# A locally compact Hausdorff space admits a local base consisting of compact sets.

I am trying to solve the following exercise:

Show that an open subset of a compact Hausdorff space is locally compact. Use this to conclude that any locally compact Hausdorff space has a local base consisting of compact sets.

Here, a topological space is locally compact if each point possesses a compact neighbourhood, and a local base is a collection such that for any given neighbourhood of a point there is a neighbourhood in the local base that is contained in the given neighbourhood.

It is clear to me why an open subset of a compact Hausdorff space is locally compact. However, I do not see how to use this to conclude that any locally compact Hausdorff space has a local base consisting of compact sets.

I am aware of another solution of the second part of the above stated exercise (using the complete regularity of the space), but I would also like a proof that follows the method outlined in the exercise.

Any help or comment is highly appreciated.

• Let $x$ be any point, and $U$ an open neighbourhood of $x$. Since $U$ is locally compact, $x$ has a compact neighbourhood in $U$. What is the relation between the neighbourhoods of $x$ in $U$ and the neighbourhoods of $x$ in the whole space? – Daniel Fischer Feb 5 '17 at 15:29
• What is your exact definition of $X$ being locally compact? Every point has a compact neighbourhood? This will matter for the proof. In your case a local base consists of neighbourhoods of the point ( having that point in its interior), not necessarily open sets? – Henno Brandsma Feb 5 '17 at 15:33
• @HennoBrandsma. Yes, exactly. This are the definitions I am using. – jvnv Feb 5 '17 at 15:34
• Are you allowed (at this stage) to use that $X$ is a regular space? – Henno Brandsma Feb 5 '17 at 15:48
• @HennoBrandsma. Yes, that is fine. As I wrote in my question, I have in fact a proof using the regularity of the space, but as far as I can see it does not follow the outline of the proof given in the exercise. – jvnv Feb 5 '17 at 15:50

Suppose $X$ is open and $x \in O$ as $O$ is locally compact by the first fact it has a compact neighbourhood $C$ in $X$ so $x \in \operatorname{Int_O}(C)$ and $C$ compact (compactness is absolute). As $O$ is open, so is $\operatorname{Int_O}(C)$ so $C$ is still a neighbourhood of $x$ in $X$, and also still compact. So compact neighbourhoods form a local base at $x$ as we can do this for every open $O$ that contains $x$.