Prove that if bound sequence diverges > two subseq converge

  • Thread starter Thread starter FaroukYasser
  • Start date Start date
  • Tags Tags
    Bound Sequence
FaroukYasser
Messages
62
Reaction score
3

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?
 
Physics news on Phys.org
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##?
 
Last edited:
  • Like
Likes mfb
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top