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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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    A space is compact if and only if every system of closed sets having [finite intersection property](http://en.wikipedia.org/wiki/Finite_intersection_property) has non-empty intersection. This is well-known characterization of compactness, you can probably find the proof in many introductory textbooks or with little googling, e.g. [planetmath](http://planetmath.org/encyclopedia/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty.html), [Ask a Topologist](http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2004;task=show_msg;msg=1102.0001).2012-03-08

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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!2012-03-06