TITLE: 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?
QUESTION [13 upvotes]: 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?

REPLY [3 votes]: This is actually the same as proving that

for every pair of nonempty upper bounded sets of real numbers $U$ and $V$ we have $$\sup(U+V)=\sup U+\sup V$$

which is an easier exercise.
For an upper bounded (nonempty) set of positive numbers $X$, consider $X'=\{\log x:x\in X\}$ (natural logarithm). Then, due to the fact that logarithm and exponential are continuous, increasing and inverse of one another,
$$
\sup X=\exp(\sup X')
$$
Since $(AB)'=A'+B'$, we also have
\begin{align}
\sup(AB)&=\exp(\sup((AB)'))\\[4px]
&=\exp(\sup(A'+B'))\\[4px]
&=\exp(\sup A'+\sup B')\\[4px]
&=\exp(\sup A')\exp(\sup B')\\[4px]
&=(\sup A)(\sup B)
\end{align}