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A Riemannian manifold is known to be complete if and only if any of its closed bounded subset is compact.

Is there a similar criterion for Riemannian manifolds for which every bounded subset is totally bounded with respect to the geodesic distance, that is, every bounded set for the geodesic distance can be covered by finitely many geodesic balls of arbitrarily small radius?

Obviously this will happen in any complete Riemannian manifold.

There are manifolds that are bounded without being totally bounded, see https://math.stackexchange.com/q/749891/105127.

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  • $\begingroup$ What do you actually mean by similar criterion? You started with $P \iff Q$ where $P$ is complete and $Q$ is closed bounded subset is compact. Is your goal to replace $Q$ by "every bounded subset is totally bounded" and ask what is the corresponding generalization of $P$? $\endgroup$
    – Willie Wong
    Jun 19, 2020 at 14:58
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    $\begingroup$ (Minor comment: doesn't Hopf Rinow require finite dimensions?) $\endgroup$
    – Willie Wong
    Jun 19, 2020 at 15:01
  • $\begingroup$ Convexity would be enough. $\endgroup$
    – Moishe Kohan
    Jun 19, 2020 at 15:25
  • $\begingroup$ Possibly a not very useful observation: for metric spaces the property "every bounded subset is totally bounded" is equivalent to "the Cauchy-completion having the Heine-Borel property.". Hopf Rinow says that finite dimensional Riemannian manifolds is geodesically complete IFF Cauchy-complete; unfortunately the Cauchy completion of an arbitrary Riemannian manifold need not be equipped with a good Riemannian structure. $\endgroup$
    – Willie Wong
    Jun 19, 2020 at 18:48
  • $\begingroup$ @WillieWong I am wondering about a necessary and sufficient condition as fruitful as the equivalence for completeness. $\endgroup$
    – Jean Van Schaftingen
    Jun 22, 2020 at 6:44

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