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!

  • $\begingroup$ The first paragraph certainly needs to be rewritten more clearly. "Empty intersection over an index, like the one mentioned earlier" is particularly confusing. $\endgroup$ – dfeuer Aug 11 '13 at 6:19
  • $\begingroup$ You're right so I went ahead and made an edit. Hopefully it sounds more clear now. $\endgroup$ – Shant Danielian Aug 11 '13 at 6:31
  • $\begingroup$ There is no inherent contradiction in the first paragraph because you don't explicitly assume compactness. $\endgroup$ – dfeuer Aug 11 '13 at 6:37
  • $\begingroup$ I see, is better now? $\endgroup$ – Shant Danielian Aug 11 '13 at 6:41
  • $\begingroup$ 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. $\endgroup$ – dfeuer Aug 11 '13 at 6:51

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.