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I came across an exercise on compactness and, though it seems really easy, I can't get it done! I'd appreciate it if you gave me some sort of hint and not the full solution!

So, here it is:

Let $(X,τ) $ be a topological space, $\left \{F_i :i\in I \right\}$ a collection of closed subsets and $U$ an open subset of $X$ such that $\bigcap_iF_i \subseteq U$. If some $F_{i_0}$ is compact, then there are $i_1, ..., i_n \in I$ such that $\bigcap_{k=0}^{n}F_{i_k} \subseteq U$.

Thoughts:

1) I believe that the trick right here is to show that $U^{c}$ is compact, but there's nothing to lead me there.

2) Just because we are working with closed sets I believe that the F.I.P. (finite intersection property) will come in handy, but I really can't see how this is going to happen.

Thank you in advance!

1 Answers 1

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Hint: by relativizing to $F_{i_0}$, you can reduce the question to the case when $X$ itself is compact. The condition that $ \bigcap_i F_i\subseteq U$ is equivalent to $\bigcap_i(F_i\setminus U)=\emptyset$.

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    First of all, thank you for answering! Secondly, can you tell me if the following is correct? The sets $K_i:=F_{i_0}∩F_i∩U^{c},i∈I$ are closed subsets of the compact space $F_{i_0}$, so if they have the FIP, then $⋂_i K_i$ is nonempty. But, because $⋂_i K_i=∅$, we have the desired result! – 2017-01-09
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    @SotirisSimos: Yes, that seems correct. – 2017-01-09
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    @SotirisSimos: Glad you are happy. If you think your question is resolved, you should mark the preferred answer (by clicking the tick sign under the voting arrows). – 2017-01-09