Can we establish the following inequalities Let $A$ and $B$ both be sets of real numbers. Let $A \times B$ be $\{xy \mid x \in A, y \in B\}$. Can we establish the following:
$\sup A \sup B \leq \sup(A\times B)$
and
$\inf A \inf B  \geq \inf(A\times B)$
I've been trying hard to come up with counterexamples but I haven't been able to come up with any. Are those inequalities true?
 A: Choose $(a_{n})\subseteq A$, $(b_{n})\subseteq B$ such that $a_{n}\rightarrow\sup A$ and $b_{n}\rightarrow\sup B$, then $\sup(AB)\geq a_{n}b_{n}$, taking limit as $n\rightarrow\infty$, then $\sup(AB)\geq(\sup A)(\sup B)$.
If we define $0\cdot\infty=0$, it still goes through: For $\sup A=\infty$ and $\sup B=0$, then choosing $a\in A$, $a>0$, $(b_{n})\subseteq B$, $b_{n}\rightarrow 0$, then $\sup(AB)\geq ab_{n}$, taking limit as $n\rightarrow\infty$, we have $\sup(AB)\geq 0=(\sup A)(\sup B)$.
A: With many equalities in Analysis, it's usually relatively easy to get one direction directly; but the other direction is much harder to do directly, so we do any approximations we can with $\epsilon$ to get some wiggle room.
Fix any $\epsilon > 0$. By the definition of supremum there are $a \in A, b \in B$ with $\sup A - \epsilon < a $ and $\sup B - \epsilon < b $. Multiplying these inequalities we get  $\sup A \sup B + \epsilon (\epsilon - \sup A - \sup B) \le ab \le \sup( A \times B)$. Thus, since $\sup A \sup B + \epsilon (\epsilon - \sup A - \sup B) \le \sup(A \times B)$ for every $\epsilon > 0$, we have that $\sup A \sup B  \le \sup(A \times B)$ 
