Prove that $\sup (C) = \sup (A)\sup (B)$ and $\inf (C) = \inf (A)\inf (B)$ 
Let $A$ and $B$ be two nonempty bounded sets of nonnegative real numbers. Define the set $C:= \{ab: a\in A, b \in B\}$. Show that $C$ is a bounded set and that $\sup (C) = \sup (A)\sup (B)$ and that $\inf (C) = \inf (A)\inf (B)$.

I have asked the mathematical assistant center at my school but no one knows how to solve this problem. I've come here as a last result. Can someone please help me? 
 A: Let $\bar a=\sup A$ and $\bar b=\sup B$. To prove that $\bar c=\bar a \bar b$ is the supremum of $C$, you must prove two things: $\bar c$ is an upper bound for $C$ and $\bar c - \epsilon$ is not an upper bound for $C$ for any $\epsilon > 0$. 
First off, it is clear that if any of $\bar a$ or $\bar b$ equals $0$, then one of the sets $A, B$ contains only $0$ and therefore $C=\{0\}$ and $\sup C = 0 = \bar a \bar b$. 
Now we focus on the case where $\bar a,\bar b > 0$.


*

*Take any $c\in C$. You know that $c=ab$ for some $a\in A, b\in B$. Because $\bar a$ is an upper bound for $A$ and $\bar b$ for $B$ (because they are supremums), you know that $\bar a\geq a\geq 0$ and $\bar b \geq b\geq 0$, you know that $\bar c = \bar a\bar b \geq ab$, so $\bar c$ is indeed an upper bound.

*Take any $\epsilon>0$. You know that, for any $\min\{\bar a, \bar b\}>\delta>0$, because $\bar a-\delta$ is not a upper bound for $A$ (because it is a supremum),that there exists $a\in A$ such that $a>\bar a - \delta$. In the same way, you get $b\in B$ such that $b>\bar b - \delta$. You know that $ab\in C$ and also that $$ab>(\bar a - \delta)(\bar b - \delta) = \bar c - (\bar a + \bar b)\delta + \delta ^2.$$
Choosing $\delta$ small enough that $(\bar a + \bar b)\delta - \delta ^2<\epsilon,$ the equation means that $ab>\bar c - \epsilon$, proving that $\bar c = \sup C$.

A: *

*If $a\in A$ and $b\in B$ then $ab \le \sup A \sup B$ so $\sup C\le \sup A\sup B$. 


There's no need of sequences for the other inequality. In fact the following approach uses only the definition of the least upper bound property.


*

*If $a\in A^*$ and $b\in B^*$ (WLOG),then $ab \leq \sup C$


Hence $a\leq \frac{\sup C}{b}$
Hence $ \forall b \in B^*, \sup A \leq \frac{\sup C}{b}$
Hence $\forall b \in B^*,b \leq \frac{\sup C}{\sup A}$
Hence $\sup B \leq \frac{\sup C}{\sup A}$
Hence $\sup A \times \sup B \leq \sup C$


*

*Hence $$\sup A \times \sup B = \sup C$$

*The other assertion follows the same proof:
If $a\in A$ and $b\in B$ then $ab \geq \inf A \inf B$ so $\inf C\geq \inf A\inf B$. 


*

*If $a\in A^*$ and $b\in B^*$ (WLOG),then $ab \geq \inf C$


Hence $a\geq \frac{\inf C}{b}$
Hence $ \forall b \in B^*, \inf A \geq \frac{\inf C}{b}$
Hence $\forall b \in B^*,b \geq \frac{\inf C}{\inf A}$
Hence $\inf B \geq \frac{\inf C}{\inf A}$
Hence $\inf A \times \inf B \geq \inf C$


*

*Hence $$\inf A \times \inf B = \inf C$$

A: If $a\in A$ and $b\in B$ then $ab \le \sup A \sup B$ so $\sup C\le \sup A\sup B$. 
If $\langle a_n\rangle$ be a sequence on $A$ (that is, $a_n\in A$ for all $n$) that converges to $\sup A$, and $\langle b_n\rangle$ be a sequence on $B$ that converges to $\sup B$, then $a_n b_n\le \sup C$ for all $n$. Take $n\to\infty$ then we get $\sup A\sup B \le \sup C$.
A: If either $A=\{0\}$ or $B=\{0\}$, then the statement is trivial, so assume that $A\neq\{0\}$ and $B\neq\{0\}$.
On one hand, we must have $\sup{C} \le \sup{A}\sup{B}$, since for any $x\in{C}$ we have $x=ab$, where $a\in{A}$ and $b\in{B}$, and thus $x=ab\le\sup{A}\sup{B}$.
On the other hand, if we assume that $\sup{C} \lt \sup{A}\sup{B}$ then we can write: $$\frac{\sup{C}}{\sup{A}}\lt\sup{B}$$
therefore there exists some non-zero element $b\in{B}$, such that $\frac{\sup{C}}{\sup{A}}\lt b$.
We can rewrite this as $\frac{\sup{C}}{b}\lt \sup{A}$, so there is an element $a\in{A}$, such that $\frac{\sup{C}}{b}\lt a$. Thus, we get:
$$\sup{C}\lt ab\in{C}$$
which is a contradiction, and the required equality ensues.
