Let, $M=\{0\}\cup\left\{\frac{1}{n}:n \ge 1\right\}$. Define two metrics $d$ and $\rho$ on $M$ ; $d$ as usual metric and $\rho$ is defined as $$\rho \left(\frac{1}{n},\frac{1}{m}\right)=\left|\frac{1}{n}-\frac{1}{m}\right|\\\rho \left(\frac{1}{n},1\right)=\frac{1}{n},n \ge 2\\\rho \left(\frac{1}{n},0\right)=1-\frac{1}{n}, n\ge 2\\\rho(0,1)=1$$
Show that there exists a homeomorphism between $(M,d)$ and $(M,\rho)$.
First of all I'm trying to define a bijective continuous map $(M,d) \to(M,\rho)$.
Define , $f:(M,d) \to(M,\rho)$ by $f(0)=0$ and $\displaystyle f\left(\frac{1}{n}\right)=\begin{cases}\frac{1}{n+1}&\text{ , if $n$ is odd}\\\frac{1}{n-1} &\text{ , if $n$ is even}\end{cases}$. Then $f$ is clearly bijetive. I've stuck to show $f$ is continuous , although I can show that $f$ is NOT uniformly continuous , as $d(1/n,0)\to 0$ as $n\to \infty$ but $\rho(f(1/n),f(0))\to 1$ as $n\to \infty$.
Is this $f$ continuous ? If yes then how I can show it ? If NOT then define such a function which is a homeomorphism.
Hint: interchange $0$ and $1$, keeep the rest fixed. $\rho$ computes the distance to $0$ as if it is the $d$-distance to $1$ and vice versa. We even get an isometry this way. A bijective isometry is a homeomorphism.