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

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!

• Writing $C = \{ (-a)(-b) \mid -a \in -A, -b \in -B\}$ might help since then you need only deal with positive values Oct 16, 2023 at 14:52
• @George but aren't the values in $C$ positive already? Oct 16, 2023 at 14:56
• I suppose it feels like the equivalent statement $\sup C = \sup(-A) \sup(-B)$ might be easier to work with intuitively Oct 16, 2023 at 14:57
• Your sign seems to be the wrong way around - remember, multiplication by a negative number flips the signs of the inequality. Oct 16, 2023 at 15:01
• @Kyky So it becomes $\sup C \ge c = ab \le \inf A \inf B$. What do I do then? Oct 16, 2023 at 15:04

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

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

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

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

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 $$ab$$.

• This solution seems more intuitive to me than the other one proposed. Thank you!! Oct 17, 2023 at 11:25

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 $$\sup C=\inf A\inf B$$.

• This might be really dumb but I don't understand how knowing that $\sup C \ge c = ab \le \inf A \inf B$ means that $\sup C \le \inf A \inf B$. Can you explain please? Oct 16, 2023 at 15:25
• We have shown $\inf A\inf B$ is an upper bound of $C$, and $\sup C$ is the least upper bound of $C$, so it follows that $\sup C\geq \inf A\inf B$. Oct 16, 2023 at 15:29
• You mean $\sup C \le \inf A \inf B$? Oct 16, 2023 at 15:31
• Yep, that was a typo. Sorry for the confusion. Oct 16, 2023 at 15:35
• I actually got to a point similar to what you've written earlier. It went something like $\inf A \inf B + \inf A \epsilon_2 + \inf B \epsilon_1 + \epsilon_1 \epsilon_2$. I wasn't sure if I can just discard the terms that have $\epsilon$ and if they can be considered an arbitrary quantity since both $\inf A$ and $\inf B$ are constants. My question is, can I do that or not? Oct 16, 2023 at 15:38