From 868857d2764a26668e32d3e4f9764f728fb162a4 Mon Sep 17 00:00:00 2001 From: adelon <22380201+adelon@users.noreply.github.com> Date: Tue, 25 Jun 2024 16:52:32 +0200 Subject: Update equivalence.tex --- library/relation/equivalence.tex | 28 +++++++++++++++++----------- 1 file changed, 17 insertions(+), 11 deletions(-) (limited to 'library') diff --git a/library/relation/equivalence.tex b/library/relation/equivalence.tex index bda8486..0c5dbfa 100644 --- a/library/relation/equivalence.tex +++ b/library/relation/equivalence.tex @@ -221,51 +221,57 @@ \begin{definition}\label{equivalence_from_partition} - $\equivfrompartition{P} = \{(a, b)\mid a\in A, b\in A\mid \exists C\in P.\ a, b\in C\}$. + $\equivfrompartition{P}{A} = \{(a, b)\mid a\in A, b\in A\mid \exists C\in P.\ a, b\in C\}$. \end{definition} \begin{proposition}\label{equivalence_from_partition_intro} Let $P$ be a partition of $A$. Let $a,b\in A$. Suppose $a,b\in C\in P$. - Then $a\mathrel{\equivfrompartition{P}} b$. + Then $a\mathrel{\equivfrompartition{P}{A}} b$. \end{proposition} \begin{proposition}\label{equivalence_from_partition_reflexive} Let $P$ be a partition of $A$. - $\equivfrompartition{P}$ is reflexive on $A$. + $\equivfrompartition{P}{A}$ is reflexive on $A$. \end{proposition} \begin{proposition}\label{equivalence_from_partition_symmetric} Let $P$ be a partition. - $\equivfrompartition{P}$ is symmetric. + $\equivfrompartition{P}{A}$ is symmetric. \end{proposition} \begin{proof} - Follows by \cref{symmetric,equivalence_from_partition,notin_emptyset}. + Omitted. \end{proof} \begin{proposition}\label{equivalence_from_partition_transitive} Let $P$ be a partition. - $\equivfrompartition{P}$ is transitive. + $\equivfrompartition{P}{A}$ is transitive. \end{proposition} +\begin{proof} + Omitted. +\end{proof} \begin{proposition}\label{equivalence_from_partition_is_equivalence} Let $P$ be a partition of $A$. - $\equivfrompartition{P}$ is an equivalence on $A$. + $\equivfrompartition{P}{A}$ is an equivalence on $A$. \end{proposition} +\begin{proof} + Omitted. +\end{proof} \begin{proposition}\label{equivalence_from_quotient} Let $E$ be an equivalence on $A$. - Then $\equivfrompartition{\quotient{A}{E}} = E$. + Then $\equivfrompartition{\quotient{A}{E}}{A} = E$. \end{proposition} \begin{proof} - Follows by set extensionality. + Omitted. \end{proof} \begin{proposition}\label{partition_eq_quotient_by_equivalence_from_partition} Let $P$ be a partition of $A$. - Then $\quotient{A}{\equivfrompartition{P}} = P$. + Then $\quotient{A}{\equivfrompartition{P}{A}} = P$. \end{proposition} \begin{proof} - Follows by set extensionality. + Omitted. \end{proof} -- cgit v1.2.3 From 0c82b10cd3ac1787838038b4b443f79cbb1612d9 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 26 Jun 2024 13:56:47 +0200 Subject: Working at the numbers.tex --- .gitignore | 1 + library/everything.tex | 8 ++++---- library/numbers.tex | 26 ++++++++++++++++++++++++++ 3 files changed, 31 insertions(+), 4 deletions(-) (limited to 'library') diff --git a/.gitignore b/.gitignore index cd7aa0f..656e54b 100644 --- a/.gitignore +++ b/.gitignore @@ -41,6 +41,7 @@ premseldump/ haddocks/ stack.yaml.lock zf*.svg +check/ diff --git a/library/everything.tex b/library/everything.tex index 783679f..29b97b7 100644 --- a/library/everything.tex +++ b/library/everything.tex @@ -29,12 +29,12 @@ \import{topology/basis.tex} \import{topology/disconnection.tex} \import{topology/separation.tex} -\import{numbers.tex} +%\import{numbers.tex} \begin{proposition}\label{trivial} $x = x$. \end{proposition} -%\begin{proposition}\label{safe} -% Contradiction. -%\end{proposition} +\begin{proposition}\label{safe} + Contradiction. +\end{proposition} diff --git a/library/numbers.tex b/library/numbers.tex index 1837ae8..f13214d 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -1,5 +1,6 @@ \import{order/order.tex} \import{relation.tex} +\import{set/suc.tex} \section{The real numbers} @@ -15,6 +16,11 @@ $x \rmul y$ is a set. \end{signature} +%Structure TODO: +% Take as may axioms as needed - Tarski Axioms of reals +%Implement Naturals -> Integer -> rationals -> reals + + \subsection{Basic axioms of the reals} \begin{axiom}\label{reals_axiom_zero_in_reals} @@ -29,6 +35,26 @@ $\zero \neq 1$. \end{axiom} +\begin{inductive}\label{naturals_subset_reals} + Define $\naturals \subseteq \reals$ inductively as follows. + \begin{enumerate} + \item $\zero \in \naturals$. + \item If $n\in \naturals$, then $\successor{n} \in \naturals$. + \end{enumerate} +\end{inductive} + +%\begin{axiom}\label{negativ_is_set} +% $\negativ{x}$ is a set. +%\end{axiom} + +%\begin{axiom}\label{negativ_of} +% $\negativ{x} \in \reals$ iff $x\in \reals$. +%\end{axiom} +% +%\begin{axiom}\label{negativ_behavior} +% $x + \negativ{x} = \zero$. +%\end{axiom} + \begin{definition}\label{reals_definition_order_def} $x < y$ iff there exist $z \in \reals$ such that $x + (z \rmul z) = y$. \end{definition} -- cgit v1.2.3 From 7e65d40f100af326adbd4ef1d32fdd0aabc92f4b Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 2 Jul 2024 11:05:31 +0200 Subject: Definition and axioms of naturals. --- library/numbers.tex | 191 +++++++++++++++++++++++++++++++++++++--------------- 1 file changed, 135 insertions(+), 56 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index f13214d..4b5d10b 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -2,6 +2,7 @@ \import{relation.tex} \import{set/suc.tex} + \section{The real numbers} \begin{signature} @@ -16,12 +17,47 @@ $x \rmul y$ is a set. \end{signature} +\begin{axiom}\label{reals_axiom_order} + $\lt[\reals]$ is an order on $\reals$. +\end{axiom} + +\begin{abbreviation}\label{lesseq_on_reals} + $x \leq y$ iff $(x,y) \in \lt[\reals]$. +\end{abbreviation} + +\begin{abbreviation}\label{less_on_reals} + $x < y$ iff it is wrong that $y \leq x$. +\end{abbreviation} + +\begin{abbreviation}\label{greater_on_reals} + $x > y$ iff $y \leq x$. +\end{abbreviation} + +\begin{abbreviation}\label{greatereq_on_reals} + $x \geq y$ iff it is wrong that $x < y$. +\end{abbreviation} + +\begin{abbreviation}\label{is_positiv} + $x$ is positiv iff $x > \zero$. +\end{abbreviation} + + %Structure TODO: % Take as may axioms as needed - Tarski Axioms of reals %Implement Naturals -> Integer -> rationals -> reals -\subsection{Basic axioms of the reals} +\subsection{Creation of natural numbers} + +\subsubsection{Defenition and axioms} + +\begin{abbreviation}\label{inductive_set} + $A$ is an inductive set iff $\emptyset\in A$ and for all $a\in A$ we have $\suc{a}\in A$. +\end{abbreviation} + +\begin{abbreviation}\label{zero_is_emptyset} + $\zero = \emptyset$. +\end{abbreviation} \begin{axiom}\label{reals_axiom_zero_in_reals} $\zero \in \reals$. @@ -35,49 +71,101 @@ $\zero \neq 1$. \end{axiom} -\begin{inductive}\label{naturals_subset_reals} - Define $\naturals \subseteq \reals$ inductively as follows. - \begin{enumerate} - \item $\zero \in \naturals$. - \item If $n\in \naturals$, then $\successor{n} \in \naturals$. - \end{enumerate} -\end{inductive} - -%\begin{axiom}\label{negativ_is_set} -% $\negativ{x}$ is a set. -%\end{axiom} - -%\begin{axiom}\label{negativ_of} -% $\negativ{x} \in \reals$ iff $x\in \reals$. -%\end{axiom} -% -%\begin{axiom}\label{negativ_behavior} -% $x + \negativ{x} = \zero$. -%\end{axiom} - -\begin{definition}\label{reals_definition_order_def} - $x < y$ iff there exist $z \in \reals$ such that $x + (z \rmul z) = y$. +\begin{axiom}\label{one_is_suc_zero} + $\suc{\zero} = 1$. +\end{axiom} + +\begin{definition}\label{naturals} + $\naturals = \{ x \in \reals \mid \exists y \in \reals. \suc{y} = x \lor x = \zero \}$. \end{definition} -%\begin{axiom}\label{reals_axiom_order} -% $\lt[\reals]$ is an order on $\reals$. -%\end{axiom} +\begin{lemma}\label{naturals_subseteq_reals} + $\naturals \subseteq \reals$. +\end{lemma} -%\begin{abbreviation}\label{lesseq_on_reals} -% $x \leq y$ iff $(x,y) \in \lt[\reals]$. -%\end{abbreviation} +%\begin{inductive}\label{naturals_subset_reals} +% Define $\naturals \subseteq \reals$ inductively as follows. +% \begin{enumerate} +% \item $\zero \in \naturals$. +% \item If $n\in \naturals$, then $\suc{n} \in \naturals$. %Naproche translate this to for all n \in R \suc{n} \in R we want something like the definition before. +% \end{enumerate} +%\end{inductive} -\begin{abbreviation}\label{less_on_reals} - $x \leq y$ iff it is wrong that $y < x$. -\end{abbreviation} +\begin{axiom}\label{suc_eq_plus_one} + For all $n \in \naturals$ we have $\suc{n} = n + 1$. +\end{axiom} -\begin{abbreviation}\label{greater_on_reals} - $x > y$ iff $y \leq x$. +\begin{abbreviation}\label{is_a_natural_number} + $n$ is a natural number iff $n \in \naturals$. \end{abbreviation} -\begin{abbreviation}\label{greatereq_on_reals} - $x \geq y$ iff it is wrong that $x < y$. -\end{abbreviation} +\begin{axiom}\label{naturals_addition_axiom_1} + For all $n \in \naturals$ $n + \zero = \zero + n = n$. +\end{axiom} + +\begin{axiom}\label{naturals_addition_axiom_2} + For all $n, m \in \naturals$ $n + \suc{m} = \suc{n} + m = \suc{n+m}$. +\end{axiom} + +\begin{axiom}\label{naturals_mul_axiom_1} + For all $n \in \naturals$ we have $n \rmul \zero = \zero$. +\end{axiom} + +\begin{axiom}\label{naturals_mul_axiom_2} + For all $n,m \in \naturals$ we have $\suc{n} \rmul m = (n \rmul m) + m$. +\end{axiom} + +\begin{axiom}\label{addition_on_naturals} + If $x$ is a natural number and $y$ is a natural number then $x+y$ is a natural number. +\end{axiom} + + +\subsubsection{Properties and Facts natural numbers} + +\begin{proposition}\label{naturals_kommu} + For all $n,m \in \naturals$ we have $n + m = m + n$. +\end{proposition} +\begin{proof} + \begin{byCase} + \caseOf{$n = \emptyset$.} Trivial. + \caseOf{$m = \emptyset$.} Trivial. + \caseOf{$n \neq \emptyset \land m \neq \emptyset$.} Omitted. + \end{byCase} +\end{proof} + + + +\begin{lemma}\label{naturals_inductive_set} + $\naturals$ is an inductive set. +\end{lemma} + + +\begin{lemma}\label{naturals_smallest_inductive_set} + Let $A$ be an inductive set. + Then $\naturals\subseteq A$. +\end{lemma} + + +\begin{lemma}\label{naturals_is_equal_to_two_times_naturals} + $\{x+y \mid x \in \naturals, y \in \naturals \} = \naturals$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +%\begin{lemma}\label{oeder_on_naturals} +% $\lt[\reals] \inter (\naturals \times \naturals)$ is an order on $\naturals$. +%\end{lemma} + + + + + + +\subsection{The Intergers} + + + \begin{axiom}\label{reals_axiom_dense} For all $x,y \in \reals$ if $x < y$ then @@ -127,7 +215,9 @@ \begin{proposition}\label{reals_reducion_on_addition} For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$. \end{proposition} - +\begin{proof} + Omitted. +\end{proof} %\begin{signature}\label{invers_is_set} @@ -147,9 +237,7 @@ % $x \rminus \addInv{x} = \zero$. %\end{lemma} -\begin{abbreviation}\label{is_positiv} - $x$ is positiv iff $x > \zero$. -\end{abbreviation} + \begin{lemma}\label{reals_add_of_positiv} Let $x,y \in \reals$. @@ -221,7 +309,9 @@ \begin{lemma}\label{order_reals_lemma00} For all $x,y \in \reals$ such that $x > y$ we have $x \geq y$. \end{lemma} - +\begin{proof} + Omitted. +\end{proof} \begin{lemma}\label{order_reals_lemma5} For all $x,y \in \reals$ such that $x < y$ we have $x \leq y$. @@ -289,24 +379,13 @@ -\section{The natural numbers} -\begin{abbreviation}\label{def_suc} - $\successor{n} = n + 1$. -\end{abbreviation} -\begin{inductive}\label{naturals_definition_as_subset_of_reals} - Define $\nat \subseteq \reals$ inductively as follows. - \begin{enumerate} - \item $\zero \in \nat$. - \item If $n\in \nat$, then $\successor{n} \in \nat$. - \end{enumerate} -\end{inductive} -\begin{lemma}\label{reals_order_suc} - $n < \successor{n}$. -\end{lemma} + + + %\begin{proposition}\label{safe} % Contradiction. -- cgit v1.2.3 From bff76c7fabb9f2c0b9dcac915dfa68e930baf4d4 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 2 Jul 2024 21:24:44 +0200 Subject: Further Formalisation of naturals. Such as induction on naturals and addition laws. --- library/numbers.tex | 177 +++++++++++++++++++++++++++++++++++++++++----------- library/ordinal.tex | 107 +++++++++++++++---------------- 2 files changed, 195 insertions(+), 89 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 4b5d10b..b83827a 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -1,6 +1,7 @@ \import{order/order.tex} \import{relation.tex} \import{set/suc.tex} +\import{ordinal.tex} \section{The real numbers} @@ -75,24 +76,17 @@ $\suc{\zero} = 1$. \end{axiom} -\begin{definition}\label{naturals} - $\naturals = \{ x \in \reals \mid \exists y \in \reals. \suc{y} = x \lor x = \zero \}$. -\end{definition} - -\begin{lemma}\label{naturals_subseteq_reals} +\begin{axiom}\label{naturals_subseteq_reals} $\naturals \subseteq \reals$. -\end{lemma} +\end{axiom} -%\begin{inductive}\label{naturals_subset_reals} -% Define $\naturals \subseteq \reals$ inductively as follows. -% \begin{enumerate} -% \item $\zero \in \naturals$. -% \item If $n\in \naturals$, then $\suc{n} \in \naturals$. %Naproche translate this to for all n \in R \suc{n} \in R we want something like the definition before. -% \end{enumerate} -%\end{inductive} +\begin{axiom}\label{naturals_inductive_set} + $\naturals$ is an inductive set. +\end{axiom} -\begin{axiom}\label{suc_eq_plus_one} - For all $n \in \naturals$ we have $\suc{n} = n + 1$. +\begin{axiom}\label{naturals_smallest_inductive_set} + Let $A$ be an inductive set. + Then $\naturals\subseteq A$. \end{axiom} \begin{abbreviation}\label{is_a_natural_number} @@ -119,33 +113,147 @@ If $x$ is a natural number and $y$ is a natural number then $x+y$ is a natural number. \end{axiom} +\subsubsection{Natural numbers as ordinals} -\subsubsection{Properties and Facts natural numbers} -\begin{proposition}\label{naturals_kommu} - For all $n,m \in \naturals$ we have $n + m = m + n$. -\end{proposition} +\begin{lemma}\label{nat_is_successor_ordinal} + Let $n\in\naturals$. + Suppose $n\neq \emptyset$. + Then $n$ is a successor ordinal. +\end{lemma} \begin{proof} - \begin{byCase} - \caseOf{$n = \emptyset$.} Trivial. - \caseOf{$m = \emptyset$.} Trivial. - \caseOf{$n \neq \emptyset \land m \neq \emptyset$.} Omitted. - \end{byCase} + Let $M = \{ m\in \naturals \mid\text{$m = \emptyset$ or $m$ is a successor ordinal}\}$. + $M$ is an inductive set by \cref{suc_ordinal,naturals_inductive_set,successor_ordinal,emptyset_is_ordinal}. + Now $M\subseteq \naturals\subseteq M$ + by \cref{subseteq,naturals_smallest_inductive_set}. + Thus $M = \naturals$. + Follows by \cref{subseteq}. \end{proof} - +\begin{lemma}\label{nat_is_transitiveset} + $\naturals$ is \in-transitive. +\end{lemma} +\begin{proof} + Let $M = \{ m\in\naturals \mid \text{for all $n\in m$ we have $n\in\naturals$} \}$. + $\emptyset\in M$. + For all $n\in M$ we have $\suc{n}\in M$ + by \cref{naturals_inductive_set,suc}. + Thus $M$ is an inductive set. + Now $M\subseteq \naturals\subseteq M$ + by \cref{subseteq,naturals_smallest_inductive_set}. + Hence $\naturals = M$. +\end{proof} -\begin{lemma}\label{naturals_inductive_set} - $\naturals$ is an inductive set. +\begin{lemma}\label{natural_number_is_ordinal} + Every natural number is an ordinal. \end{lemma} +\begin{proof} + Follows by \cref{suc_ordinal,suc_neq_emptyset,naturals_inductive_set,nat_is_successor_ordinal,successor_ordinal,suc_ordinal_implies_ordinal}. +\end{proof} +\begin{lemma}\label{omega_is_an_ordinal} + $\naturals$ is an ordinal. +\end{lemma} +\begin{proof} + Follows by \cref{natural_number_is_ordinal,transitive_set_of_ordinals_is_ordinal,nat_is_transitiveset}. +\end{proof} -\begin{lemma}\label{naturals_smallest_inductive_set} - Let $A$ be an inductive set. - Then $\naturals\subseteq A$. +\begin{lemma}\label{omega_is_a_limit_ordinal} + $\naturals$ is a limit ordinal. \end{lemma} +\begin{proof} + $\emptyset\precedes \naturals$. + If $n\in \naturals$, then $\suc{n}\in\naturals$. +\end{proof} +\subsubsection{Properties and Facts natural numbers} + +\begin{theorem}\label{induction_principle} + Let $P$ be a set. + Suppose $P \subseteq \naturals$. + Suppose $\zero \in P$. + Suppose $\forall n \in P. \suc{n} \in P$. + Then $P = \naturals$. +\end{theorem} +\begin{proof} + Trivial. +\end{proof} + +\begin{proposition}\label{existence_of_suc} + Let $n \in \naturals$. + Suppose $n \neq \zero$. + Then there exist $n' \in \naturals$ such that $\suc{n'} = n$. +\end{proposition} +\begin{proof} + Follows by \cref{transitiveset,nat_is_transitiveset,suc_intro_self,successor_ordinal,nat_is_successor_ordinal}. +\end{proof} + +\begin{proposition}\label{suc_eq_plus_one} + Let $n \in \naturals$. + Then $\suc{n} = n + 1$. +\end{proposition} +\begin{proof} + Let $P = \{ n \in \naturals \mid n + 1 = 1 + n \}$. + $\zero \in P$. + It suffices to show that if $m \in P$ then $\suc{m} \in P$. +\end{proof} + +\begin{proposition}\label{naturals_1_kommu} + Let $n \in \naturals$. + Then $1 + n = n + 1$. +\end{proposition} + +\begin{proposition}\label{naturals_add_kommu} + For all $n \in \naturals$ we have for all $m\in \naturals$ we have $n + m = m + n$. +\end{proposition} +\begin{proof} + Fix $n \in \naturals$. + Let $P = \{ m \in \naturals \mid m + n = n + m \}$. + It suffices to show that $P = \naturals$. + $P \subseteq \naturals$. + $\zero \in P$. + It suffices to show that if $m \in P$ then $\suc{m} \in P$. +\end{proof} + +\begin{proposition}\label{naturals_add_assoc} + Suppose $n,m,k \in \naturals$. + Then $n + (m + k) = (n + m) + k$. +\end{proposition} +\begin{proof} + Let $P = \{ k \in \naturals \mid \text{for all $n',m' \in \naturals$ we have $n' + (m' + k) = (n' + m') + k$}\}$. + $P \subseteq \naturals$. + We show that $P = \naturals$. + \begin{subproof} + $\zero \in P$. + It suffices to show that for all $k \in P$ we have $\suc{k} \in P$. + Fix $k \in P$. + \begin{align*} + (n + m) + \suc{k} \\ + &= \suc{(n+m) + k} \\ + &= \suc{n + (m + k)} \\ + &= n + \suc{(m + k)} \\ + &= n + (m + \suc{k}) + \end{align*} + For all $n,m \in \naturals$ we have $(n + m) + \suc{k} = n + (m + \suc{k})$. + \end{subproof} +\end{proof} + +\begin{proposition}\label{naturals_rmul_one} + For all $n \in \naturals$ we have $n \rmul 1 = n$. +\end{proposition} + + +\begin{proposition}\label{naturals_mul_kommu} + Let $n, m \in \naturals$. + Then $n \rmul m = m \rmul n$. +\end{proposition} + +\begin{proposition}\label{naturals_rmul_assoc} + Suppose $n,m,k \in \naturals$. + Then $n \rmul (m \rmul k) = (n \rmul m) \rmul k$. +\end{proposition} + \begin{lemma}\label{naturals_is_equal_to_two_times_naturals} $\{x+y \mid x \in \naturals, y \in \naturals \} = \naturals$. \end{lemma} @@ -153,9 +261,6 @@ Omitted. \end{proof} -%\begin{lemma}\label{oeder_on_naturals} -% $\lt[\reals] \inter (\naturals \times \naturals)$ is an order on $\naturals$. -%\end{lemma} @@ -387,9 +492,9 @@ -%\begin{proposition}\label{safe} -% Contradiction. -%\end{proposition} +\begin{proposition}\label{safe} + Contradiction. +\end{proposition} \section{The integers} diff --git a/library/ordinal.tex b/library/ordinal.tex index 3063162..6f924c1 100644 --- a/library/ordinal.tex +++ b/library/ordinal.tex @@ -3,7 +3,7 @@ \import{set/powerset.tex} \import{set/regularity.tex} \import{set/suc.tex} -\import{nat.tex} +%\import{nat.tex} \section{Transitive sets} @@ -584,55 +584,56 @@ Then $\alpha\subseteq\beta$. Follows by \cref{subseteq_antisymmetric}. \end{proof} - -\subsection{Natural numbers as ordinals} - -\begin{lemma}\label{nat_is_successor_ordinal} - Let $n\in\naturals$. - Suppose $n\neq \emptyset$. - Then $n$ is a successor ordinal. -\end{lemma} -\begin{proof} - Let $M = \{ m\in \naturals \mid\text{$m = \emptyset$ or $m$ is a successor ordinal}\}$. - $M$ is an inductive set by \cref{suc_ordinal,naturals_inductive_set,successor_ordinal,emptyset_is_ordinal}. - Now $M\subseteq \naturals\subseteq M$ - by \cref{subseteq,naturals_smallest_inductive_set}. - Thus $M = \naturals$. - Follows by \cref{subseteq}. -\end{proof} - -\begin{lemma}\label{nat_is_transitiveset} - $\naturals$ is \in-transitive. -\end{lemma} -\begin{proof} - Let $M = \{ m\in\naturals \mid \text{for all $n\in m$ we have $n\in\naturals$} \}$. - $\emptyset\in M$. - For all $n\in M$ we have $\suc{n}\in M$ - by \cref{naturals_inductive_set,suc}. - Thus $M$ is an inductive set. - Now $M\subseteq \naturals\subseteq M$ - by \cref{subseteq,naturals_smallest_inductive_set}. - Hence $\naturals = M$. -\end{proof} - -\begin{lemma}\label{natural_number_is_ordinal} - Every natural number is an ordinal. -\end{lemma} -\begin{proof} - Follows by \cref{suc_ordinal,suc_neq_emptyset,naturals_inductive_set,nat_is_successor_ordinal,successor_ordinal,suc_ordinal_implies_ordinal}. -\end{proof} - -\begin{lemma}\label{omega_is_an_ordinal} - $\naturals$ is an ordinal. -\end{lemma} -\begin{proof} - Follows by \cref{natural_number_is_ordinal,transitive_set_of_ordinals_is_ordinal,nat_is_transitiveset}. -\end{proof} - -\begin{lemma}\label{omega_is_a_limit_ordinal} - $\naturals$ is a limit ordinal. -\end{lemma} -\begin{proof} - $\emptyset\precedes \naturals$. - If $n\in \naturals$, then $\suc{n}\in\naturals$. -\end{proof} +% +%\subsection{Natural numbers as ordinals} +% +%\begin{lemma}\label{nat_is_successor_ordinal} +% Let $n\in\naturals$. +% Suppose $n\neq \emptyset$. +% Then $n$ is a successor ordinal. +%\end{lemma} +%\begin{proof} +% Let $M = \{ m\in \naturals \mid\text{$m = \emptyset$ or $m$ is a successor ordinal}\}$. +% $M$ is an inductive set by \cref{suc_ordinal,naturals_inductive_set,successor_ordinal,emptyset_is_ordinal}. +% Now $M\subseteq \naturals\subseteq M$ +% by \cref{subseteq,naturals_smallest_inductive_set}. +% Thus $M = \naturals$. +% Follows by \cref{subseteq}. +%\end{proof} +% +%\begin{lemma}\label{nat_is_transitiveset} +% $\naturals$ is \in-transitive. +%\end{lemma} +%\begin{proof} +% Let $M = \{ m\in\naturals \mid \text{for all $n\in m$ we have $n\in\naturals$} \}$. +% $\emptyset\in M$. +% For all $n\in M$ we have $\suc{n}\in M$ +% by \cref{naturals_inductive_set,suc}. +% Thus $M$ is an inductive set. +% Now $M\subseteq \naturals\subseteq M$ +% by \cref{subseteq,naturals_smallest_inductive_set}. +% Hence $\naturals = M$. +%\end{proof} +% +%\begin{lemma}\label{natural_number_is_ordinal} +% Every natural number is an ordinal. +%\end{lemma} +%\begin{proof} +% Follows by \cref{suc_ordinal,suc_neq_emptyset,naturals_inductive_set,nat_is_successor_ordinal,successor_ordinal,suc_ordinal_implies_ordinal}. +%\end{proof} +% +%\begin{lemma}\label{omega_is_an_ordinal} +% $\naturals$ is an ordinal. +%\end{lemma} +%\begin{proof} +% Follows by \cref{natural_number_is_ordinal,transitive_set_of_ordinals_is_ordinal,nat_is_transitiveset}. +%\end{proof} +% +%\begin{lemma}\label{omega_is_a_limit_ordinal} +% $\naturals$ is a limit ordinal. +%\end{lemma} +%\begin{proof} +% $\emptyset\precedes \naturals$. +% If $n\in \naturals$, then $\suc{n}\in\naturals$. +%\end{proof} +% \ No newline at end of file -- cgit v1.2.3 From c580901e967c6bf0b012017a868a2c360e25370a Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Fri, 5 Jul 2024 10:42:24 +0200 Subject: Proof of 1 is identity on the naturals and begin of Disstributiv law naturals proof --- library/numbers.tex | 68 +++++++++++++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 66 insertions(+), 2 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index b83827a..113aadb 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -102,7 +102,7 @@ \end{axiom} \begin{axiom}\label{naturals_mul_axiom_1} - For all $n \in \naturals$ we have $n \rmul \zero = \zero$. + For all $n \in \naturals$ we have $n \rmul \zero = \zero = \zero \rmul n$. \end{axiom} \begin{axiom}\label{naturals_mul_axiom_2} @@ -203,6 +203,9 @@ Let $n \in \naturals$. Then $1 + n = n + 1$. \end{proposition} +\begin{proof} + Follows by \cref{suc_eq_plus_one,naturals_addition_axiom_2,naturals_addition_axiom_1,naturals_inductive_set,one_is_suc_zero}. +\end{proof} \begin{proposition}\label{naturals_add_kommu} For all $n \in \naturals$ we have for all $m\in \naturals$ we have $n + m = m + n$. @@ -240,14 +243,75 @@ \end{proof} \begin{proposition}\label{naturals_rmul_one} - For all $n \in \naturals$ we have $n \rmul 1 = n$. + For all $n \in \naturals$ we have $1 \rmul n = n$. \end{proposition} +\begin{proof} + Fix $n \in \naturals$. + \begin{align*} + 1 \rmul n \\ + &= \suc{\zero} \rmul n \\ + &= (\zero \rmul n) + n \\ + &= (\zero) + n \\ + &= n + \end{align*} +\end{proof} + +\begin{proposition}\label{naturals_add_remove_brakets} + Suppose $n,m,k \in \naturals$. + Then $(n + m) + k = n + m + k$. +\end{proposition} + +\begin{proposition}\label{natural_disstro} + Suppose $n,m,k \in \naturals$. + Then $n \rmul (m + k) = (n \rmul m) + (n \rmul k)$. +\end{proposition} +\begin{proof} + Let $P = \{n \in \naturals \mid \text{for all $m,k \in \naturals$ we have $n \rmul (m + k) = (n \rmul m) + (n \rmul k)$}\}$. + $\zero \in P$. + $P \subseteq \naturals$. + It suffices to show that for all $n \in P$ we have $\suc{n} \in P$. + Fix $n \in P$. + It suffices to show that for all $m \in \naturals$ we have $\suc{n} \rmul (m + k) = (\suc{n} \rmul m) + (\suc{n} \rmul k)$. + Fix $m \in \naturals$. + It suffices to show that for all $k \in \naturals$ we have $\suc{n} \rmul (m + k) = (\suc{n} \rmul m) + (\suc{n} \rmul k)$. + Fix $k \in \naturals$. + \begin{align*} + \suc{n} \rmul (m + k) \\ + &= (n \rmul (m + k)) + (m + k) \\% \explanation{by \cref{naturals_mul_axiom_2}}\\ + &= ((n \rmul m) + (n \rmul k)) + (m + k) \explanation{by assumption}\\ + &= ((n \rmul m) + (n \rmul k)) + m + k \explanation{by \cref{naturals_add_remove_brakets,naturals_add_kommu}}\\ + &= (((n \rmul m) + (n \rmul k)) + m) + k \explanation{by \cref{naturals_add_kommu,naturals_add_remove_brakets,naturals_add_assoc}}\\ + &= (m + ((n \rmul m) + (n \rmul k))) + k \explanation{by \cref{naturals_add_kommu}}\\ + &= ((m + (n \rmul m)) + (n \rmul k)) + k \explanation{by \cref{naturals_add_assoc}}\\ + &= (((n \rmul m) + m) + (n \rmul k)) + k \explanation{by \cref{naturals_add_kommu}}\\ + &= ((n \rmul m) + m) + ((n \rmul k) + k) \explanation{by \cref{naturals_add_assoc}}\\ + &= (\suc{n} \rmul m) + (\suc{n} \rmul k) \explanation{by \cref{naturals_mul_axiom_2}} + \end{align*} +\end{proof} \begin{proposition}\label{naturals_mul_kommu} Let $n, m \in \naturals$. Then $n \rmul m = m \rmul n$. \end{proposition} +\begin{proof} + Let $P = \{n \in \naturals \mid \forall m \in \naturals. n \rmul m = m \rmul n\}$. + $\zero \in P$. + $P \subseteq \naturals$. + It suffices to show that for all $n \in P$ we have $\suc{n} \in P$. + Fix $n \in P$. + It suffices to show that for all $m \in \naturals$ we have $\suc{n} \rmul m = m \rmul \suc{n}$. + Fix $m \in \naturals$. + $n \rmul m = m \rmul n$. + + \begin{align*} + \suc{n} \rmul m \\ + &= (n \rmul m) + m \\ + &= (m \rmul n) + m \\ + &= m + (m \rmul n) \\ + &= m \rmul \suc{n} \\ + \end{align*} +\end{proof} \begin{proposition}\label{naturals_rmul_assoc} Suppose $n,m,k \in \naturals$. -- cgit v1.2.3 From 698bc0ec8128889aae37a766130e9b193c399b9c Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 6 Jul 2024 17:55:36 +0200 Subject: Finished the formalization of naturals. --- library/numbers.tex | 119 ++++++++++++++++++++++++++++++++++++---------------- 1 file changed, 84 insertions(+), 35 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 113aadb..a185940 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -113,6 +113,14 @@ If $x$ is a natural number and $y$ is a natural number then $x+y$ is a natural number. \end{axiom} +\begin{axiom}\label{naturals_rmul_is_closed_in_n} + For all $n,m \in \naturals$ we have $(n \rmul m) \in \naturals$. +\end{axiom} + +\begin{axiom}\label{naturals_add_is_closed_in_n} + For all $n,m \in \naturals$ we have $(n + m) \in \naturals$. +\end{axiom} + \subsubsection{Natural numbers as ordinals} @@ -226,23 +234,22 @@ \begin{proof} Let $P = \{ k \in \naturals \mid \text{for all $n',m' \in \naturals$ we have $n' + (m' + k) = (n' + m') + k$}\}$. $P \subseteq \naturals$. - We show that $P = \naturals$. - \begin{subproof} - $\zero \in P$. - It suffices to show that for all $k \in P$ we have $\suc{k} \in P$. - Fix $k \in P$. - \begin{align*} - (n + m) + \suc{k} \\ - &= \suc{(n+m) + k} \\ - &= \suc{n + (m + k)} \\ - &= n + \suc{(m + k)} \\ - &= n + (m + \suc{k}) - \end{align*} - For all $n,m \in \naturals$ we have $(n + m) + \suc{k} = n + (m + \suc{k})$. - \end{subproof} + $\zero \in P$. + It suffices to show that for all $k \in P$ we have $\suc{k} \in P$. + Fix $k \in P$. + It suffices to show that for all $n' \in \naturals$ we have for all $m' \in \naturals$ we have $n' + (m' + \suc{k}) = (n' + m') + \suc{k}$. + Fix $n' \in \naturals$. + Fix $m' \in \naturals$. + \begin{align*} + (n' + m') + \suc{k} \\ + &= \suc{(n' + m') + k} \\ + &= \suc{n' + (m' + k)} \\ + &= n' + \suc{(m' + k)} \\ + &= n' + (m' + \suc{k}) + \end{align*} \end{proof} -\begin{proposition}\label{naturals_rmul_one} +\begin{proposition}\label{naturals_rmul_one_left} For all $n \in \naturals$ we have $1 \rmul n = n$. \end{proposition} \begin{proof} @@ -258,7 +265,12 @@ \begin{proposition}\label{naturals_add_remove_brakets} Suppose $n,m,k \in \naturals$. - Then $(n + m) + k = n + m + k$. + Then $(n + m) + k = n + m + k = n + (m + k)$. +\end{proposition} + +\begin{proposition}\label{naturals_add_remove_brakets2} + Suppose $n,m,k,l \in \naturals$. + Then $(n + m) + (k + l)= n + m + k + l = n + (m + k) + l = (((n + m) + k) + l)$. \end{proposition} \begin{proposition}\label{natural_disstro} @@ -271,26 +283,43 @@ $P \subseteq \naturals$. It suffices to show that for all $n \in P$ we have $\suc{n} \in P$. Fix $n \in P$. - It suffices to show that for all $m \in \naturals$ we have $\suc{n} \rmul (m + k) = (\suc{n} \rmul m) + (\suc{n} \rmul k)$. - Fix $m \in \naturals$. - It suffices to show that for all $k \in \naturals$ we have $\suc{n} \rmul (m + k) = (\suc{n} \rmul m) + (\suc{n} \rmul k)$. - Fix $k \in \naturals$. + + It suffices to show that for all $m'$ such that $m' \in \naturals$ we have for all $k' \in \naturals$ we have $\suc{n} \rmul (m' + k') = (\suc{n} \rmul m') + (\suc{n} \rmul k')$. + Fix $m' \in \naturals$. + Fix $k' \in \naturals$. + $n \in \naturals$. \begin{align*} - \suc{n} \rmul (m + k) \\ - &= (n \rmul (m + k)) + (m + k) \\% \explanation{by \cref{naturals_mul_axiom_2}}\\ - &= ((n \rmul m) + (n \rmul k)) + (m + k) \explanation{by assumption}\\ - &= ((n \rmul m) + (n \rmul k)) + m + k \explanation{by \cref{naturals_add_remove_brakets,naturals_add_kommu}}\\ - &= (((n \rmul m) + (n \rmul k)) + m) + k \explanation{by \cref{naturals_add_kommu,naturals_add_remove_brakets,naturals_add_assoc}}\\ - &= (m + ((n \rmul m) + (n \rmul k))) + k \explanation{by \cref{naturals_add_kommu}}\\ - &= ((m + (n \rmul m)) + (n \rmul k)) + k \explanation{by \cref{naturals_add_assoc}}\\ - &= (((n \rmul m) + m) + (n \rmul k)) + k \explanation{by \cref{naturals_add_kommu}}\\ - &= ((n \rmul m) + m) + ((n \rmul k) + k) \explanation{by \cref{naturals_add_assoc}}\\ - &= (\suc{n} \rmul m) + (\suc{n} \rmul k) \explanation{by \cref{naturals_mul_axiom_2}} + \suc{n} \rmul (m' + k') \\ + &= (n \rmul (m' + k')) + (m' + k') \\% \explanation{by \cref{naturals_mul_axiom_2}}\\ + &= ((n \rmul m') + (n \rmul k')) + (m' + k') \\%\explanation{by assumption}\\ + &= ((n \rmul m') + (n \rmul k')) + m' + k' \\%\explanation{by \cref{naturals_add_remove_brakets,naturals_add_kommu}}\\ + &= (((n \rmul m') + (n \rmul k')) + m') + k' \\%\explanation{by \cref{naturals_add_kommu,naturals_add_remove_brakets,naturals_add_assoc}}\\ + &= (m' + ((n \rmul m') + (n \rmul k'))) + k' \\%\explanation{by \cref{naturals_add_kommu}}\\ + &= ((m' + (n \rmul m')) + (n \rmul k')) + k' \\%\explanation{by \cref{naturals_add_assoc}}\\ + &= (((n \rmul m') + m') + (n \rmul k')) + k' \\%\explanation{by \cref{naturals_add_kommu}}\\ + &= ((n \rmul m') + m') + ((n \rmul k') + k') \\%\explanation{by \cref{naturals_add_assoc}}\\ + &= (\suc{n} \rmul m') + (\suc{n} \rmul k') %\explanation{by \cref{naturals_mul_axiom_2}} \end{align*} \end{proof} +\begin{proposition}\label{naturals_rmul_one_kommu} + For all $n \in \naturals$ we have $n \rmul 1 = n$. +\end{proposition} +\begin{proof} + Let $P = \{ n \in \naturals \mid n \rmul 1 = n\}$. + $1 \in P$. + It suffices to show that for all $n' \in P$ we have $\suc{n'} \in P$. + Fix $n' \in P$. + It suffices to show that $\suc{n'} \rmul 1 = \suc{n'}$. + \begin{align*} + \suc{n'} \rmul 1 \\ + &= n' \rmul 1 + 1 \\ + &= n' + 1 \\ + &= \suc{n'} + \end{align*} +\end{proof} -\begin{proposition}\label{naturals_mul_kommu} +\begin{proposition}\label{naturals_rmul_kommu} Let $n, m \in \naturals$. Then $n \rmul m = m \rmul n$. \end{proposition} @@ -303,12 +332,13 @@ It suffices to show that for all $m \in \naturals$ we have $\suc{n} \rmul m = m \rmul \suc{n}$. Fix $m \in \naturals$. $n \rmul m = m \rmul n$. - \begin{align*} \suc{n} \rmul m \\ &= (n \rmul m) + m \\ &= (m \rmul n) + m \\ &= m + (m \rmul n) \\ + &= (m \rmul 1) + (m \rmul n) \\ + &= m \rmul (1 + n) \\ &= m \rmul \suc{n} \\ \end{align*} \end{proof} @@ -317,13 +347,32 @@ Suppose $n,m,k \in \naturals$. Then $n \rmul (m \rmul k) = (n \rmul m) \rmul k$. \end{proposition} +\begin{proof} + Let $P = \{ n \in \naturals \mid \text{for all $m,k \in \naturals$ we have $n \rmul (m \rmul k) = (n \rmul m) \rmul k$ }\}$. + $\zero \in P$. + It suffices to show that for all $n' \in P$ we have $ \suc{n'} \in P$. + Fix $n' \in P$. + It suffices to show that for all $m' \in \naturals$ we have for all $k' \in \naturals$ we have $\suc{n'} \rmul (m' \rmul k') = (\suc{n'} \rmul m') \rmul k'$. + Fix $m' \in \naturals$. + Fix $k' \in \naturals$. + \begin{align*} + \suc{n'} \rmul (m' \rmul k') \\ + &=(n' \rmul (m' \rmul k')) + (m' \rmul k') \\ + &=((n' \rmul m') \rmul k') + (m' \rmul k') \\ + &=(k' \rmul (n' \rmul m')) + (k' \rmul m') \\ + &=k' \rmul ((n' \rmul m') + m') \\ + &=k' \rmul (\suc{n'} \rmul m') \\ + &=(\suc{n'} \rmul m') \rmul k' + \end{align*} +\end{proof} \begin{lemma}\label{naturals_is_equal_to_two_times_naturals} $\{x+y \mid x \in \naturals, y \in \naturals \} = \naturals$. \end{lemma} -\begin{proof} - Omitted. -\end{proof} + + + + -- cgit v1.2.3 From b71f135d5762f2a12bf08c71ecdcd221ed87cff0 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 6 Jul 2024 19:22:03 +0200 Subject: Formalisation of integers. --- library/numbers.tex | 123 ++++++++++++++++++++++++++++++++++++++--------- source/Syntax/Lexicon.hs | 1 + 2 files changed, 101 insertions(+), 23 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index a185940..08cbc70 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -371,19 +371,18 @@ \end{lemma} +\subsection{Axioms of the reals Part One} +We seen before that we can proof the common behavior of the naturals. +Since we now want to get furhter in a more efficient way, we introduce the basis axioms of the reals. +Such that we can introduce the integers and raitionals as smooth as possible. +\begin{axiom}\label{reals_rmul} + For all $n,m \in \reals$ we have $(n \rmul m) \in \reals$. +\end{axiom} - - - - - - - -\subsection{The Intergers} - - - +\begin{axiom}\label{reals_add} + For all $n,m \in \reals$ we have $(n + m) \in \reals$. +\end{axiom} \begin{axiom}\label{reals_axiom_dense} For all $x,y \in \reals$ if $x < y$ then @@ -423,20 +422,104 @@ such that $x < z < y$. \end{axiom} -\begin{proposition}\label{reals_disstro2} +\begin{axiom}\label{reals_disstro2} For all $x,y,z \in \reals$ $(y + z) \rmul x = (y \rmul x) + (z \rmul x)$. -\end{proposition} +\end{axiom} + + +\begin{axiom}\label{reals_reducion_on_addition} + For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$. +\end{axiom} + + + + + + + +\subsection{The Intergers} + + +\begin{axiom}\label{neg} + For all $n \in \reals$ we have $\neg{n} \in \reals$ and $n + \neg{n} = \zero$. +\end{axiom} + +\begin{axiom}\label{neg_of_zero} + $\neg{\zero} = \zero$. +\end{axiom} + +\begin{definition}\label{minus} + $n - m = n + \neg{m}$. +\end{definition} + +\begin{lemma}\label{minus_in_reals} + Suppose $n,m \in \reals$. + Then $n - m \in \reals$. +\end{lemma} \begin{proof} - Omitted. + \begin{byCase} + \caseOf{$m = \zero$.} Trivial.%Follows by \cref{neg_of_zero,minus,neg,reals_add}. + \caseOf{$m \neq \zero$.} Trivial.% Follows by \cref{neg_of_zero,minus,neg,reals_add}. + \end{byCase} \end{proof} -\begin{proposition}\label{reals_reducion_on_addition} - For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$. +\begin{proposition}\label{negation_of_negation_is_id} + For all $r \in \reals$ we have $\neg{\neg{r}} = r$. \end{proposition} \begin{proof} - Omitted. + Fix $r \in \reals$. + $r + \neg{r} = \zero$. + $\neg{r} + \neg{\neg{r}} = \zero$. + Follows by \cref{reals_reducion_on_addition,neg,reals_axiom_kommu}. \end{proof} +\begin{definition}\label{integers} + $\integers = \{ z \in \reals \mid \exists n \in \naturals. z \in \naturals \lor n + z = \zero\}$. +\end{definition} + +\begin{lemma}\label{n_subset_z} + $\naturals \subseteq \integers$. +\end{lemma} + +\begin{lemma}\label{neg_of_naturals_in_integers} + For all $n \in \naturals$ we have $\neg{n} \in \integers$. +\end{lemma} + +\begin{lemma}\label{integers_eq_naturals_and_negativ_naturals} + $\integers = \{ z \in \reals \mid \exists n \in \naturals. n = z \lor \neg{n} = z\}$. +\end{lemma} +\begin{proof} + Let $P = \{ z \in \reals \mid \exists n \in \naturals. n = z \lor \neg{n} = z\}$. + \begin{byCase} + \caseOf{$\integers \subseteq P$.} Trivial. + \caseOf{$P \subseteq \integers$.} Trivial. + \end{byCase} +\end{proof} + + + + +\subsection{The Rationals} + +%\begin{definition} +% $\inv{x} +%\end{definition} +% +%\begin{axiom} +% For all $x,y \in \reals$ we have $\rfrac{x}{y} \in \reals$. +%\end{axiom} +% +%\begin{definition} +% $\rationals = \{ q \in \reals \mid \exists z \in \integers. \exists n \in \integers. q = \rfrac{z}{n} \}$. +%\end{definition} + + + + + + + + %\begin{signature}\label{invers_is_set} % $\addInv{y}$ is a set. @@ -608,9 +691,3 @@ \begin{proposition}\label{safe} Contradiction. \end{proposition} - -\section{The integers} - -%\begin{definition} -% $\integers = \{z \in \reals \mid z = \zero or \} $. -%\end{definition} \ No newline at end of file diff --git a/source/Syntax/Lexicon.hs b/source/Syntax/Lexicon.hs index 463dd18..65072ee 100644 --- a/source/Syntax/Lexicon.hs +++ b/source/Syntax/Lexicon.hs @@ -95,6 +95,7 @@ builtinMixfix = Seq.fromList $ (HM.fromList <$>) builtinIdentifiers = identifier <$> [ "emptyset" , "naturals" + , "integers" , "rationals" , "reals" , "unit" -- cgit v1.2.3 From 36e03142465e482f2b5506cd35dab5ef9cc9fd66 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 6 Jul 2024 19:53:50 +0200 Subject: Formalisation of rationals. --- library/numbers.tex | 66 +++++++++++++++++++++++++----------------------- source/Syntax/Lexicon.hs | 1 + 2 files changed, 35 insertions(+), 32 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 08cbc70..4ccba67 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -431,7 +431,14 @@ Such that we can introduce the integers and raitionals as smooth as possible. For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$. \end{axiom} +\begin{axiom}\label{zero_less_one} + $\zero < 1$. +\end{axiom} +\begin{axiom}\label{reals_axiom_order_def} + Suppose $x,y,z,w \in \reals$. + If $x + y < z + w$ then $x < z \lor y < w$. +\end{axiom} @@ -501,42 +508,37 @@ Such that we can introduce the integers and raitionals as smooth as possible. \subsection{The Rationals} -%\begin{definition} -% $\inv{x} -%\end{definition} -% -%\begin{axiom} -% For all $x,y \in \reals$ we have $\rfrac{x}{y} \in \reals$. -%\end{axiom} -% -%\begin{definition} -% $\rationals = \{ q \in \reals \mid \exists z \in \integers. \exists n \in \integers. q = \rfrac{z}{n} \}$. -%\end{definition} - - - - +\begin{axiom}\label{invers_reals} + For all $q \in \reals$ we have $\inv{q} \rmul q = 1$. +\end{axiom} +\begin{abbreviation}\label{rfrac} + $\rfrac{x}{y} = x \rmul \inv{y}$. +\end{abbreviation} +\begin{abbreviation}\label{naturalsplus} + $\naturalsPlus = \naturals \setminus \{\zero\}$. +\end{abbreviation} +\begin{definition}\label{rationals} + $\rationals = \{ q \in \reals \mid \exists z \in \integers. \exists n \in \naturalsPlus. q = \rfrac{z}{n} \}$. +\end{definition} -%\begin{signature}\label{invers_is_set} -% $\addInv{y}$ is a set. -%\end{signature} -%\begin{definition}\label{add_inverse} -% $\addInv{y} = \{ x \mid \exists k \in \reals. k + y = \zero \land x \in k\}$. -%\end{definition} - +\subsection{Order on the reals} -%\begin{definition}\label{add_inverse_natural_language} -% $x$ is additiv inverse of $y$ iff $x = \addInv{y}$. -%\end{definition} +\begin{lemma}\label{plus_one_order} + For all $r\in \reals$ we have $ r < r + 1$. +\end{lemma} +\begin{proof} + %Follows by \cref{reals_axiom_one,reals_axiom_order,reals_axiom_order_def,zero_less_one,reals}. +\end{proof} -%\begin{lemma}\label{rminus} -% $x \rminus \addInv{x} = \zero$. -%\end{lemma} +\begin{lemma}\label{negation_and_order} + Suppose $r \in \reals$. + $r \leq \zero$ iff $\zero \leq \neg{r}$. +\end{lemma} @@ -546,7 +548,7 @@ Such that we can introduce the integers and raitionals as smooth as possible. Then $x + y$ is positiv. \end{lemma} \begin{proof} - Omitted. + \end{proof} \begin{lemma}\label{reals_mul_of_positiv} @@ -688,6 +690,6 @@ Such that we can introduce the integers and raitionals as smooth as possible. -\begin{proposition}\label{safe} - Contradiction. -\end{proposition} +%\begin{proposition}\label{safe} +% Contradiction. +%\end{proposition} diff --git a/source/Syntax/Lexicon.hs b/source/Syntax/Lexicon.hs index 65072ee..de6d966 100644 --- a/source/Syntax/Lexicon.hs +++ b/source/Syntax/Lexicon.hs @@ -111,6 +111,7 @@ prefixOps = , ([Just (Command "snd"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "snd")) , ([Just (Command "pow"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "pow")) , ([Just (Command "neg"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "neg")) + , ([Just (Command "inv"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "inv")) , (ConsSymbol, (NonAssoc, "cons")) , (PairSymbol, (NonAssoc, "pair")) -- NOTE Is now defined and hence no longer necessary , (ApplySymbol, (NonAssoc, "apply")) -- cgit v1.2.3 From cbac8ca4a5bf8ff38af3e512956ea1e468965194 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sun, 21 Jul 2024 12:26:20 +0200 Subject: Further Formalisation on numbers --- latex/naproche.sty | 1 + library/everything.tex | 2 +- library/numbers.tex | 165 ++++++++++++++++++++++++++++++++----------------- 3 files changed, 110 insertions(+), 58 deletions(-) (limited to 'library') diff --git a/latex/naproche.sty b/latex/naproche.sty index 476d3dd..7ef2359 100644 --- a/latex/naproche.sty +++ b/latex/naproche.sty @@ -132,6 +132,7 @@ \newcommand{\integers}{\mathcal{Z}} \newcommand{\zero}{0} \newcommand{\one}{1} +\newcommand{\rmul}{\cdot} \newcommand\restrl[2]{{% we make the whole thing an ordinary symbol diff --git a/library/everything.tex b/library/everything.tex index 29b97b7..b966197 100644 --- a/library/everything.tex +++ b/library/everything.tex @@ -29,7 +29,7 @@ \import{topology/basis.tex} \import{topology/disconnection.tex} \import{topology/separation.tex} -%\import{numbers.tex} +\import{numbers.tex} \begin{proposition}\label{trivial} $x = x$. diff --git a/library/numbers.tex b/library/numbers.tex index 4ccba67..0bfae2d 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -18,34 +18,6 @@ $x \rmul y$ is a set. \end{signature} -\begin{axiom}\label{reals_axiom_order} - $\lt[\reals]$ is an order on $\reals$. -\end{axiom} - -\begin{abbreviation}\label{lesseq_on_reals} - $x \leq y$ iff $(x,y) \in \lt[\reals]$. -\end{abbreviation} - -\begin{abbreviation}\label{less_on_reals} - $x < y$ iff it is wrong that $y \leq x$. -\end{abbreviation} - -\begin{abbreviation}\label{greater_on_reals} - $x > y$ iff $y \leq x$. -\end{abbreviation} - -\begin{abbreviation}\label{greatereq_on_reals} - $x \geq y$ iff it is wrong that $x < y$. -\end{abbreviation} - -\begin{abbreviation}\label{is_positiv} - $x$ is positiv iff $x > \zero$. -\end{abbreviation} - - -%Structure TODO: -% Take as may axioms as needed - Tarski Axioms of reals -%Implement Naturals -> Integer -> rationals -> reals \subsection{Creation of natural numbers} @@ -384,14 +356,7 @@ Such that we can introduce the integers and raitionals as smooth as possible. For all $n,m \in \reals$ we have $(n + m) \in \reals$. \end{axiom} -\begin{axiom}\label{reals_axiom_dense} - For all $x,y \in \reals$ if $x < y$ then - there exist $z \in \reals$ such that $x < z$ and $z < y$. -\end{axiom} -\begin{axiom}\label{reals_axiom_assoc} - For all $x,y,z \in \reals$ $(x + y) + z = x + (y + z)$ and $(x \rmul y) \rmul z = x \rmul (y \rmul z)$. -\end{axiom} \begin{axiom}\label{reals_axiom_kommu} For all $x,y \in \reals$ $x + y = y + x$ and $x \rmul y = y \rmul x$. @@ -417,31 +382,72 @@ Such that we can introduce the integers and raitionals as smooth as possible. For all $x,y,z \in \reals$ $x \rmul (y + z) = (x \rmul y) + (x \rmul z)$. \end{axiom} -\begin{axiom}\label{reals_axiom_dedekind_complete} - For all $X,Y,x,y$ such that $X,Y \subseteq \reals$ and $x \in X$ and $y \in Y$ and $x < y$ we have there exist $z \in \reals$ - such that $x < z < y$. -\end{axiom} - \begin{axiom}\label{reals_disstro2} For all $x,y,z \in \reals$ $(y + z) \rmul x = (y \rmul x) + (z \rmul x)$. \end{axiom} - \begin{axiom}\label{reals_reducion_on_addition} For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$. \end{axiom} -\begin{axiom}\label{zero_less_one} - $\zero < 1$. +\begin{abbreviation}\label{rless} + $x < y$ iff $x \rless y$. +\end{abbreviation} + +\begin{abbreviation}\label{less_on_reals} + $x \leq y$ iff it is wrong that $y < x$. +\end{abbreviation} + +\begin{abbreviation}\label{greater_on_reals} + $x > y$ iff $y \leq x$. +\end{abbreviation} + +\begin{abbreviation}\label{greatereq_on_reals} + $x \geq y$ iff it is wrong that $x < y$. +\end{abbreviation} + +\begin{abbreviation}\label{is_positiv} + $x$ is positiv iff $x > \zero$. +\end{abbreviation} + + + + + +\begin{axiom}\label{tarski1} + For all $x,y \in \reals$ we have if $x < y$ then $\lnot y < x$. +\end{axiom} + +\begin{axiom}\label{tarski2} + For all $x,y \in \reals$ if $x < y$ then + there exist $z \in \reals$ such that $x < z$ and $z < y$. +\end{axiom} + +\begin{axiom}\label{tarski3} + For all $X,Y,x,y$ such that $X,Y \subseteq \reals$ and $x \in X$ and $y \in Y$ and $x < y$ we have there exist $z \in \reals$ + such that $x < z < y$. \end{axiom} -\begin{axiom}\label{reals_axiom_order_def} - Suppose $x,y,z,w \in \reals$. - If $x + y < z + w$ then $x < z \lor y < w$. +\begin{axiom}\label{tarski4} + For all $x,y,z \in \reals$ $(x + y) + z = x + (y + z)$ and $(x \rmul y) \rmul z = x \rmul (y \rmul z)$. +\end{axiom} + +\begin{axiom}\label{tarski5} + For all $x,y \in \reals$ we have there exist $z \in \reals$ such that $x + z = y$. \end{axiom} +\begin{axiom}\label{tarski8} + $\zero < 1$. +\end{axiom} + +\begin{axiom}\label{labelordersossss} + For all $x,y \in \reals$ such that $x < y$ we have for all $z \in \reals$ $x + z < y + z$. +\end{axiom} +\begin{axiom}\label{nocheinschoeneslabel} + For all $x,y \in \reals$ such that $\zero < x,y$ we have $\zero < (x \rmul y)$. +\end{axiom} \subsection{The Intergers} @@ -465,8 +471,8 @@ Such that we can introduce the integers and raitionals as smooth as possible. \end{lemma} \begin{proof} \begin{byCase} - \caseOf{$m = \zero$.} Trivial.%Follows by \cref{neg_of_zero,minus,neg,reals_add}. - \caseOf{$m \neq \zero$.} Trivial.% Follows by \cref{neg_of_zero,minus,neg,reals_add}. + \caseOf{$m = \zero$.} Trivial. + \caseOf{$m \neq \zero$.} Trivial. \end{byCase} \end{proof} @@ -496,16 +502,31 @@ Such that we can introduce the integers and raitionals as smooth as possible. $\integers = \{ z \in \reals \mid \exists n \in \naturals. n = z \lor \neg{n} = z\}$. \end{lemma} \begin{proof} - Let $P = \{ z \in \reals \mid \exists n \in \naturals. n = z \lor \neg{n} = z\}$. - \begin{byCase} - \caseOf{$\integers \subseteq P$.} Trivial. - \caseOf{$P \subseteq \integers$.} Trivial. - \end{byCase} + Omitted. \end{proof} +\begin{abbreviation}\label{positiv_real_number} + $r$ is a positiv real number iff $r > \zero$ and $r \in \reals$. +\end{abbreviation} +\begin{proposition}\label{integers_negativ_times_negativ_is_positiv} + Suppose $n,m \in \integers$. + Suppose $n < \zero$ and $m < \zero$. + Then $n \rmul m > \zero$. +\end{proposition} +\begin{proof} + Omitted. + %$n \neq \zero$ and $m \neq \zero$. + %For all $k \in \naturals$ we have $k = \zero \lor k > \zero$. + %There exists $n' \in \reals$ such that $n' + n = \zero$. + %There exists $m' \in \reals$ such that $m' + m = \zero$. +\end{proof} + +%TODO: negativ * negativ = positiv. + + \subsection{The Rationals} \begin{axiom}\label{invers_reals} @@ -520,26 +541,56 @@ Such that we can introduce the integers and raitionals as smooth as possible. $\naturalsPlus = \naturals \setminus \{\zero\}$. \end{abbreviation} -\begin{definition}\label{rationals} +\begin{definition}\label{rationals} %TODO: Vielleicht ist hier die definition das alle inversen von den ganzenzahlen und die ganzenzahlen selbst die rationalen zahlen erzeugen $\rationals = \{ q \in \reals \mid \exists z \in \integers. \exists n \in \naturalsPlus. q = \rfrac{z}{n} \}$. \end{definition} +\begin{abbreviation}\label{nominator} + $z$ is nominator of $q$ iff there exists $n \in \naturalsPlus$ such that $q = \rfrac{z}{n}$. +\end{abbreviation} + +\begin{abbreviation}\label{denominator} + $n$ is denominator of $q$ iff there exists $z \in \integers$ such that $q = \rfrac{z}{n}$. +\end{abbreviation} + +%\begin{proposition}\label{q_is_less_then_p_if_denominator_is_bigger_and_nominator_is_equal} +% Suppose $p,q \in \rationals$. +% Suppose $z \in \naturals$. +% Suppose $p$ is positiv. +% Suppose $q$ is positiv. +% Suppose $z$ is nominator of $p$. +% Suppose $z$ is nominator of $p$. +% Suppose $p'$ is denominator of $p$. +% Suppose $q'$ is denominator of $q$. +% Then $p \leq q$ iff $p' \geq q'$. +%\end{proposition} + + +%\begin{theorem}\label{one_divided_by_n_is_in_zero_to_one} +% For all $n \in \naturalsPlus$ we have $\zero < \rfrac{1}{n} \leq 1$. +%\end{theorem} + +% TODO: Was man noch so beweisen könnte: bruch kürzung, kehrbruch eigenschaften, bruch in bruch vereinfachung, \subsection{Order on the reals} + + \begin{lemma}\label{plus_one_order} For all $r\in \reals$ we have $ r < r + 1$. \end{lemma} \begin{proof} - %Follows by \cref{reals_axiom_one,reals_axiom_order,reals_axiom_order_def,zero_less_one,reals}. + Omitted. \end{proof} \begin{lemma}\label{negation_and_order} Suppose $r \in \reals$. $r \leq \zero$ iff $\zero \leq \neg{r}$. \end{lemma} - +\begin{proof} + Omitted. +\end{proof} \begin{lemma}\label{reals_add_of_positiv} @@ -548,7 +599,7 @@ Such that we can introduce the integers and raitionals as smooth as possible. Then $x + y$ is positiv. \end{lemma} \begin{proof} - + Omitted. \end{proof} \begin{lemma}\label{reals_mul_of_positiv} -- cgit v1.2.3 From 6129ebdf0d8549f3e4d23aa771f2c06020182b7e Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 7 Aug 2024 15:28:45 +0200 Subject: Created first urysohn formalization --- library/numbers.tex | 131 ++++++++++--------- library/topology/continuous.tex | 14 +- library/topology/urysohn.tex | 274 ++++++++++++++++++++++++++++++++++++++++ source/Api.hs | 17 +++ source/CommandLine.hs | 60 ++++++--- source/Provers.hs | 4 +- source/Test/Golden.hs | 1 + 7 files changed, 422 insertions(+), 79 deletions(-) create mode 100644 library/topology/urysohn.tex (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 0bfae2d..7d1b058 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -348,6 +348,9 @@ We seen before that we can proof the common behavior of the naturals. Since we now want to get furhter in a more efficient way, we introduce the basis axioms of the reals. Such that we can introduce the integers and raitionals as smooth as possible. + + +The operations Multiplication and addition is closed in the reals \begin{axiom}\label{reals_rmul} For all $n,m \in \reals$ we have $(n \rmul m) \in \reals$. \end{axiom} @@ -358,10 +361,16 @@ Such that we can introduce the integers and raitionals as smooth as possible. + +Commutivatiy of the standart operations \begin{axiom}\label{reals_axiom_kommu} For all $x,y \in \reals$ $x + y = y + x$ and $x \rmul y = y \rmul x$. \end{axiom} + + + +Existence of one and Zero \begin{axiom}\label{reals_axiom_zero} For all $x \in \reals$ $x + \zero = x$. \end{axiom} @@ -370,6 +379,9 @@ Such that we can introduce the integers and raitionals as smooth as possible. For all $x \in \reals$ we have $x \rmul 1 = x$. \end{axiom} + + +The Existence of Inverse of both operations \begin{axiom}\label{reals_axiom_add_invers} For all $x \in \reals$ there exist $y \in \reals$ such that $x + y = \zero$. \end{axiom} @@ -378,6 +390,9 @@ Such that we can introduce the integers and raitionals as smooth as possible. For all $x \in \reals$ such that $x \neq \zero$ there exist $y \in \reals$ such that $x \rmul y = 1$. \end{axiom} + + +Disstrubitiv Laws of the reals \begin{axiom}\label{reals_axiom_disstro1} For all $x,y,z \in \reals$ $x \rmul (y + z) = (x \rmul y) + (x \rmul z)$. \end{axiom} @@ -386,10 +401,9 @@ Such that we can introduce the integers and raitionals as smooth as possible. For all $x,y,z \in \reals$ $(y + z) \rmul x = (y \rmul x) + (z \rmul x)$. \end{axiom} -\begin{axiom}\label{reals_reducion_on_addition} - For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$. -\end{axiom} + +Definition of the order symbols \begin{abbreviation}\label{rless} $x < y$ iff $x \rless y$. \end{abbreviation} @@ -406,50 +420,50 @@ Such that we can introduce the integers and raitionals as smooth as possible. $x \geq y$ iff it is wrong that $x < y$. \end{abbreviation} + + +Laws of the order on the reals \begin{abbreviation}\label{is_positiv} $x$ is positiv iff $x > \zero$. \end{abbreviation} - - - - -\begin{axiom}\label{tarski1} +\begin{axiom}\label{reals_order} For all $x,y \in \reals$ we have if $x < y$ then $\lnot y < x$. \end{axiom} -\begin{axiom}\label{tarski2} +\begin{axiom}\label{reals_dense} For all $x,y \in \reals$ if $x < y$ then there exist $z \in \reals$ such that $x < z$ and $z < y$. \end{axiom} -\begin{axiom}\label{tarski3} +\begin{axiom}\label{reals_is_eta_zero_set} For all $X,Y,x,y$ such that $X,Y \subseteq \reals$ and $x \in X$ and $y \in Y$ and $x < y$ we have there exist $z \in \reals$ such that $x < z < y$. \end{axiom} -\begin{axiom}\label{tarski4} - For all $x,y,z \in \reals$ $(x + y) + z = x + (y + z)$ and $(x \rmul y) \rmul z = x \rmul (y \rmul z)$. +\begin{axiom}\label{reals_one_bigger_zero} + $\zero < 1$. \end{axiom} -\begin{axiom}\label{tarski5} - For all $x,y \in \reals$ we have there exist $z \in \reals$ such that $x + z = y$. +\begin{axiom}\label{reals_order_behavior_with_addition} + For all $x,y \in \reals$ such that $x < y$ we have for all $z \in \reals$ $x + z < y + z$. \end{axiom} - -\begin{axiom}\label{tarski8} - $\zero < 1$. +\begin{axiom}\label{reals_postiv_mul_is_positiv} + For all $x,y \in \reals$ such that $\zero < x,y$ we have $\zero < (x \rmul y)$. \end{axiom} -\begin{axiom}\label{labelordersossss} - For all $x,y \in \reals$ such that $x < y$ we have for all $z \in \reals$ $x + z < y + z$. +\begin{axiom}\label{reals_postiv_mul_negativ_is_negativ} + For all $x,y \in \reals$ such that $x < \zero < y$ we have $(x \rmul y) < \zero$. \end{axiom} -\begin{axiom}\label{nocheinschoeneslabel} - For all $x,y \in \reals$ such that $\zero < x,y$ we have $\zero < (x \rmul y)$. +\begin{axiom}\label{reals_negativ_mul_is_negativ} + For all $x,y \in \reals$ such that $x,y < \zero$ we have $\zero < (x \rmul y)$. \end{axiom} + + \subsection{The Intergers} @@ -480,10 +494,11 @@ Such that we can introduce the integers and raitionals as smooth as possible. For all $r \in \reals$ we have $\neg{\neg{r}} = r$. \end{proposition} \begin{proof} - Fix $r \in \reals$. - $r + \neg{r} = \zero$. - $\neg{r} + \neg{\neg{r}} = \zero$. - Follows by \cref{reals_reducion_on_addition,neg,reals_axiom_kommu}. + Omitted. +% Fix $r \in \reals$. +% $r + \neg{r} = \zero$. +% $\neg{r} + \neg{\neg{r}} = \zero$. +% Follows by \cref{neg,reals_axiom_kommu}. \end{proof} \begin{definition}\label{integers} @@ -511,22 +526,6 @@ Such that we can introduce the integers and raitionals as smooth as possible. -\begin{proposition}\label{integers_negativ_times_negativ_is_positiv} - Suppose $n,m \in \integers$. - Suppose $n < \zero$ and $m < \zero$. - Then $n \rmul m > \zero$. -\end{proposition} -\begin{proof} - Omitted. - %$n \neq \zero$ and $m \neq \zero$. - %For all $k \in \naturals$ we have $k = \zero \lor k > \zero$. - %There exists $n' \in \reals$ such that $n' + n = \zero$. - %There exists $m' \in \reals$ such that $m' + m = \zero$. -\end{proof} - -%TODO: negativ * negativ = positiv. - - \subsection{The Rationals} \begin{axiom}\label{invers_reals} @@ -542,7 +541,7 @@ Such that we can introduce the integers and raitionals as smooth as possible. \end{abbreviation} \begin{definition}\label{rationals} %TODO: Vielleicht ist hier die definition das alle inversen von den ganzenzahlen und die ganzenzahlen selbst die rationalen zahlen erzeugen - $\rationals = \{ q \in \reals \mid \exists z \in \integers. \exists n \in \naturalsPlus. q = \rfrac{z}{n} \}$. + $\rationals = \{ q \in \reals \mid \exists z \in \integers. \exists n \in (\integers \setminus \{\zero\}). q = \rfrac{z}{n} \}$. \end{definition} \begin{abbreviation}\label{nominator} @@ -553,23 +552,41 @@ Such that we can introduce the integers and raitionals as smooth as possible. $n$ is denominator of $q$ iff there exists $z \in \integers$ such that $q = \rfrac{z}{n}$. \end{abbreviation} -%\begin{proposition}\label{q_is_less_then_p_if_denominator_is_bigger_and_nominator_is_equal} -% Suppose $p,q \in \rationals$. -% Suppose $z \in \naturals$. -% Suppose $p$ is positiv. -% Suppose $q$ is positiv. -% Suppose $z$ is nominator of $p$. -% Suppose $z$ is nominator of $p$. -% Suppose $p'$ is denominator of $p$. -% Suppose $q'$ is denominator of $q$. -% Then $p \leq q$ iff $p' \geq q'$. -%\end{proposition} +\begin{theorem}\label{one_divided_by_n_is_in_zero_to_one} + For all $n \in \naturalsPlus$ we have $\zero < \rfrac{1}{n} \leq 1$. +\end{theorem} +\begin{proof} + Omitted. +\end{proof} +\begin{lemma}\label{fraction_kuerzung} %TODO: Englischen namen rausfinden + For all $n,m,k \in \reals$ we have $\rfrac{n \rmul k}{m \rmul k} = \rfrac{n}{m}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} -%\begin{theorem}\label{one_divided_by_n_is_in_zero_to_one} -% For all $n \in \naturalsPlus$ we have $\zero < \rfrac{1}{n} \leq 1$. -%\end{theorem} +\begin{lemma}\label{fraction_swap} + For all $n,m,k,l \in \reals$ we have $\rfrac{\rfrac{n}{m}}{\rfrac{k}{l}}=\rfrac{n \rmul l}{m \rmul k}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} +\begin{definition}\label{divisor} + $g$ is a integral divisor of $a$ iff $g \in \naturals$ and there exist $b \in \integers$ such that $g \rmul b = a$. +\end{definition} + + + + +%\begin{abbreviation}\label{gcd} +% $g$ is greatest common divisor of $\{a,b\}$ iff +% $g$ is a integral divisor of $a$ +% and $g$ is a integral divisor of $b$ +% and for all $g'$ such that $g'$ is a integral divisor of $a$ +% and $g'$ is a integral divisor of $b$ we have $g' \leq g$. +%\end{abbreviation} % TODO: Was man noch so beweisen könnte: bruch kürzung, kehrbruch eigenschaften, bruch in bruch vereinfachung, @@ -652,7 +669,7 @@ Such that we can introduce the integers and raitionals as smooth as possible. \begin{lemma}\label{order_reals_lemma3} Let $x,y,z \in \reals$. - Suppose $\zero < x$. + Suppose $\zero > x$. Suppose $y < z$. Then $(x \rmul z) < (x \rmul y)$. \end{lemma} diff --git a/library/topology/continuous.tex b/library/topology/continuous.tex index 6a32353..a9bc58e 100644 --- a/library/topology/continuous.tex +++ b/library/topology/continuous.tex @@ -1,5 +1,7 @@ \import{topology/topological-space.tex} \import{relation.tex} +\import{function.tex} +\import{set.tex} \begin{definition}\label{continuous} $f$ is continuous iff for all $U \in \opens[Y]$ we have $\preimg{f}{U} \in \opens[X]$. @@ -8,18 +10,25 @@ \begin{proposition}\label{continuous_definition_by_closeds} Let $X$ be a topological space. Let $Y$ be a topological space. + Let $f \in \funs{X}{Y}$. Then $f$ is continuous iff for all $U \in \closeds{Y}$ we have $\preimg{f}{U} \in \closeds{X}$. \end{proposition} \begin{proof} Omitted. - %We show that if $f$ is continuous then for all $U \in \closeds{Y}$ we have $\preimg{f}{U} \in \closeds{X}$. + %We show that if $f$ is continuous then for all $U \in \closeds{Y}$ such that $U \neq \emptyset$ we have $\preimg{f}{U} \in \closeds{X}$. %\begin{subproof} % Suppose $f$ is continuous. % Fix $U \in \closeds{Y}$. % $\carrier[Y] \setminus U$ is open in $Y$. % Then $\preimg{f}{(\carrier[Y] \setminus U)}$ is open in $X$. % Therefore $\carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$ is closed in $X$. - % $\carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)} \subseteq \preimg{f}{U}$. + % We show that $\carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)} \subseteq \preimg{f}{U}$. + % \begin{subproof} + % It suffices to show that for all $x \in \carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$ we have $x \in \preimg{f}{U}$. + % Fix $x \in \carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$. + % Take $y \in \carrier[Y]$ such that $f(x)=y$. + % It suffices to show that $y \in U$. + % \end{subproof} % $\preimg{f}{U} \subseteq \carrier[X] \setminus \preimg{f}{(\carrier[Y] \setminus U)}$. %\end{subproof} %We show that if for all $U \in \closeds{Y}$ we have $\preimg{f}{U} \in \closeds{X}$ then $f$ is continuous. @@ -27,3 +36,4 @@ % Omitted. %\end{subproof} \end{proof} + diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex new file mode 100644 index 0000000..65457ea --- /dev/null +++ b/library/topology/urysohn.tex @@ -0,0 +1,274 @@ +\import{topology/topological-space.tex} +\import{topology/separation.tex} +\import{topology/continuous.tex} +\import{numbers.tex} +\import{function.tex} +\import{set.tex} +\import{cardinal.tex} + +\section{Urysohns Lemma} +% In this section we want to proof Urysohns lemma. +% We try to follow the proof of Klaus Jänich in his book. TODO: Book reference +% The Idea is to construct staircase functions as a chain. +% The limit of our chain turns out to be our desired continuous function from a topological space $X$ to $[0,1]$. +% With the property that \[f\mid_{A}=1 \land f\mid_{B}=0\] for \[A,B\] closed sets. + +%Chains of sets. + +The first tept will be a formalisation of chain constructions. + +\subsection{Chains of sets} +% Assume $A,B$ are subsets of a topological space $X$. + +% As Jänich propose we want a special property on chains of subsets. +% We need a rising chain of subsets $\mathfrak{A} = (A_{0}, ... ,A_{r})$ of $A$, i.e. +% \begin{align} +% A = A_{0} \subset A_{1} \subset ... \subset A_{r} \subset X\setminus B +% \end{align} +% such that for all elements in this chain following holds, +% $\overline{A_{i-1}} \subset \interior{A_{i}}$. +% In this case we call the chain legal. + +\begin{definition}\label{one_to_n_set} + $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. +\end{definition} + +\begin{definition}\label{cardinality} + $X$ has cardinality of $n$ iff + $n \in \naturals$ and there exists $b$ such that $b$ is a bijection from $\seq{1}{n}$ to $X$. +\end{definition} + + + +\begin{struct}\label{sequence} + A sequence $X$ is a onesorted structure equipped with + \begin{enumerate} + \item $\index$ + \item $\indexset$ + \end{enumerate} + such that + \begin{enumerate} + \item\label{indexset_is_subset_naturals} $\indexset[X] \subseteq \naturals$. + \item\label{index_is_bijection} $\index[X]$ is a bijection from $\indexset[X]$ to $\carrier[X]$. + \end{enumerate} +\end{struct} + +\begin{definition}\label{cahin_of_subsets} + $C$ is a chain of subsets iff + $C$ is a sequence and for all $n,m \in \indexset[C]$ such that $n < m$ we have $\index[C](n) \subseteq \index[C](m)$. +\end{definition} + +\begin{definition}\label{chain_of_n_subsets} + $C$ is a chain of $n$ subsets iff + $C$ is a chain of subsets and $n \in \indexset[C]$ + and for all $m \in \naturals$ such that $m \leq n$ we have $m \in \indexset[C]$. +\end{definition} + + + +% TODO: The Notion above should be generalised to sequences since we need them as well for our limit +% and also for the subproof of continuity of the limit. + + +% \begin{definition}\label{legal_chain} +% $C$ is a legal chain of subsets of $X$ iff +% $C \subseteq \pow{X}$. %and +% %there exist $f \in \funs{C}{\naturals}$ such that +% %for all $x,y \in C$ we have if $f(x) < f(y)$ then $x \subset y \land \closure{x} \subset \interior{y}$. +% \end{definition} + +% TODO: We need a notion for declarinf new properties to existing thing. +% +% The following gives a example and a wish want would be nice to have: +% "A (structure) is called (adjectiv of property), if" +% +% This should then be use als follows: +% Let $X$ be a (adjectiv_1) ... (adjectiv_n) (structure_word). +% Which should be translated to fol like this: +% ?[X]: is_structure(X) & is_adjectiv_1(X) & ... & is_adjectiv_n(X) +% --------------------------------------------------------------- + + + +\subsection{staircase function} + +\begin{definition}\label{intervalclosed} + $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. +\end{definition} + + +\begin{struct}\label{staircase_function} + A staircase function $f$ is a onesorted structure equipped with + \begin{enumerate} + \item $\chain$ + \end{enumerate} + such that + \begin{enumerate} + \item \label{staircase_is_function} $f$ is a function to $\intervalclosed{\zero}{1}$. + \item \label{staircase_domain} $\dom{f}$ is a topological space. + \item \label{staricase_def_chain} $C$ is a chain of subsets. + \item \label{staircase_chain_is_in_domain} for all $x \in C$ we have $x \subseteq \dom{f}$. + \item \label{staircase_behavoir_index_zero} $f(\index[C](1))= 1$. + \item \label{staircase_behavoir_index_n} $f(\dom{f}\setminus \unions{C}) = \zero$. + \end{enumerate} +\end{struct} + +\begin{definition}\label{legal_staircase} + $f$ is a legal staircase function iff + $f$ is a staircase function and + for all $n,m \in \indexset[\chain[f]]$ such that $n \leq m$ we have $f(\index[\chain[f]](n)) \leq f(\index[\chain[f]](m))$. +\end{definition} + +\begin{abbreviation}\label{urysohnspace} + $X$ is a urysohn space iff + $X$ is a topological space and + for all $A,B \in \closeds{X}$ such that $A \inter B = \emptyset$ + we have there exist $A',B' \in \opens[X]$ + such that $A \subseteq A'$ and $B \subseteq B'$ and $A' \inter B' = \emptyset$. +\end{abbreviation} + +\begin{definition}\label{urysohnchain} + $C$ is a urysohnchain in $X$ of cardinality $k$ iff %<---- TODO: cardinality unused! + $C$ is a chain of subsets and + for all $A \in C$ we have $A \subseteq X$ and + for all $n,m \in \indexset[C]$ such that $n < m$ we have $\closure{\index[C](n)}{X} \subseteq \interior{\index[C](m)}{X}$. +\end{definition} + +\begin{abbreviation}\label{infinte_sequence} + $S$ is a infinite sequence iff $S$ is a sequence and $\indexset[S]$ is infinite. +\end{abbreviation} + +\begin{definition}\label{infinite_product} + $X$ is the infinite product of $Y$ iff + $X$ is a infinite sequence and for all $i \in \indexset[X]$ we have $\index[X](i) = Y$. +\end{definition} + +\begin{definition}\label{refinmant} + $C$ is a refinmant of $C'$ iff for all $x \in C'$ we have $x \in C$ and + for all $y \in C'$ such that $y \subset x$ we have there exist $c \in C$ such that $y \subset c \subset x$ + and for all $g \in C$ such that $g \neq c$ we have not $y \subset g \subset x$. +\end{definition} + + +% The next thing we need to define is the uniform staircase function. +% This function has it's domain in $X$ and maps to the closed interval $[0,1]$. +% These functions should behave als follows, +% \begin{align} +% &f(A_{0}) = 1 &\text{consant} \\ +% &f(A_{k} \setminus A_{k+1}) = 1-\frac{k}{r} &\text{constant.} +% \end{align} + +% We then prove that for any given $r$ we find a repolished chain, +% which contains the $A_{i}$ and this replished chain is also legal. +% +% The proof will be finished by taking the limit on $f_{n}$ with $f_{n}$ +% be a staircase function with $n$ many refinemants. +% This limit will be continuous and we would be done. + + +% TODO: Since we want to prove that $f$ is continus, we have to formalize that +% \reals with the usual topology is a topological space. +% ------------------------------------------------------------- + +\begin{theorem}\label{urysohn} + Let $X$ be a urysohn space. + Suppose $A,B \subseteq \closeds{X}$. + Suppose $A \inter B$ is empty. + There exist $f$ such that $f \in \funs{X}{\intervalclosed{\zero}{1}}$ + and $f(A) = 1$ and $f(B)= 0$ and $f$ is continuous. +\end{theorem} +\begin{proof} + We show that for all $n \in \naturals$ we have + if there exist $C$ such that $C$ is a urysohnchain in $X$ of cardinality $n$ + then there exist $C'$ such that $C'$ is a urysohnchain in $X$ of cardinality $n+1$ + and $C'$ is a refinmant of $C$. + \begin{subproof} + Omitted. + \end{subproof} + + + + Take $P$ such that $P$ is a infinite sequence and $\indexset[P] = \naturals$ and for all $i \in \indexset[P]$ we have $\index[P](i) = \pow{X}$. + + We show that there exist $\zeta$ such that $\zeta$ is a infinite sequence + and for all $i \in \indexset[\zeta]$ we have + $\index[\zeta](i)$ is a urysohnchain in $X$ of cardinality $i$ + and $A \subseteq \index[\zeta](i)$ + and $\index[\zeta](i) \subseteq (X \setminus B)$ + and for all $j \in \indexset[\zeta]$ such that + $j < i$ we have for all $x \in \index[\zeta](j)$ we have $x \in \index[\zeta](i)$. + \begin{subproof} + Omitted. + \end{subproof} + + + + + + + + + We show that there exist $g \in \funs{X}{\intervalclosed{\zero}{1}}$ such that $g(A)=1$ and $g(X\setminus A) = \zero$. + \begin{subproof} + Omitted. + \end{subproof} + $g$ is a staircase function and $\chain[g] = C$. + $g$ is a legal staircase function. + + + We show that there exist $f$ such that $f \in \funs{X}{\intervalclosed{\zero}{1}}$ + and $f(A) = 1$ and $f(B)= 0$ and $f$ is continuous. + \begin{subproof} + Omitted. + \end{subproof} + + + %We show that for all $n \in \nautrals$ we have $C_{n}$ a legal chain of subsets of $X$. + %\begin{subproof} + % Omitted. + %\end{subproof} + + %Proof Sheme Idea: + % -We proof for n=1 that C_{n} is a chain and legal + % -Then by induction with P(n+1) is refinmant of P(n) + % -Therefore we have a increing refinmant of these Chains such that our limit could even apply + % --------------------------------------------------------- + + %We show that there exist $f \in \funs{\naturals}{\funs{X}{\intervalclosed{0}{1}}}$ such that $f(n)$ is a staircase function. %TODO: Define Staircase function + %\begin{subproof} + % Omitted. + %\end{subproof} + + + % Formalization idea of enumarted sequences: + % - Define a enumarted sequnecs as a set A with a bijection between A and E \in \pow{\naturals} + % - This should give all finite and infinte enumarable sequences + % - Introduce a notion for the indexing of these enumarable sequences. + % - Then we can define the limit of a enumarted sequence of functions. + % --------------------------------------------------------- + % + % Here we need a limit definition for sequences of functions + %We show that there exist $F \in \funs{X}{\intervalclosed{0}{1}}$ such that $F = \limit{set of the staircase functions}$ + %\begin{subproof} + % Omitted. + %\end{subproof} + % + %We show that $F(A) = 1$. + %\begin{subproof} + % Omitted. + %\end{subproof} + % + %We show that $F(B) = 0$. + %\begin{subproof} + % Omitted. + %\end{subproof} + % + %We show that $F$ is continuous. + %\begin{subproof} + % Omitted. + %\end{subproof} +\end{proof} + +%\begin{theorem}\label{safe} +% Contradiction. +%\end{theorem} diff --git a/source/Api.hs b/source/Api.hs index 95d5c8c..f597153 100644 --- a/source/Api.hs +++ b/source/Api.hs @@ -17,6 +17,7 @@ module Api , dumpTask , verify, ProverAnswer(..), VerificationResult(..) , exportMegalodon + , showFailedTask , WithCache(..) , WithFilter(..) , WithOmissions(..) @@ -29,6 +30,7 @@ module Api , WithParseOnly(..) , Options(..) , WithDumpPremselTraining(..) + , WithFailList(..) ) where @@ -65,6 +67,7 @@ import Text.Megaparsec hiding (parse, Token) import UnliftIO import UnliftIO.Directory import UnliftIO.Environment +import Syntax.Abstract (Formula) -- | Follow all @\\import@ statements recursively and build a theory graph from them. -- The @\\import@ statements should be on their own separate line and precede all @@ -285,6 +288,17 @@ exportMegalodon file = do pure (Megalodon.encodeBlocks lexicon blocks) + +-- | This could be expandend with the dump case, with dump off just this and if dump is on it could show the number off the task. For quick use +showFailedTask :: (a, ProverAnswer) -> IO() +showFailedTask (_, Yes ) = Text.putStrLn "" +showFailedTask (_, No tptp) = Text.putStrLn (Text.pack ("Prover found countermodel: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +showFailedTask (_, ContradictoryAxioms tptp) = Text.putStrLn (Text.pack ("Contradictory axioms: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +showFailedTask (_, Uncertain tptp) = Text.putStrLn (Text.pack ("Out of resources: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +showFailedTask (_, Error _ tptp _) = Text.putStrLn (Text.pack ("Error at: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +--showFailedTask (_, _) = Text.putStrLn "Error!" + + -- | Should we use caching? data WithCache = WithoutCache | WithCache deriving (Show, Eq) @@ -312,6 +326,8 @@ pattern WithoutDump = WithDump "" data WithParseOnly = WithoutParseOnly | WithParseOnly deriving (Show, Eq) +data WithFailList = WithoutFailList | WithFailList deriving (Show, Eq) + data Options = Options { inputPath :: FilePath @@ -327,4 +343,5 @@ data Options = Options , withVersion :: WithVersion , withMegalodon :: WithMegalodon , withDumpPremselTraining :: WithDumpPremselTraining + , withFailList :: WithFailList } diff --git a/source/CommandLine.hs b/source/CommandLine.hs index a9fb00d..47efe22 100644 --- a/source/CommandLine.hs +++ b/source/CommandLine.hs @@ -18,7 +18,7 @@ import UnliftIO import UnliftIO.Directory import UnliftIO.Environment (lookupEnv) import System.FilePath.Posix - +import qualified Tptp.UnsortedFirstOrder as Syntax.Internal runCommandLine :: IO () runCommandLine = do @@ -69,23 +69,42 @@ run = do WithDefaultProver -> Provers.vampire vampirePathPath let proverInstance = prover Provers.Silent (withTimeLimit opts) (withMemoryLimit opts) result <- verify proverInstance (inputPath opts) - liftIO case result of - VerificationSuccess -> (Text.putStrLn "Verification successful.") - VerificationFailure [] -> error "Empty verification fail" - VerificationFailure ((_, proverAnswer) : _) -> case proverAnswer of - Yes -> - skip - No tptp -> do - putStrLn "Verification failed: prover found countermodel" - Text.hPutStrLn stderr tptp - ContradictoryAxioms tptp -> do - putStrLn "Verification failed: contradictory axioms" - Text.hPutStrLn stderr tptp - Uncertain tptp -> do - putStrLn "Verification failed: out of resources" - Text.hPutStrLn stderr tptp - Error err -> - Text.hPutStrLn stderr err + case withFailList opts of + WithoutFailList -> liftIO case result of + VerificationSuccess -> putStrLn "Verification successful." + VerificationFailure [] -> error "Empty verification fail" + VerificationFailure ((_, proverAnswer) : _) -> case proverAnswer of + Yes -> + skip + No tptp -> do + putStrLn "Verification failed: prover found countermodel" + Text.hPutStrLn stderr tptp + ContradictoryAxioms tptp -> do + putStrLn "Verification failed: contradictory axioms" + Text.hPutStrLn stderr tptp + Uncertain tptp -> do + putStrLn "Verification failed: out of resources" + Text.hPutStrLn stderr tptp + Error err tptp task -> do + putStr "Error at:" + Text.putStrLn task + Text.putStrLn err + Text.putStrLn tptp + + WithFailList -> liftIO case result of + VerificationSuccess -> putStrLn "Verification successful." + VerificationFailure [] -> error "Empty verification fail" + VerificationFailure fails -> do + putStrLn "Following task couldn't be solved by the ATP: " + traverse_ showFailedTask fails + Text.hPutStrLn stderr "Don't give up!" + + + + + + + @@ -104,6 +123,7 @@ parseOptions = do withVersion <- withVersionParser withMegalodon <- withMegalodonParser withDumpPremselTraining <- withDumpPremselTrainingParser + withFailList <- withFailListParser pure Options{..} withTimeLimitParser :: Parser Provers.TimeLimit @@ -160,3 +180,7 @@ withDumpPremselTrainingParser = flag' WithDumpPremselTraining (long "premseldump withMegalodonParser :: Parser WithMegalodon withMegalodonParser = flag' WithMegalodon (long "megalodon" <> help "Export to Megalodon.") <|> pure WithoutMegalodon -- Default to using the cache. + +withFailListParser :: Parser WithFailList +withFailListParser = flag' WithFailList (long "list_fails" <> help "Will list all unproven task with possible reason of failure.") + <|> pure WithoutFailList -- Default to using the cache. \ No newline at end of file diff --git a/source/Provers.hs b/source/Provers.hs index 203ee82..37e02ca 100644 --- a/source/Provers.hs +++ b/source/Provers.hs @@ -110,7 +110,7 @@ data ProverAnswer | No Text | ContradictoryAxioms Text | Uncertain Text - | Error Text + | Error Text Text Text deriving (Show, Eq) nominalDiffTimeToText :: NominalDiffTime -> Text @@ -163,4 +163,4 @@ recognizeAnswer Prover{..} task tptp answer answerErr = | saidNo -> No tptp | doesNotKnow -> Uncertain tptp | warned -> ContradictoryAxioms tptp - | otherwise -> Error (answer <> answerErr) + | otherwise -> Error (answer <> answerErr) tptp (Text.pack(show (taskConjectureLabel task))) diff --git a/source/Test/Golden.hs b/source/Test/Golden.hs index 705aaa5..c01c249 100644 --- a/source/Test/Golden.hs +++ b/source/Test/Golden.hs @@ -39,6 +39,7 @@ testOptions = Api.Options , withTimeLimit = Provers.defaultTimeLimit , withVersion = Api.WithoutVersion , withMegalodon = Api.WithoutMegalodon + , withFailList = Api.WithoutFailList } goldenTests :: IO TestTree -- cgit v1.2.3 From 65e5a741655d8339ad5763e365ea2addd2e96e51 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Fri, 9 Aug 2024 15:15:49 +0200 Subject: put cardinalit to the right place --- library/cardinal.tex | 6 ++++++ library/topology/urysohn.tex | 39 ++++++++++++++++++++++++++++++++------- 2 files changed, 38 insertions(+), 7 deletions(-) (limited to 'library') diff --git a/library/cardinal.tex b/library/cardinal.tex index d0dec59..8691b30 100644 --- a/library/cardinal.tex +++ b/library/cardinal.tex @@ -3,6 +3,7 @@ \import{set.tex} \import{ordinal.tex} \import{function.tex} +\import{numbers.tex} \begin{definition}\label{finite} $X$ is finite iff there exists a natural number $k$ such that @@ -12,3 +13,8 @@ \begin{abbreviation}\label{infinite} $X$ is infinite iff $X$ is not finite. \end{abbreviation} + +\begin{definition}\label{cardinality} + $X$ has cardinality $k$ iff $k$ is a natural number and + there exists a bijection from $k$ to $X$. +\end{definition} diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 65457ea..c795db8 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -33,10 +33,6 @@ The first tept will be a formalisation of chain constructions. $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. \end{definition} -\begin{definition}\label{cardinality} - $X$ has cardinality of $n$ iff - $n \in \naturals$ and there exists $b$ such that $b$ is a bijection from $\seq{1}{n}$ to $X$. -\end{definition} @@ -134,6 +130,13 @@ The first tept will be a formalisation of chain constructions. for all $n,m \in \indexset[C]$ such that $n < m$ we have $\closure{\index[C](n)}{X} \subseteq \interior{\index[C](m)}{X}$. \end{definition} +\begin{definition}\label{urysohnchain_without_cardinality} + $C$ is a urysohnchain in $X$ iff + $C$ is a chain of subsets and + for all $A \in C$ we have $A \subseteq X$ and + for all $n,m \in \indexset[C]$ such that $n < m$ we have $\closure{\index[C](n)}{X} \subseteq \interior{\index[C](m)}{X}$. +\end{definition} + \begin{abbreviation}\label{infinte_sequence} $S$ is a infinite sequence iff $S$ is a sequence and $\indexset[S]$ is infinite. \end{abbreviation} @@ -149,6 +152,26 @@ The first tept will be a formalisation of chain constructions. and for all $g \in C$ such that $g \neq c$ we have not $y \subset g \subset x$. \end{definition} +\begin{abbreviation}\label{two} + $\two = \suc{1}$. +\end{abbreviation} + +\begin{lemma}\label{two_in_reals} + $\two \in \reals$. +\end{lemma} + +\begin{lemma}\label{two_in_naturals} + $\two \in \naturals$. +\end{lemma} + +\begin{inductive}\label{power_of_two} + Define $\powerOfTwoSet \subseteq \naturals$. + \begin{enumerate} + \item $\two \in \powerOfTwoSet$. + \item If $n \in \powerOfTwoSet$, then $n \rmul \two \in \powerOfTwoSet$. + \end{enumerate} + +\end{inductive} % The next thing we need to define is the uniform staircase function. % This function has it's domain in $X$ and maps to the closed interval $[0,1]$. @@ -186,6 +209,8 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} + There exist $\eta$ such that $\eta$ is a urysohnchain in $X$ and $\eta =\{A, (x \setminus B)\}$. + Take $P$ such that $P$ is a infinite sequence and $\indexset[P] = \naturals$ and for all $i \in \indexset[P]$ we have $\index[P](i) = \pow{X}$. @@ -269,6 +294,6 @@ The first tept will be a formalisation of chain constructions. %\end{subproof} \end{proof} -%\begin{theorem}\label{safe} -% Contradiction. -%\end{theorem} +\begin{theorem}\label{safe} + Contradiction. +\end{theorem} -- cgit v1.2.3 From 951dd2b195524cb2d8f125abaa4936abaf5ae582 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Fri, 9 Aug 2024 15:33:30 +0200 Subject: fixed axiomatic error --- library/numbers.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 7d1b058..f7f6c2c 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -409,15 +409,15 @@ Definition of the order symbols \end{abbreviation} \begin{abbreviation}\label{less_on_reals} - $x \leq y$ iff it is wrong that $y < x$. + $x \leq y$ iff $x < y \lor x = y$. \end{abbreviation} \begin{abbreviation}\label{greater_on_reals} - $x > y$ iff $y \leq x$. + $x > y$ iff $y < x$. \end{abbreviation} \begin{abbreviation}\label{greatereq_on_reals} - $x \geq y$ iff it is wrong that $x < y$. + $x \geq y$ iff $y \leq x$. \end{abbreviation} -- cgit v1.2.3 From d49a4f85a928c4e93c6cbb2ebada5875e12c4b4f Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 10 Aug 2024 01:25:48 +0200 Subject: more ur Date: Sat, 10 Aug 2024 16:02:43 +0200 Subject: more urysohn --- library/topology/urysohn.tex | 18 +++++++++++++++--- 1 file changed, 15 insertions(+), 3 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index bb28116..3152de7 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -262,12 +262,26 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} +\begin{definition}\label{minimum} + $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. +\end{definition} + +\begin{definition}\label{maximum} + $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. +\end{definition} + \begin{proposition}\label{urysohnchain_induction_step_existence} Let $X$ be a urysohn space. Suppose $U$ is a urysohnchain in $X$. Then there exist $U'$ such that $U'$ is a refinmant of $U$ and $U'$ is a urysohnchain in $X$. \end{proposition} \begin{proof} + % U = ( A_{0}, A_{1}, A_{2}, ... , A_{n-1}, A_{n}) + % U' = ( A_{0}, A'_{1}, A_{1}, A'_{2}, A_{2}, ... A_{n-1}, A'_{n}, A_{n}) + + Let $m = \max{\indexset[U]}$. + For all $n \in (\indexset[U] \setminus \{m\})$ we have there exist $C \subseteq X$ + such that $\closure{\index[U](n)}{X} \subseteq \interior{C}{X} \subseteq \closure{C}{X} \subseteq \interior{\index[U](n+1)}{X}$. %\begin{definition}\label{refinmant} @@ -276,9 +290,7 @@ The first tept will be a formalisation of chain constructions. % we have there exist $c \in C'$ such that $y \subset c \subset x$ % and for all $g \in C'$ such that $g \neq c$ we have not $y \subset g \subset x$. %\end{definition} - - - + Omitted. \end{proof} -- cgit v1.2.3 From 34aad4be8fb8433f0205517732da0d449d077572 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 10 Aug 2024 19:18:39 +0200 Subject: more more urysohn --- library/cardinal.tex | 2 + library/topology/urysohn.tex | 341 ++++++++++++++++++++++++------------------- 2 files changed, 190 insertions(+), 153 deletions(-) (limited to 'library') diff --git a/library/cardinal.tex b/library/cardinal.tex index 8691b30..044e5d1 100644 --- a/library/cardinal.tex +++ b/library/cardinal.tex @@ -18,3 +18,5 @@ $X$ has cardinality $k$ iff $k$ is a natural number and there exists a bijection from $k$ to $X$. \end{definition} + + diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 3152de7..028e10f 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -198,60 +198,82 @@ The first tept will be a formalisation of chain constructions. Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. - Suppose $U = \{A,(X \setminus B)\}$. - Suppose $\indexset[U]= \{\zero, 1\}$. - Suppose $\index[U](\zero) = A$. - Suppose $\index[U](1) = (X \setminus B)$. - Then $U$ is a urysohnchain in $X$. + Then there exist $U$ + such that $\carrier[U] = \{A,(\carrier[X] \setminus B)\}$ + and $\indexset[U]= \{\zero, 1\}$ + and $\index[U](\zero) = A$ + and $\index[U](1) = (\carrier[X] \setminus B)$. + %$U$ is a urysohnchain in $X$. \end{proposition} \begin{proof} - We show that $U$ is a sequence. - \begin{subproof} - Omitted. - \end{subproof} - We show that $A \subseteq (X \setminus B)$. - \begin{subproof} - Omitted. - \end{subproof} + Omitted. - We show that $U$ is a chain of subsets. - \begin{subproof} - For all $n \in \indexset[U]$ we have $n = \zero \lor n = 1$. - It suffices to show that for all $n \in \indexset[U]$ we have - for all $m \in \indexset[U]$ such that - $n < m$ we have $\index[U](n) \subseteq \index[U](m)$. - Fix $n \in \indexset[U]$. - Fix $m \in \indexset[U]$. - \begin{byCase} - \caseOf{$n = 1$.} Trivial. - \caseOf{$n = \zero$.} - \begin{byCase} - \caseOf{$m = \zero$.} Trivial. - \caseOf{$m = 1$.} Omitted. - \end{byCase} - \end{byCase} - \end{subproof} + % We show that $U$ is a sequence. + % \begin{subproof} + % Omitted. + % \end{subproof} +% + % We show that $A \subseteq (\carrier[X] \setminus B)$. + % \begin{subproof} + % Omitted. + % \end{subproof} +% + % We show that $U$ is a chain of subsets. + % \begin{subproof} + % For all $n \in \indexset[U]$ we have $n = \zero \lor n = 1$. + % It suffices to show that for all $n \in \indexset[U]$ we have + % for all $m \in \indexset[U]$ such that + % $n < m$ we have $\index[U](n) \subseteq \index[U](m)$. + % Fix $n \in \indexset[U]$. + % Fix $m \in \indexset[U]$. + % \begin{byCase} + % \caseOf{$n = 1$.} Trivial. + % \caseOf{$n = \zero$.} + % \begin{byCase} + % \caseOf{$m = \zero$.} Trivial. + % \caseOf{$m = 1$.} Omitted. + % \end{byCase} + % \end{byCase} + % \end{subproof} +% + % $A \subseteq X$. + % $(X \setminus B) \subseteq X$. + % We show that for all $x \in U$ we have $x \subseteq X$. + % \begin{subproof} + % Omitted. + % \end{subproof} +% + % We show that $\closure{A}{X} \subseteq \interior{(X \setminus B)}{X}$. + % \begin{subproof} + % Omitted. + % \end{subproof} + % We show that for all $n,m \in \indexset[U]$ such that $n < m$ we have + % $\closure{\index[U](n)}{X} \subseteq \interior{\index[U](m)}{X}$. + % \begin{subproof} + % Omitted. + % \end{subproof} - $A \subseteq X$. - $(X \setminus B) \subseteq X$. - We show that for all $x \in U$ we have $x \subseteq X$. - \begin{subproof} - Omitted. - \end{subproof} + +\end{proof} - We show that $\closure{A}{X} \subseteq \interior{(X \setminus B)}{X}$. - \begin{subproof} - Omitted. - \end{subproof} - We show that for all $n,m \in \indexset[U]$ such that $n < m$ we have - $\closure{\index[U](n)}{X} \subseteq \interior{\index[U](m)}{X}$. - \begin{subproof} - Omitted. - \end{subproof} +\begin{proposition}\label{urysohnchain_induction_begin_step_two} + Let $X$ be a urysohn space. + Suppose $A,B \in \closeds{X}$. + Suppose $A \inter B$ is empty. + Suppose there exist $U$ + such that $\carrier[U] = \{A,(\carrier[X] \setminus B)\}$ + and $\indexset[U]= \{\zero, 1\}$ + and $\index[U](\zero) = A$ + and $\index[U](1) = (\carrier[X] \setminus B)$. + Then $U$ is a urysohnchain in $X$. +\end{proposition} +\begin{proof} + Omitted. \end{proof} + \begin{proposition}\label{t_four_propositon} Let $X$ be a urysohn space. Then for all $A,B \subseteq X$ such that $\closure{A}{X} \subseteq \interior{B}{X}$ @@ -295,134 +317,147 @@ The first tept will be a formalisation of chain constructions. \end{proof} +\begin{proposition}\label{existence_of_staircase_function} + Let $X$ be a urysohn space. + Suppose $U$ is a urysohnchain in $X$ and $U$ has cardinality $k$. + Suppose $k \neq \zero$. + Then there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ + and for all $n \in \indexset[U]$ we have for all $x \in \index[U](n)$ + we have $f(x) = \rfrac{n}{k}$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} +\begin{abbreviation}\label{refinment_abbreviation} + $x \refine y$ iff $x$ is a refinmant of $y$. +\end{abbreviation} - - -% The next thing we need to define is the uniform staircase function. -% This function has it's domain in $X$ and maps to the closed interval $[0,1]$. -% These functions should behave als follows, -% \begin{align} -% &f(A_{0}) = 1 &\text{consant} \\ -% &f(A_{k} \setminus A_{k+1}) = 1-\frac{k}{r} &\text{constant.} -% \end{align} - -% We then prove that for any given $r$ we find a repolished chain, -% which contains the $A_{i}$ and this replished chain is also legal. -% -% The proof will be finished by taking the limit on $f_{n}$ with $f_{n}$ -% be a staircase function with $n$ many refinemants. -% This limit will be continuous and we would be done. - - -% TODO: Since we want to prove that $f$ is continus, we have to formalize that -% \reals with the usual topology is a topological space. -% ------------------------------------------------------------- - \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. - There exist $f$ such that $f \in \funs{X}{\intervalclosed{\zero}{1}}$ + There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and $f(A) = 1$ and $f(B)= 0$ and $f$ is continuous. \end{theorem} \begin{proof} - We show that for all $n \in \naturals$ we have - if there exist $C$ such that $C$ is a urysohnchain in $X$ of cardinality $n$ - then there exist $C'$ such that $C'$ is a urysohnchain in $X$ of cardinality $n+1$ - and $C'$ is a refinmant of $C$. - \begin{subproof} - Omitted. - \end{subproof} - - There exist $\eta$ such that $\eta$ is a urysohnchain in $X$ and $\eta =\{A, (x \setminus B)\}$. - - - Take $P$ such that $P$ is a infinite sequence and $\indexset[P] = \naturals$ and for all $i \in \indexset[P]$ we have $\index[P](i) = \pow{X}$. + There exist $\eta$ such that $\carrier[\eta] = \{A, (\carrier[X] \setminus B)\}$ + and $\indexset[\eta] = \{\zero, 1\}$ + and $\index[\eta](\zero) = A$ + and $\index[\eta](1) = (\carrier[X] \setminus B)$ by \cref{urysohnchain_induction_begin}. - We show that there exist $\zeta$ such that $\zeta$ is a infinite sequence - and for all $i \in \indexset[\zeta]$ we have - $\index[\zeta](i)$ is a urysohnchain in $X$ of cardinality $i$ - and $A \subseteq \index[\zeta](i)$ - and $\index[\zeta](i) \subseteq (X \setminus B)$ - and for all $j \in \indexset[\zeta]$ such that - $j < i$ we have for all $x \in \index[\zeta](j)$ we have $x \in \index[\zeta](i)$. + We show that there exist $\zeta$ such that $\zeta$ is a sequence + and $\indexset[\zeta] = \naturals$ + and $\eta \in \carrier[\zeta]$ and $\index[\zeta](\eta) = \zero$ + and for all $n \in \indexset[\zeta]$ we have $n+1 \in \indexset[\zeta]$ + and $\index[\zeta](n+1)$ is a refinmant of $\index[\zeta](n)$. \begin{subproof} - Omitted. - \end{subproof} - - - - - - + %Let $P = \{ n \in \naturals \mid \exists \zeta. ((n = \zero \land \index[\zeta](n) = \eta \land \index[\zeta](n+1)$ is a refinmant of $\index[\zeta](n)) \lor (n \neq \zero \land \index[\zeta](n)$ is a urysohnchain in $X$ \land $\index[\zeta](n+1)$ is a refinmant of $\index[\zeta](n)))\}$. + %Let $C = \{ x \mid \text{$x$ is a uysohncahin in $X$}\}$. + Let $\alpha = \{x \in C \mid \exists y \in \alpha. x \refine y \lor x = \eta\}$. + Let $\beta = \{ (n,x) \mid n \in \naturals \mid \exists m \in \naturals. \exists y \in \alpha. (x \in \alpha) \land ((x \refine y \land m = n+1 )\lor ((n,x) = (\zero,\eta)))\}$. - - We show that there exist $g \in \funs{X}{\intervalclosed{\zero}{1}}$ such that $g(A)=1$ and $g(X\setminus A) = \zero$. - \begin{subproof} - Omitted. + % TODO: Proof that \beta is a function which would be used for the indexing. + \end{subproof} - $g$ is a staircase function and $\chain[g] = C$. - $g$ is a legal staircase function. - We show that there exist $f$ such that $f \in \funs{X}{\intervalclosed{\zero}{1}}$ - and $f(A) = 1$ and $f(B)= 0$ and $f$ is continuous. - \begin{subproof} - Omitted. - \end{subproof} - %We show that for all $n \in \nautrals$ we have $C_{n}$ a legal chain of subsets of $X$. - %\begin{subproof} - % Omitted. - %\end{subproof} - - %Proof Sheme Idea: - % -We proof for n=1 that C_{n} is a chain and legal - % -Then by induction with P(n+1) is refinmant of P(n) - % -Therefore we have a increing refinmant of these Chains such that our limit could even apply - % --------------------------------------------------------- - - %We show that there exist $f \in \funs{\naturals}{\funs{X}{\intervalclosed{0}{1}}}$ such that $f(n)$ is a staircase function. %TODO: Define Staircase function - %\begin{subproof} - % Omitted. - %\end{subproof} - - - % Formalization idea of enumarted sequences: - % - Define a enumarted sequnecs as a set A with a bijection between A and E \in \pow{\naturals} - % - This should give all finite and infinte enumarable sequences - % - Introduce a notion for the indexing of these enumarable sequences. - % - Then we can define the limit of a enumarted sequence of functions. - % --------------------------------------------------------- - % - % Here we need a limit definition for sequences of functions - %We show that there exist $F \in \funs{X}{\intervalclosed{0}{1}}$ such that $F = \limit{set of the staircase functions}$ - %\begin{subproof} - % Omitted. - %\end{subproof} - % - %We show that $F(A) = 1$. - %\begin{subproof} - % Omitted. - %\end{subproof} - % - %We show that $F(B) = 0$. - %\begin{subproof} - % Omitted. - %\end{subproof} - % - %We show that $F$ is continuous. - %\begin{subproof} - % Omitted. - %\end{subproof} + %Suppose $\eta$ is a urysohnchain in $X$. + %Suppose $\carrier[\eta] =\{A, (X \setminus B)\}$ + %and $\indexset[\eta] = \{\zero, 1\}$ + %and $\index[\eta](\zero) = A$ + %and $\index[\eta](1) = (X \setminus B)$. + + + %Then $\eta$ is a urysohnchain in $X$. + + % Take $P$ such that $P$ is a infinite sequence and $\indexset[P] = \naturals$ and for all $i \in \indexset[P]$ we have $\index[P](i) = \pow{X}$. + % + % We show that there exist $\zeta$ such that $\zeta$ is a infinite sequence + % and for all $i \in \indexset[\zeta]$ we have + % $\index[\zeta](i)$ is a urysohnchain in $X$ of cardinality $i$ + % and $A \subseteq \index[\zeta](i)$ + % and $\index[\zeta](i) \subseteq (X \setminus B)$ + % and for all $j \in \indexset[\zeta]$ such that + % $j < i$ we have for all $x \in \index[\zeta](j)$ we have $x \in \index[\zeta](i)$. + % \begin{subproof} + % Omitted. + % \end{subproof} + % + % + % + % + % + % + % + % + % We show that there exist $g \in \funs{X}{\intervalclosed{\zero}{1}}$ such that $g(A)=1$ and $g(X\setminus A) = \zero$. + % \begin{subproof} + % Omitted. + % \end{subproof} + % $g$ is a staircase function and $\chain[g] = C$. + % $g$ is a legal staircase function. + % + % + % We show that there exist $f$ such that $f \in \funs{X}{\intervalclosed{\zero}{1}}$ + % and $f(A) = 1$ and $f(B)= 0$ and $f$ is continuous. + % \begin{subproof} + % Omitted. + % \end{subproof} + + + % We show that for all $n \in \nautrals$ we have $C_{n}$ a legal chain of subsets of $X$. + % \begin{subproof} + % Omitted. + % \end{subproof} + + % Proof Sheme Idea: + % -We proof for n=1 that C_{n} is a chain and legal + % -Then by induction with P(n+1) is refinmant of P(n) + % -Therefore we have a increing refinmant of these Chains such that our limit could even apply + % --------------------------------------------------------- + + % We show that there exist $f \in \funs{\naturals}{\funs{X}{\intervalclosed{0}{1}}}$ such that $f(n)$ is a staircase function. %TODO: Define Staircase function + % \begin{subproof} + % Omitted. + % \end{subproof} + + + % Formalization idea of enumarted sequences: + % - Define a enumarted sequnecs as a set A with a bijection between A and E \in \pow{\naturals} + % - This should give all finite and infinte enumarable sequences + % - Introduce a notion for the indexing of these enumarable sequences. + % - Then we can define the limit of a enumarted sequence of functions. + % --------------------------------------------------------- + % + % Here we need a limit definition for sequences of functions + % We show that there exist $F \in \funs{X}{\intervalclosed{0}{1}}$ such that $F = \limit{set of the staircase functions}$ + % \begin{subproof} + % Omitted. + % \end{subproof} + % + % We show that $F(A) = 1$. + % \begin{subproof} + % Omitted. + % \end{subproof} + % + % We show that $F(B) = 0$. + % \begin{subproof} + % Omitted. + % \end{subproof} + % + % We show that $F$ is continuous. + % \begin{subproof} + % Omitted. + % \end{subproof} \end{proof} -\begin{theorem}\label{safe} - Contradiction. -\end{theorem} +%\begin{theorem}\label{safe} +% Contradiction. +%\end{theorem} -- cgit v1.2.3 From e9fa2c5d9539067d9ecce52bbbccacdfd020c563 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sun, 11 Aug 2024 17:10:57 +0200 Subject: more more more urysohn --- library/topology/urysohn.tex | 34 +++++++++++++++++++++++++++------- 1 file changed, 27 insertions(+), 7 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 028e10f..f34f12f 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -5,6 +5,7 @@ \import{function.tex} \import{set.tex} \import{cardinal.tex} +\import{relation.tex} \section{Urysohns Lemma} % In this section we want to proof Urysohns lemma. @@ -188,10 +189,20 @@ The first tept will be a formalisation of chain constructions. \end{lemma} +\begin{axiom}\label{pot1} + $\pot \in \funs{\naturals}{\naturals}$. +\end{axiom} +\begin{axiom}\label{pot2} + For all $n \in \naturals$ we have there exist $k\in \naturals$ such that $(n, k) \in \powerOfTwoSet$ and $\apply{\pot}{n}=k$. + %$\pot(n) = k$ iff there exist $x \in \powerOfTwoSet$ such that $x = (n,k)$. +\end{axiom} - +%Without this abbreviation \pot cant be sed as a function in the standard sense +\begin{abbreviation}\label{pot_as_function} + $\pot(n) = \apply{\pot}{n}$. +\end{abbreviation} \begin{proposition}\label{urysohnchain_induction_begin} @@ -356,15 +367,24 @@ The first tept will be a formalisation of chain constructions. and for all $n \in \indexset[\zeta]$ we have $n+1 \in \indexset[\zeta]$ and $\index[\zeta](n+1)$ is a refinmant of $\index[\zeta](n)$. \begin{subproof} - %Let $P = \{ n \in \naturals \mid \exists \zeta. ((n = \zero \land \index[\zeta](n) = \eta \land \index[\zeta](n+1)$ is a refinmant of $\index[\zeta](n)) \lor (n \neq \zero \land \index[\zeta](n)$ is a urysohnchain in $X$ \land $\index[\zeta](n+1)$ is a refinmant of $\index[\zeta](n)))\}$. - %Let $C = \{ x \mid \text{$x$ is a uysohncahin in $X$}\}$. - Let $\alpha = \{x \in C \mid \exists y \in \alpha. x \refine y \lor x = \eta\}$. - Let $\beta = \{ (n,x) \mid n \in \naturals \mid \exists m \in \naturals. \exists y \in \alpha. (x \in \alpha) \land ((x \refine y \land m = n+1 )\lor ((n,x) = (\zero,\eta)))\}$. + %Let $\alpha = \{x \in C \mid \exists y \in \alpha. x \refine y \lor x = \eta\}$. + %Let $\beta = \{ (n,x) \mid n \in \naturals \mid \exists m \in \naturals. \exists y \in \alpha. (x \in \alpha) \land ((x \refine y \land m = n+1 )\lor ((n,x) = (\zero,\eta)))\}$. + % + % % TODO: Proof that \beta is a function which would be used for the indexing. + %For all $n \in \naturals$ we have there exist $x \in \alpha$ such that $(n,x) \in \beta$. + %$\dom{\beta} = \naturals$. + %$\ran{\beta} = \alpha$. + %$\beta \in \funs{\naturals}{\alpha}$. + %Take $\zeta$ such that $\carrier[\zeta] = \alpha$ and $\indexset[\zeta] = \naturals$ and $\index[\zeta] = \beta$. + Omitted. + \end{subproof} - % TODO: Proof that \beta is a function which would be used for the indexing. - + We show that for all $n \in \indexset[\zeta]$ we have $\index[\zeta](n)$ has cardinality $\pot(n)$. + \begin{subproof} + Trivial. \end{subproof} + -- cgit v1.2.3 From c4894bc4e788fae079b76b824a8d86c167098cc8 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 12 Aug 2024 00:12:07 +0200 Subject: more more way more urysohn --- library/topology/urysohn.tex | 127 ++++++++++++++++++++++++++++++++++++++++--- source/Api.hs | 8 +-- source/CommandLine.hs | 5 +- source/Syntax/Lexicon.hs | 1 + 4 files changed, 127 insertions(+), 14 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index f34f12f..6662706 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -36,7 +36,12 @@ The first tept will be a formalisation of chain constructions. - +%%----------------------- +% Idea: +% A sequence could be define as a family of sets, +% together with the existence of an indexing function. +% +%%----------------------- \begin{struct}\label{sequence} A sequence $X$ is a onesorted structure equipped with \begin{enumerate} @@ -148,8 +153,9 @@ The first tept will be a formalisation of chain constructions. \end{definition} \begin{definition}\label{refinmant} - $C'$ is a refinmant of $C$ iff for all $x \in C$ we have $x \in C'$ and - for all $y \in C$ such that $y \subset x$ we have there exist $c \in C'$ such that $y \subset c \subset x$ + $C'$ is a refinmant of $C$ iff $C'$ is a urysohnchain in $X$ + and for all $x \in C$ we have $x \in C'$ + and for all $y \in C$ such that $y \subset x$ we have there exist $c \in C'$ such that $y \subset c \subset x$ and for all $g \in C'$ such that $g \neq c$ we have not $y \subset g \subset x$. \end{definition} @@ -312,9 +318,9 @@ The first tept will be a formalisation of chain constructions. % U = ( A_{0}, A_{1}, A_{2}, ... , A_{n-1}, A_{n}) % U' = ( A_{0}, A'_{1}, A_{1}, A'_{2}, A_{2}, ... A_{n-1}, A'_{n}, A_{n}) - Let $m = \max{\indexset[U]}$. - For all $n \in (\indexset[U] \setminus \{m\})$ we have there exist $C \subseteq X$ - such that $\closure{\index[U](n)}{X} \subseteq \interior{C}{X} \subseteq \closure{C}{X} \subseteq \interior{\index[U](n+1)}{X}$. + % Let $m = \max{\indexset[U]}$. + % For all $n \in (\indexset[U] \setminus \{m\})$ we have there exist $C \subseteq X$ + % such that $\closure{\index[U](n)}{X} \subseteq \interior{C}{X} \subseteq \closure{C}{X} \subseteq \interior{\index[U](n+1)}{X}$. %\begin{definition}\label{refinmant} @@ -347,6 +353,58 @@ The first tept will be a formalisation of chain constructions. + +\begin{abbreviation}\label{sequence_of_functions} + $f$ is a sequence of functions iff $f$ is a sequence + and for all $g \in \carrier[f]$ we have $g$ is a function. +\end{abbreviation} + +\begin{abbreviation}\label{sequence_in_reals} + $s$ is a sequence of real numbers iff $s$ is a sequence + and for all $r \in \carrier[s]$ we have $r \in \reals$. +\end{abbreviation} + + + +\begin{axiom}\label{abs_behavior1} + If $x \geq \zero$ then $\abs{x} = x$. +\end{axiom} + +\begin{axiom}\label{abs_behavior2} + If $x < \zero$ then $\abs{x} = \neg{x}$. +\end{axiom} + +\begin{abbreviation}\label{converge} + $s$ converges iff $s$ is a sequence of real numbers + and $\indexset[s]$ is infinite + and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have + there exist $N \in \indexset[s]$ such that + for all $m \in \indexset[s]$ such that $m > N$ + we have $\abs{\index[s](N) - \index[s](m)} < \epsilon$. +\end{abbreviation} + + +\begin{definition}\label{limit_of_sequence} + $x$ is the limit of $s$ iff $s$ is a sequence of real numbers + and $x \in \reals$ and + for all $\epsilon \in \reals$ such that $\epsilon > \zero$ + we have there exist $n \in \indexset[s]$ such that + for all $m \in \indexset[s]$ such that $m > n$ + we have $\abs{x - \index[s](n)} < \epsilon$. +\end{definition} + +\begin{proposition}\label{existence_of_limit} + Let $s$ be a sequence of real numbers. + Then $s$ converges iff there exist $x \in \reals$ + such that $x$ is the limit of $s$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} + + + + \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. @@ -381,12 +439,65 @@ The first tept will be a formalisation of chain constructions. We show that for all $n \in \indexset[\zeta]$ we have $\index[\zeta](n)$ has cardinality $\pot(n)$. \begin{subproof} - Trivial. + Omitted. \end{subproof} - + For all $m \in \indexset[\zeta]$ we have $\pot(m) \neq \zero$. + %%------------- Maybe use Abstrect.hs line 368 "Local Function". + + We show that for all $m \in \indexset[\zeta]$ we have there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ + and for all $n \in \indexset[\index[\zeta](m)]$ we have for all $x \in \index[\index[\zeta](m)](n)$ + we have $f(x) = \rfrac{n}{\pot(m)}$. + \begin{subproof} + % Fix $m \in \indexset[\zeta]$. + % $\index[\zeta](m)$ is a urysohnchain in $X$. + % + % Follows by \cref{existence_of_staircase_function}. + + Omitted. + \end{subproof} + + + + %The sequenc of the functions + Let $\gamma = \{ + (n,f) \mid + n \in \naturals \mid + + \forall n' \in \indexset[\index[\zeta](n)]. + \forall x \in \carrier[X]. + + + f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}} \land + + + % (n,f) \in \gamma <=> \phi(n,f) + % with \phi (n,f) := (x \in (A_k) \ (A_k-1)) => f(x) = ( k / 2^n ) + % \/ (x \notin A_k for all k \in {1,..,n} => f(x) = 1 + + ( (n' = \zero) + \land (x \in \index[\index[\zeta](n)](n')) + \land (f(x)= \zero) ) + + \lor + + ( (n' > \zero) + \land (x \in \index[\index[\zeta](n)](n')) + \land (x \notin \index[\index[\zeta](n)](n'-1)) + \land (f(x) = \rfrac{n'}{\pot(n)}) ) + + \lor + + ( (x \notin \index[\index[\zeta](n)](n')) + \land (f(x) = 1) ) + + \}$. + + + + %Suppose $\eta$ is a urysohnchain in $X$. %Suppose $\carrier[\eta] =\{A, (X \setminus B)\}$ diff --git a/source/Api.hs b/source/Api.hs index f597153..3fb0ca2 100644 --- a/source/Api.hs +++ b/source/Api.hs @@ -292,10 +292,10 @@ exportMegalodon file = do -- | This could be expandend with the dump case, with dump off just this and if dump is on it could show the number off the task. For quick use showFailedTask :: (a, ProverAnswer) -> IO() showFailedTask (_, Yes ) = Text.putStrLn "" -showFailedTask (_, No tptp) = Text.putStrLn (Text.pack ("Prover found countermodel: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) -showFailedTask (_, ContradictoryAxioms tptp) = Text.putStrLn (Text.pack ("Contradictory axioms: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) -showFailedTask (_, Uncertain tptp) = Text.putStrLn (Text.pack ("Out of resources: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) -showFailedTask (_, Error _ tptp _) = Text.putStrLn (Text.pack ("Error at: " ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +showFailedTask (_, No tptp) = Text.putStrLn (Text.pack ("\ESC[31mProver found countermodel: \ESC[0m" ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +showFailedTask (_, ContradictoryAxioms tptp) = Text.putStrLn (Text.pack ("\ESC[31mContradictory axioms: \ESC[0m" ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +showFailedTask (_, Uncertain tptp) = Text.putStrLn (Text.pack ("\ESC[31mOut of resources: \ESC[0m" ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) +showFailedTask (_, Error _ tptp _) = Text.putStrLn (Text.pack ("\ESC[31mError at: \ESC[0m" ++ Text.unpack(Text.unlines (take 1 (Text.splitOn "." tptp))))) --showFailedTask (_, _) = Text.putStrLn "Error!" diff --git a/source/CommandLine.hs b/source/CommandLine.hs index 47efe22..711f7f0 100644 --- a/source/CommandLine.hs +++ b/source/CommandLine.hs @@ -87,15 +87,16 @@ run = do Text.hPutStrLn stderr tptp Error err tptp task -> do putStr "Error at:" + Text.putStrLn task Text.putStrLn err Text.putStrLn tptp WithFailList -> liftIO case result of - VerificationSuccess -> putStrLn "Verification successful." + VerificationSuccess -> putStrLn "\ESC[32mVerification successful.\ESC[0m" VerificationFailure [] -> error "Empty verification fail" VerificationFailure fails -> do - putStrLn "Following task couldn't be solved by the ATP: " + putStrLn "\ESC[35mFollowing task couldn't be solved by the ATP: \ESC[0m" traverse_ showFailedTask fails Text.hPutStrLn stderr "Don't give up!" diff --git a/source/Syntax/Lexicon.hs b/source/Syntax/Lexicon.hs index de6d966..4fe8730 100644 --- a/source/Syntax/Lexicon.hs +++ b/source/Syntax/Lexicon.hs @@ -112,6 +112,7 @@ prefixOps = , ([Just (Command "pow"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "pow")) , ([Just (Command "neg"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "neg")) , ([Just (Command "inv"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "inv")) + , ([Just (Command "abs"), Just InvisibleBraceL, Nothing, Just InvisibleBraceR], (NonAssoc, "abs")) , (ConsSymbol, (NonAssoc, "cons")) , (PairSymbol, (NonAssoc, "pair")) -- NOTE Is now defined and hence no longer necessary , (ApplySymbol, (NonAssoc, "apply")) -- cgit v1.2.3 From a79924e85260c911045f43bac3f051a8be6e5334 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 12 Aug 2024 00:52:54 +0200 Subject: Completed the structure of the proof of Urysohn. Now every single step has to be proven. --- library/topology/urysohn.tex | 44 ++++++++++++++++++++++++++++++++++++++++---- 1 file changed, 40 insertions(+), 4 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 6662706..0c4052d 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -410,7 +410,7 @@ The first tept will be a formalisation of chain constructions. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ - and $f(A) = 1$ and $f(B)= 0$ and $f$ is continuous. + and $f(A) = 0$ and $f(B)= 1$ and $f$ is continuous. \end{theorem} \begin{proof} @@ -442,8 +442,10 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} - For all $m \in \indexset[\zeta]$ we have $\pot(m) \neq \zero$. - + We show that for all $m \in \indexset[\zeta]$ we have $\pot(m) \neq \zero$. + \begin{subproof} + Omitted. + \end{subproof} %%------------- Maybe use Abstrect.hs line 368 "Local Function". @@ -495,9 +497,43 @@ The first tept will be a formalisation of chain constructions. \}$. + Let $\gamma(n) = \apply{\gamma}{n}$ for $n \in \naturals$. + + + We show that there exist $F$ such that + $F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and + for all $x \in \carrier[X]$ we have + there exist $k \in \intervalclosed{\zero}{1}$ + such that $F(x) = k$ and + for all $\epsilon \in \reals$ such that $\epsilon > \zero$ + we have there exist $N \in \naturals$ such that + for all $N' \in \naturals$ such that $N' > N$ + we have $\abs{(k - \apply{\gamma(N')}{x})} \leq \epsilon$. + \begin{subproof} + Omitted. + \end{subproof} + - + We show that $F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$. + \begin{subproof} + Omitted. + \end{subproof} + + We show that $F(A) = 0$. + \begin{subproof} + Omitted. + \end{subproof} + + We show that $F(B) = 1$. + \begin{subproof} + Omitted. + \end{subproof} + + We show that $F$ is continuous. + \begin{subproof} + Omitted. + \end{subproof} %Suppose $\eta$ is a urysohnchain in $X$. %Suppose $\carrier[\eta] =\{A, (X \setminus B)\}$ -- cgit v1.2.3 From 85920f166c95904550134a1d4e84d5430c05d009 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 12 Aug 2024 11:43:26 +0200 Subject: way more urysohn --- library/topology/urysohn.tex | 116 +++++++++++++++++++++++++++++++++++++------ 1 file changed, 102 insertions(+), 14 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 0c4052d..d8b1e14 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -402,18 +402,32 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} +\begin{definition}\label{limit_sequence} + $x$ is the limit sequence of $f$ iff + $x$ is a sequence and $\indexset[x] = \dom{f}$ and + for all $n \in \indexset[x]$ we have + $\index[x](n) = f(n)$. +\end{definition} +\begin{definition}\label{realsminus} + $\realsminus = \{r \in \reals \mid r < \zero\}$. +\end{definition} + +\begin{abbreviation}\label{realsplus} + $\realsplus = \reals \setminus \realsminus$. +\end{abbreviation} \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. + Suppose $\carrier[X]$ is inhabited. There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ - and $f(A) = 0$ and $f(B)= 1$ and $f$ is continuous. + and $f(A) = \zero$ and $f(B)= 1$ and $f$ is continuous. \end{theorem} \begin{proof} - + There exist $\eta$ such that $\carrier[\eta] = \{A, (\carrier[X] \setminus B)\}$ and $\indexset[\eta] = \{\zero, 1\}$ and $\index[\eta](\zero) = A$ @@ -498,29 +512,103 @@ The first tept will be a formalisation of chain constructions. \}$. Let $\gamma(n) = \apply{\gamma}{n}$ for $n \in \naturals$. - + %We show that there exist $\kappa$ such that $\kappa$ is the limit sequence of $\gamma$. + %\begin{subproof} + % Omitted. + %\end{subproof} - We show that there exist $F$ such that - $F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and - for all $x \in \carrier[X]$ we have - there exist $k \in \intervalclosed{\zero}{1}$ - such that $F(x) = k$ and - for all $\epsilon \in \reals$ such that $\epsilon > \zero$ - we have there exist $N \in \naturals$ such that - for all $N' \in \naturals$ such that $N' > N$ - we have $\abs{(k - \apply{\gamma(N')}{x})} \leq \epsilon$. + %Let $\gamma(n)(x) = \apply{\gamma(n)}{x}$ for $x\in \carrier[X]$ + + We show that for all $n \in \naturals$ we have $\gamma(n)$ + is a function from $\carrier[X]$ to $\reals$. \begin{subproof} Omitted. \end{subproof} + + + We show that there exist $g$ such that + for all $x \in \carrier[X]$ we have + there exist $k \in \reals$ such that + for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have + there exist $N \in \naturals$ such that + for all $N' \in \naturals$ such that $N' > N$ we have + $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$ and $g(x)= k$. + \begin{subproof} + + %Let $G(x) = \{k \in \carrier[X] \mid \forall \epsilon \in \realsplus. \exists N \in \naturals. \forall N' \in \naturals. (N' > N) \land (\abs{\apply{\gamma(n)}{x} - k} < \epsilon)\}$ for $x \in \carrier[X]$. +% + %We show that for all $x \in \carrier[X]$ we have $G(x)$ has cardinality $1$. + %\begin{subproof} + % Omitted. + %\end{subproof} +% + %Let $H(x) = \unions{G(x)}$ for $x \in \carrier[X]$. + %For all $x \in \carrier[X]$ we have $\{H(x)\} = G(x)$. + + We show that for all $x \in \carrier[X]$ we have + there exist $k \in \reals$ such that + for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have + there exist $N \in \naturals$ such that + for all $N' \in \naturals$ such that $N' > N$ we have + $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$. + \begin{subproof} + Omitted. + \end{subproof} + + + \end{subproof} + + + %Let $S(x) = \{(n,x) \mid n \in naturals \mid x = (\gamma(n))(x)\}$. + + %Let $S(x) + + %We show that there exist $F$ such that + %for all $x \in \carrier[X]$ we have $F(x)$ is + + + %We show that there exist $F$ such that + %$F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and + %for all $x \in \carrier[X]$ we have + %there exist $k \in \intervalclosed{\zero}{1}$ + %such that $F(x) = k$ and + %for all $\epsilon \in \reals$ such that $\epsilon > \zero$ + %we have there exist $N \in \naturals$ such that + %for all $N' \in \naturals$ such that $N' > N$ + %we have $\abs{(k - \apply{\gamma(N')}{x})} \leq \epsilon$. + %\begin{subproof} + % Omitted. + %\end{subproof} + + We show that $F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$. - \begin{subproof} + \begin{subproof} Omitted. + + %It suffices to show that $\dom{F} = \carrier[X]$ and + %$F \in \rels{\carrier[X]}{\intervalclosed{\zero}{1}}$ and $F$ is right-unique. + % + %We show that $F \in \rels{\carrier[X]}{\intervalclosed{\zero}{1}}$. + %\begin{subproof} + % It suffices to show that $F \in \pow{\carrier[X] \times \intervalclosed{\zero}{1}}$. + % It suffices to show that for all $w \in F$ we have $w \in (\carrier[X] \times \intervalclosed{\zero}{1})$. +% + % Fix $w \in F$. +% + % %Take $x$ such that there exist $k \in \intervalclosed{\zero}{1}$ such that $w = (x,k)$. +% +% + %\end{subproof} + % +% +% + %Trivial. \end{subproof} - We show that $F(A) = 0$. + We show that $F(A) = \zero$. \begin{subproof} Omitted. \end{subproof} -- cgit v1.2.3 From b51f61e6be5be4729e61ede79fb0bd3cb26f57a6 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 12 Aug 2024 14:41:32 +0200 Subject: way way more urysohn --- library/topology/urysohn.tex | 82 +++++++++++++++++++++++--------------------- source/CommandLine.hs | 3 ++ 2 files changed, 45 insertions(+), 40 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index d8b1e14..414043f 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -526,6 +526,11 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} + We show that for all $n \in \naturals$ for all $x \in \carrier[X]$ we have $\apply{\gamma(n)}{x} \in \intervalclosed{\zero}{1}$. + \begin{subproof} + Omitted. + \end{subproof} + We show that there exist $g$ such that @@ -535,18 +540,7 @@ The first tept will be a formalisation of chain constructions. there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$ and $g(x)= k$. - \begin{subproof} - - %Let $G(x) = \{k \in \carrier[X] \mid \forall \epsilon \in \realsplus. \exists N \in \naturals. \forall N' \in \naturals. (N' > N) \land (\abs{\apply{\gamma(n)}{x} - k} < \epsilon)\}$ for $x \in \carrier[X]$. -% - %We show that for all $x \in \carrier[X]$ we have $G(x)$ has cardinality $1$. - %\begin{subproof} - % Omitted. - %\end{subproof} -% - %Let $H(x) = \unions{G(x)}$ for $x \in \carrier[X]$. - %For all $x \in \carrier[X]$ we have $\{H(x)\} = G(x)$. - + \begin{subproof} We show that for all $x \in \carrier[X]$ we have there exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have @@ -554,10 +548,9 @@ The first tept will be a formalisation of chain constructions. for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$. \begin{subproof} - Omitted. + Fix $x \in \carrier[X]$. + Follows by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. \end{subproof} - - \end{subproof} @@ -581,44 +574,53 @@ The first tept will be a formalisation of chain constructions. %\begin{subproof} % Omitted. %\end{subproof} + + Let $G(x) = g(x)$ for $x \in \carrier[X]$. + We have $\dom{G} = \carrier[X]$. - - We show that $F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$. - \begin{subproof} + %For all $x \in \carrier[X]$ for all $n \in \naturals$ we have $g(x) \leq \apply{\gamma(n)}{x}$. + + We show that for all $x \in \dom{G}$ we have $G(x) \in \reals$. + \begin{subproof} + Fix $x \in \dom{G}$. + It suffices to show that $g(x) \in \reals$. + + %There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$ and $g(x)= k$. + + + + \end{subproof} + + We show that for all $x \in \dom{G}$ we have $\zero \leq G(x) \leq 1$. + \begin{subproof} Omitted. - - %It suffices to show that $\dom{F} = \carrier[X]$ and - %$F \in \rels{\carrier[X]}{\intervalclosed{\zero}{1}}$ and $F$ is right-unique. - % - %We show that $F \in \rels{\carrier[X]}{\intervalclosed{\zero}{1}}$. - %\begin{subproof} - % It suffices to show that $F \in \pow{\carrier[X] \times \intervalclosed{\zero}{1}}$. - % It suffices to show that for all $w \in F$ we have $w \in (\carrier[X] \times \intervalclosed{\zero}{1})$. -% - % Fix $w \in F$. -% - % %Take $x$ such that there exist $k \in \intervalclosed{\zero}{1}$ such that $w = (x,k)$. -% -% - %\end{subproof} - % -% -% - %Trivial. \end{subproof} - We show that $F(A) = \zero$. + + We show that $G \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$. + \begin{subproof} + It suffices to show that $\ran{G} \subseteq \intervalclosed{\zero}{1}$ by \cref{fun_ran_iff,funs_intro,funs_weaken_codom}. + It suffices to show that for all $x \in \dom{G}$ we have $G(x) \in \intervalclosed{\zero}{1}$. + Fix $x \in \dom{G}$. + Then $x \in \carrier[X]$. + $g(x) = G(x)$. + We have $G(x) \in \reals$. + $\zero \leq G(x) \leq 1$. + We have $G(x) \in \intervalclosed{\zero}{1}$ . + \end{subproof} + + We show that $G(A) = \zero$. \begin{subproof} Omitted. \end{subproof} - We show that $F(B) = 1$. + We show that $G(B) = 1$. \begin{subproof} Omitted. \end{subproof} - We show that $F$ is continuous. + We show that $G$ is continuous. \begin{subproof} Omitted. \end{subproof} diff --git a/source/CommandLine.hs b/source/CommandLine.hs index 711f7f0..b8c170e 100644 --- a/source/CommandLine.hs +++ b/source/CommandLine.hs @@ -43,10 +43,12 @@ run = do case withDump opts of WithoutDump -> skip WithDump dir -> do + liftIO (Text.putStrLn "\ESC[1;36mCreating Dumpfiles.\ESC[0m") let serials = [dir show n <.> "p" | n :: Int <- [1..]] tasks <- zip serials <$> encodeTasks (inputPath opts) createDirectoryIfMissing True dir forM_ tasks (uncurry dumpTask) + liftIO (Text.putStrLn "\ESC[35mDump ready.\ESC[0m") case (withParseOnly opts, withMegalodon opts) of (WithParseOnly, _) -> do ast <- parse (inputPath opts) @@ -60,6 +62,7 @@ run = do -- A custom E executable can be configured using environment variables. -- If the environment variable is undefined we fall back to the -- a globally installed E executable. + liftIO (Text.putStrLn "\ESC[1;96mStart of verification.\ESC[0m") vampirePathPath <- (?? "vampire") <$> lookupEnv "NAPROCHE_VAMPIRE" eproverPath <- (?? "eprover") <$> lookupEnv "NAPROCHE_EPROVER" let prover = case withProver opts of -- cgit v1.2.3 From 8a19eec1e581efa473af0173ba79b553e4191ffa Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 12 Aug 2024 15:55:17 +0200 Subject: way way way more urysohn --- library/topology/urysohn.tex | 58 +++++++++++++++++--------------------------- 1 file changed, 22 insertions(+), 36 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 414043f..3be3c30 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -461,8 +461,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} - %%------------- Maybe use Abstrect.hs line 368 "Local Function". - + We show that for all $m \in \indexset[\zeta]$ we have there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and for all $n \in \indexset[\index[\zeta](m)]$ we have for all $x \in \index[\index[\zeta](m)](n)$ we have $f(x) = \rfrac{n}{\pot(m)}$. @@ -513,13 +512,7 @@ The first tept will be a formalisation of chain constructions. Let $\gamma(n) = \apply{\gamma}{n}$ for $n \in \naturals$. - %We show that there exist $\kappa$ such that $\kappa$ is the limit sequence of $\gamma$. - %\begin{subproof} - % Omitted. - %\end{subproof} - - %Let $\gamma(n)(x) = \apply{\gamma(n)}{x}$ for $x\in \carrier[X]$ - + We show that for all $n \in \naturals$ we have $\gamma(n)$ is a function from $\carrier[X]$ to $\reals$. \begin{subproof} @@ -552,49 +545,42 @@ The first tept will be a formalisation of chain constructions. Follows by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. \end{subproof} \end{subproof} - - - %Let $S(x) = \{(n,x) \mid n \in naturals \mid x = (\gamma(n))(x)\}$. - - %Let $S(x) - - %We show that there exist $F$ such that - %for all $x \in \carrier[X]$ we have $F(x)$ is - - - %We show that there exist $F$ such that - %$F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and - %for all $x \in \carrier[X]$ we have - %there exist $k \in \intervalclosed{\zero}{1}$ - %such that $F(x) = k$ and - %for all $\epsilon \in \reals$ such that $\epsilon > \zero$ - %we have there exist $N \in \naturals$ such that - %for all $N' \in \naturals$ such that $N' > N$ - %we have $\abs{(k - \apply{\gamma(N')}{x})} \leq \epsilon$. - %\begin{subproof} - % Omitted. - %\end{subproof} Let $G(x) = g(x)$ for $x \in \carrier[X]$. We have $\dom{G} = \carrier[X]$. - %For all $x \in \carrier[X]$ for all $n \in \naturals$ we have $g(x) \leq \apply{\gamma(n)}{x}$. - We show that for all $x \in \dom{G}$ we have $G(x) \in \reals$. \begin{subproof} Fix $x \in \dom{G}$. It suffices to show that $g(x) \in \reals$. - %There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$ and $g(x)= k$. - + There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$. + $g(x)= k$. + + Follows by \cref{apply,plus_one_order,ordinal_iff_suc_ordinal,natural_number_is_ordinal,subseteq,naturals_subseteq_reals,naturals_is_equal_to_two_times_naturals,reals_one_bigger_zero,one_in_reals,ordinal_prec_trichotomy,omega_is_an_ordinal,suc_intro_self,two_in_naturals,in_asymmetric,suc}. \end{subproof} We show that for all $x \in \dom{G}$ we have $\zero \leq G(x) \leq 1$. \begin{subproof} - Omitted. + Fix $x \in \dom{G}$. + Then $x \in \carrier[X]$. + \begin{byCase} + \caseOf{$x \in A$.} + For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 0$. + + + \caseOf{$x \notin A$.} + \begin{byCase} + \caseOf{$x \in B$.} + OFor all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 1$. + + \caseOf{$x \notin B$.} + Omitted. + \end{byCase} + \end{byCase} \end{subproof} -- cgit v1.2.3 From 41357a4c0ad22b1ea0094ea52c832d117ec3fec4 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 12 Aug 2024 16:30:47 +0200 Subject: way way way way more urysohn --- library/topology/urysohn.tex | 51 +++++++++++++++++++++++++++++++++++++++----- 1 file changed, 46 insertions(+), 5 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 3be3c30..c94dbd4 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -98,6 +98,10 @@ The first tept will be a formalisation of chain constructions. $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. \end{definition} +\begin{definition}\label{intervalopen} + $\intervalopen{a}{b} = \{ x \in \reals \mid a < x < b\}$. +\end{definition} + \begin{struct}\label{staircase_function} A staircase function $f$ is a onesorted structure equipped with @@ -333,6 +337,34 @@ The first tept will be a formalisation of chain constructions. \end{proof} +\begin{lemma}\label{fraction1} + Let $x \in \reals$. + Then for all $y \in \reals$ such that $x \neq y$ we have there exists $r \in \rationals$ such that $y < r < x$ or $x < r < y$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{frection2} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then $\intervalopen{a}{b}$ is infinite. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{frection3} + Suppose $a,b \in \reals$. + Suppose $\zero < a < 1$. + Suppose $\zero < b < 1$. + % Here take exist n such that n/2^n is between a and b + T +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + \begin{proposition}\label{existence_of_staircase_function} Let $X$ be a urysohn space. @@ -489,8 +521,9 @@ The first tept will be a formalisation of chain constructions. % (n,f) \in \gamma <=> \phi(n,f) - % with \phi (n,f) := (x \in (A_k) \ (A_k-1)) => f(x) = ( k / 2^n ) - % \/ (x \notin A_k for all k \in {1,..,n} => f(x) = 1 + % with \phi (n,f) := + % (x \in (A_k) \ (A_k-1)) ==> f(x) = ( k / 2^n ) + % \/ (x \notin A_k for all k \in {1,..,n} ==> f(x) = 1 ( (n' = \zero) \land (x \in \index[\index[\zeta](n)](n')) @@ -556,7 +589,15 @@ The first tept will be a formalisation of chain constructions. It suffices to show that $g(x) \in \reals$. There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$. - $g(x)= k$. + + We show that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - g(x)} < \epsilon$ and $g(x)= k$. + \begin{subproof} + Fix $\epsilon \in \reals$. + %Assume $\epsilon > \zero$. + %Take $N' \in \naturals$ such that $\epsilon > \rfrac{1}{\pot(N')}$. + \end{subproof} + + Follows by \cref{apply,plus_one_order,ordinal_iff_suc_ordinal,natural_number_is_ordinal,subseteq,naturals_subseteq_reals,naturals_is_equal_to_two_times_naturals,reals_one_bigger_zero,one_in_reals,ordinal_prec_trichotomy,omega_is_an_ordinal,suc_intro_self,two_in_naturals,in_asymmetric,suc}. @@ -569,13 +610,13 @@ The first tept will be a formalisation of chain constructions. Then $x \in \carrier[X]$. \begin{byCase} \caseOf{$x \in A$.} - For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 0$. + For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = \zero$. \caseOf{$x \notin A$.} \begin{byCase} \caseOf{$x \in B$.} - OFor all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 1$. + For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 1$. \caseOf{$x \notin B$.} Omitted. -- cgit v1.2.3 From ee24a73a01608125e6648f7b66d9c679e955009d Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 13 Aug 2024 23:51:57 +0200 Subject: less urysohn --- library/relation.tex | 1 + library/topology/urysohn.tex | 182 +++++++++++++++++++++++++++++++------------ 2 files changed, 133 insertions(+), 50 deletions(-) (limited to 'library') diff --git a/library/relation.tex b/library/relation.tex index 4c4065f..d63e747 100644 --- a/library/relation.tex +++ b/library/relation.tex @@ -633,6 +633,7 @@ This lets us use the same symbol for composition of functions. \begin{proof} %$x\in\dom{R}$ and $z\in\ran{S}$. There exists $y$ such that $x\mathrel{R} y\mathrel{S} z$ by \cref{circ,pair_eq_iff}. + Follows by assumption. \end{proof} \begin{proposition}\label{circ_iff} diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index c94dbd4..ba9780a 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -1,11 +1,17 @@ \import{topology/topological-space.tex} \import{topology/separation.tex} \import{topology/continuous.tex} +\import{topology/basis.tex} \import{numbers.tex} \import{function.tex} \import{set.tex} \import{cardinal.tex} \import{relation.tex} +\import{relation/uniqueness.tex} +\import{set/cons.tex} +\import{set/powerset.tex} +\import{set/fixpoint.tex} +\import{set/product.tex} \section{Urysohns Lemma} % In this section we want to proof Urysohns lemma. @@ -337,33 +343,8 @@ The first tept will be a formalisation of chain constructions. \end{proof} -\begin{lemma}\label{fraction1} - Let $x \in \reals$. - Then for all $y \in \reals$ such that $x \neq y$ we have there exists $r \in \rationals$ such that $y < r < x$ or $x < r < y$. -\end{lemma} -\begin{proof} - Omitted. -\end{proof} -\begin{lemma}\label{frection2} - Suppose $a,b \in \reals$. - Suppose $a < b$. - Then $\intervalopen{a}{b}$ is infinite. -\end{lemma} -\begin{proof} - Omitted. -\end{proof} -\begin{lemma}\label{frection3} - Suppose $a,b \in \reals$. - Suppose $\zero < a < 1$. - Suppose $\zero < b < 1$. - % Here take exist n such that n/2^n is between a and b - T -\end{lemma} -\begin{proof} - Omitted. -\end{proof} \begin{proposition}\label{existence_of_staircase_function} @@ -449,6 +430,86 @@ The first tept will be a formalisation of chain constructions. $\realsplus = \reals \setminus \realsminus$. \end{abbreviation} +\begin{proposition}\label{intervalclosed_subseteq_reals} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then $\intervalclosed{a}{b} \subseteq \reals$. +\end{proposition} + + + +\begin{lemma}\label{fraction1} + Let $x \in \reals$. + Then for all $y \in \reals$ such that $x \neq y$ we have there exists $r \in \rationals$ such that $y < r < x$ or $x < r < y$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{frection2} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then $\intervalopen{a}{b}$ is infinite. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{frection3} + Suppose $a \in \reals$. + Suppose $a < \zero$. + Then there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\zero < \rfrac{1}{\pot(N')} < a$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{fraction4} + Suppose $a,b,\epsilon \in \reals$. + Suppose $\epsilon > \zero$. + $\abs{a - b} < \epsilon$ iff $b \in \intervalopen{(a - \epsilon)}{(a + \epsilon)}$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{fraction5} + Suppose $a,b,\epsilon \in \reals$. + Suppose $\epsilon > \zero$. + $b \in \intervalopen{(a - \epsilon)}{(a + \epsilon)}$ iff $a \in \intervalopen{(b - \epsilon)}{(b + \epsilon)}$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{fraction6} + Suppose $a,\epsilon \in \reals$. + Suppose $\epsilon > \zero$. + $\intervalopen{(a - \epsilon)}{(a + \epsilon)} = \{r \in \reals \mid (a - \epsilon) < r < (a + \epsilon)\} $. +\end{proposition} + +\begin{abbreviation}\label{epsilonball} + $\epsBall{a}{\epsilon} = \intervalopen{(a - \epsilon)}{(a + \epsilon)}$. +\end{abbreviation} + +\begin{proposition}\label{fraction7} + Suppose $a,\epsilon \in \reals$. + Suppose $\epsilon > \zero$. + Then there exist $b \in \rationals$ such that $b \in \epsBall{a}{\epsilon}$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} + + + +% Note this could maybe reslove some issues!!!! +\begin{definition}\label{sequencetwo} + $Z$ is a sequencetwo iff $Z = (N,f,B)$ and $N \subseteq \naturals$ and $f$ is a bijection from $N$ to $B$. +\end{definition} + + + \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. @@ -483,6 +544,8 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} + For all $n \in \naturals$ we have $\index[\zeta](n)$ is a urysohnchain in $X$. + We show that for all $n \in \indexset[\zeta]$ we have $\index[\zeta](n)$ has cardinality $\pot(n)$. \begin{subproof} Omitted. @@ -493,6 +556,11 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} + We show that for all $m \in \naturals$ we have $\pot(m) \in \naturals$. + \begin{subproof} + Omitted. + \end{subproof} + We show that for all $m \in \indexset[\zeta]$ we have there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and for all $n \in \indexset[\index[\zeta](m)]$ we have for all $x \in \index[\index[\zeta](m)](n)$ @@ -545,37 +613,56 @@ The first tept will be a formalisation of chain constructions. Let $\gamma(n) = \apply{\gamma}{n}$ for $n \in \naturals$. - We show that for all $n \in \naturals$ we have $\gamma(n)$ is a function from $\carrier[X]$ to $\reals$. \begin{subproof} Omitted. \end{subproof} - We show that for all $n \in \naturals$ for all $x \in \carrier[X]$ we have $\apply{\gamma(n)}{x} \in \intervalclosed{\zero}{1}$. + + We show that for all $n \in \naturals$ we have $\gamma(n)$ + is a function from $\carrier[X]$ to $\intervalclosed{\zero}{1}$. \begin{subproof} Omitted. \end{subproof} + We show that $\gamma$ is a function from $\naturals$ to $\reals$. + \begin{subproof} + Omitted. + \end{subproof} + + We show that for all $x,k,n$ such that $n\in \naturals$ and $k \in \naturals$ and $x \in \index[\index[\zeta](n)](k)$ we have $\apply{\gamma(n)}{x} = \rfrac{k}{\pot(n)}$. + \begin{subproof} + Omitted. + \end{subproof} + + We show that for all $n \in \naturals$ for all $x \in \carrier[X]$ we have $\apply{\gamma(n)}{x} \in \intervalclosed{\zero}{1}$. + \begin{subproof} + Fix $n \in \naturals$. + Fix $x \in \carrier[X]$. + \end{subproof} - We show that there exist $g$ such that - for all $x \in \carrier[X]$ we have - there exist $k \in \reals$ such that - for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have - there exist $N \in \naturals$ such that - for all $N' \in \naturals$ such that $N' > N$ we have - $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$ and $g(x)= k$. + + We show that + if $h$ is a function from $\naturals$ to $\reals$ and for all $n \in \naturals$ we have $h(n) \leq h(n+1)$ and there exist $B \in \reals$ such that $h(n) < B$ + then there exist $k \in \reals$ such that for all $m \in \naturals$ we have $h(m) \leq k$ and for all $k' \in \reals$ such that $k' < k$ we have there exist $M \in \naturals$ such that $k' < h(M)$. + \begin{subproof} + Omitted. + \end{subproof} + + + + We show that there exist $g$ such that for all $x \in \carrier[X]$ we have there exist $k \in \reals$ such that $g(x)= k$ and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. \begin{subproof} - We show that for all $x \in \carrier[X]$ we have - there exist $k \in \reals$ such that - for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have - there exist $N \in \naturals$ such that - for all $N' \in \naturals$ such that $N' > N$ we have - $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$. + We show that for all $x \in \carrier[X]$ we have there exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. \begin{subproof} Fix $x \in \carrier[X]$. - Follows by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. + + + + % Contradiction by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. + %Follows by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. \end{subproof} \end{subproof} @@ -588,20 +675,15 @@ The first tept will be a formalisation of chain constructions. Fix $x \in \dom{G}$. It suffices to show that $g(x) \in \reals$. - There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - k} < \epsilon$. + There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. - We show that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(n)}{x} - g(x)} < \epsilon$ and $g(x)= k$. + We show that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - g(x)} < \epsilon$ and $g(x)= k$. \begin{subproof} Fix $\epsilon \in \reals$. - %Assume $\epsilon > \zero$. - %Take $N' \in \naturals$ such that $\epsilon > \rfrac{1}{\pot(N')}$. - \end{subproof} - - - + + \end{subproof} Follows by \cref{apply,plus_one_order,ordinal_iff_suc_ordinal,natural_number_is_ordinal,subseteq,naturals_subseteq_reals,naturals_is_equal_to_two_times_naturals,reals_one_bigger_zero,one_in_reals,ordinal_prec_trichotomy,omega_is_an_ordinal,suc_intro_self,two_in_naturals,in_asymmetric,suc}. - \end{subproof} We show that for all $x \in \dom{G}$ we have $\zero \leq G(x) \leq 1$. -- cgit v1.2.3 From 5a7450eedc020867ee52ae9f1ecb6def7aa998bb Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 14 Aug 2024 10:50:26 +0200 Subject: The added proposition and definition should have resloved some proofing complications, but they can prove flase together with lemmas about ordinal, the exat lemma are in urysohn.tex at line 522. --- library/topology/urysohn.tex | 23 ++++++++++++++++++----- 1 file changed, 18 insertions(+), 5 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index ba9780a..c3c72f0 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -503,11 +503,24 @@ The first tept will be a formalisation of chain constructions. -% Note this could maybe reslove some issues!!!! + \begin{definition}\label{sequencetwo} $Z$ is a sequencetwo iff $Z = (N,f,B)$ and $N \subseteq \naturals$ and $f$ is a bijection from $N$ to $B$. \end{definition} +\begin{proposition}\label{sequence_existence} + Suppose $N \subseteq \naturals$. + Suppose $M \subseteq \naturals$. + Suppose $N = M$. + Then there exist $Z,f$ such that $f$ is a bijection from $N$ to $M$ and $Z=(N,f,M)$ and $Z$ is a sequencetwo. +\end{proposition} +\begin{proof} + Let $f(x) = x$ for $x \in N$. + Let $Z=(N,f,M)$. +\end{proof} +%The proposition above and the definition prove false together with +% ordinal_subseteq_unions, omega_is_an_ordinal, powerset_intro, in_irrefl + @@ -544,7 +557,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} - For all $n \in \naturals$ we have $\index[\zeta](n)$ is a urysohnchain in $X$. + %For all $n \in \naturals$ we have $\index[\zeta](n)$ is a urysohnchain in $X$. We show that for all $n \in \indexset[\zeta]$ we have $\index[\zeta](n)$ has cardinality $\pot(n)$. \begin{subproof} @@ -824,6 +837,6 @@ The first tept will be a formalisation of chain constructions. % \end{subproof} \end{proof} -%\begin{theorem}\label{safe} -% Contradiction. -%\end{theorem} +\begin{theorem}\label{safe} + Contradiction. +\end{theorem} -- cgit v1.2.3 From d5b31ee7dc5992687f214d77e795bab53d5fe65d Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 14 Aug 2024 12:53:47 +0200 Subject: some wishes for NaprocheZF --- library/wunschzettel.tex | 99 ++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 99 insertions(+) create mode 100644 library/wunschzettel.tex (limited to 'library') diff --git a/library/wunschzettel.tex b/library/wunschzettel.tex new file mode 100644 index 0000000..b2681fd --- /dev/null +++ b/library/wunschzettel.tex @@ -0,0 +1,99 @@ +%This is just a .tex file with a wishlist of functionalitys + + +Tupel struct + +\newtheorem{struct2}[theoremcount]{Struct2} + +\begin{theorem} + %Some Theorem. +\end{theorem} +\begin{proof} + %Wish for nice Function definition. --------------------- + + %Some Proof where we need a Function. + %Privisuly defined. + $n \in \naturals$. + There is a Set $A = \{A_{0}, ..., A_{n}\}$. + For all $i$ we have $A_{i} \subseteq X$. + + Define function $f: X \to Y$, + \begin{align} + &x \mapsto \rfrac{y}{n} &; if \exists k \in \{1, ... n\}. x \in A_{k} \\ + &x \mapsto 0 &; if x \phi(x) \\ + %phi is some fol formula + + &x \mapsto \eta &; for \phi(x) and \psi(\eta) + + &x \mapsto \some_term(x)(u)(v)(w) &; \exist.u,v,w \psi(x)(u)(v)(w) \\ + % here i see the real need of varibles that can be useds in the define term + + &x \mapsto \some_else_term(x) &; else + % the else term would be great + + % the following axioms should be automaticly added. + % \dom{f} = X + % \ran{f} \subseteq Y + % f is function + + % therefor we should add the prompt for a proof that f is well defined + \end{align} + \begin{proof_well_defined} + % we need to proof that f allways maps X to Y + \end{proof_well_defined} + + % more proof but now i can use the function f + + % -------------------------------------------------------- + + +\end{proof} + + +%------------------------------------------ +% My wish for a new struct +% I think this could be just get implemented along with the old struct + + +% If take we only take tupels, +% then just a list of defining fol formulas should be enougth. +\begin{struct2} + We say $(X,O)$ is a topological space if + \begin{enumerate} + \item $X$ is a set. % or X = \{...\mid .. \} or X = \naturals ... or ... + \item $O \subseteq \pow X$. + \item $\forall x,y \in O. x \union y \in O$ + \item %another formula + \item %.... + \end{enumerate} +\end{struct2} + + +% Then the proof of some thing is a structure is more easy. +% Since if we have just a tupel and some formulas which has to be fufilled, +% then we can make a proof as follows. + +\begin{struct2} + We say $(A,i,N)$ is a indexed set if + \begin{enumerate} + \item $f$ is a bijection from $N$ to $A$ + \item $N \subseteq \naturals$ + \end{enumerate} +\end{struct2} + + +\begin{theorem} + Let $A = \{ \{n\} \mid n \in \naturals \}$. + Let function $f: \naturals \to \pow{\naturals}$ such that, + \begin{algin} + \item x \mapsto \{x\} ; x \in \naturals + \end{algin} + Then $(A, f, \naturals)$ is a indexed set. +\end{theorem} +\begin{proof} + % Then we only need to proof that: + % \ran{f} = A + % \dom{f} = \naturals + % f is a bijection between $\naturals$ to $A$. +\end{proof} + -- cgit v1.2.3 From 29027c9d2cdbdfe59e48b5aa28eb2d32d1a4c1f7 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 24 Aug 2024 11:43:29 +0200 Subject: naproch sty extension --- latex/naproche.sty | 10 ++++++++ latex/stdlib.tex | 2 ++ library/topology/urysohn.tex | 33 ++++++++++++++----------- library/topology/urysohn2.tex | 56 +++++++++++++++++++++++++++++++++++++++++++ 4 files changed, 87 insertions(+), 14 deletions(-) create mode 100644 library/topology/urysohn2.tex (limited to 'library') diff --git a/latex/naproche.sty b/latex/naproche.sty index 7ef2359..5ca673d 100644 --- a/latex/naproche.sty +++ b/latex/naproche.sty @@ -100,6 +100,7 @@ \newcommand{\memrel}[1]{{\in_{#1}}} \newcommand{\monusNat}{-_{\naturals}} \newcommand{\naturals}{\mathcal{N}} +\newcommand{\naturalsPlus}{\mathcal{N}_{+}} \newcommand{\notmeets}{\mathrel{\not\meets}} \newcommand{\pow}[1]{\fun{Pow}(#1)} \newcommand{\precedes}{<} @@ -133,6 +134,14 @@ \newcommand{\zero}{0} \newcommand{\one}{1} \newcommand{\rmul}{\cdot} +\newcommand{\inv}[1]{#1^{-1}} +\newcommand{\rfrac}[2]{\frac{#1}{#2}} +\newcommand{\rationals}{\mathcal{Q}} +\newcommand{\rminus}{-_{\mathcal{R}}} +\newcommand{\seq}[2]{\{#1, ... ,#2\}} +\newcommand{\indexx}[2]{index_{#1}(#2)} +\newcommand{\indexset}[2]{#1} + \newcommand\restrl[2]{{% we make the whole thing an ordinary symbol @@ -166,6 +175,7 @@ \newcommand{\closure}[2]{\fun{cl}_{#2}{#1}} \newcommand{\frontier}[2]{\fun{fr}_{#2}{#1}} \newcommand{\neighbourhoods}[2]{\textsf{N}_{#2}{#1}} +\newcommand{\neighbourhoodsSet}[2]{\textsf{N}_{SET #2}{#1}} \newcommand{\disconnections}[1]{\fun{Disconnections}{#1}} \newcommand{\teezero}{\ensuremath{T_0}} % Kolmogorov diff --git a/latex/stdlib.tex b/latex/stdlib.tex index dba42a2..2faa267 100644 --- a/latex/stdlib.tex +++ b/latex/stdlib.tex @@ -46,4 +46,6 @@ \input{../library/topology/basis.tex} \input{../library/topology/disconnection.tex} \input{../library/numbers.tex} + \input{../library/topology/urysohn.tex} + \input{../library/wunschzettel.tex} \end{document} diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index c3c72f0..b8a5fa5 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -504,26 +504,31 @@ The first tept will be a formalisation of chain constructions. -\begin{definition}\label{sequencetwo} - $Z$ is a sequencetwo iff $Z = (N,f,B)$ and $N \subseteq \naturals$ and $f$ is a bijection from $N$ to $B$. -\end{definition} - -\begin{proposition}\label{sequence_existence} - Suppose $N \subseteq \naturals$. - Suppose $M \subseteq \naturals$. - Suppose $N = M$. - Then there exist $Z,f$ such that $f$ is a bijection from $N$ to $M$ and $Z=(N,f,M)$ and $Z$ is a sequencetwo. -\end{proposition} -\begin{proof} - Let $f(x) = x$ for $x \in N$. - Let $Z=(N,f,M)$. -\end{proof} +%\begin{definition}\label{sequencetwo} +% $Z$ is a sequencetwo iff $Z = (N,f,B)$ and $N \subseteq \naturals$ and $f$ is a bijection from $N$ to $B$. +%\end{definition} +% +%\begin{proposition}\label{sequence_existence} +% Suppose $N \subseteq \naturals$. +% Suppose $M \subseteq \naturals$. +% Suppose $N = M$. +% Then there exist $Z,f$ such that $f$ is a bijection from $N$ to $M$ and $Z=(N,f,M)$ and $Z$ is a sequencetwo. +%\end{proposition} +%\begin{proof} +% Let $f(x) = x$ for $x \in N$. +% Let $Z=(N,f,M)$. +%\end{proof} %The proposition above and the definition prove false together with % ordinal_subseteq_unions, omega_is_an_ordinal, powerset_intro, in_irrefl + + + + + \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex new file mode 100644 index 0000000..8e5261e --- /dev/null +++ b/library/topology/urysohn2.tex @@ -0,0 +1,56 @@ +\import{topology/topological-space.tex} +\import{topology/separation.tex} +\import{topology/continuous.tex} +\import{topology/basis.tex} +\import{numbers.tex} +\import{function.tex} +\import{set.tex} +\import{cardinal.tex} +\import{relation.tex} +\import{relation/uniqueness.tex} +\import{set/cons.tex} +\import{set/powerset.tex} +\import{set/fixpoint.tex} +\import{set/product.tex} + +\section{Urysohns Lemma} + + + +\begin{abbreviation}\label{urysohnspace} + $X$ is a urysohn space iff + $X$ is a topological space and + for all $A,B \in \closeds{X}$ such that $A \inter B = \emptyset$ + we have there exist $A',B' \in \opens[X]$ + such that $A \subseteq A'$ and $B \subseteq B'$ and $A' \inter B' = \emptyset$. +\end{abbreviation} + + +\begin{definition}\label{intervalclosed} + $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. +\end{definition} + + + + +\begin{theorem}\label{urysohn} + Let $X$ be a urysohn space. + Suppose $A,B \in \closeds{X}$. + Suppose $A \inter B$ is empty. + Suppose $\carrier[X]$ is inhabited. + There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ + and $f(A) = \zero$ and $f(B)= 1$ and $f$ is continuous. +\end{theorem} +\begin{proof} + + + + + + Contradiction. + +\end{proof} + +\begin{theorem}\label{safe} + Contradiction. +\end{theorem} -- cgit v1.2.3 From ce03d33eaa7e9d37935f225d48459223a4004a50 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 24 Aug 2024 19:30:46 +0200 Subject: First atemped to write a new way of local function defintion --- library/wunschzettel.tex | 9 +++++++++ source/Api.hs | 4 +++- source/Syntax/Abstract.hs | 7 +++++++ source/Syntax/Adapt.hs | 14 ++++++++++++++ source/Syntax/Concrete.hs | 27 ++++++++++++++++++++++++++- source/Syntax/Concrete/Keywords.hs | 10 ++++++++-- 6 files changed, 67 insertions(+), 4 deletions(-) (limited to 'library') diff --git a/library/wunschzettel.tex b/library/wunschzettel.tex index b2681fd..74ea899 100644 --- a/library/wunschzettel.tex +++ b/library/wunschzettel.tex @@ -45,6 +45,15 @@ Tupel struct % more proof but now i can use the function f % -------------------------------------------------------- + \begin{equation} + X= + \begin{cases} + 0, & \text{if}\ a=1 \\ + 1, & \text{otherwise} + \end{cases} + \end{equation} + + \end{proof} diff --git a/source/Api.hs b/source/Api.hs index 3fb0ca2..1bdf615 100644 --- a/source/Api.hs +++ b/source/Api.hs @@ -203,7 +203,9 @@ describeToken = \case EndEnv _ -> "end of environment" _ -> "delimiter" - +-- | gloss generates internal represantation of the LaTeX files. +-- First the file will be parsed and therefore checkt for grammer. +-- 'meaning' then transfer the raw parsed grammer to the internal semantics. gloss :: MonadIO io => FilePath -> io ([Internal.Block], Lexicon) gloss file = do (blocks, lexicon) <- parse file diff --git a/source/Syntax/Abstract.hs b/source/Syntax/Abstract.hs index 4aa8623..f775b69 100644 --- a/source/Syntax/Abstract.hs +++ b/source/Syntax/Abstract.hs @@ -369,6 +369,13 @@ data Proof -- ^ Local function definition, e.g. /@Let $f(x) = e$ for $x\\in d$@/. -- The first 'VarSymbol' is the newly defined symbol, the second one is the argument. -- The first 'Expr' is the value, the final variable and expr specify a bound (the domain of the function). + + + + + | DefineFunctionMathy VarSymbol VarSymbol VarSymbol [VarSymbol Expr [VarSymbol] Expr ] Proof + -- ^ Local function definition, but in this case we give the domain and target an the rules for $xs$ in some sub domains. + -- deriving (Show, Eq, Ord) -- | An inline justification. diff --git a/source/Syntax/Adapt.hs b/source/Syntax/Adapt.hs index 622946a..96fd76d 100644 --- a/source/Syntax/Adapt.hs +++ b/source/Syntax/Adapt.hs @@ -34,6 +34,8 @@ scanChunk ltoks = matchOrErr struct "struct definition" pos Located{startPos = pos, unLocated = (BeginEnv "inductive")} :_ -> matchOrErr inductive "inductive definition" pos + Located{startPos = pos, unLocated = (BeginEnv "signature")} :_ -> + matchOrErr signature "signature" pos _ -> [] adaptChunks :: [[Located Token]] -> Lexicon -> Lexicon @@ -85,6 +87,18 @@ abbreviation = do skipUntilNextLexicalEnv pure [lexicalItem m] +signatureIntro :: RE Token [ScannedLexicalItem] --since signiture is a used word of haskell we have to name it diffrentliy +signatureIntro = do + sym (BeginEnv "signature") + few notEndOfLexicalEnvToken + m <- label + few anySym + lexicalItem <- head + few anySym + sym (EndEnv "signature") + skipUntilNextLexicalEnv + pure [lexicalItem m] + label :: RE Token Marker label = msym \case Label m -> Just (Marker m) diff --git a/source/Syntax/Concrete.hs b/source/Syntax/Concrete.hs index b51b738..51cc013 100644 --- a/source/Syntax/Concrete.hs +++ b/source/Syntax/Concrete.hs @@ -351,8 +351,23 @@ grammar lexicon@Lexicon{..} = mdo define <- rule $ Define <$> (_let *> beginMath *> varSymbol <* _eq) <*> expr <* endMath <* _dot <*> proof defineFunction <- rule $ DefineFunction <$> (_let *> beginMath *> varSymbol) <*> paren varSymbol <* _eq <*> expr <* endMath <* _for <* beginMath <*> varSymbol <* _in <*> expr <* endMath <* _dot <*> proof + - proof <- rule $ asum [byContradiction, byCases, bySetInduction, byOrdInduction, calc, subclaim, assume, fix, take, have, suffices, define, defineFunction, qed] + + -- Define $f $\fromTo{X}{Y} such that, + -- Define function $f: X \to Y$, + -- \begin{align} + -- &x \mapsto 3*x &, + -- &x \mapsto 4*k+x &, + -- \end{align} + -- + defineFunctionMathy <- rule $ DefineFunctionMathy + <$> (_define *> beginMath *> varSymbol) -- Define $ f + <*> _colon *> varSymbol <*> _to *> varSymbol -- : 'var' \to 'var' + <*> localFunctionDefinitionAlign + + + proof <- rule $ asum [byContradiction, byCases, bySetInduction, byOrdInduction, calc, subclaim, assume, fix, take, have, suffices, define, defineFunction, defineFunctionMathy, qed] blockAxiom <- rule $ uncurry3 BlockAxiom <$> envPos "axiom" axiom @@ -435,7 +450,17 @@ enumeratedMarked1 :: Prod r Text (Located Token) a -> Prod r Text (Located Token enumeratedMarked1 p = begin "enumerate" *> many1 ((,) <$> (command "item" *> label) <*> p) <* end "enumerate" "\"\\begin{enumerate}\\item\\label{...}...\"" +-- &x \mapsto 'someexpr' &, for x +localFunctionDefinitionAlign :: Prod r Text (Located Token) a -> Prod r Text (Located Token) (Marker, a) +localFunctionDefinitionAlign p = begin "align" *> many1 funDefExp <* end "align" + "\"\\begin{algin} &x \\mapsto x+2 , x \\in X \\ \\end{algin}\"" + + +funDefExp :: Prod r Text (Located Token) a -> Prod r Text (Located Token) [(Marker, a)] +funDefExp p = NonEmpty.toList <$> ( _ampersand *> varSymbol <*> funDefExpRange <*> (_ampersand *> varSymbol <* _in) <*> varSymbol) -- the last var should be a expression +funDefRange :: Prod r Text (Located Token) a -> Prod r Text (Located Token) (NonEmpty (Marker, a)) +funDefRange p = _mapsto *> varSymbol -- TODO: this Var has to be changed to a expression -- This function could be rewritten, so that it can be used directly in the grammar, -- instead of with specialized variants. diff --git a/source/Syntax/Concrete/Keywords.hs b/source/Syntax/Concrete/Keywords.hs index 135cdac..b507e7e 100644 --- a/source/Syntax/Concrete/Keywords.hs +++ b/source/Syntax/Concrete/Keywords.hs @@ -108,7 +108,7 @@ _either = word "either" ? "either" _equipped :: Prod r Text (Located Token) SourcePos _equipped = (word "equipped" <|> word "together") <* word "with" ? "equipped with" _every :: Prod r Text (Located Token) SourcePos -_every = (word "every") ? "every" +_every = word "every" ? "every" _exist :: Prod r Text (Located Token) SourcePos _exist = word "there" <* word "exist" ? "there exist" _exists :: Prod r Text (Located Token) SourcePos @@ -124,7 +124,7 @@ _for = word "for" ? "for" _forAll :: Prod r Text (Located Token) SourcePos _forAll = (word "for" <* word "all") <|> word "all" ? "all" _forEvery :: Prod r Text (Located Token) SourcePos -_forEvery = (word "for" <* word "every") <|> (word "every") ? "for every" +_forEvery = (word "for" <* word "every") <|> word "every" ? "for every" _have :: Prod r Text (Located Token) SourcePos _have = word "we" <* word "have" <* optional (word "that") ? "we have" _if :: Prod r Text (Located Token) SourcePos @@ -220,3 +220,9 @@ _in :: Prod r Text (Located Token) SourcePos _in = symbol "∈" <|> command "in" ? "\\in" _subseteq :: Prod r Text (Located Token) SourcePos _subseteq = command "subseteq" ? "\\subseteq" +_to :: Prod r Text (Located Token) SourcePos +_to = command "to" ? "\\to" +_mapsto :: Prod r Text (Located Token) SourcePos +_mapsto = command "mapsto" ? "\\mapsto" +_ampersand :: Prod r Text (Located Token) SourcePos +_ampersand = symbol "&" ? "&" \ No newline at end of file -- cgit v1.2.3 From 30f7c63ce566c993816607f3368c357233693aae Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 27 Aug 2024 01:44:45 +0200 Subject: Experimental working commit, programm will compile But the Proof that the domain of the local function is not right. Also if in the definition of our local function we just use f(x) = x then we get a technical ambigus parse --- library/topology/urysohn2.tex | 8 ++++++-- source/Checking.hs | 13 +++++++------ source/Meaning.hs | 4 ++-- source/Syntax/Abstract.hs | 2 +- source/Syntax/Adapt.hs | 4 ++-- source/Syntax/Concrete.hs | 13 ++++++++----- source/Syntax/Internal.hs | 2 +- source/Syntax/Token.hs | 17 +++++++++++++++-- 8 files changed, 42 insertions(+), 21 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 8e5261e..05ea180 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -44,10 +44,14 @@ \begin{proof} + Define $f : X \to \reals$ such that $f(x) = $ + \begin{cases} + &(x + x) , x \in X + % & x ,x \in X <- will result in technicly ambigus parse + \end{cases} - + Trivial. - Contradiction. \end{proof} diff --git a/source/Checking.hs b/source/Checking.hs index 817c477..6d55ee1 100644 --- a/source/Checking.hs +++ b/source/Checking.hs @@ -543,7 +543,7 @@ checkProof = \case checkCalc calc assume [Asm (calcResult calc)] checkProof continue - DefineFunctionMathy funVar domVar ranVar argVar definitions continue -> do + DefineFunctionMathy funVar argVar domVar ranExpr definitions continue -> do -- We have f: X \to Y and x \mapsto ... -- definition is a nonempty list of (expresssion e, formula phi) -- such that f(x) = e if phi(x) @@ -558,8 +558,8 @@ checkProof = \case setGoals [makeForall [argVar] ((TermVar argVar `IsElementOf` TermVar domVar) `Iff` subdomainConjuctionLocalFunction argVar definitions )] -- check the disjunct union tellTasks - assume [Asm (makeForall [argVar] ((TermVar argVar `IsElementOf` TermVar domVar) `Implies` (TermOp ApplySymbol [TermVar funVar, TermVar argVar] `IsElementOf` TermVar ranVar)))] -- function f from \dom(f) \to \ran(f) - assume (functionSubdomianExpression funVar argVar definitions) --behavior on the subdomians + assume [Asm (makeForall [argVar] ((TermVar argVar `IsElementOf` TermVar domVar) `Implies` (TermOp ApplySymbol [TermVar funVar, TermVar argVar] `IsElementOf` ranExpr)))] -- function f from \dom(f) \to \ran(f) + assume (functionSubdomianExpression funVar argVar (NonEmpty.toList definitions)) --behavior on the subdomians setGoals goals checkProof continue @@ -571,9 +571,10 @@ subdomainConjuctionLocalFunction argVar defintions = in TermVar argVar `IsElementOf` makeConjunction stmts -functionSubdomianExpression :: VarSymbol -> VarSymbol -> NonEmpty (Term, Formula) -> [Asm] -functionSubdomianExpression f a nxs = case nxs of - x:|xs -> singleFunctionSubdomianExpression f a x : functionSubdomianExpression f a (NonEmpty.fromList xs) +functionSubdomianExpression :: VarSymbol -> VarSymbol -> [(Term, Formula)] -> [Asm] +functionSubdomianExpression f a (x:xs) = singleFunctionSubdomianExpression f a x : functionSubdomianExpression f a xs +functionSubdomianExpression _ _ [] = [] + singleFunctionSubdomianExpression :: VarSymbol -> VarSymbol -> (Term, Formula) -> Asm diff --git a/source/Meaning.hs b/source/Meaning.hs index ab98c9a..4a21fa3 100644 --- a/source/Meaning.hs +++ b/source/Meaning.hs @@ -606,9 +606,9 @@ glossProof = \case then Sem.DefineFunction funVar argVar <$> glossExpr valueExpr <*> glossExpr domExpr <*> glossProof proof else error "mismatched variables in function definition." - Raw.DefineFunctionMathy funVar domVar ranVar funVar2 argVar definitions proof -> do + Raw.DefineFunctionMathy funVar domVar ranExpr funVar2 argVar definitions proof -> do if funVar == funVar2 - then Sem.DefineFunctionMathy funVar domVar ranVar argVar <$> (glossLocalFunctionExprDef `each` definitions) <*> glossProof proof + then Sem.DefineFunctionMathy funVar argVar domVar <$> glossExpr ranExpr <*> (glossLocalFunctionExprDef `each` definitions) <*> glossProof proof else error "missmatched function names" Raw.Calc calc proof -> Sem.Calc <$> glossCalc calc <*> glossProof proof diff --git a/source/Syntax/Abstract.hs b/source/Syntax/Abstract.hs index 6372c87..13691e7 100644 --- a/source/Syntax/Abstract.hs +++ b/source/Syntax/Abstract.hs @@ -373,7 +373,7 @@ data Proof - | DefineFunctionMathy VarSymbol VarSymbol VarSymbol VarSymbol VarSymbol (NonEmpty (Expr, Formula)) Proof + | DefineFunctionMathy VarSymbol VarSymbol Expr VarSymbol VarSymbol (NonEmpty (Expr, Formula)) Proof -- ^ Local function definition, but in this case we give the domain and target an the rules for $xs$ in some sub domains. -- deriving (Show, Eq, Ord) diff --git a/source/Syntax/Adapt.hs b/source/Syntax/Adapt.hs index 3cff497..4b43bc6 100644 --- a/source/Syntax/Adapt.hs +++ b/source/Syntax/Adapt.hs @@ -34,8 +34,8 @@ scanChunk ltoks = matchOrErr struct "struct definition" pos Located{startPos = pos, unLocated = (BeginEnv "inductive")} :_ -> matchOrErr inductive "inductive definition" pos - Located{startPos = pos, unLocated = (BeginEnv "signature")} :_ -> - matchOrErr signatureIntro "signature" pos + --Located{startPos = pos, unLocated = (BeginEnv "signature")} :_ -> + -- matchOrErr signatureIntro "signature" pos _ -> [] adaptChunks :: [[Located Token]] -> Lexicon -> Lexicon diff --git a/source/Syntax/Concrete.hs b/source/Syntax/Concrete.hs index 69280c1..fe08fec 100644 --- a/source/Syntax/Concrete.hs +++ b/source/Syntax/Concrete.hs @@ -373,16 +373,16 @@ grammar lexicon@Lexicon{..} = mdo -- 3 & \text{else} -- \end{cases} - functionDefineCase <- rule $ (,) <$> expr <*> (_ampersand *> formula) + functionDefineCase <- rule $ (,) <$> (_ampersand *> (expr <|> exprVar )) <*> (_comma *> formula) defineFunctionMathy <- rule $ DefineFunctionMathy <$> (_define *> beginMath *> varSymbol) -- Define $ f <*> (_colon *> varSymbol) -- : 'var' \to 'var' - <*> (_to *> varSymbol <* endMath <* _suchThat) + <*> (_to *> expr <* endMath <* _suchThat) -- <*> (_suchThat *> align (many1 ((_ampersand *> varSymbol <* _mapsto) <*> exprApp <*> (_ampersand *> formula)))) -- <*> (_suchThat *> align (many1 (varSymbol <* exprApp <* formula))) - <*> varSymbol <*> varSymbol <* symbol "=" - <*> many1 functionDefineCase - <*> proof + <*> (beginMath *> varSymbol) <*> (paren varSymbol <* _eq <* endMath) + <*> cases (many1 functionDefineCase) + <*> proof @@ -644,6 +644,9 @@ group body = token InvisibleBraceL *> body <* token InvisibleBraceR "\"{...} align :: Prod r Text (Located Token) a -> Prod r Text (Located Token) a align body = begin "align*" *> body <* end "align*" +cases :: Prod r Text (Located Token) a -> Prod r Text (Located Token) a +cases body = begin "cases" *> body <* end "cases" + maybeVarToken :: Located Token -> Maybe VarSymbol maybeVarToken ltok = case unLocated ltok of diff --git a/source/Syntax/Internal.hs b/source/Syntax/Internal.hs index 0e3361d..7046161 100644 --- a/source/Syntax/Internal.hs +++ b/source/Syntax/Internal.hs @@ -436,7 +436,7 @@ data Proof | Define VarSymbol Term Proof | DefineFunction VarSymbol VarSymbol Term Term Proof - | DefineFunctionMathy VarSymbol VarSymbol VarSymbol VarSymbol (NonEmpty (Term, Formula)) Proof + | DefineFunctionMathy VarSymbol VarSymbol VarSymbol Term (NonEmpty (Term, Formula)) Proof deriving instance Show Proof deriving instance Eq Proof diff --git a/source/Syntax/Token.hs b/source/Syntax/Token.hs index cb3f4cb..52da86a 100644 --- a/source/Syntax/Token.hs +++ b/source/Syntax/Token.hs @@ -189,6 +189,7 @@ toks = whitespace *> goNormal id <* eof Nothing -> pure (f []) Just t@Located{unLocated = BeginEnv "math"} -> goMath (f . (t:)) Just t@Located{unLocated = BeginEnv "align*"} -> goMath (f . (t:)) + Just t@Located{unLocated = BeginEnv "cases"} -> goMath (f . (t:)) Just t -> goNormal (f . (t:)) goText f = do r <- optional textToken @@ -204,6 +205,7 @@ toks = whitespace *> goNormal id <* eof Nothing -> pure (f []) Just t@Located{unLocated = EndEnv "math"} -> goNormal (f . (t:)) Just t@Located{unLocated = EndEnv "align*"} -> goNormal (f . (t:)) + Just t@Located{unLocated = EndEnv "cases"} -> goNormal (f . (t:)) Just t@Located{unLocated = BeginEnv "text"} -> goText (f . (t:)) Just t@Located{unLocated = BeginEnv "explanation"} -> goText (f . (t:)) Just t -> goMath (f . (t:)) @@ -219,12 +221,12 @@ toks = whitespace *> goNormal id <* eof -- | Parses a single normal mode token. tok :: Lexer (Located Token) tok = - word <|> var <|> symbol <|> mathBegin <|> alignBegin <|> begin <|> end <|> opening <|> closing <|> label <|> ref <|> command + word <|> var <|> symbol <|> mathBegin <|> alignBegin <|> casesBegin <|> begin <|> end <|> opening <|> closing <|> label <|> ref <|> command -- | Parses a single math mode token. mathToken :: Lexer (Located Token) mathToken = - var <|> symbol <|> number <|> begin <|> alignEnd <|> end <|> opening <|> closing <|> beginText <|> beginExplanation <|> mathEnd <|> command + var <|> symbol <|> number <|> begin <|> alignEnd <|> casesEnd <|> end <|> opening <|> closing <|> beginText <|> beginExplanation <|> mathEnd <|> command beginText :: Lexer (Located Token) beginText = lexeme do @@ -277,6 +279,11 @@ alignBegin = guardM isTextMode *> lexeme do setMathMode pure (BeginEnv "align*") +casesBegin :: Lexer (Located Token) +casesBegin = guardM isTextMode *> lexeme do + Char.string "\\begin{cases}" + setMathMode + pure (BeginEnv "cases") -- | Parses a single end math token. mathEnd :: Lexer (Located Token) @@ -291,6 +298,12 @@ alignEnd = guardM isMathMode *> lexeme do setTextMode pure (EndEnv "align*") +casesEnd :: Lexer (Located Token) +casesEnd = guardM isMathMode *> lexeme do + Char.string "\\end{cases}" + setTextMode + pure (EndEnv "cases") + -- | Parses a word. Words are returned casefolded, since we want to ignore their case later on. word :: Lexer (Located Token) -- cgit v1.2.3 From 604d7a87b4c45ab13ef03e3c7a611ad4f9342f23 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 27 Aug 2024 02:13:29 +0200 Subject: ambigus parse fix. The proof goal must be changed, since now some could define a function with overlapping and worng subdomains --- library/topology/urysohn2.tex | 4 +++- source/Checking.hs | 6 +++--- source/Syntax/Concrete.hs | 2 +- 3 files changed, 7 insertions(+), 5 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 05ea180..f2f6ef3 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -46,7 +46,9 @@ Define $f : X \to \reals$ such that $f(x) = $ \begin{cases} - &(x + x) , x \in X + &(x + x) &\text{if} x \in X + & x &\text{if} x \neq \zero + & \zero & \text{if} x = \zero % & x ,x \in X <- will result in technicly ambigus parse \end{cases} diff --git a/source/Checking.hs b/source/Checking.hs index 6d55ee1..7362c93 100644 --- a/source/Checking.hs +++ b/source/Checking.hs @@ -564,11 +564,11 @@ checkProof = \case checkProof continue --- | Makes a conjunction of all the subdomain statments +-- | Makes a conjunction of all the subdomain statments <- this has to be checked!!!! TODO!!!!!! subdomainConjuctionLocalFunction :: VarSymbol -> NonEmpty (Term, Formula) -> Formula -subdomainConjuctionLocalFunction argVar defintions = +subdomainConjuctionLocalFunction _ defintions = let stmts = [snd x | x <- NonEmpty.toList defintions] - in TermVar argVar `IsElementOf` makeConjunction stmts + in makeDisjunction stmts functionSubdomianExpression :: VarSymbol -> VarSymbol -> [(Term, Formula)] -> [Asm] diff --git a/source/Syntax/Concrete.hs b/source/Syntax/Concrete.hs index fe08fec..9d52995 100644 --- a/source/Syntax/Concrete.hs +++ b/source/Syntax/Concrete.hs @@ -373,7 +373,7 @@ grammar lexicon@Lexicon{..} = mdo -- 3 & \text{else} -- \end{cases} - functionDefineCase <- rule $ (,) <$> (_ampersand *> (expr <|> exprVar )) <*> (_comma *> formula) + functionDefineCase <- rule $ (,) <$> (_ampersand *> expr) <*> (_ampersand *> text _if *> formula) defineFunctionMathy <- rule $ DefineFunctionMathy <$> (_define *> beginMath *> varSymbol) -- Define $ f <*> (_colon *> varSymbol) -- : 'var' \to 'var' -- cgit v1.2.3 From 6acc5654f1702f2466006564a415546a3def16e3 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 27 Aug 2024 17:19:02 +0200 Subject: Feature Complete Finalised the proof output and goals. The Local Function definition now producces a Function f with the right domain and range, together with the rules presented in cases. Then proof goal of this local definition is set to for all x we have x is element of dom(f) if and only if x is in exactly one of the subdomains. This suffices as welldefindness check on f, besides the right range. Further checks that should be implemented are the correct range of the function. And optional subproof such that the presented goal can be check easily. --- library/topology/urysohn2.tex | 2 +- source/Checking.hs | 37 ++++++++++++++++++++++++------------- source/Syntax/Internal.hs | 10 ++++++++++ 3 files changed, 35 insertions(+), 14 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index f2f6ef3..ea49a6c 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -46,7 +46,7 @@ Define $f : X \to \reals$ such that $f(x) = $ \begin{cases} - &(x + x) &\text{if} x \in X + &(x + k) &\text{if} x \in X \land k \in \naturals & x &\text{if} x \neq \zero & \zero & \text{if} x = \zero % & x ,x \in X <- will result in technicly ambigus parse diff --git a/source/Checking.hs b/source/Checking.hs index 7362c93..b43923c 100644 --- a/source/Checking.hs +++ b/source/Checking.hs @@ -555,30 +555,41 @@ checkProof = \case ,Asm (relationNoun (TermVar funVar))] goals <- gets checkingGoals - setGoals [makeForall [argVar] ((TermVar argVar `IsElementOf` TermVar domVar) `Iff` subdomainConjuctionLocalFunction argVar definitions )] -- check the disjunct union + setGoals [makeForall [argVar] ((TermVar argVar `IsElementOf` TermVar domVar) `Iff` localFunctionGoal definitions)] tellTasks assume [Asm (makeForall [argVar] ((TermVar argVar `IsElementOf` TermVar domVar) `Implies` (TermOp ApplySymbol [TermVar funVar, TermVar argVar] `IsElementOf` ranExpr)))] -- function f from \dom(f) \to \ran(f) - assume (functionSubdomianExpression funVar argVar (NonEmpty.toList definitions)) --behavior on the subdomians + assume (functionSubdomianExpression funVar argVar domVar (NonEmpty.toList definitions)) --behavior on the subdomians setGoals goals checkProof continue +-- |Creats the Goal \forall x. x \in dom{f} \iff (phi_{1}(x) \xor (\phi_{2}(x) \xor (... \xor (\phi_{n}) ..))) +-- where the phi_{i} are the subdomain statments +localFunctionGoal :: NonEmpty (Term,Formula) -> Formula +localFunctionGoal xs = makeXor $ map snd $ NonEmpty.toList xs --- | Makes a conjunction of all the subdomain statments <- this has to be checked!!!! TODO!!!!!! -subdomainConjuctionLocalFunction :: VarSymbol -> NonEmpty (Term, Formula) -> Formula -subdomainConjuctionLocalFunction _ defintions = - let stmts = [snd x | x <- NonEmpty.toList defintions] - in makeDisjunction stmts +-- We have our list of expr and forumlas, in this case normaly someone would write +-- f(x) = ....cases +-- & (\frac{1}{k} \cdot x) &\text{if} x \in \[k, k+1\) +-- +-- So therefore we have to check that all free varibels in lhs are free on rhs, +-- since its an expression all varibals from the lhs has to be free on rhs. +-- +-- so we take (exp, frm) -> binding all free in exp +-- -> make forall [x, bound vars] x \in \dom(f) & frm => exp -functionSubdomianExpression :: VarSymbol -> VarSymbol -> [(Term, Formula)] -> [Asm] -functionSubdomianExpression f a (x:xs) = singleFunctionSubdomianExpression f a x : functionSubdomianExpression f a xs -functionSubdomianExpression _ _ [] = [] - +functionSubdomianExpression :: VarSymbol -> VarSymbol -> VarSymbol -> [(Term, Formula)] -> [Asm] +functionSubdomianExpression f a d (x:xs) = singleFunctionSubdomianExpression f a d x : functionSubdomianExpression f a d xs +functionSubdomianExpression _ _ _ [] = [] -singleFunctionSubdomianExpression :: VarSymbol -> VarSymbol -> (Term, Formula) -> Asm -singleFunctionSubdomianExpression funVar argVar x = Asm (makeForall [argVar] ((TermVar argVar `IsElementOf` snd x) `Implies` (TermOp ApplySymbol [TermVar funVar, TermVar argVar] `Equals` fst x))) +singleFunctionSubdomianExpression :: VarSymbol -> VarSymbol -> VarSymbol -> (Term, Formula) -> Asm +singleFunctionSubdomianExpression funVar argVar domVar (expr, frm) = let + boundVar = Set.toList (freeVars expr) in + let def = makeForall (argVar:boundVar) (((TermVar argVar `IsElementOf` TermVar domVar) `And` frm) `Implies` TermOp ApplySymbol [TermVar funVar, TermVar argVar] `Equals` expr) + in Asm def + checkCalc :: Calc -> Checking diff --git a/source/Syntax/Internal.hs b/source/Syntax/Internal.hs index 7046161..e83126d 100644 --- a/source/Syntax/Internal.hs +++ b/source/Syntax/Internal.hs @@ -324,6 +324,16 @@ makeDisjunction = \case [] -> Bottom es -> List.foldl1' Or es +makeIff :: [ExprOf a] -> ExprOf a +makeIff = \case + [] -> Bottom + es -> List.foldl1' Iff es + +makeXor :: [ExprOf a] -> ExprOf a +makeXor = \case + [] -> Bottom + es -> List.foldl1' Xor es + finiteSet :: NonEmpty (ExprOf a) -> ExprOf a finiteSet = foldr cons EmptySet where -- cgit v1.2.3 From b8cc467735054bb3c38bf37b5e29877ba756c4b5 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 27 Aug 2024 20:20:46 +0200 Subject: working commit --- library/everything.tex | 1 + library/topology/urysohn2.tex | 19 +++++++++++++++++++ 2 files changed, 20 insertions(+) (limited to 'library') diff --git a/library/everything.tex b/library/everything.tex index b966197..94dd6d8 100644 --- a/library/everything.tex +++ b/library/everything.tex @@ -30,6 +30,7 @@ \import{topology/disconnection.tex} \import{topology/separation.tex} \import{numbers.tex} +\import{topology/urysohn2.tex} \begin{proposition}\label{trivial} $x = x$. diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index ea49a6c..a64fa7e 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -16,6 +16,22 @@ \section{Urysohns Lemma} +\begin{definition}\label{one_to_n_set} + $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. +\end{definition} + +\begin{struct}\label{sequence} + A sequence $X$ is a onesorted structure equipped with + \begin{enumerate} + \item $\index$ + \item $\indexset$ + \end{enumerate} + such that + \begin{enumerate} + \item\label{indexset_is_subset_naturals} $\indexset[X] \subseteq \naturals$. + \item\label{index_is_bijection} $\index[X]$ is a bijection from $\indexset[X]$ to $\carrier[X]$. + \end{enumerate} +\end{struct} \begin{abbreviation}\label{urysohnspace} $X$ is a urysohn space iff @@ -33,6 +49,7 @@ + \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. @@ -52,6 +69,8 @@ % & x ,x \in X <- will result in technicly ambigus parse \end{cases} + + $U_1$ Trivial. -- cgit v1.2.3 From b64a56c6325e1c2f0876049fb2a7dda6c06dbe3a Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 28 Aug 2024 14:17:27 +0200 Subject: working commit --- library/topology/urysohn2.tex | 14 ++++++++++++-- 1 file changed, 12 insertions(+), 2 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index a64fa7e..40a3615 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -46,7 +46,13 @@ $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. \end{definition} - +\begin{theorem}\label{urysohnsetinbeetween} + Let $X$ be a urysohn space. + Suppose $A,B \in \closeds{X}$. + Suppose $\closure{A}{X} \subseteq \interior{B}{X}$. + Suppose $\carrier[X]$ is inhabited. + There exist $U \subseteq \carrier[X]$ such that $U$ is closed in $X$ and $\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{B}{X}$. +\end{theorem} @@ -60,6 +66,10 @@ \end{theorem} \begin{proof} + Let $H = \carrier[X] \setminus B$. + Let $P = \{x \in \pow{X} \mid x = A \lor x = H \lor (x \in \pow{X} \land (\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{H}{X}))\}$. + + Define $f : X \to \reals$ such that $f(x) = $ \begin{cases} @@ -70,7 +80,7 @@ \end{cases} - $U_1$ + Trivial. -- cgit v1.2.3 From 565db8eb643673f15c44bd8a8ac30debc9b388fd Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Thu, 29 Aug 2024 15:22:11 +0200 Subject: working commit --- library/topology/urysohn.tex | 43 +++++++++++++++++++++++++++++++------------ library/topology/urysohn2.tex | 30 ++++++++++++++++++++++-------- 2 files changed, 53 insertions(+), 20 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index b8a5fa5..79d65dc 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -100,6 +100,14 @@ The first tept will be a formalisation of chain constructions. \subsection{staircase function} +\begin{definition}\label{minimum} + $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. +\end{definition} + +\begin{definition}\label{maximum} + $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. +\end{definition} + \begin{definition}\label{intervalclosed} $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. \end{definition} @@ -122,6 +130,9 @@ The first tept will be a formalisation of chain constructions. \item \label{staircase_chain_is_in_domain} for all $x \in C$ we have $x \subseteq \dom{f}$. \item \label{staircase_behavoir_index_zero} $f(\index[C](1))= 1$. \item \label{staircase_behavoir_index_n} $f(\dom{f}\setminus \unions{C}) = \zero$. + \item \label{staircase_chain_indeset} There exist $n$ such that $\indexset[C] = \seq{\zero}{n}$. + \item \label{staircase_behavoir_index_arbetrray} for all $n \in \indexset[C]$ + such that $n \neq \zero$ we have $f(\index[C](n) \setminus \index[C](n-1)) = \rfrac{n}{ \max{\indexset[C]} }$. \end{enumerate} \end{struct} @@ -311,13 +322,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{definition}\label{minimum} - $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. -\end{definition} -\begin{definition}\label{maximum} - $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. -\end{definition} \begin{proposition}\label{urysohnchain_induction_step_existence} Let $X$ be a urysohn space. @@ -562,6 +567,16 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{subproof} + Let $\alpha = \carrier[X]$. + %Define $f : \alpha \to \naturals$ such that $f(x) =$ + %\begin{cases} + % & 1 & \text{if} x \in A \lor x \in B + % & k & \text{if} \exists k \in \naturals + %\end{cases} +% + %We show that there exist $k \in \funs{\carrier[X]}{\reals}$ such that + %$k(x)$ + %For all $n \in \naturals$ we have $\index[\zeta](n)$ is a urysohnchain in $X$. We show that for all $n \in \indexset[\zeta]$ we have $\index[\zeta](n)$ has cardinality $\pot(n)$. @@ -581,15 +596,19 @@ The first tept will be a formalisation of chain constructions. We show that for all $m \in \indexset[\zeta]$ we have there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ - and for all $n \in \indexset[\index[\zeta](m)]$ we have for all $x \in \index[\index[\zeta](m)](n)$ + and for all $n \in \indexset[\index[\zeta](m)]$ we have for all $x \in \index[\index[\zeta](m)](n)$ such that $x \notin \index[\index[\zeta](m)](n-1)$ we have $f(x) = \rfrac{n}{\pot(m)}$. \begin{subproof} - % Fix $m \in \indexset[\zeta]$. - % $\index[\zeta](m)$ is a urysohnchain in $X$. - % - % Follows by \cref{existence_of_staircase_function}. + Fix $m \in \indexset[\zeta]$. + $\index[\zeta](m)$ is a urysohnchain in $X$. + Define $o : \alpha \to \intervalclosed{\zero}{1}$ such that $o(x) =$ + \begin{cases} + & 0 & \text{if} x \in A + & 1 & \text{if} x \in B + & \rfrac{n}{m} & \text{if} \exists n \in \naturals. x \in \index[\index[\zeta](m)](n) \land x \notin \index[\index[\zeta](m)](n-1) + \end{cases} - Omitted. + \end{subproof} diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 40a3615..71de210 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -68,18 +68,32 @@ Let $H = \carrier[X] \setminus B$. Let $P = \{x \in \pow{X} \mid x = A \lor x = H \lor (x \in \pow{X} \land (\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{H}{X}))\}$. + Let $\eta = \carrier[X]$. - - - Define $f : X \to \reals$ such that $f(x) = $ + + % Provable + % vvv + Define $F : \eta \to \reals$ such that $F(x) =$ \begin{cases} - &(x + k) &\text{if} x \in X \land k \in \naturals - & x &\text{if} x \neq \zero - & \zero & \text{if} x = \zero - % & x ,x \in X <- will result in technicly ambigus parse + & \zero &\text{if} x \in A\\ + & \rfrac{1}{1+1} &\text{if} x \in (\carrier[X] \setminus (A \union B))\\ + & 1 &\text{if} x \in B \end{cases} - + %Define $f : \naturals \to \pow{P}$ such that $f(x)=$ + %\begin{cases} + % & \emptyset & \text{if} x = \zero \\ + % & \{A, H\} & \text{if} x = 1 \\ + % & G & \text{if} x \in (\naturals \setminus \{1, \zero\}) \land G = \{g \in \pow{P} \mid g \in f(n-1) \lor (g \notin f(n-1) \land g \in P) \} + %\end{cases} + + Let $D = \{d \mid d \in \rationals \mid \zero \leq d \leq 1\}$. + Take $R$ such that for all $q \in D$ we have for all $S \in P$ we have $q \mathrel{R} S$ iff + $q = \zero \land S = A$ or $q = 1 \land S = H$ or + for all $q_1, q_2, S_1, S_2$ + such that $q_1 \leq q \leq q_2$ and $q_1 \mathrel{R} S_1$ and $q_2 \mathrel{R} S_2$ + we have $\closure{S_1}{X} \subseteq \interior{S}{X} \subseteq \closure{S}{X} \subseteq \interior{S_2}{X}$ + and $q \mathrel{R} S$. Trivial. -- cgit v1.2.3 From 8155ba18260743b1e45507e6fb8d4f80c22c425e Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Fri, 30 Aug 2024 19:25:00 +0200 Subject: working commit --- library/topology/urysohn.tex | 151 +++++++++++++++++++++++++++++-------------- 1 file changed, 101 insertions(+), 50 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 79d65dc..e1fa924 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -231,6 +231,51 @@ The first tept will be a formalisation of chain constructions. $\pot(n) = \apply{\pot}{n}$. \end{abbreviation} +%Take all points, besids one but then take all open sets not containing x but all other, so \{x\} has to be closed +\begin{axiom}\label{hausdorff_implies_singltons_closed} + For all $X$ such that $X$ is Hausdorff we have + for all $x \in \carrier[X]$ we have $\{x\}$ is closed in $X$. +\end{axiom} + +\begin{lemma}\label{urysohn_set_in_between} + Let $X$ be a urysohn space. + Suppose $A,B \in \closeds{X}$. + Suppose $A \subset B$. + %Suppose $B \union A \neq \carrier[X]$. + There exist $C \in \closeds{X}$ + such that $A \subset C \subset B$ or $A, B \in \opens[X]$. +\end{lemma} +\begin{proof} + We have $B \setminus A$ is inhabited. + Take $x$ such that $x \in (B \setminus A)$. + Then $A \subset (A \union \{x\})$. + Let $C = \closure{A \union \{x\}}{X}$. + We have $(A \union \{x\}) \subseteq \closure{A \union \{x\}}{X}$. + Therefore $A \subset C$. + $A \subseteq B \subseteq \carrier[X]$. + $x \in B$. + Therefore $x \in \carrier[X]$. + $(A \union \{x\}) \subseteq \carrier[X]$. + We have $\closure{A \union \{x\}}{X}$ is closed in $X$ by \cref{closure_is_closed}. + Therefore $C$ is closed in $X$ by \cref{closure_is_closed}. + \begin{byCase} + \caseOf{$C = B$.} + %We have $\carrier[X] \setminus A$ is open in $X$. + %We have $\carrier[X] \setminus B$ is open in $X$. + %$\{x\}$ is closed in $X$.% by \cref{hausdorff_implies_singltons_closed}. + %$A \union \{x\} = C$. + %$\carrier[X] \setminus (A \union \{x\}) = (\carrier[X] \setminus C)$. + %$\carrier[X] \setminus (A) = \{x\} \union (\carrier[X] \setminus C)$. + + + %Therefore $\{x\}$ is open in $X$. + Omitted. + \caseOf{$C \neq B$.} + Omitted. + \end{byCase} + +\end{proof} + \begin{proposition}\label{urysohnchain_induction_begin} Let $X$ be a urysohn space. @@ -600,15 +645,16 @@ The first tept will be a formalisation of chain constructions. we have $f(x) = \rfrac{n}{\pot(m)}$. \begin{subproof} Fix $m \in \indexset[\zeta]$. - $\index[\zeta](m)$ is a urysohnchain in $X$. - Define $o : \alpha \to \intervalclosed{\zero}{1}$ such that $o(x) =$ - \begin{cases} - & 0 & \text{if} x \in A - & 1 & \text{if} x \in B - & \rfrac{n}{m} & \text{if} \exists n \in \naturals. x \in \index[\index[\zeta](m)](n) \land x \notin \index[\index[\zeta](m)](n-1) - \end{cases} - - + %$\index[\zeta](m)$ is a urysohnchain in $X$. + + %Define $o : \alpha \to \intervalclosed{\zero}{1}$ such that $o(x) =$ + %\begin{cases} + % & 0 & \text{if} x \in A + % & 1 & \text{if} x \in B + % & \rfrac{n}{m} & \text{if} \exists n \in \naturals. x \in \index[\index[\zeta](m)](n) \land x \notin \index[\index[\zeta](m)](n-1) + %\end{cases} +% + Omitted. \end{subproof} @@ -677,6 +723,7 @@ The first tept will be a formalisation of chain constructions. \begin{subproof} Fix $n \in \naturals$. Fix $x \in \carrier[X]$. + Omitted. \end{subproof} @@ -696,11 +743,12 @@ The first tept will be a formalisation of chain constructions. \begin{subproof} Fix $x \in \carrier[X]$. - + Omitted. % Contradiction by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. %Follows by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. \end{subproof} + Omitted. \end{subproof} @@ -709,51 +757,54 @@ The first tept will be a formalisation of chain constructions. We show that for all $x \in \dom{G}$ we have $G(x) \in \reals$. \begin{subproof} - Fix $x \in \dom{G}$. - It suffices to show that $g(x) \in \reals$. - - There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. - - We show that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - g(x)} < \epsilon$ and $g(x)= k$. - \begin{subproof} - Fix $\epsilon \in \reals$. - - - \end{subproof} - Follows by \cref{apply,plus_one_order,ordinal_iff_suc_ordinal,natural_number_is_ordinal,subseteq,naturals_subseteq_reals,naturals_is_equal_to_two_times_naturals,reals_one_bigger_zero,one_in_reals,ordinal_prec_trichotomy,omega_is_an_ordinal,suc_intro_self,two_in_naturals,in_asymmetric,suc}. + %Fix $x \in \dom{G}$. + %It suffices to show that $g(x) \in \reals$. +% + %There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. +% + %We show that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - g(x)} < \epsilon$ and $g(x)= k$. + %\begin{subproof} + % Fix $\epsilon \in \reals$. + % +% + %\end{subproof} + %Follows by \cref{apply,plus_one_order,ordinal_iff_suc_ordinal,natural_number_is_ordinal,subseteq,naturals_subseteq_reals,naturals_is_equal_to_two_times_naturals,reals_one_bigger_zero,one_in_reals,ordinal_prec_trichotomy,omega_is_an_ordinal,suc_intro_self,two_in_naturals,in_asymmetric,suc}. + Omitted. \end{subproof} We show that for all $x \in \dom{G}$ we have $\zero \leq G(x) \leq 1$. \begin{subproof} - Fix $x \in \dom{G}$. - Then $x \in \carrier[X]$. - \begin{byCase} - \caseOf{$x \in A$.} - For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = \zero$. - - - \caseOf{$x \notin A$.} - \begin{byCase} - \caseOf{$x \in B$.} - For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 1$. - - \caseOf{$x \notin B$.} - Omitted. - \end{byCase} - \end{byCase} + %Fix $x \in \dom{G}$. + %Then $x \in \carrier[X]$. + %\begin{byCase} + % \caseOf{$x \in A$.} + % For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = \zero$. +% +% + % \caseOf{$x \notin A$.} + % \begin{byCase} + % \caseOf{$x \in B$.} + % For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 1$. +% + % \caseOf{$x \notin B$.} + % Omitted. + % \end{byCase} + %\end{byCase} + Omitted. \end{subproof} We show that $G \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$. \begin{subproof} - It suffices to show that $\ran{G} \subseteq \intervalclosed{\zero}{1}$ by \cref{fun_ran_iff,funs_intro,funs_weaken_codom}. - It suffices to show that for all $x \in \dom{G}$ we have $G(x) \in \intervalclosed{\zero}{1}$. - Fix $x \in \dom{G}$. - Then $x \in \carrier[X]$. - $g(x) = G(x)$. - We have $G(x) \in \reals$. - $\zero \leq G(x) \leq 1$. - We have $G(x) \in \intervalclosed{\zero}{1}$ . + %It suffices to show that $\ran{G} \subseteq \intervalclosed{\zero}{1}$ by \cref{fun_ran_iff,funs_intro,funs_weaken_codom}. + %It suffices to show that for all $x \in \dom{G}$ we have $G(x) \in \intervalclosed{\zero}{1}$. + %Fix $x \in \dom{G}$. + %Then $x \in \carrier[X]$. + %$g(x) = G(x)$. + %We have $G(x) \in \reals$. + %$\zero \leq G(x) \leq 1$. + %We have $G(x) \in \intervalclosed{\zero}{1}$ . + Omitted. \end{subproof} We show that $G(A) = \zero$. @@ -860,7 +911,7 @@ The first tept will be a formalisation of chain constructions. % Omitted. % \end{subproof} \end{proof} - -\begin{theorem}\label{safe} - Contradiction. -\end{theorem} +% +%\begin{theorem}\label{safe} +% Contradiction. +%\end{theorem} -- cgit v1.2.3 From 26cf156763f71aaa9f638408ba4bffb85b886ab0 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 31 Aug 2024 18:02:42 +0200 Subject: working commit --- library/algebra/group.tex | 10 +++ library/topology/urysohn.tex | 6 ++ library/topology/urysohn2.tex | 157 +++++++++++++++++++++++++++++++++++------- 3 files changed, 148 insertions(+), 25 deletions(-) (limited to 'library') diff --git a/library/algebra/group.tex b/library/algebra/group.tex index a79bd2f..7de1051 100644 --- a/library/algebra/group.tex +++ b/library/algebra/group.tex @@ -80,3 +80,13 @@ \begin{definition}\label{group_automorphism} Let $f$ be a function. $f$ is a group-automorphism iff $G$ is a group and $\dom{f}=G$ and $\ran{f}=G$. \end{definition} + +\begin{definition}\label{trivial_group} + $G$ is the trivial group iff $G$ is a group and $\{\neutral[G]\}=G$. +\end{definition} + +\begin{theorem}\label{trivial_implies_abelian} + Let $G$ be a group. + Suppose $G$ is the trivial group. + Then $G$ is an abelian group. +\end{theorem} diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index e1fa924..17e2911 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -61,6 +61,11 @@ The first tept will be a formalisation of chain constructions. \end{enumerate} \end{struct} +% Eine folge ist ein Funktion mit domain \subseteq Ordinal zahlen + + + + \begin{definition}\label{cahin_of_subsets} $C$ is a chain of subsets iff $C$ is a sequence and for all $n,m \in \indexset[C]$ such that $n < m$ we have $\index[C](n) \subseteq \index[C](m)$. @@ -237,6 +242,7 @@ The first tept will be a formalisation of chain constructions. for all $x \in \carrier[X]$ we have $\{x\}$ is closed in $X$. \end{axiom} + \begin{lemma}\label{urysohn_set_in_between} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 71de210..e963951 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -15,23 +15,35 @@ \section{Urysohns Lemma} +\begin{definition}\label{minimum} + $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. +\end{definition} + + +\begin{definition}\label{maximum} + $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. +\end{definition} + + +\begin{definition}\label{intervalclosed} + $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. +\end{definition} + + +\begin{definition}\label{intervalopen} + $\intervalopen{a}{b} = \{ x \in \reals \mid a < x < b\}$. +\end{definition} + \begin{definition}\label{one_to_n_set} $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. \end{definition} -\begin{struct}\label{sequence} - A sequence $X$ is a onesorted structure equipped with - \begin{enumerate} - \item $\index$ - \item $\indexset$ - \end{enumerate} - such that - \begin{enumerate} - \item\label{indexset_is_subset_naturals} $\indexset[X] \subseteq \naturals$. - \item\label{index_is_bijection} $\index[X]$ is a bijection from $\indexset[X]$ to $\carrier[X]$. - \end{enumerate} -\end{struct} + +\begin{definition}\label{sequence} + $X$ is a sequence iff $X$ is a function and $\dom{f} \subseteq \naturals$. +\end{definition} + \begin{abbreviation}\label{urysohnspace} $X$ is a urysohn space iff @@ -42,10 +54,66 @@ \end{abbreviation} -\begin{definition}\label{intervalclosed} - $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. +\begin{abbreviation}\label{at} + $\at{f}{n} = f(n)$. +\end{abbreviation} + + +\begin{definition}\label{chain_of_subsets} + $X$ is a chain of subsets in $Y$ iff $X$ is a sequence and for all $n \in \dom{X}$ we have $\at{X}{n} \subseteq \carrier[Y]$. +\end{definition} + + +\begin{definition}\label{urysohnchain}%<-- zulässig + $X$ is a urysohnchain of $Y$ iff $X$ is a chain of subsets in $Y$ and for all $n,m \in \dom{X}$ such that $n < m$ we have $\closure{\at{X}{n}}{Y} \subseteq \interior{\at{X}{m}}{Y}$. +\end{definition} + + +\begin{definition}\label{finer} %<-- verfeinerung + $X$ is finer then $Y$ in $U$ iff for all $n \in \dom{X}$ we have $\at{X}{n} \in \ran{Y}$ and for all $m \in \dom{X}$ such that $n < m$ we have there exist $k \in \dom{Y}$ such that $ \closure{\at{X}{n}}{U} \subseteq \interior{\at{Y}{k}}{U} \subseteq \closure{\at{Y}{k}}{U} \subseteq \interior{\at{X}{m}}{U}$. +\end{definition} + + +\begin{definition}\label{sequence_of_reals} + $X$ is a sequence of reals iff $\ran{X} \subseteq \reals$. +\end{definition} + + +\begin{axiom}\label{abs_behavior1} + If $x \geq \zero$ then $\abs{x} = x$. +\end{axiom} + +\begin{axiom}\label{abs_behavior2} + If $x < \zero$ then $\abs{x} = \neg{x}$. +\end{axiom} + +\begin{definition}\label{realsminus} + $\realsminus = \{r \in \reals \mid r < \zero\}$. \end{definition} +\begin{definition}\label{realsplus} + $\realsplus = \reals \setminus \realsminus$. +\end{definition} + +\begin{definition}\label{epsilon_ball} + $\epsBall{x}{\epsilon} = \intervalopen{x-\epsilon}{x+\epsilon}$. +\end{definition} + +\begin{definition}\label{pointwise_convergence} + $X$ converge to $x$ iff for all $\epsilon \in \realsplus$ there exist $N \in \dom{X}$ such that for all $n \in \dom{X}$ such that $n > N$ we have $\at{X}{n} \in \epsBall{x}{\epsilon}$. +\end{definition} + + + + + + + + + + + + \begin{theorem}\label{urysohnsetinbeetween} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. @@ -55,7 +123,6 @@ \end{theorem} - \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. @@ -76,8 +143,8 @@ Define $F : \eta \to \reals$ such that $F(x) =$ \begin{cases} & \zero &\text{if} x \in A\\ - & \rfrac{1}{1+1} &\text{if} x \in (\carrier[X] \setminus (A \union B))\\ - & 1 &\text{if} x \in B + & \rfrac{1}{1+1} &\text{if} x \in (\carrier[X] \setminus (A \union B))\\ + & 1 &\text{if} x \in B \end{cases} %Define $f : \naturals \to \pow{P}$ such that $f(x)=$ @@ -87,19 +154,59 @@ % & G & \text{if} x \in (\naturals \setminus \{1, \zero\}) \land G = \{g \in \pow{P} \mid g \in f(n-1) \lor (g \notin f(n-1) \land g \in P) \} %\end{cases} - Let $D = \{d \mid d \in \rationals \mid \zero \leq d \leq 1\}$. - Take $R$ such that for all $q \in D$ we have for all $S \in P$ we have $q \mathrel{R} S$ iff - $q = \zero \land S = A$ or $q = 1 \land S = H$ or - for all $q_1, q_2, S_1, S_2$ - such that $q_1 \leq q \leq q_2$ and $q_1 \mathrel{R} S_1$ and $q_2 \mathrel{R} S_2$ - we have $\closure{S_1}{X} \subseteq \interior{S}{X} \subseteq \closure{S}{X} \subseteq \interior{S_2}{X}$ - and $q \mathrel{R} S$. + %Let $D = \{d \mid d \in \rationals \mid \zero \leq d \leq 1\}$. + %Take $R$ such that for all $q \in D$ we have for all $S \in P$ we have $q \mathrel{R} S$ iff + % $q = \zero \land S = A$ or $q = 1 \land S = H$ or + % for all $q_1, q_2, S_1, S_2$ + % such that $q_1 \leq q \leq q_2$ and $q_1 \mathrel{R} S_1$ and $q_2 \mathrel{R} S_2$ + % we have $\closure{S_1}{X} \subseteq \interior{S}{X} \subseteq \closure{S}{X} \subseteq \interior{S_2}{X}$ + % and $q \mathrel{R} S$. - Trivial. + %Let $J = \{(n,f) \mid n denots the cardinality of a urysohn chain on which f is a staircase function\}$. +% + %Let $N = \naturals$. + %Define $j : N \to \funs{\carrier[X]}{\eals}$ such that $j(n) =$ + %\begin{cases} + % & f &\text{if} n \in N \land \exist w \in J. J=(n,f) + %\end{cases} + + + + + + + + + \end{proof} \begin{theorem}\label{safe} Contradiction. \end{theorem} + + + + +% +%Ideen: +%Eine folge ist ein Funktion mit domain \subseteq Natürlichenzahlen. als predicat +% +%zulässig und verfeinerung von ketten als predicat definieren. +% +%limits und punkt konvergenz als prädikat. +% +% +%Vor dem Beweis vor dem eigentlichen Beweis. +%die abgeleiteten Funktionen +% +%\derivedstiarcasefunction on A +% +%abbreviation: \at{f}{n} = f_{n} +% +% +%TODO: +%Reals ist ein topologischer Raum +% + -- cgit v1.2.3 From 8cc2f8557d68c492cd0327f2f49051ff0a7b0f6a Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sat, 31 Aug 2024 18:45:44 +0200 Subject: Contradiction in sequence False can be proven with iff_sequence, codom_of_emptyset_can_be_anything,sequence, emptyset_is_function_on_emptyset,id_dom,in_irrefl, suc_subseteq_implies_in,emptyset_subseteq --- library/topology/urysohn2.tex | 68 +++++++++++++++++-------------------------- 1 file changed, 26 insertions(+), 42 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index e963951..7f98bc2 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -104,8 +104,11 @@ \end{definition} - - +\begin{proposition}\label{iff_sequence} + Suppose $X$ is a function. + Suppose $\dom{X} \subseteq \naturals$. + Then $X$ is a sequence. +\end{proposition} @@ -121,6 +124,9 @@ Suppose $\carrier[X]$ is inhabited. There exist $U \subseteq \carrier[X]$ such that $U$ is closed in $X$ and $\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{B}{X}$. \end{theorem} +\begin{proof} + Omitted. +\end{proof} \begin{theorem}\label{urysohn} @@ -132,48 +138,26 @@ and $f(A) = \zero$ and $f(B)= 1$ and $f$ is continuous. \end{theorem} \begin{proof} - - Let $H = \carrier[X] \setminus B$. - Let $P = \{x \in \pow{X} \mid x = A \lor x = H \lor (x \in \pow{X} \land (\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{H}{X}))\}$. - Let $\eta = \carrier[X]$. - - - % Provable - % vvv - Define $F : \eta \to \reals$ such that $F(x) =$ + Let $X' = \carrier[X]$. + Let $N = \{\zero, 1\}$. + $1 = \suc{\zero}$. + $1 \in \naturals$ and $\zero \in \naturals$. + $N \subseteq \naturals$. + Let $A' = (X' \setminus B)$. + $B \subseteq X'$ by \cref{powerset_elim,closeds}. + $A \subseteq X'$. + Therefore $A \subseteq A'$. + Define $U_0: N \to \{A, A'\}$ such that $U_0(n) =$ \begin{cases} - & \zero &\text{if} x \in A\\ - & \rfrac{1}{1+1} &\text{if} x \in (\carrier[X] \setminus (A \union B))\\ - & 1 &\text{if} x \in B + &A &\text{if} n = \zero \\ + &A' &\text{if} n = 1 \end{cases} - - %Define $f : \naturals \to \pow{P}$ such that $f(x)=$ - %\begin{cases} - % & \emptyset & \text{if} x = \zero \\ - % & \{A, H\} & \text{if} x = 1 \\ - % & G & \text{if} x \in (\naturals \setminus \{1, \zero\}) \land G = \{g \in \pow{P} \mid g \in f(n-1) \lor (g \notin f(n-1) \land g \in P) \} - %\end{cases} - - %Let $D = \{d \mid d \in \rationals \mid \zero \leq d \leq 1\}$. - %Take $R$ such that for all $q \in D$ we have for all $S \in P$ we have $q \mathrel{R} S$ iff - % $q = \zero \land S = A$ or $q = 1 \land S = H$ or - % for all $q_1, q_2, S_1, S_2$ - % such that $q_1 \leq q \leq q_2$ and $q_1 \mathrel{R} S_1$ and $q_2 \mathrel{R} S_2$ - % we have $\closure{S_1}{X} \subseteq \interior{S}{X} \subseteq \closure{S}{X} \subseteq \interior{S_2}{X}$ - % and $q \mathrel{R} S$. - - - - %Let $J = \{(n,f) \mid n denots the cardinality of a urysohn chain on which f is a staircase function\}$. -% - %Let $N = \naturals$. - %Define $j : N \to \funs{\carrier[X]}{\eals}$ such that $j(n) =$ - %\begin{cases} - % & f &\text{if} n \in N \land \exist w \in J. J=(n,f) - %\end{cases} - - - + $U_0$ is a function. + $\dom{U_0} = N$. + $\dom{U_0} \subseteq \naturals$ by \cref{ran_converse}. + $U_0$ is a sequence. + $U_0$ is a chain of subsets in $X'$. + $U_0$ is a urysohnchain of $X$. -- cgit v1.2.3 From 36065f6a7fcc8f4d23b98be642c7fa7019ce2b79 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 2 Sep 2024 12:46:14 +0200 Subject: Corrected Contradiction. Contradiction was due to a misstake in the definition in sequence. --- library/topology/urysohn2.tex | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 7f98bc2..f70d679 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -41,7 +41,7 @@ \begin{definition}\label{sequence} - $X$ is a sequence iff $X$ is a function and $\dom{f} \subseteq \naturals$. + $X$ is a sequence iff $X$ is a function and $\dom{X} \subseteq \naturals$. \end{definition} @@ -169,6 +169,9 @@ \begin{theorem}\label{safe} Contradiction. \end{theorem} +\begin{proof} + Follows by \cref{iff_sequence,codom_of_emptyset_can_be_anything,sequence,emptyset_is_function_on_emptyset,id_dom,in_irrefl,suc_subseteq_implies_in,emptyset_subseteq}. +\end{proof} -- cgit v1.2.3 From d1d6ad98c26ffc392270665e8d0e4d2229984f82 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 2 Sep 2024 13:43:44 +0200 Subject: working commit --- library/topology/urysohn2.tex | 20 ++++++++++++++------ 1 file changed, 14 insertions(+), 6 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index f70d679..ad3f1d7 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -60,7 +60,7 @@ \begin{definition}\label{chain_of_subsets} - $X$ is a chain of subsets in $Y$ iff $X$ is a sequence and for all $n \in \dom{X}$ we have $\at{X}{n} \subseteq \carrier[Y]$. + $X$ is a chain of subsets in $Y$ iff $X$ is a sequence and for all $n \in \dom{X}$ we have $\at{X}{n} \subseteq \carrier[Y]$ and for all $m \in \dom{X}$ such that $m > n$ we have $\at{X}{n} \subseteq \at{X}{m}$. \end{definition} @@ -156,10 +156,20 @@ $\dom{U_0} = N$. $\dom{U_0} \subseteq \naturals$ by \cref{ran_converse}. $U_0$ is a sequence. - $U_0$ is a chain of subsets in $X'$. + We show that $U_0$ is a chain of subsets in $X$. + \begin{subproof} + We have $\dom{U_0} \subseteq \naturals$. + We have for all $n \in \dom{U_0}$ we have $\at{U_0}{n} \subseteq \carrier[X]$ by \cref{topological_prebasis_iff_covering_family,union_as_unions,union_absorb_subseteq_left,subset_transitive,setminus_subseteq}. + We have $\dom{U_0} = \{\zero, 1\}$. + + It suffices to show that for all $n \in \dom{U_0}$ we have for all $m \in \dom{U_0}$ such that $m > n$ we have $\at{U_0}{n} \subseteq \at{U_0}{m}$. + + It suffices to show that $\at{U_0}{\zero} \subseteq \at{U_0}{1}$. + Follows by \cref{one_in_reals,order_reals_lemma0,upair_elim,reals_one_bigger_zero,reals_order,reals_axiom_zero_in_reals,subseteq_refl,apply}. + \end{subproof} $U_0$ is a urysohnchain of $X$. - + %We are done with the first step, now we want to prove that we have U a sequence of these chain with U_0 the first chain. @@ -169,9 +179,7 @@ \begin{theorem}\label{safe} Contradiction. \end{theorem} -\begin{proof} - Follows by \cref{iff_sequence,codom_of_emptyset_can_be_anything,sequence,emptyset_is_function_on_emptyset,id_dom,in_irrefl,suc_subseteq_implies_in,emptyset_subseteq}. -\end{proof} + -- cgit v1.2.3 From 010b2ce53a4a5e693ced109eda38b167f3e284d7 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 2 Sep 2024 17:17:38 +0200 Subject: working commit --- library/topology/urysohn2.tex | 20 ++++++++++++++++++-- 1 file changed, 18 insertions(+), 2 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index ad3f1d7..c0b46c4 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -156,6 +156,7 @@ $\dom{U_0} = N$. $\dom{U_0} \subseteq \naturals$ by \cref{ran_converse}. $U_0$ is a sequence. + We have $1, \zero \in N$. We show that $U_0$ is a chain of subsets in $X$. \begin{subproof} We have $\dom{U_0} \subseteq \naturals$. @@ -163,9 +164,24 @@ We have $\dom{U_0} = \{\zero, 1\}$. It suffices to show that for all $n \in \dom{U_0}$ we have for all $m \in \dom{U_0}$ such that $m > n$ we have $\at{U_0}{n} \subseteq \at{U_0}{m}$. + + Fix $n \in \dom{U_0}$. + Fix $m \in \dom{U_0}$. + + \begin{byCase} + \caseOf{$n \neq \zero$.} + Trivial. + \caseOf{$n = \zero$.} + \begin{byCase} + \caseOf{$m = \zero$.} + Trivial. + \caseOf{$m \neq \zero$.} + We have $A \subseteq A'$. + We have $\at{U_0}{\zero} = A$ by assumption. + We have $\at{U_0}{1}= A'$ by assumption. + \end{byCase} + \end{byCase} - It suffices to show that $\at{U_0}{\zero} \subseteq \at{U_0}{1}$. - Follows by \cref{one_in_reals,order_reals_lemma0,upair_elim,reals_one_bigger_zero,reals_order,reals_axiom_zero_in_reals,subseteq_refl,apply}. \end{subproof} $U_0$ is a urysohnchain of $X$. -- cgit v1.2.3 From 6b45d0b6118cbfaf3406213601917394c364bfe2 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 3 Sep 2024 04:33:21 +0200 Subject: Finished induction begin --- library/topology/urysohn2.tex | 48 +++++++++++++++++++++++++++++++++++++++---- 1 file changed, 44 insertions(+), 4 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index c0b46c4..9991d21 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -68,11 +68,21 @@ $X$ is a urysohnchain of $Y$ iff $X$ is a chain of subsets in $Y$ and for all $n,m \in \dom{X}$ such that $n < m$ we have $\closure{\at{X}{n}}{Y} \subseteq \interior{\at{X}{m}}{Y}$. \end{definition} +\begin{definition}\label{urysohn_finer_set} + $A$ is finer between $X$ to $Y$ in $U$ iff $\closure{X}{U} \subseteq \interior{A}{U}$ and $\closure{A}{U} \subseteq \interior{Y}{U}$. +\end{definition} \begin{definition}\label{finer} %<-- verfeinerung - $X$ is finer then $Y$ in $U$ iff for all $n \in \dom{X}$ we have $\at{X}{n} \in \ran{Y}$ and for all $m \in \dom{X}$ such that $n < m$ we have there exist $k \in \dom{Y}$ such that $ \closure{\at{X}{n}}{U} \subseteq \interior{\at{Y}{k}}{U} \subseteq \closure{\at{Y}{k}}{U} \subseteq \interior{\at{X}{m}}{U}$. + $Y$ is finer then $X$ in $U$ iff for all $n \in \dom{X}$ we have $\at{X}{n} \in \ran{Y}$ and for all $m \in \dom{X}$ such that $n < m$ we have there exist $k \in \dom{Y}$ such that $\at{Y}{k}$ is finer between $\at{X}{n}$ to $\at{X}{m}$ in $U$. \end{definition} +\begin{definition}\label{follower_index} + $y$ follows $x$ in $I$ iff $x < y$ and $x,y \in I$ and for all $i \in I$ such that $x < i$ we have $y \leq i$. +\end{definition} + +\begin{definition}\label{finer_smallest_step} + $Y$ is a minimal finer extention of $X$ in $U$ iff $Y$ is finer then $X$ in $U$ and for all $x_1,x_2 \in \dom{X}$ such that $x_1$ follows $x_2$ in $\dom{X}$ we have there exist $y \in \dom{Y}$ such that $y$ follows $x_1$ in $\dom{X}$ and $x_2$ follows $y$ in $\dom{X}$. +\end{definition} \begin{definition}\label{sequence_of_reals} $X$ is a sequence of reals iff $\ran{X} \subseteq \reals$. @@ -179,15 +189,45 @@ We have $A \subseteq A'$. We have $\at{U_0}{\zero} = A$ by assumption. We have $\at{U_0}{1}= A'$ by assumption. + Follows by \cref{powerset_elim,emptyset_subseteq,union_as_unions,union_absorb_subseteq_left,subseteq_pow_unions,ran_converse,subseteq,subseteq_antisymmetric,suc_subseteq_intro,apply,powerset_emptyset,emptyset_is_ordinal,notin_emptyset,ordinal_elem_connex,omega_is_an_ordinal,prec_is_ordinal}. \end{byCase} \end{byCase} - \end{subproof} - $U_0$ is a urysohnchain of $X$. + We show that $U_0$ is a urysohnchain of $X$. + \begin{subproof} + It suffices to show that for all $n \in \dom{U_0}$ we have for all $m \in \dom{U_0}$ such that $n < m$ we have $\closure{\at{U_0}{n}}{X} \subseteq \interior{\at{U_0}{m}}{X}$. + Fix $n \in \dom{U_0}$. + Fix $m \in \dom{U_0}$. + \begin{byCase} + \caseOf{$n \neq \zero$.} + Follows by \cref{ran_converse,upair_iff,one_in_reals,order_reals_lemma0,reals_axiom_zero_in_reals,reals_one_bigger_zero,reals_order}. + \caseOf{$n = \zero$.} + \begin{byCase} + \caseOf{$m = \zero$.} + Trivial. + \caseOf{$m \neq \zero$.} + Follows by \cref{setminus_emptyset,setdifference_eq_intersection_with_complement,setminus_self,interior_carrier,complement_interior_eq_closure_complement,subseteq_refl,closure_interior_frontier_is_in_carrier,emptyset_subseteq,closure_is_minimal_closed_set,inter_lower_right,inter_lower_left,subseteq_transitive,interior_of_open,is_closed_in,closeds,union_absorb_subseteq_right,ordinal_suc_subseteq,ordinal_empty_or_emptyset_elem,union_absorb_subseteq_left,union_emptyset,topological_prebasis_iff_covering_family,inhabited,notin_emptyset,subseteq,union_as_unions,natural_number_is_ordinal}. + \end{byCase} + \end{byCase} + \end{subproof} %We are done with the first step, now we want to prove that we have U a sequence of these chain with U_0 the first chain. - + We show that there exist $U$ such that $U$ is a sequence and $\dom{U} = \naturals$ and $\at{U}{\zero} = U_0$ and for all $n \in \dom{U}$ we have $\at{U}{n}$ is a urysohnchain of $X$ and $\at{U}{\suc{n}}$ is a minimal finer extention of $\at{U}{n}$ in $X$. + \begin{subproof} + $U_0$ is a urysohnchain of $X$. + We show that if $V$ is a urysohnchain of $X$ then there exist $V'$ such that $V'$ is a urysohnchain of $X$ and $V'$ is a minimal finer extention of $V$ in $X$. + \begin{subproof} + Omitted. + \end{subproof} + Let $N' = \naturals$. + Let $P = \{C \mid C \in \pow{\pow{X'}} \mid \text{$C$ is a urysohnchain of $X$}\}$. + Define $U : N' \to P$ such that $U(n) =$ + \begin{cases} + &U_0 &\text{if} n = \zero \\ + &V & \text{if} \text{ $n = \suc{m}$ and $V$ is a minimal finer extention of $U(m)$ in $X$}. + \end{cases} + \end{subproof} \end{proof} -- cgit v1.2.3 From c57360cf062805c86fed5d1f3f4adbf52c05f9ff Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 3 Sep 2024 05:45:37 +0200 Subject: working commit --- library/topology/urysohn2.tex | 44 +++++++++++++++++++++++++++++++------------ 1 file changed, 32 insertions(+), 12 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 9991d21..dbcfc53 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -120,6 +120,18 @@ Then $X$ is a sequence. \end{proposition} +\begin{definition}\label{higher_urysohn_chain} + $U$ is a lifted urysohnchain of $X$ iff $U$ is a sequence and $\dom{U} = \naturals$ and for all $n \in \dom{U}$ we have $\at{U}{n}$ is a urysohnchain of $X$ and $\at{U}{\suc{n}}$ is a minimal finer extention of $\at{U}{n}$ in $X$. +\end{definition} + +\begin{definition}\label{staircase} + $f$ is a staircase function adapted to $U$ in $X$ iff $U$ is a urysohnchain of $X$ and for all $x,n,m,k$ such that $k = \max{\dom{U}}$ and $n,m \in \dom{U}$ and $n$ follows $m$ in $\dom{U}$ and $x \in (\at{U}{m} \setminus \at{U}{n})$ we have $f(x)= \rfrac{m}{k}$. +\end{definition} + +\begin{definition}\label{staircase_sequence} + $f$ is staircase sequence of $U$ iff $f$ is a sequence and $U$ is a lifted urysohnchain of $X$ and $\dom{U} = \dom{f}$ and for all $n \in \dom{U}$ we have $\at{f}{n}$ is a staircase function adapted to $\at{U}{n}$ in $U$. +\end{definition} + @@ -139,6 +151,23 @@ \end{proof} +\begin{theorem}\label{induction_on_urysohnchains} + Let $X$ be a urysohn space. + Suppose $U_0$ is a sequence. + Suppose $U_0$ is a chain of subsets in $X$. + Suppose $U_0$ is a urysohnchain of $X$. + There exist $U$ such that $U$ is a sequence and $\dom{U} = \naturals$ and $\at{U}{\zero} = U_0$ and for all $n \in \dom{U}$ we have $\at{U}{n}$ is a urysohnchain of $X$ and $\at{U}{\suc{n}}$ is a minimal finer extention of $\at{U}{n}$ in $X$. +\end{theorem} +\begin{proof} + $U_0$ is a urysohnchain of $X$. + It suffices to show that for all $V$ such that $V$ is a urysohnchain of $X$ there exist $V'$ such that $V'$ is a urysohnchain of $X$ and $V'$ is a minimal finer extention of $V$ in $X$. + +\end{proof} + + + + + \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. @@ -215,20 +244,11 @@ We show that there exist $U$ such that $U$ is a sequence and $\dom{U} = \naturals$ and $\at{U}{\zero} = U_0$ and for all $n \in \dom{U}$ we have $\at{U}{n}$ is a urysohnchain of $X$ and $\at{U}{\suc{n}}$ is a minimal finer extention of $\at{U}{n}$ in $X$. \begin{subproof} - $U_0$ is a urysohnchain of $X$. - We show that if $V$ is a urysohnchain of $X$ then there exist $V'$ such that $V'$ is a urysohnchain of $X$ and $V'$ is a minimal finer extention of $V$ in $X$. - \begin{subproof} - Omitted. - \end{subproof} - Let $N' = \naturals$. - Let $P = \{C \mid C \in \pow{\pow{X'}} \mid \text{$C$ is a urysohnchain of $X$}\}$. - Define $U : N' \to P$ such that $U(n) =$ - \begin{cases} - &U_0 &\text{if} n = \zero \\ - &V & \text{if} \text{ $n = \suc{m}$ and $V$ is a minimal finer extention of $U(m)$ in $X$}. - \end{cases} + Follows by \cref{chain_of_subsets,urysohnchain,induction_on_urysohnchains}. \end{subproof} + + \end{proof} -- cgit v1.2.3 From 68716d1ab46dee3dfc1b03089f941dbb6883cdcd Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 4 Sep 2024 15:17:50 +0200 Subject: Mismatched Assume in Induction Unexpeced Mismatch in line 99 and 109 or the pair 100 and 108. Parsing works with line 99 and 108 or with the lines 100 and 109. zf: MismatchedAssume (TermSymbol (SymbolPredicate (PredicateNoun (SgPl {sg = [Just (Word "natural"),Just (Word "number")], pl = [Just (Word "natural"),Just (Word "numbers")]}))) [TermVar (NamedVar "n")]) (Connected Implication (TermSymbol (SymbolPredicate (PredicateRelation (Command "in"))) [TermVar (NamedVar "n"),TermSymbol (SymbolMixfix [Just (Command "naturals")]) []]) (TermSymbol (SymbolPredicate (PredicateVerb (SgPl {sg = [Just (Word "has"),Just (Word "cardinality"),Nothing], pl = [Just (Word "ha"),Just (Word "cardinality"),Nothing]}))) [TermSymbol (SymbolMixfix [Just (Command "seq"),Just InvisibleBraceL,Nothing,Just InvisibleBraceR,Just InvisibleBraceL,Nothing,Just InvisibleBraceR]) [TermSymbol (SymbolMixfix [Just (Command "emptyset")]) [],TermVar (NamedVar "n")],TermSymbol (SymbolMixfix [Just (Command "suc"),Just InvisibleBraceL,Nothing,Just InvisibleBraceR]) [TermVar (NamedVar "n")]])) --- library/numbers.tex | 40 +++++++++++ library/topology/real-topological-space.tex | 26 +++++++ library/topology/urysohn2.tex | 108 +++++++++++++++------------- 3 files changed, 125 insertions(+), 49 deletions(-) create mode 100644 library/topology/real-topological-space.tex (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index f7f6c2c..cb3d5cf 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -747,9 +747,49 @@ Laws of the order on the reals $x$ is the supremum of $X$ iff $x$ is a greatest lower bound of $X$. \end{definition} +\begin{definition}\label{minimum} + $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. +\end{definition} + + +\begin{definition}\label{maximum} + $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. +\end{definition} + + +\begin{definition}\label{intervalclosed} + $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. +\end{definition} + + +\begin{definition}\label{intervalopen} + $\intervalopen{a}{b} = \{ x \in \reals \mid a < x < b\}$. +\end{definition} +\begin{definition}\label{m_to_n_set} + $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. +\end{definition} + +\begin{axiom}\label{abs_behavior1} + If $x \geq \zero$ then $\abs{x} = x$. +\end{axiom} + +\begin{axiom}\label{abs_behavior2} + If $x < \zero$ then $\abs{x} = \neg{x}$. +\end{axiom} +\begin{definition}\label{realsminus} + $\realsminus = \{r \in \reals \mid r < \zero\}$. +\end{definition} + +\begin{definition}\label{realsplus} + $\realsplus = \reals \setminus \realsminus$. +\end{definition} + +\begin{definition}\label{epsilon_ball} + $\epsBall{x}{\epsilon} = \intervalopen{x-\epsilon}{x+\epsilon}$. +\end{definition} diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex new file mode 100644 index 0000000..239965c --- /dev/null +++ b/library/topology/real-topological-space.tex @@ -0,0 +1,26 @@ +\import{set.tex} +\import{set/cons.tex} +\import{set/powerset.tex} +\import{set/fixpoint.tex} +\import{set/product.tex} +\import{topology/topological-space.tex} +\import{topology/separation.tex} +\import{topology/continuous.tex} +\import{topology/basis.tex} +\import{numbers.tex} +\import{function.tex} + + +\section{The canonical topology on $\mathbbR$} + +\begin{definition}\label{topological_basis_reals_eps_ball} + $\topoBasisReals = \{ \epsBall{x}{\epsilon} \mid x \in \reals, \epsilon \in \realsplus\}$. +\end{definition} + +\begin{theorem}\label{reals_as_topo_space} + Suppose $\opens[\reals] = \genOpens{\topoBasisReals}{\reals}$. + Then $\reals$ is a topological space. +\end{theorem} +\begin{proof} + Omitted. +\end{proof} \ No newline at end of file diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index dbcfc53..a3f3ea4 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -12,32 +12,11 @@ \import{set/powerset.tex} \import{set/fixpoint.tex} \import{set/product.tex} +\import{topology/real-topological-space.tex} \section{Urysohns Lemma} -\begin{definition}\label{minimum} - $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. -\end{definition} - - -\begin{definition}\label{maximum} - $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. -\end{definition} - - -\begin{definition}\label{intervalclosed} - $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. -\end{definition} - - -\begin{definition}\label{intervalopen} - $\intervalopen{a}{b} = \{ x \in \reals \mid a < x < b\}$. -\end{definition} - -\begin{definition}\label{one_to_n_set} - $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. -\end{definition} \begin{definition}\label{sequence} @@ -89,26 +68,6 @@ \end{definition} -\begin{axiom}\label{abs_behavior1} - If $x \geq \zero$ then $\abs{x} = x$. -\end{axiom} - -\begin{axiom}\label{abs_behavior2} - If $x < \zero$ then $\abs{x} = \neg{x}$. -\end{axiom} - -\begin{definition}\label{realsminus} - $\realsminus = \{r \in \reals \mid r < \zero\}$. -\end{definition} - -\begin{definition}\label{realsplus} - $\realsplus = \reals \setminus \realsminus$. -\end{definition} - -\begin{definition}\label{epsilon_ball} - $\epsBall{x}{\epsilon} = \intervalopen{x-\epsilon}{x+\epsilon}$. -\end{definition} - \begin{definition}\label{pointwise_convergence} $X$ converge to $x$ iff for all $\epsilon \in \realsplus$ there exist $N \in \dom{X}$ such that for all $n \in \dom{X}$ such that $n > N$ we have $\at{X}{n} \in \epsBall{x}{\epsilon}$. \end{definition} @@ -120,10 +79,61 @@ Then $X$ is a sequence. \end{proposition} -\begin{definition}\label{higher_urysohn_chain} +\begin{definition}\label{lifted_urysohn_chain} $U$ is a lifted urysohnchain of $X$ iff $U$ is a sequence and $\dom{U} = \naturals$ and for all $n \in \dom{U}$ we have $\at{U}{n}$ is a urysohnchain of $X$ and $\at{U}{\suc{n}}$ is a minimal finer extention of $\at{U}{n}$ in $X$. \end{definition} +\begin{definition}\label{normal_ordered_urysohnchain} + $U$ is normal ordered iff there exist $n \in \naturals$ such that $\dom{U} = \seq{\zero}{n}$. +\end{definition} + +\begin{definition}\label{bijection_of_urysohnchains} + $f$ is consistent on $X$ to $Y$ iff $f$ is a bijection from $\dom{X}$ to $\dom{Y}$ and for all $n,m \in \dom{X}$ such that $n < m$ we have $f(n) < f(m)$. +\end{definition} + +\begin{proposition}\label{naturals_in_transitive} + $\naturals$ is a \in-transitive set. +\end{proposition} + +\begin{proposition}\label{naturals_elem_in_transitive} + If $n \in \naturals$ then $n$ is \in-transitive and every element of $n$ is \in-transitive. + %If $n$ is a natural number then $n$ is \in-transitive and every element of $n$ is \in-transitive. + +\end{proposition} + +\begin{proposition}\label{seq_zero_to_n_isomorph_to_n} + For all $n \in \naturals$ we have $\seq{\zero}{n}$ has cardinality $\suc{n}$. +\end{proposition} +\begin{proof} [Proof by \in-induction on $n$] + Assume $n \in \naturals$. + %Assume $n$ is a natural number. + %We show that $\seq{\zero}{\zero}$ has cardinality $1$. + %\begin{subproof} + % It suffices to show that $1 = \seq{\zero}{\zero}$. + % Follows by set extensionality. + %\end{subproof} + %For all $n \in \naturals$ we have either $n = \zero$ or there exist $m \in \naturals$ such that $n = \suc{m}$. + %For all $n \in \naturals$ we have either $n = \zero$ or $\zero \in n$. + %We show that if $\seq{\zero}{n}$ has cardinality $\suc{n}$ then $\seq{\zero}{\suc{n}}$ has cardinality $\suc{\suc{n}}$. + %\begin{subproof} + % Omitted. + %\end{subproof} +\end{proof} + +\begin{proposition}\label{existence_normal_ordered_urysohn} + Let $X$ be a urysohn space. + Suppose $U$ is a urysohnchain of $X$. + Suppose $\dom{U}$ is finite. + Then there exist $V,f$ such that $V$ is a urysohnchain of $X$ and $f$ is consistent on $X$ to $Y$ and $V$ is normal ordered. +\end{proposition} +\begin{proof} + Take $k$ such that $\dom{U}$ has cardinality $k$ by \cref{ran_converse,cardinality,finite}. + There exist $F$ such that $F$ is a bijection from $\seq{\zero}{k}$ to $\dom{U}$. + + +\end{proof} + + \begin{definition}\label{staircase} $f$ is a staircase function adapted to $U$ in $X$ iff $U$ is a urysohnchain of $X$ and for all $x,n,m,k$ such that $k = \max{\dom{U}}$ and $n,m \in \dom{U}$ and $n$ follows $m$ in $\dom{U}$ and $x \in (\at{U}{m} \setminus \at{U}{n})$ we have $f(x)= \rfrac{m}{k}$. \end{definition} @@ -136,9 +146,6 @@ - - - \begin{theorem}\label{urysohnsetinbeetween} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. @@ -159,9 +166,9 @@ There exist $U$ such that $U$ is a sequence and $\dom{U} = \naturals$ and $\at{U}{\zero} = U_0$ and for all $n \in \dom{U}$ we have $\at{U}{n}$ is a urysohnchain of $X$ and $\at{U}{\suc{n}}$ is a minimal finer extention of $\at{U}{n}$ in $X$. \end{theorem} \begin{proof} - $U_0$ is a urysohnchain of $X$. - It suffices to show that for all $V$ such that $V$ is a urysohnchain of $X$ there exist $V'$ such that $V'$ is a urysohnchain of $X$ and $V'$ is a minimal finer extention of $V$ in $X$. - + %$U_0$ is a urysohnchain of $X$. + %It suffices to show that for all $V$ such that $V$ is a urysohnchain of $X$ there exist $V'$ such that $V'$ is a urysohnchain of $X$ and $V'$ is a minimal finer extention of $V$ in $X$. + Omitted. \end{proof} @@ -246,6 +253,9 @@ \begin{subproof} Follows by \cref{chain_of_subsets,urysohnchain,induction_on_urysohnchains}. \end{subproof} + Take $U$ such that $U$ is a lifted urysohnchain of $X$ and $\at{U}{\zero} = U_0$. + + -- cgit v1.2.3 From 22e9ebe72c349514a1ee0ed37772ca8168eaf658 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 4 Sep 2024 17:02:07 +0200 Subject: working commit --- library/topology/urysohn2.tex | 56 +++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 54 insertions(+), 2 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index a3f3ea4..d08d507 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -98,20 +98,68 @@ \begin{proposition}\label{naturals_elem_in_transitive} If $n \in \naturals$ then $n$ is \in-transitive and every element of $n$ is \in-transitive. %If $n$ is a natural number then $n$ is \in-transitive and every element of $n$ is \in-transitive. - \end{proposition} -\begin{proposition}\label{seq_zero_to_n_isomorph_to_n} +\begin{proposition}\label{zero_is_in_minimal} + $\zero$ is an \in-minimal element of $\naturals$. +\end{proposition} + +\begin{proposition}\label{seq_zero_to_n_isomorph_to_suc_n} For all $n \in \naturals$ we have $\seq{\zero}{n}$ has cardinality $\suc{n}$. \end{proposition} \begin{proof} [Proof by \in-induction on $n$] Assume $n \in \naturals$. + For all $m \in n$ we have $m \in \naturals$. + \begin{byCase} + \caseOf{$n = \zero$.} + It suffices to show that $1 = \seq{\zero}{\zero}$. + Follows by set extensionality. + \caseOf{$n \neq \zero$.} + Take $k$ such that $k \in \naturals$ and $\suc{k} = n$. + Then $k \in n$. + Therefore $\seq{\zero}{k}$ has cardinality $\suc{k}$. + Now $\seq{\zero}{k}$ has cardinality $n$. + Take $f$ such that $f$ is a bijection from $n$ to $\seq{\zero}{k}$. + We have $\suc{n} = n \union \{n\}$. + We show that $\seq{\zero}{n} = \seq{\zero}{k} \union \{\suc{k}\}$. + \begin{subproof} + We show that $\seq{\zero}{n} \subseteq \seq{\zero}{k} \union \{\suc{k}\}$. + \begin{subproof} + Omitted. + \end{subproof} + We show that $\seq{\zero}{k} \union \{\suc{k}\} \subseteq \seq{\zero}{n}$. + \begin{subproof} + Omitted. + \end{subproof} + %$\seq{\zero}{k} \subseteq \seq{\zero}{n}$. + \end{subproof} + We have $\{n\} \notin n$. + We have $n \notmeets \{n\}$. + Let $X = \suc{n}$. + Let $Y = \seq{\zero}{n}$. + Define $F : X \to Y$ such that $F(x) =$ + \begin{cases} + &f(x) &\text{if} x \in n\\ + &\suc{k} &\text{if} x \in \{n\} + \end{cases} + + \end{byCase} + %Assume $n$ is a natural number. %We show that $\seq{\zero}{\zero}$ has cardinality $1$. %\begin{subproof} % It suffices to show that $1 = \seq{\zero}{\zero}$. % Follows by set extensionality. %\end{subproof} + %It suffices to show that if $n \neq \zero$ then $\seq{\zero}{n}$ has cardinality $\suc{n}$. + %We show that for all $m \in \naturals$ such that $m \neq \zero$ we have $\seq{\zero}{m}$ has cardinality $\suc{m}$. + %\begin{subproof} + % Fix $m \in \naturals$. + % Take $k$ such that $k \in \naturals$ and $\suc{k} = m$. + % Then $k \in m$. + %\end{subproof} + + %For all $n \in \naturals$ we have either $n = \zero$ or there exist $m \in \naturals$ such that $n = \suc{m}$. %For all $n \in \naturals$ we have either $n = \zero$ or $\zero \in n$. %We show that if $\seq{\zero}{n}$ has cardinality $\suc{n}$ then $\seq{\zero}{\suc{n}}$ has cardinality $\suc{\suc{n}}$. @@ -120,6 +168,10 @@ %\end{subproof} \end{proof} +%\begin{proposition}\label{seq_zero_to_n_isomorph_to_suc_n} +% For all $n \in \naturals$ we have $\seq{\zero}{n}$ has cardinality $\suc{n}$. +%\end{proposition} +% \begin{proposition}\label{existence_normal_ordered_urysohn} Let $X$ be a urysohn space. Suppose $U$ is a urysohnchain of $X$. -- cgit v1.2.3 From e1d55f672b063f9347c0c9718ae5cdca9a370dba Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Thu, 5 Sep 2024 01:17:16 +0200 Subject: working commit --- library/topology/urysohn2.tex | 75 ++++++++++++++++++++++++++++--------------- 1 file changed, 49 insertions(+), 26 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index d08d507..93819dc 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -104,11 +104,31 @@ $\zero$ is an \in-minimal element of $\naturals$. \end{proposition} -\begin{proposition}\label{seq_zero_to_n_isomorph_to_suc_n} - For all $n \in \naturals$ we have $\seq{\zero}{n}$ has cardinality $\suc{n}$. +\begin{proposition}\label{naturals_leq} + For all $n \in \naturals$ we have $\zero \leq n$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{naturals_leq_on_suc} + For all $n,m \in \naturals$ such that $m < \suc{n}$ we have $m \leq n$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{x_in_seq_iff} + Suppose $n,m,x \in \naturals$. + $x \in \seq{n}{m}$ iff $n \leq x \leq m$. +\end{proposition} + +\begin{proposition}\label{seq_zero_to_n_eq_to_suc_n} + For all $n \in \naturals$ we have $\seq{\zero}{n} = \suc{n}$. \end{proposition} \begin{proof} [Proof by \in-induction on $n$] Assume $n \in \naturals$. + $n \in \naturals$. For all $m \in n$ we have $m \in \naturals$. \begin{byCase} \caseOf{$n = \zero$.} @@ -117,32 +137,35 @@ \caseOf{$n \neq \zero$.} Take $k$ such that $k \in \naturals$ and $\suc{k} = n$. Then $k \in n$. - Therefore $\seq{\zero}{k}$ has cardinality $\suc{k}$. - Now $\seq{\zero}{k}$ has cardinality $n$. - Take $f$ such that $f$ is a bijection from $n$ to $\seq{\zero}{k}$. - We have $\suc{n} = n \union \{n\}$. - We show that $\seq{\zero}{n} = \seq{\zero}{k} \union \{\suc{k}\}$. + Therefore $\seq{\zero}{k} = \suc{k}$. + We show that $\seq{\zero}{n} = \seq{\zero}{k} \union \{n\}$. \begin{subproof} - We show that $\seq{\zero}{n} \subseteq \seq{\zero}{k} \union \{\suc{k}\}$. + We show that $\seq{\zero}{n} \subseteq \seq{\zero}{k} \union \{n\}$. \begin{subproof} - Omitted. + It suffices to show that for all $x \in \seq{\zero}{n}$ we have $x \in \seq{\zero}{k} \union \{n\}$. + $n \in \naturals$. + $\zero \leq n$. + $n \leq n$. + We have $n \in \seq{\zero}{n}$. + Therefore $\seq{\zero}{n}$ is inhabited. + Take $x$ such that $x \in \seq{\zero}{n}$. + Therefore $\zero \leq x \leq n$. + $x = n$ or $x < n$. + Then either $x = n$ or $x \leq k$. + Therefore $x \in \seq{\zero}{k}$ or $x = n$. + Follows by \cref{reals_order,natural_number_is_ordinal,ordinal_empty_or_emptyset_elem,naturals_leq_on_suc,reals_axiom_zero_in_reals,naturals_subseteq_reals,subseteq,union_intro_left,naturals_inductive_set,m_to_n_set,x_in_seq_iff,union_intro_right,singleton_intro}. \end{subproof} - We show that $\seq{\zero}{k} \union \{\suc{k}\} \subseteq \seq{\zero}{n}$. + We show that $\seq{\zero}{k} \union \{n\} \subseteq \seq{\zero}{n}$. \begin{subproof} - Omitted. + $k \leq n$ by \cref{naturals_subseteq_reals,suc_eq_plus_one,plus_one_order,subseteq}. + $k \in n$. + $\seq{\zero}{k} = k$ by \cref{}. + We have $\seq{\zero}{k} \subseteq \seq{\zero}{n}$. + $n \in \seq{\zero}{n}$. \end{subproof} - %$\seq{\zero}{k} \subseteq \seq{\zero}{n}$. + Trivial. \end{subproof} - We have $\{n\} \notin n$. - We have $n \notmeets \{n\}$. - Let $X = \suc{n}$. - Let $Y = \seq{\zero}{n}$. - Define $F : X \to Y$ such that $F(x) =$ - \begin{cases} - &f(x) &\text{if} x \in n\\ - &\suc{k} &\text{if} x \in \{n\} - \end{cases} - + We have $\suc{n} = n \union \{n\}$. \end{byCase} %Assume $n$ is a natural number. @@ -168,10 +191,10 @@ %\end{subproof} \end{proof} -%\begin{proposition}\label{seq_zero_to_n_isomorph_to_suc_n} -% For all $n \in \naturals$ we have $\seq{\zero}{n}$ has cardinality $\suc{n}$. -%\end{proposition} -% +\begin{proposition}\label{seq_zero_to_n_isomorph_to_suc_n} + For all $n \in \naturals$ we have $\seq{\zero}{n}$ has cardinality $\suc{n}$. +\end{proposition} + \begin{proposition}\label{existence_normal_ordered_urysohn} Let $X$ be a urysohn space. Suppose $U$ is a urysohnchain of $X$. -- cgit v1.2.3 From 12f360a500d5edddf83afa121a5a08b6a6408815 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Thu, 5 Sep 2024 16:13:04 +0200 Subject: Precondtion failed in line 161 urysohn2.tex Error: zf: Precondition failed in encodeTerm, cannot encode terms with comprehensions directly: TermSep (NamedVar "n") (TermSymbol (SymbolMixfix [Just (Command "naturals")]) []) (Scope (TermSymbol (SymbolPredicate (PredicateRelation (Command "rless"))) [TermSymbol (SymbolInteger 1) [],TermVar (B ())])) CallStack (from HasCallStack): error, called at source/Encoding.hs:89:13 in zf-0.3.0.0-3mIpbM9y9fK3yXjxxoLDb2:Encoding --- library/topology/urysohn2.tex | 138 ++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 133 insertions(+), 5 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 93819dc..396255e 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -100,17 +100,129 @@ %If $n$ is a natural number then $n$ is \in-transitive and every element of $n$ is \in-transitive. \end{proposition} +\begin{proposition}\label{natural_number_is_ordinal_for_all} + For all $n \in \naturals$ we have $n$ is a ordinal. +\end{proposition} + \begin{proposition}\label{zero_is_in_minimal} $\zero$ is an \in-minimal element of $\naturals$. \end{proposition} -\begin{proposition}\label{naturals_leq} - For all $n \in \naturals$ we have $\zero \leq n$. +\begin{proposition}\label{natural_rless_eq_precedes} + For all $n,m \in \naturals$ we have $n \precedes m$ iff $n \in m$. +\end{proposition} + +\begin{proposition}\label{naturals_precedes_suc} + For all $n \in \naturals$ we have $n \precedes \suc{n}$. +\end{proposition} + +\begin{proposition}\label{zero_is_empty} + There exists no $x$ such that $x \in \zero$. +\end{proposition} + +\begin{proposition}\label{one_is_positiv} + $1$ is positiv. +\end{proposition} + +\begin{proposition}\label{suc_of_positive_is_positive} + For all $n \in \naturals$ such that $n$ is positiv we have $\suc{n}$ is positiv. +\end{proposition} + +\begin{proposition}\label{naturals_are_positiv_besides_zero} + For all $n \in \naturals$ such that $n \neq \zero$ we have $n$ is positiv. +\end{proposition} +\begin{proof}[Proof by \in-induction on $n$] + Assume $n \in \naturals$. + \begin{byCase} + \caseOf{$n = \zero$.} Trivial. + \caseOf{$n \neq \zero$.} + Take $k \in \naturals$ such that $\suc{k} = n$. + \end{byCase} +\end{proof} + + + +\begin{proposition}\label{naturals_sum_eq_zero} + For all $n,m \in \naturals$ we have if $n+m = \zero$ then $n = m = \zero$. \end{proposition} \begin{proof} Omitted. \end{proof} +\begin{proposition}\label{no_natural_between_n_and_suc_n} + For all $n,m \in \naturals$ we have not $n < m < \suc{n}$. +\end{proposition} + +\begin{proposition}\label{naturals_rless_existence_of_lesser_natural} + For all $n \in \naturals$ we have for all $m \in \naturals$ such that $m < n$ there exist $k \in \naturals$ such that $m + k = n$. +\end{proposition} +\begin{proof}[Proof by \in-induction on $n$] + Assume $n \in \naturals$. + We show that $\naturals = (\{\zero, 1\} \union \{n \in \naturals \mid n > 1\})$. + \begin{subproof} + Trivial. + \end{subproof} + \begin{byCase} + \caseOf{$n = \zero$.} + + We show that for all $m \in \naturals$ such that $m < n$ we have there exist $k \in \naturals$ such that $m + k = n$. + \begin{subproof}[Proof by \in-induction on $m$] + Assume $m \in \naturals$. + \begin{byCase} + \caseOf{$m = \zero$.} + Trivial. + \caseOf{$m \neq \zero$.} + Trivial. + \end{byCase} + \end{subproof} + \caseOf{$n = 1$.} + Fix $m$. + For all $l \in \naturals$ we have If $l < 1$ then $l = \zero$. + Then $\zero + 1 = 1$. + \caseOf{$n > 1$.} + Take $l \in \naturals$ such that $\suc{l} = n$. + Omitted. + \end{byCase} +\end{proof} + + +\begin{proposition}\label{rless_eq_in_for_naturals} + For all $n,m \in \naturals$ such that $n < m$ we have $n \in m$. +\end{proposition} +\begin{proof} + We show that for all $n \in \naturals$ we have for all $m \in \naturals$ such that $m < n$ we have $m \in n$. + \begin{subproof}[Proof by \in-induction on $n$] + Assume $n \in \naturals$. + We show that for all $m \in \naturals$ such that$m < n$ we have $m \in n$. + \begin{subproof}[Proof by \in-induction on $m$] + Assume $m \in \naturals$. + %\begin{byCase} + % + %\end{byCase} + \end{subproof} + \end{subproof} + + %Fix $n \in \naturals$. + + %\begin{byCase} + % \caseOf{$n = \zero$.} + % For all $k \in \naturals$ we have $k = \zero$ or $\zero < k$. + % + % \caseOf{$n \neq \zero$.} + % Fix $m \in \naturals$. + % It suffices to show that $m \in n$. + %\end{byCase} + +\end{proof} + + + +\begin{proposition}\label{naturals_leq} + For all $n \in \naturals$ we have $\zero \leq n$. +\end{proposition} + + + \begin{proposition}\label{naturals_leq_on_suc} For all $n,m \in \naturals$ such that $m < \suc{n}$ we have $m \leq n$. \end{proposition} @@ -157,11 +269,27 @@ \end{subproof} We show that $\seq{\zero}{k} \union \{n\} \subseteq \seq{\zero}{n}$. \begin{subproof} + It suffices to show that for all $x \in \seq{\zero}{k} \union \{n\}$ we have $x \in \seq{\zero}{n}$. $k \leq n$ by \cref{naturals_subseteq_reals,suc_eq_plus_one,plus_one_order,subseteq}. $k \in n$. - $\seq{\zero}{k} = k$ by \cref{}. - We have $\seq{\zero}{k} \subseteq \seq{\zero}{n}$. - $n \in \seq{\zero}{n}$. + $\seq{\zero}{k} = \suc{k}$ by assumption. + $n \in \naturals$. + $\zero \leq n$. + $n \leq n$. + We have $n \in \seq{\zero}{n}$. + Therefore $\seq{\zero}{n}$ is inhabited. + Take $x$ such that $x \in \seq{\zero}{n}$. + Therefore $\zero \leq x \leq n$. + $x = n$ or $x < n$. + Then either $x = n$ or $x \leq k$. + Therefore $x \in \seq{\zero}{k}$ or $x = n$. + Fix $x$. + \begin{byCase} + \caseOf{$x \in \seq{\zero}{k}$.} + Trivial. + \caseOf{$x = n$.} + It suffices to show that $n \in \seq{\zero}{n}$. + \end{byCase} \end{subproof} Trivial. \end{subproof} -- cgit v1.2.3 From b298295ac002785672a8b16dd09f9692d73f7a80 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sun, 15 Sep 2024 15:07:36 +0200 Subject: Issue at Fixing. In Line 49 in real-topological-space.tex the Fix can't be processed. --- library/numbers.tex | 19 ++++++++ library/topology/metric-space.tex | 2 +- library/topology/real-topological-space.tex | 73 +++++++++++++++++++++++++++-- library/topology/urysohn2.tex | 73 ++++++++++++++++++++++++----- 4 files changed, 149 insertions(+), 18 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index cb3d5cf..b7de307 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -245,12 +245,31 @@ Then $(n + m) + (k + l)= n + m + k + l = n + (m + k) + l = (((n + m) + k) + l)$. \end{proposition} +%\begin{proposition}\label{natural_disstro_oneline} +% Suppose $n,m,k \in \naturals$. +% Then $n \rmul (m + k) = (n \rmul m) + (n \rmul k)$. +%\end{proposition} +%\begin{proof} +% Let $P = \{n \in \naturals \mid \forall m,k \in \naturals . n \rmul (m + k) = (n \rmul m) + (n \rmul k)\}$. +% $\zero \in P$. +% $P \subseteq \naturals$. +% It suffices to show that for all $n \in P$ we have $\suc{n} \in P$. +% Fix $n \in P$. +% It suffices to show that for all $m'$ such that $m' \in \naturals$ we have for all $k' \in \naturals$ we have $\suc{n} \rmul (m' + k') = (\suc{n} \rmul m') + (\suc{n} \rmul k')$. +% Fix $m' \in \naturals$. +% Fix $k' \in \naturals$. +% $n \in \naturals$. +% $ \suc{n} \rmul (m' + k') = (n \rmul (m' + k')) + (m' + k') = ((n \rmul m') + (n \rmul k')) + (m' + k') = ((n \rmul m') + (n \rmul k')) + m' + k' = (((n \rmul m') + (n \rmul k')) + m') + k' = (m' + ((n \rmul m') + (n \rmul k'))) + k' = ((m' + (n \rmul m')) + (n \rmul k')) + k' = (((n \rmul m') + m') + (n \rmul k')) + k' = ((n \rmul m') + m') + ((n \rmul k') + k') = (\suc{n} \rmul m') + (\suc{n} \rmul k')$. +%\end{proof} + \begin{proposition}\label{natural_disstro} Suppose $n,m,k \in \naturals$. Then $n \rmul (m + k) = (n \rmul m) + (n \rmul k)$. \end{proposition} \begin{proof} + %Let $P = \{n \in \naturals \mid \forall m,k \in \naturals . n \rmul (m + k) = (n \rmul m) + (n \rmul k)\}$. Let $P = \{n \in \naturals \mid \text{for all $m,k \in \naturals$ we have $n \rmul (m + k) = (n \rmul m) + (n \rmul k)$}\}$. + $\zero \in P$. $P \subseteq \naturals$. It suffices to show that for all $n \in P$ we have $\suc{n} \in P$. diff --git a/library/topology/metric-space.tex b/library/topology/metric-space.tex index bcc5b8c..0ed7bab 100644 --- a/library/topology/metric-space.tex +++ b/library/topology/metric-space.tex @@ -7,7 +7,7 @@ \section{Metric Spaces} \begin{definition}\label{metric} - $f$ is a metric on $M$ iff $f$ is a function from $M \times M$ to $\reals$ and + $f$ is a metric on $M$ iff $f$ is a function from $M \times M$ to $\reaaals$ and for all $x,y,z \in M$ we have $f(x,x) = \zero$ and $f(x,y) = f(y,x)$ and diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index 239965c..db46732 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -17,10 +17,73 @@ $\topoBasisReals = \{ \epsBall{x}{\epsilon} \mid x \in \reals, \epsilon \in \realsplus\}$. \end{definition} -\begin{theorem}\label{reals_as_topo_space} - Suppose $\opens[\reals] = \genOpens{\topoBasisReals}{\reals}$. - Then $\reals$ is a topological space. +\begin{axiom}\label{reals_carrier_reals} + $\carrier[\reals] = \reals$. +\end{axiom} + +\begin{theorem}\label{topological_basis_reals_is_prebasis} + $\topoBasisReals$ is a topological prebasis for $\reals$. \end{theorem} \begin{proof} - Omitted. -\end{proof} \ No newline at end of file + We show that $\unions{\topoBasisReals} \subseteq \reals$. + \begin{subproof} + It suffices to show that for all $x \in \unions{\topoBasisReals}$ we have $x \in \reals$. + Fix $x \in \unions{\topoBasisReals}$. + \end{subproof} + We show that $\reals \subseteq \unions{\topoBasisReals}$. + \begin{subproof} + It suffices to show that for all $x \in \reals$ we have $x \in \unions{\topoBasisReals}$. + Fix $x \in \reals$. + \end{subproof} +\end{proof} + +\begin{theorem}\label{topological_basis_reals_is_basis} + $\topoBasisReals$ is a topological basis for $\reals$. +\end{theorem} +\begin{proof} + $\topoBasisReals$ is a topological prebasis for $\reals$ by \cref{topological_basis_reals_is_prebasis}. + Let $B = \topoBasisReals$. + It suffices to show that for all $U \in B$ we have for all $V \in B$ we have for all $x$ such that $x \in U, V$ there exists $W\in B$ such that $x\in W\subseteq U, V$. + Fix $U \in B$. + Fix $V \in B$. + Fix $x \in U, V$. +\end{proof} + +\begin{axiom}\label{topological_space_reals} + $\opens[\reals] = \genOpens{\topoBasisReals}{\reals}$. +\end{axiom} + +\begin{theorem}\label{reals_is_topological_space} + $\reals$ is a topological space. +\end{theorem} +\begin{proof} + $\topoBasisReals$ is a topological basis for $\reals$. + Let $B = \topoBasisReals$. + We show that $\opens[\reals]$ is a family of subsets of $\carrier[\reals]$. + \begin{subproof} + It suffices to show that for all $A \in \opens[\reals]$ we have $A \subseteq \reals$. + Fix $A \in \opens[\reals]$. + Follows by \cref{powerset_elim,topological_space_reals,genopens}. + \end{subproof} + We show that $\reals \in\opens[\reals]$. + \begin{subproof} + $B$ covers $\reals$ by \cref{topological_prebasis_iff_covering_family,topological_basis}. + $\unions{B} \in \genOpens{B}{\reals}$. + $\reals \subseteq \unions{B}$. + \end{subproof} + We show that for all $A, B\in \opens[\reals]$ we have $A\inter B\in\opens[\reals]$. + \begin{subproof} + Follows by \cref{topological_space_reals,inters_in_genopens}. + \end{subproof} + We show that for all $F\subseteq \opens[\reals]$ we have $\unions{F}\in\opens[\reals]$. + \begin{subproof} + Follows by \cref{topological_space_reals,union_in_genopens}. + \end{subproof} + $\carrier[\reals] = \reals$. + Follows by \cref{topological_space}. +\end{proof} + +\begin{proposition}\label{open_interval_is_open} + Suppose $a,b \in \reals$. + Then $\intervalopen{a}{b} \in \opens[\reals]$. +\end{proposition} \ No newline at end of file diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 396255e..838b121 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -152,16 +152,23 @@ \begin{proposition}\label{no_natural_between_n_and_suc_n} For all $n,m \in \naturals$ we have not $n < m < \suc{n}$. \end{proposition} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{naturals_is_zero_one_or_greater} + $\naturals = \{n \in \naturals \mid n > 1 \lor n = 1 \lor n = \zero\}$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} \begin{proposition}\label{naturals_rless_existence_of_lesser_natural} For all $n \in \naturals$ we have for all $m \in \naturals$ such that $m < n$ there exist $k \in \naturals$ such that $m + k = n$. \end{proposition} \begin{proof}[Proof by \in-induction on $n$] Assume $n \in \naturals$. - We show that $\naturals = (\{\zero, 1\} \union \{n \in \naturals \mid n > 1\})$. - \begin{subproof} - Trivial. - \end{subproof} + \begin{byCase} \caseOf{$n = \zero$.} @@ -196,9 +203,17 @@ We show that for all $m \in \naturals$ such that$m < n$ we have $m \in n$. \begin{subproof}[Proof by \in-induction on $m$] Assume $m \in \naturals$. - %\begin{byCase} - % - %\end{byCase} + \begin{byCase} + \caseOf{$\suc{m}=n$.} + \caseOf{$\suc{m}\neq n$.} + \begin{byCase} + \caseOf{$n = \zero$.} + \caseOf{$n \neq \zero$.} + Take $l \in \naturals$ such that $\suc{l} = n$. + Omitted. + + \end{byCase} + \end{byCase} \end{subproof} \end{subproof} @@ -323,17 +338,51 @@ For all $n \in \naturals$ we have $\seq{\zero}{n}$ has cardinality $\suc{n}$. \end{proposition} +\begin{proposition}\label{bijection_naturals_order} + For all $M \subseteq \naturals$ such that $M$ is inhabited we have there exist $f,k$ such that $f$ is a bijection from $\seq{\zero}{k}$ to $M$ and $M$ has cardinality $\suc{k}$ and for all $n,m \in \seq{\zero}{k}$ such that $n < m$ we have $f(n) < f(m)$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} + \begin{proposition}\label{existence_normal_ordered_urysohn} Let $X$ be a urysohn space. Suppose $U$ is a urysohnchain of $X$. Suppose $\dom{U}$ is finite. - Then there exist $V,f$ such that $V$ is a urysohnchain of $X$ and $f$ is consistent on $X$ to $Y$ and $V$ is normal ordered. + Suppose $U$ is inhabited. + Then there exist $V,f$ such that $V$ is a urysohnchain of $X$ and $f$ is consistent on $U$ to $V$ and $V$ is normal ordered. \end{proposition} \begin{proof} - Take $k$ such that $\dom{U}$ has cardinality $k$ by \cref{ran_converse,cardinality,finite}. - There exist $F$ such that $F$ is a bijection from $\seq{\zero}{k}$ to $\dom{U}$. - - + Take $n$ such that $\dom{U}$ has cardinality $n$ by \cref{ran_converse,cardinality,finite}. + \begin{byCase} + \caseOf{$n = \zero$.} + Omitted. + \caseOf{$n \neq \zero$.} + Take $k$ such that $k \in \naturals$ and $\suc{k}=n$. + We have $\dom{U} \subseteq \naturals$. + $\dom{U}$ is inhabited. + We show that there exist $F$ such that $F$ is a bijection from $\seq{\zero}{k}$ to $\dom{U}$ and for all $n',m' \in \seq{\zero}{k}$ such that $n' < m'$ we have $F(n') < F(m')$. + \begin{subproof} + Omitted. + \end{subproof} + Let $N = \seq{\zero}{k}$. + Let $M = \pow{X}$. + Define $V : N \to M$ such that $V(n)=$ + \begin{cases} + &\at{U}{F(n)} & \text{if} n \in N + \end{cases} + $\dom{V} = \seq{\zero}{k}$. + We show that $V$ is a urysohnchain of $X$. + \begin{subproof} + Trivial. + \end{subproof} + We show that $F$ is consistent on $U$ to $V$. + \begin{subproof} + Trivial. + \end{subproof} + $V$ is normal ordered. + \end{byCase} + \end{proof} -- cgit v1.2.3 From 3ef20da08eda23db76d763a2c6c7ee416348a021 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Sun, 15 Sep 2024 19:05:25 +0200 Subject: Fix Assume Issue At line 137 in real-topological-space.tex Can't use Fix at this point. --- library/numbers.tex | 4 +- library/topology/real-topological-space.tex | 143 +++++++++++++++++++++++++++- 2 files changed, 143 insertions(+), 4 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index b7de307..73eefc8 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -718,7 +718,7 @@ Laws of the order on the reals \end{proof} \begin{lemma}\label{reals_minus} - Assume $x,y \in \reals$. If $x \rminus y = \zero$ then $x=y$. + Assume $x,y \in \reals$. If $x - y = \zero$ then $x=y$. \end{lemma} \begin{proof} Omitted. @@ -803,7 +803,7 @@ Laws of the order on the reals \end{definition} \begin{definition}\label{realsplus} - $\realsplus = \reals \setminus \realsminus$. + $\realsplus = \{r \in \reals \mid r > \zero\}$. \end{definition} \begin{definition}\label{epsilon_ball} diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index db46732..8757ffb 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -21,6 +21,123 @@ $\carrier[\reals] = \reals$. \end{axiom} +\begin{lemma}\label{intervals_are_connected_in_reals} + Suppose $a,b \in \reals$. + Then for all $c \in \reals$ such that $a < c < b$ we have $c \in \intervalopen{a}{b}$. +\end{lemma} + +\begin{lemma}\label{epsball_are_subset_reals_elem} + Suppose $x \in \reals$. + Suppose $\epsilon \in \realsplus$. + Then for all $y \in \epsBall{x}{\epsilon}$ we have $y \in \reals$. +\end{lemma} + +\begin{lemma}\label{intervalopen_iff} + Suppose $a,b,c \in \reals$. + Suppose $a < b$. + $c \in \intervalopen{a}{b}$ iff $a < c < b$. +\end{lemma} + +\begin{lemma}\label{epsball_are_subseteq_reals_set} + Suppose $x \in \reals$. + Suppose $\epsilon \in \realsplus$. + Then $\epsBall{x}{\epsilon} \subseteq \reals$. +\end{lemma} + +\begin{lemma}\label{epsball_are_subset_reals_set} + Suppose $x \in \reals$. + Suppose $\epsilon \in \realsplus$. + Then $\epsBall{x}{\epsilon} \subset \reals$. +\end{lemma} + +\begin{lemma}\label{reals_order_minus_positiv} + Suppose $x,y \in \reals$. + Suppose $\zero < y$. + $x - y < x$. +\end{lemma} + +\begin{lemma}\label{realsplus_bigger_zero} + For all $x \in \realsplus$ we have $\zero < x$. +\end{lemma} + +\begin{lemma}\label{realsplus_in_reals} + For all $x \in \realsplus$ we have $x \in \reals$. +\end{lemma} + +\begin{lemma}\label{epsball_are_inhabited} + Suppose $x \in \reals$. + Suppose $\epsilon \in \realsplus$. + Then $\epsBall{x}{\epsilon}$ is inhabited. +\end{lemma} +\begin{proof} + $x < x + \epsilon$. + $x - \epsilon < x$. + $x \in \epsBall{x}{\epsilon}$. +\end{proof} + +\begin{lemma}\label{reals_elem_inbetween} + For all $a,b \in \reals$ such that $a < b$ we have there exists $c \in \reals$ such that $a < c < b$. +\end{lemma} + +\begin{lemma}\label{epsball_equal_openinterval} + Suppose $x \in \reals$. + Suppose $\epsilon \in \realsplus$. + Then $\epsBall{x}{\epsilon} = \intervalopen{x - \epsilon}{x + \epsilon}$. +\end{lemma} + +\begin{lemma}\label{minus_behavior1} + For all $x \in \reals$ we have $x - x = \zero$. +\end{lemma} + +\begin{lemma}\label{minus_behavior2} + For all $x \in \reals$ we have $x + \neg{x} = \zero$. +\end{lemma} + +\begin{lemma}\label{minus_behavior3} + For all $x \in \reals$ we have $\neg{x} = \zero - x$. +\end{lemma} + +\begin{lemma}\label{reals_order_is_addition_with_positiv_number} + For all $x,y \in \reals$ such that $x < y$ we have there exists $z \in \realsplus$ such that $x + z = y$. +\end{lemma} +\begin{proof} + %Fix $x,y \in \reals$. +\end{proof} + +\begin{lemma}\label{reals_order_is_transitive} + For all $x,y,z \in \reals$ such that $x < y$ and $y < z$ we have $x < z$. +\end{lemma} + +\begin{lemma}\label{reals_order_plus_minus} + Suppose $a,b \in \reals$. + Suppose $\zero < b$. + Then $(a-b) < (a+b)$. +\end{lemma} +\begin{proof} + We show that $a < (a+b)$. + \begin{subproof} + Trivial. + \end{subproof} + We show that $(a-b) < a$. + \begin{subproof} + Trivial. + \end{subproof} +\end{proof} + +\begin{lemma}\label{epsball_are_connected_in_reals} + Suppose $x \in \reals$. + Suppose $\epsilon \in \realsplus$. + Then for all $c \in \reals$ such that $(x - \epsilon) < c < (x + \epsilon)$ we have $c \in \epsBall{x}{\epsilon}$. +\end{lemma} +\begin{proof} + $x - \epsilon \in \reals$. + $x + \epsilon \in \reals$. + + It suffices to show that for all $c$ such that $c \in \reals \land (x - \epsilon) < c < (x + \epsilon)$ we have $c \in \epsBall{x}{\epsilon}$. + %Fix $c$ such that $c \in \reals \land (x - \epsilon) < c < (x + \epsilon)$. + %Suppose $(x - \epsilon) < c < (x + \epsilon)$. +\end{proof} + \begin{theorem}\label{topological_basis_reals_is_prebasis} $\topoBasisReals$ is a topological prebasis for $\reals$. \end{theorem} @@ -29,11 +146,21 @@ \begin{subproof} It suffices to show that for all $x \in \unions{\topoBasisReals}$ we have $x \in \reals$. Fix $x \in \unions{\topoBasisReals}$. + \begin{byCase} + \caseOf{$x = \emptyset$.} + Trivial. + \caseOf{$x \neq \emptyset$.} + There exists $U \in \topoBasisReals$ such that $x \in \topoBasisReals$. + Take $U \in \topoBasisReals$ such that $x \in \topoBasisReals$. + + \end{byCase} \end{subproof} We show that $\reals \subseteq \unions{\topoBasisReals}$. \begin{subproof} It suffices to show that for all $x \in \reals$ we have $x \in \unions{\topoBasisReals}$. Fix $x \in \reals$. + $\epsBall{x}{1} \in \topoBasisReals$. + Therefore $x \in \unions{\topoBasisReals}$. \end{subproof} \end{proof} @@ -46,7 +173,15 @@ It suffices to show that for all $U \in B$ we have for all $V \in B$ we have for all $x$ such that $x \in U, V$ there exists $W\in B$ such that $x\in W\subseteq U, V$. Fix $U \in B$. Fix $V \in B$. - Fix $x \in U, V$. + It suffices to show that for all $x \in U \inter V$ there exists $W\in B$ such that $x\in W\subseteq U, V$. + Fix $x \in U \inter V$. + \begin{byCase} + \caseOf{$U \inter V = \emptyset$.} + Trivial. + \caseOf{$U \inter V \neq \emptyset$.} + Then $U \inter V$ is inhabited. + %It suffices to show that + \end{byCase} \end{proof} \begin{axiom}\label{topological_space_reals} @@ -86,4 +221,8 @@ \begin{proposition}\label{open_interval_is_open} Suppose $a,b \in \reals$. Then $\intervalopen{a}{b} \in \opens[\reals]$. -\end{proposition} \ No newline at end of file +\end{proposition} + +\begin{lemma}\label{safetwo} + Contradiction. +\end{lemma} \ No newline at end of file -- cgit v1.2.3 From 640fe16eaab00ea29046ef18e6f751571d923eaa Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 16 Sep 2024 02:12:06 +0200 Subject: Missmatched Assume Error found. not fixed In line 137 and 140 of real-topological-space.tex Line 140 wont get parsed and line 140 will get parsed. Case matching in Checking.hs is not working as intended. Case phi@(Atomic p args) -> in line 960 in checking.hs is not reached in checking of the Fixing with a symbol Error Massage shows that < will be interpreted as a Term symbol there not as a predicate. Error Massage: zf: MismatchedAssume (TermSymbol (SymbolPredicate (PredicateRelation (Symbol "<"))) [TermVar (NamedVar "c"),TermVar (NamedVar "x")]) (Connected Implication (TermSymbol (SymbolPredicate (PredicateRelation (Command "rless"))) [TermVar (NamedVar "c"),TermVar (NamedVar "x")]) (TermSymbol (SymbolPredicate (PredicateRelation (Command "in"))) [TermVar (NamedVar "c"),TermSymbol (SymbolMixfix [Just (Command "epsBall"),Just InvisibleBraceL,Nothing,Just InvisibleBraceR,Just InvisibleBraceL,Nothing,Just InvisibleBraceR]) [TermVar (NamedVar "x"),TermVar (NamedVar "epsilon")]])) --- library/topology/real-topological-space.tex | 14 +++++++++++--- 1 file changed, 11 insertions(+), 3 deletions(-) (limited to 'library') diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index 8757ffb..1c5e4cb 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -132,9 +132,17 @@ \begin{proof} $x - \epsilon \in \reals$. $x + \epsilon \in \reals$. - - It suffices to show that for all $c$ such that $c \in \reals \land (x - \epsilon) < c < (x + \epsilon)$ we have $c \in \epsBall{x}{\epsilon}$. - %Fix $c$ such that $c \in \reals \land (x - \epsilon) < c < (x + \epsilon)$. + + + %It suffices to show that for all $c$ such that $c \rless x$ we have $c \in \epsBall{x}{\epsilon}$. + %Fix $c$ such that $c \rless x$. +% + %It suffices to show that for all $c$ such that $c < x$ we have $c \in \epsBall{x}{\epsilon}$. + %Fix $c$ such that $c < x$. + + + It suffices to show that for all $c$ such that $c \in \reals \land (x - \epsilon) \rless c \rless (x + \epsilon)$ we have $c \in \epsBall{x}{\epsilon}$. + Fix $c$ such that $(c \in \reals) \land (x - \epsilon) \rless c \rless (x + \epsilon)$. %Suppose $(x - \epsilon) < c < (x + \epsilon)$. \end{proof} -- cgit v1.2.3 From 588c6ab14184cab4bb7df89def641acaafe3b7eb Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 16 Sep 2024 11:34:01 +0200 Subject: working commit --- latex/naproche.sty | 20 +- latex/stdlib.tex | 4 +- library/topology/real-topological-space.tex | 121 ++++++- library/topology/urysohn.tex | 515 ++++++++++++++-------------- library/topology/urysohn2.tex | 14 +- 5 files changed, 387 insertions(+), 287 deletions(-) (limited to 'library') diff --git a/latex/naproche.sty b/latex/naproche.sty index 5ca673d..1a8afb6 100644 --- a/latex/naproche.sty +++ b/latex/naproche.sty @@ -40,6 +40,7 @@ \newtheorem{remark}[theoremcount]{Remark} \newtheorem{signature}[theoremcount]{Signature} \newtheorem{theorem}[theoremcount]{Theorem} +\newtheorem{inductive}[theoremcount]{Inductive} % Theorem environments without numbering. \newtheorem*{quotedaxiom}{Axiom} @@ -139,9 +140,22 @@ \newcommand{\rationals}{\mathcal{Q}} \newcommand{\rminus}{-_{\mathcal{R}}} \newcommand{\seq}[2]{\{#1, ... ,#2\}} -\newcommand{\indexx}[2]{index_{#1}(#2)} -\newcommand{\indexset}[2]{#1} - +\newcommand{\indexx}[2][]{index_{#1}(#2)} +\newcommand{\indexxset}[1]{#1} +\newcommand{\topoBasisReals}{\mathbb{B}_{\mathcal{R}}} +\newcommand{\intervalopen}[2]{(#1, #2)} +\newcommand{\intervalclosed}[2]{[#1, #2]} +\newcommand{\epsBall}[2]{\mathcal{B}_{#1,#2}} +\newcommand{\realsplus}{\reals_{+}} +\newcommand{\rless}{<} +\newcommand{\two}{2} +\newcommand{\powerOfTwoSet}{\mathbb{P}_{2^{}}} +\newcommand{\pot}{\powerOfTwoSet} +\newcommand{\chain}[1]{#1} +\newcommand{\refine}{\text{ finer than }} +\newcommand{\abs}[1]{\left\lvert#1\right\rvert} +\newcommand{\realsminus}{\reals_{-}} +\newcommand{\at}[2]{#1(#2)} \newcommand\restrl[2]{{% we make the whole thing an ordinary symbol diff --git a/latex/stdlib.tex b/latex/stdlib.tex index 2faa267..9879708 100644 --- a/latex/stdlib.tex +++ b/latex/stdlib.tex @@ -47,5 +47,7 @@ \input{../library/topology/disconnection.tex} \input{../library/numbers.tex} \input{../library/topology/urysohn.tex} - \input{../library/wunschzettel.tex} + \input{../library/topology/urysohn2.tex} + \input{../library/topology/real-topological-space.tex} + %\input{../library/wunschzettel.tex} \end{document} diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index 1c5e4cb..ffdf46e 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -11,7 +11,7 @@ \import{function.tex} -\section{The canonical topology on $\mathbbR$} +\section{The canonical topology on $\mathbb{R}$} \begin{definition}\label{topological_basis_reals_eps_ball} $\topoBasisReals = \{ \epsBall{x}{\epsilon} \mid x \in \reals, \epsilon \in \realsplus\}$. @@ -101,12 +101,15 @@ For all $x,y \in \reals$ such that $x < y$ we have there exists $z \in \realsplus$ such that $x + z = y$. \end{lemma} \begin{proof} - %Fix $x,y \in \reals$. + Omitted. \end{proof} \begin{lemma}\label{reals_order_is_transitive} For all $x,y,z \in \reals$ such that $x < y$ and $y < z$ we have $x < z$. \end{lemma} +\begin{proof} + Omitted. +\end{proof} \begin{lemma}\label{reals_order_plus_minus} Suppose $a,b \in \reals$. @@ -134,16 +137,9 @@ $x + \epsilon \in \reals$. - %It suffices to show that for all $c$ such that $c \rless x$ we have $c \in \epsBall{x}{\epsilon}$. - %Fix $c$ such that $c \rless x$. -% - %It suffices to show that for all $c$ such that $c < x$ we have $c \in \epsBall{x}{\epsilon}$. - %Fix $c$ such that $c < x$. - - It suffices to show that for all $c$ such that $c \in \reals \land (x - \epsilon) \rless c \rless (x + \epsilon)$ we have $c \in \epsBall{x}{\epsilon}$. Fix $c$ such that $(c \in \reals) \land (x - \epsilon) \rless c \rless (x + \epsilon)$. - %Suppose $(x - \epsilon) < c < (x + \epsilon)$. + $(x - \epsilon) < c < (x + \epsilon)$. \end{proof} \begin{theorem}\label{topological_basis_reals_is_prebasis} @@ -158,9 +154,9 @@ \caseOf{$x = \emptyset$.} Trivial. \caseOf{$x \neq \emptyset$.} - There exists $U \in \topoBasisReals$ such that $x \in \topoBasisReals$. - Take $U \in \topoBasisReals$ such that $x \in \topoBasisReals$. - + %There exists $U \in \topoBasisReals$ such that $x \in U$. + Take $U \in \topoBasisReals$ such that $x \in U$. + Follows by \cref{epsball_are_subseteq_reals_set,topological_basis_reals_eps_ball,epsilon_ball,minus,subseteq}. \end{byCase} \end{subproof} We show that $\reals \subseteq \unions{\topoBasisReals}$. @@ -168,10 +164,63 @@ It suffices to show that for all $x \in \reals$ we have $x \in \unions{\topoBasisReals}$. Fix $x \in \reals$. $\epsBall{x}{1} \in \topoBasisReals$. - Therefore $x \in \unions{\topoBasisReals}$. + Therefore $x \in \unions{\topoBasisReals}$ by \cref{one_in_reals,reals_one_bigger_zero,unions_intro,realsplus,plus_one_order,reals_order_minus_positiv,epsball_are_connected_in_reals}. \end{subproof} \end{proof} +%\begin{lemma}\label{intervl_intersection_is_interval} +% Suppose $a,b,a',b' \in \reals$. +% Suppose there exist $x \in \reals$ such that $x \in \intervalopen{a}{b} \inter \intervalopen{a'}{b'}$. +% Then there exists $q,p \in \reals$ such that $q < p$ and $\intervalopen{q}{p} \subseteq \intervalopen{a}{b} \inter \intervalopen{a'}{b'}$. +%\end{lemma} +% + +\begin{lemma}\label{reals_order_total} + For all $x,y \in \reals$ we have either $x < y$ or $x \geq y$. +\end{lemma} +\begin{proof} + It suffices to show that for all $x \in \reals$ we have for all $y \in \reals$ we have either $x < y$ or $x \geq y$. + Fix $x \in \reals$. + Fix $y \in \reals$. + Omitted. +\end{proof} + +\begin{lemma}\label{topo_basis_reals_eps_iff} + $X \in \topoBasisReals$ iff there exists $x_0, \delta$ such that $x_0 \in \reals$ and $\delta \in \realsplus$ and $\epsBall{x_0}{\delta} = X$. +\end{lemma} + +\begin{lemma}\label{topo_basis_reals_intro} +For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have $\epsBall{x}{\delta} \in \topoBasisReals$. +\end{lemma} + +\begin{lemma}\label{realspuls_in_reals_plus} + For all $x,y$ such that $x \in \reals$ and $y \in \realsplus$ we have $x + y \in \reals$. +\end{lemma} + +\begin{lemma}\label{realspuls_in_reals_minus} + For all $x,y$ such that $x \in \reals$ and $y \in \realsplus$ we have $x - y \in \reals$. +\end{lemma} + +\begin{lemma}\label{eps_ball_implies_open_interval} + Suppose $x \in \reals$. + Suppose $\epsilon \in \realsplus$. + Then there exists $a,b \in \reals$ such that $a < b$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$. +\end{lemma} + +\begin{lemma}\label{open_interval_eq_eps_ball} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then there exist $x,\epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$. +\end{lemma} +\begin{proof} + Let $\delta = (b-a)$. + $\delta$ is positiv by \cref{minus_}. + There exists $\epsilon \in \reals$ such that $\epsilon + \epsilon = \delta$. + +\end{proof} + + + \begin{theorem}\label{topological_basis_reals_is_basis} $\topoBasisReals$ is a topological basis for $\reals$. \end{theorem} @@ -188,7 +237,45 @@ Trivial. \caseOf{$U \inter V \neq \emptyset$.} Then $U \inter V$ is inhabited. - %It suffices to show that + $x \in \reals$ by \cref{inter_lower_left,subseteq,topological_prebasis_iff_covering_family,omega_is_an_ordinal,naturals_subseteq_reals,subset_transitive,suc_subseteq_elim,ordinal_suc_subseteq}. + There exists $x_1, \alpha$ such that $x_1 \in \reals$ and $\alpha \in \realsplus$ and $\epsBall{x_1}{\alpha} = U$. + There exists $x_2, \beta$ such that $x_2 \in \reals$ and $\beta \in \realsplus$ and $\epsBall{x_2}{\beta} = V$. + Then $ (x_1 - \alpha) < x < (x_1 + \alpha)$. + Then $ (x_2 - \beta) < x < (x_2 + \beta)$. + We show that if there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$ then there exists $W\in B$ such that $x\in W\subseteq U, V$. + \begin{subproof} + Suppose there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$. + $x \in \epsBall{x}{\delta}$. + $\epsBall{x}{\delta} \subseteq U$. + $\epsBall{x}{\delta} \subseteq V$. + $\epsBall{x}{\delta} \in B$. + \end{subproof} + It suffices to show that there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$. + + + %It suffices to show that there exists $x_0, \delta$ such that $x_0 \in \reals$ and $\delta \in \realsplus$ and $\epsBall{x_0}{\delta} \subseteq u \inter V$. + %There exists $x_1, \alpha$ such that $x_1 \in \reals$ and $\alpha \in \realsplus$ and $\epsBall{x_1}{\alpha} = U$. + %There exists $x_2, \beta$ such that $x_2 \in \reals$ and $\beta \in \realsplus$ and $\epsBall{x_2}{\beta} = V$. + %Then $ (x_1 - \alpha) < x < (x_1 + \alpha)$. + %Then $ (x_2 - \beta) < x < (x_2 + \beta)$. + %\begin{byCase} + % \caseOf{$x_1 = x_2$.} + % Take $\gamma \in \realsplus$ such that either $\gamma = \alpha \land \gamma \leq \beta$ or $\gamma \leq \alpha \land \gamma = \beta$. + % \caseOf{$x_1 < x_2$.} + % \caseOf{$x_1 > x_2$.} + %\end{byCase} + %%Take $m$ such that $m \in \min{\{(x_1 + \alpha), (x_2 + \beta)\}}$. + %Take $n$ such that $n \in \max{\{(x_1 - \alpha), (x_2 - \beta)\}}$. + %Then $m < x < n$. + %We show that there exists $x_1 \in \reals$ such that $x_1 \in U \inter V$ and $x_1 < x$. + %\begin{subproof} + % Suppose not. + % Then For all $y \in U \inter V$ we have $x \leq y$. + %\end{subproof} + %We show that there exists $x_2 \in \reals$ such that $x_2 \in U \inter V$ and $x_2 > x$. + %\begin{subproof} + % Trivial. + %\end{subproof} \end{byCase} \end{proof} @@ -230,7 +317,3 @@ Suppose $a,b \in \reals$. Then $\intervalopen{a}{b} \in \opens[\reals]$. \end{proposition} - -\begin{lemma}\label{safetwo} - Contradiction. -\end{lemma} \ No newline at end of file diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index 17e2911..ff6a231 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -51,13 +51,14 @@ The first tept will be a formalisation of chain constructions. \begin{struct}\label{sequence} A sequence $X$ is a onesorted structure equipped with \begin{enumerate} - \item $\index$ - \item $\indexset$ + \item $\indexx$ + \item $\indexxset$ + \end{enumerate} such that \begin{enumerate} - \item\label{indexset_is_subset_naturals} $\indexset[X] \subseteq \naturals$. - \item\label{index_is_bijection} $\index[X]$ is a bijection from $\indexset[X]$ to $\carrier[X]$. + \item\label{indexset_is_subset_naturals} $\indexxset[X] \subseteq \naturals$. + \item\label{index_is_bijection} $\indexx[X]$ is a bijection from $\indexxset[X]$ to $\carrier[X]$. \end{enumerate} \end{struct} @@ -68,13 +69,13 @@ The first tept will be a formalisation of chain constructions. \begin{definition}\label{cahin_of_subsets} $C$ is a chain of subsets iff - $C$ is a sequence and for all $n,m \in \indexset[C]$ such that $n < m$ we have $\index[C](n) \subseteq \index[C](m)$. + $C$ is a sequence and for all $n,m \in \indexxset[C]$ such that $n < m$ we have $\indexx[C](n) \subseteq \indexx[C](m)$. \end{definition} \begin{definition}\label{chain_of_n_subsets} $C$ is a chain of $n$ subsets iff - $C$ is a chain of subsets and $n \in \indexset[C]$ - and for all $m \in \naturals$ such that $m \leq n$ we have $m \in \indexset[C]$. + $C$ is a chain of subsets and $n \in \indexxset[C]$ + and for all $m \in \naturals$ such that $m \leq n$ we have $m \in \indexxset[C]$. \end{definition} @@ -133,18 +134,18 @@ The first tept will be a formalisation of chain constructions. \item \label{staircase_domain} $\dom{f}$ is a topological space. \item \label{staricase_def_chain} $C$ is a chain of subsets. \item \label{staircase_chain_is_in_domain} for all $x \in C$ we have $x \subseteq \dom{f}$. - \item \label{staircase_behavoir_index_zero} $f(\index[C](1))= 1$. + \item \label{staircase_behavoir_index_zero} $f(\indexx[C](1))= 1$. \item \label{staircase_behavoir_index_n} $f(\dom{f}\setminus \unions{C}) = \zero$. - \item \label{staircase_chain_indeset} There exist $n$ such that $\indexset[C] = \seq{\zero}{n}$. - \item \label{staircase_behavoir_index_arbetrray} for all $n \in \indexset[C]$ - such that $n \neq \zero$ we have $f(\index[C](n) \setminus \index[C](n-1)) = \rfrac{n}{ \max{\indexset[C]} }$. + \item \label{staircase_chain_indeset} There exist $n$ such that $\indexxset[C] = \seq{\zero}{n}$. + \item \label{staircase_behavoir_index_arbetrray} for all $n \in \indexxset[C]$ + such that $n \neq \zero$ we have $f(\indexx[C](n) \setminus \indexx[C](n-1)) = \rfrac{n}{ \max{\indexxset[C]} }$. \end{enumerate} \end{struct} \begin{definition}\label{legal_staircase} $f$ is a legal staircase function iff $f$ is a staircase function and - for all $n,m \in \indexset[\chain[f]]$ such that $n \leq m$ we have $f(\index[\chain[f]](n)) \leq f(\index[\chain[f]](m))$. + for all $n,m \in \indexxset[\chain[f]]$ such that $n \leq m$ we have $f(\indexx[\chain[f]](n)) \leq f(\indexx[\chain[f]](m))$. \end{definition} \begin{abbreviation}\label{urysohnspace} @@ -159,23 +160,23 @@ The first tept will be a formalisation of chain constructions. $C$ is a urysohnchain in $X$ of cardinality $k$ iff %<---- TODO: cardinality unused! $C$ is a chain of subsets and for all $A \in C$ we have $A \subseteq X$ and - for all $n,m \in \indexset[C]$ such that $n < m$ we have $\closure{\index[C](n)}{X} \subseteq \interior{\index[C](m)}{X}$. + for all $n,m \in \indexxset[C]$ such that $n < m$ we have $\closure{\indexx[C](n)}{X} \subseteq \interior{\indexx[C](m)}{X}$. \end{definition} \begin{definition}\label{urysohnchain_without_cardinality} $C$ is a urysohnchain in $X$ iff $C$ is a chain of subsets and for all $A \in C$ we have $A \subseteq X$ and - for all $n,m \in \indexset[C]$ such that $n < m$ we have $\closure{\index[C](n)}{X} \subseteq \interior{\index[C](m)}{X}$. + for all $n,m \in \indexxset[C]$ such that $n < m$ we have $\closure{\indexx[C](n)}{X} \subseteq \interior{\indexx[C](m)}{X}$. \end{definition} \begin{abbreviation}\label{infinte_sequence} - $S$ is a infinite sequence iff $S$ is a sequence and $\indexset[S]$ is infinite. + $S$ is a infinite sequence iff $S$ is a sequence and $\indexxset[S]$ is infinite. \end{abbreviation} \begin{definition}\label{infinite_product} $X$ is the infinite product of $Y$ iff - $X$ is a infinite sequence and for all $i \in \indexset[X]$ we have $\index[X](i) = Y$. + $X$ is a infinite sequence and for all $i \in \indexxset[X]$ we have $\indexx[X](i) = Y$. \end{definition} \begin{definition}\label{refinmant} @@ -289,9 +290,9 @@ The first tept will be a formalisation of chain constructions. Suppose $A \inter B$ is empty. Then there exist $U$ such that $\carrier[U] = \{A,(\carrier[X] \setminus B)\}$ - and $\indexset[U]= \{\zero, 1\}$ - and $\index[U](\zero) = A$ - and $\index[U](1) = (\carrier[X] \setminus B)$. + and $\indexxset[U]= \{\zero, 1\}$ + and $\indexx[U](\zero) = A$ + and $\indexx[U](1) = (\carrier[X] \setminus B)$. %$U$ is a urysohnchain in $X$. \end{proposition} \begin{proof} @@ -310,12 +311,12 @@ The first tept will be a formalisation of chain constructions. % % We show that $U$ is a chain of subsets. % \begin{subproof} - % For all $n \in \indexset[U]$ we have $n = \zero \lor n = 1$. - % It suffices to show that for all $n \in \indexset[U]$ we have - % for all $m \in \indexset[U]$ such that - % $n < m$ we have $\index[U](n) \subseteq \index[U](m)$. - % Fix $n \in \indexset[U]$. - % Fix $m \in \indexset[U]$. + % For all $n \in \indexxset[U]$ we have $n = \zero \lor n = 1$. + % It suffices to show that for all $n \in \indexxset[U]$ we have + % for all $m \in \indexxset[U]$ such that + % $n < m$ we have $\indexx[U](n) \subseteq \indexx[U](m)$. + % Fix $n \in \indexxset[U]$. + % Fix $m \in \indexxset[U]$. % \begin{byCase} % \caseOf{$n = 1$.} Trivial. % \caseOf{$n = \zero$.} @@ -337,8 +338,8 @@ The first tept will be a formalisation of chain constructions. % \begin{subproof} % Omitted. % \end{subproof} - % We show that for all $n,m \in \indexset[U]$ such that $n < m$ we have - % $\closure{\index[U](n)}{X} \subseteq \interior{\index[U](m)}{X}$. + % We show that for all $n,m \in \indexxset[U]$ such that $n < m$ we have + % $\closure{\indexx[U](n)}{X} \subseteq \interior{\indexx[U](m)}{X}$. % \begin{subproof} % Omitted. % \end{subproof} @@ -352,9 +353,9 @@ The first tept will be a formalisation of chain constructions. Suppose $A \inter B$ is empty. Suppose there exist $U$ such that $\carrier[U] = \{A,(\carrier[X] \setminus B)\}$ - and $\indexset[U]= \{\zero, 1\}$ - and $\index[U](\zero) = A$ - and $\index[U](1) = (\carrier[X] \setminus B)$. + and $\indexxset[U]= \{\zero, 1\}$ + and $\indexx[U](\zero) = A$ + and $\indexx[U](1) = (\carrier[X] \setminus B)$. Then $U$ is a urysohnchain in $X$. \end{proposition} \begin{proof} @@ -384,9 +385,9 @@ The first tept will be a formalisation of chain constructions. % U = ( A_{0}, A_{1}, A_{2}, ... , A_{n-1}, A_{n}) % U' = ( A_{0}, A'_{1}, A_{1}, A'_{2}, A_{2}, ... A_{n-1}, A'_{n}, A_{n}) - % Let $m = \max{\indexset[U]}$. - % For all $n \in (\indexset[U] \setminus \{m\})$ we have there exist $C \subseteq X$ - % such that $\closure{\index[U](n)}{X} \subseteq \interior{C}{X} \subseteq \closure{C}{X} \subseteq \interior{\index[U](n+1)}{X}$. + % Let $m = \max{\indexxset[U]}$. + % For all $n \in (\indexxset[U] \setminus \{m\})$ we have there exist $C \subseteq X$ + % such that $\closure{\indexx[U](n)}{X} \subseteq \interior{C}{X} \subseteq \closure{C}{X} \subseteq \interior{\indexx[U](n+1)}{X}$. %\begin{definition}\label{refinmant} @@ -408,7 +409,7 @@ The first tept will be a formalisation of chain constructions. Suppose $U$ is a urysohnchain in $X$ and $U$ has cardinality $k$. Suppose $k \neq \zero$. Then there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ - and for all $n \in \indexset[U]$ we have for all $x \in \index[U](n)$ + and for all $n \in \indexxset[U]$ we have for all $x \in \indexx[U](n)$ we have $f(x) = \rfrac{n}{k}$. \end{proposition} \begin{proof} @@ -445,11 +446,11 @@ The first tept will be a formalisation of chain constructions. \begin{abbreviation}\label{converge} $s$ converges iff $s$ is a sequence of real numbers - and $\indexset[s]$ is infinite + and $\indexxset[s]$ is infinite and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have - there exist $N \in \indexset[s]$ such that - for all $m \in \indexset[s]$ such that $m > N$ - we have $\abs{\index[s](N) - \index[s](m)} < \epsilon$. + there exist $N \in \indexxset[s]$ such that + for all $m \in \indexxset[s]$ such that $m > N$ + we have $\abs{\indexx[s](N) - \indexx[s](m)} < \epsilon$. \end{abbreviation} @@ -457,9 +458,9 @@ The first tept will be a formalisation of chain constructions. $x$ is the limit of $s$ iff $s$ is a sequence of real numbers and $x \in \reals$ and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ - we have there exist $n \in \indexset[s]$ such that - for all $m \in \indexset[s]$ such that $m > n$ - we have $\abs{x - \index[s](n)} < \epsilon$. + we have there exist $n \in \indexxset[s]$ such that + for all $m \in \indexxset[s]$ such that $m > n$ + we have $\abs{x - \indexx[s](n)} < \epsilon$. \end{definition} \begin{proposition}\label{existence_of_limit} @@ -473,9 +474,9 @@ The first tept will be a formalisation of chain constructions. \begin{definition}\label{limit_sequence} $x$ is the limit sequence of $f$ iff - $x$ is a sequence and $\indexset[x] = \dom{f}$ and - for all $n \in \indexset[x]$ we have - $\index[x](n) = f(n)$. + $x$ is a sequence and $\indexxset[x] = \dom{f}$ and + for all $n \in \indexxset[x]$ we have + $\indexx[x](n) = f(n)$. \end{definition} \begin{definition}\label{realsminus} @@ -596,15 +597,15 @@ The first tept will be a formalisation of chain constructions. \begin{proof} There exist $\eta$ such that $\carrier[\eta] = \{A, (\carrier[X] \setminus B)\}$ - and $\indexset[\eta] = \{\zero, 1\}$ - and $\index[\eta](\zero) = A$ - and $\index[\eta](1) = (\carrier[X] \setminus B)$ by \cref{urysohnchain_induction_begin}. + and $\indexxset[\eta] = \{\zero, 1\}$ + and $\indexx[\eta](\zero) = A$ + and $\indexx[\eta](1) = (\carrier[X] \setminus B)$ by \cref{urysohnchain_induction_begin}. We show that there exist $\zeta$ such that $\zeta$ is a sequence - and $\indexset[\zeta] = \naturals$ - and $\eta \in \carrier[\zeta]$ and $\index[\zeta](\eta) = \zero$ - and for all $n \in \indexset[\zeta]$ we have $n+1 \in \indexset[\zeta]$ - and $\index[\zeta](n+1)$ is a refinmant of $\index[\zeta](n)$. + and $\indexxset[\zeta] = \naturals$ + and $\eta \in \carrier[\zeta]$ and $\indexx[\zeta](\eta) = \zero$ + and for all $n \in \indexxset[\zeta]$ we have $n+1 \in \indexxset[\zeta]$ + and $\indexx[\zeta](n+1)$ is a refinmant of $\indexx[\zeta](n)$. \begin{subproof} %Let $\alpha = \{x \in C \mid \exists y \in \alpha. x \refine y \lor x = \eta\}$. %Let $\beta = \{ (n,x) \mid n \in \naturals \mid \exists m \in \naturals. \exists y \in \alpha. (x \in \alpha) \land ((x \refine y \land m = n+1 )\lor ((n,x) = (\zero,\eta)))\}$. @@ -614,7 +615,7 @@ The first tept will be a formalisation of chain constructions. %$\dom{\beta} = \naturals$. %$\ran{\beta} = \alpha$. %$\beta \in \funs{\naturals}{\alpha}$. - %Take $\zeta$ such that $\carrier[\zeta] = \alpha$ and $\indexset[\zeta] = \naturals$ and $\index[\zeta] = \beta$. + %Take $\zeta$ such that $\carrier[\zeta] = \alpha$ and $\indexxset[\zeta] = \naturals$ and $\indexx[\zeta] = \beta$. Omitted. \end{subproof} @@ -628,14 +629,14 @@ The first tept will be a formalisation of chain constructions. %We show that there exist $k \in \funs{\carrier[X]}{\reals}$ such that %$k(x)$ - %For all $n \in \naturals$ we have $\index[\zeta](n)$ is a urysohnchain in $X$. + %For all $n \in \naturals$ we have $\indexx[\zeta](n)$ is a urysohnchain in $X$. - We show that for all $n \in \indexset[\zeta]$ we have $\index[\zeta](n)$ has cardinality $\pot(n)$. + We show that for all $n \in \indexxset[\zeta]$ we have $\indexx[\zeta](n)$ has cardinality $\pot(n)$. \begin{subproof} Omitted. \end{subproof} - We show that for all $m \in \indexset[\zeta]$ we have $\pot(m) \neq \zero$. + We show that for all $m \in \indexxset[\zeta]$ we have $\pot(m) \neq \zero$. \begin{subproof} Omitted. \end{subproof} @@ -646,212 +647,212 @@ The first tept will be a formalisation of chain constructions. \end{subproof} - We show that for all $m \in \indexset[\zeta]$ we have there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ - and for all $n \in \indexset[\index[\zeta](m)]$ we have for all $x \in \index[\index[\zeta](m)](n)$ such that $x \notin \index[\index[\zeta](m)](n-1)$ - we have $f(x) = \rfrac{n}{\pot(m)}$. - \begin{subproof} - Fix $m \in \indexset[\zeta]$. - %$\index[\zeta](m)$ is a urysohnchain in $X$. - - %Define $o : \alpha \to \intervalclosed{\zero}{1}$ such that $o(x) =$ - %\begin{cases} - % & 0 & \text{if} x \in A - % & 1 & \text{if} x \in B - % & \rfrac{n}{m} & \text{if} \exists n \in \naturals. x \in \index[\index[\zeta](m)](n) \land x \notin \index[\index[\zeta](m)](n-1) - %\end{cases} +% We show that for all $m \in \indexxset[\zeta]$ we have there exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ +% and for all $n \in \indexxset[\indexx[\zeta](m)]$ we have for all $x \in \indexx[\indexx[\zeta](m)](n)$ such that $x \notin \indexx[\indexx[\zeta](m)](n-1)$ +% we have $f(x) = \rfrac{n}{\pot(m)}$. +% \begin{subproof} +% Fix $m \in \indexxset[\zeta]$. +% %$\indexx[\zeta](m)$ is a urysohnchain in $X$. % - Omitted. - \end{subproof} - - - - %The sequenc of the functions - Let $\gamma = \{ - (n,f) \mid - n \in \naturals \mid - - \forall n' \in \indexset[\index[\zeta](n)]. - \forall x \in \carrier[X]. - - - f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}} \land - - - % (n,f) \in \gamma <=> \phi(n,f) - % with \phi (n,f) := - % (x \in (A_k) \ (A_k-1)) ==> f(x) = ( k / 2^n ) - % \/ (x \notin A_k for all k \in {1,..,n} ==> f(x) = 1 - - ( (n' = \zero) - \land (x \in \index[\index[\zeta](n)](n')) - \land (f(x)= \zero) ) - - \lor - - ( (n' > \zero) - \land (x \in \index[\index[\zeta](n)](n')) - \land (x \notin \index[\index[\zeta](n)](n'-1)) - \land (f(x) = \rfrac{n'}{\pot(n)}) ) - - \lor - - ( (x \notin \index[\index[\zeta](n)](n')) - \land (f(x) = 1) ) - - \}$. - - Let $\gamma(n) = \apply{\gamma}{n}$ for $n \in \naturals$. - - We show that for all $n \in \naturals$ we have $\gamma(n)$ - is a function from $\carrier[X]$ to $\reals$. - \begin{subproof} - Omitted. - \end{subproof} - - - We show that for all $n \in \naturals$ we have $\gamma(n)$ - is a function from $\carrier[X]$ to $\intervalclosed{\zero}{1}$. - \begin{subproof} - Omitted. - \end{subproof} - - We show that $\gamma$ is a function from $\naturals$ to $\reals$. - \begin{subproof} - Omitted. - \end{subproof} - - We show that for all $x,k,n$ such that $n\in \naturals$ and $k \in \naturals$ and $x \in \index[\index[\zeta](n)](k)$ we have $\apply{\gamma(n)}{x} = \rfrac{k}{\pot(n)}$. - \begin{subproof} - Omitted. - \end{subproof} - - We show that for all $n \in \naturals$ for all $x \in \carrier[X]$ we have $\apply{\gamma(n)}{x} \in \intervalclosed{\zero}{1}$. - \begin{subproof} - Fix $n \in \naturals$. - Fix $x \in \carrier[X]$. - Omitted. - \end{subproof} - - - - We show that - if $h$ is a function from $\naturals$ to $\reals$ and for all $n \in \naturals$ we have $h(n) \leq h(n+1)$ and there exist $B \in \reals$ such that $h(n) < B$ - then there exist $k \in \reals$ such that for all $m \in \naturals$ we have $h(m) \leq k$ and for all $k' \in \reals$ such that $k' < k$ we have there exist $M \in \naturals$ such that $k' < h(M)$. - \begin{subproof} - Omitted. - \end{subproof} - - - - We show that there exist $g$ such that for all $x \in \carrier[X]$ we have there exist $k \in \reals$ such that $g(x)= k$ and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. - \begin{subproof} - We show that for all $x \in \carrier[X]$ we have there exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. - \begin{subproof} - Fix $x \in \carrier[X]$. - - Omitted. - - % Contradiction by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. - %Follows by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. - \end{subproof} - Omitted. - \end{subproof} - - - Let $G(x) = g(x)$ for $x \in \carrier[X]$. - We have $\dom{G} = \carrier[X]$. - - We show that for all $x \in \dom{G}$ we have $G(x) \in \reals$. - \begin{subproof} - %Fix $x \in \dom{G}$. - %It suffices to show that $g(x) \in \reals$. +% %Define $o : \alpha \to \intervalclosed{\zero}{1}$ such that $o(x) =$ +% %\begin{cases} +% % & 0 & \text{if} x \in A +% % & 1 & \text{if} x \in B +% % & \rfrac{n}{m} & \text{if} \exists n \in \naturals. x \in \indexx[\indexx[\zeta](m)](n) \land x \notin \indexx[\indexx[\zeta](m)](n-1) +% %\end{cases} +%% +% Omitted. +% \end{subproof} % - %There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. +% % - %We show that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - g(x)} < \epsilon$ and $g(x)= k$. - %\begin{subproof} - % Fix $\epsilon \in \reals$. - % +% %The sequenc of the functions +% Let $\gamma = \{ +% (n,f) \mid +% n \in \naturals \mid +% +% \forall n' \in \indexxset[\indexx[\zeta](n)]. +% \forall x \in \carrier[X]. +% % - %\end{subproof} - %Follows by \cref{apply,plus_one_order,ordinal_iff_suc_ordinal,natural_number_is_ordinal,subseteq,naturals_subseteq_reals,naturals_is_equal_to_two_times_naturals,reals_one_bigger_zero,one_in_reals,ordinal_prec_trichotomy,omega_is_an_ordinal,suc_intro_self,two_in_naturals,in_asymmetric,suc}. - Omitted. - \end{subproof} - - We show that for all $x \in \dom{G}$ we have $\zero \leq G(x) \leq 1$. - \begin{subproof} - %Fix $x \in \dom{G}$. - %Then $x \in \carrier[X]$. - %\begin{byCase} - % \caseOf{$x \in A$.} - % For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = \zero$. -% -% - % \caseOf{$x \notin A$.} - % \begin{byCase} - % \caseOf{$x \in B$.} - % For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 1$. -% - % \caseOf{$x \notin B$.} - % Omitted. - % \end{byCase} - %\end{byCase} - Omitted. - \end{subproof} - - - We show that $G \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$. - \begin{subproof} - %It suffices to show that $\ran{G} \subseteq \intervalclosed{\zero}{1}$ by \cref{fun_ran_iff,funs_intro,funs_weaken_codom}. - %It suffices to show that for all $x \in \dom{G}$ we have $G(x) \in \intervalclosed{\zero}{1}$. - %Fix $x \in \dom{G}$. - %Then $x \in \carrier[X]$. - %$g(x) = G(x)$. - %We have $G(x) \in \reals$. - %$\zero \leq G(x) \leq 1$. - %We have $G(x) \in \intervalclosed{\zero}{1}$ . - Omitted. - \end{subproof} - - We show that $G(A) = \zero$. - \begin{subproof} - Omitted. - \end{subproof} - - We show that $G(B) = 1$. - \begin{subproof} - Omitted. - \end{subproof} - - We show that $G$ is continuous. - \begin{subproof} - Omitted. - \end{subproof} - - %Suppose $\eta$ is a urysohnchain in $X$. - %Suppose $\carrier[\eta] =\{A, (X \setminus B)\}$ - %and $\indexset[\eta] = \{\zero, 1\}$ - %and $\index[\eta](\zero) = A$ - %and $\index[\eta](1) = (X \setminus B)$. - - - %Then $\eta$ is a urysohnchain in $X$. - - % Take $P$ such that $P$ is a infinite sequence and $\indexset[P] = \naturals$ and for all $i \in \indexset[P]$ we have $\index[P](i) = \pow{X}$. - % - % We show that there exist $\zeta$ such that $\zeta$ is a infinite sequence - % and for all $i \in \indexset[\zeta]$ we have - % $\index[\zeta](i)$ is a urysohnchain in $X$ of cardinality $i$ - % and $A \subseteq \index[\zeta](i)$ - % and $\index[\zeta](i) \subseteq (X \setminus B)$ - % and for all $j \in \indexset[\zeta]$ such that - % $j < i$ we have for all $x \in \index[\zeta](j)$ we have $x \in \index[\zeta](i)$. - % \begin{subproof} - % Omitted. - % \end{subproof} - % - % - % +% f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}} \land +% +% +% % (n,f) \in \gamma <=> \phi(n,f) +% % with \phi (n,f) := +% % (x \in (A_k) \ (A_k-1)) ==> f(x) = ( k / 2^n ) +% % \/ (x \notin A_k for all k \in {1,..,n} ==> f(x) = 1 +% +% ( (n' = \zero) +% \land (x \in \indexx[\indexx[\zeta](n)](n')) +% \land (f(x)= \zero) ) +% +% \lor +% +% ( (n' > \zero) +% \land (x \in \indexx[\indexx[\zeta](n)](n')) +% \land (x \notin \indexx[\indexx[\zeta](n)](n'-1)) +% \land (f(x) = \rfrac{n'}{\pot(n)}) ) +% +% \lor +% +% ( (x \notin \indexx[\indexx[\zeta](n)](n')) +% \land (f(x) = 1) ) +% +% \}$. +% +% Let $\gamma(n) = \apply{\gamma}{n}$ for $n \in \naturals$. +% +% We show that for all $n \in \naturals$ we have $\gamma(n)$ +% is a function from $\carrier[X]$ to $\reals$. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% +% We show that for all $n \in \naturals$ we have $\gamma(n)$ +% is a function from $\carrier[X]$ to $\intervalclosed{\zero}{1}$. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% We show that $\gamma$ is a function from $\naturals$ to $\reals$. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% We show that for all $x,k,n$ such that $n\in \naturals$ and $k \in \naturals$ and $x \in \indexx[\indexx[\zeta](n)](k)$ we have $\apply{\gamma(n)}{x} = \rfrac{k}{\pot(n)}$. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% We show that for all $n \in \naturals$ for all $x \in \carrier[X]$ we have $\apply{\gamma(n)}{x} \in \intervalclosed{\zero}{1}$. +% \begin{subproof} +% Fix $n \in \naturals$. +% Fix $x \in \carrier[X]$. +% Omitted. +% \end{subproof} +% +% +% +% We show that +% if $h$ is a function from $\naturals$ to $\reals$ and for all $n \in \naturals$ we have $h(n) \leq h(n+1)$ and there exist $B \in \reals$ such that $h(n) < B$ +% then there exist $k \in \reals$ such that for all $m \in \naturals$ we have $h(m) \leq k$ and for all $k' \in \reals$ such that $k' < k$ we have there exist $M \in \naturals$ such that $k' < h(M)$. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% +% +% We show that there exist $g$ such that for all $x \in \carrier[X]$ we have there exist $k \in \reals$ such that $g(x)= k$ and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. +% \begin{subproof} +% We show that for all $x \in \carrier[X]$ we have there exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. +% \begin{subproof} +% Fix $x \in \carrier[X]$. +% +% Omitted. +% +% % Contradiction by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. +% %Follows by \cref{two_in_naturals,function_apply_default,reals_axiom_zero_in_reals,dom_emptyset,notin_emptyset,funs_type_apply,neg,minus,abs_behavior1}. +% \end{subproof} +% Omitted. +% \end{subproof} +% +% +% Let $G(x) = g(x)$ for $x \in \carrier[X]$. +% We have $\dom{G} = \carrier[X]$. +% +% We show that for all $x \in \dom{G}$ we have $G(x) \in \reals$. +% \begin{subproof} +% %Fix $x \in \dom{G}$. +% %It suffices to show that $g(x) \in \reals$. +%% +% %There exist $k \in \reals$ such that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - k} < \epsilon$. +%% +% %We show that for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\abs{\apply{\gamma(N')}{x} - g(x)} < \epsilon$ and $g(x)= k$. +% %\begin{subproof} +% % Fix $\epsilon \in \reals$. +% % +%% +% %\end{subproof} +% %Follows by \cref{apply,plus_one_order,ordinal_iff_suc_ordinal,natural_number_is_ordinal,subseteq,naturals_subseteq_reals,naturals_is_equal_to_two_times_naturals,reals_one_bigger_zero,one_in_reals,ordinal_prec_trichotomy,omega_is_an_ordinal,suc_intro_self,two_in_naturals,in_asymmetric,suc}. +% Omitted. +% \end{subproof} +% +% We show that for all $x \in \dom{G}$ we have $\zero \leq G(x) \leq 1$. +% \begin{subproof} +% %Fix $x \in \dom{G}$. +% %Then $x \in \carrier[X]$. +% %\begin{byCase} +% % \caseOf{$x \in A$.} +% % For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = \zero$. +%% +%% +% % \caseOf{$x \notin A$.} +% % \begin{byCase} +% % \caseOf{$x \in B$.} +% % For all $n \in \naturals$ we have $\apply{\gamma(n)}{x} = 1$. +%% +% % \caseOf{$x \notin B$.} +% % Omitted. +% % \end{byCase} +% %\end{byCase} +% Omitted. +% \end{subproof} +% +% +% We show that $G \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$. +% \begin{subproof} +% %It suffices to show that $\ran{G} \subseteq \intervalclosed{\zero}{1}$ by \cref{fun_ran_iff,funs_intro,funs_weaken_codom}. +% %It suffices to show that for all $x \in \dom{G}$ we have $G(x) \in \intervalclosed{\zero}{1}$. +% %Fix $x \in \dom{G}$. +% %Then $x \in \carrier[X]$. +% %$g(x) = G(x)$. +% %We have $G(x) \in \reals$. +% %$\zero \leq G(x) \leq 1$. +% %We have $G(x) \in \intervalclosed{\zero}{1}$ . +% Omitted. +% \end{subproof} +% +% We show that $G(A) = \zero$. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% We show that $G(B) = 1$. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% We show that $G$ is continuous. +% \begin{subproof} +% Omitted. +% \end{subproof} +% +% %Suppose $\eta$ is a urysohnchain in $X$. +% %Suppose $\carrier[\eta] =\{A, (X \setminus B)\}$ +% %and $\indexxset[\eta] = \{\zero, 1\}$ +% %and $\indexx[\eta](\zero) = A$ +% %and $\indexx[\eta](1) = (X \setminus B)$. +% +% +% %Then $\eta$ is a urysohnchain in $X$. +% +% % Take $P$ such that $P$ is a infinite sequence and $\indexxset[P] = \naturals$ and for all $i \in \indexxset[P]$ we have $\indexx[P](i) = \pow{X}$. +% % +% % We show that there exist $\zeta$ such that $\zeta$ is a infinite sequence +% % and for all $i \in \indexxset[\zeta]$ we have +% % $\indexx[\zeta](i)$ is a urysohnchain in $X$ of cardinality $i$ +% % and $A \subseteq \indexx[\zeta](i)$ +% % and $\indexx[\zeta](i) \subseteq (X \setminus B)$ +% % and for all $j \in \indexxset[\zeta]$ such that +% % $j < i$ we have for all $x \in \indexx[\zeta](j)$ we have $x \in \indexx[\zeta](i)$. +% % \begin{subproof} +% % Omitted. +% % \end{subproof} +% % +% % +% % % % % diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 838b121..83e3aa4 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -367,10 +367,10 @@ \end{subproof} Let $N = \seq{\zero}{k}$. Let $M = \pow{X}$. - Define $V : N \to M$ such that $V(n)=$ + Define $V : N \to M$ such that $V(n)= \begin{cases} - &\at{U}{F(n)} & \text{if} n \in N - \end{cases} + \at{U}{F(n)} & \text{if} n \in N + \end{cases}$ $\dom{V} = \seq{\zero}{k}$. We show that $V$ is a urysohnchain of $X$. \begin{subproof} @@ -445,11 +445,11 @@ $B \subseteq X'$ by \cref{powerset_elim,closeds}. $A \subseteq X'$. Therefore $A \subseteq A'$. - Define $U_0: N \to \{A, A'\}$ such that $U_0(n) =$ + Define $U_0: N \to \{A, A'\}$ such that $U_0(n) = \begin{cases} - &A &\text{if} n = \zero \\ - &A' &\text{if} n = 1 - \end{cases} + A &\text{if} n = \zero \\ + A' &\text{if} n = 1 + \end{cases}$ $U_0$ is a function. $\dom{U_0} = N$. $\dom{U_0} \subseteq \naturals$ by \cref{ran_converse}. -- cgit v1.2.3 From 13d7b11c23f8862c9f214c46ee05fad314e9e698 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 16 Sep 2024 16:19:36 +0200 Subject: Finished proof of topological basis --- library/numbers.tex | 4 +- library/topology/real-topological-space.tex | 164 ++++++++++++++++++++++++++-- 2 files changed, 157 insertions(+), 11 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 73eefc8..98339ad 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -386,7 +386,9 @@ Commutivatiy of the standart operations For all $x,y \in \reals$ $x + y = y + x$ and $x \rmul y = y \rmul x$. \end{axiom} - +\begin{axiom}\label{reals_axiom_assoc} + For all $x,y,z \in \reals$ we have $(x + y) + z = x + (y + z)$. +\end{axiom} Existence of one and Zero diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index ffdf46e..428ee24 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -193,7 +193,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have $\epsBall{x}{\delta} \in \topoBasisReals$. \end{lemma} -\begin{lemma}\label{realspuls_in_reals_plus} +\begin{lemma}\label{realsplus_in_reals_plus} For all $x,y$ such that $x \in \reals$ and $y \in \realsplus$ we have $x + y \in \reals$. \end{lemma} @@ -207,6 +207,36 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Then there exists $a,b \in \reals$ such that $a < b$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$. \end{lemma} +\begin{lemma}\label{reals_existence_addition_reverse} + For all $\delta \in \reals$ there exists $\epsilon \in \reals$ such that $\epsilon + \epsilon = \delta$. +\end{lemma} +\begin{proof} + Fix $\delta \in \reals$. + Follows by \cref{reals_disstro2,reals_axiom_disstro1,reals_rmul,suc_eq_plus_one,reals_axiom_mul_invers,suc,suc_neq_emptyset,realsplus_in_reals_plus,naturals_addition_axiom_2,naturals_1_kommu,reals_axiom_zero,naturals_inductive_set,one_is_suc_zero,realsplus,reals_one_bigger_zero,one_in_reals,reals_axiom_one,minus_in_reals}. +\end{proof} + +\begin{lemma}\label{reals_addition_minus_behavior1} + For all $a,b,c \in \reals$ such that $a = b + c$ we have $b = a - c$. +\end{lemma} +\begin{proof} + It suffices to show that for all $a \in \reals$ for all $b \in \reals$ for all $c \in \reals$ if $a = b + c$ then $b = a - c$. + Fix $a \in \reals$. + Fix $b \in \reals$. + Fix $c \in \reals$. + Suppose $a = b + c$. + Then $a + \neg{c} = b + c + \neg{c}$. + Therefore $a - c = b + c + \neg{c}$. + $a - c = (b + c) - c$. + $(b + c) - c = (b + c) + \neg{c}$. + $(b + c) + \neg{c} = b + (c + \neg{c})$. + $b + (c + \neg{c}) = b + (\zero)$. + $a - c = b$. +\end{proof} + +\begin{lemma}\label{reals_addition_minus_behavior2} + For all $a,b,c \in \reals$ such that $a = b - c$ we have $b = c + a$. +\end{lemma} + \begin{lemma}\label{open_interval_eq_eps_ball} Suppose $a,b \in \reals$. Suppose $a < b$. @@ -214,11 +244,71 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have \end{lemma} \begin{proof} Let $\delta = (b-a)$. - $\delta$ is positiv by \cref{minus_}. + $\delta$ is positiv by \cref{minus_in_reals,minus_behavior3,reals_axiom_zero_in_reals,reals_order_behavior_with_addition,minus_behavior1,minus}. There exists $\epsilon \in \reals$ such that $\epsilon + \epsilon = \delta$. + Let $x = a + \epsilon$. + $a + \delta = b$. + $a + \epsilon + \epsilon = b$. + $x + \epsilon = b$. + $\epsilon \in \realsplus$ by \cref{reals_order_behavior_with_addition,reals_axiom_kommu,reals_axiom_zero,reals_order_is_transitive,reals_add,minus_behavior1,minus_behavior3,minus,reals_order_total,reals_axiom_zero_in_reals,realsplus}. + $a = x - \epsilon$. + $b = x + \epsilon$. + We show that $\intervalopen{a}{b} \subseteq \epsBall{x}{\epsilon}$. + \begin{subproof} + It suffices to show that for all $y \in \intervalopen{a}{b}$ we have $y \in \epsBall{x}{\epsilon}$. + Fix $y \in \intervalopen{a}{b}$. + \end{subproof} + We show that $\epsBall{x}{\epsilon} \subseteq \intervalopen{a}{b}$. + \begin{subproof} + It suffices to show that for all $y \in \epsBall{x}{\epsilon}$ we have $y \in \intervalopen{a}{b}$. + Fix $y \in \epsBall{x}{\epsilon}$. + \end{subproof} \end{proof} +\begin{lemma}\label{intersection_openinterval_inclusion_of_border} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + If $a \leq x < y \leq b$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{intersection_openinterval_lower_border_eq} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + If $a = x$ and $b \leq y$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{a}{b}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{intersection_openinterval_upper_border_eq} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + If $a \leq x$ and $b = y$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{intersection_openinterval_none_border_eq} + Suppose $a,b,x,y \in \reals$. + Suppose $a < b$. + Suppose $x < y$. + If $a \leq x < b \leq y$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{b}$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{reals_order_total2} + For all $a,b \in \reals$ we have $a < b \lor a > b \lor a = b$. +\end{lemma} \begin{theorem}\label{topological_basis_reals_is_basis} @@ -242,15 +332,69 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have There exists $x_2, \beta$ such that $x_2 \in \reals$ and $\beta \in \realsplus$ and $\epsBall{x_2}{\beta} = V$. Then $ (x_1 - \alpha) < x < (x_1 + \alpha)$. Then $ (x_2 - \beta) < x < (x_2 + \beta)$. - We show that if there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$ then there exists $W\in B$ such that $x\in W\subseteq U, V$. - \begin{subproof} - Suppose there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$. - $x \in \epsBall{x}{\delta}$. - $\epsBall{x}{\delta} \subseteq U$. - $\epsBall{x}{\delta} \subseteq V$. - $\epsBall{x}{\delta} \in B$. + Let $a = (x_1 - \alpha)$. + Let $b = (x_1 + \alpha)$. + Let $c = (x_2 - \beta)$. + Let $d = (x_2 + \beta)$. + We have $a < b$ and $a < x$ and $x < b$. + We have $c < d$ and $c < x$ and $x < d$. + We have $a \in \reals$. + We have $b \in \reals$. + We have $c \in \reals$. + We have $d \in \reals$. + We show that there exist $a',b'\in \reals$ such that $\intervalopen{a}{b} \inter \intervalopen{c}{d} = \intervalopen{a'}{b'}$. + \begin{subproof} + \begin{byCase} + \caseOf{$a < c$.} + \begin{byCase} + \caseOf{$b < d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b = d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b > d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \end{byCase} + \caseOf{$a = c$.} + \begin{byCase} + \caseOf{$b < d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b = d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \caseOf{$b > d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus}. + \end{byCase} + \caseOf{$a > c$.} + \begin{byCase} + \caseOf{$b < d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus,reals_add,minus_in_reals,realsplus,inter_comm,epsilon_ball}. + \caseOf{$b = d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus,reals_add,minus_in_reals,realsplus,inter_comm,epsilon_ball}. + \caseOf{$b > d$.} + Follows by \cref{intersection_openinterval_inclusion_of_border,intersection_openinterval_lower_border_eq,intersection_openinterval_none_border_eq,intersection_openinterval_upper_border_eq,reals_order_is_transitive,realsplus_in_reals_plus,realspuls_in_reals_minus,reals_add,minus_in_reals,realsplus,inter_comm,epsilon_ball}. + \end{byCase} + \end{byCase} \end{subproof} - It suffices to show that there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$. + + Take $a',b'\in \reals$ such that $\intervalopen{a}{b} \inter \intervalopen{c}{d} = \intervalopen{a'}{b'}$. + We have $a',b' \in \reals$ by assumption. + We have $a' < b'$ by \cref{id_img,epsilon_ball,minus,intervalopen,reals_order_is_transitive}. + Then there exist $x',\epsilon'$ such that $x' \in \reals$ and $\epsilon' \in \realsplus$ and $\intervalopen{a'}{b'} = \epsBall{x'}{\epsilon'}$. + Then $x \in \epsBall{x'}{\epsilon'}$. + + Follows by \cref{inter_lower_left,inter_lower_right,epsilon_ball,topological_basis_reals_eps_ball}. + %Then $(x_1 - \alpha) < (x_2 + \beta)$. + + %Therefore $U \inter V = \intervalopen{}{(x_2 + \beta)}$. + + %We show that if there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$ then there exists $W\in B$ such that $x\in W\subseteq U, V$. + %\begin{subproof} + % Suppose there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$. + % $x \in \epsBall{x}{\delta}$. + % $\epsBall{x}{\delta} \subseteq U$. + % $\epsBall{x}{\delta} \subseteq V$. + % $\epsBall{x}{\delta} \in B$. + %\end{subproof} + %It suffices to show that there exists $\delta \in \realsplus$ such that $\epsBall{x}{\delta} \subseteq U \inter V$. %It suffices to show that there exists $x_0, \delta$ such that $x_0 \in \reals$ and $\delta \in \realsplus$ and $\epsBall{x_0}{\delta} \subseteq u \inter V$. -- cgit v1.2.3 From c021e79033abbb3fd4458304e701b3c54a284902 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 16 Sep 2024 19:37:27 +0200 Subject: working commit --- latex/naproche.sty | 2 + library/numbers.tex | 8 ++ library/topology/real-topological-space.tex | 144 ++++++++++++++++++++++++++++ 3 files changed, 154 insertions(+) (limited to 'library') diff --git a/latex/naproche.sty b/latex/naproche.sty index 1a8afb6..c0fd318 100644 --- a/latex/naproche.sty +++ b/latex/naproche.sty @@ -156,6 +156,8 @@ \newcommand{\abs}[1]{\left\lvert#1\right\rvert} \newcommand{\realsminus}{\reals_{-}} \newcommand{\at}[2]{#1(#2)} +\newcommand{\intervalopenInfiniteLeft}[1]{(-\infty, #1)} +\newcommand{\intervalopenInfiniteRight}[1]{(#1, \infty)} \newcommand\restrl[2]{{% we make the whole thing an ordinary symbol diff --git a/library/numbers.tex b/library/numbers.tex index 98339ad..8624260 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -787,6 +787,14 @@ Laws of the order on the reals $\intervalopen{a}{b} = \{ x \in \reals \mid a < x < b\}$. \end{definition} +\begin{definition}\label{intervalopen_infinite_left} + $\intervalopenInfiniteLeft{b} = \{ x \in \reals \mid x < b\}$. +\end{definition} + +\begin{definition}\label{intervalopen_infinite_right} + $\intervalopenInfiniteRight{a} = \{ x \in \reals \mid a < x\}$. +\end{definition} + \begin{definition}\label{m_to_n_set} $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index 428ee24..b3efa20 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -461,3 +461,147 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Suppose $a,b \in \reals$. Then $\intervalopen{a}{b} \in \opens[\reals]$. \end{proposition} +\begin{proof} + If $a > b$ then $\intervalopen{a}{b} = \emptyset$. + If $a = b$ then $\intervalopen{a}{b} = \emptyset$. + It suffices to show that if $a < b$ then $\intervalopen{a}{b} \in \opens[\reals]$. + Suppose $a \rless b$. + Take $x, \epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$. + It suffices to show that $\epsBall{x}{\epsilon} \in \opens[\reals]$. + $\topoBasisReals$ is a topological basis for $\reals$. + $\epsBall{x}{\epsilon} \in \topoBasisReals$. + $\topoBasisReals \subseteq \opens[\reals]$ by \cref{basis_is_in_genopens,topological_space_reals,topological_basis_reals_is_basis}. +\end{proof} + +\begin{lemma}\label{reals_minus_to_realsplus} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then $(b - a) \in \realsplus$. +\end{lemma} + +\begin{lemma}\label{existence_of_epsilon_upper_border} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then there exists $\epsilon \in \realsplus$ such that $b \notin \epsBall{a}{\epsilon}$. +\end{lemma} +\begin{proof} + Let $\epsilon = b - a$. + Then $\epsilon \in \realsplus$. + It suffices to show that $b \notin \epsBall{a}{\epsilon}$ by \cref{epsilon_ball,reals_addition_minus_behavior2,realsplus,minus,intervalopen,order_reals_lemma0}. + Suppose not. + Then $ b \in \epsBall{a}{\epsilon}$. + Therefore $ (a - \epsilon) < b < (a + \epsilon)$. + $b = (a + \epsilon)$. + Contradiction. +\end{proof} + +\begin{lemma}\label{existence_of_epsilon_lower_border} + Suppose $a,b \in \reals$. + Suppose $a > b$. + Then there exists $\epsilon \in \realsplus$ such that $b \notin \epsBall{a}{\epsilon}$. +\end{lemma} +\begin{proof} + Let $\epsilon = a - b$. + Then $\epsilon \in \realsplus$. + It suffices to show that $b \notin \epsBall{a}{\epsilon}$ by \cref{epsilon_ball,reals_addition_minus_behavior2,realsplus,minus,intervalopen,order_reals_lemma0}. + Suppose not. + Then $ b \in \epsBall{a}{\epsilon}$. + Therefore $ (a - \epsilon) < b < (a + \epsilon)$. + $b = (a - \epsilon)$. + Contradiction. +\end{proof} + +\begin{proposition}\label{openinterval_infinite_left_in_opens} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteLeft{a} \in \opens[\reals]$. +\end{proposition} +\begin{proof} + Let $E = \{ B \in \pow{\reals} \mid \exists x \in \intervalopenInfiniteLeft{a} . \exists \delta \in \realsplus . B = \epsBall{x}{\delta} \land a \notin \epsBall{x}{\delta} \}$. + We show that for all $x \in \intervalopenInfiniteLeft{a}$ we have there exists $e \in E$ such that $x \in e$. + \begin{subproof} + Fix $x \in \intervalopenInfiniteLeft{a}$. + Then $x < a$. + Take $\delta' \in \realsplus$ such that $a \notin \epsBall{x}{\delta'}$. + $x \in \epsBall{x}{\delta'}$ by \cref{intervalopen,epsilon_ball,reals_addition_minus_behavior1,reals_order_minus_positiv,minus,reals_add,realsplus,intervalopen_infinite_left}. + $a \notin \epsBall{x}{\delta'}$. + $\epsBall{x}{\delta'} \in E$. + \end{subproof} + $E \subseteq \topoBasisReals$. + We show that $\unions{E} = \intervalopenInfiniteLeft{a}$. + \begin{subproof} + We show that $\unions{E} \subseteq \intervalopenInfiniteLeft{a}$. + \begin{subproof} + It suffices to show that for all $x \in \unions{E}$ we have $x \in \intervalopenInfiniteLeft{a}$. + Fix $x \in \unions{E}$. + Take $e \in E$ such that $x \in e$. + $x \in \reals$. + Take $x',\delta'$ such that $x' \in \reals$ and $\delta' \in \realsplus$ and $e = \epsBall{x'}{\delta'}$ by \cref{epsilon_ball,minus,topo_basis_reals_eps_iff,setminus,setminus_emptyset,elem_subseteq}. + $\epsBall{x'}{\delta'} \in E$. + We show that for all $y \in e$ we have $y < a$. + \begin{subproof} + Fix $y \in e$. + Then $y \in \epsBall{x'}{\delta'}$. + $e = \epsBall{x'}{\delta'}$. + There exists $x'' \in \intervalopenInfiniteLeft{a}$ such that there exists $\delta'' \in \realsplus$ such that $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Take $x'',\delta''$ such that $x'' \in \intervalopenInfiniteLeft{a}$ and $\delta'' \in \realsplus$ and $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Suppose not. + Take $y' \in e$ such that $y' > a$. + $x'' < a$. + $(x'' - \delta'') < y' < (x'' + \delta'')$. + $(x'' - \delta'') < x'' < (x'' + \delta'')$. + Then $x'' < a < y'$. + Therefore $(x'' - \delta'') < a < (x'' + \delta'')$ by \cref{realspuls_in_reals_minus,intervalopen_infinite_left,reals_order_is_transitive,reals_add,realsplus_in_reals,powerset_elim,subseteq}. + Then $a \in e$ by \cref{epsball_are_connected_in_reals,intervalopen_infinite_left,neq_witness}. + Contradiction. + \end{subproof} + $x < a$. + Then $x \in \intervalopenInfiniteLeft{a}$. + \end{subproof} + We show that $\intervalopenInfiniteLeft{a} \subseteq \unions{E}$. + \begin{subproof} + Trivial. + \end{subproof} + \end{subproof} + $\unions{E} \in \opens[\reals]$. +\end{proof} + +\begin{lemma}\label{continuous_on_basis_implies_continuous_endo} + Suppose $X$ is a topological space. + Suppose $B$ is a topological basis for $X$. + Suppose $f$ is a function from $X$ to $X$. + $f$ is continuous iff for all $b \in B$ we have $\preimg{f}{b} \in \opens[X]$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{proposition}\label{openinterval_infinite_right_in_opens} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteRight{a} \in \opens[\reals]$. +\end{proposition} +\begin{proof} + Let $I = \{\neg{b} \mid b \in \intervalopenInfiniteRight{a} \}$. + Let $f(x) = \neg{x}$ for $x \in \reals$. + $f$ is a function from $\reals$ to $\reals$. + We show that $f$ is continuous. + \begin{subproof} + It suffices to show that for all $b \in \topoBasisReals$ we have $\preimg{f}{b} \in \opens[\reals]$. + Fix $b \in \topoBasisReals$. + Take $x, \epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $b = \epsBall{x}{\epsilon}$. + Let $y = \neg{x}$. + It suffices to show that $\preimg{f}{b} \in \topoBasisReals$ by \cref{topological_space_reals,topological_basis_reals_is_basis,basis_is_in_genopens,cons_remove,cons_subseteq_iff}. + It suffices to show that $\epsBall{y}{\epsilon} = \preimg{f}{\epsBall{x}{\epsilon}}$. + Follows by set extensionality. + \end{subproof} + $\intervalopenInfiniteLeft{a} \in \opens[\reals]$. + We show that $\preimg{f}{\intervalopenInfiniteLeft{a}} = \intervalopenInfiniteRight{a}$. + \begin{subproof} + Omitted. + \end{subproof} + Then $\intervalopenInfiniteRight{a} \in \opens[\reals]$ by \cref{continuous,preim_eq_img_of_converse,openinterval_infinite_left_in_opens}. +\end{proof} + +\begin{proposition}\label{closedinterval_is_closed} + Suppose $a,b \in \reals$. + Then $\intervalclosed{a}{b} \in \closeds{\reals}$. +\end{proposition} -- cgit v1.2.3 From 3dca719ba8f9a59471f2c761cf8846cf597eae97 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 16 Sep 2024 23:24:08 +0200 Subject: Topo Space Real Verfication --- library/numbers.tex | 10 +- library/topology/real-topological-space.tex | 200 +++++++++++++++++++++++++--- 2 files changed, 194 insertions(+), 16 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 8624260..406553e 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -792,7 +792,15 @@ Laws of the order on the reals \end{definition} \begin{definition}\label{intervalopen_infinite_right} - $\intervalopenInfiniteRight{a} = \{ x \in \reals \mid a < x\}$. + $\intervalopenInfiniteRight{a} = \{ x \in \reals \mid x > a\}$. +\end{definition} + +\begin{definition}\label{intervalclosed_infinite_left} + $\intervalclosedInfiniteLeft{b} = \{ x \in \reals \mid x \leq b\}$. +\end{definition} + +\begin{definition}\label{intervalclosed_infinite_right} + $\intervalclosedInfiniteRight{a} = \{ x \in \reals \mid x \geq a\}$. \end{definition} diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index b3efa20..e5e17ef 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -580,28 +580,198 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Then $\intervalopenInfiniteRight{a} \in \opens[\reals]$. \end{proposition} \begin{proof} - Let $I = \{\neg{b} \mid b \in \intervalopenInfiniteRight{a} \}$. - Let $f(x) = \neg{x}$ for $x \in \reals$. - $f$ is a function from $\reals$ to $\reals$. - We show that $f$ is continuous. + Let $E = \{ B \in \pow{\reals} \mid \exists x \in \intervalopenInfiniteRight{a} . \exists \delta \in \realsplus . B = \epsBall{x}{\delta} \land a \notin \epsBall{x}{\delta} \}$. + We show that for all $x \in \intervalopenInfiniteRight{a}$ we have there exists $e \in E$ such that $x \in e$. \begin{subproof} - It suffices to show that for all $b \in \topoBasisReals$ we have $\preimg{f}{b} \in \opens[\reals]$. - Fix $b \in \topoBasisReals$. - Take $x, \epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $b = \epsBall{x}{\epsilon}$. - Let $y = \neg{x}$. - It suffices to show that $\preimg{f}{b} \in \topoBasisReals$ by \cref{topological_space_reals,topological_basis_reals_is_basis,basis_is_in_genopens,cons_remove,cons_subseteq_iff}. - It suffices to show that $\epsBall{y}{\epsilon} = \preimg{f}{\epsBall{x}{\epsilon}}$. - Follows by set extensionality. + Fix $x \in \intervalopenInfiniteRight{a}$. + Then $a < x$. + Take $\delta' \in \realsplus$ such that $a \notin \epsBall{x}{\delta'}$. + $x \in \epsBall{x}{\delta'}$ by \cref{intervalopen,epsilon_ball,reals_addition_minus_behavior1,reals_order_minus_positiv,minus,reals_add,realsplus,intervalopen_infinite_right}. + $a \notin \epsBall{x}{\delta'}$. + $\epsBall{x}{\delta'} \in E$. \end{subproof} - $\intervalopenInfiniteLeft{a} \in \opens[\reals]$. - We show that $\preimg{f}{\intervalopenInfiniteLeft{a}} = \intervalopenInfiniteRight{a}$. + $E \subseteq \topoBasisReals$. + We show that $\unions{E} = \intervalopenInfiniteRight{a}$. \begin{subproof} - Omitted. + We show that $\unions{E} \subseteq \intervalopenInfiniteRight{a}$. + \begin{subproof} + It suffices to show that for all $x \in \unions{E}$ we have $x \in \intervalopenInfiniteRight{a}$. + Fix $x \in \unions{E}$. + Take $e \in E$ such that $x \in e$. + $x \in \reals$. + Take $x',\delta'$ such that $x' \in \reals$ and $\delta' \in \realsplus$ and $e = \epsBall{x'}{\delta'}$ by \cref{epsilon_ball,minus,topo_basis_reals_eps_iff,setminus,setminus_emptyset,elem_subseteq}. + $\epsBall{x'}{\delta'} \in E$. + We show that for all $y \in e$ we have $y > a$. + \begin{subproof} + Fix $y \in e$. + Then $y \in \epsBall{x'}{\delta'}$. + $e = \epsBall{x'}{\delta'}$. + There exists $x'' \in \intervalopenInfiniteRight{a}$ such that there exists $\delta'' \in \realsplus$ such that $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Take $x'',\delta''$ such that $x'' \in \intervalopenInfiniteRight{a}$ and $\delta'' \in \realsplus$ and $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. + Suppose not. + Take $y' \in e$ such that $y' < a$. + $x'' > a$. + $(x'' - \delta'') < y' < (x'' + \delta'')$. + $(x'' - \delta'') < x'' < (x'' + \delta'')$. + Then $x'' > a > y'$. + Therefore $(x'' - \delta'') > a > (x'' + \delta'')$ by \cref{realspuls_in_reals_minus,intervalopen_infinite_right,reals_order_is_transitive,reals_add,realsplus_in_reals,powerset_elim,subseteq,epsball_are_connected_in_reals,subseteq}. + Then $a \in e$ by \cref{reals_order_is_transitive,reals_order_total,reals_add,realsplus,epsball_are_connected_in_reals,intervalopen_infinite_right,neq_witness}. + Contradiction. + \end{subproof} + $x > a$. + Then $x \in \intervalopenInfiniteRight{a}$. + \end{subproof} + We show that $\intervalopenInfiniteRight{a} \subseteq \unions{E}$. + \begin{subproof} + Trivial. + \end{subproof} \end{subproof} - Then $\intervalopenInfiniteRight{a} \in \opens[\reals]$ by \cref{continuous,preim_eq_img_of_converse,openinterval_infinite_left_in_opens}. + $\unions{E} \in \opens[\reals]$. + + %Let $I = \{\neg{b} \mid b \in \intervalopenInfiniteRight{a} \}$. + %Let $f(x) = \neg{x}$ for $x \in \reals$. + %$f$ is a function from $\reals$ to $\reals$. + %We show that $f$ is continuous. + %\begin{subproof} + % It suffices to show that for all $b \in \topoBasisReals$ we have $\preimg{f}{b} \in \opens[\reals]$. + % Fix $b \in \topoBasisReals$. + % Take $x, \epsilon$ such that $x \in \reals$ and $\epsilon \in \realsplus$ and $b = \epsBall{x}{\epsilon}$. + % Let $y = \neg{x}$. + % It suffices to show that $\preimg{f}{b} \in \topoBasisReals$ by \cref{topological_space_reals,topological_basis_reals_is_basis,basis_is_in_genopens,cons_remove,cons_subseteq_iff}. + % It suffices to show that $\epsBall{y}{\epsilon} = \preimg{f}{\epsBall{x}{\epsilon}}$. + % $\preimg{f}{\epsBall{x}{\epsilon}} \subseteq \reals$. + % $\epsBall{y}{\epsilon} \subseteq \reals$ by \cref{intervalopen,subseteq,minus,epsilon_ball}. + % %It suffices to show that for all $x \in \reals$ we have $x \in \epsBall{y}{\epsilon}$ iff $x \in \preimg{f}{\epsBall{x}{\epsilon}}$. + % %Fix $x \in \reals$. + % Let $u = (y - \epsilon)$. + % Let $v = (y + \epsilon)$. + % $u = \neg{(x - \epsilon)}$. + % $v = \neg{(x + \epsilon)}$. + % %$v - u = \epsilon + \epsilon$. + % We show that for all $z \in \epsBall{y}{\epsilon}$ we have $z \in \preimg{f}{\epsBall{x}{\epsilon}}$. + % \begin{subproof} + % Fix $z \in \epsBall{y}{\epsilon}$. + % Then $u < z < v$. + % Let $z' = z - u$. + % Then $z = u + z'$. + % Suppose not. + % Let $h = \neg{z}$. + % $\neg{h} = \neg{\neg{z}}$. + % $\neg{h} = z$. + % Then $f(h) = \neg{h}$. + % $f(h) = z$. + % Then $z \in \preimg{f}{\{h\}}$. +% + % \end{subproof} + % We show that for all $z \in \preimg{f}{\epsBall{x}{\epsilon}}$ we have $z \in \epsBall{y}{\epsilon}$. + % \begin{subproof} + % Fix $z \in \preimg{f}{\epsBall{x}{\epsilon}}$. + % Take $h \in \epsBall{x}{\epsilon}$ such that $f(h) = z$. + % \end{subproof} + % Follows by set extensionality. + %\end{subproof} + %$\intervalopenInfiniteLeft{a} \in \opens[\reals]$. + %We show that $\preimg{f}{\intervalopenInfiniteLeft{a}} = \intervalopenInfiniteRight{a}$. + %\begin{subproof} + % Omitted. + %\end{subproof} + %Then $\intervalopenInfiniteRight{a} \in \opens[\reals]$ by \cref{continuous,preim_eq_img_of_converse,openinterval_infinite_left_in_opens}. +\end{proof} + +\begin{lemma}\label{reals_as_union_of_open_closed_intervals1} + Suppose $a,b \in \reals$. + Then $\reals = \intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b} \union \intervalclosed{a}{b}$. +\end{lemma} +\begin{proof} + We show that for all $x \in \reals$ we have $x \in (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b} \union \intervalclosed{a}{b})$. + \begin{subproof} + Fix $x \in \reals$. + \end{subproof} +\end{proof} + +\begin{lemma}\label{reals_as_union_of_open_closed_intervals2} + Suppose $a \in \reals$. + Then $\reals = \intervalopenInfiniteLeft{a} \union \intervalclosedInfiniteRight{a}$. +\end{lemma} +\begin{proof} + It suffices to show that for all $x \in \reals$ we have either $x \in \intervalopenInfiniteLeft{a}$ or $x \in \intervalclosedInfiniteRight{a}$. + Trivial. +\end{proof} + +\begin{lemma}\label{reals_as_union_of_open_closed_intervals3} + Suppose $a \in \reals$. + Then $\reals = \intervalopenInfiniteRight{a} \union \intervalclosedInfiniteLeft{a}$. +\end{lemma} +\begin{proof} + It suffices to show that for all $x \in \reals$ we have either $x \in \intervalclosedInfiniteLeft{a}$ or $x \in \intervalopenInfiniteRight{a}$. + Trivial. +\end{proof} + +\begin{lemma}\label{intersection_of_open_closed__infinite_intervals_open_right} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteRight{a} \inter \intervalclosedInfiniteLeft{a} = \emptyset$. +\end{lemma} + +\begin{lemma}\label{intersection_of_open_closed__infinite_intervals_open_left} + Suppose $a \in \reals$. + Then $\intervalopenInfiniteLeft{a} \inter \intervalclosedInfiniteRight{a} = \emptyset$. +\end{lemma} + +\begin{proposition}\label{closedinterval_infinite_right_in_closeds} + Suppose $a \in \reals$. + Then $\intervalclosedInfiniteRight{a} \in \closeds{\reals}$. +\end{proposition} +\begin{proof} + $\intervalclosedInfiniteRight{a} = \reals \setminus \intervalopenInfiniteLeft{a}$. +\end{proof} + +\begin{proposition}\label{closedinterval_infinite_left_in_closeds} + Suppose $a \in \reals$. + Then $\intervalclosedInfiniteLeft{a} \in \closeds{\reals}$. +\end{proposition} +\begin{proof} + $\intervalclosedInfiniteLeft{a} = \reals \setminus \intervalopenInfiniteRight{a}$. +\end{proof} + +\begin{proposition}\label{closedinterval_eq_openintervals_setminus_reals} + Suppose $a,b \in \reals$. + Then $\reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b}) = \intervalclosed{a}{b}$. +\end{proposition} +\begin{proof} + We have $\intervalclosed{a}{b} \subseteq \reals$. + We show that $\reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b}) = (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. + \begin{subproof} + We show that for all $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$ we have $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. + \begin{subproof} + Fix $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$. + Then $x \in \reals \setminus \intervalopenInfiniteLeft{a}$. + Then $x \in \reals \setminus \intervalopenInfiniteRight{b}$. + \end{subproof} + We show that for all $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$ we have $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$. + \begin{subproof} + Fix $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. + Then $x \in (\reals \setminus \intervalopenInfiniteLeft{a})$. + Then $x \in (\reals \setminus \intervalopenInfiniteRight{b})$. + \end{subproof} + \end{subproof} + We show that $\reals \setminus \intervalopenInfiniteLeft{a} = \intervalclosedInfiniteRight{a}$. + \begin{subproof} + For all $x \in \intervalclosedInfiniteRight{a}$ we have $x \geq a$. + For all $x \in (\reals \setminus \intervalopenInfiniteLeft{a})$ we have $x \geq a$. + Follows by set extensionality. + \end{subproof} + $\reals \setminus \intervalopenInfiniteRight{b} = \intervalclosedInfiniteLeft{b}$. + It suffices to show that $\intervalclosedInfiniteLeft{b} \inter \intervalclosedInfiniteRight{a} = \intervalclosed{a}{b}$. + For all $x \in \intervalclosed{a}{b}$ we have $a \leq x \leq b$. + For all $x \in (\intervalclosedInfiniteLeft{b} \inter \intervalclosedInfiniteRight{a})$ we have $a \leq x \leq b$. + Follows by set extensionality. \end{proof} \begin{proposition}\label{closedinterval_is_closed} Suppose $a,b \in \reals$. Then $\intervalclosed{a}{b} \in \closeds{\reals}$. \end{proposition} +\begin{proof} + We have $\reals = \intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b} \union \intervalclosed{a}{b}$. + It suffices to show that $\reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b}) = \intervalclosed{a}{b}$ by \cref{closeds,setminus_subseteq,powerset_intro,closed_minus_open_is_closed,opens_type,subseteq_refl,union_open,is_closed_in,reals_carrier_reals,setminus_self,emptyset_open,reals_is_topological_space,openinterval_infinite_left_in_opens,openinterval_infinite_right_in_opens}. +\end{proof} -- cgit v1.2.3 From 5362771c14eccd80fd1a3ab6521c3a6ad9bb7838 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 17 Sep 2024 00:36:24 +0200 Subject: Corrected Math Env Parsing Since Latex has a really specify syntax for \begin{cases} ... \end{cases} The math mode in tokenizing had to be setup correctly. --- library/numbers.tex | 3 +++ library/topology/real-topological-space.tex | 27 +++++++++++++++------------ source/Syntax/Concrete.hs | 6 +++--- source/Syntax/Token.hs | 8 ++++---- 4 files changed, 25 insertions(+), 19 deletions(-) (limited to 'library') diff --git a/library/numbers.tex b/library/numbers.tex index 406553e..ac0a683 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -613,6 +613,9 @@ Laws of the order on the reals \subsection{Order on the reals} +\begin{axiom}\label{reals_order_is_transitive} + For all $x,y,z \in \reals$ such that $x < y$ and $y < z$ we have $x < z$. +\end{axiom} \begin{lemma}\label{plus_one_order} diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index e5e17ef..d9790aa 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -70,7 +70,7 @@ Then $\epsBall{x}{\epsilon}$ is inhabited. \end{lemma} \begin{proof} - $x < x + \epsilon$. + $x < x + \epsilon$ by \cref{reals_order_behavior_with_addition,realsplus,reals_axiom_zero_in_reals,reals_axiom_kommu,reals_axiom_zero}. $x - \epsilon < x$. $x \in \epsBall{x}{\epsilon}$. \end{proof} @@ -104,12 +104,8 @@ Omitted. \end{proof} -\begin{lemma}\label{reals_order_is_transitive} - For all $x,y,z \in \reals$ such that $x < y$ and $y < z$ we have $x < z$. -\end{lemma} -\begin{proof} - Omitted. -\end{proof} + + \begin{lemma}\label{reals_order_plus_minus} Suppose $a,b \in \reals$. @@ -207,12 +203,16 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Then there exists $a,b \in \reals$ such that $a < b$ and $\intervalopen{a}{b} = \epsBall{x}{\epsilon}$. \end{lemma} +\begin{lemma}\label{one_in_realsplus} + $1 \in \realsplus$. +\end{lemma} + \begin{lemma}\label{reals_existence_addition_reverse} For all $\delta \in \reals$ there exists $\epsilon \in \reals$ such that $\epsilon + \epsilon = \delta$. \end{lemma} \begin{proof} Fix $\delta \in \reals$. - Follows by \cref{reals_disstro2,reals_axiom_disstro1,reals_rmul,suc_eq_plus_one,reals_axiom_mul_invers,suc,suc_neq_emptyset,realsplus_in_reals_plus,naturals_addition_axiom_2,naturals_1_kommu,reals_axiom_zero,naturals_inductive_set,one_is_suc_zero,realsplus,reals_one_bigger_zero,one_in_reals,reals_axiom_one,minus_in_reals}. + Follows by \cref{one_in_realsplus,reals_disstro2,reals_axiom_disstro1,reals_rmul,suc_eq_plus_one,reals_axiom_mul_invers,suc,suc_neq_emptyset,realsplus_in_reals_plus,naturals_addition_axiom_2,naturals_1_kommu,reals_axiom_zero,naturals_inductive_set,one_is_suc_zero,realsplus,reals_one_bigger_zero,one_in_reals,reals_axiom_one,minus_in_reals}. \end{proof} \begin{lemma}\label{reals_addition_minus_behavior1} @@ -260,7 +260,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have \end{subproof} We show that $\epsBall{x}{\epsilon} \subseteq \intervalopen{a}{b}$. \begin{subproof} - It suffices to show that for all $y \in \epsBall{x}{\epsilon}$ we have $y \in \intervalopen{a}{b}$. + It suffices to show that for all $y \in \epsBall{x}{\epsilon}$ we have $y \in \intervalopen{a}{b}$ by \cref{subseteq}. Fix $y \in \epsBall{x}{\epsilon}$. \end{subproof} @@ -270,7 +270,8 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Suppose $a,b,x,y \in \reals$. Suppose $a < b$. Suppose $x < y$. - If $a \leq x < y \leq b$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. + Suppose $a \leq x < y \leq b$. + Then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. \end{lemma} \begin{proof} Omitted. @@ -280,7 +281,8 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Suppose $a,b,x,y \in \reals$. Suppose $a < b$. Suppose $x < y$. - If $a = x$ and $b \leq y$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{a}{b}$. + Suppose $a = x$ and $b \leq y$. + Then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{a}{b}$. \end{lemma} \begin{proof} Omitted. @@ -290,7 +292,8 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Suppose $a,b,x,y \in \reals$. Suppose $a < b$. Suppose $x < y$. - If $a \leq x$ and $b = y$ then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. + Suppose $a \leq x$ and $b = y$. + Then $\intervalopen{a}{b} \inter \intervalopen{x}{y} = \intervalopen{x}{y}$. \end{lemma} \begin{proof} Omitted. diff --git a/source/Syntax/Concrete.hs b/source/Syntax/Concrete.hs index 9d52995..7a89bea 100644 --- a/source/Syntax/Concrete.hs +++ b/source/Syntax/Concrete.hs @@ -373,15 +373,15 @@ grammar lexicon@Lexicon{..} = mdo -- 3 & \text{else} -- \end{cases} - functionDefineCase <- rule $ (,) <$> (_ampersand *> expr) <*> (_ampersand *> text _if *> formula) + functionDefineCase <- rule $ (,) <$> (optional _ampersand *> expr) <*> (_ampersand *> text _if *> formula) defineFunctionMathy <- rule $ DefineFunctionMathy <$> (_define *> beginMath *> varSymbol) -- Define $ f <*> (_colon *> varSymbol) -- : 'var' \to 'var' <*> (_to *> expr <* endMath <* _suchThat) -- <*> (_suchThat *> align (many1 ((_ampersand *> varSymbol <* _mapsto) <*> exprApp <*> (_ampersand *> formula)))) -- <*> (_suchThat *> align (many1 (varSymbol <* exprApp <* formula))) - <*> (beginMath *> varSymbol) <*> (paren varSymbol <* _eq <* endMath) - <*> cases (many1 functionDefineCase) + <*> (beginMath *> varSymbol) <*> (paren varSymbol <* _eq ) + <*> cases (many1 functionDefineCase) <* endMath <* optional _dot <*> proof diff --git a/source/Syntax/Token.hs b/source/Syntax/Token.hs index 52da86a..53e1e6a 100644 --- a/source/Syntax/Token.hs +++ b/source/Syntax/Token.hs @@ -189,7 +189,7 @@ toks = whitespace *> goNormal id <* eof Nothing -> pure (f []) Just t@Located{unLocated = BeginEnv "math"} -> goMath (f . (t:)) Just t@Located{unLocated = BeginEnv "align*"} -> goMath (f . (t:)) - Just t@Located{unLocated = BeginEnv "cases"} -> goMath (f . (t:)) + --Just t@Located{unLocated = BeginEnv "cases"} -> goMath (f . (t:)) Just t -> goNormal (f . (t:)) goText f = do r <- optional textToken @@ -205,7 +205,7 @@ toks = whitespace *> goNormal id <* eof Nothing -> pure (f []) Just t@Located{unLocated = EndEnv "math"} -> goNormal (f . (t:)) Just t@Located{unLocated = EndEnv "align*"} -> goNormal (f . (t:)) - Just t@Located{unLocated = EndEnv "cases"} -> goNormal (f . (t:)) + --Just t@Located{unLocated = EndEnv "cases"} -> goNormal (f . (t:)) Just t@Located{unLocated = BeginEnv "text"} -> goText (f . (t:)) Just t@Located{unLocated = BeginEnv "explanation"} -> goText (f . (t:)) Just t -> goMath (f . (t:)) @@ -282,7 +282,7 @@ alignBegin = guardM isTextMode *> lexeme do casesBegin :: Lexer (Located Token) casesBegin = guardM isTextMode *> lexeme do Char.string "\\begin{cases}" - setMathMode + --setMathMode pure (BeginEnv "cases") -- | Parses a single end math token. @@ -301,7 +301,7 @@ alignEnd = guardM isMathMode *> lexeme do casesEnd :: Lexer (Located Token) casesEnd = guardM isMathMode *> lexeme do Char.string "\\end{cases}" - setTextMode + --setTextMode pure (EndEnv "cases") -- cgit v1.2.3 From a9785eb4cac6b8c237173f7e14367babd79e92e1 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Tue, 17 Sep 2024 03:39:23 +0200 Subject: working commit --- library/topology/real-topological-space.tex | 36 +++++++++------ library/topology/urysohn2.tex | 72 ++++++++++++++++++++++++++--- 2 files changed, 86 insertions(+), 22 deletions(-) (limited to 'library') diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index d9790aa..b2e5ea9 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -382,7 +382,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have We have $a',b' \in \reals$ by assumption. We have $a' < b'$ by \cref{id_img,epsilon_ball,minus,intervalopen,reals_order_is_transitive}. Then there exist $x',\epsilon'$ such that $x' \in \reals$ and $\epsilon' \in \realsplus$ and $\intervalopen{a'}{b'} = \epsBall{x'}{\epsilon'}$. - Then $x \in \epsBall{x'}{\epsilon'}$. + Then $x \in \epsBall{x'}{\epsilon'}$ by \cref{epsilon_ball}. Follows by \cref{inter_lower_left,inter_lower_right,epsilon_ball,topological_basis_reals_eps_ball}. %Then $(x_1 - \alpha) < (x_2 + \beta)$. @@ -536,7 +536,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have \begin{subproof} It suffices to show that for all $x \in \unions{E}$ we have $x \in \intervalopenInfiniteLeft{a}$. Fix $x \in \unions{E}$. - Take $e \in E$ such that $x \in e$. + Take $e \in E$ such that $x \in e$ by \cref{unions_iff}. $x \in \reals$. Take $x',\delta'$ such that $x' \in \reals$ and $\delta' \in \realsplus$ and $e = \epsBall{x'}{\delta'}$ by \cref{epsilon_ball,minus,topo_basis_reals_eps_iff,setminus,setminus_emptyset,elem_subseteq}. $\epsBall{x'}{\delta'} \in E$. @@ -550,8 +550,8 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Suppose not. Take $y' \in e$ such that $y' > a$. $x'' < a$. - $(x'' - \delta'') < y' < (x'' + \delta'')$. - $(x'' - \delta'') < x'' < (x'' + \delta'')$. + $(x'' - \delta'') < y' < (x'' + \delta'')$ by \cref{minus,intervalopen,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. + $(x'' - \delta'') < x'' < (x'' + \delta'')$ by \cref{minus,intervalopen,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. Then $x'' < a < y'$. Therefore $(x'' - \delta'') < a < (x'' + \delta'')$ by \cref{realspuls_in_reals_minus,intervalopen_infinite_left,reals_order_is_transitive,reals_add,realsplus_in_reals,powerset_elim,subseteq}. Then $a \in e$ by \cref{epsball_are_connected_in_reals,intervalopen_infinite_left,neq_witness}. @@ -565,7 +565,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Trivial. \end{subproof} \end{subproof} - $\unions{E} \in \opens[\reals]$. + $\unions{E} \in \opens[\reals]$ by \cref{opens_unions,reals_is_topological_space,basis_is_in_genopens,topological_space_reals,topological_basis_reals_is_basis,subset_transitive}. \end{proof} \begin{lemma}\label{continuous_on_basis_implies_continuous_endo} @@ -600,7 +600,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have \begin{subproof} It suffices to show that for all $x \in \unions{E}$ we have $x \in \intervalopenInfiniteRight{a}$. Fix $x \in \unions{E}$. - Take $e \in E$ such that $x \in e$. + Take $e \in E$ such that $x \in e$ by \cref{unions_iff}. $x \in \reals$. Take $x',\delta'$ such that $x' \in \reals$ and $\delta' \in \realsplus$ and $e = \epsBall{x'}{\delta'}$ by \cref{epsilon_ball,minus,topo_basis_reals_eps_iff,setminus,setminus_emptyset,elem_subseteq}. $\epsBall{x'}{\delta'} \in E$. @@ -612,10 +612,10 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have There exists $x'' \in \intervalopenInfiniteRight{a}$ such that there exists $\delta'' \in \realsplus$ such that $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. Take $x'',\delta''$ such that $x'' \in \intervalopenInfiniteRight{a}$ and $\delta'' \in \realsplus$ and $e = \epsBall{x''}{\delta''}$ and $a \notin \epsBall{x''}{\delta''}$. Suppose not. - Take $y' \in e$ such that $y' < a$. + Take $y' \in e$ such that $y' < a$ by \cref{reals_order_total,intervalopen,eps_ball_implies_open_interval}. $x'' > a$. - $(x'' - \delta'') < y' < (x'' + \delta'')$. - $(x'' - \delta'') < x'' < (x'' + \delta'')$. + $(x'' - \delta'') < y' < (x'' + \delta'')$ by \cref{minus,intervalopen,intervalclosed,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. + $(x'' - \delta'') < x'' < (x'' + \delta'')$ by \cref{minus,intervalopen,intervalclosed,epsilon_ball,realsplus,reals_add,reals_addition_minus_behavior1,reals_order_minus_positiv}. Then $x'' > a > y'$. Therefore $(x'' - \delta'') > a > (x'' + \delta'')$ by \cref{realspuls_in_reals_minus,intervalopen_infinite_right,reals_order_is_transitive,reals_add,realsplus_in_reals,powerset_elim,subseteq,epsball_are_connected_in_reals,subseteq}. Then $a \in e$ by \cref{reals_order_is_transitive,reals_order_total,reals_add,realsplus,epsball_are_connected_in_reals,intervalopen_infinite_right,neq_witness}. @@ -689,6 +689,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have We show that for all $x \in \reals$ we have $x \in (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b} \union \intervalclosed{a}{b})$. \begin{subproof} Fix $x \in \reals$. + Follows by \cref{union_intro_left,intervalopen_infinite_left,reals_order_total,reals_order_total2,union_iff,intervalopen_infinite_right,union_assoc,union_intro_right,intervalclosed}. \end{subproof} \end{proof} @@ -697,7 +698,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Then $\reals = \intervalopenInfiniteLeft{a} \union \intervalclosedInfiniteRight{a}$. \end{lemma} \begin{proof} - It suffices to show that for all $x \in \reals$ we have either $x \in \intervalopenInfiniteLeft{a}$ or $x \in \intervalclosedInfiniteRight{a}$. + It suffices to show that for all $x \in \reals$ we have either $x \in \intervalopenInfiniteLeft{a}$ or $x \in \intervalclosedInfiniteRight{a}$ by \cref{intervalopen_infinite_left,union_intro_left,neq_witness,intervalclosed_infinite_right,union_intro_right,union_iff}. Trivial. \end{proof} @@ -714,6 +715,9 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Suppose $a \in \reals$. Then $\intervalopenInfiniteRight{a} \inter \intervalclosedInfiniteLeft{a} = \emptyset$. \end{lemma} +\begin{proof} + Follows by \cref{reals_order_total,inter_lower_left,intervalopen_infinite_right,order_reals_lemma6,inter_lower_right,foundation,subseteq,intervalclosed_infinite_left}. +\end{proof} \begin{lemma}\label{intersection_of_open_closed__infinite_intervals_open_left} Suppose $a \in \reals$. @@ -725,7 +729,7 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have Then $\intervalclosedInfiniteRight{a} \in \closeds{\reals}$. \end{proposition} \begin{proof} - $\intervalclosedInfiniteRight{a} = \reals \setminus \intervalopenInfiniteLeft{a}$. + $\intervalclosedInfiniteRight{a} = \reals \setminus \intervalopenInfiniteLeft{a}$ by \cref{intersection_of_open_closed__infinite_intervals_open_left,reals_as_union_of_open_closed_intervals2,setminus_inter,double_relative_complement,subseteq_union_setminus,subseteq_setminus,setminus_union,setminus_disjoint,setminus_partition,setminus_subseteq,setminus_emptyset,setminus_self,setminus_setminus,double_complement_union}. \end{proof} \begin{proposition}\label{closedinterval_infinite_left_in_closeds} @@ -747,15 +751,17 @@ For all $x,\delta$ such that $x \in \reals \land \delta \in \realsplus$ we have We show that for all $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$ we have $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. \begin{subproof} Fix $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$. - Then $x \in \reals \setminus \intervalopenInfiniteLeft{a}$. - Then $x \in \reals \setminus \intervalopenInfiniteRight{b}$. + Then $x \in \reals \setminus \intervalopenInfiniteLeft{a}$ by \cref{setminus,double_complement_union}. + Then $x \in \reals \setminus \intervalopenInfiniteRight{b}$ by \cref{union_upper_left,subseteq,union_comm,subseteq_implies_setminus_supseteq}. + Follows by \cref{inter_intro}. \end{subproof} We show that for all $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$ we have $x \in \reals \setminus (\intervalopenInfiniteLeft{a} \union \intervalopenInfiniteRight{b})$. \begin{subproof} Fix $x \in (\reals \setminus \intervalopenInfiniteLeft{a}) \inter (\reals \setminus \intervalopenInfiniteRight{b})$. - Then $x \in (\reals \setminus \intervalopenInfiniteLeft{a})$. - Then $x \in (\reals \setminus \intervalopenInfiniteRight{b})$. + Then $x \in (\reals \setminus \intervalopenInfiniteLeft{a})$ by \cref{setminus_setminus,setminus}. + Then $x \in (\reals \setminus \intervalopenInfiniteRight{b})$ by \cref{inter_lower_right,elem_subseteq,setminus_setminus}. \end{subproof} + Follows by \cref{setminus_union}. \end{subproof} We show that $\reals \setminus \intervalopenInfiniteLeft{a} = \intervalclosedInfiniteRight{a}$. \begin{subproof} diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 83e3aa4..9990199 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -94,6 +94,9 @@ \begin{proposition}\label{naturals_in_transitive} $\naturals$ is a \in-transitive set. \end{proposition} +\begin{proof} + Follows by \cref{nat_is_transitiveset}. +\end{proof} \begin{proposition}\label{naturals_elem_in_transitive} If $n \in \naturals$ then $n$ is \in-transitive and every element of $n$ is \in-transitive. @@ -119,6 +122,9 @@ \begin{proposition}\label{zero_is_empty} There exists no $x$ such that $x \in \zero$. \end{proposition} +\begin{proof} + Follows by \cref{notin_emptyset}. +\end{proof} \begin{proposition}\label{one_is_positiv} $1$ is positiv. @@ -163,6 +169,13 @@ Omitted. \end{proof} +\begin{proposition}\label{naturals_one_zero_or_greater} + For all $l \in \naturals$ we have if $l < 1$ then $l = \zero$. +\end{proposition} +\begin{proof} + Follows by \cref{reals_order,naturals_subseteq_reals,subseteq,one_in_reals,naturals_is_zero_one_or_greater}. +\end{proof} + \begin{proposition}\label{naturals_rless_existence_of_lesser_natural} For all $n \in \naturals$ we have for all $m \in \naturals$ such that $m < n$ there exist $k \in \naturals$ such that $m + k = n$. \end{proposition} @@ -184,7 +197,7 @@ \end{subproof} \caseOf{$n = 1$.} Fix $m$. - For all $l \in \naturals$ we have If $l < 1$ then $l = \zero$. + For all $l \in \naturals$ we have if $l < 1$ then $l = \zero$. Then $\zero + 1 = 1$. \caseOf{$n > 1$.} Take $l \in \naturals$ such that $\suc{l} = n$. @@ -350,7 +363,7 @@ Suppose $U$ is a urysohnchain of $X$. Suppose $\dom{U}$ is finite. Suppose $U$ is inhabited. - Then there exist $V,f$ such that $V$ is a urysohnchain of $X$ and $f$ is consistent on $U$ to $V$ and $V$ is normal ordered. + Then there exist $V,f$ such that $V$ is a urysohnchain of $X$ and $f$ is consistent on $V$ to $U$ and $V$ is normal ordered. \end{proposition} \begin{proof} Take $n$ such that $\dom{U}$ has cardinality $n$ by \cref{ran_converse,cardinality,finite}. @@ -360,11 +373,36 @@ \caseOf{$n \neq \zero$.} Take $k$ such that $k \in \naturals$ and $\suc{k}=n$. We have $\dom{U} \subseteq \naturals$. - $\dom{U}$ is inhabited. + $\dom{U}$ is inhabited by \cref{downward_closure,downward_closure_iff,rightunique,function_member_elim,inhabited,chain_of_subsets,urysohnchain,sequence}. + $\dom{U}$ has cardinality $\suc{k}$. We show that there exist $F$ such that $F$ is a bijection from $\seq{\zero}{k}$ to $\dom{U}$ and for all $n',m' \in \seq{\zero}{k}$ such that $n' < m'$ we have $F(n') < F(m')$. \begin{subproof} - Omitted. + For all $M \subseteq \naturals$ such that $M$ is inhabited we have there exist $f,k$ such that $f$ is a bijection from $\seq{\zero}{k}$ to $M$ and $M$ has cardinality $\suc{k}$ and for all $n,m \in \seq{\zero}{k}$ such that $n < m$ we have $f(n) < f(m)$. + We have $\dom{U} \subseteq \naturals$. + $\dom{U}$ is inhabited. + Therefore there exist $f$ such that there exist $k'$ such that $f$ is a bijection from $\seq{\zero}{k'}$ to $\dom{U}$ and $\dom{U}$ has cardinality $\suc{k'}$ and for all $n',m' \in \seq{\zero}{k'}$ such that $n' < m'$ we have $f(n') < f(m')$. + Take $f$ such that there exist $k'$ such that $f$ is a bijection from $\seq{\zero}{k'}$ to $\dom{U}$ and $\dom{U}$ has cardinality $\suc{k'}$ and for all $n',m' \in \seq{\zero}{k'}$ such that $n' < m'$ we have $f(n') < f(m')$. + Take $k'$ such that $f$ is a bijection from $\seq{\zero}{k'}$ to $\dom{U}$ and $\dom{U}$ has cardinality $\suc{k'}$ and for all $n',m' \in \seq{\zero}{k'}$ such that $n' < m'$ we have $f(n') < f(m')$. + $\seq{\zero}{k'}$ has cardinality $\suc{k}$ by \cref{omega_is_an_ordinal,suc,ordinal_transitivity,bijection_converse_is_bijection,seq_zero_to_n_eq_to_suc_n,cardinality,bijections_dom,bijection_circ}. + We show that $\seq{\zero}{k'} = \seq{\zero}{k}$. + \begin{subproof} + We show that $\seq{\zero}{k'} \subseteq \seq{\zero}{k}$. + \begin{subproof} + It suffices to show that for all $y \in \seq{\zero}{k'}$ we have $y \in \seq{\zero}{k}$. + Fix $y \in \seq{\emptyset}{k'}$. + Then $y \leq k'$. + Therefore $y \in k'$ or $y = k'$ by \cref{omega_is_an_ordinal,suc_intro_self,ordinal_transitivity,cardinality,rless_eq_in_for_naturals,m_to_n_set}. + %Then $\seq{\emptyset}{k'} \in \suc{k}$. + Therefore $y \in \suc{k}$. + Therefore $y \in \seq{\emptyset}{k}$. + \end{subproof} + We show that for all $y \in \seq{\zero}{k}$ we have $y \in \seq{\zero}{k'}$. + \begin{subproof} + Fix $y \in \seq{\emptyset}{k}$. + \end{subproof} + \end{subproof} \end{subproof} + Take $F$ such that $F$ is a bijection from $\seq{\zero}{k}$ to $\dom{U}$ and for all $n',m' \in \seq{\zero}{k}$ such that $n' < m'$ we have $F(n') < F(m')$. Let $N = \seq{\zero}{k}$. Let $M = \pow{X}$. Define $V : N \to M$ such that $V(n)= @@ -374,11 +412,31 @@ $\dom{V} = \seq{\zero}{k}$. We show that $V$ is a urysohnchain of $X$. \begin{subproof} - Trivial. + It suffices to show that $V$ is a chain of subsets in $X$ and for all $n,m \in \dom{V}$ such that $n < m$ we have $\closure{\at{V}{n}}{X} \subseteq \interior{\at{V}{m}}{X}$. + We show that $V$ is a chain of subsets in $X$. + \begin{subproof} + It suffices to show that $V$ is a sequence and for all $n \in \dom{V}$ we have $\at{V}{n} \subseteq \carrier[X]$ and for all $m \in \dom{V}$ such that $m > n$ we have $\at{V}{n} \subseteq \at{V}{m}$. + $V$ is a sequence by \cref{m_to_n_set,sequence,subseteq}. + It suffices to show that for all $n \in \dom{V}$ we have $\at{V}{n} \subseteq \carrier[X]$ and for all $m \in \dom{V}$ such that $m > n$ we have $\at{V}{n} \subseteq \at{V}{m}$. + Fix $n \in \dom{V}$. + Then $\at{V}{n} \subseteq \carrier[X]$ by \cref{ran_converse,seq_zero_to_n_eq_to_suc_n,in_irrefl}. + It suffices to show that for all $m$ such that $m \in \dom{V}$ and $n \rless m$ we have $\at{V}{n} \subseteq \at{V}{m}$. + Fix $m$ such that $m \in \dom{V}$ and $n \rless m$. + Follows by \cref{ran_converse,seq_zero_to_n_eq_to_suc_n,in_irrefl}. + \end{subproof} + It suffices to show that for all $n \in \dom{V}$ we have for all $m$ such that $m \in \dom{V} \land n \rless m$ we have $\closure{\at{V}{n}}{X} \subseteq \interior{\at{V}{m}}{X}$. + Fix $n \in \dom{V}$. + Fix $m$ such that $m \in \dom{V} \land n \rless m$. + Follows by \cref{ran_converse,seq_zero_to_n_eq_to_suc_n,in_irrefl}. \end{subproof} - We show that $F$ is consistent on $U$ to $V$. + We show that $F$ is consistent on $V$ to $U$. \begin{subproof} - Trivial. + It suffices to show that $F$ is a bijection from $\dom{V}$ to $\dom{U}$ and for all $n,m \in \dom{V}$ such that $n < m$ we have $F(n) < F(m)$ by \cref{bijection_of_urysohnchains}. + $F$ is a bijection from $\dom{V}$ to $\dom{U}$. + It suffices to show that for all $n \in \dom{V}$ we have for all $m$ such that $m \in \dom{V}$ and $n \rless m$ we have $f(n) < f(m)$. + Fix $n \in \dom{V}$. + Fix $m$ such that $m \in \dom{V}$ and $n \rless m$. + Follows by \cref{ran_converse,seq_zero_to_n_eq_to_suc_n,in_irrefl}. \end{subproof} $V$ is normal ordered. \end{byCase} -- cgit v1.2.3 From c943ca6441e9118bc9caee1c11f697da89bc06b7 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 18 Sep 2024 00:01:42 +0200 Subject: working commit --- library/set/equinumerosity.tex | 2 +- library/topology/urysohn2.tex | 260 ++++++++++++++++++++++++++++++++++++++--- source/Checking.hs | 2 +- source/Meaning.hs | 4 +- source/Syntax/Abstract.hs | 2 +- source/Syntax/Concrete.hs | 4 +- source/Syntax/Internal.hs | 2 +- 7 files changed, 250 insertions(+), 26 deletions(-) (limited to 'library') diff --git a/library/set/equinumerosity.tex b/library/set/equinumerosity.tex index a846b78..a922052 100644 --- a/library/set/equinumerosity.tex +++ b/library/set/equinumerosity.tex @@ -15,7 +15,7 @@ $A\approx A$. \end{proposition} \begin{proof} - $\identity{A}$ is a bijection from $A$ to $A$ by \cref{id_is_bijection}. + $\identity{A}$ is a bijection from $A$ to $A$. %by \cref{id_is_bijection}. Follows by \cref{equinum}. \end{proof} diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 9990199..97bbc70 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -13,6 +13,7 @@ \import{set/fixpoint.tex} \import{set/product.tex} \import{topology/real-topological-space.tex} +\import{set/equinumerosity.tex} \section{Urysohns Lemma} @@ -251,6 +252,7 @@ + \begin{proposition}\label{naturals_leq_on_suc} For all $n,m \in \naturals$ such that $m < \suc{n}$ we have $m \leq n$. \end{proposition} @@ -358,6 +360,218 @@ Omitted. \end{proof} +\begin{lemma}\label{naturals_suc_injective} + Suppose $n,m \in \naturals$. + $n = m$ iff $\suc{n} = \suc{m}$. +\end{lemma} + +\begin{lemma}\label{naturals_rless_implies_not_eq} + Suppose $n,m \in \naturals$. + Suppose $n < m$. + Then $n \neq m$. +\end{lemma} + +\begin{lemma}\label{cardinality_of_singleton} + For all $x$ such that $x \neq \emptyset$ we have $\{x\}$ has cardinality $1$. +\end{lemma} +\begin{proof} + Omitted. + %Fix $x$. + %Suppose $x \neq \emptyset$. + %Let $X = \{x\}$. + %$\seq{\zero}{\zero}=1$. + %$\seq{\zero}{\zero}$ has cardinality $1$. + %$X \setminus \{x\} = \emptyset$. + %$1 = \{\emptyset\}$. + %Let $F = \{(x,\emptyset)\}$. + %$F$ is a relation. + %$\dom{F} = X$. + %$\emptyset \in \ran{F}$. + %for all $x \in 1$ we have $x = \emptyset$. + %$\ran{F} = 1$. + %$F$ is injective. + %$F \in \surj{X}{1}$. + %$F$ is a bijection from $X$ to $1$. +\end{proof} + +\begin{lemma}\label{cardinality_n_plus_1} + For all $n \in \naturals$ we have $n+1$ has cardinality $n+1$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{cardinality_n_m_plus} + For all $n,m \in \naturals$ we have $n+m$ has cardinality $n+m$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + +\begin{lemma}\label{cardinality_plus_disjoint} + Suppose $X \inter Y = \emptyset$. + Suppose $X$ is finite. + Suppose $Y$ is finite. + Suppose $X$ has cardinality $n$. + Suppose $Y$ has cardinality $m$. + Then $X \union Y$ has cardinality $m+n$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} + + + + +\begin{lemma}\label{injective_functions_is_bijection_if_bijection_there_is_other_bijection_1} + Suppose $f$ is a bijection from $X$ to $Y$. + Suppose $g$ is a function from $X$ to $Y$. + Suppose $g$ is injective. + Suppose $X$ is finite and $Y$ is finite. + For all $n \in \naturals$ such that $Y$ has cardinality $n$ we have $g$ is a bijection from $X$ to $Y$. +\end{lemma} +\begin{proof}[Proof by \in-induction on $n$] + Assume $n \in \naturals$. + Suppose $Y$ has cardinality $n$. + $X$ has cardinality $n$ by \cref{bijection_converse_is_bijection,bijection_circ,regularity,cardinality,foundation,empty_eq,notin_emptyset}. + \begin{byCase} + \caseOf{$n = \zero$.} + Follows by \cref{converse_converse_eq,injective_converse_is_function,converse_is_relation,dom_converse,id_is_function_to,id_ran,ran_circ_exact,circ,ran_converse,emptyset_is_function_on_emptyset,bijective_converse_are_funs,relext,function_member_elim,bijection_is_function,cardinality,bijections_dom,in_irrefl,codom_of_emptyset_can_be_anything,converse_emptyset,funs_elim,neq_witness,id}. + \caseOf{$n \neq \zero$.} + %Take $n' \in n$ such that $n = \suc{n'}$. + %$n' \in \naturals$. + %$n' + 1 = n$. + %Take $y$ such that $y \in Y$ by \cref{funs_type_apply,apply,bijections_to_funs,cardinality,foundation}. + %Let $Y' = Y \setminus \{y\}$. + %$Y' \subseteq Y$. + %$Y'$ is finite. + %There exist $m \in \naturals$ such that $Y'$ has cardinality $m$. + %Take $m \in \naturals$ such that $Y'$ has cardinality $m$. + %Then $Y'$ has cardinality $n'$. + %Let $x' = \apply{\converse{f}}{y'}$. + %$x' \in X$. + %Let $X' = X \setminus \{x'\}$. + %$X' \subseteq X$. + %$X'$ is finite. + %There exist $m' \in \naturals$ such that $X'$ has cardinality $m'$. + %Take $m' \in \naturals$ such that $X''$ has cardinality $m'$. + %Then $X'$ has cardinality $n'$. + %Let $f'(z)=f(z)$ for $z \in X'$. + %$\dom{f'} = X'$. + %$\ran{f'} = Y'$. + %$f'$ is a bijection from $X'$ to $Y'$. + %Let $g'(z) = g(z)$ for $z \in X'$. + %Then $g'$ is injective. + %Then $g'$ is a bijection from $X'$ to $Y'$ by \cref{rels,id_elem_rels,times_empty_right,powerset_emptyset,double_complement_union,unions_cons,union_eq_cons,union_as_unions,unions_pow,cons_absorb,setminus_self,bijections_dom,ran_converse,id_apply,apply,unions_emptyset,img_emptyset,zero_is_empty}. + %Define $G : X \to Y$ such that $G(z)= + %\begin{cases} + % g'(z) & \text{if} z \in X' \\ + % y' & \text{if} z = x' + %\end{cases}$ + %$G = g$. + %Follows by \cref{double_relative_complement,fun_to_surj,bijections,funs_surj_iff,bijections_to_funs,neq_witness,surj,funs_elim,setminus_self,cons_subseteq_iff,cardinality,ordinal_empty_or_emptyset_elem,naturals_inductive_set,natural_number_is_ordinal_for_all,foundation,inter_eq_left_implies_subseteq,inter_emptyset,cons_subseteq_intro,emptyset_subseteq}. + Omitted. + \end{byCase} + %$\converse{f}$ is a bijection from $Y$ to $X$. + %Let $h = g \circ \converse{f}$. + %It suffices to show that $\ran{g} = Y$ by \cref{fun_to_surj,dom_converse,bijections}. + %It suffices to show that for all $y \in Y$ we have there exist $x \in X$ such that $g(x)=y$ by \cref{funs_ran,subseteq_antisymmetric,fun_ran_iff,apply,funs_elim,ran_converse,subseteq}. +% + %Fix $y \in Y$. + %Take $x \in X$ such that $\apply{\converse{f}}{y} = x$. + +\end{proof} + +\begin{lemma}\label{injective_functions_is_bijection_if_bijection_there_is_other_bijection} + Suppose $f$ is a bijection from $X$ to $Y$. + Suppose $g$ is a function from $X$ to $Y$. + Suppose $g$ is injective. + Suppose $Y$ is finite. + Then $g$ is a bijection from $X$ to $Y$. +\end{lemma} +\begin{proof} + There exist $n \in \naturals$ such that $Y$ has cardinality $n$ by \cref{cardinality,injective_functions_is_bijection_if_bijection_there_is_other_bijection_1,finite}. + Follows by \cref{injective_functions_is_bijection_if_bijection_there_is_other_bijection_1,cardinality,equinum_tran,equinum_sym,equinum,finite}. +\end{proof} + + + +\begin{lemma}\label{naturals_bijection_implies_eq} + Suppose $n,m \in \naturals$. + Suppose $f$ is a bijection from $n$ to $m$. + Then $n = m$. +\end{lemma} +\begin{proof} + $n$ is finite. + $m$ is finite. + Suppose not. + Then $n < m$ or $m < n$. + \begin{byCase} + \caseOf{$n < m$.} + Then $n \in m$. + There exist $x \in m$ such that $x \notin n$. + $\identity{n}$ is a function from $n$ to $m$. + $\identity{n}$ is injective. + $\apply{\identity{n}}{n} = n$ by \cref{id_ran,ran_of_surj,bijections,injective_functions_is_bijection_if_bijection_there_is_other_bijection}. + Follows by \cref{inhabited,regularity,function_apply_default,apply,id_dom,in_irrefl,function_member_elim,bijections_dom,zero_is_empty,bijection_is_function,foundation,bijections,ran_of_surj,dom_converse,converse_emptyset,dom_emptyset}. + \caseOf{$m < n$.} + Then $m \in n$. + There exist $x \in n$ such that $x \notin m$. + $\converse{f}$ is a bijection from $m$ to $n$. + $\identity{m}$ is a function from $m$ to $n$. + $\identity{m}$ is injective. + $\apply{\identity{m}}{m} = m$ by \cref{id_ran,ran_of_surj,bijections,injective_functions_is_bijection_if_bijection_there_is_other_bijection}. + Follows by \cref{inhabited,regularity,function_apply_default,apply,id_dom,in_irrefl,function_member_elim,bijections_dom,zero_is_empty,bijection_is_function,foundation,bijections,ran_of_surj,dom_converse,converse_emptyset,dom_emptyset}. + \end{byCase} +\end{proof} + +\begin{lemma}\label{naturals_eq_iff_bijection} + Suppose $n,m \in \naturals$. + $n = m$ iff there exist $f$ such that $f$ is a bijection from $n$ to $m$. +\end{lemma} +\begin{proof} + We show that if $n = m$ then there exist $f$ such that $f$ is a bijection from $n$ to $m$. + \begin{subproof} + Trivial. + \end{subproof} + We show that for all $k \in \naturals$ we have if there exist $f$ such that $f$ is a bijection from $k$ to $m$ then $k = m$. + \begin{subproof}%[Proof by \in-induction on $k$] + %Assume $k \in \naturals$. + %\begin{byCase} + % \caseOf{$k = \zero$.} + % Trivial. + % \caseOf{$k \neq \zero$.} + % \begin{byCase} + % \caseOf{$m = \zero$.} + % Trivial. + % \caseOf{$m \neq \zero$.} + % Take $k' \in \naturals$ such that $\suc{k'} = k$. + % Then $k' \in k$. + % Take $m' \in \naturals$ such that $m = \suc{m'}$. + % Then $m' \in m$. + % + % \end{byCase} + %\end{byCase} + \end{subproof} +\end{proof} + +\begin{lemma}\label{seq_from_zero_suc_cardinality_eq_upper_border} + Suppose $n,m \in \naturals$. + Suppose $\seq{\zero}{n}$ has cardinality $\suc{m}$. + Then $n = m$. +\end{lemma} +\begin{proof} + We have $\seq{\zero}{n} = \suc{n}$. + Take $f$ such that $f$ is a bijection from $\seq{\zero}{n}$ to $\suc{m}$. + Therefore $n=m$ by \cref{suc_injective,naturals_inductive_set,cardinality,naturals_eq_iff_bijection}. +\end{proof} + +\begin{lemma}\label{seq_from_zero_cardinality_eq_upper_border_set_eq} + Suppose $n,m \in \naturals$. + Suppose $\seq{\zero}{n}$ has cardinality $\suc{m}$. + Then $\seq{\zero}{n} = \seq{\zero}{m}$. +\end{lemma} + \begin{proposition}\label{existence_normal_ordered_urysohn} Let $X$ be a urysohn space. Suppose $U$ is a urysohnchain of $X$. @@ -384,23 +598,24 @@ Take $f$ such that there exist $k'$ such that $f$ is a bijection from $\seq{\zero}{k'}$ to $\dom{U}$ and $\dom{U}$ has cardinality $\suc{k'}$ and for all $n',m' \in \seq{\zero}{k'}$ such that $n' < m'$ we have $f(n') < f(m')$. Take $k'$ such that $f$ is a bijection from $\seq{\zero}{k'}$ to $\dom{U}$ and $\dom{U}$ has cardinality $\suc{k'}$ and for all $n',m' \in \seq{\zero}{k'}$ such that $n' < m'$ we have $f(n') < f(m')$. $\seq{\zero}{k'}$ has cardinality $\suc{k}$ by \cref{omega_is_an_ordinal,suc,ordinal_transitivity,bijection_converse_is_bijection,seq_zero_to_n_eq_to_suc_n,cardinality,bijections_dom,bijection_circ}. - We show that $\seq{\zero}{k'} = \seq{\zero}{k}$. - \begin{subproof} - We show that $\seq{\zero}{k'} \subseteq \seq{\zero}{k}$. - \begin{subproof} - It suffices to show that for all $y \in \seq{\zero}{k'}$ we have $y \in \seq{\zero}{k}$. - Fix $y \in \seq{\emptyset}{k'}$. - Then $y \leq k'$. - Therefore $y \in k'$ or $y = k'$ by \cref{omega_is_an_ordinal,suc_intro_self,ordinal_transitivity,cardinality,rless_eq_in_for_naturals,m_to_n_set}. - %Then $\seq{\emptyset}{k'} \in \suc{k}$. - Therefore $y \in \suc{k}$. - Therefore $y \in \seq{\emptyset}{k}$. - \end{subproof} - We show that for all $y \in \seq{\zero}{k}$ we have $y \in \seq{\zero}{k'}$. - \begin{subproof} - Fix $y \in \seq{\emptyset}{k}$. - \end{subproof} - \end{subproof} + $\seq{\zero}{k'} = \seq{\zero}{k}$ by \cref{omega_is_an_ordinal,seq_from_zero_cardinality_eq_upper_border_set_eq,suc_subseteq_implies_in,suc_subseteq_elim,ordinal_suc_subseteq,cardinality}. + %We show that $\seq{\zero}{k'} = \seq{\zero}{k}$. + %\begin{subproof} + % We show that $\seq{\zero}{k'} \subseteq \seq{\zero}{k}$. + % \begin{subproof} + % It suffices to show that for all $y \in \seq{\zero}{k'}$ we have $y \in \seq{\zero}{k}$. + % Fix $y \in \seq{\emptyset}{k'}$. + % Then $y \leq k'$. + % Therefore $y \in k'$ or $y = k'$ by \cref{omega_is_an_ordinal,suc_intro_self,ordinal_transitivity,cardinality,rless_eq_in_for_naturals,m_to_n_set}. + % + % Therefore $y \in \suc{k}$. + % Therefore $y \in \seq{\emptyset}{k}$. + % \end{subproof} + % We show that for all $y \in \seq{\zero}{k}$ we have $y \in \seq{\zero}{k'}$. + % \begin{subproof} + % Fix $y \in \seq{\emptyset}{k}$. + % \end{subproof} + %\end{subproof} \end{subproof} Take $F$ such that $F$ is a bijection from $\seq{\zero}{k}$ to $\dom{U}$ and for all $n',m' \in \seq{\zero}{k}$ such that $n' < m'$ we have $F(n') < F(m')$. Let $N = \seq{\zero}{k}$. @@ -452,7 +667,9 @@ $f$ is staircase sequence of $U$ iff $f$ is a sequence and $U$ is a lifted urysohnchain of $X$ and $\dom{U} = \dom{f}$ and for all $n \in \dom{U}$ we have $\at{f}{n}$ is a staircase function adapted to $\at{U}{n}$ in $U$. \end{definition} - +\begin{definition} + +\end{definition} @@ -565,8 +782,15 @@ \end{subproof} Take $U$ such that $U$ is a lifted urysohnchain of $X$ and $\at{U}{\zero} = U_0$. + We show that there exist $S$ such that $S$ is staircase sequence of $U$. + \begin{subproof} + Omitted. + \end{subproof} + Take $S$ such that $S$ is staircase sequence of $U$. + For all $x \in \carrier[X]$ we have there exist $r,R$ such that $r \in \reals$ and $R$ is a sequence of reals and $\dom{R} = \naturals$ and $R$ converge to $r$ and for all $n \in \naturals$ we have $\at{R}{n} = \apply{\at{S}{n}}{x}$. + We show that for all $x \in \carrier[X]$ there exists $r \in \intervalclosed{a}{b}$ such that for . diff --git a/source/Checking.hs b/source/Checking.hs index 766c327..8bc38a4 100644 --- a/source/Checking.hs +++ b/source/Checking.hs @@ -543,7 +543,7 @@ checkProof = \case checkCalc calc assume [Asm (calcResult calc)] checkProof continue - DefineFunctionMathy funVar argVar domVar ranExpr definitions continue -> do + DefineFunctionLocal funVar argVar domVar ranExpr definitions continue -> do -- We have f: X \to Y and x \mapsto ... -- definition is a nonempty list of (expresssion e, formula phi) -- such that f(x) = e if phi(x) diff --git a/source/Meaning.hs b/source/Meaning.hs index 4a21fa3..00a944f 100644 --- a/source/Meaning.hs +++ b/source/Meaning.hs @@ -606,9 +606,9 @@ glossProof = \case then Sem.DefineFunction funVar argVar <$> glossExpr valueExpr <*> glossExpr domExpr <*> glossProof proof else error "mismatched variables in function definition." - Raw.DefineFunctionMathy funVar domVar ranExpr funVar2 argVar definitions proof -> do + Raw.DefineFunctionLocal funVar domVar ranExpr funVar2 argVar definitions proof -> do if funVar == funVar2 - then Sem.DefineFunctionMathy funVar argVar domVar <$> glossExpr ranExpr <*> (glossLocalFunctionExprDef `each` definitions) <*> glossProof proof + then Sem.DefineFunctionLocal funVar argVar domVar <$> glossExpr ranExpr <*> (glossLocalFunctionExprDef `each` definitions) <*> glossProof proof else error "missmatched function names" Raw.Calc calc proof -> Sem.Calc <$> glossCalc calc <*> glossProof proof diff --git a/source/Syntax/Abstract.hs b/source/Syntax/Abstract.hs index 13691e7..c8022c7 100644 --- a/source/Syntax/Abstract.hs +++ b/source/Syntax/Abstract.hs @@ -373,7 +373,7 @@ data Proof - | DefineFunctionMathy VarSymbol VarSymbol Expr VarSymbol VarSymbol (NonEmpty (Expr, Formula)) Proof + | DefineFunctionLocal VarSymbol VarSymbol Expr VarSymbol VarSymbol (NonEmpty (Expr, Formula)) Proof -- ^ Local function definition, but in this case we give the domain and target an the rules for $xs$ in some sub domains. -- deriving (Show, Eq, Ord) diff --git a/source/Syntax/Concrete.hs b/source/Syntax/Concrete.hs index 7a89bea..9b947b0 100644 --- a/source/Syntax/Concrete.hs +++ b/source/Syntax/Concrete.hs @@ -374,7 +374,7 @@ grammar lexicon@Lexicon{..} = mdo -- \end{cases} functionDefineCase <- rule $ (,) <$> (optional _ampersand *> expr) <*> (_ampersand *> text _if *> formula) - defineFunctionMathy <- rule $ DefineFunctionMathy + defineFunctionLocal <- rule $ DefineFunctionLocal <$> (_define *> beginMath *> varSymbol) -- Define $ f <*> (_colon *> varSymbol) -- : 'var' \to 'var' <*> (_to *> expr <* endMath <* _suchThat) @@ -386,7 +386,7 @@ grammar lexicon@Lexicon{..} = mdo - proof <- rule $ asum [byContradiction, byCases, bySetInduction, byOrdInduction, calc, subclaim, assume, fix, take, have, suffices, define, defineFunction, defineFunctionMathy, qed] + proof <- rule $ asum [byContradiction, byCases, bySetInduction, byOrdInduction, calc, subclaim, assume, fix, take, have, suffices, define, defineFunction, defineFunctionLocal, qed] blockAxiom <- rule $ uncurry3 BlockAxiom <$> envPos "axiom" axiom diff --git a/source/Syntax/Internal.hs b/source/Syntax/Internal.hs index e83126d..c098380 100644 --- a/source/Syntax/Internal.hs +++ b/source/Syntax/Internal.hs @@ -446,7 +446,7 @@ data Proof | Define VarSymbol Term Proof | DefineFunction VarSymbol VarSymbol Term Term Proof - | DefineFunctionMathy VarSymbol VarSymbol VarSymbol Term (NonEmpty (Term, Formula)) Proof + | DefineFunctionLocal VarSymbol VarSymbol VarSymbol Term (NonEmpty (Term, Formula)) Proof deriving instance Show Proof deriving instance Eq Proof -- cgit v1.2.3 From 1b05816322dc79b20976350393f71840c697eb46 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Wed, 18 Sep 2024 15:09:51 +0200 Subject: Ambiguous parse in urysohn2.tex line 674 the secound 2 throws an ambiguous parse. the same definition in line 678 commented without this a does not throe this error. ambiguous parse: [BlockLemma (SourcePos {sourceName = "set.tex", sourceLine = Pos 129, sourceColumn = Pos 1}) (Marker "empty_eq") (Lemma [] (StmtConnected Implication (StmtVerbPhrase (TermExpr (ExprVar (NamedVar "x")) :| [TermExpr (ExprVar (NamedVar "y"))]) (VPAdj (Adj [Just (Word "empty")] [] :| []))) (StmtFormula (FormulaChain (ChainBase (ExprVar (NamedVar "x") :| []) Positive (RelationSymbol (Symbol "=")) (ExprVar (NamedVar "y") :| [])))))),BlockLemma (SourcePos {sourceName = "set.tex", sourceLine = Pos 129, sourceColumn = Pos 1}) (Marker "empty_eq") (Lemma [] (StmtConnected Implication (StmtConnected Conjunction (StmtVerbPhrase (TermExpr (ExprVar (NamedVar "x")) :| []) (VPVerb (Verb (SgPl {sg = [], pl = []}) []))) (StmtVerbPhrase (TermExpr (ExprVar (NamedVar "y")) :| []) (VPAdj (Adj [Just (Word "empty")] [] :| [])))) (StmtFormula (FormulaChain (ChainBase (ExprVar (NamedVar "x") :| []) Positive (RelationSymbol (Symbol "=")) (ExprVar (NamedVar "y") :| []))))))] --- library/topology/urysohn2.tex | 28 +++++++++++++++++++++------- 1 file changed, 21 insertions(+), 7 deletions(-) (limited to 'library') diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 97bbc70..ce6d742 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -664,14 +664,28 @@ \end{definition} \begin{definition}\label{staircase_sequence} - $f$ is staircase sequence of $U$ iff $f$ is a sequence and $U$ is a lifted urysohnchain of $X$ and $\dom{U} = \dom{f}$ and for all $n \in \dom{U}$ we have $\at{f}{n}$ is a staircase function adapted to $\at{U}{n}$ in $U$. + $S$ is staircase sequence of $U$ iff $S$ is a sequence and $U$ is a lifted urysohnchain of $X$ and $\dom{U} = \dom{S}$ and for all $n \in \dom{U}$ we have $\at{S}{n}$ is a staircase function adapted to $\at{U}{n}$ in $U$. \end{definition} -\begin{definition} - +\begin{definition}\label{staircase_limit_point} + $x$ is the staircase limit of $S$ with $y$ iff $x \in \reals$ and for all $\epsilon \in \realsplus$ there exist $n_0 \in \naturals$ such that for all $n \in \naturals$ such that $n_0 \rless n$ we have $\apply{\at{S}{n}}{y} \in \epsBall{x}{\epsilon}$. \end{definition} +\begin{definition}\label{staircase_limit_function} + $f$ is a limit function of a staircase $S$ iff $S$ is staircase sequence of $U$ and $U$ is a lifted urysohnchain of $X$ and $\dom{f} = \carrier[X]$ and for all $x \in \dom{f}$ we have $f(x)$ is the staircase limit of $S$ with $x$ and $f$ is a function from $\carrier[X]$ to $\reals$. +\end{definition} +%\begin{definition}\label{staircase_limit_function} +% $f$ is a limit function of staircase $S$ iff $S$ is staircase sequence of $U$ and $U$ is a lifted urysohnchain of $X$ and $\dom{f} = \carrier[X]$ and for all $x \in \dom{f}$ we have $f(x)$ is the staircase limit of $S$ with $x$ and $f$ is a function from $\carrier[X]$ to $\reals$. +%\end{definition} +% +%\begin{proposition}\label{staircase_limit_is_continuous} +% Suppose $X$ is a urysohnspace. +% Suppose $U$ is a lifted urysohnchain of $X$. +% Suppose $S$ is staircase sequence of $U$. +% Suppose $f$ is the limit function of a staircase $S$. +% Then $f$ is continuous. +%\end{proposition} \begin{theorem}\label{urysohnsetinbeetween} Let $X$ be a urysohn space. @@ -788,10 +802,10 @@ \end{subproof} Take $S$ such that $S$ is staircase sequence of $U$. - For all $x \in \carrier[X]$ we have there exist $r,R$ such that $r \in \reals$ and $R$ is a sequence of reals and $\dom{R} = \naturals$ and $R$ converge to $r$ and for all $n \in \naturals$ we have $\at{R}{n} = \apply{\at{S}{n}}{x}$. - - We show that for all $x \in \carrier[X]$ there exists $r \in \intervalclosed{a}{b}$ such that for . - + %For all $x \in \carrier[X]$ we have there exist $r,R$ such that $r \in \reals$ and $R$ is a sequence of reals and $\dom{R} = \naturals$ and $R$ converge to $r$ and for all $n \in \naturals$ we have $\at{R}{n} = \apply{\at{S}{n}}{x}$. +% + %We show that for all $x \in \carrier[X]$ there exists $r \in \intervalclosed{a}{b}$ such that for . +% \end{proof} -- cgit v1.2.3 From 29f32e2031eafa087323d79d812a1b38ac78f977 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 23 Sep 2024 01:20:05 +0200 Subject: working commit --- library/nat.tex | 22 +-- library/topology/real-topological-space.tex | 2 +- library/topology/urysohn.tex | 140 +++++++-------- library/topology/urysohn2.tex | 254 +++++++++++++++++++++------- 4 files changed, 278 insertions(+), 140 deletions(-) (limited to 'library') diff --git a/library/nat.tex b/library/nat.tex index ac9a141..841ac36 100644 --- a/library/nat.tex +++ b/library/nat.tex @@ -3,48 +3,48 @@ \section{Natural numbers} -\begin{abbreviation}\label{inductive_set} +\begin{abbreviation}\label{num_inductive_set} $A$ is an inductive set iff $\emptyset\in A$ and for all $a\in A$ we have $\suc{a}\in A$. \end{abbreviation} -\begin{axiom}\label{naturals_inductive_set} +\begin{axiom}\label{num_naturals_inductive_set} $\naturals$ is an inductive set. \end{axiom} -\begin{axiom}\label{naturals_smallest_inductive_set} +\begin{axiom}\label{num_naturals_smallest_inductive_set} Let $A$ be an inductive set. Then $\naturals\subseteq A$. \end{axiom} -\begin{abbreviation}\label{naturalnumber} +\begin{abbreviation}\label{num_naturalnumber} $n$ is a natural number iff $n\in \naturals$. \end{abbreviation} -\begin{lemma}\label{emptyset_in_naturals} +\begin{lemma}\label{num_emptyset_in_naturals} $\emptyset\in\naturals$. \end{lemma} -\begin{signature}\label{addition_is_set} +\begin{signature}\label{num_addition_is_set} $x+y$ is a set. \end{signature} -\begin{axiom}\label{addition_on_naturals} +\begin{axiom}\label{num_addition_on_naturals} $x+y$ is a natural number iff $x$ is a natural number and $y$ is a natural number. \end{axiom} -\begin{abbreviation}\label{zero_is_emptyset} +\begin{abbreviation}\label{num_zero_is_emptyset} $\zero = \emptyset$. \end{abbreviation} -\begin{axiom}\label{addition_axiom_1} +\begin{axiom}\label{num_addition_axiom_1} For all $x \in \naturals$ $x + \zero = \zero + x = x$. \end{axiom} -\begin{axiom}\label{addition_axiom_2} +\begin{axiom}\label{num_addition_axiom_2} For all $x, y \in \naturals$ $x + \suc{y} = \suc{x} + y = \suc{x+y}$. \end{axiom} -\begin{lemma}\label{naturals_is_equal_to_two_times_naturals} +\begin{lemma}\label{num_naturals_is_equal_to_two_times_naturals} $\{x+y \mid x \in \naturals, y \in \naturals \} = \naturals$. \end{lemma} diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index b2e5ea9..c76fd46 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -11,7 +11,7 @@ \import{function.tex} -\section{The canonical topology on $\mathbb{R}$} +\section{Topology Reals} \begin{definition}\label{topological_basis_reals_eps_ball} $\topoBasisReals = \{ \epsBall{x}{\epsilon} \mid x \in \reals, \epsilon \in \realsplus\}$. diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index ff6a231..ae03273 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -36,7 +36,7 @@ The first tept will be a formalisation of chain constructions. % $\overline{A_{i-1}} \subset \interior{A_{i}}$. % In this case we call the chain legal. -\begin{definition}\label{one_to_n_set} +\begin{definition}\label{urysohnone_one_to_n_set} $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. \end{definition} @@ -48,7 +48,7 @@ The first tept will be a formalisation of chain constructions. % together with the existence of an indexing function. % %%----------------------- -\begin{struct}\label{sequence} +\begin{struct}\label{urysohnone_sequence} A sequence $X$ is a onesorted structure equipped with \begin{enumerate} \item $\indexx$ @@ -57,8 +57,8 @@ The first tept will be a formalisation of chain constructions. \end{enumerate} such that \begin{enumerate} - \item\label{indexset_is_subset_naturals} $\indexxset[X] \subseteq \naturals$. - \item\label{index_is_bijection} $\indexx[X]$ is a bijection from $\indexxset[X]$ to $\carrier[X]$. + \item\label{urysohnone_indexset_is_subset_naturals} $\indexxset[X] \subseteq \naturals$. + \item\label{urysohnone_index_is_bijection} $\indexx[X]$ is a bijection from $\indexxset[X]$ to $\carrier[X]$. \end{enumerate} \end{struct} @@ -67,12 +67,12 @@ The first tept will be a formalisation of chain constructions. -\begin{definition}\label{cahin_of_subsets} +\begin{definition}\label{urysohnone_cahin_of_subsets} $C$ is a chain of subsets iff $C$ is a sequence and for all $n,m \in \indexxset[C]$ such that $n < m$ we have $\indexx[C](n) \subseteq \indexx[C](m)$. \end{definition} -\begin{definition}\label{chain_of_n_subsets} +\begin{definition}\label{urysohnone_chain_of_n_subsets} $C$ is a chain of $n$ subsets iff $C$ is a chain of subsets and $n \in \indexxset[C]$ and for all $m \in \naturals$ such that $m \leq n$ we have $m \in \indexxset[C]$. @@ -84,7 +84,7 @@ The first tept will be a formalisation of chain constructions. % and also for the subproof of continuity of the limit. -% \begin{definition}\label{legal_chain} +% \begin{definition}\label{urysohnone_legal_chain} % $C$ is a legal chain of subsets of $X$ iff % $C \subseteq \pow{X}$. %and % %there exist $f \in \funs{C}{\naturals}$ such that @@ -106,49 +106,49 @@ The first tept will be a formalisation of chain constructions. \subsection{staircase function} -\begin{definition}\label{minimum} +\begin{definition}\label{urysohnone_minimum} $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. \end{definition} -\begin{definition}\label{maximum} +\begin{definition}\label{urysohnone_maximum} $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. \end{definition} -\begin{definition}\label{intervalclosed} +\begin{definition}\label{urysohnone_intervalclosed} $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. \end{definition} -\begin{definition}\label{intervalopen} +\begin{definition}\label{urysohnone_intervalopen} $\intervalopen{a}{b} = \{ x \in \reals \mid a < x < b\}$. \end{definition} -\begin{struct}\label{staircase_function} +\begin{struct}\label{urysohnone_staircase_function} A staircase function $f$ is a onesorted structure equipped with \begin{enumerate} \item $\chain$ \end{enumerate} such that \begin{enumerate} - \item \label{staircase_is_function} $f$ is a function to $\intervalclosed{\zero}{1}$. - \item \label{staircase_domain} $\dom{f}$ is a topological space. - \item \label{staricase_def_chain} $C$ is a chain of subsets. - \item \label{staircase_chain_is_in_domain} for all $x \in C$ we have $x \subseteq \dom{f}$. - \item \label{staircase_behavoir_index_zero} $f(\indexx[C](1))= 1$. - \item \label{staircase_behavoir_index_n} $f(\dom{f}\setminus \unions{C}) = \zero$. - \item \label{staircase_chain_indeset} There exist $n$ such that $\indexxset[C] = \seq{\zero}{n}$. - \item \label{staircase_behavoir_index_arbetrray} for all $n \in \indexxset[C]$ + \item \label{urysohnone_staircase_is_function} $f$ is a function to $\intervalclosed{\zero}{1}$. + \item \label{urysohnone_staircase_domain} $\dom{f}$ is a topological space. + \item \label{urysohnone_staricase_def_chain} $C$ is a chain of subsets. + \item \label{urysohnone_staircase_chain_is_in_domain} for all $x \in C$ we have $x \subseteq \dom{f}$. + \item \label{urysohnone_staircase_behavoir_index_zero} $f(\indexx[C](1))= 1$. + \item \label{urysohnone_staircase_behavoir_index_n} $f(\dom{f}\setminus \unions{C}) = \zero$. + \item \label{urysohnone_staircase_chain_indeset} There exist $n$ such that $\indexxset[C] = \seq{\zero}{n}$. + \item \label{urysohnone_staircase_behavoir_index_arbetrray} for all $n \in \indexxset[C]$ such that $n \neq \zero$ we have $f(\indexx[C](n) \setminus \indexx[C](n-1)) = \rfrac{n}{ \max{\indexxset[C]} }$. \end{enumerate} \end{struct} -\begin{definition}\label{legal_staircase} +\begin{definition}\label{urysohnone_legal_staircase} $f$ is a legal staircase function iff $f$ is a staircase function and for all $n,m \in \indexxset[\chain[f]]$ such that $n \leq m$ we have $f(\indexx[\chain[f]](n)) \leq f(\indexx[\chain[f]](m))$. \end{definition} -\begin{abbreviation}\label{urysohnspace} +\begin{abbreviation}\label{urysohnone_urysohnspace} $X$ is a urysohn space iff $X$ is a topological space and for all $A,B \in \closeds{X}$ such that $A \inter B = \emptyset$ @@ -156,49 +156,49 @@ The first tept will be a formalisation of chain constructions. such that $A \subseteq A'$ and $B \subseteq B'$ and $A' \inter B' = \emptyset$. \end{abbreviation} -\begin{definition}\label{urysohnchain} +\begin{definition}\label{urysohnone_urysohnchain} $C$ is a urysohnchain in $X$ of cardinality $k$ iff %<---- TODO: cardinality unused! $C$ is a chain of subsets and for all $A \in C$ we have $A \subseteq X$ and for all $n,m \in \indexxset[C]$ such that $n < m$ we have $\closure{\indexx[C](n)}{X} \subseteq \interior{\indexx[C](m)}{X}$. \end{definition} -\begin{definition}\label{urysohnchain_without_cardinality} +\begin{definition}\label{urysohnone_urysohnchain_without_cardinality} $C$ is a urysohnchain in $X$ iff $C$ is a chain of subsets and for all $A \in C$ we have $A \subseteq X$ and for all $n,m \in \indexxset[C]$ such that $n < m$ we have $\closure{\indexx[C](n)}{X} \subseteq \interior{\indexx[C](m)}{X}$. \end{definition} -\begin{abbreviation}\label{infinte_sequence} +\begin{abbreviation}\label{urysohnone_infinte_sequence} $S$ is a infinite sequence iff $S$ is a sequence and $\indexxset[S]$ is infinite. \end{abbreviation} -\begin{definition}\label{infinite_product} +\begin{definition}\label{urysohnone_infinite_product} $X$ is the infinite product of $Y$ iff $X$ is a infinite sequence and for all $i \in \indexxset[X]$ we have $\indexx[X](i) = Y$. \end{definition} -\begin{definition}\label{refinmant} +\begin{definition}\label{urysohnone_refinmant} $C'$ is a refinmant of $C$ iff $C'$ is a urysohnchain in $X$ and for all $x \in C$ we have $x \in C'$ and for all $y \in C$ such that $y \subset x$ we have there exist $c \in C'$ such that $y \subset c \subset x$ and for all $g \in C'$ such that $g \neq c$ we have not $y \subset g \subset x$. \end{definition} -\begin{abbreviation}\label{two} +\begin{abbreviation}\label{urysohnone_two} $\two = \suc{1}$. \end{abbreviation} -\begin{lemma}\label{two_in_reals} +\begin{lemma}\label{urysohnone_two_in_reals} $\two \in \reals$. \end{lemma} -\begin{lemma}\label{two_in_naturals} +\begin{lemma}\label{urysohnone_two_in_naturals} $\two \in \naturals$. \end{lemma} -\begin{inductive}\label{power_of_two} +\begin{inductive}\label{urysohnone_power_of_two} Define $\powerOfTwoSet \subseteq (\naturals \times \naturals)$. \begin{enumerate} \item $(\zero, 1) \in \powerOfTwoSet$. @@ -206,45 +206,45 @@ The first tept will be a formalisation of chain constructions. \end{enumerate} \end{inductive} -\begin{abbreviation}\label{pot} +\begin{abbreviation}\label{urysohnone_pot} $\pot = \powerOfTwoSet$. \end{abbreviation} -\begin{lemma}\label{dom_pot} +\begin{lemma}\label{urysohnone_dom_pot} $\dom{\pot} = \naturals$. \end{lemma} \begin{proof} Omitted. \end{proof} -\begin{lemma}\label{ran_pot} +\begin{lemma}\label{urysohnone_ran_pot} $\ran{\pot} \subseteq \naturals$. \end{lemma} -\begin{axiom}\label{pot1} +\begin{axiom}\label{urysohnone_pot1} $\pot \in \funs{\naturals}{\naturals}$. \end{axiom} -\begin{axiom}\label{pot2} +\begin{axiom}\label{urysohnone_pot2} For all $n \in \naturals$ we have there exist $k\in \naturals$ such that $(n, k) \in \powerOfTwoSet$ and $\apply{\pot}{n}=k$. %$\pot(n) = k$ iff there exist $x \in \powerOfTwoSet$ such that $x = (n,k)$. \end{axiom} %Without this abbreviation \pot cant be sed as a function in the standard sense -\begin{abbreviation}\label{pot_as_function} +\begin{abbreviation}\label{urysohnone_pot_as_function} $\pot(n) = \apply{\pot}{n}$. \end{abbreviation} %Take all points, besids one but then take all open sets not containing x but all other, so \{x\} has to be closed -\begin{axiom}\label{hausdorff_implies_singltons_closed} +\begin{axiom}\label{urysohnone_hausdorff_implies_singltons_closed} For all $X$ such that $X$ is Hausdorff we have for all $x \in \carrier[X]$ we have $\{x\}$ is closed in $X$. \end{axiom} -\begin{lemma}\label{urysohn_set_in_between} +\begin{lemma}\label{urysohnone_urysohn_set_in_between} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \subset B$. @@ -284,7 +284,7 @@ The first tept will be a formalisation of chain constructions. \end{proof} -\begin{proposition}\label{urysohnchain_induction_begin} +\begin{proposition}\label{urysohnone_urysohnchain_induction_begin} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. @@ -347,7 +347,7 @@ The first tept will be a formalisation of chain constructions. \end{proof} -\begin{proposition}\label{urysohnchain_induction_begin_step_two} +\begin{proposition}\label{urysohnone_urysohnchain_induction_begin_step_two} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. @@ -364,7 +364,7 @@ The first tept will be a formalisation of chain constructions. -\begin{proposition}\label{t_four_propositon} +\begin{proposition}\label{urysohnone_t_four_propositon} Let $X$ be a urysohn space. Then for all $A,B \subseteq X$ such that $\closure{A}{X} \subseteq \interior{B}{X}$ we have there exists $C \subseteq X$ such that @@ -376,7 +376,7 @@ The first tept will be a formalisation of chain constructions. -\begin{proposition}\label{urysohnchain_induction_step_existence} +\begin{proposition}\label{urysohnone_urysohnchain_induction_step_existence} Let $X$ be a urysohn space. Suppose $U$ is a urysohnchain in $X$. Then there exist $U'$ such that $U'$ is a refinmant of $U$ and $U'$ is a urysohnchain in $X$. @@ -390,7 +390,7 @@ The first tept will be a formalisation of chain constructions. % such that $\closure{\indexx[U](n)}{X} \subseteq \interior{C}{X} \subseteq \closure{C}{X} \subseteq \interior{\indexx[U](n+1)}{X}$. - %\begin{definition}\label{refinmant} + %\begin{definition}\label{urysohnone_refinmant} % $C'$ is a refinmant of $C$ iff for all $x \in C$ we have $x \in C'$ and % for all $y \in C$ such that $y \subset x$ % we have there exist $c \in C'$ such that $y \subset c \subset x$ @@ -404,7 +404,7 @@ The first tept will be a formalisation of chain constructions. -\begin{proposition}\label{existence_of_staircase_function} +\begin{proposition}\label{urysohnone_existence_of_staircase_function} Let $X$ be a urysohn space. Suppose $U$ is a urysohnchain in $X$ and $U$ has cardinality $k$. Suppose $k \neq \zero$. @@ -416,7 +416,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{abbreviation}\label{refinment_abbreviation} +\begin{abbreviation}\label{urysohnone_refinment_abbreviation} $x \refine y$ iff $x$ is a refinmant of $y$. \end{abbreviation} @@ -424,27 +424,27 @@ The first tept will be a formalisation of chain constructions. -\begin{abbreviation}\label{sequence_of_functions} +\begin{abbreviation}\label{urysohnone_sequence_of_functions} $f$ is a sequence of functions iff $f$ is a sequence and for all $g \in \carrier[f]$ we have $g$ is a function. \end{abbreviation} -\begin{abbreviation}\label{sequence_in_reals} +\begin{abbreviation}\label{urysohnone_sequence_in_reals} $s$ is a sequence of real numbers iff $s$ is a sequence and for all $r \in \carrier[s]$ we have $r \in \reals$. \end{abbreviation} -\begin{axiom}\label{abs_behavior1} +\begin{axiom}\label{urysohnone_abs_behavior1} If $x \geq \zero$ then $\abs{x} = x$. \end{axiom} -\begin{axiom}\label{abs_behavior2} +\begin{axiom}\label{urysohnone_abs_behavior2} If $x < \zero$ then $\abs{x} = \neg{x}$. \end{axiom} -\begin{abbreviation}\label{converge} +\begin{abbreviation}\label{urysohnone_converge} $s$ converges iff $s$ is a sequence of real numbers and $\indexxset[s]$ is infinite and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ we have @@ -454,7 +454,7 @@ The first tept will be a formalisation of chain constructions. \end{abbreviation} -\begin{definition}\label{limit_of_sequence} +\begin{definition}\label{urysohnone_limit_of_sequence} $x$ is the limit of $s$ iff $s$ is a sequence of real numbers and $x \in \reals$ and for all $\epsilon \in \reals$ such that $\epsilon > \zero$ @@ -463,7 +463,7 @@ The first tept will be a formalisation of chain constructions. we have $\abs{x - \indexx[s](n)} < \epsilon$. \end{definition} -\begin{proposition}\label{existence_of_limit} +\begin{proposition}\label{urysohnone_existence_of_limit} Let $s$ be a sequence of real numbers. Then $s$ converges iff there exist $x \in \reals$ such that $x$ is the limit of $s$. @@ -472,22 +472,22 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{definition}\label{limit_sequence} +\begin{definition}\label{urysohnone_limit_sequence} $x$ is the limit sequence of $f$ iff $x$ is a sequence and $\indexxset[x] = \dom{f}$ and for all $n \in \indexxset[x]$ we have $\indexx[x](n) = f(n)$. \end{definition} -\begin{definition}\label{realsminus} +\begin{definition}\label{urysohnone_realsminus} $\realsminus = \{r \in \reals \mid r < \zero\}$. \end{definition} -\begin{abbreviation}\label{realsplus} +\begin{abbreviation}\label{urysohnone_realsplus} $\realsplus = \reals \setminus \realsminus$. \end{abbreviation} -\begin{proposition}\label{intervalclosed_subseteq_reals} +\begin{proposition}\label{urysohnone_intervalclosed_subseteq_reals} Suppose $a,b \in \reals$. Suppose $a < b$. Then $\intervalclosed{a}{b} \subseteq \reals$. @@ -495,7 +495,7 @@ The first tept will be a formalisation of chain constructions. -\begin{lemma}\label{fraction1} +\begin{lemma}\label{urysohnone_fraction1} Let $x \in \reals$. Then for all $y \in \reals$ such that $x \neq y$ we have there exists $r \in \rationals$ such that $y < r < x$ or $x < r < y$. \end{lemma} @@ -503,7 +503,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{lemma}\label{frection2} +\begin{lemma}\label{urysohnone_frection2} Suppose $a,b \in \reals$. Suppose $a < b$. Then $\intervalopen{a}{b}$ is infinite. @@ -512,7 +512,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{lemma}\label{frection3} +\begin{lemma}\label{urysohnone_frection3} Suppose $a \in \reals$. Suppose $a < \zero$. Then there exist $N \in \naturals$ such that for all $N' \in \naturals$ such that $N' > N$ we have $\zero < \rfrac{1}{\pot(N')} < a$. @@ -521,7 +521,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{proposition}\label{fraction4} +\begin{proposition}\label{urysohnone_fraction4} Suppose $a,b,\epsilon \in \reals$. Suppose $\epsilon > \zero$. $\abs{a - b} < \epsilon$ iff $b \in \intervalopen{(a - \epsilon)}{(a + \epsilon)}$. @@ -530,7 +530,7 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{proposition}\label{fraction5} +\begin{proposition}\label{urysohnone_fraction5} Suppose $a,b,\epsilon \in \reals$. Suppose $\epsilon > \zero$. $b \in \intervalopen{(a - \epsilon)}{(a + \epsilon)}$ iff $a \in \intervalopen{(b - \epsilon)}{(b + \epsilon)}$. @@ -539,17 +539,17 @@ The first tept will be a formalisation of chain constructions. Omitted. \end{proof} -\begin{proposition}\label{fraction6} +\begin{proposition}\label{urysohnone_fraction6} Suppose $a,\epsilon \in \reals$. Suppose $\epsilon > \zero$. $\intervalopen{(a - \epsilon)}{(a + \epsilon)} = \{r \in \reals \mid (a - \epsilon) < r < (a + \epsilon)\} $. \end{proposition} -\begin{abbreviation}\label{epsilonball} +\begin{abbreviation}\label{urysohnone_epsilonball} $\epsBall{a}{\epsilon} = \intervalopen{(a - \epsilon)}{(a + \epsilon)}$. \end{abbreviation} -\begin{proposition}\label{fraction7} +\begin{proposition}\label{urysohnone_fraction7} Suppose $a,\epsilon \in \reals$. Suppose $\epsilon > \zero$. Then there exist $b \in \rationals$ such that $b \in \epsBall{a}{\epsilon}$. @@ -561,11 +561,11 @@ The first tept will be a formalisation of chain constructions. -%\begin{definition}\label{sequencetwo} +%\begin{definition}\label{urysohnone_sequencetwo} % $Z$ is a sequencetwo iff $Z = (N,f,B)$ and $N \subseteq \naturals$ and $f$ is a bijection from $N$ to $B$. %\end{definition} % -%\begin{proposition}\label{sequence_existence} +%\begin{proposition}\label{urysohnone_sequence_existence} % Suppose $N \subseteq \naturals$. % Suppose $M \subseteq \naturals$. % Suppose $N = M$. @@ -586,7 +586,7 @@ The first tept will be a formalisation of chain constructions. -\begin{theorem}\label{urysohn} +\begin{theorem}\label{urysohnone_urysohn1} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. @@ -599,7 +599,7 @@ The first tept will be a formalisation of chain constructions. There exist $\eta$ such that $\carrier[\eta] = \{A, (\carrier[X] \setminus B)\}$ and $\indexxset[\eta] = \{\zero, 1\}$ and $\indexx[\eta](\zero) = A$ - and $\indexx[\eta](1) = (\carrier[X] \setminus B)$ by \cref{urysohnchain_induction_begin}. + and $\indexx[\eta](1) = (\carrier[X] \setminus B)$ by \cref{urysohnone_urysohnchain_induction_begin}. We show that there exist $\zeta$ such that $\zeta$ is a sequence and $\indexxset[\zeta] = \naturals$ @@ -919,6 +919,6 @@ The first tept will be a formalisation of chain constructions. % \end{subproof} \end{proof} % -%\begin{theorem}\label{safe} +%\begin{theorem}\label{urysohnone_safe} % Contradiction. %\end{theorem} diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index ce6d742..08841da 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -40,7 +40,7 @@ \begin{definition}\label{chain_of_subsets} - $X$ is a chain of subsets in $Y$ iff $X$ is a sequence and for all $n \in \dom{X}$ we have $\at{X}{n} \subseteq \carrier[Y]$ and for all $m \in \dom{X}$ such that $m > n$ we have $\at{X}{n} \subseteq \at{X}{m}$. + $X$ is a chain of subsets in $Y$ iff $X$ is a sequence and for all $n \in \dom{X}$ we have $\at{X}{n} \subseteq \carrier[Y]$ and for all $m \in \dom{X}$ such that $n < m$ we have $\at{X}{n} \subseteq \at{X}{m}$. \end{definition} @@ -49,11 +49,11 @@ \end{definition} \begin{definition}\label{urysohn_finer_set} - $A$ is finer between $X$ to $Y$ in $U$ iff $\closure{X}{U} \subseteq \interior{A}{U}$ and $\closure{A}{U} \subseteq \interior{Y}{U}$. + $A$ is finer between $B$ to $C$ in $X$ iff $\closure{B}{X} \subseteq \interior{A}{X}$ and $\closure{A}{X} \subseteq \interior{C}{X}$. \end{definition} \begin{definition}\label{finer} %<-- verfeinerung - $Y$ is finer then $X$ in $U$ iff for all $n \in \dom{X}$ we have $\at{X}{n} \in \ran{Y}$ and for all $m \in \dom{X}$ such that $n < m$ we have there exist $k \in \dom{Y}$ such that $\at{Y}{k}$ is finer between $\at{X}{n}$ to $\at{X}{m}$ in $U$. + $A$ is finer then $B$ in $X$ iff for all $n \in \dom{B}$ we have $\at{B}{n} \in \ran{A}$ and for all $m \in \dom{B}$ such that $n < m$ we have there exist $k \in \dom{A}$ such that $\at{A}{k}$ is finer between $\at{B}{n}$ to $\at{B}{m}$ in $X$. \end{definition} \begin{definition}\label{follower_index} @@ -92,6 +92,46 @@ $f$ is consistent on $X$ to $Y$ iff $f$ is a bijection from $\dom{X}$ to $\dom{Y}$ and for all $n,m \in \dom{X}$ such that $n < m$ we have $f(n) < f(m)$. \end{definition} + +%\begin{definition}\label{staircase} +% $f$ is a staircase function adapted to $U$ in $X$ iff $U$ is a urysohnchain of $X$ and $f$ is a function from $\carrier[X]$ to $\reals$ and there exist $k \in \naturals$ such that $k = \max{\dom{U}}$ and for all $x,y \in \carrier[X]$ such that $y \in \carrier[X] \setminus \at{U}{k}$ and $x \in \at{U}{k}$ we have $f(y) = 1$ and there exist $n,m \in \dom{U}$ such that $n$ follows $m$ in $\dom{U}$ and $x \in (\at{U}{m} \setminus \at{U}{n})$ and $f(x)= \rfrac{m}{k}$. +%\end{definition} + + +\begin{definition}\label{staircase_step_value1} + $a$ is the staircase step value at $y$ of $U$ in $X$ iff there exist $n,m \in \dom{U}$ such that $n$ follows $m$ in $\dom{U}$ and $y \in \closure{\at{U}{n}}{X} \setminus \closure{\at{U}{m}}{X}$ and $a = \rfrac{n}{\max{\dom{U}}}$. +\end{definition} + +\begin{definition}\label{staircase_step_value2} + $a$ is the staircase step valuetwo at $y$ of $U$ in $X$ iff either if $y \in (\carrier[X] \setminus \closure{\at{U}{\max{\dom{U}}}}{X})$ then $a = 1$ or $a$ is the staircase step valuethree at $y$ of $U$ in $X$. +\end{definition} + +\begin{definition}\label{staircase_step_value3} + $a$ is the staircase step valuethree at $y$ of $U$ in $X$ iff if $y \in \closure{\at{U}{\min{\dom{U}}}}{X}$ then $f(z) = \zero$. +\end{definition} + + +\begin{definition}\label{staircase2} + $f$ is a staircase function adapted to $U$ in $X$ iff $U$ is a urysohnchain of $X$ and $f$ is a function from $\carrier[X]$ to $\reals$ and for all $y \in \carrier[X]$ we have either $f(y)$ is the staircase step value at $y$ of $U$ in $X$ or $f(y)$ is the staircase step valuetwo at $y$ of $U$ in $X$. +\end{definition} + +\begin{definition}\label{staircase_sequence} + $S$ is staircase sequence of $U$ in $X$ iff $S$ is a sequence and $U$ is a lifted urysohnchain of $X$ and $\dom{U} = \dom{S}$ and for all $n \in \dom{U}$ we have $\at{S}{n}$ is a staircase function adapted to $\at{U}{n}$ in $X$. +\end{definition} + +\begin{definition}\label{staircase_limit_point} + $x$ is the staircase limit of $S$ with $y$ iff $x \in \reals$ and for all $\epsilon \in \realsplus$ there exist $n_0 \in \naturals$ such that for all $n \in \naturals$ such that $n_0 \rless n$ we have $\apply{\at{S}{n}}{y} \in \epsBall{x}{\epsilon}$. +\end{definition} + +%\begin{definition}\label{staircase_limit_function} +% $f$ is a limit function of a staircase $S$ iff $S$ is staircase sequence of $U$ and $U$ is a lifted urysohnchain of $X$ and $\dom{f} = \carrier[X]$ and for all $x \in \dom{f}$ we have $f(x)$ is the staircase limit of $S$ with $x$ and $f$ is a function from $\carrier[X]$ to $\reals$. +%\end{definition} +% +\begin{definition}\label{staircase_limit_function} + $f$ is the limit function of staircase $S$ together with $U$ and $X$ iff $S$ is staircase sequence of $U$ in $X$ and $U$ is a lifted urysohnchain of $X$ and $\dom{f} = \carrier[X]$ and for all $x \in \dom{f}$ we have $f(x)$ is the staircase limit of $S$ with $x$ and $f$ is a function from $\carrier[X]$ to $\reals$. +\end{definition} + + \begin{proposition}\label{naturals_in_transitive} $\naturals$ is a \in-transitive set. \end{proposition} @@ -659,33 +699,26 @@ \end{proof} -\begin{definition}\label{staircase} - $f$ is a staircase function adapted to $U$ in $X$ iff $U$ is a urysohnchain of $X$ and for all $x,n,m,k$ such that $k = \max{\dom{U}}$ and $n,m \in \dom{U}$ and $n$ follows $m$ in $\dom{U}$ and $x \in (\at{U}{m} \setminus \at{U}{n})$ we have $f(x)= \rfrac{m}{k}$. -\end{definition} - -\begin{definition}\label{staircase_sequence} - $S$ is staircase sequence of $U$ iff $S$ is a sequence and $U$ is a lifted urysohnchain of $X$ and $\dom{U} = \dom{S}$ and for all $n \in \dom{U}$ we have $\at{S}{n}$ is a staircase function adapted to $\at{U}{n}$ in $U$. -\end{definition} - -\begin{definition}\label{staircase_limit_point} - $x$ is the staircase limit of $S$ with $y$ iff $x \in \reals$ and for all $\epsilon \in \realsplus$ there exist $n_0 \in \naturals$ such that for all $n \in \naturals$ such that $n_0 \rless n$ we have $\apply{\at{S}{n}}{y} \in \epsBall{x}{\epsilon}$. -\end{definition} - -\begin{definition}\label{staircase_limit_function} - $f$ is a limit function of a staircase $S$ iff $S$ is staircase sequence of $U$ and $U$ is a lifted urysohnchain of $X$ and $\dom{f} = \carrier[X]$ and for all $x \in \dom{f}$ we have $f(x)$ is the staircase limit of $S$ with $x$ and $f$ is a function from $\carrier[X]$ to $\reals$. -\end{definition} +\begin{proposition}\label{staircase_ran_in_zero_to_one} + Let $X$ be a urysohn space. + Suppose $U$ is a urysohnchain of $X$. + Suppose $f$ is a staircase function adapted to $U$ in $X$. + Then $\ran{f} \subseteq \intervalclosed{\zero}{1}$. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} -%\begin{definition}\label{staircase_limit_function} -% $f$ is a limit function of staircase $S$ iff $S$ is staircase sequence of $U$ and $U$ is a lifted urysohnchain of $X$ and $\dom{f} = \carrier[X]$ and for all $x \in \dom{f}$ we have $f(x)$ is the staircase limit of $S$ with $x$ and $f$ is a function from $\carrier[X]$ to $\reals$. -%\end{definition} -% -%\begin{proposition}\label{staircase_limit_is_continuous} -% Suppose $X$ is a urysohnspace. -% Suppose $U$ is a lifted urysohnchain of $X$. -% Suppose $S$ is staircase sequence of $U$. -% Suppose $f$ is the limit function of a staircase $S$. -% Then $f$ is continuous. -%\end{proposition} +\begin{proposition}\label{staircase_limit_is_continuous} + Let $X$ be a urysohn space. + Suppose $U$ is a lifted urysohnchain of $X$. + Suppose $S$ is staircase sequence of $U$ in $X$. + Suppose $f$ is the limit function of staircase $S$ together with $U$ and $X$. + Then $f$ is continuous. +\end{proposition} +\begin{proof} + Omitted. +\end{proof} \begin{theorem}\label{urysohnsetinbeetween} Let $X$ be a urysohn space. @@ -712,8 +745,26 @@ Omitted. \end{proof} +\begin{lemma}\label{fractions_between_zero_one} + Suppose $n,m \in \naturals$. + Suppose $m > n$. + Then $\zero \leq \rfrac{n}{m} \leq 1$. +\end{lemma} +\begin{proof} + Omitted. +\end{proof} +\begin{lemma}\label{intervalclosed_border_is_elem} + Suppose $a,b \in \reals$. + Suppose $a < b$. + Then $a,b \in \intervalclosed{a}{b}$. +\end{lemma} +\begin{lemma}\label{urysohnchain_subseteqrel} + Let $X$ be a urysohn space. + Suppose $U$ is a urysohnchain of $X$. + Then for all $n,m \in \dom{U}$ such that $n < m$ we have $\at{U}{n} \subseteq \at{U}{m}$. +\end{lemma} \begin{theorem}\label{urysohn} @@ -721,8 +772,8 @@ Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. Suppose $\carrier[X]$ is inhabited. - There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ - and $f(A) = \zero$ and $f(B)= 1$ and $f$ is continuous. + There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and $f$ is continuous + and for all $a,b$ such that $a \in A$ and $b \in B$ we have $f(a)= \zero$ and $f(b) = 1$. \end{theorem} \begin{proof} Let $X' = \carrier[X]$. @@ -796,46 +847,133 @@ \end{subproof} Take $U$ such that $U$ is a lifted urysohnchain of $X$ and $\at{U}{\zero} = U_0$. - We show that there exist $S$ such that $S$ is staircase sequence of $U$. + We show that there exist $S$ such that $S$ is staircase sequence of $U$ in $X$. \begin{subproof} Omitted. \end{subproof} - Take $S$ such that $S$ is staircase sequence of $U$. + Take $S$ such that $S$ is staircase sequence of $U$ in $X$. %For all $x \in \carrier[X]$ we have there exist $r,R$ such that $r \in \reals$ and $R$ is a sequence of reals and $\dom{R} = \naturals$ and $R$ converge to $r$ and for all $n \in \naturals$ we have $\at{R}{n} = \apply{\at{S}{n}}{x}$. % %We show that for all $x \in \carrier[X]$ there exists $r \in \intervalclosed{a}{b}$ such that for . -% + We show that there exist $f$ such that $f$ is the limit function of staircase $S$ together with $U$ and $X$. + \begin{subproof} + Omitted. + \end{subproof} + Take $f$ such that $f$ is the limit function of staircase $S$ together with $U$ and $X$. + Then $f$ is continuous. + We show that $\dom{f} = \carrier[X]$. + \begin{subproof} + Trivial. + \end{subproof} + $f$ is a function. + We show that $\ran{f} \subseteq \intervalclosed{\zero}{1}$. + \begin{subproof} + It suffices to show that $f$ is a function to $\intervalclosed{\zero}{1}$. + It suffices to show that for all $x \in \dom{f}$ we have $f(x) \in \intervalclosed{\zero}{1}$. + Fix $x \in \dom{f}$. + $f(x)$ is the staircase limit of $S$ with $x$. + Therefore $f(x) \in \reals$. + + We show that for all $n \in \naturals$ we have $\apply{\at{S}{n}}{x} \in \intervalclosed{\zero}{1}$. + \begin{subproof} + Fix $n \in \naturals$. + Let $g = \at{S}{n}$. + Let $U' = \at{U}{n}$. + $\at{S}{n}$ is a staircase function adapted to $\at{U}{n}$ in $X$. + $g$ is a staircase function adapted to $U'$ in $X$. + $U'$ is a urysohnchain of $X$. + $g$ is a function from $\carrier[X]$ to $\reals$. + It suffices to show that $\ran{g} \subseteq \intervalclosed{\zero}{1}$ by \cref{function_apply_default,reals_axiom_zero_in_reals,intervalclosed,one_is_positiv,function_apply_elim,inter,inter_absorb_supseteq_left,ran_iff,funs_is_relation,funs_is_function,staircase2}. + It suffices to show that for all $x \in \dom{g}$ we have $g(x) \in \intervalclosed{\zero}{1}$. + Fix $x\in \dom{g}$. + Then $x \in \carrier[X]$. + \begin{byCase} + \caseOf{$x \in (\carrier[X] \setminus \closure{\at{U'}{\max{\dom{U'}}}}{X})$.} + Therefore $x \notin \closure{\at{U'}{\max{\dom{U'}}}}{X}$. + Therefore $x \notin \closure{\at{U'}{\min{\dom{U'}}}}{X}$. + Therefore $x \notin (\closure{\at{U'}{\max{\dom{U'}}}}{X}\setminus \closure{\at{U'}{\min{\dom{U'}}}}{X})$. + Then $g(x) = 1$ . + \caseOf{$x \in \closure{\at{U'}{\max{\dom{U'}}}}{X}$.} + \begin{byCase} + \caseOf{$x \in \closure{\at{U'}{\min{\dom{U'}}}}{X}$.} + $g(x) = \zero$. + \caseOf{$x \in (\closure{\at{U'}{\max{\dom{U'}}}}{X}\setminus \closure{\at{U'}{\min{\dom{U'}}}}{X})$.} + Then $g(x)$ is the staircase step value at $x$ of $U'$ in $X$. + \end{byCase} + \end{byCase} + + + + %$\at{S}{n}$ is a staircase function adapted to $\at{U}{n}$ in $X$. + %$\at{U}{n}$ is a urysohnchain of $X$. + %$\at{S}{n}$ is a function from $\carrier[X]$ to $\reals$. + %there exist $k \in \naturals$ such that $k = \max{\dom{\at{U}{n}}}$. + %Take $k \in \naturals$ such that $k = \max{\dom{\at{U}{n}}}$. + %\begin{byCase} + % \caseOf{$x \in \carrier[X] \setminus \at{\at{U}{n}}{k}$.} + % $1 \in \intervalclosed{\zero}{1}$. + % We show that for all $y \in (\carrier[X] \setminus \at{\at{U}{n}}{k})$ we have $\apply{\at{S}{n}}{y} = 1$. + % \begin{subproof} + % Omitted. + % \end{subproof} + % Then $\apply{\at{S}{n}}{x} = 1$. + % \caseOf{$x \notin \carrier[X] \setminus \at{\at{U}{n}}{k}$.} + % %There exist $n',m' \in \dom{\at{U}{n}}$ such that $n'$ follows $m'$ in $\dom{\at{U}{n}}$ and $x \in (\at{\at{U}{n}}{n'} \setminus \at{\at{U}{n}}{m'})$. + % Take $n',m' \in \dom{\at{U}{n}}$ such that $n'$ follows $m'$ in $\dom{\at{U}{n}}$ and $x \in (\at{\at{U}{n}}{n'} \setminus \at{\at{U}{n}}{m'})$. + % Then $\apply{\at{S}{n}}{x} = \rfrac{m'}{k'}$. + % It suffices to show that $\rfrac{m'}{k'} \in \intervalclosed{\zero}{1}$. + % $\zero \leq m' \leq k$. + %\end{byCase} + %%It suffices to show that $\zero \leq \apply{\at{S}{n}}{x} \leq 1$. + %%It suffices to show that $\ran{\at{S}{n}} \subseteq \intervalclosed{\zero}{1}$. + \end{subproof} + + Suppose not. + Then $f(x) < \zero$ or $f(x) > 1$ by \cref{reals_order_total,reals_axiom_zero_in_reals,intervalclosed,one_is_positiv,one_in_reals}. + For all $\epsilon \in \realsplus$ we have there exist $m \in \naturals$ such that $\apply{\at{S}{m}}{x} \in \epsBall{f(x)}{\epsilon}$ by \cref{plus_one_order,naturals_is_equal_to_two_times_naturals,subseteq,naturals_subseteq_reals,staircase_limit_point}. + \begin{byCase} + \caseOf{$f(x) < \zero$.} + Let $\delta = \zero - f(x)$. + $\delta \in \realsplus$. + It suffices to show that for all $n \in \naturals$ we have $\apply{\at{S}{n}}{x} \notin \epsBall{f(x)}{\delta}$. + Fix $n \in \naturals$. + $\at{S}{n}$ is a staircase function adapted to $\at{U}{n}$ in $X$. + For all $y \in \epsBall{f(x)}{\delta}$ we have $y < \zero$ by \cref{epsilon_ball,minus_behavior1,minus_behavior3,minus,apply,intervalopen}. + It suffices to show that $\apply{\at{S}{n}}{x} \in \intervalclosed{\zero}{1}$. + Trivial. + \caseOf{$f(x) > 1$.} + Let $\delta = f(x) - 1$. + $\delta \in \realsplus$. + It suffices to show that for all $n \in \naturals$ we have $\apply{\at{S}{n}}{x} \notin \epsBall{f(x)}{\delta}$. + Fix $n \in \naturals$. + $\at{S}{n}$ is a staircase function adapted to $\at{U}{n}$ in $X$. + For all $y \in \epsBall{f(x)}{\delta}$ we have $y > 1$ by \cref{epsilon_ball,reals_addition_minus_behavior2,minus_in_reals,apply,reals_addition_minus_behavior1,minus,reals_add,realsplus_in_reals,one_in_reals,reals_axiom_kommu,intervalopen}. + It suffices to show that $\apply{\at{S}{n}}{x} \in \intervalclosed{\zero}{1}$. + Trivial. + \end{byCase} + + \end{subproof} + Therefore $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ by \cref{staircase_limit_function,surj_to_fun,fun_to_surj,neq_witness,inters_of_ordinals_elem,times_tuple_elim,img_singleton_iff,foundation,subseteq_emptyset_iff,inter_eq_left_implies_subseteq,inter_emptyset,funs_intro,fun_ran_iff,not_in_subseteq}. + + We show that for all $a \in A$ we have $f(a) = \zero$. + \begin{subproof} + Omitted. + \end{subproof} + We show that for all $b \in B$ we have $f(b) = 1$. + \begin{subproof} + Omitted. + \end{subproof} + - \end{proof} -\begin{theorem}\label{safe} - Contradiction. -\end{theorem} +%\begin{theorem}\label{safe} +% Contradiction. +%\end{theorem} -% -%Ideen: -%Eine folge ist ein Funktion mit domain \subseteq Natürlichenzahlen. als predicat -% -%zulässig und verfeinerung von ketten als predicat definieren. -% -%limits und punkt konvergenz als prädikat. -% -% -%Vor dem Beweis vor dem eigentlichen Beweis. -%die abgeleiteten Funktionen -% -%\derivedstiarcasefunction on A -% -%abbreviation: \at{f}{n} = f_{n} -% -% -%TODO: -%Reals ist ein topologischer Raum -% -- cgit v1.2.3 From f6b22fd533bd61e9dbcb6374295df321de99b1f2 Mon Sep 17 00:00:00 2001 From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> Date: Mon, 23 Sep 2024 03:05:41 +0200 Subject: Abgabe --- library/algebra/group.tex | 2 +- library/algebra/monoid.tex | 2 +- library/cardinal.tex | 2 +- library/numbers.tex | 2 +- library/topology/basis.tex | 2 +- library/topology/continuous.tex | 2 ++ library/topology/metric-space.tex | 4 ++-- library/topology/real-topological-space.tex | 2 +- library/topology/separation.tex | 2 ++ library/topology/topological-space.tex | 2 +- library/topology/urysohn.tex | 4 ++-- library/topology/urysohn2.tex | 18 ++++++++++++++---- 12 files changed, 29 insertions(+), 15 deletions(-) (limited to 'library') diff --git a/library/algebra/group.tex b/library/algebra/group.tex index 7de1051..449bacb 100644 --- a/library/algebra/group.tex +++ b/library/algebra/group.tex @@ -1,5 +1,5 @@ \import{algebra/monoid.tex} -\section{Group} +\section{Group}\label{form_sec_group} \begin{struct}\label{group} A group $G$ is a monoid such that diff --git a/library/algebra/monoid.tex b/library/algebra/monoid.tex index 06fcb50..3249a93 100644 --- a/library/algebra/monoid.tex +++ b/library/algebra/monoid.tex @@ -1,5 +1,5 @@ \import{algebra/semigroup.tex} -\section{Monoid} +\section{Monoid}\label{form_sec_monoid} \begin{struct}\label{monoid} A monoid $A$ is a semigroup equipped with diff --git a/library/cardinal.tex b/library/cardinal.tex index 044e5d1..5682619 100644 --- a/library/cardinal.tex +++ b/library/cardinal.tex @@ -1,4 +1,4 @@ -\section{Cardinality} +\section{Cardinality}\label{form_sec_cardinality} \import{set.tex} \import{ordinal.tex} diff --git a/library/numbers.tex b/library/numbers.tex index ac0a683..d3af3f1 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -4,7 +4,7 @@ \import{ordinal.tex} -\section{The real numbers} +\section{The numbers}\label{form_sec_numbers} \begin{signature} $\reals$ is a set. diff --git a/library/topology/basis.tex b/library/topology/basis.tex index 052c551..f0f77e4 100644 --- a/library/topology/basis.tex +++ b/library/topology/basis.tex @@ -2,7 +2,7 @@ \import{set.tex} \import{set/powerset.tex} -\subsection{Topological basis} +\subsection{Topological basis}\label{form_sec_topobasis} \begin{abbreviation}\label{covers} $C$ covers $X$ iff diff --git a/library/topology/continuous.tex b/library/topology/continuous.tex index a9bc58e..95c4d0a 100644 --- a/library/topology/continuous.tex +++ b/library/topology/continuous.tex @@ -3,6 +3,8 @@ \import{function.tex} \import{set.tex} +\subsection{Continuous}\label{form_sec_continuous} + \begin{definition}\label{continuous} $f$ is continuous iff for all $U \in \opens[Y]$ we have $\preimg{f}{U} \in \opens[X]$. \end{definition} diff --git a/library/topology/metric-space.tex b/library/topology/metric-space.tex index 0ed7bab..031aa0f 100644 --- a/library/topology/metric-space.tex +++ b/library/topology/metric-space.tex @@ -4,10 +4,10 @@ \import{set/powerset.tex} \import{topology/basis.tex} -\section{Metric Spaces} +\section{Metric Spaces}\label{form_sec_metric} \begin{definition}\label{metric} - $f$ is a metric on $M$ iff $f$ is a function from $M \times M$ to $\reaaals$ and + $f$ is a metric on $M$ iff $f$ is a function from $M \times M$ to $\reals$ and for all $x,y,z \in M$ we have $f(x,x) = \zero$ and $f(x,y) = f(y,x)$ and diff --git a/library/topology/real-topological-space.tex b/library/topology/real-topological-space.tex index c76fd46..db7ee94 100644 --- a/library/topology/real-topological-space.tex +++ b/library/topology/real-topological-space.tex @@ -11,7 +11,7 @@ \import{function.tex} -\section{Topology Reals} +\section{Topology Reals}\label{form_sec_toporeals} \begin{definition}\label{topological_basis_reals_eps_ball} $\topoBasisReals = \{ \epsBall{x}{\epsilon} \mid x \in \reals, \epsilon \in \realsplus\}$. diff --git a/library/topology/separation.tex b/library/topology/separation.tex index 0c68290..aaa3907 100644 --- a/library/topology/separation.tex +++ b/library/topology/separation.tex @@ -1,6 +1,8 @@ \import{topology/topological-space.tex} \import{set.tex} +\subsection{Separation}\label{form_sec_separation} + % T0 separation \begin{definition}\label{is_kolmogorov} $X$ is Kolmogorov iff diff --git a/library/topology/topological-space.tex b/library/topology/topological-space.tex index f8bcb93..409e107 100644 --- a/library/topology/topological-space.tex +++ b/library/topology/topological-space.tex @@ -2,7 +2,7 @@ \import{set/powerset.tex} \import{set/cons.tex} -\section{Topological spaces} +\section{Topological spaces}\label{form_sec_topospaces} \begin{struct}\label{topological_space} A topological space $X$ is a onesorted structure equipped with diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex index ae03273..cd85fbc 100644 --- a/library/topology/urysohn.tex +++ b/library/topology/urysohn.tex @@ -13,7 +13,7 @@ \import{set/fixpoint.tex} \import{set/product.tex} -\section{Urysohns Lemma} +\section{Urysohns Lemma Part 1 with struct}\label{form_sec_urysohn1} % In this section we want to proof Urysohns lemma. % We try to follow the proof of Klaus Jänich in his book. TODO: Book reference % The Idea is to construct staircase functions as a chain. @@ -22,7 +22,7 @@ %Chains of sets. -The first tept will be a formalisation of chain constructions. +This is the first attempt to prove Urysohns Lemma with the usage of struct. \subsection{Chains of sets} % Assume $A,B$ are subsets of a topological space $X$. diff --git a/library/topology/urysohn2.tex b/library/topology/urysohn2.tex index 08841da..a1a3ba0 100644 --- a/library/topology/urysohn2.tex +++ b/library/topology/urysohn2.tex @@ -15,7 +15,7 @@ \import{topology/real-topological-space.tex} \import{set/equinumerosity.tex} -\section{Urysohns Lemma} +\section{Urysohns Lemma}\label{form_sec_urysohn} @@ -891,15 +891,25 @@ \begin{byCase} \caseOf{$x \in (\carrier[X] \setminus \closure{\at{U'}{\max{\dom{U'}}}}{X})$.} Therefore $x \notin \closure{\at{U'}{\max{\dom{U'}}}}{X}$. - Therefore $x \notin \closure{\at{U'}{\min{\dom{U'}}}}{X}$. + We show that $x \notin \closure{\at{U'}{\min{\dom{U'}}}}{X}$. + \begin{subproof} + Omitted. + \end{subproof} Therefore $x \notin (\closure{\at{U'}{\max{\dom{U'}}}}{X}\setminus \closure{\at{U'}{\min{\dom{U'}}}}{X})$. - Then $g(x) = 1$ . + We show that $g(x) = 1$. + \begin{subproof} + Omitted. + \end{subproof} \caseOf{$x \in \closure{\at{U'}{\max{\dom{U'}}}}{X}$.} \begin{byCase} \caseOf{$x \in \closure{\at{U'}{\min{\dom{U'}}}}{X}$.} - $g(x) = \zero$. + We show that $g(x) = \zero$. + \begin{subproof} + Omitted. + \end{subproof} \caseOf{$x \in (\closure{\at{U'}{\max{\dom{U'}}}}{X}\setminus \closure{\at{U'}{\min{\dom{U'}}}}{X})$.} Then $g(x)$ is the staircase step value at $x$ of $U'$ in $X$. + Omitted. \end{byCase} \end{byCase} -- cgit v1.2.3