Show that metrics d_1,d_2 are equivalent

  • Thread starter Ratpigeon
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  • #1
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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) [itex]\subset[/itex] 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 [itex]\in[/itex] B_(d_1)(x,epsilon) and
a not [itex]\in[/itex] 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)[itex]\subset[/itex] I, but I'm not sure how.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
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I see. Apparently I didn't know what equivalence of metric means. I was thinking of the equivalence between the two topologies.
 
  • #6
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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...
 

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