Showing that $\sup(A\cdot B) = \sup(A)\cdot\sup(B)$ for $A$ and $B$ subsets of non-negative reals

Question

While $A \cdot B=\{x \cdot y \mid x \in A, y \in B\}$, show that, for $A$, $B \subseteq [0,\infty)$, $$\sup(A \cdot B)= \sup(A) \cdot \sup(B)$$

My demonstration:

First:

$$\forall a\in A, a\le\sup(A)$$ $$\forall b\in B, b\le\sup(B)$$

so, $$\begin{align} a\cdot b\le\sup(A)\cdot\sup(B) &\implies a\cdot b \le \sup(A \cdot B)\le\sup(A)\cdot\sup(B) \\ &\implies \sup(A\cdot B)\le\sup(A)\cdot\sup(B) \tag{1} \end{align}$$

Second: Using $(1)$, while $b\ne0$, $$a \le \frac{sup(A\cdot B)}{b} \implies \sup(A)\le \frac{\sup(A\cdot B)}{b}$$

so $$\begin{align} b\le\frac{\sup(A\cdot B)}{\sup(A)} &\implies \sup(B)\le \frac{\sup(A\cdot B)}{\sup(A)} \\ &\implies \sup(A)\cdot\sup(B)\le \sup(A\cdot B) \tag{2} \end{align}$$

Then $(1)$ and $(2)$ imply $$\sup(A \cdot B)= \sup(A) \cdot \sup(B) \tag{3}$$

Is the demonstration correct? Can the case $b = 0$ be discarded for being trivial?

Answer 1

It is basically correct. But you should have started your proof by saying that the cases in which $A=\{0\}$ or $B=\{0\}$ are trivial and that you will assume that $A,B\neq\{0\}$.