Post History

Let $L:=(\sup A)(\sup B)$. The statement then can be rephrased as $\sup A\cdot B = L$. By definition of "supremum", one needs to show the following two things, 

- $L$ is an upper bound for $A\cdot B$; 

- $L$ is the smallest upper bound for $A\cdot B$. 

The first statement is very easy to prove: since for every $a\in A$ and every $b\in B$, one has $a\le \sup A$ and $b\le \sup B$, which immediately implies that $ab\le (\sup A)(\sup B)$.

To prove the second statement, assume that $L'$ is an upper bound of $A\cdot B$. We want to show that $L\le L'$. But the assumption of $L'$, we have for every $a\in A$, 

$$\text{for every } b\in B: ab\le L'$$

which is equivalent to 

$$\text{for every } b\in B: b\le \frac{1}{a}\cdot L'$$

So it follows that $\sup B\le L'/a$. Thus for every $a\in A$, one has

$$
a\le \frac{L'}{\sup B}
$$
and thus $\sup(A)\le \frac{L'}{\sup B}$, which implies that $L\le L'$.