Show that metrics d_1,d_2 are equivalent

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In summary, the two metrics are equivalent if and only if for all epsilon>0, there exists delta>0 such that B_(d_1)(x,epsilon) \subset B_(d_2)(x,delta) and vice versa.
  • #1
Ratpigeon
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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.
 
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  • #2
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
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
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
I see. Apparently I didn't know what equivalence of metric means. I was thinking of the equivalence between the two topologies.
 
  • #6
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...
 

FAQ: Show that metrics d_1,d_2 are equivalent

What does it mean for metrics d_1 and d_2 to be equivalent?

When two metrics, d_1 and d_2, are equivalent, it means that they produce the same topology on a given set X. This means that they define the same notions of convergence, continuity, and open sets.

How can I show that two metrics, d_1 and d_2, are equivalent?

To show that two metrics are equivalent, you can use the definition of equivalence, which states that for every open set U in the topology generated by d_1, there exists an open set V in the topology generated by d_2 such that U = V, and vice versa. You can also show that the identity map from (X, d_1) to (X, d_2) is a homeomorphism, which preserves the topologies.

What is the importance of showing that two metrics are equivalent?

Showing that two metrics are equivalent is important because it allows us to use different metrics to define the same notions of convergence, continuity, and open sets. This can be especially useful in situations where one metric may be easier to work with than the other.

Can two metrics, d_1 and d_2, be equivalent but not uniformly equivalent?

Yes, it is possible for two metrics to be equivalent but not uniformly equivalent. This means that the identity map from (X, d_1) to (X, d_2) may not be uniformly continuous, even though it is a homeomorphism. This is because uniform continuity is a stronger condition than continuity.

Are equivalent metrics unique?

No, equivalent metrics are not unique. There can be multiple metrics that are equivalent to a given metric. For example, if d is a metric on a set X, then any bi-Lipschitz equivalent metric will also be equivalent to d. However, there can only be one metric that is uniformly equivalent to a given metric.

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