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I have a question about these metric spaces.

I know that totally bounded -> bounded. And this link gives that totally bounded -> separable. http://www.proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Separable

So I have that totally bounded -> bounded and separable

But can we go the opposite way, that is can we say for metric spaces that: bounded and separable -> totally bounded?

I know that we can not say separable -> totally bounded, since R is separable, but not totally bounded. And I assume that we can not generally say that bounded -> totally bounded, because then we would have no use for the definition of totally bounded. But if we include both, do we then have a totally bounded metric space?

UPDATE: I got a fast answer that the answer was no, thanks! Then my follow up is this: I really hope you can help me with the follow up aswell.

Would it help if you put in complete aswell? The reason I am asking is because I have read that a subset A for R^n is compact if it is closed and bounded. Genereally a subset A would be compact iff it is closed, totally bounded and complete. So a subset A of R^n that is bounded must then also be totally bounded? What is it with R^n that makes bounded->totally bounded?

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    $\begingroup$ Take a countably infinite set with the discrete metric. $\endgroup$ – David Mitra Jan 7 '14 at 17:48
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Consider the space $X = \{e_k\}_{k=0}^\infty \subset l_2$ with the $l_2$ norm ($e_k$ is all zeros except for a one in the $k$'th position).

$X$ is clearly separable and bounded, but for any $\epsilon<\sqrt{2}$ has no finite $\epsilon$-net, hence is not totally bounded.

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  • $\begingroup$ Thanks, would it help if you put in complete aswell? The reason I am asking is because I have read that a subset A for R^n is compact if it is closed and bounded. Genereally a subset A would be compact if it is closed, totally bounded and complete. So a subset A of R^n that is bounded must then also be totally bounded. Why is it with R^n that makes bounded->totally bounded? $\endgroup$ – user119615 Jan 7 '14 at 17:42
  • $\begingroup$ No, the example $X$ is a closed set in $l_2$, hence complete. One thing that makes $\mathbb R^n$ work this way is local compactness. $\endgroup$ – GEdgar Jan 7 '14 at 18:54
  • $\begingroup$ In fact: (complete + totally bounded) <-> (compact), in a metric space. $\endgroup$ – GEdgar Jan 7 '14 at 18:55
  • $\begingroup$ Ah thanks, so I guess "seperable+complete+bounded->totally bounded" is very false then. So I guess it is very difficult to find conditions so that "bounded->totally bounded" holds?, and we are just lucky that it holds in Rn? $\endgroup$ – user119615 Jan 7 '14 at 19:52
  • $\begingroup$ I believe clarification that $X$ is an infinite dimensional space would clarify the text; e.g., $X = \{e_k\}_{k\in\mathbb{N}}$ (clarifying that $e_k$ is a map $\mathbb{N} \to \{0, 1\}$ such that $e_k(n) = 1$ if $k = n$, and $e_k(n) = 0$) $\endgroup$ – Anakhand Feb 18 at 15:58

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