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Cauchy Sequence Problem Update

  1. Dec 6, 2008 #1
    This is what it is in my book, i dont know how it went from saying that the difference is 2e and then using e/2 . The part that confuses me , is why the book outlines (1) to then use it in the proof, it looks that (1) doesnt even matter because since e > 0 then e/2 will do just fine.

    (1) Suppose { a_n } converges to A. Choose e > 0, THere is a positive integer N such that, if
    n, m >= N , then A - e < a_n < A + e and A - e < a_m < A + e
    Thus for all n, m >= N we find a_n ∈ ( A - e , A + e ) and
    a_m ∈ ( A - e , A +e ) . the set ( A - e, A +e ) is an interval of length 2e ,
    hence the difference between a_n, and a_m is less then 2e

    we will now state a theorem , the proof of which we have just outlined


    THEOREM :Every convergent sequence is a cauchy sequence

    Proof: Suppose {a_n} converges to A, choose e > 0 , then e/2 > 0

    there is a positive integer N such that n>=N implies abs[a_n - A ] < e/2

    by (1) the difference between a_n , a_m was less than twice the original choice of e

    Now if m , n >= N , then abs[a_n - A ] < e/2 and abs[a_m - A] < e/2 , hence

    abs[a_n - a_m ] = abs[ a_n - A + A - a_m ] <= abs[a_n - A ] + abs[ A - a_m ]
    = abs[a_n - A ] + abs[ a_m - A ] < e/2 + e/2 = e
    thus {a_n} is cauchy
     
  2. jcsd
  3. Dec 6, 2008 #2
    You want your conclusion to be |a_n - a_m| < e to satisfy the definition of a Cauchy sequence. It is not always easy to get e by itself in the inequality right away. (1) is rough work, after which you divide all expressions containing e by 2 to get the proof.

    If you understand the proof, then don't worry about (1) so much. Looking at other epsilon proofs will help.
     
  4. Dec 7, 2008 #3
    I think i am going to follow your advise Mutton, (1) seems not to be connected to the proof.
    In the proof if e > 0 then it must be true for e / 2 , and e/2 is used because at the end the answer is e

    Thanks again Mutton.
     
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