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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.

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  • $\begingroup$ Have you taken a look at this answer: math.stackexchange.com/a/1325286/318467 ? $\endgroup$ Nov 20, 2019 at 0:17
  • $\begingroup$ @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. $\endgroup$
    – Math1000
    Nov 20, 2019 at 0:21
  • $\begingroup$ 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. $\endgroup$ Nov 20, 2019 at 0:25
  • $\begingroup$ 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 $x$". $\endgroup$
    – Math1000
    Nov 20, 2019 at 0:29

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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 $U\cap V=\emptyset$.

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  • $\begingroup$ Thanks for the answer. Just one question - I don't see where you used the Hausdorff property. Is it that compact Hausdorff spaces are regular, not just compact spaces? $\endgroup$
    – Math1000
    Nov 20, 2019 at 3:13
  • $\begingroup$ Ah sorry yes, I got tricked by habit of using the French terminology. By compact I mean Hausdorff compact. $\endgroup$ Nov 20, 2019 at 3:33
  • $\begingroup$ It makes sense now. Thanks again! $\endgroup$
    – Math1000
    Nov 20, 2019 at 3:34

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