Product of sets and supremum 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?
 A: I provide the answer in a more general setting, which gives $\sup(A + B) = \sup A + \sup B$ as well.

Theorem: Let $G$ be a bi-ordered group with least upper bound property. Then for any $A, B\subseteq G$, we have
$$
\sup(AB) = (\sup A)(\sup B)\text.
$$

Proof. "$\le$": We show that the RHS is an upper bound of $AB$. Let $x\in A$ and $y\in B$. Then $x\le\sup A$ and $y\le \sup B$. Due to bi-invariance of $\le$, we have that $xy\le(\sup A)(\sup B)$, as required.
"$\ge$": Let $x\in A$ and $y\in B$. Then $xy\in AB$ and hence $xy\le \sup(AB)$, which implies $x\le \sup(AB)y^{-1}$. Since $x\in A$ was arbitrary, we have $\sup A\le\sup(AB)y^{-1}$, which gives, again using bi-invariance, $y\le (\sup A)^{-1}\sup(AB)$. Again, $y\in B$ was arbitrary, and hence $\sup B\le (\sup A)^{-1}\sup(AB)$, or $(\sup A)(\sup B)\le\sup(AB)$, as required.
Now, we just note that $(\mathbb R, +)$ and $(\mathbb R^+, \cdot)$ are bi-ordered (in fact abelian) groups (under the usual order). And boom! You get the additive and multiplicative facts in a single shot!

Note that the set of nonzero reals does not form an ordered group under multiplication. This gives a clearer picture of why we concern ourselves with only positive reals in the multiplicative case.
A: 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 $\sup AB = \alpha \beta$.
