Finding the supremum of a set composed of two other sets 
Question
Let $A$, $B$ be two nonempty sets of real numbers with suprema $\sup{(A)}=\alpha$, and $\sup{(B)}=\beta$, respectively. Define the set $AB = \{ab\,\colon\, a\in A, b\in B\}$. Show that if the elements of $A$ and $B$ are positive then $\sup{(AB)}=\alpha\beta$. 

To show $\sup{(AB)} = \alpha\beta$ we must show $\forall\varepsilon>0$, $\exists a\in A$,  $\exists b\in B$ such that $\alpha\beta-\varepsilon<ab\le\alpha\beta$
$$\text{We know }\begin{cases}
\sup{(A)}=\alpha\implies \forall\varepsilon'>0,\, \exists a\in A \text{ s.t. } \alpha-\varepsilon'<a\le\alpha. \\
\sup{(B)}=\beta\implies \forall\varepsilon'>0,\, \exists b\in B \text{ s.t. } \beta-\varepsilon'<b\le\beta.
\end{cases}$$
Multiplying these we get $$(\alpha-\varepsilon')(\beta-\varepsilon')<ab\le\alpha\beta.$$
$$\iff \alpha\beta - (\alpha+\beta)\varepsilon' + (\varepsilon')^{2}<ab\le\alpha\beta.$$
So can we take $\varepsilon = (\alpha+\beta)\varepsilon' - (\varepsilon')^{2}$ leaving us with $$\alpha\beta-\varepsilon<ab\le\alpha\beta.$$ thus $\sup{(AB)} = \alpha\beta$?

The only problem I can think of at the moment is are we sure $\varepsilon>0$. Note $$\varepsilon=(\alpha+\beta)\varepsilon' - (\varepsilon')^{2}>0$$ $$\iff \varepsilon'(\alpha+\beta-\varepsilon')>0$$ $$\iff \varepsilon'(\varepsilon'-\alpha-\beta)<0$$ $$\iff 0<\varepsilon'<\alpha+\beta.$$ Now this is a contradiction since we chose $\varepsilon'>0$ but not less than $\alpha+\beta$. So evidently my working is incorrect, can someone tell me what's wrong?
 A: This shouldn't be that confusing, $\varepsilon>0$ is given and you want to find some $\varepsilon'>0$ for which we have $(\alpha+\beta)\varepsilon'-(\varepsilon')^2<\varepsilon$ because then $\alpha\beta-\varepsilon<\alpha\beta-(\alpha+\beta)\varepsilon'+(\varepsilon')^2$.
You can find such $\varepsilon'$ just from the fact $(\alpha+\beta)\varepsilon'-(\varepsilon')^2$ gets arbitrarily small as you decrease $\varepsilon'$.

There's also a nicer, but trickier way:
$$\alpha\beta-ab=\alpha\beta-a\beta+(a\beta-ab)=(\alpha-a)\beta+a(\beta-b)<\frac\varepsilon{2\beta}\beta+\alpha\frac\varepsilon{2\alpha}=\varepsilon$$
by choosing $\varepsilon'<\frac12\min(\frac\varepsilon\alpha,\frac\varepsilon\beta)$.
A: Your solution is fundamentally correct.  I would refer to the problem you're (keenly) noting here as a red herring.  If we happen to choose $\varepsilon' > \alpha + \beta$, then certainly $\varepsilon < 0$, but this is not a problem:
Clearly if $\varepsilon' > \alpha$, then $(\alpha - \varepsilon' , \alpha)$ contains an element of $A$, since the elements of $A$ are positive.  So you don't need to employ your algebraic argument in this case (or in the case that $\varepsilon' > \beta$).  Using this you can then add the assumption that $\varepsilon' < \alpha + \beta$ to the beginning of your proof above.
