Understanding the proof of $\sup(AB)=\sup A\cdot \sup B$.

Question

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 $\sup A>0$ and $\sup B>0$.

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 $0<\varepsilon<\sup A\cdot \sup B$.

Then for $\varepsilon_1=\frac{\varepsilon}{2\sup A}>0$ and $\varepsilon_2=\frac{\varepsilon}{2\sup B}>0$ there exists $a\in A$ and $b\in B$ such that $a>\sup A-\varepsilon_1$ and $b>\sup B-\varepsilon_2$. $\Rightarrow ab>(\sup A-\varepsilon_1)(\sup B-\varepsilon_2)=\sup A\sup B -\varepsilon +\frac{\varepsilon^2}{4\sup A\sup B}>\sup A \sup B-\varepsilon$

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 $$\frac{\varepsilon\cdot\sup A}{2\sup B}+\frac{\varepsilon\cdot \sup B}{2\sup A}=\varepsilon$$?

Answer 1

I think that it should be $(\sup A-\epsilon_2)(\sup B-\epsilon_1)$, then the answer falls out.

Answer 2

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 $e>0$, take $ab-e<ab$, so $b-\frac e a<b$, since $b$ is supreme of $B$ so there exist some $u\in B$ such that $b-\frac e a<u$. That is $ab-e<au$ so that is $ab-\frac e u<a$ and since $a$ is $\sup A$ so there exist some $v\in A$ such that $ab-\frac e u<v$ that is, $ab-e<uv$, since $vu\in AB$, so there exist an element $uv$ in $AB$ such that $ab-e<uv$. Hence $\sup(AB) =ab=\sup A\sup B$.