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Every compact set is totally bounded, but can we say that every totally bounded set is compact?

I'm a beginner in metric space. My thinking is that a totally bounded set behaves like a finite set and in some sense it is small. So it is very much like a compact set.

Someone please help me to clear my doubts. Thanks.

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    $\begingroup$ I believe the open ball is totally bounded but it is not compact. $\endgroup$ – Yanko Dec 8 '18 at 13:37
  • $\begingroup$ Oh! yes, I didn't think of that. But one more question why are we taking open balls of arbitrary diameter? Why don't we take closed balls of arbitrary diameter? $\endgroup$ – Arjun Banerjee Dec 8 '18 at 13:41
  • $\begingroup$ closed balls are compact... (In $\mathbb{R}^n$ at least) $\endgroup$ – Yanko Dec 8 '18 at 13:41
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    $\begingroup$ @ArjunBanerjee You ask what happens if we change the definition of totally bounded by replacing open balls with closed balls? In this case the open ball is still totally bounded but it is not compact. $\endgroup$ – Yanko Dec 8 '18 at 13:47
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    $\begingroup$ @ArjunBanerjee It doesn't really matter, we have $B(x_0,r/2) \subset B[x_0,r] \subset B(x_0,2r)$ anyway. Of course, here I used $B[x_0,r]$ to denote a closed ball of radius $r$. $\endgroup$ – BigbearZzz Dec 8 '18 at 13:48
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No, that is not enough (but almost).

Consider the set $(0,1)$ in the metric space $\Bbb R$, it is totally bounded but not compact. However, it is well known that totally boundedness & completeness is equivalent to compactness. You can read more about that here.

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