1
$\begingroup$

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.

$\endgroup$
8
  • 1
    $\begingroup$ 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 $\tau_1=\tau_2$. $\endgroup$
    – Santana Afton
    Sep 17, 2018 at 14:25
  • $\begingroup$ Topologies are sets, try to prove the double inclusion. $\endgroup$
    – Marco Lecci
    Sep 17, 2018 at 14:25
  • 1
    $\begingroup$ 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". $\endgroup$
    – Torsten Schoeneberg
    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 $\tau_2\subseteq\tau_1$. $\endgroup$
    – Walt van Amstel
    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) $\;B_{d_2}(x,r_2)\subset B_{d_1}(x,r).$ $\endgroup$
    – DanielWainfleet
    Sep 18, 2018 at 4:49

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.