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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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2 Answers 2

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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Just be careful considering two cases: $\epsilon\ge 1$ and $0<\epsilon<1$ when you go one direction. –  Vadim Apr 20 '12 at 0:02
    
I'm afraid I don't entirely understand yet. Could you explicitly show one way? –  Mike C. Apr 20 '12 at 1:24
    
$\sigma(\rho)$ is a strictly increasing function, hence, bijective and has inverse on the range. A $\sigma$-ball at $x$ of radius $\epsilon$ is the set of points $y$ such that $\sigma(x,y)<\epsilon$, which corresponds to points $y$ such that $\rho(x,y)<\delta$ for some $\delta$, i.e. a $\rho$-ball. Once again, just be careful enough with large values for $\epsilon$ (for example, $\epsilon\ge 1$ if $\rho$ goes to infinity). –  Vadim Apr 20 '12 at 2:19
    
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. –  Vadim Apr 20 '12 at 2:23
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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 $\rho(x,y)<\delta$.

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