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 $\sup \paren {A + B} \le \sup A + \sup B$.
$\blacksquare$