\section{Regular Spaces}
\label{section:regularspaces}

\begin{definition}[Regular Space]
\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 fundamental 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}[Proof {{\cite[Proposition 1.4.11]{Bourbaki}}}. ]
    $(1) \Rightarrow (2)$: 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 \autoref{definition:hausdorff}.

    Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of \autoref{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, \autoref{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}

\begin{proposition}
\label{proposition:second-countable-regular}
    Let $X$ be a second countable regular space, then $X$ is normal. 
\end{proposition}
\begin{proof}
    Let $\cb \subset 2^X$ be a base for $X$ and $A, B \subset X$ be disjoint closed sets. For each $x \in A$ and $y \in B$, there exists $U_x, V_y \in \cb$ such that $x \in \ol{U_x} \subset B^c$ and $y \in \ol{V_y} \subset B^c$. Since $\cb$ is countable, let $\seq{U_n}$ and $\seq{V_n}$ be enumerations of $\bracs{U_x|x \in A}$ and $\bracs{V_y|y \in B}$, respectively. 

    For each $n \in \natp$, let
    \[
    U_n' = U_n \setminus \bigcup_{j = 1}^n \ol{V_j} \quad V_n' = V_n \setminus \bigcup_{j = 1}^n \ol{V_j}
    \]

    then $U_n'$ and $V_n'$ are both open. Let
    \[
    U = \bigcup_{n \in \natp}U_n' \quad V = \bigcup_{n \in \natp}V_n'
    \]

    then $U \in \cn_X(A)$ and $V \in \cn_X(B)$. For each $m, n \in \natp$ with $m \le n$, $V_n \cap \bigcup_{j = 1}^n U_j = \emptyset$, so $U_m \cap V_n = \emptyset$. Likewise, if $m \ge n$, then $U_m \cap V_n = \emptyset$ as well. Therefore $U \cap V = \emptyset$. 
\end{proof}