# Equivalent distances define same topology

I have to prove that

equivalent distances define same topology.

I know there are similar questions, so please don't have a go at me but I am still confused and they don't answer it in the way I have been taught.

If distances are equivalent then there exist an $\alpha$ and $\beta$ more than zero such that $$\alpha d_1(x,y) \leq d_2(x,y) \leq \beta d_1(x,y)$$

• What is the way you were taught? Do you want to use balls or sequences or something else? // You didn't use $\beta$ in the definition of equivalence of metrics. // To get $d_1$ and $d_2$ use d_{1} and d_{2}. – Martin May 5 '13 at 1:00
• The above is my definition for equivalent distances... i'm assuming that i have an $\epsilon \greater 0$ and a $\delta \greater 0$ – andwil May 5 '13 at 1:02
• Show every $d_1$-ball is an open $d_2$-set and vice-versa. – Neal May 5 '13 at 1:06
Assume that $U$ is $d_1$-open. It means that for any $x$ there is an $r>0$ such that $B_x^{d_1}(r)\subseteq U$. But we have $B_x^{d_2}(r/\beta)\subseteq B_x^{d_1}(r)$ by assumption.
The other direction goes similarly, using $\alpha$.