# Homework Help: Prove that if bound sequence diverges > two subseq converge

1. Sep 20, 2015

### FaroukYasser

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. Sep 20, 2015

### 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?

3. Sep 21, 2015

### sebasvargasl

As mfb said; use compactness.

4. Sep 21, 2015

### Ray Vickson

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