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Rubin's Theorem 2.34 says that "Compact subsets of metric spaces are closed." But I'm wondering should it also be true that, compact subsets of metric spaces are closed and bounded at the same time? The closeed-ness comes from Rudin's proof, and the bounded-ness comes from the fact that for a compact subset $E$, we can take the finite open cover and than this open cover is what bounds $E$? Discussions are welcomed! Thank you!!!

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    $\begingroup$ @KprimeX The Heine-Borel theorem doesn't apply to arbitrary metric spaces. It doesn't even apply to $\mathbb{R}^n$ if you choose a different metric. $\endgroup$ – Matt Samuel Nov 15 '15 at 2:48
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Yes, this is correct. To be a bit more precise, if $E$ is compact, you can fix any point $x$ and consider the open balls $B_n(x)$ for each $n\in\mathbb{N}$. These are an open cover of $E$, so there is a finite subcover. It follows that there exists an $N$ such that $d(x,y)<N$ for all $y\in E$, so $E$ is bounded.

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  • $\begingroup$ Closed and bounded is necessary but not always sufficient for compactness.(It depends on the space.) Another way to see that an unbounded set $S$ is not compact is that for any point $p$ in the space, $\{B_d(p,n) :n\in N\}$ is an open cover of $S$ with no finite sub-cover. $\endgroup$ – DanielWainfleet Nov 15 '15 at 8:09
  • $\begingroup$ Your "other way" is exactly the same as what I wrote... $\endgroup$ – Eric Wofsey Nov 15 '15 at 17:07
  • $\begingroup$ It was late for me.I didn't intend any criticism. $\endgroup$ – DanielWainfleet Nov 15 '15 at 17:23

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