Do the metrics $d$ and $\frac{d}{1+d}$ induce the same uniformity?

Question

Let $d_1$ and $d_2$ be two metrics on the same set $M$. Then $d_1$ and $d_2$ are called uniformly equivalent if the identity maps $i:(M,d_1)\rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)\rightarrow(M,d_1)$ are uniformly continuous. Now this textbook gives the following exercise:

Given any metric space $(M,d)$, show that the metric $\rho=\frac{d}{1+d}$ is always uniformly equivalent to $d$[.]

My question is, is the result of the exercise correct? Because two metrics are uniformly equivalent if and only if they induce the same uniformity, and if two metrics induce the same uniformity then they have the same bounded sets. Yet all sets are bounded with respect to $\frac{d}{1+d}$, whereas all sets need not be bounded with respect to $d$.

Where am I going wrong?

Answer 1

The textbook you linked distinguishes between $3$ types of equivalence:

•strong equivalence

•uniform equivalence &

•equivalence.

Evidently the condition that the identity map and its inverse be uniformly continuous corresponds to uniform equivalence. But strong equivalence is stronger, and is the one for which bounded sets are preserved.

As pointed out in the text, $\rho\le C\sigma$ may fail under uniform equivalence. In particular, it may fail when $\sigma =\frac{\rho}{1+\rho}$.

Answer 2

Yes, the exercise is correct, and $\mathcal{U}(d) = \mathcal{U}(d')$ where $d'=\frac{d}{1+d}$. In the latter uniformity all subsets of $X$ are $d'$-bounded but the $\mathcal{U}(d')$-bounded sets are not always the same as the $d'$-bounded sets. $d$ and $d'$ induce the same uniform structure (and thus the same uniform-bounded and totally-bounded bornologies) but not the same metric bornologies.