Assume ZF together with some choice principle needed to make the choice in the following recursion that constructs a free ultrafilter on $\Bbb{N}$

  • $\mathcal{F}_0$ is the filter of co-finite subsets of $\Bbb{N}$.
  • For any ordinal $\beta$ assume $\mathcal{F}_{\beta}$ has already been constructed and define $\mathcal{F}_{\beta+1}$ as follows. If $\mathcal{F}_\beta$ is already an ultrafilter let $\mathcal{F}_{\beta+1} = \mathcal{F}_{\beta}$. Otherwise there is some $C\subseteq \Bbb{N}$ such that $C\not\in\mathcal{F}_\beta$ and $C^c\not\in\mathcal{F}_\beta$ where $C^c = \Bbb{N}\setminus C$. It is possible to show that $C,C^c$ and both infinite and furthermore $C\cap F, C^c\cap F$ are both infinite for every $F\in\mathcal{F}_\beta$. So choose one of $D=C$ or $D=C^c$ using the presumed choice machinary, and let $\mathcal{F}_{\beta+1}=\{A\subseteq \Bbb{N}:\exists F\in\mathcal{F}_\beta\; (F\cap D\subseteq A)\}$.
  • For a non-zero limit ordinal $\alpha$, if $\mathcal{F}_\beta$ has been constructed for all $\beta < \alpha$, let $\mathcal{F}_\alpha = \bigcup_{\beta<\alpha} \mathcal{F}_\beta$.

Finally, define the length of the above recursion as the least ordinal $\beta$ such that $\mathcal{F}_\beta = \mathcal{F}_{\beta+1}$.

I'm wondering what is known about ordinals $\lambda$ that can be a length of this process. Since each recursion step corresponding to the transition $\beta\to\beta+1$ chooses a distinct member of $\mathcal{P}(\Bbb{N})$, then $\lambda<\kappa$ where $\kappa$ is the Hartogs number of $2^{\aleph_0}$. My questions are

  1. Is the length $\lambda$ always a limit ordinal or can there be a 'last step' that transitions from $\lambda-1$ to $\lambda$? Is the length always a cardinal?
  2. What is the smallest $\lambda$ that can be a length?
  3. Is the supremum of all possible lengths equal to $\kappa$ or can it be strictly smaller?

Regarding the 3rd question, if $\nu$ is the supremum over all possible lengths and $\nu<\kappa$, then again, taking into account that the recursion chooses a distinct subset of $\Bbb{N}$ at each incremental step, the recursion can construct a number of ultrafilters which is at most the number of sequences from $\nu$ to $2^{\aleph_0}$, but this is $$(2^{\aleph_0})^{|\nu|} = 2^{\aleph_0 |\nu|} = 2^{|\nu|}$$ and I'd like to say that since $\nu<\kappa$ then $|\nu| < 2^{\aleph_0}$ unless $2^{\aleph_0}$ is well-orderable, and then $2^{|\nu|} < 2^{2^{\aleph_0}}$ so the process can't generate all free ultrafilters on $\Bbb{N}$. However, the deduction $|\nu| < 2^{\aleph_0}\rightarrow 2^{|\nu|} < 2^{2^{\aleph_0}}$ is probably incorrect, and it's possible that $2^{\aleph_0}$ is well-orderable, so $|\nu|=2^{\aleph_0}$.

What's already known about this?

  • 1
    $\begingroup$ Let me point out that there is no particular reason to expect this sort of recursive construction to work in a situation where $\cal P(\omega)$ cannot be well-ordered. Recursion works hand-in-hand with well-orders, whereas choice principles like the Boolean Prime Ideal theorems simply assert the existence of an object. I can imagine there can be a situation where no ultrafilter base is well-orderable, but ultrafilters exist nonetheless (although I don't have any concrete example right now). $\endgroup$
    – Asaf Karagila
    Jul 1, 2023 at 18:19
  • $\begingroup$ This is related, although I think not quite identical to, the cardinal characterstic $\mathfrak{u}$; see this MO discussion. $\endgroup$ Jul 1, 2023 at 18:19
  • $\begingroup$ @AsafKaragila Maybe $L(\mathbb{R})[\mathcal{U}]$ assuming large cardinals? $\endgroup$ Jul 1, 2023 at 18:20
  • $\begingroup$ @Noah: Probably. I'd reckon the Cohen model would be an example, too. Or that at the very least there is a LC-free example. $\endgroup$
    – Asaf Karagila
    Jul 1, 2023 at 18:32
  • 1
    $\begingroup$ My point is that the assumption that this process even yields all the ultrafilters (and under the assumption that these exist, of course) is already somehow suspect to well ordering the continuum. $\endgroup$
    – Asaf Karagila
    Jul 2, 2023 at 12:04

1 Answer 1


Is the length $\lambda$ always a limit ordinal or can there be a 'last step' that transitions from $\lambda-1$ to $\lambda$? Is the length always a cardinal?

The length can be a successor ordinal. Indeed, if $F$ and $G$ are two distinct nonprincipal ultrafilters then the only proper filters that strictly extend the filter $F\cap G$ are $F$ and $G$ (this follows immediately from Stone duality, for instance). So you can start by a recursion that generates the filter $F\cap G$, and then it will take only one more step to reach the ultrafilter $F$, since you just need to adjoin any element of $F\setminus G$.

What is the smallest $\lambda$ that can be a length?

Equivalently, you are asking for the smallest (well-orderable) cardinality of a subset of $\mathcal{P}(\mathbb{N})$ which generates a nonprincipal ultrafilter (since given such a set, you can well-order it and then take a recursion that adjoins its elements in that order). This is a cardinal characteristic of the continuum known as the ultrafilter number $\mathfrak{u}$. Its value is independent of ZFC; for instance, MA implies $\mathfrak{u}=2^{\aleph_0}$ but it is also consistent to have $\mathfrak{u}<2^{\aleph_0}$ (see here for more details and a reference).

  1. Is the supremum of all possible lengths equal to $\kappa$ or can it be strictly smaller?

It must be equal to $\kappa$. There exists a family $I$ of $2^{\aleph_0}$ elements of $\mathcal{P}(\mathbb{N})/\mathrm{fin}$ which are independent, i.e. they freely generate a Boolean subalgebra (see here for instance). This family can then be adjoined in any order to generate a filter, and by independence none of these generators will be redundant. So for any ordinal $\alpha<\kappa$ (i.e. for any well-ordering of a subset of $I$), this gives a recursion of length $\alpha$ to generate a filter, which can then be extended further to generate an ultrafilter.


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