Union of Bounded Above Real Subsets is Bounded Above

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

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

Then $A \cup B$ is also bounded above.

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

Then by definition $A$ and $B$ both have an upper bound $U_A$ and $U_B$ respectively.

Suppose $U_A \le U_B$.

Then:
 * $\forall a \in A: a \le U_B$

and also, by definition:
 * $\forall b \in B: b \le U_B$

and so $U_B$ is an upper bound for $A$.

Otherwise, suppose $U_A > U_B$.

Then:
 * $\forall b \in B: b \le U_A$

and also, by definition:
 * $\forall a \in A: a \le U_A$

Let $x \in A \cup B$.

Then from the above, either $x \le U_A$ or $x \le U_B$.

So either $U_A$ or $U_B$ is an upper bound for $A \cup B$.

Hence, by definition, $A \cup B$ is bounded above by either $U_A$ or $U_B$.