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.


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.