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Question
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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 $\sup{(AB)}=\alpha\beta$. 

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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 $\sup{(AB)} = \alpha\beta$?

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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?