Proving Limit Infs and Sups to Solving Lim Inf, Lim Sup Proof Problems

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Homework Help Overview

The discussion revolves around proving the relationship between the limit inferior and limit superior of a sequence, specifically the assertion that lim inf Sn = -lim sup (-Sn). Participants are exploring concepts related to sequences and their limits in the context of real analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss their understanding of limit inferior and limit superior, with some attempting to relate these concepts to the properties of sequences. There are questions about how to rigorously apply definitions and theorems regarding limits, particularly in the context of negative sequences.

Discussion Status

Some participants have offered insights and clarifications regarding the definitions of lim inf and lim sup, while others are seeking further guidance on how to rigorously complete their proofs. There is an ongoing exploration of the relationships between bounds and limits, with no explicit consensus reached on the proof itself.

Contextual Notes

Participants are grappling with the definitions and properties of limit inferior and limit superior, and there are indications of uncertainty regarding the application of these concepts in their proofs. The discussion reflects a mix of attempts to apply theoretical knowledge and the need for deeper understanding of the underlying principles.

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Homework Statement


prove lim inf Sn= - Lim Sup (-Sn)


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The Attempt at a Solution



I know how to run the proof when dealing with the sup and inf. But the limsup and liminf is another story. I want to do it the same way by saying that limsup (-SN) >= Sn but i know I cannot say that. From what I gather all I can say that the Sups >= Limsup. But that does not help me relate it to the sequence itself. Need a push to get me going.
 
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Sn is a sequence of numbers, right?
So remember, liminf (Sn)= lim n->infinity (inf{Sm: m>=n})
Now, you know that inf{Sm: m>=n} = -sup{-Sm: m>=n}.
That's my push
 
I need more of a nudge

I am with you on that inf(Sn : n> N} = -sup{-Sn: n >= N} Though I do not think I can say lim as n goes to infinity and finish it off. I know as n goes to invintiy that lim sup <= -sup{-Sn: n >= N} and inf{Sm: m>=n} <= Lim inf. I was thinking this would run similar to the inf Sn = -sup Sn. So I would have to prove -sup Sn is a lowerbound and also the greatest lower bound.
 
I did some reading and slepted on it so how about the following

Let S0 be the lim sup of -Sn. Thus as n -> infinity -Sn <= S0. By the order postulates it follows that -S0 <= Sn making, -S0 a lowerbound for Sn as n -> infinity. Now assume L >= -S0, but L <= Sn, making L also a lowerbound. By order postulates -L >= -Sn as n-> infinity making L an upperbound. By definition of the suprenum, S0 <= -L but by the order postualtes L<= -S0 which is a contradiction -S0 = lim inf of Sn.


Is the basic idea for the lim sup and lim inf is that it forms a box where the limit is allowed to be, and as n ---> infinty Sn <= lim Sup and same logic for the lim inf. Also the infs are an increasing sequence and sups are decreasing and essential squeeze the box to get the limit value if it exists?

Thanks in advance
 
yeah, you're right in your description of the basic idea of liminf and limsup.

Though I do not think I can say lim as n goes to infinity and finish it off.

Why not?

Let S0 be the lim sup of -Sn. Thus as n -> infinity -Sn <= S0.

That's not completely rigorous. Note that it could be the case that for all n, -Sn>S0. For example, let Sn=-1/n, so Sn=1/n. Then S0 = limsup -Sn = lim -Sn = 0, even though for all n, -Sn=1/n>0.

So you really do need to talk about limits.
 

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