Adjusted citation formats. Moved citation off of named theorems if possible.
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Bokuan Li
2026-03-19 23:58:16 -04:00
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so $V = \bigcup_{j = 1}^n V_{x_j} \in \cn^o(K)$ is precompact.
\end{proof}
\begin{lemma}[Urysohn's Lemma (LCH), {{\cite[Lemma 4.32]{Folland}}}]
\begin{lemma}[Urysohn's Lemma (LCH)]
\label{lemma:lch-urysohn}
Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $f \in C_c(X; [0, 1])$ such that $\supp{f} \subset U$.
\end{lemma}
\begin{proof}
\begin{proof}[Proof {{\cite[Lemma 4.32]{Folland}}}. ]
By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
\[
K \subset V \subset \ol{V} \subset W \subset \ol{W} \subset U

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\section{Regular Spaces}
\label{section:regularspaces}
\begin{definition}[Regular Space, {{\cite[Proposition 1.4.11]{Bourbaki}}}]
\begin{definition}[Regular Space]
\label{definition:regular}
Let $X$ be a topological space, then the following are equivalent:
\begin{enumerate}
@@ -10,7 +10,7 @@
\end{enumerate}
If $X$ is a T1 space such that the above holds, then $X$ is \textbf{regular}.
\end{definition}
\begin{proof}
\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.