Totally bounded subset in complete metric space implies compact?

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.

• I believe it should read "closed and totally bounded" Oct 3, 2013 at 22:30

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.
• @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 $R$. Oct 3, 2013 at 22:51