This is might be too easy, but just to make sure I am on the right track.
Prove that if sup $S \in S \Rightarrow$ sup $S =$ max $S$
By definition, the maximum $max$ of a set $S$ is the number that is greater or equal to all the elements of $S$. The supremum $sup$, if in the set, must be greater or equal to all numbers in the set, therefore $sup S = max S$.
Is this correct? Am I missing something?
I think you need to prove it rigorously.
$x\leq $ sup $S$ for all $x\in S$. Also $x\leq$ max $S$ for all $x\in S$.
Since sup $S\in S, $ sup $S\leq$ max $S$. Since always max $S\in S,$ max $S\leq $ sup $S$.
So, max $S=$sup $S$
It's certainly correct, technically. Stylistically, it feels to me like it's been in the reverse of the logical order. That is to say, the logic seems to flow better as:
The supremum is defined to be greater than or equal to every element in $S$. Given that $s=\sup S$ is a member of $S$, it is thus an element of $S$ greater than or equal to any given element of $S$. This is the definition of the maximum, hence $s$ is the maximum of $S$ as well.
We're trying to conclude something about the maximum given something about the supremum, so it makes sense to start by talking about the supremum.
We can prove the reverse as well, that is, if sup S = max S ⇒ sup S ∈ S. Assume the opposite, that is, sup S ∉ S. Hence, x < sup S for all x ∈ S. Also, x ≤ max S for all x ∈ S. As max S ∈ S always, Hence, max S < sup S. This contradicts our assumption that sup S = max S. Hence, our assumption was wrong. Hence, sup S ∈ S. Hence, sup S = max S ⇒ sup S ∈ S. So finally, sup S ∈ S ⇔ sup S = max S.
