I'm having trouble fully understanding the proof of this statement.

Suppose $X$ is a locally compact Hausdorff topological space. Then if $K$ is a compact subset of $X$ and $O$ is any open subset of $X$ containing $K$, $\exists$ an open subset $U$ of $X$ such that $K \subseteq U \subseteq \overline{U} \text{ compact} \subseteq O$, with $\overline{U}$ being the closure of $U$.

Here is how I was taught the proof goes:

Let $K \subseteq X$ be compact, and let $O \subseteq X$ be open such that $K \subseteq O$. Then since $O$ is open, $O^{c}$ is closed. But $O^{c}$ closed and $K$ compact implies we can find $U, V$ open such that $K \subseteq U$ and $O^{c} \subseteq V$ with $U \cap V = \emptyset$ (this is because locally compact Hausdorff topological spaces are regular). But if $O^{c} \subseteq V$, then $V^{c} \subseteq O$.

Since $V$ is open, $V^{c}$ is closed. Also, $U \cap V = \emptyset \implies U \subseteq V^{c}$. Since $V^{c}$ is closed, we have the closure of $U$, $\overline{U}$, is contained in $V^{c}$. So, we have $K \subseteq U \subseteq \overline{U} \subseteq O$. We can assume $\overline{U}$ is compact, and so we are done.

First question: Is there an easy way to prove that a locally compact Hausdorff topological space is regular? I wasn't able to do so on my own.

Second question: Why can we assume $\overline{U}$ is compact at the end? Does it have anything to do with the space being locally compact?

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    $\begingroup$ Corollary 3.4 is a proof showing that a LCHS is regular. It doesn't look too bad. Sorry I can't be more help, my topology is pretty rusty. Also for what it's worth, compact subsets of Hausdorff spaces are always compact. books.google.com/… $\endgroup$ – graydad Sep 2 '14 at 5:00
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    $\begingroup$ @user166967 Thanks for the reference. I couldn't find much when I tried googling it myself. Also, did you mean "compact subsets of Hausdorff spaces are always closed"? If so, this doesn't imply closed subsets are necessarily compact. :( $\endgroup$ – layman Sep 2 '14 at 5:04
  • $\begingroup$ @user166967 that is exactly what I meant haha. It may not be the result we need, but it could help us get there. $\endgroup$ – graydad Sep 2 '14 at 5:05
  • $\begingroup$ @layman, could you explain how is that being in a regular space helps you to find open subsets $U$ and $V$ such that one contain a compact and other a closed and $U\cap V=\emptyset$ ? $\endgroup$ – Isa May 28 at 20:14
  • $\begingroup$ AFAIK, the regular space allows you to separate a point and a closed set. $\endgroup$ – Isa May 28 at 20:16

First question: In this link, you can find your desired proof. It invokes the one point compactification and I think that it is a standard one.

I would like to note that, in your case, where you are trying to separate a compact set from a closed set, the proof can be done without invoking regularity (at least, it does not use it explicitly). For the proof, take a look in Theorem 2.7. of Rudin's book.

Second question: Yes, it is related with locally compactness. Indeed, once $K$ is compact, you can find a open neighbourhood of $K$ with compact closure. Indeed, for each $p\in K$, there is $p\in U_p$ open with $\overline{U}_p$ compact. The Family $\{U_p\}_{p\in K}$ is an open cover of $K$, thus, we can assume that $$K\subset \bigcup_{i=1}^N U_{p_i},$$

for some finite $N$, therefore, $$K\subset \bigcup_{i=1}^N \overline{U_{p_i}},$$

which is compact.

  • $\begingroup$ Yes, but is this compact set you found contained within $V^{c}$? $\endgroup$ – layman Sep 2 '14 at 13:53
  • $\begingroup$ Also, is $\bigcup \limits_{i = 1}^{N} U_{p_{i}}$ contained in $V^{c}$? It seems to me that the $U$'s you found are independent of the $U$ and $V$ I picked such that $U \subseteq V^{c}$. $\endgroup$ – layman Sep 2 '14 at 13:55
  • $\begingroup$ That's not the argument. I will leave to you, to think more on the problem. After some thought, come here and I will help you again. $\endgroup$ – Tomás Sep 2 '14 at 13:56
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    $\begingroup$ Well ok, however, be advised that if you have thinked a little bit on the problem, you might have solved it by your own. Consider your $U$ in your proof and my finite collection $\cup_{i=1}^N U_{p_i}$. What can you say about the set $$U\cap \left(\bigcup_{i=1}^N U_{p_i}\right)$$ $\endgroup$ – Tomás Sep 2 '14 at 14:09
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    $\begingroup$ Yes, that's the idea and that's why in your proof, it can be assumed that $\overline{U}$ is compact. $\endgroup$ – Tomás Sep 2 '14 at 15:36

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