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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, 2016 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, 2016 at 18:45
  • $\begingroup$ I have just seen, that xyzzyz posted the same answer before me. $\endgroup$
    – user178826
    Oct 10, 2016 at 18:46
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    $\begingroup$ 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. $\endgroup$ Oct 13, 2016 at 16:14

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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.

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    $\begingroup$ 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, 2016 at 18:49
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A metric space is proper iff (by definition) all closed balls are compact. https://en.wikipedia.org/wiki/Metric_space#Locally_compact_and_proper_spaces

All proper spaces are locally compact (see the link above) (hence "if" is true in your claim) and complete (see the link below): https://en.wikipedia.org/wiki/Glossary_of_Riemannian_and_metric_geometry#P

So any incomplete locally compact metric space is a counter-example to "only if".

Moreover, as mentioned Tsemo Aristide's answer, any non-compact metric space, even a proper one, has the same topology as some improper metric space.

A normed space X is proper iff it is locally bounded (iff it is finite-dimensional).

This last claim follows from Theorem 1.22 of Rudin: Functional Analysis. That says that locally compact topological vector spaces are finite-dimensional, hence equivalent to some $\mathbb K^n$, by Theorem 1.21, where $\mathbb K$ is your scalar field (i.e., $\mathbb R$ or $\mathbb C$). $\mathbb K^n$ being proper is well known.

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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.

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