Prove that $O[f(n)] \cdot O[g(n)] = O[f(n) \cdot g(n)]$, knowing that $O[g(n)] = \left\{ f(n) \mid \exists\ c,n_0 > 0\ :\ 0 \leq f(n) \leq c \cdot g(n)\ \forall\ n \geq n_0 \right\}$
I don't really know what to do here. I tried to use the definition to relate the first part of the equality with the second:
Be $x(n) \in O[f(n)] \cdot O[g(n)]$, $\implies 0 \leq x(n) \leq cf(n)$ and $0 \leq x(n) \leq cg(n)\ \forall\ n \geq n_0$.
But I got stuck trying to find a connection between the two inequalities.
Assume that we have $x(n)\in O[f(n)]\cdot O[g(n)]$. This means that there exists $y(n)\in O[f(n)]$ and $z(n)\in O[g(n)]$ and $c_1, n_{1}, c_2, n_2>0$ such that $x(n)=y(n)\cdot z(n)$ and $0\leq y(n)\leq c_1 f(n)$ and $0\leq z(m)\leq c_2 g(m)$ for all $n\geq n_1$, $m\geq n_2$. Take $n_0=\max\{n_1,n_2\}$. Then \begin{align*} x(n)=y(n)z(n)\leq c_1 c_2 f(n)g(n) \end{align*} for all $n\geq n_0$. So $x(n)$ satisfies the condition of $O[f(n)g(n)]$ with $n_{0}$ and constant $c_1 c_2$. So $O[f(n)]\cdot O[g(n)]\subseteq O[f(n)g(n)]$. I'll leave you to complete the reverse inclusion.