Every isometry is a bilipschitz equivalence and every bilipschitz equivalence is a quasi-isometry, and two metrics on one set are Lipschitz equivalent exactly when the identity is a bilipschitz equivalence between them

Given: The hypotheses of the Statement.

A map is a bilipschitz embedding when $c^{-1}d(x,x^{\mathrm{\prime }})\le d(f(x),f(x^{\mathrm{\prime }}))\le cd(x,x^{\mathrm{\prime }})$ for some $c>0$, and a bilipschitz equivalence when it is a bijective such map with bilipschitz inverse (Bilipschitz embeddings and bilipschitz equivalences of metric spaces).

A map is $(L,C)$-coarse Lipschitz when $d(f(x),f(x^{\mathrm{\prime }}))\le Ld(x,x^{\mathrm{\prime }})+C$, and an $(L,C)$-quasi-isometric embedding when in addition $L^{-1}d(x,x^{\mathrm{\prime }})-C\le d(f(x),f(x^{\mathrm{\prime }}))$ (Coarse Lipschitz maps and quasi-isometric embeddings).

A subset is coarsely dense when every point of the space is within a fixed distance of it, and a quasi-isometry is a coarse Lipschitz map admitting a coarse Lipschitz quasi-inverse (Coarsely dense subsets, quasi-inverses and quasi-isometries).

Isometric embedding and isometry. A function $f:X\to Y$ is an isometric embedding if (Isometry, isometric embedding, and the subspace metric on a subset).

• $d$ and $d^{\mathrm{\prime }}$ are topologically equivalent if they have the same metric topology: ${\mathcal{T}}_d={\mathcal{T}}_{d^{\mathrm{\prime }}}\mathrm{.}$ - $d$ and $d^{\mathrm{\prime }}$ are uniformly equivalent if for every real $\epsilon >0$ there are reals $\delta >0$ and ${\delta }^{\mathrm{\prime }}>0$ such that, for all $x,y\in X$, $d(x,y)<\delta ⟹d^{\mathrm{\prime }}(x,y)<\epsilon \text{and}{d}^{\mathrm{\prime }}(x,y)<{\delta }^{\mathrm{\prime }}⟹d(x,y)<\epsilon \mathrm{.}$ - $d$ and $d^{\mathrm{\prime }}$ are Lipschitz equivalent if there are reals $\alpha ,\beta >0$ with $\alpha d(x,y)\le d^{\mathrm{\prime }}(x,y)\le \beta d(x,y)\text{for all }x,y\in X\mathrm{.}$ (Topologically, uniformly and Lipschitz equivalent metrics on a set).

If $d$ and $d^{\mathrm{\prime }}$ are Lipschitz equivalent, they are uniformly equivalent. (Lipschitz equivalence implies uniform equivalence implies topological equivalence).