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diff --git a/library/topology/separation.tex b/library/topology/separation.tex index 535a51f..3268bb9 100644 --- a/library/topology/separation.tex +++ b/library/topology/separation.tex @@ -145,3 +145,46 @@ $x\in U$ 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 $V$.
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