Supremum of Sum equals Sum of Suprema

Theorem
Let $A$ and $B$ be non-empty real sets.

Let $A$ and $B$ have supremums.

Let $A\text{+}B$ be $\left\{{x + y: x \in A, y \in B}\right\}$.

Then $\sup \left({A\text{+}B}\right)$ exists and:


 * $\sup \left({A\text{+}B}\right) = \sup A + \sup B$.

Proof
We have:


 * $x \le \sup A$ for an arbitrary $x$ in $A$


 * $y \le \sup B$ for an arbitrary $y$ in $B$

Adding these inequalities, we get:


 * $x + y \le \sup A + \sup B$

The number $x + y$ is an arbitrary element of $A\text{+}B$ as $x$ and $y$ are arbitrary elements of $A$ and $B$ respectively.

Therefore, $\sup A + \sup B$ is an upper bound for $A\text{+}B$.

$A\text{+}B$ is non-empty as $A$ and $B$ are non-empty.

Accordingly, $A\text{+}B$ has a supremum by the Continuum Property.

We have $\sup \left({A\text{+}B}\right) \le \sup A + \sup B$ as $\sup A + \sup B$ is an upper bound for $A\text{+}B$.

Accordingly, either:


 * $\sup \left({A\text{+}B}\right) < \sup A + \sup B$

or:


 * $\sup \left({A\text{+}B}\right) = \sup A + \sup B$.

Suppose that:


 * $\sup \left({A\text{+}B}\right) < \sup A + \sup B$.

Let $\epsilon = \sup A + \sup B - \sup \left({A\text{+}B}\right)$.

We note that $\epsilon > 0$.

Since $\sup A$ is the least upper bound of $A$, there is an element $x$ in $A$ such that:


 * $x > \sup A - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close

Since $\sup B$ is the least upper bound of $B$, there is an element $y$ in $B$ such that:


 * $y > \sup B - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close

Adding these inequalities, we get:

which is impossible since the number $x + y$ is an element of $A\text{+}B$ as $x \in A$ and $y \in B$.

We have found that:


 * $\sup \left({A\text{+}B}\right) < \sup A + \sup B$ is not true.

Therefore,
 * $\sup \left({A\text{+}B}\right) = \sup A + \sup B$ as $\sup \left({A\text{+}B}\right) \le \sup A + \sup B$.