$\sup(A+B) = \sup(A) + \sup(B)$

Question

This question has been asked before but if someone could check my proof - it appears different then other answers I have read - I would appreciate it. Thanks in advance.

Suppose $A+B = \lbrace a + b : a \in A, b\in B \rbrace$. And suppose $\sup(A), \sup(B)$ exist. Prove $\sup(A+B)=\sup(A) + \sup(B)$.

To keep things brief suppose I have already shown that $\sup(A) + \sup(B)$ is an upper bound of $A+B$. Now suppose $\exists$ an upper bound $y$ of $A+B$ such that $y < \sup(A) + \sup(B)$. Hence $y - \sup(A) < \sup(B) \implies\exists b \in B$ such that $y - \sup(A) < b \leq \sup(B) \implies y - b < \sup(A) \implies\exists a \in A$ such that $y - b < a \leq \sup(A) \implies y < a+b \in A+B \implies y$ is not an upper bound of $A+B$. Hence $\sup(A)+\sup(B) = \sup(A+B)$.

Answer 1

While your proof is fine, this is one of the instances where a proof by contradiction I think is more confusing than the more direct proof. In particular, a direct proof here would say that for any $\epsilon > 0$, there exists an $a \in A$ and $b \in B$ such that $a > \sup(A) - \epsilon/2$ and $b > \sup(B) - \epsilon/2$. Then, there is an element in $A + B$ that is arbitrarily close to $\sup(A) + \sup(B)$, which shows that $\sup(A + B) \geq \sup(A) + \sup(B)$.

Answer 2

Yes, the proof looks good. As far as presentation goes, I would probably add the following line at the end:

$y < a + b \implies y$ is not an upper bound of $A+B$. This is a contradiction. Hence for all $y$, if $y$ is an upper bound of $A+B$, we must have $\sup A + \sup B \le y$. It follows that $\sup A + \sup B = \sup(A+B)$.