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 ? 
 A: Let $(a_n)_n$ be a Cauchy sequence w.r.t $\sigma$. Fix $\varepsilon > 0$.
The function $x \mapsto \frac{x}{1-x}$ is continuous so there exists $\delta > 0$ such that $|x| < \delta \implies \left|\frac{x}{1-x}\right| < \varepsilon$. For example, you can take $\delta = \frac{\varepsilon}{1+\varepsilon}$.
Since $(a_n)_n$ is Cauchy, there exists $n_0\in\mathbb{N}$ such that $m,n\ge n_0 \implies \sigma(a_m, a_n) < \delta$. Hence, for $m,n \ge n_0$ we have
$$d(a_m,a_n) = \frac{\sigma(a_m, a_n)}{1-\sigma(a_m, a_n)} < \varepsilon$$
Therefore, the sequence $(a_n)_n$ is Cauchy w.r.t $d$. Since $d$ is complete, there exists $a \in X$ such that $a_n \xrightarrow{n\to\infty} a$ w.r.t. $d$.
Now $\sigma(a_n, a) \le d(a_n, a) \xrightarrow{n\to\infty} 0$ so $a_n \xrightarrow{n\to\infty} a$ w.r.t. $\sigma$.
We conclude that $(X, \sigma)$ is a complete metric space.
A: One would start with the definition of "complete" metric space- a metric space is "complete" if and only if every Cauchy sequence converges.  If $\{a_n\}$ is a Cauchy sequence with the "$\sigma$" metric then $\sigma(a_n, a_m)= \frac{d(a_n, a_m)}{1+ d(a_n, a_m)}$ goes to 0 as m and n go to infinity independently.  What does that tell you about $d(a_n, a_m)$?
A: 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.
