Does topological equivalence of metrics imply strong equivalence?

Question

I know that if $(X,d_1)$ and $(X,d_2)$ are metric spaces and for some positive constants $a,b$ , $ad_1(x,y) \le d_2(x,y) \le b d_2(x,y) $ for every $x,y$ in $X$ , then a subset $A$ of $X$ is $d_1$ open iff $d_2$ open that is $(X,d_1)$ and $(X,d_2)$ are topologically equivalent . But is the converse true , that is if $(X,d_1)$ and $(X,d_2)$ are topologically equivalent then does there exist some positive constants $a,b$ such that , $ad_1(x,y) \le d_2(x,y) \le b d_2(x,y) $ for every $x,y$ in $X$ ?

Answer 1

No. Let $X=(0,1], d_1=|x-y|,d_2=|\frac{1}{x}-\frac{1}{y}|$, you can check they are topologically equivalent. But we don't have strong equivalence as $x,y\to 0$