diff options
| -rw-r--r-- | library/numbers.tex | 41 | ||||
| -rw-r--r-- | library/topology/basis.tex | 51 | ||||
| -rw-r--r-- | library/topology/metric-space.tex | 77 | ||||
| -rw-r--r-- | library/topology/separation.tex | 80 | ||||
| -rw-r--r-- | library/topology/topological-space.tex | 270 |
5 files changed, 387 insertions, 132 deletions
diff --git a/library/numbers.tex b/library/numbers.tex index 5dd06da..2451730 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -4,8 +4,6 @@ \section{The real numbers} -%TODO: Implementing Notion for negativ number such as -x. - %TODO: %\inv{} für inverse benutzen. Per Signatur einfüheren und dann axiomatisch absicher %\cdot für multiklikation verwenden. @@ -17,11 +15,11 @@ \end{signature} \begin{signature} - $x + y$ is a set. + $x \add y$ is a set. \end{signature} \begin{signature} - $x \times y$ is a set. + $x \rmul y$ is a set. \end{signature} \begin{axiom}\label{one_in_reals} @@ -58,7 +56,7 @@ \end{axiom} \begin{axiom}\label{reals_axiom_order_def} - $x < y$ iff there exist $z \in \reals$ such that $\zero < z$ and $x + z = y$. + $x < y$ iff there exist $z \in \reals$ such that $\zero < z$ and $x \add z = y$. \end{axiom} \begin{lemma}\label{reals_one_bigger_than_zero} @@ -67,11 +65,11 @@ \begin{axiom}\label{reals_axiom_assoc} - For all $x,y,z \in \reals$ $(x + y) + z = x + (y + z)$ and $(x \times y) \times z = x \times (y \times z)$. + For all $x,y,z \in \reals$ $(x \add y) \add z = x \add (y \add z)$ 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 $x \times y = y \times x$. + For all $x,y \in \reals$ $x \add y = y \add x$ and $x \rmul y = y \rmul x$. \end{axiom} \begin{axiom}\label{reals_axiom_zero_in_reals} @@ -79,32 +77,32 @@ \end{axiom} \begin{axiom}\label{reals_axiom_zero} - For all $x \in \reals$ $x + \zero = x$. + For all $x \in \reals$ $x \add \zero = x$. \end{axiom} \begin{axiom}\label{reals_axiom_one} - For all $x \in \reals$ $1 \neq \zero$ and $x \times 1 = x$. + For all $x \in \reals$ $1 \neq \zero$ and $x \rmul 1 = x$. \end{axiom} \begin{axiom}\label{reals_axiom_add_invers} - For all $x \in \reals$ there exist $y \in \reals$ such that $x + y = \zero$. + For all $x \in \reals$ there exist $y \in \reals$ such that $x \add y = \zero$. \end{axiom} \begin{axiom}\label{reals_axiom_mul_invers} - For all $x \in \reals$ such that $x \neq \zero$ there exist $y \in \reals$ such that $x \times y = 1$. + For all $x \in \reals$ such that $x \neq \zero$ there exist $y \in \reals$ such that $x \rmul y = 1$. \end{axiom} \begin{axiom}\label{reals_axiom_disstro1} - For all $x,y,z \in \reals$ $x \times (y + z) = (x \times y) + (x \times z)$. + For all $x,y,z \in \reals$ $x \rmul (y \add z) = (x \rmul y) \add (x \rmul z)$. \end{axiom} \begin{proposition}\label{reals_disstro2} - For all $x,y,z \in \reals$ $(y + z) \times x = (y \times x) + (z \times x)$. + For all $x,y,z \in \reals$ $(y \add z) \rmul x = (y \rmul x) \add (z \rmul x)$. \end{proposition} \begin{proposition}\label{reals_reducion_on_addition} - For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$. + For all $x,y,z \in \reals$ if $x \add y = x \add z$ then $y = z$. \end{proposition} \begin{axiom}\label{reals_axiom_dedekind_complete} @@ -116,20 +114,20 @@ \begin{lemma}\label{order_reals_lemma1} For all $x,y,z \in \reals$ such that $\zero < x$ if $y < z$ - then $(y \times x) < (z \times x)$. + then $(y \rmul x) < (z \rmul x)$. \end{lemma} \begin{lemma}\label{order_reals_lemma2} For all $x,y,z \in \reals$ such that $\zero < x$ if $y < z$ - then $(x \times y) < (x \times z)$. + then $(x \rmul y) < (x \rmul z)$. \end{lemma} \begin{lemma}\label{order_reals_lemma3} For all $x,y,z \in \reals$ such that $x < \zero$ if $y < z$ - then $(x \times z) < (x \times y)$. + then $(x \rmul z) < (x \rmul y)$. \end{lemma} \begin{lemma}\label{o4rder_reals_lemma} @@ -145,17 +143,13 @@ \end{lemma} \begin{axiom}\label{reals_axiom_minus} - For all $x \in \reals$ $x - x = \zero$. + For all $x \in \reals$ $x \rmiuns x = \zero$. \end{axiom} \begin{lemma}\label{reals_minus} - Assume $x,y \in \reals$. If $x - y = \zero$ then $x=y$. + Assume $x,y \in \reals$. If $x \rmiuns y = \zero$ then $x=y$. \end{lemma} -%\begin{definition}\label{reasl_supremum} %expaction "there exists" after \mid -% $\rsup{X} = \{z \mid \text{ $z \in \reals$ and for all $x,y$ such that $x \in X$ and $y,x \in \reals$ and $x < y$ we have $z \leq y$ }\}$. -%\end{definition} - \begin{definition}\label{upper_bound} $x$ is an upper bound of $X$ iff for all $y \in X$ we have $x > y$. \end{definition} @@ -185,7 +179,6 @@ \end{definition} \begin{lemma}\label{infimum_unique} - %Let $x,y \in \reals$ and let $X$ be a subset of $\reals$. If $x$ is a greatest lower bound of $X$ and $y$ is a greatest lower bound of $X$ then $x = y$. \end{lemma} diff --git a/library/topology/basis.tex b/library/topology/basis.tex index 6fa9fbd..bd7d15a 100644 --- a/library/topology/basis.tex +++ b/library/topology/basis.tex @@ -56,24 +56,52 @@ $\emptyset \in \genOpens{B}{X}$. \end{lemma} -\begin{lemma}\label{all_is_in_genopens} + + +\begin{lemma}\label{union_in_genopens} Assume $B$ is a topological basis for $X$. - $X \in \genOpens{B}{X}$. + Assume $F\subseteq \genOpens{B}{X}$. + Then $\unions{F}\in\genOpens{B}{X}$. \end{lemma} \begin{proof} - $B$ covers $X$ by \cref{topological_prebasis_iff_covering_family,topological_basis}. - $\unions{B} \in \genOpens{B}{X}$. - $X \subseteq \unions{B}$. + We have $\unions{F} \in \pow{X}$. + + Show for all $x\in \unions{F}$ there exists $W \in B$ + such that $x\in W$ and $W \subseteq \unions{F}$. + \begin{subproof} + Fix $x \in \unions{F}$. + There exists $V \in F$ such that $x \in V$ by \cref{unions_iff}. + $V \in \genOpens{B}{X}$. + There exists $W \in B$ such that $x \in W \subseteq V$. + Then $W \subseteq \unions{F}$. + \end{subproof} + Then $\unions{F}\in\genOpens{B}{X}$ by \cref{genopens}. \end{proof} -\begin{lemma}\label{union_in_genopens} +\begin{lemma}\label{basis_is_in_genopens} Assume $B$ is a topological basis for $X$. - For all $F\subseteq \genOpens{B}{X}$ we have $\unions{F}\in\genOpens{B}{X}$. + $B \subseteq \genOpens{B}{X}$. \end{lemma} \begin{proof} - Omitted. + We show for all $V \in B$ $V \in \genOpens{B}{X}$. + \begin{subproof} + Fix $V \in B$. + For all $x \in V$ $x \in V \subseteq V$. + $V \subseteq X$ by \cref{topological_prebasis_iff_covering_family,topological_basis}. + $V \in \pow{X}$. + $V \in \genOpens{B}{X}$. + \end{subproof} \end{proof} +\begin{lemma}\label{all_is_in_genopens} + Assume $B$ is a topological basis for $X$. + $X \in \genOpens{B}{X}$. +\end{lemma} +\begin{proof} + $B$ covers $X$ by \cref{topological_prebasis_iff_covering_family,topological_basis}. + $\unions{B} \in \genOpens{B}{X}$. + $X \subseteq \unions{B}$. +\end{proof} \begin{lemma}\label{inters_in_genopens} Assume $B$ is a topological basis for $X$. @@ -92,16 +120,13 @@ Then $x\in A,C$. There exists $V' \in B$ such that $x \in V' \subseteq A$ by \cref{genopens}. There exists $V'' \in B$ such that $x \in V''\subseteq C$ by \cref{genopens}. - There exists $W \in B$ such that $x \in W$ and $W \subseteq V'$ and $W \subseteq V''$. + There exists $W \in B$ such that $x \in W$ and $W \subseteq V'$ and $W \subseteq V''$ by \cref{topological_basis}. Show $W \subseteq (A\inter C)$. \begin{subproof} - %$W \subseteq v'$ and $W \subseteq V''$. For all $y \in W$ we have $y \in V'$ and $y \in V''$. \end{subproof} \end{subproof} - %Therefore for all $A, C, x$ such that $A \in \genOpens{B}{X}$ and $C \in \genOpens{B}{X}$ and $x \in (A \inter C)$ we have there exists $W \in B$ - %such that $x\in W$ and $W \subseteq (A\inter C)$. - $(A\inter C) \in \genOpens{B}{X}$. + $(A\inter C) \in \genOpens{B}{X}$ by \cref{genopens}. \end{proof} diff --git a/library/topology/metric-space.tex b/library/topology/metric-space.tex index 1c6a0ca..bcc5b8c 100644 --- a/library/topology/metric-space.tex +++ b/library/topology/metric-space.tex @@ -21,44 +21,16 @@ -%\begin{definition}\label{induced_topology} -% $O$ is the induced topology of $d$ in $M$ iff -% $O \subseteq \pow{M}$ and -% $d$ is a metric on $M$ and -% for all $x,r,A,B,C$ -% such that $x \in M$ and $r \in \reals$ and $A,B \in O$ and $C$ is a family of subsets of $O$ -% we have $\openball{r}{x}{d} \in O$ and $\unions{C} \in O$ and $A \inter B \in O$. -%\end{definition} - -%\begin{definition} -% $\projcetfirst{A} = \{a \mid \exists x \in X \text{there exist $x \i } \}$ -%\end{definition} - \begin{definition}\label{set_of_balls} $\balls{d}{M} = \{ O \in \pow{M} \mid \text{there exists $x,r$ such that $r \in \reals$ and $x \in M$ and $O = \openball{r}{x}{d}$ } \}$. \end{definition} -%\begin{definition}\label{toindsas} -% $\metricopens{d}{M} = \{O \in \pow{M} \mid \text{ -% $d$ is a metric on $M$ and -% for all $x,r,A,B,C$ -% such that $x \in M$ and $r \in \reals$ and $A,B \in O$ and $C$ is a family of subsets of $O$ -% we have $\openball{r}{x}{d} \in O$ and $\unions{C} \in O$ and $A \inter B \in O$. -% } \}$. -% -%\end{definition} - \begin{definition}\label{metricopens} $\metricopens{d}{M} = \genOpens{\balls{d}{M}}{M}$. \end{definition} -\begin{theorem} - Let $d$ be a metric on $M$. - $M$ is a topological space. -\end{theorem} - @@ -93,13 +65,6 @@ \end{lemma} -%\begin{definition}\label{lenght_of_interval} %TODO: take minus if its implemented -% $\lenghtinterval{x}{y} = r$ -%\end{definition} - - - - \begin{lemma}\label{metric_implies_topology} @@ -108,45 +73,3 @@ \end{lemma} - - - -%\begin{struct}\label{metric_space} -% A metric space $M$ is a onesorted structure equipped with -% \begin{enumerate} -% \item $\metric$ -% \end{enumerate} -% such that -% \begin{enumerate} -% \item \label{metric_space_d} $\metric[M]$ is a function from $M \times M$ to $\reals$. -% \item \label{metric_space_distence_of_a_point} $\metric[M](x,x) = \zero$. -% \item \label{metric_space_positiv} for all $x,y \in M$ if $x \neq y$ then $\zero < \metric[M](x,y)$. -% \item \label{metric_space_symetrie} $\metric[M](x,y) = \metric[M](y,x)$. -% \item \label{metric_space_triangle_equation} for all $x,y,z \in M$ $\metric[M](x,y) < \metric[M](x,z) + \metric[M](z,y)$ or $\metric[M](x,y) = \metric[M](x,z) + \metric[M](z,y)$. -% \item \label{metric_space_topology} $M$ is a topological space. -% \item \label{metric_space_opens} for all $x \in M$ for all $r \in \reals$ $\{z \in M \mid \metric[M](x,z) < r\} \in \opens$. -% \end{enumerate} -%\end{struct} - -%\begin{definition}\label{open_ball} -% $\openball{r}{x}{M} = \{z \in M \mid \metric(x,z) < r\}$. -%\end{definition} - -%\begin{proposition}\label{open_ball_is_open} -% Let $M$ be a metric space,let $r \in \reals $, let $x$ be an element of $M$. -% Then $\openball{r}{x}{M} \in \opens[M]$. -%\end{proposition} - - - - - - -%TODO: - Basis indudiert topology lemma -% - Offe Bälle sind basis - -% Was danach kommen soll bleibt offen, vll buch oder in proof wiki -% Trennungsaxiom, - -% Notaionen aufräumen damit das gut gemercht werden kann. - diff --git a/library/topology/separation.tex b/library/topology/separation.tex index f70cb50..3268bb9 100644 --- a/library/topology/separation.tex +++ b/library/topology/separation.tex @@ -73,14 +73,29 @@ \begin{proposition}\label{teeone_implies_singletons_closed} Let $X$ be a \teeone-space. - Then for all $x\in\carrier[X]$ we have $\{x\}$ is closed in $X$. + Assume $x \in \carrier[X]$. + Then $\{x\}$ is closed in $X$. \end{proposition} \begin{proof} - Omitted. - % TODO - % Choose for every y distinct from x and open subset U_y containing y but not x. - % The union U of all the U_y is open. - % {x} is the complement of U in \carrier[X]. + Let $V = \{ U \in \opens[X] \mid x \notin U\}$. + For all $y \in \carrier[X]$ such that $x \neq y$ there exist $U \in \opens[X]$ such that $x \notin U \ni y$. + For all $y \in \carrier[X]$ such that $y \neq x$ there exists $U \in V$ such that $y \in U$. + + $\unions{V} \in \opens[X]$. + For all $y \in \carrier[X]$ such that $x \neq y$ we have $y \in \unions{V}$. + We show that $\carrier[X] \setminus \{x\} = \unions{V}$. + \begin{subproof} + We show that for all $y \in \carrier[X] \setminus \{x\}$ we have $y \in \unions{V}$. + \begin{subproof} + Fix $y \in \carrier[X] \setminus \{x\}$. + $y \neq x$. + $y \in \carrier[X]$. + $y \in \unions{V}$. + \end{subproof} + For all $y \in \unions{V}$ we have $y \notin \{x\}$. + For all $y \in \unions{V}$ we have $y \in \carrier[X] \setminus \{x\}$. + Follows by set extensionality. + \end{subproof} \end{proof} % % Conversely, if \{x\} is open, then for any y distinct from x we can use @@ -120,5 +135,56 @@ Then $X$ is a \teeone-space. \end{proposition} \begin{proof} - Omitted. % TODO + We show that for all $x,y\in\carrier[X]$ such that $x\neq y$ + there exist $U, V\in\opens[X]$ such that + $U\ni x\notin V$ and $V\ni y\notin U$. + \begin{subproof} + $X$ is hausdorff. + For all $x,y\in\carrier[X]$ such that $x\neq y$ + there exist $U, V\in\opens[X]$ such that + $x\in U$ and $y\in V$ and $U$ is disjoint from $V$. + \end{subproof} +\end{proof} + +\begin{definition}\label{regular} + $X$ is regular iff for all $C,p$ such that $p \in \carrier[X]$ and $p \notin C \in \closeds{X}$ we have there exists $U,C \in \opens[X]$ such that $p \in U$ and $C \subseteq V$ and $U \inter V = \emptyset$. +\end{definition} + +\begin{definition}\label{regular_space} + $X$ is a regular space iff $X$ is a topological space and $X$ is regular. +\end{definition} + + +\begin{definition}\label{teethree} + $X$ is \teethree\ iff $X$ is regular and $X$ is \teezero\ . +\end{definition} + +\begin{definition}\label{teethree_space} + $X$ is a \teethree-space iff $X$ is a topological space and $X$ is \teethree\ . +\end{definition} + +\begin{proposition}\label{teethree_space_is_teetwo_space} + Let $X$ be a \teethree-space. + Then $X$ is a \teetwo-space. +\end{proposition} +\begin{proof} + For all $x,y \in \carrier[X]$ such that $x \neq y$ we have $x \notin \{y\}$. + It suffices to show that $X$ is hausdorff. + It suffices to show that for all $x \in \carrier[X]$ we have for all $y \in \carrier[X]$ such that $y \neq x$ we have there exist $U,V \in \opens[X]$ such that $x\in U$ and $y \in V$ and $U$ is disjoint from $V$. + Fix $x \in \carrier[X]$. + It suffices to show that for all $y \in \carrier[X]$ such that $y \neq x$ we have there exist $U,V \in \opens[X]$ such that $x\in U$ and $y \in V$ and $U$ is disjoint from $V$. + Fix $y \in \carrier[X]$. + + %There exist $U' \in \opens[X]$ such that $x\in U'\not\ni y$ or $x\notin U'\ni y$ by \cref{}. + %There exist $C \in \closeds{X}$ such that $y \in C \not\ni X$. + We show that there exist $U,V,C$ such that $U,V \in \opens[X]$ and $C\in \closeds{X}$ and $x \in U$ and $y \in C \subseteq V$ and $U$ is disjoint from $V$. + \begin{subproof} + Omitted. + \end{subproof} + $y \in V$. + Follows by assumption. \end{proof} + +% for all $x,y\in\carrier[X]$ such that $x\neq y$ +% there exist $U, V\in\opens[X]$ such that +% $x\in U$ and $y\in V$ and $U$ is disjoint from $V$.
\ No newline at end of file diff --git a/library/topology/topological-space.tex b/library/topology/topological-space.tex index 55bc253..73b7ec5 100644 --- a/library/topology/topological-space.tex +++ b/library/topology/topological-space.tex @@ -161,12 +161,21 @@ \begin{proposition}\label{carrier_is_closed} Let $X$ be a topological space. - Then $\emptyset$ is closed in $X$. + Then $\carrier[X]$ is closed in $X$. \end{proposition} \begin{proof} $\carrier[X]\setminus \carrier[X] = \emptyset$. + Follows by \cref{emptyset_open,is_closed_in}. \end{proof} +\begin{proposition}\label{opens_minus_closed_is_open} + Let $X$ be a topological space. + Suppose $A, B \subseteq \carrier[X]$. + Suppose $A$ is open in $X$. + Suppose $B$ is closed in $X$. + Then $A \setminus B$ is open in $X$. +\end{proposition} + \begin{definition}[Closed sets]\label{closeds} $\closeds{X} = \{ A\in\pow{\carrier[X]}\mid\text{$A$ is closed in $X$}\}$. \end{definition} @@ -216,31 +225,270 @@ Follows by \cref{inters_singleton,closure}. \end{proof} +\begin{proposition}\label{subseteq_inters_iff_to_right} + Let $A,F$ be sets. + Suppose $A \subseteq \inters{F}$. + Then for all $X \in F$ we have $A \subseteq X$. +\end{proposition} + + +\begin{proposition}\label{subseteq_of_all_then_subset_of_union} + Let $A,F$ be sets. + Suppose $F$ is inhabited. %TODO: Remove!! + Suppose for all $X \in F$ we have $A \subseteq X$. + Then $A \subseteq \unions{F}$. +\end{proposition} +\begin{proof} + There exist $X \in F$ such that $X \subseteq \unions{F}$. + $A \subseteq X \subseteq \unions{F}$. +\end{proof} + + + +\begin{proposition}\label{subseteq_inters_iff_to_left} + Let $A,F$ be sets. + Suppose $F$ is inhabited. % TODO:Remove!! + Suppose for all $X \in F$ we have $A \subseteq X$. + Then $A \subseteq \inters{F}$. +\end{proposition} +\begin{proof} + \begin{byCase} + \caseOf{$A = \emptyset$.}Trivial. + \caseOf{$A \neq \emptyset$.} + $F$ is inhabited. + It suffices to show that for all $a \in A$ we have $a \in \inters{F}$. + Fix $a \in A$. + For all $X \in F$ we have $a \in X$. + $A \subseteq \unions{F}$. + $a \in \unions{F}$. + \end{byCase} +\end{proof} + +\begin{proposition}\label{subseteq_inters_iff_new} + Suppose $F$ is inhabited. + $A \subseteq \inters{F}$ iff for all $X \in F$ we have $A \subseteq X$. +\end{proposition} + +\begin{proposition}\label{set_is_subseteq_to_closure_of_the_set} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + $A \subseteq \closure{A}{X}$. +\end{proposition} +\begin{proof} + \begin{byCase} + \caseOf{$A = \emptyset$.} + Trivial. + \caseOf{$A \neq \emptyset$.} + We show that $\carrier[X] \in \closures{A}{X}$. + \begin{subproof} + $\carrier[X]$ is closed in $X$. + $\carrier[X] \in \pow{\carrier[X]}$. + \end{subproof} + $\closures{A}{X}$ is inhabited. + For all $A' \in \closures{A}{X}$ we have $A \subseteq A'$. + Therefore $A \subseteq \inters{\closures{A}{X}}$. + \end{byCase} +\end{proof} + +\begin{proposition}\label{complement_of_closure_of_complement_of_x_subseteq_x} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + $(\carrier[X] \setminus \closure{(\carrier[X]\setminus A)}{X}) \subseteq A$. +\end{proposition} +\begin{proof} + It suffices to show that for all $x \in (\carrier[X] \setminus \closure{(\carrier[X]\setminus A)}{X})$ we have $x \in A$. + If $x \in \carrier[X]\setminus A$ then $x \in \closure{(\carrier[X]\setminus A)}{X}$. +\end{proof} + +\begin{proposition}\label{complement_of_closed_is_open} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + Suppose $A$ is closed in $X$. + Then $\carrier[X] \setminus A$ is open in $X$. +\end{proposition} + +\begin{proposition}\label{complement_of_open_is_closed} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + Suppose $A$ is open in $X$. + Then $\carrier[X] \setminus A$ is closed in $X$. +\end{proposition} + +\begin{proposition}\label{intersection_of_closed_is_closed} + Let $X$ be a topological space. + Suppose $A, B \subseteq \carrier[X]$. + Suppose $A$ are closed in $X$. + Suppose $B$ are closed in $X$. + Then $A \inter B$ is closed in $X$. +\end{proposition} +\begin{proof} + $\carrier[X] \setminus A, \carrier[X] \setminus B \in \opens[X]$. + $(\carrier[X] \setminus A) \union (\carrier[X] \setminus B) \in \opens[X]$. + $A \inter B = \carrier[X] \setminus ((\carrier[X] \setminus A) \union (\carrier[X] \setminus B))$. +\end{proof} + +\begin{proposition}\label{union_of_closed_is_closed} + Let $X$ be a topological space. + Suppose $A, B \subseteq \carrier[X]$. + Suppose $A$ are closed in $X$. + Suppose $B$ are closed in $X$. + Then $A \union B$ is closed in $X$. +\end{proposition} +\begin{proof} + $\carrier[X] \setminus A, \carrier[X] \setminus B \in \opens[X]$. + $(\carrier[X] \setminus A) \inter (\carrier[X] \setminus B) \in \opens[X]$. + $A \union B = \carrier[X] \setminus ((\carrier[X] \setminus A) \inter (\carrier[X] \setminus B))$. +\end{proof} + +\begin{proposition}\label{closed_minus_open_is_closed} + Let $X$ be a topological space. + Suppose $A, B \subseteq \carrier[X]$. + Suppose $A$ is open in $X$. + Suppose $B$ is closed in $X$. + Then $B \setminus A$ is closed in $X$. +\end{proposition} + + + +\begin{proposition}\label{intersection_of_closed_is_closed_infinite} + Let $X$ be a topological space. + Suppose $F \subseteq \pow{\carrier[X]}$. + Suppose for all $A \in F$ we have $A$ is closed in $X$. + Suppose $F$ is inhabited. + Then $\inters{F}$ is closed in $X$. +\end{proposition} +\begin{proof} + Let $F' = \{Y \in \pow{\carrier[X]} \mid \text{there exists $C \in F$ such that $Y = \carrier[X] \setminus C$ }\} $. + For all $Y \in F'$ we have $Y$ is open in $X$. + $\unions{F'}$ is open in $X$. + $\unions{F'}, \inters{F} \subseteq \carrier[X]$. + We show that $\inters{F} = \carrier[X] \setminus (\unions{F'})$. + \begin{subproof} + We show that for all $a \in \inters{F}$ we have $a \in \carrier[X] \setminus (\unions{F'})$. + \begin{subproof} + Fix $a \in \inters{F}$. + $a \in \carrier[X]$. + For all $A \in F$ we have $a \in A$. + For all $A \in F$ we have $a \notin (\carrier[X] \setminus A)$. + Then $a \notin \unions{F'}$. + Therefore $a \in \carrier[X] \setminus (\unions{F'})$. + \end{subproof} + We show that for all $a \in \carrier[X] \setminus (\unions{F'})$ we have $a \in \inters{F}$. + \begin{subproof} + \begin{byCase} + \caseOf{$\inters{F} = \emptyset$.} + $F$ is inhabited. + Take $U$ such that $U \in F$. + Let $F'' = F \setminus \{U\}$. + There exist $U' \in F'$ such that $U' = \carrier[X] \setminus U$. + \caseOf{$\inters{F} \neq \emptyset$.} + Fix $a \in \carrier[X] \setminus (\unions{F'})$. + $a \in \carrier[X]$. + $a \notin \unions{F'}$. + For all $A \in F'$ we have $a \notin A$. + For all $A \in F'$ we have $a \in (\carrier[X] \setminus A)$. + For all $A \in F'$ there exists $Y \in F$ such that $Y = (\carrier[X] \setminus A)$. + For all $Y \in F $ there exists $A \in F'$ such that $a \in Y = (\carrier[X] \setminus A)$. + For all $Y \in F$ we have $a \in Y$. + $\inters{F}$ is inhabited. + Therefore $a \in \inters{F}$. + \end{byCase} + \end{subproof} + Follows by set extensionality. + \end{subproof} + Follows by \cref{complement_of_open_is_closed}. +\end{proof} + +\begin{proposition}\label{closure_is_closed} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + Then $\closure{A}{X}$ is closed in $X$. +\end{proposition} +\begin{proof} + \begin{byCase} + \caseOf{$\closure{A}{X} = \emptyset$.} + Trivial. + \caseOf{$\closure{A}{X} \neq \emptyset$.} + $\closures{A}{X}$ is inhabited. + $\closures{A}{X} \subseteq \pow{\carrier[X]}$. + For all $B \in \closures{A}{X}$ we have $B$ is closed in $X$. + $\inters{\closures{A}{X}}$ is closed in $X$. + \end{byCase} +\end{proof} + -\begin{proposition}%[Complement of interior equals closure of complement] -\label{complement_interior_eq_closure_complement} + +\begin{proposition}\label{closure_is_minimal_closed_set} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + For all $Y \in \closeds{X}$ such that $A \subseteq Y$ we have $\closure{A}{X} \subseteq Y$. +\end{proposition} +\begin{proof} + Follows by \cref{closure,closeds,inters_subseteq_elem,closures}. +\end{proof} + +\begin{proposition}\label{complement_interior_eq_closure_complement} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. $\carrier[X]\setminus\interior{A}{X} = \closure{(\carrier[X]\setminus A)}{X}$. \end{proposition} \begin{proof} - Omitted. % Use general De Morgan's Law in Pow|X| + We show that for all $x \in \carrier[X]\setminus\interior{A}{X}$ we have $x \in \closure{(\carrier[X]\setminus A)}{X}$. + \begin{subproof} + Fix $x \in \carrier[X]\setminus\interior{A}{X}$. + Suppose not. + $x \notin \closure{(\carrier[X]\setminus A)}{X}$. + $x \in \carrier[X]$. + $(\carrier[X] \setminus \closure{(\carrier[X]\setminus A)}{X}) \inter \closure{(\carrier[X]\setminus A)}{X} = \emptyset$. + $x \in A$. + $x \in A \inter (\carrier[X] \setminus \closure{(\carrier[X]\setminus A)}{X})$. + $\carrier[X] \setminus \closure{(\carrier[X]\setminus A)}{X} \in \opens[X]$. + Contradiction. + \end{subproof} + $\carrier[X]\setminus\interior{A}{X} \subseteq \closure{(\carrier[X]\setminus A)}{X}$. + $\closure{(\carrier[X]\setminus A)}{X} \subseteq \carrier[X]\setminus\interior{A}{X}$. \end{proof} + + \begin{definition}[Frontier]\label{frontier} $\frontier{A}{X} = \closure{A}{X}\setminus\interior{A}{X}$. \end{definition} +\begin{proposition}\label{closure_interior_frontier_is_in_carrier} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + Then $\closure{A}{X}, \interior{A}{X}, \frontier{A}{X} \subseteq \carrier[X]$. +\end{proposition} + +\begin{proposition}\label{frontier_is_closed} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. + Then $\frontier{A}{X}$ is closed in $X$. +\end{proposition} +\begin{proof} + $\closure{A}{X}\setminus\interior{A}{X}$ is closed in $X$ by \cref{closure_interior_frontier_is_in_carrier,closure_is_closed,interior_is_open,closed_minus_open_is_closed}. +\end{proof} + +\begin{proposition}\label{setdifference_eq_intersection_with_complement} + Suppose $A,B \subseteq C$. + Then $A \setminus B = A \inter (C \setminus B)$. +\end{proposition} -\begin{proposition}%[Frontier as intersection of closures] -\label{frontier_as_inter} + +\begin{proposition}\label{frontier_as_inter} + Let $X$ be a topological space. + Suppose $A \subseteq \carrier[X]$. $\frontier{A}{X} = \closure{A}{X} \inter \closure{(\carrier[X]\setminus A)}{X}$. \end{proposition} \begin{proof} - % TODO - Omitted. - %$\frontier{A}{X} = \closure{A}{X}\setminus\interior{A}{X}$. % by frontier (definition) - %$\closure{A}{X}\setminus\interior{A}{X} = \closure{A}{X}\inter(\carrier[X]\setminus\interior{A}{X})$. % by setminus_eq_inter_complement - %$\closure{A}{X}\inter(\carrier[X]\setminus\interior{A}{X}) = \closure{A}{X} \inter \closure{(\carrier[X]\setminus A)}{X}$. % by complement_interior_eq_closure_complement + \begin{align*} + \frontier{A}{X} \\ + &= \closure{A}{X}\setminus\interior{A}{X} \\ + &= \closure{A}{X} \inter (\carrier[X] \setminus \interior{A}{X}) \explanation{by \cref{setdifference_eq_intersection_with_complement,closure_interior_frontier_is_in_carrier}}\\ + &= \closure{A}{X} \inter \closure{(\carrier[X]\setminus A)}{X} \explanation{by \cref{complement_interior_eq_closure_complement}} + \end{align*} \end{proof} \begin{proposition}\label{frontier_of_emptyset} |