Proving the maximum of a set is unique

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SUMMARY

The discussion centers on proving the uniqueness of the maximum of a set S, specifically that if m and n are both maxima of S, then m must equal n. The proof utilizes the properties of supremum, establishing that if m ≠ n, contradictions arise regarding upper bounds. The uniqueness of the supremum is referenced as a critical theorem that supports the argument. The proof is confirmed to be valid, despite concerns about circular reasoning and detail.

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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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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!
 
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!
 

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