Fixed definition of LCH spaces.

src/topology/main/lch.tex

@@ -13,7 +13,7 @@
     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}.
+    (1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact. By \autoref{proposition:compact-closed}, $K$ is closed. Let $U \in \cn^o(x)$ be open, then $K \setminus U$ is closed. By \autoref{proposition:compact-extensions}, $K \setminus U$ is compact. Since $X$ is Hausdorff, for each $y \in K \setminus U$, there exists $U_y \in \cn(x)$ and $V_y \in \cn^o(y)$ open such that $U_y \cap V_y = \emptyset$. As $K \setminus U$ is compact, there exists a $L \subset K \setminus U$ finite such that $\bigcup_{y \in L}V_y \supset K \setminus U$. In which case, $K \setminus \bigcup_{y \in L}V_y$ is compact by \autoref{proposition:compact-extensions}, and a neighbourhood of $x$ because it contains $\bigcap_{y \in L}U_y$. 
 
     (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}