Geometric condition for Dependent Choice

We provide a geometric condition which characterises when the Principle of Dependent Choice holds in a Fraenkel--Mostowski--Specker permutation model. This condition is a slight weakening of requiring the filter of groups to be closed under countable intersections. We show that this condition holds nontrivially in a new permutation model we call"the nowhere dense model"and we study its extensions to uncountable cardinals as well.


Introduction
Permutation models are models of set theory with atoms. 1 These models are used often to prove basic independence results related to the Axiom of Choice.The technique was originally developed by Fraenkel, improved upon by Mostowski by introducing the concept of supports, and the modern presentation is due to Specker using filters of subgroups. 2 With these models it is relatively easy to construct models where the Axiom of Choice fails, and using a myriad of transfer theorems we can translate some of these results to the context of Zermelo-Fraenkel.Regardless of their intended use, permutation models are still an interesting way to exhibit the tension between symmetries and definablity on the one hand and the Axiom of Choice on the other.
We rarely want to violate the Axiom of Choice so much that it is become completely useless.Indeed, often times the goal is to preserve a fragment of the Axiom of Choice.One of the important fragments is the Principle of Dependent Choice (DC), which has many equivalents and reformulations throughout mathematics (e.g. the Baire Category Theorem), and can be succinctly stated as "Every tree without maximal nodes has an infinite chain".
Reading the standard proof separating Dependent Choice from other principles (see Chapter 8 of [3]) it is easy to conjecture a permutation model satisfies DC if and only if the filter of subgroups is countably complete.In this work we show that this is not quite the right notion, and we can in fact weaken the completeness to what we call "shift completeness" of the filter.

Preliminaries
Let us recall the basics of permutation models, following [3].The underlying theory we are working with is ZFA, i.e.ZF together with atoms.For simplicity we will also always assume that the collection of atoms is a set A. Models of this theory have a natural von Neumann like hierarchy where instead of starting with the empty set, we start with the set of atoms A and then successively build powersets 1 Traditionally either by weakening Extensionality (to allow "multiple empty sets") or Regularity (to allow Quine atoms, i.e. x = {x}).

2
For a more detailed historical review see the remarks at the end of Chapter 4 in [3].and unions at limit steps.Whenever π is a permutation of A, π recursiveley extends via this hierarchy to the whole universe by stipulating that π(x) = {π(y) : y ∈ x}.Now let G be a group of permutations of A. A filter of subgroups of G is a non-empty set F of subgroups of G that is closed under supergroups and finite intersections.It is called normal if it is also closed under conjugation by elements of G, i.e. for every H ∈ F and π ∈ G, πHπ −1 ∈ F. We say that F is κ-complete, for a cardinal κ, if whenever γ < κ and In the context of some fixed G and a normal filter F of subroups of G, we say that a set x is symmetric if there is a group H ∈ F such that π(x) = x for every π ∈ H.In other words, sym(x) := {π ∈ G : π(x) = x} ∈ F. Moreover we say that x is hereditarily symmetric if every element of the transitive closure tcl({x} ∪ x) is symmetric.The class of hereditarily symmetric sets is then denoted HS.
We will say that HS is the permutation model obtained from G and F .Definition 2. Let F be a normal filter of subgroups of G.We say that F is shift complete if for any decreasing sequence ).To improve the readability, we will always use K n to denote the shifted sequence and we will use σ n to denote the composition Assuming DC, it is true that F is shift complete if and only if it has a basis which is shift complete (requiring that n∈ω K n contains a basis element, in this case).This is similar to the case of normality, although we need to rely on a modicum of choice since we need to choose infinitely many basis elements at once.

Definition 3.
Let A be a collection of atoms, let G be group of permutations of A and F be a normal filter of subgroups of G. Then the essential subfilter of F is the filter generated by {sym(x) : x ∈ HS}.
It is easy to see that the essential subfilter is itself normal and produces the same permutation model, and in typical constructions F is itself already essential.Proposition 4. Assuming DC holds in V , if F is shift complete, then the essential subfilter is shift complete.
Proof.Let H n : n ∈ ω be a sequence in the essential subfilter, and choose x n ∈ HS such that sym(x n ) = H n for every n ∈ ω.Using the shift completeness of F , there is and n∈ω K n is in the essential subfilter, as wanted.

Dependent Choice is Essentially shift completeness
Main Theorem.Assume DC holds in V and let F be a normal filter of subgroups of G. Then the following are equivalent: (1) The essential subfilter of F is shift complete.
Proof.(1) implies (2): Let (T, ≤) ∈ HS be a tree without a maximal element.Since DC holds in V , there is an increasing sequence s n : n ∈ ω in T .Let H 0 := sym(T, ≤) and H n+1 := sym(T, ≤, s i : i ≤ n ) for every n ∈ ω.Then each H n is in the essential subfilter of F and they form a decreasing sequence.According to (1), there are π n ∈ G such that K := n∈ω K n ∈ F. Further let t n := σ n (s n ), for every n ∈ ω.Then t n ∈ HS and Moreover, as (s n , s n+1 ) ∈ ≤ and σ n+1 ∈ H 0 = sym(≤), we have that The last equality follows since π n+1 ∈ K n+1 ⊆ sym(t n ).Thus t n : n ∈ ω is increasing in T and t n : n ∈ ω ∈ HS.
(2) implies ( 1): Let H n : n ∈ ω be in the essential subfilter.Using DC in V , for every n ∈ ω, let x n ∈ HS be such that, without loss of generality, sym(x n ) = H n+1 .
Define T := {π( we have that T ∈ HS, and as a tree ordered by inclusion T has no maximal elements.Thus, by (2), there is a branch Proof.For n = 0 this is trivial, as K 0 = H 0 and π 0 = τ 0 ∈ H 0 .For n ≥ 1, since Proof.Let π ∈ K and n ∈ ω be arbitrary.Then π ∈ K 0 and for n ≥ 1, since , it suffices to show that π fixes τ n−1 (x n−1 ).This is similar to the previous claim.
Since K is in the essential subfilter of F , this completes the proof.

Nowhere dense model
We consider the rational numbers, Q, with their linear order as our structure.The group G is the group of order automorphisms (i.e.order preserving bijections).For any subset E ⊆ Q, we let fix(E) := {π ∈ G : π ↾ E = id}.Let F be the filter generated by {fix(E) : E ⊆ Q is nowhere dense}.It is not hard to see that any singleton is nowhere dense, so the filter is certainly not countably complete.
We claim that F is shift complete.Let E n be an increasing sequence of nowhere dense subsets of (Q, <).Let I n : n ∈ ω enumerate all open intervals of Q.We will recursively define automorphisms π n and non-empty intervals J n ⊆ I n such that (1) Lemma 5. Let E be a nowhere dense set and J = i<n (a i −ε i , b i +ε i ) be a disjoint union of open intervals for some ε i > 0. There is an automorphism, π ∈ fix(Q \ J), and take π to be the composition of these automorphisms, since the intervals are disjoint this composition is commutative.For readability let us omit i from the subscript, as we are working on each of the intervals separately.
Since E is nowhere dense, we can find (c, Since E 1 is nowhere dense we can apply the lemma to obtain π 0 ∈ fix(E 0 ) such that π 0 "E 1 ∩ (a 0 − ε0 4 , b 0 + ε0 4 ) = ∅.Suppose that we have defined J i = (a i , b i ) and π i for i < n such that for some It is important to note that the sequence of ε i may be taken to be different at each step.We want to find and the previously chosen intervals, by perhaps shrinking the ε i even more in order to apply the lemma to J = i≤n (a i − ε i , b i + ε i ) and σ n "E n+1 .Certainly, since σ n "E n is nowhere dense that requirement is easy to fulfil; if we cannot fulfil the second requirement, then we can choose J n to be contained in one of the J i for i < n and we can apply the lemma.This ensures that n∈ω σ n "E n ∩ n∈ω J n = ∅ as wanted.

Generalised versions of Dependent Choice and shift completeness
We can generalise DC to higher cardinals in the following way.We say that a tree T is κ-closed if every chain of order type < κ has an upper bound.Then DC κ states that every κ-closed tree has a chain of order type κ or a maximal element.It is not hard to verify, in this formulation, that if λ < κ, then DC κ implies DC λ holds as well.We write DC <κ to denote DC λ holds for all λ < κ.In the case where κ = λ + this is just DC λ , and if κ is singular, then DC <κ implies DC κ .However, for inaccessible cardinals DC <κ is indeed weaker than DC κ .Definition 6.Let δ be an infinite ordinal.Then we say that F is δ-shift complete if for any sequence H α : α < γ with γ < δ in F , there are σ α ∈ G, for α < γ, such that (1) α<γ K α ∈ F, where It is not hard to see that shift-complete as we previously defined is ω + 1-shift complete.
Lemma 7. Let F be a filter of subgroups of G.
Proof.For simplicity, we assume that γ = |γ|.To see that F is γ-complete, let H α : α < β be in F , where β < γ.Next extend this sequence arbitrarily to H α : α < γ , for example by repeating G after β.Let σ α and K α be as in the definition of shift complete, for each α ≤ γ.Then we have that α<β K α ∈ F. On the other hand, for each α < β, σ The proof in the case of γ + 2-shift completeness is similar, but we can now use a sequence of length γ to begin with, thus proving that F is γ + -complete.
We saw with the nowhere dense model that the above theorem is the best we can get, since it is possible to get a filter of groups that is ω + 1-shift complete, but not ω 1 -complete.
It is a standard observation that if F is κ-complete and DC <κ holds in V , then DC <κ holds in HS.The proof of this observation actually shows more, it shows that HS is closed under γ-sequences for any γ < κ.Indeed, if X = {x α | α < γ} ⊆ HS, then α<γ sym(x α ) = H ∈ F, and it is not hard to check that H fixes X pointwise, so X ∈ HS.
Proof.Suppose that (T, ≤) ∈ HS is a κ-closed tree with no maximal elements.We have that T is κ-closed in V , since there is some γ < κ and a chain in T of order type γ which is not in HS.But by the lemma above, F is κ-complete, so this is impossible.Thus there is a branch s α : α < κ ∈ V in T .Now we proceed exactly as in the proof of Theorem 3 to get a shifted branch in HS.
Corollary.Assuming DC <κ holds in V , if F is κ-shift complete, then DC <κ holds in HS.

Open questions
From the work of Blass in [1] and [2] we know that there is a complete characterisation of when the Boolean Prime Ideal theorem holds in a permutation model.The property used by Blass is called a "Ramsey filter", but upon deeper inspection it seems to have a strong finitary nature which is at odds with the infinitary nature of shift completeness.Question 9. Is there a natural example of a permutation model where the filter which is both Ramsey and shift complete?It seems somewhat unlikely that κ + 1-shift completeness will be equivalent to DC κ holding in HS for uncountable κ.It seems reasonable to expect that HS might not be closed under γ-sequences for all γ < κ, but DC κ still holds there.
Question 10.What is the "correct" generalisation of shift completeness which does not imply κ-completeness?