Variation of omega definition The same question has been asked here


  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 $\overset{\infty}{\Omega}$


The solution given in the link hints that we can negate the definition of $ \overset{\infty}{\Omega}$ which gives us the definition of big O. 

But negation of the definition of $ {\Omega}$ also gives us the same result, doesn't it?

I cannot wrap my head on how the condition of infinitely many integers makes difference rather than saying for all n greater than certain $n_0$.

Please help me understand this.
When you negate $\Omega$ you get $\stackrel{\infty}{O}$, the analog of $\stackrel{\infty}{\Omega}$. I suggest writing the definition of $\Omega$ as a first-order formula and then negating it.

To see the difference between $\Omega$ and $\stackrel\infty\Omega$, consider the functions $f(n) = n$ and $$g(n) = \begin{cases} 1 &amp; \text{$n$ even,} \\ n^2 &amp; \text{$n$ odd.} \end{cases}$$
You can check that $f(n) \neq O(g(n))$ and $f(n) \neq \Omega(g(n))$, whereas $f(n) = \stackrel\infty O(g(n))$ and $f(n) = \stackrel\infty\Omega(g(n))$.