In CLRS, one question defines $\Omega^\infty(g(n))$ such that: $$f(n) = \Omega^\infty (g(n)) \iff \exists c \in \mathbb{R}^+ : \text{for infinitely many integers } n, 0 \leq cg(n) \leq f(n) \ .$$

My question: What does it mean by "infinitely many integers"? Does it mean $\{n : n \in \mathbb{Z}\}$ or does it just mean some infinite subset of $\mathbb{Z}$ (say, for instance, every odd positive integer)?

If the latter is true, then why is $\Omega^\infty$ even useful? I'm struggling to find out what knowing the behaviour of $f(n)$ for some arbitrary infinite scattering of integers could possibly achieve.

If the former is true, why do we care about what a function $f(n)$ does for small $n$ (i.e. for $n < n_0$)? Isn't the whole point of asymptotics that we're looking for the behaviour of functions as they approach infinity?

Can someone give a bit of context to the definition of $\Omega ^\infty$, because at the moment, it just seems a very contrived definition.


1 Answer 1


The phrase "for infinitely many integers" means "for an infinite set of integers". This strange definition is what you get when you negate little $o$. That is, for any two functions $f,g \geq 0$, either $f = o(g)$ or $f = \Omega^\infty(g)$. In some cases $f = \Omega^\infty(g)$ is enough to derive certain consequences, and so is a useful property.

  • $\begingroup$ I can't see the difference between $\Omega$ and $\Omega^\infty$ here. $\endgroup$
    – rus9384
    Commented Jul 8, 2017 at 23:09
  • 1
    $\begingroup$ Common $\Omega$ holds for all $n \n_0$. In contrast, $\Omega^\infty$ could hold only for even $n$, for example. $\endgroup$ Commented Jul 9, 2017 at 4:32

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