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.

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!!

share|improve this question
    
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

1 Answer 1

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?

share|improve this answer
    
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

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.