Question: Let $$d(x,y)=\left\lvert \dfrac{x}{1+\lvert x\rvert}-\dfrac{y}{1+\lvert y\rvert}\right\rvert$$ for all $x,y\in \mathbb{R}$. How is the topology induced by $d$ related to the usual topology on $\mathbb{R}$?
The usual topology on $\mathbb{R}$ mentioned in the question is the topology induced by $d_1(x,y)=\lvert x-y\rvert$ for all $x,y\in \mathbb{R}$.
What I did? I have proved that $(\mathbb{R},d)$ is not complete becuase the sequence defined as $x_n=n$ for all $n\in \mathbb{N}$ is a Cauchy sequence but does not converge to a number. Because of that, they are not the same topology. I believe, topology $(\mathbb{R},d_1)$ contains the topology $(\mathbb{R},d)$.
If $x$ and $y$ have the same sign, then I can tell $d(x,y)\leq d_1(x,y)$. My job is easy if open subset $U$ of $(\mathbb{R},d)$ does not contains $0$, $$\forall x\in U, \exists \epsilon>0, d(x,y)<\epsilon\implies y\in U$$ If we define $\epsilon_x=\min{\left\{\left\lvert\frac{x}{2}\right\rvert,\epsilon\right\}}>0$, then for every $y\in \mathbb{R}$ satisfying $d_1(x,y)<\epsilon_x$, $x$ and $y$ will have same sign, and $$d(x,y)\leq d_1(x,y)<\epsilon_x\leq \epsilon\implies y\in U$$ Hence, $U$ is open in the topology incuded by $d_1$ on $\mathbb{R}$. However, how can I obtain the same result if $U$ contains $0$? For elements of $U$ which are not $0$, there is nothing changed, but when $x=0\in U$, we cannot apply exactly same method since $\epsilon_x=0$. I will be glad if you help for $x=0$ case or correct my mistakes if there are.
