Proving the maximum of a set is unique

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The discussion centers on proving that if m and n are both maxima of a set S, then m must equal n. The proof begins by assuming m ≠ n and demonstrating that this leads to contradictions regarding the properties of upper bounds and the supremum of the set. It highlights that if n is greater than m, it would contradict the uniqueness of the supremum, while if n is less than m, it would contradict n being a maximum. The author expresses concern about the proof being circular or lacking detail, while also noting that the statement seems intuitively obvious. Ultimately, the proof effectively confirms the uniqueness of the maximum in a set.
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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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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