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Assume X is a Hausdorff space. Is the following statement about local compactnes true? (Is there a proof or counterexample?)

$X$ Hausdorf loc cpt, $\forall V$cpt $\subseteq X: (U\cap V)$ cpt $\Rightarrow U$ closed.

(the opposite statement is true and easy to prove:

$X$ Hausdorf loc cpt, $U$ closed $\Rightarrow$ $\forall V$cpt $\subseteq X: (U\cap V)$ cpt.

)

Apass.Jack
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georgy_dunaev
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  • I think that's true and you can prove it via a contrapositive. Assume $U$ is not closed. Then there exists a limit point in $x \in X$ that is not in $U$. Take a relative compact neighborhood of $x$ and let $V$ be the closure of this neighborhood. Then $U\cap V$ is not empty but also can't be compact since it has a limit point that is not in $U\cap V$. I didn't check all the details there. It's just a rough idea. – Maik Pickl Dec 15 '22 at 18:50

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Maik Pickl outlined an approach. The following proof follows this post roughly.

Fix $x\in X\setminus U$.

$X$ is locally compact means there is an open set $O_x$ and a compact set $C_x$ such that $x\in O_x\subseteq C_x$. So $U\cap C_x$ is compact, hence, as a Hausdorff space, closed in $C_x$. We have $$x\in O_x\setminus U=O_x\setminus(U\cap C_x)$$ is open in $O_x$, hence also open in $X$. We have found an open neighborhood of $x$ in $X$ disjoint from $U$.

Since $x$ was an arbitrary point of $X\setminus U$, $U$ must be closed.

Apass.Jack
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