# Rudin's proof about closed subsets of compact sets

The theorem, and proof presented in Rudin is:

Theorem: closed subsets of compact sets are compact

Proof: Suppose $$F \subset K \subset X$$, $$F$$ is closed (relative to $$X$$), and $$K$$ is compact. Let $$\\{V_{\alpha}\\}$$ be an open cover of $$F$$. If $$F_c$$ is adjoined to $$\\{V_{\alpha}\\}$$, we obtain an open cover $$\Omega$$ of $$K$$. Since $$K$$ is compact, there is a finite subcollection $$\Phi$$ of $$\Omega$$ which covers $$K$$, and hence $$F$$. If $$F^c$$ is a member of $$\Phi$$, we may remove it from $$\Phi$$ and still retain an open cover of $$F$$. We have thus shown that a finite subcollection of $$\\{V_{\alpha}\\}$$ covers $$F$$.

I read wrong, and thought the open cover $$\\{ V_{\alpha}\\}$$ was of $$K$$. Now, my question is. Why cant we just use that finite cover to proof $$F$$ as compact?

I know it doesnt make sense because, for example, $$(0,1)$$ is not compact but is a subset of $$[0,1]$$ which is compact. I would like to understand formally what is the problem with using the same finite subcover in $$K$$ and $$F$$.

Thank you

• Because not every cover of F is a cover of K. Commented Sep 20, 2023 at 5:04
• @Andrew Yeah but, I mean, the other way. A finite cover of K would also cover any of it's subsets. That's what I understand, and I don't know if there is something wrong with that reasoning.
– DAGO
Commented Sep 20, 2023 at 5:20
• Okay and how does that prove F is compact? Commented Sep 20, 2023 at 5:23
• @Andrew I just realized my fault and understood your comment. Sorry. I didn't thought of covers of F that do not cover K. So i was just confused about the need of taking another one from the original cover of K.
– DAGO
Commented Sep 20, 2023 at 5:32

We know that the $$V_\alpha$$ cover $$F$$, but we cannot be sure that they cover the bigger $$K$$. This is why we have to adjoin $$F_c$$. Then we obtain an open cover of $$X$$ which in particular covers $$K$$.