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Cauchy sequence proof.

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Assume [itex] x_n [/itex] and [itex] y_n [/itex] are Cauchy sequences.
    Give a direct argument that [itex] x_n+y_n [/itex] is Cauchy.
    That does not use the Cauchy criterion or the algebraic limit theorem.
    A sequence is Cauchy if for every [itex] \epsilon>0 [/itex] there exists an
    [itex] N\in \mathbb{N} [/itex] such that whenever [itex] m,n\geq N [/itex]
    it follows that [itex] |a_n-a_m|< \epsilon [/itex]
    3. The attempt at a solution
    Lets call [itex] x_n+y_n=c_n [/itex]
    now we want to show that [itex] |c_m-c_n|< \epsilon [/itex]
    Lets assume for the sake of contradiction that
    [itex] c_m-c_n> \epsilon [/itex]
    so we would have
    [itex]|x_m+y_m-x_n-y_n|> \epsilon [/itex]
    [itex] x_m> \epsilon+y_n-y_m [/itex]
    since [itex] y_n>y_m [/itex]
    and we know that [itex] x_m< \epsilon [/itex]
    so this is a contradiction and the original statement must be true.
     
  2. jcsd
  3. Mar 4, 2012 #2

    Office_Shredder

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    It's always good to be careful with wording. We want to show that if n and m are big enough, that this inequality holds.

    This is probably not true, especially since you haven't even said what n and m are besides arbitrary numbers!

    This is also probably not true since there's no reason to think the limit is zero

    To get the contradiction you're going to want to use the triangle inequality on |(xn-xm)+(yn-ym)|
     
  4. Mar 4, 2012 #3
    ok thanks for your response.
    So I take [itex] |(x_n-x_m)+(y_n-y_m)| \leq |x_n-x_m|+|y_n-y_m| [/itex]
    lets assume that [itex] |x_n-x_m|+|y_n-y_m| > \epsilon [/itex]
    Im going to rewrite it as [itex] A+B> \epsilon [/itex]
    so now we have [itex] A> \epsilon -B [/itex]
    Can I just say this since we know that [itex] A< \epsilon [/itex] and [itex]B< \epsilon [/itex] since [itex] \epsilon [/itex] can be any number bigger than zero, then both of these values should be less than [itex] \frac{\epsilon}{2} [/itex]
    therefore [itex] A+B< \epsilon [/itex]
    I have a feeling my last step is not ok
     
    Last edited: Mar 4, 2012
  5. Mar 4, 2012 #4

    Office_Shredder

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    This is the crux of the argument. It's not the whole proof of course - A and B aren't always that small. Feel free to post a full proof if you want it checked over for errors
     
  6. Mar 4, 2012 #5
    Ok , Am I thinking about this in the right way.
     
  7. Mar 4, 2012 #6

    Office_Shredder

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    Well, I can't read your mind but the part I quoted is the basis of the a correct argument for the proof. You just have to add the fact that these are sequences - A and B aren't always that small, but as long as n and m are big enough they are (by definition of the x's and the y's forming Cauchy sequences)
     
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