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Given two metric spaces $(X_1,d_1),(X_2,d_2)$ one can define $d_{GH}(X_1,X_2)$---the Gromov Hausdorff distance between them. It appears to be $0$ iff $X_1$ and $X_2$ are isometric. One can therefore form the space $\mathcal{X}$ of all (classes of) compact metric spaces. I would like to know whether there is some model for this space i.e. metric space $Y$ which is homeomorphic (or better isometric) to $\mathcal{X}$.

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  • $\begingroup$ Since a compact metric space has at most the cardinality of the continuum, one could simply pick one isometric copy of each compact metric space in which the points are real numbers and this set with the Gromov-Hausdorff distance is a proper metric space. $\endgroup$ – Michael Greinecker Mar 26 '14 at 22:16
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    $\begingroup$ @MichaelGreinecker: I think the question is whether $\mathcal{X}$ is homeomorphic/isometric to some more familiar or more easily constructed metric space. $\endgroup$ – Nate Eldredge Mar 26 '14 at 23:52
  • $\begingroup$ Is it at least known that the space of compact metric spaces modulo isometry is separable? This reminds me of a similar result of Pisier asserting that for each $n\geqslant 2$ the space of $n$-dimensional operator spaces, $O_n$, is non-separable. I am wondering if one could embed $O_n$ in your space. $\endgroup$ – Tomek Kania Oct 18 '14 at 20:54
  • $\begingroup$ I believe the set of finite metric spaces with pairwise rational distances is dense in the Gromov-Hausdorff space of compact metric spaces. $\endgroup$ – Noah Schweber Jul 27 '15 at 19:11
  • $\begingroup$ @NoahSchweber Yes, that looks correct. You can post it as an answer. $\endgroup$ – nbarto Oct 4 '18 at 11:48

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