38 lines
2.4 KiB
TeX
38 lines
2.4 KiB
TeX
\section{Regular Spaces}
|
|
\label{section:regularspaces}
|
|
|
|
\begin{definition}[Regular Space, {{\cite[Proposition 1.4.11]{Bourbaki}}}]
|
|
\label{definition:regular}
|
|
Let $X$ be a topological space, then the following are equivalent:
|
|
\begin{enumerate}
|
|
\item For each $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U \in \cn(x)$ and $V \in \cn(A)$ such that $U \cap V = \emptyset$.
|
|
\item For each $x \in X$, the closed neighbourhoods of $x$ forms a fundamantal system of neighbourhoods at $x$.
|
|
\end{enumerate}
|
|
If $X$ is a T1 space such that the above holds, then $X$ is \textbf{regular}.
|
|
\end{definition}
|
|
\begin{proof}
|
|
$(1) \Rightarrow (2)$: Let $U \in \cn^o(x)$, then $U^c$ is closed with $x \not\in U^c$ by (V1). Thus there exists $V \in \cn^o(U^c)$ such that $x \not\in V$, so $V^c \in \cn(x)$ is closed with $V^c \subset U$.
|
|
|
|
$(2) \Rightarrow (1)$: Since $X$ is Hausdorff, $X$ is T1 as well. Let $x \in X$ and $A \subset X$ closed such that $x \not\in A$, then $A^c \in \cn(x)$. So there exists $K \in \cn(x)$ closed such that $x \in K \subset A^c$. Thus $K \in \cn(x)$ and $K^c \in \cn(A)$ are the desired neighbourhoods.
|
|
\end{proof}
|
|
|
|
\begin{theorem}[{{\cite[Theorem 1.8.1]{Bourbaki}}}]
|
|
\label{theorem:regularextension}
|
|
Let $X$ be a topological space, $Y$ be a regular space, $A \subset X$, and $f \in C(A; Y)$, then the following are equivalent:
|
|
\begin{enumerate}
|
|
\item There exists $F \in C(X; Y)$ such that $F|_A = f$.
|
|
\item For each $x \in X$, $\lim_{y \to x, y \in A}f(y)$ exists.
|
|
\end{enumerate}
|
|
\end{theorem}
|
|
\begin{proof}
|
|
$(\Rightarrow)$: Let $x \in X$ and $\fF = \bracs{U \cap A| U \in \cn(x)}$, then $f(\fF) = F(\fF)$. By continuity of $F$, $F(\fF)$ converges to $F(x)$, so $\lim_{y \to x, x \in A}f(x)$ exists.
|
|
|
|
$(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_A = f$ by (4) of \ref{definition:hausdorff}.
|
|
|
|
Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of \ref{definition:regular}, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^o(x)$ such that $f(U \cap A) \subset V$. In which case, \ref{lemma:openneighbourhood} implies that $U \in \cn^o(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
|
|
\[
|
|
F(y) = \lim_{\substack{z \to y \\ z \in A}}f(z) \in \ol{f(U \cap A)} \subset V
|
|
\]
|
|
as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.
|
|
\end{proof}
|