Relationship between Sup and Limsup of Sequences

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Discussion Overview

The discussion centers on the relationship between the supremum (sup) and the limit superior (limsup) of sequences, specifically in the context of countably infinite sets. Participants explore definitions, conditions under which these concepts are equal, and provide examples to illustrate their points.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants propose that limsup equals sup if and only if there is a subsequence converging to sup.
  • Others argue that in general, sup is always greater than or equal to limsup, and both are only well-defined if the sequence is bounded above.
  • One participant defines limsup as the largest value for which there is a subsequence converging to it, describing it as the largest limit point of the sequence.
  • A specific example is discussed where for the sequence 1, 1/2, 1/3, ..., the sup is 1 while the limsup is 0, illustrating the concepts in practice.
  • Another participant introduces the idea of superior and inferior numbers as a clarifying concept, referencing a book for further context.

Areas of Agreement / Disagreement

Participants express some agreement on the definitions and relationships between sup and limsup, but there are also nuances and conditions that remain contested, particularly regarding the circumstances under which they are equal.

Contextual Notes

The discussion includes assumptions about boundedness and the nature of subsequences, which are not fully resolved. The implications of these assumptions on the definitions and relationships discussed are acknowledged but not clarified.

AxiomOfChoice
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So if you have a countably infinite set \{ x_n \} and consider also the sequence (x_n), what's the relationship between \sup \{ x_n \} and \limsup x_n?
 
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limsup will equal sup if and only there is a subsequence converging to sup. In general sup is always larger or equal to limsup. Both are only well-defined if the sequence is bounded above.

limsup is the largest value for which there is a subsequence converging to it. In other words it's the largest limit point of the sequence. I may be mistaken, feel free to correct me if I'm wrong.
 
Last edited:
disregardthat said:
limsup will equal sup if and only there is a subsequence converging to sup. In general sup is always larger or equal to limsup. Both are only well-defined if the sequence is bounded above.

limsup is the largest value for which there is a subsequence converging to it. In other words it's the largest limit point of the sequence. I may be mistaken, feel free to correct me if I'm wrong.

No, this definitely makes sense. I suppose in the case of the sequence 1, 1/2, 1/3, \ldots, we have \sup\limits_n x_n = 1 (since 1 is certainly the least upper bound), but \limsup\limits_{n\to \infty} x_n = 0 (since 0 is the only limit point of this set). Thanks!
 
AxiomOfChoice said:
No, this definitely makes sense. I suppose in the case of the sequence 1, 1/2, 1/3, \ldots, we have \sup\limits_n x_n = 1 (since 1 is certainly the least upper bound), but \limsup\limits_{n\to \infty} x_n = 0 (since 0 is the only limit point of this set). Thanks!


That seems about right, and no problem.
 
I found it very clarifying to introduce the idea of a superior number & an inferior number,
as is done in this book. Just have a look at the page above the one that comes up in the link.
 

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