MHB Is the Supremum of the Sum of Two Sets Equal to the Sum of their Suprema?

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The discussion centers on determining the relationship between the supremum of the sum of two sets of real numbers, A and B, and the sum of their individual suprema. It is proposed that the supremum of the set C, defined as C = {x+y: x ∈ A, y ∈ B}, equals the sum of the suprema of A and B, expressed as sup C = sup A + sup B. The argument establishes that sup A + sup B serves as an upper bound for C. To prove the equality, it is suggested to show that for any ε > 0, there exists an element in C that exceeds sup A + sup B - ε, thus contradicting the assumption that sup A + sup B - ε is an upper bound for C. The discussion emphasizes the need for a constructive approach to find such an element in C.
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Let $A$ and $B$ be sets of real numbers and write $$C=\{x+y:x\in A,y\in B\}.$$ Find a relation among $\sup A$, $\sup B$, and $\sup C$.

My attempt:

I'm assuming the answer is $\sup C=\sup A+\sup B$.

$x\le \sup A\ \forall x\in A$
$y\le \sup B\ \forall y\in B$
$\implies x+y\le \sup A+\sup B$ $\forall x\in A,\ y\in B$

So, $\sup A+\sup B$ is an upper bound for $C$.

Suppose $\sup C\neq\sup A+\sup B$.

$\implies \exists l<\sup A+\sup B$ such that $x+y\le l$ $\forall x\in A,\ y\in B$

$\implies x\le l-y\ \forall x\in A,\ y\in B$

So, $l-y$ is an upper bound of $A\ \forall y\in B$

I feel that I'm on the right track but I don't know how to get a contradiction. Any suggestions?
 
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Prove that $\sup A+\sup B-\varepsilon$ is not an upper bound of $C$ for any $\varepsilon>0$. For this find a $c\in C$ such that $c>\sup A+\sup B-\varepsilon$.
 
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