I already know that there are some answers to this problem, but in my case I want to solve a particular question, related to my show process.
Let $A$, $B$ sets of real positive numbers, bounded above. Let $a=sup(A)$, $b=sup(B)$, and $C:=\{x\cdot y | x\in A, y\in B\}$, show that $a\cdot b = sup(C)$.
For any $x \in A$ y $y \in B$, by supreme definition: $$0 \le x \le a \quad \forall x \in A$$ $$0 \le y \le b \quad \forall y \in B$$ As $x,y,a$ y $b$ are positive reals, it is true that: $$x\cdot y \le a\cdot b$$ Note that $x\cdot y \in C$, so $a\cdot b$ is an upper bound of $C$, thus being $C \neq \emptyset$ and bounded above, the supreme of $C$ exists (Completeness Axiom). Now we need to show that $a\cdot b$ is the minumum upper bound of $C$. So $a\cdot b = sup(C)$ and $a\cdot b \in C$, then $sup(C)\le a\cdot b$.
Let $\epsilon > 0$, thus by the approximation property of Suprema, we know that exists $x_0 \in A$ y $y_0 \in B$ such that: $$a - \epsilon < x_0 \le a$$ $$b - \epsilon < y_0 \le b$$ Multipliying the inequalities, we get: $$(a - \epsilon)(b - \epsilon) < x_0 y_0 \le ab$$
And this is the problem I have, because $\epsilon$ can be greater than $a$ or $b$, and it implies that $a - \epsilon<0$ or $b - \epsilon<0$. But I need to get and expresion like: $$ab - \epsilon \le x_{0} y_{0} \le sup(C)$$ In order to show that $a\cdot b = sup(C)$.
I thank everyone who can help me.