Prove that if bound sequence diverges > two subseq converge

In summary: No, what are they?Limit superior and limit inferior are two concepts that are used when dealing with extremum and infimum. They are used to help make sure that two sequences are converging to the same limit.
  • #1
FaroukYasser
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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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  • #2
You cannot guarantee that your subsequences have more than one element.

Hint: What can you know about limit points of your bound sequence?
 
  • #3
As mfb said; use compactness.
 
  • #4
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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Related to Prove that if bound sequence diverges > two subseq converge

What does it mean for a bound sequence to diverge?

A bound sequence is a sequence of numbers that is limited or "bounded" by some finite value. A divergent bound sequence is one in which the terms of the sequence get larger and larger without ever reaching a specific limit.

What is a subsequence?

A subsequence is a sequence that is obtained by removing some terms from a given sequence. For example, the subsequence of {1,2,3,4,5} could be {2,4}.

How can we prove that a bound sequence diverging implies that two subsequnces must converge?

To prove this, we can use the Bolzano-Weierstrass theorem, which states that every bounded sequence has a convergent subsequence. Since the bound sequence is diverging, the terms of the sequence must be getting larger and larger. This means that at least one of the subsequnces we create by removing terms from the bound sequence must have a limit, and thus, must converge.

Can we have more than two convergent subsequnces if the bound sequence diverges?

Yes, it is possible to have more than two convergent subsequnces if the bound sequence diverges. However, the proof for this statement is more complex and requires a deeper understanding of mathematical concepts such as the Cantor set and uncountable sets.

What are some real-world applications of this concept?

This concept is commonly used in the field of numerical analysis, where it is used to analyze the convergence of algorithms and methods for solving equations. It is also used in physics and engineering to study the behavior of physical systems, as well as in economics and finance to analyze trends and predict future outcomes.

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