Supremum of set of products of elements of two sets

Question

Let $A, B \subseteq \mathbb R$ and $C = \{ab\mid a\in A, b \in B\}$. Show $\sup C \geq \sup A \sup B$.

I don't really get the point here. I understand it intuitively - if $A$ and $B$ contain big negative numbers, the supremum of $C$ gets big since the negative signs disappear but the suprema of $A$ and $B$ may be small if $A$ and $B$ only contain small positive numbers.

If $A = \{-3,-2,0,1\}, B = \{-5,0,1\}$, then $\sup A = 1, \sup B = 1$ and $\sup C = (-3)(-5) = 15$. So in this case, the supremum would be bigger. But I don't know how to prove this formally...

Answer 1

HINT: $$ \forall a\in A\,\forall b\in B\,a\leq\sup A\text{ and }\,b\leq\sup B\text{ and }ab\leq \sup C. $$

Answer 2

Case 1: $\sup A = 0$ or $\sup B = 0$ (wolog $\sup A = 0$).

If $\sup C < \sup A \sup B = 0=\sup A$ then for all $a\in A, b\in B$ then $ab < 0$. So $a \ne 0$ but $a < 0$ so $b > 0$. So for any $b\in B; b> 0$ we can find an $a \in A$ so that $\frac {\sup C}b< a < 0$ so $\sup C < ab < 0$ . But $ab \in C$ so that's a contradiction.

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Case 2: $\sup A < 0; \sup B < 0$.

Trivial: If $a \le \sup A < 0$ and $b \le \sup B < 0$ then $ab \in C$ so $\sup C \ge ab \ge \sup A\sup B$.

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Case 3: $\sup A > 0$ and $\sup B > 0$

If $0 < k < \sup A\sup B$ then $\frac k{\sup B} < \sup A$. So there is a $k' \in A$ so that $\frac k{\sup B} < k' \le \sup A$.

And $\frac k{k'}< \sup B$ so there is $\overline k \in B$ so that $\frac k{k'} <\overline k\le \sup B$. So $k <k\overline k\in C$. So $k$ can't be an upperbound of $C$. So either $\sup A\sup B$ is a least upper bound or not an upper bound at all[*].

So $\sup A \sup B \le \sup C$.

[*] $\sup A\sup B$ need not be an upper bound if we can have negative elements in both sets with absolute values that exceep the suprememums of the set. If $a<0$ and $|a|>\sup A$ and $b< 0$ and $|b| > \sup B$ then $ab >\sup A\sup B$.

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Case 4: one supremum is positive and the other negative (wolog $\sup A < 0$ and $\sup B >0$ )

If $\sup C < \sup A\sup B < 0$ then $\frac {\sup C}{\sup B} < \sup A < 0$ and we can find an $a\in A$ so that $\frac {\sup C}{\sup B} < a$. As $\sup B > 0$ there is a $b \in B$ so that $0 < b \le \sup B$ so $\frac 1b \ge \frac 1 {\sup B}$ and $\frac {\sup C}b \le \frac {\sup C}{\sup B} < a < 0$.

So $\sup C < ab < 0$. But $ab \in C$ and that is a contradiction.

Answer 3

We assume that $\sup A=:\sigma_A$ and $\sup B=:\sigma_B$ are finite. Let an $\epsilon>0$ be given. Since multiplication is continuous at the point $(\sigma_A,\sigma_B)\in{\mathbb R}^2$ there is a $\delta>0$ such that $$|a-\sigma_A|<\delta \ \ \wedge\ \ |b-\sigma_B|<\delta\qquad\Rightarrow \qquad ab>\sigma_A\sigma_B-\epsilon\ .$$ By definition of the supremum we can choose here $a\in A$ and $b\in B$. This implies $$\sigma_C\geq ab>\sigma_A\sigma_B-\epsilon\ ,$$ and as $\epsilon>0$ was arbitrary we can conclude that $\sigma_C\geq\sigma_A\sigma_B$.