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Given two metric spaces $(X,d_X), (Y, d_Y)$, a bi-Lipschitz map $f:X \to Y$ and a finite set of points $\{x_1, \ldots, x_n\} \in X$. Consider in addition, that $X$ is a vector space over $\mathbb{R}$, concretely some $l^p$ space. Under which conditions is it possible to find points $\{y_1, \ldots, y_n\} \in Y$ such that for each pair of points $x_i,x_j$ we have $d_X(x_i, x_j) = d_Y(y_i, y_j)$?

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    $\begingroup$ Rephrased: let $X,Y$ be bilipschitz homeomorphic metric spaces. Under what condition is it true that $X$ and $Y$ are "finitely isometric"? (where "finitely isometric" means that every finite subset of $X$ is isometric to some finite subset of $Y$?). The question sounds a bit vague, what do you except other than trivial restatements or a list of spaces that are both bilipschitz and finitely isometric? $\endgroup$ – YCor Dec 18 '15 at 14:35
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    $\begingroup$ @YCor I would guess that some connectedness assumptions might help, so that continuity arguments become possible. The intention of the question could be that the points $y_i$ are not too far away from $f(x_i)$ - maybe the OP wants to clarify this? I agree that the question is a bit too vague, but is it really so bad that it has to be closed? $\endgroup$ – Sebastian Goette Dec 20 '15 at 12:04
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    $\begingroup$ Well, if you only assume that $X$ is an $\ell^p$-space and $Y$ is any metric space, you have an endless list of trivial counterexamples... and even when $Y$ is also a Banach space. $\endgroup$ – YCor Dec 21 '15 at 10:47
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    $\begingroup$ @YCor I disagree with your statement. For example the Banach-space-theoretical version of the question asked by CAT0 which I suggest in mathoverflow.net/questions/227073 is of interest. Possibly other, more metric-geometrical versions are also of interest. I would suggest to re-open this question. $\endgroup$ – Mikhail Ostrovskii Dec 27 '15 at 9:43
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    $\begingroup$ @YCor I do not see an obvious answer to the question of CAT0 either, if we restate it as: Let $Y$ be a metric space bilipschitz-equivalent to $\ell_p$, $1<p<\infty$. Is it true that any finite subset of $\ell_p$ is isometric to a subset of $Y$? (There are much stronger chances to construct a counterexample than in my question, but I do not see an obvious counterexample in this case, too.) $\endgroup$ – Mikhail Ostrovskii Dec 27 '15 at 16:35
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This can go very wrong in totally disconnected spaces.

For instance, if $X$ and $Y$ are both finite, one can adjust the distances between points via the bi-Lipschitz map so that every collection of points gets their distances changed.

Even further, you can let $X$ be the rationals and let $Y$ be constructed by recursively moving each point of $X$ to a point in a distinct coset of the rationals, moving each point by only $1/2^n$, say. This is another example of a pair of spaces where you can never map a collection of points isometricly.

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