Let $A$ and $B$ be two non-empty bounded subsets of $\mathbb{R}$. Write $A + B = \{x + y : x \in A, y \in B\}$. Show that $\sup(A + B) = \sup A + \sup B$.
This is a problem from an assignment from my analysis course. the two definitions I know of supremum is that it is the least upper bound and also for all $\epsilon >0 \sup(S)-\epsilon < a$ for some $a$ belonging to the set $S$.
I am unable to prove this using the above definitions (though the second one is actually a theorem that can be proven from the previous definition). I have already tried writing the definition of supremum and saying that for all $\epsilon >0$, there exists $a+b$ belonging to $A+B$ such that $\sup(A+B)<a+b$ But I am unable to proceed further.
Can anyone please provide a right approach to solve this problem and what lemmas must be proven.
You wrote in the comments that you see why $supA+supB$ is an upper bound of the set $A+B$. Now you just have to prove it is the least upper bound. Alright, so fix $\epsilon>0$. Then you know there exists $a\in A$ such that $a>supA-\frac{\epsilon}{2}$ and there exists $b\in B$ such that $b>supB-\frac{\epsilon}{2}$. So then $a+b\in A+B$ and $a+b>supA-\frac{\epsilon}{2}+supB-\frac{\epsilon}{2}=supA+supB-\epsilon$. Hence anything smaller than $supA+supB$ is not an upper bound of $A+B$.