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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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Are the $C_\alpha$ compact? If not, what are the $C_{\alpha n}$'s exactly? Are they subsets of some $C_\alpha$ or so? – Asaf Karagila Nov 30 '12 at 0:00
It still doesn’t quite make sense. Is $C_{\alpha 0}$ one of the sets in the family, one that just happens to be compact? And what do the sets $C_{\alpha 1},\dots,C_{\alpha n}$ have to do with $C_{\alpha 0}$? – Brian M. Scott Nov 30 '12 at 0:20
$ C_{\alpha 0}$ belongs to the family, and sorry didn´t write well the end: $ C_{\alpha 0} \bigcap C_{\alpha 1} \bigcap ... C_{\alpha n} \subset U$ – vic Nov 30 '12 at 0:34
That was my guess, but I wanted to be sure. – Brian M. Scott Nov 30 '12 at 0:42

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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I have some doubts: K is compact beacause $ K \subset U (C_{\alpha_0} / C_\alpha \alpha \in A) $ and $ C_{\alpha_0} $ is compact so $ C_{\alpha_0} $ is covered by a subcover finite and so K ?? – vic Dec 3 '12 at 2:02
@vic: $K$ is compact simply because it’s a closed subset of the compact set $C_{\alpha_0}$. – Brian M. Scott Dec 3 '12 at 2:06
abd the contradiction is beacuse $\bigcap_{\alpha\in A\setminus\{\alpha_0\}}C_\alpha\ne\varnothing$ is contained in U and $K=C_{\alpha_0}\setminus U$ ?? – vic Dec 3 '12 at 2:11
@vic: That’s exactly right. – Brian M. Scott Dec 3 '12 at 2:12
Thank you!!! :D – vic Dec 3 '12 at 2:31

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