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Cauchy sequences is my proof correct?

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Let (xn)n[itex]\in[/itex]ℕ and (yn)n[itex]\in[/itex]ℕ be Cauchy sequences of real numbers.

    Show, without using the Cauchy Criterion, that if zn=xn+yn, then (zn)n[itex]\in[/itex]ℕ is a Cauchy sequence of real numbers.
    2. Relevant equations



    3. The attempt at a solution
    Here's my attempt at a proof:

    Let (xn) and (yn) be Cauchy sequences. Let (zn) be a sequence and let zn=xn+yn.

    Since (xn) and (yn) are Cauchy, [itex]\exists[/itex]N[itex]\in[/itex]ℕ such that,
    |xn-xm|<ε/2, and
    |yn-ym|<ε/2 for n,m≥N.

    Let n,m≥N and let zn,zm[itex]\in[/itex](zn).
    Then,
    |zn-zm|=|xn-xm|+|yn-ym|
    <ε/2+ε/2=ε.
    Therefore,
    |zn-zm|<ε for all n,m≥N and hence, (zn) is a Cauchy sequence of real numbers.

    Is this correct?
    Any input is appreciated.

    Thanks.
     
  2. jcsd
  3. Oct 16, 2012 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    Your first step should be an inequality and not an equality, other than that seems ok, you just need to take the maximum of N1 and N2 where these indices are the ones for which

    [tex]\forall n,m \geq N_1 \ |x_n-x_m|\leq \epsilon/2[/tex]
    [tex]\forall n,m \geq N_2 \ |y_n-y_m| \leq \epsilon /2[/tex]
     
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