Supremum of the product of sets

Question

Let $A, B$ be subsets of positive real numbers that are bounded above, and let $A\cdot B=\{ a b : a\in A, b\in B\}$.
Show that $$ \sup (A\cdot B) = \sup A \sup B. $$

Proof:

This is obvious. It is clear that for the set $A \cdot B$, $\sup A \sup B$ is an upper bound, as $\alpha\beta \geq ab$, $\forall ab \in A \cdot B$. Now we just need show that $\alpha \beta$ is the smallest such bound. Let $\epsilon_1 > 0$, Let $\epsilon_2 = \alpha \epsilon_1 + \beta \epsilon_1 + \epsilon_1^2$. It's clear that $a > \alpha - \epsilon_1$ and $b> \beta-\epsilon_1$ for some $a \in A$ and some $b \in B$ Thus $(\alpha-\epsilon_1)(\beta-\epsilon_1) = \alpha\beta - \epsilon_2 < ab$ for some $ab \in A \cdot B$. Ergo, $\alpha \beta$ is the smallest such upper bound and thus $$ \sup (A\cdot B) = \sup A \sup B $$

Is this a valid proof? I've been debating this with a friend of mine for the past half and hour..

Answer 1

$\def\epsilon{\varepsilon}$I find the logic rather confusing in your proof. My problem is that you need to show that $A\cdot B$ contains an element greater than $\alpha\beta-\epsilon$ for every $\epsilon>0$. That is, given $\epsilon>0$, you need to find $a\in A$ and $b\in B$ in terms of $\epsilon,\alpha$ and $\beta$, such that $$ab>\alpha\beta-\epsilon\ .$$ You say "it's clear that $a>\alpha-\epsilon_1$", but it's not at all clear to me, since you have not said what $a$ is!

You could try something like this. Let $\epsilon>0$. Since $\alpha,\beta$ are the suprema of $A,B$ respectively, there exist $a\in A$ and $b\in B$ such that $$a>\alpha-\frac{\epsilon}{2\beta}\ ,\quad b>\beta-\frac{\epsilon}{2\alpha}\ ;$$ Then $$ab>\alpha\beta-\frac\epsilon2-\frac\epsilon2+\frac{\epsilon^2}{4\alpha\beta} >\alpha\beta-\epsilon\ .$$ Note that if $\epsilon$ is large, these inequalities could involve negative numbers and may no longer be correct. This is "not a problem" since we are really interested only in small $\epsilon$; but if you want a really watertight proof you would have to write it down more carefully.

Answer 2

Your idea is correct. But I think you'd better state the relation of $\epsilon_2$ and $\epsilon_1$ because you want to show $\forall \epsilon>0, \exists ab>\alpha\beta-\epsilon$, but you have proved $\forall \epsilon_1>0, \exists ab>\alpha\beta-\epsilon_2$ . Does it guarantee $ \epsilon_2<\epsilon_1$? Moreover, once you get $a>c,b>d$, you need to make sure $c,d>0$ before getting $ab>cd$.

You can improve it like this.

Let $\alpha=\sup A, \beta=\sup B>0, \epsilon_1 > 0$, Let $\epsilon_2 = \min(\frac{\alpha}{2},\frac{\beta}{2},\frac{\epsilon_1}{\alpha+\beta+1})>0$.

By definition of $\sup$, $\alpha - \epsilon_2$ is not an upper bounded of $A$, hence $\exists a > \alpha - \epsilon_2\bf{>0}$ for some $a\in A$. Similarly $\exists b> \beta-\epsilon_2\bf{>0}$ for some $b\in B$. Thus $$ab>(\alpha-\epsilon_2)(\beta-\epsilon_2) = \alpha\beta - (\alpha+\beta)\epsilon_2+\epsilon_2^2 \\>\alpha\beta - (\alpha+\beta)\epsilon_2>\alpha\beta-\epsilon_1$$

Answer 3

For any $\epsilon > 0$, and $\forall m, n \in \mathbb{N}$, $\exists a_m \in A, \exists b_n \in B$ such that:

$\alpha - \dfrac{1}{m} < a_m$, and $\beta - \dfrac{1}{n} < a_n$. Now choose $m, n$ such that: $n > \dfrac{3\alpha}{\epsilon},m > \dfrac{3\beta}{\epsilon}, mn > \dfrac{3}{\epsilon}$. Then:

$\alpha\cdot \beta - a_mb_n < \left(a_m+\dfrac{1}{m}\right)\left(b_n+\dfrac{1}{n}\right) -a_mb_n = \dfrac{a_m}{n} + \dfrac{b_n}{m} + \dfrac{1}{mn} < \dfrac{\alpha}{n} + \dfrac{\beta}{m} + \dfrac{\epsilon}{3} < \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} = \epsilon$. This shows $\text{Sup(AB)} = \text{SupA}\cdot \text{SupB}$