# Theorem 29.2 - Munkres's Topology

Theorem 29.2. Let $X$ be a Hausdorff space. Then $X$ is locally compact if and only if given $x \in X$, and given a neighborhood $U$ of $x$,there is a neighborhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$.

Proof: Clearly this new formulation implies local compactness; the set $C = \overline{V}$ is the desired compact set containing a neighborhood of $x$. To prove the converse, suppose $X$ is locally compact, let $x$ be a point of $X$ and let $U$ be a neighborhood of $x$. Take the one-point compactification $Y$ of $X$, and let $C$ be the set $Y - U$. Then $C$ is closed in $Y$, so that $C$ is a compact subspace of $Y$. Apply Lemma 26.4 to choose disjoint open sets $V$ and $W$ containing $x$ and $C$, respectively. Then the closure $\overline{V}$ of $V$ in $Y$ is compact, furthermore, $\overline{V}$ is disjoint from $C$, so that $\overline{V} \subset U$, as desired.

I didn't understand how exactly the fact of $V$ and $W$ are disjoint opent sets containing $x$ and $C$, respectively imply that $\overline{V}$ of $V$ in $Y$ is compact, someone can help me? Thanks in advance!

The set $\overline{V}$ is closed, and since $Y$ is compact, closed subsets of $Y$ are compact by Theorem 26.2 in Munkres.
The subspace $\overline{V}$ is compact since $Y$ is a compact space, and since closed subspaces of compact spaces are compact, you have the result.