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Math Help - Analysis Proof Involving Cauchy Sequences

  1. #1
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    Analysis Proof Involving Cauchy Sequences

    So I teach high school math, but I love to study math in my spare time. I'm charging my way through an analysis book, trying to solve all of the problems. Been looking at this one for a couple of days now and I need to get un-stuck. It's clear to me that lim_{n\rightarrow \infty}\{a_n-b_n\}=a-b and that lim_{n\rightarrow \infty}\{a_n-b_n\}\leq0 since a_n\leq b_n, but I'm stuck on proving it. Any help appreciated. Thanks!

    Here is the problem:
    Suppose that \{a_n\} and \{b_n\} are both Cauchy sequences and that a_n\leq b_n for each n. Prove that lim_{n\rightarrow\infty}a_n\leq lim_{n\rightarrow\infty}b_n
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  2. #2
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    Re: Analysis Proof Involving Cauchy Sequences

    Quote Originally Posted by UNLVRich View Post
    Here is the problem:
    Suppose that \{a_n\} and \{b_n\} are both Cauchy sequences and that a_n\leq b_n for each n. Prove that lim_{n\rightarrow\infty}a_n\leq lim_{n\rightarrow\infty}b_n
    Because \mathbb{R} is complete we know that \left( {{a_n}} \right) \to a\;\& \;\left( {{b_n}} \right) \to b~.

    All you need to show is that a\le b.

    Have you (can you) shown that if \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0~?

    If you can then let c_n=b_n-a_n~.
    Thanks from topsquark and UNLVRich
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  3. #3
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    Re: Analysis Proof Involving Cauchy Sequences

    Quote Originally Posted by Plato View Post
    Have you (can you) shown that if \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0~?
    I've boiled it down to being able to show \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0, but there is something that I'm missing. I'm not sure how to show that.
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  4. #4
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    Re: Analysis Proof Involving Cauchy Sequences

    c_n is a snake that crawls arbitrarily close to c. On the other hand, if c < 0, then there is a finite distance between c and 0. Eventually the distance between c_n and c becomes smaller than this distance. What happens then?
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  5. #5
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    Re: Analysis Proof Involving Cauchy Sequences

    Quote Originally Posted by UNLVRich View Post
    I've boiled it down to being able to show \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0, but there is something that I'm missing. I'm not sure how to show that.
    I take you at your word that you are self studying basic analysis. Here is very useful concept.
    First, "almost all" means all but a finite collection.

    So if (c_n)\to c then almost all of the terms are 'near' c.
    As was pointed out in reply #4, if c<0 that would impossible if \forall n[c_n\ge 0].
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  6. #6
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    Re: Analysis Proof Involving Cauchy Sequences

    Hi,
    Maybe this will help. Try proof by contradiction. Namely, assume c < 0. Then in the definition of limit take \epsilon={-c\over2}. Then there is N such that c_N-c\leq|c_N-c|<\epsilon. But this says c_N<c+\epsilon={c\over2}<0, a contradiction.
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  7. #7
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    Re: Analysis Proof Involving Cauchy Sequences

    Thanks to johng, Plato and emakarov. I will meditate further on this.
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