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Complete metric space

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    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.


    3. 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.
     
  2. jcsd
  3. Nov 23, 2008 #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)?
     
  4. Nov 23, 2008 #3
    I'm guessing it is yes. Because we have M' that we can multiply the sequence by?
     
  5. Nov 24, 2008 #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?
     
  6. Nov 24, 2008 #5
    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...?
     
  7. Nov 24, 2008 #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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