Earlier I found an extended definition of Big-O that turned out to be Little-o. With that in mind, how would we prove that for $f,g$ that map naturals to positive reals, if $g \in o(f)$ then $f\not \in O(g)$.
The clue is that $f$ and $g$ map to positive reals.
Recall that the definition of little-oh is:
$$g \in o(f) : \forall c \in \mathbb{R^+}, \exists n_0 \in \mathbb{R^+}, \forall n \in \mathbb{N}, n \geq n_0 \implies g(n) \leq cf(n).$$
$f\in O(g)$ implies that $\dfrac{f}{g}$ is bounded for $n\geq n_0$ where $n_0\in\mathbb{N}$. On the other hand, $g(n)\leq cf(n)\Leftrightarrow \dfrac{1}{c}\leq\dfrac{f(n)}{g(n)}$. You can choose $c$ arbitrarily small to make $\dfrac{f}{g}$ arbitrarily large for $n\geq n_1$ where $n_1\in\mathbb{N}$. In other words, $\dfrac{f}{g}$ is not bounded as $n$ grows.