Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
$\{X\setminus P\}_{P\in B}$ is an open cover of $X$. –  Davide Giraudo Oct 18 '12 at 11:35
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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