Complete metric space

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  • #1
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Homework Statement


We say that two metrics d, d' on a space S are equivalent if each "dominates" the other in the following sense: there exist constants M, M'>0 such that
d'(x,y)<=M' d(x,y) and d(x,y)<=M d'(x,y) for all x,y in S.

If metrics d, d' are equivalent, prove that (S,d) is complete<==>(S,d') is a complete metric space.


The Attempt at a Solution


If (S,d) is complete then every Cauchy sequence in S converges to a limit in S. I want to go somewhere along the lines of saying that multiplying by a constant will not change its convergence.....am I going along the right lines here? I don't really know how else to go about this.
 

Answers and Replies

  • #2
morphism
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Let's do the => direction first. Take a cauchy sequence {x_n} in (S,d'). Is {x_n} cauchy in (S,d)?
 
  • #3
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I'm guessing it is yes. Because we have M' that we can multiply the sequence by?
 
  • #4
HallsofIvy
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And, of course, a sequence convertes if and only if "given [itex]\epsilon> 0[/itex], there exist N such that ...

Given [itex]\epsilon> 0[/itex] in d', what does that tell you about [itex]\epsilon/M'[/itex] in d?
 
  • #5
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And, of course, a sequence convertes if and only if "given [itex]\epsilon> 0[/itex], there exist N such that ...

Given [itex]\epsilon> 0[/itex] in d', what does that tell you about [itex]\epsilon/M'[/itex] in d?

To finish the first part of your response, "there exists N such that..."
is the end of that "n larger than N will be equal to some epsilon>0"?

For the second part, epsilon/M' will be some small value in d...?
 
  • #6
HallsofIvy
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If [itex]d(x_n, L)< \epsilon[/tex] for all n> N, then [itex]d'(x_n,L)= M'\epsilon[/itex] and conversely.
 

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