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This question already has an answer here:

For the closed part, I just noted that it's a compact subspace of a Hausdorff space and therefore it's closed. For the bounded part, I know intuitively that since every open cover has a finite subcover, I just have to take the largest ball that includes all these covers, but I don't know how to write it rigorously. I know they'd all fit a large ball...

I also must find a metric space in which not every closed subspace is compact. Which is an example of a metric space in which not every closed is compact? Because once I know that, I could just take the metric $0,1$

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marked as duplicate by user228113, Claude Leibovici, user91500, E. Joseph, астон вілла олоф мэллбэрг Nov 17 '16 at 9:45

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  • $\begingroup$ Are you unable to use the general Heine-Borel theorem? $\endgroup$ – A.Riesen Nov 17 '16 at 1:14
  • $\begingroup$ Well, in $\Bbb R$ there are closed subsets which are not compact. For instance $\Bbb R$ itself (since it's not bounded). $\endgroup$ – user228113 Nov 17 '16 at 1:14
  • $\begingroup$ Compact imply finite subcover for every open cover. Then for the cover $\bigcup \Bbb B(x,\epsilon)$ for all $x$ in the compact set this imply that the set is bounded. $\endgroup$ – Masacroso Nov 17 '16 at 1:20
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    $\begingroup$ @G.Sassatelli but I needed a bounded one :c $\endgroup$ – Guerlando OCs Nov 17 '16 at 1:23
  • $\begingroup$ @GuerlandoOCs You did not mention it, though. Pick your favourite closed interval of $\Bbb Q$. $\endgroup$ – user228113 Nov 17 '16 at 1:28
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An insightful cover to use:

Let $S$ be the compact set. Pick any point $x \in S$. Now consider the cover $\{B(x, n) \ | \ n \in \mathbb{N} \}$, where $B(x, n)$ denotes the open ball centered at $x$ of radius $n$. Notice that $S$ is contained inside the largest $B(x, k)$ in the finite subcover this cover admits.

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  • $\begingroup$ You're saying that there will be a finite subcover that's just a ball, and therefore the set is bounded because the ball contains the compact entirely? $\endgroup$ – Guerlando OCs Nov 17 '16 at 1:23
  • $\begingroup$ Indeed @GuerlandoOCs. The union of everything in the finite subcover will simply be the largest ball in the finite subcover. $\endgroup$ – Kaj Hansen Nov 17 '16 at 1:25

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