# Bi-Lipschitz maps and equivalent metrics

Suppose $X$ is a non empty set and let $d_1, d_2$ be two metric on $X$, so we have two metric spaces over the same set $X$: $(X,d_1), (X,d_2).$

Suppose that there exists a surjective bi-Lipschitz map $\phi: (X,d_1)\to (X,d_2)$. Can we conclude that $d_1$ and $d_2$ are equivalent?

• what definition do you use for two metrics are equivalent? They generate the same topology or simply $$cd_1(x,y)\leq d_2(x,y) \leq Cd_1(x,y).$$ – Xiao Sep 24 '15 at 14:47
• They generate the same topology – batman Sep 24 '15 at 15:58

Take $X=\mathbb{R}$ with its usual metric $d_1(x,y)=|x-y|$. Take $\phi : X \to X$ to be a horrible, awful, very bad bijection. Do your worst. Make it discontuous everywhere, etc. etc. Define $d_2(x,y) = d_1(\phi(x),\phi(y))$. Then I see no reasonable definition of equivalence of metrics under which one would say that $d_1$ and $d_2$ are equivalent.
ADDED: With the definition of equivalence given by the OP in comments, the metrics $d_1$ and $d_2$ are not equivalent, indeed the identity map is not even continuous.
• Unless your definition of equivalence has a built-in invariance under bijections of $X$. In other words, your definition of equivalence would have to be formatted as follows: – Lee Mosher Sep 24 '15 at 15:10
• - Two metrics $d_1(x,y) : X \times X \to [0,\infty)$ and $d_2(x,y) : X \times X \to [0,\infty)$ are equivalent if there exists a bijection $\phi : X \to X$ such that the functions $d_1(\phi(x),\phi(y)) : X \times X \to [0,\infty)$ and $d_2(x,y) : X \times X \to [0,\infty)$ are related by [RELATION]. – Lee Mosher Sep 24 '15 at 15:10
• $(X, d_1)$ and $(X, d_2)$ would even be isometric in this case. – Cloudscape Sep 24 '15 at 15:48
• @LeeMosher: Thanks for your answer. My definition of equivalence is that the identity maps $I:(X,d_1)\to (X,d_2)$ and $I:(X,d_2)\to(X,d_1)$ are continuous. – batman Sep 25 '15 at 7:18