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Let $X$ be a locally compact Hausdorff space, and $A \subset X$ closed. I want to show that $X - A$ is locally compact.

I have found a proof here: Open subspaces of locally compact Hausdorff spaces are locally compact, but since I am not familiar with the notion of regularity I am looking for another explanation for the (apparent) fact that:

for any $x \in X-A$, I can find an open $V$ such that $x \in V$ and $\overline{V} \subset X-A$.

My efforts so far: Fix $x \in X - A$, then since $X$ is locally compact, $x$ has a compact neighborhood $B\subset X$. So there is a $U$ open in $X$ such that $x \in U \subset B$. $A$ is closed, so $X-A$ is open, therefore $U \cap (X-A)$ is open. This way, I was trying to construct a compact neighborhood of $x$ in $X-A$, but I can't figure out how. Any help is much appreciated!

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    $\begingroup$ Just to clarify: Are you looking for a proof that locally compact Hausdorff spaces are regular? Or trying to show that open subsets of locally compact Hausdorff spaces are locally compact without using the notion of regular? $\endgroup$
    – graydad
    Jan 26, 2015 at 18:02
  • $\begingroup$ The latter: I would like to show it without the notion of regular. Thanks for clarifying that! $\endgroup$
    – linx
    Jan 26, 2015 at 18:04
  • $\begingroup$ Thanks! I'll try to think of a way around it. Would you be opposed to proving that locally compact Hausdorff spaces are regular though? If you complete that proof, then the proof in the link you provided would make sense, and you'd get to add the notion of "regular" spaces to your Topology arsenal. It doesn't show up a ton, but I do remember "regular" spaces showing up multiple times in proofs about compactness. $\endgroup$
    – graydad
    Jan 26, 2015 at 18:06
  • $\begingroup$ Not opposed, but I haven't been taught about regularity, but the exercise is in the book, so I'm curious what the solution can be. $\endgroup$
    – linx
    Jan 26, 2015 at 18:11
  • $\begingroup$ I'm not sure I follow; aren't $X\setminus A$ and $B$ open in $X$, so there union should be open? $\endgroup$
    – graydad
    Jan 26, 2015 at 18:22

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Suppose $U$ is an open subset of a locally compact Hausdorff space $X$, $K\subset U$ and $K$ is compact. Then there is an open set $V$ with compact closure such that $$K\subset V\subset \bar{V}\subset U.$$

This is Theorem 2.7 in Rudin's Real and Complex analysis. To prove it first you need to show (the easy) fact that in a Hausdorff space compact subsets and points can be separated. More precisely, if $K\subset X$ is compact and $p\in X\setminus K$, then there exist disjoint open sets $V, W$ containing $K$ and $p$, respectively.

For your purposes you need a special case of this theorem where $K$ is a point in the open set $U$.

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  • $\begingroup$ Thank you, this is very clear. I now remember that we indeed learned about this theorem you mention, that disjoint compact sets can be separated in a Hausdorff space. Wouldn't a singleton be closed in $X$ Hausdorff, and therefore compact? $\endgroup$
    – linx
    Jan 27, 2015 at 10:32
  • $\begingroup$ Could I say $x$ is closed, $X\setminus K$ is closed, so both are compact. So there must exist disjoint open sets $U,V$ such that $x \in U$, $X\setminus K \subset V$. then by [math.stackexchange.com/questions/108852/…, the closure of $U$ is disjoint from $V$, and therefore a compact neighborhood of $x$ in $X\setminus K$. $\endgroup$
    – linx
    Jan 27, 2015 at 10:51

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