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Let $A$ and $B$ be nonempty sets of positive real numbers that are bounded above. Also let $AB = \{ab: a \in A, b \in B \}$. Prove that $AB$ is bounded above and $\sup(AB) = (\sup A) (\sup B)$.

So $\sup A$ and $\sup B$ exist by completeness. An upper bound for $AB$ is $(\sup A)(\sup B)$. Let $\alpha = \sup A$ and $\beta = \sup B$. We want to show that if $c$ is an upper bound for $AB$ then $\alpha \beta \leq c$. For $a \in A$, $ab \leq c$ for all $b \in B$. So $c/b$ is an upper bound for $A$. Thus $\alpha \leq c/b$. It follows that $\alpha \beta \leq c$.

Is this correct?

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    I think you should add some more details between the last two sentences since it's unclear whether you are mixing $\beta$ and $b$ or not.2011-06-21

1 Answers 1

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Let $\alpha = \sup A$ and $\beta = \sup B$.

For every $a\in A$ and $b \in B$ we have

$ab \leq \sup_{b\in B} a b = a \beta \leq \sup_{a\in A} a \beta = \alpha \beta,$

so we have $\sup AB \leq \alpha\beta$.

Now let $(a_n)_{n\in \mathbb N} \subset A$ and $(b_n)_{n \in \mathbb N} \subset B$ be sequences such that $a_n \to \alpha$ and $b_n \to \beta$ as $n \to \infty$.

It is then clear, that $(a_nb_n)_{n \in \mathbb N} \subset AB$ and $a_nb_n \to \alpha\beta$ as $n \to \infty$, so $\sup AB \geq \alpha\beta$ and therefor we have $\sup AB = \alpha \beta$.

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    Again by rearrangement we get $b\leq c/\alpha$ for all $b\in B$, so $\beta\leq c/\alpha$, hence $\alpha\beta\leq c$.2011-06-21