Proving the Least Upper Bound Property: A Mathematical Inquiry

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Homework Help Overview

The discussion revolves around proving the Least Upper Bound property in the context of real analysis. The original poster presents a statement regarding a nonempty set A and its upper bound x, exploring the conditions under which x is the least upper bound.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of x being an upper bound and the conditions necessary for it to be the least upper bound. There is an exploration of the relationship between upper bounds and the existence of elements in A that approach x.

Discussion Status

Some participants are attempting to clarify the proof structure, particularly regarding the reverse direction of the proof. There is an ongoing exploration of potential contradictions that arise if an upper bound z exists below y, with no consensus reached on the completion of the proof.

Contextual Notes

Participants note the importance of epsilon (E) in the proof and the need to demonstrate that no number less than y can serve as an upper bound for A. There are indications of typos in the posts that may affect the clarity of the arguments presented.

ssayan3
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Least Upper Bound proof...

Homework Statement


Suppose A is a nonempty set that has x as an upper bound. Prove that x is the least upper bound of the set A iff for any E>0 there exists a y in A such that y>x-E


Homework Equations


None


The Attempt at a Solution


The forward where you assume that x is the least upper bound is very easy, but I'm having some trouble proving the reverse...

This is what I have so far...

Let x be an upper bound of A, and choose a point z in A.
If x is an upper bound of A, then x+z is also an upper bound.
 
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To prove the reverse, you are given that x is an upper bound for A having the property:

If \epsilon &gt; 0 there is a y in A satisfying x - \epsilon < y

You have to show that no number z < y is an upper bound for A. What problem would arise if there was such a number z?

[Edit] Sorry, there is a typo. The last paragraph should have read:

You have to show that no number z < x is an upper bound for A. What problem would arise if there was such a number z?
 
Last edited:


Hmm... if there were such a number z, then y could not be the least upper bound...

Could the proof go something like this?:

Choose arbitrary E>0, and let y be an upper bound of A

Suppose z is an upper bound of A, and y>z>y-E.

y is not the lub

does this finish the proof?
 


ssayan3 said:
Hmm... if there were such a number z, then y could not be the least upper bound...

Could the proof go something like this?:

Choose arbitrary E>0, and let y be an upper bound of A

Suppose z is an upper bound of A, and y>z>y-E.

y is not the lub

does this finish the proof?

No. Sorry, but I had a typo which I have corrected. Read my reply and try again.
 

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