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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!

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

2 Answers 2

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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 $ab$.

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  • $\begingroup$ This solution seems more intuitive to me than the other one proposed. Thank you!! $\endgroup$
    – Marin
    Oct 17, 2023 at 11:25
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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 $\sup C=\inf A\inf B$.

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  • $\begingroup$ 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? $\endgroup$
    – Marin
    Oct 16, 2023 at 15:25
  • $\begingroup$ 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 $\sup C\geq \inf A\inf B$. $\endgroup$ Oct 16, 2023 at 15:29
  • $\begingroup$ You mean $\sup C \le \inf A \inf B$? $\endgroup$
    – Marin
    Oct 16, 2023 at 15:31
  • $\begingroup$ Yep, that was a typo. Sorry for the confusion. $\endgroup$ Oct 16, 2023 at 15:35
  • $\begingroup$ 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? $\endgroup$
    – Marin
    Oct 16, 2023 at 15:38

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