Prove that: $|P(\mathbb{R}) \setminus P(\mathbb{R}\setminus \mathbb{Z})| = 2^{2^{\aleph_0}}$
It is obvious that: $P(\mathbb{R}) \setminus P(\mathbb{R}\setminus \mathbb{Z}) \subseteq P(\mathbb{R})$ so: $|P(\mathbb{R}) \setminus P(\mathbb{R}\setminus \mathbb{Z})| \le 2^{2^{\aleph_0}}$
But I don't know how to prove that it is also $\ge 2^{2^{\aleph_0}}$.