1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Daniel Fischer Feb 5 '17 at 15:29
  • $\begingroup$ 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? $\endgroup$ – Henno Brandsma Feb 5 '17 at 15:33
  • $\begingroup$ @HennoBrandsma. Yes, exactly. This are the definitions I am using. $\endgroup$ – jvnv Feb 5 '17 at 15:34
  • $\begingroup$ Are you allowed (at this stage) to use that $X$ is a regular space? $\endgroup$ – Henno Brandsma Feb 5 '17 at 15:48
  • $\begingroup$ @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. $\endgroup$ – jvnv Feb 5 '17 at 15:50
1
$\begingroup$

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 $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.