Locally compact metric space I'm trying to prove that a metric space is locally compact iff every closed ball is compact, using the more general definition that applies to Hausdorff spaces, that every point has a compact neighbourhood.
So call $X$ my space. The only non trivial thing to prove is that every closed ball is compact, assuming $X$ is locally compact. So consider $N$ a compact neighbourhood of some $x\in X$. Then as a neighbourhood, it contains $B(x,r)$ for some $r$. So it contains $\bar{B}(x,r/2)$. This is closed inside $N$ which is compact, so it's also compact. So I've proven that at any point there is a compact closed neighbourhood ball. Surely it's not too hard to prove all the bigger closed balls are compact ?

If $(X,d)$ is a metric space, then $d'(x,y) = d(x,y)/(1+d(x,y))$ defines a new metric having the same (open or closed) balls as $(X,d)$. But then $X$ is the ball of radius $1$ centered anywhere. Hence if, in addition, $X$ is locally compact but not compact, the metric space $(X,d')$ admits a closed non compact ball.
