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Cauchy sequence; I need some help

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]x_{n}[/tex] be a Cauchy sequence. Suppose that for every [tex]\epsilon>0[/tex] there is [tex] n > \frac{1}{\epsilon}[/tex] such that [tex]|x_{n}| < \epsilon[/tex]. Prove that [tex]x_{n} \rightarrow 0[/tex].


    2. Relevant equations



    3. The attempt at a solution

    My problem with the question is I do not understand it.

    if,

    [tex]|x_{n}| < \epsilon[/tex] when [tex] n > \frac{1}{\epsilon}[/tex] ;



    Doesn't that mean that [tex]Lim_{n\rightarrow infinity} x_{n} = 0 ?[/tex]

    In which case [tex]x_{n} \rightarrow 0[/tex] because if the limit as n goes to infinity is zero, then the terms can be arbitrarily brought close to zero as n gets large enough.



    What exactly is the point of me knowing that [tex]x_{n}[/tex] is a Cauchy sequence ?
    It doesn't seem like I even need to know that it is a Cauchy sequence. After-all, every convergent sequence is a cauchy sequence.




    Am I missing something ?
     
  2. jcsd
  3. Sep 1, 2010 #2

    Office_Shredder

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    You only know that there is one n such that [tex]|x_n|<\epsilon[/tex]. You don't know that for example [tex]|x_{n+1}|<\epsilon[/tex]

    If we take away the Cauchy condition then the sequence (1,0,1,0,1,0,1,0,...) would fit the criterion
     
  4. Sep 1, 2010 #3
    Sorry I don't follow.

    Doesn't the inequality hold for every epsilon?

    And given that n is not a specific number can't I just make n to be n+1 ?

    And given your sequence, if [tex] x_{n} = 1[/tex] the inequality would not hold.

    I am a bit lost. Please explain further.
     
  5. Sep 1, 2010 #4

    Office_Shredder

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    Ok, let's continue with my example.

    Let's pick [tex]\epsilon=\frac{1}{100,000}[/tex]. Then we need to find n larger than 100,000 such that [tex]x_n<\epsilon[/tex]. Well that's easy, n=100,002 works, since we know that's zero.

    The point is that it doesn't say there exists N, such that for all n>N, |xn|<epsilon, it only says there exists ONE n for which |xn|<epsilon

    Basically, the statement of the problem is that a SUBSEQUENCE of xn converges to zero
     
  6. Sep 1, 2010 #5
    Okay I understand the problem now. I will think more about the problem and request more help if I need to. Thanks for now. I'll repost if i don't get anywhere.
     
  7. Sep 2, 2010 #6
    Okay, I thought about the question and I am still not sure what to do.

    Now that I know a sub-sequence in [tex]x_{n}[/tex] converges. So this means the following are true:

    1) The Cauchy sequence converges since we have found a convergent subsequence.

    2) The Cauchy sequence converges to 0 also , since the terms of the Cauchy sequence can brought arbitrarily close to each other. This is a Lemma in my book anyway; I assume I can use it.
    3) since the Cauchy sequence converges to zero this means [tex]x_{n} \rightarrow 0[/tex]


    How exactly do I go about proving this mathematically ?
     
  8. Sep 2, 2010 #7
    Can someone please point me in the right direction?!
     
  9. Sep 2, 2010 #8

    Office_Shredder

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    You basically have the right idea. If you want to get the sequence within [tex]\epsilon[/tex], pick N so that the elements of your subsequence are within [tex]\frac{\epsilon}{2}[/tex] and the elements of your sequence are within [tex] \frac{\epsilon}{2}[/tex] of each other when the indices are larger than N and see what you can do
     
  10. Sep 2, 2010 #9
    UThe elements of the sequence are within epsilon of what? 0? Are you suggesting some sort of triangle inequality ? Something like
    [tex]|x_{n}-x_{m}|\leq |x_{n}|+|x_{m}|< \frac{ \epsilon_{0}}{2}} + \frac{\epsilon_{0}}{2}[/tex].
    EDIT

    I am not sure how to get the terms in the subsequence close to those of the sequence. Can you help me with this?
     
    Last edited: Sep 2, 2010
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