Proving the maximum of a set is unique

In summary, the student's proof of the equality of m and n for a set of maxima is different from the answer key. However, the proof is still valid, and the student can stop here.
  • #1
stripes
266
0

Homework Statement



Suppose that m and n are both maxima of a set S. Prove that m = n.

Homework Equations



None

The Attempt at a Solution



My proof seems much different than the answer key. Here is mine:

Suppose m ≠ n. Take m = max S. We know that m = sup S since m is an upperbound for S and that m ∈ S. If n>m, then n is an upperbound for S. n must be the least upperbound since we know n ∈ S. But we originally said m = sup S, and we know sup S is unique (from a previous theorem). This is a contradiction, so n is not greater than m. If n<m, then n is not an upperbound for S. But we know that n = max S, which means that n is the least upperbound for S, a contradiction. So n is not less than m. We have that m must equal n.

The question is trivially obvious, which is why I think it's a little tough to prove...we can just say it's obvious. But it's hard to not do that...if you know what I'm saying. Anyways I just feel like my proof might be circular or not detailed enough. Please let me know if there is anything I could do better, thank you all in advance!
 
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  • #2
stripes said:

Homework Statement



Suppose that m and n are both maxima of a set S. Prove that m = n.

Homework Equations



None

The Attempt at a Solution



My proof seems much different than the answer key. Here is mine:

Suppose m ≠ n. Take m = max S. We know that m = sup S since m is an upperbound for S and that m ∈ S. If n>m, then n is an upperbound for S. n must be the least upperbound since we know n ∈ S. But we originally said m = sup S, and we know sup S is unique (from a previous theorem).
If you have, in fact, previously proved that the supremum of a set is unique, and are allowed to use that theorem, you can stop here!

This is a contradiction, so n is not greater than m. If n<m, then n is not an upperbound for S. But we know that n = max S, which means that n is the least upperbound for S, a contradiction. So n is not less than m. We have that m must equal n.

The question is trivially obvious, which is why I think it's a little tough to prove...we can just say it's obvious. But it's hard to not do that...if you know what I'm saying. Anyways I just feel like my proof might be circular or not detailed enough. Please let me know if there is anything I could do better, thank you all in advance!
You say there is a different proof in the answer key? I'll bet it's exactly like the proof of the uniqueness of the supremum!
 
  • #3
That's what I thought...I could stop right there. Originally I DID, but as I was typing it up, I thought I should include it, as it seemed necessary. But thank you so much anyways!
 

What does it mean to prove the maximum of a set is unique?

To prove the maximum of a set is unique means to show that there is only one element in the set that has the highest value.

Why is it important to prove the maximum of a set is unique?

Proving the maximum of a set is unique is important because it ensures that there is no ambiguity or confusion about which element has the highest value in the set.

What is the process for proving the maximum of a set is unique?

The process for proving the maximum of a set is unique involves demonstrating that there is only one element in the set that has the highest value, either through mathematical equations or logical reasoning.

Can the maximum of a set be proven to be unique for all sets?

No, the maximum of a set cannot be proven to be unique for all sets. There may be cases where multiple elements in a set have the same maximum value, making it impossible to prove uniqueness.

What are some common methods for proving the maximum of a set is unique?

Common methods for proving the maximum of a set is unique include using the definition of a maximum element, using the order properties of the elements in the set, and using contradiction or proof by contradiction.

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