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Given $A\subseteq (-\infty , -1)$ and $B\subseteq (1 ,+\infty)$, and $C = \{ ab :\, a\in A,~ b\in B\}$.

In addition, $A$ and $B$ have at least 2 elements and are bounded.

I believe that $\sup C = (\sup A)(\inf B)$.

How do I prove it?

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up vote 3 down vote accepted

For all $a\in A$ and $b\in B$ $$ -a\geq -\sup A \geq 0\qquad b\geq \inf B\geq 0 $$ hence $$ -ab\geq -\sup A\inf B\qquad\Leftrightarrow\quad ab\leq \sup A\inf B. $$ So $$ \sup AB\leq \sup A\inf B. $$

To get the reverse inequality, take sequences $a_n$ in $A$ and $b_n$ in $B$ such that $$ \lim a_n=\sup A\quad \lim b_n=\inf B. $$ For all $n$, we have $\sup AB\geq a_nb_n$, so $$ \sup AB\geq \lim a_nb_n=\sup A\inf B. $$

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Sorry, I don't quite get the reverse inequality. How did you deduce that sup(AB) ≥ lim( – Sapphire Mar 5 '13 at 13:48
@Sapphire For all $n$, $a_nb_n\leq \sup AB$. Then pass to the limit: $\lim a_nb_n\leq \sup AB$. Edited. – 1015 Mar 5 '13 at 13:53
Can I use the same method as the 1st part to deduce the reverse inequality. (i.e. a ≤ supA and -b ≤ -infB will give me -ab ≤ -supA.infB and hence ab ≥ supA.infB)? – Sapphire Mar 5 '13 at 14:09
No you can't. Note that both equation you gave are with nonpositive numbers. If you multiply them, this reverses the inequality and you get first part again. – 1015 Mar 5 '13 at 14:35

The number $\sup C$ is characterized by

  • $c\leq \sup C$ for all $c\in C$.
  • There exists a sequence $(c_n)_{n\geq 1}\subseteq C$ such that $\lim_{n\to\infty}c_n=\sup C$.

Show that the number $\sup A\cdot \inf B$ satisfies these two properties using that $\sup A$ satisfies the two properties with $C$ replaced by $A$ and $\inf B$ satisfies the two properties with $\geq$ instead of $\leq$ and $C$ replaced by $B$.

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