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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?

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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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