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Proposition. If $K$ is a compact metric space, $F \subset K$, and $F$ is closed, then $F$ is compact.

Proof. Suppose $F$ is a closed subset of the compact set $K$. If $\mathcal{G} = \{G_\alpha\}$ is an open cover of $F$, then $\mathcal{G}' = \mathcal{G} \cup \{F^c\}$ will be an open cover of $K$. We are given that $K$ is compact, so let $\mathcal{H}$ be a finite subcover of $\mathcal{G}'$ for $K$. If $F^c$ is one of the open sets in $\mathcal{H}$, omit it. The resulting finite subcover of $\mathcal{G}$ will be an open cover of $F$.

I have a question. So I understand that we can throw out $F^c$, as it's disjoint from $F$. But how do we know that a finite subcover of $\mathcal{G}' = \mathcal{G} \cup \{F^c\}$ (which for all purposes and intents, just consider $\mathcal{G}$) for $K$ is actually an open cover of $F$? How do we know we aren't missing any part of $F$, i.e. there's a part of $F$ that hasn't been covered?

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    $\begingroup$ It covers $K$ and $F$ is a subset of $K$ ... $\endgroup$ – user251257 Aug 17 '16 at 23:16
  • $\begingroup$ Any cover of $K$ is also a cover of $F$ since $F$ is a subset of $K$. And the elements of the cover just need to be open in $K$ for it to be an open cover for $F$, since we're proving that $F$ is compact in $K$. $\endgroup$ – florence Aug 17 '16 at 23:18
  • $\begingroup$ I understand the proof, but I also want to ask if we can argue that since $F$ is closed and is a subset of compact set $K$ which is bounded, then $F$ must also be bounded. Hence, $F$ is also compact....? $\endgroup$ – TheLast Cipher Jul 22 '18 at 6:16
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Basically what this proof does is it shows that given an open cover $U$ of $F$, we can extend $U$ to cover all of $K$ by adding the open set $F^c$ (this is possible since $F$ is closed). Then since this new open cover covers all of $K$, it must have a finite subcover by compactness, call it $U' \subseteq U \cup \left\{F^c \right\}$. Then $U'$ covers $F$ since $F \subseteq K$, and we can toss out $F^c$ to get an open cover of $F$ that is a finite subcover of $U$. Since $U$ was arbitrary, the claim follows.

Note that we didn't use any particular properties of metric spaces here, this actually holds in any compact topological space $K$.

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  • $\begingroup$ I understand the proof, but I also want to ask if we can argue that since $F$ is closed and is a subset of compact set $K$ which is bounded, then $F$ must also be bounded. Hence, $F$ is also compact....? $\endgroup$ – TheLast Cipher Jul 22 '18 at 6:17
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    $\begingroup$ Closed and bounded does not imply compact in a general metric space. $\endgroup$ – M10687 Jul 22 '18 at 6:21
  • $\begingroup$ But if we are to restrict to sets in $\Bbb{R}$, we can? thanks!! $\endgroup$ – TheLast Cipher Jul 22 '18 at 6:24
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    $\begingroup$ Yes, in $\mathbb{R}^n$ this would be ok, but I would personally avoid it since you'd be implicitly using a theorem that is much too powerful for this situation, namely Heine-Borel. $\endgroup$ – M10687 Jul 22 '18 at 6:26
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This follows from the fact $F \subset K$.

In other words, if $\mathcal{G}$ is an open cover for $K$, then $$K \subset \bigcup_{\alpha\in A} G_{\alpha} $$

and thus $\mathcal{G}$ must also be an open cover for $F$, since $$F \subset K, \quad K \subset \bigcup_{\alpha\in A} G_{\alpha} $$ $$\implies F \subset \bigcup_{\alpha\in A} G_{\alpha}$$ In other words, subset inclusion is transitive.

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