# Metrics and Comparative Distances

Suppose $d_1$ and $d_2$ are metrics on a set M. I'm trying to find some insight to a condition where for all $w, x, y, z \in M$ then $d_1(w, x) > d_1(y, z) \iff d_2(w, x) > d_2(y, z)$, in other words, distance comparison is the same in the two metrics.

That such conditions exist is clear: take for example $d_2 = x.d_1$ where $x$ is constant. So this is a sufficient condition, but is it necessary ?

So far I have tried to show that this would follows if the metrics are strongly equivalent, but could only get as far as $d_1(w, x) > d_1(y, z) \implies \gamma d_2(w, x) > d_2(y, z)$ where $\gamma$ is a constant $\ge 1$. $\gamma = \beta / \alpha$ in the relation of strong equivalence $\alpha d_2() \le d_1() \le \beta d_2()$. Nor could I prove the converse that preserving comparison of distances would imply strongly equivalent metrics.

Google search didn't help: any assistance would be appreciated, especially if expressed in basic terms.

• From "where $\gamma$ is a constant..." till the end of that sentence it's not understandable (e.g., what is $=$ sign doing there?). Please, clarify what you are trying to state. – MASL Sep 17 '15 at 10:20
• @MASL Hopefully clearer now ? – Tom Collinge Sep 17 '15 at 10:24

Well it isn't true. Take $M=\{0,1,2\}$ and let $d_1(x,y) = |(x-y)^2|$ and $d_2(x,y) = |(x-y)^3|$, then $d_1$ and $d_2$ fulfils the requirements, but they are not proportional.
If you weaken the statement, that $d_2 = f(d_1)$ then it's enough. Since the requirement implies that $d_1(w,x) = d_2(y,z)\Leftrightarrow d_2(w,x) = d_2(y,z)$ we can assign values for $f(d_1(x,y))$ in a unique way (then we can extend $f$ in regions outside the range of $d_1$ if wanted).
• Thanks: that establishes that a linear relation between the metrics is not necessary and suggest that a relationship $d_2 = f(d_1)$ where $f$ is any positive monotonic function is sufficient: to complete the story, is such a condition necessary ? – Tom Collinge Sep 17 '15 at 11:29