## Completeness of the club filter without AC

Let $\kappa$ be a regular cardinal. If I understand correctly, the proof that the intersection of $<\kappa$ many club subsets of $\kappa$ is a club does not require AC. However, the proof that the club filter on $\kappa$ is $\kappa$-complete does ostensibly require AC because given a sequence of sets in the club filter, we need to pick out a sequence of clubs that they contain in order to show that their intersection is in the club filter. So I'd like to know, in the absence of AC, how strong is the assertion that the club filter on $\kappa$ is $\kappa$-complete? Does it imply $AC_{\kappa}$, or is it weaker? If weaker, then how much weaker?

Is there any well-known model of ZF without AC in which there is a proper class or better of regular cardinals $\kappa$ such that the club filter on $\kappa$ is $\kappa$-complete?

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This is not much of an answer, but it may be useful:

a. It is an open problem (asked by Woodin) whether ZF + DC suffices to prove that there is a regular cardinal larger than $\omega_1$. This is why, presently, we cannot verify whether ZF + DC suffices to prove that for a proper class of regular cardinals $\kappa$ we have that the club filter on $\kappa$ is $\kappa$-complete.

The best result towards Woodin's question that I am aware of is in Apter-Henle-Jackson, "The calculus of partition sequences, changing cofinalities, and a question of Woodin". Trans. Amer. Math. Soc. 352 (2000), no. 3, 969–1003.

In the paper it is shown that:

If ZF + AD is consistent, then so is ZF + DC + $\omega_1$ is the only regular uncountable cardinal $\le\aleph_{\omega_1 +1}$.

The argument uses a mixture of Radin forcing techniques and AD results, and it is not clear how far it can be generalized. The ideal result from their techniques would be that, starting with $L({\mathbb R})$ and assuming determinacy, there is a forcing extension with the same value of $\Theta$ and where DC holds and $\omega_1$ is the only regular uncountable cardinal below $\Theta$. It seems to me that this would still be a long way from a positive answer to Woodin's question.

b. If determinacy holds then, for example, $L({\mathbb R})$ is a model of DC where your question has a positive answer if we replace club with the weaker notion of $\omega$-club. To see this, given a cardinal $\kappa$, let $W_\kappa$ be the collection of subsets of $\kappa$ that are $\omega$-club in $\kappa$. This means that $A\subseteq\kappa$ is in $W_\kappa$ iff $A$ is unbounded in $\kappa$ and contains all its limit points of countable cofinality.

The argument goes back to results on the $\omega$-club filter that can be found in Kleinberg's nice little book "Infinitary combinatorics and the axiom of determinateness."

The partition relation $$\kappa\to(\kappa)^\omega_{\lt\kappa}$$ means the following: Suppose that $\lambda\lt\kappa$ and that $f:[\kappa]^\omega\to\lambda$. (Here, if $X\subseteq\kappa$, by ${}[X]^\omega$ we denote the collection of subsets of $X$ of order type $\omega$.) Then there is a subset $A$ of $\kappa$ of type $\kappa$ with $f$ constant on $[A]^\omega$. (We say that $A$ is homogeneous for $f$.)

This relation is false for any $\kappa$ in the presence of the axiom of choice. If $\kappa$ satisfies such a relation we say it is a strong partition cardinal. One can prove that strong partition cardinals are measurable, and this (essentially) gives the result you ask.

Suppose $\kappa\to(\kappa)^\omega_{\lt\kappa}$. Then $W_\kappa$ is an ultrafilter. To see this, let $A\subseteq\kappa$, and consider $f:[\kappa]^\omega\to2$ given by $f(p)=0$ iff $\sup(p)\in A$. Suppose that $C$ is homogeneous for $f$ of type $\kappa$. Denote by $(C)$ the set of suprema of $\omega$-sequences of elements of $C$. This is an $\omega$-club. If $f$ takes always the value 0 on $[C]^\omega$, then $(C)\subseteq A$. Else, $(C)\subseteq A^c$.

In fact, $W_\kappa$ is $\kappa$-complete. In effect, suppose $\{A_\alpha\mid\alpha\lt\lambda\}\subseteq W_\kappa$, where $\lambda\lt\kappa$. We need to check that $\bigcap_\alpha A_\alpha\in W_\kappa$. For this, define $f:[\kappa]^\omega\to\lambda$ as follows: Let $p\in[\kappa]^\omega$. If $\sup(p)\in\bigcap_\alpha A_\alpha$, let $f(p)=0$. Else, let $f(p)=\alpha+1$ where $alpha$ is least such that $\sup(p)\notin A_\alpha$. One argues that if $C$ is homogeneous of type $\kappa$, then $f$ always takes the value 0 on $[C]^\omega$, and $(C)$ is an $\omega$-club contained in $\bigcap_\alpha A_\alpha$. (Furthermore, $W_\kappa$ is normal, by an elaboration of this argument.)

The connection with determinacy is the following: In Kechris-Kleinberg-Moschovakis-Woodin, "The Axiom of Determinacy, strong partition properties and nonsingular measures". In Kechris-Martin-Moschovakis eds., Cabal Seminar 77–79. Proceedings, Caltech-UCLA Logic Seminar 1977–79. Lecture Notes in Mathematics 839. Berlin, Springer-Verlag 1981, pp. 77–99, it is shown that under AD+DC, the strong partition cardinals are unbounded below $\Theta$. (DC is not needed here if we assume the stronger determinacy axiom AD${}^+$.)

This gives that, under determinacy, in $L({\mathbb R})$, the regular cardinals $\kappa$ where $W_\kappa$ is $\kappa$-complete are unbounded below $\Theta$. But $L({\mathbb R})$ is a (symmetric) extension of its inner model HOD, which is a model of choice, by a forcing that adds a subset $A$ of $\Theta$, $L({\mathbb R})=HOD[A]$. Then one can use an analysis of the forcing poset (essentially, Vopenka) and the result above to see that, above $\Theta$, we have that if $\kappa$ is regular then $W_\kappa$ is again $\kappa$-complete.

Under AD, $L({\mathbb R})$ is a model of AD${}^+$, and the argument above generalizes to models of AD${}^+$ of the form $L({\mathcal P}({\mathbb R}))$, so there are plenty of natural models where the $\omega$-club version of your question has a positive answer. I may be wrong, but I suspect that the result for the club filters should follow by soft arguments from the result for the filters $W_\kappa$.

(If this is indeed the case, then your question of how much choice follows from the assumption of completeness of the filters is: Not even DC. In models of AD${}^+$, a weak version of DC known as DC${}_{\mathbb R}$ holds. I do not know if you can recover this version, but I doubt it.)

Of course, this begs the question of how more general this argument is. But currently, the only known models where we have (many) strong partition cardinals are models of determinacy. (And it is an interesting difficult open problem to see whether there are other models or even models with unboundedly many such cardinals.)

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 Thanks for comment (1). I don't know so much about ZF without AC. No regular cardinals above $\omega_1$ -- that's kind of disturbing. I'm not sure whether my modification works better than what I had previously. – Norman Lewis Perlmutter Nov 24 2010 at 1:57 As for comment (2), it is interesting, thanks. – Norman Lewis Perlmutter Nov 24 2010 at 2:00 @Norman : I've added some details. – Andres Caicedo Nov 25 2010 at 7:00