# Question on the proof that a closed subset of a compact set is compact

I was reading the proof here on the claim that a closed subset of a compact set is compact, which reads:

Say F ⊂ K ⊂ X where F is closed and K is compact. Let $\{V_α\}$ be an open cover of F. Then $F^c$ is a trivial open cover of $F^c$. Consequently { $F^c$} ∪ $\{V_α\}$ is an open cover of K. By compactness of K it has a finite sub-cover – which gives us a finite sub-cover of F.

The proof has to add $\{F^c\}$ to an open cover of of F so that it covers K. What about the open cover $\{V_α\}$ without $\{F^c\}$? How can one be sure that it has a finite subcover for F since it is not necessarily true that $\{V_α\}$ covers K?

UPDATE

I want to put the question differently. How do I know that there is not an open cover that covers F but not K?

• But when you add the set $F^{c}$, that's just adding one more set. – quasi Nov 6 '17 at 7:48
• I misread your statement. But you are sure, because as quasi has remarked,you have just added one set, so you can subtract it as well (if the process of going to a finite subcover hasn't done that for you already). – B. Pasternak Nov 6 '17 at 7:49
• @quasi The open cover without $\{F^c\}$ should have a finite subcover as well, right? – user1691278 Nov 6 '17 at 7:49
• @B.Pasternak I thought you need to show that every open cover has a finite subcover. The open cover $\{V_α\}$ without $\{F^c\}$ should have a finite subcover too. How can I show that? – user1691278 Nov 6 '17 at 7:50
• @quasi I updated the question. – user1691278 Nov 6 '17 at 7:54

How do I know that there is not an open cover that covers $F$ but not $K$?

You do not know. For example if $F = K$, then every open cover of $F$ also covers $K$. But this is none of your business.

I guess I understand your confusion now. Let me spell out the proof.

Your assumption is that $\{ V_\alpha\}_\alpha$ covers $F$. So $\{ V_\alpha\}_\alpha \cup F^c$ covers $K$. From there you can use the assumption that $K$ is compact and pick a finite subcover. The finite subcover looks like

$$\{ V_{\alpha_1}, V_{\alpha_2} , \cdots ,V_{\alpha_k}, F^c\}.$$

This covers $K$ and in particular covers $F$. But now we can take $F^c$ away: we know by definition of $F^c$ that $F\cap F^c$ is empty anyway. Thus

$$\{ V_{\alpha_1}, V_{\alpha_2} , \cdots ,V_{\alpha_k}\}$$

covers $F$ and is a finite subcover of $\{V_\alpha\}_\alpha$.

• Well said. Thank you! – user1691278 Nov 6 '17 at 8:07

You don't have to know that $V_{alpha}$ doesn't cover $K$, if it happens to cover $K$ then still adding $F^C$ wouldn't alter anything - the resulting cover would still cover $K$.

Now of course in the subcover of $K$ including $F^C$ we must know that the subcover excluding $F^C$ would cover $F^C$, but this is rather obvious. If $x\in F$ then it would be in at least one of the set in the subcover, but it can't be in $F^C$ by definition so it must be in at least another of the sets in the subcover.

• Well, $V_\alpha$ has to produce a finite subcover for F without having $F^C$ added to it, right? Assuming that $V_\alpha$ is an open cover for F but not K. – user1691278 Nov 6 '17 at 8:01
• @user1691278 Yes, the finite subcover of $K$ will produce a subcover of $F$ by removing $F^C$ - I've added a comment about that in my answer. – skyking Nov 6 '17 at 8:20

The idea of the proof is that, according to the definition of compactness, every open cover for $K$ has a finitie subcover. Now we found an open cover for $K$, containing only one set more than the open cover for $F$, we conclude that this set has a finite subcover. Since $F^c \not\subset F$, we know that there exists a subset of the finite subcover for $K$ also covering $F$, that does not contain $F^c$, and is finite since it contains as most as many sets as the finite cover for $K$.

To answer your update, there exist open covers for $F$ not covering $K$, but by construction, adding $F^c$ makes suren the open cover also covers $K$. That makes the possibility you're asking about irrelevant.