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Let $X$ be a compact set. Suppose $V$ be a collection of set of closed set $P$ where each $P$ are closed set in $X$ and any intersection of finite subcollection of $V$ is nonempty. then $\bigcap_{P\in V} P$ is also non empty. I tried to use contradiction to prove it but cannot get a contradiction. Any method is fine. Thx

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$\{X\setminus P\}_{P\in B}$ is an open cover of $X$. – Davide Giraudo Oct 18 '12 at 11:35

We have that $\bigcap_{P\in V}P=\emptyset$ is equivalent to $\bigcup_{P\in V}P^C=X$, which amounts to the family of sets of the form $P^C$ forming an open cover of $X$. If there is a finite family $F$ such that $\bigcup_{P\in F}P^C=X$, then we also have $\bigcap_{P\in F}=\emptyset$, which means that the intersection of finitely many elements of $V$ is empty. But such a family $F$ exists by compactness.

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