Show that metrics d_1,d_2 are equivalent

[Ratpigeon]

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.

 

[sunjin09]

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.

 

[Ratpigeon]

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.

 

[Ratpigeon]

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.

 

[sunjin09]

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

 

[Ratpigeon]

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...