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Homework Help: Every Cauchy sequence of real numbers converges

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    ra6.JPG

    I understand everything except the last two lines. I am really confused about the last two lines of the proof. (actually I was never able to fully understand it since my first year calculus)
    I agree that if ALL three of the conditions n≥N, k≥K, and nk≥N are satisfied, then the last line is true. But why does the condition n≥N alone imply that the last line is true? Does n≥N guarantee that k≥K, and nk≥N?
    Why is it true that n≥N => |an-an_k| < ε/2 ?
    And why is it true that n≥N => |an_k-L| < ε/2 ?
    Can somebody kindly explain the last two lines of the proof in more detail?

    2. Relevant equations
    N/A

    3. The attempt at a solution
    N/A

    Any help is much appreciated!
     
    Last edited: Jan 17, 2010
  2. jcsd
  3. Jan 17, 2010 #2

    HallsofIvy

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    Knowing that [itex]n\ge N[/itex] doesn't tell you anything about k. That was why the proof says "Pick any [itex]k\ge K[/itex]" k is chosen to be larger than K. That has nothing to do with n. What the proof is really saying is "K is a fixed number so certainly there exist k> K. Also N is a fixed number and nk is a sequence that increases without bound so there exist numbers in the subsequence such that nk> N. Of all the nk> N, choose one for which k is also > K.

     
  4. Jan 17, 2010 #3
    But at the end, we are supposed to prove that:
    an->L
    i.e. for all ε>0, there exists N such that n≥N => |an-L| < ε ?
    (note that here only n appears, there is nothing about k)

    Looking at the definition of an->L above, all we have to do is to construct an N that works. Just like every ε-limit proof, we have to find an N that works. And if we can find such an N, then the only restriction should be n≥N and nothing else, but in this proof we also have other restrictions k≥K and nk≥N which does not even appear in |an-L| < ε . How come? Please help...I really don't understand :(

    Do we need all three conditions (n≥N, k≥K, and nk≥N) to be simultaneously satisfied in order for |an-L| < ε to hold??
     
  5. Jan 17, 2010 #4

    vela

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    All you have to show is that given an [itex]\epsilon[/itex], an N exists. It doesn't really matter how you found it; you just have to show it exists.
     
  6. Jan 17, 2010 #5
    But what is that "N" in this case?
     
  7. Jan 17, 2010 #6

    vela

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    It says what N is in the fifth sentence of the proof.
     
  8. Jan 17, 2010 #7
    So the exact same N there would work at the end?
     
    Last edited: Jan 17, 2010
  9. Jan 17, 2010 #8

    vela

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    Yes. That's the point of the rest of the proof.
     
  10. Jan 18, 2010 #9
    I've seen another proof of this theorem:

    Given ε>0.
    There exists N s.t. m,n≥N => |an-am|<ε/2
    There exists M s.t. k≥M => |ank-L|<ε/2 and also M≥N.
    Take N'=max{N,nM}
    Then if n>N',
    |an-L|≤|an-anM|+|anM-L|<ε/2+ε/2=ε
    =====================

    Is this proof right or wrong? If it's right, I have the following questions:

    1) Why is M≥N?

    2) Why should we take N'=max{N,nM}?

    Does anyone have any idea?


    (I increased the font size to make the subscripts legible)
     
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