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Is there a complete measurable metric space $(X,d)$ for which its $p$-Wasserstein space $W(X)$ is isometrically isomorphic to $(X,d)$ for some $p \in [1,\infty]$?

Note that there is a canonical non-surjective isometric map from $X$ to $W(X)$, but this seems to not necessarily contradict the existence of an isometric isomorphism.

The intuition why the intial question could be true is that one can iterate taking Wasserstein spaces, i.e., one can take $W(W(X)),\ldots,W(\ldots(W(X))),\ldots$, each canonically embedding to the next one. Then, one can take union of all of them in a suitable sense, and then metric completion.

Is there anything known about the space resulting from this construction, e.g. if one starts with the two-point space $X$?

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  • $\begingroup$ There are the two trivial answers to the first question, the empty space and the space consisting of a single point. It feels like everything after that can probably be ruled out by a cardinality argument, though maybe for extremely large spaces there might not be enough ways to construct measures anymore. $\endgroup$
    – mlk
    Commented Apr 18 at 7:34
  • $\begingroup$ Thank you for your reply! I also had this feeling that the Wasserstein space must be "bigger", but cardinality is not useful as every separable metric space has cardinality at most of the continuum. Moreover, the Wasserstein space of a separable complete metric space is separable and complete again. So the question is, what is the right concept of a metric space beeing "bigger" than another? $\endgroup$ Commented Apr 18 at 11:33
  • $\begingroup$ Okay, thinking about it, the notion of "bigger" does indeed not work except for maybe something small enough like countable $X$. However $C([0,1])$ is separable and embeds isometrically into $W(C([0,1]))$, but as this is also separable, the Banach-Mazur theorem tells us that it in turn embeds isometrically back into $C([0,1])$. So they are of the same size for all intents and purposes. $\endgroup$
    – mlk
    Commented Apr 19 at 8:11

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