Polished A-A and added new lines for broken enumerates.

src/topology/main/lch.tex

@@ -8,7 +8,8 @@
         \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}
+    \end\{enumerate\}
+
     If the above holds, then $X$ is a \textbf{locally compact Hausdorff (LCH)} space.
 \end{definition}
 \begin{proof}
@@ -76,6 +77,35 @@
     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:
@@ -92,7 +122,8 @@
         \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}
+    \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
@@ -154,7 +185,8 @@
     \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}
+    \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
@@ -238,12 +270,4 @@
     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}
 
-\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}
-