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If $X$ is compact Hausdorf and $A$ is closed than $Y=A^c$ is locally compact

I attempted this question but I am not sure of my solution and whether it can be simplified if correct.

Take $y\in Y$, as $X$ is compact it is locally compact and there is a compact set $y \in K \subseteq X$, I want to show that $\tilde K = K\cap Y$ is also compact. Let $\tilde K\subseteq \bigcup_\alpha (U_\alpha \cap Y)$ be a covering of $\tilde K$, then we can add to this covering $K\cap A$ and then intersect with $Y$ to get the empty set: $\tilde K \subseteq (\bigcup_\alpha U_\alpha \cup (K\cap A)) \cap Y$. Now I can look at the expression before taking the intersection with $Y$, since $K\cap A$ is closed in a regular space (since it is compact Hausdorf), there is an open set $K\cap A\subseteq U$. Since $K\subseteq \bigcup_\alpha U_\alpha \cup U$ is a an open cover there is a finite subcover $K\subseteq \bigcup _iU_i$ and $\tilde K \subseteq (\bigcup_i U_i) \cap Y$.

Is this correct?

Omer Paz
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  • You cannot show this like that, because $K$ can be simply chosen as $X$. And then $\tilde K=Y$ which does not have to be compact. The idea here is that $Y$ is open. And since $X$ is locally compact Hausdorff, then one of the equivalent conditions is that $X$ has a local basis of compact neighbourhoods. In particular there is open $V$ such that $y\in V$ and $\overline{V}\subseteq Y$ and $\overline{V}$ is compact. This can be probably found in Spanier or somewhere. – freakish Jun 26 '23 at 18:03
  • Since $X$ is compact and $A$ is closed, $A$ is compact. If $y\in Y$, by the lemma here, there are open sets $U, V$ such that $A\subseteq U, y\in V, U\cap V=\emptyset$. Let $C=X\setminus U$. Then $C$ is closed, hence compact, and $y\in V\subseteq C\subseteq Y$. So $C$ is a compact neighborhood of $y$. – Chad K Jun 26 '23 at 18:39

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