# Supremum of Sum equals Sum of Suprema

## Theorem

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

Let $A + B$ be $\set {x + y: x \in A, y \in B}$.

Let either $A$ and $B$ have suprema or $A + B$ have a supremum.

Then all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist and:

- $\sup \paren {A + B} = \sup A + \sup B$

## Proof

Assume first that $A$ and $B$ have suprema.

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 $A + B$ as $x$ and $y$ are arbitrary elements of $A$ and $B$ respectively.

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

$A + B$ is non-empty as $A$ and $B$ are non-empty.

Accordingly, $A + B$ has a supremum by the Continuum Property.

Next, assume that $\sup \paren {A + B}$ has a supremum.

We need to prove that $A$ and $B$ have suprema.

Let $y$ be a point in $B$.

We have:

\(\ds x + y\) | \(\le\) | \(\ds \sup \paren {A + B}\) | for every $x$ in $A$ as $x + y$ is a point in $A + B$ | |||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\le\) | \(\ds \sup \paren {A + B} - y\) |

Therefore, $A$ has an upper bound as $x$ is an arbitrary point in $A$.

Also, we have that $A$ is non-empty.

Accordingly, $A$ has a supremum by the Continuum Property.

A similar argument gives that $B$ has a supremum.

So, we have shown that all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist.

We proceed to show that $\sup \paren {A + B} = \sup A + \sup B$.

We have $\sup \paren {A + B} \le \sup A + \sup B$ as $\sup A + \sup B$ is an upper bound for $A + B$.

Accordingly, either:

- $\sup \paren {A + B} < \sup A + \sup B$

or:

- $\sup \paren {A + B} = \sup A + \sup B$.

Aiming for a contradiction, suppose:

- $\sup \paren {A + B} < \sup A + \sup B$.

Let $\epsilon = \sup A + \sup B - \sup \paren {A + B}$.

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:

\(\ds x + y\) | \(>\) | \(\ds \sup A - \frac \epsilon 2 + \sup B - \frac \epsilon 2\) | ||||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x + y\) | \(>\) | \(\ds \sup A + \sup B - \epsilon\) | |||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x + y\) | \(>\) | \(\ds \sup A + \sup B - \paren {\sup A + \sup B - \sup \paren {A + B} }\) | definition of $\epsilon$ | ||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x + y\) | \(>\) | \(\ds \sup \paren {A + B}\) |

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

We have found that:

- $\sup \paren {A + B} < \sup A + \sup B$ is not true.

Therefore:

- $\sup \paren {A + B} = \sup A + \sup B$ as $\sup \paren {A + B} \le \sup A + \sup B$.

$\blacksquare$