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Easy cauchy sequence prob

  1. Oct 14, 2005 #1
    How would one prove that the sum of 2 cauchy sequences is cauchy? I said let e>0 and take 2 arbitrary cauchy sequences then
    |Sn - St|<e/2 whenever n,t>N1 and |St - Sm|<e/2 whenever t,m >N2.


    |Sn - Sm|=|Sn - St + St - Sm|<= |Sn - St|+|St - Sm|< e/2 + e/2 <= e

    So n,m>max{N1, N2} imples |Sn - Sm|<e thus cauchy

    Am I close or way off?
  2. jcsd
  3. Oct 14, 2005 #2
    bump ttt bump
  4. Oct 15, 2005 #3
    The method looks perfect to me.

    However, a slight issue with your subscripts. When I did this problem I only needed two subscripts. Observe:

    Take two arbitrary Cauchy sequences [itex]\{x_n\}[/itex] and [itex]\{y_n\}[/itex]. Then for any [itex]\epsilon > 0[/itex] there exists [itex]N_1 \in \mathbb{N}[/itex] such that for any [itex]n,m \geq N_1[/itex] we have

    |x_n - x_m| < \frac{\epsilon}{2}

    and there also exists an [itex]N_2 \in \mathbb{N}[/itex] such that for any [itex]n,m \geq N_2[/itex] we have

    |y_n - y_m| < \frac{\epsilon}{2}

    Then, as you said, choose [itex]N = \max\{N_1,N_2\}[/itex]. Then for any [itex]n,m \geq N[/itex] we have

    [tex]|(x_n + y_n) - (x_m - y_m)| = |(x_n - x_m) + (y_n - y_m)| \leq |x_n-x_m| + |y_n - y_m| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon[/tex]

    Therefore [itex]\{x_n + y_n\}[/itex] is a Cauchy sequence. [itex]\square[/itex]
  5. Oct 15, 2005 #4
    thank you!
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