TITLE: Totally bounded and closed implies compact??
QUESTION [10 upvotes]: Is there a fault in exercice 9.3.1 b) from Analysis by Zorich? The exercice asks to prove that a subset of a metric space is compact if and only if it is totally bounded and closed. But I have a counterexample for it: Consider the open unit ball $B(0,1)$ in $\mathbb{R}^n$ as a metric space itself. Then it is closed in itself and totally bounded, but not compact. 
Am I right?

REPLY [6 votes]: $B(0,1)$ is closed in the space $B(0,1)$ endowed with the subspace topology of $\mathbb{R}^n$. But this is true for every $S \subset \mathbb{R}^n$ if we endow $S$ with the subspace topology.
The requirement is that the set is closed in $\mathbb{R}^n$, not in itself.
Moreover, in a generic tolopogical space X, given $A \subset X$, the equivalence " $A$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $X$ is complete. In this case every closed subspace of $X$ is also complete.