TITLE: Lower density of {primes} times themselves
QUESTION [6 upvotes]: We say that a set $A\subseteq \mathbb{N}$ has lower density 0 if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 0.$$
Given $A,B\subseteq \mathbb{N}$ we set $A\cdot B = \{a\cdot b: a\in A, b\in B\}$. The set $A^n$ for $n\in\mathbb{N}$ is defined inductively in the obvious manner.
Let $P$ be the set of prime numbers in $\mathbb{N}$. What is the smallest positive $m_0\in\mathbb{N}$ such that the lower density of $P^{m_0}$ is $>0$? And what is the lower density of $P^{m_0}$?

REPLY [2 votes]: The following results are from the paper Ivan Niven, The asymptotic density of sequences. Bull. Amer. Math. Soc., 57(6):420-434, 1951. I changed the notation to denote upper asymptotic density by $\overline d(A)$ and asymptotic density by $d(A)$. (Which seems to be used more frequently nowadays than the notation $\delta_2(A)$ and $\delta(A)$ from that paper.)

For any prime $p$ let $A_p$ denote the subset $A_p=\{n\in A; p\mid n, p^2\nmid n\}$.
Theorem 1. If $\{p_i\}$ is a set of primes such that $\sum\frac1{p_i}=\infty$, then $\overline d(A)\le \sum\overline d(A_{p_i})$ for any $A$.
Corollary 1. If a set of primes $\{p_i\}$ we have $d(A_{p_i})=0$ for every $i$, and if $\sum p_i^{-1}=\infty$ then $d(A)=0$.
Corollary 2. For any fixed $k$, if $\{p_i\}$ is a set of primes such that $\sum\frac1{p_i}=\infty$, and if $A$ is any set whose members are divisible by at most $k$ of these primes to the first degree, then $d(A)=0$.

From Corollary 2 it follows that $d(P^{m_0})=0$ for each $m_0$. (In the other words, Corollary 2 implies that the set of all numbers having at most $m_0$ prime factors has density zero. Niven also mentions in connection with this weaker result the paper Willy Feller, Erhard Tornier: Mengentheoretische Untersuchung von Eigenschaften der Zahlenreihe, Mathematische Annalen 1933, Volume 107, Issue 1, pp 188-232.)