Any Cauchy sequence converges.

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    Cauchy Sequence
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Homework Help Overview

The discussion revolves around the properties of Cauchy sequences, specifically focusing on the concepts of limit inferior (lim inf) and limit superior (lim sup) in relation to convergence. Participants are exploring the implications of bounded sequences and the relationships between subsequential limits and the overall limit of the sequence.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of inequalities related to lim inf and lim sup, particularly how these limits relate to the bounds defined by a sequence. There is also a request for clarification on the definitions of lim sup and lim inf.

Discussion Status

Some participants are actively seeking clarification on the inequalities presented, while others are reflecting on the implications of having equal lim sup and lim inf. There is an acknowledgment of the need for clearer definitions and understanding of the concepts involved.

Contextual Notes

There are indications that the discussion is influenced by the formatting of the original posts, which some participants found challenging to read. Additionally, there is a focus on ensuring that the definitions of key terms are understood correctly.

peripatein
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Hello,

My instructor, whilst trying to prove that liminf of sequence a_n = limsup of sequence a_n = A,
_
wrote that since we know that a_n0-ε<an<a_n0+ε → a_n0-ε ≤ A ≤ A ≤ a_n0+ε.
Why is that true? I mean, how do we know that if a sequence is bounded then its lim inf (i.e. the lowest amongst the limits of its subsequences) and lim sup (the highest amongst the limits of its subsequences; do forgive me for potential misnomers) are indeed within that area bounded by a_n0-ε and a_n0+ε?
I hope you may assit in clarifying this. Thank you!
 
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The inequality did not come out right. The second A (following a_n0-epsilon <= A_) should be upper A (or, if I understand it aright, lim sup.
 
peripatein said:
Hello,

My instructor, whilst trying to prove that liminf of sequence a_n = limsup of sequence a_n = A,
_
wrote that since we know that a_n0-ε<an<a_n0+ε → a_n0-ε ≤ A ≤ A ≤ a_n0+ε.
Why is that true? I mean, how do we know that if a sequence is bounded then its lim inf (i.e. the lowest amongst the limits of its subsequences) and lim sup (the highest amongst the limits of its subsequences; do forgive me for potential misnomers) are indeed within that area bounded by a_n0-ε and a_n0+ε?
I hope you may assit in clarifying this. Thank you!
This is quite hard to read. You'll get much better response if you take the trouble to make it legible, e.g. by writing it out in latex.
 
Hello,

My instructor, whilst trying to prove that liminf of sequence an = limsup of sequence an = A, stated that since we know that an0-ε<an<an0+ε \Rightarrow an0-ε ≤ A ≤ A-- ≤ an0+ε.

Why is that true? I mean, how do we know that if a sequence is bounded then its lim inf (i.e. the lowest amongst the limits of its subsequences) and lim sup (the highest amongst the limits of its subsequences; do forgive me for potential misnomers) are indeed within that neighbourhood bounded by an0-ε and an0+ε?
I hope you may assit in clarifying this. Thank you!
 
Following haruspex's advice, the content is now far more legible. May someone please help calrifying this issue? Why is that inequality valid given that the sequence is bounded?
 
What are your definitions for "limsup" and "liminf" of a sequence?
 
I have explained that above, in parenthesis.
 
Thanks. For some reason I didn't notice that before. So if limsup= liminf= A, you know that all subsequential limits are less than or equal to A and larger than or equal to A. Well, a subsequential limit cannot be both "less than A" and "larger than A" can it? So this means that all subsequential limits are equal to A which means that the sequence itself converges to A.
 

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