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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.

Math1000
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  • Have you taken a look at this answer: https://math.stackexchange.com/a/1325286/318467 ? – Captain Lama Nov 20 '19 at 00:17
  • @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. – Math1000 Nov 20 '19 at 00:21
  • 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. – Captain Lama Nov 20 '19 at 00:25
  • 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 $x$". – Math1000 Nov 20 '19 at 00:29

1 Answers1

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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 $U\cap V=\emptyset$.

Captain Lama
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