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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.

Xander Henderson
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phil demeanour
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  • Perhaps the statement holds only for $A, B \subset \mathbb R_+$. – Jimmy R. Mar 23 '15 at 21:40
  • AB is the set of products ab with a from A and b from B. Your counterexample works fine, since you get AB=(1,100) which does not have MN=1 as an upper bound. – coffeemath Mar 23 '15 at 21:45

1 Answers1

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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 $N = \max\{ -\inf(B), \sup(B) \}$.

Xander Henderson
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