Analysis , sequences, limits, supremum explaination needed

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.
 
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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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