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I'm reading some stuff related to Gromov-Hausdorff distance. I have a question out of curiosity:

Does the isometry classes of compact metric spaces form a set $\mathcal{M}$ (but not a proper class)?

As written in Dmitri Burago's book A Course in Metric Geometry, it should be yes because he states that $(\mathcal{M},d_{GH})$ is a metric space. Is there a proof? Maybe some embedding theorems can be applied?

Thanks in advance!

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Any compact metric space can be embedded in $[0,1]^{\mathbb{N}}$, so indeed the isometry classes form a set (precisely: the set of isometry classes of subspaces of $[0,1]^{\mathbb{N}}$).

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    $\begingroup$ Slight quibble: the collection of isometry classes itself isn't even a class, it's a "hyperclass" (or "2-class" or "conglomerate" or ...). However, there are only "set-many" isometry classes; precisely, there is a set consisting of exactly one space from each isometry class. $\endgroup$ Commented Apr 25 at 20:40

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