Prove the completeness of the following metric space We know that if $(X,d)$ is a metric space then $\sigma=\frac{d}{1+d}$ is also a metric on $X$. 
If $(X,d)$ is a complete metric space then how to prove that $(X,\sigma)$ is also a complete metric space ? 

To get the triangle inequality, note that the map $t \mapsto \frac{t}{t+1}$ is increasing. We have the triangle inequality for the metric $d$ already, so
\begin{equation*}
    \begin{split}
        &\quad \quad \quad d(x,z) \leq d(x,y) + d(y,z) \\
        &\implies \frac{d(x,z)}{1+ d(x,z)} \leq \frac{d(x,y)}{1+ d(x,y)} + \frac{d(y,z)}{1+ d(y,z)} \\
    \end{split}
\end{equation*}
Then symmetry follows for free and if the above metric vanishes, the numerator must vanish and again since $d$ is a metric this would force $x=y$.
The completeness also follows for free, since a sequence that is Cauchy under the above metric must be Cauchy in the original metric.
