Prove that if bound sequence diverges > two subseq converge

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Homework Help Overview

The problem involves proving that if a bounded sequence \(\{X_a\}\) is divergent, then there exist two subsequences that converge to different limits. The subject area pertains to real analysis, specifically the properties of sequences and convergence.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to construct two subsequences based on bounds and monotonicity, questioning the correctness of their reasoning. Some participants raise concerns about the guarantee of subsequences having more than one element and suggest considering limit points of the bounded sequence.

Discussion Status

The discussion is ongoing, with participants exploring different approaches, such as the concepts of limit superior and limit inferior. There is no explicit consensus yet, but hints and suggestions for further exploration have been provided.

Contextual Notes

Participants are discussing the implications of boundedness and divergence in sequences, as well as the relevance of compactness in the context of subsequences.

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Homework Statement


Prove that if a bound sequence ##\left\{ { X }_{ a } \right\} ## is divergent then there are two sub sequences that converge to different limits.

Homework Equations


None.

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

Hint: What can you know about limit points of your bound sequence?
 
As mfb said; use compactness.
 
FaroukYasser said:

Homework Statement


Prove that if a bound sequence ##\left\{ { X }_{ a } \right\} ## is divergent then there are two sub sequences that converge to different limits.
FaroukYasser said:

Homework Statement


Prove that if a bound sequence ##\left\{ { X }_{ a } \right\} ## is divergent then there are two sub sequences that converge to different limits.

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##?
 
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