$X$ 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.

  • 1
    $\begingroup$ 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

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 $\cap_{1}^{n}F_{a_{k}} \cap U^{c} \subset \cap_{1}^{n}F_{a_{k}} \subset \cap_{a}F_{a}.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.