# supremum of a set

We have the following: $A,B,C$ are sets.

$C = \{ab: a \in A, b \in B\}$.

What is the relationship between $\sup(C),\sup(A)$, and $\sup(B)$?.

Is it: $$\sup(C) \le \sup(A) \sup(B)\;,$$ and why?.

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Are $A,B$, and $C$ sets of real numbers? Can they contain negative numbers? If so, what happens if $A=B=\{x\in\Bbb R:x<0\}$? What are $\sup A,\sup B$, and $\sup C$ in that case? –  Brian M. Scott Apr 26 '12 at 14:34
so the A,B musst be then real not negative. –  Jooan Hussain Apr 26 '12 at 14:46
Sets in the question? Hooray, must be [set-theory]! –  Asaf Karagila Apr 26 '12 at 15:49

If $A$ and $B$ are sets of non-negative real numbers, your conjecture is correct. Let $x=\sup A$ and $y=\sup B$. Then for any $c\in C$ there are $a\in A$ and $b\in B$ such that $c=ab$, and since $a,b\ge 0$, $ab\le xy$. That is, $c\le xy$ for every $c\in C$, so $\sup C\le xy=\sup A\sup B$.

But we can say more. Suppose for now that $0<x,y<\infty$, and let $\epsilon<\min\{x,y\}$ be positive. Then \begin{align*}(x-\epsilon)(y-\epsilon)&=xy-(x+y)\epsilon+\epsilon^2\\ &>xy-\epsilon(x+y)\;. \end{align*}

Now $x=\sup A$, so there is an $a_\epsilon\in A\cap(x-\epsilon,x]$. Similarly, there is a $b_\epsilon\in B\cap(y-\epsilon,y]$, and clearly $xy-\epsilon(x+y)<a_\epsilon b_\epsilon\le xy$. Since $x+y$ is a positive real, by taking $\epsilon$ small enough we can make $\epsilon(x+y)$ as small as we like. Thus, we can find products $a_\epsilon b_\epsilon\in C$ as close to $xy$ as we like, and it follows that $xy=\sup C$.

It’s easy to check that this is also the case when one of $x$ and $y$ is $0$ and the other is finite, and when one of $x$ and $y$ is infinite and the other is positive. When one is $0$ and the other is infinite, $\sup C=0$. Thus, if we (perhaps somewhat arbitrarily) define $0\cdot\infty=0$, we can say that $$\sup(AB)=\sup A \sup B$$ when $A$ and $B$ are sets of non-negative real numbers, where $$AB=\{ab:a\in A\text{ and }b\in B\}\;.$$

As you can see from the example in the comments, matters are much more complicated when $A$ and $B$ are allowed to contain negative numbers.

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Suppose for simplicity that $A$ and $B$ are finite. For all $a \in A$, we have $a \leq \max A$. For all $b \in B$, we have $b \leq \max B$. What can you say about $ab$ (assuming $A,B\subseteq \mathbb{R}_{\geq 0}$)? What can you deduce about $\max C$?

You can also think about the extreme case $A = \{a\}$, $B = \{b\}$, $C = \{ab\}$. What does it imply about the inequality?

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I think you mean $C=\{ab\}$ for the extreme case. –  Cameron Buie Apr 26 '12 at 15:01