Finding isometries

Question

I was wondering if for a given set of points $x_1,...,x_n$, $n > 2$, and two metric spaces $(X,d_1),(Y,d_2)$, $X \neq Y$, it s possible to find a continuous mapping $T: X \rightarrow Y $, such that $d_1(x_i,x_j) = d_2(T(x_i),T(x_j))$ for all pairs of indices $(i,j)$. I'm sure there s something around in the optimisation literature but I cant seem to find any keywords; maybe some optimal transport related topic. Any help or pointers would be greatly appreciated.

Answer 1

No, this isn't always possible. Here is an extreme counterexample.

Let $X$ be the real line but under the discrete metric $d_1$. Let $Y$ be the real line under the ordinary metric $d_2$. Take 3 distinct points $x_1, x_2$ and $x_3$ in $X$. Suppose you could find a continuous map $T: X \to Y$ preserving the distances between all the $x_i$.

Since $X$ has the discrete metric this means that $d_2(T(x_i), T(x_j))=1$ whenever $i \neq j$. Suppose, without loss of generality, that the $x_i$ are indexed so that $T(x_1) < T(x_2) < T(x_3)$. This is clearly impossible since now $T(x_1)$ and $T(x_3)$ are now distance 2 apart.