Showing two metrics are equivalent. Let $(X,d)$ be a metric space. Define
$$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$$
(do you know the name of this metric?)
Show that the metrics $d$ and $d_1$ are equivalent.
Edited: Captain Lama pointed out that I was looking for strong equivalence which is not there. So how do I show they are equivalent?
 A: Consider a point $p\in X$, and let $U^1$ be any $d_1$-neighborhood of $p$. Then there is an $\epsilon>0$ with $U_\epsilon ^1(p)\subset U^1$. Since $d_1(x,y)<d(x,y)$ it follows that
$$U_\epsilon(p)=\{x\>|\>d(x,p)<\epsilon\}\subset\{x\>|\>d_1(x,p)<\epsilon\}=U_\epsilon ^1(p)\subset U^1\ .$$
This shows that $U^1$ is  a neighborhood of $p$ with respect to $d$ as well.
Conversely: Consider a point $p\in X$, and let $U$ be any $d$-neighborhood of $p$. Then there is a positive $\epsilon<1$ with $U_\epsilon(p)\subset U$. Since
$$d(x,y)={d_1(x,y)\over 1-d_1(x,y)}\leq 2d_1(x,y)$$
when $d_1(x,y)<{1\over2}$ it follows that
$$U^1_{\epsilon/2}(p)=\{x\>|\>d_1(x,p)<\epsilon/2\}\subset \{x\>|\>d(x,p)<\epsilon\}=U_\epsilon (p)\subset U\ .$$
This shows that $U$ is  a neighborhood of $p$ with respect to $d_1$ as well.
Altogether we have proven that $(X,d)$ and $(X,d_1)$ possess the same open sets.
A: In order to show that two spaces, $X,Y$ are topologically equivalent it is enough to prove that the identity function is continuous in both directions $X\to Y$ and $Y\to X$ (so the inverse image of every open set of $Y$ is an open set of $X$ and reciprocally).
A function $f$ from $(X,d)\to (X,d_1)$ is continuous if for all sequence $x_n\to x$ in $(X,d)$ it is verify that
$\forall \epsilon >0$ there exist $\delta>0$ such that $d(x_n-x)<\delta\Rightarrow d_1(f(x_n)-f(x))<\epsilon$
This is clearly verified in both directions when $f(x)=x$ because of the equality
$$d_1(x,y)=\frac {d(x,y)}{1+d(x,y)}$$
Thus, the two metrics define the same topology.
