Let $(X,d_1)$ be a metric space. Assume there are $x,y \in X$, such that $$d_1(x,z)\leq d_1(y,z) \quad \forall z \in X\setminus \lbrace x,y\rbrace. \quad (*) $$

I am trying to show that if one defines another metric $d_2$ on X, this property is preserved, i.e.

$$(*) \implies d_2(x,z)\leq d_2(y,z) \quad \forall z \in X\setminus \lbrace x,y\rbrace $$

I am not sure if it is actually possible to prove this in general. I guess one should rather focus on properties, that $d_2$ has to fullfil in order to preserve $(*)$. I tried it with topollogically equivalent metrics but so far i couldn't come up with a proof.

Any suggestion towards proving this hypothesis would be helpful.


This is not true. Consider $X=\mathbb R$ and define $d_1$ to be the discrete metric and $d_2$ to be the Euclidian metric. Then the first inequality is trivially satisfied for all $x,y\in X$, where the last one is not. Take, for instance, $x=0,y=3,z=2$. Then $|x-z|=2>1=|y-z|$.

  • $\begingroup$ Ok I guess this does not hold in many spaces. The space i am considering is actually a graph G=(X,E) with $d_1$ as the geodesic distance. I have checked that if $d_2$ is the resistance distance then the property holds. But I can not find any characteristic of G or $d_1$ and $d_2$ explaining that. I posted the question more general but I guess this could be a unique feature of graphs. $\endgroup$ – munky Feb 24 '15 at 20:41

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