Let $X$ be a set. Then we can define at least three equivalence relations on the set of metrics on $X$. We say that two metrics $d_1$ and $d_2$ are topologically equivalent if the identity maps $i:(X,d_1)\rightarrow(X,d_2)$ and $i^{-1}:(X,d_2)\rightarrow(X,d_1)$ are continuous. We say that $d_1$ and $d_2$ are uniformly equivalent if $i$ and $i^{-1}$ are uniformly continuous. And we say that $d_1$ and $d_2$ are strongly equivalent if there exists constants $\alpha,\beta>0$ such that $\alpha d_1(x,y)\leq d_2(x,y)\leq\beta d_1(x,y)$ for all $x,y\in X$.

We can take equivalence classes of metrics under each of these equivalence relations. Now two metrics are topologically equivalent if and only if they induce the same topology on $X$, so we can identify equivalence classes under topological equivalence with topologies on $X$. And two metrics are uniformly equivalent if and only if they induce the same uniformity on $X$, so we can identify equivalence classes under uniform equivalence with uniformities on $X$. But my question is, what structures can we identify equivalence classes under strong equivalence with? To put it another way, two metrics are strongly equivalent if and only if they have the same ... what?

One thing worth noting is that if two metrics are strongly equivalent, then they’re both uniformly equivalent and they have the same bounded sets. Or in fancier language, they induce both the same uniformity and the same bornology. But the converse is not true; uniformly equivalent metrics with the same bounded sets need not be strongly equivalent. So that means there must be some structure on top of the uniformity and bornology that is preserved by strong equivalence.

Note: This was posted on Mathematics Stack Exchange with no answer.

lipeomorphism. So if two spaces are related like this, we could say that they arelipeomorphic. $\endgroup$ – Gerald Edgar Jan 7 at 20:51