Definitions of bounds: this.
Proof
Let \(A, B \subseteq \mathbb{R}\) be nonempty and bounded above. By the Axiom of Completeness, \(\sup A\) and \(\sup B\) both exist. Since \(\sup A\) and \(\sup B\) both exist, then the set \(\{\sup A, \sup B\}\) is nonempty and bounded above. Therefore, \(\sup (A \cup B)\) exists by the Axiom of Completeness.
We claim that that \(\sup (A \cup B) = \max (\{\sup A, \sup B\})\). [Why? because the maximum of the set is the supremum of the set (proof)] Without the loss of generality, suppose that \(\sup A \geq \sup B\). If that’s the case, then \(\sup (\{\sup A, \sup B\}) = \sup A\). for this is straight forward). So now the goal is to prove that \(\sup (A \cup B) = \sup A\). To show this, we will verify the two conditions of the least upper bound definition.
First, we want to show that \(\sup A\) is an upper bound on \(A \cup B\). We know that \(a \leq \sup A\) for any \(a \in A\). We also know that \(b \leq \sup B\) for any \(b \in B\). But \(\sup B < \sup A\), so \(b \leq \sup A\). Therefore, \(\sup A\) is an upper bound on \(A \cup B\).
Next, we want to show that \(\sup A\) is the least upper bound of \(A \cup B\). So suppose it wasn’t and suppose that \(u = \sup(A \cup B)\) be. Then, \(u \leq \sup A\) by the definition of the least upper bound. But \(\sup A\) is the least upper bound for \(A\). So there must exist an element \(a \in A\) such that \(a > u\). This is a contradiction since we said that \(u\) is the least upper bound on \(A \cup B\). Therefore, \(\sup A\) must be the least upper bound on \(A \cup B\). \(\blacksquare\)