# Prove that a nonempty finite contains its Supremum

1. Sep 30, 2012

### inverse

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

Last edited: Sep 30, 2012
2. Sep 30, 2012

### SteveL27

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?

3. Sep 30, 2012

### mathman

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.

4. Sep 30, 2012

### SteveL27

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.

5. Sep 30, 2012

### inverse

this is an exercise in a book of mathematical analysis, as there are no solutions wondered to know what is the correct ration.

6. Sep 30, 2012

### SteveL27

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.

7. Oct 1, 2012

### inverse

thank you very much to everyone who have helped me :)