Show that every locally compact Hausdorff space is regular.

Question

I have read through answers on this site and they all seem to rely on using the one-point compactification to prove that locally compact Hausdorff spaces are in fact completely regular. I would like to instead prove it more directly from the definition, using this lemma:

Let $X$ be a Hausdorff space. Then $X$ is locally compact iff for any given $x\in X$ and a neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\overline V$ is compact and $\overline V\subset U$.

Now, we must show that given $x\in X$ and $A$ closed in $X$ with $x\notin A$, that there exist disjoint neighborhoods $U$ and $V$ that contain $x$ and $A$, respectively.

I am not sure how to proceed. I know that compact sets in a Hausdorff space are closed, and that seems important in the proof, but I don't know how to use that fact. Any hints would be appreciated.

Answer 1

I'm just reformulating the answer I linked to.

Take $W$ an open neighborhood of $x$ such that $K=\overline{W}$ is compact (which exists according to your lemma). Define $B=K\cap A$, a closed subset of $K$. Now we can use the fact that compact spaces are regular: in $K$, we can find disjoint open subsets $U'$ and $V'$ with $x\in U'$ and $B\subset V'$.

Then we can take $U=U'\cap W$ and $V=V'\cup (X\setminus K)$: we have $x\in U$, $A\subset V$ and $U\cap V=\emptyset$.

Answer 2

Note that in a Hausdorff space, compacts are closed. In fact, Hausdorff may be weakened to the property KC in this theorem.

So each point of a locally compact space has local neighborhood basis of compact sets, which is a local neighborhood basis of closed sets.

Each point having a local neighborhood basis of closed sets is equivalent to regular: for each point and an open neighborhood, there is an open set whose closure lies within that open neighborhood.

(Bonus: all locally compact regular spaces are completely regular. But you cannot improve this to get normality.)