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

  • $\begingroup$ Because not every cover of F is a cover of K. $\endgroup$
    – Andrew
    Commented Sep 20, 2023 at 5:04
  • $\begingroup$ @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. $\endgroup$
    – DAGO
    Commented Sep 20, 2023 at 5:20
  • $\begingroup$ Okay and how does that prove F is compact? $\endgroup$
    – Andrew
    Commented Sep 20, 2023 at 5:23
  • $\begingroup$ @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. $\endgroup$
    – DAGO
    Commented Sep 20, 2023 at 5:32

1 Answer 1


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 $K$.


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