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 ?

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    $\begingroup$ This is not true. Consider $\mathbb{R}$ with a metric $d(x, y) = min(1, |x - y|)$. It generates the regular topology on $\mathbb{R}$, but the closed ball $B(0, 2) = \mathbb{R}$ is clearly not compact. $\endgroup$ – xyzzyz Oct 10 '16 at 18:42
  • $\begingroup$ This is not true. For instance, let us consider in $R$ the distance $d(x,y):=|x-y|\wedge 1$. This distant induces the usual topology. However, any ball with radious grater than $1$ is all the space, which is not compact. $\endgroup$ – user178826 Oct 10 '16 at 18:45
  • $\begingroup$ I have just seen, that xyzzyz posted the same answer before me. $\endgroup$ – user178826 Oct 10 '16 at 18:46
  • $\begingroup$ The correct theorem is "if $X$ is a complete locally compact geodesic metric space then every closed metric ball in $X$ is compact. This is a form of the classical Hopf-Rinow theorem from Riemannian geometry. A metric space is geodesic if between any pair of points there is a path whose length is the distance between the points. $\endgroup$ – Moishe Kohan Oct 13 '16 at 16:14

Consider $R^2-\{(0,0)\}$ endowed with the canonical metric, it is locally compact. But $B((0,1);2)$ is not compact.

But the result is true if $X$ is endowed with a norm.

  • $\begingroup$ Hum... you're right. Ok so is there a better way to summarise local compactness for the (not so) special case of metric spaces ? "There is a compact closed ball centered at each point" ? $\endgroup$ – James Well Oct 10 '16 at 18:49

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