I want to prove the following result:
Theorem $\quad $ If $A$ and $B$ are bounded sets of real numbers, then
\begin{gather*}\tag{$\star$}
\sup(A\cdot B)=\max\{\sup A\cdot\sup B, \sup A\cdot\inf B, \inf A\cdot\sup B, \inf A\cdot\inf B\}.
\end{gather*}
Although in the thread Show that sup(A⋅B)=max{supA⋅supB,supA⋅infB,infA⋅supB,infA⋅infB}, some suggestions are given, but I have trouble in understanding them. So
I have tried to prove as follows. Please check whether it is right or wrong. Note that $AB$ is defined by $$AB=\{z\in\mathbb{R}\mid \exists x\in A, y\in B: z=xy\}.$$
In order to prove this theorem, we need the following lemma.
Lemma $\quad $
Let $A$ and $B$ be nonempty sets of nonnegative real numbers. Suppose that $A$ and $B$ are bounded above. Then
\begin{gather*}
\sup AB=\sup A\cdot\sup B.
\end{gather*}
Proof.
Let $A\subset [0,+\infty)$ and $B\subset [0,+\infty),$ and $A,B$ nonempty, bounded above. Put $a=\sup A, b=\sup B, c=\sup AB.$ Since $A\subset [0,+\infty)$ and $B\subset [0,+\infty),$ we see that $0$ is a lower bound of both $A$ and $B,$ so $a, b, c$ are nonnegative. Let $z\in AB.$ Then there are $x\in A, y\in B$ such that $z=xy.$ Since $0\leq x\leq a$ and $0\leq y\leq b,$ we have $xy\leq ab.$ So $z\leq ab.$ By arbitrariness of $z,$ we deduce that $ab$ is an upper bound of $AB.$ Hence $c\leq ab.$ Note that $c\geq0.$ If $c<ab,$ then we have $a>0$ and $b>0.$ It follows that $\frac{c}{b}<a=\sup A.$ Hence there exists $x_1\in A$ such that $\frac{c}{b}<x_1.$ So $x_1>0.$ Then we deduce that $\frac{c}{x_1}<b=\sup B.$ So there exists $y_1\in B$ such that $\frac{c}{x_1}<y_1.$ From $x_1>0$ it follows that $c<x_1y_1.$ But $x_1y_1\in AB,$ so we have $\sup AB=c<x_1y_1\in AB,$ which is absurd. Hence we have $c\geq ab.$ And thus we conclude that $\sup AB=c=ab=\sup A\cdot \sup B.$ $\qquad\Box$
Proof of the theorem.
let $A$ and $B$ be bounded sets of real numbers. Put
\begin{gather*}
a=\sup A, \quad a'=\inf A,\quad b=\sup B,\quad b'=\inf B, \quad c=\sup AB.
\end{gather*}
We shall prove that $c=\max\{ab,a'b,ab',a'b'\}.$ And we plan to prove this statement by cases.
(i) Case $a'\geq 0, b'\geq 0.$ Thus, for $x\in A$ and $y\in B,$ we have $a\geq x\geq a'\geq 0, b\geq y\geq b'\geq 0,$ so $A\subset [0,+\infty), B\subset [0,+\infty),$ and $a'b\leq ab, ab'\leq ab, a' b'\leq ab,$ which implies that $\max\{ab,a'b,ab',a'b'\}=ab.$ By the above Lemma , we have $\sup AB=\sup A\cdot \sup B=ab.$ Hence in this case the desired equality $(\star)$ holds.
(ii) $a'\geq 0, b'<0.$ In this case we have for all $x\in A,$ $x\geq a'\geq 0,$ and so $x\geq 0.$ We consider two sub-cases.
(ii.1) $b>0.$ In this case $0\leq a'\leq a, b'<0<b,$ so $a'b\leq ab, ab'\leq a'b'\leq 0\leq ab,$ which gives that
$\max\{ab,a'b,ab',a'b'\}=ab.$
Let $z\in AB,$ then there exists $x\in A, y\in B$ such that $z=xy.$ Since $x\geq 0,$ $xy\leq xb\leq ab.$ And so $c\leq ab.$ If $a=0,$ then we deduce that $a'=0,$ and so $A=\{0\}=AB,$ which shows the desired result is obvious. If $a>0,$ then, for every $0<\epsilon<\min\{a,b\},$ we have $a-\epsilon>0, b-\epsilon>0.$ Hence there exist $x_1\in A, y_1\in B$ such that $x_1>a-\epsilon >0, y_1>b-\epsilon>0,$ which implies that $x_1y_1>(a-\epsilon)(b-\epsilon)=ab-\epsilon(a+b-\epsilon).$ Since $\lim_{\epsilon\to 0+}\epsilon(a+b-\epsilon)=0,$ we have, by the order preserving property of limit, $x_1y_1\geq ab,$ and so $c\geq ab.$ It follows that $c=ab.$ Hence we have proved, in this case, that $\sup AB=c=ab=\max\{ab,a'b,ab',a'b'\}.$
(ii.2) $b\leq 0.$ Then $a'b'\leq a'b$ and $ab'\leq ab\leq a'b.$ So $\max\{ab,a'b,ab',a'b'\}=a'b.$ Let $z\in AB,$ then there are $x\in A, y\in B$ such that $z=xy\leq xb\leq a'b,$ which implies that $c\leq a'b.$ On the other hand, for every $\epsilon >0,$ there exists $x_2\in A, y_2\in B,$ such that $0\leq x_2<a'+\epsilon, 0\geq b\geq y_2>b-\epsilon,$ it follows that $x_2y_2\geq (a'+\epsilon)y_2>(a'+\epsilon)(b-\epsilon)=a'b+\epsilon(b-a'-\epsilon),$ which implies that $x_2y_2\geq a'b.$ Thus $c\geq a'b.$ As a result, $c=a'b=\max\{ab,a'b,ab',a'b'\}.$
(iii) $a'<0, b'\geq 0.$ Swapping $A$ and $B,$ and using Case (ii), the desired result follows.
(iv) $a'<0, b'<0.$ There are four sub-cases.
(iv.1) $a> 0, b>0.$ That is, $a'<0<a, b'<0<b.$ Because in this case $a'b<0<a'b'\leq \max\{ab,a'b'\}$ and $ab'<0<ab\leq \max\{ab,a'b'\},$ we deduce that $\max\{ab,a'b,ab',a'b'\}=\max\{ab,a'b'\}.$
Then, for any $x\in A$ and $y\in B,$ we have $xy\leq \max\{ab, a'b'\}.$ Assume first that $\max\{ab, a'b'\}=a'b'.$ For sufficiently small $\epsilon>0,$ such that $a'+\epsilon<0$ and $b'+\epsilon<0,$ there exist $x^*\in A$ and $y^*\in B$ for which $x^*<a'+\epsilon<0$ and $y^*<b '+\epsilon<0.$ This gives $x^*y^*>(a'+\epsilon)(b'+\epsilon)=a'b'+\epsilon(a'+b'+\epsilon).$
Hence $\sup AB=c\geq x^*y^*\geq a'b'.$ Therefore $a'b'$ is the least upper bound of $A\cdot B.$ In the case where $\max\{ab, a'b'\}=ab,$ for sufficiently small $\epsilon>0,$ such that
$a-\epsilon>0$ and $b-\epsilon>0,$
there exists $x_1\in A$ and $y_1\in B$ such that $x_1>a-\epsilon>0, y_1>b-\epsilon>0.$ This gives $x_1y_1>(a-\epsilon)(b-\epsilon)=ab-a\epsilon-\epsilon(b-\epsilon)=ab-\epsilon\big(a+b-\epsilon\big).$ Since
$\lim_{\epsilon\to 0+}\epsilon\big(
a+b-\epsilon\big)=0,$ the order preserving property of limit gives that $x_1y_1\geq ab,$ and hence $c\geq ab.$ Therefore $\sup (AB)=c=ab=\max\{ab,a'b,ab',a'b'\}.$ Therefore we have proved that, in this sub-case, $\sup AB=\max\{ab,ab',a'b,a'b'\}.$
(iv.2) $a>0, b\leq 0.$ Then $ab\leq 0\leq a'b\leq a'b', ab'\leq 0\leq a'b',$ which gives that $\max\{ab,a'b,ab',a'b'\}=a'b'.$
In this case, for every $y\in B,$ $y\leq 0.$ Let $z\in AB,$ then there exist $x\in A, y\in B$ such that $z=xy.$ If $x\geq 0,$ then since $a'< x\leq a,$ so $z=xy\leq a'y\leq a'b';$ while if $x<0,$ then $z=xy\leq a'y\leq a'b'.$ Hence $a'b'$ is an upper bound of $AB.$ Since $a'<0$ and $b'<0,$ for sufficiently small $\epsilon>0,$
such that $a'+\epsilon<0$ and $b'+\epsilon<0,$
there exist $x_1\in A, y_1\in B$ such that $x_1<a'+\epsilon<0, y_1<b'+\epsilon<0,$ which implies that
$x_1y_1>(a'+\epsilon)(b'+\epsilon)=a'b'+\epsilon(a'+b'+\epsilon)\to a'b', $ as $\epsilon\to 0+0.$ Hence $x_1y_1\geq a'b'.$ Thus $c=\sup AB\geq x_1y_1\geq a'b'.$ Therefore we have $\sup AB=c=a'b'=\max\{ab,a'b,ab',a'b'\}.$
(iv.3) $a\leq 0, b>0.$ Similar to sub-case (iv.2), or you can argue by simply swapping $A$ and $B.$
(iv.4) $a\leq 0, b\leq 0.$ Because in this case $a,a',b,b'$ are non-positive, $ab\leq a'b\leq a'b',$ and $ab'\leq a'b',$ we deduce that $\{ab,a'b,ab',a'b'\}=a'b'.$
For every $x\in A, y\in B,$ we have $a'\leq x\leq a\leq 0, b'\leq y\leq b\leq 0,$ so
$xy\leq a'b'.$ From this it follows that $a'b'$ is an upper bound of $AB.$ Let $\epsilon>0$ be sufficiently small such that
$a'+\epsilon<0$ and $b'+\epsilon<0.$
Then there exist $x_1\in A, y_1\in B$ such that $x_1<a'+\epsilon<0, y_1<b'+\epsilon<0.$ So $x_1y_1>(a'+\epsilon)(b'+\epsilon)=a'b'+\epsilon(a'+b'+\epsilon). $ It follows that $c\geq a'b'.$ As a consequence, we have $c=a'b'=\max\{ab,a'b,ab',a'b'\}.$
$\qquad\Box$
Best Answer
Given a set $A \subset \mathbb{R}$, define $A^-=A \cap (-\infty, 0]$ and $A^+=A \cap (0, +\infty)$. Of course, $A=A^- \cup A^+$.
Observe that $$AB=(A^- \cup A^+) \cdot (B^- \cup B^+)$$ $$=(A^- B^-) \cup (A^- B^+) \cup (A^+ B^-) \cup (A^+ B^+)$$
So $\sup(AB)=\max(\sup(A^- B^-), \ \sup(A^- B^+), \ \sup(A^+ B^-), \ \sup(A^+ B^+))$. Here, I'm considering $\sup \emptyset = -\infty$.
Suppose first that $A^-, A^+, B^-$ and $B^+$ are all nonempty.
Then $$ \sup A \sup B = \sup A^+ \sup B^+ =^{lemma} \sup A^+ B^+$$ $$ \sup A \inf B = \sup A^+ \inf B^- = -\sup A^+ \sup (-B^-)=^{lemma} - \sup(-A^+ B^-) = \inf(A^+ B^-) \le \sup(A^+ B^-)$$ $$ \sup B \inf A = \sup B^+ \inf A^- = -\sup B^+ \sup (-A^-)=^{lemma} - \sup(-B^+ A^-) = \inf(B^+ A^-) \le \sup(B^+ A^-)$$ $$ \inf A \inf B = \inf A^- \inf B^- = \sup(- A^-) \sup (-B^-) =^{lemma} \sup A^- B^-$$
Consider $S= \max(\sup A \sup B, \ \sup A \inf B, \ \sup B \inf A, \ \inf A \inf B)$. Then $S \le \sup(AB)$. Since it is clear that $S \ge \sup(AB)$, it follows that $S = \sup(AB)$.
The other cases can be dealt with, similarly.