Do not understand why log n = O(n^c) (for any c>0)

Problem Detail: 

Can anyone help me understand this equation?

$\log (n) = O(n^c)$ (for any $c>0$)

Does it mean that $O(\log (n)) < O(n^c)$ (for any $c>0$)?

Added:

Please also prove that $\log (n) = O(n^c)$ is true.

Asked By : Xin

Answered By : Luke Mathieson

It's not actually an equation $f = O(g)$ is a lazy shorthand that should be written $f \in O(g)$. So if you look back at the definition of $O$, you should be able to see what $\log n \in O(n^{c})$ for any $c > 0$ means:

For every $c > 0$, there exists $n_{0} \geq 0$ and $k \geq 0$ such that $\log n \leq k\cdot n^{c}$ for all $n \geq n_{0}$.

Always remember that $O(\cdot)$ describes a set, $O(\log n) < O(n^{c})$ doesn't actually make sense (unless you make up a special meaning for $<$, but then no-one will know what you're talking about). You could say $O(\log n) = O(n^{c})$, as equality for sets has a understood meaning, though of course this statement in particular would be false.

As John Kugelman points out in the comments below, normal set relations do make sense, so $O(f) \subset O(g)$, $O(f) \subseteq O(g)$, etc., make sense.

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Question Source : http://cs.stackexchange.com/questions/41588