Merge pull request #2 from Simon-Kor/main

Merge (finally)

1 files changed, 165 insertions, 6 deletions

diff --git a/library/topology/separation.tex b/library/topology/separation.tex
index f70cb50..aaa3907 100644
--- a/library/topology/separation.tex
+++ b/library/topology/separation.tex
@@ -1,5 +1,7 @@
 \import{topology/topological-space.tex}
+\import{set.tex}
 
+\subsection{Separation}\label{form_sec_separation}
 
 % T0 separation
 \begin{definition}\label{is_kolmogorov}
@@ -73,14 +75,30 @@
 
 \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 $x \notin U \ni y$ by \cref{carrier_open,teeone}.
+    %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 +138,146 @@
     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{is_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{abbreviation}\label{regular_space}
+    $X$ is a regular space iff $X$ is a topological space and $X$ is regular.
+\end{abbreviation}
+
+
+\begin{abbreviation}\label{teethree}
+    $X$ is \teethree\ iff $X$ is regular and $X$ is \teezero\ .
+\end{abbreviation}
+
+\begin{abbreviation}\label{teethree_space}
+    $X$ is a \teethree-space iff $X$ is a topological space and $X$ is \teethree\ .
+\end{abbreviation}
+
+\begin{proposition}\label{teethree_implies_closed_neighbourhood_in_open}
+    Let $X$ be a topological space.
+    Suppose $X$ is inhabited.
+    Suppose $X$ is \teethree\ .
+    For all $U \in \opens[X]$ we have for all $x \in U$ we have there exist $N \in \neighbourhoods{x}{X}$ such that $N \subseteq U$ and $N$ is closed in $X$.    
+\end{proposition}
+\begin{proof}
+    Omitted.
+    %%Suppose $X$ is regular and kolmogorov.
+    %Fix $U \in \opens[X]$.
+    %Fix $x \in U$.
+    %Let $C = \carrier[X] \setminus U$.
+    %Then $C \in \closeds{X}$.
+    %$x \notin C$.
+    %$x \in \carrier[X]$.
+    %We show that there exists $A,B \in \opens[X]$ such that $x \in B$ and $C \subseteq A$ and $A \inter B = \emptyset$.     
+    %We show that $B \subseteq (\carrier[X] \setminus A)$.
+    %$(\carrier[X] \setminus A) \subseteq (\carrier[X] \setminus (\carrier[X] \setminus U))$.
+    %$(\carrier[X] \setminus (\carrier[X] \setminus U)) = U$.
+    %$x \in B \subseteq (\carrier[X] \setminus A) \subseteq U$.
+    %Let $N = (\carrier[X] \setminus A)$.
+    %Then $N \in \closeds{X}$ and $x \in N$ and $N \subseteq U$.
+    %$N \in \neighbourhoods{x}{X}$.
+\end{proof}
+
+\begin{proposition}\label{teethree_iff_each_closed_is_intersection_of_its_closed_neighborhoods}
+    Let $X$ be a topological space.
+    Suppose $X$ is inhabited.
+    $X$ is \teethree\ iff for all $H \in \closeds{X}$ such that $F = \{ N \in \neighbourhoodsSet{H}{X} \mid N \in \closeds{X}\}$ we have $H = \inters{F}$.    
+\end{proposition}
+\begin{proof}
+    We show that if $X$ is \teethree\ then for all $H \in \closeds{X}$ such that $F = \{ N \in \neighbourhoodsSet{H}{X} \mid N \in \closeds{X}\}$ we have $H = \inters{F}$.
+    \begin{subproof}
+        %For all $U \in \opens[X]$ we have for all $x \in U$ we have there exist $N \in \neighbourhoods{x}{X}$ such that $N \subseteq U$ and $N$ is closed in $X$.
+        Omitted.
+    \end{subproof}
+
+    We show that if for all $H \in \closeds{X}$ such that $F = \{ N \in \neighbourhoodsSet{H}{X} \mid N \in \closeds{X}\}$ we have $H = \inters{F}$ then $X$ is \teethree\ .
+    \begin{subproof}
+        Omitted.
+    \end{subproof}
 \end{proof}
+
+
+\begin{proposition}\label{teethree_iff_closed_neighbourhood_in_open}
+    Let $X$ be a topological space.
+    Suppose $X$ is inhabited.
+    $X$ is \teethree\ iff for all $U \in \opens[X]$ we have for all $x \in U$ we have there exist $N \in \neighbourhoods{x}{X}$ such that $N \subseteq U$ and $N$ is closed in $X$.    
+\end{proposition}
+\begin{proof}
+    Omitted.
+    %Follows by \cref{teethree_iff_each_closed_is_intersection_of_its_closed_neighborhoods,teethree_implies_closed_neighbourhood_in_open}.
+\end{proof}
+
+
+\begin{proposition}\label{teethree_space_is_teetwo_space}
+    Let $X$ be a \teethree-space.
+    Suppose $X$ is inhabited.
+    Then $X$ is a \teetwo-space.
+\end{proposition}
+\begin{proof}
+    Omitted.
+\end{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 $y \in \carrier[X]$.
+
+
+    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}
+        There exist $C' \in \closeds{X}$ such that $x \in \carrier[X]$ and $x \notin C' \in \closeds{X}$ and there exists $U',V' \in \opens[X]$ such that $x \in U'$ and $C' \subseteq V'$ and $U' \inter V' = \emptyset$.
+        There exists $U',V' \in \opens[X]$ such that $x \in U'$ and $C' \subseteq V'$ and $U' \inter V' = \emptyset$.
+        $U'$ is disjoint from $V'$.
+        $x \in U'$.
+        $x \notin C' \subseteq V'$.
+        $U',V' \in \opens[X]$.
+        $C' \in \closeds{X}$.
+        We show that there exist $K \in \closeds{X}$ such that $x \notin K \ni y$.
+        \begin{subproof}
+            $X$ is Kolmogorov.
+            For all $x',y'\in\carrier[X]$ such that $x'\neq y'$ there exist $H\in\opens[X]$ such that $x'\in H\not\ni y'$ or $x'\notin H\ni y'$.
+            we show that there exist $H \in \opens[X]$ such that $x \notin H \ni y$ or $y \notin H \ni x$.
+            \begin{subproof}
+                Omitted.
+            \end{subproof}
+            $H \subseteq \carrier[X]$ by \cref{opens_type,subseteq}.
+            Since $\carrier[X] \ni x \notin H$ or $\carrier[X] \ni y \notin H$, we have there exist $c \in H$. 
+            Then $H \neq \carrier[X]$.
+            Since $y \in H$ or $x \in H$, we have $H \neq \emptyset$.
+            Let $K = \carrier[X] \setminus H$.
+            $K$ is inhabited.
+            $K \in \closeds{X}$ by \cref{complement_of_open_is_closed}.
+            $x \notin K \ni y$ or $y \notin K \ni x$.
+            \begin{byCase}
+                \caseOf{$y \in K$.} Trivial.
+                \caseOf{$y \notin K$.}
+                Then there exist $U'',V'' \in \opens[X]$ such that $x \in U''$ and $K \subseteq V''$ and $U'' \inter V'' = \emptyset$ by \cref{is_regular}.
+                Let $K' = \carrier[X] \setminus U''$.
+                $x \in K'$.
+                $K' \in \closeds{X}$.
+            \end{byCase}
+
+
+        \end{subproof}
+
+        Follows by assumption.
+    \end{subproof}
+    $y \in V$ by assumption.
+    Follows by assumption.
+
+    
+%\end{proof}