0
$\begingroup$

Can you please let me know if my proof is correct? I would like to make sure that I've understood the concept of open covers and subcoverings correctly.

Proof:

Suppose $\Pi = \bigcup\limits_{k=1}^n H_k$ is a finite union of $n$ compact sets. Then there is a finite subcovering for every open covering of $H_k$. In particular, if $\Psi_k = \{V_{\alpha}\}_{\alpha\in A}^{(k)}$ is an open covering of $H_k$ then $\Psi_{k_0}=\{V_{\alpha}\}_{\alpha\in A_0}^{(k)}$ (where $A_0$ is a finite subset of $A$) is a subcovering of $H_k$. Consider $\Xi = \bigcup\limits_{k=1}^n\Psi_{k_0}$, the union of all subcoverings of $H_k$ (for $1\le k\le n$). Then $X$ is an open subcovering of $\Pi$, which implies that $\Pi$ is compact.

$\endgroup$
  • $\begingroup$ No. You have to start with an open cover of the union. BTW, it suffices to consider only two compact sets. Always simplify in order to understand thigs better. $\endgroup$ – Friedrich Philipp Mar 9 '16 at 2:32
  • $\begingroup$ @FriedrichPhilipp: Do you mean I have to start with an open cover for $\Pi$? But I have to derive that there is this open cover by arguing first that there is an open cover for each $H_k$. $\endgroup$ – sequence Mar 9 '16 at 2:35
  • $\begingroup$ You need to show if you have any open cover of $\Pi$, there is a finite subcover. That's what it means for $\Pi$ to be compact. You aren't allowed to start with open covers of the $H_k$. $\endgroup$ – D_S Mar 9 '16 at 2:38
  • $\begingroup$ @D_S: But how do I know that I have an open cover of $\Pi$? I only know that I have open covers for $H_k$, because I know that they are compact. It's that which I need to prove that $\Pi$ has an open covering. I think I'm totally lost now. $\endgroup$ – sequence Mar 9 '16 at 2:42
  • $\begingroup$ The whole space is an open cover for $\Pi$. $\endgroup$ – Friedrich Philipp Mar 9 '16 at 2:43
2
$\begingroup$

Start with an open cover $(U_\alpha)_{\alpha\in A}$ of $K_1\cup K_2$. Now, is it an open cover of $K_1$? Yes, fine. So, take a finite subcover. Do the same for $K_2$ and take the union of the two finite subcovers (where you cancel those in the second one that already are in the covering of $K_1$).

$\endgroup$
  • $\begingroup$ How do we know that an open cover of $K_1 \cup K_2$ exists? $\endgroup$ – sequence Mar 9 '16 at 2:50
  • $\begingroup$ The existence of a specific open cover for $K_1 \cup K_2$ is irrelevant here (if you want one, take just the whole topological space, that's an open cover by one open set). You need to show that if you begin with any open cover of $K_1 \cup K_2$, then it can be reduced to a finite subcover. $\endgroup$ – D_S Mar 9 '16 at 2:53
  • $\begingroup$ It's not like you can prove compactness by finding an open cover of $K_1 \cup K_2$ and reducing that thing to a finite subcover. $\endgroup$ – D_S Mar 9 '16 at 2:54
  • $\begingroup$ Correct, since all subcovers are supposed to contain a finite one. $\endgroup$ – Friedrich Philipp Mar 9 '16 at 2:56
  • 1
    $\begingroup$ Yes, that's fine. Only one thing: $K_1\subset\{U_\alpha\}$ does not make sense. You mean $K_1\subset\bigcup_\alpha U_\alpha$. $\endgroup$ – Friedrich Philipp Mar 9 '16 at 3:24

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.