In a book it is written that by writing $f(n)=\overset{\infty}{\Omega}(g(n))$ we mean that there exists a positive constant $c$ such that $f(n) \ge c\times g(n) \geq 0$ for infinitely many integers $n$. Whereas a Big Omega definition says that there exist positive constants $c$ and $n_0$ such that $0 \leq c\times g(n) \leq f(n)$ for all $n \geq n_0$.
My question is why is it not right to say that these definitions are equal when for the Big Omega definition there is a $n \geq n_0$ condition for $n$, which in my opinion tells that the definition must be true for infinitely many $n$'s that are just greater or equal to $n_0$.
Suppose $f(n)=1$ and $g(n)=0$ when $n$ is even. Suppose $f(n)=0$ and $g(n)=1$ when $n$ is odd. With $c=1,$ the sets $\{n\in \Bbb N: f(n)\ge cg(n)\ge 0\}$ and $\{n\in \Bbb N: g(n)\ge cf(n)\}$ are both infinite. But we do NOT have $f=O(g)$ nor $g=O(f).$
The Landau "big-$O$" notation is only concerned with the absolute values of the functions, and may be used when the variable $n$ can take non-integer values, and may be used when the variable approaches $r\in \Bbb R.\,$
If it is understood that $n$ must be an integer then "$f(n)=O(g(n))$ as $n\to \infty$" means $$\exists n'\in \Bbb N\, \exists c>0\,\forall n\in \Bbb N \,(n>n'\implies |f(n)|\le c|g(n)|\,).$$
If the variable can take any real value, it is customary to use $x,$ not $n$. So "$f(x)=O(g(x))$ as $x \to\infty$" means $$\exists x'\in \Bbb R^+\,\exists c>0\,\forall x\in \Bbb R\,(x>x'\implies |f(x)|\le c|g(x)|\,).$$
If $r\in \Bbb R$ then "$f(x)=O(g(x))$ as $x\to r$" means $$\exists \delta >0 \, \exists c>0 \,\forall x\in \Bbb \,(0<|x-r|<\delta \implies |f(x)|\le c|g(x)|\,).$$ For example $(\sin x)^2/x= O(x)$ as $x\to 0.$ In some contexts $x$ may be restricted to values only $>r.$