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Set $X$ to be a compact topological space. Let ${V_\alpha}$ be a system of the closed subsets of $X$ where $\bigcap\limits_{\alpha\in ℤ}{V_\alpha}≠\emptyset$ ($\alpha$ is finite). Show $\bigcap\limits_{∞}{V_\alpha}≠\emptyset$.

I'm having a terrible time getting to grips with topology after just starting to study it- hopefully I'll cross the bridge and get it soon! But this question has me truly stumped.

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Which definition of compactness are you working with? (This is one of them.) –  Qiaochu Yuan Mar 6 '12 at 19:20
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Prove the contrapositive using the "open cover definition" of compactness and DeMorgan's laws. This is all that's needed. –  David Mitra Mar 6 '12 at 19:21
    
"I'm having a terrible time getting to grips with topology after just starting to study it" - would you expect anything different from what is probably the first really abstract topic one learns in mathematics? –  Asaf Karagila Mar 6 '12 at 19:49
    
@AsafKaragila Not particularly surprised, just hate the feeling! Like many others I'm quite shocked by the transition to graduate level- didn't expect such a great increase in difficulty compared to everything else ever studied! –  D.G.S Mar 6 '12 at 19:52
    
@D.G.S Don't worry, what you're feeling is normal. –  Gunnar Magnusson Mar 6 '12 at 21:13
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Hint: towards proving the contrapositive, suppose the collection of closed sets $\{V_\alpha \mid\alpha\in A\}$ has the property that any finite subcollection has non-empty intersection but $\bigcap\limits_{\alpha\in A} V_\alpha=\emptyset$. Consider the open cover $\{V_\alpha^C\mid \alpha\in A \}$ of $X$ (this is an open cover of $X$ since $X=\emptyset^C=(\bigcap\limits_{\alpha\in A} V_\alpha)^C=\bigcup\limits_{\alpha\in A} V_\alpha^C$). Can this have a finite subcover?

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Thank you very much for your help. Checking definitions of covers at the moment and shall work through this! –  D.G.S Mar 6 '12 at 19:53
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