Finitely isometrically persistent metric spaces

Question

The goal of this question is to develop further the discussion initiated in Under which conditions is it possible to find points with same distances under bi-Lipschitz map. The mentioned question was closed because the author stated the question somewhat vaguely, but I think that the corresponding direction deserves attention.

Definition: We say that a metric space $X$ is finitely isometrically persistent if for any metric space $Y$ which is bilipschitz equivalent to $X$ (this means that there exists a bijective Lipschitz map $F:X\to Y$ such that $F^{-1}$ is also Lipschitz) and any finite set $A\subset X$ there is a subset $B\subset Y$ such that $A$ and $B$ (with the induced metrics) are isometric.

It is easy to find numerous examples of spaces which are not finitely isometrically persistent.

Question: Does there exist an infinite separable metric space which is finitely isometrically persistent?

Restricting the category of metric spaces and (possibly also) the category of allowed maps $F:X\to Y$ we get many questions, some of which are definitely interesting. Some of them were asked in the question mentioned above, Is $\ell_p$ $(1<p<\infty)$ finitely isometrically distortable?, and Isometric embeddings of finite subsets of $\ell_2$ into infinite-dimensional Banach spaces.

One can also consider suitably defined $\varepsilon$-isometric finite persistence. Many of the interesting and important results in the theory of Banach spaces and theory of metric embeddings can be stated in these terms. An interesting counterexample (trees, Theorem 1.12 in Mendel-Naor, JEMS, 15 (2013), 287-337) is also known.

Inserted after the question was answered: The answer of Nik Weaver (see below) seems to show that in the study of this question we have to restrict ourselves, for example, to Banach spaces and linear isomorphisms (one can try something in between this and the general metric category).

Answer 1

There is no such space with more than one point. Let $X$ be any metric space and suppose there exist $x,y,z \in X$ such that $$a = d(x,y) > {\rm max}(d(x,z), x(y,z)) = b.$$ That is, there is a triangle in which one side is longer than the other two. Define $\omega: [0,\infty) \to [0,\infty)$ by $$\omega(t) = \begin{cases} t&0 \leq t \leq b\cr m(t-b) + b&b \leq t \end{cases}$$ where $m = \frac{a-b}{2b}$, and let $d'$ be a new metric on $X$ defined by $d' = \omega\circ d$.

Let $Y$ be the set $X$ equipped with this new metric. It is clear that the identity map is bi-Lipschitz. However, there are no points $x',y',z' \in Y$ satisfying $d'(x',y') = d(x,y)$, $d'(x',z') = d(x,z)$ and $d'(y',z') = d(y,z)$. For distances $\leq b$ do not change, so the three conditions force $d(x',z') = d(x,z)$ and $d(y',z') = d(y,z)$, and this implies $d(x',y') \leq d(x',z') + d(y',z') \leq 2b$, which means that $$d'(x',y') \leq \omega(2b) = \frac{a-b}{2} + b = \frac{a+b}{2} < a = d(x,y).$$

Thus, there is a space $Y$ which is bi-Lipschitz equivalent to $X$ but does not have a triangle congruent to $\overline{xyz}$.

---

Editing to include the case when no triangle has a strictly longest side. The argument is actually a bit different. Suppose $X$ has this property, choose any two distinct $x,y \in X$, and let $a = d(x,y)$. Then let $d' = \omega\circ d$ where $$\omega(t) = \begin{cases} t&0 \leq t < a\cr t + 1&a \leq t. \end{cases}$$ The only thing to check is that $d'$ is a metric; granting this, it is clearly bi-Lipschitz equivalent to $d$ (distances change by at most a factor of $\frac{a+1}{a}$) and does not contain a two-element set isometric to $\{x,y\} \subseteq X$. The only issue is the triangle inequality: we need $r \leq s + t$ whenever $r$, $s$, and $t$ are the sides of a triangle in $X'$. Since this inequality holds with respect to $d$, a moment's thought shows that the only threat is if $r$ increases but $s$ and $t$ do not. But that could only happen if $r > {\rm max}(s,t)$, contradiction.

Answer 2

Although this is not a direct reply to your question, I think that it is in the spirit of the question. I want to mention the paper of Falconer and Marsh, On the Lipschitz equivalence of Cantor sets, Mathematika 39 (1992), 223-233.

Copying the abstract: "We show that under certain circumstances quasi self-similar fractals of equal Hausdorff dimensions that are homeomorphic to Cantor sets are equivalent under Hölder bijections of exponents arbitrarily close to 1. By setting up algebraic invariants for strictly self-similar sets, we show that such sets are not, in general, equivalent under Lipschitz bijections."

It goes without saying that it is unlikely that the techniques in the paper extend to invariant sets of nonlinear nature, although of course they apply verbatim at the level of topological entropy.