How can I show that $\sup(AB)\geq\sup A\sup B$ for $A,B\subset\mathbb{R}$ where $A\cup B$ is positive and bounded? The question is based on the following exercise in real analysis:


  Assume that $A,B\subset{\Bbb R}$ are both bounded and $x&gt;0$ 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 $0&lt;a\leq\sup A$ and $0&lt;b\leq\sup B$ 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 $\epsilon &gt;0$, $\sup AB-\epsilon \geq \sup A\sup B$. If one uses the definition of supremum, one has the estimates that for every $\epsilon&gt;0$,
$$
\sup A-\epsilon\leq a, \quad \sup B-\epsilon\leq b
$$
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?
There's a simple way to prove the result. For all $0 &lt; a\in A$, $0 &lt; b\in B$ we have:
$$\sup(AB)\geq ab\iff\frac{1}{a}\sup(AB)\geq b$$
hence $B$ is bounded above by $\frac{1}{a}\sup(AB)$ so
$$\frac{1}{a}\sup(AB)\geq \sup B\iff \frac{1}{\sup B}\sup(AB)\geq a$$
hence $A$ is bounded above by $\frac{1}{\sup B}\sup(AB)$ so
$$\frac{1}{\sup B}\sup(AB)\geq \sup A\iff \sup(AB)\geq \sup(A)\sup(B)$$
I think the key here is that all x are positive, and the sets are bounded so the required $\mathrm{sup}$'s are finite. Let $a \in A$. Then $a \leq \mathrm{sup}A$. Let $b \in B$. Then $ab \leq  (\mathrm{sup}A)b \leq (\mathrm{sup}A) (\mathrm{sup}B)$, using the fact that $b \leq \mathrm{sup}B$.