$\sup C = \inf A \inf B$ for $A,B$ bounded sets of negative reals and $C$ products of $A$ and $B$

Question

Given that A and B are bounded sets of negative real numbers, prove that if the set $C = \{ ab | a \in A, b \in B\}$ then $\sup C = \inf A \inf B$. I understand why this is true intuitively but the proof doesn't feel quite right. Here's what I've written so far:

$\forall a \in A : a \ge \inf A$ and $\forall b \in B : b \ge \inf B$, we have $\forall c \in C : \sup C \ge c = ab \ge \inf A \inf B$.

I know that I have to now prove the opposite inequality, i.e. $\sup C \le \inf A \inf B$. I also know that I should probably use the property $\forall \epsilon \gt 0, \exists a_0 \in A : \inf A + \epsilon \gt a_0$. The problem is each time I try to use it I can't figure out how to end the argument. Please help, thanks!

Answer 1

Let $\alpha = \inf A, \beta= \inf B$

$\forall \epsilon_1 >0, \exists a \in A, \alpha<a<\alpha+\epsilon_1. $

$\forall \epsilon_2 >0, \exists b \in B, \beta <b<\beta+\epsilon_2. $

These are due to a corollary of the continuum hypothesis of the reals. Every least upper bound can be approximated arbitrarily closely by an element of the set. Ditto greatest lower bounds.

$\alpha$ and $\beta$ must both be negative since they are smaller than any of the negative numbers in their respective sets.

$\beta\alpha+\beta \epsilon_1< a \beta < \alpha \beta$

$\alpha \beta + \alpha \epsilon_2<\alpha b< \alpha \beta$

$\alpha b + \epsilon_1 b < ab<\alpha b$

$\beta a + \epsilon_2 a < ab < a\beta$

By the transitive property we, now know that $\alpha \beta$ is an upper bound of $C$. Since, e.g., $ab< \alpha b<\alpha \beta$.

$\beta\alpha+ \beta \epsilon_1 + a \epsilon_2< \beta a + \epsilon_2 a< ab< \alpha b<\alpha \beta$

$\beta\alpha+ \beta \epsilon_1 + a \epsilon_2<ab<\alpha \beta$

We now have an equivalent statement for $ab$ to that before mentioned corollary of the continuum hypothesis. Combined, this means $\alpha \beta$ is not just an upper bound, but the least upper bound of $ab$.

Answer 2

As OP mentioned, they actually proved $ab\leq\inf A\inf B$. I'll prove the opposite in this post, by showing that there exist $a\in A, b\in B$ such that $ab\geq\inf A\inf B-\varepsilon$ for any $\varepsilon>0$.

Let $\delta=-\frac{\varepsilon}{\inf A+\inf B}>0$. We know we can find $a, b$ such that $a-\inf A, b-\inf B<\delta$.

Next, we find

\begin{align*} ab&>(\inf A+\delta)(\inf B+\delta) \\ &=\inf A\inf B+(A+B)\delta+\delta^2 \\ &>\inf A\inf B+(A+B)\delta \\ &=\inf A\inf B-\varepsilon \end{align*}

As a result, $\sup C\geq\inf A\inf B$. Combined with what OP has shown, we find $\sup C=\inf A\inf B$.