Problem
Assume $A \subset \{x|x \geq 0, x \in \mathbb{R}\}$ and $B \subset \{y|y \geq 0, y \in \mathbb{R}\}$. $A,B$ are both nonempty and bounded. Define $AB=\{xy|x \in A,y \in B\}.$ Prove $\sup AB=\sup A \sup B.$
Proof
If $\sup A\sup B=0$, then $A=\{0\}$ or $B=\{0\}$. The statement holds trivially. Now, we assume $\sup A\sup B \neq 0$, namely, $\sup A,\sup B>0$.
$\forall x \in A, y \in B：0 \leq x \leq \sup A, 0 \leq y \leq \sup B$. Then $xy \leq \sup A \sup B,$ which shows $\sup A \sup B$ is a upper bound for all $xy.$ Thus $$\sup AB \leq \sup A \sup B.\tag1$$ $\forall \varepsilon \in (0,\min\{\sup A,\sup B\}), \exists x_0 \in A, y_0 \in B:$ $$x_0>\sup A-\varepsilon>0;$$ $$y_0>\sup B-\varepsilon>0.$$ Hence \begin{align*} x_0y_0&>\sup A\sup B-(\sup A+\sup B)\varepsilon+\varepsilon^2\\&>\sup A\sup B-(\sup A+\sup B)\varepsilon\\&=\sup A\sup B-\varepsilon',\tag2 \end{align*} where $\varepsilon'=(\sup A+\sup B)\varepsilon.$ From $(1)(2)$, we may claim $\sup AB=\sup A \sup B.$