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i have this trouble, but no idea how to star, so I need some help!

Let X be a Hausdorff space, and let $\{C_\alpha| \alpha \in A \}$ a family of closed subsets of $X$ such that $\bigcap C_\alpha \neq \emptyset$. Let U and open that contains $ \bigcap C_\alpha $. Prove that for each $C_{\alpha0} $ compact exist $C_{\alpha1},C_{\alpha2},..., C_{\alpha n},$ such that $ C_{\alpha1} \bigcap C_{\alpha2} \bigcap ... \bigcap C_{\alpha n} \subset U$?

Thank you!!

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    That was my guess, but I wanted to be sure.2012-11-30

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HINT: Suppose that $C_{\alpha_0}$ is compact for some $\alpha_0\in A$. Suppose, to get a contradiction, that $\left(C_{\alpha_0}\cap\bigcap_{\alpha\in F}C_\alpha\right)\setminus U\ne\varnothing\tag{1}$ for each finite $F\subseteq A$. For each $\alpha\in A\setminus\{\alpha_0\}$ let $V_\alpha=X\setminus C_\alpha$, and let $\mathscr{V}=\big\{V_\alpha:\alpha\in A\setminus\{\alpha_0\}\big\}$. Finally, let $K=C_{\alpha_0}\setminus U$. Then $K$ is compact, and

$\left(C_{\alpha_0}\cap\bigcap_{\alpha\in F}C_\alpha\right)\setminus U=(C_{\alpha_0}\setminus U)\cap\bigcap_{\alpha\in F}C_\alpha=K\setminus\bigcup_{\alpha\in F}V_\alpha\;,$

so no finite subset of $\mathscr{V}$ covers $K$, and therefore $\mathscr{V}$ does not cover $K$. That is, $K\nsubseteq\bigcup\mathscr{V}$, and therefore $K\cap\bigcap_{\alpha\in A\setminus\{\alpha_0\}}C_\alpha\ne\varnothing\;.$

Do you see why this is a contradiction?

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    @vic: You’re welcome.2012-12-03