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We know that from axiom of choice (or just BPIT) we can deduce ultrafilter lemma, which states that every filter can be extended to an ultrafilter. From this lemma we can derive the existence of at least 2 different ultrafilters (example: take an ultrafilter containing set of even numbers and ultrafilter containing set of odd numbers. They are distinct, as otherwise it would contain a set together with its complement).

I was wondering if it is possible (without assuming ultrafilter lemma) that there exist exactly one ultrafilter on $\omega$, or maybe that from one ultrafilter we can always construct another? My exact question is: Is it consistent with ZF that there exists exactly one ultrafilter on $\omega$?

dankness
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Wojowu
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    From any permutation $\pi$ of $\omega$ we will be able to construct other (nonprincipal) ultrafilters from a given ultrafilter $\mathcal{U}$: take $\mathcal{V} = { \pi [A] : A \in \mathcal{U} }$. If $\pi[A] \notin \mathcal{U}$ for some $A \in \mathcal{U}$, then $\mathcal{V} \neq \mathcal{U}$, although in some sense $\mathcal{U}$ and $\mathcal{V}$ are "the same" (at least they are "trivially isomorphic"). – user642796 May 03 '14 at 16:20

1 Answers1

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If there exists one free ultrafilter on $\omega$, then there are at least $2^{\aleph_0}$.

This appears as an exercise in Herrlich's The Axiom of Choice, section 4.3.

There he hints that one should use the fact that there exists an almost disjoint family of sets of integers of cardinality $2^{\aleph_0}$ which is true without the axiom of choice (note, we don't require it to be maximal, just almost disjoint).

Suppose that $\{A_i\mid i\in I\}$ is such family, and $\cal U$ is a free ultrafilter on $\omega$, then by a bijection we can generate $\cal U_i$ which is a free ultrafilter on $A_i$, and note that $\{X\mid X\cap A_i\in\cal U_i\}$ is a free ultrafilter on $\omega$. Moreover, if $i\neq j$ then $\cal U_i\neq U_j$ for obvious reasons (e.g. $A_i\in\cal U_i$ but not in any other $\cal U_j$).

Asaf Karagila
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