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I came across an exercise that asked to characterize compactness in terms of closed sets. This is what I came up with:

Claim: $X$ is compact $\Leftrightarrow$ for every set of closed sets $\{C_\alpha\}$ with $\cap_{\alpha}C_\alpha=\emptyset$ has a finite subset $\{C_{\alpha_1},\dots,C_{\alpha_n}\}$ s.t. $C_{\alpha_1}\cap\dots\cap C_{\alpha_n}=\emptyset$.

But apparently the correct characterization is: $X$ is compact $\Leftrightarrow$ any collection of closed sets with the finite intersection property has non-empty intersection.

That seems like a much more complicated way to do it. So my question is, what's wrong with my characterization? Is mine wrong? If so, where's the mistake in the following proof? If mine is not wrong, why don't I find this characterization anywhere, while I find the other version everywhere? Thank you for any help!

Proof of Claim: $(\Rightarrow)$ Suppose $X$ is compact. Let $\{C_\alpha\}$ be a collection of closed sets s.t. $\cap_{\alpha}C_\alpha=\emptyset$. Then $\{C_\alpha^c\}$ is a collection of open sets and $\cup_\alpha C_\alpha^c=(\cap C_\alpha)^c=\emptyset^c=X$. Thus $\{C_\alpha^c\}$ is an open cover of $X$. Since $X$ is compact there is a finite subcover $C_{\alpha_1}^c, \dots, C_{\alpha_n}^c$. Since $C_{\alpha_1}^c\cup \cdots\cup C_{\alpha_n}^c=X$, $\emptyset=X^c=(C_{\alpha_1}^c\cup \cdots\cup C_{\alpha_n}^c)^c=C_{\alpha_1}\cap \cdots\cap C_{\alpha_n}$.

$(\Leftarrow)$ Let $\{U_\alpha\}$ be an open cover of $X$. Then $\{U_\alpha^c\}$ is a collection of closed sets s.t. $\cap_\alpha U_\alpha^c=(\cup U_\alpha)^c=X^c=\emptyset$. Thus $\exists$ finite subset $\{U_{\alpha_1}^c,\dots,U_{\alpha_n}^c\}$ s.t. $U_{\alpha_1}^c \cap\dots \cap U_{\alpha_n}^c=\emptyset$. But then $(U_{\alpha_1}^c \cap\dots \cap U_{\alpha_n}^c)^c=U_{\alpha_1} \cup\dots \cup U_{\alpha_n}=X$. So $\{U_\alpha\}$ has a finite subcover.

Gregory Grant
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    You characterization is right. The other characterization is usually more useful so that's why you've found that one and not yours. – Daniel Apr 05 '15 at 20:20
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    Nothing is wrong with your characterization; it’s trivially equivalent to the other one by taking the contrapositive. – Brian M. Scott Apr 05 '15 at 20:21
  • @DanielEscudero Thank you. Strange to me that this is not easy to find written down anywhere. – Gregory Grant Apr 05 '15 at 20:24
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    @GregoryGrant You're welcome. As others have said, your characterizarion doesn't require the proof you gave. – Daniel Apr 05 '15 at 20:25
  • @BrianM.Scott Thank you. But I don't see how it is the contrapositive. Wouldn't the contrapositive be $X$ is not compact $\Leftrightarrow$ $\exists$ an open cover with no finite subcover. – Gregory Grant Apr 05 '15 at 20:26
  • @DanielEscudero Why doesn't it require proof? It's almost something I can see in my head but I think it's better to write it down. As I said above, I don't see how this is the contrapositive. – Gregory Grant Apr 05 '15 at 20:27
  • It seems to prove the equivalence you need DeMorgan's law. You wouldn't need another theorem if it was just the straight contra-positive, no? – Gregory Grant Apr 05 '15 at 20:28
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    @GregoryGrant It's not the contrapositive. It's the same claim in terms of closed sets: Just take complements in your definition of compactness. – Daniel Apr 05 '15 at 20:28
  • @DanielEscudero Thank you, that's what I thought. Ok I think I'm totally straight on this now! – Gregory Grant Apr 05 '15 at 20:29
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    @Gregory: It’s the contrapositive of the condition, not of the theorem. You have: if $\bigcap\mathscr{C}=\varnothing$, then some finite subset of $\mathscr{C}$ has empty intersection. The contrapositive of that is: if $\mathscr{C}$ has the f.i.p., then $\bigcap\mathscr{C}\ne\varnothing$. In other words, for all practical purposes your version is identical to the other one. – Brian M. Scott Apr 05 '15 at 20:30
  • @BrianM.Scott Got it. BTW, I was reading carefully an answer you left in 2013:

    http://math.stackexchange.com/questions/427374/why-is-the-inverse-image-of-a-compact-set-under-a-special-sort-of-function-compa

    Do you think that could be proven more easily using this other "contra-positive" characterization?

     – Gregory Grant Apr 05 '15 at 20:33
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    @Gregory: I don’t see any obvious way in which that would simplify the argument. – Brian M. Scott Apr 05 '15 at 20:38
  • @BrianM.Scott Ok thank you. I ran across that exercise in an undergraduate textbook called "Basic Topology" in chapter 3. So it seems unlikely the author had that proof in mind, especially because he never mentioned the FIP, though he did ask the reader to characterize compactness in terms of closed sets. Your proof is beautiful and inspiring, I just wonder what the author had in mind that's all. Thanks again for your help! – Gregory Grant Apr 05 '15 at 20:42

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