We have a standard metric $d_1(x,y) := |x-y|$ on $\mathbb{R}$, and another metric $d_2(x,y) := |f(x)-f(y)|$ on $\mathbb{R}$, where $f(x)=\frac{x}{1+|x|}$. I want to show these two metrics induce the same topology on $\mathbb{R}$, but $d_2$ is not complete.
I have no idea about the way to prove "metrics induce the same topology". A hint is to show the identity $id: (\mathbb{R},d_1) \to (\mathbb{R},d_2)$ is a homeomorphism, but I don't understand why the homeomorphic property implies the same topology.
In addition, I know $d_2$ is not complete just by taking a specific sequence like $(x_n = n)_{n \in \mathbb{N}}$, but I want to learn a general proof rather than to use a counter-example.
Probably it's clearer to consider an intermediate space, namely the range of $f$, that is the open interval $(-1,1)$, with its usual metric, and prove that $f$ is a homeomorphism $(\Bbb R, d_1)\to (-1,1)$.
This can be done by directly calculating the inverse of $f$, separating the cases $x\ge0$ and $x<0 $.
The same $f$ as $(\Bbb R, d_2)\to(-1,1)$ is isometric, by definition of $d_2$, hence it is also a homeomorphism.
Consequently, $id:(\Bbb R, d_1)\to(\Bbb R, d_2)\ =f^{-1}\circ f$ is also a homeomorphism.
That means that it's continuous, so $d_2$-open sets are also $d_1$-open, and its inverse is continuous, so $d_1$-open sets are also $d_2$-open.