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I want to show that given a metric space $\left(X,d\right)$ the metric $\rho=\frac{d}{d+1}$ on $X$ induces the same topology on $X$ as $d$. It suffices to show that any open ball in $\left(X,d\right)$ is open in $\left(X,\rho\right)$ and vice versa. What I'm stuck on is finding a $\delta>0$ such that given $y\in B_{d}\left(x,\varepsilon\right)$ I will get $y\in B_{\rho}\left(y,\delta\right)\subseteq B_{d}\left(x,\varepsilon\right)$. This will give me that every open ball in $ \left(X,d\right)$ is open in $\left(X,\rho\right)$ .

Help would be appreciated.

Serpahimz
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  • Another way using the lemma that "if $X$ has a metric $d$, then the topology induced by $d$ is the coarsest topology relative to which the function $d$ is continuous" can be found here. The lemma is from the exercise 3 of section 20 of Munkres's Topology (2nd edition). – hengxin Feb 06 '14 at 15:18

2 Answers2

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Note that for any metric $\rho$, we have always that $\rho\geq 0$, so that $$\frac{\rho}{1+\rho}\leq \rho$$

so one direction is immediate. On the other hand, this tells us that given $\epsilon$ and $\rho(x,y)<\varepsilon $; we will have to take $\frac{\rho}{1+\rho}$ much smaller to get it inside the ball. We want $$\frac{\rho(x,y)}{1+\rho(x,y)}<\delta$$ to imply that $$\rho(x,y)<\epsilon$$ Since all is positive

$$\begin{align} \frac{{\rho (x,y)}}{{1 + \rho (x,y)}} <& \delta \\ \rho (x,y) <& \delta + \delta \rho (x,y) \\ \left( {1 - \delta } \right)\rho (x,y) <& \delta \\ \rho (x,y) <& \frac{\delta }{{1 - \delta }} \end{align}$$

so $$\frac{{\rho (x,y)}}{{1 + \rho (x,y)}} < \delta\implies \rho (x,y) < \frac{\delta }{{1 - \delta }}$$

The inverse of $x\mapsto \frac{x}{1-x}$ is $x \mapsto \frac{x}{{1 + x}}$ so if we take $\delta = \frac{\varepsilon }{{1 + \varepsilon }}$ we will have$$\frac{{\rho (x,y)}}{{1 + \rho (x,y)}} < \delta \Rightarrow \rho (x,y) < \varepsilon $$ as desired.

Pedro
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Try $\delta=\varepsilon/(1+\varepsilon)$.

Did
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