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!
Best Answer
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.