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**gordo151091** Hey, this is the problem

$\displaystyle A,B \subset \mathbb{R}$ A,B Bounded. Let

$\displaystyle AB :=\left \{ ab:a \in A, b \in B \right \}$. Proof $\displaystyle sup(A)sup(B)=sup(AB)$

I proofed that $\displaystyle xy \leq sup(A)sup(B) \forall x \in A, y \in B$ ( this is false as pointed out by GJA). So by definition $\displaystyle sup(A)sup(B)$ bounds $\displaystyle AB$. I want to show that indeed $\displaystyle sup(A)sup(B)=sup(AB)$ by showing that any there's no element in $\displaystyle AB$ lower than $\displaystyle sup(a)sup(b)$ than bounds the set. I trying to make the proof by contradiction, by contradicting the fact that theres a element in etheir set $\displaystyle A$ or $\displaystyle B$ that bounds it and is lower than its sup.

Any hints?

Thank you.