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Suppose $(X,d)$ is a countably infinite set endowed with a metric $d$ that satisfies the following condition:

Every finite subset of $X$ with the induced metric is isometric to a subset of some finite-dimensional euclidean space.

Does it follow that $X$ is isometric to a subset of Hilbert space $H = \ell^2(\mathbb Z_+)$ ?

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  • $\begingroup$ It's true ($X$ need not be finite if you allow arbitrary Hilbert spaces). A (semi)metric space is isometrically embeddable into a Hilbert space iff the square of its distance is conditionally negative definite, and the latter condition is finitary. $\endgroup$
    – YCor
    Commented Feb 5 at 23:15
  • $\begingroup$ Alternatively, it follows from a standard ultraproduct procedure (using that the metric ultraproduct of Hilbert spaces is Hilbert). But of course this is essentially a nonstandard restatement of Christian Remling's answer. $\endgroup$
    – YCor
    Commented Feb 5 at 23:18

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When embedding $f_n:\{ x_1, \ldots , x_{n+1}\}\to\mathbb R^n\cong \ell^2(\{ 1, \ldots ,n\})$, we can assume that $f_n(x_j)\in \ell^2(\{ 1, \ldots , j-1\})$ (and $f_n(x_1)=0$), by giving $\ell^2$ a suitable basis. By a diagonal procedure and compactness, we find a subsequence $n_k$ such that $f(x_j):=\lim f_{n_k}(x_j)$ exists for all $j\ge 1$, and this function works as the desired isometry.

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  • $\begingroup$ That appears to show that the entire embedding lies in the increasing union of all finite-dimensional euclidean spaces (so that every point in the final embedding has only finitely many nonzero coordinates). $\endgroup$
    – Daniel Asimov
    Commented Feb 5 at 18:58
  • $\begingroup$ @DanielAsimov: This is not an additional property but rather the general claim: every countable subset $y_1, y_2, \ldots \in\ell^2$ satisfies $y_j\in L(e_1,\ldots , e_j)$ if we choose the ONB $e_j$ appropriately (apply Gram-Schmidt to the $y_j$). $\endgroup$
    – Christian Remling
    Commented Feb 5 at 19:53

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