We say that $X$ is locally compact if for every $x\in X$ and every open $V$ with $x\in V$, there exists $K$ compact such that $x\in \text{Int}(K)\subseteq K\subseteq V$. Prove that if $X$ is locally compact, $U$ is open and $F$ is closed then $U\cap F$ is locally compact.

My question is: do you think this could be false if $X$ is not Hausdorff? The above problem came in an exam and it just looks to me that $X$ must be Hausdorff, so the problem would be solved in the following natural way.

Take $x\in U\cap F$ and $V$ open in $U\cap F$, say $V=V'\cap U\cap F$ with $V'$ open in $X$. Then there exists $K$ compact such that $x\in \text{Int}(K)\subseteq K\subseteq V'\cap U$. If $X$ were Hausdorff, $F\cap K$ would be compact because $K$ would be closed, and then $K\cap F$ is a compact set that shows $U\cap F$ is locally compact.

What do you think?


Let $x \in U \cap F$. It follows that there is some compact $K_1$ such that $x \in int(K_1) \subseteq K_1 \subseteq U$. Let $N \cap (U \cap F)$ be a neighborhood of $x$ in the subspace topology of $U \cap F$. It follows there is some compact $K_2$ such that $x \in int(K_2) \subseteq K_2 \subseteq N \cap int(K_1)$.

Finish the proof (if you can from here) and answer your question yourself: Could this be false if $X$ is not Hausdorff?

Hope this helps.

  • 1
    $\begingroup$ Thank you. I don't understand everything yet. You still want some compact $K$ in $U\cap F$. Would you mean to take $K_2\cap F$? In some part you need to intersect with $F$, or am I wrong? $\endgroup$ – JonSK Jan 23 '16 at 15:50
  • $\begingroup$ Yeeeeup. You're very correct. Now you need to explain why $K_2 \cap F$ is compact and then why it proves that $U \cap F$ is locally compact. $\endgroup$ – Logician6 Jan 23 '16 at 16:12
  • 1
    $\begingroup$ Oh, I see. $K_2\cap F$ is a closed subset of $K_2$ with its subspace topology, hence $K_2\cap F$ is compact. $\endgroup$ – JonSK Jan 23 '16 at 16:42
  • $\begingroup$ Yup. And that's the last assumption put in motion and basically finishes the proof. Make sure to make it clear in your set theory arguments though that $K_2 \cap F$ is sufficiently contained in the neighborhood $N \cap (U \cap F)$. I left it to you to reason through that one. $\endgroup$ – Logician6 Jan 23 '16 at 16:56

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.