# Metric space and compactness

Prove that if in a metric space all closed balls are compact, a subset is compact if and only if it is closed and bounded.

Attempt: If all closed balls are compact, then there is a converging subsequence, so the subset is closed and bounded since the subsequence converges. Is that right? How do I show the other direction?

Suppose $C$ is closed and bounded set in $X$, then due to boundedness $\exists R>0$ such that $C\subset [-R,R]$. But $[-R,R]$ is compact as per hypothesis and any closed subset of a compact space is itself compact.
• I proved that if $C$ is closed and bounded then it is compact. For other way, suppose that $C$ is compact and consider the open cover $\{B(x_0,n)|n\in \mathbb{N}\}$ where $x_0\in X$ is fixed. It admits a finite subcover and so $C$ is bounded. To show $C$ is closed show that $X/C$ is open. – wanderer Mar 19 '14 at 4:47
• @wanderer what does $C\subset [-R,R]$ mean here? – Ittay Weiss Mar 19 '14 at 7:56
• @IttayWeiss : I meant to say that given any closed and bounded set $C$ we can find $R>0$ such that $C$ is contained in the closed ball $B[0,R]$. actually $B[0,R]$ is bit larger ball (we can get a better $R$ with suitable choice of the centre of the ball). – wanderer Mar 19 '14 at 8:15