Understanding the proof of $\sup(AB)=\sup A\cdot \sup B$. I need help understanding this:
Let $A, B\subseteq [0,\infty)$ be bounded from above and non-empty sets. Let $AB=\{ab, a\in A, b\in B\}$ Prove that $\sup(AB)=\sup A\cdot \sup B$.

If $\sup A=0 \Rightarrow A=\{0\}$ so $AB=\{0\}$ so the statement is true.
Suppose $\sup A&gt;0$ and $\sup B&gt;0$.
Then for $a\in A$ and $b\in B$: $ab\leq \sup A\cdot \sup B$. $\Rightarrow \sup A\cdot \sup B$ is an upper bound of $AB$.
Let $0&lt;\varepsilon&lt;\sup A\cdot \sup B$.
Then for  $\varepsilon_1=\frac{\varepsilon}{2\sup A}&gt;0$ and $\varepsilon_2=\frac{\varepsilon}{2\sup B}&gt;0$ there exists $a\in A$ and $b\in B$ such that $a&gt;\sup A-\varepsilon_1$ and $b&gt;\sup B-\varepsilon_2$.
$\Rightarrow ab&gt;(\sup A-\varepsilon_1)(\sup B-\varepsilon_2)=\sup A\sup B -\varepsilon +\frac{\varepsilon^2}{4\sup A\sup B}&gt;\sup A \sup B-\varepsilon$

I don't understand this part:
$$(\sup A-\varepsilon_1)(\sup B-\varepsilon_2)=\sup A\sup B -\varepsilon +\frac{\varepsilon^2}{4\sup A\sup B}$$
As far as I know: $$(\sup A-\varepsilon_1)(\sup B-\varepsilon_2)=\sup A\cdot \sup B -\frac{\varepsilon\cdot\sup A}{2\sup B}-\frac{\varepsilon\cdot \sup B}{2\sup A}+\frac{\varepsilon^2}{4\sup A\sup B}$$
Why is $$\frac{\varepsilon\cdot\sup A}{2\sup B}+\frac{\varepsilon\cdot \sup B}{2\sup A}=\varepsilon$$?
I think that it should be $(sup A-\epsilon_2)(sup B-\epsilon_1)$, then the answer falls out.
As $AB=\{xy:x\in A, y\in B\}$, $A$ and $B$ are subsets of positive real numbers such that $\sup A$ and $\sup B$ exists.
Now Assume that $\sup A=a$ and $\sup B=b$.
Then it is easy to see that
$xy\le ab$, for all $xy\in AB$.
So $ab$ is an upper bound for $AB$.
Now let $e&gt;0$, take $ab-e&lt;ab$,
so $b-\frac e a&lt;b$,
since $b$ is supreme of $B$ so there exist some $u\in B$ such that $b-\frac e a&lt;u$.
That is $ab-e&lt;au$ so that is
$ab-\frac e u&lt;a$ and
since $a$ is $\sup A$ so there exist some $v\in A$ such that
$ab-\frac e u&lt;v$ that is,
$ab-e&lt;uv$,
since $vu\in AB$, so there exist an element $uv$ in $AB$ such that
$ab-e&lt;uv$.
Hence $\sup(AB) =ab=\sup A\sup B$.