Alphabeta Math
RemarkSession-authored (Fable 5 assisted) sources checked 2026-07-26 not proved here
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

The Feferman-Levy model: the reals as a countable union of countable sets

Statement

If ZF is consistent, then ZF is consistent with all of the following holding simultaneously:

  • R\mathbb{R} is a countable union of countable sets;
  • the first uncountable ordinal ω1\omega_1 is singular, indeed cf(ω1)=ω\mathrm{cf}(\omega_1) = \omega;
  • consequently "a countable union of countable sets is countable" fails, and ω1\omega_1 is not regular.

Feferman and Levy (1963) obtain this by collapsing: starting from a model of ZFC, force with the finite-support product that makes each n\aleph_n of the ground model countable, for nNn \in \mathbb{N}, and take the symmetric submodel with finite supports. The ground model's ω\aleph_\omega becomes the new ω1\omega_1, and it is the supremum of the countably many ordinals n\aleph_n, each now countable, so its cofinality is ω\omega. The reals of the extension are the union over nn of the reals added at stage nn, and each of those layers is countable in the extension.

Note what does not fail: ω1\omega_1 still exists, and R\mathbb{R} is still uncountable. A countable union of countable sets is being exhibited whose union is uncountable, which is possible exactly because no enumeration of the layers can be chosen uniformly.

Remarks

Depends on

Used by

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Sources