Added the Singer representation theorem.

src/topology/main/lch.tex

@@ -237,3 +237,13 @@
 
     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}
+
+