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I have come across this problem in my book which I have not solved.

F is a metric space,$\mathbb{F}$ is a set of all closed subsets of the close set $F$. $\tilde{F}$ is a set of all collections $\{F_{\alpha}~|~\alpha\in A\} \subseteq\mathbb{F} $,such that for any $\{F_{\alpha_1},F_{\alpha_2},\ldots,F_{\alpha_n}\} \subseteq \tilde{F} $, $\bigcap_{i=1}^{n}F_{\alpha_i}\neq \varnothing$.Prove that $F$ is compact if and only if for any $\{F_{\alpha}\}$, $\bigcap_{\alpha \in A} F_{\alpha}\neq \varnothing$.

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What is your definition of compactness. What conditions are imposed on $F$? You say it is a "close set", by which I assume you mean a "closed set". Closed set contained in what? Is $F$ a metric space, or just a topological space? – Alex Becker Sep 16 '12 at 3:45
Hint: a collection of closed sets has empty intersection iff their complements are an open cover. – Nate Eldredge Sep 16 '12 at 4:58
up vote 3 down vote accepted

First, either you’ve miscopied the problem, or the person who wrote it misused some of the notation rather badly. If $\Bbb F$ is the set of all closed subsets of the closed set $F$, then every member of $\Bbb F$ is a subset of $F$; $\Bbb F$ does not contain any collections of closed sets. $\tilde F$ is actually the set of all collections $\{F_\alpha:\alpha\in A\}\subseteq\Bbb F$ (note the subset relation, not the membership relation) satisfying a certain condition. That condition says that the collection is centred, or has the finite intersection property: $\{F_\alpha:\alpha\in A\}\in\tilde F$ if and only if

  1. $\{F_\alpha:\alpha\in A\}\subseteq\Bbb F$, and
  2. for every finite $\{\alpha_1,\dots,\alpha_n\}\subseteq A$, $\bigcup\limits_{k=1}^nF_{\alpha_k}\ne\varnothing$.

Your problem is to prove that $F$ is compact iff $\bigcap_{\alpha\in A}F_\alpha\ne\varnothing$ for every family $\{F_\alpha:\alpha\in A\}\in\tilde F$. In other words, $F$ is compact if and only if every centred family of closed subsets of $F$ has non-empty intersection.

This of course requires that you show two things: if $F$ is compact, then every centred family of closed subsets of $F$ has non-empty intersection, and if every centred family of closed subsets of $F$ has non-empty intersection, then $F$ is compact. The same basic observations are used to prove both directions:

Let $\mathscr{U}$ be a family of open subsets of $F$, and let $\mathscr{F}=\{F\setminus U:U\in\mathscr{U}\}$.

  1. $\mathscr{U}$ covers $F$ if and only if $\bigcap\mathscr{F}=\varnothing$.
  2. If $\mathscr{U}$ covers $F$, $\mathscr{U}$ has a finite subcover if and only if $\mathscr{F}$ is not centred (or if you prefer, is not in $\tilde F$).

Prove these, and you’re well on your way.

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