Prove or Disprove: the product of the upper bounds is an upper bound of the Minkowski product Prove or find a counterexample to the following statement:

If $A$, $B \subseteq \mathbb{R}$ are nonempty, $M$ is an upper bound for $A$ and $N$ is an upper bound for $B$, then $MN$ is an upper bound for
  $AB := \{ab \mid a \in A, b \in B\}$

Now I've gone through my notes and had a look at my textbook and I figure  that a  counterexample for is easily found if you take both $A$ and $B$ to be subsets of the negative numbers something along the lines of: let $A = B = (-10, -1)$ and $M = N = -1$. If anybody maybe could explain how to go about this question would be great. I think I'm getting stuck on the notation namely is $AB$ the Cartesian product I'm not too sure.
 A: So that this question has an answer:
The definition of the set $AB$ is as in the question, i.e.
$$ AB := \{ ab \mid a\in A, b\in B\}. $$
Here $ab$ is simply multiplication in $\mathbb{R}$.  This set is not the Cartesian product, with the the collection of all ordered pairs $(a,b)$ with $a\in A$ and $b\in B$.  I have heard this set referred to as the Minkowski product of two sets (in analogy to the Minkowski sum or difference of two sets, but I am not entirely certain how standard that term is.  In any event, the set $AB$ consists of all products of numbers $a\in A$ and $b \in B$.  For example if $A = \{ -1, 0, 1\}$ and $B = \{ 1, 2, 3\}$, then
\begin{align}
AB
&= \{ -1(1), -1(2), -1(3), 0(1), 0(2), 0(3), 1(1), 1(2), 1(3) \} \\
&= \{ -3, -2, -1, 0, 1, 2, 3 \}.
\end{align}
Your counterexample looks good (coffeemath gives the argument in a comment), though I would like to suggest what I think is an even easier counterexample:  take $A = B = \{ -1 \}$.  Note that we may take $M = N = 0$ to be an upper bound of both $A$ and $B$ (we are not asked to take $M$ and $N$ to be least upper bounds, so why make life hard for ourselves?).  Then
$$ AB = \{ 1 \}
\qquad\text{and}\qquad
MN = 0. $$
But $0$ not an upper bound of the set $\{1\}$, so we have a counter-example.
As Jimmy R. points out, you might want to consider what hypotheses will make this statement true.  The suggestion that $A,B \subseteq (0,\infty)$ seems reasonable.  You could probably extend that to all of $\mathbb{R}$ by considering $M = \max\{ -\inf(A), \sup(A) \}$ and $N = \max\{ -\inf(B), \sup(B) \}$.
