Show that $f(n) = O(g(n))$ or $f(n) = \overset{\infty}{\Omega}(g(n))$
If you consider the following functions defined for $x\in \mathbb{R}$:
$f(x) = \left\{\begin{array}{rl}1 &amp; \text{if }x\in\mathbb{N}\\x&amp;\text{otherwise}\end{array}\right.$ and $g(x) = \left\{\begin{array}{rl}x &amp; \text{if }x\in\mathbb{N}\\1&amp;\text{otherwise}\end{array}\right.$
Then you have:

neither $f \in O(g)$, because whatever the constant $C$, $x\in]C,+\infty[\setminus \mathbb{N}\Rightarrow f(x)  &gt; Cg(x)$;
nor $f\in \overset{\infty}{\Omega}(g)$ because whatever the constant $C$, $n\in]\frac{1}{C},+\infty[\cap\mathbb{N}\Rightarrow f(n)&lt;Cg(n)$ (so there cannot be infinitely many integers so that $f(x) \geq Cg(n)$).

So the question here implicitly supposes that the big Oh notation considers that $n\in\mathbb{N}$ (and I think that's why the chosen notation is $n$).