Rank Preserving Maps between Metric Spaces

Question

Consider two Metric Spaces $(U, d_{U})$ and $(V, d_{V})$.

Here let

$f: U \rightarrow V$

Then $f$ is an isometry between if

$\forall u_{i}, u_{j} \in U$

$d_{U}(u_{i},u_{j}) = d_{V}(f(u_{i}), f(u_{j}))$

I'm interested in studying a different type of mapping in which the ranking of distances between a set element are preserved, i.e.

Let $u_{T}, u_{1}, u_{2}, \dots u_{n} \in U$ and let

$d_{U}(u_{T}, u_{1}) \leq d_{U}(u_{T}, u_{2}) \leq \cdots \leq d_{U}(u_{T}, u_{1})$

Then I'm interested in studying maps between $U$ and $V$ such that this is preserved, i.e.

$g:U \rightarrow W$

$d_{U}(u_{T}, u_{1}) \leq d_{U}(u_{T}, u_{2}) \leq \cdots \leq d_{U}(u_{T}, u_{1}) \rightarrow d_{W}(g(u_{T}), g(u_{1})) \leq d_{W}(g(u_{T}), g(u_{2})) \leq \cdots \leq d_{W}(g(u_{T}), g(u_{1}))$

Is this studied at all? and if so, where should I go to find more information about it?

Answer 1

The chain of inequalities forces equality to hold because the last term in the inequality is the same as the first. Your condition is equivalent to $f$ being isometric.