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If $X$ is locally compact Hausdorff, then $X$ is regular.

Proof: Suppose that $x$ is some point not in the closed set $C$. Then $x$ is in the open set $C^c$, and $X$ being LCH implies there exists an open set $U$ such that $x \in U \subseteq \overline {U} \subseteq C^c$, where $\overline{U}$ is compact. Taking the complement, we get $C \subseteq \overline{U}^c$, and obviously $U \cap \overline{U}^c = \emptyset$.

Something about this seems fishy, especially since I never used the fact that $U$ has compact closure.

For completeness, we know:

Theorem 29.2: Let $X$ be Hausdorff. Then $X$ is locally compact if and only if given $x\in X$, and given a neighborhood $U$ of $X$ there is a neighborhood $V$ of $x$ such that $\overline V$ is compact and $\overline V\subseteq U.$

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  • $\begingroup$ How does LCH imply the existence of such a $U$? (You must use at least some compactness in that, since not all Hausdorff spaces are regular.) $\endgroup$ – Daniel Fischer Jan 17 '18 at 17:13
  • $\begingroup$ How do we know that $\overline{U}\subseteq C^c$ just because $\overline U$ is compact? All we know is that there is an open set $V$ containing $X$ such that $\overline{V}$ is compact. Setting $U=V\cap C^c$ gives that $\overline{U}$ is also compact, but it doesn't show that $\overline{U}\subseteq C^c.$ $\endgroup$ – Thomas Andrews Jan 17 '18 at 17:26
  • $\begingroup$ But you are correct that it is enough to show such a $U$ exists. $\endgroup$ – Thomas Andrews Jan 17 '18 at 17:30
  • $\begingroup$ @DanielFischer Theorem 29.2: "Let $X$ be Hausdorff. Then $X$ is locally compact if and only if given $x \in X$, and given a neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$." From this theorem, I get my set $U$ such that $\overline{U} \subseteq C^c$, but I figured that $\overline{U}$ being compact would be used in the proof somewhere. $\endgroup$ – user193319 Jan 17 '18 at 17:30
  • $\begingroup$ Then everything interesting happened in the proof of 29.2. That theorem says that a Hausdorff space is locally compact if and only if the compact neighbourhoods form a neighbourhood basis at each point. A Hausdorff space is regular if and only if the closed neighbourhoods form a neighbourhood basis at each point (depending on the used nomenclature, that may not need the Hausdorffness restriction). Since compact subsets are closed in Hausdorff spaces, the implication is direct. $\endgroup$ – Daniel Fischer Jan 17 '18 at 18:30
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Given that you know Theorem 29.2, then what you are using for local compactness is just that $U$ exists such that $\overline{U}\subseteq C^c.$

While you aren't using that $\overline U$ is compact, it is not true in general that $U$ exists if you don't have local compactness of $X.$

Indeed, "regular" can be seen as equivalent to:

Given any $x\in X$ and any neighborhood $V$ of $x$, there is a neighborhood $U$ of $X$ such that $\overline{U}\subseteq V.$

So Theorem 29.2 is stating that, for Hausdorff spaces, local compactness is a stronger form of this definition, with the added requirement that $\overline{U}$ is compact. The proof of 29.2 contains all the hard work.


(Correction: Contrary to my claim in prior versions of this document, this can all proceed "at $x$," rather than requiring global properties. Here's the outline.)

Interestingly, if we define "Hausdorff," "locally compact," and "regular" at a point as:

$X$ is "locally compact at $x$" if there is some open neighborhood $U$ of $x$ such that $\overline{U}$ is compact.

and:

$X$ is "regular at $x$" if, for any closed set $C$ not containing $x$, there exists disjoint open sets $U,V$ such that $x\in U, C\subseteq V.$

and:

$X$ is "Hausdorff at $x$" if, for any $y\neq x$ there are disjoint open sets $U,V$ with $x\in U$ and $y\in V.$

Then we get the following version of $29.2$:

Theorem 29.2-L: Given that $X$ is Hausdorff at $x$, then $X$ is locally compact at $x$ if and only if for any open neighborhood $V$ of $x$, we have an open neighborhood $U$ of $x$ such that $\overline{U}$ is compact and $\overline{U}\subseteq V.$

Proof: Let $W$ be the neighborhood of $x$ such that $\overline{W}$ is compact. Let $U_1=W\cap V.$ Let $C=\overline{U_1}\cap V^c$, which is a closed subset of the compact $\overline U$ and hence compact.

For each $y\in C$, find a neighborhood $W_y$ of $y$ such that $x\not\in\overline{W_y}.$ (You can find such $W_y$ because $X$ is Hausdorff at $x.$)

Then $\{W_y\}_{y\in C}$ is an open cover of $C$, so there are $y_1,\dots, y_n$ such that $W_{y_1}\cup\cdots\cup W_{y_n}$ covers $C$.

Now let $U=U_1\setminus\left(\overline{W_{y_1}}\cup\cdots\cup\overline{W_{y_n}}\right).$

We know that $U$ is open and contains $x$ and is a subset of $W$, so $\overline{U}$ must be compact.

The last step is to prove that $\overline{U}\subseteq V.$ But $\overline{U}$ is contained in the closed set $\overline{U_1}\setminus\left(W_{y_1}\cup\cdots\cup W_{y_n}\right)$ which can't contain any elements of $V^c$, since any element in $\overline{U_1}\cap V^c=C$ is necessarily in some $W_{y_i}.$

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