Topologically, uniformly and Lipschitz equivalent metrics on a set

Let $X$ be a set and let $d$ and $d^{\mathrm{\prime }}$ both be metrics on $X$ (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric). Note that the underlying set is the same; nothing below compares metrics on different sets.

• $d$ and $d^{\mathrm{\prime }}$ are topologically equivalent if they have the same metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement): ${\mathcal{T}}_d={\mathcal{T}}_{d^{\mathrm{\prime }}}\mathrm{.}$
• $d$ 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 $\alpha d(x,y)\le d^{\mathrm{\prime }}(x,y)\le \beta d(x,y)\text{for all }x,y\in X\mathrm{.}$

What the middle condition says in words. It is the statement that both identity maps $\mathrm{i}\mathrm{d}:(X,d)\to (X,d^{\mathrm{\prime }})$ and $\mathrm{i}\mathrm{d}:(X,d^{\mathrm{\prime }})\to (X,d)$ are uniformly continuous: the same $\delta$ works at every pair of points, not merely at each point separately as in Continuity of a map between metric spaces, at a point and globally, in the $\epsilon$-$\delta$ form. Uniform continuity has no definition of its own at this point in the reading order, so the condition is written out in full above; a later page defines it, and until then this write-out is what earlier pages quote.

Each of the three is an equivalence relation on the metrics on $X$. Reflexivity is immediate ($\delta =\epsilon$, and $\alpha =\beta =1$); symmetry is built into the statements, the uniform one being symmetric by construction and the Lipschitz one because $\alpha d\le d^{\mathrm{\prime }}\le \beta d$ gives ${\beta }^{-1}{d}^{\mathrm{\prime }}\le d\le {\alpha }^{-1}{d}^{\mathrm{\prime }}$; and transitivity follows by composing the $\delta$s and multiplying the constants.

• The three are ranked, and the ranking is proved, not assumed: Lipschitz equivalence implies uniform equivalence implies topological equivalence (Lipschitz equivalence implies uniform equivalence implies topological equivalence). Neither implication reverses, and the witnesses live on the companion page.
• Naming forks in the literature. Many texts say strongly equivalent for what is called Lipschitz equivalent here, and many say simply equivalent for what is called topologically equivalent here. This library always writes the qualifier, so that no statement depends on which convention a reader brings. A few texts define topological equivalence by "the identity is a homeomorphism", which is the same condition (Metric continuity characterisations, with countable choice for the sequential converse).
• Topological equivalence preserves exactly the topological notions: open, closed, closure, interior, boundary, convergence of sequences, continuity of maps into and out of the space. It does not preserve boundedness, diameters or the Lipschitz constants, and FALSE: boundedness of a metric space is determined by its topology records the first of those failures.