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 ++++++++++++++++++++++++++++++++++++++++--- 1 file changed, 119 insertions(+), 8 deletions(-) (limited to 'library/topology/urysohn.tex') 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 $\carrier[\eta] =\{A, (X \setminus B)\}$ -- cgit v1.2.3