2
$\begingroup$

I came across the following proposition in my Complex Analysis book, stated as "A set $K \subseteq X$ is compact iff every collection $\mathcal{F}$ of closed subsets of $K$ with the f.i.p. has $\cap_{F\in \mathcal{F}} F\neq \emptyset$", where the finite intersection property is that any finite subset of the set has nonempty intersection. The forward direction of the proof was given, but the converse was left as an exercise. I struggled to prove the converse and finally resorted to looking up the proof and found this: Compact spaces and closed sets (finite intersection property). This statement is slightly different, in that it talks about a metric space, rather than a subset of a metric space.

My approach was to assume that $K$ had the given property, but was not compact. Then there was some open cover of $K$ with no finite subcover, say $\mathcal{F}$. Then let $\mathcal{G}=\{F^C: F\in \mathcal{F}\}$. In order to use the original assumption, I would need to intersect each $G\in \mathcal{G}$ with $K$ to get subsets of $K$. However, unless I prove that $K$ is closed, I would also need to use the closures of these intersections, which stops being useful. If I had that $K$ was closed, that the $\mathcal{G}$ has the f.i.p. is easy, since $\mathcal{F}$ has no finite subcover, and then I can use the hypothesis to show that all of the sets in $\mathcal{G}$ (intersected with $K$) share some common point which must also be in $K$, which contradicts $\mathcal{F}$ being a cover of $K$. However, I'm not sure how to get that $K$ is closed. Since I have to prove $K$ is compact, it is closed, but I'm not sure how to get that $K$ is closed from the given property. Is proving $K$ to be closed the best way to complete this proof?

$\endgroup$
  • $\begingroup$ You cannot deduce that $K$ is closed because if the space is not Hausdorff it need not be closed and the equivalence holds anyway. $\endgroup$ – Matt Samuel Jun 12 '16 at 23:21
3
$\begingroup$

What you’re missing is that ‘$A$ is a closed subset of $K$’ means that there is a closed set $H$ in $X$ such that $A=K\cap H$. That is, by closed subset of $K$ the author means a subset of $K$ that is closed in the relative (i.e., subspace) topology on $K$. Thus, each of your sets $K\cap F^C$ is a closed subset of $K$ in the required sense, and the fact that their intersection is non-empty gives you the desired contradiction and shows that $K$ is compact. So long as $X$ is a Hausdorff space you can then conclude further that $K$ is closed in $X$ and hence that the sets $F^C$ were closed in $X$ as well, but this is not necessary to the proof that $K$ is compact.

$\endgroup$
  • $\begingroup$ Thank you, I suspected that was the case. I'm new to topology, so when I see "closed subset", I assume in whichever topology the proposition is given, in this case, $X$. $\endgroup$ – Kevin Long Jun 12 '16 at 23:38
  • $\begingroup$ @Kevin: You’re welcome. That’s a natural assumption, and there are times when it will be correct; writers aren’t always careful, and context can be tricky when you’re beginning. $\endgroup$ – Brian M. Scott Jun 12 '16 at 23:40

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.