Suppose \(A, B \subseteq \mathbf{R}\) are nonempty and bounded above. Find a formula for \(\sup (A \cup B)\) and prove that it is correct.


For the definitions of an upper bound and the least upper bound of a set see this.

Proof:

Let \(A, B \subseteq \mathbf{R}\) be nonempty and bounded above. By the Axiom of Completeness, \(\sup A\) and \(\sup B\) both exist. We claim that that \(\sup (A \cup B) = \sup (\{\sup A, \sup B\})\). Since \(\sup A\) and \(\sup B\) both exist, then the set \(\{\sup A, \sup B\}\) is nonempty and bounded above. Therefore, \(\sup (A \cup B)\) also exists by the Axiom of Completeness.

Without the loss of generality, suppose that \(\sup A \geq \sup B\). If that’s the case, then \(\sup (\{\sup A, \sup B\}) = \sup A\). (Why? because the maximum of the set is the supremum of the set. The proof for this is straight forward). So now the goal is to prove that \(\sup (A \cup B) = \sup A\). To do this, we will prove that

  • \(\sup (A \cup B) \geq \sup A\)
  • \(\sup (A \cup B) \leq \sup A\)

For the first inequality, we want to prove that \(\sup (A \cup B)\) is an upper bound on the set \(A\). First, \(\sup(A \cup B)\) is an upper bound on the set \(A \cup B\) (by the definition of the least upper bound). But we know that \(A \subseteq A \cup B\). Therefore, \(\sup(A \cup B)\) is an upper bound on the set \(A\) as well. Second, we know that \(\sup A\) must be less than or equal to any other upper bound by the definition of the least upper bound. Therefore \(\sup A \leq \sup(A \cup B)\).

For the second inequality, suppose for the sake of contradiction that it is not the case that \(\sup (A \cup B) \leq \sup A\). So \(\sup (A \cup B) > \sup A\). This means that \(\sup A\) is not an upper bound on the set \(A \cup B\). Therefore, there must exists some element \(b \in B\) such that \(b > \sup A\). But we know by the definition of the least upper bound that \(\sup B \geq b\). So now we have,

$$ \begin{align*} \sup A < b \leq \sup B \end{align*} $$

This is a contradiction since we assumed that \(\sup A > \sup B\) and therefore we must have \(\sup (A \cup B) \leq \sup A\) as required. \(\blacksquare\)

References: