1

I have to show that the metric $e\left(x,y\right)=\frac{d\left(x,y\right)}{1+d\left(x,y\right)}$ and $d\left(x,y\right)$ generate the same topology. I've already shown that $B_{e}\left(x,\frac{\epsilon}{1+\epsilon}\right)\subseteq B_{d}\left(x,\epsilon\right)$, but I'm having trouble finding an $r>0$ such that $B_{d}\left(x,r\right)\subseteq B_{e}\left(x,\epsilon\right)$. I know that this question already exists, but the ones I have seen I had trouble understanding.

I started with if $e\left(x,y\right)<\epsilon$, then $d\left(x,y\right)<\frac{\epsilon}{1-\epsilon}$, which is not less than $\epsilon$. I have also started with if $e\left(x,y\right)<\frac{\epsilon}{1+\epsilon}$, then $d\left(x,y\right)<\epsilon$, which is not less than $\frac{\epsilon}{1+\epsilon}$. So, I'm having trouble finding $r>0$ such that $B_{d}\left(x,r\right)\subseteq B_{e}\left(x,\epsilon\right)$.

1 Answers1

2

Hint. If $\epsilon\geq 1$ then $B_{e}\left(x,\epsilon\right)$ is the whole space because $\frac{d\left(x,y\right)}{1+d\left(x,y\right)}<1\leq \epsilon$ for any $x,y$.

If $0<\epsilon< 1$ then $$\frac{d\left(x,y\right)}{1+d\left(x,y\right)}<\epsilon\Leftrightarrow d(x,y)<\frac{\epsilon}{1-\epsilon}. $$

Robert Z
  • 147,345