Prove that for any function $f : \mathbb{N} \to \mathbb{R}^+$ that there exists a function $g : \mathbb{N} \to \mathbb{R}^{\geq 0}$ so that $g \not \in O(f)$ or in other words $$\forall c, n_0 \in \mathbb{R^+}, \exists n \in \mathbb{N}, n \geqslant n_0 \land g(n) > cf(n).$$
I believe we would have to consider cases for $g$, but don't know how to begin. I tried $g(n) = f(n) + 1$ (if $n$ is even) or $g(n) = 0$ (if $n$ is odd). But then we have to prove $$\forall c, n_0 \in \mathbb{R^+}, \exists n \in \mathbb{N}, n \geqslant n_0 \land g(n) > cf(n).$$ Which doesn't seem right because I would try to force $n$ so that $g(n) = f(n)+1$, so in this case we have to prove $$\forall c, n_0 \in \mathbb{R^+}, \exists n \in \mathbb{N}, n \geqslant n_0 \land f(n)+1 > cf(n),$$ which seems false...
$g(n)=nf(n)$ has this property since there is always an integer $n$ such that $n \geq n_0$ and $n >c$. .