Given a metric d, I showed that d/(1+d) is a metric. Now I want to verify whether it preserves open sets. If a set U is open in (X,d), then is it open in (X,d')? Let $x \in U$, let r>0 such that $B(x,r)\subset U$. I tried letting r'=r/(1+r) and r'=r(1-r) as well as a few other likely candidates, But I can't show that in general openness is conserved. However, I also can't find a counter example.
Any help would be appreciated
Let $B(x,r) := \{y\in X:\ d(y,x) < r\}$ and $B'(x,\rho) := \{y\in X:\ \varphi(d(y,x)) < \rho\}$, where $\varphi(t) := t/(1+t)$, $t\geq 0$.
It is enough to show that, for every $x\in X$ and $r>0$, there exists $\rho > 0$ such that $B(x,r) \supset B'(x,\rho)$ and viceversa.
One inclusion follows immediately observing that $B(x,r) = B'(x, \varphi(r))$.
The other inclusion follows observing that $\varphi$ is an increasing bijection between $[0,+\infty)$ and $[0,1)$, so that $$ B'(x,\rho) = \begin{cases} B(x, \rho(1-\rho)), &\text{if}\ \rho\in [0,1),\\ X, &\text{if}\ \rho \geq 1. \end{cases} $$