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Cauchy Sequences Triangle Inequality.

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    assuming an and bn are cauchy, use a triangle inequality argument to show that cn=
    | an-bn| is cauchy

    2. Relevant equations

    an is cauchy iff for all e>0, there is some natural N, m,n>=N-->|an-am|<e

    3. The attempt at a solution
    I am currently trying to work backwards on this one, since we know
    for some m,n>=N where N is a natural
    Thus, through the triangle inequality this is >=|am-an+bn-bm|=|(bn-an)-(bm-am)|>=|bn-an|-|bm-am|.

    Am I going completely wrong somewhere, because I am getting very stuck here. To anyone who answers, please only hints, not full answers (I am sure this is customary). Again, many thanks.
  2. jcsd
  3. Feb 10, 2009 #2

    Tom Mattson

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    Two quick comments:

    I would have said that [itex]|a_n-a_m|< \epsilon / 2[/itex] and [itex]|b_n-b_m|< \epsilon /2[/itex] just to get [itex]|(a_n-b_n)-(a_m-b_m)|<\epsilon[/itex].

    Here's a nitpicky thing that your professor might correct (mine would have). Once [itex]\epsilon >0[/itex] is given, there's no reason to think that the same [itex]N[/itex] will work for both sequences. So I would have said that there exists [itex]N_1[/itex] for one sequence and [itex]N_2[/itex] for the other, such that blah blah blah... Then I would have let [itex]N=max\{N_1,N_2\}[/itex].
  4. Feb 10, 2009 #3
    Yeah, my instructor does require this also. I can't believe the anguish this is causing me, It must be a very simple problem, and no one asked about it in class. Right now, I am supposed to be reading a history book, but this problem is back there bugging me.
  5. Feb 10, 2009 #4
    Not to imply that anyone read this incorrectly, but the problem is cn=abs(an-bn) not just an-bn. Just entered my head that this might not have been emphasized.
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