TITLE: Is a closed and bounded subset of any complete metric space compact?
QUESTION [6 upvotes]: We know that every closed and bounded subset of $\Bbb{R}$ is compact. The proof proceeds by bifurcating $[a,b]$, and then using the property that in a complete metric space the infinite intersection of closed and bounded sets contains one point. 
I was wondering if this can be extended to any complete metric space. Is a closed and bounded set of any complete metric space compact? If one were to extrapolate the proof given above for this situation, how do you bifurcate a complete metric space? 
Also, assume that the complete metric space is ordered. Can you still bifurcate it? How do you find the mid-point?
Thanks in advance!

REPLY [6 votes]: André’s example is probably the simplest; here’s a fairly general technique for constructing such examples. If $\langle X,d\rangle$ is any metric space, the function
$$d\,':X\times X\to\Bbb R:\langle x,y\rangle\mapsto\min\{d(x,y),1\}$$
is a metric on $X$ that generates the same topology as $d$. If $\langle X,d\rangle$ is complete but not compact, $\langle X,d\,'\rangle$ provides a counterexample: every subset of $X$ is $d\,'$-bounded, so $X$ is a closed, bounded subset that is not compact. Since $\Bbb R$ with the usual metric is complete but not compact, it provides one such example.