# Hausdorff space in which each point has a compact neighbourhood is locally compact

Could you help me prove the following fact? I've been trying to prove it and I've searched for a hint in Englking's book, but I haven't come up with anything:

If $X$ is a Hausdorff space and each $x \in X$ has at least one compact neighbourhood, then $X$ is locally compact.

$U$ is a neighbourhood of $x \in X$ $\iff$ $\exists V$ - open $: x \in V \subset U$.

$X$ is locally compact $\iff$ each $x \in X$ has a basis of compact neighbourhoods.

• What is your definition of locally compact? Sep 7 '13 at 17:14
• I've just added my definition of locally compact space. Sep 7 '13 at 17:16
• Are you allowed to use that compact Hausdorff spaces are regular? Sep 7 '13 at 17:18
• Yes. Even that they're normal :) I know that if $X$ is regulat, then each $x \in X$ has a basis of closed neighbourhoods. But does that mean that they're compact? Sep 7 '13 at 17:23

## 2 Answers

Actually, a stronger result is true: If each point has a compact Hausdorff neighborhood, then $X$ is locally compact.

Let $x\in X$, $U$ a neighborhood of $x$, and $K$ a compact Hausdorff neighborhood of $x$. Since $K$ is regular as a subspace of $X$ and $U\cap K$ is a neighborhood of $x$ in $K,$ there is a neighborhood $C\subseteq U\cap K$ of $x$ in $K$ (and thus also in $X$ since $K$ is a neighborhood of $x$) which is closed in $K$, thus compact. This proves that there are arbitrary small compact neighborhoods around $x$.

• Ok, now I understand. Thank you very much. Sep 7 '13 at 17:26
• It's missing that as $C$ is a neighborhood of $x$ in $K$ it follows that $C$ is a neighborhood of $x$ in $X$ (since $K$ was a neighborhood of $x$ in $X$). Nov 4 '16 at 19:11
• @AlexanderFrei: Yes, I couldn't be bothered to be that verbose when I wrote my answer, and I guessed that the reader would make that implication themselves. Nevertheless, I've added it to the post now. Nov 5 '16 at 14:33

A tricky one, indeed. I already had that once. Here we go :

We suppose that every point of $X$ has at least one compact neighborhood.

Let $x\in X$ and let $V$ be an open neighborhood of $x$. We want to show that there exists a compact neighborhood $W$ of $x$ with $W \subset V$.

By hypothesis, $x$ has a compact neighborhood, let's call it $C$. Let's take the subspace topology for $C$, so we have that $C$ is a compact Hausdorff space.

Let's now take $U = C \cap V$. And now we have that in $C$, $U$ is an open neighborhood of $x$. We know that $C$ is locally compact since it's compact and Hausdorff.

Thus, in $C$ we have a compact neighborhood $W$ of $x$ with $W \subset U$, and then also $W \subset V$.

In the end, we must check that $W$ is a neighborhood of $x$ in $X$, too. But since a neighborhood in a neighborhood is a neighborhood, we have it directly.

• The proof is circular as it uses the hypothesis as argument: "... is locally compact since it's compact and Hausdorff." Nov 4 '16 at 16:12
• @AlexanderFrei: I disagree, I'm proving the minimal result the Op asked for, using one of the first equivalent definition of being locally compact one sees when one learns about it. My hypothesis is as follows : X is Hausdorff and each x in X has a compact neighborhood. Note that X is not supposed compact.
– Lery
Nov 12 '16 at 9:49
• Ahhh I think I got it: The idea is to reduce the problem to compact Hausdorff space. There one easily checks local compactness. In return this extends to the ambient space as the restricted space is a neighborhood in the ambient space. Nov 12 '16 at 15:40
• Up-vote for the nicely strategic approach! (+1) Nov 12 '16 at 15:42