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Show that a locally compact Hausdorff space $(X,\tau)$ is regular.

I have already shown that a compact Hausdorff space is regular.

My textbook proposes 2 methodes, but I get stuck at both. The first method looks the most elegant, but how can I continue?

Compactification method.

Consider the compactification $(X_\infty, \tau_\infty)$ where $X_\infty = X\sqcup \infty$ and $\tau_\infty = \tau \cup \{ X_\infty \setminus K: K \text{ is a compact subset of } X\}$

Then $(X_\infty, \tau_\infty)$ is regular.

Let $x\in X\setminus F, F\subseteq X$ closed in $\tau$. Then $x\in X_\infty$ and $F\sqcup \{ \infty\}$ is closed in $\tau_\infty$. By regularity:

$$(\exists U_\infty, V_\infty \in \tau_\infty)(x\in U_\infty, (F\sqcup \{ \infty\} ) \subseteq V_\infty, U_\infty\cap V\infty = \varnothing)$$

Since $\infty \in V_\infty$ then $U_\infty \in \tau$.

How can I prove $V_\infty \setminus \{ \infty\} \in \tau?$ Any pointers?

Direct method:

Consider $x\in X\setminus F, F\subseteq X$ closed in $\tau$.

We are on the lookout for $(U, V\in \tau)$ such that $U\cap V = \varnothing, x\in U, F\subseteq V$.

Since $x\in X$ there (exists $K\subseteq X$ compact)($K$ is a neighbourhood of $x$)

Consider the compact subspace $(K, \tau_K)$ of $(X,\tau)$ which is Hausdorff. Then $K\cap F$ is closed in $\tau_K$ which makes $(K,\tau_K)$ regular. And then $(\exists U_K, V_K \in \tau_K)(x\in U_K, (K\cap F)\subseteq V_K, U_K\cap V_K = \varnothing)$.

But how can I expand this $V_K$, such that $F\subseteq V$?

Martin Sleziak
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dietervdf
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  • Do you know that a compact subspace of a Hausdorff space is closed? And do you know that a closed subspace of a compact space is compact? – Stefan Hamcke Jun 05 '15 at 19:50
  • @StefanHamcke Yes I do, I which of the methods could that be used? – dietervdf Jun 05 '15 at 19:52
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    Okay, then you know that $X\setminus K$ is open. So you can take $V=V_K\cup X\setminus K$ – Stefan Hamcke Jun 05 '15 at 19:53
  • Note that $U_K$ is already a neighborhood of $x$ since it is open relative to the neighborhood $K$ of $x$. – Stefan Hamcke Jun 05 '15 at 19:55
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    @StefanHamcke I see it, thanks! Any ideas for the first method? Can I say $(X, \tau)$ is a relative topology to $(X_\infty, \tau_\infty)$? Then $V_\infty \cap X$ would be be in $\tau$. – dietervdf Jun 05 '15 at 19:58
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    That $V_\infty\setminus{\infty}\in\tau$ follows immediately from the fact that $X\in\tau_\infty$. – Brian M. Scott Jun 05 '15 at 19:58
  • @BrianM.Scott Offcourse! It's so obvious when you see it. A bit frustrating though, I often fail to see the missing link. If anyone would turn this into an answer I'd gladly accept it. – dietervdf Jun 05 '15 at 20:00
  • @StefanHamcke One final question. Since $U_K, V_K\in \tau_K$ then $U_K = \tilde U \cap K$ where $\tilde U\in \tau$ and $V_K = \tilde V \cap K$ where $\tilde V \in \tau$. But why can we be sure $\tilde U\cap \tilde V = \varnothing?$ I ask since I need to work with sets of $\tau$. – dietervdf Jun 05 '15 at 20:05
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    Note that $\tilde U\cap K$ is a neighborhood of $x$ as it contains the open set $int K\cap\tilde U$. So you can assume that $\tilde U=U_K$. – Stefan Hamcke Jun 05 '15 at 20:10

4 Answers4

18

Suppose $X$ is a Hausdorff space which is locally compact, meaning that every point has a compact neighborhood. Consider a closed set $F\subseteq X$ and a point $x\in X\setminus F.$ Let $K$ be a compact neighborhood of $x.$

Since $X$ is Hausdorff, and since $F\cap K$ is compact and $x\notin F\cap K,$ there are disjoint open sets $U,V$ such that $x\in U$ and $F\cap K\subseteq V.$

Now $x\in\operatorname{int}K$ (since $K$ is a neighborhood of $x$), and $K$ is closed (since $K$ is compact and $X$ is Hausdorff).

Thus we have disjoint open sets $U_0=U\cap\operatorname{int}K$ and $V_0=V\cup(X\setminus K)$ with $x\in U_0$ and $F\subseteq V_0.$

bof
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  • Why $F\subseteq V_0$? Do we have $F\subseteq V$? – Sigur Jun 02 '16 at 12:11
  • @Sigur: Denote $A=F\cap K$, then $F=(F-A)\cup A$, notice $A$ contained in $V$, and $F-A$=$F-K$ contained entirely in $X-K$, hence: $F=(F-A)\cup A\subseteq V\cup (X-K) = V_0$ – Daniel Jun 03 '17 at 10:23
  • @bof Why is $F\cap K$ compact? – Ruochan Liu Jul 12 '21 at 23:10
  • @Larry $K$ is compact and $F$ is closed. In general, the intersection of a compact set with a closed set is compact. – bof Jul 12 '21 at 23:26
  • @bof Thank you for your response. To clarify, when you say "in general", does that mean always or not? – Ruochan Liu Jul 12 '21 at 23:50
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    @Larry It means always, in any topological space $X$. This is an easy consequence of definitions. If $\mathcal U$ is an open cover of $F\cap K$, then $\mathcal U\cup{X\setminus F}$ is an open cover of $K$. If $\mathcal V$ is a finite subcollection of $\mathcal U\cup{X\setminus F}$ which covers $K$, then $\mathcal V\cap\mathcal U$ is a finite subcollection of $\mathcal U$ which covers $F\cap K$. – bof Jul 13 '21 at 02:56
  • $F\cap K$ could well be empty, which would make the argument pointless, no? $V$ need not necessarily exist disjoint from $U$ if the intersection is empty – FShrike Jan 27 '22 at 12:28
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For the first approach, note that $X\in\tau_\infty$, which immediately yields $V_\infty\setminus\{\infty\}\in\tau$.

Brian M. Scott
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8

The second approach can be made clearer (for my tastes, which do not have to agree with yours) by using the English language more.

A simple reformulation is that you are looking for a closed neighbourhood of $x$ that does not meet $F$. Now $x$ has a compact neighbourhood $K$ and a closed neighbourhood $W_K$ in $K$ that does not meet $F$. Since $K$ is a neighbourhood of $x$, $W_K$ is also a neighbourhood of $x$ in $X$, and since $K$ is closed, $W_K$ is also closed in $X$.

(The closed neighbourhood $W_K$ can be chosen as the closure of $U_K$ in your formulation of the proof.)

Carsten S
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  • If you edit your post I will remove the downvote; I intended to upvote but misclicked, and realise now only too late as my vote has been locked in. Your answer was useful to me! – FShrike Jan 27 '22 at 13:53
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In the compactification method, just use that every subspace $Y$ of a regular space $Z$ is regular, since the intersections of the respective neighborhoods of $x$ and $A$ with $Y$ give the desired neighborhoods of $x$ and $A\cap Y$ in $Y$.

In the second method, having found disjoint neighborhoods $U_K, V_K$ of $x$ and $F\cap K$ in $K$, you can take $U=U_K$ and $V=V_K\cup X\setminus K$, which are then disjoint neighborhoods of $x$ and $F$ in $X$.

Note that a locally compact Hausdorff space $X$ is not only regular, but even completely regular. Using the one-point compactification $\hat X$, the proof is easy. You can simply use the fact that $\hat X$ is normal and regular, and thus completely regular, and that this property is hereditary.
A direct proof without compactifications is possible, but a bit more cumbersome.

Stefan Hamcke
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