Supremum of Union of Bounded Above Sets of Real Numbers

Theorem
Let $A$ and $B$ be sets of real numbers.

Let $A$ and $B$ both be bounded above.

Then:
 * $\map \sup {A \cup B} = \max \set {\sup A, \sup B}$

where $\sup$ denotes the supremum.

Proof
Let $A$ and $B$ both be bounded above.

By the Continuum Property, $A$ and $B$ both admit a supremum.

Let $x \in A \cup B$.

Then either $x \le \sup A$ or $x \le \sup B$ by definition of supremum.

Hence:
 * $x \le \max \set {\sup A, \sup B}$

and so $\max \set {\sup A, \sup B}$ is certainly an upper bound of $A \cup B$.

It remains to be shown that $\max \set {\sup A, \sup B}$ is the smallest upper bound.

there exists $m \in \R$ such that:
 * $m < \max \set {\sup A, \sup B}$

and:
 * $\forall x \in A \cup B: x \le m$

, let $\sup A \ge \sup B$.

Then:
 * $\max \set {\sup A, \sup B} = \sup A$

and so:
 * $m < \sup A$

But then by definition of supremum:


 * $\exists a \in A: a > m$

and so $m$ is not an upper bound of $A \cup B$.

This contradicts our assumption that $m$ is a supremum of $A \cup B$.

It follows by Proof by Contradiction that $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.

The same argument shows, mutatis mutandis, that if $\sup A \le \sup B$, $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.