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I am reading the book Elements of Functional analysis by Kolmogorov and Fomin. In chapter 2, section 16 on compact metric spaces the author poses the following theorem which he demonstrates afterwards:

"A necessary and sufficient condition that a subset M of a complete metric space R be compact is that M be totally bounded."

I have been able to easily prove necessity by negation. Nevertheless I find the proof on sufficiency provided by the author unsatisfactory since I cannot see the way in which completeness of the whole space R would imply convergence of a cauchy sequence in M to a point in M. Take for instance any open ball of radius epsilon in an euclidean space. The ball is totally bounded and the space is complete, nevertheless the ball is clearly not compact.

Thank you in advance for your time, any advice will be greatly appreciated

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  • $\begingroup$ I believe it should read "closed and totally bounded" $\endgroup$ – Will Jagy Oct 3 '13 at 22:30
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You are correct. The result should read:

A necessary and sufficient condition that a closed subset $M$ of a complete metric space $R$ be compact is that $M$ be totally bounded.

If $M$ is compact, then of course $M$ will be closed, and if $M$ is a closed subset of a complete space, then $M$ is complete as well.

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  • $\begingroup$ thank you very much. I have realized that if we ask for M to be complete in itself the theorem also follows $\endgroup$ – Arturios Oct 3 '13 at 22:50
  • $\begingroup$ @Arturios: You’re very welcome. Yes, asking for $M$ to be complete in the metric inherited from $R$ is equivalent to asking $M$ to be closed in $R$. $\endgroup$ – Brian M. Scott Oct 3 '13 at 22:51

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