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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?

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Hint: note that $\sigma=\frac{\rho}{1+\rho}=1-\frac{1}{1+\rho}$ is strictly increasing on $[0;+\infty)$, therefore, \sigma<\epsilon implies \rho<\delta (for some $\delta$) and vice versa.

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    If this does not help, draw a graph of the function $\sigma(\rho)$ and try to understand why \sigma<\epsilon implies \rho<\delta for some delta, and vice versa.2012-04-20
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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<1 since: \begin{align*} \sigma(x,y)=\frac{\rho(x,y)}{1+\rho(x,y)}< \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<\varepsilon<1 by choosing $\delta=\frac{\varepsilon}{1-\varepsilon}$ we see that \sigma(x,y)<\varepsilon implies \rho(x,y)<\delta.