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I am studying Turan's proof of the fact that "almost all" $n$ have "very close to" $\ln \ln n$ prime factors. The statement of the theorem is:

Let $\omega (n) \rightarrow \infty$ arbitrarily slowly. Then the number of $x$ in $\{1, 2, ..., n \}$ such that $$|\nu(x)-\ln\ln n|>\omega(n)\sqrt{\ln\ln n}$$ is $o(n)$. Here $\nu(x)$ is the number of distinct prime factors of $x$.

What does the first line (in bold) mean?

JDF
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    Can you give more context? At this point, I would assume this just means "Let $\omega(n)$ be a function satisfying $\omega(n) \to \infty$", with "arbitrarily slowly" just being a hint that there are no assumptions on the growth rate. – Nate Eldredge Sep 20 '16 at 20:10
  • Without context, the statement presumably means nothing more than $\omega\to\infty$, with no conditions on its growth rate. – anomaly Sep 20 '16 at 20:12
  • @NateEldredge, I've edited and given the whole statement. – JDF Sep 20 '16 at 20:12
  • I think it means for any value M we can find an n where $\omega (n) $ is more or less equal to M. In other $\omega $ doesn't "quickly" jump from $\omega (n) $ being "reasonably average in size" to $\omega (n+1) $ big "really really huge". That's what I think it means. – fleablood Sep 20 '16 at 20:15
  • I haven't even understood what $\omega(n)$ refers to. Does it mean I am taking a function $f(n)$ which is $\omega(n)$ and which goes to infinity arbitrarily slowly? – JDF Sep 20 '16 at 20:17
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    Maybe a way to formally state it is f (x)- > infinity but f'(x) - > 0. f '(0) is the derivative which is "the rate of change". If f'(x) - > 0 then we can find points as n gets large where the rate of change is arbitrarily small. I.e the function approaches infinity "arbitrarily slowly". – fleablood Sep 20 '16 at 20:21
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    Yeah... take 2^x ,x^2, x, root(x) and ln (x). These all approach infinity but some "do it faster" . We want to compare how fast |v (x) - ln ln x)| approaches infinity compared to how fast root( ln ln x) approaches infinity. We tweak them into more or less the same range by scaling the root (ln ln x) function by multiplying it by a function w that also approaches infinity but at a rate that makes the rate of the two sides relatively comparable. – fleablood Sep 20 '16 at 20:36
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    I don't think $\omega(n)$ is meant as Landau notation here, if that's what you're wondering; it's just an arbitrary function named $\omega$. – Nate Eldredge Sep 20 '16 at 22:55

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