Let $A$ and $B$ be nonempty sets of positive real numbers that are bounded above. Also let $AB = \{ab: a \in A, b \in B \}$. Prove that $AB$ is bounded above and $\sup(AB) = (\sup A) (\sup B)$.

So $\sup A$ and $\sup B$ exist by completeness. An upper bound for $AB$ is $(\sup A)(\sup B)$. Let $\alpha = \sup A$ and $\beta = \sup B$. We want to show that if $c$ is an upper bound for $AB$ then $\alpha \beta \leq c$. For $a \in A$, $ab \leq c$ for all $b \in B$. So $c/b$ is an upper bound for $A$. Thus $\alpha \leq c/b$. It follows that $\alpha \beta \leq c$.

Is this correct?