$(X,d)$ 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.
 A: You can see that $d' = {d \over 1+ d} \le {d \over 1} = d$, hence a $d$-Cauchy sequence is a $d'$-Cauchy sequence.
Note that if $d'(x,y) \neq 1$, then since $d'(x,y) = { d(x,y) \over 1 + d(x,y) }$, then $d(x,y) = { d'(x,y) \over 1 - d'(x,y) }$.
Now suppose $x_n$ is a $d'$-Cauchy sequence, and choose $N$ such that if $m,n \ge N$, then $d'(x_n,x_m) \le {1 \over 2}$. Then for $m,n \ge N$, we have
$d(x,y) \le  2 d'(x,y)$ and so the sequence is a $d$-Cauchy sequence.
If $(X,d)$ is complete, and $x_n$ is $d$-Cauchy, then there is some $x$ such that $d(x,x_n) \to 0$. Since $d' \le d$, it follows that $d'(x,x_n) \to 0$ and hence $(X,d')$ is complete.
Similarly, if $(X,d')$ is complete, and $x_n$ is $d'$-Cauchy, then there is some $x$ such that $d'(x,x_n) \to 0$. Choose $N$ such that if $n \ge N$
then $d'(x,x_n) \le {1 \over 2}$, then we have $d(x,x_n) \le 2 d'(x,x_n)$ and
so $d(x,x_n) \to 0$ and hence $(X,d)$ is complete.
A: 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.
A: HINT: If $d(x,y)\ge\epsilon$, then
$$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}=1-\frac1{1+d(x,y)}\ge 1-\frac1{1+\epsilon}=\frac{\epsilon}{1+\epsilon}\;.$$
If $d'(x,y)\ge\epsilon$, then $\epsilon<1$, and
$$\frac{d(x,y)}{1+d(x,y)}\ge\epsilon\;,$$
so $d(x,y)\ge\epsilon+\epsilon d(x,y)$, and 
$$d(x,y)\ge\frac{\epsilon}{1-\epsilon}\;.$$
Use these to show that if a sequence is not Cauchy in one metric, then it’s not Cauchy in the other metric, and use the fact (that I suspect you’ve already proved) that $d$ and $d'$ are topologically equivalent.
