How to prove a ball is open in different metric spaces So, I was given, $\rho: X$ x $X \to \mathbb{R}$ defined by $\rho(x,y)={d(x,y)\over d(x,y)+1}$. 
How do I prove that for any $x \in X$ and any $\epsilon>0$, $B_{\rho}(x,\epsilon)$ is open in $(X,d)$
This is what I did: let $y\in B_{\rho}(x,\epsilon)$, there exists $\delta>0$ such that $\forall z, d(y,z)<\delta$. Let $\epsilon=2\delta$. 
I need to show that $B_d(y,\delta)\subset B_{\rho}(x,\epsilon)$
Let $z\in B_d(y,\delta)$, then 
$d(x,z)<d(x,y)+d(y,z)<2\delta=\epsilon$
I did not use the defined metric $\rho$ at all, so my guess is I did something wrong. 
I would appreiate it if you can give me some idea, thank you so much.

Fix $x\in X$ and $\epsilon>0$. Take $y\in B_\rho (x,\epsilon)$. We have to find $\delta>0$ such that $$B_d (y,\delta)\subset B_\rho (x,\epsilon)$$
First observe that the function $\frac{t}{1+t}$, for $t\geq 0$ is increasing. Now, note that for $z\in B_d(y,\delta)$ 
\begin{eqnarray}
 \rho(x,z) &\leq& \rho(x,y)+\rho(y,z)      \nonumber \\
   &=& \rho(x,y)+\frac{d(y,z)}{1+d(y,z)} \nonumber \\
   &<& \rho(x,y)+\delta\\
  &&
\end{eqnarray}
From the last inequality, you can see that it is sufficient to take $\delta=\epsilon-\rho(x,y)$.
