TITLE: Does the Prime Number Theorem have anything to do with Erdos-Kac law or vice versa?
QUESTION [10 upvotes]: The prime number theorem says on average we can find $\frac n{\log n}$ primes of magnitude $n$.
Erdos-Kac law state a typical number of magnitude $n$ has $\log\log n$ primes.
Somehow the fact $e^{\log\log n}=\log n$ seems to make both facts related. 
How related are these two facts in number theory?

Let me put it in this way.
Suppose we somehow naturally choose a subset of natural numbers where Erdos-Kac law applies but with average factors as $f(\log\log n)$ for some $f(x)=o(x)$ then how many primes of magnitude $n$ can we expect in this subset? Say for instance $f(\log\log n)=(\log\log n)^\alpha$ or $f(\log\log n)=(\log\log\log n)^{1/\alpha}$ where $0<\alpha\ll1$ holds?
I think such sets should typically have $\frac{n}{2^{f(\log\log n)}}\gg\frac{n}{\log n}$ primes.

REPLY [10 votes]: The right context of your question is probably the area of Beurling primes. An arithmetic semigroup is a semigroup $(G,\cdot)$ together with a norm $\|\cdot\|:G\rightarrow[1,\infty)$, such that $\|gh\|=\|g\|\|h\|$ and for all $x$ we have that $N(x) = \#\{g\in G: \|g\|\leq x\}$ is finite. Define $\pi(x)$ as the number of indecompsable elements of $G$ with norm $\leq x$. Beurling proved that $N(x)=cx+\mathcal{O}(\frac{x}{x^{3/2+\delta}})$ implies the prime number theorem in the form $\pi(x)\sim\frac{x}{\log x}$, and the condition is best possible in the sense that for every $\delta>0$ there are groups satisfying $N(x)=cx+\mathcal{O}(\frac{x}{x^{3/2-\delta}})$ and $\pi(x)\not\sim\frac{x}{\log x}$.
The theory of Beurling primes has two motivations: On one hand it has applications to algebra and logic, see the books "Abstract analytic number theory" by Knopfmacher and "Number theoretic densities and logical limit laws" by Burris. On the other hand it serves as a context in which "equivalence" of statements about primes can be made precise.
In this context your question becomes: Is the set of arithmetic semigroups satisfying the prime number theorem equal to the set of semigroups satisfying Erdos-Kac? Erdos-Kac is essentially equivalent to $\sum_{p\leq x}\frac{1}{p}\sim\log\log x$, which is obviously implied by the prime number theorem, but Pollack ( http://pollack.uga.edu/beurling.pdf Wayback Machine ) showed that $\sum_{p\leq x}\frac{1}{p}\sim\log\log x$ is equivalent to the statement $\zeta_G(s)\sim\frac{A}{s-1}$, where $\zeta_G(s)=\sum_{g\in G}\|g\|^{-s}$. In particular all semigroups with $N(x)\sim Ax$ satisfy Erdos-Kac, while the counterexamples found by Beurling show that Erdos-Kac is strictly weaker then the prime number theory.