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 $|\mathcal{P}(\mathbb{N})| > \aleph_\omega$?
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2 Answers
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It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$. More generally Easton has shown that the only constraints on exponentiation of regular cardinals is that it be monotonic and satisfy $\mathrm{cof}(2^\kappa) > \kappa$. For singular cardinals, the matter is much more subtle. For instance one of Shelah's celebrated results is that if $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega} < \aleph_{\omega_4}$ (and no, that $4$ is not a misprint).
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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.)
