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Is it true that for some function $f:N\to N$, a positive function, then either one of these conditions hold. First is that there is some $a>0$, such that $f(n)\in O(n^a)$. Or Second, there is some $b>1$ such that $f(n)\in\Omega(b^n)$.

So what I understand from this is that $O$ is like the upper bound and $\Omega$ is the lower bound. So now, if we have some positive function say $f(n)=\log(n)$, then one of these conditions always holds? Doesn't the first condition hold for $\log$? Is there a way to prove this for any positive function?

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    $\log n$ is not $\mathbb N\to\mathbb N$ ... – 2017-01-31
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    @HenningMakholm Oh that's true. Then how would I go about proving this? – 2017-01-31

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No, that is not true -- a counterexample would be $$ f(n) = \begin{cases} 1 & n\text{ odd} \\ 2^n & n\text{ even} \end{cases}$$

With a bit more work one can construct strictly increasing examples too, such as this recurrence: $$ f(0)=1 \qquad f(n+1) = \begin{cases} 2^n & \text{if }f(n)\le 2n \\ f(n)+1 & \text{otherwise} \end{cases} $$

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    Umm. Hold on. I meant to ask a slightly different question. Should I ask a new one? – 2017-01-31
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    @user19289: Yes, it's better to ask a new question once the original has been answered. (Unless my recent addition answers what you really meant to ask, of course :-) – 2017-01-31
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    Yes, i'll do that – 2017-01-31