Can we say that if a sequence in $(X,\frac d{1+d})$ is cauchy, then is also in $(X,d)$?

Question

If $(X,d)$ is a metric space and $\{x_n\}$ a cauchy sequence in $(X,d)$, then for every $\epsilon>0$ there is an $N\in \mathbb N$ such that for all $m,n\ge N$, $d(x_m,x_n)<\epsilon$. Now since for all $x,y\in X$, $\frac{d(x,y)}{1+d(x,y)}\le d(x,y)$ we can conclude that for all $n,m\ge N$, $\frac{d(x_m,x_n)}{1+d(x_m,x_n)}<\epsilon$. That is, $\{x_n\}$ is also a cauchy sequence in $(X,\frac d{1+d})$.

However I'm not sure if the converse is also true. Can we say such a thing? How?

Answer 1

The converse is true. For $\{x_n\}$ Cauchy in $(X, \frac{d}{1+d})$ and $\epsilon > 0$, choose $N$ such that $m, n\geq N$ implies $$\frac{d(x_m, x_n)}{1+d(x_m, x_n)} < \frac{\epsilon}{1+\epsilon}$$ As $x\mapsto \frac{x}{1+x}$ is strictly increasing on $[0, \infty)$, this implies $d(x_m, x_n) < \epsilon$.

Answer 2

Hint: If $\epsilon \in (0,1)$ and $\frac{d}{1+d} < \epsilon$, then $d < \frac{\epsilon}{1-\epsilon}$.