# 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?

If $M$ is an inner model of a forcing extension $V[G]$ of a model $V$ of Choice via a partial order of cardinality $\kappa$ in $V$, then for every regular cardinal $\lambda$ of $V$ greater than $\kappa$, every club subset of $\lambda$ in $V[G]$ will contain one from $V$. So, in $M$, the club filter on $\lambda$ will be $\lambda$-complete for any such $\lambda$, as $V[G]$ can carry out the usual argument with the sets from $M$, and then shrink the result to a set in $V$. A similar situation holds for models of ZF which have set-forcing extensions satisfying Choice. There are many such models of each type in which AC fails, for instance the Feferman-Levy model, in which $\omega_{1}$ is singular.

Andres's answer is more interesting, of course.

• Don't you have to choose the sets from $V$? – Asaf Karagila Feb 18 '15 at 8:54
• I think you take your collection (in $M$) of less than $\lambda$ many subsets of $\lambda$, each containing a club subset of $\lambda$, let $V[G]$ pick the clubs and intersect them, and shrink the intersection to a set in $V$, which is contained in $M$. – Paul Larson Feb 19 '15 at 3:47

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

a. It is an open problem (asked by Woodin) whether $\mathsf{ZF} + \mathsf{DC}$ suffices to prove that there is a regular cardinal larger than $\omega_1$. This is why, presently, we cannot verify whether $\mathsf{ZF} + \mathsf{DC}$ 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

Arthur W. Apter, James M. Henle, and Stephen C. Jackson. The calculus of partition sequences, changing cofinalities, and a question of Woodin, Trans. Amer. Math. Soc., 352 (3), (2000), 969–1003. MR1695015 (2000j:03063).

In the paper it is shown that:

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

The argument uses a mixture of Radin forcing techniques and $\mathsf{AD}$ 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 $\mathsf{DC}$ 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 $\mathsf{DC}$ 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

Eugene M. Kleinberg. Infinitary combinatorics and the axiom of determinateness. Lecture Notes in Mathematics, Vol. 612. Springer-Verlag, Berlin-New York, 1977. MR0479903 (58 #109).

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. One can call a cardinals $\kappa$ satisfying such a relation a partition cardinal. And one can prove that 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 $\bigcap_\alpha A_\alpha\in W_\kappa$, the trouble being that without choice we cannot simply argue by picking $\omega$-clubs contained in the $A_\alpha$ and proceed to intersect them. 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: See

Alexander S. Kechris, Eugene M. Kleinberg, Yiannis N. Moschovakis, and W. Hugh Woodin. The axiom of determinacy, strong partition properties and nonsingular measures. In Cabal Seminar 77–79 (Proc. Caltech-UCLA Logic Sem., 1977–79), Alexander S. Kechris, Donald A. Martin, and Yiannis N. Moschovakis, eds., pp. 75–99, Lecture Notes in Math., 839, Springer, Berlin-New York, 1981. MR0611168 (83f:03047). Reprinted in Games, scales, and Suslin cardinals. The Cabal Seminar. Vol. I, Alexander S. Kechris, Benedikt Löwe and John R. Steel, eds. Lecture Notes in Logic, 31. Association for Symbolic Logic, Chicago, IL; Cambridge University Press, Cambridge, 2008. MR2463612 (2010b:03003).

In this paper, it is shown that under $\mathsf{AD}+\mathsf{DC}$, the strong partition cardinals are unbounded below $\Theta$. ($\mathsf{DC}$ is not needed here if we assume the stronger determinacy axiom $\mathsf{AD}^+$.)

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 $\mathsf{HOD}$, 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, the Vopěnka algebra) and the result above to see that, above $\Theta$, we have that if $\kappa$ is regular then $W_\kappa$ is again $\kappa$-complete.

Under $\mathsf{AD}$, $L({\mathbb R})$ is a model of $\mathsf{AD}^+$, and the argument above generalizes to models of $\mathsf{AD}^+$ 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 $\mathsf{DC}$. In models of $\mathsf{AD}^+$, a weak version of $\mathsf{DC}$ known as $\mathsf{DC}_{\mathbb R}$ holds. I do not know if you can recover this version, but I doubt it.)

For an introduction to $\mathsf{AD}^+$, I suggest

Andrés Eduardo Caicedo, and Richard Ketchersid. A trichotomy theorem in natural models of $\mathsf{AD}^+$. In Set theory and its applications. Papers from the Annual Boise Extravaganzas (BEST) held in Boise, ID, 1995–2010. Liljana Babinkostova, Andrés E. Caicedo, Stefan Geschke, and Marion Scheepers, eds., pp. 227–258, Contemp. Math., 533, Amer. Math. Soc., Providence, RI, 2011. MR2777751 (2012i:03146).

Of course, this raises 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.)

• 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 '10 at 1:57
• As for comment (2), it is interesting, thanks. – Norman Lewis Perlmutter Nov 24 '10 at 2:00
• @Norman : I've added some details. – Andrés E. Caicedo Nov 25 '10 at 7:00
• Wait, why is the result by Woodin is anything less than immediate? Assume $\sf ZF+DC+AD$, then just run a Prikry forcing to make $\omega_2$ singular. – Asaf Karagila Sep 22 '13 at 8:53
• (@AsafKaragila I'm afraid I don't understand your comment. In particular, I do not know what result by Woodin you are referring to.) – Andrés E. Caicedo Sep 22 '13 at 16:46