I am trying to prove that:
For nonempty subsets of the positive reals $A,B$, both of which are bounded above, define
$$A \cdot B = \{ab \mid a \in A, \; b \in B\}.$$
Prove that $\sup(A \cdot B) = \sup A \cdot \sup B$.
Here is what I have so far.
Let $A, B \subset \mathbb{R}^+$ be nonempty and bounded above, so $\sup A$ and $\sup B$ exist by the least-upper-bound property of $\mathbb{R}$. For any $a \in A$ and $b \in B$, we have
$$ab \leq \sup A \cdot b \leq \sup A \cdot \sup B.$$
Hence, $A \cdot B$ is by bounded above by $\sup A \cdot \sup B$. Since $A$ and $B$ are nonempty, $A \cdot B$ is nonempty by construction, so $\sup(A \cdot B)$ exists. Furthermore, since $\sup A \cdot \sup B$ is an upper bound of $A \cdot B$, by the definition of the supremum, we have
$$\sup(A \cdot B) \leq \sup A \cdot \sup B.$$
It suffices to prove that $\sup(A \cdot B) \geq \sup A \cdot \sup B$.
I cannot figure out the other half of this. A trick involving considering $\sup A – \epsilon$ and $\sup B – \epsilon$ for some $\epsilon > 0$ and establishing that $\sup(A \cdot B) < \sup A \cdot \sup B + \epsilon$ did not seem to work, though it did in the additive variant of this proof. I haven't anywhere used the assumption that $A$ and $B$ are contained in the positive real numbers, and it seems to me that this assumption must be important, probably as it pertains to inequality sign, so I assume that at some point I will need to multiply inequalities by some positive number. I cannot seem to get a good start on this, though. A hint on how to get started on this second half would be very much appreciated.
Best Answer
If $\varepsilon>0$, take $a\in A,b\in B$ such that $\sup A-\varepsilon<a$ and $\sup B-\varepsilon<b$. Then it is
$$(\sup A-\varepsilon)\cdot(\sup B-\varepsilon)<ab\leq\sup(A\cdot B) $$
So, $$(\sup A-\varepsilon)\cdot(\sup B-\varepsilon)<\sup(A\cdot B) $$ is true for any $\varepsilon>0$. What happens if you let $\varepsilon\to0^+$?