Variations of Omega and Omega infinity

Some authors define $\Omega$ in a slightly different way: let’s use
$ \overset{\infty}{\Omega}$
(read “omega infinity”) for this alternative definition. We say that $f(n) = \overset{\infty}{\Omega}(g(n))$
if there exists a positive constant $c$ such that $f(n) \geq c\cdot g(n) \geq 0$ for infinitely many integers $n$, whereas the usual $\Omega$ requires that this holds for all integers greater than a certain $n_0$.

Show that for any two functions $f(n)$ and $g(n)$ that are asymptotically nonnegative,
either $f(n) = O(g(n))$ or $f(n)= \overset{\infty}{\Omega}(g(n))$ or both, whereas this is not true if we use $\Omega$ in place of $\overset{\infty}{\Omega}$.

I am trying learn Algorithms. But I am unable to prove this. Can the experts help me ?

Hint: If $f(n) \notin \overset{\infty}{\Omega}(g(n))$ and $g(n)$ is asymptotically non-negative, then for all positive constants $c$, $f(n) \leq c \cdot g(n)$ for large enough $n$. This follows by ignoring the condition $c \cdot g(n) \geq 0$ and negating the definition of $f(n) \in \overset{\infty}{\Omega}(g(n))$. In fact, this way you get the stronger result that either $f(n) \in \overset{\infty}{\Omega}(g(n))$ or $f(n) \in o(g(n))$ (but not both).

Further hint: You can start by showing that the negation of "$P(n)$ for infinitely many $n$" is "$\lnot P(n)$ for large enough $n$".

StackExchange Computer Science Q#12147, answer score: 6