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Prove that if bound sequence diverges > two subseq converge

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that if a bound sequence ##\left\{ { X }_{ a } \right\} ## is divergent then there are two sub sequences that converge to different limits.

    2. Relevant equations
    None.

    3. The attempt at a solution

    Ok so I am not sure if my attempt for a solution is correct or not, but I have no ideas except this one.

    Since ##\left\{ { X }_{ a } \right\} ## is bound, Let ##x## be the highest lower bound and ##y## be the lowest upper bound.

    ##\Longrightarrow \quad x<\left\{ { X }_{ a } \right\} <y\quad ,\quad x<y##

    Next, Let ##\left\{ { X }_{ \alpha _{ k } } \right\} ## be a sub sequence of ##\left\{ { X }_{ a } \right\} ## where

    ##\frac { x+y }{ 2 } <\left\{ { X }_{ \alpha _{ k } } \right\} <y\quad and\quad { X }_{ \alpha _{ 1 } }\quad \le { \quad X }_{ \alpha _{ 2 } }\le { \quad X }_{ \alpha _{ 3 } }\quad \le ...##

    Since ##\left\{ { X }_{ \alpha _{ k } } \right\} ## is an increasing sequence and is bounded then it converges.

    Assume ##\lim _{ x\longrightarrow \infty }{ { X }_{ \alpha _{ k } } } ={ L }_{ 1 }\quad (*)##

    Next, Let ##\left\{ { X }_{ b_{ k } } \right\} ## be a sub sequence of ##\left\{ { X }_{ a } \right\} ## where

    ##x<\left\{ { X }_{ b_{ k } } \right\} <\frac { x+y }{ 2 } \quad and\quad X_{ { b }_{ 1 } }\quad \ge \quad X_{ { b }_{ 2 } }\quad \ge \quad X_{ { b }_{ 3 } }\quad \ge \quad ...##

    Since ##\left\{ { X }_{ b_{ k } } \right\} ## is a decreasing sequence and is bounded then it converges:

    Assume ##\lim _{ x\longrightarrow \infty }{ { X }_{ b_{ k } } } ={ L }_{ 2 }\quad (**)##

    ##(*)-(**)\quad =\quad { L }_{ 1 }-{ L }_{ 2 }\quad =\quad \lim _{ x\longrightarrow \infty }{ { X }_{ \alpha _{ k } } } -\lim _{ x\longrightarrow \infty }{ { X }_{ b_{ k } } } ## , Since ##{ X }_{ \alpha _{ k } }>{ X }_{ b_{ k } }## for all ##\alpha _{ k },b_{ k }##, then by the order rule ##{ L }_{ 1 }-{ L }_{ 2 }\quad >\quad 0\quad \Longleftrightarrow \quad { L }_{ 1 }\quad >\quad { L }_{ 2 }##, Therefore the two sub sequences converge to two different limits. Q.E.D

    Any idea if what i wrote is correct or not? if not, any idea on how I can approach this?
     
  2. jcsd
  3. Sep 20, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    You cannot guarantee that your subsequences have more than one element.

    Hint: What can you know about limit points of your bound sequence?
     
  4. Sep 21, 2015 #3
    As mfb said; use compactness.
     
  5. Sep 21, 2015 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    Have you ever heard of the concepts of limit superior and limit inferior, that is, ##\limsup_{n \to \infty} X_n## and ##\liminf_{n \to \infty} X_n##?
     
    Last edited: Sep 22, 2015
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