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

[alexmahone]

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?

 

[Evgeny.Makarov]

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 $c>\sup A+\sup B-\varepsilon$.

 

[math771]

To get a contradiction, we can use the fact that $\sup A$ and $\sup B$ are the least upper bounds for $A$ and $B$, respectively. This means that for any $\epsilon>0$, there exists $x\in A$ and $y\in B$ such that $\sup A-\epsilon<x$ and $\sup B-\epsilon<y$.

Now, if $\sup C<\sup A+\sup B$, then we can choose $\epsilon$ small enough such that $\sup A-\epsilon<x$ and $\sup B-\epsilon<y$, which would mean that $x+y<\sup A+\sup B$. This contradicts the assumption that $x+y\leq \sup A+\sup B$ for all $x\in A$ and $y\in B$. Therefore, $\sup C\geq \sup A+\sup B$.

On the other hand, if $\sup C>\sup A+\sup B$, then there exists $z\in C$ such that $z>\sup A+\sup B$. Since $z=x+y$ for some $x\in A$ and $y\in B$, we have $x+y>\sup A+\sup B$. This means that either $x>\sup A$ or $y>\sup B$. But this is a contradiction since $\sup A$ and $\sup B$ are the least upper bounds for $A$ and $B$, respectively. Therefore, $\sup C\leq \sup A+\sup B$.

Combining the two inequalities, we get $\sup A+\sup B\leq \sup C\leq \sup A+\sup B$, which implies $\sup C=\sup A+\sup B$. This completes the proof.