Feferman 1965: ZF does not prove that a free ultrafilter on the naturals exists

If ZF is consistent, then ZF does not prove that there is a free (that is, non-principal) ultrafilter on $\mathbb{N}$.

Feferman (1965), using Cohen's forcing, produces a model of ZF in which every ultrafilter on $\mathbb{N}$ is principal. The forcing adjoins countably many Cohen reals $x_n \subseteq \mathbb{N}$ by finite partial functions $\mathbb{N} \times \mathbb{N} \to \{0,1\}$, but the symmetry group is not a group of permutations of the indices: it is the group of automorphisms obtained by flipping the generic bits on an arbitrary set of coordinates, with supports the finite subsets of $\mathbb{N}$. A hereditarily symmetric name for an ultrafilter on $\mathbb{N}$ has such a finite support, and for any index $n$ outside that support there is a flip that fixes the name while replacing $x_n$ by a set differing from it on a cofinite set. The purported ultrafilter would then have to contain both, which forces it to be principal.

Consequence, and this is the form the library needs. The ultrafilter lemma (UL), that every filter on a set extends to an ultrafilter, produces a free ultrafilter on $\mathbb{N}$ from the filter of cofinite sets. So, if ZF is consistent, UL is not a theorem of ZF, and neither is its equivalent, the Boolean prime ideal theorem.

• This is not the basic Cohen model. The distinction is load-bearing. The basic Cohen model uses the same forcing but the group of permutations of the indices, and by Halpern and Lévy 1971: the Boolean prime ideal theorem does not imply the Axiom of Choice ‡ the Boolean prime ideal theorem, hence the ultrafilter lemma, holds there, so the basic Cohen model does contain free ultrafilters on $\mathbb{N}$. Only the flip-symmetric model above kills them. Reading Feferman's model as "the Cohen model" would put two items on this page in direct contradiction.

• Not proved in this library. Forcing and symmetric models are not developed here.

• What would prove it. The forcing track named in Cohen 1963: ZF does not prove the Axiom of Choice ‡, plus the symmetry argument above: no hereditarily symmetric name can decide the membership of every subset of $\mathbb{N}$ in a purported ultrafilter.

• Why it matters here. What the ultrafilter lemma costs: a choice principle strictly weaker than AC ↗ cites exactly this result for its first bullet, "UL is not a theorem of ZF", which is one of the two poles that locate UL strictly between ZF and the Axiom of Choice; the other pole is Halpern and Lévy 1971: the Boolean prime ideal theorem does not imply the Axiom of Choice ‡. It is also what makes FALSE, once the ultrafilter lemma is available: every ultrafilter is principal ↗ an unusual item: that statement is false in ZFC, since The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter ↗ refutes it, and yet it is consistent with ZF, so the refutation genuinely consumes a choice principle and cannot be made choice-free. The strengthening from $\mathbb{N}$ to arbitrary sets is Blass 1977: a model of ZF with no free ultrafilter on any set ‡.

• Conditional discipline. The statement is relative to the consistency of ZF. This library never says "free ultrafilters do not exist"; it says that ZF alone cannot produce one, and that Ultrafilter ↗ is satisfied non-principally only once a choice principle is available.