Permutation Models for Set TheoryAn essay submitted for Part III of the Mathematical Tripos

Clive Newstead

Robinson College, University of Cambridge

Friday 3rd May 2013

Permutation Models for Set Theory

Contents

1 Introduction 1

1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . 5

2 The Boolean prime ideal theorem 6

2.1 Some equivalences . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Set theory with atoms 16

3.1 Fraenkel–Mostowski permutation models . . . . . . . . . . . 18

3.2 Parameter modification . . . . . . . . . . . . . . . . . . . . . 19

3.3 Normal filter–normal ideal duality . . . . . . . . . . . . . . . 21

4 Mostowski’s ordered model 25

5 Topological groups and their actions 28

5.1 Extremely amenable groups . . . . . . . . . . . . . . . . . . 28

5.2 Ramsey filters . . . . . . . . . . . . . . . . . . . . . . . . . . 31

6 The correspondence 33

6.1 Post mortem . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

References 40

1 Introduction

Permutation models of set theory were devised by Abraham Fraenkel in the1920s to prove the independence of the axiom of choice from other axioms.They were found very useful in the subsequent decades for establishing non-implications—for instance, that the ordering principle does not imply theaxiom of choice—but soon fell out of use for this purpose when Paul Cohenintroduced the forcing method in the 1960s. The reason for this is thatpermutation models require a weakening of the axiom of extensionality toallow for the existence of atoms (empty objects which are not sets; seeSection 3), whereas forcing allows one to work entirely within the moreconventional set theory ZFC.

Nonetheless, the methods used in the study of permutation models live on.Permutation methods underlie forcing via Boolean-valued models, nominalsets in computer science (e.g. [18]), and a proposed proof due to RandallHolmes that Quine’s New Foundations set theory is consistent relative toZF, a question which has been open for seventy-six years.

The focus of this essay is Andreas Blass’s recent result [4] that permuta-tion models in which the Boolean prime ideal theorem holds but the axiomof choice fails correspond with nontrivial extremely amenable topologicalgroups with small open subgroups. The reader is not expected to under-stand the statement of this result at this stage; the necessary theory will bedeveloped throughout the essay.

Blass’s result is surprising for a number of reasons. Firstly, it draws froma diverse range of distinct fields of mathematics, most notably set theory,group theory, combinatorics and topology. Secondly, the theory of amenablegroups, more specifically of extremely amenable groups, was not contrivedin any way to prove results in set theory: the question of the existence ofnontrivial extremely amenable groups was asked by Theodore Mitchell [16]in a paper about topological semigroups and answered by Wojciech Hererand Jens Christensen [8] five years later in a measure theory paper. Furtherexamples were given by Vladimir Pestov [17] in a paper about topologicaldynamics.

What I hope to achieve with this essay is to give a self-contained expositionof Blass’s theorem, proving (almost) all results in full and building up allthe necessary grounding. In Section 2 the Boolean prime ideal theorem isexplored in depth: we establish how strong a statement it is relative toother weakenings of the axiom of choice and prove that it is equivalent toa number of other assertions in mathematics. Set theory with atoms andFraenkel–Mostowski permutation models are introduced in Section 3 andtheir basic theory explored, including a digression to discuss normal filtersand normal ideals in greater depth. Section 4 discusses a particular example

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of a permutation model, namely Mostowski’s ordered model, which will serveto illustrate the main theorem of the essay. After a digression in Section 5to build up the group theory and topology needed to understand the otherhalf of the correspondence, we will be ready to state the result in full detailand prove it in Section 6.

1.1 Preliminaries

This part of the essay is devoted to briskly introducing some concepts whichunderpin the rest of the essay—particularly those outside of set theory. Abasic grounding of sentential logic, first-order logic and axiomatic set theoryis assumed, such as that found in [6] or [12], as is an elementary knowledgeof group theory and general topology.

Partially ordered sets

A partially ordered set P = 〈P,≤〉 is a set P together with a reflexive,transitive, antisymmetric binary relation ≤. I will abuse notation slightlyby writing x ∈ P to mean x ∈ P , and omit reference to the underlying set.The strict part of the order relation ≤ is denoted by <. The discrete orderis defined by p ≤ q → p = q for all p, q ∈ P.

P is totally ordered if for each x, y ∈ P either x ≤ y or y ≤ x. The orderextension principle is the assertion that every partial order can be extendedto a total order. The ordering principle is the assertion that every set canbe endowed with a total order. A subset C ⊆ P is a chain if it is totallyordered by the restriction of ≤ to C.

P is wellordered if it is totally ordered and every nonempty subset of P hasa <-least member. The wellordering principle is the assertion that every setcan be wellordered, or equivalently that for every set x there is an ordinalαx and an injective function x→ αx.

A function f : P → P′ is order-preserving if for all x, y ∈ P, x ≤ y impliesf(x) ≤ f(y), order-reversing if x ≤ y implies f(y) ≤ f(x) and monotonic ifit is order-preserving or order-reversing. It is an order-isomorphism if it isbijective and both f and f−1 are order-preserving, in which case we writeP ∼= P′. An order-automorphism of P is an order-isomorphism P→ P.

A lattice L is a partially ordered set with all finite joins (suprema, ∨) andmeets (infima, ∧). In particular, every lattice has a least element 0 =

∨∅

and a greatest element 1 =∧

∅. A lattice is complete if it has all joins. Inthis essay, all lattices will be assumed to be nontrivial, i.e. 0 6= 1. For a pairx, y ∈ L write x ∨ y =

∨{x, y} and x ∧ y =

∧{x, y}.

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A lattice L is distributive if for any x, y, z ∈ L we have

x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) and x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z)

Lemma 1.1 If L is a distributive lattice then there exists a set X and alattice L of subsets of X such that L ∼= L. �

A distributive lattice B is a Boolean algebra if every element x ∈ B has acomplement ¬x ∈ B, which is unique with the properties

x ∧ ¬x = 0, x ∨ ¬x = 1, ¬¬x = x

The dual algebra Bop of a Boolean algebra B has the same underlying setbut with reversed order.

The axiom of choice and Zermelo–Fraenkel set theory

The axiom of choice (AC) is the assertion that every set of nonempty setshas a choice function; that is, if I is a set and {xα : α ∈ I} is a set ofnonempty sets then there is a function f : I →

⋃α∈I xα such that f(α) ∈ xα

for each α ∈ I.

Zorn’s lemma is the assertion that if P is a partially ordered set all of whosechains have an upper bound, then P has a maximal element.

Lemma 1.2 The axiom of choice is equivalent to both the wellordering prin-ciple and Zorn’s lemma. �

ZF denotes Zermelo–Fraenkel set theory, ZFC denotes ZF with AC appendedas an axiom, and ZFA denotes a modification of ZF to be defined in Section3. We will work inside ZF(A) wherever possible; choice is assumed to hold inthe ‘real world’ (the metatheory) and will be mentioned explicitly whereverused.

A formula φ is independent of a theory T if T 0 φ and T 0 ¬φ, and isindependent of another formula ψ in the presence of T if T 0 ψ → φ.Similarly, φ and ψ are equivalent in T if T ` φ ↔ ψ. I omit reference to Twhen the independence or equivalence holds in each of ZF, ZFC and ZFA.

P(x) denotes the powerset of a set x. The transitive closure of a set x isdefined by trcl(x) =

⋃{x,⋃x,⋃⋃

x, . . . }. If φ is a predicate with one freevariable then H(φ) is the class of hereditarily φ sets, defined by

x ∈ H(φ)↔ φ(x) ∧ ∀y(y ∈ trcl(x)→ φ(y))

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The symbols N, Z, Q, R denote the sets of natural numbers, integers, ra-tional numbers and real numbers, respectively; we take 0 to be a naturalnumber. Ordinals are taken to be von Neumann ordinals, i.e. sets α whichare wellordered by ∈ and such that β ∈ α → β ⊆ α. The successor ofan ordinal α is α + 1 = α ∪ {α}; ordinals not of this form are called limitordinals. The class of all ordinals is denoted by Ord. We adopt the usualhabit of identifying finite ordinals with natural numbers.

I will use arrow notation f→, f← for the direct image and inverse image ofa function f . Specifically, if f : X → Y is a function, A ⊆ X and B ⊆ Y ,then

f→(A) = {f(x) : x ∈ A} and f←(B) = {x ∈ X : f(x) ∈ B}

It is important to distinguish between f(A) and f→(A) whenever A∪{A} ⊆X, especially if f is defined on the entire set-theoretic universe, because f(A)and f→(A) may not be equal.

General topology

A (topological) space is a set X equipped with a topology O(X), which isa lattice of subsets of X closed under arbitrary joins (unions) and finitemeets (intersections). Elements of O(X) are called open sets and theircomplements in X are called closed sets. If O(X) is a topology on X andB ⊆ O(X) then O(X) is generated by B, and B is a base for O(X), if O(X)is the closure of B under arbitrary unions.

The discrete topology on a space X is P(X). If {Xi : i ∈ I} is a set oftopological spaces and X =

∏i∈I Xi = {(f : I → Xi) : f(i) ∈ Xi} then

define πi : X → Xi by πi(f) = f(i); the product topology on X is thetopology generated by sets of the form π−1

i (U) for U open in Xi.

A space X is Hausdorff if for any distinct points x, y ∈ X there are disjointopen sets U, V with x ∈ U and y ∈ V . X is compact if whenever {Uα : α ∈I} is a set of open sets with

⋃α∈I Uα = X, there is a finite subset J ⊆ I

such that⋃α∈J Uα = X.

Lemma 1.3 If X is a compact Hausdorff space and x, y ∈ X are distinctthen there exists a continuous map f : X → [0, 1] such that f(x) = 0 andf(y) = 1. �

Groups and group actions

If G is a group then Sub(G) denotes the lattice of subgroups of G, whereH ∧K = H ∩K and H ∨K is the subgroup generated by the elements of

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H ∪K. H ≤ G denotes that H is a subgroup of G, and H E G denotes thatH is a normal subgroup of G.

If X is a set, then a left action of G on X will be denoted by Gy X. Thiswill be formalised as a map G × X → X given by 〈g, x〉 7→ g · x, whichsatisfies 1 · x = x and (gh) · x = g · (h · x) for all g, h ∈ G and x ∈ X. Theorbit of an element x ∈ X will be denoted by

G(x) = {g · x : g ∈ G}

1.2 Acknowledgements

First and foremost, I am extremely grateful to Thomas Forster. He is re-sponsible for nurturing my interest in logic and set theory, and for sparkingmy interest in permutation models through the Fraenkel–Mostowski permu-tation models reading group which he organised in Michaelmas Term 2012.I have enjoyed indulging in numerous discussions with him about logic andset theory, and I owe him a great debt for his undying support over the lasttwo years.

I would also like to thank Andreas Blass for being kind enough to point mein the direction of his paper [4], from which the most important result ofthe essay is drawn, and for responding to my questions [5].

Finally I would like to thank the members of the Fraenkel–Mostowski modelsreading group and the con(NF) discussion group, especially Philipp Klepp-mann, who provided me with a paper [15] on which Section 3.3 is based.

A complete list of papers that I have read and from which ideas have beendrawn can be found in the list of references on page 41. I have drawn mostheavily from Thomas Jech’s book The Axiom of Choice [10] and AndreasBlass’s papers [3] and [4], from which I learnt the material on which theessay is based. Many of the proofs in the essay are adapted directly fromthese references. The results whose proofs are my own are Lemmas 2.1, 4.1and 6.6; Propositions 2.3, 2.5, 2.13, 3.1, 3.6, 3.7∗, 5.1∗ and 5.2; and Theorem6.7. Those marked with ∗ are my own results.

My construction of ZFA adopts the convention that the collection of atomsneedn’t form a set, even though for the purposes of this essay it will always bea set. My definitions of mess and consistency principle (page 10) generalisethose usually found in the literature (e.g. [10]); Proposition 2.6 is generalisedaccordingly.

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2 The Boolean prime ideal theorem

Let L be a distributive lattice and let I ⊆ L be a nonempty subset. Considerthe following properties that I may, or may not, have:

(a) I is closed downwards: if y ∈ I and x ≤ y then x ∈ I;

(b) I is closed under pairwise joins: if x, y ∈ I then x ∨ y ∈ I;

(c) I is proper: I 6= L;

(d) L− I is closed under joins: if x ∧ y ∈ I then x ∈ I or y ∈ I.

(e) I is maximal: if I ⊆ J and J satisfies (a)–(c) then I = J .

Then I is:

� an ideal if it satisfies (a)–(b);

� a proper ideal if it satisfies (a)–(c);

� a prime ideal if it satisfies (a)–(d);

� a maximal ideal if it satisfies (a)–(c) and (e).

Dually, there are notions of filter, proper filter, prime filter and ultrafilter,which are obtained by swapping ∨ and ∧, and replacing ≤ by ≥, in (a)–(e)above. For example, a filter is an upwards-closed set which is closed underpairwise meets. Note that, for an ideal I, (c) is equivalent to the assertionthat 1 6∈ I.

For a Boolean algebra B, there is another property that I may have:

(∗) For all b ∈ B, either b ∈ I or ¬b ∈ I, but not both.

Lemma 2.1 Let B be a Boolean algebra. A proper ideal I ⊆ B is prime ifand only if it is maximal. Furthermore, this holds if and only if (∗) holds.

Proof We prove that primality (d) and maximality (e) are both equivalentto (∗) in the presence of properness (a)–(c).

(d) ↔ (∗). If I is prime and b ∈ B then b ∧ ¬b = 0 ∈ I and so b ∈ I or¬b ∈ I by (d), so (∗) holds. Conversely, if (∗) holds and a, b ∈ B with a 6∈ Iand b 6∈ I then ¬a ∈ I and ¬b ∈ I. Then ¬a ∨ ¬b = ¬(a ∧ b) ∈ I by (b), soa ∧ b 6∈ I by (c), and so I is prime.

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(e) ↔ (∗). Let I be a proper ideal and suppose b ∈ B is a witness to thefailure of (∗), so that b 6∈ I and ¬b 6∈ I. We must have b < 1 because¬1 = 0 ∈ I. Let

I+ = I ∪ {a ∈ B : a ≤ b} ∪ {a ∨ b : a ∈ I}

It is plain to see that I+ is closed downwards and under joins. The onlyways in which it could fail to be proper are if we had either 1 ≤ b and henceb = 1, which is impossible, or 1 = a ∨ b for some a ∈ I, in which case

a = a ∨ 0 = a ∨ (b ∧ ¬b) = (a ∨ b) ∧ (a ∨ ¬b) = 1 ∧ (a ∨ ¬b) = a ∨ ¬b

But then ¬b ≤ a, and hence ¬b ∈ I, contradicting our assumption. So I+

is a proper ideal strictly containing I, and I is not maximal. Conversely,suppose (∗) holds and that J is a proper ideal with I ⊆ J . If b ∈ J but b 6∈ Ithen ¬b ∈ I, so ¬b ∈ J and hence 1 = b ∨ ¬b ∈ J contradicting propernessof J ; hence I = J and so I is maximal.

Hence (d) ↔ (e) ↔ (∗). And if any of these hold and b ∈ I then ¬b 6∈ I,else b ∨ ¬b = 1 ∈ I, contradicting maximality. �

The Boolean prime ideal theorem (BPIT) is the assertion that any properideal in a Boolean algebra can be extended to a prime ideal. In fact thisstatement is equivalent to an apparently weaker one:

Lemma 2.2 The Boolean prime ideal theorem is equivalent to the assertionthat every Boolean algebra has a prime ideal.

Proof The latter statement is implied by BPIT since {0} is an ideal in anyBoolean algebra B and any ideal must contain 0.

Conversely, suppose every Boolean algebra has a prime ideal and let I bean ideal in a Boolean algebra B. Define a relation ∼ on B by

a ∼ b if and only if (a ∧ ¬b) ∨ (¬a ∧ b) ∈ I

Then ∼ is an equivalence relation. It is reflexive since 0 ∈ I, it is clearlysymmetric, and it is transitive since if a ∼ b and b ∼ c then by closure underunions we have

(a ∧ ¬b) ∨ (¬a ∧ b) ∨ (b ∧ ¬c) ∨ (¬b ∧ c) ∈ I

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and so we obtain

(a ∧ ¬c) ∨ (¬a ∧ c) = (a ∧ b ∧ ¬c) ∨ (a ∧ ¬b ∧ ¬c)∨ (¬a ∧ b ∧ c) ∨ (¬a ∧ ¬b ∧ c)

≤ (a ∧ b ∧ ¬c) ∨ (a ∧ ¬b ∧ ¬c) ∨ (a ∧ ¬b ∧ c)∨ (¬a ∧ b ∧ c) ∨ (¬a ∧ ¬b ∧ c) ∨ (¬a ∧ b ∧ ¬c)

= (a ∧ ¬b) ∧ (c ∨ ¬c) ∨ (¬a ∧ b) ∧ (c ∨ ¬c)∨ (b ∧ ¬c) ∧ (a ∨ ¬a) ∨ (¬b ∧ c) ∧ (a ∨ ¬a)

= (a ∧ ¬b) ∨ (¬a ∧ b) ∨ (b ∧ ¬c) ∨ (¬b ∧ c) ∈ I

Hence (a ∧ ¬c) ∨ (¬a ∧ c) ∈ I by downwards-closure, and so a ∼ c. So ∼ istransitive, and hence an equivalence relation.

Define operations on the set B/∼ of ∼-equivalence classes pointwise; thenB/∼ is a Boolean algebra, and hence contains a prime ideal Q say. But then

P = {b ∈ B : [b]∼ ∈ Q}

is a prime ideal in B containing I. �

Proposition 2.3 The axiom of choice implies the Boolean prime ideal the-orem.

Proof Suppose the axiom of choice holds and let B be a Boolean algebra.

The partially ordered set of proper ideals in B is chain-complete; this can beeasily verified by noting that the supremum of a chain is simply its union,which is proper. Let I be any proper ideal in B. By Zorn’s lemma, which isequivalent to the axiom of choice by Lemma 1.2, there is a proper ideal Pcontaining I which is ⊆-maximal amongst all proper ideals of B; i.e. it is amaximal ideal, and hence is prime by Lemma 2.1. �

In fact, there is a chain of implications

axiom of choice→ BPIT→ order-extension principle→ ordering principle

The first implication is Proposition 2.3, the second follows immediately fromTheorem 2.4 and Lemma 2.7 below, and the third is an application of theorder extension principle to sets with the discrete ordering. We will see inSection 4 that in Mostowski’s ‘ordered’ model of ZFA the ordering principleholds but the axiom of choice does not. As such, Mostowski’s ordered modelmust exhibit the failure of one of these reverse implications. It is strikingthat the axiom of choice fails so weakly in this model that BPIT still holds,thus proving that BPIT is independent from the axiom of choice in thepresence of the ZFA axioms.

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2.1 Some equivalences

In order to prove this independence result and the main result of the paper,we must venture through numerous areas of mathematics and logic, each ofwhich lends itself to different ways of thinking. The Boolean prime idealtheorem, as stated, is not much use for many of our purposes. As such,the remainder of this section is devoted to proving that BPIT is equivalentto a number of different statements. Namely, we introduce the consistencyprinciple (CP), the ultrafilter lemma (UL), the sentential compactness theo-rem (SCT) and the almost maximal ideal theorem (AMIT), and prove resultsworking up to the following theorem.

Theorem 2.4 The following are equivalent:

� Boolean prime ideal theorem

� Consistency principle

� Almost maximal ideal theorem

� Ultrafilter lemma

� Sentential compactness theo-rem

Proof The propositions in the remainder of Section 2 give the followingchain of implications:

BPIT(2.5)−−→ UL

(2.6)−−→ CP(2.8)−−→ SCT

(2.12)−−−→ AMIT(2.13)−−−→ BPIT

so the equivalences are established. �

For example, to prove ZFA 0 BPIT → AC we will instead prove ZFA 0CP → AC (Theorem 4.7), and to prove, for a particular model M of ZFA,that M � BPIT is equivalent to a certain condition we shall prove that thecondition is implied by SCT and implies AMIT (Theorem 6.5).

Ultrafilter lemma

Ultrafilters are dual to maximal ideals in the sense that U is an ultrafilter ina Boolean algebra B if and only if it is a maximal ideal in the dual algebraBop.

The ultrafilter lemma (UL) is the assertion that any proper filter on aBoolean algebra can be extended to an ultrafilter.

Proposition 2.5 The Boolean prime ideal theorem implies the ultrafilterlemma.

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Proof Let B be a Boolean algebra and F ⊆ B be a proper filter. Then F isa proper ideal in Bop, so it can be extended to a prime ideal U in Bop. ByLemma 2.1, U is a maximal ideal in Bop, and so U is an ultrafilter in B. �

Consistency principle

We will now reformulate BPIT in combinatorial terms. A mess on a latticeL is a pair M = 〈M, I〉, where I is an ideal in L and M is a set of pairs〈x, y〉 with x ∈ I and y ≤ x, such that:

� For all x ∈ I there exists y ≤ x with 〈x, y〉 ∈M .

� If 〈x, y〉 ∈M and z ∈ I then 〈z, y ∧ z〉 ∈M .

We say 〈x, y〉 and 〈x′, y′〉 are compatible if x ∧ y′ = x′ ∧ y. If c ∈ L is anarbitrary element and M is a mess on L then c is consistent with M if〈x, c∧ x〉 ∈M for each x ∈ I. When L = P(X) for some set X and I is theideal of finite subsets of X, we identifyM with M and sayM is a mess onX.

A complete lattice F is called a frame if for all sets I and all x, yi ∈ F (i ∈ I)

x ∧∨i∈I

yi =∨i∈I

x ∧ yi and x ∨∧i∈I

yi =∧i∈I

x ∨ yi

Note in particular that all power sets are frames.

The consistency principle (CP) is the assertion that if F is a frame andM = 〈M, I〉 is a mess on F, then there is an element c ∈ F which isconsistent with M.

Proposition 2.6 The ultrafilter lemma implies the consistency principle.

Proof Let M = 〈M, I〉 be a mess on a frame F. Let A be the set of allfunctions j : J → I with J ⊆ I such that

(a) j(x) ≤ x and 〈x, j(x)〉 ∈M for all x ∈ J ;

(b) If x, y ∈ J then 〈x, j(x)〉 and 〈y, j(y)〉 are compatible.

For each x ∈ I and y ≤ x, define

Ax = {(j : J → I) ∈ A : x ∈ J}

Bx,y = {(j : J → I) ∈ A : x ∈ J and j(x) = y}

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Let F be the filter generated by {Ax : x ∈ I}. Each Ax is nonempty: ifx ∈ I then there exists y ≤ x with 〈x, y〉 ∈M ; then the function j : {x} → Igiven by j(x) = y lies in Ax. Furthermore, if x, x′ ∈ I then Ax ∩ Ax′ 6= ∅.To see this, choose y, y′ such that 〈x, y〉 ∈M and 〈x′, y′〉 ∈M . By property(b) of messes, 〈y′, y ∧ y′〉 ∈ M and so 〈x, x ∧ y ∧ y′〉 = 〈x, y ∧ y′〉 ∈ M , andlikewise 〈x′, y ∧ y′〉 ∈ M . Define j : {x, x′} → I by j(x) = j(x′) = y ∧ y′.Then

x′ ∧ j(x) = x′ ∧ y ∧ y′ = y ∧ y′ = x ∧ y ∧ y′ = x ∧ j(x′)So 〈x, j(x)〉 and 〈x′, j(x′)〉 are compatible, and j ∈ Ax ∩ Ax′ . Hence F is aproper filter.

By the ultrafilter lemma, there is an ultrafilter U on A extending F . Now,for each x ∈ I, we can write Ax as a disjoint union:

Ax = Bx,y1 ∪ · · · ∪Bx,ym

where 〈x, y1〉, . . . , 〈x, ym〉 exhausts all elements of M whose first componentis x. Hence there exists x∗ ≤ x with 〈x, x∗〉 ∈M such that Bx,x∗ ∈ U . Thisx∗ is unique: the Bx,yi are disjoint, so if more than one were to lie in Uthen so would their intersection, and then we’d have ∅ ∈ U , contradictingproperness. Then for any x, z ∈ I, Bx,x∗ ∩Bz,z∗ 6= ∅ by properness of U . Bydefinition of A, 〈x, x∗〉 and 〈z, z∗〉 are compatible, so z ∧ x∗ = x ∧ z∗.

Define c =∨x∈I x∗. Then for any z ∈ I,

z ∧ c =∨x∈I

z ∧ x∗ =∨x∈I

x ∧ z∗ = z∗

Hence 〈z, z ∧ c〉 ∈M , and c is consistent with M. �

Sentential compactness theorem

Our viewpoint now turns to that of logic; in particular, sentential logic.Let P be a set of sentential variables and Σ be a set of sentences over P .The sentential compactness theorem (SCT) states that if each finite subsetΣ0 ⊆ Σ has a model then Σ has a model.

Lemma 2.7 The sentential compactness theorem implies the order exten-sion principle.

Proof Let P = 〈P,≤〉 be a partially ordered set. Define sentential variablesJp ≤t qK for p, q ∈ P encoded, say, as elements of P × P . Let

Σ = {Jp ≤t qK ∧ Jq ≤t rK⇒ Jp ≤t rK : p, q, r ∈ P} ← transitivity

∪ {Jp ≤t qK ∧ Jq ≤t pK⇒ Jp = qK : p, q ∈ P} ← symmetry

∪ {Jp ≤t qK ∨ Jq ≤t pK : p, q ∈ P} ← totality

∪ {Jp ≤t qK : p, q ∈ P such that p ≤ q} ← extension

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Let Σ0 ⊆ Σ be finite, and let P0 ⊆ P be the finite subset of P consisting ofall elements appearing in sentences in Σ0. It follows from the finite axiomof choice that P0 has a total ordering extending the restriction to P0 of ≤,and hence Σ0 has a model. By the sentential compactness theorem, Σ hasa model, which is to say that P has a total ordering ≤t extending ≤. �

Proposition 2.8 The consistency principle implies the sentential compact-ness theorem.

Proof Let P be a set of sentential variables and Σ be a set of sentencesover P . Suppose each Σ0 ⊆ Σ has a model vΣ0 : P → {>,⊥}. Let PΣ0 ⊆ Pbe the set of sentential variables appearing in formulae in Σ0. For F ⊆ PΣ0

let F> = {p ∈ F : vΣ0(p) = >}. Define

M = {〈F,E〉 : Σ0 ⊆ Σ finite, F ⊆ PΣ0 , E ⊆ F>}

Each PΣ0 must be finite since Σ0 is finite and each formula in Σ0 contains onlyfinitely many variables, so each F and E are certainly finite. Furthermore,for each F ⊆ PΣ0 , 〈F, F>〉 ∈ M; and if 〈F,E〉 ∈ M and G ⊆ PΣ0 thenE∩G ⊆ G> and so 〈G,E∩G〉 ∈ M. HenceM is a mess on P , and so thereis some subset C ⊆ P which is consistent with M. Define v : P → {>,⊥}by

v(p) =

{> if p ∈ C⊥ if p 6∈ C

Then for each finite Σ0 ⊆ Σ and F ⊆ PΣ0 , 〈F,C ∩F 〉 ∈ M, so C ∩F ⊆ F>.Hence v is a model of Σ. �

Almost maximal ideal theorem

Let L be a nontrivial distributive lattice (with 0 and 1) and let I ⊆ L be anideal. The dual I∗ of I is given by

I∗ = {v ∈ L : u ∨ v = 1 for some u ∈ I}

Lemma 2.9 I is proper if and only if I ∩ I∗ = ∅

Proof If u ∈ I ∩ I∗ then u ∨ v = 1 for some v ∈ I and hence 1 ∈ I, soI is not proper. Conversely if 1 ∈ I then I ⊆ I∗ since 1 ∨ v = 1 and soI ∩ I∗ 6= ∅. �

12

We say that I is an almost maximal ideal if it is prime and its dual ismaximal, in the sense that if J ⊆ I and J∗ = I∗ then J = I. Moreconcisely, for an ideal I ⊆ L define

I = {u ∈ L : v ∈ I∗ whenever u ∨ v = 1}

Then I is almost maximal if it is prime and I = I.

The almost maximal ideal theorem (AMIT) states that every nontrivial dis-tributive lattice has an almost maximal ideal.

To prove that AMIT is equivalent to BPIT we first need to build up somemachinery. We will use the sentential compactness theorem along the way.

First we fix some notation. Let L be a nontrivial distributive lattice with0 and 1. Let P be the set of sentential variables Jx ∨ y ∈ IK, where I ⊆ Lis a proper ideal and x, y ∈ L with x ∨ y ∈ I. This can be formalised byencoding Jx∨y ∈ IK as 〈I, x, y〉, for example. We define a set Σ of sentencesover P as follows. Let X be the set of all triples 〈I, F, f〉 where

� I ⊆ L is a proper ideal;

� F ⊆ L is a finite subset with∨F ∈ I∗;

� f : F → L is a function with x ∧ f(x) ∈ I for each x ∈ F .

For 〈I, F, f〉 ∈ X define

ξ(I, F, f) =∨x∈F

¬Jx ∨ f(x) ∈ IK, η(I, F, f) =∨x∈F

Jf(x) ∨ x ∈ IK

and let

Σ = {ξ(I, F, f) : 〈I, F, f〉 ∈ X} ∪ {η(I, F, f) : 〈I, F, f〉 ∈ X}

Lemma 2.10 The sentential compactness theorem implies that Σ is consis-tent.

Proof In any sentence in Σ, each variable refers to just one ideal. LetΣ0 ⊆ Σ be finite, and assume that the only variables that appear in sentencesin Σ0 refer to the ideal I. If each Σ0 of this form has a model then so doall finite subsets of Σ since any finite subset will be a finite union of (finite)sets of this form.

Let Q = {〈x, y〉 : Jx∨y ∈ IK appears in a sentence in Σ0}. Then Q is finite,and by definition of Σ we have u =

∨(x,y)∈Q(x ∧ y) ∈ I. But then

u =∧C⊆Q

∨〈x,y〉∈C

x ∨∨

〈x,y〉∈Q−C

y

13

Since I is proper, u 6∈ I∗, and hence there exists a subset C ⊆ Q such that∨〈x,y〉∈C

x ∨∨

〈x,y〉∈Q−C

y 6∈ I∗

Fix such a subset C. Define a valuation v0 : P → {>,⊥} by declaring

v0(Jx ∨ y ∈ IK) =

{> if 〈x, y〉 ∈ C⊥ otherwise

and let v0 be the (unique) extension of the valuation v0 to L(P ).

Now∨F1 6∈ I∗ and

∨F2 6∈ I∗, where

F1 = {x ∈ L : 〈x, y〉 ∈ C for some y ∈ L}

F2 = {y ∈ L : 〈x, y〉 ∈ Q− C for some x ∈ L}Since

∨F ∈ I∗, we must have F * F1 and F * F2, and so in particular

there exist x0, y0 ∈ F such that 〈x0, f(x0)〉 6∈ C and 〈f(y0), y0〉 6∈ Q− C.

If ξ(I, F, f) ∈ Σ0 then 〈x0, f(x0)〉 ∈ Q. Since 〈x0, f(x0)〉 6∈ C, we havev0(Jx0 ∨ f(x0) ∈ IK) = ⊥, and so v0(ξ(I, F, f)) = >. Likewise if η(I, F, f) ∈Σ0 then 〈f(y0), y0〉 ∈ Q, and since 〈f(y0), y0〉 6∈ Q− C we have v0(Jf(y0) ∨y0 ∈ IK) = >, and so v0(η(I, F, f)) = >.

Thus v→0 (Σ0) = {>}, so v0 is a model of Σ0. By the above remark, wemay now apply the sentential compactness theorem to deduce that Σ has amodel. �

Denote the model of Σ by v : P → {>,⊥}. For a proper ideal I ⊆ L define

V1(I) = {x ∈ L : v(Jx ∨ y ∈ IK) = > for some y ∈ L}

V2(I) = {y ∈ L : v(Jx ∨ y ∈ IK) = > for some x ∈ L}And for i = 1, 2 let V +

i (I) be the ideal generated by I ∪ Vi(I).

Lemma 2.11 V +1 (I) and V +

2 (I) are proper ideals.

Proof Let F ⊆ V1(I) be a finite subset. For each x ∈ F , there exists y ∈ Lsuch that v(Jx ∨ y ∈ IK) = >, so let f(x) be some such y. (The axiom ofchoice is not required here since F is finite.) Then

v

(∨x∈F

¬Jx ∨ f(x) ∈ IK

)= ⊥

and hence∨F 6∈ I∗, since if

∨F ∈ I∗ then

∨x∈F ¬Jx ∨ f(x) ∈ IK ∈ Σ.

Since no finite subset of V1(I) has a join in I∗, V +1 (I) is proper.

The proof that V +2 (I) is proper is almost identical. �

14

Proposition 2.12 The sentential compactness theorem implies the almostmaximal ideal theorem.

Proof Define a nondecreasing transfinite sequence of proper ideals of L asfollows:

� I0 = {0};

� Iα+1 =

V +

1 (Iα) if Iα ( V +1 (Iα)

V +2 (Iα) if Iα = V +

1 (Iα) ( V +2 (Iα)

I otherwise

;

� Iλ =⋃α<λ Iα if λ is a limit ordinal.

This sequence is well-defined since the partially ordered set of proper idealsis chain-complete, so that unions of proper ideals are proper, and by Lemma2.11 the ideals indexed by successor ordinals are proper.

The sequence 〈Iα : α ∈ Ord〉 must eventually be constant with value A

say, where A is a proper ideal. Then A = A, so it suffices to prove that A isprime so that A will be an almost maximal ideal.

Let x, y ∈ L with x ∧ y ∈ A. Then Jx ∨ y ∈ AK ∈ P . If v(Jx ∨ y ∈ AK) = >then

x ∈ V1(A) ⊆ V +1 (A) = A

so x ∈ A. If v(Jx ∨ y ∈ AK) = ⊥ then

y ∈ V2(A) ⊆ V +2 (A) = A

so y ∈ A. Hence A is almost maximal. �

At long last we are able to prove the final implication, from which Theorem2.4 will follow.

Proposition 2.13 The almost maximal ideal theorem implies the Booleanprime ideal theorem.

Proof Let B be a Boolean algebra. Then B is in particular a distributivelattice, so has an almost maximal ideal by AMIT. But an almost maximalideal is prime by definition, so B has a prime ideal. �

15

3 Set theory with atoms

Proposition 3.1 There are no nontrivial ∈-automorphisms of the ZF(C)universe V.

Proof Let f : V → V be an automorphism. If x is a set and f(y) = y forall y ∈ x then f(x) = {f(y) : y ∈ x} = {y : y ∈ x} = x. Therefore f istrivial by ∈-induction. �

The key idea in set theory with atoms is to construct a universe whichdoes admit nontrivial ∈-automorphisms. This is achieved by allowing foratoms, also known in the literature as urelemente. Atoms are objects withno set-theoretic structure; they are distinct from sets and are empty in thesense that they contain no elements, but they may be elements of sets. Theresulting theory, ZFA, is obtained from the axioms of ZF by making thefollowing modifications:

� Introduce a unary predicate symbol A, where A(x) will denote ‘x isan atom’.

� Introduce an axiom of atoms, which states that all atoms are empty:

∀x(A(x)→ ∀y y 6∈ x)

� Modify the empty set axiom to state that there is a set which is empty:

∃x(¬A(x) ∧ ∀y y 6∈ x)

� Modify the axiom of extensionality so that it refers only to sets:

∀x∀y(¬A(x) ∧ ¬A(y)→ (∀u(u ∈ x↔ u ∈ y)→ x = y))

This ensures that the atoms do not all collapse onto the empty set.

We will often abuse notation by writing A to denote the class {x : A(x)}and x ∈ A to denote A(x). For the purposes of this essay we will make afurther modification:

� Introduce an atom-set axiom, which states that there is a set of atoms:

∃x∀y(y ∈ x↔ A(y))

If we did not do this then, without further tampering with the axioms,problems would arise. To illustrate this, consider the powerset axiom:

∀x∃y∀z(∀w(w ∈ z → w ∈ x)→ z ∈ y)

16

For any set x this produces some y ⊇ A ∪ P(x), since all atoms z satisfy∀w(w ∈ z → w ∈ x). So if A were a proper class then this axiom wouldproduce a set y containing A, and the axiom of separation would give thatA = {z ∈ y : A(z)} was a set. . . a contradiction! Instead of declaring thatA be a set we could modify the powerset axiom to state

∀x∃y∀z(¬A(z) ∧ ∀w(w ∈ z → w ∈ x)→ z ∈ y)

However, we shall not at any point require that A be a proper class andrequiring that it be a set means that we can leave the remaining axiomsunchanged.

We can make sense of the notion of rank in ZFA in much the same wayas we can in ZF. For a set (or atom) S, define a cumulative hierarchy{Pα(S) : α ∈ Ord} by P0(S) = S and

Pα+1(S) = P(Pα(S)), Pλ(S) =⋃α<λ

Pα(S), P∞(S) =⋃

α∈Ord

Pα(S)

where α, λ denote ordinals and λ is a nonzero limit ordinal.

Let Vα = Pα(A), V = P∞(A), Kα = Pα(∅) and K = P∞(∅).

Figure 1: The cumulative hierarchy with atoms

Proposition 3.2 V is the ZFA universe.

Proof Suppose there is some set x with x 6∈ V. Let y ∈ trcl(x) be ∈-leastsuch that y 6∈ V. Then for all z ∈ y we must have z ∈ V, and hence thereis a least ordinal αz with z ∈ Vαz+1. But then if α =

⋃{αz + 1 : z ∈ y}

we must have z ∈ Vα for each z ∈ y, and so y ∈ Vα+1, and hence y ∈ V,contradicting our assumption. �

The rank of a set x ∈ V is the least ordinal α for which x ∈ Vα+1. ByProposition 3.2, every set has a rank. Furthermore, K is a model of ZF,

17

called the kernel of V, and the sets in K are called pure sets. A picture ofthe ZFA universe is shown in Figure 1.

More generally, if M is a model of ZFA then M∩K is the kernel of M. Notealso that K = H(¬A), or in other words, the pure sets are precisely thosewhich are hereditarily not atoms.

3.1 Fraenkel–Mostowski permutation models

The set of permutations A → A forms a group Aut(A) under composition.Every f ∈ Aut(A) extends to a unique ∈-automorphism of V defined re-cursively on sets by f(x) = f→(x) = {f(y) : y ∈ x}. Provided |A| > 1,nontrivial ∈-automorphisms of V will exist, in stark contrast to Proposition3.1.

We will exploit this new symmetry to pass to submodels of V consistingof those sets which are suitably symmetric. By controlling what we meanby ‘suitably symmetric’ we will be able to prove independence results withrelatively little effort. For instance, suppose ψ is a formula and φ is aformula of the form ∀x∃yφ0(x, y). We will obtain ZFA 0 ψ → φ by passingto a submodel in which ψ holds but in which there is a suitably symmetricset x but no suitably symmetric y satisfying φ0(x, y).

For a set x and a subgroup G ≤ Aut(A), the (setwise) stabilizer of x by Gis

stabG(x) = {f ∈ Aut(A) : f(x) = x}and the fixator of x by G, sometimes called its elementwise stabilizer, is

fixG(x) = {f ∈ Aut(A) : f(y) = y for all y ∈ x} =⋂y∈x

stabG(y)

Note that fixG(x) E stabG(x) ≤ G.

A normal filter F (of subgroups of G) is a filter in Sub(G) which is closedunder conjugations, i.e. if H ∈ F and f ∈ G then f−1Hf ∈ F .

We say a set x is (F -)symmetric if stabG(x) ∈ F . The class of hereditarily F -symmetric sets is denoted by M(A,G,F) and called a (Fraenkel–Mostowski)permutation model of set theory.

Theorem 3.3 M = M(A,G,F) is a transitive model of ZFA, K ⊆ M andA ∈M.

Proof Since M is defined hereditarily it is a transitive class. Also, G actstrivially on K, as can be proved by ∈-induction: if f(y) = y for all y ∈ x

18

thenf(x) = {f(y) : y ∈ x} = {y : y ∈ x} = x

so K ⊆M. And stabG(A) = G since G is a group of permutations of A.

It remains to prove that M is a model of ZFA. A proof of this fact, involvingthe notions of almost universality and Godel operations, can be found in [10,Ch. 4, §2]. �

3.2 Parameter modification

Although a permutation model M is determined by some parameters A, Gand F , these parameters are not determined by the model itself. Later on itwe will need to impose certain conditions on these parameters without anychange to the model—in this subsection we prove that this is possible.

Asymmetric atoms

For a given normal filter F , it may be the case that some atoms are not F -symmetric. Since the permutation model only sees that which is symmetric,if any atoms are asymmetric then they will be forgotten when passing fromV to M and it will be as if they never existed at all in the universe. This ismade more precise in Proposition 3.4 below.

Proposition 3.4 Let M = M(A,G,F) be a permutation model. Then thereexist a subset A′ ⊆ A of atoms, a group G′ of automorphisms of A′ and anormal filter F ′ of subgroups of G′, such that

(i) M ⊆ V(A′) = P∞(A′);

(ii) M = M(A′, G′,F ′);

(iii) Each a ∈ A′ is F ′-symmetric;

(iv) Each H ∈ F ′ is the stabilizer of some x ∈M.

Proof Let A′ be the collection of all F -symmetric atoms. If a ∈ A − A′

then stabG(a) 6∈ F and so a 6∈ M; so in particular if x ∈ V − V(A′) thenx 6∈M. Hence M ⊆ V(A′). This proves (i). Define

N = {f ∈ G : f(a) = a for all a ∈ A′} E G

Then G′ = G/N acts as a group of permutations on A′, and

F/N = {H/(N ∩H) : H ∈ F}

19

is a normal filter of subgroups of G′ and M(A′, G/N,F/N) = M(A,G,F).

For H ≤ G write H ′ = H/N ∩H, and define

F ′ = {H ′ ∈ F/N : H ′ = stabG′(x) for some x ∈M}

Then F ′ is a normal filter:

Upwards closure. Suppose H ′ ≤ K ′ ≤ G′ and H = stabG′(x) ∈ F/N forsome x ∈M. Then f(x) lies in M for each f ∈ G′, because if y ∈ x then

stabG′(f(y)) = f stabG′(y)f−1 ∈ F/N

In particular, z = K ′(x) ∈ M. Clearly K ′ ⊆ stabG′(z). Conversely, iff ∈ stabG′(z) then fk1(x) = k2(x) for some k1, k2 ∈ K ′. Hence fk1k

−12 ∈ H ′

and so f ∈ H ′k2k−11 ⊆ K ′. So K ′ = stabG′(z).

Closure under intersections. If H ′ = stabG′(x) and K ′ = stabG′(y) liein F ′ then H ′ ∩K ′ = stabG′(〈x, y〉) ∈ F ′.

Closure under conjugation. If H ′ = stabG′(x) ∈ F ′ then

fH ′f−1 = stabG′(f(x)) ∈ F ′

Now since F ′ ⊆ F/N we have M(A′, G′,F ′) ⊆M(A′, G/N,F/N) = M. Thereverse inclusion is immediate from how we defined F ′, so we have equality.Thus (ii) is proved, and (iii) and (iv) follow immediately. �

The situation of having asymmetric atoms is avoided in some books (e.g.[7], [10]) by encoding into the definition of a normal filter that stabilizers ofatoms lie in the filter. However, to avoid any dependence on A in our defini-tion of a normal filter, we will overcome this anomaly by using Proposition3.4.

Shrinking

Given a group G, a normal filter F on G and a subgroup H ∈ F , define theshrink FH of F to H by

FH = {K ∈ F : K ≤ H} = F ∩ Sub(H)

We can shrink the filter giving rise to it without changing the model. In-tuitively this is because if a group stabilizes some set then so do all of itssubgroups.

Proposition 3.5 Let A be a set of atoms, G a group of permutations ofA and F a normal filter on G. Then M(A,G,F) = M(A,H,FH) for anyH ∈ F .

20

Proof If stabH(x) ∈ FH then stabH(x) ∈ F , and hence stabG(x) ∈ F sincestabH(x) ≤ stabG(x) and F is upwards-closed. Conversely, if stabG(x) ∈ Fthen stabH(x) = stabG(x) ∩ H ∈ F since H ∈ F by assumption; but thenstabH(x) ∈ FH since stabH(x) ≤ H.

Since stabG(x) ∈ F if and only if stabH(x) ∈ FH , it follows immediatelythat x ∈M(A,G,F) if and only if x ∈M(A,H,FH). �

Quotienting out triviality

The final modification we will make later on regards the filter. If the inter-section of all the subgroups in F acts trivially on the sets in M then wemay quotient out by this triviality without changing the model.

Proposition 3.6 Let N =⋂F . Then N E G, and M(A,G/N,F/N) =

M(A,G,F).

Proof Let g ∈ G and n ∈ N . Then n ∈ H for all H ∈ F , so gng−1 ∈ gHg−1

for all H ∈ F . By normality, gng−1 ∈ H for all H ∈ F , so gng−1 ∈ N .Hence N E G. In fact, N E H for all H ∈ F , so in particular N E stabG(x)for all x ∈M. Since N acts trivially on x, we have

stabG/N(x) = stabG(x)/N

so x ∈M(A,G,F) if and only if x ∈M(A,G/N,F/N). �

3.3 Normal filter–normal ideal duality

In constructing permutation models it will prove useful to be able to charac-terise the normal filter F in terms of subsets of A, whenever this is possible.This provides a more convenient setting for working with the wellorderingprinciple, which is equivalent to the axiom of choice by Lemma 1.2. To thisend, we say that a nonempty collection I of subsets of A is a normal idealif it is an ideal of subsets of A which is closed under the action of G, i.e.if S ∈ I and f ∈ G then f(S) ∈ I. In this way, normal filters give rise tonormal ideals and vice versa, and these operations are almost but not quitemutually inverse, in a sense explained below.

If I is a normal ideal of subsets of A then define

I∗ = {H ≤ G : fixG(S) ≤ H for some S ∈ I}

This is a normal filter of subgroups of G. Conversely, if A is a set, G is agroup of permutations of A and H ≤ G then define the H-fixed subset

AH = {a ∈ A : f(a) = a for all f ∈ H} ⊆ A

21

If F is a normal filter of subgroups of G then define

F∗ = {S ⊆ A : S ⊆ AH for some H ∈ F}

This is a normal ideal of subsets of A.

In passing we note the following result. A Galois connection between par-tially ordered sets P and P′ is a pair 〈f, g〉 of monotone maps f : P → P′and g : P′ → P such that, for all x ∈ P and y ∈ P′, f(x) ≤ y if and only ifx ≤ g(y).

Proposition 3.7 The assignments S 7→ fixG(S) and H 7→ AH define aGalois connection between P(A) and Sub(G).

Proof Let H ≤ G and S ⊆ A. Then S ⊆ AH if and only if every elementof H fixes each element of S, which occurs if and only if fixG(S) ≤ H.Furthermore, if K ≤ H and T ⊆ S then AK ⊇ AH and fixG(T ) ≥ fixG(S),so the assignments are monotonic. �

If I is a normal ideal on A then x ∈M(A,G, I∗) if and only if there is someS ∈ I with fixG(S) ≤ stabG(x). We call such a set S a support for x.

Lemma 3.8 A set x ∈ M = M(A,G,F) has a wellordering in M if andonly if fixG(x) ∈ F .

Proof The existence of a wellorder of x in M is equivalent to the existenceof an ordinal α and an injection h : x → α. For each y ∈ x, h(y) ∈ αlies in the kernel, so is fixed by G. Thus for all f ∈ G and y ∈ x we havef(〈y, h(y)〉) = 〈f(y), h(y)〉. Since h is injective, we thus have h(f) = f ifand only if f(y) = y for each y ∈ x. Therefore f ∈ stabG(h) if and only iff ∈ fixG(x), and so stabG(h) = fixG(x). �

This fact will be easy to exploit in order to force that certain sets have nowellordering in a given permutation model.

Fraenkel’s basic model

Fraenkel’s basic model is a very simple permutation model that illustrateshow useful the duality between normal filters and normal ideals is. Let Abe a countably infinite set of atoms, G be the group of all permutations ofA and I be the set of all finite subsets of A. It is clear that I is a normalideal, so we obtain a model M = M(A,G, I∗) of ZFA in the usual way.

22

Theorem 3.9 M � ¬AC

Proof Note first that A ∈ M, since stabG(A) = G ∈ I∗ and each atomis symmetric by definition of I∗. Now if S ∈ I then AfixG(S) = S since ifx, y ∈ A − S then the tranposition of x and y lies in fixG(S). So if S ∈ Iwith fixG(S) ≤ fixG(A) then

A = AfixG(A) ⊆ AfixG(S) = S

This contradicts finiteness of S, so fixG(A) 6∈ I∗, and A admits no wellorder-ing in M by Lemma 3.8. �

Corollary 3.10 The axiom of choice is independent from the axioms ofZFA. �

Before moving on to a more interesting example of a permutation model, wewill explore the duality between normal filters and normal ideals further. InFraenkel’s basic model, it was the fact that the normal filter arose from anideal that allowed us to break the axiom of choice so easily, but in generalnormal filters may not arise in this way.

Proposition 3.11

(i) Given a normal filter F on a group G, F ⊇ F∗∗ and F∗∗∗ = F∗.

(ii) Given a normal ideal I on a set A, I ⊆ I∗∗ and I∗∗∗ = I∗.

Proof

(i) Let H ∈ F∗∗. Then there exists S ∈ F∗ with fixG(S) ≤ H, and for thisS there is some K ∈ F with S ⊆ AK . But then K ≤ fixG(S) ≤ H,and so H ∈ F by upwards-closure.

The first part of (ii) applied to F∗ gives F∗ ⊆ F∗∗∗. If S ∈ F∗∗∗ thenfixG(S) ≤ H for some H ∈ F∗∗; but then H ∈ F since F∗∗ ⊆ F , soS ∈ F∗ and F∗ = F∗∗∗.

(ii) Let S ∈ I, then fixG(S) ∈ I∗, so AfixG(S) ∈ I∗∗. But S ⊆ AfixG(S), soS ∈ I∗∗ by downwards-closure.

The first part of (i) applied to I∗ gives I∗∗∗ ⊆ I∗. If H ∈ I∗ thenfixG(S) ≤ H for some S ∈ I; but then S ∈ I∗∗ since I ⊆ I∗∗, soH ∈ I∗∗∗ and I∗ = I∗∗∗. �

23

A normal filter F (resp. normal ideal I) is reflexive if F = F∗∗ (resp.I = I∗∗). Note that if a normal filter (resp. ideal) F is irreflexive then theredoes not exist a normal ideal (resp filter) I such that F = I∗. The followingtwo examples demonstrate that Proposition 3.11 is the best we can hopefor, in that irreflexive normal filters and ideals exist.

Example Let G be the group of all permutation of Z acting on R by f(n+δ) = f(n) + δ for all f ∈ G, n ∈ Z and 0 ≤ δ < 1; that is, G acts on R bypermuting intervals of the form [n, n + 1) for n ∈ Z. Let I be the normalideal of all finite subsets of R. Then [0, 1) ∈ I∗∗ since [0, 1) = RfixG({0}), but[0, 1) is infinite, so [0, 1) 6∈ I. Hence I 6= I∗∗.

Example For X ⊆ Z let SX denote the group of permutations of X whichfix all but finitely many points and AX denote the subgroup of SX consistingof all even permutations.

F = {SZ−E : E ⊆ Z finite} ∪ {AZ−E : E ⊆ Z finite}

Write G = SZ, and note that SZ−E = fixG(E) for all E ⊆ Z. Now F is closedunder intersections, since

SZ−E ∩ SZ−F = SZ−(E∪F )

SZ−E ∩ AZ−F = AZ−(E∪F )

AZ−E ∩ AZ−F = AZ−(E∪F )

and it is closed upwards by elementary results in group theory. Closure underconjugation follows from the fact that fSZ−Ef

−1 = SZ−f(E) and fAZ−Ef−1 =

AZ−f(E). Hence F is a normal filter. However AZ 6∈ F∗∗ since fixG(ZAZ) = Gand fixG(ZfixG({n})) = fixG({n}), both of which contain odd permutations.Hence F 6= F∗∗.

24

4 Mostowski’s ordered model

By introducting an order structure on the set of atoms, Mostowski’s orderedmodel allows us to refine the result obtained from Fraenkel’s model to provenot just that the axiom of choice is independent from the axioms of ZFA(Lemma 4.1), but in fact that the axiom of choice is independent from theordering principle in the presence of the ZFA axioms (Theorem 4.4). Witha bit more work, we will see that the Boolean prime ideal theorem alsoholds in Mostowski’s ordered model (Theorem 4.7), therefore proving thatthe axiom of choice is independent from the Boolean prime ideal theorem inthe presence of the ZFA axioms.

Let A denote the set of atoms, which is assumed to be countably infinite,and let <A be an ordering on A such that 〈A,<A〉 is order-isomorphic to〈Q, <〉. Let G be the group of all <A-preserving permutations of A and letI be the normal ideal of all finite subsets of A. Again, we get an inducedpermutation model M = M(A,G, I∗).

Lemma 4.1 M � A has no wellordering.

Proof Let S be a finite subset of A. Let h : Q→ A be an order-isomorphismand let S = h←(S) ⊆ Q. Suppose that max(S) ≤ 0; otherwise replace f by

f −max(S). Define f : Q→ Q by

f(x) =

{x if x ≤ 0

2x if x ≥ 0

Then f = h ◦ f ◦ h−1 is a nontrivial <A-preserving permutation of A whichfixes each s ∈ S. �

So the axiom of choice fails in M. In order to prove the independence of theordering principle from the axiom of choice, we need to find a way of linearlyordering each set in M. To do so, some preliminary results are needed.

Lemma 4.2 Each x ∈ M has a least support Sx, and the class of all pairs〈x, Sx〉 for x ∈M is I∗-symmetric.

Proof Let S and T be supports of x, and let U = S ∩ T .

Clearly fixG(S), fixG(T ) ≤ fixG(U). Moreover, if f ∈ fixG(U) then we canfind g1, . . . , gn ∈ fixG(S) ∪ fixG(T ) with gn ◦ · · · ◦ g1 = f . Intuitively, thisis done as follows. Partition A − U into |U | + 1 sets in the obvious way.

25

Taking each partition in turn, alternately slide the elements of S and T inthe partition along (by elements of fixG(T ) and fixG(S), respectively) untilthey and all the points between them lie at their images under f . This canbe done in finitely many steps since S ∪ T is finite. It follows that fixG(U)is equal to the group generated by fixG(S) and fixG(T ) and hence U is asupport of x.

This implies that any finite intersection of supports is a support. Now if Sis the collection of all supports of x then there is some finite subcollectionS ′ ⊆ S such that

⋂S ′ =

⋂S = Sx; for instance, pick an arbitrary S ∈ S

and let S ′ = S ∩ P(S). Hence x has a least support.

To prove the last part, note that if f ∈ G, S ∈ I and x ∈M then

fixG(f(S)) = f fixG(S)f−1 and stabG(f(x)) = f stabG(x)f−1

so Sx is the least support of x if and only if f(Sx) is the least support off(x). �

Lemma 4.3 There is a symmetric injective function-class ι : M→ Ord×I.

Proof The orbits G(x) = {f(x) : f ∈ G} are pairwise disjoint and sym-metric, so can be enumerated by ordinals, and furthermore this enumerationis symmetric. Let αx be the ordinal assigned to the orbit G(x), and defineι(x) = 〈αx, Sx〉. Then ι is symmetric, since f(Sx) = Sf(x) and f(x) ∈ G(x)so αf(x) = αx. Furthermore, ι is injective: if ι(x) = ι(y) then Sx = Sy andy ∈ G(x). Let y = f(x) for f ∈ G. Then Sx = Sf(x) = f(Sx), and so f ∈ Sx.But that means that f(x) = x, so y = x. �

Theorem 4.4 M � Every set has a linear ordering.

Proof Note first that <A is symmetric, since by definition each f ∈ Grespects <A. Furthermore, I can be ordered lexicographically since it is aset of finite subsets of a linearly ordered set; hence Ord×I can be orderedlexicographically (by <Ord×I , say). Hence we can define the relation-class<M on M by x <M y if and only if ι(x) <Ord×I ι(y). Then <M is a linearordering of M, and so descends to a linear ordering of each set x ∈ M byrestriction. �

Corollary 4.5 The axiom of choice is independent from the ordering prin-ciple in the presence of the axioms of ZFA. �

Now we will prove that the Boolean prime ideal theorem holds in M. It mayseem strange to do this because this result is stronger than Theorem 4.4;however, its proof is far more involved and uses different methods.

26

By Theorem 2.4 it suffices to prove that the consistency principle holds inM, and by Lemma 1.1 we need only consider messes on sets. Given a binarymess M ∈ M on a set X ∈ M, define M and X as follows. Let SM be theleast support of M and, for each x ∈ X, let

O(x) = fixG(SM)(x) = {f(x) : f ∈ fixG(SM)}

X = O→(X) = {O(x) : x ∈ X}Since fixG(SM) ≤ stabG(X) we have O(x) ⊆ X for each x ∈ X, and hence

X ⊆ P(X).

Given a finite subset F ⊆ X, a subset E ⊆ F , and a finite subset P ⊆⋃F ,

we say P admits E if 〈P,O←(E)〉 ∈ M. Let M be the collection of all pairs

〈F,E〉 with F ⊆ X finite and E ⊆ F such that every subset of⋃F admits

E.

Lemma 4.6 M is a mess on X.

Idea of proof A full proof can be found in [7, Theorem 7.16] or [10, The-orem 7.1], with a slight change in language: pairs 〈F,E〉 correspond withfinite partial functions g : X ⇁ 2 with F = dom(g) and E = g←(1).

It suffices to prove that for each finite F ⊆ X there is a subset E ⊆ Fwith 〈F,E〉 ∈ M. The proof splits into the cases where SM is empty andnonempty and appeals to combinatorial arguments about n-sets, i.e. setsthat can be partitioned into a union of n sets of the form f→i (P0) for fi ∈ Gand P0 fixed, and types of n-sets, i.e. equivalence classes of n-sets under therelation of differing by an element of G. �

Theorem 4.7 The consistency principle holds in M.

Proof Let M ∈ M be a mess on a set X ∈ M. Working outside of M—inparticular, assuming the axiom of choice holds in our metatheory—we canfind a subset C ⊆ X which is consistent with M. Now define

C = {x ∈ X : O(x) ∈ C} ⊆ X

Then C is symmetric, so C ∈M, and C must be consistent with M. �

Corollary 4.8 The Boolean prime ideal theorem is independent from theaxiom of choice in the presence of the axioms of ZFA.

Proof M � ¬AC by Lemma 4.1 and M � CP by Theorem 4.7. By Theorem2.4, CP is equivalent to BPIT, and so M � BPIT + (¬AC). �

27

5 Topological groups and their actions

This short section builds up some theory necessary to state and prove themain result of the essay.

5.1 Extremely amenable groups

A G-set is a set X equipped with an action G y X. A topological group isa topological space G equipped with a group structure whose operations

(−) · (−) : G×G→ G, (−)−1 : G→ G

are continuous with respect to the topology.

A neighbourhood base N of x ∈ X is a collection of subsets of G whichcontain x, such that if x ∈ U ⊆ G then U is open if and only if N ⊆ U forsome N ∈ N . A topological group G has small open subgroups if and onlyif every open neighbourhood of the identity contains an open subgroup, i.e.the open subgroups form a neighbourhood base of 1.

For a topological group G and a normal filter F of subgroups of G, define

F † = {U ⊆ G : ∀g ∈ U ∃H ∈ F gH ⊆ U}

Conversely, for a topology O on G, define

O† = {H ≤ G : H ∈ O} = {O-open subgroups of G}

Proposition 5.1 F † is a topology and O† is a normal filter. MoreoverF †† = F , and G has small open subgroups if and only if O ⊆ O††.

Proof Trivially ∅ ∈ F † and G ∈ F †. Let U, V ∈ F †. If g ∈ U ∩ V thenthere exist H,K ≤ G such that gH ⊆ U and gK ⊆ V . Then gH ∩ gK =g(H∩K) ⊆ U∩V , and H∩K is a subgroup, so U∩V ∈ F †. Let {Ui : i ∈ I}be a collection of subsets of G, all of which we may assume to be nonempty,and let U =

⋃i∈I Ui. If g ∈ U then g ∈ Uj for some j ∈ I. Then there exists

H ≤ G such that gH ⊆ Uj; so gH ⊆ U , so I ∈ F †. Hence F † is a topologyon G.

O† is nonempty since G is an open subgroup. It is closed under pairwiseintersections since if H and K are open subgroups then H ∩K is an opensubgroup. It is closed upwards since if K ≤ H ≤ G with K open thenH =

⋃h∈H hK and each coset hK is open by continuity of the action of

G on itself, so H is open. Likewise, if g ∈ G and H ≤ G is open thengHg−1 ≤ G is open by continuity. Hence O† is a normal filter.

28

If H ∈ F then hH ⊆ H for all h ∈ H, so H ∈ F ††. Conversely if K ∈ F ††then there is some H ∈ F with 1H = H ⊆ K, so K ∈ F by upwards-closure.

Now O†† is precisely the set of all subsets U ⊆ G such that for all g ∈ Uthere is some H ≤ G with gH ⊆ U . This occurs if and only if H ⊆ g−1U .Now U is open if and only if g−1U is open, so O ⊆ O†† if and only if G hassmall open subgroups. �

Recall from Proposition 3.6 that without changing our model we may re-place G by G/N where N =

⋂F . Together with the following result, this

means that we may replace G by a quotient of G and thus assume that Gis Hausdorff.

Proposition 5.2 Let G be a topological group with topology O and supposethat G has small open subgroups. Then

⋂O† = {1} if and only if G is

Hausdorff.

Proof If 1 6= g ∈⋂O† then g ∈ H for each H ∈ O†. In particular, if

U ⊆ G is open and contains 1 then g ∈ U , so 1 and g cannot be separatedby open sets, so G is not Hausdorff.

Conversely, suppose G is Hausdorff. Then for each g ∈ G there exists anopen subset Ug ⊆ G containing 1 but not g, and hence there exists an opensubgroup Hg not containing g. Then

⋂O† ⊆

⋂g∈GHg = {1}. �

If G is a topological group and X is a topological space then an actionGy X is continuous if it is continuous as a map G×X → X of topologicalspaces. A topological space X equipped with a continuous G-action is calleda G-space.

A compact Hausdorff G-space is called a G-flow. A group G is extremelyamenable if for each G-flow X there is a point x∗ ∈ X for which g · x∗ = x∗

for all g ∈ G, called a fixed point of the G-flow.

Extremely amenable groups are at one end of our correspondence; as such weneed to make sure no damage is done when passing to a Hausdorff quotientusing Propositions 3.6 and 5.2.

Proposition 5.3 Let G be a topological group with topology O and writeN =

⋂O†. Then G has small open subgroups if and only if G/N does.

Moreover, if G has small open subgroups, then G is extremely amenable ifand only if G/N is extremely amenable.

29

Proof The quotient topology on G/N is precisely O/N = {U/N : U ∈ O},where we take the topological quotient. Furthermore, O ⊆ O†† if and only ifO/N ⊆ O††/N = (O/N)††. By Proposition 5.1, G has small open subgroupsif and only if G/N does.

Let X be a G-flow. For each open V ⊆ X, the set

U(V ) = {g ∈ G : g→(V ) ⊆ V }

is open in G, and hence contains some H ∈ O†. In particular, N ⊆ U(V ),so for all n ∈ N and open V ⊆ X, n→(V ) ⊆ V . Since X is Hausdorff, Nacts trivially on X.

Thus there is an exact correspondence between actions ofG onX and actionsof G/N on X given by (gN) · x = g · x. In particular, x∗ is a fixed pointfor G y X if and only if it is a fixed point for the corresponding actionG/N y X. �

Lemma 5.4 A G-flow X has a fixed point if and only if for each n ∈ N,continuous f : X → Rn, ε > 0 and finite subset F ⊆ G, there exists x ∈ Xwith

‖f(x)− f(g · x)‖ ≤ ε for all g ∈ Fwhere ‖−‖ denotes the Euclidean norm on Rn.

Proof (⇒) This direction is trivial by setting x = x∗.

(⇐) Given f, ε, F as in the statement of the lemma, put

Af,ε,F = {x ∈ X : ‖f(x)− f(g · x)‖ ≤ ε for all g ∈ F}

Then each Af,ε,F is closed, and hence compact since X is compact.

Now if {Afj ,εj ,Fj: 1 ≤ j ≤ m} is any finite collection of such sets, with

fj : X → Rnj for each j, then putting

f = (f1, . . . , fn) : X → Rn1+···+nm , ε = min1≤j≤m

εj, F =m⋃j=1

Fj

gives ∅ 6= Af ,ε,F ⊆n⋂j=1

Afj ,εj ,Fj; that is, the intersection of any finite collec-

tion of the sets Af,ε,F is nonempty.

By compactness the intersection A of all the Af,ε,F is nonempty. Let x∗ ∈ Aand suppose it is not a fixed point. Choose g ∈ G with x∗ 6= g · x∗. ByLemma 1.3 there is a continuous map f : X → R with f(x∗) = 0 andf(g · x∗) = 1; but this contradicts x∗ ∈ A.

So any point in A is fixed. In particular, A 6= ∅, so a fixed point exists. �

30

5.2 Ramsey filters

A G-set X has the Ramsey property if for every 2-colouring c : X → 2 andevery finite F ⊆ X there exists g ∈ G such that g ·F is monochromatic, i.e.c(g · x) = c(g · x′) for all x, x′ ∈ F . In this case we call the action of G on Xa Ramsey action and we call X a Ramsey G-set.

A subgroup H ≤ G is a Ramsey subgroup of G if the set G/H of left-cosetsof H in G is a Ramsey G-set under the usual action g · (xH) = (gx)H. Tospell it out, this means that for every 2-colouring c : G/H → 2 and everyfinite set F of cosets of H in G, some G-translation g ·F is monochromatic.

We say a normal filter F of subgroups of a group G is a Ramsey filter (aswitnessed by H) if it contains a subgroup H for which whenever K ≤ Hwith K ∈ F , K is a Ramsey subgroup of H.

When we come to prove the correspondence between models and groupsin Section 6, our strategy will be to go via Ramsey filters. It will becomeapparent that our definition is quite tight and for our purposes it will beconvenient to have a weaker definition. The following proposition says wecan do that at no cost.

Lemma 5.5 Fix k ≥ 2. A G-set X has the Ramsey property if and only iffor every finite subset F ⊆ X there exists a finite subset Y ⊆ X such that,whenever c : Y → k is a k-colouring of Y , there exists g ∈ G with g ·F ⊆ Yand g · F is monochromatic.

Proof Denote the latter statement, for a given k, by R(k).

(⇒) is clear: suppose that X has the Ramsey property; then since a 2-colouring is a k-colouring, we may simply take Y = g · F and consider therestriction of said colouring to Y .

(⇐) is proved by induction on k, using a compactness argument in the basecase. Take Jx gets colour iK for x ∈ X and i ∈ 2 as sentential variables,encoded as elements of X × 2 say, and for each 2-colouring c : X → 2 definea valuation vc : X × 2→ {>,⊥} by

vc(Jx gets colour iK) =

{> if c(x) = i

⊥ otherwise

Let Σ be the set of sentences of the form

Jx gets colour 0K ∨ Jx gets colour 1K

¬(Jx gets colour 0K ∧ Jx gets colour 1K)

31

¬

(∧x∈F

Jg · x gets colour 0K ∨∧x∈F

Jg · x gets colour 1K

)for x ∈ X and g ∈ G. In turn, these mean that each x ∈ X has a colour,that no x ∈ X has two colours, and that no G-translation g · F of F ismonochromatic.

Let F be a finite subset that witnesses the failure of R(2). Then any finitesubset of Σ has a model vc for some colouring c, and hence Σ has a modelby compactness. But this is precisely the statement that F witnesses thefailure of the Ramsey property.

Now suppose k > 2 and that that R(k′) holds for all 2 ≤ k′ < k. Fix afinite subset F ⊆ X. We may choose Y that satisfies R(k− 1) using F ; andwithout loss of generality Y ⊇ F (if not, replace it by Y ∪ F ). By R(2)with Y in the place of F , there exists a finite subset Z ⊇ Y such that every2-colouring of Z makes some subset of the form g · Y monochromatic.

Let c : Z → k be a k-colouring, and define c0 : Z → 2 by

c0(z) =

{0 if c(z) 6= k − 1

1 if c(z) = k − 1

By assumption there is some g ∈ G such that c0 is constant on g · Y .

If c→0 (g · Y ) = {1} then c→(g · Y ) = {k − 1}, so in particular c is constanton g · F and we’re done.

If not then c→0 (g · Y ) = {0} and c→(g · Y ) ⊆ k − 1, so there is a (k − 1)-colouring c1 : Y → k − 1 defined by c1(y) = c0(g · y). By R(k − 1) there issome g′ ∈ G such that g′ · F ⊆ Y and c1 is constant on g′ · F ; but then c isconstant on (gg′) · F . �

From now on, the Ramsey property will interchangeably refer to the Ramseyproperty as originally defined and the statement R(k) for an arbitrary k asin Lemma 5.5.

32

6 The correspondence

We are now ready to state the long-promised correspondence. Its proof re-lies on two theorems (6.2 and 6.5), whose proofs occupy the remainder ofthe section. Throughout this section, G will be a topological group of per-mutations of a set A of atoms, F will be the normal filter of open subgroupsof G, and M = M(A,G,F) will be the corresponding permutation model ofZFA.

Theorem 6.1 Suppose G is Hausdorff and has small open subgroups, thatall atoms are F-symmetric and that each H ∈ F stabilizes some set inM. Then M � BPIT + (¬AC) if and only if G is nontrivial and extremelyamenable (with a slight caveat).

Proof

G is nontrivial andextremely amenable

(6.2)←→F is a Ramsey

filter of subgroupsof G and {1} 6∈ F

(6.5)←→ M � BPIT + (¬AC)

The caveat is made precise in Theorems 6.2 and 6.5 and discussed in Section6.1. �

Theorem 6.2 Suppose G is Hausdorff and has small open subgroups. Thenthe following are equivalent:

(i) G is nontrivial and extremely amenable;

(ii) F is a Ramsey filter as witnessed by G, and {1} 6∈ F .

ProofSuppose (i) holds, so that G is nontrivial and extremely amenable. LetH ≤ G be an open subgroup, c : G/H → 2 be a 2-colouring of G/H, andA ⊆ G/H be a finite subset.

G acts on Y = 2G/H by (g · p)(xH) = p(g−1xH) for p ∈ Y , g ∈ G andxH ∈ G/H, and this action turns Y into a G-flow. Note also that thecolouring c lies in Y . Let X = G(c) be the closure in Y of the orbit of cunder the action of G. Then X is a G-flow and there is a fixed point p∗ ∈ X.

For any g ∈ G we have p∗(gH) = (g−1 · p∗)(H) = p∗(H). But G actstransitively on G/H, so p∗ : G/H → 2 is constant, say p∗(gH) = i for allgH ∈ G/H, where i = 0 or 1.

33

Since p∗ ∈ G(c), there is a g ∈ G for which (g · c)(aH) = p∗(aH) = i for allaH ∈ A. Hence H is a Ramsey subgroup of G, so (ii) holds.

Now suppose that (ii) holds. Then F satisfies the corresponding definition ofa Ramsey filter (witnessed by G) referring to right-cosets H/G rather thanleft-cosets G/H, where the action Gy H/G is given by g ·Hh = H(hg−1).

Let X be a G-flow, f : X → Rn be a continuous map, ε > 0 and F ⊆ G bea finite subset. Let H ⊆ G be an open neighbourhood of 1 ∈ G for which

‖f(x)− f(h · x)‖ ≤ ε

3for all h ∈ H and x ∈ X

Since G has small open subgroups, we can suppose without loss of generalitythat H is a subgroup of G.

Since f→(X) ⊆ Rn is compact, we can partition it into m sets A0, . . . , Am−1

of diameter ≤ ε3. Fix x0 ∈ X, and for each 0 ≤ i < k let

Ki = {g ∈ G : f(g · x0) ∈ Ai}

andHi = {hk : h ∈ H, k ∈ Ki} =

⋃k∈Ki

Hk

We can identify Hi with the subset {Hk : k ∈ Ki} of H/G. Also,⋃iHi =

G/H, and so there is an m-colouring c : G/H → m such that c−1({i}) ⊆ Hi.But then, since F is a Ramsey filter, there is 0 ≤ i < m and g ∈ G with(F ∪ {1})g ⊆ Hi. Set x = g · x0 for this g.

For y ∈ F , choose h ∈ H with hyg ∈ Ki. Then f(hyg · x0) = f(hy · x) ∈ Ai.Then ‖f(hy · x)− f(y · x)‖ ≤ ε

3, and so f(y · x) lies in the ε

3-neighbourhood

of Ai. But 1 ∈ F , so ‖f(x)− f(y · x)‖ ≤ ε.

It follows from Lemma 5.4 that X has a fixed point for G y X, and so (i)holds. �

To prove Theorem 6.5 we introduce some notation and prove some prelimi-nary results. The idea is similar to that of Theorem 2.12.

Suppose now that each subgroup in F stabilizes some set in M; this ispossible by Proposition 3.4. Fix a subgroup H ∈ F .

Let K,L ∈ F be subgroups of H and consider H/L as a K-set. Suppose Fis a finite subset of some K-orbit T ⊆ H/L which is a counterexample tothe Ramsey property, in the sense that there is a finite subset Y ⊆ T suchthat for all k ∈ K, Y can be partitioned into two sets Y> and Y⊥, neitherwholly containing kF .

34

For such F,K,L define the 〈F,K,L〉-orbit X〈F,K,L〉 to be some set isomor-phic (as an H-set) to H/L which is disjoint from X〈F ′,K′,L′〉 for 〈F,K,L〉 6=〈F ′, K ′, L′〉. For a given F , let F be the image of F under the isomorphismH/L ∼= X〈F,K,L〉.

LetX =

⋃〈F,K,L〉

X〈F,K,L〉

with H-action induced by the isomorphisms X〈F,K,L〉 ∼= H/L. Without lossof generality, X is a set in M: if not then it is isomorphic to one, since everysubgroup in F stabilizes some set, and the action G y M transfers acrossthe isomorphism.

Consider X as a set of variables and let Σ ⊆ L(X) be the set of sentencesof the form

ξ(F,K,L, h) =∨x∈F

¬hx, η(F,K,L, h) =∨x∈F

hx

for X〈F,K,L〉 as above and h ∈ H. Then Σ ∈M.

Lemma 6.3 If Σ is consistent (in M) then F has the Ramsey property.

Proof Let v : X → {>,⊥} be a model of X in M, and let K ∈ F stabilizev. We may assume that K ≤ H by replacing K by K ∩ H if necessary,since K ∩H certainly stabilizes v if K does. Let L ∈ F be a subgroup withL ≤ H; we show that L is a Ramsey subgroup of H.

Suppose not, and let L be a counterexample. Let F ⊆ K/L ⊆ H/Lbe a finite subset, such that each finite subset Y ⊆ F has a partitionY = Y> ∪ Y⊥ with gF * Y>, Y⊥ for all g ∈ G. Then X〈F,K,L〉 ∈ X andξ(F,K,L, 1), η(F,K,L, 1) ∈ Σ. Hence v is not constant on F . But all thevariables in F lie in the same K-orbit, so this is impossible since v is fixedby K.

This is the required contradiction. �

We now prove a very suggestive result, bearing in mind that we can’t as-sume the sentential compactness theorem and therefore can’t prove that Σis consistent.

Lemma 6.4 Every finite subset of Σ has a model in M.

Proof Let Σ0 ⊆ Σ be finite. Each sentence in Σ involves only one elementfrom a single H-translation hT of the K-orbit T , so we may assume without

35

loss of generality that Σ0 consists entirely of sentences all of whose variablescome from such hT . In fact, we may take h = 1. We can then recover Σ0 byleft-multiplication by h, and piece together the valuations (in a sense madeprecise) to obtain a model for a general finite subset of Σ.

Let Y = {x ∈ T : x appears in Σ0}. By definition, there exists a partitionY = Y> ∪ Y⊥ of Y with hF * Y> and hF * Y⊥. Define a valuationv : X → {>,⊥} by

v(x) =

> if x ∈ Y>⊥ if x ∈ Y⊥∗ if x 6∈ Y

where ∗ denotes an arbitrary value which does not matter. This allows usto piece together the valuations for the general case.

Now, v defines a model of Σ0, and by the above reasoning it follows thatevery finite subset of Σ has a model. �

Theorem 6.5 Suppose all atoms are F-symmetric and each H ∈ F occursas a stabilizer of some set in M. Then the following are equivalent:

(i) M � BPIT;

(ii) F is a Ramsey filter of subgroups of G.

Moreover, M � AC if and only if {1} ∈ F .

Proof By Theorem 2.4, it suffices to prove

SCT holds in M (a)−→ F is a Ramsey filter(b)−→ AMIT holds in M

We use all notation as above.

(a) Suppose M � SCT; then Σ has a model by Lemma 6.4. By Lemma 6.3,for an arbitrary H ∈ F we can find K ≤ H such that every subgroupof K is a Ramsey subgroup. But then the set M of such Ks for eachH ∈ F forms a basis for F , so that F has the Ramsey property.

(b) Conversely, suppose that F is a Ramsey filter, and let B be the basisof F consisting of those K ∈ F arising in the manner described above.Fix a (nontrivial) distributive lattice L in M, and find some K ∈ Bstabilizing L.

Using the axiom of choice in the meta-theory, we can find a maximalideal I ⊆ L with K ≤ stabG(I). Then I is F -symmetric, so lies in M,

and by maximality I = I. It remains to prove that I is prime.

36

Suppose not. Then there exist x, y ∈ L − I with x ∧ y ∈ I. Then theideal generated by

I ∪ {kx : k ∈ K}

is K-invariant, so lies in M, and properly contains I. Replacing x byy yields the same result for I ∪ {ky : k ∈ K}. Hence there existk1, . . . , kr ∈ K with

k1x ∨ · · · ∨ krx ∈ I∗ and k1y ∨ · · · ∨ kry ∈ I∗

Let L = stabK(〈x, y〉). Then L ∈ F and L ≤ K, and since F is aRamsey subgroup, X = orbK(〈x, y〉) ∼= K/H has the Ramsey property.

Let F = {k1〈x, y〉, . . . , kr〈x, y〉} and from the Ramsey property obtaina finite set

Y = {y1〈x, y〉, . . . , ys〈x, y〉}

where y1, . . . , ys ∈ K are such that whenever Y is partitioned into twopieces, one of the pieces contains kF for some k ∈ K.

For C ⊆ {1, 2, . . . , s} let zC =(∧

i∈C yix ∨∧j 6∈C yjx

). Then

z =∧

C⊆{1,2,...,s}

zC =s∨i=1

(yix ∨ yiy) =s∨i=1

yi · x ∧ y ∈ I

Fix a subset C ⊆ {1, 2, . . . , s}. This determines a partition Y = Y>∪Y⊥,where

Y> = {yi〈x, y〉 : i ∈ C} and Y⊥ = {yj〈x, y〉 : j 6∈ C}

By choice of Y , there exists k ∈ K with either kF ⊆ Y>, in which case,

zC ≥∨i∈C

yix ≥r∨j=1

kkjx ∈ kI∗ = I∗

or kF ⊆ Y⊥, in which case

zC ≥∨i 6∈C

yix ≥r∨j=1

kkjy ∈ kI∗ = I∗

But then z =∧C⊆{1,...,s} zC ∈ I∗. But we saw above that z ∈ I, so

I ∩ I∗ = ∅, contradicting Lemma 2.9.

We thus have (a)↔(b). Now if {1} ∈ F then fixG(x) ∈ F for each x ∈ Mby upwards-closure, so every set in M has a wellorder in M by Lemma 3.8.Conversely, if every set can be wellordered then fixG(A) = {1} ∈ F . ByLemma 1.2, the axiom of choice holds in M if and only if {1} ∈ F . �

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6.1 Post mortem

The ‘caveat’

As alluded to in its statement, the correspondence in Theorem 6.1 is notquite exact. Both the hypotheses and the specification of the normal filterF differ between Theorems 6.2 and 6.5. The hypotheses differ in that:

� Theorem 6.2 requires that G be Hausdorff and that it have small opensubgroups;

� Theorem 6.5 requires that all atoms be symmetric and that each sub-group in the filter F stabilise some set in M.

Fortunately this has no implications for our correspondence. The require-ment that G be Hausdorff corresponds exactly with our requirement thatall atoms be symmetric. If it weren’t Hausdorff then Propositions 5.2 and5.3 give that some quotient G/N is Hausdorff—and is extremely amenableif and only if G is—and Proposition 3.6 tells us that the model M is unaf-fected by this modification. Proposition 3.4 takes care of the requirementthat each subgroup in F stabilise some set in the model.

The specification of the normal filter F , however, does affect the exactnessof the correspondence. Specifically:

� In Theorem 6.2, the Ramsey property of F must be witnessed by G;

� In Theorem 6.5, the Ramsey property of F can be witnessed by anysubgroup of G.

By Proposition 3.5, if the Ramsey property in Theorem 6.5 is witnessed byH ≤ G then we can replace F by its shrink FH without changing the model,and then Theorem 6.2 tells us that H is extremely amenable. Note that thistells us nothing about the extreme amenability or otherwise of G.

Mostowski’s ordered model revisited

Recall that in Mostowski’s ordered model the set A of atoms was equippedwith a linear order <A making 〈A,<A〉 order-isomorphic to 〈Q, <〉, a groupG of all <A-automorphisms of 〈A,<A〉, and a normal filter F generated bythe fixators of finite subsets of A. We endow G with the product topologyinherited from AA in the obvious way, which in turn is given the producttopology from the discrete topology on A.

Lemma 6.6 G ∼= Aut(Q, <) has small open subgroups.

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Proof Define maps πa : G → A for a ∈ A by πa(f) = f(a). The discretetopology on A is generated by its singletons, so since inverse images preserveunions and intersections, the product topology O on G is generated by finiteintersections of sets of the form π←a ({b}) for a, b ∈ A.

If 1 ∈ π←a ({b}) then a = b and π←a ({b}) = fixG({a}). For a finite subsetF ⊆ A, ⋂

a∈F

fixG({a}) = fixG(F ) ≤ G

and so the fixators of finite subsets of A are a neighbourhood base for 1. Inparticular, if U ∈ O then fixG(F ) ⊆ U for some finite F ⊆ A. �

Theorem 6.7 G ∼= Aut(Q, <) is extremely amenable.

Proof By Lemma 6.6, G has small open subgroups, and certainly everyatom is symmetric and every subgroup of G lying in F stabilizes some setin M. By Theorems 4.1 and 4.7, M 2 AC and M � BPIT, so we may applyTheorem 6.5 to get that F is a Ramsey filter of subgroups of G and {1} 6∈ F .

Working through the proof of Theorem 6.5 with H = G we see that K = Gworks as a witness to the Ramsey property on F , so that by Theorem 6.2G is extremely amenable. �

A direct proof that Aut(Q, <) is extremely amenable can be found in [17].

Further research

Andreas Blass provides some open questions in [4] concerning Ramsey filters;for instance, if G is a group, H is a Ramsey subgroup of G and K is a Ramseysubgroup of H, is K a Ramsey subgroup of G?

Other avenues of thought include the following:

� Is there an interesting Galois theory surrounding Proposition 3.7?

� What properties do the models of ZFA+ (¬AC) +BPIT correspondingwith other groups have?

� Given a formula φ implied by the axiom of choice, is there a class ofgroups satisfying a certain condition that correspond with models ofZFA + (¬AC) + φ?

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[3] Andreas Blass. Prime ideals yield almost maximal ideals. FundamentaMathematicae, 127(1):57–66, 1987.

[4] Andreas Blass. Partitions and permutation groups. ContemporaryMathematics, 558:453–466, 2011.

[5] Andreas Blass. Models of ZFA corresponding exactlywith a particular class of groups. MathOverflow, 2013.http://mathoverflow.net/questions/119673 (version: 2013-01-23).

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[15] Philipp Kleppmann. Duals of filters and ideals. Unpublishedmanuscript, 2013.

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[18] Andrew M. Pitts. Nominal sets and their applications. Lectures atMidland Graduate School, University of Nottingham, 2011.

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