1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    If [itex]d(x_n, L)< \epsilon[/tex] for all n> N, then [itex]d'(x_n,L)= M'\epsilon[/itex] and conversely.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook