For prime number $p \in \mathbb{P}$, define a countably infinite subset $A_p \subset \mathbb{N}$
$A_p = \{p^n \mid n \in \mathbb{N} + 1\}$
For example, $A_2 = \{2^n \mid n \in \mathbb{N} + 1\}$.
It is known that
- the cardinality of this example set is countable infinity: $|A_2| = \aleph_0$
- the natural density of this example set is zero: $d(A_2) = 0$.
- finite additivity is satisfied: $d(A_2 \cup A_3) = d(A_2) + d(A_3) = 0$.
However, imagine $\bigcup_{p \in \mathbb{P}} A_p$ (Note that $A_p \cap A_q = \emptyset$ for any $p \ne q \in \mathbb{P}$).
$\bigcup_{p \in \mathbb{P}} A_p = \{2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,\cdots\}$
It seems dense in natural number.
Is the density of this countable union really zero: $d(\bigcup_{p \in \mathbb{P}} A_p) = 0$?
Is there any possibility that there is a infinite-superadditivity in natural density: $d(\bigcup_{p \in \mathbb{P}} A_p) \ge \sum_{p \in \mathbb{P}} d(A_p) = 0$?
