TITLE: Equivalent distances define same topology
QUESTION [5 upvotes]: 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 $$\alpha d_1(x,y) \leq d_2(x,y) \leq \beta d_1(x,y)$$
Please help me

REPLY [3 votes]: 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 $\alpha$.