Is this simple calculus 1 proof true?

Question

What's up math experts, yesterday I saw a question related to calculus 1 in which they asked for a proof for the following: Let A and B be two bounded subsets of R. A ⊂ B ⇒ sup(A) ≤ sup(B). The moment I saw it, reasoning by absurd came to my mind. So my personnal proof would be: let's assume that the assertion A ⊂ B ⇒ sup(A) > sup(B) is true and we find a contradiction: sup (A) > sup (B), since sup(B) is the smallest upper bound of B and sup(A) strictly greater than sup(B) then this means that sup(A) which can belong to set A does not belong to set B and this is a contradiction because A is included in B . so we have shown by contradiction that the assertion A ⊂ B ⇒ sup(A) ≤ sup(B) is true. Many teachers prooved the same assertion with direct reasoning but I preffered the absurd reasoning. what do you think about it? is my proof true? thanks already.

Answer 1

Your reasoning is on the right track but slightly flawed. We don't want to assume the if/then statement you chose above. Instead, we assume both parts of that statement are true (rather than one implies the other):

A ⊂ B and sup(A) > sup(B).

From there, your proof by contradiction is valid, if a bit imprecise.