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Let $X$ be a locally compact Hausdorff space. Let $K \subset X$ be a compact subspace, and $U \subset X$ an open set, such that $K \subset U$.

Can we find an open set $V \subset X$ such that $K \subset V \subset U$ and $\overline V$ is compact?

I know that for all $x \in U$ there is an open set $V_x \ni x$ with compact closure contained in $U$. But the closure of $\bigcup_{x \in K} V_x$ doesn't have to be compact.

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The set $\cup_{x\in K} V_x$ that you mention is almost what we want. Since $K$ is compact and $\{V_x\}_{x\in K}$ is an open cover of $K$, we can choose a finite subcover. That is, we can choose $x_1,\ldots,x_n\in K$ so that \begin{equation} V = \bigcup_{i=1}^n V_{x_i} \end{equation} covers $K$. But we can also check that \begin{equation} \overline{V} = \bigcup_{i=1}^n \overline{V}_{x_i} \end{equation} is compact (and contained in $U$).

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