$\sup_{x\in A}x \sup_{y\in B}y=\sup_{x\in A,y\in B}xy$ Let $A$ and $B$ be two sets of nonnegtive numbers. Prove that

$\sup_{x\in A}x \sup_{y\in B}y=\sup_{x\in A,y\in B}xy$.

Thanks for your help.
 A: If $x \in A$, then $x \le \sup A$. Similarly, $y \le \sup B$. Since $x,y \ge 0$, we have $xy \le \sup A \sup B$. It follows that $\sup_{x \in A, y \in B} x y \le \sup A \sup B$. 
In the other direction, if $x \in A, y \in B$, then $xy \le \sup_{x' \in A, y' \in B} x' y'$. Then $\sup_{y \in B} x y = x \sup B \le \sup_{x' \in A, y' \in B} x' y'$. Since this is true for all $x \in A$, and both $x\ge0$ and $\sup B \ge 0$, we have $\sup_{x \in A} x \sup B =\sup A \sup B \le\sup_{x' \in A, y' \in B} x' y'$.
A: This may help:
For each real number $M $:
$$\sup_{x\in A}x \sup_{y\in B}y\le M$$
$$\leftrightarrow (\forall x\in A)(x \sup_{y\in B}y\le M)$$
$$\leftrightarrow (\forall x\in A)(\forall y\in B)(xy\le M)$$
$$\leftrightarrow (\forall (x,y)\in A\times B)(xy\le M)$$
$$\leftrightarrow \sup_{(x,y)\in A\times B} xy\le M$$
Now try 
$$M=\sup_{x\in A}x \sup_{y\in B}y$$
and
$$M=\sup_{(x,y)\in A\times B} xy$$


  
*
  
*if $a\ge 0$ and for each $x\in A$, $f(x)\in \Bbb R$ then we have:
  $$(\forall x\in A)( af(x)\le M)\leftrightarrow a\sup_{x\in A} f(x)\le M$$
  

