Is a closed and bounded subset of any complete metric space compact? 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! 
 A: 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.
A: Let $X$ be an infinite set. Define a metric $d$ on $X$ by $d(x,y) = 0$ if $x = y$, and $d(x,y) = 1$ if $x \neq y$. This is known as the discrete metric. This metric makes $X$ a complete metric space.
Let $E$ be any infinite subset of $X$. Then $E$ is closed and bounded, but not compact.
Remark: If we assume in addition that the set is totally bounded, the usual proof can be carried out.
Therefore closed and bounded subset of a complete metric space need not be compact.
