Proving $d_1=\frac{d(x,y)}{1+d(x,y)}$ is a metric equivalent with $d$, the Euclidean metric Given a set $X$, define a metric by $d_1=\frac{d(x,y)}{1+d(x,y)}$ for all $x,y\in X$. I want to show that $d_1$ is a metric equivalent with $d$, the standard Euclidean metric. Another way of writing $d_1$ would be 
$$d_1=\frac{|x-y|}{1+|x-y|}.$$
Here is my attempt so far:
Given a set $U\subseteq X$ which is open in the $d$ metric and $x\in U$, there exists an $\varepsilon>0$ such that $B_{d}(x,\varepsilon)\subseteq U$. We want to show that there exists a $\delta>0$ such that $B_{d_1}(x,\delta)\subseteq U$, and one way to do that is to show that there exists a $\delta>0$ such that $B_{d_1}(x,\delta)\subseteq B_{d}(x,\varepsilon).$
Given any $y\in B_{d}(x,\varepsilon)$, we have $d(x,y)<\varepsilon$. Since $d_1(x,y)\leq d(x,y)$, we know $d_1(x,y)< \varepsilon$, so that $y\in B_{d_1}(x,\varepsilon)$. Therefore $B_{d}(x,\varepsilon)\subseteq B_{d_1}(x,\varepsilon)$. But this is the opposite of what I want, isn't it?
Can someone help me complete this proof and help me understand what I am doing wrong? Thank you.
 A: It seems like you are using this theorem:

Metrics $d_1$ and $d_2$ are equivalent $\Leftrightarrow$ for every
  open ball centered at $x\in X$ according to one metric,
  $B_{d_1}(x,r_1)$, there exists an open ball centered at $x$ according
  to the other metric, $B_{d_2}(x,r_2)$, such that
  $B_{d_1}(x,r_1)\subseteq B_{d_2}(x,r_2)$ and the other way round.

You showed one part of this criterion- for every $x\in X$ and radius $r>0$ we have a corresponding radius (which happens to be the same) such that $B_{d}(x,r)\subseteq B_{d_1}(x,r)$.
What you were missing is that we should show both parts. So we now have to show that for every $x\in X$ and radius $r_1>0$ there is a corresponding radius $r_2>0$ such that $B_{d_1}(x,r_1)\subseteq B_{d}(x,r_2)$. In other words, because of the definition of an open ball, given $r_1>0$ we should find an $r_2>0$ such that $d_1(x,y)< r_1$ implies $d(x,y)< r_2$.
$d_1=\frac{d(x,y)}{1+d(x,y)}$ implies the opposite relation $d=\frac{d_1(x,y)}{1-d_1(x,y)}$. So given $d_1(x,y)< r_1$ we have $d=\frac{d_1(x,y)}{1-d_1(x,y)}< \frac{r_1}{1-r_1}$ (make sure you understand why this is true). In other words, we found $r_2=\frac{r_1}{1-r_1}$ which assures $B_{d_1}(x,r_1)\subseteq B_{d}(x,r_2)$ for every $x\in X$ and $r_1>0$ given.
Hence, $d$ and $d_1$ are equivalent metrics on $X$.
