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I'm reading Intro to Topology from Mendelson.

The entire problem statement is,

Prove that $X$ is compact if and only if for each family $\lbrace F_\alpha\rbrace_{\alpha\in I}$ of closed subsets of $X$ that has the finite intersection property, we have $\bigcap_{\alpha\in I} F_\alpha\neq\varnothing$.

My attempt at the proof is as follows:

First assume that $X$ is compact. For the sake of contradiction, suppose that there exists a family $\{ F_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ with FIP such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$. Since $X$ is compact if $\lbrace F_\alpha\rbrace_{\alpha\in I}$ any family of closed sets such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$, then there exists a finite set of indices $\lbrace \alpha_1,\dots,\alpha_n\rbrace$ such that $\bigcap\limits_{i=1}^n F_{\alpha_i}=\varnothing$. Yet, since $\lbrace F_\alpha\rbrace_{\alpha\in I}$ has the finite intersection property, for every finite index of $J\subset I$, $\bigcap\limits_{\alpha\in I} F_\alpha\neq\varnothing$.

Suppose now that $X$ is not compact, that is, there exists an open cover $\lbrace U_\alpha\rbrace_{\alpha\in I}$ of $X$ with no finite subcover. Consider the set $F_\alpha=C(U_\alpha).$ Then $\lbrace F_\alpha\rbrace_{\alpha\in I}$ is a collection of closed subsets of $X$ with the finite intersection property and were $\bigcap_{\alpha\in I} F_\alpha=\varnothing.$

Thanks for any feedback!

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The first paragraph certainly needs to be rewritten more clearly. "Empty intersection over an index, like the one mentioned earlier" is particularly confusing. –  dfeuer Aug 11 '13 at 6:19
    
You're right so I went ahead and made an edit. Hopefully it sounds more clear now. –  Shant Danielian Aug 11 '13 at 6:31
    
There is no inherent contradiction in the first paragraph because you don't explicitly assume compactness. –  dfeuer Aug 11 '13 at 6:37
    
I see, is better now? –  Shant Danielian Aug 11 '13 at 6:41
    
I made a few edits, mostly minor. Note that \{\} does the same thing as \lbrace\rbrace with fewer characters, and that \bigcap is usually more readable with subscripts/superscripts than is \cap. The last sentence of the first paragraph is still completely unreadable. –  dfeuer Aug 11 '13 at 6:51

1 Answer 1

up vote 1 down vote accepted

Although your proof goes well, but it needs more explanation. Here is an attempt.

Suppose that $X$ is compact and let $\{F_\alpha:\alpha\in I\}$ be a family of closed subsets of $X$ with $FIP$. On contrary, suppose that $\cap_{\alpha\in I}F_\alpha=\emptyset$. Then $\cup_{\alpha\in I}F_\alpha^c=X$. As $X$ is compact, so there is a finite subset $J$ of $I$ such that $\cup_{\alpha\in J}F_\alpha^c=X$ and consequently $\cap_{\alpha\in J}F_\alpha^c=\emptyset$, which is a contradiction to $FIP$.

Conversely, let $\{U_\alpha:\alpha\in I\}$ be an open cover of $X$. Now put $F_\alpha=U_\alpha^c$. If $X$ is not compact then the family $\{F_\alpha:\alpha\in I\}$, of closed subsets of $X$, has $FIP$ with $\cap_{\alpha\in I}F_\alpha=\emptyset$, which contradicts our hypothesis.

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Thanks for the response and cleaning up the proof considerably. –  Shant Danielian Aug 11 '13 at 8:49

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