# Proving a metric space on a set X

Assume that $$(X,d)$$ is a metric space, define $$\displaystyle \rho(x,y)= \frac{d(x,y)}{1+d(x,y)}$$ for all $$x,y \in X.$$

a. Show $$(X,\rho)$$ is a metric.

b. A sequence $$(x)_{n \geq}$$ in $$X$$ converges to $$p$$ in the metric space $$(X,d)$$ iff it converges to $$p$$ in $$(X,\rho.)$$

My attempt:

a. I have a problem with showing the tringle inequality \begin{align*} \rho(x,z)=\frac{d(x,z)}{1+d(x,z)}& \leq \frac{d(x,y)}{1+d(x,z)}+\frac{d(y,z)}{1+d(x,z)} \end{align*} I could not figure the way to show that.

b. I am good with the first implication. The problem is when I assume $$(x_n)$$ converges to $$p$$ in $$(X, \rho)$$, and want to show this sequence converges in $$(X,d).$$ Here is my attempt:

Assume for the sake of a contradiction that this sequence does not converge in $$(X,d)$$. By this, there exists an $$\epsilon_0>0$$ such that for all $$N \in \mathbb{Z^+}$$, we can find $$n_{_{N}} \geq N$$ such that $$d(x_{n_{_N}},p)> \epsilon_0$$. Now by the assumption, for $$\epsilon=\frac{\epsilon_0}{1+\epsilon_0}$$ there exists $$N' \in \mathbb{Z}^+$$ such that for all $$n \geq N'$$, $$\rho(x_n,p)<\frac{\epsilon_0}{1+\epsilon_0}$$. However, by our hypothesis, for this particular $$N'$$, there exists $$n_{_{N'}} \geq N'$$ such that $$d(x_{n_{_{N'}}},p)>\epsilon_0$$, so $$\frac{\epsilon_0}{1+d(x_{n_{_{N'}}},p)} < \frac{d(x_{n_{_{N'}}},p)}{1+d(x_{n_{_{N'}}},p)}=\rho(x_{n_{_{N'}}},p)<\frac{\epsilon_0}{1+\epsilon_0}$$ what we conclude is $$\displaystyle \frac{\epsilon_0}{1+d(x_{n_{_{N'}}},p)} < \frac{\epsilon_0}{1+\epsilon_0}$$ which is true since $$d(x_{n_{N'}},p)>\epsilon_0$$, I am geting stuck here.

I will appreciate any help or hint for that.

a) $$\frac x {1+x}$$ is an incersing function on $$[0,\infty)$$
b) $$\frac {a+b} {1+a+b} \leq \frac a {1+a} +\frac b {1+b}$$ for all R$$a, b \geq 0$$.
For the second part you only need the fact that $$\frac x {1+x} <\epsilon$$ iff $$x<\frac {\epsilon} {1-\epsilon}$$ provided $$0 <\epsilon <1$$ (and $$x <\epsilon$$ iff $$\frac x {1+x} <\frac {\epsilon} {1+\epsilon}$$.