Show that $\sup (A\cdot B)=\max\{\sup A\cdot\sup B, \sup A\cdot\inf B,\inf A\cdot\sup B,\inf A\cdot\inf B\}$

Question

Given nonempty subsets $A$ and $B$ of positive real numbers, define $$A\cdot B=\{z=x\cdot y:x\in A,\,y\in B \}$$ show that if $A$ and $B$ are bounded sets of real numbers, then $$\sup(A\cdot B)=\max\{\sup A\cdot\sup B, \sup A\cdot\inf B,\inf A\cdot\sup B,\inf A\cdot\inf B\}$$

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For this question, I know that $\sup(A\cdot B)=\sup(A)\cdot\sup(B)$. Do I need to compare all elements in the set then show $\sup(A\cdot B)=\sup(A)\cdot\sup(B)$? Am I on the right way? If not, can anyone give a hit or a suggestion to start? Thanks.

Answer 1

For any $x\in A, y\in B$, there is $$ \inf{A}\leqslant x\leqslant \sup{A} \hspace{4 mm} \text{and} \hspace{4 mm} \inf{B}\leqslant y\leqslant \sup{B} $$ So $$ xy\leqslant\max\{\sup A\cdot\sup B, \sup A\cdot\inf B,\inf A\cdot\sup B,\inf A\cdot\inf B\} $$ And $$ xy\geqslant\min\{\sup A\cdot\sup B, \sup A\cdot\inf B,\inf A\cdot\sup B,\inf A\cdot\inf B\} $$ Thus $$ \sup{A\cdot B}\leqslant\max\{\sup A\cdot\sup B, \sup A\cdot\inf B,\inf A\cdot\sup B,\inf A\cdot\inf B\} $$ And $$ \inf{A\cdot B}\geqslant\min\{\sup A\cdot\sup B, \sup A\cdot\inf B,\inf A\cdot\sup B,\inf A\cdot\inf B\} $$

Answer 2

Let the inf and sup of $A$ be $a_1,a_2$ and of $B$ be $b_1,b_2.$ Now let $R$ denote the rectangle $R=[a_1,a_2]\times [b_1,b_2],$ and we seek the sup of $p=xy$ over $R.$

We first assume that $A,B$ are the entire closed intervals between their respective inf and sup, and say later why this is OK. Note first that the only critical point of $p$ is at the origin. If this critical point happens to lie interior to $R$ then each of $a_2,b_2$ are positive, and since they are sups, there are sequences approaching them, so that the sup we seek is at least $a_2b_2>0,$ which means the sup of $p$ is positive, so the maximum of $p$ on $R$ does not occur at the critical point of $p$.

So the max of $p$ must occur at a boundary point of $R,$ and focusing on any of the four edges of $R$ then shows $p$ cannot have its max at an interior point of an edge of $R.$ This leaves only the four vertices of $R$ as candidates for the max of $p$ over $R.$ So by the usual method of maximizing a two variable differentiable function over a compact domain, the maximum of $p$ over $R$ is the greatest product among $a_1b_1,\ a_1b_2,\ a_2b_1,\ a_2b_2$ which in our notation is the desired supremum of $p.$ Call this amount $M.$

Slight correction: If an edge of $R$ happens to be along the $x$ or $y$ axis, then $p$is constant ($p=0$) on that edge. So above, instead of saying the max of $p$ cannot occur at an interior point of an edge, I should have said more correctly something like "The max of $p$ is taken on at an endpoint of the edge (and perhaps at interior points, if the edge is along an axis)."

Now of course the given sets $A,B$ need not be intervals, but by the above argument the product $p$ taken at any point $(x,y) \in A \times B$ will be at most $M$. We can be sure the supremum is actually equal to $M$ since whichever "corner" of the rectangle is chosen, there are sequences in $A,B$ approaching the coordinates of that corner.