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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. –  Apostolos Jun 21 '11 at 16:42

1 Answer 1

up vote 7 down vote accepted

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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How could I salvage my proof? –  Damien Jun 21 '11 at 18:35
    
@Damien: Your proof is correct, but ( the presentation of ) your use of $\alpha$ and $\beta$, is rather puzzling. While creativity surely is a good thing to cultivate, try to refine your proofs to present the arguments in a more clear way. It will help you correct your proofs by yourself, make them more accessible and easier to remember. Note how the second half of your proof reads like this: Fix an element $a$ of $A$. For every $b\in B$ we have $ab\leq c$. By rearrangement we get $a\leq c/b$ for all $b\in B$. Because $a$ was arbitrary we have $α\leq c/b$ for all $b\in B$. –  Alexander Thumm Jun 21 '11 at 21:34
    
Again by rearrangement we get $b\leq c/\alpha$ for all $b\in B$, so $\beta\leq c/\alpha$, hence $\alpha\beta\leq c$. –  Alexander Thumm Jun 21 '11 at 21:35

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