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Convergence on metric

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that lim[itex]_{n} p_{n}= p[/itex] iff the sequence of real numbers {d{p,p[itex]_{n}[/itex]}} satisfies lim[itex]_{n}[/itex]d(p,p[itex]_{n}[/itex])=0

    2. Relevant equations



    3. The attempt at a solution
    I think I can get the first implication. If [itex]lim_{n} p_{n}[/itex]= p, then we know that d(p,p[itex]_{n}[/itex]) = d(p[itex]_{n}[/itex],p) <[itex] \epsilon[/itex]. Then given [itex]\epsilon[/itex] > 0 and some N, for n>N we have |d{p,p[itex]_{n}[/itex]-0|<d{p,p[itex]_{n} = d(p_{n},p) < \epsilon[/itex].

    I'm having a little trouble with the backwards implication, do I just do what I did up above but backwards sorta? Or should I pick some p[itex]_{n}[/itex] and show that it converges to 0, like 1/n or something.
     
  2. jcsd
  3. Sep 16, 2011 #2

    CompuChip

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    Homework Helper

    You are very much on the right track, but your "proofs" are still quite sloppy and it's hard to see whether you're reasoning circularly here.
    For example,
    Actually, you know that for any [itex]\epsilon > 0[/itex] there is an N such that this is true for all n > N.

    Maybe it helps if you first write out exactly what you need to prove, in the form:

    For all [itex]\epsilon > 0[/itex], I need to prove that ...(if there exists an N such that ... there also exists an N' such that ...)... , and that (the other implication).
     
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