# Show that metrics d_1,d_2 are equivalent

## Homework Statement

Show that two metrics d_1,d_2 are equivalent iff for all epsilon>0, exists delta>0 such that
B_(d_1)(x,epsilon) $\subset$ B_(d_2)(x,delta) and vice versa (Where B_(d_1)(x, epsilon) is the open ball on the metric d_1 around x with radius epsilon.

## Homework Equations

pretty much what is in the first part

## The Attempt at a Solution

This is what I tried:
Let B_(d_1)(x,epsilon)=I, which is an open interval;

Assume
a $\in$ B_(d_1)(x,epsilon) and
a not $\in$ B_(d_2)(x,deltamax)
I'm trying to show this leads to a contradiction of deltamax being the largest delta that gives B_(d_2)(x,delta)$\subset$ I, but I'm not sure how.

## Answers and Replies

For the if part, all you have to show is an open set under d1 is also an open set under d2. Considering definition of open set in a metric space, i.e., each point of the set has an open ball contained in the set, this is almost obvious. The only if part is even more obvious following definition.

I have to show - insofar as I understand - that if they are equivalent metrics, then for any open ball for d1, there is an identical open ball for d2 ( because B_d1(x, epsilon) subset of B_d2(x,delta) and b_d2(x,delta) subset of B_d1(x,epsilon), which is more restrictive than the solution you suggested.
But thank you for your help.

I have to show - insofar as I understand - that if they are equivalent metrics, then for any open ball for d1, there is an identical open ball for d2 ( because B_d1(x, epsilon) subset of B_d2(x,delta) and b_d2(x,delta) subset of B_d1(x,epsilon), which is more restrictive than the solution you suggested.
But thank you for your help.

I see. Apparently I didn't know what equivalence of metric means. I was thinking of the equivalence between the two topologies.

The lecturer explained the problem after class - the notation had been unclear, and I'd misunderstood the problem. I've got it out now. But thanks - your suggestion was actually the right answer...