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 as we know it $\mathbb{N}$.

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 $P^{m_0}$?

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 $P^{m_0}$?