Analysis , sequences, limits, supremum explaination needed

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Homework Help Overview

The discussion revolves around the concepts of sequences, limits, and the supremum in the context of real analysis. The original poster seeks clarification on a proof related to constructing a nondecreasing sequence that converges to the supremum of a bounded nonempty subset of real numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the construction of a monotone increasing sequence and its convergence to the supremum. Questions arise regarding the role of the term 1/n and its relation to the definition of supremum and convergence.

Discussion Status

Participants are actively engaging with the proof and exploring its implications. Some have provided insights into the completeness property and its relevance to the discussion, while others express confusion about specific elements of the proof.

Contextual Notes

There is mention of the completeness property and its significance in the context of supremum, as well as the use of ε in relation to the sequence elements. The original poster and others are navigating through the definitions and implications without reaching a consensus on all points.

retspool
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Analysis , sequences, limits, supremum explanation needed :(

So i have a question and the answer as well, but i will need some explanation.

here is the Question

Let S be a bounded nonempty subset of R and suppose supS ∉S . Prove that there is a
nondecreasing sequence (Sn) of points in S such that lim Sn =SupS .

Answer Proof

Since supS ∉ S , there exist Sn ∈ S for all n ∈N such that Sn > S - 1/n
.
Hence limSn = supS and (Sn) is a nondecreasing sequence.
 
Last edited:
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What are you not clear about? Is it the proof ?

Edit

Just saw your edit. :-)

The idea of the proof is constructing a monotone increasing sequence that converges to the suprema.

Using completeness and the definition of sup S we know that there exist an element of S such that
s_0 > supS - e.

Basically, the statement means that if you go a little bit to the left of the suprema of a set you encounter points of your set that are to the right of that number on the number line.

This is intuitive giving that the supS is the LEAST upper bound. That is, any other upper bound must the greater that supS.
 
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Yes this is the proof.

i don't how it makes sense :(
 


Please review my edit and post if you still need help :-).
 


Thanks


I get that they are trying to locate a supremum using its defn, but the 1/n throws me off.

is -(1/n) just an ε through which the statement Sn > S - 1/n
Satisfies the definition of both a supremem and convergence of {Sn} to a real number?

I never thought that we would have to use completeness.
But using it does seems more appropriate.

I thought of using the property liminfSn = limSn = limSupSn, but i guess the approach is incorrect
 


Yes, that is right 1/n is pretty much your epsilon.

The reason for using 1/n is that works nicely for specifying the elements of the sequence.

I.e
x_1 > supS -1
x_2 > supS -1/2
etc

Whenever you see supA you know completeness is used. You don't need to use limsup and liminf here.

Is everything clear now ?
 


Crystal.

Thanks a ton.!
 

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