Results 1 to 7 of 7
Like Tree5Thanks
  • 2 Post By Plato
  • 1 Post By emakarov
  • 1 Post By Plato
  • 1 Post By johng

Thread: Analysis Proof Involving Cauchy Sequences

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    17

    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 $\displaystyle lim_{n\rightarrow \infty}\{a_n-b_n\}=a-b$ and that $\displaystyle lim_{n\rightarrow \infty}\{a_n-b_n\}\leq0$ since $\displaystyle a_n\leq b_n$, but I'm stuck on proving it. Any help appreciated. Thanks!

    Here is the problem:
    Suppose that $\displaystyle \{a_n\}$ and $\displaystyle \{b_n\}$ are both Cauchy sequences and that $\displaystyle a_n\leq b_n$ for each n. Prove that $\displaystyle lim_{n\rightarrow\infty}a_n\leq lim_{n\rightarrow\infty}b_n$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,734
    Thanks
    2810
    Awards
    1

    Re: Analysis Proof Involving Cauchy Sequences

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

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

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

    If you can then let $\displaystyle c_n=b_n-a_n~.$
    Thanks from topsquark and UNLVRich
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    17

    Re: Analysis Proof Involving Cauchy Sequences

    Quote Originally Posted by Plato View Post
    Have you (can you) shown that if $\displaystyle \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0~?$
    I've boiled it down to being able to show $\displaystyle \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0$, but there is something that I'm missing. I'm not sure how to show that.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790

    Re: Analysis Proof Involving Cauchy Sequences

    $\displaystyle c_n$ is a snake that crawls arbitrarily close to $\displaystyle c$. On the other hand, if $\displaystyle c < 0$, then there is a finite distance between $\displaystyle c$ and 0. Eventually the distance between $\displaystyle c_n$ and $\displaystyle c$ becomes smaller than this distance. What happens then?
    Thanks from UNLVRich
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,734
    Thanks
    2810
    Awards
    1

    Re: Analysis Proof Involving Cauchy Sequences

    Quote Originally Posted by UNLVRich View Post
    I've boiled it down to being able to show $\displaystyle \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0$, 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 $\displaystyle (c_n)\to c$ then almost all of the terms are 'near' $\displaystyle c$.
    As was pointed out in reply #4, if $\displaystyle c<0$ that would impossible if $\displaystyle \forall n[c_n\ge 0].$
    Thanks from UNLVRich
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    1,141
    Thanks
    475

    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 $\displaystyle \epsilon={-c\over2}$. Then there is N such that $\displaystyle c_N-c\leq|c_N-c|<\epsilon$. But this says $\displaystyle c_N<c+\epsilon={c\over2}<0$, a contradiction.
    Thanks from UNLVRich
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2011
    Posts
    17

    Re: Analysis Proof Involving Cauchy Sequences

    Thanks to johng, Plato and emakarov. I will meditate further on this.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof involving convergent sequences.
    Posted in the Advanced Math Topics Forum
    Replies: 9
    Last Post: Sep 17th 2012, 04:59 PM
  2. Analysis Proof involving compositon f o g.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 13th 2010, 10:14 AM
  3. Replies: 4
    Last Post: Nov 16th 2008, 01:40 PM
  4. Replies: 2
    Last Post: Oct 4th 2008, 04:57 PM
  5. Replies: 1
    Last Post: Apr 8th 2008, 07:00 PM

Search Tags


/mathhelpforum @mathhelpforum