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 $x>0$ 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?
 A: There's a simple way to prove the result. For all $0 < a\in A$, $0 < b\in B$ 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
$$\frac{1}{\sup B}\sup(AB)\geq \sup A\iff \sup(AB)\geq \sup(A)\sup(B)$$
A: 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}
A: 
I was trying to show that for every $ϵ>0$, $\sup AB−ϵ≥\sup A \sup B$.....

In fact you have to prove: $\hspace{3mm}\sup AB + ϵ≥\sup A \sup B$, $\hspace{3mm}\forallϵ>0$ and this is actually what you do so I suppose you are aware that instead $-\epsilon$ you need 
$+\epsilon$
A: What you've done is basically enough to finish. You want to show that for any $\delta > 0$, $\sup A \sup B \le \sup AB + \delta$. Choose $\epsilon$ in the final inequality you have written so that $\epsilon (a+b)+ \epsilon^2 \le \delta$. You'll notice that this is equivalent to a quadratic inequality $\epsilon^2 + (a+b) \epsilon - \delta \le 0$ which is clearly negative for $\epsilon < \frac{-a-b+\sqrt{(a+b)^2-4 \delta}}{2}$, so any positive value less than this will work. If you want a particular choice that will make things easy, try $\epsilon = \frac12 \min \{ \frac\delta{a+b}, \sqrt{\delta} \}$.
A: Correct me if I'm wrong but I think you can take limits from both sides and let $\epsilon \to 0$ because of the following reason:
If $ \forall x\in X: f(x) < g(x)$ and $\displaystyle  \lim_{x \to a} f(x)$ and $\displaystyle \lim_{x \to a} g(x)$ exist then $\displaystyle  \lim_{x \to a} f(x) \leq \lim_{x \to a} g(x)$. 
However, I think there is an easier way of proving this:
Let $\alpha = \sup{A}$ and $\beta = \sup{B}$. Clearly $\alpha\cdot\beta$ is an upper bound for the set $AB$. Because $\forall x \in A, \forall y \in B: 0<x \leq \alpha , 0<y \leq \beta \implies x\cdot y\leq \alpha\cdot\beta$.
Now suppose that $\gamma$ is an arbitrary upper bound for $AB$. Then $\forall x\in A, \forall y \in B: x\cdot y \leq \gamma $
Therefore since $y>0$ we have $\displaystyle x \leq \frac{\gamma}{y}$. If we think of y as being constant, this implies that $\displaystyle \alpha=\sup{A} \leq \frac{\gamma}{y}$.
Therefore $\displaystyle y \leq \frac{\gamma}{\alpha}$ because $\alpha=\sup{A}>0$. This implies that $\displaystyle \beta=\sup{B} \leq \frac{\gamma}{\alpha}$. We conclude that $\alpha\cdot\beta \leq \gamma$. And this means that $\sup{A}\cdot\sup{B} \leq \gamma$. Therefore $\sup{A}\cdot\sup{B}$ is the least upper bound of $AB$ and the proof is completed.
A: Thank you for all the previous answers. I would like to try to continue the approach I wrote in the question. 
I've shown that for every $\epsilon>0$,
$$
\sup A\sup B\leq (a+\epsilon)(b+\epsilon)=ab+\epsilon(a+b)+\epsilon^2\leq \sup AB+\epsilon (a+b)+\epsilon^2
$$
for some $a\in A,\ b\in B$. If I go on with this, then 
$$
\sup A\sup B\leq (a+\epsilon)(b+\epsilon)=\sup AB+\epsilon(2M)+\epsilon^2  
$$
where $M$ is such that $x<M$ for all $x\in A\cup B$. Such $M$ exists by the assumption of boundedness of $A$ and $B$. Since $\epsilon(2M)+\epsilon^2$ can be arbitrarily "small", which is not hard to show, we are done.
A: Sorry to reopen such an old question. But I think I have an alternative solution. 
We begin first by stating that $ \sup A \ge a \gt 0$ and $\sup B \ge b \gt 0$ for any $a \in A$ and $b \in B$. 
Suppose to the contrary that $\sup(AB)\lt \sup A\sup B$. Then $\dfrac {\sup(AB)}{\sup B} \lt \sup A$. This implies that there is $a \in A$ such that $\dfrac {\sup(AB)}{\sup B} \lt a \implies \dfrac{\sup(AB)}{a} \lt \sup B$ which implies there is $b \in B$ such that $\dfrac{\sup(AB)}{a} \lt b \implies \sup(AB) \lt ab$ for some $a\in A$ and $b\in B$ leading to a contradiction.  
A: 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 $b \leq \mathrm{sup}B$.
