Metric Space Open Sets. Let $(X, \rho)$ be a metric. I've shown $\sigma(s,t) = \frac{\rho(s,t)}{1 + \rho(s,t)}$ is also a metric on $X$. 
I'm having trouble showing that the open sets defined by the metric $\rho$ are the same as the open sets defined by $\sigma$. I know I must show that an open ball in the $\rho$ metric is an open set in the $\sigma$ metric, and that an open ball in the $\sigma$ metric is an open set in the $\rho$ metric. Any hints or advice? 

The other direction is immediate, because for all $x,y\in X$ we have
\begin{align*}
\sigma(x,y)=\frac{\rho(x,y)}{1+\rho(x,y)}\leq \frac{\rho(x,y)}{1+0}=\rho(x,y).
\end{align*}
On the other hand $\sigma&lt1$ since:
\begin{align*}
\sigma(x,y)=\frac{\rho(x,y)}{1+\rho(x,y)}&lt \frac{1+\rho(x,y)}{1+\rho(x,y)}=1.
\end{align*}
So by rearranging the terms in the definition of $\sigma$ we get (which is well defined by the previous remark):
\begin{align*}
\rho(x,y)=\frac{\sigma(x,y)}{1-\sigma(x,y)}
\end{align*}
Thus for any $0&lt\varepsilon&lt1$ by choosing $\delta=\frac{\varepsilon}{1-\varepsilon}$ we see that $\sigma(x,y)&lt\varepsilon$ implies $\rho(x,y)&lt\delta$.
