Topologically equivalent metric 
Show that in $\mathbb{R}$ the distance
$d'(x,y)=\left|\frac{x}{1+|x|} - \frac{y}{1+|y|} \right|$ is topologically equivalent to the usual metric in $\mathbb{R}$, $d(x,y)=|x-y|$ But $(\mathbb{R},d')$ is not complete.

Im not being able to show that they are topologically equivalent. I've tried to show that if $A \subset \mathbb{R}$ is open with $d$ then it should be open with $d'$. However, when dealing with the $\delta$ and $\varepsilon$, I don't know how to proceed. Any hint is more than welcome!
 A: One way of playing with this is to note that $d'(x,y) = d(f(x),f(y))$ where $f(t) = t/(1+|t|)$ maps $\mathbb{R}\to (-1,1)$.  In other words, $d'$ is the standard metric on the interval $(-1,1)$ pulled back by the homeomorphism $f$. So this problem boils down to showing that $f$ is continuous and continuously invertible.
A: Show that an open ball wrt d is an open ball wrt d' (not necessarily of the same radius), and vice versa. Then, since the open balls form a basis for the topology, the topologies are the same.
A: Here's what you will do:
First you will show that $d'$ is a finer metric that $d$ on $\mathbb R$.
To do it, first pick a point $x \in \mathbb R$ and let $\varepsilon >0$ be a fixed positive number.
Now, you want to find $\varepsilon'>0$ ( which may depend on $\varepsilon$ and $x$ ) satisfying
$$B_{d'}(x,\varepsilon') \subseteq B_d(x,\varepsilon) \tag{1}$$
So, you will say that for $\varepsilon'= f(\varepsilon,x)$ the relation (1) holds.
Then similarly, you will show that $d$ is a finer metric than $d'$ on $\mathbb R$ and this will complete the proof that these metrics are topologically equivalent.
