Showing that $d$ and $d'$ generate the same topology

Question

Let $(X, d)$ be a metric space. Show that $d': X \times X \to \mathbb{R}$ given by $$ d'(x, y) = \frac{d(x, y)}{1 + d(x, y)} $$ defines a distance on $X$ that generates the same topology.

I´ve already proved that $d'$ defines a distance and that every open ball with respect to $d$ is contained in every open ball with respect to $d'$, because for any $x, y \in X$ $$ d'(x, y) = \frac{d(x, y)}{1 + d(x, y)} \leq d(x, y). $$ Now to show the reverse inclusion, let $r > 0$ be a real number and note that for every $x, y \in X$ such that $$ d'(x, y) < r $$ we have $$ d(x, y) < r + r d(x, y). $$ That is $$ d(x, y) < \frac{r}{1-r}. $$ My guess is that $r' = r / (1+r)$ will do the job but I´m making myself a bit of a mess with the arithmetic. I need a bit of help to finish the proof.

Answer 1

A little correction:

and that every open ball with respect to $d$ is contained in every open ball with respect to $d'$

You mean

and that every open ball with respect to $d$ is contained in some open ball with respect to $d'$

Anyway, take the open ball around $x_0$ of radius $r$ in metric $d^\prime$. Then, for any $x$ in this ball we have $$\frac{d(x,x_0)}{1+d(x,x_0)}<r$$ which rearranges to $$(1-r)d(x,x_0)<r$$ If $r<1$, then you are done. Otherwise since $d^\prime(\cdot,\cdot)$ is always $<1$, the $r$-ball is just the $1$-ball. The key point is that you can throw away the $1$-ball from the basis of the topology, since it is the full space $X$ (adding it does not create any new open sets as it can only generate $X$).

Hope this helps. :)

Answer 2

What you can do is show that the identity mapping $\left( X,d \right) \longrightarrow \left( X,d' \right)$ is an homeomorphism. That would mean that both distances are equivalent hence they define the same topology. The identity mapping $\left( X,d \right) \longrightarrow \left( X,d' \right)$ is $1-$Lipschitz hence continuous. Let $\epsilon > 0$ and choose $\delta<\min \lbrace 1,\epsilon \rbrace$. It follows that

$$d'(x,y)<\delta \Longrightarrow d(x,y)<\epsilon$$

Thus the identity mapping $\left( X,d' \right) \longrightarrow \left( X,d \right)$ is continuous, therefore the identity mapping $\left( X,d \right) \longrightarrow \left( X,d' \right)$ is an homeomorphism.