\section{Locally Compact Hausdorff Spaces}
\label{section:lch}

\begin{definition}[Locally Compact Hausdorff Space]
\label{definition:lch}
    Let $X$ be a Hausdorff space, then the following are equivalent:
    \begin{enumerate}
        \item For any $x \in X$, there exists $K \in \cn(x)$ compact.
        \item For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of compact sets.
        \item For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of precompact sets.
    \end{enumerate}

    If the above holds, then $X$ is a \textbf{locally compact Hausdorff (LCH)} space.
\end{definition}
\begin{proof}
    (1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact and $U \in \cn(x)$, then $\overline{U \cap K}$ is closed. By \autoref{proposition:compact-closed}, $K$ itself is closed, so $\overline{U \cap K} \subset K$ is a closed subset of a compact set, and compact by \autoref{proposition:compact-extensions}.

    (2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By \autoref{proposition:compact-closed}, $K$ is closed, so $\overline{K^o} \subset K$ is compact by \autoref{proposition:compact-extensions}.
\end{proof}

\begin{lemma}
\label{lemma:lch-compact-neighbour}
    Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exits $V \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$.
\end{lemma}
\begin{proof}
    For each $x \in K$, there exists $V_x \in \cn^o(x)$ be precompact such that $x \in V_x \subset \overline{V_x} \subset U$ by (3) of \autoref{definition:lch}. Since $K$ is compact, there exists $\seqf{x_j} \subset K$ such that
    \[
    K \subset \bigcup_{j = 1}^n V_{x_j} \subset U
    \]

    By \autoref{proposition:closure-finite-union},
    \[
    \ol{\bigcup_{j = 1}^n V_{x_j}} = \bigcup_{j = 1}^n \overline{V_{x_j}} \subset U
    \]

    so $V = \bigcup_{j = 1}^n V_{x_j} \in \cn^o(K)$ is precompact.
\end{proof}

\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}[Proof {{\cite[Lemma 4.32]{Folland}}}. ]
    By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
    \[
    K \subset V \subset \ol{V} \subset W \subset \ol{W} \subset U
    \]

    As $\ol{W}$ is compact, it is normal by \autoref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \autoref{proposition:compact-closed}.

    By \hyperref[Urysohn's lemma]{lemma:urysohn}, there exists $f \in C(\ol{V}; [0, 1])$ such that $f|_K = 1$ and $f|_{\ol{W} \setminus V} = 0$. Let
    \[
    F: X \to [0, 1] \quad x \mapsto \begin{cases}
    f(x) &x \in W \\
    0 &x \in X \setminus \ol{V}
    \end{cases}
    \]

    then by the \hyperref[gluing lemma for continuous functions]{lemma:gluing-continuous}, $F \in C_c(X; [0, 1])$ with $F|_{K} = 1$ and $\supp{f} \subset U$.
\end{proof}

\begin{theorem}[Tietze Extension Theorem (LCH)]
\label{theorem:lch-tietze}
    Let $X$ be a LCH space, $K \subset X$ be compact, $U \in \cn^o(K)$, and $f \in C(K; \real)$, then there exists $F \in C_c(U; \real)$ such that $F|_K = f$.
\end{theorem}
\begin{proof}
    By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$. As $\ol{W}$ is compact, it is normal by \autoref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \autoref{proposition:compact-closed}.

    By the \hyperref[Tietze extension theorem]{theorem:tietze}, there exists $F \in C(\ol{W}; \real)$ such that $F|_K = f$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\eta \in C_c(X; [0, 1])$ such that $\eta|_K = 1$ and $\supp{\eta} \subset V$. In which case, define
    \[
    \ol F: X \to \real \quad x \mapsto \begin{cases}
        \eta(x) \cdot f(x) &x \in V \\
        0 &x \in X \setminus \supp{\eta}
    \end{cases}
    \]

    then by the \hyperref[gluing lemma for continuous functions]{lemma:gluing-continuous}, $\ol F \in C_c(X; \real)$ with $\ol F|_K = F|_K = f$ and $\supp{F} \subset \supp{\eta} \subset V \subset U$.
\end{proof}

\begin{proposition}
\label{proposition:lch-compactly-generated}
    Let $X$ be a LCH space, then:
    \begin{enumerate}
        \item $X$ is compactly generated. 
        \item For any uniform space $Y$, $C(X; Y) \subset Y^X$ is closed with respect to the compact-open topology. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Let $U \subset X$ such that $U \cap K$ is open in $K$ for all $K \subset X$ compact. For any $x \in U$, there exists a compact neighbourhood $K \in \cn(x)$. In which case, $U \supset U \cap K \in \cn(x)$, so $U \in \cn(x)$ for all $x \in U$. By \autoref{lemma:openneighbourhood}, $U$ is open. 

    (2): By \autoref{proposition:compact-uniform-open}, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$. Since $X$ is compactly generated, $C(X; Y) \subset Y^X$ is closed with respect to the compact-open topology by \autoref{corollary:uniform-limit-continuous-generated}. 
\end{proof}


\begin{proposition}
\label{proposition:lch-product}
    Let $\seqi{X}$ be a family of LCH spaces. If $X_i$ is compact for all but finitely many $i \in I$, then $X = \prod_{i \in I}X_i$ is a LCH space. 
\end{proposition}
\begin{proof}
    By \autoref{proposition:product-hausdorff}, $\prod_{i \in I}X_i$ is Hausdorff. Let $x \in \prod_{i \in I}X_i$ and $i \in I$. If $X_i$ is not compact, let $U_i \in \cn_{X_i}(\pi_i(x))$ be compact. Otherwise, let $U_i = X_i$. Let $U = \prod_{i \in I}U_i$, then since $U_i \ne X_i$ for only finitely many $i \in I$, $U \in \cn_X(x)$. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $U$ is compact. Therefore $X$ is locally compact. 
\end{proof}


\subsection{Paracompactness and LCH Spaces}
\label{subsection:lch-paracompact}



\begin{proposition}[{{\cite[Proposition 4.39]{Folland}}}]
\label{proposition:lch-sigma-compact}
    Let $X$ be a LCH space, then the following are equivalent:
    \begin{enumerate}
        \item $X$ is $\sigma$-compact.
        \item There exists an exhaustion of $X$ by compact sets.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1) $\Rightarrow$ (2): Let $\seq{K_n} \subset 2^X$ be compact such that $\bigcup_{n \in \natp}K_n = X$, and $U_0 = \emptyset$.

    Assume inductively that $\bracs{U_j}_0^n$ has been constructed such that:
    \begin{enumerate}
        \item[(a)] For each $0 \le k \le n$, $U_k$ is a precompact open set.
        \item[(b)] For each $0 \le k < n$, $\overline{U_k} \subset U_{k+1}$.
        \item[(c)] For each $1 \le k \le n$, $U_k \supset \bigcup_{j = 1}^k K_j$.
    \end{enumerate}

    By \autoref{lemma:lch-compact-neighbour}, there exists $U_{n+1} \in \cn^o(\overline{U_n} \cup K_{n+1})$ precompact. In which case, by (c),
    \[
    U_{n+1} \supset \ol{U_n} \cup K_{n+1} \supset \bigcup_{j = 1}^n K_j \cup K_{n+1} = \bigcup_{j = 1}^{n+1}K_j
    \]

    Thus $\bracs{U_j}_0^{n+1}$ satisfies (a), (b), and (c), and $\seq{U_n}$ is an exhaustion of $X$ by compact sets.
\end{proof}

\begin{proposition}[{{\cite[Proposition 4.41]{Folland}}}]
\label{proposition:lch-partition-of-unity}
    Let $X$ be a LCH space, $K \subset X$ be compact, and $\seqi{U}$ be an open cover of $K$, then there exists a $C_c$ partition of unity on $K$ subordinate to $\seqi{U}$.
\end{proposition}
\begin{proof}
    Since $K$ is compact, assume without loss of generality that $\seqi{U} = \seqf{U_j}$.

    For every $x \in K$, there exists $1 \le j \le n$ and $N_x \in \cn(x)$ compact such that $x \in N_x \subset U_j$. By compactness of $K$, there exists $\seqf[m]{x_j} \subset K$ such that $K = \bigcup_{j = 1}^m N_{x_j}$.

    For each $1 \le j \le n$, let
    \[
    F_j = \bigcup_{\substack{1 \le k \le m \\ N_{x_k} \subset U_j}}N_{x_k}
    \]

    then $F_j \subset U_j$ is compact, and $\bigcup_{j = 1}^n F_j \supset K$.

    By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{f_j} \subset C_c(X; [0, 1])$ such that for each $1 \le j \le n$, $f_j|_{F_j} = 1$, and $\supp{f_j} \subset U_j$.

    By Urysohn's lemma again, there exists $f_{j + 1} \in C(X; [0, 1])$ such that $f_{j+1}|_{K} = 0$ and $\bracs{f_{j+1} = 0} \subset \bigcup_{j = 1}^n \supp{f_j}$. Let $F = \sum_{j = 1}^{n+1}f_j$, then $F(x) > 0$ for all $x \in X$. For each $1 \le j \le n$, let $g_j = f_j/F$, then $g_j \in C_c(X; [0, 1])$ with $\supp{g_j} \subset U_j$. In addition, since $f_{j+1}|_K = 0$,
    \[
    \sum_{j = 1}^n g_j|_K = \frac{\sum_{j = 1}^n f_j}{F} = \frac{\sum_{j = 1}^n f_j}{\sum_{j = 1}^n f_j} = 1
    \]

    Therefore $\seqf{g_j}$ is the desired partition of unity.
\end{proof}

\begin{lemma}
\label{lemma:lch-locally-finite-precompact-refine}
    Let $X$ be a LCH space and $\ce \subset 2^X$ be a locally finite precompact open cover of $X$, then there exists locally finite precompact open covers $\bracs{F_E}_{E \in \ce}, \bracs{G_E}_{E \in \ce} \subset 2^X$ such that for each $E \in \ce$, $F_E \subset \ol{F_E} \subset E \subset \ol{E} \subset G_E$.
\end{lemma}
\begin{proof}
    $(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracs{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by \autoref{lemma:locally-finite-compact}. Let
    \[
    F_E = \bigcup_{\substack{F \in \ce} \\ F \cap \ol E \ne \emptyset}F
    \]

    then $F_E \in \cn(\ol{E})$ is precompact.

    Let $N \subset X$ and $E \in \ce$. If $N \cap F_E \ne \emptyset$, then there exists $F \in \ce$ such that $N \cap F \ne \emptyset$ and $F \cap \ol{E} \ne \emptyset$. Thus
    \[
    \bracs{E \in \ce|N \cap F_E \ne \emptyset} \subset \bigcup_{\substack{F \in \ce \\ F \cap N \ne \emptyset}}\bracs{E \in \ce|F \cap \ol{E} \ne \emptyset} \subset \bigcup_{\substack{F \in \ce \\ F \cap N \ne \emptyset}}\bracs{E \in \ce|\ol{F} \cap \ol{E} \ne \emptyset}
    \]

    By \autoref{lemma:locally-finite-closure}, $\bracsn{\ol E|E \in \ce}$ is also locally finite. Hence for every $F \in \ce$, $\bracsn{E \in \ce|F \cap \ol{E} \ne \emptyset}$ is finite.

    Let $x \in X$, then there exists $N \in \cn(x)$ such that $\bracs{F \in \ce|N \cap F \ne \emptyset}$ is finite. In which case, $\bracs{E \in \ce|N \cap F_E \ne \emptyset}$ is finite as well. Therefore $\bracs{F_E}_{E \in \ce}$ is locally finite.

    $(\bracs{G_E}_{E \in \ce})$: For each $x \in X$, there exists $E \in \ce$ and $N_x \in \cn^o(x)$ precompact with $x \in N_x \subset \ol{N_x} \subset E$.

    For any $E \in \ce$, $\ol{E}$ is compact, so there exists $X_E \subset X$ finite such that
    \begin{enumerate}
        \item[(a)] $\ol{E} \subset \bigcup_{x \in X_E}N_x$.
        \item[(b)] For every $x \in X_E$, $N_x \cap E \ne \emptyset$.
    \end{enumerate}

    Let $X_{\ce} = \bigcup_{E \in \ce}X_\ce$, and for each $E \in \ce$, let
    \[
    G_E = \bigcup_{\substack{x \in X_\ce \\ \ol{N_x} \subset E}}N_x
    \]

    then $\bracs{G_E}_{E \in \ce}$ is an open cover of $X$. Since $G_E \subset E$ for all $E \in \ce$, $\bracs{G_E}_{E \in \ce}$ is locally finite.

    It remains to show that $\ol{G_E} \subset E$. Let $x \in X_F$ such that $N_x \subset E$, then $N_x \cap F \ne \emptyset$. Since $N_x \subset E$, $E \cap F \ne \emptyset$. Thus
    \[
    \bracsn{x \in X_\ce|\ol{N_x} \subset E} \subset \bigcup_{\substack{F \in \ce \\ E \cap F \ne \emptyset}}X_F \subset \bigcup_{\substack{F \in \ce \\ \ol E \cap F \ne \emptyset}}X_F
    \]

    is finite by \autoref{lemma:locally-finite-compact}, so
    \[
    \ol{G_E} = \bigcup_{\substack{x \in X_\ce \\ \ol{N_x} \subset E}}\ol N_x \subset E
    \]

    by \autoref{proposition:closure-finite-union}.
\end{proof}

\begin{proposition}
\label{proposition:lch-paracompact}
    Let $X$ be a LCH space, then the following are equivalent:
    \begin{enumerate}
        \item $X$ is paracompact.
        \item There exists a locally finite precompact open cover $\cf$ of $X$.
        \item For any open cover $\mathcal{U}$ of $X$, there exists a locally finite refinement $\mathcal{V}$ of $\mathcal{U}$ consisting of precompact open sets.
        \item For any open cover $\mathcal{U}$ of $X$, there exists locally finite refinements $\seqi{V}, \seqi{W} \subset 2^X$ of $\mathcal{U}$ consisting of precompact open sets such that $\ol{W_i} \subset V_i$ for all $i \in I$.
        \item For any open cover $\mathcal{U}$ of $X$, there exists a $C_c(X; [0, 1])$ partition of unity subordinate to it.
        \item $X$ admits a $C_c(X; [0, 1])$ partition of unity.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1) $\Rightarrow$ (2): For each $x \in X$, there exists a precompact open neighbourhood $U_x \in \cn^o(x)$. Since $\bracs{U_x| x \in X}$ is an open cover of $X$, there exists a locally finite refinement $\mathcal{V}$. For each $V \in \mathcal{V}$, there exists $x \in X$ such that $V \subset U_x$. In which case, $\ol{V} \subset \ol{U_x}$ is compact.

    (2) $\Rightarrow$ (3): Let $\cf \subset 2^X$ be a locally finite open cover of $X$ consisting of precompact open sets. By \autoref{lemma:lch-locally-finite-precompact-refine}, there exists a locally finite open cover $\bracs{G_F}_{F \in \cf}$ of $X$ consisting of precompact open sets such that $\ol{F} \subset G_F$ for all $F \in \cf$.

    For each $F \in \cf$, let
    \[
    \mathcal{U}_F = \bracs{U \cap G_F|U \in \mathcal{U}}
    \]

    then $\mathcal{U}_F$ is a precompact open cover of $\ol{F}$. By compactness of $\ol{F}$, there exists $\mathcal{V}_F \subset \mathcal{U}_F$ finite such that $\ol{F} \subset \bigcup_{V \in \mathcal{V}_F}V$.

    Let $\mathcal{V} = \bigcup_{F \in \cf}\mathcal{V}_F$, then $\mathcal{V}$ is a precompact open cover of $X$. For any $x \in X$, there exists $N \in \cn(x)$ such that $\bracs{F \in \cf|N \cap G_F}$ is finite. Thus
    \[
    \bracs{V \in \mathcal{V}| N \cap V} \subset \bigcup_{\substack{F \in \cf \\ N \cap G_F \ne \emptyset}}\mathcal{V}_F
    \]

    is finite, and $\mathcal{V}$ is locally finite.

    (3) $\Rightarrow$ (4): By \autoref{lemma:lch-locally-finite-precompact-refine}.

    (4) $\Rightarrow$ (5): Let $\seqi{V}, \seqi{W} \subset 2^X$ be locally finite refinements of $\mathcal{U}$ consisting of precompact open sets such that for each $i \in I$, $\ol{W_i} \subset V_i$.

    By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqi{f} \in C_c(X; [0, 1])$ such that for each $i \in I$, $f_i|_{\ol{W_i}} = 1$ and $\supp{f_i} \subset V_i$.

    Let $F = \sum_{i \in I}f_i$. For each $x \in X$, there exists $N_x \in \cn^o(x)$ such that $\bracs{i \in I|N_x \cap V_i \ne \emptyset}$ is finite. In which case,
    \[
    F|_{N_x} = \sum_{\substack{i \in I \\ N_x \cap V_i \ne \emptyset}}f_i|_{N_x}
    \]

    thus $F|_{N_x} \in C(N_x; \real)$. By \autoref{lemma:gluing-continuous}, $F \in C(X; \real)$.

    Since $\seqi{W}$ is an open cover of $X$, $F(x) > 0$ for all $x \in X$. For each $i \in I$, let $g_i = f_i/F$, then $g_i \in C_c(X; [0, 1])$ with $\supp{g_i} = \supp{f_i} \subset W_i$. For any $x \in X$, there exists $N_x \in \cn^o(x)$ such that $\bracs{i \in I|N_x \cap W_i \ne \emptyset}$ is finite. In which case, $\bracs{i \in I|0 < g_i|_{N_x}}$ is also finite. Thus $\seqi{g}$ is a $C_c$ partition of unity subordinate to $\mathcal{U}$.

    (5) $\Rightarrow$ (1): Let $\mathcal{U}$ be an open cover of $X$ and $\seqi{f} \subset C_c(X; [0, 1])$ subordinate to $\mathcal{U}$. For each $i \in I$, let $V_i = \bracs{f_i > 0}$, then $\seqi{V}$ is a locally finite refinement of $\mathcal{U}$.

    (5) $\Rightarrow$ (6): Take $\mathcal{U} = \bracs{X}$.

    (6) $\Rightarrow$ (2): Let $\seqi{f} \subset C_c(X; [0, 1])$ be a partition of unity. For each $i \in I$, let $V_i = \bracs{f_i > 0}$, then $\seqi{V}$ is a locally finite precompact open cover of $\mathcal{U}$.
\end{proof}

\begin{proposition}
\label{proposition:lch-sigma-paracompact}
    Let $X$ be a $\sigma$-compact LCH space, then $X$ is paracompact.
\end{proposition}
\begin{proof}
    By \autoref{proposition:lch-sigma-compact}, there exists an exhaustion $\seq{U_n} \subset 2^X$ of $X$ by precompact open sets. Denote $U_0 = \emptyset$. For each $n \in \natp$, let $V_n = U_{n+1} \setminus \ol{U_{n-1}}$.

    Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_n \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n} \setminus \ol{U_{n - 2}} = V_{n-1}$. If $n = 1$, then $x \in U_{2} = V_1$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_m \cap V_n \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of \autoref{proposition:lch-paracompact}, $X$ is paracompact.
\end{proof}