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

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    $\begingroup$ Consider the map $$\phi \colon \mathbb{R} \to (-1,1);\quad \phi(x) = \frac{x}{1+\lvert x\rvert}.$$ Show it is a homeomorphism. $\endgroup$ – Daniel Fischer Nov 24 '13 at 14:24
  • $\begingroup$ I can prove your map is a homeomorphism, and $d_{2}(x,y) = d(\phi(x), \phi(y))$, but then why $\tau_{2}$ is finer than $\tau_{1}$??? $\endgroup$ – le duc quang Nov 24 '13 at 14:39

Let's break it down like this:

  1. 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.

  2. 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 $X$, $Y$, $f$, $d$, $\tau$, $\tau_{f^\ast d}$.

  • $\begingroup$ perfect. Thanks so much Daniel. $\endgroup$ – le duc quang Nov 25 '13 at 1:23

Or you can show that if $d_1(x_n,x)\to 0<=>d_2(x_n,x)\to 0$. It is more easy sometimes to use convergance,besides topology was invented for convergance. To give you a hint i'll show the more easy part ($=>$).

Let $d_1(x_n,x)=|x_n-x|\to 0$ then $$d_2(x_n,x)=\frac {x_n}{1+|x_n|}-\frac {x}{1+|x|}=\frac {(x_n-x)+(|x|x_n-|x_n|x)}{(1+|x_n|)(1+|x|)}$$. We have that $|x_n-x|\to 0$ and $x_n\to x=>|x_n|\to |x|$ and thus $d_2(x_n,x)\to 0$. Your turn:)

  • $\begingroup$ Can you explain why if $d_{1}(x_{n}, x) \rightarrow 0 <=> d_{2}(x_{n}, x) \rightarrow 0$, then 2 topologies are the same. This statement seems quite helpful in many case? Thanks $\endgroup$ – le duc quang Nov 25 '13 at 1:26
  • $\begingroup$ Yes $d1(x_n,x)→0<=>d2(x_n,x)→0$ which means that they produce the same open sets. to prove this,is not difficult but you need to write down some things( i cannot think any faster way),but think of it that they produce the same structure (this is algebraic term). Or each open set can be described equivalenlty by these two metrics. $\endgroup$ – Haha Nov 25 '13 at 10:11
  • $\begingroup$ I don't think it's right Dimitris. With your condition, I think we can only state that 2 topologies have the same Cauchy sequence, so if one is complete, then the other is complete, too. But how can they be the same? $\endgroup$ – le duc quang Nov 25 '13 at 12:23
  • $\begingroup$ No,it's quite right my friend:). Use it in these problems and you will remember me:) $\endgroup$ – Haha Nov 25 '13 at 22:17

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