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Is there a nonprincipal ultrafilter $\omega$ on $\mathbb N$ such that for any metric space $M$ there is an isometry $$(M^\omega)^\omega\to M^\omega?$$ (In other words, the $\omega$-power of $\omega$-power of $M$ is isometric to the $\omega$-power of $M$.)

Comments.

  • You may assume that $M$ has finite diameter, otherwise $M^\omega$ is an $\infty$-metric space; that is, distance between points might take infinite value.

  • I want to thank Will Brian for pointing to this post of Andreas Blass. It solves my problem negatively, at least if one assumes some natural condition on the isometry.

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    $\begingroup$ I presume $M^\omega$ is the "standard part" of the usually-defined ultrapower, right (namely, take the subset of the full ultrapower consisting of points a finite distance from the image of some element of $M$, and then quotient out by the infinitesimal distance relation)? $\endgroup$ Jul 5, 2022 at 16:53
  • $\begingroup$ @NoahSchweber, I added a comment about it. $\endgroup$ Jul 5, 2022 at 20:55
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    $\begingroup$ You still need to mod out by infinitesimal distance - otherwise, it's not a genuine metric space. For example, consider (the points corresponding to) the constant sequence $(0,0,0,...)$ and the harmonic sequence $(1,{1\over 2},{1\over 3},{1\over 4},...)$ in the ultrapower of $[0,1]$ (with the usual metric). $\endgroup$ Jul 5, 2022 at 21:06
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    $\begingroup$ I don't know the answer to this question, but I think this thread is relevant: mathoverflow.net/questions/324254/…. If I understand correctly, Andreas' answer there blocks a plausible way to proving a positive answer to your question. $\endgroup$
    – Will Brian
    Jul 5, 2022 at 21:16
  • $\begingroup$ @NoahSchweber ultralimit is usually defined as the corresponding metric space to the constructed pseudometric space; that is, it is a metric on equivalence classes defined by $x\sim y$ iff $|x-y|=0$. $\endgroup$ Jul 6, 2022 at 11:06

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