# If $X$ is locally compact Hausdorff, then $X$ is regular.

The following are from Munkres' book on 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 $\overline{V} \subseteq U$.

Lemma 31.1 a. Let $X$ be a topological space. Let one-point sets in $X$ be closed. $X$ is regular if and only if given a point $x$ of $X$ and a neighborhood $U$ of x, there is a neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.

Doesn't the $(\implies)$ part of theorem 29.2 and the $(\impliedby)$ part of lemma 31.1 a together prove that locally compact Hausdorff spaces are regular? I'm a little bit suspicious because this seems an easy solution, I might be missing something.

• Why do I need to show that compact Hausdorff spaces are regular? – Kurome May 19 '16 at 14:36
• You’re right: I didn’t read carefully enough and was using a slightly different characterization of local compactness. It’s fine as is. – Brian M. Scott May 19 '16 at 14:40

## 1 Answer

In fact, X is completely regular: Let X be locally compact Hausdorff. Then, there exists a compact Hausdorff space Y containing X as a subspace. Now since Y is normal, Y is completely regular, which implies that X is completely regular, because it is a subspace of a completely regular space.