Problem about the proof of $\sup(AB) \le \sup(A)\sup(B)$

Question

It is an old question. But what I concern is, in many proofs of this result, many argue that

For all $a \in A, b\in B$. Since $$a\le \sup(A)\quad and\quad b\le \sup(B)$$ Then, as $ab\le \sup(A)\sup(B)$, $\sup(A)\sup(B)$ is an upper bound of the set $AB$.

I think it is OK up to this step.

But then people argue that,

Consequently $$\sup(AB)\le \sup(A)\sup(B)$$

How this conclusion comes? An Upper bound $M$ only ensures for all $s\le S$, $s\le M$. But $\sup(AB)$ may not in the set $AB$.

Consider $S=\{x: 0\lt x\lt 1\}$, then we hae $\sup(S)$ not in $S$.

Answer 1

The statement is of course false in general: the supremum of $[-1,1]$ is $1$, the supremum of $[-2,1]$ is $1$, whereas the supremum of the product set is $2$.

If we assume that $A$ and $B$ only consist of positive numbers, then the result is true. You just need to show that, for each $a\in A$ and $b\in B$, $ab\le \sup(A)\sup(B)$, which follows from $a\le\sup(A)$ and $b\le\sup(B)$.

So $\sup(A)\sup(B)$ is an upper bound for $AB$, and therefore $\sup(AB)\le\sup(A)\sup(B)$, because $\sup(AB)$ is the least upper bound for $AB$.

Answer 2

So we have $ab \le \sup(A) \sup(B)$ for $a \in A$ and $b \in B$.
Now, $a$ and $b$ where arbitrary elements, and $AB = \{ ab \mid a \in A , b \in B\}$, so we can take the supremum ob both sides which gives $$\sup (AB) = \sup_{ab \in AB} \{ab \} \le \sup_{ab \in AB} \{ \sup (A) \sup(B) \}.$$ However, the RHS is the supremum over a set with only one element, so $$\sup_{ab \in AB} \{ \sup (A) \sup(B) \} = \sup (A) \sup(B).$$ And so we have $$\sup(AB) \le \sup(A) \sup(B).$$

An other way to see $\sup(AB) \le \sup(A) \sup(B)$ is that we have that $\sup(A) \sup(B)$ is a upper bound of $AB$, and $\sup(AB)$ is the least upper bound, which means that for any upper bound $M$ of $AB$, $\sup (AB) \le M$.
So as $\sup(A)\sup(B)$ is an upper bound of $AB$, we have $$\sup(AB) \le \sup(A) \sup(B).$$

Furthermore, I would like to note that $a \le \sup(A)$ and $ b \le \sup(B)$ does not mean $ab \le \sup(A) \sup(B)$ in general. If we want to conclude $ab \le \sup(A) \sup(B)$, we need that $0 \le \inf (A)$ and $0 \le \inf(B)$, thus $A$ and $B$ exists only of non-negative numbers.