Prove that $d_{1}$ and $d_{2}$ induce the same topology 
Suppose $d_{1}(x,y) = |x-y|$, $d_{2}(x, y) = |\phi(x) - \phi(y)|$, where $\phi(x) = \frac{x}{1 + |x|}$. Prove that $d_{1}$ and $d_{2}$ are metrics on $\mathbb{R}$ which induce the same topology.

It's easy to prove that $d_{1}$ and $d_{2}$ are metrics. Call the topologies induced by them $\tau_{1}$ and $\tau_{2}$. I can prove that $\tau_{1}$ is finer than $\tau_{2}$, but I can't prove the inverse. Can any body please help me. Thanks
 A: Let's break it down like this:


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*Given a metric space $(Y,d)$, a set $X$, and a bijection $f\colon X \to Y$, the pull-back of $d$ via $f$, $f^\ast d \colon X\times X \to [0,\infty);\; f^\ast d(a,b) = d(f(a),f(b))$ is a metric on $X$. Thus $f\colon (X,f^\ast d) \to (Y,d)$ is an isometry, in particular a homeomorphism.

*If, in the situation above, $X$ carries a prior topology $\tau$, we have two topologies on $X$, the original $\tau$, and the one induced by the pull-back of $d$, let's call it $\tau_{f^\ast d}$. We have $\tau = \tau_{f^\ast d}$ if and only if $\operatorname{id} \colon (X,\tau) \to (X,\tau_{f^\ast d})$ is a homeomorphism. Since the composition of homeomorphisms is a homeomorphism, and we know that $f \colon (X,\tau_{f^\ast d}) \to (Y,d)$ is a homeomorphism, that is the case if and only if $f\circ \operatorname{id} \colon (X,\tau) \to (Y,d)$ is a homeomorphism.
It remains to identify $X$, $Y$, $f$, $d$, $\tau$, $\tau_{f^\ast d}$.
A: Or you can show that if $d_1(x_n,x)\to 0\iff d_2(x_n,x)\to 0$. It is more easy sometimes to use convergence,besides topology was invented for convergence. To give you a hint i'll show the more easy part ($\implies$).
Let $d_1(x_n,x)=|x_n-x|\to 0;$ then \begin{align}d_2(x_n,x)&=\frac {x_n}{1+|x_n|}-\frac {x}{1+|x|}\\&=\frac {(x_n-x)+(|x|x_n-|x_n|x)}{(1+|x_n|)(1+|x|)}.\end{align} We have that $|x_n-x|\to 0$ and $x_n\to x\implies |x_n|\to |x|$ and thus $d_2(x_n,x)\to 0.$
