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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.

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Fore triangle inequality verify these two facts:

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$a, b \geq 0$.

Both are simple algebraic manipulations.

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 $\frac x {1+x} <\frac {\epsilon} {1+\epsilon}$.

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