TITLE: Is it possible for P(N) to be larger than Aleph_omega?
QUESTION [5 upvotes]: I have seen a proof that $|\mathcal{P}(\mathbb{N})| \neq \aleph_\omega$ using the fact that $\aleph_\omega$ is the union of countably many smaller cardinals, while $|\mathcal{P}(\mathbb{N})|$ is not. Is it consistent with ZFC that $|\mathcal{P}(\mathbb{N})| > \aleph_\omega$?

REPLY [13 votes]: Yes.  In fact, the only requirement on $2^{\aleph_0}$ is that $cf(2^{\aleph_0})>\aleph_0$.  (Cohen's argument can make the power set any regular cardinal, and I think it requires at most small modifications to handle the general case.)