$(X,d)$ is a complete metric space iff $(X,d')$ is a complete metric space. I recently got asked this question on an exam and I wasn't able to give a solution. 
If $(X,d)$ is a metric space and we define
 $$d'(x,y)= \frac {d(x,y)}{1+d(x,y)}$$
(I've already proved that $(X,d')$ is a metric space.)
Prove that $(X,d)$ is a complete metric space iff $(X,d')$ is a complete metric space.

Consider the function given by $f(x)= \frac{x}{x+1}$ and study its graph paying attention to:


*

*The boundedness of $f$ on $\mathbb{R}_{0}^{+}$

*The monotonicity of $f$

*The concavity of $f$


With this in mind convince yourself about these equalities of balls:


*

*$B(x_0,r) = B '(x_0, \frac{r}{1+r})$ for all $x_0$ in $X$ and $0 \leq r$

*$B'(x_0,\rho) = B(x_0, \frac{\rho}{1-\rho})$ for all $x_0$ in $X$ and
$0 \leq \rho < 1$


Because of these equalities the two metrics define the same balls, hence the same open sets, hence the same topologies. Now observe that $\frac{r}{1+r} \rightarrow 0$ when $r \rightarrow 0$ and that $\frac{\rho}{1-\rho} \rightarrow 0$ when $\rho \rightarrow 0$. It follows that $(X,d)$ and $(X,d')$ are metric spaces with the same underlying topology and $(X,d)$ is complete if and only if $(X,d')$ is.
