(dis)prove: M compact if and only if every closed ball in M in compact.

Question

I was wondering if my counter-example was ok:

Consider M=(0,1) with the standard metric on the real numbers. We have that M is not compact. Consider B, a closed ball with centre x and radius r in M. We know that B is compact iff B is closed and bounded. Now B is closed (I have proven this before) and certainly B is contained in the open ball with centre x and radius r+1. Thus shown that the statement is false.

If you have any feedback I would be grateful.

Answer 1

This doesn’t actually work, because in the space $M$ the closed ball of radius $\frac12$ centred at $\frac12$ is $M$ itself, which is not compact. Similarly, the closed ball of radius $\frac14$ centred at $\frac18$ is $\left(0,\frac38\right]$, which is not compact. If you replace $M$ by $\Bbb R$, however, you get a genuine example: now all of the closed balls are closed intervals in $\Bbb R$ and as such are compact, but $\Bbb R$ is not.

Answer 2

Take the set of real numbers, $\Bbb{R}$. Every closed interval in real line is closed and bounded even compact but $\Bbb{R}$ is not compact.