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

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SUMMARY

The supremum of the sum of two sets of real numbers, denoted as $\sup C$, is equal to the sum of their individual suprema, $\sup A + \sup B$. The reasoning follows from the definitions of supremum and upper bounds. For any $x \in A$ and $y \in B$, it holds that $x + y \leq \sup A + \sup B$, establishing that $\sup A + \sup B$ is an upper bound for the set $C = \{x + y : x \in A, y \in B\}$. To prove that $\sup C$ cannot be less than $\sup A + \sup B$, one must show that for any $\varepsilon > 0$, there exists an element in $C$ that exceeds $\sup A + \sup B - \varepsilon$.

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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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