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


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

  • $\begingroup$ But how does that follow from $a>\sup A-\varepsilon_1$ and $b>\sup B-\varepsilon_2$? $\endgroup$ – lmc Jan 8 '17 at 12:41
  • $\begingroup$ if a> b and c>d, multiply 1st inequality by c, giving ac>bc, multiply second inequality by b, giving bc>bd. By transitivity ac>bd $\endgroup$ – matt Jan 8 '17 at 14:17

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