For $A, B \subset \mathbb{R}^+$, $\sup(A \cdot B) = \sup A \sup B$. 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. 
 A: 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 $\varepsilon\to0^+$?
A: Here's a completely different idea. (I've left only the straightforward proof of Lemma 1 to be filled in.)
Lemma 1. If $E$ is a nonempty subset of $\mathbb{R}$ that is bounded above, and $c > 0,$ then
$$
\sup cE = c\sup E.
$$
Lemma 2. If $(E_i)_{i \in I}$ is a nonempty family of nonempty subsets of $\mathbb{R}$ that are all bounded above, then
\begin{equation}
\label{3712256:eq:1}\tag{1}
\sup\bigcup_{i \in I}E_i = \sup\{\sup E_i : i \in I\},
\end{equation}
in the sense that if either supremum exists, then so does the other, and they are equal.
Proof.
If the supremum on the left of \eqref{3712256:eq:1} exists, then it is an upper bound for $\bigcup_{i \in I}E_i,$ therefore also an upper bound for $E_j$ for all $j \in I.$ Therefore, $\sup E_j$ exists for all $j \in I$ (we assumed this anyway), and
$$
\sup E_j \leqslant \sup\bigcup_{i \in I}E_i \, \text{ for all } j \in I,
$$
therefore the set $\{\sup E_j : j \in I\}$ is bounded above, and
$$
\sup\{\sup E_j : j \in I\} \leqslant \sup\bigcup_{i \in I}E_i.
$$
So the supremum on the right of \eqref{3712256:eq:1} also exists, and is bounded above by the supremum on the left.
Conversely, if the supremum on the right of \eqref{3712256:eq:1} exists, then it is an upper bound for $\sup E_i,$ therefore also an upper bound for $E_i,$ for all $i \in I.$ Therefore, it is an upper bound for $\bigcup_{i \in I}E_i.$ Therefore, the supremum on the left of \eqref{3712256:eq:1} also exists, and
$$
\sup\bigcup_{i \in I}E_i \leqslant \sup\{\sup E_i : i \in I\}.
$$
We have now shown that if either side of \eqref{3712256:eq:1} is defined, then so is the other; and we have proved an inequality in both directions, therefore the two sides are equal. $\ \square$
Now, with Lemma 2 doing all the hard work for us, the proof is straightforward:
\begin{align*}
\sup AB & = \sup\bigcup_{a \in A}aB \\
& = \sup\{\sup aB : a \in A \} & \text{by Lemma 2} \\
& = \sup\{a\sup B : a \in A \} & \text{by Lemma 1} \\
& = \sup(A\sup B) \\
& = (\sup A)(\sup B). & \text{by Lemma 1}
\end{align*}
A: Hint:
Rather than $\sup A - \varepsilon$ and $\sup B - \varepsilon,$ subtract appropriate multiples of $\varepsilon$ from $\sup A, \sup B$ respectively. You'll need to assume that $\varepsilon$ isn't too big.
Full proof:
[I'm sorry, I can't get the wretched spoiler mechanism to work, so I'm afraid you'll have to avert your eyes!]
Let $s = \sup A > 0,$ and $t = \sup B > 0.$
You have already proved that $\sup AB \leqslant st.$
For every $\varepsilon$ such that $\varepsilon > 0$ and $\varepsilon < 2st,$ there exist $a \in A$ and $b \in B$ such that
\begin{align*}
a & > s - \frac\varepsilon{2t} > 0, \\
b & > t - \frac\varepsilon{2s} > 0.
\end{align*}
Therefore
$$
ab > \left(s - \frac\varepsilon{2t}\right)\left(t - \frac\varepsilon{2s}\right) =
st - \frac\varepsilon2 - \frac\varepsilon2 + \frac{\varepsilon^2}{4st} > st - \varepsilon.
$$
Therefore $\sup AB \geqslant st,$ therefore $\sup AB = st = (\sup A)(\sup B).$
