# Coinciding open sets

I'm given two distances that are defined on some metric space and I need to show that open sets and Cauchy sequences coincide for the two distances. What does this mean? I'm avoiding giving details on purpose, but a nice definition would help :)

I found that for two distances to be equivalent we need $\exists$ $0<c_1<c_2<\infty$ s.t. $c_1< d_1/d_2< c_2$, but in my particular case I can't find a particular lower bound.

-

What you wrote in your last paragraph is strong equivalence of metrics. What you have to actually show is weak equivalence, that is that the two metrics define the same topology.

Without knowing more about your metrics, you need to show: If $U$ is open in the sense of $d_1$, then it is open in the sense of $d_2$. It suffices to show that for every open ball in the sense of $d_1$ and every point $x$ in that ball, there is a $d_2$-ball around $x$ contained in the given $d_1$-ball (and vice versa). The statement about Cauchy sequences will then follow immediately.

-
So open sets are equivalent in both metrics? –  Simon Sehayek Nov 2 '12 at 23:17
@Atreyu: Yes, according to your vague question, that is exactly the first part what you said that you had to show. –  Hagen von Eitzen Nov 3 '12 at 21:24
Is this equivalent to showing if $d_1 (x,y)< \epsilon$ then this implies $\exists$ $\delta$ s.t. $d_2 (x,y)< \delta$ and vice-versa, or is this just a statement about open sets mapping to open sets? I mean it seems to me the way you stated it that in the condition I just mentioned $\epsilon = \delta$ (for both inclusions), or else they do not coincide. –  Simon Sehayek Nov 4 '12 at 5:28

You are probably referring to equivalent metrics. Two metrics $d_1$ and $d_2$ on a set $X$ are equivalent if and only if there exist positive constants $A$ and $B$ such that for all $x,y \in X$: $$A d_2(x,y) \leq d_1(x,y) \leq B d_2(x,y).$$

Then its easy to see that $O$ is open in $(X,d_1)$ iff it is open in $(X,d_2)$ and $(x_n)$ is Cauchy in $(X,d_1)$ iff it is Cauchy in $(X,d_2)$.

-