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$X$ is a compact space, $U \subseteq X$ an open set of $X$ and $\mathcal{F}$ a family of closed sets of $X$ that $\cap \mathcal{F} \subseteq U$. Prove that exists $\mathcal{F'}\subseteq\mathcal{F}$ finite that $\cap \mathcal{F'}\subseteq U$.

I'm stuck in that question. I think that I will find those closed sets with the compactness but I don't know how.

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    $\begingroup$ Hint: consider the complements of the closed sets in the family $\mathcal F$, together with the open set $U$. Together, these form an open cover for $X$. But $X$ is compact, so... $\endgroup$ – Kenny Wong Jun 24 '17 at 13:33
  • $\begingroup$ Thank you, this solved my problem :) $\endgroup$ – Ettore Moura Jun 24 '17 at 14:06
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To formally prove it, let $\{ F_{a} \}$ be the given family of closed sets. Then $(\cap_{a}F_{a})^{c} = \cup_{a}F_{a}^{c} \supset U^{c}$ by assumption. So $\cup_{a}F_{a}^{c} \cup U,$ which $\supset (U^{c} \cup U) = X$ is an open cover of $X$, and hence $X$ compact by assumption implies there are some $a_{1},\dots, a_{n}$ such that $\cup_{1}^{n}F_{a_{k}}^{c} \cup U \supset X$. Note that the set $\cap_{1}^{n}F_{a_{k}} \cap U^{c}$ is closed in $X$. Note that $\cap_{1}^{n}F_{a_{k}} \cap U^{c} \subset \cap_{1}^{n}F_{a_{k}} \subset \cap_{a}F_{a}.$

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