Is the cardinal $2^{\aleph_0}$ regular, working in ZFC?
2026-04-08 02:31:49.1775615509
Is the cardinal $2^{\aleph_0}$ regular, working in ZFC?
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Not necessarily - all ZFC can prove is that the cofinality of $2^{\aleph_0}$ is uncountable. (That $cof(2^{\aleph_0})>\omega$ follows from König's theorem; that this is the only restriction provable from ZFC is due to Solovay, and a huge strengthening of this result is due to Easton.) It is consistent, for example, that $$2^{\aleph_0}=\aleph_{\omega_1},$$ which is of course singular.