Prove that a nonempty finite contains its Supremum

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The discussion revolves around proving that a nonempty finite subset S of the real numbers ℝ contains its supremum. Participants are exploring the properties of finite sets in relation to their supremum and discussing the implications of finiteness versus infiniteness in this context.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Some participants attempt to establish that a finite set must contain its maximum element, suggesting that this relates to the existence of the supremum. Others question the implications of finiteness, particularly why the argument does not hold for infinite sets. There are inquiries about the meaning of subsets being "less than or equal to ℝ" and the nature of comparisons needed to find a maximum.

Discussion Status

The discussion is active, with participants providing insights and raising questions about the original poster's reasoning. Some guidance has been offered regarding the existence of maximum elements in finite sets, and there is an acknowledgment of the need for clarity in distinguishing between finite and infinite cases.

Contextual Notes

Participants note that the problem is framed as an exercise in mathematical analysis, emphasizing the lack of provided solutions and the need for a rigorous proof based on real number axioms.

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Prove that a nonempty finite S\,\subseteq\,\mathbb{R} contains its Supremum.

If S is a finite subset of ℝ less than or equal to ℝ, then ∃ a value "t" belonging to S such that t ≥ s where s ∈ S.

This is the only way I see to prove it, I hope your help :))

Regards
 
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inverse said:
Prove that a nonempty finite S\,\subseteq\,\mathbb{R} contains its Supremum.

If S is a finite subset of R less than or equal to IR, then ∃ a value "t" belonging to S such that t ≥ s where s ∈ S.

What does it mean for a subset S to be less than or equal to ℝ? [Markup note: to the right there's a box full of handy symbols. You can just click on ℝ and you don't need LaTeX]

And why doesn't your proof work is S is infinite? Why doesn't t exist in that case? What is is about the finite case that makes this work?
 
If S is finite, the largest member can be found in a finite number of steps using pairwise comparisons.

When S is infinite, this may not be possible.
 
mathman said:
If S is finite, the largest member can be found in a finite number of steps using pairwise comparisons.

When S is infinite, this may not be possible.

That's true, but hardly on point for the OP's question. By your reasoning, we'd be hard-pressed to show that [0,1] contains its sup, since computationally we'd have to do uncountably many comparisons.

OP is asking a real analysis question, not a computability theory question.
 
this is an exercise in a book of mathematical analysis, as there are no solutions wondered to know what is the correct ration.
 
inverse said:
this is an exercise in a book of mathematical analysis, as there are no solutions wondered to know what is the correct ration.

The answer is that a finite set of reals always contains its max. If you need to prove that from the real number axioms you can note that for any two distinct reals a and b, either a < b or b < a. In either case the max exists.

[Oh! LOL now I understand mathman's remark. It went over my head the first time. My apologies!]

So you can prove that the max of two numbers exists; and by induction you can prove that the max of n numbers exists. That was mathman's point.

Just look at examples. Any finite set of real numbers has a largest element.
 
thank you very much to everyone who have helped me :)
 

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