Prove that $\sup(AB) = \inf{A} \cdot \inf{B}$ 
If $A,B \subseteq (-\infty,0]$, prove that $\sup(AB) = \inf{A} \cdot \inf{B}$.

For this question you can assume that if $A,B \subseteq [0,\infty)$ then $\sup(AB) = \sup{A} \sup{B}$ as that was an earlier question I did and that $\inf(AB) = \inf{A} \inf{B}$. Thus, it suffices to prove that $\sup(-A\cdot -B) = \inf{-A} \cdot \inf{-B}$ and the result will follow.
 A: For all $a \in A$ and $b \in B$ we have $$\inf A \le a \;\;\; \text{ and } \;\;\; \inf B \le b.$$Since these are non-positive numbers we have $$ab \le \inf A \inf B$$ which gives us $$\sup(AB) \le \inf A \inf B.$$
For all $a \in A$ and $b \in B$ we have (for $0 \not \in B$ and $\inf A \ne 0$ ---- These cases are trivial.) $$ab \le \sup(AB)$$ $$\implies a \ge \sup(AB)b^{-1}$$ $$\implies \inf A \ge \sup(AB)b^{-1}$$ $$\implies b\inf A \le \sup(AB)$$ $$\implies b \ge \sup (AB)(\inf A)^{-1}$$ $$\implies \inf B \ge \sup (AB)(\inf A)^{-1}$$ $$\implies \inf A \inf B \le \sup(AB).$$
A: Let $C = -A = \{-x|x \in A\}; D = -B = \{-x| x \in B\}$.
Then $AB = \{xy|x \in A, y \in B\} = \{(-x)(-y)|x \in A, y \in B\}=\{(-x)(-y)|-x \in C, -y \in D\} = CD$.
$\inf A \le a; \forall a \in A$ so $-inf A \ge -a; \forall a \in A$ so $-\inf A \ge -a; \forall -a \in C$.  So $-\inf A$ is an upper bound of $C$. 
If $c < -\inf A$ then $-c > \inf A$ so $-c$ is not a lowever bound of A so there is a $d \in A$ so that $-c > d > \inf A$.  So $c < -d < -\inf A$.  But $-d \in C$ so $c$ is not an upper bound of $C$.  So $-\inf A = \sup C$.
Likewise $-\inf B = \sup C$
Now you have already proven $\sup CD = \sup C \sup D$.
So $\sup AB = \sup CD = \sup C \sup D = -\inf A -\inf B = \inf A \inf B$
