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?
Best Answer
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.