# Show that every locally compact Hausdorff space is regular.

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.

• Have you taken a look at this answer: math.stackexchange.com/a/1325286/318467 ? Nov 20, 2019 at 0:17
• @CaptainLama I don't really follow that answer. It talks about "compact neighborhoods" and "closed neighborhoods." To me, a "neighborhood" is an open set. It also doesn't seem to show the desired result at all. Nov 20, 2019 at 0:21
• A neighborhood of $x$ is a set $V$ such that there is an open subset $U\subset V$ with $x\in U$. If you really have trouble understanding the answer I linked I can try to reformulate it if you want. Nov 20, 2019 at 0:25
• Yes, that is the more general definition of neighborhood. I am following Munkres' convention of using "neighborhood" as "open neighborhood," i.e. "$U$ is an open set containing $x$" is equivalent to "$U$ is a neighborhood of $x$". Nov 20, 2019 at 0:29

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 $$U\cap V=\emptyset$$.