Prove that a nonempty finite contains its Supremum

In summary, a nonempty finite subset S of the real numbers always contains its supremum, or largest element. This can be proven by using pairwise comparisons to find the maximum of a finite set, and by induction for larger sets. This is a result from real analysis and can be shown using the real number axioms.
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inverse
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Prove that a nonempty finite [itex]S\,\subseteq\,\mathbb{R}[/itex] 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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  • #2
inverse said:
Prove that a nonempty finite [itex]S\,\subseteq\,\mathbb{R}[/itex] 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?
 
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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.
 
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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.
 
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this is an exercise in a book of mathematical analysis, as there are no solutions wondered to know what is the correct ration.
 
  • #6
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.
 
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thank you very much to everyone who have helped me :)
 

FAQ: Prove that a nonempty finite contains its Supremum

What does it mean for a set to contain its Supremum?

A set containing its Supremum means that the Supremum, or the least upper bound, is actually an element of the set.

How do you prove that a nonempty finite set contains its Supremum?

This can be proven by showing that the Supremum is an element of the set and that it is the smallest element that is greater than or equal to all other elements in the set.

Can a nonempty finite set contain multiple Supremums?

No, a nonempty finite set can only contain one Supremum. This is because the Supremum is defined as the smallest element that is greater than or equal to all other elements in the set.

Why is it important to prove that a set contains its Supremum?

Proving that a set contains its Supremum is important because it shows that the set has a well-defined maximum element. This can be useful in various mathematical proofs and can also help in determining the behavior of functions and sequences.

Is it possible for an infinite set to contain its Supremum?

Yes, an infinite set can also contain its Supremum. However, the proof for this may be more complex and may involve concepts such as limits and convergence.

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