Product of two sets

Question

Let A and B two bonded sets from $R$ i want to prove that the set $AB=\{ab,a\in A,b\in B\}$ is bounded

Let $c\in AB$ then there exists $a\in A$ and $b\in B$ such that $c=ab$ then $\inf(A)\times\inf(B)\leq c\leq \sup(A)\sup(B)$

Is this always true? or one of sets must be in $R^+$ or the two?

Answer 1

Your claim is not true in general. For example, consider $ A = \{ 1, -1 \}, B = \{ 2, 1 \}$.
We have $\inf (A) \times \inf (B) = -1 \times 1 = -1 > -2 \in AB $.

If both of the sets are positive, then your claim is true. Can you prove this?

---

To prove the general statement that you want, show that if $ A \subset [ -x, x ] $ and $ B \subset [ -y, y ]$, then prove that $ AB \subset [ -xy, xy ] $.

Answer 2

The inequalities you have written are not true in general: Take $A=\{0,1\}, B=\{-1\}$ and $c=0$.

If $A$ and $B$ are bounded then there exist $M,N \in (0,\infty)$ such that $|a| \leq M$ for all $a \in A$ and $|b|\leq N$ for all $b \in B$. If $c=ab$ with $a \in A$ and $b \in B$ the $|c|=|a||b| \leq MN$, so $C$ is bounded.

Answer 3

For your flaw see the answer of Calvin Lin.

Hint:

Find a positive constant $c$ such that $|a|\leq c$ and $|b|\leq c$ for every $a\in A$ and $b\in B$.

Now prove that $-c^2\leq ab\leq c^2$ for every $a\in A$ and $b\in B$.