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Cauchy sequence proof , help

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Every convergent sequence is a cauchy sequence

    2. Relevant equations

    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

    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

    3. The attempt at a solution

    the part where i have trouble understanding this proof is , where does the e/2 comes from?
    in other words how does e/2 appears in the proof, how can i represent it in the real number graph?
    if the answer is this " the difference between a_n , a_m was less than twice the original choice of e " i would like to see the arithmetic that produces e/2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 6, 2008 #2
    The sentence "the difference between a_n , a_m was less than twice the original choice of e " is false; get rid of it.

    In the first sentence, e is fixed.

    The convergence of [tex]\{a_n\}[/tex] to A means that for any [tex]\varepsilon > 0[/tex], there exists an N such that [tex]n \ge N[/tex] implies [tex]|a_n - A| < \varepsilon[/tex].

    So we can pick any [tex]\varepsilon[/tex]. In the proof, [tex]\varepsilon = e/2[/tex] is picked. [tex]\varepsilon = e/4[/tex] could have been picked and it would still work. [tex]\varepsilon = 123[/tex] could have been picked and it wouldn't work.
     
  4. Dec 6, 2008 #3
    the sentence " the difference between a_n , a_m was less than twice the original choice of e " is false; get rid of it. " is not false , look , a_n and a_m are in the set ( A - e , A + e ) , and this is an interval of length 2e , and in the proof the choice of e/2 was not a coincidence it was because ( A +e ) - ( A -e ) = 2e , i dont know if this is right, but it comes straight from my book
     
  5. Dec 6, 2008 #4
    In this proof, a_n and a_m are in (A - e/2, A + e/2) because |a_n - A| < e/2 and |a_m - A| < e/2.

    The goal was to show that |a_n - a_m| < e, not 2e. See the second last line of the proof.
     
  6. Dec 6, 2008 #5
  7. Dec 6, 2008 #6
    how can e/4 work ?

    could you please give me your version of this cauchy sequence ( explaining the e/2 's or 2e's . My book is confusing me more.

    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
     
    Last edited: Dec 6, 2008
  8. Dec 6, 2008 #7
    Just replace e/2 in the proof with e/4. Then near the end, e/4 + e/4 = e/2 < e.
     
  9. Dec 7, 2008 #8
    Wait, in the original cauchy proof, e/2 + e/ 2 is e , but at the end it becomes
    e < e ?? isnt this false

    | a_n - a_m | < e , we have to satisfy this
    with e /2 we reach e < e ??
     
  10. Dec 7, 2008 #9
    Nowhere is e < e implied. |a_n - a_m| is less than or equal to the sum of two expressions. Each expression is less than e/2, so their sum is less than e.

    |a_n - a_m| [tex]\le[/tex] |a_n - A| + |a_m - A| < e/2 + e/2 = e
     
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