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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.

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    $\begingroup$ To prove that two topologies $\tau_1,\tau_2$ are the same, you should show that every open set in $\tau_1$ is open in $\tau_2$, and that every open set in $\tau_2$ is open in $\tau_1$; this shows that $\tau_1\subseteq\tau_2$ and $\tau_2\subseteq\tau_1$, so $\tau_1=\tau_2$. $\endgroup$ Sep 17, 2018 at 14:25
  • $\begingroup$ Topologies are sets, try to prove the double inclusion. $\endgroup$ Sep 17, 2018 at 14:25
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    $\begingroup$ Regarding the final paragraph: A "general proof" for what statement? A counterexample is a very good proof that something does not have a property of the form "all x are y". $\endgroup$ Sep 17, 2018 at 14:30
  • $\begingroup$ Your hint to prove that the identity map is a homeomorphism is useful: Since the identity is clearly bijective it only remains to verify that the map is both ways continuous. But pre-images are preserved under continuous maps. Since the map under consideration is the identity map, this would mean that $\tau_1\subseteq\tau_2$ and $\tau_2\subseteq\tau_1$. $\endgroup$ Sep 17, 2018 at 14:48
  • $\begingroup$ Two metrics $d_1, d_2$ on a set $R$ are equivalent (i.e. generate the same topology) iff , for any $x\in R$ and any $r>0$ there exists $r_1>0$ and $r_2>0$ such that (i) $\;B_{d_1}(x,r_1)\subset B_{d_2}(x,r),$ and (ii) $\;B_{d_2}(x,r_2)\subset B_{d_1}(x,r).$ $\endgroup$ Sep 18, 2018 at 4:49

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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.

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