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?.
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?.
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
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)
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.
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?