# $\sup (A\cdot B)=\sup A \cdot \sup B$ [duplicate]

I would like to prove that $\sup (A\cdot B)=\sup A \cdot \sup B$, where $A,B$ are sets of positive real numbers. I've already proved $\sup (A+B)=\sup A +\sup B$ and $\sup (c\cdot A)=c\cdot \sup A$$(c\gt 0)$ easily without any problems. However, I'm stuck in the multiplication $A\cdot B$ case.

I need help.

Thanks a lot

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## marked as duplicate by user7530, Asaf Karagila, draks ..., Henry T. Horton, Davide GiraudoFeb 16 '13 at 23:16

Duplicate of math.stackexchange.com/q/46738/264 – Zev Chonoles Feb 16 '13 at 22:41
@user42912 This might help you. – Git Gud Feb 16 '13 at 22:42

If $A=\{-1,+1\}$, and $B=\{-2,1\}$, then $\sup A = 1$, $\sup B = 1$, but $\sup A \cdot B = 2$.

If $A,B \subset [0,\infty)$, then if $a \in A, b \in B$, we have $ab \leq \sup A \sup B$, hence $\sup A \cdot B \leq \sup A \sup B$.
Furthermore, since $\sup A \cdot B = \sup_{a \in A} \sup_{b \in B} ab$, we have $a b \leq \sup A \cdot B$ for all $a \in A, b \in B$. Since $a,b \geq 0$, we have $a \sup B \leq \sup A \cdot B$, and hence $\sup A \sup B \leq \sup A \cdot B$.