The question is based on the following exercise in real analysis:
Assume that $A,B\subset{\Bbb R}$ are both bounded and $x>0$ for all $x\in A\cup B$. Show that $$\sup(AB)=\sup A\sup B$$ where $$AB:=\{ab\in{\Bbb R}:a\in A, b\in B\}.$$
Since $0<a\leq\sup A$ and $0<b\leq\sup B$ for all $a\in A$ and $b\in B$, we have$$ab\leq\sup A\sup B$$for all $ab\in AB$ which implies that $\sup AB\leq\sup A\sup B$. I have trouble with another direction:$$\sup AB\geq\sup A\sup B$$I was trying to show that for every $\epsilon >0$, $\sup AB-\epsilon \geq \sup A\sup B$. If one uses the definition of supremum, one has the estimates that for every $\epsilon>0$,$$\sup A-\epsilon\leq a, \quad \sup B-\epsilon\leq b$$for some $a\in A,\ b\in B$. It follows that$$\sup A\sup B\leq (a+\epsilon)(b+\epsilon)=ab+\epsilon(a+b)+\epsilon^2\leq \sup AB+\epsilon (a+b)+\epsilon^2$$which seems quite close to what I want. How can I go on?