Let $X$ be complete finite dimensional Alexandrov space with curvature bounded from below. Is it true that any two points can be connected by a shortest path? If this is not true in general, it it true under some assumptions on $X$? (E.g. this is certainly true for $X$ being a complete smooth Riemannian manifold.)
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3$\begingroup$ If $(X,d)$ is a complete, locally-compact path metric space, then any two points can be joined by a minimizing geodesic (using the Arzela-Ascoli and lowersemicontinuity of length). Finite-dimensional complete Alexandrov spaces are locally compact. $\endgroup$– JHMApr 11, 2015 at 18:28
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$\begingroup$ @J.Martel: Do you need just local compactness or that any closed ball is compact? (in any case, you are right since complete finite dimensional Alexandrov spaces do have this property, as I learned.) $\endgroup$– asvApr 11, 2015 at 18:33
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The answer is "yes", see e.g., p. 114 in [Plaut, Spaces of Wald-Berestovskii Curvature, http://link.springer.com.sci-hub.org/article/10.1007/BF02921569] - which is available online. It is claimed: "... If X is locally compact, Ascoli's theorem can be used to obtain the existence of minimal curves between all pairs of points ..."
Possibly, the other reference might be Y. Burago, M. Gromov, G. Perelman; Alexandrov's spaces with curvatures bounded from below I. Uspeki Mat. Nauk, 47 (1992), pp. 3–51. - which I did not find online
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3$\begingroup$ Well, this is true for any locally compact lenght metric space (see Hopf--Rinow theorem). You need to show that finite dimensional Alexandrov spaces are locally compact. The later is proved already in Burago-Gromov-Perelman. $\endgroup$ Apr 12, 2015 at 3:08