Extensions of the Axiom of Determinacy

Home , Almost, Axiom of determinacy, Bijection, Countable set, Disjoint union, Hereditarily finite set

Paul B. Larson

September 22, 2021 2 Contents

0.1 Introduction ...... 5 0.2 Notation ...... 7 0.3 Prerequisites ...... 9 0.4 Forms of Choice ...... 9 0.5 Partial orders ...... 9

I Preliminaries 11

1 The Axiom of Determinacy 13 1.1 Turing determinacy ...... 15

2 The Wadge hierarchy 19 2.1 Lipschitz determinacy ...... 19 2.2 Lipschitz degrees and Wadge degrees ...... 21 2.3 Pointclasses ...... 26 2.4 Universal sets ...... 28 2.5 The cardinal Θ ...... 33

3 Coding Lemmas 37

4 Properties of pointclasses 43 4.1 Separation and reduction ...... 43 4.2 The prewellordering property ...... 48 4.3 Prewellorderings and wellfounded relations ...... 52 4.4 Closure under wellordered unions ...... 54

5 Strong Partition Cardinals 57 5.1 The Martin Conditions ...... 59

6 Suslin sets and Uniformization 63 6.1 Uniformization ...... 68 6.2 The Kunen-Martin property ...... 69 6.3 A pointclass fact ...... 69 6.4 The Solovay sequence ...... 70

3 4 CONTENTS

II AD+ 73

7 <Θ-determinacy 75

8 Cone measures 81 8.1 Measurable cardinals from cone measures ...... 85 8.2 The degree order on sets of ordinals ...... 86 8.3 Pointed trees ...... 91 8.4 Coding ultrapowers ...... 92 8.5 Forcing with positive sets ...... 93

9 ∞-Borel sets 97 9.1 Local ∞-Borel codes ...... 103

10 Vopˇenka algebras 107 10.1 Codes for projections, and Uniformization ...... 111 10.2 Vopˇenka algebras and ∞-Borel sets ...... 116 10.3 ∞-Borel subsets of P(δ) ...... 120

11 Applications 125 11.1 Strong ∞-Borel codes ...... 125 11.2 Producing strong ∞-Borel codes ...... 127 11.3 ∞-Borel representations from Uniformization ...... 130 11.4 Closure of the Suslin cardinals ...... 132

12 Scales from Uniformization 137 12.1 Ordinal determinacy in the codes ...... 137 12.2 Boundedness for ∞-Borel relations ...... 141 12.3 Becker’s argument ...... 142

13 ADR 149 13.1 Weakly homogeneous trees ...... 150 13.2 Normal measures ...... 153

13.3 AD implies ADR if all sets are Suslin ...... 157 14 Questions 161 0.1. INTRODUCTION 5

This draft includes notes to myself, in footnotes. These should probably be ignored. Comments and corrections are more than welcome, although some parts are obviously very rough and these may not be worth commenting on, since they’ll probably we rewritten. Chapters 1, 2 and 3 should be essentially in their ﬁnal form, so please let me know when you see issues there. Ideally, I’ll remember to post frequent updates.

0.1 Introduction

The Axiom of Determinacy (AD) is the statement that all length-ω integer games of perfect information are determined. The beginning of Chapter 1 contains a more precise deﬁnition, but we expect the reader to be familiar with the classical theory of determinacy, as found in [7, 9], for instance. The axiom AD+ is a generalization, due to W. Hugh Woodin, of the Axiom of Determinacy. We deﬁne it as the conjunction of three statements.

0.1.1 Deﬁnition. The axiom AD+ is the conjunction of the following three statements.

1. DCR 2. Every subset of ωω is ∞-Borel.

3. (<Θ-Determinacy) For all λ < Θ, every A ⊆ ωω and every continuous function π : ωλ → ωω, π−1(A) is determined (as a subset of ωλ).

The ∞-Borel sets are deﬁned in Deﬁnition 9.0.1 for subsets of 2ω, and in Remark 9.0.8 for subsets of ωω. The notion of continuity in <Θ-Determinacy refers to the discrete topology on λ, not the interval topology. The restriction of <Θ-Determinacy to the case where π is the identity function on ωω is exactly AD. It is an open question whether AD implies any or all of the parts of AD+, and in fact whether AD plus any two parts of AD+ implies the third. It also + open whether ADR implies AD . The issue in this case is whether ADR implies <Θ-Determinacy, as DCR is easily seen to follow from ADR, and the second part + of AD follows from ADR (moreover, from AD + Uniformization) by Theorem 11.3.2. If M ⊆ N are models of AD with the same reals, and every set of reals in M is Suslin in N, then M |= AD+.1 In fact, this is the context which the axiom AD+ was designed to describe; its original name was “within scales”. Here is a brief2 outline of what we aim to prove is this book (all due to Woodin):

+ • In L(R), AD implies AD .

1Reference the relevant theorems. 2and incomplete 6 CONTENTS

• If AD+ holds then every inner model containing the reals satisﬁes AD+.

• If AD+ holds then the set of Suslin cardinals is closed below Θ. This follows from Theorem 11.4.1.3

• Assuming DCR and <Θ-Determinacy, ADR is equivalent to the assertion that the Solovay sequence has limit length. This is shown in Section ??.4

+ 2 • AD implies Σ1 reﬂection into the Suslin, co-Suslin (i.e., hom) sets

• AD+ implies that the ultrapower of V by the Turing measure is wellfounded.

ω • AD + DCR implies that for all ordinals χ, every subset of ω in L(P(χ)) is ∞-Borel.

Theorem 0.1.2 (Woodin). If <Θ-Determinacy and Uniformization hold, then every subset of ωω is Suslin.

Proof. By Theorem 11.3.2, the hypotheses give that every subset of ωω is ∞- Borel. The theorem then follows from Corollary 10.1.7 and Theorem 11.2.1.

The ﬁrst sentence of the following theorem is due to Woodin. The implication from (4) to (1) under DC combines results of Woodin and Becker (see Chapter 12)

Theorem 0.1.3. Each of the following statements implies the ones below it, and the ﬁrst two statements are equivalent. If DC holds, then all four statements are equivalent.

1. AD + “every subset of ωω is Suslin”

2. AD+ + “every subset of ωω is Suslin”

3. ADR 4. AD + Uniformization

Proof. To see that (3) implies (4), note that ADR implies Uniformization, via a game in which each player plays once. That (2) implies (1) follows from applying Ordinal Determinacy in the case λ = ω. As discussed in Chapter 6, Suslin sets can be uniformized, and Uniformization implies DCR. The implication from (1) to (2) then follows from Theorems 5.0.2 and 7.0.3, and Remark 6.0.9. Theorem 13.0.1 says that (1) implies (3). That (4) implies (1) under DC is Theorem 12.3.1 (the coﬁnality of Θ being uncountable is the relevant consequence of DC). 5

3Get closure below the supremum just from AD; comment on the reversal. 4 Does this need DCR? 5I should also introduce this theorem properly. 0.2. NOTATION 7

ω Theorem 0.1.4. Assume that AD + DCR holds. Let Γ be the set of A ⊆ ω for + which L(A, R) |= AD . Then

+ L(Γ, R) |= AD ,

ω 6 and, if Γ 6= P(ω ), then L(Γ, R) |= DC + ADR. Proof78 The following theorem follows from part (1) of Corollary 10.3.7.

ω Theorem 0.1.5. Assuming AD+DCR, a subset of ω is ∞-Borel if and only if there is a set of ordinals S such that A ∈ L(S, R). The following is Theorem 11.4.8.

+ Theorem 0.1.6. If AD holds, then ADR fails if and only if there is a set of ordinals T such that L(P(R)) = L(T, R). + Theorem 0.1.7. Assume that V = L(P(R)) and that AD holds. Then for all A ⊆ ωω, either V = L(A, R), or A# exists.9

Theorem 0.1.8. If AD and DCR hold, then for every ordinal χ, every subset of ωω inL (P(χ)) is ∞-Borel.

Part I of the book collects various facts (primarily from the Cabal seminar) which will be needed in Part II. A natural way to read the book would be to start at the beginning of Part II and refer back to Part I as needed. This book has a great deal of overlap with many sources, notably [6, 18, 19].

0.2 Notation

We reserve the symbol R for the real line, which is never used directly. However, we use traditional notation such as ADR, DCR, L(R) and so on, as these terms ω are equivalent to more relevant forms such as ADωω , DCωω and L(ω ). We informally use the word “real” to mean an element of the Baire space ωω. We use the following notational conventions.

• We write Ord for the class of ordinals.

• We write XY to mean the set of functions from Y to X, and, when γ is <γ S α an ordinal X to mean α∈γ X .

6 ω What is the proof? Let χB be the ∞-Borel threshhold. Then P(ω ) ∩ L(P(χB)) is the ∞-Borel sets. Look at a limit of Suslin cardinals; all sets are ¡ Borel, so also Suslin? 7 ω Let χB be the universally Baire threshold. Then P(ω ) ∩ L(P(χB)) = ∞-Borel. 8Also : let λ be a limit of Suslin cardinals. All sets are < λ-Borel, so also Suslin (what did mean by this?? 9 Another statement to consider : The axiom ADR implies that the sharp of each subset of ωω exists. 8 CONTENTS

• For an ordinal α and a set of ordinals X,[X]α denotes the collection of subsets of X of ordertype α, and [X]<α denotes the collection of subsets of X of ordertype less than α.

• Given a set X, we write ∃X and ∀X for existential and universal quantiﬁ- cation over X, respectively.

• A preorder on a set is a binary relation which is reﬂexive and transitive. A preorder is wellfounded if every nonempty subset of its domain has a minimal element.

• If ≤ is a wellfounded preorder on a set X, the canonical rank function rank≤ associated to ≤ assigns to each x ∈ X the least ordinal α such α > rank≤(y) for all y ∈ X such that y ≤ x and x 6≤ y.

• In any preorder, especially the natural order on the ordinals, the strict supremum of a set (if it exists) is the least element in the domain of the order which is strictly greater than every element of the set.

• We let ≤Gö be the Gödelorder, i.e., the order on pairs of ordinals defined as follows : (α, β) ≤Gö (δ, γ) if any of the following holds:

– max{α, β} < max{δ, γ}; – max{α, β} = max{δ, γ} and α < δ; – max{α, β} = max{δ, γ}, α = δ and β ≤ γ.

• Given ordinals α and β, we write ≺α, β for the ordinal ordertype of the set of predecessors of (α, β) in the order ≤G¨o.

• For a set X, we let TC(X) denote the transitive closure of X.

• Given a cardinal κ, H(κ) denotes the collection of sets of hereditary cardinality less than κ, i.e., the set of x for which | TC(x)| < κ.

We sometimes write HF for H(ℵ0). The members of HF are said to be hereditarily ﬁnite. A set x ⊆ HF is semi-recursive if x is Σ1-deﬁnable over HF, and recursive if x and HF \ x are both semi-recursive. Given A, B ⊆ HF, we say that A is Turing reducible to B, and write A ≤Tu B, to mean that there is a semi-recursive function f : HF → HF such that A = f −1[B]. We say that A is Turing equivalent to B, and write A ≡Tu B, to mean that A ≤Tu B and B ≤Tu A. We also sometimes write HC for H(ℵ1). The members of HC are said to be hereditarily countable. We say that a set x ∈ ωω HC-codes a set y ∈ HC if (ω, {(n, m) ∈ ω × ω : x(2n3m) = 0}) is isomorphic to (TC({y}), ∈). 0.3. PREREQUISITES 9

0.3 Prerequisites

We expect the reader to be familiar with Zermelo-Fraenkel set theory, G¨odel numbering of formulas, relative constructibility, ordinal deﬁnability, inner models of the form HODX , ultrapowers and forcing. We do not expect the reader to be familiar with everything in the books [7, 9, 16, 27, 30], but they make good references.

0.4 Forms of Choice

Our base theory in this book is Zermelo-Fraenkel set theory (ZF). Additional axioms will be stated as used. Although we will sometimes consider models of the Axiom of Choice (AC), our main interest is in models of the Axiom of Determinacy (AD), which contradicts AC. Weak forms of AC can (and do) hold in models of AD, however. Given a set X, the principle of Dependent Choice for X (DCX ) is the statement that whenever T ⊆ <ωX is a tree (i.e., a subset of <ωX closed under initial segments) with the property that every element of T has a proper extension in T , there exists an inﬁnite path through T . The principle of Countable Choice for X (CCX ) is the statement that for all countable sets Y , if hAy : y ∈ Y i is a sequence of nonempty subsets of X, then there exists a function f : Y → X such that f(y) is in Ay, for each y ∈ Y . The principle CCX follows immediately from DCX . A classical argument (due to Mycielski; see Remark 1.0.2) shows that CCR follows from AD. Whether or not DCR follows from AD is an open question. Note however that if DCR holds, then any inner model containing P(ω) satisﬁes DCR. + The following theorem is part of the result that, in L(R), AD implies AD . 10 Theorem 0.4.1 (Kechris). Assuming V =L(R), AD implies DCR.

The axiom of Dependent Choice (DC) asserts that DCX holds for every set X. Similarly, the axiom of Countable Choice (CC) asserts that CCX holds for every set X. 0.4.2 Remark. We make frequent use of the standard fact11 that if every set is deﬁnable from a member of X, then DCX implies DC.

0.5 Partial orders

We list here some classical partial orders which are used as forcing notions throughout the book.

• Col(κ, X), where κ is an inﬁnite cardinal and X is a set. Conditions are functions f : α → X, where α < κ. The order is extension.

10citation 11citation? 10 CONTENTS

• Col∗(κ, X), where κ is an inﬁnite cardinal and X is a set. Conditions are injective functions f : α → X, where α < κ. The order is extension.

• Given a set X consisting of infinite subsets of ω, the classical almost- disjoint coding forcing for X (due to Jensen and Solovay [8]) consists of pairs (a, B), where a is a finite subset of ω and B is a finite subset of X, with the order (a, B) ≤ (c, D) if – c is either the emptyset or a ∩ (max(c) + 1), – D ⊆ B and – (a \ c) ∩ r = ∅ for all r ∈ D.

This partial order is c.c.c. and adds a subset of ω having finite intersection with each member of X and infinite intersection with each element of P(ω) not contained mod-finite in the union of a finite subset of X. Part I

Preliminaries

Chapter 1

The Axiom of Determinacy

Given a set X, a set A ⊆ Xω is determined (as a subset of Xω) if there is a function π : X<ω → X such that one of the two following statements holds. ω 1. For every x ∈ X , if x(2n) = π(x2n) holds for all n ∈ ω, then x ∈ A. ω 2. For every x ∈ X , if x(2n + 1) = π(x(2n + 1)) hold for all n ∈ ω, then x 6∈ A. ω We let ADX denote the statement that every subset of X is determined (as a ω subset of X ). The Axiom of Determinacy (AD) is the statement ADω. 1.0.1 Remark. Let X and Y be sets. The following assertions can be easily veriﬁed.

• If there is an injection from X to Y , then ADY implies ADX .

• If X is wellorderable and there is a surjection from X to Y , then ADX implies ADY .

• If AD holds, then so does ADX for each countable set X. It is convenient and conventional to rephrase determinacy in terms of games. ω A set A ⊆ X corresponds to a game GA between players I and II, who alternate picking members of X, with I winning if and only if the induced member of Xω is in A.

I x(0) x(2) x(4) ... II x(1) x(3) ...

The game GA; I wins if and only if x is in A.

<ω A function π : X → X is then said to be a strategy in the game GA.A function π as in case (1) above is a winning strategy for player I; in case (2) it

13 14 CHAPTER 1. THE AXIOM OF DETERMINACY is a winning strategy for player II. If σ is a strategy and x is in Xω, then we write σ ∗ x for combined output of the two players when I plays according to σ and II plays x, that is, the unique y ∈ Xω such that

• y(2n + 1) = x(n) for all n ∈ ω;

• y(n) = σ(yn) for all even n ∈ ω. Similarly, we write x ∗ σ for the unique y ∈ Xω such that

• y(2n) = x(n) for all n ∈ ω;

• y(n) = σ(yn) for all odd n ∈ ω. The statement that a set A ⊆ Xω is determined can then be rephrased as assert- ing the existence of a strategy σ such that one of the two following statements holds.

• For every x ∈ Xω, σ ∗ x is in A. • For every x ∈ Xω, x ∗ σ is not in A. 1.0.2 Remark. We list four classical consequences of AD, the details of which are presented in Chapter 33 of [7] and Sections 27 and 28 of [9]. An abbreviated history of determinacy axioms can be found in [21].

1. Jan Mycielski observed that AD implies CCR. To see this, let Ai (i ∈ ω) be nonempty subsets of ωω, and consider the game where I plays i ∈ ω and then II must list the values of a member of Ai. 2. Morton Davis proved that AD implies that every uncountable subset of ωω contains a perfect set. To see this, consider a set A ⊆ ωω, and the game where players I and II collaborate to build an x ∈ ωω, with player II choosing individual digits as usual, but player I allowed to play ﬁnite sequences from ω, with I winning if the concatenation of these moves is in A. Player II has a winning strategy if and only if A is countable, and a winning strategy for I induces a perfect subset of A. It follows from ZF that if every uncountable subset of ωω contains a perfect set then there is ω no injection from ω1 into ω (such an injection would give a wellordering ω ω of ω in ordertype ω1). The nonexistence of an injection from ω1 into ω ℵ0 (which we will denote by writing ℵ1 6≤ 2 ) is equivalent to the assertion that for any inner model M satisfying Choice (equivalently, for all models of the form L[S], for S a subset of ω1), and any countable ordinal α, P(α)∩M is countable. Since ZF implies the existence of partition of P(ω)

into ℵ1 many sets, this shows that the statement ADω1 is inconsistent with ZF. 3. Stefan Banach proved that AD implies that every subset of ωω (similarly, every subset of 2ω) has the property of Baire, i.e., that for each A ⊆ ωω there exists an open U ⊆ ωω such that the symmetric diﬀerence A4U is 1.1. TURING DETERMINACY 15

meager. To see this, fix a bijection π : ω → ω<ω and, given A ⊆ ωω, let B be the set of x ∈ ωω for which the concatenation of the values π(x(i)) (i ∈ ω) is in A. If player I was a winning strategy in GB, then A is relatively comeager in some open set; if player II does, then A is meager. Wac law Sierpińskithat nonpricipal ultrafilters on ω (considered as subsets of 2ω) do not have the property of Baire. It follows that nonpricipal ultrafilters on ω don’t exist under AD, and from this that, under AD every nonprincipal ultrafilter on any set is countably complete.

4. Robert Solovay proved AD implies that the club ﬁlter on ω1 is an ultraﬁlter. This gives another proof that (under AD) there is no injection from ω1 into P(ω). We present a proof that AD implies the measurabilty of ω1 (not Solovay’s original proof) in Remark 1.1.7.

Given Γ ⊆ P(ωω), we will write Baire(Γ) for the assertion that every element of Γ has the property of Baire.

1.1 Turing determinacy

In this section we prove Martin’s theorem that under AD the cone measure on the Turing degrees is an ultrafilter. Although it may not be necessary for what follows, we prove the theorem for the finest equivalence relation for which Martin’s original proof applies. In practice we will often use the theorem with equivalence relations coarser than Turing equivalence, as discussed in this section. <ω ω Fixing an enumeration hσn : n < ωi of ω , we can associate to each x ∈ ω <ω a function (strategy) πx : ω → ω defined by the formula πx(σn) = x(n). Let even: ωω → ωω be the function defined by letting even(y)(n) = y(2n) for each n ∈ ω and let odd: ωω → ωω be the function defined by letting odd(y)(n) = y(2n + 1) for each n ∈ ω. Given x ∈ ωω, let

∗ ω ω ∗ • Ix : ω → ω be the function deﬁned by letting Ix(y) = (πy ∗ x) (i.e., the result of playing x for player II against the strategy πy for player I);

∗ ω ω ∗ • IIx : ω → ω be the function deﬁned by letting IIx(y) = (x ∗ πy) (i.e., the result of playing x for player I against the strategy πy for player II);

ω • Fx be the smallest set of functions on ω which is closed under composition ∗ and contains the identity function, even, odd and the functions If(x) and ∗ IIf(x) for each f ∈ Fx. ω Let ≤Ma be the reflexive binary relation on ω defined by setting y ≤Ma x if y = f(x) for some f ∈ Fx. Observe that if y ≤Ma x then Fy ⊆ Fx; it follows from this that ≤Ma is transitive. The functions even and odd can be used to show that ≤Ma is countably directed. We let ≡Ma denote the equivalence relation ≤Ma ∩ ≥Ma. The relations ≡Tu (Turing equivalence) and ≡Ma are instances of the following definition. 16 CHAPTER 1. THE AXIOM OF DETERMINACY

1.1.1 Deﬁnition. An ordered equivalence relation is a pair (E, ≤E) where E is an equivalence relation on a set X and ≤E is a partial order on X such that E =≤E ∩ ≥E. We say that (E, ≤E) is an ordered equivalence relation on X. We write [x]E for the E-equivalence class of x, and CE for the set of E-equivalence classes (or E-classes). ω ω Given an ordered equivalence relation (E, ≤E) on ω , and x ∈ ω , we deﬁne the upward ≤E-cone of x to be UE(x) = {[y]E : y ≥E x} and the downward ≤E-cone of x to be DE(x) = {[y]E : y ≤E x}. In each case we say that x is a base for the corresponding cone. Since we are typically interested in upward cones, we often write “cone” for “upward cone”. The ≤E-cone measure is

{A ⊆ {[x]E : x ∈ X} : ∃x ∈ X UE(x) ⊆ A}.

We will write [x]Ma, CMa,[x]Tu and CTu instead of the more cumbersome forms using ≡Ma and ≡Tu. We also write µMa and µTu for the corresponding cone measures, which are ultraﬁlters under AD. We call µTu the Turing measure. A Turing cone is an upward cone for ≤Tu.

Theorem 1.1.2 (Martin). Suppose that AD holds. Then µMa is a countably complete ultraﬁlter.

Proof. Since AD implies CCR, µMa is countably complete. It suﬃces then to show that each subset of CMa either contains or is disjoint from a cone. Fix ∗ ω A ⊆ CMa and consider the game GA∗ , where A is the set of x ∈ ω such that ω [x]Ma is in A. Let π be a winning strategy (for either player), and let x ∈ ω ω be such that π = πx. Let y ∈ ω be such that y ≥Ma x. If π is a winning strategy for player I, then [πx ∗ y]Ma is in A, and [y]Ma = [πx ∗ y]Ma (one ∗ direction of the equivalence uses the function Iy ; the other uses the function odd). If π is a winning strategy for player II, then [y ∗ πx]Ma is not in A, and [y]Ma = [y ∗ πx]Ma. We will typically be considering equivalence relations satisfying the following deﬁnition, with ≤Ma or ≤Tu in the role of F , especially those of the form of the last part in Example 1.1.4.

1.1.3 Deﬁnition. Given two ordered equivalence relations (E, ≤E) and (F, ≤F ) on the same underlying set X, say that (E, ≤E) is as thick as (F, ≤F ) if, for all x, y in X,

• if x ≤F y then x ≤E y;

• if x ≤E y, then for some z ∈ [y]E, x ≤F z. Every ordered equivalence relation is trivially as thick as itself. We leave it as an exercise to check that the “as thick as” relation is transitive. Given the ﬁrst condition in the deﬁnition of “as thick as”, the second condition is equivalent to saying that for all x ∈ X,

UE(x) = {[y]E : x ≤F y}. 1.1. TURING DETERMINACY 17

If CCR holds and (E, ≤E) and (F, ≤F ) are ordered equivalence relations, with F a subset of E and F countably directed, then E is also countably directed. 1.1.4 Example. The following are examples of ordered equivalence relations which are as thick as (≡Ma, ≤Ma).

ω • (≡Tu, ≤Tu), restricted to ω . • For any set S, the equivalence relation L(S, x) = L(S, y), under the order x ∈ L(S, y).

• For any set S, the equivalence relation HODS∪{x} = HODS∪{y}, under the order x ∈ HODS∪{y}.

1.1.5 Remark. Suppose that (E, ≤E) and (F, ≤F ) are ordered equivalence relations on a set X,(E, ≤E) is as thick as (F, ≤F ), and µF is an ultrafilter on the set of F -classes. Then µE is an ultrafilter on the set of E-classes. To see this, let A be a set of E-classes, and let B be the set of F -classes contained in a ω member of A. Since µF is an ultrafilter, we can fix an x ∈ ω such that UF (x) is either contained in or disjoint from B. Since UE(x) = {[y]E : y ≥F x}, UE(x) is either contained in or disjoint from A. Remark 1.1.5 and Theorem 1.1.2 give the following.

Corollary 1.1.6. If AD holds and (E, ≤E) is an ordered equivalence relation ω on ω which is as thick as ≡M, then µE is a countably complete ultraﬁlter on CE.

We let Turing Determinacy (TD) denote the statement that the ≤Tu-cone measure is an ultraﬁlter on the ≡Tu-classes. A recent result of Peng and Yu

[28] shows that TD implies CCR. It is open whether TD implies DCR, or whether either of TD and TD + DCR implies AD. Woodin has shown that TD + DCR + V =L(R) implies AD [3]. The following remark gives (from TD) Solovay’s theorem that, assuming AD, ω1 is measurable. It also shows that TD ℵ0 implies that ℵ1 6≤ 2 .

ω 1.1.7 Remark. Let (E, ≤E) be an ordered equivalence relation on ω such that each downward cone is countable, and suppose that µE is a countably ω complete ultrafilter on CE. For each α < ω1, let Xα be the set of x ∈ ω which HC-code α, in the sense of Section 0.2. For each σ ∈ CE, let ασ be sup{α < ω1 : ∃x ∈ Xα ∃y ∈ σ x ≤E y}. Let U be the set of A ⊆ ω1 for which {σ ∈ CE : ασ ∈ A} ∈ µE. Then U is a countably complete ultrafilter on ω1. The following example gives a curious consequence of the Turing measure being an ultrafilter. Similar arguments will appear in Chapters 8, 10 and 11.

ω 1.1.8 Example. Let S be a set of ordinals, let ≤S be the binary relation on ω given by the binary relation x ∈ L[S, y], let ≡S be the corresponding equivalence relation and let µS be the ≡S-cone measure. Assume that µS is an ultraﬁlter. ℵ0 ω It follows then by Remark 1.1.7 that ℵ1 6≤ 2 . For each x ∈ ω , there exist y 18 CHAPTER 1. THE AXIOM OF DETERMINACY and z in ωω such that y is generic for Sacks forcing over L[S, x] and z codes a wellordering of ω in ordertype ωL[S,x]. By standard facts about Sacks forcing S 1 [29], n∈ω{(2n, x(n)), (2n + 1, y(n))} is an immediate ≤S-successor of x. It ω follows then that the set of ≤S-successors contains a set in µS. Let f : ω → ω1 L[S,x] be the S-invariant function deﬁned by setting f(x) to be ω1 . Since Sacks ω ω forcing preserves ω1, we have that for every x ∈ ω , there exist y, z in ω both ≤S-above x, such that f(x) = f(y) and f(x) < f(z). Chapter 2

The Wadge hierarchy

2.1 Lipschitz determinacy

Given sets A, B ⊆ ωω, we say that A is Wadge reducible to B (and write ω ω −1 A ≤Wa B) if there is a continuous function f : ω → ω such that A = f [B]. ω ω ω A function f : ω → ω is Lipschitz if, for all x, y ∈ ω and n ∈ ω, if xn = yn, then f(x)n = f(y)n. We say that A is Lipschitz reducible to B (and write ω ω −1 A ≤L B) if there is a Lipschitz function f : ω → ω such that A = f [B]. Since Lipschitz functions are continuous, A ≤L B implies A ≤W B.

−1 ω −1 ω 2.1.1 Remark. Since A = f [B] implies that (ω \A) = f [ω \B], A ≤W B ω ω ω ω implies (ω \ A) ≤W (ω \ B) and A ≤L B implies (ω \ A) ≤L (ω \ B).

The relations ≤W and ≤L are easily seen to be preorders, that is, reﬂexive and transitive. We write =Wa and =Li respectively for the equivalence relations ≤W ∩ ≥Wa and ≤L ∩ ≥Li, whose equivalence classes are respectively ω called Wadge classes and Lipschitz classes. Given A ⊆ ω , we write [A]Wa for the Wadge class of A, and [A]Li for the Lipschitz class of A. We say that a Wadge class or Lipschitz class is selfdual if it contains a pair of complements (in which case it is closed under complements, by Remark 2.1.1), and nonselfdual otherwise. We write x

2.1.2 Deﬁnition. Let A, B be subsets of ωω. We write A B for the set of hxi : i ∈ ωi for which hx2i : i ∈ ωi ∈ A if and only if hx2i+1 : i ∈ ωi ∈/ B.

ω The game GA B can then be seen as a game where I plays x ∈ ω , II plays y ∈ ωω, and II wins if and only if (x ∈ A ⇔ y ∈ B). We call this the Lipschitz game for the pair (A, B).

2.1.3 Remark. The relation A ≤L B is equivalent to the assertion that player II has a winning strategy in the game GA B; it also follows from player I having a winning strategy in GB (ωω \A).

19 20 CHAPTER 2. THE WADGE HIERARCHY

This observation leads to the following fundamental fact.

Theorem 2.1.4 (Wadge). Let A and B be subsets of ωω. If player I has a ω winning strategy in GA B, then (ω \ B) ≤L A. If player II has a winning strategy in GA B, then A ≤L B. Wadge’s Theorem (along with Remarks 2.1.1 and 2.1.3) has the following consequences. Proposition 2.1.5 (which can be proved directly from Wadge’s Theorem) shows that the ordering on the Lipschitz degrees induced by ≤Li ω is almost linear, the exceptions being pairs of the form [A]Li,[ω \ A]Li, for nonselfdual classes [A]Li. The conclusion of the proposition is sometimes called the Semi-Linear Ordering Principle for Lipschitz maps.

Proposition 2.1.5. Let A and B be subsets of ωω such that A B and B A ω are both determined. If A 6≤L B and B 6≤L A, then A =Li (ω \ B). Proposition 2.1.6. Let A and B be subsets of ωω such that the sets B A and ω B (ω \ A) are both determined. If A

ω ω Proof. In the case of B (ω \A), one gets otherwise that A

Proposition 2.1.7. Let A and B be subsets of ωω such that A B and B A are both determined. If A

ω Proof. Since B 6≤W A, B 6≤L A, so by Proposition 2.1.5, either A =Li (ω \ B) or A ≤L B. The former is impossible, as then A ≤W B would imply that ω ω (ω \ B) ≤W (ω \ A) and thus B ≤W A, giving a contradiction.

We let Lipschitz Determinacy be the statement that A B is determined for all subsets A, B of ωω. It is not hard to see that Lipschitz Determinacy implies

CCR (see [20], for instance; the proof there combines the proof of Theorem 2.5.4 with1). By Proposition 2.1.7, the following theorem shows implies the corresponding version for

Proof. Suppose towards a contradiction that such a sequence hAi : i < ωi does 0 1 exist. Applying Proposition 2.1.6 and CCR we can ﬁx winning strategies fi , fi ω (i ∈ ω) for player I in the games GAi Ai+1 and GAi (ω \Ai+1), respectively. ω ω For each x ∈ 2, deﬁne yi(x) ∈ ω (i ∈ ω) by letting yi(x)(k) be

x(i) fi (hyi(x)(0), yi+1(x)(0), . . . , yi(x)(k − 1), yi+1(x)(k − 1)i).

1the strategy selection argument 2.2. LIPSCHITZ DEGREES AND WADGE DEGREES 21

Note that the values yi(x)(k) are defined recursively in k, for all x and i simultaneously. For each fixed x, each yi(x) is the set of values given by the strategy x(i) fi when playing against the moves for Player II listed by yi+1(x). It follows that for each i ∈ ω, if x(i) = 0 then yi(x) ∈ Ai ⇔ yi+1(x) ∈/ Ai+1, and if x(i) = 1 then yi(x) ∈ Ai ⇔ yi+1(x) ∈ Ai+1. ω Let Z be the set of x ∈ 2 for which y0(x) ∈ A0. It suffices to show that whenever x, x0 ∈ 2ω differ at exactly one point, x ∈ Z if and only if x0 6∈ Z, since no subset of 2ω with the property of Baire can have this property. First observe that if x, x0 ∈ ω2 and i ∈ ω are such that x(j) = x0(j) for all 0 0 j ≥ i, then yi(x) = yi(x ). If i0 is the unique i ∈ ω such that x(i) 6= x (i), then 0 0 yi0+1(x) = yi0+1(x ). Since x(i0) 6= x (i0), it follows that yi0 (x) ∈ Ai0 if and 0 0 only if yi0 (x ) 6∈ Ai0 . Since x(i) = x (i) for all i < i0, it follows that yi(x) ∈ Ai 0 if and only if yi(x ) 6∈ Ai for all such i. Recall that a preorder is wellfounded if every nonempty subset of its domain has a minimal element. The additional assumption of DCR derives the wellfoundedness of

ω Corollary 2.1.9 (Martin). If DCR+Lipschitz Determinacy+Baire(P(ω )) holds, then

Proof. If

2 As noted in Section 0.4, it is open whether AD implies DCR. Moreover, it is an open question whether DCR is needed for Corollary 2.1.9. The Lipschitz rank LR(A) (respectively, Wadge rank WR(A)) of a set A ⊆ ωω is recursively deﬁned to the be the least ordinal greater than the Lipschitz ω (Wadge) rank of every B ⊆ ω with B

WR(A) is deﬁned. In Part II of the book we will usually assume AD + DCR, so this notation will not be needed, since in this case W = P(ωω). By Proposition 2.1.5, for each ordinal α, the subsets of ωω of Lipschitz rank (Wadge rank) α (if there are any) consist either of a single Lipschitz (Wadge) class or a pair of Lipschitz (Wadge) classes corresponding to complements. Theorem 2.2.8 says more about the relationship between the two sets of classes.

2.2 Lipschitz degrees and Wadge degrees

Following [34], we give some more details on the structure of the Lipschitz and Wadge hierarchies, using Lipschitz Determinacy and the assumption that all sets of reals have the Baire property, but not DCR. We begin by noting that ∅ and 2 Woodin has shown that one gets a model of ADR from a counterexample? 22 CHAPTER 2. THE WADGE HIERARCHY

ωω are the only subsets of ωω of Lipschitz or Wadge rank 0, and that they are Wadge (and thus Lipschitz) inequivalent. We write s_x for the concatenation of s and x, where s is a finite sequence and x is either finite or infinite. Given A ⊆ ωω and i ∈ ω, we write A(i) for _ _ (i) {hii x : x ∈ A} and A(i) for {x : hii x ∈ A}. Note that (A )(i) = A, while (i) (A(i)) = {x ∈ A : x(0) = i}. (i) Proposition 2.2.1 shows (among other things) that [A ]Li is least Lipschitz class above [A]Li when A is selfdual. .

ω Proposition 2.2.1 (ZF+Lipschitz Determinacy). Let A ⊆ ω be such that [A]Li is selfdual, and let i be an element of ω. Then

(i) (i) 1. [A ]Li is selfdual, and [A ]Li is the ≤L-least Lipschitz class above [A]Li;

2. A(i)

_ (i) Proof. The function f(x) = hii x shows that A ≤L A and A(i) ≤L A. To (i) ω ω see that A

f(x) = hx(1), x(2), x(3),...i.

(i) ω ω It follows that no such f witnesses that A ≤L (ω \ A) or (ω \ A) ≤L A(i), which is enough since [A]Li is selfdual. (i) Using the assumption that A is selfdual, it is easy to that [A ]Li is selfdual. ω Finally, if C ⊆ ω is such that A

ω (i) 2.2.2 Remark. For any A ⊆ ω and i ∈ ω,[A]Wa = [A ]Wa. For the direction left open by Proposition 2.2.1, consider the function on ωω which removes the ﬁrst coordinate of its input.

2.2.3 Remark. An argument similar to the proof of Proposition 2.2.1 shows S (i) that [ i∈ω A ]Li is also the ≤L-least Lipschitz class above [A]Li (again assum- (j) ing Lipschitz Determinacy). It is also not hard to show directly that A =Li S (i) i∈ω A for each j ∈ ω.

2.2.4 Remark. Let π : ω × ω → ω be a bijection, and suppose that Ai (i ∈ ω) are subsets of ωω. Let C ⊆ ωω be the set of functions of the form

hπ(i, x(0)), x(1), x(2),...i for i ∈ ω and x ∈ Ai. Then Ai ≤L C for all i ∈ ω. If CCR holds, then for all ω D ⊆ ω , if Ai ≤L D for all i ∈ ω then C ≤L D. If {Ai : i ∈ ω} does not have a ≤L-maximal element, then Ai

If Lipschitz Determinacy holds, then [C]Li is nonselfdual exactly in the case where {Ai : i ∈ ω} has a ≤L-maximal element whose Lipschitz class is nonselfdual. To see this, note ﬁrst of all that [C]Li is clearly selfdual in the case where {Ai : i ∈ ω} has a ≤L-maximal element whose Lipschitz class is selfdual, as [C]Li is equal to this class (similarly, if {Ai : i ∈ ω} has a ≤W-maximal element, then [C]Wa is equal to this class). In the remaining case, {[Ai]Li : i ∈ ω} has the same supremum as

ω {[Ai]Li : i ∈ ω} ∪ {[ω \ Ai]Li : i ∈ ω}.

ω Running the construction of C above with the set {Ai : i ∈ ω}∪{ω \Ai : i ∈ ω} clearly gives a selfdual Lipschitz class (equal to [C]Li). This argument gives the following facts, for any A ⊆ ωω.

ω • If [A]Li is nonselfdual, then the pair [A]Li, [ω \ A]Li has a ≤L-least upper bound, and this upper bound is selfdual.

• If [A]Li is the ≤L-supremum of a countable set of Lipschitz classes strictly below it, and Lipschitz Determinacy holds, then [A]Li is selfdual (Proposi- tion 2.2.5 below gives the converse, for non-successor classes).

From Proposition 2.2.1 and Remarks 2.2.2 and 2.2.4 it follows that for all ω A ⊆ ω such that [A]Li is selfdual, the ﬁrst ω1 many Lipschitz classes above [A]Li are all selfdual and contained in [A]Wa. Proposition 2.2.5 shows that the nonselfdual Lipschitz classes are exactly those whose Lipschitz rank is either 0 or an ordinal of uncountable coﬁnality. First we prove the following fact about sets of the form A(i), which is dual to Proposition 2.2.1.

ω Proposition 2.2.5 (ZF+Lipschitz Determinacy). Let A ⊆ ω be such that [A]Li is selfdual, and [A]Li is not the ≤L-least Lipschitz class above any other class. Then [A]Li is the ≤L-supremum of a countable set of Lipschitz classes strictly below it.

Proof. Let π : ω × ω → ω be a bijection. For each i ∈ ω, let Bi be the set of sequences of the form hπ(n, x(0)), x(1), x(2),...i such that either n is even and hi, x(0), x(1), x(2),...i is in A, or n is odd and hi, x(0), x(1), x(2),...i is not in A. Then by Remark 2.2.4 [Bi]Li is the least ω upper bound of the pair {[A(i)]Li, [(ω \ A)(i)]Li} and Bi is selfdual. By Propo- sition 2.2.1, and the assumption that [A]Li is not the ≤L-least Lipschitz class above any other class, we have that Bi

Let g : <ωω → ω be such that for all n ∈ ω \ 2,

g(hi0, . . . , ini) = hπ(i0, i0), π(0, i1), i2, . . . , ini. Let g∗ : ωω → ωω be the Lipschitz function induced by g. Let us see that g∗ ω ∗ witnesses that A ≤L D. Fix x ∈ ω , and let g (x) have the form

hπ(i0, y(0)), y(1), y(2),...i

ω for some i0 ∈ ω and some y ∈ ω . Then i0 = x(0) = y(0), y(1) = π(0, x(1)) and y(i) = x(i) for all i ∈ ω \ 2. We want to see that x ∈ A if and only if g∗(x) ∈ D. ∗ Now, g (x) is in D if and only if y ∈ Ci0 , which in turn happens if and only if hπ(0, x(1)), x(2), x(3),...i is in Bi0 , which happens if and only if x is in A. The following proposition completes the analysis of which Wadge classes are selfdual classes, as well as the relationship between the Lipschitz classes and the Wadge classes. Theorem 2.2.6 (Steel, Van Wesep). Suppose that

Lipschitz Determinacy + Baire(P(ωω))

ω holds. For all A ⊆ ω , if [A]Wa is selfdual, then so is [A]Li.

Proof. Suppose towards a contradiction that [A]Wa is selfdual but [A]Li is not. By Lipschitz Determinacy, if [A]Li is not selfdual, then player I wins GA (ωω \A). Let σ be a strategy witnessing this. Consider the following game between players I and II (which is called the Wadge game). For each i ∈ ω, player I plays a value x(i), and player II either passes or plays a value y(j), for j the least k ∈ i + 1 for which a value for y(k) has not yet been chosen. If at the end of the game there is a j ∈ ω for which y(j) has not been chosen, then II loses. Otherwise II wins if and only ω if x ∈ A ⇔ y 6∈ A. The statement that A ≤W (ω \ A) is equivalent to the existence of a winning strategy for II. Let τ be such a strategy. n For each positive n ∈ ω and each sequencec ¯ = hcm : m < ni ∈ 3 , there is a unique (possibly partial) function sc¯ on n×ω satisfying the following conditions.

• For each m < n, the set of i for which (m, i) ∈ dom(sc¯) is an ordinal αm ∈ ω + 1. We let tm be the function with domain αm such that tm(i) = sc¯(m, i) for all i < αm.

• The largest m < n for which αm > 0 is the largest m < n for which cm = 2, and for this m, αm = 1 and sc¯(m, 0) = σ(hi).

• For each m < n − 1 such that cm = 2, αm = αm+1 + 1, and for each i < αm, sc¯(m, i) is the response given by σ when player II plays tmi.

• For each m < n − 1 such that cm = 0, αm = αm+1, and tm = tm+1.

• For each m < n − 1 such that cm = 1, tm is the longest sequence of nonpassing moves made by τ in response to tm+1. 2.2. LIPSCHITZ DEGREES AND WADGE DEGREES 25

Choose integers ik (k ∈ ω) so that i0 = 0, ik+1 > ik + 1 for all k ∈ ω and, n for each sequencec ¯ = hcm : m < ik+1i in 3 , if cm = 2 for all

m ∈ ik+1 \{ip : p ≤ k}, then the corresponding value αj is at least k for each j ≤ ik. ω x For each x ∈ 2 and each m ∈ ω, let cm be x(k) if m = ik for some x k ∈ ω, and let cm be 2 otherwise. For each such x there is a unique sequence x ω ω hym : m ∈ ωi ∈ (ω ) such that for each m ∈ ω,

x x • if cm = 2, then ym is the sequence produced by σ when player II plays x ym+1;

x x x • if cm = 0, then ym = ym+1;

x x • if cm = 1, then ym is the sequence produced by τ when player I plays x ym+1. Then as in the proof of Theorem 2.1.8, if x, x0 ∈ 2ω and k ∈ ω are such that 0 x x0 x(p) = x (p) for all p > k, then ym = ym for all m > ik. So again, if we let Z x be the set of x for which y0 ∈ A, Z cannot have the property of Baire, since whenever x and x0 disagree at exactly one point, exactly one of them will be in Z.

Theorem 2.2.6 has the following corollary.

Corollary 2.2.7. Suppose that Lipschitz Determinacy+Baire(P(ωω)) holds. Let ω A be a subset of ω . If [A]Li is nonselfdual, then [A]Li = [A]Wa.

Proof. Supposing otherwise, ﬁx B ∈ [A]Wa \ [A]Li. By Wadge’s Theorem (The- ω ω orem 2.1.4), either (ω \ A) ≤L B or (ω \ B) ≤L A. Each of these implies that [A]Wa is selfdual, giving a condradiction by Theorem 2.2.6.

Summarizing, we have the following.

Theorem 2.2.8. Suppose that Lipschitz Determinacy + Baire(P(ωω)) holds.

1. The minimal Wadge classes consist of the singletons {∅} and {ωω}.

2. Each selfdual Wadge class contains ℵ1 many selfdual Lipschitz classes, and these are ordered in ordertype ω1 by ≤L. 3. Each nonselfdual Wage class is equivalent to the corresponding Lipschitz class, and has a selfdual class as an immediate successor.

4. A Wadge class which is neither minimal nor a successor is selfdual if and only if it is the supremum of a countable set of classes strictly below it.

As we show in Proposition 2.5.4, a straightforward diagonal argument (from [31]) shows that there is no largest Wadge degree, if Lipschitz Determinacy holds. 26 CHAPTER 2. THE WADGE HIERARCHY

2.2.9 Remark. The material in this section does not give a deﬁnition for the least Wadge class above a given selfdual class, or show (without assuming DCR) that such a class exists. Adding DCR to the hypotheses of Theorem 2.2.8, Proposition 2.2.5 gives that each seldfual Wadge class [A]Wa has a pair of nonselfdual classes as immediate successors, the Wadge classes corresponding to the

2.3 Pointclasses

The notion of Wadge reducibility naturally generalizes to other topological spaces. In general, one could say that for any pair of topological spaces X and Y , and any sets A ⊆ X and B ⊆ Y , that A ≤W B if there is a continuous function f : X → Y such that A = f −1[B]. We will be concerned only with topological spaces of the form X1 × · · · × Xn for some positive n ∈ ω, where at ω ω least one Xi is ω , and each Xi is either ω or ω. Let X be the collection of such spaces; these spaces are all homeomorphic with ωω. The notions of Wadge reducibilty and Wadge rank carry over naturally to subsets of spaces in X . We use the word pointclass to denote a collection of subsets of spaces in X .A boldface pointclass is a pointclass closed under continuous preimages, i.e., the union of an initial segment of the Wadge hierarchy. ω Given a pointclass Γ, we write o(Γ) for the ordertype of (Γ ∩ P(ω ), ≤W), identifying this with the corresponding ordinal in the case that the restriction ω of ≤W to Γ ∩ P(ω ) is wellfounded. In this case o(Γ) is the supremum of the Wadge ranks of the members of Γ. <ω ω Given a ﬁnite sequence s ∈ ω, we let [s] = {x ∈ ω | x|s| = s}.A basic open interval of a space X1 × · · · Xn in X is a product of the form a1 × · · · × an, where, for each i ∈ {1, . . . , n},

<ω ω • ai is of the form [s], for some s ∈ ω, if Xi = ω ;

• ai is either ∅, ω or {m} for some m ∈ ω if Xi = ω.

We say that continuous functions f : X1 × · · · × Xn → Y1 × · · · × Ym between spaces in X is recursive if the set of pairs of basic open intervals U ⊆ X, V ⊆ Y for which f[U] ⊆ V is recursive (i.e., ∆1-deﬁnable over HF). A lightface pointclass is a pointclass closed under preimages of continuous functions which are recursive. Under these deﬁnitions boldface pointclasses are also lightface. Given a set A ⊆ X (for some X ∈ X ), we write Aˇ for X \ A. Given a pointclass Γ, we write Γˇ for {Aˇ : A ∈ Γ}. We say that a pointclass Γ is selfdual if Γ = Γ;ˇ otherwise it is nonselfdual. Observe that Γ is a boldface pointclass if and only if Γˇ is.

1 2.3.1 Example. The collection of analytic subsets of spaces in X (i.e., Σ∼1) is 1 1 a nonselfdual boldface pointclass, as are the projective pointclasses Σ∼n and Π∼n 1 for all n ∈ ω. The pointclasses and ∆∼n are boldface and selfdual. 2.3. POINTCLASSES 27

The following is an immediate, but useful, corollary of Theorem 2.1.4 (and Remark 2.1.1). Corollary 2.3.2. Assume that Lipschitz Determinacy+Baire(P(ωω)) holds. Let Γ be a nonselfdual boldface pointclass, and suppose that Λ is a boldface pointclass properly containing Γ. Then Γˇ ⊆ Λ. A member A of a pointclass Γ is complete for Γ (or Γ-complete) if every member of Γ is a continuous preimage of (e.g., Wadge below) A. Proposition 2.3.3. Suppose that Lipschitz Determinacy + Baire(P(ωω)) holds. If Γ is a boldface pointclass, then every member of Γ \ Γˇ is Γ-complete. Proof. Fix a set A ∈ P(ωω) ∩ (Γ \ Γ)ˇ .

Then [A]L is nonselfdual, so [A]L = [A]W , by Corollary 2.2.7. Suppose towards a contradiction that there is a B ∈ Γ such that B 6≤W A. Then, by Theorem ω 2.1.4, A ≤W (ω \ B), contradicting our assumption that A 6∈ Γ.ˇ We write ωω × X for the set of X ∈ X of the form

ω ω × X1 × · · · × Xn,

ω ω for some X1,...,Xn (all equal to either ω or ω ). Given X ∈ ω × X and ω A ⊆ X, we write ∃ω A for the set

ω {(x1, . . . , xn): ∃x0 ∈ ω (x0, . . . , xn) ∈ A},

ω and ∀ω A for the set

ω {(x1, . . . , xn): ∀x0 ∈ ω (x0, . . . , xn) ∈ A}.

ω Given a pointclass Γ, we write ∃ω Γ for

ω {∃ω A : A ∈ Γ ∧ ∃X ∈ ωω × X A ⊆ X}

ω and ∀ω Γ for ω {∀ω A : A ∈ Γ ∧ ∃X ∈ ωω × X A ⊆ X}. Similarly, for any ordinal δ, we write S Γ for the collection of sets of the form S δ α<δ Aα, where each Aα is in Γ, and the Aα’s are all subsets of the same element of X . ω ω ω ω A pointclass Γ is ∃ω -closed if ∃ω Γ ⊆ Γ, and ∀ω -closed if ∀ω Γ ⊆ Γ.3

ω Proposition 2.3.4 (ZF + CC ). If Γ is an ∃ω -closed boldface pointclass with R ω a complete set, then Γ is closed under countable unions. If ∆ is a selfdual ∃ω - closed boldface pointclass and o(∆) has uncountable coﬁnality, then ∆ is closed under countable unions and countable intersections. 3We need to avoid trivialities like the Wadge class of the emptyset. Ideally, our usage is consistent. 28 CHAPTER 2. THE WADGE HIERARCHY

ω ω Proof. We prove the ﬁrst part ﬁrst. Let A ⊆ ω be Γ-complete, and let Bi ⊆ ω

(i ∈ ω) be elements of Γ. Applying CCR, for each i, let fi be a continuous −1 ω ω function such that Bi = fi [A]. Let C be {(x, y) ∈ ω × ω : y ∈ Bx(0)} and −1 define g : ω × ω → ω by setting g(x, y) to be fx(0)(y). Then C = g [A], so S ω C ∈ Γ. Since i∈ω Bi = {y : ∃x ∈ ω (x, y) ∈ C}, we are done. For the second part, the assumptions imply that every countable subset of ωω ∆ is contained in a boldface pointclass Γ0 having a complete set. Then ∃ Γ0 is contained in ∆ and satisfies the assumptions of the first part. 2.3.5 Remark. The proof of Proposition 2.3.4 shows the following, without the assumption of CCR ω • If Γ is an ∃ω -closed boldface pointclass with a complete set, then Γ is closed under unions.

ω • If ∆ is a selfdual ∃ω -closed boldface pointclass and o(∆) has uncountable coﬁnality, then ∆ is closed under unions and intersections. 2.3.6 Remark. Let Γ be a boldface pointclass. If Γ is closed under unions, ω ω then so are ∃ω Γ and ∀ω Γ (the latter case can be veriﬁed by coding a pair of reals with a single real). Similarly, if Γ is closed under countable unions, then ω ω so are ∃ω Γ and ∀ω Γ.

2.4 Universal sets

Given an integer n ∈ ω \ 2, a set A ⊆ (ωω)n is universal for a pointclass Γ (or Γ-universal) if A ∈ Γ, and for every B ⊆ (ωω)n−1 in Γ there is an x ∈ ωω such that B = {(y1, . . . , yn−1) | (x, y1, . . . , yn−1) ∈ A}

(we call this set Ax). If Γ is closed under Wadge-equivalence, and n is in ω \ 2, then there exists a universal subset of (ωω)n in Γ if and only if there is a universal subset of (ωω)2 in Γ.

<ω 2.4.1 Example. Fixing an enumeration hσn : n ∈ ωi of ω , the set of (x, y) ω 2 −1 in (ω ) such that y ∈ [σn] for some n ∈ x [{0}] is a universal open set. It follows that there exist a universal closed set C ⊆ (ωω)3, and universal analytic and coanalytic subsets of (ωω)2. The usual diagonal argument shows that selfdual lightface pointclasses do not have universal sets. Proposition 2.4.2. If ∆ is a selfdual lightface pointclass, then ∆ does not have a universal set. Proof. Given a set A ⊆ (ωω)2 in ∆, let B be the set of x ∈ ωω for which ω (x, x) 6∈ A. Then for any x ∈ ω , x ∈ B if and only if x 6∈ Ax. Nonselfdual boldface pointclasses have universal sets, under suitable determinacy assumptions. 2.4. UNIVERSAL SETS 29

Theorem 2.4.3. Suppose that Lipschitz Determinacy+Baire(P(ωω)) holds. If Γ is a boldface pointclass, then Γ has a universal set if and only if it is nonselfdual. Proof. The selfdual case follows from Proposition 2.4.2. For the other case, suppose that Γ is nonselfdual, and ﬁx a set

A ∈ P(ωω) ∩ (Γ \ Γ)ˇ .

Then [A]L is nonselfdual, so [A]L = [A]W , by Corollary 2.2.7, and A is Γ- complete by Proposition 2.3.3. Fix a recursive bijection ρ: ω<ω → ω. For each ω ω ω x ∈ ω , deﬁne fx : ω → ω by setting fx(y)(n) to be x(ρ(y(n + 1))). ω ω Then each fx is Lipschitz, and each Lipschitz function from ω to ω is ω ω 2 equal to fx for some x ∈ ω . Now let U be the set of (x, y) ∈ (ω ) such that fx(y) ∈ A. Then U =Wa A, and U is Γ-universal. 2.4.4 Remark. Suppose that Γ is a boldface pointclass and U ⊆ (ωω)n is Γ-universal, for some n ∈ ω \ 3. Then

ω n−2 ω {(z1, . . . , zn−2) ∈ (ω ) : ∃y ∈ ω (x, y, z1, . . . , zn−2) ∈ U}

ω is universal for ∃ω Γ and

ω n−2 ω {(z1, . . . , zn−2) ∈ (ω ) : ∀y ∈ ω (x, y, z1, . . . , zn−2) ∈ U}

ω is universal for ∀ω Γ. It follows from Theorem 2.4.3 that if Γ is nonselfdual, ω ω then so are ∃ω Γ and ∀ω Γ (although either or both of these pointclasses could be equal to Γ). In order to prove the Moschovakis Coding Lemma (in Chapter 3) we will need sequences of universal sets in diﬀerent dimensions which are suitably coherent.

2.4.5 Deﬁnition. Let U¯ = hUn : n ∈ ω \{0}i be such that each Un is a subset of (ωω)n+1.

• The sequence U¯ has the s-m-n property if for each pair of positive integers ω n+1 ω n < m, there exists a continuous sm,n :(ω ) → ω such that, for all ω x, y1, . . . , ym ∈ ω , (x, y1, . . . , ym) ∈ Um

if and only if (sm,n(x, y1, . . . , yn), yn+1, . . . , ym) ∈ Um−n.

• The sequence U¯ has the recursion property (with respect to a pointclass Γ) if for each n ∈ ω \{0} and each

A ∈ P((ωω)n+1) ∩ Γ,

ω ω there exists an x ∈ ω such that for all y1, . . . , yn ∈ ω ,

(x, y1, . . . , yn) ∈ Un

if and only if (x, y1, . . . , yn) ∈ A. 30 CHAPTER 2. THE WADGE HIERARCHY

We will omit the phrase “with respect to Γ” when talking about sequences U¯ with the recursion property, since Γ is recoverable from U¯. Similarly, when we say that hUn : n ∈ ω \{0}i is a sequence of sets with the s-m-n property, we ω n+1 will mean that each Un is a subset of (ω ) . 2.4.6 Remark. The recursion property defined above of universal set U is a fixed point property. By the universality of U there is, for each set A in the ω ω definition, and each x ∈ ω , a z ∈ ω such that Ax = Un,z. The point of the definition is that there is some x such that Ax = Un,x. Theorems 2.4.7 and 2.4.8 below show that if Γ is a boldface pointclass with a universal set, then there exists a sequence of Γ-universal set with the s-m-n and recursion properties. The statement of Theorem 2.4.7, and its proof, are taken from [6].4

Theorem 2.4.7. If Γ is a boldface pointclass with a universal set then there exists a sequence of Γ-universal sets with the s-m-n property.

ω ω n Proof. Fix homeomorphisms πn : ω → (ω ) for each n ∈ ω \ 2, and let ω ω πm,n : ω → ω (n < m ∈ ω \ 2) be such that

πm(x) = (πm,0(x), . . . , πm,m−1(x))

ω −1 ω for all x ∈ ω , so that πm,n(πm (x0, . . . , xm−1)) = xn for all x0, . . . , xm−1 ∈ ω . ω 2 Let U ⊆ (ω ) be a universal set for Γ. For each n ∈ ω \{0}, let Un be the set ω n+1 of (x, y1, . . . , yn) ∈ (ω ) such that

−1 (π2,0(x), πn+1(π2,1(x), y1, . . . , yn)) ∈ U.

ω n We check first that each Un is Γ-universal. Fix n ∈ ω \{0} and A ⊆ (ω ) ω ω in Γ. We want to find an x ∈ ω such that Un,x = A. Fixing any z ∈ ω , we −1 have that πn+1[{(z, y1, . . . , yn):(y1, . . . , yn) ∈ A}] is in Γ. Since U is universal, ω −1 there is a w ∈ ω such that Uw = πn+1[{(z, y1, . . . , yn):(y1, . . . , yn) ∈ A}]. Let −1 ω x = π2 (w, z). Then for all y1, . . . , yn ∈ ω ,(x, y1, . . . , yn) is in Un if and only −1 if (w, πn+1(z, y1, . . . , yn)) is in U, which holds if and only if (y1, . . . , yn) is in A. To check that the s-m-n property holds, fix m > n in ω. Let W be the set of w ∈ ωω for which

−1 (u(w), πm+1(v(w), r1(w), . . . , rn(w), t1(w), . . . , tm−n(w))) ∈ U, where

• u(w) = π2,0(πn+1,0(πm−n+1,0(w)));

• v(w) = π2,1(πn+1,0(πm−n+1,0(w)));

• ri(w) = πn+1,i(πm−n+1,0(w)) for i ∈ {1, . . . , n};

4But to whom are they due? 2.4. UNIVERSAL SETS 31

• tj(w) = πm−n+1,j(w) for j ∈ {1, . . . , m − n}.

ω Then W ≤W U, and, as U is universal for Γ, there exists a z ∈ ω such that Uz = W . ω n+1 ω Deﬁne sm,n :(ω ) → ω by setting

−1 −1 sm,n(x, y1, . . . , yn) = π2 (z, πn+1(x, y1, . . . , yn))

ω ω for all x, y1, . . . , yn ∈ ω . Now ﬁx x, y1, . . . , ym ∈ ω . Then (x, y1, . . . , ym) ∈ Um if and only if −1 (π2,0(x), πm+1(π2,1(x), y1, . . . , ym)) ∈ U, and (sm,n(x, y1, . . . , yn), yn+1, . . . , ym) ∈ Um−n if and only if

−1 (π2,0(sm,n(x, y1, . . . , yn)), πm−n+1(π2,1(sm,n(x, y1, . . . , yn)), yn+1, . . . , ym)) ∈ U.

Now, π2,0(sm,n(x, y1, . . . , yn)) = z and

−1 π2,1(sm,n(x, y1, . . . , yn)) = πn+1(x, y1, . . . , yn).

Since Uz = W , we have that

−1 −1 (z, πm−n+1(πn+1(x, y1, . . . , yn), yn+1, . . . , ym)) ∈ U

−1 −1 if and only if πm−n+1(πn+1(x, y1, . . . , yn), yn+1, . . . , ym) ∈ W , which, letting w −1 −1 be πm−n+1(πn+1(x, y1, . . . , yn), yn+1, . . . , ym), holds if and only if

−1 (u(w), πm+1(v(w), r1(w), . . . , rn(w), t1(w), . . . , tm−n(w))) ∈ U, which holds if and only if

−1 (π2,0(x), πm+1(π2,1(x), y1, . . . , ym)) ∈ U, since

• u(w) = π2,0(x),

• v(w) = π2,1(x),

• ri(w) = yi for i ∈ {1 . . . , n} and

• tj(w) = yn+i for j ∈ {1, . . . , m − n}.

Finally, we observe that a sequence of Γ-universal sets with the s-m-n property has the U¯ has the recursion property.

Theorem 2.4.8. If Γ is a pointclass and U¯ = hUn : n ∈ ω \{0}i is a sequence of Γ-universal sets with the s-m-n property, then U¯ has the recursion property. 32 CHAPTER 2. THE WADGE HIERARCHY

Proof. Fix n ∈ ω \{0} and A ∈ P((ωω)n+1) ∩ Γ. Applying the assump- ω tion that Un+1 is Γ-universal, let y ∈ ω be such that for all w, z1, . . . , zn ω in ω ,(y, w, z1, . . . , zn) ∈ Un+1 if and only if (s(w, w), z1, . . . , zn) ∈ A, where ω 2 ω s:(ω ) → ω witnesses the s-m-n property for U¯ in the role of sn+1,1. Then ω for all w, z1, . . . , zn ∈ ω ,

(s(y, w), z1, . . . , zn) ∈ Un if and only if (y, w, z1, . . . , zn) ∈ Un+1 if and only if (s(w, w), z1, . . . , zn) ∈ A. Then x = s(y, y) is as desired. Theorem 2.4.9 below is known as the Recursion Theorem. A function ω ω 1 1 ω ω f : ω → ω is in∼ Σ1 if it is∼ Σ1 as a subset of ω × ω (equivalently, if the 1 f-preimage of each open set is in Σ∼1. ω Theorem 2.4.9 (Kleene). Suppose that Γ is a ∃ω -closed pointclass and that U¯ = hUn : n ∈ ω \{0}i is a sequence of Γ-universal sets with the recursion ω ω 1 ω property. If f : ω → ω is in ∼Σ1, then there is an x ∈ ω such that U2,x = U2,f(x).

ω Proof. Since Γ is ∃ω -closed, the set

ω 3 A = {(x, y, z) ∈ (ω ) :(f(x), y, z) ∈ U2}

ω is in Γ. Then for each x ∈ ω , Ax = U2,f(x). Applying the recursion property, ω we get an x ∈ ω such that U2,x = Ax. Using a recursive bijection π : ω → ω × ω, we can associate to each y ∈ ω ω ω an ω-sequence h(y)n : n ∈ ωi of elements of ω by setting (y)n(m) to be y(π−1(n, m)). Loosely following 7D.7 (page 430) of [27], we say that a set B ⊆ X 1 ω (for some X in X ) is pos-Σ∼1(A) (for some A ⊆ ω ) if ω B = {x ∈ X : ∃y ∈ ω ((∀n ∈ ω (y)n ∈ A) ∧ (x, y) ∈ S)},

1 ω 1 for some Σ1 set S ⊆ X × ω . These are essentially the sets which are Σ1 ∼ ω ∼ using a predicate for A positively. If A1,...,An are subsets of ω , we write 1 1 pos-Σ∼1(A1,...,An) for pos-Σ∼1(A), where A is the disjoint union of A1,...,An, i.e., the set of reals of the form hπ(i, x(0)), x(1), x(2),...i for x ∈ Ai and i ∈ {1, . . . , n} (we call this set A1 ⊕ · · · ⊕ An).

ω 1 2.4.10 Remark. For any X ∈ X and A ⊆ ω , A is in pos-Σ∼1(A), and 1 ωω pos-Σ∼1(A) is a boldface pointclass closed under ∃ , ∧ and ∨. The existence of 1 universal sets for Σ∼1 (as shown in Example 2.4.1) implies that each pointclass 1 of the form pos-Σ∼1(A) has a universal set. 2.5. THE CARDINAL Θ 33

ω 1 Given A ⊆ ω ,Σ∼1(A) the collection of subsets of spaces in X which can be 1 deﬁned by a Σ∼1 formula in a predicate for A. This is the same as the pointclass 1 ω 1 pos-Σ∼1(A, ω \ A). Similarly, we write Σ∼1(A1,...,An) for

1 ω ω pos-Σ∼1(A1,...,An, ω \ A1, . . . , ω \ An).

1 1 Given a positive n ∈ ω,Π∼n(A) is the set of complements of sets in Σ∼1(A), and 1 1 ∼Σn+1(A) is the set of continuous images of sets in Π∼n(A). We say that a set B is projective in A if it is in S Σ1 (A). We say that a pointclass ∆ is n∈ω ∼n+1 projectively closed if for each A ∈ ∆ ∩ P(ωω), every set projective in A is in ∆.

¯ 1 2.4.11 Remark. Let U = hUn : n < ωi be a sequence of universal sets for∼ Σ1 ω with the s-m-n property. For any A ⊆ ω , let Un(A) be the set

ω n+1 ω {x ∈ (ω ) : ∃y ∈ ω ((∀n ∈ ω (y)n ∈ A) ∧ (x, y) ∈ Un+1)}.

1 Then each Un(A) is universal for pos-Σ∼1(A). Furthermore, if the functions sm,n (n < m < ω) witness the s-m-n property for U¯, then for all n < m < ω, function sm+1,n witnesses the s-m-n property for U¯(A) = hUn(A): n < ωi for m and n. ω ω Similarly, given a function f : ω → ω and n ∈ ω \{0}, if Un,x = Un,f(x), then ω Un,x(A) = Un(A)x = Un(A)f(x) = Un,f(x)(A) for all A ⊆ ω .

ω For U¯ as Remark 2.4.11, n ∈ ω and A1,...,Am ⊆ ω , we write Un(A1,...,Am) for Un(A1 ⊕ · · · ⊕ Am).

2.5 The cardinal Θ

This section presents results of Solovay on the height of the Wadge hierarchy, taken from [31].

2.5.1 Deﬁnition. The ordinal Θ is deﬁned to be the least ordinal which is not a surjective image of P(ω).

It follows immediately from this deﬁnition that Θ is a cardinal.

2.5.2 Remark. We will often making use of the fact that continuous functions on ωω can be coded by members of ωω. While there are many ways of doing this, we ﬁx one for concreteness and convenience. Recall from Example 2.4.1 that there is a universal set C ⊆ (ωω)3 for the pointclass of closed subsets of spaces in X . For each continuous function f : ωω → ωω, there is an x ∈ ωω such c that Cx = f. Let F be the set of x for which Cx is a continuous function from ω ω c c c 1 ω to ω , and for each x ∈ F let fx denote the set Cx. Then F is in Π∼1. ω k+3 Using universal closed sets Ck ⊆ (ω ) (k ∈ ω), we can in a similar fashion ﬁx for each pair (n, m) ∈ ω × ω a set F c,n,m of codes x for all continuous functions c,n,m ω n ω m fx :(ω ) → (ω ) . 34 CHAPTER 2. THE WADGE HIERARCHY

2.5.3 Remark. The usual diagonal argument shows that it is not possible to ω ∗ ω ω have an association of each x ∈ ω to a continuous function fx : ω → ω in ∗ ω 2 such a way that the map (x, y) 7→ fx (y) is continuous on (ω ) . To see this, ω ω ∗ consider the function g : ω → ω deﬁned by setting g(x)(n) = fx (x)(n) + 1 for ω ∗ each x ∈ ω , for any such map x 7→ fx . Another diagonalization argument gives a Wadge class above a given selfdual class. Theorem 2.5.4 (Solovay). If Lipschitz Determinacy holds, then there a function ω ω ω j : P(ω ) → P(ω ) such that, for all A ⊆ ω , A Wa A as in Remark 2.2.4 (recall that [A]Wa = [A]Li in this case, by Theorem 2.2.6. c c −1 In the selfdual case, let j(A) be the set of x ∈ F such that x 6∈ (fx) [A]. 2.5.5 Remark. The proof of Theorem 2.5.4 shows that if A ⊆ ωω and ∆ is the smallest selfdual boldface pointclass containing A, then

• if [A]Wa is nonselfdual, then [j(A)]Wa = ∆;

• if [A]Wa is selfdual, and ∆ contains a universal closed set, then j(A) is in ω ∀ω ∆;

1 ωω • if [A]Wa is selfdual and∼ Π1 ⊆ ∆, then j(A) is in ∃ ∆. Recall that we let W denote the set of A ⊆ ωω for which the Wadge rank WR(A) is deﬁned. The proof of Theorem 2.5.4 does not show how high the Wadge rank of j(A) is relative to A. In particular, we have the following question (one could ask a similar question about the limit stages of the construction in the proof of Theorem 2.5.9). 2.5.6 Question. Assuming AD, must j map W into W? The main result of this section is the following fact. One direction is proved in Proposition 2.5.8, the other in Theorem 2.5.9. Theorem 2.5.7 (Solovay). If Lipschitz Determinacy holds then Θ is the supremum of the set of ordinals γ for which there exists a

hAα : α < γi.

ω If one assumes in addition that P(ω ) = W (i.e., that

(if there exists such an α, and 0 otherwise). It follows that γ < Θ. It shows moreover that for each A ∈ W, WR(A) < Θ. Proposition 2.5.8. Assume that Lipschitz Determinacy holds, let A be a subset of ωω, and let ∆ be the smallest selfdual boldface pointclass containing A and F c. Let ≤ be the order on F c deﬁned by setting x ≤ y if and only if

c −1 c −1 (fx) [A] ≤W (fy ) [A].

c ω Then (F , ≤) is isomorphic to ({B ⊆ ω : B ≤W A}, ≤W), and ≤ is in ω ω ∃ω ∀ω ∆.

c ω Proof. That (F , ≤) is isomorphic to ({B ⊆ ω : B ≤W A}, ≤W) is immediate ω from the deﬁnitions. That ≤ is in ∀ω ∆ follows noting that x ≤ y if and only if {x, y} ⊆ F c and there exists a u ∈ ωω such that for all z, w, v, t in ωω, if c c c w = fx(z), v = fu(z) and t = fy (v), then w ∈ A if and only if t ∈ A (that is, c −1 c c −1 z ∈ (fx) [A] if and only if fu(z) ∈ (fy ) [A]).

Finally, we use Theorem 2.5.4 to building

ω ω j(π[{(x, y) ∈ ω × ω | g(y) < α and x ∈ Ag(y)}]), where j is as in Theorem 2.5.4.

ω 2.5.10 Remark. Let ∆ be a ∃ω -closed selfdual pointclass containing F c, let S ω γ be a ordinal such that γ ∆ ⊆ ∆ and let f : ω → γ be a surjection such that f −1[{α}] ∈ ∆ for all α < γ. The proof of Theorem 2.5.9, along with Remark 2.5.5, shows that ∆ contains a set whose Wadge rank is at least γ (or undeﬁned). It follows that if P(ωω) = W then Θ is the supremum of the Wadge ranks of the elements of P(ωω).

Corollary 2.5.11 (Solovay). Suppose that Lipschitz Determinacy holds, and that WR(A) is deﬁned for every A ⊆ ωω. Then Θ = {WR(A) | A ⊆ ωω}. 36 CHAPTER 2. THE WADGE HIERARCHY Chapter 3

Coding Lemmas

This chapter presents the Moschovakis Coding Lemma, which is one of the fundamental theorems of the theory of determinacy. Among other things the Coding Lemma can be used to map ωω onto the powerset of any ordinal below Θ (see Corollary 3.0.2). Theorem 3.0.1 is the basic form of the lemma. Theorem 3.0.3 is a uniform version, which will be needed in Chapter 5. We follow [19]. Section 7D of [27] also contains proofs of the theorems in this section. Given a strategy σ : ω<ω → ω, and x ∈ ωω, we let σ ◦ x (similarly, x ◦ σ) be the sequence of moves played by player I (II) when he plays according to σ and player II (I) plays x. Theorem 3.0.1 (The Coding Lemma; Moschovakis). Assume that AD holds. Let

• X be a subset of ωω; • Z be a subset of X × ωω; • f be a function from X to the ordinals;

2 • =f be {(y, z) ∈ X : f(y) = f(z)};

2 •

For each y ∈ X, let [y]f denote {z ∈ X : f(y) = f(z)}. Then there exists a 1 pos-∼Σ1(=f ,

Proof. It suﬃces to consider the case where the range of f is an ordinal γ, and, proving the theorem by induction, we may assume that it holds for all 1 smaller ordinals. As pos-Σ∼1(=f , limit ordinal. Applying Theorem 2.4.7 and Remark 2.4.11, let U¯ = hUn : n ∈ ω \{0}i be a sequence of universal 1 ω sets for pos-Σ∼1(=f ,

37 38 CHAPTER 3. CODING LEMMAS

1. U2,x ⊆ Z;

ω ω 2. for all y ∈ X, if Z ∩ ([y]f × ω ) 6= ∅ then U2,x ∩ ([y]f × ω ) 6= ∅.

ω ω Let Y = {x ∈ ω : U2,x ⊆ Z}. For each x ∈ ω , let αx be the least value f(y) for any y ∈ X witnessing the failure of item (2) for x, if there exists such a y; otherwise, let αx = γ. Consider the game between players I and II where ω ω player I builds x1 ∈ ω , player II builds x2 ∈ ω , and I wins if and only if x1 ∈ Y and either x2 6∈ Y or αx1 ≥ αx2 . We will show that a winning strategy for either player gives the desired conclusion. First, suppose that σ is a winning strategy for player I. Then for all x ∈ ω ωω, U ⊆ Z. The set S U is in pos-Σ1(= , < ), which is ∃ω - 2,σ◦x x∈ωω 2,σ◦x ∼1 f f closed. By the assumption that the theorem holds for all ordinals smaller than γ, there exists for each α < γ an x ∈ Y such that αx ≥ α. It follows then that S x∈ωω U2,σ◦x is as desired. Now suppose that τ is a winning strategy for player II. For each y ∈ X, let ω 4 [

[ ω U2,h1(x) = (U2,h0(x,w)◦τ ∩ ([w]f × ω )). (3.2) w∈X

ω By Theorem 2.4.9, there exists an x1 ∈ ω such that U2,x1 = U2,h1(x1). We show now that x1 satisﬁes conditions (1) and (2). For condition (1), suppose toward a contradiction that the condition fails, and consider (y, z) ∈ U2,x1 \ Z with f(y) minimal. Since U2,x1 = U2,h1(x1), there exists a w ∈ X such that

ω (y, z) ∈ U2,h0(x1,w)◦τ ∩ ([w]f × ω ).

Then f(y) = f(w) and, by equation (3.1), for all (a, b) ∈ U2,h0(x1,w),(a, b) is in

U2,x1 and f(a) < f(w). By the minimality of f(y), it follows that (a, b) ∈ Z and therefore that h0(x1, w) ∈ Y . Since τ is a winning strategy for II, h0(x1, w)◦τ ∈ Y , which means that (y, z) ∈ Z, giving a contradiction.

For condition (2), suppose toward a contradiction that αx1 < γ. Let y ∈ X be such that f(y) = αx1 . By equation (3.1), the fact that x1 is in Y and the 39

deﬁnition of αx1 , h0(x1, y) ∈ Y , and αh0(x1,y) = αx1 . Since τ is a winning strategy for II, αh0(x1,y)◦τ > αh0(x1,y) = αx1 , which is impossible, as

ω (U2,h0(x1,y)◦τ ∩ ([y]f × ω )) ⊆ U2,x1 , by equation (3.2) and the choice of x1. This completes the proof.

ω 1 As there is a surjection from ω to∼ Σ1, Theorem 3.0.1 gives the following immediate corollary. Corollary 3.0.2. If AD holds, then each of the following hold. 1. For each λ < Θ there is a surjection from P(ω) to P(λ). 2. Θ is a limit cardinal. We now prove a uniform version of Theorem 3.0.1 which is used in the proof of Theorem 5.1.3, which in turn is used to prove that the strong partition cardinals are coﬁnal below Θ. Given a sequence hUn : n < ωi of universal sets 1 ω for Σ∼1, and an A ⊆ ω , we let hUn(A): n < ωi be as in Remark 2.4.11. Theorem 3.0.3 (The Uniform Coding Lemma). Assume that AD holds. Let ¯ 1 • U = hUn : n < ωi be a sequence of universal sets for ∼Σ1 with the s-m-n property;

• X be a subset of ωω; • Z be a subset of X × ωω; • f be a function from X to the ordinals.

For each y ∈ X, let [y]f denote {z ∈ X : f(y) = f(z)}, let [

ω Cy = ω \ ([y]f ∪ [

Then there exists an x ∈ ωω such that for all y ∈ X,

ω 1. U2,x([y]f ,Cy) ⊆ Z ∩ ([y]f × ω ),

ω 2. U2,x([y]f ,Cy) 6= ∅ if and only if Z ∩ ([y]f × ω ) 6= ∅. Proof. It suﬃces to consider the case where the range of f is an ordinal γ, and, proving the theorem by induction, we may assume that it holds for all smaller 1 ω ordinals. As Σ∼1 is closed under unions and contains all ﬁnite subsets of ω , we may assume that γ is a limit ordinal. Let Y be the set

ω ω {x ∈ ω : ∀y ∈ XU2,x([y]f ,Cy) ⊆ Z ∩ ([y]f × ω )}.

ω For each x ∈ ω , let αx be the least value f(y) for a y ∈ X witnessing the failure of conclusion (2) of the theorem, with respect to x, if there exists such 40 CHAPTER 3. CODING LEMMAS

a y; otherwise, let αx = γ. Consider the game between players I and II where ω ω player I builds x1 ∈ ω , player II builds x2 ∈ ω , and I wins if and only if x1 ∈ Y and either x2 6∈ Y or αx1 ≥ αx2 . We will show that a winning strategy for either player gives the desired conclusion. ω First, suppose that σ is a winning strategy for player I. Let zσ ∈ ω be such S ω that U2,zσ = x∈ωω U2,σ◦x. Then for all P1,P2 ⊆ ω (in particular, in the case P1 = [y]f , P2 = Cy, for some y ∈ X), [ U2,zσ (P1,P2) = U2,σ◦x(P1,P2). x∈ωω

Since σ is a winning strategy for I, zσ is in Y . By our assumption that the theorem holds for all ordinals smaller than γ, there exist for each α < γ an x ∈ Y such that αx ≥ α. It follows then that zσ is as desired. Now suppose that τ is a winning strategy for player II. Fix a recursive bijection π : ω × ω → ω. For each i ∈ {1, 2}, let Qi be the set of pairs (y, r) ∈ ωω × ωω such that (r)0 = hπ(i, y(0)), y(1), y(2),...i 1 (this corresponds to how we deﬁned the class pos-Σ∼1(A1,A2) in terms of our 1 0 ω ω deﬁnition of pos-Σ∼1(A) in Section 2.4). Then for any P,P ⊆ ω and z ∈ ω , ω 0 • z ∈ P if and only if (z, r) ∈ Q1 holds for some r ∈ ω with (r)0 ∈ P ⊕ P .

0 ω 0 • z ∈ P if and only if (z, r) ∈ Q2 holds for some r ∈ ω with (r)0 ∈ P ⊕ P .

ω − ω − For each r ∈ ω , let r ∈ ω be such that (r )n = (r)n+1 for all n ∈ ω. ω ω 5 Let a0 ∈ ω be such that U5,a0 is the set of (x, y, z, w, r) ∈ (ω ) for which − ω 3 ω (x, z, w, r ) ∈ U3 and (y, r) ∈ Q2. Let s5,2 :(ω ) → ω be a continuous function witnessing the s-m-n property of U¯ for n = 2 and m = 5, and let ω 2 ω h0 :(ω ) → ω be deﬁned by setting h0(x, y) = s5,2(a0, x, y). Then for all (x, y) ∈ ωω × ωω, all P,P 0 ⊆ ωω and all (z, w) ∈ ωω × ωω,

0 (z, w) ∈ U2,h0(x,y)(P,P ) if and only if

ω 0 ∃r ∈ ω ((∀n ∈ ω (r)n ∈ P ⊕ P ) ∧ (h0(x, y), z, w, r) ∈ U3) if and only if

ω 0 ∃r ∈ ω ((∀n ∈ ω (r)n ∈ P ⊕ P ) ∧ (a0, x, y, z, w, r) ∈ U5) if and only if

ω 0 − ∃r ∈ ω ((∀n ∈ ω (r)n ∈ P ⊕ P ) ∧ (x, z, w, r ) ∈ U3 ∧ (y, r) ∈ Q2) if and only if 0 0 (z, w) ∈ U2,x(P,P ) ∧ y ∈ P . 41

That is, we have the following fact (**) : for all (x, y) ∈ ωω × ωω, and all 0 ω 0 0 P,P ⊆ ω (in particular, sets of the form Cy), U2,h0(x,y)(P,P ) is U2,x(P,P ) if y is in P 0 and ∅ otherwise. ω ω 4 Now let a1 ∈ ω be such that U4,a1 is the set of (x, z, w, r) ∈ (ω ) such ω − that, for some y ∈ ω ,(z, w, r ) ∈ U3,h0(x,y)◦τ and (y, r) ∈ Q1 holds. Let ω 2 ω s4,1 :(ω ) → ω be a continuous function witnessing the s-m-n property of ω ω U¯ for n = 1 and m = 3, and let h1 : ω → ω be deﬁned by setting h1(x) = s4,1(a1, x). Then for all x ∈ ωω, all P,P 0 ⊆ ωω and all (z, w) ∈ ωω × ωω,

0 (z, w) ∈ U2,h1(x)(P,P ) if and only if

ω 0 ∃r ∈ ω ((∀n ∈ ω (r)n ∈ P ⊕ P ) ∧ (h1(x), z, w, r) ∈ U3) if and only if

ω 0 ∃r ∈ ω ((∀n ∈ ω (r)n ∈ P ⊕ P ) ∧ (a1, x, z, w, r) ∈ U4) if and only if

ω 0 ω − ∃r ∈ ω ((∀n ∈ ω (r)n ∈ P ⊕P )∧∃y ∈ ω ((z, w, r ) ∈ U3,h0(x,y)◦τ ∧(y, r) ∈ Q1)) if and only if 0 ∃y ∈ P (z, w) ∈ U2,h0(x,y)◦τ (P,P ). Then for all x ∈ ωω and all P,P 0 ⊆ ωω,

0 [ 0 U2,h1(x)(P,P ) = U2,h0(x,y)◦τ (P,P ). (3.3) y∈P

ω By Theorem 2.4.9 and Remark 2.4.11, there exists an x1 ∈ ω such that 0 0 0 ω U2,x1 (P,P ) = U2,h1(x1)(P,P ) for all P,P ⊆ ω . We show now that x1 satisﬁes both conditions in the conclusion of the theorem. For condition (1) (i.e., the ∗ assertion that x1 ∈ Y ), ﬁx y ∈ X with

∗ ∗ ω ∗ U2,x1 ([y ]f ,Cy ) 6⊆ Z ∩ ([y ]f × ω ) and f(y∗) minimal. Since

∗ ∗ ∗ ∗ U2,x1 ([y ]f ,Cy ) = U2,h1(x1)([y ]f ,Cy ),

0 ∗ there exists by equation (3.3) a y ∈ [y ]f such that

∗ ∗ ω 0 ∗ U2,h0(x1,y )◦τ ([y ]f ,Cy ) 6⊆ Z ∩ ([y ]f × ω ).

0 We want to see that h0(x1, y ) ∈ Y , since then, as τ is a winning strategy for 0 II, we will have that h0(x1, y ) ◦ τ ∈ Y , giving a contradiction. For each y ∈ X, we have by (**) that

ω 0 U2,h0(x1,y )([y]f ,Cy) = U2,x1 ([y]f ,Cy) ⊆ Z ∩ ([y]f × ω ) 42 CHAPTER 3. CODING LEMMAS

0 ∗ ∗ 0 if f(y) < f(y ), by the minimality of f(y ), and U2,h0(x1,y )([y ]f ,Cy) = ∅ ω 0 otherwise. In either case, U2,h0(x1,y )([y]f ,Cy) ⊆ Z ∩ ([y]f × ω ), which shows 0 that h0(x1, y ) ∈ Y as desired. ∗ For condition (2), suppose toward a contradiction that αx1 < γ. Let y ∈ X ∗ be such that f(y ) = αx1 . By (**), the fact that x1 is in Y and the deﬁnition ∗ ∗ of αx1 , h0(x1, y ) ∈ Y , and αh0(x1,y ) = αx1 . Since τ is a winning strategy for ∗ ∗ II, αh0(x1,y )◦τ > αh0(x1,y ) = αx1 , which is impossible, as

∗ ∗ ∗ ∗ ∗ (U2,h0(x1,y )◦τ ([y ]f ,Cy ) ⊆ U2,x1 ([y ]f ,Cy ), by equation (3.3) and the choice of x1. This completes the proof. Chapter 4

Properties of pointclasses

In this chapter we develop various properties of pointclasses, including closure and separation properties. Our immediate goal is preparing for the proof of the existence of strong partition cardinal in Chapter 5. Some of the material in this chapter will also be used in Part II of the book (in Remark 9.0.15, for instance).

4.1 Separation and reduction

Given a pointclass Γ, and disjoint subsets A and B of the same space in X , we say that A and B are Γ-separable if there exists a set C ∈ Γ ∩ Γˇ such that A ⊆ C and B ∩ C = ∅ (and Γ-inseparable otherwise). We generally use this property only with selfdual pointclasses ∆. We say that a pointclass Γ satisﬁes the separation property if all pairs of disjoint subsets of ωω in Γ are (Γ ∩ Γ)-ˇ separable. We write Sep(Γ) to indicate that Γ has the separation property. We follow [33]. Let us say that a function f : ωω → ωω is strongly Lipschitz ω if for all x, y ∈ ω and n ∈ ω, if xn = yn, then f(x)(n) = f(y)(n). Strategies for Player I in Lipschitz games induce strongly Lipschitz functions. The following lemma is used in the proof of Theorem 4.1.2.

Lemma 4.1.1 (ZF + AD; Steel). Let Γ be a boldface pointclass and let (A0,A1) ω be a Γ-inseparable pair of subsets of ω . Let B0 and B1 be disjoint subsets of ωω, both in in Γ or both in Γˇ. Then there is a strongly Lipschitz map f so that f[B0] ⊆ A0 and f[B1] ⊆ A1. Proof. Let ∆ = Γ ∩ Γ.ˇ Consider the game where I plays x, II plays y, and I wins if y ∈ B0 implies x ∈ A0, and y ∈ B1 implies x ∈ A1. It suffices to see that II cannot have a winning strategy. Supposing that σ were a winning strategy for II, let g be the function x 7→ x ◦ σ (this notation was defined at −1 the beginning of Chapter 3). Then g is continuous, and g [B0] is in ∆, since ω −1 −1 −1 ω \ g [B0] = g [B1]. Furthermore, g [B0] separates A0 and A1. Theorem 4.1.2 (Steel). Assume that AD holds. If Γ is a nonselfdual boldface pointclass, then at least one of Γ and Γˇ satisfies the separation property.

43 44 CHAPTER 4. PROPERTIES OF POINTCLASSES

Proof. Suppose toward a contradiction that the pairs (A0,A1) and (C0,C1) form counterexamples to Sep(Γ) and Sep(Γ),ˇ respectively. By Lemma 4.1.1, there is a strongly Lipschitz function f mapping A0 into C0 and A1 into C1. −1 −1 Let B0 = f [C0] and let B1 = f [C1]. Then B0 and B1 are disjoint and both in Γ,ˇ and A0 ⊆ B0 and A1 ⊆ B1. Applying Lemma 4.1.1 again, there exist strongly Lipschitz functions f0, f1 and f2 such that

• f0[A0] ⊆ A1;

• f0[A1] ⊆ A0;

• f1[A0] ⊆ A0; ω • f1[ω \ B0] ⊆ A1;

• f2[A1] ⊆ A1; ω • f2[ω \ B1] ⊆ A0. ω Since f0, f1 and f2 are all strongly Lipschitz, for each r ∈ 3 there is a r ω r r unique sequence hxn : n < ωi of elements of ω such that xn = fr(n)(xn+1) for all n ∈ ω. Let 3ω have its induced topology as a subspace of ωω. By Baire(P(ωω)), every subset of 3ω has the property of Baire, in particular the set ω r C = {r ∈ 3 : x0 6∈ A0 ∪ A1}. Suppose first that C is nonmeager. By the Baire property, we may fix an <ω ω sar s ∈ 3 such that {r ∈ 3 : x0 6∈ A0 ∪ A1} is comeager. However, for each <ω ω r sar s ∈ 3 and r ∈ 3 , if x0 ∈ A0 ∪ A1 then x0 ∈ A0 ∪ A1. It follows then that r ω x0 6∈ A0 ∪ A1, for comeagerly many r ∈ 3 . Since B0 ∩ B1 = ∅, we may fix ∗ r ω an i ∈ {0, 1} such that x0 6∈ Bi∗ , for nonmeagerly many r ∈ 3 . Inspecting ω r f0, f1 and f2, we see that for each i ∈ {0, 1} and each r ∈ 3 , if x0 6∈ Bi hi+1iar r then x0 ∈ A1−i. It follows then that x0 ∈ A0 ∪ A1 for nonmeagerly many r ∈ 3ω, which gives a contradiction. Suppose on the other hand that C is comeager. Choose i∗ ∈ {0, 1} so that ω r <ω ω sar {r ∈ 3 : x0 ∈ Ai∗ } is nonmeager. Choose s ∈ 3 so that {r ∈ 3 : x0 ∈ ω r sah0iar Ai∗ } is comeager. For any r ∈ 3 such that x0 ∈ Ai∗ , x0 ∈ A1−i∗ . sar ω It follows that x0 ∈ A1−i∗ for nonmeagerly many r ∈ 3 , giving another contradiction. 4.1.3 Remark. Theorems 5.2 and 5.3 of [34] show that if AD holds then for any nonselfdual pointclass Γ, at most one of Γ and Γˇ has the separation property. A pointclass Γ is said to have the reduction property if for all A, B ⊆ ωω in Γ, there exist disjoint A0,B0 ∈ Γ such that A0 ⊆ A, B0 ⊆ B and A0 ∪B0 = A∪B. In this case we say that the sets A0 and B0 reduce the sets A and B. We write Red(Γ) to indicate that the pointclass Γ has the reduction property. It follows easily from the definitions that Red(Γ) implies Sep(Γ).ˇ Along with Remark 4.1.3, this shows that, under AD, Red(Γ) and Sep(Γ) cannot both hold for a nonselfdual pointclass. The following gives a direct proof.1 1where is it from? 4.1. SEPARATION AND REDUCTION 45

Theorem 4.1.4. Assume that Wadge Determinacy + Baire(P(ωω)) holds. If Γ is a nonselfdual boldface pointclass and Red(Γ) holds, then Sep(Γ) does not hold.

Proof. By Theorem 2.4.3 there exist a Γ-universal set U ⊆ (ωω)2. Let π : ωω → ω 2 ω (ω ) be a continuous bijection, and let π0, π1 be the functions on ω such that ω ω π(x) = (π0(x), π1(x)) for all x ∈ ω . Let A be the set of x ∈ ω such that ω 0 (π0(x), x) ∈ U and let B be the set of x ∈ ω such that (π1(x), x) ∈ U. Let A and B0 in Γ witness the reduction property for A and B. Suppose now that C ∈ Γ ∩ Γˇ witnesses the separation property for A0 and B0 (i.e., C contains A0 and is disjoint from B0). Let x ∈ ωω be such that ω 0 Uπ0(x) = ω \ C and Uπ1(x) = C. If x ∈ C, then x ∈ B \ A, so x ∈ B , giving a contradiction. Similarly, if x 6∈ C, then x ∈ A \ B, so x ∈ A0, giving another contradiction.

4.1.5 Remark. Theorems 4.1.2 and 4.1.4 together imply, assuming AD, that if Γ is a nonselfdual boldface pointclass, then at most one of Γ and Γˇ has the reduction property. Theorem 4.1.6, which is Theorem 5.5 of [34], shows that if in addition Γ is closed under ﬁnite unions and intersections, then the separation property holds for exactly one of Γ and Γ,ˇ and the reduction property holds for the other.

Theorem 4.1.6 (Van Wesep). If Γ is a boldface pointclass closed under ﬁnite intersections, and Sep(Γ)ˇ holds, then Red(Γ) holds.

The proof of Theorem 4.1.6 (which we give below) uses the following lemma.

Lemma 4.1.7. Suppose that AD holds, and that Γ is a boldface pointclass closed under ﬁnite intersections. If Red(Γ) does not hold, then for all A, B in P(ωω)∩Γˇ there exist C,D ∈ Γ such that

• A \ B ⊆ C \ D;

• B \ A ⊆ D \ C;

• C ∪ D = ωω. Proof. Let (E,F ) be subsets of ωω witnessing ¬Red(Γ), and ﬁx A, B in P(ωω)∩ Γ.ˇ Consider the game G(E, F, A, B) in which player I produces x ∈ ωω, player II produces y ∈ ωω and player II wins if and only if the following conditions are met:

• y ∈ E ∪ F ;

• if x ∈ A \ B then y ∈ E \ F ;

• if x ∈ B \ A then y ∈ F \ E. Suppose toward a contradiction that σ is a winning strategy for player I. Let E00 = {y ∈ ωω : σ ◦ y 6∈ A} and let F 00 = {y ∈ ωω : σ ◦ y 6∈ B}. Then we have the following: 46 CHAPTER 4. PROPERTIES OF POINTCLASSES

• E ∪ F ⊆ E00 ∪ F 00;

• (E ∪ F ) ∩ (E00 ∩ F 00) = ∅;

• E \ F ⊆ E00;

• F \ E ⊆ F 00;

• E00,F 00 ∈ Γ. Let E0 = E ∩ E00 and let F 0 = F ∩ F 00. Then E0 and F 0 are disjoint sets in Γ, and E ∪ F = E0 ∪ F 0. This contradicts the assumption that E and F witness the failure of Red(Γ). It follows then that there is a winning strategy τ for player II. Let C = {x ∈ ωω : x ◦ τ ∈ E} and let D = {x ∈ ωω : x ◦ τ ∈ F }. Then C and D are as desired.

Proof of Theorem 4.1.6. Supposing that Red(Γ) fails, we show that Red(Γ)ˇ holds, contradicting Theorem 4.1.4. Let A and B be subsets of ωω in Γ. Let C and D be as given by Lemma 4.1.7. Applying Sep(Γ),ˇ let E ∈ Γ ∩ Γˇ be such that ωω \ C ⊆ E and E ∩ (ωω \ D) = ∅. Let C0 = C \ E and let D0 = D ∩ E. Then

D0 ∩ (A \ B) = C0 ∩ (B \ A) = C0 ∩ D0 = ∅ and A∪B ⊆ C0 ∪D0, so A\D0 and B \C0 satisfy the conclusion of the reduction property with respect to A and B.

The following result (which we need for Theorem 6.0.7) is sometimes called the 0th Periodicity Theorem. Recall that x ≤Tu y and x ≡Tu y refer to Turing reducibility and were deﬁned in Section 0.2.

Theorem 4.1.8 (ZF + AD; Kechris). 2 Suppose that Γ is a boldface pointclass closed under countable intersections and countable unions, and that Red(Γ) holds.

ω ω • If ∃ω Γ ⊆ Γ, then Red(∀ω Γ) holds.

ω ω • If ∀ω Γ ⊆ Γ, then Red(∃ω Γ) holds.3 Proof. Let A and B be subsets of (ωω)2 in Γ. For the ﬁrst part we will ﬁnd a reduction for A∗ = {x : ∀y ∈ ωω (x, y) ∈ A} and B∗ = {x : ∀y ∈ ωω (x, y) ∈ B}. Let 0 ω 2 A = {(x, z) ∈ (ω ) : ∀y ≤Tu z (x, y) ∈ A} and let 0 ω 2 B = {(x, z) ∈ (ω ) : ∀y ≤Tu z (x, y) ∈ B}.

2and more than AD? 3We may need a lot of basic stuﬀ about the reduction property, like its relation to PWO, and whether selfdual classes can satisfy it. 4.1. SEPARATION AND REDUCTION 47

Since Γ is closed under countable intersections, A0 and B0 are in Γ; ﬁx sets C ⊆ A0 and D ⊆ B0 reducing them. Let

0 ω 2 0 0 C = {(x, z) ∈ (ω ) : ∃z ≡Tu z (x, z ) ∈ C} and let 0 ω 2 0 0 D = {(x, z) ∈ (ω ) : ∀z ≡Tu z (x, z ) ∈ D}. Then C0 and D0 are in Γ, since Γ is closed under countable intersections and countable unions, and they reduce A0 and B0. ω Let GA(x) (for x ∈ ω ) be the game where players I and II collaborate to build z ∈ ωω, where I wins if and only if (x, z) ∈ C0, and let E be the set of ω x for which I has a winning strategy in GA(x). Let GB(x) (for x ∈ ω ) be the game where players I and II collaborate to build z ∈ ωω, where I wins if and only if (x, z) ∈ D0, and let F be the set of x for which I has a winning strategy ωω in GB(x). Then E and F are in ∀ Γ (via the assertion that player II does ω not have a winning strategy, and the assumption that ∃ω Γ ⊆ Γ), and they are disjoint, since C0 and D0 are. To see that E ad F reduce A∗ and B∗, ﬁx x ∈ ωω and suppose that at least ω 0 ω one of Ax and Bx is all of ω . In the former case, Ax = ω , and in the latter 0 ω ω 0 0 case Bx = ω . Then for each z ∈ ω ,(x, z) is either in C or D , and one of the 0 0 sets {[z]Tu :(x, z) ∈ C } and {[z]Tu :(x, z) ∈ D } contains a Turing cone. Then player I has a winning strategy in one of the games GA(x) and GB(x). For the second part (redirecting our labels) we ﬁnd a reduction for A∗ = {x : ∃y ∈ ωω (x, y) ∈ A} and B∗ = {x : ∃y ∈ ωω (x, y) ∈ B}. Let

0 ω 2 A = {(x, z) ∈ (ω ) : ∃y ≤Tu z (x, y) ∈ A} and let 0 ω 2 B = {(x, z) ∈ (ω ) : ∃y ≤Tu z (x, y) ∈ B}. Since Γ is closed under countable unions, A0 and B0 are in Γ, so there exist C and D in Γ reducing them. Let

To see that E and F reduce A∗ and B∗, ﬁx x ∈ ωω and suppose that at least 0 one of Ax and Bx is nonempty. Then again one of the sets {[z]Tu :(x, z) ∈ C } 0 and {[z]Tu :(x, z) ∈ D } contains a Turing cone. Then player I has a winning strategy in one of the games GA(x) and GB(x).

4.2 The prewellordering property

Given a subset P of a space in X , a norm on P is a function from P to the ordinals. A norm is regular if its range is an ordinal. We write |φ| for the range of a regular norm φ. A prewellordering on a set P is a binary relation ≤ on which is reﬂexive, transitive, total (so for all x, y ∈ X, at least one of x ≤ y and y ≤ x holds) and wellfounded. Equivalently (via the canonical rank function for a wellfounded preorder), a prewellordering on a set P is a set of the form {(x, y) ∈ P : f(x) ≤ f(y)}, where f is a (regular) norm on P . When f is regular, the range of f is the length of the prewellordering. Given a pointclass Γ, we let δ(Γ) be the supremum of the lengths of the prewellorderings in Γ ∩ Γ.ˇ Given a pointclass Γ, a space X in X and a set P ⊆ X, a Γ-norm on P is norm φ on P for which each of the following sets are in Γ : ∗ • ≤φ, the set of pairs (x, y) ∈ X × X such that x ∈ P and either y 6∈ P or φ(x) ≤ φ(y);

∗ • <φ, the set of pairs (x, y) ∈ X × X such that x ∈ P and either y 6∈ P or φ(x) < φ(y). 4.2.1 Remark. Suppose that Γ is a boldface pointclass, and φ is a Γ-norm on a subset P of a space X in X . Let Y be a space in X and let f : Y → X be a continuous function. Then φ ◦ f is a Γ-norm on f −1[P ]. In particular, if P is Γ-complete, and P is the domain of a Γ-norm, then every member of Γ is the domain of some Γ-norm. 4.2.2 Remark. Let Γ be a pointclass, let P be an element of Γ contained in a space X in X . Let φ be a Γ-norm on P . For points x, y in X, the negation of ∗ x ≤φ y is equivalent to the statement x 6∈ P ∨ (y ∈ P ∧ φ(y) < φ(x))

∗ and the negation of x <φ y is equivalent to the statement x 6∈ P ∨ (y ∈ P ∧ φ(y) ≤ φ(x)). 4.2.3 Remark. It follows from Remark 4.2.2 that if Γ is a boldface pointclass, φ is a Γ norm on a set P and z ∈ P , then {(x, y): φ(x) ≤ φ(y) < φ(z)} is a prewellordering in Γ ∩ Γ.ˇ Lemma 4.2.4. Suppose that Γ is boldface pointclass closed under unions and universal quantiﬁcation over ωω, and that φ is a Γ-norm on a set P ∈ Γ \ Γˇ. Let Q be a subset of P in Γˇ. Then there exists a y ∈ P such that φ(x) < φ(y) for all x ∈ Q. 4.2. THE PREWELLORDERING PROPERTY 49

Proof. Supposing otherwise, P ∈ Γ,ˇ as y ∈ P if and only if there exists an x ∈ Q ∗ such that x <φ y fails.

A pointclass Γ has the prewellordering property if every member of Γ has a Γ-norm. We write PWO(Γ) to mean that Γ has the prewellordering property.

Theorem 4.2.5. If Γ is a boldface pointclass closed under unions and satisfying the prewellordering property, then Red(Γ) holds.

ω ω ω Proof. Let A and B be subsets of ω in Γ. For each i ∈ 2, let fi : ω → ω be ω such that, for each x ∈ ω , fi(x)(0) = i and fi(x)(n + 1) = x(n) for all n ∈ ω. 0 ω ∗ Let φ be a Γ-norm on f0[A] ∪ f1[B]. Let A = {x ∈ ω : f0(x) ≤φ f1(x)} and 0 ω ∗ 0 0 let B = {x ∈ ω : f1(x) <φ f0(x)}. Then A and B reduce A and B.

1 4.2.6 Remark. The pointclass∼ Π1 has the prewellordering property (see 4B.1 of [27]). Adapting an argument of Novikov, Moschovakis showed that if Γ is a ω ω boldface pointclass closed under ∀ω and PWO(Γ) holds, then PWO(∃ω Γ) also holds. The First Periodicity Theorem (due to Martin and Moschovakis) says ω that, assuming AD, if Γ is a boldface pointclass closed under ∃ω and PWO(Γ) ω holds, then PWO(∀ω Γ) also holds. See 4B.1 and 6B.3 of [27].4

4.2.7 Remark. Suppose that Γ is a boldface pointclass, and P is an element of Γ \ Γ,ˇ and φ is a Γ-norm on P with range γ. Then φ gives rise to a γ-sequence of sets in Γ ∩ Γˇ whose union is P . In particular, Γˇ is not closed under γ-unions.

ω 2 Given a set A ⊆ ω , a∼ Σ1(A) set is a set of the form

ω n ω {(x1, . . . , xn) ∈ (ω ) : ∃B ⊆ ω φ(A, B, x1, . . . , xn, y)}, where the quantiﬁers in φ range over the hereditarily countable sets, and y is an ω element of ω . If φ(A, B, x1, . . . , xn, y) holds for suitable A, B x1, . . . , xn and ω y, we say that the formula ∃B ⊆ ω φ(A, B, x1, . . . , xn, y) is witnessed by B. 2 2 The pointclass∼ Σ1(A) is the smallest boldface pointclass containing each Σ∼1(A) subset of ωω.

ω 2 4.2.8 Remark. A set C ⊆ ω is Σ∼1(A) if, for some first order formula φ and some y ∈ ωω, C is the set of x ∈ ωω for which there is an ω-structure M containing ωω ∪{A} such that (M,E) |= φ(A, y, x), where E is the ∈-relation of 2 M. Modifying this formulation one can define a universal Σ∼1(A) set from the set of (n, x, y) ∈ ω × ωω × ωω for which there exists an ω-structure M containing ωω ∪ {A} and a set T (naturally coded by a set of reals) such that T is the theory of M (in parameters from M, using the Gödelnumbering for first order formulas) and n is the Gödelnumber of a formula φ such that M |= φ(A, x, y). 2 2 It follows Σ1(A) is nonselfdual. One can also verify in this way that Σ1(A) is ∼ ω ∼ closed under ∃ω and ∧.

4Get the reference right. These references are from the new version of the book. 50 CHAPTER 4. PROPERTIES OF POINTCLASSES

2 ˇ 2 2 We write Π1(A) for ∼Σ1(A) (the class of complements of members of∼ Σ1(A)), and ∆2 for Σ2(A) ∩ Π2(A). We write δ 2(A) for δ(Σ2(A)). Proposition 2.5.8 ∼1 ∼1 ∼1 ∼1 ∼1 2 shows that for each A ∈ W, ∼δ 1(A) is at least the Wadge rank of A. 4.2.9 Remark. Assuming that W = P(ωω), for any set A ⊆ ωω the pointclass 2 2 Σ1(A) has the prewellordering property. To see this, fix a Σ1(A) set X of the form {x ∈ ωω : ∃B ⊆ ωω ψ(A, B, x)}. Orders ≤∗ and <∗ witnessing that X has the prewellordering property can be defined using the Wadge rank of the witness B. That is, define x ≤∗ y to hold if there exists a B ⊆ ωω such that ψ(A, B, x) holds, and for every continuous ω ω −1 −1 function f : ω → ω such that f [B]

• for all α < β < ρ, f(α) ⊆ f(β); S • for all limit ordinals β < ρ, f(β) = α<β f(α); S • F = α<ρ f(α). Define φ: F → ρ by setting φ(x) to be the least α < ρ such that x ∈ f(α). Then S ω ∗ φ is a ρ ∆-norm on F . To see this, note first that, for all x, y in ω , x <φ y if and only if there exists an α < ρ such that x ∈ f(α) and y 6∈ f(α), and this set is in S ∆. Similarly, x ≤∗ y if and only if there exists an α < ρ such that ρ φ S x ∈ f(α + 1) and y 6∈ f(α), and this set is also in ρ ∆. 4.2.11 Remark. Assuming that Wadge Determinacy and Baire(P(ωω)) hold, Theorem 4.2.10 implies that if ∆ is a selfdual boldface pointclass containing a S countable Wadge-cofinal subset, then ω ∆ has the prewellordering property. This holds in particular whenever ∆ is the pointclass of sets projective in a fixed subset of ωω. The following theorem shows that at most one member of a complementary pair of nonselfdual bolface pointclasses can have the prewellordering property. Theorem 4.2.12. If Γ is a boldface pointclass with a universal set, then Γ and Γˇ do not both have the prewellordering property. 4.2. THE PREWELLORDERING PROPERTY 51

Proof. Let U ⊆ ωω × ωω be a universal Γ-set, and let φ be a Γ-norm on U. ω ω ω Let π : ω → ω × ω be a recursive bijection, and let π0 and π1 be such that ω π(x) = (π0(x), π1(x)) for all x ∈ ω . Let B be the set of (x, y) such that

∗ (π0(x), y) <φ (π1(x), y), and let C be the set of (x, y) such that

∗ (π1(x), y) <φ (π0(x), y).

Then B and C are disjoint and in Γ. Let ρ: ω × 2 → ω be bijection. For each i ∈ s, let ρ∗ be the function on ω ω ∗ ω × ω deﬁned by setting ρi (x, y) to be (hπ(x(0), i), x1, x2,...i, y). Let P be the set ∗ ω ω ∗ ω ω (ρ0[(ω × ω ) \ B] ∪ ρ1[(ω × ω ) \ C]. Then P is in Γ.ˇ Let ψ be a Γ-normˇ on P . Let E be the set of (x, y) ∈ ωω × ωω such that ∗ ∗ ∗ ρ0(x, y) <ψ ρ1(x, y). Since B ∩ C = ∅, E is also the set of (x, y) such that

∗ ∗ ∗ ¬(ρ1(x, y) ≤ψ ρ0(x, y)), so E is in ∆. We derive a contradiction to Theorem 2.4.2 by showing that E is universal ω for ∆. Fixing D ∈ ∆, let x be such that ω \ D = Ux0 and D = Ux1 . Then ω ω ω \ D = Bx and D = Cx, since, for all y ∈ ω ,

y ∈ D ⇔ ((x1, y) ∈ U ∧ (x0, y) 6∈ U).

Finally, for each y ∈ ωω,

(x, y) ∈ C ⇒ (x, y) ∈ E and (x, y) ∈ B ⇒ (x, y) 6∈ E.

It follows that Ex = D.

Finally, the following theorem shows that for certain pointclasses, every witness to the prewellordering property has the same length. Part of the proof of Theorem 4.2.13 is reused for Theorem 4.3.3.

ω Theorem 4.2.13 (Moschovakis). Suppose that Γ is a ∀ω -closed boldface pointclass with a universal set which is closed under unions, and let hUn : n ∈ ω\{0}i be sequence of Γ-universal sets with the recursion property. Then for every n ∈ ω \{0} and every regular Γ-norm φ on Un, |φ| = δ(Γ). 52 CHAPTER 4. PROPERTIES OF POINTCLASSES

Proof. Fix n ∈ ω, and let φ be a Γ-norm on Un. Let ≺ be a strict prewellordering on a set Y ⊆ (ωω)n, with ≺∈ Γ.ˇ Since any two spaces in X are homeomorphic, it suffices to show that the rank of ≺ is at most |φ|. Let Q be the set of ω ω n ω n ∗ (x, y) ∈ ω × (ω ) such that for all z ∈ (ω ) , z ≺ y implies (x, z) <φ (x, y). ∗ ω Then Q ∈ Γ. By Theorem 2.4.8, there exists an x ∈ ω such that Un,x∗ = Qx∗ . Then one can verify by induction on the ≺-rank of y that for all y ∈ Y , ω n ∗ ∗ y ∈ Un,x∗ and, for all z ∈ (ω ) , z ≺ y implies φ(x , z) < φ(x , y). To do this, fix y ∈ Y , and suppose that for all z ∈ Y with z ≺ y, z is in Un,x∗ and, for all ω n ∗ ∗ w ∈ (ω ) w ≺ z implies φ(x , w) < φ(x , z). To see that y is in Un,x∗ , we need ω n ∗ ∗ ∗ ω n to see that for all z ∈ (ω ) , if z ≺ y then (x , z) <φ (x , y). Fixing z ∈ (ω ) ∗ ∗ ∗ such that z ≺ y, have that the failure of (x , z) <φ (x , y) implies that either ∗ ∗ ∗ ∗ ∗ (x , z) 6∈ Un or (x , y) ∈ Un and φ(x , y) ≤ φ(x , z)). Since (x , z) is in Un, we ∗ ω n have that (x , y) is in Un. Finally, fix once again a z ∈ (ω ) such that z ≺ y. ∗ ∗ ∗ ∗ ∗ Since (x , y) ∈ Un, we have that (x , z) <φ (x , y), and since (x , y) is in Un, this means that φ(x∗, z) < φ(x∗, y), as desired. 4.2.14 Remark. It follows from Theorem 4.2.13 and Remark 4.2.1 and 4.2.3 ω that if Wadge determinacy holds, Γ is a ∀ω -closed boldface pointclass with a universal set which is closed under unions, P is in Γ\Γˇ and φ is a regular Γ-norm on P , then |φ| = δ(Γ).

4.3 Prewellorderings and wellfounded relations

Theorems 4.3.1 and 4.3.2 connect the length of prewellorderings in a bolface pointclass Γ with the closure of Γ under wellfounded unions. Theorem 4.3.1 (Martin). Assume that AD holds. Let Γ be a nonselfdual bold- ω face pointclass closed under ∃ω and ∧. Let ρ be an ordinal. Suppose that there S exists a prewellordering in Γ of length ρ. Then ρ Γ ⊆ Γ. S Proof. Fix f : ρ → Γ, and let F = α<ρ f(α). We want to see that F is in Γ. Let be a prewellordering in Γ of length ρ. Let X be the domain of ≺, and for each x ∈ X let αx denote the -rank of x. By Theorem 2.4.3, we may let U ⊆ (ωω)2 be a Γ-universal set. Let R be the set of (x, y) ∈ X × ωω for which Uy = f(αx). By the Coding Lemma (Theorem 3.0.1), there is a set S ⊂ R 1 ωω in pos-Σ∼1() (which is contained in Γ, since Γ is ∃ -closed) such that for all ω ω x ∈ ω , if Rx 6= ∅ then Sx 6= ∅. Then for all z ∈ ω , z ∈ F if and only if there exist x ∈ X and y ∈ ωω such that S(x, y) and U(y, z). It follows that F is in Γ. Theorem 4.3.2 (Martin). Assume that AD holds. Let Γ be a nonselfdual ω pointclass closed under ∀ω and ∨, and assume that Γ has the prewellordering property. Let ∆ = Γ ∩ Γˇ. Then ∆ is closed under < δ(Γ)-length unions and intersections. Proof. Since ∆ is closed under complements, it suﬃces to see that it is closed under unions of length less than δ(Γ). Supposing otherwise, let ρ < δ(Γ) be 4.3. PREWELLORDERINGS AND WELLFOUNDED RELATIONS 53

S minimal such that there is exists a function f : ρ → ∆ with α<ρ f(α) 6∈ ∆. Since ∆ is closed under unions, ρ is a limit ordinal. By Theorem 4.2.10, the S S ˇ pointclass ρ ∆ has the prewellordering property. By Theorem 4.3.1, ρ ∆ = Γ. We then have a contradiction to Theorem 4.2.12. A binary relation ≺ on a set X is said to be wellfounded if there is a function f : X → Ord such that f(x) < f(y) whenever x ≺ y. The least ordinal containing the range of such a function is said to the rank of ≺. The following theorem relates the lengths of prewellorderings in certain pointclasses to the ranks of the wellfounded relations in the same class. Theorem 4.3.3 (Moschovakis). Suppose that

Wadge Determinacy + Baire(P(ωω))

ω holds. Let Γ be a nonselfdual boldface pointclass closed under ∀ω and ∨, and suppose that PWO(Γ) holds. Then any wellfounded relation in Γˇ has length less than some prewellordering in Γ ∩ Γˇ.

Proof. Let U¯ = hUn : n < ωi be a sequence of universal sets for Γ with the s-m-n property. Let φ be a regular Γ-norm on U2. Then the range of φ is some δ ≤ δ(Γ), by Remark 4.2.3. Let ≺ be a wellfounded relation in Γˇ on a set X ∈ Γ.ˇ We will show that the rank of ≺ is at most δ. This will suffice, since there are wellfounded relations in Γˇ whose rank is more than that of ≺ (one more, for instance), and the argument will apply to these relations also. Let A be the set of (x, y) ∈ ωω × ωω such that, for all z ∈ X, if z ≺ y then ∗ ω (x, z) <φ (x, y). By Theorem 2.4.8, there is an x ∈ ω such that Ax = U2,x. ω We verify first by induction on the ≺-rank of y ∈ ω that (x, y) ∈ U2. First note that if (x, y) 6∈ U2 and (x, z) is in U2 for all z ≺ y, then (x, y) is in A, which implies that in fact (x, y) is in U2. It follows then that for all y, z ∈ X, if z ≺ y then φ(x, z) < φ(x, y), as desired. In conjunction with Theorem 4.3.5, the following theorem shows that the 2 supremum of the Wadge ranks of the sets in a pointclass of the form Σ∼1(A) (assuming that the Wadge hierarchy is wellfounded) is a regular cardinal. Theorem 4.3.4. Assume that AD holds. Let Γ be a nonselfdual boldface point- ω class closed under ∃ω and ∧. Then the supremum of the ranks of the wellfounded relations in Γ is a regular cardinal.5 Proof. Let κ be the supremum of the ranks of the wellfounded relations in Γ. Clearly, κ is a limit ordinal. Let ρ be the cofinality of κ, suppose that ρ < κ, and let f : ρ → κ be cofinal. Let ≺ be a wellfounded relation in Γ, on a set X ⊆ ωω, of rank ρ, and let be the non-strict part of ≺. For each x ∈ X, let αx be the ≺-rank of x. Applying Theorem 2.4.3, let U ⊆ (ωω)3 be a universal Γ-set, ω and let R be the set of (x, y) ∈ X × ω for which Uy a wellfounded relation of 1 length at least f(αx). By Theorem 3.0.1, there exists a pos-Σ∼1()-set A ⊆ R 5To whom is this due? 54 CHAPTER 4. PROPERTIES OF POINTCLASSES

such that, for all x ∈ X, if Rx 6= ∅ then Ax 6= ∅. Then A is in Γ. Now deﬁne the relation < on (ωω)3 by setting (x, y, z) < (a, b, c) if and only if (x, y) ∈ A, (x, y) = (a, b) and (y, z, c) ∈ U. Then < is a wellfounded relation in Γ of rank κ, giving a contradiction.

A projective algebra is a selfdual boldface pointclass containing all closed subsets of spaces in X and closed under real quantiﬁcation. The following is a weak version of a theorem from [15].

Theorem 4.3.5. Suppose that AD holds, and that W = P(ωω). Let Γ be a non- ω selfdual boldface pointclass closed under ∀ω and ∨, and suppose that PWO(Γ) holds. Let ∆ = Γ ∩ Γˇ, and suppose that ∆ is a projective algebra. Then the following three ordinals are equivalent:

• o(∆);

• δ(Γ);

• The supremum of the ranks of the wellfounded relations in ∆. Proof. Let κ be the supremum of the ranks of the wellfounded relations in ∆. That δ(Γ) ≤ κ follows immediately; that δ(Γ) = κ follows from Theorem 4.3.3. That o(∆) ≤ δ(Γ) follows from Proposition 2.5.8. That δ(Γ) ≤ o(∆) follows from Theorems 2.5.9 and 4.3.2, and Remark 2.5.10.

4.4 Closure under wellordered unions

We include an application of the material in Section 4.2 and 4.3 which will be used in Chapter 6.6 The following is extracted from the proof of Lemma 2.18 of [6]

Lemma 4.4.1. Suppose that AD holds, and that Γ is a nonselfdual boldface ω pointclass closed under ∃ω and ﬁnite intersections, but not wellordered unions. ωω S Let Γ1 be ∃ Γˇ and let κ be the least ordinal γ such that Γ 6⊆ Γ. Then S S γ κ Γ1 = Γ1, from which it follows that κ Γ = Γ1. ˇ S S ωω Proof. By Corollary 2.3.2, Γ ⊆ κ Γ. Since κ Γ is closed under ∃ , the ωω ˇ S pointclass ∃ Γ (which we will call Γ1) is also contained in κ Γ. We will show ﬁrst there is a prewellordering of length κ in Γ1. From this and Theorem 4.3.1 S it will follow that κ Γ1 ⊆ Γ1, from which the rest of the lemma follows. ω ˇ ω 2 1 Let A ⊆ ω be any set in Γ \ Γ. Let S ⊆ (ω ) be a universal Σ∼1 set, and ω ωω let B = {x ∈ ω : Sx ⊆ A}. Since Γˇ is closed under ∀ and ∨, B ∈ Γ,ˇ so S B can be written as α<κ Bα, where each Bα is in Γ. For each α < κ, let A = {y ∈ ωω : ∃x ∈ B y ∈ S }. Then A = S A , and every Σ1 subset of α α x α<κ α ∼1 ω A (being Sx for some x ∈ ω ) is contained in some Aα.

6this has to be improved 4.4. CLOSURE UNDER WELLORDERED UNIONS 55

Let U ⊆ (ωω)2 be a universal Γ set. Let G be the game where I plays x ∈ ωω, II plays y and z, and II wins if and only if x 6∈ A or [ ∃α > |x|(Uy = Aα ∧ z ∈ Aα \ Aβ), β<α

1 where |x| denotes the least γ < κ such that x ∈ Aγ . Since every∼ Σ1 subset of A is contained in some Aα, I cannot have a winning strategy in G. Let τ be a ω winning strategy for II. Given x ∈ ω , let τy(x) and τz(x) denote, respectively, the y and z parts of the response to x (played by I) given by τ. Deﬁne the ω binary relation ≺ on ω by setting x0 ≺ x1 if and only if {x0, x1} ⊆ A and

τz(x1) 6∈ Uτy (x0). Then for all x0, x1 ∈ A, x0 ≺ x1 if and only if, for some α < κ, τz(x0) ∈ Aα and τz(x1) 6∈ Aα. It follows that ≺ is a strict prewellordering in Γˇ of length κ. The relation of being in the same ≺-class is then the intersection of a relation in Γ with one in Γ,ˇ and therefore in Γ1.

Theorem 4.4.2. Suppose that AD holds, and that W = P(ωω). Suppose that S S ∆ is a projective algebra such that ω ∆ 6⊆ ∆. Then ω ∆ is nonselfdual, has the prewellordering property and is closed under wellordered unions.

S ω Proof. Since ∆ is a projective algebra and ω ∆ 6⊆ ∆, there exist Bi ⊆ ω (i ∈ ω) which are collectively Wadge coﬁnal in ∆. Let π : ωω → (ωω)ω be ω a continuous bijection, and let πi (i ∈ ω) be the functions on ω such that ω π(x) = hπi(x): i ∈ ωi for all x ∈ ω . Recalling our coding in Remark2.5.2 of c c ω continuous functions fx by x ∈ F ⊆ ω , for each i ∈ ω let Ci be the set of ω 2 c c (x, y) ∈ (ω ) such that πi(x) ∈ F and fπ (x)(y) ∈ Bi. Since ∆ is a projective S i algebra, each Ci is in ∆. Let C = Ci. We claim that C is universal for S i∈ω S the pointclass ω ∆. To verify this, consider a set i∈ω Ai, where each Ai is in c ∆. Applying CCR, we may ﬁnd Ei and xi (i ∈ ω) such that each xi is in F , each E is (f c )−1[B ] and S A = S E . Letting g : ωω → (ωω)2 be the i xi i i∈ω i i∈ω i −1 S −1 function sending y to (π (hxi : i ∈ ωi), y), we have that Ei = g [C]. S i∈ω Since ω ∆ has a universal set, it is nonselfdual, by Proposition 2.4.2. S ωω Let Γ = ω ∆. Then Γ is closed under ∃ and ∧. It follows now by Theorem 4.2.10 that Γ has the prewellordering property. By the First Periodicity ω Theorem (see Remark 4.2.6), ∀ω Γ does as well. We will now show (extracting the relevant part of the proof of Lemma 2.18 of [6]) that Γ is closed under wellordered unions. ωω Supposing otherwise, let Γ1 = ∃ Γˇ and let κ be the least ordinal γ such S S S that Γ 6⊆ Γ. By Lemma 4.4.1, Γ = Γ1 = Γ1. Let ∆1 = Γ1 ∩ Γˇ1. γ κ κ S Let ρ be the least ordinal γ (necessarily at most κ) such that ∆1 6⊆ ∆1. S γ S Since ρ ∆1 properly contains ∆1 and is contained in Γ1,Γ1 = ρ ∆1. It follows from Theorem 4.2.10 that Γ1 has the prewellordering property. Since Γˇ1 has the prewellordering propery (as observed above) this contradicts Theorem 4.2.12.

Lemma 2.4 of [15]. 56 CHAPTER 4. PROPERTIES OF POINTCLASSES

ω Proposition 4.4.3. Assume7. Let Γ be a ∃ω -closed boldface pointclass which ω is closed under countable unions and intersections but not ∀ω -closed. If Red(Γ) holds, then Γ is closed under wellordered unions.8 Proof. Working toward a contradiction, let κ be the least ordinal γ such that for some f : γ → Γ, S f(α) 6∈ Γ. Then κ is a regular uncountable cardinal. α<γ S Fix such a function f, and let F = α<κ f(α). By Remark 2.2.4 and Theorem S ˇ ωω ˇ 2.2.6, the boldface pointclass κ Γ contains both Γ and Γ. Letting Γ1 = ∃ Γ, S ωω we have also that Γ ⊆ Γ1 ⊆ Γ, in the latter case because Γ is ∃ -closed. κ S Letting ∆1 = Γ1 ∩ Γˇ1, we have that Γ ⊆ ∆1, so ∆1 contains Γ1, which S κ S properly contains ∆1. Let ρ be minimal such that ρ ∆1 6⊆ ∆1. Then ρ ∆1 has the prewellordering property, by Proposition 4.2.10. S S 9 By Lemma 4.4.1, Γ1 = κ Γ1. As above it equals ρ ∆1. Let δ1 be the supremum of the ranks of the wellfounded relations in Γ1. By Proposition 4.3.3, δ1 is at most the supremum of the lengths of the prewellorderings in ∆1. It follows that ρ ≥ δ1, since otherwise, by Proposition 4.2.10, Γ1 has the prewellordering property, contradicting the fact that Γˇ1has the reduction property, by the 0th Periodicity theorem (Theorem 4.1.8). Let δ be the supremum of the ranks of the wellfounded relations in Γ. Let U ⊆ (ωω)3) be a Γ-universal set, and let W be the set of x ∈ ωω for which ω 2 Ux is a wellfounded relation. Then the relation < on (ω ) defined by setting (x, y) < (w, z) if and only if x = w, x ∈ W and (y, z) ∈ Ux is in Γ1, which shows (since Γ1 is closed under unions which implies that δ1 is a limit ordinal) that 10 δ < δ1. It follows that κ > δ. S Now let ≺ be a wellfounded relation in Γ1. Since Γ1 ⊆ Γ, we can write S κ ≺ as α<κ ≺α, where each ≺α is in Γ. By the minimality of κ we may assume that α < β < κ implies that ≺α⊆≺β. For each x in the field of ≺, and each α < κ, let gx(α) be 0 if x is not in the field of ≺α, and its rank in ≺α otherwise. Then each function gx is nondecreasing. Since δ < κ and κ is regular, each gx has an eventually constant value ξx. If x ≺ y, then ξx < ξy, which means that the rank of ≺ is at most δ, showing that δ1 ≤ δ, giving a contradiction.

7at least Wadge Determinacy and Baire(P(ωω)) 8You may have to say/show that Red follows from PWO, or something else. 9We need to see that PWO implies reduction. 10this argument appears in Kechris ’74 Chapter 5

Strong Partition Cardinals

Given an ordinal α and a set X of ordinals, [X]α denotes the collection of subsets δ of X of ordertype α. Given ordinals α, β, δ and γ the formula α → (β)γ asserts that for every partition of [α]δ into γ many pieces, there exists an X ∈ [α]β such that [X]δ is contained in one piece. We write α → (β)δ when γ = 2. A strong κ partition cardinal is a regular uncountable cardinal κ such that κ → (κ)µ holds for all µ < κ.

5.0.1 Remark. Some authors define a strong partition cardinal using the os- tensibly weaker property κ → (κ)κ. We do not know whether ZF implies that these two definitions are equivalent. However, if κ is a regular uncountable car- κ µ dinal such that κ → (κ) , then ∀µ < κ(κ → (κ)µ) holds. This can be seen, for instance, by considering, given an f :[κ]µ → µ, the function f 0 :[κ]κ → 2 defined by setting f 0(X) = 0 if and only if f(a) = f(b), where a is the initial segment of X of ordertype µ, and b is the initial segment of X \ a of ordertype µ µ. The property ∀µ < κ(κ → (κ)µ) is enough for our application of strong partition cardinals in Theorem 7.0.3.

Our aim in this chapter is to prove the following theorem.

Theorem 5.0.2. ([12]) If AD holds then the strong partition cardinals are co- ﬁnal below Θ.

2 We do not know whether AD implies that cardinals of the form ∼δ 1(A) are strong partition cardinals. One problem is that we do not know how to show that 2 the pointclass∼ Σ1(A) has the prewellordering property without some additional assumption. In Remark 4.2.9 we show this using the additional assumption that the Wadge hierarchy is wellfounded. A second problem is that we do know 2 ωω whether AD implies that Σ∼1(A) is ∀ -closed, although it is in many natural models of AD. Adding these two properties as hypotheses, we get the following.

Theorem 5.0.3. If AD holds and W = P(ωω), then for each A ⊆ ωω for which ω Σ2(A) is ∀ω -closed, δ 2(A) is a strong partition cardinal. ∼1 ∼1

57 58 CHAPTER 5. STRONG PARTITION CARDINALS

2 To get around these issues, we relativize Σ∼1(A) to the inner model L(A, R). 2 We write loc-Σ∼1(A) for 2 L(A,R) ∼Σ1(A) , loc-∆2(A) for ∆2(A)L(A,R) and loc-δ 2(A) for δ 2(A)L(A,R). As with Σ2(A), each ∼1 ∼1 ∼1 ∼1 ∼1 2 ωω pointclass loc-Σ∼1(A) is nonselfdual, has a universal set and closed under ∃ and ∧.

2 5.0.4 Remark. To see that the pointclasses loc-Σ∼1(A) satisfy the prewellorder- ω 2 ing property (without assuming that W = P(ω )), ﬁx a Σ1(A) set X of the form

{x ∈ ωω : L(A, ωω) |= ∃B ⊆ ωω ψ(A, B, x)}.

Orders ≤∗ and <∗ witnessing that X has the prewellordering property can be ω defined using the least ordinal α such that Lα(A, ω ) contains a set B as in the definition of X. The Solovay Basis Theorem (Theorem 5.0.5) shows that this 2 gives a loc-Σ∼1(A)-norm on X. The Solovay Basis Theorem (Theorem 5.0.5 below) says that for each A ⊆ ω 2 ω ω , loc-∆∼1(A) is Σ1-elementary in P(ω ) ∩ L(A, R). The proof of the theorem, which we leave to the reader, follows from the fact that in models of the form L(A, R), every set is ordinal definable from A and a member of ωω, which implies that every set has an elementary submodel which is the surjective image of ωω. This means that there is a minimal (relative to a parameter from ωω) 2 witness to each true Σ∼1(A) statement, and that that minimal witness is in 2 L(A,R) 2 loc-∆∼1(A) . In particular, membership in loc-Σ∼1(A) sets is witnessed by 2 loc-∆∼1(A) sets. Using the Solovay Basis Theorem, it is not hard to show that 2 ωω pointclasses of the form loc-Σ∼1(A) are ∀ -closed. Theorem 5.0.5 (Solovay Basis Theorem). Let A be a subset of ωω and let x and y be an element of ωω. Let φ be a formula whose quantifiers range over the hereditarily countable sets. If there exists a set B in P(ωω) ∩ L(A, R) such that 2 φ(A, B, x, y) holds, then there exists such a set B in loc-∼∆1(A). The following gives Theorem 5.0.2.

2 Theorem 5.0.6. If AD holds then loc-∼δ 1(A) is a strong partition cardinal, for each A ⊆ ωω.

Our proofs of Theorems 5.0.3 and 5.0.2 diverge at just one point, establishing that each of the relevant pointclasses satisﬁes the prewellordering property (see Section 4.2). The prewellordering property is part of a very deep and general theory of boldface pointclasses under AD, but we are mostly interested in taking the shortest path to proving the existence of strong partition cardinals. Readers interested in this general theory are directed to [13, 14] for an introduction. Much of this chapter is taken from [6] and conversations with its author. 5.1. THE MARTIN CONDITIONS 59

5.1 The Martin Conditions

We complete the proof of Theorem 5.0.6 (so also Theorem 5.0.2) by presenting an argument of Martin connecting closure properties of pointclasses with partition properties of their associated ordinals. The following definition (a slightly modified version of Definition 2.33 on page 1783 of [6]) is due to Martin; the properties that it lists are known as the Martin conditions.

5.1.1 Deﬁnition. Let λ ≤ κ be ordinals. We say that κ is λ-reasonable if there ω exist a non-selfdual boldface pointclass Γ closed under ∀ω and a function

φ: ωω → P(λ × κ) such that, letting ∆ = Γ ∩ Γ,ˇ

1. for each function F : λ → κ there is an x ∈ ωω such that φ(x) = F ;

2. for each pair (β, γ) ∈ λ × κ, the set

ω Rβ,γ = {x ∈ ω : {η < κ :(β, η) ∈ φ(x)} = {γ}}

is in ∆;

ωω 3. for each β < λ and each A in ∃ ∆ ∩ P(Rβ) there exists a γ0 < κ such that [ A ⊆ Rβ,γ ,

γ<γ0 S where Rβ = {Rβ,γ : γ < κ}.

Given a function φ and sets Rβ (β < λ) as in Deﬁnition 5.1.1, φ(x) is a T function from λ to κ for each x ∈ β<λ Rβ. In Theorem 5.1.2 and 5.1.3 below, we assume that Γ has the prewellordering property, which may make these theorems less than optimal. Versions without this additional hypothesis appear in [6].1

Theorem 5.1.2 (Martin). Suppose that AD holds. Let Γ be a nonselfdual bol- ω face pointclass closed under ∀ω and ∨, and assume that Γ has the prewellordering property. Let ∆ = Γ ∩ Γˇ, and suppose that ∆ is a projective algebra. Let κ be δ(Γ) and suppose that κ is a regular cardinal. Let λ be an ordinal such that ω · λ ≤ κ. If Γ witnesses that κ is ω · λ-reasonable, then κ → (κ)λ.

Proof. Let φ, Rβ,γ (β < λ, γ < κ) and Rβ (β < λ) be as in Deﬁnition 5.1.1, with respect to ω · λ, κ and Γ. Fix P :[κ]λ → 2. Consider the integer game in which player I plays the values of x ∈ ωω and player II plays the values of y ∈ ωω.

1Really? 60 CHAPTER 5. STRONG PARTITION CARDINALS

I x(0) x(1) x(2) ... II y(0) y(1) ...

If there is a least ordinal β < ω · λ such that {x, y} 6⊆ Rβ, then II wins provided that x 6∈ Rβ. Otherwise, letting fx,y : λ → κ be deﬁned by setting fx,y(β) to be

sup{max(φ(x)(η), φ(y)(η)) : η < ω · (β + 1)},

II wins if and only if P (fx,y) = 1. Suppose first that τ is a winning strategy for player II. Fix for this para- T S graph a pair (β, γ) in (ω · λ) × κ. Let Sβ,γ = δ≤β η≤γ Rδ,η. Then Sβ,γ is in ∆ by Theorem 4.3.2. Therefore, the set Sβ,γ ◦ τ = {x ◦ τ : x ∈ Sβ,γ } is in ∆, T as ∆ is a projective algebra. We have that Sβ,γ ⊆ δ≤β Rβ, which implies that Sβ,γ ◦ τ ⊆ Rβ. Letting θ(β, γ) be sup{φ(y)(β): y ∈ Sβ,γ ◦ τ}, we have by part (3) of Definition 5.1.1 that θ(β, γ) < κ. Let C0 be the set of elements of κ closed under θ, and C be any cofinal subset of C0 (for the present theorem it suffices to work with C0, but the fact that the rest of the argument works with any cofinal C ⊆ C0 will be used later in this section). Let C0 be the set of γ ∈ C for which ot(γ ∩ C) has the form η + ω, for some ordinal η. Suppose now that F : λ → C0 is increasing; we want to see that P (F ) = 1. Applying part 1 of Definition reasonabledef, let x be such that φ(x) is an increasing function, with range contained in C, such that F (β) = sup{φ(x)(δ): δ < ω · (β + 1)} for all β < ω · λ. Let y = x ◦ τ; then y T is in β<ω·λ Rβ. It suffices to see that φ(y)(β) ≤ φ(x)(β + 1) for all β < ω · λ, since then F = fx,y, so P (F ) = 1. Fix β < ω · λ. We have that x is in Sβ,φ(x)(β), so y is in Sβ,φ(x)(β) ◦ τ. Then φ(y)(β) ≤ θ(β, φ(x)(β)) < φ(x)(β + 1), since φ(x) is increasing and φ(x)(β + 1), being in C, is closed under θ. A winning strategy for player I can be used as a winning strategy for player II in the game where the values of P are reversed. It follows that if there is a winning strategy for player I, then there is homogeneous set for value 0.

As Remarks 4.2.8 and 5.0.4 and Theorems 4.3.3, 4.3.4 and 4.3.5 together 2 show that the pointclasses loc-Σ∼1(A) satisfy the hypotheses of Theorem 5.1.3 2 (and Remark 4.2.9 shows that the pointclasses∼ Σ1(A) do, if the Wadge hierarchy is assumed to be wellfounded), Theorem 5.1.3 gives the versions of Theorems 5.0.3 and 5.0.6 for the weaker notion of strong partition cardinal discussed in Remark 5.0.1.

Theorem 5.1.3. Assume that AD holds. Let Γ be a nonselfdual bolface point- ω class with the prewellordering property, closed under ∀ω and ∨. Let ∆ = Γ∩Γˇ, and suppose that ∆ is a projective algebra. Let κ be δ(Γ) and suppose that κ is a regular cardinal. Then κ → (κ)κ. 5.1. THE MARTIN CONDITIONS 61

Proof. By Theorem 5.1.2, it suﬃces to see that κ is ω·κ-reasonable. By Theorem 4.3.5, o(∆) = δ(Γ). By Theorem 4.2.13 and Remark 4.2.3, there exist a set X ⊆ ωω in Γ \ Γˇ and a surjection f : X → κ such that, for each x ∈ X, the set

{y ∈ X : f(y) ≤ f(x)} is in ∆. Let, for each x ∈ X,

[x]f = {y ∈ X : f(y) = f(x)},

[< x]f = {y ∈ X : f(y) < f(x)} and ω Cx = ω \ ([x]f ∪ [< x]f ). By Theorem 4.3.2, each of these sets is in ∆. ¯ 1 Fix a sequence U = hUn : n < ω \{0}i of∼ Σ1-universal sets with the s-m-n property. Deﬁne φ: ωω → P(κ×κ) by setting, for each z ∈ ωω and (β, γ) ∈ κ×κ, (β, γ) ∈ φ(z) if and only if there exist x, y in X with

• f(x) = β,

• f(y) = γ and

• (x, y) ∈ U2,z([x]f ,Cx).

• for all (v, w) ∈ U2,z([x]f ,Cx), f(w) = γ. The Uniform Coding Lemma (Theorem 3.0.3) implies that part (1) of Deﬁ- nition 5.1.1 is satisﬁed. To see this, given F : κ → κ, let ZF be

{(x, y) ∈ X × X : F (f(x)) = f(y)}.

By the Uniform Coding Lemma, there is a z ∈ ωω such that, for all x ∈ X,

ω 1. U2,z([x]f ,Cx) ⊆ ZF ∩ ([x]f × ω ),

ω 2. U2,z([x]f ,Cx) 6= ∅ if and only if ZF ∩ ([x]f × ω ) 6= ∅. It follows from this that φ(z) = F . To see that part (2) of Definition 5.1.1 is satisfied, fix (β, γ) ∈ κ × κ. By the ω definition of Rβ,γ , for each z ∈ ω , z ∈ Rβ,γ implies (β, γ) ∈ φ(z). The reverse implication follows from last condition on the definition of φ. It suffices then to see that {z ∈ ωω :(β, γ) ∈ φ(z)} is in ∆. This can be seen by fixing x∗, y∗ ∈ X such that f(x∗) = β and f(y∗) = γ, and applying the fact that ∆ is a projective algebra. ωω For part (3), fix β < κ and A ⊆ Rβ with A in ∃ ∆ (which is the same as ∆). Fix x∗ ∈ X with f(x∗) = β. Let D be the set of y ∈ ωω for which there ∗ ∗ exist z ∈ A and x ∈ [x ]f with (x, y) ∈ U2,z([x ]f ,Cx∗ ). Then D ⊆ X and D ∈ ∆, so f[D] is a bounded subset of κ, by Lemma 4.2.4. 62 CHAPTER 5. STRONG PARTITION CARDINALS

We now finally finish the proofs of Theorems 5.0.3 and 5.0.6. Given ordinals µ, κ, let us define (µ, κ)-club directedness to be the statement that if hCα : α < µi is a sequence of nonempty sets of club subsets of κ, then there is a club subset of κ contained in at least one member of each Cα. The proof of Theorem 5.1.2 (i.e., the fact that the proof worked with any cofinal C ⊆ C0) shows that, for each λ µ < κ, the statement κ → (κ)µ follows from the hypotheses of that theorem, if one adds to the hypotheses the statement that (µ, κ)-club directedness holds. We derive this latter property from the Martin conditions and the Moschovakis Coding Lemma. Theorem 5.1.4. Suppose that AD holds. Let Γ be a nonselfdual bolface point- ω class closed under ∀ω and ∨, and assume that Γ has the prewellordering property. Let κ a regular uncountable cardinal. If Γ witnesses that κ is ω · κ- reasonable, then, for all µ < κ, (µ, κ)-club directedness holds.

Proof. Let φ, Rβ,γ (β < λ, γ < κ) and Rβ (β < λ) be as in Deﬁnition 5.1.1, with respect to ω · κ, κ and Γ. Fix µ < κ, and let hCα : α < µi be a sequence S of nonempty sets of club subsets of κ. Let X be α<µ R0,α, and let f : X → µ be such that x ∈ R0,f(x) for all x ∈ X. By Theorem 4.3.2, the corresponding relations =f and

Then the closure points of g form a club subset of κ contained in at least one member of each Cα. Chapter 6

Suslin sets and Uniformization

Recall that a tree T on a set X is a set of finite sequences from X which is closed under initial segments. We write [T ] for the set of infinite sequences whose finite initial segments are all in T . If T is a tree on a product of two sets X × Y , and <ω s is a finite sequence of elements of X, then Ts denotes the set of t ∈ Y for ω S which (s, t) ∈ T . For each f ∈ X , we write Tf for n∈ω Tfn. The projection ω of T , p[T ], is the set of f ∈ x for which Tf is illfounded, i.e., for which there exists a g ∈ Y ω such that (f, g) ∈ [T ] (identifying, for notational simplicity, pairs of sequences with sequences of pairs). If T is an illfounded tree on a set X, and ≤ is a wellordering of X, the ω leftmost branch of T relative to ≤ (lb≤(T )) is the unique f ∈ X such that, for each n ∈ ω, f(n) is the ≤-least member of X for which the tree

<ω _ {σ ∈ X :(fn) σ ∈ T } is illfounded. When X is an ordinal or a ﬁnite product of ordinals we use the ordinal ordering or the corresponding lexicographical ordering for ≤ and write lb(T ).

6.0.1 Deﬁnition. Given ordinals γ and η, a set A ⊆ ηω is γ-Suslin if there exists a tree T ⊆ (η × γ)<ω such that A = p[T ]. A set A ⊆ ηω is Suslin if it is γ-Suslin for some ordinal γ.

6.0.2 Remark. Every Suslin subset of ωω is γ-Suslin for some γ < Θ. To see this, suppose that α is an ordinal and T is tree on ω × α, and observe that the ω set {lb(Tx)(n): x ∈ p[T ], n ∈ ω} is a surjective image of ω , and therefore has ordertype less than Θ. Alternately one can use the fact that (assuming only ZF) if β is an ordinal with T ∈ Lβ(T, R), then Lβ(T, R) has an elementary submodel which contains ωω and is a surjective image of ωω.

Given an ordinal γ, we write Sγ for the smallest boldface pointclass contain- ω S ing each γ-Suslin subsets of ω and S<γ for α<γ Sα. Then Sγ and S<γ are ω boldface pointclasses closed under ∃ω , intersections and unions.

63 64 CHAPTER 6. SUSLIN SETS AND UNIFORMIZATION

6.0.3 Remark. If γ is less than Θ, and AD holds, then, by the Coding Lemma, CCP(γ) holds, so Sγ is closed under countable intersections and countable unions. The following definition is due to Kechris. 6.0.4 Definition. A cardinal κ is Suslin if it is infinite and there exists A ⊆ ωω which is κ-Suslin but not γ-Suslin for any γ < κ.

In symbols, κ is Suslin if and only if Sκ \S<κ 6= ∅. By Remark 6.0.2, every Suslin cardinal is less than Θ. By Remark 6.0.3, AD implies that a limit of Suslin cardinals of countable coﬁnality is a Suslin cardinal if it is below Θ (more is true, as well shall see in Section 11.4). The following theorem will be used in the proof of Theorem 6.0.7 below. Theorem 6.0.5. Assume that AD holds, and let Γ be a boldface selfdual point- ω class closed under ∃ω , and let κ be an inﬁnite cardinal such that there exists a ω prewellordering of ω of length κ in Γ. Then Sκ ⊆ Γ.

ω Proof. We ﬁx an A ∈ Sκ ∩ P(ω ) and show that A ∈ Γ. Replacing κ with a smaller cardinal if necessary we may assume that A is not λ-Suslin for any λ < κ. Let be a prewellordering of ωω in Γ of length κ. Using a bijection between κ × ω and ω if necessary, we may ﬁx a tree T ⊆ (ω × κ)<ω such that A = p[T ] and such that there is a function u: κ → T such that for each α ∈ κ, u(α) is the unique node (s, t) of T for which α is the last value taken by t. Let f : X → κ be the function which sends each member of X to its -rank. Let π : ωω → (ω<ω × (ωω)<ω) be a recursive bijection. Let Z be the set of ω (x, y) ∈ X × ω such that π(y) is a tuple (s, z0, . . . , z|s|−1) for some nonempty s ∈ ω<ω, such that

• s is the ﬁrst coordinate of u(f(x))),

• for all positive n < |s|, sn is the ﬁrst coordinate of u(f(zn−1))). By the Coding Lemma, there is a set R ⊆ Z such that for all x, if there exists a y with (x, y) ∈ Z, then there exist (x0, y0) ∈ R with f(x) = f(y). Then A is the ω ω ω set of w ∈ ω for which there exist hqi : i ∈ ωi ∈ (ω ) such that for all n ∈ ω there exist (x, y) ∈ R such that, letting π(y) = (s, z0, . . . , z|s|−1), s = wn and, for each m < n, f(zm) = f(qm). This shows that A is in Γ. 6.0.6 Deﬁnition. Let γ be an ordinal and let T be a tree on ω × γ. The minimization of T is the tree

{(xn, lb(Tx)n): x ∈ p[T ], n ∈ ω}. We say that T is minimal if it is equal to its minimization and γ-full if

{h(0, α)i : α < γ} ⊆ T.

We note that the minimization of a tree T as in Deﬁnition 6.0.6 is minimal and has the same projection as T . 65

Theorem 6.0.7 (Kechris [10]). Assume that AD holds.1 If κ is a Suslin cardi- 2 nal, then Sκ is not closed under complements.

Proof. Assume toward a contradiction that Sκ is closed under complements. ωω Since Sκ is (always) ∃ -closed, it follows from this assumption that it is also ωω ∀ -closed. Since Sκ is closed under countable unions, it follows from Remark 4.2.11 that it then has the prewellordering property. By Theorem 4.3.5 and the assumption that Sκ is closed under complements, the supremum of the Wadge ranks of the members of Sκ is the same as the supremum of the lengths of the prewellorderings in Sκ. Call this ordinal λ. ω 1 1 For each A ⊆ ω in Sκ,Σ∼1(A) and Σ∼2(A) are properly contained in Sκ. By Remark 4.1.5 and Theorems 4.1.8 and 4.4.3, one of these two pointclasses (for each A) is closed under wellordered unions. It follows then that for each ρ < λ, any welloredered union of sets from Sκ of Wadge rank less than λ is in Sκ. ω 1 1 Let A be a subset of ω in Sκ\S<κ and let Γ be the member of {∼Σ1(A), ∼Σ2(A)} which is closed under wellordered unions. Then S<κ is contained in Γ, since every member of S<κ is Wadge-reducible to A. If the cofinality of κ were uncountable, it would follow that Sκ ⊆ Γ, giving a contradiction. To see this, let <ω T be a tree on ω ×κ, and for each α < κ, let T α denote T ∩(ω ×α) . If κ has S uncountable cofinality, then p[T ] = α<κ p[T α], and the latter is a wellordered union of sets in S<κ. To complete the proof, we show that κ has uncountable cofinality. Since Sκ is closed under countable unions, λ has uncountable cofinality. It suffices then to show that λ = κ. By Theorem 6.0.5, λ ≤ κ. To show that κ ≤ λ, let T be a minimal tree on ω × κ projecting to A. Then T has cardinality κ. For each σ ∈ T , let Aσ be the set of x ∈ A for which σ is an initial segment of (x, lb(Tx). For each ρ < λ, let T ρ be the set of σ ∈ T for which the Wadge rank of Aσ is less than ρ. For each n ∈ ω and ρ < λ, let

6.0.8 Remark. If Wadge Determinacy and Baire(P(ωω)) hold, and κ < λ are Suslin cardinals, then, by Corollary 2.3.2, Sˇκ ⊆ Sλ. It follows (under the same assumption) that if κ is a limit of Suslin cardinals, then S<κ is a projective algebra. If AD holds, λ is a Suslin cardinal and ∆ is a projective algebra S contained in Sλ, then Sλ contains ω ∆ (see Remark 6.0.3). If in addition ∆ is not closed under countable unions then, by Theorem 4.4.2, every wellordered union of sets in ∆ is in Sλ.

1And more, e.g., W = P(ωω)? 2 ω Does this argument show that for every A ∈ Sκ \S<κ, ω \ A 6∈ Sκ? 66 CHAPTER 6. SUSLIN SETS AND UNIFORMIZATION

6.0.9 Remark. By the Moschovakis Coding Lemma, if AD holds then for each κ < Θ is there is a subset of ωω which is not κ-Suslin. It follows that if every subset of ωω is Suslin, then the Suslin cardinals are coﬁnal in Θ. The converse is shown in Theorem 6.0.13 below. Lemma 6.0.11 below shows that if λ is a Suslin cardinal, then there exists a strictly ⊆-increasing λ-sequence of Suslin sets, induced by a single tree on ω × λ.3 2 Let <` be the following version of the lexicographical order on ﬁnite se- 2 quences of pairs of ordinals : (s0, t0) <` (s1, t1) if and only if there exists an i ∈ ω such that the following hold:

• (s0i, t0i) = (s1i, t1i);

• s0(i) < s1(i) ∨ (s0(i) = s1(i) ∧ t0(i) < t1(i)). If A is a set of pairwise incompatible sequences, and n ∈ ω is such that each 2 member of A has length at most n, then the restriction of <` to A wellorders A. Lemma 6.0.10. Let γ and η be ordinals, let T be a minimal tree on ω × γ, and let A ⊆ T be an antichain such that

• every element of [T ] intersects A; 2 • the restriction of <` to A has ordertype η. Then there is a minimal η-full tree T 0 on ω × max{γ, η}, deﬁnable from A and T , such that p[T ] = p[T 0].

2 Proof. Enumerate A in <` -order as h(sα : tα): α < ηi. For each α < η, let

• Tα be the set of (s, t) ∈ T which are compatible with (sα, tα); <ω <ω • fα : γ → max{γ, η} be deﬁned by setting

– |fα(t)| = |t|;

– for all i ∈ |t| ∩ |tα|, fα(t)(i) = α;

– for all i ∈ |t| \ |tα|, fα(t)(i) = t(i);

0 • Tα be {(s, fα(t)) : (s, t) ∈ Tα}. 0 S 0 0 Let T = α<η Tα. Then T is η-full, and since p[Tα] = p[Tα] for all α < η, and every element of [T ] intersects A, p[T 0] = p[T ]. To see that T 0 is minimal, fix 0 0 0 0 0 0 (s , t ) ∈ T and α < η such that (s , t ) ∈ Tα. It suffices to consider the case 0 0 0 where |s | ≥ |sα|. Let (s, t) ∈ Tα be such that s = s and t = fα(t). Since T is minimal, there exists an x ∈ p[Tα] such that (s, t) is an initial segment of 2 (x, lb(Tx)). Applying the definition of <` we have that for all β < α, x 6∈ p[Tβ], 0 0 0 0 0 so x 6∈ p[Tβ]. Then lb(Tx) = lb(Tα,x), and t is an initial segment of lb(Tx) as desired. 3Irrelevant question : does the conclusion of Lemma 6.0.11 imply that λ is a Suslin cardinal? 67

Lemma 6.0.11. If λ is a Suslin cardinal, there is a minimal λ-full tree on ω × λ.

ω Proof. Let T0 be a tree on ω × λ projecting to a subset of ω which is not γ-Suslin for any γ < λ, and such that each member of p[T0] takes value 0 at 0. Let T1 be the minimization of T0. Since T0 witnesses that λ is Suslin, |T1| = λ.

{(xn, lb(T0,x)n): x ∈ p[T0], n ∈ ω}. 2 Then |T1| = λ and T1 is minimal. If some level of T1 has <` -ordertype at least 2 0 λ, let A be the length-λ <` -initial segment of such a level, and let T1 be the set 0 of nodes of T1 compatible with a member of A. Then T1 is minimal, and the 0 lemma follows by applying Lemma 6.0.10 with T1 and A. Suppose then that T1 does not contain such an antichain. For each n ∈ ω, let κn be the cardinality of the set of nodes of T1 of length n. Then each κn is less than λ, and supn∈ω κn = λ. We can then complete the proof by using (for each n ∈ ω) the nth level to produce a minimal κn-full tree on ω × κn, modifying the trees to make their domains disjoint, and combining these trees to make a minimal λ-full tree on ω × λ. We give the details below. 2 For each n ∈ ω, let An be the length-κn <` -initial segment of the nth level of T1, and let T1,n be the set of nodes of T1,n compatible with a member of An. Then T1,n is minimal; let T2,n be the result of applying Lemma 6.0.10 to T1,n and An. ∗ Let π : ω × ω → ω be an injection, and for each n ∈ ω let π0,n be the length- <ω ∗ preserving function on ω deﬁned by setting π0,n(s)(0) to be s(0) (in the case ∗ where s is nonempty) and π0,n(s)(i) = π(n, i) for all nonzero i ∈ dom(s). Again P ∗ for each n ∈ ω, let γn = m

∗ ∗ {(π0,n(s), π1,n(t)) : n ∈ ω, (s, t) ∈ T2,n}.

Then T2 is as desired. 6.0.12 Remark. Suppose that AD holds, λ is an ordinal and that T is a minimal λ-full tree on ω × λ. For each α < λ, let

• Tα be the set of σ ∈ T which are either empty or have (0, β), for some β < α as their ﬁrst element;

• Aα = p[Tα].

Then hAα : α < λi is a strictly ⊆-increasing sequence of λ-Suslin sets. Moreover, S for each pair α ≤ β below λ, Aα × Aβ is λ-Suslin. Let ≤= α≤β<λ Aα × Aβ. Then ≤ is a prewellordering of length λ. Let ∆ be a projective algebra containing S 4 S Sλ such that ω ∆ 6⊆ ∆. If AD holds, then, by Theorem 4.4.2, ≤ is in ω ∆.

4 By Kechris’s theorem that Sκ is not closed under complements (Theorem 6.0.7), the smallest projective algebra containing Sλ works. 68 CHAPTER 6. SUSLIN SETS AND UNIFORMIZATION

Still assuming AD, if κ is a Suslin cardinal and ∆ ⊆ Sκ (which holds by Remark 6.0.8 if there are inﬁnitely many Suslin cardinals between λ and κ), then ≤ is in Sκ, by Remark 6.0.3. Theorem 6.0.13. If AD holds, and the Suslin cardinals are coﬁnal in Θ, then every subset of ωω is Suslin.

Proof. Suppose towards a contradiction that there is a subset of ωω which is not Suslin. It follows then by Wadge Determinacy that there is a set A ⊆ ωω such that every Suslin set is Wadge reducible to A. Let ∆ be the smallest projective S algebra with A as a member. Then ω ∆ is also a proper initial segment of the Wadge hierarchy, and S ∆ 6⊆ ∆.5 It follows from the ﬁrst paragraph of ω S Remark 6.0.12 that for each Suslin cardinal κ there is a prewellordering in ω ∆ of length κ, contradicting the deﬁnition of Θ.

6.1 Uniformization

Given sets X, Y and A ⊆ X ×Y , we say that a function f : X → Y uniformizes A if dom(f) = {x ∈ X | Ax 6= ∅}, and (x, f(x)) ∈ A for all x ∈ dom(f). We say then that A is uniformized. If A ⊆ αω × βω is the projection of a tree T on α × β × γ, for some ordinals α, β and γ, then a uniformizing function for A can be constructed from T by considering T as a tree on α × (β × γ) and using the function x 7→ lb(Tx) for x ∈ p[T ]. This gives the following fact. Theorem 6.1.1. Every Suslin subset of (ωω)2 is uniformized.

We let Uniformization be the statement that every subset of (ωω)2 has a uniformizing function. Note that Uniformization easily implies DCR. 6.1.2 Deﬁnition. Given a set X, a set A ⊆ Xω is quasi-determined (as a subset of Xω) if there is a function π : X<ω → P(X) \ {∅} such that one of the two following statements holds.

ω 1. For every x ∈ X , if x(2n) ∈ π(x2n) holds for all n ∈ ω, then x ∈ A.

ω 2. For every x ∈ X , if x(2n + 1) ∈ π(x(2n + 1)) hold for all n ∈ ω, then x 6∈ A.

A function π : X<ω → P(X)\{0} is then said to be a quasi-strategy in the game GA. Such a function π as in case (1) above is a winning quasi-strategy for player I; in case (2) it is a winning quasi-strategy for player II. We let quasi-ADX denote the statement that every subset of Xω is quasi-determined, as a subset of Xω.

If X is wellorderable, then quasi-ADX and ADX are equivalent. Remark 6.1.3 shows that a form of uniformization suﬃces to prove this.

5Say why, for both parts of the sentence? 6.2. THE KUNEN-MARTIN PROPERTY 69

6.1.3 Remark. If X is a set, and every subset of X<ω × X is uniformized, then every quasi-determined game on X is determined. As quasi-ADR implies Uniformization (via a game where I plays a real, and II plays the inﬁnite string of coordinates of a partner real), quasi-ADR and ADR are equivalent.

Given sets t1, . . . , tn, we let ODt1,...,tn denote the class of sets which are definable from t1, . . . , tn and finite sequence of ordinals (i.e., which are ordinal definable from t1, . . . , tn). For each finite sequence t1, . . . , tn, there is a definable class-length wellordering of ODt1,...,tn . Theorem 6.1.4 (Kechris-Solovay). Suppose that there is a set Z such that every subset of ωω is ordinal definable from Z and a member of ωω, and that ω ω ω 2 for all x ∈ ω , ω 6⊆ ODZ,x. Then the set {(x, y) ∈ (ω ) | y 6∈ ODZ,x} is not uniformizable. Proof. Any uniformizing function would be ordinal definable from Z and some ω x ∈ ω , which would mean that f(x) ∈ ODZ,x. The following theorem will be used in the proof of Corollary 9.1.6, which says that, assuming AD, every set of reals in L(R) is ∞-Borel.6 Theorem 6.1.5 (Martin-Steel[25]). Assuming AD + V =L(R), the Suslin sets 2 are exactly the ∼Σ1 sets. 2 Theorem 6.1.4 can be used to show that, in L(R), not all Π∼1 sets are Suslin.

6.2 The Kunen-Martin property

1 ω Every Σ∼1 prewellordering of ω has countable length (see [27, 17], for instance). Applying this fact in a forcing extension by Col(ω, κ), we get the following. Theorem 6.2.1 (Kunen-Martin). Every κ-Suslin prewellordering has length less than κ+.

6.3 A pointclass fact

6Isn’t it also used to reflect Ordinal Determinacy? 7Probably none of this is needed, and we should just move the rest of this section somewhere else : Suppose that ∆ is a boldface pointclass, and that {Ai : i < ω} ⊆ ∆ is Wadge-cofinal ∗ ω in ∆. Let A be the set of x ∈ ω such that hx(1), x(2),...i ∈ Ax0 . We say that ∆ has the + 1 Kunen-Martin property if w(A) = δ1 (A). Becker [1] showed that if ∆ is a projectively closed pointclass with countable Wadge cofinality, every set in ∆ is uniformized by a set in ∆, and ∆ satisfies the Kunen-Martin property, then δ∆ is a Suslin cardinal. Woodin showed that the Kunen-Martin property followed from the other hypotheses in this implication. 8Obviously this section needs a better title 70 CHAPTER 6. SUSLIN SETS AND UNIFORMIZATION

Theorem 6.3.1. Assume that AD holds and let ∆ be a boldface selfdual point- ω class closed under ∃ω , and let κ be a cardinal less than cof(o(∆)). Then Sκ ⊆ ∆. Proof. Let A be a κ-Suslin subset of ωω. We may assume that A is not λ-Suslin for any λ < κ. By Proposition 2.5.8, we may ﬁx a prewellordering in ∆ of length κ on a set X. Using a bijection between κ × ω and ω if necessary, we may ﬁx a tree T ⊆ (ω × κ)<ω such that A = p[T ] and such that there is a function u: κ → T such that for each α ∈ κ, u(α) is the unique node (s, t) of T for which α is the last value taken by t. Let f : X → κ be the function which sends each member of X to its -rank. Let π : ωω → (ω<ω × (ωω)<ω) be a recursive bijection. Let Z be the set of (x, y) ∈ X × ωω such that π(y) is a tuple <ω (s, z0, . . . , z|s|−1) for some nonempty s ∈ ω , such that

• s is the ﬁrst coordinate of u(f(x))),

6.4 The Solovay sequence

The Solovay sequence [31] is the unique continuous sequence hθα : α ≤ βi such that

ω • θ0 is the least ordinal which is not the surjective image of ω under an OD function;

• for every ordinal α such that α + 1 ≤ β, θα+1 is the least ordinal which is ω ω not the surjective image of ω under an OD{A} function for any A ⊆ ω of Wadge rank θα;

• θβ = Θ.

We call β the length of the Solovay sequence. In L(R), θ0 = Θ, so β = 0. For ω each α ≤ β, if we let Γα be the set subsets of ω of Wadge rank less than θα, ω ω then HODΓα is a model of ZF containing ω whose subsets of ω are exactly the members of Γα.

6.4.1 Remark. If ADR holds, and the length of the Solovay sequence has uncountable coﬁnality, then L(Γα) |= ADR for club many α on the Solovay sequence. To see this, suppose that ADR holds, and consider any family {Ax : x ∈ ωω} of subsets of (ωω)ω. Consider the real game in which player I plays x and the two players play the real game with payoﬀ set Ax, with player II choosing 6.4. THE SOLOVAY SEQUENCE 71

first which player to be in the game with payoff set Ax. Player I cannot have ω a winning strategy. This shows for instance that, under ADR, for each A ⊆ ω there is a B ⊆ ωω such that every real game with payoff set Wadge below A has a winning strategy with payoff set below B.

There is an alternate deﬁnition of the Solovay sequence in which θα+1 is deﬁned to be the least ordinal which is not the surjective image of P(θα) under an OD function. The Moschovakis Coding Lemma implies that this version of 910 θα+1 is at most the one given above. In fact the two are the same Theorem 2.5.9 and 6.1.4 give the following.

ℵ0 Theorem 6.4.2. If Wadge Determinacy, Uniformization and ℵ1 6≤ 2 hold, then the length of the Solovay sequence is a limit ordinal.

6.4.3 Remark. While it is possible for nonlimit elements of the Solovay sequence to be singular11, if the length of the Solovay sequence is not a limit ordinal, then Θ is regular, since in this case there is an A ⊂ ωω such that every element of P(ωω) is ordinal deﬁnable from A and an element of ωω.

ω Solovay [31] showed that ADR + V =L(P(ω )), DC holds if and only if the length of the Solovay sequence has uncountable coﬁnality. For the easier direction, note that Wadge Determinacy + DCP(ωω ) implies that cof(Θ) is uncountable, since otherwise there exists an ω-sequence of sets of reals whose Wadge ranks are unbounded in Θ, and thereby a set of reals of Wadge rank at least Θ contracting Corollary 2.5.11.12 The following theorem gives the harder direction.

Theorem 6.4.4 (Solovay [31]). If AD + DCR holds, V =HODP(ωω ) and cof(Θ) is uncountable, then DC holds.

Proof. By Remark 0.4.2, it suffices to show that DCP(ωω ) holds. Let T be a tree on P(ωω) without terminal nodes. We want to see that T has an infinite path. For each ξ < Θ, let Tξ be the set of t ∈ T such that the range of t is contained ω in Wξ. For each t ∈ T , let w(t) be the least Wadge rank of a set A ⊆ ω such that t_hAi is in T . First suppose that for each ξ < Θ, w[Tξ] is bounded in Θ. Define s:Θ → Θ by setting s(ξ) to be the supremum of w[Tξ]. Then since cof(Θ) > ℵ0 there is a ζ < Θ closed under w. Then Tζ is a subtree of T without terminal nodes, and since DCR holds we have that Tζ contains an infinite path, so we are done.

We continue the proof under the assumption that for some ξ∗ < Θ, w[Tξ∗ ] ω is coﬁnal in Θ. Since Tξ∗ is a surjective image of ω , the ordertype of w[Tξ∗ ]

9 assuming V = HODP(ωω ) or something like this? find the proof. the argument uses θα-Borel codes? 10 W(A) < θα implies that δω(A) < θα? 11moreover, nonlimit members of the Solovay sequence below Θ have cofinality ω, see... 12 Maybe we need to say something about ultrapowers by µTu, and even say what µTu is. We need write µTu everywhere. We might also point out that part of this argument is used later to show that DCR follows from the wellfoundedness of the Turing ultrapower. 72 CHAPTER 6. SUSLIN SETS AND UNIFORMIZATION must be less than Θ. It follows that Θ is singular. Let λ be its cofinality and let c: λ → Θ be increasing, continuous and cofinal. Toward a contradiction, suppose that T does not contain an infinite path. ω Then each Tξ is also wellfounded. Let s: ω → λ be a surjection. For each ω 0 ω 0 x ∈ ω , let s (x) = sup{s(y): y ∈ ω , y ≤T x}. Then s is Turing-invariant. 0 ω Since cof(λ) > ℵ0, s (x) < λ for each x ∈ ω . For each t ∈ T , let rt be ω the function on ω defined by setting rt(x) to be 0 if some member of t has 0 Wadge rank at least s (x), and the rank of t in Ts0(x) otherwise. Then each rt 0 is Turing invariant, and rt < rt0 on a Turing cone whenever t < t in T . Let

R = {rt : t ∈ T } and for each α < λ let Rα = {rt : t ∈ Tc(α)}. Since DCR holds and each Tξ is wellfounded, each ultrapower Rα/µT is wellfounded. Since T is illfounded, R/µT is illfounded. Since R is an increasing λ-union of the sets Rα, and λ has uncountable coﬁnality, we get a contradiction by picking an increasing sequence of αn’s such that each Rαn+1 /µT contains a strict lower bound for Rαn+1 /µT. Part II

AD+

Chapter 7

<Θ-determinacy

Let λ be an ordinal, and let π : λ<ω → ω<ω be such that s ⊆ t implies π(s) ⊆ π(t) and dom(s) = dom(π(s)) (we say that a function with this property is ω extension-preserving). Given A ⊆ ω , let Gπ,A be the game in which players I and II alternate choosing ordinals αi ∈ λ (i ∈ ω), where I wins if and only if S 1 n∈ω π(hα0, . . . , αni) ∈ A.

I α0 α2 α4 ... II α1 α3 ...

The game Gπ,A

We let λ-Determinacy denote the statement that for all π : λ<ω → ω<ω as above, Gπ,A is determined. Given an ordinal γ, we let < γ-Determinacy denote the statement that λ-Determinacy holds for all λ < γ. 7.0.1 Remark. We are primarily interested in <Θ-Determinacy, which is usually called Ordinal Determinacy. In the usual definition of Ordinal Determinacy one has an ordinal λ, a continuous function f : λω → ωω (with respect to the discrete topology on λ and the corresponding product topology on λω) and a payoff set A ⊆ ωω. The corresponding game is a defined as above, except that I wins if and only if f(hαi : i < ωi) ∈ A. The difference is that the usual definition allows continuous functions, whereas the function mapping runs of the game into ωω in our definition is Lipschitz. We leave it to the reader to verify that the two definitions are equivalent.2

7.0.2 Remark. Consider the game of length ω in which player I plays α ∈ ω1 (and then makes no other moves) and then player II plays (digit by digit) an

1The result in the KKWM paper gives determinacy for Suslin games on ordinals. This gives ordinal determinacy, so maybe we should prove it. 2Say more?

75 76 CHAPTER 7. <Θ-DETERMINACY element x of 2ω, with II winning if and only if x codes (under some fixed coding) a wellordering of ω of ordertype α. By part (2) of Remark 1.0.2, if AD holds ω 3 then there is no injection from ω1 into ω , so this game is not determined. The Lusin-Sierpińskiorder (sometimes called the Brouwer-Kleene order) is the ordering on wellordered-sequences of ordinals defined by setting s

Theorem 7.0.3 (Moschovakis, Woodin). Assume that AD + DCR holds. Sup- pose that κ < δ are inﬁnite cardinals, and that δ is regular cardinal such that

µ ∀µ < δ(δ → (δ)µ) holds. Let A ⊆ ωω be such that A and ωω \ A are both κ-Suslin. Let λ < δ <ω <ω be an ordinal, and let π : λ → ω be extension-preserving. Then Gπ,A is determined. Proof. Fix trees T and S on ω × κ such that A = p[T ] and ωω \ A = p[S]. <ω For each u ∈ ω , recalling that Tu denotes the set {s :(u, s) ∈ T }, let

∗ [ Tu = {Tun : n ≤ |u|} and let ∗ [ Su = {Sun : n ≤ |u|}. ∗ ∗ For each such u,

3Say more? 77

I α0 α2 α4 ... II α1, f1 α3, f3 ...

The game GT

S In G , player II plays ordinals αi ∈ λ (for i ∈ ω odd) and player I plays pairs (αi, fi) (for i ∈ ω even) such that each αi is in λ and each fi is a function in OP S . Player I wins a run of GS if and only if f ⊆ f for all i < j π(hα0,...,αii) i j in ω.

I α0, f0 α2, f2 α4, f4 ... II α1 α3 ...

The game GS

The games GT and GS are each closed. Therefore, in GT , either player I has a winning strategy, or player II has a winning quasi-strategy (not necessarily a strategy, as the set of possible moves for II may not be wellorderable), and in GS, either player II has a winning strategy, or player I has a winning quasi- strategy. It cannot be, however, that player II has a winning quasi-strategy T S τII in G and player I has a winning quasi-strategy τI in G . Supposing that such quasi-strategies existed, using DCR, the Moschovakis Coding Lemma, the fact that δ < Θ and the fact that each function fi as above can be coded by a subset of δ, we could ﬁnd h(αi, fi): i ∈ ωi such that

hα0, (α1, f1), α2, (α3, f3),...i is according to τII and

h(α0, f0), α1, (α2, f2), α3,...i S is according to τI . Then f2i would witness that π(hαi : i ∈ ωi) 6∈ A, and S i∈ω i∈ω f2i+1 would witness that π(hαi : i ∈ ωi) ∈ A, giving a contradiction. The proof is then completed by proving the following claims.

Claim 1. If I has a winning strategy in GT , then I has a winning strategy in Gπ,A. Claim 2. If II has a winning strategy in GS, then II has a winning strategy in Gπ,A. We will prove Claim 1 and leave the proof of Claim 2, which is a routine modification, to the reader (Claim 2 can also be proved directly from Claim 1 by modifying the payoff set A). Recall that for an ordinal α and a set of ordinals X,[X]α denotes the collection of subsets of X of ordertype α. For notational 78 CHAPTER 7. <Θ-DETERMINACY convenience, we identify each element of [X]α with the corresponding order- preserving function on α which enumerates it. Fix for this paragraph an ordinal α < δ. For each g ∈ [δ]ωα (for some α ≤ δ), let g∗ be the function on α defined by setting g∗(ξ) to be sup{g(ω·ξ+n): n ∈ ω} ∗ ωα for each ξ < α. For each X ⊆ δ, let X(α) denote {g : g ∈ [X] }. Let µα be the set of A ⊆ [δ]α for which C(α) ⊆ A, for some club C ⊆ δ.

α Subclaim. For each α < δ, µα is a δ-complete ultrafilter on [δ] . To prove the subclaim, fix ρ < δ and suppose that [δ]α is the union of sets ωα ∗ Aβ (β < ρ). For each β < ρ, let Bβ be the set of g ∈ [δ] for which g ∈ Aβ. ωα δ Since δ → (δ)ρ holds, there exist an unbounded H ∈ [δ] and a β < ρ such ωα that [H] ⊆ Bβ. Let C be the set of limit points of H. Then C is a club subset of δ. Given h ∈ [C]ωα, define g ∈ [H]ωα by setting g(γ) = min(H \ h(γ)) for all ∗ ∗ γ < ωα. Then h = g , which is in Aβ. It follows that C(α) ⊆ Aβ as desired. This ends the proof of the subclaim.

<ω ζu T For each u ∈ ω , let Qu be the bijection from [δ] to OPu with the ζu property that Qu(g) and g have the same range, for all g ∈ [δ] . Let νu be T {Qu[A]: A ∈ µζu }. Then νu a δ-complete ultraﬁlter on OPu . T Fix a winning strategy Σ for I in G . We deﬁne a strategy σ for I in Gπ,A. We let σ(hi) = Σ(hi). For each positive n ∈ ω, we let σ(hα0, . . . , α2n−1i), be the T unique value α such that, letting u = π(hα0, . . . , α2n−1i), for νu-many f ∈ OPu , Σ(α , (α , f T ∗ ), α , (α , f T ∗ ), . . . , α , (α ,T ∗)) = α. 0 1 u2 2 3 u4 2n−2 2n−1 u

Now suppose that hαi : i ∈ ωi is the result of a run of Gπ,A according to σ, and that I has lost, so that π(hαi : i ∈ ωi) is not in A. Let x = π(hαi : i ∈ ωi). S Then x 6∈ p[T ], so

α0, (α1, f1), α2, (α3, f3),... is according to Σ and yet losing for player I, giving a contradiction. Again, the Moschovakis Coding Lemma gives us that determinacy for continuous games on ωλ is local.4 Similarly, the Martin-Steel theorem gives us that,

4explain 79 assuming AD, <Θ-determinacy holds in L(R).5

5and explain 80 CHAPTER 7. <Θ-DETERMINACY Chapter 8

Cone measures

Given an ordered equivalence relation (E, ≤E) such that the cone measure µE is an ultraﬁlter on the set CE consisting of the E-equivalence classes, we let WFE CE CE denote the assertion that the ultrapower Ord /µE (equivalently, V /µE) is wellfounded. Given two such pairs (E, ≤E) and (F, ≤F ), if ≤F ⊆≤E then CE CF Ord /µE embeds into Ord /µF , so WFF implies WFE. CE The wellfoundedness of the ultrapower Ord /µE follows from DC plus the assumption that µE is a countably complete ultraﬁlter on CE. Recall that DC holds in models of the form L(A, R) satisfying DCR, if A is a set contained in + L(R). Woodin has shown that AD implies WFT, where T is Turing equivalence. 1

Theorem 8.0.1 below shows that DCR follows from the certain instances of WFE. Let us say that an ordered equivalence relation (E, ≤E) is locally countable ω S if for all x ∈ ω , DE(x) is countable. Recall that a ranking function on a tree T is a function ρ: T → Ord such that ρ(s) > ρ(t) whenever t is an extension of S. The existence of a ranking function for T implies that T is wellfounded; the 2 reverse implication follows from DCT .

Theorem 8.0.1 (Solovay [31]). Let (E, ≤E) be a locally countable ordered equivalence relation on ωω. If the E-cone measure is an ultraﬁlter, and the image of

ω1 in the corresponding ultrapower is wellfounded, then DCR holds. Proof. Let T be a tree of finite sequences from ωω without an infinite path. We ω will find a ranking function for T . For eachx ¯ ∈ T and each z ∈ ω , let fx¯([z]E) be 0 if the members ofx ¯ are not all in DE(z), and the rank of T DE(z) belowx ¯ otherwise (which exists, since, as DE(z) is countable, DCDE (z) holds). Then if x¯ extendsy ¯ in T , fx¯([z]E) < fy¯([z]E) for all [z]E containing bothx ¯ andy ¯. The mapx ¯ 7→ [fx¯]µE then ranks T .

1We need to get a proof of this, and ﬁgure out the general form. In particular, does it work for ≡M? 2Refer the reader somewhere for the deﬁnition of the canonical ranking function? Observe V that only the ultrapower of ω1 needs to be wellfounded?

81 82 CHAPTER 8. CONE MEASURES

A similar argument gives Theorem 8.0.2 below, which is similar to Theorem ω 4.3.5 and will be used in Section 8.1. Given a set A ⊆ ω , we let ∆ω(A) be the smallest projective algebra with A as an element, and say that the members of ∆ω(A) are projective in A. We let δω(A) be the supremum of the lengths of the prewellorderings in ∆ω(A). Theorem 8.0.2. Suppose that there exists a locally countable ordered equiva- ω lence relation (E, ≤E) on ω such that the E-cone measure is an ultraﬁlter and ω WFE holds. Then for each A ⊆ ω , δω(A) is the supremum of the ranks of the ω wellfounded transitive relations on ω in ∆ω(A).

ω Proof. It suffices to fix a wellfounded transitive relation R on ω in ∆ω(A) and show that its rank is less than δω(A). Let E be the set of E-equivalence classes, ω and let µE be the E-cone measure. Given x ∈ ω , define a function πx : E → ω1 ω ω by setting πx(e) to be the rank of Rxe, where Rx is {(y, z) ∈ ω ×ω : xRyRz}. Define the relation ≤∗ on ωω by setting x ≤∗ y if and only if

{e ∈ E : πx¯(e) ≤ πy¯(e)} ∈ µE.

∗ Since µE is an ultrafilter and WFE holds, ≤ is a prewellordering. Furthermore, ∗ ≤ is projective in A. Since R is transitive, if xRy, then for all e ∈ E, πy(e) ≤ ∗ πx(e). It follows that the rank of R is less than or equal to the length of ≤ . The following definition redefines notation used above. 8.0.3 Definition. Given a set of ordinals S, we say that

ω • ≤S is the binary relation on ω deﬁned by setting x ≤S y if and only if x ∈ L[S, y];

ω • ≡S is the equivalence relation ≤S ∩ ≥S on ω ;

ω • for each x ∈ ω ,[x]S is the ≡S-equivalence class of x;

• DS is the set of ≡S-classes; • an S-cone is a set of the form {y ∈ ωω : x ∈ L[S, y]}, for some x ∈ ωω;

• µS is the set of subsets of DS which contain an S-cone;

ω • a function f on ω is S-invariant if f(x) = f(y) whenever x ≡S y.

Recall from Section 1.1 that TD implies that µS is an ultraﬁlter on DS, for each set S ⊆ Ord. We will be primarily interested in ultrapowers of the form Q f(d)/µ , for S invariant functions f on ωω. We start by showing that d∈DS S L[S,x] the function f(x) = ω1 represents ω1 in these ultrapowers (this fact appears as Lemma 3.3 of [18]).

Theorem 8.0.4 (ZF + AD + DCR). Let S be a set of ordinals and let f be the ω L[S,x] function on ω deﬁned by setting f(x) to be ω1 . Then f represents ω1 in DS Ord /µS. 83

Proof. Since µS is a countably complete ultrafilter (by Corollary 1.1.6), each α < ω1 is represented by the function with constant value α. For each α < ω1, ω {[x]S : x ∈ ω , f(x) > α} is in µS, so [f]µs > α. ω Suppose now that g is a ≡S-invariant function on ω such that {[x]S : x ∈ ω ω , f(x) > g(x)} ∈ µS. We want to see that there is an α < ω1 such that ω {[x]S : x ∈ ω , g(x) = α} ∈ µS. By the countable completeness of µS, this ω follows from the existence of an α < ω1 such that {[x]S : x ∈ ω , g(x) ≤ α} ∈ µS, which we will show. ω Let A be the set of x ∈ ω for which there is a y ≥S x with g(y) < g(x). ω Since µS is an ultrafilter, one of {[x]S : x ∈ A} and {[x]S : x ∈ ω \ A} contains an S-cone. Since DCR holds, and the ordinals are wellfounded, it cannot be the ω former. So there is an x∗ ∈ ω such that whenever x∗ ≤S≤ x ≤S y, g(x) ≤ g(y). Let WO be the set of wellorderings of ω, and for each z ∈ WO, let |z| denote the ordinal ordertype of z. Let G be the game in which player I produces x ∈ ωω, ω II produces y ∈ ω and z ⊆ ω × ω, and II wins if and only if either x∗ 6≤S x or x ≤S y, z ∈ WO and |z| > g(y). Given a strategy σ for player I, let y ∈ ωω be such that σ ∈ L[S, y] and let z ∈ WO ∩ L[S, y] be such that |z| > g(y). Then if x is the response given by σ to y and z, and if x∗ ≤S x, x ∈ L[S, y], so g(x) ≤ g(y) < |z| and II wins the corresponding run of the game. So I cannot have a winning strategy. Then AD implies that there is a winning strategy σ for II. Denote by σy(x) and σz(x) the values y and z respectively given by σ in response to a play x for ω I. Then {σz(x): x ∈ ω } is an analytic subset of WO, so by the boundedness result mentioned in the first sentence of Section 6.2 there is a countable ordinal ω ω α greater than all the members of {|σz(x)| : x ∈ ω }. Then for all x ∈ ω with σ ∈ L[S, x], σy(x) ≡S x, so α > |σz(x)| > g(σy(x)) = g(x).

The following theorem shows, under similar assumptions, that, for an S-cone ω V of y ∈ ω , L[S, y] satisﬁes GCH below ω1 . This fact will be used in the proof of Theorem 11.2.1.

ℵ0 Theorem 8.0.5 (ZF + CCR + ℵ1 6≤ 2 ). Let S be a set of ordinals such that ω ω µS is an ultraﬁlter on DS. Then there is an x ∈ ω such that for all y ∈ ω V with x ≤S y, and all γ < ω1 which are inﬁnite cardinals in L[S, y], L[S, y] |= 2γ = γ+.

V Proof. By Remark 1.1.7, µS being an ultraﬁlter implies that ω1 is measurable. ω Since µE is an ultraﬁlter, there is an x0 ∈ ω such that either ω • L[S, y] |= CH for all y ∈ ω with x0 ∈ L[S, y] or ω • L[S, y] |= ¬CH for all y ∈ ω with x0 ∈ L[S, y].

We show that the former holds, by finding one such y ≥E x0. Let P be the set of ordered pairs from ω<ω, and let π : P → ω be a recursive bijection. Let p: ωω × ωω → P(ω) 84 CHAPTER 8. CONE MEASURES be defined by setting p(x, y) to be π[{(x ∩ n, y ∩ n): n ∈ ω}]. Then p sends distinct pairs from ωω to infinite almost disjoint subsets of ω. ∗ ω Recall that the partial order Col (ω1, ω ) consists of injections from count- ω ω able ordinals to ω , ordered by extension, and adds a bijection g : ω1 → ω ∗ ω (see Section 0.5). Let X˙ be a Col (ω1, ω )-name for the p-image of the set of ω ω ∗ ω (x, y) ∈ ω × ω such that g(x) < g(y). Let QX˙ be a Col (ω1, ω )-name for the Jensen-Solovay almost disjoint coding forcing for the realization of X˙ (see Section 0.5). Suppose that M is a transitive model of ZFC, and that (g, z) is M-generic ∗ ω M for (Col (ω1, ω ) ∗ QX˙ ) . Then g is an element of M[z] (see Remark A.0.5 of [22] for a discussion of why M[z] is a generic extension of M). Moreover, since Q is c.c.c. and has cardinality ℵ in M[g] (which satisfies CH), CH holds in X˙ g 1 M[g][z] (which is the same as M[z]). V Since ω1 is measurable, and since L[S, x0] |= AC, the partial order

∗ ω L[S,x0] (Col (ω1, ω ) ∗ QX˙ ) is countable, and there exists a pair (g, z) which is L[S, x0]-generic for this partial order. The argument just given shows that L[S, x0, z] |= CH, so any real in L[S, x0, z] coding the pair (x0, z) works as our desired y. V Again using the measurability of ω1 , P(α) ∩ L[S, x0] is countable for each V α < ω1 so there exist L[S, x0]-generic ﬁlters for each corresponding partial order Col(ω, α) (see Section 0.5 again). Since these extensions must also satisfy CH, γ + V we have that L[S, x0] |= 2 = γ for all γ < ω1 which are inﬁnite cardinals in ω L[S, x0]. The same considerations apply to each y ∈ ω such that [y]E ≥ [x0]E.

The argument used in the last paragraph of the proof of Theorem 8.0.5 also shows the following, using the fact that the partial orders of the form Col(ω, X) (for any set X) are homogeneous and therefore do not change the model HODS.

ℵ0 Theorem 8.0.6 (ZF + CCR + ℵ1 6≤ 2 ). Let S be a set of ordinals such that ω ω µS is an ultraﬁlter on DS. Then there is an x ∈ ω such that for all y ∈ ω V with x ≤S y, all inﬁnite successor cardinals of L[S, y] below ω1 are strongly L[S,y] inaccessible in HOD{S} .

ω L[S,y] Proof. For any y ∈ ω , forcing over L[S, y] with Col(ω, P(ω) ∩ HOD{S} ) produces a z ∈ ωω such that

L[S,y,z] L[S, y, z] |= |P(ω) ∩ HOD{S} | = ℵ0.

ω L[S,y] It follows that, for some x ∈ ω , L[S, y] |= |P(ω) ∩ HOD{S} | = ℵ0 holds for all y ≥S x. Consideration of the partial order Col(ω, γ) shows that for all y ≥S x V L[S,y] and all inﬁnite γ < ω1 , in L[S, y] |= |P(γ) ∩ HOD{S} | = |γ|. It follows that V for all y ≥S x, each inﬁnite successor cardinal of L[S, y] below ω1 is strongly L[S,y] inaccessible in HOD{S} . 8.1. MEASURABLE CARDINALS FROM CONE MEASURES 85

Woodin has shown that, assuming AD, for each set S of ordinals, for an ω L[S,x] L[S,x] S-cone of x ∈ ω , ω2 is a Woodin cardinal in HOD{S} (see [19], Theorem 5.40).3

8.1 Measurable cardinals from cone measures

We give here an application of Theorem 8.0.2, a weak consequence of which will be used in Sections 9.1 and 11.4.4

ω Theorem 8.1.1 (AD + DCR). For each A ⊆ ω , δω(A) is a limit of measurable cardinals.

ω Proof. Since δω(A) < Θ, there is a set C ⊆ ω of Wadge rank greater than δω(A). By the Moschovakis Coding Lemma, it suﬃces to establish the theorem ω in L(C, R), where DC, and thus WFT, holds. Given B ⊆ ω , we let ρB denote 1 ω the supremum of the ranks of the wellfounded relations in Σ∼1(B, ω \ B), and 1 ω we let ηB be the supremum of the ranks of the ∆∼1(B, ω \ B) prewellorderings. By Theorem 8.0.2, δω(A) = sup{ρB : B ∈ ∆ω(A)}. ω ω 3 1 ω Let B ⊆ ω be projective in A, let U ⊆ (ω ) be a complete Σ1(B, ω \ B)- ∗ ∼ set, and let B be the set of x for which Ux is wellfounded. Deﬁne a prewellorder- ∗ ∗ ∗ ing ≤ on B by setting x ≤ y if and only if the rank of rk(Ux) ≤ rk(Uy), where rk(Ux) denotes the rank of Ux. This order is not necessarily projective in A. Claim 1 follows from the fact that a union of a collection of pairwise disjoint wellfounded transitive relations is wellfounded and transitive.

∗ 1 ω Claim 1. If Z ⊆ B is in ∼Σ1(B, ω \ B), then {rk(Ux): x ∈ Z} is bounded in ρB.

We adapt Solovay’s original proof of the measurability of ω1 under AD to ﬁnd a measure on ρB. For each X ⊆ ρB we consider the game GX , where player ω ω I builds xi ∈ ω for even i ∈ ω and player II builds xi ∈ ω for odd i ∈ ω. We use the way of doing this illustrated in the following diagram.

I x0(0) x0(1), x2(0) x0(2), x2(1), x4(0) ... II x1(0) x1(1), x3(0) ...

The game GX .

Given xi (i ∈ ω) produced by a run of this game, the winner is decided as follows.

∗ • If there is an i ∈ ω such that xi 6∈ B , then I wins if the least such i is odd, and II wins if the least such i is even.

3check the reference. 4I think, in the latter case. 86 CHAPTER 8. CONE MEASURES

∗ ∗ • If xi ∈ B for all i ∈ ω, and there is an i ∈ ω such that ¬(xi ≤ xi+1), then I wins if the least such i is even, and II wins if the least such i is odd.

∗ ∗ • If xi ∈ B and xi ≤ xi+1 holds for all i ∈ ω, then I wins if and only if

sup{rk(Uxi ): i ∈ ω} ∈ X.

Since a winning strategy for player II in GX (for some X ⊆ ρB) can be converted into a winning strategy for player I in GρB \X , for each X ⊆ ρB, player I has a winning strategy in at least one of the games GX and GρB \X . Let W be the set of X ⊆ ρB for which I has a winning strategy. Claim 2 shows that W is a countably complete ultrafilter. Its completeness, which by the claim is at least ηB, is a measurable cardinal. The theorem them follows from Claim 2 and the fact that {ηB : B ∈ ∆ω(A)} is cofinal in δω(A) (by the definition of δω(A)).

Claim 2. W is closed under intersections of cardinal less than ηB. We ﬁnish by proving the claim, using the Moschovakis Coding Lemma plus 1 ω boundedness. Fix γ < ηB and a prewellordering ≤R in ∆1(B, ω \ B) of a set ω ∼ R ⊆ ω such that ≤R has length γ. Let =R be ≤R ∩ ≥R, and let

8.2 The degree order on sets of ordinals

Given S, T ⊆ Ord, we write S ≤D T (D for “degree”) to mean that

ω {[x]Tu : ω ∩ L[S, x] ⊆ L[T, x]} ∈ µTu.

We write S

A reflection argument shows that for any set S of ordinals there is a bounded T ⊆ Θ with S ≤D T . Ultimately we will be concerned only with the restriction of ≤D to bounded subsets of Θ. Theorem 8.2.1 shows that ≤D is a total order when µTu is a countably complete ultrafilter. As with Theorem 8.0.5, Theorem 8.2.1 was first proved by 5 Steel in [?] assuming AD and later by Woodin from TD + CCR. Woodin’s proof uses a variation of Mathias forcing.

Theorem 8.2.1. Assume TD + CCR. If S and T are sets of ordinals such ω V that ¬(T ≤D S) then Then for a Turing cone of x ∈ ω , for all γ < ω1 , P(γ) ∩ L[S, x] is a set of cardinality |γ|L[T,x] in L[T, x]. To prove Theorem 8.2.1, we use a variation of Mathias forcing relative to a nonprincipal ultrafilter U on ω. Our partial order is a slight simplification of the one used originally by Woodin, and is in fact forcing-equivalent to it. Let S be the set of nonempty finite subsets of ω and let F be the set of functions from ω to U. The domain of PU is S × F. The order on PU is defined as follows : (s0,F 0) ≤ (s, F ) if the following hold (where max(s) denotes the largest element of a set s): • s0 ∩ (max(s) + 1) = s; • for each k ∈ s0 \ s, k ∈ F (max(s ∩ k)); • for each i ∈ ω, F 0(i) ⊆ F (i).

The second condition above distinguishes PU from the usual Mathias forcing relative to an ultrafilter. This condition enables Lemmas 8.2.2 and 8.2.3 below. The corresponding facts hold for Mathias forcing, but only for Ramsey ultrafilters; the ultrafilters we use below are not Ramsey. Let us say that a function F ∈ F is refining if F (i) ⊆ F (j) ⊆ (ω \ (j + 1)) whenever j < i ∈ ω. For densely many PU -conditions (s, F ), F is refining. Given F ∈ F, say that an infinite a ⊆ ω is F -fast if, for all i ∈ a, a \ (i + 1) ⊆ F (i). If F is refining, then every infinite subset of an F -fast set is F -fast. Every infinite a ⊆ ω (in the ground model or any outer model) generates a filter Ga ⊆ PU consisting of those (s, F ) ∈ PU such that s is an initial segment of a and a \ max(s) is F -fast.

Lemma 8.2.2 (The Genericity Condition). Assume that CCR holds. Let M be a transitive model of ZF, and let U be a nonprincipal ultrafilter on ω in M. Let a be a subset of ω. Then Ga is M-generic for PU if and only if, for each F ∈ F in M, there is an i ∈ a such that a \ i is F -fast. Proof. The forward direction follows immediately from the fact that for every s ∈ S, and any two functions F,F 0 ∈ F,(s, F ) and (s, F 0) are compatible. For the reverse direction, fix a with the given property and let D be a dense open subset of PU in M. Working in M, we recursively define a ranking function ρD on S, letting ρD(s) be

5Say something like this where Theorem 8.0.5 is stated. 88 CHAPTER 8. CONE MEASURES

• 0 if there exists an F ∈ F such that (s, F ) ∈ D;

• α if ¬(ρD(s) < α) and {i ∈ ω : ρD(s ∪ {i}) < α} ∈ U;

• ω1 if ¬(ρD(s) < ω1).

Observe that, for each s ∈ S, if {i ∈ ω : ρD(s ∪ {i}) < ω1} ∈ U, then ρD(s) is at most sup{ρD(s ∪ {i}): i ∈ ω, ρD(s ∪ {i}) < ω1}, which is less than ω1. It follows that, by CCR, we can choose, for each s ∈ S a function Fs : ω → U such that

• if ρD(s) = 0, then (s, Fs) ∈ D;

• if ρD(s) < ω1, then for all i ∈ Fs(max(s)), i > max(s) and

ρD(s ∪ {i}) < ρD(s);

• if ρD(s) = ω1, then for all i ∈ Fs(max(s)), i > max(s) and

ρD(s ∪ {i}) = ω1.

∗ ∗ T Define F : ω → U by setting F (s) to be {Ft(s): t ∈ P(max(s) + 1) \ {∅}}. We claim that ρD(s) < ω1 for all s ∈ S. To see this, suppose that s ∈ S is ∗ such that ρD(s) = ω1. By the choice of Fs, and the definition of F , we have 0 0 0 0 0 that ρD(s ) = ω1 whenever (s ,F ) ∈ PU is such that (s ,F ) ≤ (s, F ). This contradicts our assumption that D is dense. Finally, there is an i ∈ a such that a \ i is F ∗-fast. Then the sequence of values hρD(a ∩ j): j ∈ ω \ ii must decrease until reaching 0, showing that Ga ∩ D 6= ∅. The Genericity Condition, plus the fact that F is refining for densely many conditions (s, F ), gives the following. Lemma 8.2.3. Let M be a transitive model of ZF, let U be a nonprincipal ultrafilter on ω in U, and suppose that a ⊆ ω is such that Ga is an M-generic 0 filter for PU . Then for any infinite a ⊆ a, Ga0 is an M-generic filter for PU .

Recall that a set is recursive if it is ∆1 over HF. Let us say that a function <ω <ω ∗ π : 2 → [ω] is diffuse if, letting π (x) denote ∪{π(ti): i < ω} for each ω ω ∗ ∗ x ∈ 2 , for all distinct x0, . . . , xn−1 ∈ 2 , if σi ∈ {π (xi), ω \ π (xi)} for each T i < n then icomplement). Given a function π : 2<ω → [ω]<ω, say that a set a ⊆ ω is π-weak if there exist infinitely many x ∈ 2ω such that at least one of a ∩ π∗(x) and a \ π∗(x) is finite. 8.2. THE DEGREE ORDER ON SETS OF ORDINALS 89

<ω <ω Lemma 8.2.4. If π : 2 → [ω] is diﬀuse, n is a positive integer and ai S (i < n) are π-weak subsets of ω, then i

We separate out one lemma which we be used again in the proof of Theorem 10.1.6.

Lemma 8.2.5. Let π : 2<ω → [ω]<ω be a recursive diﬀuse function in a model M of ZF. Let U be, in M, an ultraﬁlter on ω disjoint from the collection of M ω π-weak sets, and suppose that P(PU ) is countable. Let y be in 2 \ M, and ∗ M let b be a subset of ω. Then there exists an a ⊆ ω which is M-generic for PU such that b ∈ L[a∗, y].

1 ω Proof. By Σ1 absoluteness, for each a ∈ U, every z ∈ 2 such that at least one of a ∩ π∗(z) and a \ π∗(z) is finite is in M. It follows that π∗(y) and ω \ π∗(y) have infinite intersection with each member of U. By Lemma 8.2.2, there exists an M-generic set a such that a ∩ π∗(y) and a\π∗(y) are both infinite. To see this, enumerate all refining functions F : ω → U T in M as hFi : i ∈ ωi. Let a be {ni : i ∈ ω} where each ni ∈ j

The end of the proof of Theorem 8.2.1 uses following theorem of Solovay, which will be used again in Sections 8.5 and ??. We will use the theorem only for the case where x is a set of ordinals, in which case the proof is slightly easier.

Theorem 8.2.6 (Solovay). Suppose that M ⊆ N are models of ZF, P is a partial order in M and P(P) ∩ N is countable. Suppose that x ∈ M[G] for every N-generic G ⊆ P. Then x ∈ M. Proof. It suﬃces to prove the theorem in the case where x is transitive, replacing x with TC({x}) if necessary. Since P(P)∩M is countable, P(P×P) is countable. Let (G, H) be N-generic for P × P. Arguing now by induction on the rank of x, we may assume that the members of x are all in M. Then if τ and σ are P-names in M such that τG = σH , it follows by genericity that there is some condition (p, q) ∈ G × H such that, for all y ∈ M p decidesy ˇ ∈ τ and q decides yˇ ∈ σ.

ω ω Proof of Theorem 8.2.1. Let x ∈ ω , y ∈ 2 be such that y ∈ L[S2, x] \ L[S1, x]. <ω <ω Let π : 2 → [ω] be a recursive diﬀuse function. Let U be, in L[S1, x], an L[S1,x] ultraﬁlter on ω not containing any π-weak subset of ω, and let P be PU . Let b ⊆ ω HC-code an enumeration in ordertype ω of the P-names in L[S1, x] 90 CHAPTER 8. CONE MEASURES

ω ∗ for elements of ω . By Lemma 8.2.5, there is an L[S1, x]-generic set a for P ∗ ω ∗ ∗ such that b ∈ L[a , y]. Then ω ∩ L[S1, x][a ] is a countable set in L[S2, x, a ]. ω It follows by Turing Determinacy that there is an x∗ ∈ ω such that, for ω all y ≥T x∗, ω ∩ L[S1, y] is a countable set in L[S2, y]. Applying this fact and Theorem 8.2.6 to the Col(ω, γ)-extension of any such model L[S2, y], we see that L[S2,y] 6 for all such y, P(γ) ∩ L[S1, y] is a set of cardinality |γ| in L[S2, y].

8.2.7 Remark. It follows from the last paragraph of the proof of Theorem 8.2.1 ω that if S and T are sets of ordinals such that S ≤D T , then, for all x ∈ ω in a Turing cone witnessing this fact, every uncountable regular cardinal of L[T, x] V below ω1 is strongly inaccessible in L[S, x]. 8.2.8 Remark. Just as in the last paragraph of the proof of Theorem 8.2.1, consideration of the partial order Col(ω, α) shows that if S and T sets of ordinals ω such that S ≡D T , then for all x ∈ ω in a cone witnessing this face, and all V γ < ω1 , P(γ) ∩ L[S, x] = P(γ) ∩ L[T, x]. Theorem 8.2.9 lists some consequences of Theorem 8.2.1. Parts (1a) and (2a) follow from Theorems 8.2.1 and 8.2.6 by a collapsing argument as in the second half of the proof of Theorem 8.0.5 (note that both parts apply to all V L[T,x] γ below ω1 , not just ω1 ). Part (1b) follows from part (1a) and Theorem 8.2.6, by another collapsing argument. Part (2b) follow from part (2a). Note that neither part of the theorem assumes (or implies) that S is in L[T, y]. A natural application of part (1b) is when S is an ∞-Borel code.

Theorem 8.2.9. Assume TD + CCR, and let S and T be sets of ordinals.

1. If x is a base of a cone witnessing that S ≤D T , then the following hold for all y ≥Tu x.

V (a) For every γ < ω1 , P(γ) ∩ L[S, y] ⊆ L[T, y]. (b) For every formula φ,

{z ∈ ωω : L[S, z] |= φ(S, z)} ∩ L[T, y] ∈ L[T, y].

ω 2. If x ∈ ω is a base of a Turing cone witnessing that S

V (a) For every γ < ω1 ,

P(γ) ∩ L[S, y] ∈ H(|γ|+)L[T,y].

L[T,y] (b) ω1 is a strongly inaccessible cardinal in L[S, y]. 6 That P(γ)∩L[S1, y] ⊆ L[S2, y] follows from Theorem 8.2.6. It may take redoing the proof to see that L[S2, y] contains an enumeration in ordertype γ. 8.3. POINTED TREES 91

8.3 Pointed trees

In this section we prove a result due to Martin which will be used in Sections 8.4 and 8.5. We continue to use the notation from Definition 8.0.3, and add the following. Recall that a tree is perfect if each node has an incompatible pair of extensions, and, given a tree a, that [a] denotes the set of infinite branches through a. 8.3.1 Definition. Given a set of ordinals S, we say that

• a set A ⊆ ωω is S-positive if it intersects every S-cone; • a tree a ⊆ ω<ω is S-pointed if a is perfect and, for every x ∈ [a], a ∈ L[S, x]; The following theorem, which appears in [11], shows that, assuming AD, the sets of the form [a], for a an S-pointed tree, are dense in the containment order on the S-positive subsets of ωω. Theorem 8.3.2 (Martin). Suppose that AD holds. Let S be a set of ordinals and let A be a subset of ωω. Then A ⊆ ωω is S-positive if and only if there is an S-pointed perfect tree a such that [a] ⊆ A. Proof. The reverse implication (which does not use the assumption of AD) follows from the fact that for any perfect tree a ⊆ ω<ω and any x ∈ ωω there is a y ∈ [a] such that x ∈ L[a, y]. For the forward implication, consider the game G where I and II respectively ω build x and y in ω , and II wins if and only if x ≤S y and y ∈ A. If σ is a strategy for I and y ∈ A is such that σ is in L[S, y], then y is a winning play for II against σ. This shows that if A is S-positive then I cannot have a winning strategy. Suppose then that σ is a winning strategy for II. For each s ∈ ω<ω, let σ(s) denote the sequence of (|s|-many) moves made according to σ in response to s. Let y ∈ 2ω be such that σ ∈ L[S, y], and let Y be the set of t ∈ ω<ω ∪ ωω such that t(2n) = y(n) whenever n ∈ ω and 2n ∈ dom(t). Working in L[S, y] <ω <ω we can choose for each s ∈ 2 , recursively in |s|, a ts in Y ∩ ω , in such a way that, for all s, s0 ∈ 2<ω,

0 • if |s| = |s | then |ts| = |ts0 |;

0 • if s ⊥ s then σ(ts) ⊥ σ(ts0 ). The achievability of the second condition above follows from the fact that σ is a winning strategy for II, which implies that, for any t ∈ Y , the set

{x ◦ σ : x ∈ ωω ∩ Y, t ⊆ x} is S-positive, and therefore has size at least 2. Let a be the tree of initial <ω segments of the members of {ts : s ∈ 2 }. We claim that a is as desired. To see this, ﬁx z ∈ [a] and x ∈ ωω ∩ Y such that z = x ◦ σ. Since σ is a winning strategy for II, z is in A, and x is in L[S, z]. This implies that y is in L[S, z] and therefore that a is in L[S, z] as desired. 92 CHAPTER 8. CONE MEASURES

8.4 Coding ultrapowers

This section continues to use the notation from Deﬁnitions 8.0.3 and 8.3.1, for a set of ordinals S. We work under the assumption that µS is an ultraﬁlter (which ∞ Q L[S,x] AD implies that it is, by Corollary 1.1.6) on DS, and let δS be ω2 /µS, assuming that this ultraproduct is wellfounded (as is it, assuming DC). In Remark 8.5.8 we will show (assuming AD plus the wellfoundedness of the ul- DS ∞ ∞ trapower Ord /µS) that δS ≤ Θ. In this section we will show that δS < Θ, under the additional assumption that, for some set of ordinals T , S

• y and w are in L[T, x];

• y comes before w in the constructibility order in L[T, x] using T and x as predicates.

c In the statement of Theorem 8.4.1 the set ≤S × transitive set in H(ℵ1) . Then f(x)/µS is isomorphic to a relation c which is projective in ≤S × isomorphism between

(ω, {(n, m) ∈ ω × ω : y(2n3m) = 0}) and (TC({M}), ∈) sending k to p. We call this element py,k. The following sets are projective:

0 0 2 2 • {(y, y , k, k ) ∈ E × ω : py,k = py0,k0 };

0 0 2 2 • {(y, y , k, k ) ∈ E × ω : py,k ∈ py0,k0 }; For any S-invariant h such that h(x) ∈ f(x) on an S-cone there is a k ∈ ω ω ∗ such that the set A = {x ∈ ω : p ∗ = h(x)} is S-positive. Since R h,k Rf (x),k f may not be S-invariant, there may be more than one such k. The pair (k, Ah,k) codes h on an S-cone, however, in the sense that, for an S-cone of z ∈ ωω,

h(z) = h(x) = p ∗ Rf (x),k 8.5. FORCING WITH POSITIVE SETS 93

for any x ∈ [z]S ∩ Ah,k. By Theorem 8.3.2, every S-positive set contains the (S-positive) set of paths <ω through some S-pointed tree. Let hσi : i < ωi be a recursive listing of ω , and ω let Cf be the set of pairs (k, b) ∈ ω × 2 such that ab = {σi : b(i) = 1} is an S-pointed perfect tree and, for all x, y ∈ [a ], if x ≡ y then p ∗ = p ∗ . b S Rf (x),k Rf (y),k ω For each (k, b) ∈ Cf and z ∈ ω , set h(k,b)([z]S) to be the common value of p ∗ for all x ∈ [z] ∩ [a ] (if there is such a value, and ∅ otherwise). Deﬁne Rf (x),k S b an equivalence relation ∼ on Cf by setting (k, b) ∼ (k0, b0) if h(k,b) and h(k0,b0) agree on an S-cone, and let [k, b]f denote the ∼-class of 0 0 (k, b). Deﬁne [k, b]f ∈f [k , b ]f to hold whenever

h(k,b)([z]S) ∈ h(k0,b0)([z]S) ω Q for an S-cone of z ∈ ω . Then (Cf / ∼, ∈f ) is isomorphic to f(x)/µS. Since c Cf , ∼ and the relation on Cf induced by ∈f are projective in ≤S ×

8.5 Forcing with positive sets

Given a set S consisting of ordinals, let PS be the partial order of S-positive ω subsets of ω and let SS be the partial order of S-pointed perfect trees, each ordered by ⊆. By Theorem 8.3.2, each PS condition contains the set of infinite paths through some S-pointed tree, which is in turn S-positive. It follows that a V -generic filter for either partial order generates one for the other. Q A V -generic filter G ⊆ PS induces an ultrapower x∈ωω V/G, whose elements are represented by (not necessarily S-invariant) functions in V with ω ω domain ω . The generic xG ∈ ω added by PS is represented by the identity function in this ultrapower. The following theorem shows that the corresponding ultrapower of any set of ordinals, in particular the ultrapower of S itself, is the same in the ultraproducts with respect to G and µS, where the latter ultraproduct is computed in V and the former is computed in the PS-extension.

Theorem 8.5.1. Let S be a set of ordinals such that µS is a countably complete ultrafilter on DS, and let G ⊆ PS be a V -generic filter. Let k be the map sending ω [f]S to [f]G, for each S-invariant function f in V from ω to Ord. Then k maps Q Q S Ord/µS isomorphically to Ord/G. Proof. Since each S-positive set has S-positive intersection with each S-cone, k is injective (mod µS) and order preserving. We show that it is onto. Let g : A → Ord be a function in V with S-positive domain. Define f on [ {[x]S : x ∈ A} by letting f(x) be min(g[[x]S ∩ A]). Then f is S-invariant. Let C be {x ∈ A : f(x) = g(x)}. 94 CHAPTER 8. CONE MEASURES

It suﬃces (by the genericity of G) to show that C is S-positive, since C forces ω that [g]G = [f]G = k([f]S). Supposing otherwise, let x ∈ ω be such that ω {y ∈ ω : y ≥S x} is disjoint from C. Since A is S-positive, there is a y ≥S x 0 ∼ 0 0 in A. Then there is a y =S y in A such that g(y) = f(y ). Then y ∈ C and y ≥S x, giving a contradiction. Q We let jG denote the embedding of V into x∈ωω V/G sending each element of V to the class represented by its corresponding constant function, assuming that G is a V -generic ﬁlter for PS. We let jS be the corresponding embedding Q of V into x∈ωω V/µS.

Corollary 8.5.2. If S is a set of ordinals and G ⊆ PS is a V -generic ﬁlter, then for every set of ordinals T , jG(T ) = jS(T ). Proof. If f : ωω → Ord is S-invariant, then either f(x) ∈ T for an S-cone of x or f(x) 6∈ T for an S-cone of x. In either case the same must hold on a set in G.

8.5.3 Remark. We are primarily interested here in the ultrapower Y L[S, x]/G, x∈ωω Q which by Theorem 8.5.1 and Corollary 8.5.2 is isomorphic to L[S, x]/µS. Since Q the functions used to from x∈ωω L[S, x]/G are not required to be S-invariant, the usual proof ofLo´s’sTheorem goes through, showing that whenever f1, . . . , fn ω are functions on ω in V such that each value fi(x) in the corresponding model L[S, x], and φ is an n-ary formula, Y L[S, x]/G |= φ([f1]G,..., [fn]G) x∈ωω if and only if the set of x for which L[S, x] |= φ(f1(x), . . . , fn(x)) is in G. In L[S,x] particular, for each n ∈ ω the function x 7→ ωn represents the value of ωn in the ultrapower.

DS ∞ Assuming that the ultrapower Ord /µS is wellfounded, we let S be the set of ordinals represented by the function with constant value S. Given A ⊆ ωω, we say that S computes A on a cone if A ∩ L[S, x] ∈ L[S, x] for an S-cone of x ∈ ωω. Part (1b) of Theorem 8.2.9 shows that if S and T are sets of ordinals ω such that S ≤D T , then T computes {x ∈ ω : L[S, x] |= φ(S, x)} on a cone, for any binary formula φ.

Corollary 8.5.4. Let S be a set of ordinals such that µS is a countably complete DS ultraﬁlter on DS and Ord /µS is wellfounded. Let G ⊆ PS be a V -generic ﬁlter ω and let xG be the corresponding generic element of ω . Then the following hold.

Q ∞ 1. The ultraproduct x∈ωω L[S, x]/G is isomorphic to L[S , xG]. 8.5. FORCING WITH POSITIVE SETS 95

V ∞ 2. For all γ < jG(ω1 ) which are inﬁnite cardinals of L[S , xG], ∞ γ + L[S , xG] |= 2 = γ .

∞ V L[S ,xG] 3. ω1 = ω1 ∞ ∞ L[S ,xG] 4. δS = ω2 5. If A ⊆ ωω is such that S computes A on a cone, then the function x 7→ A ∩ L[S, x] represents a set A∗ ⊆ (ωω)V [G] such that A∗ ∩ (ωω)V = A. ω Proof. For part (1), the fact that the identity function on ω represents xG Q shows that x∈ωω L[S, x]/G is isomorphic to L[jG(S), xG]. Corollary 8.5.2 im- ∞ plies that this is the same as L[S , xG]. Part (2) follows from Remark 8.5.3 and Theorem 8.0.5. For part (3), use Theorems 8.5.1 and 8.0.4, and Remark 8.5.3. For part (4), use Theorem 8.5.1 and Remark 8.5.3. For part (5), use Theorem 8.5.1 again.

8.5.5 Remark. Let S be a set of ordinals such that µS is a countably complete DS ω ∞ ultraﬁlter on DS and Ord /µS is wellfounded. For all x, y ∈ ω , y ∈ L[S , x] if and only if y ∈ L[S, x] for an S-cone of z, i.e., if and only if y ∈ L[S, x]. So SS and SS∞ are the same partial order. The following theorem will be used in Section ??.7

DS Theorem 8.5.6 (ZF + AD). Let S be a set of ordinals such that Ord /µS is wellfounded. Suppose that A ⊆ ωω is such that S computes A on a cone. Then A ∈ L(S∞, R). ∞ Proof. We show that A ∈ L(S , R)[xG] whenever xG is V -generic for SS∞ ; the theorem then follows by Theorem 8.2.6. By Remark 8.5.5, the partial orders SS and SS∞ are the same, and by Theorem 8.3.2 they are forcing-equivalent to PS. It follows that if xG is V -generic for SS∞ then it is V -generic for PS, and ∗ ∞ ∗ ω V therefore that there is an A ∈ L[S , xG] such that A ∩ (ω ) = A. Since ω V ∞ ω (ω ) ∈ L(S , ω )[xG] we are done.

We say that S is strongly maximal (implicitly, with respect to ≤D) if it computes each subset of ωω on a cone. Theorem 8.5.7 follows from Theorem 8.5.6 and part (1b) of Theorem 8.2.9.

Theorem 8.5.7. Assume that AD holds, S is a set of ordinals, and the µS- ultrapower of the ordinals is wellfounded. If S is a ≤D-maximal set of ordinals, then every ∞-Borel set is in L(S∞, R). By Theorem 9.1.4, the ∞-Borel sets are all ∞-Borel in L(S∞, R), assuming the hypotheses of Theorem 8.5.7.8

7Or maybe Theorem 8.5.7. Do we need both? 8Do we need this? : Similarly AD holds and A ⊆ ωω is not ∞-Borel, then every ∞-Borel set is ∞-Borel in L(A, R) by Theorem 9.1.4, and one can apply the theorem in L(S, A, R) assuming (in addition) only AD. In this case, however, the S∞ in the statement of the theorem is as computed in L(S, A, R). 96 CHAPTER 8. CONE MEASURES

8.5.8 Remark. Suppose that AD holds. Let S be a set of ordinals such that DS the corresponding ultrapower Ord /µS is wellfounded. Let xG be V -generic ω for SS. The partial order SS is a surjective image of ω (in V ), so forcing V ∞ with SS preserves Θ as a cardinal. Since L[S , xG] is an inner model of ∞ V ∞ L[S ,xG] V V V [G], Θ is a cardinal in L[S , xG]. Since ω1 = ω1 < Θ , it must be ∞ V L[S ,xG] ∞ that Θ ≥ ω2 , which is equal to δS , by Theorem 8.5.4. Similarly, it DS follows from Corollary 8.5.4 that if Ord /µS is wellfounded and S is strongly maximal then each prewellordering of ωω in V is a subset of a prewellordering ∞ ω L[S ,xG] ∞ ∞ of (ω ) in L[S , xG], which since L[S , xG] satisﬁes CH implies that ∞ ∞ 9 δS ≥ Θ. If in addition AD holds then δS = Θ.

Wellfoundedness of the µS-ultrapower follows from AD + DC, and also from AD+, by Theorem ??.10

9Have we said why? Compare with Uniformization? 10To be proved in Part III. Is this really the best place for this remark? Chapter 9

∞-Borel sets

For an infinite cardinal κ and an infinite ordinal δ, we let Lκ,δ be the language with propositional variables Pα (α ∈ δ), closed under negation and wellordered conjunctions and disjunctions of cardinality less than κ. Formally, we require the conjunctions and disjunctions to be indexed by sets of ordinals. As the conjunctions and disjunctions are wellordered, we can associate to each sentence in Lκ,δ a unique code (in P(κ), if δ ≤ κ) via a definable (injective) pairing function ≺ ·, · on the ordinals (as defined in Section 0.2), and write φS for the sentence coded by the set S. Fixing one such coding, we let

• φ{≺0,α } be Pα for each α ∈ δ,

• φ{≺1,α :α∈S} be ¬φS, and V • φ{≺2+α,β :α<γ,β∈Sα} be α<γ φSα .

We let L∞,δ be the class of sentences which are in Lκ,δ for some infinite δ cardinal κ. Each x ∈ 2 can be thought of as an L∞,δ-structure, where, for each α ∈ δ, Pα is interpreted as true if and only if x(α) = 1. For a sentence φ δ of L∞,δ, we let Aφ be the set of x ∈ 2 such that x |= φ, and say that φ is an ∞-Borel code (or a κ-Borel code, if φ ∈ Lκ,δ) for Aφ. We also say that a set of ordinals S is an ∞-Borel code or κ-Borel code if the corresponding formula φS is. 9.0.1 Definition. Given a cardinal κ, a subset of 2δ is κ-Borel if it is equal to Aφ for some φ ∈ Lκ,δ, and <κ-Borel if it is γ-Borel for some cardinal γ < κ.A subset of 2δ is ∞-Borel if it is κ-Borel for some infinite cardinal κ. 9.0.2 Remark. For each infinite cardinal κ, the class of κ-Borel subsets of 2ω is closed under continuous preimages, and thus Wadge reducibility. To see this, ω ω fix φ ∈ Lκ,ω and a continuous function f : 2 → 2 . To express “f(x) |= φ” it suffices to find, for each n ∈ ω, an expression for “f(x)(n) = 1”, since then one −1 can form a ψ with Aψ = f [Aφ] by replacing each instance of each Pn in φ with the corresponding expression. The existence of these expressions follows from the continuity of f (and, in the case κ = ω, the compactness of 2ω).

97 98 CHAPTER 9. ∞-BOREL SETS

The following example shows that Suslin subsets of 2ω are ∞-Borel (this also follows easily from item (2) of Theorem 9.0.6 below). Given a tree T and a node t ∈ T , we let Tt (for this example) be the set of s such that the concatenation t_s is in T . We define rank(T ) to be 0 if T is empty or T = {∅}. Otherwise, rank(T ) is the strict supremum of the set {rank(Tt): t ∈ T, |t| = 1}. 9.0.3 Example. Let T be a tree on (γ × 2), for some ordinal γ. Given a node ω 0 0 <ω <ω (s, t) ∈ T and an x ∈ 2 , let Tx,s,t denote the set of (s , t ) ∈ γ × 2 for which s_s0 is an initial segment of x and (s_s0, t_t0) ∈ T . For such a T one can + α ∗ define recursively on α ≤ |γ| formulas φ(s,t) ∈ L|γ|+,ω, simultaneously for each ω (s, t) ∈ γ × 2, such that A α is the set of x ∈ 2 for which rank(T ) = α. φ(s,t) x,s,t V α The projection of T is then defined by α<|γ|+ ¬φh∅,∅i. 9.0.4 Remark. There is in general no method for picking an ∞-Borel code for each ∞-Borel set. To see this, let E0 be the equivalence relation of mod-finite agreement on 2ω, and suppose that every subset of 2ω has the property of Baire. If F were a function picking for each E0-class a code for the class, then by the Kuratowski-Ulam theorem (which among other things implies that if every set of reals has the property of Baire then every wellordered union of meager sets is meager) comeagerly many reals would have the same code for their class. 9.0.5 Remark. The collection of ∞-Borel subsets of 2ω is not necessarily closed under wellfounded unions. For example, let {xα : α < ω1} be mutually generic Cohen reals over L, and consider the model L(ωω,F ), where ωω is the set of all reals in the forcing extension, and F is the function taking α < ω1 to the -class of x . The set S F (α) cannot be ∞-Borel in this model, even E0 α α<ω1 though it is a wellordered union of countable sets. To see this, note first that ∞-Borel code for this set would be ordinal definable from F and finitely many reals. These finitely many reals would appear in L[{xα : α < β}] for some β < ω1. The homogeneity of the rest of the forcing would then imply that the ∞-Borel code for the union would be in this model, which is impossible, since L for instance if ω1 is countable it is possible to build two generic filters for the remainder of the forcing such that some real x is equal to xβ in one extension and not a member of any F (α) in the other. (This example was constructed by the author in collaboration with Trevor Wilson, although it was very likely noticed earlier.) Theorem 9.0.6 gives a number of convenient reformulations of the notion of ∞-Borel set. Item (2) below is especially useful. Given ordinals α and δ, x ∈ 2δ and C ⊆ αω × 2δ, let G(α, x, C) be the length-ω game on α such that player I wins if (β,~ x) ∈ C, where β~ is the sequence produced by the run of the game. The notion of closure in parts (3) and (4) below refers to the product topology on both αω and 2δ. Theorem 9.0.6 (ZF). Let δ be an ordinals. The following are equivalent, for a set A ⊆ 2δ. 1. A is ∞-Borel. 99

2. For some set of ordinals S and some ﬁrst-order formula θ,

A = {x ∈ 2ω : L[S, x] |= θ(S, x)}.

3. For some ordinal α and some clopen C ⊆ αω × 2δ, A is the set of x ∈ 2ω for which Player I has a winning strategy in G(α, x, C). 4. For some ordinal α and some closed C ⊆ αω × 2δ, A is the set of x ∈ 2ω for which Player I has a winning strategy in G(α, x, C). Proof. The implication from (1) to (2) follows from the fact that sentences in L∞,δ can be coded by sets of ordinals, and the deﬁnability of the relation x |= φ. For the reverse direction, for each ordinal α, let

∗ • Lα be the extension of the ﬁrst-order language of set theory given by adding predicate symbols S˙,x ˙ and constant symbols β˙ for each β ∈ α and

∗ • Tα denote the collection of Lα sentences of the form

∀x ... ∀x ((x = Xβ1 ∧ ... ∧ x = Xβn ) → ψ) 1 n 1 φ1 n φn where

– each βi is in α; – each φ is a unary formula in the corresponding L∗ ; i βi – each expression of the form x = Xβi denotes the formula saying that i φi xi is the set of sets satisfying φi in the structure

(L [S,˙ x˙]; S,˙ x,˙ δ˙ (δ ∈ d ), ∈), β˙i i ˙ where di is the set of symbols from {δ : δ < βi} appearing in φi; ∗ – ψ is a Lα formula with free variables x1, . . . , xn (so each sentence of ∗ Lα is itself an element of Tα).

Since each expression of the form xi = Xφi above asserts that xi is defined by φi, each sentence in Tα holds in the same structures as the corresponding sentence ∃x ... ∃x (x = Xβ1 ∧ ... ∧ x = Xβn ∧ ψ). 1 n 1 φ1 n φn We refer to this below as the existential form of the sentence. Working (in L[S]) by recursion on α, we describe a procedure which asso- δ ciates to each sentence θ ∈ Tα a sentence ρθ,α of L|α|+,δ such that for all x ∈ 2 , ˙ Lα[S, x] |= θ (with the symbols β (β < α), S˙ andx ˙ given their natural interpre- tations) if and only if x |= ρθ,α. For finite α this follows from the fact that the ∗ Lα-theory of (Lα[S, x]; S, x, β (β ∈ α), ∈) depends only on xα (as S is fixed) so the desired ρθ,α can consist of a disjunction of conjunctions describing xα exactly (and since there is a canonical finite set of such sentences our procedure can pick the least one in a suitable ordering). 100 CHAPTER 9. ∞-BOREL SETS

For the limit and successor cases, we induct on the complexity of ψ in the representation of the members of Tα given above. For limit α, the induction hypothesis gives us the desired sentences ρθ,α when ψ is a ∆0 formula, and the steps corresponding to ∧, ∨ and ¬ are handled by combining the formulas ρθ,α in the same manner (to see that this works in the case of ∨ and ¬, use the fact that each sentence in Tα has an equivalent existential form). If ψ has the 0 form Qxψ , for Q either ∀ or ∃, we let ρθ,α be the conjunction (when Q = ∀) or the disjunction (when Q = ∃) of all the sentences ρ β , where β < α, φ is θφ ,α ∗ β a unary formula in Lβ and θφ is formed from θ by adding ∀x to the beginning β ∗ of θ and x = Xφ at the appropriate place. The point here is that α × Lα is wellorderable, and each set in the range of the quantifier Q is defined by some S ∗ formula in β<α Lβ. The successor step from α to α + 1 is similar, except that, given a sentence θ ∈ Tα+1 as above, with ψ a ∆0 formula, we must deal with the possibility that some of the formulas φi define subsets of Lα[S, x], so that we cannot simply apply the induction hypothesis. Each such Tα+1 sentence θ is equivalent to 0 a sentence θ in Tα, formed by moving the introduction of the corresponding variables and their definitions into the ψ part of θ0. For instance, if x = Xα i φi appears in θ, then it does not appear and θ0, and ψ0 (the ψ part of θ0) asserts about the set of things satisfying φi in ˙ (Lα˙ [S,˙ x˙]; S,˙ x,˙ δ (δ ∈ di), ∈) what ψ says about xi (this involves rewriting the atomic formulas of ψ). The formula ψ may also need to be rewritten to deal with the predicates for S and x, resulting (if α < δ) in formulas ψ0 and ψ1 to accommodate the two possible values of x(α). Letting θ0 and θ1 be the two rewritten forms of θ, we can let

ρθ,α+1 be (Pα ∧ ρθ0,α) ∨ ((¬Pα) ∧ ρθ1,α). The rest of this step is the same as the limit case. To complete the proof, ﬁx an ordinal γ > δ ∪ sup(S) such that for all x ∈ 2δ, Lγ [S, x] |= θ(S, x) if and only if L[S, x] |= θ(S, x), and note that there is an δ element of Tγ which, for each x ∈ 2 , holds in the expanded version of Lγ [S, x] if and only if L[S, x] |= θ(S, x). That (3) implies (4) is immediate. The implication from (4) to (2) follows from the absoluteness of the existence of winning strategies in closed games (via a ranking function, as in [4]). To get from (1) to (3), consider a witness φ to (1) as a tree. Players I and II choose a path from the root of the tree to a terminal node, with I choosing at disjunctions, II choosing at conjunctions, and I winning if and only if x(γ) = 1, for Pγ the terminal node reached by the run of the game. If x |= φ, then I can play to maintain that x satisﬁes each sub-sentence visited during the run of the game, and conversely for II.

∗ 9.0.7 Remark. The construction of the L∞,δ-sentence in the implication from (2) to (1) in the proof of Theorem 9.0.6 is carried out in L[S], aside from the choice of γ. 101

9.0.8 Remark. Part (2) of Theorem 9.0.6 makes sense for subsets of spaces of the form ωδ and their products (or sets of subsets of any set in L), so we will say that a set A ⊆ (ωδ)n (for some n ∈ ω and some ordinal δ) is ∞-Borel is there exists a set S of ordinals such that A = {x ∈ (ωδ)n : L[S, x] |= φ(S, x)}, for some binary formula φ. One could also revise the infinitary language L∞,δ above to δ describe (for instance) subsets of ω , starting with variables Pα,m (α ∈ δ, m ∈ ω) interpreted so that x |= Pα,m if and only if x(α) = m. Under this formulation everything goes through as above (trivially, as we are merely relabeling the set ω of propositional variables). In this sense, however, ω is ℵ1-Borel as a subset of ω n P(ω × ω), but not ℵ0-Borel. When we describe subsets of (ω ) in this revised sense (i.e., when we say that some set of ordinals is an ∞-Borel code for a subset of (ωω)n) then we restrict our attention to κ-Borel sets for uncountable κ. 9.0.9 Definition. Given a set of ordinals S and a binary formula φ from the first order language of set theory, we say that the pair (S, φ) is an ∞-Borel ∗ code for the set {x ∈ (ωδ)n : L[S, x] |= φ(S, x)}. If in addition S is a subset of an uncountable ordinal γ, we say that the pair (S, φ) is a γ-Borel ∗ code for the set

δ n {x ∈ (ω ) : Lγ [S, x] |= φ(S, x)}.

9.0.10 Remark. The proof of Theorem 9.0.6 shows that, for any uncountable cardinal γ, every γ-Borel subset of 2ω has a γ+-Borel∗ code, and every subset of 2ω which has a γ-Borel∗ code is γ+-Borel. 9.0.11 Remark. If A ⊆ ωω is ∞-Borel, then this is witnessed by a pair (S, φ) with S a bounded subset of Θ. This can be shown by taking a continuum-sized ω elementary submodel (containing ω ) of a suitable model of the form Lα(T, R), where (T, φ) is a given ∞-Borel code for A. Since every element of Lα(T, R) is ω deﬁnable in Lα(T, R) from T , an ordinal and an element of ω , such a submodel can be built assuming only ZF.

ℵ0 9.0.12 Remark. If A is ∞-Borel and ℵ1 6≤ 2 then A has the property of Baire and is Lebesgue measurable. To see this for the property of Baire, let P be Cohen forcing and letc ˙ be the canonical P-name for the generic Cohen real. Suppose ∗ ℵ0 ω that (S, φ) is an ∞-Borel code for A. Since ℵ1 6≤ 2 and L[S] |= AC, ω ∩L[S] is countable, so the set of reals which are Cohen-generic over L[S] is comeager. 0 For each such real x, let gx be the filter for whichc ˙gx = x. Let A be the set of x which are Cohen-generic over L[S] for which there is a condition p ∈ gx forcing that L[S, c˙] |= φ(S, c˙). Then A0 is Borel, and the symmetric difference A4A0 is contained in the set of reals which are not Cohen-generic over L[S]. For Lebesgue measurability the argument is the same, using random forcing instead of Cohen forcing. One can use the same argument with Mathias forcing to show that A (under the same assumptions) satisfies the Ramsey property. Unlike the cases of the Baire property and Lebesgue measurability, it is still unknown whether AD alone implies that every subset of 2ω has the Ramsey property. 102 CHAPTER 9. ∞-BOREL SETS

9.0.13 Example. One can define, by recursion on α, L∞,ω-sentences φT,α, for ω all (set-sized) trees T on ω × Ord such that AφT,α is the set of x ∈ ω for which <ω the Tx = {t ∈ Ord :(x|t|, t) ∈ T } has rank at most α. Starting with the standard tree on ω × ω whose projection is the set of (codes for, under some bijection π : ω → ω × ω) illfounded linear orders on ω, this operation gives a sequence of Lℵ1,0-sentences hφα : α < ω1i such that each Aφα is the set of codes for wellorders of ω of ordertype α. 9.0.14 Remark. It follows from Theorem 9.0.6 that Suslin subsets of 2ω are ∞-Borel. Theorem 6.1.1 says that Suslin sets can be uniformized. The proof of the theorem shows that if γ is an ordinal and T is a tree on ω × γ such that p[T ] is nonempty, then p[T ] has a member in L[T ]. However, it is possible to have φ ∈ L∞,0 such that Aφ is nonempty in V but empty in L[φ]. For instance, there is a φ in L such that Aφ is the set of Cohen reals over L. Another way to see this is to note that consistently there are ∞-Borel subsets of (ωω)2 which are not uniformized (for instance, in L(R), assuming AD, by Corollary 9.1.6, Theorem 6.1.4) whereas if S were an infinity Borel code for a set A ⊆ (ωω)2 and L[S, x] ∩ A where nonempty for each x, then a uniformizing function could be found by picking the constructibly least member of each set L[S, x] ∩ A relative to S and x. Given a cardinal κ, we say that a subset of topological space X is weakly κ-Borel if it is in the smallest collection of subsets of X containing the closed sets and closed under unions and intersections of size less than κ. A set is weakly ∞-Borel if it is weakly κ-Borel for some cardinal κ. Some authors call these sets “∞-Borel” and use “effectively ∞-Borel” for the notion of ∞-Borel used here.1 9.0.15 Remark. Suppose that AD holds, and that ∆ is a projective algebra S which is not closed under countable unions. By Theorem 4.4.2, ω has the prewellordering property. By the first Periodicity Theorem (see Remark 4.2.6), ωω S ωω S 0 ˇ ∀ ω ∆ does as well. Let Γ = ∀ ω ∆ and let ∆ = Γ ∩ Γ. By Theorem 4.3.2, every weakly δ(∆)0-Borel subset of ωω is in ∆0. We will make use of the following lemma in Section 11.4.2 Lemma 9.0.16. Assume that AD holds, and suppose that κ is a limit of Suslin cardinals and that κ has uncountable cofinality. Then every set which is <κ- Borel is also <κ-Suslin. Proof. We make use of Remark 6.0.3 and the comments immediately before. Let γ < κ be a limit of Suslin cardinals of countable cofinality. By Remark 6.0.8, S<γ is a projective algebra which is not closed under countable unions. ωω S 0 ˇ Letting Γ = ∀ ω S<γ and ∆ = Γ ∩ Γ, we have by Remark 9.0.15 that every 0 <γ-Borel set is in ∆ . Since Sγ is closed under countable unions, it contains S 0 ωω ω S<γ . If γ > γ is a Suslin cardinal greater than γ, then Sγ0 is ∃ -closed 0 and contains the complement of every set in Sγ , so ∆ ⊆ Sγ0 . 1Say something more explicit about the difference? 2The lemma is not optimal. 9.1. LOCAL ∞-BOREL CODES 103

9.1 Local ∞-Borel codes

ω n Given a set A ⊆ (ω ) (for some n ∈ ω) let δA denote the supremum of the lengths of the prewellorderings P such that either P ≤W A or P ≤W Aˇ. Recalling from Remark 2.5.2 that F c,1,b is the set of x ∈ ωω such that c,1,n ω ω n fx is a continuous function from ω to (ω ) , let ≤A be the relation on (ωω)3 defined by setting (x , y , z ) ≤ (x , y , z ) if x = x ,(f c,1,n)−1[A] is a 0 0 0 A 1 1 1 0 1 x0 prewellordering and ((y , z ), (y , z )) ∈ (f c,1,n)−1[A]. Then ≤ is a wellfounded 0 0 1 1 x0 A relation projective in A, and the rank of ≤A is δA. If µT is an ultrafilter on the Turing degrees, then, by Theorem 8.0.2, δA < δω(A). By Theorem 8.1.1, assuming AD + DCR, there is a regular cardinal in the interval (δA, δω(A)). ∗ Let δA be the least such regular cardinal. In this section we show (assuming ∗ + AD + DCR) that if A is ∞-Borel then it is (δA) -Borel. We define the rank rk(φ) of a sentence in L∞,0 by setting

• rk(Pn) = 0 for each propositional variable Pn;

• rk(¬φ) = rk(φ) for all φ ∈ L∞,0; W V • rk( α∈Y φα) = rk( α∈Y φα) = sup{rk(φα) + 1 : α ∈ Y } for all sets Y ⊆ Ord and all formulas φα (α ∈ Y ) in L∞,0,

∗ For each φ ∈ L∞,0 we deﬁne the formula φ ∈ L∞,0 by recursion on the rank of φ as follows.

∗ • Pn = Pn, for each each propositional variable Pn;

∗ ∗ • (¬φ) = ¬φ , for all φ ∈ L∞,0;

• for all sets Y ⊆ Ord, and all L∞,0-sentences φα (α ∈ Y ), W ∗ W ∗ S – ( α∈Y φα) = α∈X φα, where X = {α ∈ Y : Aφα 6⊆ β∈α∩Y Aφβ }; V ∗ V ∗ T – ( α∈Y φα) = α∈X φα, where X = {α ∈ Y : Aφα 6⊇ β∈α∩Y Aφβ }, ∗ ∗∗ ∗ Then for all φ ∈ L∞,0, Aφ = Aφ∗ , rk(φ) ≥ rk(φ ) and φ = φ . We deﬁne the following ordinals, all of which are at most Θ (and which, by Theorem 9.1.7, which in turn uses a deeper theorem of Woodin to be proved 3 later, are all the same under AD + DCR):

• χB, the least χ such that all ∞-Borel sets are χ-Borel;

• κB, the supremum of the lengths of wellordered sequences of ∞-Borel codes for nonempty disjoint sets;

• λB, the supremum of the lengths of ∞-Borel prewellorderings.

• ρB, the supremum of the Wadge ranks of ∞-Borel sets.

Theorem 9.1.1. Assuming AD + DCR, χB ≤ κB ≤ λB. 3!!! 104 CHAPTER 9. ∞-BOREL SETS

Proof. To see that χB ≤ κB, let φ be any sentence in Lκ,0. For any disjunction W ∗ W α<γ ψα in φ the sentences ψα ∧ ¬( β<α ψβ) are ∞-Borel codes for nonempty disjoint sets. A similar remark applies to the conjunctions. It follows that φ∗ is in LκB,0. To see that κB ≤ λB, note that one can convert a sequence of ∞-Borel codes for nonempty disjoint sets to an ∞-Borel code for a prewellordering in a uniform fashion by means of uniformly coding the product. Consider hφα : α < γi with the Aφα ’s disjoint and nonempty. For α ≤ β take _ Aφα × Aφβ . α≤β

If each φα is in Lξ,0, the resulting code is in Lξ∪γ+,0.

4 ω We observed above that ρB = Θ if and only if all subsets of ω are ∞-Borel. The same is true for λB (as, as well shall see, for all four ordinals).

Theorem 9.1.2. If Wadge determinacy and DCR hold, then λB = Θ if and only if every subset of ωω is ∞-Borel. Proof. By deﬁnition, Θ is the supremum of the lengths of the prewellorderings ω ω of ω , so if every subset of ω is ∞-Borel then λB = Θ. For the other direction, suppose that A ⊆ ωω is not ∞-Borel and that P is an ∞-Borel prewellorder. Then A 6≤W P as the ∞-Borel sets form an initial segment of the Wadge hierar- ω chy. Thus either P ≤W A or P ≤W ω \A, and in either case |P | < δA < Θ.

Theorem 9.1.3. Assuming AD+DCR, the following statements are equivalent.

• At least one of χB, κB, λB and ρB is Θ. • All subsets of ωω are ∞-Borel

• χB = κB = λB = ρB = Θ Proof. All that remains to be shown is that if all subsets of ωω are ∞-Borel then χB = Θ. This follows from the Moschovakis Coding Lemma, which implies that for each χ < Θ there is a surjection from ωω to P(χ), and therefore one from ωω to the χ-Borel sets. An argument similar to the proof of Theorem 9.1.2 (given in the following ∗ + proof) shows that (δA) ≥ χB for every ∞-Borel set A. ω Theorem 9.1.4. Suppose that AD + DCR holds, and let A ⊆ 2 be ∞-Borel. ∗ + Then A is (δA) -Borel. ∗ Proof. If δA ≥ χB we are done. Supposing otherwise, using a sentence φ in ∗ L of minimal rank such that φ 6∈ L ∗ , we may ﬁnd (using the regularity of ∞,0 δA,0 ∗ ∗ δ ) a sequence hφ : α < δ i, consisting of L ∗ sentences, such that the sets A α A δA,0

Aφα are nonempty and pairwise disjoint. Then the prewellordering built in the

4where? assuming what? 9.1. LOCAL ∞-BOREL CODES 105

∗ + second half of the proof of Theorem 9.1.1 is (δA) -Borel, and, as it has length greater than δA, it is not Wadge below either A or Aˇ. By Wadge Determinacy, ∗ + then, A is Wadge below ≤S, which means that A is (δA) -Borel, by Remark 9.0.2.

ω Suppose (for this paragraph) that AD + DCR holds, and let A ⊆ 2 be ω ∗ ∞-Borel. Let B ⊆ ω have Wadge rank δA. By Proposition 2.5.8, there is a c prewellordering of F of length δA which is projective in B. By the Moschovakis ∗ Coding Lemma, every subset of δA is coded (relative to this prewellordering) by a set of reals projective in B. Putting this all together with Theorem 9.1.4 gives the following corollaries.

ω Corollary 9.1.5. Suppose that AD + DCR holds, and let A ⊆ 2 be ∞-Borel. Then A = A for some φ ∈ L ∗ + ∩ L(A, ). φ (δA) ,0 R Corollary 10.3.7 below is a stronger version of the following corollary, and its proof does not use Theorem 6.1.5.

Corollary 9.1.6. Suppose that AD holds. Then, in L(R), every subset of 2ω is ∞-Borel.

Proof. We work in L(R), noting that the hypothesis imply that AD and DCR hold there (the latter by Theorem 0.4.1. Supposing that some subset of 2ω is not ∞-Borel, there exists an ordinal γ such that

• in Lγ (R), Θ (as deﬁned in Lγ (R)) is a limit cardinal; ω 1 • for every A ∈ P(ω ) ∩ Lγ (R), Σ∼1(A) ⊆ Lγ (R); ω • for some A ∈ P(2 ) ∩ Lγ (R), A is not ∞-Borel in Lγ (R). 1 Since L 2 (R) is Σ1-elementary in L(R), there is a γ0 <∼ δ 2 satisfying these ∼δ 1 properties. By the Moschovakis Coding Lemma, Lγ0 (R) is correct about cardinals below its version of Θ. By Theorem 6.1.5, all sets in L 1 (R) are Suslin, ∼δ 2 which by Theorem 9.1.4 implies that in Lγ0 (R) all sets are ∞-Borel, giving a contradiction.

We conclude this section by showing that the four ordinals deﬁned above are 5 all the same under AD + DCR.

Theorem 9.1.7. Assume AD + DCR. Then

L(P(χB )) χB = κB = λB = ρB = Θ .

Proof. By Theorem 9.1.3, the theorem holds in the case where all subsets of ω ω ω are ∞-Borel. Suppose then that some A ⊆ ω is not ∞-Borel. Since DCR holds, L(A, R) satisﬁes DC. Since the ∞-Borel sets form an initial segment 5Using Hugh’s theorem that all sets in L(P(λ)) are ∞-Borel, which isn’t in the book yet, and which will appear at a later point anyway. 106 CHAPTER 9. ∞-BOREL SETS of the Wadge hierarchy, they are all in L(A, R). By Theorem 9.1.4, every ∞- Borel subset of ωω is ∞-Borel in L(A, R). By the Moschovakis Coding Lemma, L(A,R) L(A, R) contains all bounded subsets of its Θ. It follows that χB = χB , L(A,R) L(A,R) λB = λB and ρB = ρB . ω By Theorem ?? (applied in L(A, R)), every subset of ω in L(P(χB)) is ∞-Borel.6 It follows then that the ∞-Borel sets are exactly the subsets of ωω L(P(χB)) in L(P(χB)), all of which are ∞-Borel in L(P(χB)). Moreover, χB = χB , L(P(χB)) L(P(χB)) λB = λB and ρB = ρB . The theorem then follows from Theorems 9.1.1 (applied in V ) and 9.1.3 (applied in L(P(χB))).

6Maybe change the reference once we prove it? Chapter 10

Vopˇenka algebras

Recall that for a set X, ODX is the class of all sets which are definable from a finite sequence from X ∪ Ord. We let HODX denote the class of all sets in ODX whose transitive closures are contained in ODX . When the transitive closure of {X} is contained in ODX , HODX is a model of ZF; when in addition a wellordering of X exists in ODX , HODX is a model of ZFC. One can find an introduction to HOD on pages 194-195 of [7], including a proof that it satisfies 1 2 ZFC. The presentation there relativizes to HODX . Vopˇenka’s theorem (see page 249 of [7]) says that every set of ordinals is set- generic over HOD. Theorem 10.0.1 below is a relativized version of this fact. Relativizing to the sets X and Z there requires only minor modifications of the proof. The rest of this chapter contains variations of Vopˇenka’s theorem due to Woodin. Suppose that X and Z are sets with Z ∈ ODX . Let κ be the least cardinal Vλ λ such that X ⊆ Vλ and every subset of P(Z) in ODX is in ODX . Let PZ,X be the set of triples (n, x,¯ α¯) such that

• n is the Gödelnumber of a ternary formula φ, • x¯ is a finite sequence of elements of X, • α¯ is a finite sequence of elements of κ and

• the set Bn,x,¯ α¯ = {C ⊆ Z : Vκ |= φ(C, x,¯ α¯)} is nonempty.

The set PZ,X is in HODX , as is the reﬂexive and transitive ordering ≤Z,X on ¯ PZ,X deﬁned by setting (n, x,¯ α¯) ≤Z,X (m, y,¯ β) if Bn,x,¯ α¯ ⊆ Bm,y,¯ β¯. Let ≡Z,X be the induced equivalence relation on PZ,X , and for each (n, x,¯ α¯) ∈ PZ,X , let [n, x,¯ α¯]Z,X denote the corresponding equivalence class. We let VZ,X (the Vopˇenka algebra for X on Z) be the partial order whose domain is the set of ≡Z,X -classes of PZ,X , with the order inherited from ≤Z,X . We write VZ when X = ∅. 1Refer the reader to Kunen also for the basics of HOD? 2check

107 108 CHAPTER 10. VOPENKAˇ ALGEBRAS

Theorem 10.0.1. Let X be a set with TC({X}) ⊆ ODX , and let γ be an ordinal. For each A ⊆ γ, there is a HODX -generic ﬁlter G ⊆ Vγ,X such that

HODX∪{A} = HODX [G].

Proof. Fix A ⊆ γ. If D is a subset of Pγ,X and an element of HODX , then [ {Bn,x,¯ α¯ :(n, x,¯ α¯) ∈ D} is ODX . It follows that if this union is not all of P(γ), then there is a triple ¯ (m, y,¯ β) in Pγ,X such that [ Bm,y,¯ β¯ = P(γ) \ {Bn,x,¯ α¯ :(n, x,¯ α¯) ∈ D}.

From this it follows that the set of equivalences classes of triples (n, x,¯ α¯) ∈ Pγ,X for which A ∈ Bn,x,¯ α¯ is a HODX -generic filter for Vγ,X . Let GA denote this filter. The definition just given shows that GA is in HODX∪{A} and therefore that HODX∪{GA} is contained in HODX∪{A}. The set K consisting of those pairs (δ, (n, x,¯ α¯)) ∈ γ × Pγ,X for which

Bn,x,¯ α¯ = {C ⊆ γ : δ ∈ C} is also a member of HODX . Since A is equal to

{δ : ∃ [n, x,¯ α¯]Z,X ∈ GA (δ, (n, x,¯ α¯)) ∈ K},

it follows that A is in HODX [GA]. Since HODX [GA] ⊆ HODX∪{GA}, it follows also that A is in HODX∪{GA}, HODX∪{A} = HODX∪{GA} and that each of these models contains HODX [GA]. To see that HODX∪{A} ⊆ HODX [GA], it suffices to see that every subset of <ω X × Ord in HODX∪{A} is in HODX [GA], since, in HODX∪{A}, every set is a surjective image of X<ω × ζ, for some ordinal ζ. Fix then an ordinal ζ and a set <ω Q ⊆ X × ζ in HODX∪{A}. Fix a quaternary formula φ, a finite tuplex ¯ from X and a finite set of ordinalsα ¯ such that Q = {q ∈ X<ω × ζ : φ(q, x,¯ α,¯ A)}. Let T be

¯ <ω {(q, (n, y,¯ β)) ∈ (X × ζ) × Pγ,X : Bn,y,¯ β¯ = {C ⊆ γ : φ(q, x,¯ α,¯ C)}}.

Then T is in HODX , and Q, which is ¯ ¯ {q : ∃ [n, y,¯ β]γ,X ∈ GA (q, (n, y,¯ β)) ∈ T }, is in HODX [GA]. The proof of the following theorem, due to Woodin, uses the proof of Theo- rem 10.0.1. Theorem 10.0.2. Let S be a set of ordinals, and suppose that V = L(A), for some transitive set A. Then HOD{S} = L[B], for some set of ordinals B. 109

Before proving Theorem 10.0.2, we prove a lemma (a variation of Theorem 8.2.6), and prove the theorem in the case V = L[A], for A a set of ordinals.

Lemma 10.0.3. Let M1 and M2 be transitive models of ZFC, with M1 ⊆ M2. Suppose that P is a partial order in M1, G ⊆ P is an M2-generic ﬁlter and M2 ⊆ M1[G]. Then M1 = M2.

Proof. Since M1 and M2 are models of ZFC, it is enough to see that every set of ordinals in M2 is in M1. Letting X be a set of ordinals in M2, we have that X = τG, for some P-name τ in M1. Since G is M2-generic, this means that some condition in P decides all of τ, so X is in M1. Now, suppose that V = L[A], where A is a subset of an ordinal γ. Since Choice holds in HOD{S}, there exist in HOD{S} an ordinal η and bijection π : Vγ,{S} → η. Let ≤γ be the partial order on η induced by π. Let K be the set of pairs (δ, (n, x,¯ α¯)) ∈ γ × Pγ,{S} such that Bn,x,¯ α¯ = {C ⊆ γ : δ ∈ C} (i.e., the set K from the proof of Theorem 10.0.1 with {S} as X), and let Kγ be the π-image of K, that is,

{(δ, π([(n, x,¯ α¯)]γ,{S})) : (δ, (n, x,¯ α¯)) ∈ K}.

Then ≤γ and Kγ are in HOD{S}. By Theorem 10.0.1, A is Vγ,{S}-generic over HOD{S}, via the generic ﬁlter GA. Then π[GA] is HOD{S}-generic for ≤γ , and A is in L[≤γ ,Kγ ][GA]. Then

V = L[A] = HOD{S}[GA] = L[≤γ ,Kγ ][π[GA]], which implies that HOD{S} = L[≤γ ,Kγ ], by Lemma 10.0.3, with L[≤γ ,Kγ ] as M1 and HOD{S} as M2.

We introduce a new partial order to deal with the general case. Recall that, given an infinite set Z, Col∗(ω, Z) is the partial order of finite partial injections from ω to Z (each with domain some n ∈ ω), ordered by extension. Given a ∗ S filter G ⊆ Col (ω, Z), let aG = {(i, j) ∈ ω × ω : g(i) ∈ g(j)}, where g = G. Then aG is a subset of ω × ω, and, if Z is transitive, V [G] = V [aG]. As above, suppose that X and Z are sets (with Z infinite) such that Z is in Vλ ODX , and let κ be the least cardinal λ such that X ⊆ Vλ, Z is in ODX and ∗ Vλ ω every subset of P(Col (ω, Z)) in ODX is in ODX . Let PZ,X be the set of tuples (n, m, x,¯ α¯) such that

• n ∈ ω,

• m is the G¨odelnumber of a ternary formula φ,

• x¯ is a ﬁnite sequence of elements of X,

• α¯ is a ﬁnite sequence of elements of κ and 110 CHAPTER 10. VOPENKAˇ ALGEBRAS

∗ • the set Bn,m,x,¯ α¯ = {p ∈ Col (ω, Z): Vκ |= φ(p, x,¯ α¯)} is nonempty, and all members of Bn,m,x,¯ α¯ have domain n.

ω Given (n, m, x,¯ α¯) ∈ PZ,X and k ≤ n, let Bn,m,x,¯ α¯k denote the set

{pk : p ∈ Bn,m,x,¯ α¯}.

ω This set is evidently also in HODX . The set PZ,X is in HODX , as is the reﬂexive ω ω and transitive ordering ≤Z,X on PZ,X deﬁned by setting

ω ¯ (n, m, x,¯ α¯) ≤Z,X (k, j, y,¯ β) if n ≥ k and Bn,m,x,¯ β¯k ⊆ Bk,j,y,¯ β¯. ω ω ω Let ≡Z,X be the induced equivalence relation on PZ,X , and let [n, m, x,¯ α¯]Z,X ω ω denote the ≡Z,X -class of a tuple (n, m, x,¯ α¯). Let VZ,X be the partial order ω ω whose domain is the set of ≡Z,X -classes of PZ,X , with the order inherited from ω ω ≤Z,X . Each injection f : ω → Z deﬁnes a ﬁlter Hf on VZ,X , consisting of the ω ω ≡Z,X -classes of those (n, m, x,¯ α¯) ∈ PZ,X for which fn is in Bn,m,x,¯ α¯.

Lemma 10.0.4. For any set X with TC({X}) ⊆ ODX , and any infinite set ∗ S Z in ODX , if G ⊆ Col (ω, Z) is a V -generic filter, and g = G, then Hg is ω VZ,X -generic over HODX , and ag is in HODX [Hg]. ω Proof. To see that Hg is VZ,X -generic over HODX , let D be a dense open subset ω ∗ 0 of VZ,X in HODX , and let p be a condition in Col (ω, Z). Let D be the set of ω ω (n, m, x,¯ α¯) ∈ PZ,X with [(n, m, x,¯ α¯)]Z,X ∈ D. Applying the genericity of G, it suffices to find a condition p0 ≤ p and a tuple (n, m, x,¯ α¯) in D0 such that p0 is in Bn,m,x,¯ α¯. Let np be the domain of p. The set

[ 0 {Bn,m,x,¯ α¯np :(n, m, x,¯ α¯) ∈ D , n ≥ np}, being in ODX , must be the set of injections from np to Z, since otherwise the ¯ complement of this set has the form Bnp,j,y,¯ β¯ for some j and β, and we get a 0 contradiction by considering a (n, m, x,¯ α¯) ∈ D with B n ⊆ B ¯. n,m,x,¯ α¯ p np,j,y,¯ β It follows that there exist p0 ≤ p and (n, m, x,¯ α¯) as desired. As in the proof of Theorem 10.0.1, let K be the set of pairs

ω ((i, j), (n, m, x,¯ α¯)) ∈ (ω × ω) × PZ,X such that {i, j} ⊆ n and p(i) ∈ p(j) for all p ∈ Bn,m,x,¯ α¯. Then again K is in HODX , and, as

ω ag = {(i, j) ∈ ω × ω : ∃ [n, m, x,¯ α¯]Z,X ∈ Hg ((i, j), (n, m, x,¯ α¯)) ∈ K}, ag is in HODX [Hg]. Theorem 10.0.2 now follows. 10.1. CODES FOR PROJECTIONS, AND UNIFORMIZATION 111

Proof of Theorem 10.0.2. We may assume that A = Vα, for some inﬁnite ordinal α (so that A is ordinal deﬁnable). In HOD{S}, there exist an ordinal γ and ω bijection π : VA,{S} → γ. Let ≤γ be the partial order on γ induced by π. Let K be the set of pairs

ω ((i, j), (n, m, x,¯ α¯)) ∈ (ω × ω) × PA,{S} such that {i, j} ⊆ n and p(i) ∈ p(j) for all p ∈ Bn,m,x,¯ α¯ (i.e., the set K from the proof of Lemma 10.0.4 with A as Z and {S} as X), and let Kγ be the π-image of K, that is,

ω {((i, j), π([(n, m, x,¯ α¯)]A,{S})) : ((i, j), (n, m, x,¯ α¯)) ∈ K}. ∗ S Let G0 ⊆ Col (ω, A) be a V -generic ﬁlter and let g = G0. Then π[Hg] is a HOD{S}-generic ﬁlter for ≤γ . As in the proof of Lemma 10.0.4, ag is in the model L[≤γ ,Kγ ][π[Hg]], and therefore so is A. Applying Lemma 10.0.3 with L[≤γ ,Kγ ] as M1, HOD{S} as M2, ≤γ as P and π[Hg] as G, we get that L[≤γ ,Kγ ] = HOD{S}. As the model L[≤γ ,Kγ ] has the form L[B], for some set of ordinals B, we are done.

10.1 Codes for projections, and Uniformization

This section uses the results of Chapter 8, using the notation ≤S, µS, DS, etc. introduced in Definition 8.0.3. Theorem 1.1.2 shows that AD implies that, for each set of ordinals S, µS is an ultrafilter on the corresponding set DS; if CCR holds (as it does under AD) then µS is countably complete. If there exists any set of ordinals S for which µS is a countably complete ultrafilter on DS, then V ℵ0 ω1 is measurable (see Remark 1.1.7), so ℵ1 6≤ 2 . ω S For each set S of ordinals, and each x ∈ ω , let Qx denote the definably L[S,x] L[S,x] least poset on the ordinals isomorphic in HODS to the poset Vω,S (via the L[S,x] S definability order on HODS , say), and let Kx (also a set of pairs of ordinals) denote the corresponding version of the set K from the proof of Theorem 10.0.1 S S (i.e., relabeled as in the proof of Theorem 10.0.2 to refer to Qx ). Note that Qx depends only on [x]S. We will use this notation throughout this section, and in other places.

Theorem 10.1.2 below shows that if DCR holds and µS is an ultraﬁlter for each set of ordinals S then the collection of ∞-Borel sets of reals is projectively closed. We start with a lemma whose hypothesis is weaker (and appeal to the ℵ0 V fact that ℵ1 6≤ 2 is equivalent to the assertion that ω1 is strongly inaccessible in every inner model satisfying Choice).

ℵ0 Lemma 10.1.1. Let S be a set of ordinals, and suppose that ℵ1 6≤ 2 . Let φ be a ternary formula and let B denote the set

{x ∈ ωω : ∃y ∈ ωω L[S, x, y] |= φ(S, x, y)}.

Then following are equivalent, for each x ∈ ωω. 112 CHAPTER 10. VOPENKAˇ ALGEBRAS

1. x ∈ B

ω S S 2. For some w ∈ ω , for all z ≥S w, L[S, Qz ,Kz , x] |= “there is some QS ω p ∈ Col(ω, 2 z ) forcing that there exists a y ∈ ω such that L[S, x, y] |= φ(S, x, y).”

ω S S QS 3. For some z ∈ ω , L[S, Qz ,Kz , x] |= “there is some p ∈ Col(ω, 2 z ) forcing that there exists a y ∈ ωω such that L[S, x, y] |= φ(S, x, y).”

Proof. That (2) implies (3) is immediate. To see that (3) ⇒ (1), fix one such S 2ℵ0 L[S,z] V z. The cardinality of Qz in L[S, z] is at most (2 ) , which is below ω1 , V V as ω1 is strongly inaccessible in L[S, z]. Since ω1 is strongly inaccessible in S S L[S, Qz ,Kz , x], S S S P(Qz ) ∩ L[S, Qz ,Kz , x] S S S is countable, and L[S, Qz ,Kz , x]-generic filters for Qz exist below each condi- QS S tion. Since forcing with Col(ω, 2 z ) adds a generic filter for Qz , statement (1) follows. ω ω For (1) ⇒ (2), fix y0 ∈ ω such that L[S, x, y0] |= φ(S, x, y0) and z ∈ ω such that z ≥S y0 and z ≥S x. Letting Gz ⊆ Qz be the filter induced L[S,z] by z as in the proof of Theorem 10.0.1, Gz is HODS -generic. It follows S S S S that Gz is Qz -generic over L[S, Qz ,Kz ] and (using Kz ) that z is a mem- S S S S S S ber of L[S, Qz ,Kz ][Gz]. Since x is in L[S, Qz ,Kz ][Gz], L[S, Qz ,Kz , x] is a S S S generic extension of L[S, Qz ,Kz], and L[S, Qz ,Kz ][Gz] is a generic extension S S |QS | L[S,QS ,KS ] of L[S, Qz ,Kz , x] by a partial order of cardinality at most (2 z ) z z S S QS in L[S, Qz ,Kz , x]. This partial order regularly embeds into Col(ω, 2 z ) in S S 3 L[S, Qz ,Kz , x]. This gives (2).

If DC holds in L(S, R) (as it does if DCR holds), then for each set S of ordinals the ultraproduct DS V /µS

(formed using all S-invariant functions in L(S, R)) is wellfounded in L(S, R). We ∞ ∞ ∞ Q let S , QS and KS be the sets in z∈ωω V/µS represented by the functions S S z 7→ S and z 7→ Qz and z 7→ Kz , respectively.

Theorem 10.1.2. Assume that DCR holds. Let S be a set of ordinals, and suppose that µS is an ultraﬁlter. Let φ be a ternary formula, let S be a set of ordinals, and let B be

{x ∈ ωω : ∃y ∈ ωω L[S, x, y] |= φ(S, x, y)}.

Then there exist a set of ordinals T in OD{S} and a binary formula ψ such that

B = {x ∈ ωω : L[T, x] |= ψ(T, x)}.

3Explain regularly embeds, and cite McAloon? 10.1. CODES FOR PROJECTIONS, AND UNIFORMIZATION 113

V Proof. By DCR, µS is countably complete. Since µS is an ultrafilter, ω1 is a ℵ0 measurable cardinal, which implies that ℵ1 6≤ 2 (as in Remark 1.1.7). We work in L(S, R), which satisfies DC by the assumption that V satisfies DCR. Since DC holds in L(S, R), for each x ∈ ωω the ultraproduct

Y S S L[S, Qz ,Kz , x]/µS z∈ωω is wellfounded in L(S, R). Furthermore, for each x ∈ ωω,

∞ ∞ ∞ Y S S L[S ,QS ,KS , x] = L[S, Qz ,Kz , x]/µS z∈ωω

∞ ∞ ∞ and, by Lemma 10.1.1, x is in B if and only if L[S ,QS ,KS , x] satisﬁes the Q∞ statement “there is some p ∈ Col(ω, 2 S ) forcing that

∃y ∈ ωω L[S∞, x, y] |= φ(S∞, x, y).”

∞ ∞ ∞ Then we can let T be set of the ordinals coding the triple (S ,QS ,KS ) under some ﬁxed coding in L of triples of ordinals by ordinals, and let ψ be the corresponding version of the statement just given.

The following is an immediate consequence of Theorem 10.1.2.

ω <ω Theorem 10.1.3. If TD+DCR holds, then the set of ∞-Borel subsets of (ω ) is projectively-closed.

10.1.4 Remark. Since T can be taken to be in OD{S} in the conclusion of Theorem 10.1.2, TD + DCR implies that there is an ordinal deﬁnable class-sized function (minimizing in the ordinal deﬁnability order from S) taking in ∞-Borel∗ codes (S, φ) for subsets of (ωω)2 and returning the correspoding ∞-Borel∗ codes for {x ∈ ωω : ∃y ∈ ωω L[S, x, y] |= φ(S, x, y)}.

Theorem 10.1.6 below is a uniformization result derived from Lemma 10.1.1 ω and Theorem 10.1.2. Given sets B and x, we let σB,x denote ω ∩ OD{B,x}. ω 2 ω ω Recall that for a set A ⊆ (ω ) and y ∈ ω , Ay denotes the set {z ∈ ω : ωω ω (y, z) ∈ A}, and that ∃ A denotes the set {y ∈ ω : Ay 6= ∅}. Such a set can ω be uniformized if there is a function f : ∃ω A → ωω such that (y, f(y)) ∈ A for ω all y ∈ ∃ω A. Recall also that, for x, y ∈ ωω, x ⊕ y denotes the element z of ωω such that, for all n ∈ ω, z(2n) = x(n) and z(2n + 1) = y(n). The proof of Theorem 10.1.6 goes through the following observation.

Lemma 10.1.5. Suppose that A and B are sets, with A ⊆ (ωω)2. Suppose that ω for a Turing cone of x ∈ ωω, for all y ∈ L[x] ∩ ∃ω A,

Ay ∩ σB,x 6= ∅.

Then A can be uniformized. 114 CHAPTER 10. VOPENKAˇ ALGEBRAS

Proof. Let x0 be a base for a cone witnessing the relevant assumption of the ωω lemma. Given y in ∃ A, deﬁne f(y) to be the OD{B,x0,y}-least z ∈ σB,x0⊕y with (y, z) ∈ A.

ℵ0 Theorem 10.1.6. Suppose that TD + DCR + ℵ1 6≤ 2 holds, and that (S, φ) ∗ ω 2 ∞ ∞ is an ∞-Borel code for a set A ⊆ (ω ) . Let B be a set, and let S , QS ∞ and KS be as computed in HODωω ∪{S,B}. Suppose that, for a Turing cone of ω ω ∞ ∞ ∞ x ∈ ω , ω ∩ σB,x 6⊆ L[S ,QS ,KS , x]. Then A can be uniformized.

ω Proof. We prove that for a Turing cone of x ∈ ωω, for all y ∈ ∃ω A ∩ L[S, x],

Ay ∩ σB,x 6= ∅,

ω and apply Lemma 10.1.5. Let x0 ∈ ω be such that for all x ≥T x0,

ω ∞ ∞ ∞ ω ∩ L[S ,QS ,KS , x] 6⊆ σB,x.

Applying TD, suppose toward a contradiction that for a Turing cone of x there ωω exists a y ∈ ∃ A ∩ L[S, x] such that Ay ∩ σB,x = ∅. Let x1 ≥T x0 be a base for a Turing cone witnessing this. By Lemma 10.1.1, for every y ∈ ωω,

∞ ∞ ∞ ∞ Col(ω,QS ) L[S ,QS ,KS , y] |= “V |= ∃z L[S, y, z] |= φ(S, y, z)”.

As in Section 8.2, let π be a recursive diﬀuse function and let U be, in the model ∞ ∞ ∞ 4 L[S ,QS ,KS , x1], an ultraﬁlter on ω not containing any π-weak set. Let P ∞ ∞ ∞ L[S ,QS ,KS ,x1] denote PU , again as in Section 8.2. Fix

∞ ∞ ∞ t ∈ σB,x1 \ L[S ,QS ,KS , x1].

∞ ∞ ∞ For any L[S ,QS ,KS , x1]-generic

g ⊆ P, and any ωω ∞ ∞ ∞ y ∈ ∃ A ∩ L[S ,QS ,KS , x1][g], ∞ ∞ ∞ ∞ Col(ω,QS ) ω L[S ,QS ,KS , x1][g] |= “V |= ∃z ∈ ω L[S, y, z] |= φ(S, y, z)”, ∞ ∞ ∞ ∞ ∞ ∞ since L[S ,QS ,KS , x1][g] is an outer model of L[S ,QS ,KS , y]. We have that

∞ ∞ ∞ Y S S L[S ,QS ,KS , x1] = L[S, Qx ,Kx , x1]/µS.

ℵ0 ∞ ∞ ∞ ω Since ℵ1 6≤ 2 , L[S ,QS ,KS , x1] ∩ P(P(ω )) is countable, and there is an ω x2 ≥T x1 in ω such that, for all x ≥T x2,

∞ ∞ ∞ ω S S ω 1. L[S ,QS ,KS , x1] ∩ P(P(ω )) = L[S, Qx ,Kx , x1] ∩ P(P(ω )), 4We may have to go back and isolate a lemma. 10.1. CODES FOR PROJECTIONS, AND UNIFORMIZATION 115

S S 2. P ∈ L[S, Qx ,Kx , x1], ∞ ∞ ∞ ω S S 3. all P-names in L[S ,QS ,KS , x1] for elements of ω are in L[S, Qx ,Kx , x1], S S ω 4. 1P forces over L[S, Qx ,Kx , x1] that for all y ∈ ω ,

Col(ω,QS ) ω V x |= ∃z ∈ ω L[S, y, z] |= φ(S, y, z).

∗ ∞ ∞ ∞ By Lemma 8.2.5, there exists an a ⊆ ω which is L[S ,QS ,KS , x1]-generic for P such that there is a w ∈ ωω ∩L[a∗, t] which HC-codes an ω-sequence listing all the members of P( × QS × ω) in L[S, QS ,KS , x ]. P x2 x2 x2 1 Let y be an element of

ωω ∞ ∞ ∞ ∃ A ∩ L[S ,QS ,KS , x1 ⊕ g]. The following claim ﬁnishes the proof of the theorem.

Claim. Ay ∩ σB,x1⊕g 6= ∅.

We prove the claim. Since t ∈ σB,x1⊕g and w ∈ L[g, t] there exists a L[S, QS ,KS , x ][g]-generic ﬁlter G ⊆ Col(ω, QS ) in OD . By item (4) x2 x2 1 x2 {B,x1,g} from the choice of x2, there is a z ∈ ωω ∩ L[S, QS ,KS , x ][g][G] x2 x2 1 such that L[S, y, z] |= φ(S, y, z). Since

ωω ∩ L[S, QS ,KS , x ][g][G] ⊆ σ , x2 x2 1 B,x1⊕g we are done.

In Section 8.2 we deﬁned, for sets S, T of ordinals, S ≤D T to mean that ωω ∩ L[S, x] ⊆ L[T, x] for a Turing cone of x. Theorem 10.1.6 shows that if TD, ℵ0 ω 2 DCR and ℵ1 6≤ 2 hold, and A is an ∞-Borel subset of (ω ) with an ∞-Borel code which is not ≤D-maximal, then A can be uniformized. The converse also holds.5

ℵ0 ω Corollary 10.1.7. Assume TD + DCR + ℵ1 6≤ 2 , and that every subset of ω is ∞-Borel. Then Uniformization is equivalent to the assertion that ≤D has no greatest element. Proof. The reverse direction is Theorem 10.1.6. The forward direction requires only that every subset of ωω is ∞-Borel. Let S be a set of ordinals, and let f : ωω → ωω uniformize the set {(x, y) ∈ (ωω)2 : y 6∈ L[S, x]}. Let T be an inﬁnity Borel code for {(x, i, j):(i, j) ∈ f(x)} via formula φ. Then for all (x, i, j), (i, j) ∈ f(x) if and only if L[T, x] |= φ(T, x, i, j), which shows that L[T, x] is closed under f for all x.67 5We seem to be saying the same thing twice, or contradicting ourselves. 6Maybe we have to say something to make this notation oﬃcial? 7Corollary again : Assuming Turing Determinacy plus DC, if A is ∞-Borel and x ∈ F (x) (?) then F (x) ⊆ H∞[t] x → ωω ∩ L[t, x] (?) 116 CHAPTER 10. VOPENKAˇ ALGEBRAS

10.2 Vopˇenka algebras and ∞-Borel sets

Given a set Y consisting of ordinals, the Vopˇenka algebra can also be varied to use ∞-Borel codes in OD{Y } to refer to sets. As before, we deﬁne one-step and ω-sequence versions. We ﬁx a cardinal κ such that for each n ∈ ω and each ω n A ⊆ (ω ) , if there exist a set of ordinals S in OD{Y } and a formula φ such that A = {x¯ ∈ (ωω)n : L[S, x¯] |= φ(S, x¯)} then there exist a set of ordinals S in Vκ ∩ OD{Y } and a formula ψ such that ω n A = {x¯ ∈ (ω ) : Lκ[S, x¯] |= ψ(S, x¯)}.

Let P∞,Y be the set of pairs (m, S) such that • m is the G¨odelnumber of a binary formula φ,

• S ∈ Vκ ∩ OD{Y } is a set of ordinals, ω • the set Bm,S = {x ∈ ω : Lκ[S, x] |= φ(S, x)} is nonempty.

Then again we have an induced ordering ≤∞,Y on P∞,Y deﬁned by setting

(m, S) ≤∞,Y (j, T ) if Bm,S ⊆ Bj,T . Let ≡∞,Y be the induced equivalence relation on P∞,Y . Let V∞,Y be the partial order whose domain is the set of ≡∞,Y -classes of P∞,Y , with the order inherited from ≤∞,Y . ω For the sequence version, let P∞,Y be the set of triples (n, m, S) such that • n ∈ ω, • m is the G¨odelnumber of a binary formula φ,

• S ∈ Vκ ∩ OD{Y } is a set of ordinals, ω ω n • the set Bn,m,S = {x¯ ∈ (ω ) : L[S, x¯] |= φ(S, x¯)} is nonempty. ω ω Then again we have an induced ordering ≤∞,Y on P∞,Y deﬁned by setting ω (n, m, S) ≤∞,Y (k, j, T ) ω ω ω if n ≥ k and {x¯k :x ¯ ∈ Bn,m,S} ⊆ Bk,j,T . Let ≡∞,Y be the induced equivalence ω ω relation on P∞,Y . Let V∞,Y be the partial order whose domain is the set of ω ω ω ≡∞,Y -classes of P∞,Y , with the order inherited from ≤∞,Y .

10.2.1 Remark. By Theorem 10.1.2, for each bounded S ⊆ κ in HOD{Y } and each ternary formula φ, there exist a bounded T ⊆ κ and a binary formula φ such that {x ∈ ωω : ∃y ∈ ωω L[S, x, y] |= φ(S, x, y)} = {x ∈ ωω : L[T, x] |= ψ(T, x)}. ω 10.2.2 Remark. Let (n, m, S) be an element of P∞,Y , and let φ be the formula with G¨odelnumber m. Let n0 be an element of ω \ n. Then

ω n0 ω n0 {x¯ ∈ (ω ) :x ¯n ∈ Bn,m,S} = {x¯ ∈ (ω ) : Lκ(S, x¯) |= φ(S, x¯n)}, ω so this set has the form Bn0,k,S for some k ∈ ω, and Bn0,k,S ≤∞,Y Bn,m,S. Moreover, there is a recursive function (not depending on S) sending each such tuple (n, m, n0) to a corresponding value k. 10.2. VOPENKAˇ ALGEBRAS AND ∞-BOREL SETS 117

10.2.3 Remark. The first paragraph of the proof of Theorem 10.0.1 adapts ω to prove that for each x ∈ ω , the set Gx consisting of those [(m, S)]∞,{Y } ∈ P∞,{Y } such that x ∈ Bm,S is a HOD{Y }-generic filter. This also follows from ω Lemma 10.3.5 below, using the fact that the first coordinate of a V∞,{Y }-generic sequence is generic for V∞,{Y }. For each i ∈ ω, x(i) is the unique j ∈ ω such that [(∅, mi,j)]∞,{Y } ∈ Gx, where mi,j is the Gödelnumber of the formula y(i) = j. It follows then that HOD{Y }[x] = HOD{Y }[Gx], since Gx is the set of [(m, S)]∞,Y ∈ V∞,Y such that Lκ[S, x] |= φ(S, x) (where φ is the formula with Gödelnumber m), and this can be computed in HOD{Y }[x]. Since every ω condition in V∞,{Y } is a member of Gx for some x ∈ ω , every HOD{Y }-generic filter G ⊆ V∞,{Y } is equal to Gx, for x = {(i, j) : [(∅, mi,j)]∞,{Y } ∈ G}. ω 10.2.4 Remark. Since the first coordinate of a V∞,{Y }-generic sequence is ˙ ω generic for V∞,{Y }, there is a V∞,{Y }-name Q such that V∞,{Y } is forcing- ˙ equivalent to V∞,{Y } ∗Q, via a map which carries the generic real for V∞,{Y } to ω the first coordinate of the generic sequence produced by forcing with V∞,{Y }. ω ω As with the partial order VZ,X above, each injection f : ω → ω defines a ω ω ω filter Hf on V∞,Y , consisting of the ≡∞,Y -classes of those (n, m, S) ∈ P∞,Y for ω which fn is in Bn,m,S. Lemma 10.3.5 is the version of Lemma 10.0.4 for V∞,Y . Theorem 10.1.2 and Remark 10.3.2 are used in the proof of the lemma. The remainder of this section uses the notation µS and DS introduced in Definition 8.0.3, for S a set of ordinals. Recall from Remark 1.1.5 that if µ∅ is a countably complete ultrafilter on D∅ (i.e., the constructibility degrees), then, for each set S ⊆ Ord, µS is a countably complete ultrafilter on DS.

Lemma 10.2.5. Suppose that DCR holds, and that µ∅ is an ultraﬁlter on D∅. ∗ ω S ω If G ⊆ Col (ω, ω ) is V -generic, and g = G, then Hg is V∞,Y -generic over HOD{Y }, and g is in HOD{Y }[Hg]. ω Proof. Let D ⊆ V∞,Y be a dense set in HOD{Y }, and let p be a condition in ∗ ω Col (ω, ω ). Let np be the domain of p. By strengthening the conditions in D if necessary, we may assume, applying Remark 10.3.2, that each element of D ω ω has the form [(n, m, S)]∞,Y for some (n, m, S) ∈ P∞,Y with n ≥ np. Let D0 ω ω be the set of (n, m, S) ∈ P∞,Y with [(n, m, S)]∞,Y ∈ D. To show that Hg is 0 HOD{Y }-generic, we want to ﬁnd a condition p ≤ p and a (n, m, S) in D0 such 0 that p is in Bn,m,S. By Remark 10.3.1 there is an OD{Y } function associating each (n, m, S) ∈ D0 to a tuple (np, k, T ) such that • k ∈ ω,

• T ∈ Vκ ∩ OD{Y } and

• Bn,m,Snp = Bnp,k,T .

Since HOD{Y } is a model of Choice, and the translation between witnesses to parts (1) and (2) of Theorem 9.0.6 is uniformly deﬁnable,8 there exist by the

8say more. is there a reﬂection property of κ involved? 118 CHAPTER 10. VOPENKAˇ ALGEBRAS choice of κ a set T∗ ∈ Vκ ∩ OD{Y } and a binary formula ψ (with G¨odelnumber k∗) such that

ω np {x¯ ∈ (ω ) : ∃(n, m, S) ∈ D0 x¯ ∈ Bn,m,Snp}

ω np is equal to {x¯ ∈ (ω ) : Lκ[T∗, x¯] |= ψ(T∗, x¯)}. Furthermore, Bnp,k∗,T∗ must ω be the set of all injections from np to ω , since otherwise the complement of this set has the form Bnp,j,R for some j ∈ ω and R ∈ Vκ ∩ OD{Y }, and we get a contradiction by considering a (n, m, S) ∈ D0 with

{x¯np :x ¯ ∈ Bn,m,S} ⊆ Bnp,j,R.

There exists then a triple (n, m, S) ∈ D0 such that p ∈ Bn,m,Snp, as desired. This gives the genericity of Hg. To see that g is in HOD{Y }[Hg], as in the proof of Theorem 10.0.1, let K be the set of pairs

ω ((i, j, k), (n, m, S)) ∈ (ω × ω × ω) × P∞,Y such that i < n and p(i)(j) = k for all p ∈ Bn,m,S. Then again K is in HOD{Y }, ω and K can be used to deﬁne a V∞,Y -name τ ∈ HOD{Y } such that for every ∗ ω Col (ω, ω )-generic function g over V , τHg = g. The following is the main theorem of this section.9

Theorem 10.2.6. Suppose that DCR holds, and that µ∅ is an ultraﬁlter on D∅. Let Y be a set of ordinals such that V =L(Y, R). Then for each OD{Y } set ω A ⊆ ω there exist an OD{Y } set S of ordinals and a binary formula φ such that A = {x ∈ ωω : L[S, x] |= φ(S, x)}.

Proof. Leta ¯ be a finite set of ordinals, let ψ be a ternary formula, and let A be the set of x ∈ ωω for which ψ(¯a, x, Y ) holds. By Theorem 10.0.2, we may fix a set B contained in the ordinals such that HOD{Y } = L[B]. ω By Remark 10.3.4, V∞,{Y } is forcing-equivalent to an iteration of the form ˙ ˙ ω V∞,{V } ∗Q, for some V∞,{Y }-name Q. By Remark 10.3.3, for every x ∈ ω there is a HOD{Y }-generic filter Gx ⊆ V∞,{Y } such that HOD{Y }[Gx] = HOD{Y }[x]. ω ˙ Let τ be the V∞,{Y }-name from the proof of Lemma 10.3.5. Let R be a ω ˙ V∞,{Y }-name in HOD{Y } for the range of the realization of τ. Let R∗ be the ˙ ˙ V∞,{Y } ∗ Q-name induced by R and a map witnessing the forcing-equivalence ˙ ω of V∞,{Y } ∗ Q with V∞,{Y } which carries the generic real for V∞,{Y } to the ω first coordinate of the generic sequence produced by forcing with V∞,{Y }, as in

9It doesn’t show that the two versions of the Vopˇenka algebra are equivalent. Is the hypotheses right/necessary? Do we say somewhere that µ∅ being an ultraﬁlter implies it for all sets of ordinals? 10.2. VOPENKAˇ ALGEBRAS AND ∞-BOREL SETS 119

ω ˙ Remark 10.3.4. We want to see that for each x ∈ ω , all generic QGx -extensions ˙ of HOD{Y }[x] agree about whether or not φ(¯α, x, Y ) holds in L(Y, R∗). By ˙ Lemma 10.3.5, R is the realization of R∗ in one of these extensions. The theorem will then follow, with S a set of ordinals coding B, Y ,α ¯ and R˙ ∗. If this were not the case there would be x ∈ ωω and (n, m, S), (n0, m0,S0) in ω P∞,Y such that

• x ∈ Bn,m,S ∩ Bn0,m0,S0 ω ˇ ˙ ˇ ˇ • [(n, m, S)]∞,Y “L(Y, R) |= φ(α,¯ τ(0), Y )”; 0 0 0 ω ˇ ˙ ˇ ˇ • [(n , m ,S )]∞,Y “L(Y, R) |= ¬φ(α,¯ τ(0), Y )”; Let p: n + n0 → ωω and π : ω → ω be such that

• p(0) = x; • π(0) = 0;

0 0 • π(n + n ) is a permutation of n + n ; • π(i) = i for all i ≥ n + n0;

• pn ∈ Bn,m,S; 0 0 ω 0 • the function p : n → ω defined by setting p (i) = p(π(i)) is in Bn0,m0,S0 . Let g : ω → ωω be the union of a V -generic filter for Col∗(ω, ωω) containing p, and let g0 : ω → ωω be defined by setting g0(i) = g(π(i)) for all i ∈ ω. 0 ∗ ω ˙ ˙ Then g is also the union of a V -generic filter for Col (ω, ω ), RHg = RHg0 and g(0) = g0(0), giving a contradiction. Part (1) of the following corollary appears as Theorem 1.9 in [2].10

Corollary 10.2.7. Suppose that DCR holds, and that µ∅ is an ultraﬁlter on D∅. Let Y be a set of ordinals such that V =L(Y, R). Then the following hold. 1. Every subset of ωω is ∞-Borel.

ω 2. HOD{Y,x} = HOD{Y }[x] for all x ∈ ω .

the HOD{Y }-generic filter induced by x. By Theorem 10.0.1 it suffices to show that Gx is in HOD{Y }[x]. Let κ0 be the cardinal κ used in the definition of Pω,{Y } and, for each 0 (n, x,¯ α¯) ∈ Pω,{Y } let Bn,x,¯ α¯ be the set Bn,x,¯ α¯ from this definition. Similarly, let κ1 be the cardinal κ used in the definition of P∞,{Y }, and, for each (m, S) ∈ 1 P∞,{Y } let Bm,S be the set Bm,S from that definition. By Theorem 10.3.6, for 0 1 each (n, x,¯ α¯) ∈ Pω,{Y } there is a pair (m, S) ∈ P∞,{Y } such that Bn,x,¯ α¯ = Bm,S. Furthermore, the set of pairs ((n, x,¯ α¯), (m, S)) in Pω,{Y } × P∞,{Y } such that 0 1 1 Bn,x,¯ α¯ = Bm,S is in HOD{Y }. The set of (m, S) ∈ P∞,{Y } such that x ∈ Bm,S is in HOD{Y }[x]. It follows that Gx, which is the set of (n, x,¯ α¯) such that 0 x ∈ Bn,x,¯ α¯, is in HOD{Y }[x].

10.3 ∞-Borel subsets of P(δ)

In this section we vary the Vopˇenka algebra again, using ∞-Borel11 for subsets of P(δ), for some ordinal δ. As before, we let Y be a set of ordinals. We ﬁx a cardinal κ such that for each n ∈ ω and each A ⊆ (2δ)n, if there exist a set δ n of ordinals S in OD{Y } and a formula φ such that A = {x¯ ∈ (2 ) : L[S, x¯] |= φ(S, x¯)} then there exist a set of ordinals S in Vκ ∩OD{Y } and a formula ψ such that ω n A = {x¯ ∈ (ω ) : Lκ[S, x¯] |= ψ(S, x¯)}. δ Let P∞,Y be the set of pairs (m, S) such that • m is the G¨odelnumber of a binary formula φ,

• S ∈ Vκ ∩ OD{Y } is a set of ordinals,

δ • the set Bm,S = {x ∈ 2 : Lκ[S, x] |= φ(S, x)} is nonempty.

δ Then again we have an induced ordering ≤δ,Y on P∞,Y deﬁned by setting

(m, S) ≤δ,Y (j, T )

δ if Bm,S ⊆ Bj,T . Let ≡δ,Y be the induced equivalence relation on P∞,Y . Let δ δ V∞,Y be the partial order whose domain is the set of ≡δ,Y -classes of P∞,Y , with the order inherited from ≤δ,Y . δ,ω For the sequence version, let P∞,Y be the set of triples (n, m, S) such that • n ∈ ω,

• m is the G¨odelnumber of a binary formula φ,

• S ∈ Vκ ∩ OD{Y } is a set of ordinals,

ω δ n • the set Bn,m,S = {x¯ ∈ (2 ) : L[S, x¯] |= φ(S, x¯)} is nonempty.

11or * codes 10.3. ∞-BOREL SUBSETS OF P(δ) 121

δ,ω δ,ω Then again we have an induced ordering ≤Y on P∞,Y deﬁned by setting

ω (n, m, S) ≤∞,Y (k, j, T )

ω ω δ,ω if n ≥ k and {x¯k :x ¯ ∈ Bn,m,S} ⊆ Bk,j,T . Let ≡Y be the induced equivalence δ,ω δ,ω relation on P∞,Y . Let VY be the partial order whose domain is the set of δ,ω δ,ω δ,ω ≡Y -classes of P∞,Y , with the order inherited from ≤Y .

10.3.1 Remark. By Theorem 10.1.2, for each bounded S ⊆ κ in HOD{Y } and each ternary formula φ, there exist a bounded T ⊆ κ and a binary formula φ such that {x ∈ ωω : ∃y ∈ ωω L[S, x, y] |= φ(S, x, y)} = {x ∈ ωω : L[T, x] |= ψ(T, x)}.

ω 10.3.2 Remark. Let (n, m, S) be an element of P∞,Y , and let φ be the formula with G¨odelnumber m. Let n0 be an element of ω \ n. Then

ω n0 ω n0 {x¯ ∈ (ω ) :x ¯n ∈ Bn,m,S} = {x¯ ∈ (ω ) : Lκ(S, x¯) |= φ(S, x¯n)},

ω so this set has the form Bn0,k,S for some k ∈ ω, and Bn0,k,S ≤∞,Y Bn,m,S. Moreover, there is a recursive function (not depending on S) sending each such tuple (n, m, n0) to a corresponding value k.

10.3.3 Remark. The first paragraph of the proof of Theorem 10.0.1 adapts ω to prove that for each x ∈ ω , the set Gx consisting of those [(m, S)]∞,{Y } ∈ P∞,{Y } such that x ∈ Bm,S is a HOD{Y }-generic filter. This also follows from ω Lemma 10.3.5 below, using the fact that the first coordinate of a V∞,{Y }-generic sequence is generic for V∞,{Y }. For each i ∈ ω, x(i) is the unique j ∈ ω such that [(∅, mi,j)]∞,{Y } ∈ Gx, where mi,j is the Gödelnumber of the formula y(i) = j. It follows then that HOD{Y }[x] = HOD{Y }[Gx], since Gx is the set of [(m, S)]∞,Y ∈ V∞,Y such that Lκ[S, x] |= φ(S, x) (where φ is the formula with Gödelnumber m), and this can be computed in HOD{Y }[x]. Since every ω condition in V∞,{Y } is a member of Gx for some x ∈ ω , every HOD{Y }-generic filter G ⊆ V∞,{Y } is equal to Gx, for x = {(i, j) : [(∅, mi,j)]∞,{Y } ∈ G}.

ω 10.3.4 Remark. Since the ﬁrst coordinate of a V∞,{Y }-generic sequence is ˙ ω generic for V∞,{Y }, there is a V∞,{Y }-name Q such that V∞,{Y } is forcing- ˙ equivalent to V∞,{Y } ∗Q, via a map which carries the generic real for V∞,{Y } to ω the ﬁrst coordinate of the generic sequence produced by forcing with V∞,{Y }.

ω ω As with the partial order VZ,X above, each injection f : ω → ω defines a ω ω ω filter Hf on V∞,Y , consisting of the ≡∞,Y -classes of those (n, m, S) ∈ P∞,Y for ω which fn is in Bn,m,S. Lemma 10.3.5 is the version of Lemma 10.0.4 for V∞,Y . Theorem 10.1.2 and Remark 10.3.2 are used in the proof of the lemma. The remainder of this section uses the notation µS and DS introduced in Definition 8.0.3, for S a set of ordinals. Recall from Remark 1.1.5 that if µ∅ is a countably complete ultrafilter on D∅ (i.e., the constructibility degrees), then, for each set S ⊆ Ord, µS is a countably complete ultrafilter on DS. 122 CHAPTER 10. VOPENKAˇ ALGEBRAS

Lemma 10.3.5. Suppose that DCR holds, and that µ∅ is an ultraﬁlter on D∅. ∗ ω S ω If G ⊆ Col (ω, ω ) is V -generic, and g = G, then Hg is V∞,Y -generic over HOD{Y }, and g is in HOD{Y }[Hg]. ω Proof. Let D ⊆ V∞,Y be a dense set in HOD{Y }, and let p be a condition in ∗ ω Col (ω, ω ). Let np be the domain of p. By strengthening the conditions in D if necessary, we may assume, applying Remark 10.3.2, that each element of D ω ω has the form [(n, m, S)]∞,Y for some (n, m, S) ∈ P∞,Y with n ≥ np. Let D0 ω ω be the set of (n, m, S) ∈ P∞,Y with [(n, m, S)]∞,Y ∈ D. To show that Hg is 0 HOD{Y }-generic, we want to ﬁnd a condition p ≤ p and a (n, m, S) in D0 such 0 that p is in Bn,m,S. By Remark 10.3.1 there is an OD{Y } function associating each (n, m, S) ∈ D0 to a tuple (np, k, T ) such that • k ∈ ω,

• T ∈ Vκ ∩ OD{Y } and

• Bn,m,Snp = Bnp,k,T .

Since HOD{Y } is a model of Choice, and the translation between witnesses to parts (1) and (2) of Theorem 9.0.6 is uniformly deﬁnable,12 there exist by the choice of κ a set T∗ ∈ Vκ ∩ OD{Y } and a binary formula ψ (with G¨odelnumber k∗) such that

ω np {x¯ ∈ (ω ) : ∃(n, m, S) ∈ D0 x¯ ∈ Bn,m,Snp}

{x¯np :x ¯ ∈ Bn,m,S} ⊆ Bnp,j,R.

12say more. is there a reﬂection property of κ involved? 13It doesn’t show that the two versions of the Vopˇenka algebra are equivalent. Is the hypotheses right/necessary? Do we say somewhere that µ∅ being an ultraﬁlter implies it for all sets of ordinals? 10.3. ∞-BOREL SUBSETS OF P(δ) 123

Theorem 10.3.6. Suppose that DCR holds, and that µ∅ is an ultrafilter on D∅. Let Y be a set of ordinals such that V =L(Y, R). Then for each OD{Y } set ω A ⊆ ω there exist an OD{Y } set S of ordinals and a binary formula φ such that A = {x ∈ ωω : L[S, x] |= φ(S, x)}. Proof. Leta ¯ be a finite set of ordinals, let ψ be a ternary formula, and let A be the set of x ∈ ωω for which ψ(¯a, x, Y ) holds. By Theorem 10.0.2, we may fix a set B contained in the ordinals such that HOD{Y } = L[B]. ω By Remark 10.3.4, V∞,{Y } is forcing-equivalent to an iteration of the form ˙ ˙ ω V∞,{V } ∗Q, for some V∞,{Y }-name Q. By Remark 10.3.3, for every x ∈ ω there is a HOD{Y }-generic filter Gx ⊆ V∞,{Y } such that HOD{Y }[Gx] = HOD{Y }[x]. ω ˙ Let τ be the V∞,{Y }-name from the proof of Lemma 10.3.5. Let R be a ω ˙ V∞,{Y }-name in HOD{Y } for the range of the realization of τ. Let R∗ be the ˙ ˙ V∞,{Y } ∗ Q-name induced by R and a map witnessing the forcing-equivalence ˙ ω of V∞,{Y } ∗ Q with V∞,{Y } which carries the generic real for V∞,{Y } to the ω first coordinate of the generic sequence produced by forcing with V∞,{Y }, as in ω ˙ Remark 10.3.4. We want to see that for each x ∈ ω , all generic QGx -extensions ˙ of HOD{Y }[x] agree about whether or not φ(¯α, x, Y ) holds in L(Y, R∗). By ˙ Lemma 10.3.5, R is the realization of R∗ in one of these extensions. The theorem will then follow, with S a set of ordinals coding B, Y ,α ¯ and R˙ ∗. If this were not the case there would be x ∈ ωω and (n, m, S), (n0, m0,S0) in ω P∞,Y such that

• x ∈ Bn,m,S ∩ Bn0,m0,S0 ω ˇ ˙ ˇ ˇ • [(n, m, S)]∞,Y “L(Y, R) |= φ(α,¯ τ(0), Y )”; 0 0 0 ω ˇ ˙ ˇ ˇ • [(n , m ,S )]∞,Y “L(Y, R) |= ¬φ(α,¯ τ(0), Y )”; Let p: n + n0 → ωω and π : ω → ω be such that • p(0) = x; • π(0) = 0; 0 0 • π(n + n ) is a permutation of n + n ; • π(i) = i for all i ≥ n + n0;

Corollary 10.3.7. Suppose that DCR holds, and that µ∅ is an ultraﬁlter on D∅. Let Y be a set of ordinals such that V =L(Y, R). Then the following hold. 1. Every subset of ωω is ∞-Borel.

ω 2. HOD{Y,x} = HOD{Y }[x] for all x ∈ ω .

Proof. For part (1), every set in L(Y, R) is definable from Y , a finite set of ordinals and a finite subset of ωω. It follows then that for each A ⊆ ωω in ω ω ω L(Y, R), there exist an OD{Y }-set B ⊆ ω × ω and an x ∈ ω such that ω A = {y ∈ ω :(x, y) ∈ B}. By Theorem 10.3.6, there exist an OD{Y } set S of ordinals and a ternary formula φ such that B = {(x, y) ∈ ωω : L[S, x, y] |= φ(S, x, y)}. It is easy to find a set T of ordinals coding S and x in some simple way, and a binary formula φ0 such that A = {y ∈ ωω : L[T, y] |= φ0(T, y)}. ω For part (2), fix x ∈ ω . Clearly, HOD{Y }[x] is contained in HOD{Y,x}. To show the reverse inclusion, let Gx ⊆ Vω,Y (the ordinary Vopˇenka algebra) be the HOD{Y }-generic filter induced by x. By Theorem 10.0.1 it suffices to show that Gx is in HOD{Y }[x]. Let κ0 be the cardinal κ used in the definition of Pω,{Y } and, for each 0 (n, x,¯ α¯) ∈ Pω,{Y } let Bn,x,¯ α¯ be the set Bn,x,¯ α¯ from this definition. Similarly, let κ1 be the cardinal κ used in the definition of P∞,{Y }, and, for each (m, S) ∈ 1 P∞,{Y } let Bm,S be the set Bm,S from that definition. By Theorem 10.3.6, for 0 1 each (n, x,¯ α¯) ∈ Pω,{Y } there is a pair (m, S) ∈ P∞,{Y } such that Bn,x,¯ α¯ = Bm,S. Furthermore, the set of pairs ((n, x,¯ α¯), (m, S)) in Pω,{Y } × P∞,{Y } such that 0 1 1 Bn,x,¯ α¯ = Bm,S is in HOD{Y }. The set of (m, S) ∈ P∞,{Y } such that x ∈ Bm,S is in HOD{Y }[x]. It follows that Gx, which is the set of (n, x,¯ α¯) such that 0 x ∈ Bn,x,¯ α¯, is in HOD{Y }[x]. Chapter 11

Applications

11.1 Strong ∞-Borel codes

Given a sentence φ of L∞,0 and a set of ordinals σ, we deﬁne the sentence φσ recursively on the complexity of φ as follows:

• Pnσ = Pn, for each n ∈ ω;

• (¬φ)σ = ¬(φσ); V V • ( α<γ φα)σ = α∈γ∩σ(φασ); W W • ( α<γ φα)σ = α∈γ∩σ(φασ). 11.1.1 Remark. If δ is an inﬁnite cardinal, S is a subset of δ coding a sentence ω φS in Lδ,0, X ⊆ 2 and Z is an elementary submodel of Lδ+ (X,S), then, for all x ∈ X, x ∈ A if and only if x ∈ A . It does not follow, however, that φS φS (Z∩γ) A ⊆ A . One way to see this is to note that, since (¬φ) σ = ¬(φ σ), φS (Z∩γ) φS A ⊆ A and A ⊆ A implies that A is Borel. Another is to note φσ φ (¬φ)σ ¬φ φ that an intersection of uncountably many sets typically does not contain the intersection of any countable subcollection.

11.1.2 Deﬁnition. Suppose that δ is an inﬁnite cardinal and that φ is a sentence in Lδ,0. Associate to φ a game Gφ,δ on δ, where I and II collaborate to build a countable set σ ⊆ δ, and I wins if Aφσ ⊆ Aφ. We say that φ is a strong ∞-Borel code (or strong δ-Borel code) for Aφ if I has a winning strategy in Gφ,δ. If S is the set of ordinals such that φ = φS, we also say that S is a strong 1 ∞-Borel code or strong δ-Borel code for Aφ.

1There is an alternate deﬁnition, which requires a measure µ.

11.1.3 Deﬁnition (Strong ∞-Borel code). Suppose that A ⊆ ωω, S is a strong ∞-Borel code ∼ if (suppose S ⊆ δ) the set of σ ∈ Pℵ1 (δ) such that sσ = collapse(σ(σ ∩ S)) deﬁnes a Borel subset of A is of measure 1 for µ, where µ is the measure on Pℵ1 (δ).

125 126 CHAPTER 11. APPLICATIONS

Note that every element of Lℵ1,0 is a strong ∞-Borel code.

11.1.4 Remark. For any ordinal δ and any φ ∈ Lδ,0, δ-Determinacy implies the determinacy of the game Gφ,δ. One way to see this involves continuously associating to a run of the game (producing a countable σ ⊆ δ) an element of ωω coding the transitive collapse of an elementary submodel of Lδ+ [S] containing ¯ σ and, letting S be the image of S under this transitive collapse, asking if AφS¯ (which is the same as AφS σ) is a subset of AφS . Alternately, a run of the game builds a surjection from ω to σ, which can be used to continuously build a real coding φσ (starting from the last conjunction, for instance).

Theorem 11.1.5. If δ is an infinite cardinal and φ ∈ Lδ,0 is a strong ∞-Borel code, then Aφ is δ-Suslin. ω Proof. Fix a winning strategy τ for player I in Gφ,δ. Then for all x ∈ ω , x ∈ Aφ if and only if there is a countable σ ⊆ δ produced by a run of GS where I plays according to τ and x ∈ Aφσ. To see this, note first of all that, since a τ is winning strategy for player I, if there is a play against τ producing σ with x in the corresponding set Aφσ, then x ∈ A. For the other direction, fix S ⊆ δ such that φ = φS. Then for each x ∈ A there exists a σ resulting from a run of Gφ,δ in which I plays according to τ such that, for some countable Z ≺ Lδ+ [S, x], x ∈ Z and Z ∩ δ = σ, in which case x ∈ Aφσ. There is a tree T on ω × δ × ω whose branches are the triples (x, f, y) where ω ω ω x ∈ ω , f ∈ δ is a run of Gφ,δ where I plays according to τ, and y ∈ ω codes the transitive collapse M of a countable elementary submodel Z of Lδ+ [S, x] with x ∈ Z and Z ∩ δ = range(f) such that, letting S¯ be the image of S under the transitive collapse of Z, x |= φS¯. Then Aφ = p[T ]. Theorem 11.1.6 below shows that strong ∞-Borel codes are preserved under disjunctions, given the appropriate form of ordinal determinacy. Note, however, that strong ∞-Borel codes for non-Borel sets are never preserved under negation, since for any pair of strategies for player I in games Gφ,δ and G¬φ,δ respectively there exists a countable σ ⊆ δ resulting from runs of the corresponding game against each strategy. Both runs being winning for player I would show that Aφ is Borel, as in Remark 11.1.1. Similarly, there exists a wellordered conjunction of strong ∞-Borel codes which is not a strong ∞-Borel code. To see this, note that if κ has uncountable cofinality, and hφα : αi is a sequence of Lℵ1,0-sentences for a strictly ⊆-decreasing sequence of sets, then each φα is a strong-∞-Borel V code but α<κ φα is not. Such a sequence can be shown to exist (with κ = ω1) by Example 9.0.13, or alternately by applying the ∗-operation from Section 9.1 to any ∞-Borel code for a non-Borel set.

Theorem 11.1.6. Assume that DCR holds. Let κ be an inﬁnite cardinal below Θ such that κ-Determinacy holds, and let hφα : α < κi be a sequence of Lκ,0 W sentences which are strong ∞-Borel codes. Then α<κ φα is a strong ∞-Borel code. Proof. We can work in L(A, R) for some set A ⊆ ωω of Wadge rank greater W than κ, and use the fact that DC holds in L(A, R). Let φ = α<κ φα. It suﬃces 11.2. PRODUCING STRONG ∞-BOREL CODES 127

to see that player II cannot have a winning strategy in Gφ,κ. Fixing a strategy τ for player II, we can ﬁnd by DC a countable σ ⊆ κ and winning strategies ρα for player I in the games Gφα,κ (α ∈ σ) such that σ is the result of a run of Gφ,κ where II has played by τ and also, for each α ∈ σ, the result of a run of Gφα,κ where player I has played by ρα. Then [ [ Aφσ = Aφασ ⊆ Aφα ⊆ Aφ, α∈σ α∈σ showing that I wins this run of the game. In Section 11.2 we see how to convert ∞-Borel codes into strong ∞-Borel codes.

11.2 Producing strong ∞-Borel codes

In this section we use the terms µS and DS from Deﬁnition 8.0.3, for S a set ∞ Q L[S,x] of ordinals. Recall from Section 8.4 that δS denotes ω2 /µS. Recall also from Theorem 11.1.5 that subsets of ωω with strong ∞-Borel codes are Suslin. The following theorem produces strong ∞-Borel codes. Theorem 11.2.1. Let S be a set of ordinals. Suppose that the following hold.

• µS is an ultraﬁlter on DS.

DS ∞ 2 • The ultrapower Ord /µS is wellfounded, and δS < Θ. ∞ • δS -Determinacy. Then if S is an ∞-Borel code for a set A ⊆ ωω, then A has a strong ∞-Borel ∞ 3 code which is contained in δS and deﬁnable from S.

ℵ0 4 Proof. Since ω-Determinacy (i.e., AD) holds, ℵ1 6≤ 2 . Then by Theorem ω ω 8.0.5 we have that for some x0 ∈ ω and all y ∈ ω such that [y]S ≥S [x]S, V GCH holds in L[S, y] below ω1 (which is strongly inaccessible in L[S, y], since ℵ0 ℵ1 6≤ 2 ). For each such y, it follows that the partial order V∞,S (deﬁned at the beginning of Section 10.2) as computed in L[S, y], being isomorphic to ω the subset relation restricted to a subset of P(ω ), has cardinality at most ℵ2 in L[S, y]. Furthermore, since antichains in this partial order correspond to ω L[S,y] pairwise disjoint subsets of P(ω ), V∞,S is ℵ2-c.c. in L[S, y], and the set of L[S,y] all antichains of V∞,S has cardinality ℵ2 in L[S, y]. In the proof (2) ⇒ (1) of Theorem 9.0.6, the translation from ∞-Borel∗ codes to ∞-Borel codes is ∗ L[S,y] ordinal deﬁnable. It follows that for each ∞-Borel code in HOD{S} there is

2Maybe we can weaken this, either by restricting to an inner model, or an initial segment of the ordinals. 3Doesn’t the ﬁrst hypothesis follow from the third? And shouldn’t we state a corollary relating the theorem to Uniformization. Should we say what the code is? 4give a reference? 128 CHAPTER 11. APPLICATIONS

L[S,y] ω an ∞-Borel code in HOD{S} defining the same subset of ω in L[S, y], and furthermore that the set of pairs of codes (of the two types) defining the same ω L[S,y] subset of ω is also in HOD{S} . In Section 9.1 we defined the relation φ 7→ φ∗ on ∞-Borel codes, and noted (for instance in the proof of Theorem 9.1.4) that if κ is a regular cardinal and there does not exist a κ-sequence of ∞-Borel codes for nonempty (pairwise) ∗ disjoint sets, then each resulting formula φ must be in Lκ,0. This means that ∗ L[S,y] every set of reals in L[S, y] with an ∞-Borel code in HOD{S} (i.e., every set of L[S,y] L[S,y] reals corresponding to a condition in V∞,S ) has an ∞-Borel code in HOD{S} L[S,y] which is a bounded subset of ω2 . Furthermore, starting with propositional variables Pn,m (n, m ∈ ω, corresponding to the assertion z(n) = m), we can, working by recursion on the complexity of formulas in LL[S,y] ∩ HODL[S,y], ℵ2,0 {S} L[S,y] associate to each such formula ψ a condition pψ ∈ V∞,S in such a way that

L[S,y] • for all n, m ∈ ω, pPn,m is the condition in V∞,S corresponding to the set of z ∈ ωω such that z(n) = m;

for ψ ∈ LL[S,y] ∩ HODL[S,y], p is the complement of p in L[S,y]; • ℵ2,0 {S} ¬ψ ψ V∞,S

for all γ < ωL[S,y] and all ψ ∈ LL[S,y] ∩ HODL[S,y] of form W ψ , p is • 2 ℵ2,0 {S} α<γ α ψ L[S,y] the supremum in V∞,S of {pψα : α < γ};

for all γ < ωL[S,y] and all ψ ∈ LL[S,y] ∩ HODL[S,y] of form V ψ , p is • 2 ℵ2,0 {S} α<γ α ψ L[S,y] the inﬁmum in V∞,S of {pψα : α < γ}.

∗ L[S,y] Since every set of reals in L[S, y] with an ∞-Borel code in HOD{S} has an L[S,y] L[S,y] ∞-Borel code in HOD{S} which is a bounded subset of ω2 , the range of L[S,y] our association ψ 7→ pψ will be all of V∞,S . There is an S-invariant function ω y L[S,y] choosing for each y ∈ ω a condition qA in V∞,S corresponding in L[S, y] to an ∞-Borel∗ code for A ∩ L[S, y]. It follows that we can choose (for each y ≥ x , in an S-invariant fashion) a formula χ ∈ LL[S,y] ∩ HODL[S,y] (coded S 0 y ℵ3,0 {S} L[S,y] by a bounded set Ty ∈ P(ω2) as in the ﬁrst paragraph of Chapter 9) such ω L[S,y] that for all z ∈ ω (in V ), if z |= χy then z generates a HOD{S} -generic ﬁlter L[S,y] y for V∞,S below qA, so z is in A. DS Now we consider the ultrapower V /µS, which we have assumed to be 5 wellfounded. Let T be the set of ordinals represented by the function y 7→ Ty. We want to see that T is a strong ∞-Borel code for A. We have that T is a ∞ subset of δS , which by assumption is less than Θ.

5 I think we need only DCR in this case, since we can assume that we are working in a model of the form L(S, ωω,B) for some B ⊆ ωω. Check! Recall that by Solovay’s theorem we get DCR from this assumption. 11.2. PRODUCING STRONG ∞-BOREL CODES 129

∞ Since δ -Determinacy holds, the game G ∞ from Definition 11.1.2 is de- S φT ,δS termined (see Remark 11.1.4). We want to see then that player II does not have a winning strategy. Suppose towards a contradiction that Σ is such a ω strategy. For each y ∈ ω such that y ≥ x0, let Ry be the tree of attempts to ∞ find a countable σ ⊆ δS resulting from a play of GφT ,δ according to Σ, and an isomorphism between φTy and φT σ. If any of these trees is illfounded, then we have a contradiction. If not, then (since there is in each case a wellordering of Ry definable from Σ, T and Ty) we can find (in an S-invariant fashion) ranking functions ρy on each such tree Ry. Let j be the map sending each element of V DS to the element of V /µS represented by the corresponding constant function. ∗ ∗ DS In addition, let R and ρ be the elements of V /µS represented by the maps y 7→ Ry and y 7→ ρy. Then in the wellfounded model

Y L[S,T,Σ,y] HOD{S,T,Σ} /µS,

∗ ∞ ρ is a ranking function on the tree of attempts to a ﬁnd countable σ ⊆ j(δS ) resulting from a play of j(G ∞ ) according to j(Σ), and an isomorphism be- φT ,δS ∞ tween φT and φj(T )σ. Since j[δS ] is closed under j(Σ), and T is isomorphic to ∞ ∗ j[T ], which is j(T )j[δS ], there is a nonempty subset of R without terminal nodes, corresponding to those nodes where the isomorphism is contained in j, ∞ ∗ and the run of the game is contained in j[δS ]. It follows that R is illfounded, giving the desired contradiction. Suppose now that A ⊆ ωω is ∞-Borel, and that S is an ∞-Borel code for A ∞ which is not ≤D-maximal. By Theorem 8.4.1, then, δS < Θ. Theorem 11.2.1 says that S induces a strong ∞-Borel code for A, which, by Theorem 11.1.5, implies that A is Suslin. Theorems 11.1.5 and Theorem 11.2.2 give the following. ω Theorem 11.2.2. If <Θ-Determinacy and DCR hold, then every subset of ω which has an ∞-Borel code which is not ≤D-maximal is Suslin. In addition, one gets the following equivalences under AD+. Theorem 11.2.3. Assuming AD+, the following are equivalent.

1. There is no ≤D-maximal set of ordinals. 2. Every subset of ωω is Suslin.

3. ADR 4. Uniformization Proof. By Theorem 11.2.2, (1) implies (2). Theorem 13.0.1 gives that (2) implies (3). The implication (3) ⇒ (4) is discussed in the proof of Theorem 0.1.3. The equivalence of (4) and (1) follows from Corollary 10.1.7.

Recall from Theorem 6.4.2 that if Wadge Determinacy+Uniformization+ℵ1 6≤ 2ℵ0 holds, then the length of the Solovay sequence is a limit ordinal. A version of the converse holds assuming AD+. 130 CHAPTER 11. APPLICATIONS

Theorem 11.2.4. If AD+ holds and the Solovay sequence has limit length, then every subset of ωω is Suslin.

Proof. By Theorem 11.2.3, it suffices to show that there is not ≤D maximal set of ordinals. If there were such a set, then a reflection argument shows that there is such a set S which is a bounded subset of Θ.6 By the Moschovakis Coding Lemma, S is in L(A, R), for some A ⊆ ωω. Theorem 8.5.7 gives that every subset of ωω is then in L(S∞, R), and Theorem 11.2.1 gives that every subset of ωω is ordinal definable from A and a real, which implies that Θ is the least member of the Solovay sequence above the Wadge rank of A, contradicting our hypothesis that the Solovay sequence has limit length.

Proof. This follows from Theorem 8.4.1 and 11.2.1.7

11.3 ∞-Borel representations from Uniformization

The main theorem of this section is Theorem 11.3.2, which says that if AD and Uniformization hold then all sets of reals are ∞-Borel. Theorem 11.3.1 is a slightly stronger version, in which the assumption of AD is replaced with some of its consequences. Theorem 11.3.1 is Theorem 5.10 of [18], whose presentation we follow closely. Before beginning the proof of Theorem 11.3.1 we introduce some terminology for treating generic ﬁlters over countable structures in terms of descriptive set theory. Given a partial order P, a nice P-name for an element of ωω is a set τ such that

• each element of τ is a pair (p, (n, m)), where p ∈ P and n, m ∈ ω;

• for each p ∈ P and each n ∈ ω, there exist q ≤ p and m ∈ ω such that (q, (n, m)) ∈ τ;

• for all p, p0 ∈ P and all n, m, m0 ∈ ω, if (p, (n, m)) and (p0, (n, m0)) are both in τ, and p and p0 are compatible, then m = m0.

Given a nice P-name τ and a set g ⊆ τ, we let τg be the set of pairs (n, m) for which there exists a p ∈ g with (p, (n, m)) ∈ τ. Given a set X, we say that a filter g ⊆ P is D-generic if it intersects each dense subset of P in X. We say that a set Y consisting of filters on P(P) is comeager if there is a countable set X, ∗ such that Y contains every X-generic filter contained in P. We write p τ ∈ A ∗ to mean that for a comeager set of filters g ⊆ P, τg is in A, and define p τ 6∈ A similarly. We let DA be the set of p ∈ such that p ∗ τ ∈ A or p ∗ τ 6∈ A. P,τ P If P is countable, and every subset of ωω has the property of Baire, then each set of the form DA is dense in . P,τ P

6explain? maybe this is already in here somewhere. 7Say more. Maybe the theorem should assume AD+? Then you can’t have a maximal degree notion so it seems ok. Or is there some way to get ∞-Borel codes? 11.3. ∞-BOREL REPRESENTATIONS FROM UNIFORMIZATION 131

Given A ⊆ ωω, the term relation for A is the class TR(A) consisting of those triples (P, p, τ) such that P is a poset, τ is a nice P-name for an element of ωω ∗ ˙ and p τ ∈ A. We write tA,P for the set of (p, τ) for which (P, p, τ) ∈ TR(A). ˙ Note that tA,P is a P-name, as is every subset of it. If P is a countable partial order and T is a countable set of nice -names for reals and each set DA (τ ∈ T ) P P,τ is dense, then, for comeagerly many filters g ⊆ P and all τ ∈ T , τg ∈ A if and ˙ only if, for some p ∈ g,(p, τ) ∈ tA,P. For any set M we write t˙M for t˙ ∩ M. A transitive set M is said to be A,P A,P weakly-(A, )-closed if t˙M ∈ M (applying the definition to the set of names P A,P of the formx ˇ (x ∈ ωω ∩ M) we see that this implies that A ∩ M ∈ M). We say that M is strongly-(A, )-closed if t˙M ∈ M and, for all M-generic filters P A,P g ⊆ , t˙M = A ∩ M[g] (here were are referring to existing M-generic filters, P A,P,g not filters existing in a forcing extension of V ). We say that M is weakly (or strongly) A-closed if it is weakly (or strongly) (A, P)-closed for all partial orders P ∈ M. Theorem 11.3.1. Suppose that Uniformization + Baire(P(ωω)) holds and that ω 8 ω there is a normal fine measure on Pℵ1 (ω ). Then every subset of ω is ∞- Borel. Proof. Fix A ⊆ ωω. Let B be the set of (x, y) ∈ (ωω)2 such that • x HC-codes a pair (P, τ) where P is a countable poset and τ is a nice P-name for an element of ωω; • y HC-codes a countable set D consisting of dense open subsets of P, with DA ∈ D, such that for any D-generic filter g ⊆ , P,τ P ∗ – τg ∈ A if and only if ∃p ∈ g p τ ∈ A, and ∗ – τg 6∈ A if and only if ∃p ∈ g p τ 6∈ A. The conclusion of the condition on y is equivalent to : for all D-generic filters g ⊆ , τ ∈ t˙ if and only if τ ∈ A. By the density of the sets DA P g A,P,g g P,τ (mentioned above), the assumption that every set of reals has the property of Baire implies that for each x ∈ ωω coding a pair (P, τ) as above, there is a y ∈ ωω such that (x, y) ∈ B. Let f be a function uniformizing B, and F ⊆ ωω code f as follows : x ∈ F if and only if, letting x0 ∈ ωω be such that x0(n) = x(n + 2) for all n ∈ ω, 0 0 ω x ∈ dom(f) and f(x )(x(0)) = x(1). For each σ ∈ Pℵ1 (ω ), let Nσ be

Lω1 [TR(A), TR(F ), σ], and let Mσ be Nσ HOD{TR(A),TR(F )}.

Then the models Nσ and Mσ are weakly A-closed and weakly F -closed, and since ω1 is measurable in V (see the last sentence of Remark 1.1.7) they are models of ZFC. 8Do you really need normality? Does AD + Uniformization give a normal ﬁne measure?

Look at the ADR section. 132 CHAPTER 11. APPLICATIONS

ω Claim. For each σ ∈ Pℵ1 (ω ), Mσ is strongly A-closed.

ω To prove the claim, fix σ ∈ Pℵ1 (ω ) and a partial order P ∈ Mσ. Let C be ω the set of nice P-names in Mσ for elements of ω . Let Q be Col(ω, P), and for ω each τ ∈ C, letz ˙τ be a nice Q-name for an element of ω coding the pair (P, τ). Mσ Since Mσ is weakly F -closed, t˙ is in Mσ. Let E be a countable set of F,Q dense open subsets of Q, containing each dense open subset of Q from Mσ and having the property that for each E-generic h ⊆ Q and all nice names ρ ∈ Mσ ω Mσ for elements of ω , ρh ∈ F if and only if, for some p ∈ h,(p, ρ) ∈ t˙ . Applying F,Q this to the collection of nice Q-names ρ for which it is forced by 1Q that (for some τ ∈ C) ρ(n + 2) =z ˙τ (n) for all n ∈ ω, we get that for any E-generic filter ω h ⊆ Q, and for each τ ∈ C,z ˙τ,h ∈ ω and f(z ˙τ,h) is in Mσ[h]. Fix such a h, and let Dh be the collection of dense open subsets of P coded by the members of {f(z ˙τ,h): τ ∈ C}. Let g ⊆ P be an Mσ[h]-generic filter. Then g is Dh-generic. It follows then that t˙M = A ∩ M [g]. Since M [h][g] = M [g][h] (i.e., (g, h) is generic for A,P,g σ σ σ ∗ ), it follows that t˙M = A ∩ M [g] holds for any M -generic filter g ⊆ , P Q A,P,g σ σ P as desired. This ends the proof of the claim.

We complete the proof working in the inner model L(TR(A), TR(F ), µ, ωω), ω where µ is a normal fine measure on Pℵ1 (ω ). Since Uniformization holds, DCR holds. It follows that L(TR(A), TR(F ), µ, ωω) satisfies DC, and, since µ is Q countably complete, ω Mσ/µ (which we will call M∞) is wellfounded Pℵ1 (ω ) ω ω when computed in L(TR(A), TR(F ), µ, ω ). For each σ ∈ Pℵ1 (ω ), let Pσ be Mσ (V∞,∅) , as defined in Section 10.2. Using the ordinal-definability order, we Mσ can pick in each Nσ a set of ordinals Sσ coding the set of conditions in (V∞,∅) which force that the generic real will be in the realization of t˙M (for instance, A,Pσ either by listing the relevant pairs from P∞,∅, or converting to a sentence in Mσ L∞,0 using Theorem 9.0.6). Since each real in Nσ is Mσ-generic for (V∞,∅) , ∗ we have that, in Nσ, Sσ is an ∞-Borel code (or an ∞-Borel code, depending on which method of coding we choose) for A ∩ Nσ. Let S∞ be the µ-ultraproduct of the sets Sσ. Since µ is fine, and since A is the µ-ultraproduct of the sets ∗ A ∩ σ, we have that S∞ is an ∞-Borel code (or ∞-Borel -code) for A.

Putting together Theorems 1.1.2 and 11.3.1 with part (3) of Remark 1.0.2, we have the following.

Theorem 11.3.2. If AD + Uniformization holds then every subset of ωω is ∞- Borel.

11.4 Closure of the Suslin cardinals

Recall from Chapter 6 that a cardinal κ is Suslin if there is a set A ⊆ ωω which is κ-Suslin but not γ-Suslin for any γ < κ. In this section we prove Theorem 11.4.1 below, on the closure of the set of Suslin cardinals. Corollary 11.4.6 gives 11.4. CLOSURE OF THE SUSLIN CARDINALS 133 that under AD+ this set is closed below Θ. Again, we follow the presentation in [18]. 910

Theorem 11.4.1 (Steel-Woodin11). Suppose that

1. DCR holds, 2. κ < Θ is a limit of Suslin cardinals,

3. κ-Determinacy holds and that

4. hSα : α < κi is a sequence of bounded subsets of κ which are ∞-Borel codes for pairwise disjoint subsets of 2ω.

Then κ is a Suslin cardinal.

It follows from AD alone (or, more directly part (1) of Corollary 3.0.2 and the fact that AD implies CCR; see part (1) of Remark 1.0.2) that if κ < Θ is a limit of an ω-sequence of Suslin cardinals, then κ is a Suslin cardinal. We assume then for the rest of this section that cof(κ) is uncountable, in which case (since we are assuming AD) the <κ-Borel sets and the <κ-Suslin sets coincide, by Lemma 9.0.16.

11.4.2 Remark. The map φ 7→ φ∗ from Section 9.1 shows that a sequence hSα : α < κi as in hypothesis (4) of Theorem 11.4.1 exists if there is an ∞- Borel set which is not <κ-Borel (an ∞-Borel set which is not κ-Borel gives this immediately, but such a sequence can also be extracted from a κ-Borel set which is not <κ-Borel, since the corresponding formula φ∗ will contain conjunctions or disjunctions of coﬁnally many cardinalities below κ).12 If κ < Θ, then this in turn follows from either the assumption that every subset of 2ω is ∞-Borel, or the assumption that κ is a Suslin cardinal and there is a Suslin cardinal above κ, along with Lemma 9.0.16.

ω Let µS and [y]S (for y ∈ ω and S a set of ordinals) be as deﬁned in Section 8.4. Since assuming AD, µS is an ultraﬁlter on the set of S-degrees, for each such S, by Corollary 1.1.6. In this section we will use only the case S = ∅. While this is not necessary for the present argument, we note that the two following lemmas imply the corresponding version where ∅ is replaced with an arbitrary set of ordinals.

9Various thoughts, idle and otherwise. There is something about the equivalence, given AD, of AD+ and something about the Suslin cardinals. Is Θ-Determinacy impossible? What is the relationship between the Wadge rank of a set of reals, and the least cardinal for which it is Suslin, if any? 10This should also give the earlier result that AD implies that the Suslin cardinals are closed below their sup. Also, a nice consequence of the theorem is that if A is Suslin then L(A, R) satisﬁes AD+, assuming AD? 11?? 12Add explanation? Do you need that a limit of Suslin cardinals is a limit of regular cardinals? And do Remark 6.0.8 and Theorem 8.1.1 show this? 134 CHAPTER 11. APPLICATIONS

Lemma 11.4.3. Assume that AD holds. If κ is a limit of Suslin cardinals of Q L[T,x] uncountable coﬁnality, then for all bounded T ⊆ κ, ω1 /µ∅ < κ.

c Proof. By Lemma 9.0.16, every <κ-Borel set (e.g., ≤∅ ×

Lemma 11.4.4. Assume that AD holds. If κ is a limit of Suslin cardinals of Q L[S,y] 13 uncountable coﬁnality, then for each bounded S ⊆ κ, ω2 /µ∅ < κ. Proof. The set of pairs (x, y) ∈ ωω such that x ∈ L[S, y] is <κ-Borel, so its complement is as well. By Lemma 9.0.16, the complement is <κ-Suslin. The leftmost branch construction for uniformizing Suslin sets then shows that there is a bounded T ⊆ κ such that ωω ∩ L[S, y] is a proper subset of ωω ∩ L[T, y] for all y ∈ ωω (one could also use Corollary 10.1.7 for this step). It follows from L[S,y] L[T,y] Theorem 8.2.1 then that the set of [y]∅ for which ω2 < ω1 is in µ∅. The lemma then follows from Lemma 11.4.3.

The proof of Theorem 11.4.1 is completed by appealing to Theorems 11.1.6 and 11.2.1.

Proof of Theorem 11.4.1. We assume that κ has uncountable cofinality, as the countable cofinality case was dealt with just after the statement of the theorem. ω For each α < κ, let Aα be the subset of 2 coded by Sα. Using the coding at the beginning of Chapter 8, we may find Sα,β (α ≤ β < κ), bounded subsets of κ, such that each Sα,β is an ∞-Borel code for the set Aα × Aβ. Lemma ∗ ∗ 11.4.4 and Theorem 11.2.1 give a sequence hSα,β : α ≤ β < κi, where each Sα,β is a bounded subset of κ and a strong ∞-Borel code for the corresponding set S Aα ×Aβ. Let ≤= α≤β<κ Aα ×Aβ and observe that this is a prewellordering of length κ. Theorem 11.1.6 then shows that ≤ is κ-Suslin. By the Kunen-Martin Theorem (Theorem 6.2.1), ≤ cannot be <κ-Suslin.

Recalling Remark 11.4.2, Theorem 11.4.1 has the following consequences.14

Corollary 11.4.5. If AD holds, then the set of Suslin cardinals is closed below its supremum.

Corollary 11.4.6. If AD+ holds, then the set of Suslin cardinals is closed below Θ.

13Maybe you need to worry about wellfoundedness issues in this section, and in the AD+ and not ADR section. 14?? Well, the AD+ result is an immediate consequence, but the AD result seems to require some work. Richard’s paper says that AD implies κ-determinacy for κ below the supremum of the Suslin cardinals. 11.4. CLOSURE OF THE SUSLIN CARDINALS 135

The reverse direction of the following theorem is beyond the scope of this book.15 See Section 7 of [18] for a proof.16

+ Theorem 11.4.7. Suppose that AD + DCR holds. Then AD holds if and only if the set of Suslin cardinals is a closed subset of Θ. By Theorems 11.4.1 and 13.0.1, the context of the following theorem is ex- + 17 actly the case where AD holds and ADR fails. Theorem 11.4.8. Assume that AD+ holds and that there is a largest Suslin cardinal. Then there is a set of ordinals S such that P(ωω) is contained in L(S, R). Proof. By the Moschovakis Coding Lemma, if there is a largest Suslin cardinal, then some subset of ωω is not Suslin. It follows by Theorem 11.2.3 that there is a ≤D-maximal set of ordinals. From and Theorem 8.5.7 it follows that there is a set S of ordinals such that P(ωω) is contained in L(S, R).

15Really? 16? 17Maybe this has all been said already. 136 CHAPTER 11. APPLICATIONS Chapter 12

Scales from Uniformization

In this chapter we will prove Theorem 12.3.1, which says that AD + DC + Uniformization implies that every subset of ωω is Suslin. The proof is a combi- nation of arguments of Becker, Harrington, Kechris, Steel and Woodin.

12.1 Ordinal determinacy in the codes

The proof of Theorem 12.3.1 uses a weaker notion of strong ∞-Borel code (introduced in Section 12.2) where the players play real number codes for ordinals. The determinacy of the associated game follows from AD, by Theorem 12.1.1, which is a variation of the Harrington-Kechris theorem [5] on the determinacy (under AD) of certain real games where the payoff set depends only on the rank of the reals played in a fixed prewellordering. In this chapter, we will say that a sequence P¯ = hPi : i ∈ ωi of subsets of ω ω ω ω is good if, for each i ∈ ω, every subset of ω × ω which is projective in Pi is 1 uniformized by a function which is Σ1 in P (without parameters). Recall that ω 1 ω for any set P ⊆ ω , there is a universal Σ1(Pi+1) set U ⊆ ω × ω such that 1 ω every Σ1(P ) subset of ω equal to {x :(n, x) ∈ U} for some n ∈ ω. Moreover, such a set can be defined from P , so in particular for any sequence hPi : i ∈ ωi ω of subsets of ω there is a sequence hUi : i ∈ ωi such that each Ui is a universal 1 ω Σ1(Pi) subset of ω × ω . We will use a special case of this fact below. ω We say that a sequence h≤i: i < ωi of prewellorderings of 2 is suitable if ω there exists a good sequence hPi : i ∈ ωi of subsets of ω such that, for all i, ≤i is projective in Pi. ω Now suppose that Q = h≤i: i ∈ ωi is a sequence of prewellorderings of 2 , and, for each i ∈ ω, let γi be the length of ≤i and πi be the associated rank Q function. Given A ⊆ i∈ω γi, we let G(Q, A) be the game where players I and ω II successively pick xi ∈ 2 and Player I wins if and only if hπi(xi): i ∈ ωi ∈ A. Theorem 12.1.1 is a version of the Harrington-Kechris theorem from [5]. The proof we give is the Harrington-Kechris proof, as modified by Woodin.

137 138 CHAPTER 12. SCALES FROM UNIFORMIZATION

Theorem 12.1.1. Suppose that AD holds. Let Q = h≤i: i < ωi be a suitable ω sequence of prewellorderings of 2 , and for each i ∈ ω let γi be the length of ≤i. Q Then for each A ⊆ i∈ω γi the game G(Q, A) is determined. The rest of this section is a proof of Theorem 12.1.1, so we assume AD throughout. We fix a suitable sequence Q = h≤i: i < ωi of prewellorderings of ω ω 2 , as witnessed by sets Pi ⊆ ω (i ∈ ω). For each i ∈ ω we let γi be the length Q of ≤i and πi be the corresponding rank function. Fix A ⊆ i∈ω γi. We say that a subtree T of ω<ω is 2-perfect if each node of T has a unique shortest pair of incompatible extensions. There is then a natural bijection be- ω 1 tween [T ] and 2 . Using the uniform existence of universal Σ1(P ) sets, we 1 ω can fix hTi : i ∈ ωi such that, for each i ∈ ω, every Σ1(Pi) function from 2 to the set of ω-sequences of 2-perfect subtrees of ω<ω is equal to the function x 7→ Ti(n, x), for some n ∈ ω (fixing and suppressing some coding of subtrees of ω<ω by elements of ωω). The proof of Lemma 12.1.2 uses the Kuratowski-Ulam theorem1 (which implies that if every subset of ωω has the property of Baire, then every wellordered union of meager sets is meager) and a diagonalization along the lines of the recursion theorem (which results in the double use of m in the formula Ti+1(m, y)m).

Lemma 12.1.2. For each i ∈ ω and each H : ω × 2ω × ωω → ωω projective in ω Pi there exists an m ∈ ω such that for each y ∈ 2 , πi(H(m, y, z)) is the same for all paths z through Ti+1(m, y)m.

Proof. Let B be the set of (m, y, T ) such that (m, y) ∈ ω × 2ω, T ⊆ ω<ω is a 2-perfect tree and πi(H(m, y, z)) is the same for all z ∈ [T ]. Then B is projective in Pi, and, by Kuratowksi-Ulam, for each (m, y) there is a T such that (m, y, T ) ∈ B. 1 ω Since hPj : j ∈ ωi is good, there is a Σ1(Pi+1) function g on ω×2 such that, for all (m, y) ∈ ω × ωω,(m, y, g(m, y)) ∈ B. Moreover, the function sending ω 1 each y ∈ 2 to the sequence hg(n, y): n ∈ ωi is Σ1(Pi+1), so for some m ∈ ω ω ω we have that g(n, y) = Ti+1(m, y)n for all (n, y) ∈ ω × 2 . Then for all y ∈ 2 , Ti+1(m, y)m = g(m, y), so πi(H(m, y, z)) is the same for all paths z through Ti+1(m, y)m.

We next define an integer game whose determinacy (which follows from AD) implies the determinacy of G(Q, A). We fix (and suppress) a recursive bijection between 2ω and 2ω ×ω ×2ω ×ωω, assigning a 4-tuple to each element of 2ω in such a way that the first three coordinates of the output are continuous in the input. For each i ∈ ω, we let ω ω Fi be the function on ω × 2 × ω which on input (n, w, x) returns the 4-tuple ω coded by x as a path through Ti(n, w)n (i.e., assigned to the element of 2 which is the image of x under the natural bijection between [Ti(n, w)n] and ω ω 2 ). For all i, n ∈ ω and w ∈ 2 , then, the function sending x to Fi(n, w, x) is a bijection between 2ω and 2ω × ω × 2ω × ωω. Given (n, w, u, m, y, z) ∈

1citation 12.1. ORDINAL DETERMINACY IN THE CODES 139

ω ω ω ω ∗ ω×2 ×2 ×ω×2 ×ω , we let Fi (n, w, u, m, y, z) be the unique x ∈ [Ti(n, w)n] such that Fi(n, w, x) = hu, m, y, zi. ω ω We say that (n, w, x) ∈ ω×2 ×ω is i-good if x is a path through Ti(n, w)n. 0 We let G (Q, A) be the integer game in which Player I plays (n0, w0, x0) ∈ ω ω ω ω ω ×2 ×ω and Player II plays (m0, y0, z0) ∈ ω ×2 ×ω , both one coordinate per move. To determine the winner of the game, we then let, for each i ∈ ω,

(u2i, ni+1, wi+1, xi+1) be F2i(ni, wi, xi) and (u2i+1, mi+1, yi+1, zi+1) 0 be F2i+1(mi, yi, zi). Player I wins this run of G (h≤i: i ∈ ωi,A) if and only if either

• there is an i ∈ ω such that (mi, yi, zi) is not (2i + 1)-good, and such that, for all j ≤ i,(nj, wj, xj) is (2j)-good, or

• hπi(ui): i < ωi ∈ A. Since G0(Q, A) is an integer game, AD implies that it is determined. We will show that if Player I has a winning strategy in G0(Q, A) then he also has one in G(Q, A). The corresponding fact for Player II is proved by a similar, slightly simpler, argument. 2 These two facts give the determinacy of each game G(Q, A) under AD. We ﬁx a winning strategy σ for player I in G0(Q, A), and write (n, w, x) = σ(m, y, z) to mean that (n, w, x) is the result of Player I’s moves when he plays according to σ and Player II plays (m, y, z). Our desired strategy for G(Q, A) will be induced by the function E given in the following lemma.

Lemma 12.1.3. There is a function E : (2ω)<ω → (ω × 2ω × 2ω)<ω such that

1. for all t, t0 ∈ (2ω)<ω, if t ⊆ t0 then E(t) ⊆ E(t0);

2. for all t ∈ (2ω)<ω, length(E(t)) = length(t) + 1;

ω 3. for all u1, u3 . . . , u2k−1 ∈ ω , if

E(u1, u3, . . . , u2k−1) = (m0, y0, v0, m1, y1, v2, . . . , mk, yk, v2k),

then for each zk ∈ [T2k+1(mk, yk)mk ], if

∗ • for all i < k, zi = F2i+1(mi, yi, u2i+1, mi+1, yi+1, zi+1);

• (n0, w0, x0) = σ(m0, y0, z0);

• for all i ≤ k, (u2i, ni+1, wi+1, xi+1) = F2i(ni, wi, xi);

then π2i(u2i) = π2i(v2i) for all i ≤ k. 2say more? it seems that the fact for player II gives it for I, but not obviously the other way around. You could give the proof for II instead.... 140 CHAPTER 12. SCALES FROM UNIFORMIZATION

Granting the claim, we get a strategy for I in G(h≤i: i ∈ ωi,A) as follows: When II plays u1, u3 ..., I answers by v0, v2,..., where

0 0 1 1 k k E(u1, u3, . . . , u2k−1) = (m , y , v0, m , y , u2, . . . , m , y , v2k) for each k ∈ ω. We will show that this is indeed a winning strategy for player I. Suppose that II has played u2i+1 (i ∈ ω) in a run of the game and I produced v2i (i ∈ ω) following this strategy. Then we have mi and yi (i ∈ ω) as produced by E. For k k k each k ∈ ω deﬁne C0 ,C1 ,...,Ck as follows :

k • Ck is the set of z ∈ [T2k+1(mk, yk)mk ] such that 0 F2k+1(mk, yk, z) = hu2k+1, mk+1, yk+1, z i

for some z0 ∈ ωω;

k • for each i < k, Ci is the set of z ∈ [T2i+1(mi, yi)mi ] such that

0 F2i+1(mi, yi, z) = hu2i+1, mi+1, yi+1, z i

0 k for some z ∈ Ci+1.

k Then hC0 : k ∈ ωi is a ⊆-decreasing sequence of nonempty closed sets, so we may pick a z0 in their intersection and let Player II play (m0, y0, z0). Then the induced values (mi, yi, zi) according to the rules of the game agree with the values (mi, yi) given by E, are each respectively (2i + 1)-good, and the corresponding values u2i+1 are also as given. Moreover, if hn0, w0, x0i = σ(hm0, y0, z0i), then the induced values hni, wi, xii are all good. By part (3) of Lemma 12.1.3, we have for all k ∈ ω that for all i ≤ k + 1, πi(u2i) = πi(v2i). Therefore the same is true for all i ∈ ω. But since 0 I won the run of G (h≤i: i ∈ ωi,A) by playing (n0, w0, x0), he wins the run of G(h≤i: i ∈ ωi,A) by playing v2i (i ∈ ω). It remains then to prove Lemma 12.1.3. Proof of Lemma 12.1.3. We construct f(t) recursively on the length of t. We ﬁrst consider the case where t = ∅, where we have to produce m0, y0 and v0 such that, for any z ∈ [T1(m0, y0)m0 ], if hn0, w0, x0i = σ(m0, y0, z), and 0 0 0 F0(n0, w0, x0) = hu, n , w , x i, then π0(u) = π0(v0). ω Fix a surjective coding of strategies by elements of 2 , let σy be the strategy coded by y and let y0 be such that σ = σy0 . Let H be the function sending (a candidate) m ∈ ω, y ∈ 2ω and z (in [T1(m, y)m]) to the ﬁrst coordinate of F0(σy(m, y, z)). Then H is projective in P0, so Lemma 12.1.2 gives an m0 ∈ ω such that π0(H(m0, y0, z)) is the same for all z ∈ [T1(m0, y0)m0 ]. Then we can let v0 be any such value H(m0, y0, z), for instance, the value produced when z is the leftmost branch of T0(m0, y0)m0 . In the case where t 6= ∅, let t = (u1, . . . , u2k+1). We can assume

k E(u1, . . . , u2k−1) = (m0, y0, v0, . . . , m , yk, v2k) 12.2. BOUNDEDNESS FOR ∞-BOREL RELATIONS 141

is known. We need to ﬁnd mk+1, yk+1 and v2k+2 such that for each zk+1 ∈

[T2k+3(mk+1, yk+1)mk+1 ], if

• z0, z1, . . . , zk are the unique members of

[T1(m0, y0)m0 ],..., [T2k+1(mk, yk)mk ] respectively such that

F2i+1(mi, yi, zi) = hu2i+1, mi+1, yi+1, zi+1i for all i ≤ k;

• (n0, w0, x0) is the result of playing against (m0, y0, z0) according to σ;

• for all i ≤ k + 1, (u2i, ni+1, wi+1, xi+1) = F2i(ni, wi, xi); then π2k+2(u2k+2) = π2k+2(v2k+2). Consider the function which takes in candidatem ˆ 0,y ˆ0,z ˆ0, ...,m ˆ k+1,y ˆk+1, zˆk+1 (forgetting for the moment that the values of m0,y0,z0,...,mk, yk, zk have been fixed already) and returns the corresponding u2k+2 as above, using a strat- egyσ ˆ. This function is projective in P2k+2, so applying Lemma 12.1.2 (where we interpret y as coding a strategyσ ˆ and arbitrary values form ˆ 0,ˆy0,ˆz0,...,m ˆ k, yˆk,z ˆk) again we can find an mk+1 as desired, with the chosen yk+1 coding σ and our fixed values of m0,y0,z0,...,mk, yk, zk.

12.2 Boundedness for ∞-Borel relations

In Section 12.1 we considered a version of coded ordinal determinacy, where the players play elements of 2ω representing ordinals via some fixed prewellorderings. The following definition gives a version of the notion of strong ∞-Borel code relative to such a game. 12.2.1 Definition. Suppose that δ is an infinite cardinal, π : ωω → δ is a surjection and φ is a sentence in Lδ,0. Associate to φ and π a game Gφ,π on ωω, where I and II collaborate to build a countable set σ ⊆ ωω, and I wins if A ⊆ A . We say that φ is a π-strong ∞-Borel code (or π-strong δ-Borel φπ[σ] φ code) for Aφ if I has a winning strategy in Gφ,π. If S is the set of ordinals such that φ = φS, we also say that S is a π-strong ∞-Borel code or π-strong δ-Borel code for Aφ. ∞ Replacing the use of δS -Determinacy in the proof of Theorem 11.2.1 with AD, we get the following. Recall that AD implies that µS is an ultrafilter on 3 DS, for each set S of ordinals. Theorem 12.2.2. Let S be a set of ordinals. Suppose that AD holds, and the DS ∞ ultrapower Ord /µS is wellfounded, and δS < Θ. If S is an ∞-Borel code for ω ∞ a set A ⊆ ω , then A has a π-strong ∞-Borel code which is contained in δS and definable from S.

3reference? 142 CHAPTER 12. SCALES FROM UNIFORMIZATION

We use Theorem 12.2.2 to prove Theorem 12.2.3 below. The proof uses Woodin’s collapse embedding technique, which is also used in the proof of ....4 Theorem 8.4.1 and Corollary 10.1.7 together give that if Uniformization holds ∞ then for each bounded subset S of Θ, δS < Θ. Since DC holds, there are coﬁnally many κ < Θ with the property that every subset of ωω with Wadge ∞ rank less κ has an ∞-Borel code S with δS < κ. Theorem 12.2.3. Let γ < Θ be an ordinal. Suppose that, for each i ∈ ω,

ω • πi is a surjection from 2 onto some ordinal γi < γ;

• Si is a bounded subset of γ and a πi-strong ∞-Borel code for a subset of ωω × ωω. Suppose also that R = S A is a wellfounded relation. Then the rank of R i∈ω φSi is less than γ+.

Proof. For each i ∈ ω, let Σi be a strategy witnessing that Si is a πi-strong W ω ∞-Borel code. Let φ be i∈ω φSi . Let B ⊆ ω be of Wadge rank greater than γ, and such thatφ, hπi : i ∈ ωi and hΣi : i ∈ ωi are in L(B, R). Let τ be a Col(ω, γ)-name for an element of ωω coding the pair (γ, φ).5 p ω Given a condition p ∈ Col(ω, γ), let Iτ be the set of X ⊆ ω for which player I wins the game where I and II play a descending sequence of Col(ω, γ) conditions, with I winning if the induced realization of τ is in X. Let g ⊂ ω ω Col(ω, γ) be L(B, R)-generic, and let Ug be the set of X ⊆ ω for which ω \ X p is in Iτ , for some p ∈ g. Since DC holds, the ultrapower Mg = Ult(L(B, R),Ug) is wellfounded, and, since Uniformization holds, the induced embedding j from L(B, R) to Mg is elementary. Moreover, j(R) is wellfounded. The identity function represents Mg τg, so γ and φ are both in H(ℵ1) . Since γ is countable in Mg, φ codes 0 a Borel relation R in Mg which contains R. Since Col(ω, γ) has cardinality + V Mg 0 γ,(γ ) ≥ ω1 . If R is wellfounded in Mg, then we have contradiction to 1 ∼Σ1-boundedness. To complete the argument, consider for each i ∈ ω the tree of attempts to find a run of G according to j(Σi) where the induced j(φS ) j(πi)(σ) j(φSi ),j(πi) i is isomorphic to φSi . It suffices to see that each such tree has an infinite path 0 in Mg, since then R is contained in j(R). If for some i there is no such path, then this tree of attempts has a ranking function in Mg. However, each node in the tree contained in the image of j has an extension in the image of j. Consideration of the least rank of a node in the image of j shows that this is impossible.

12.3 Becker’s argument

In this section we complete the proof of the following theorem

4?? 5and more? 12.3. BECKER’S ARGUMENT 143

Theorem 12.3.1. If AD + DC + Uniformization holds, then every set of reals is Suslin. To prove the theorem, we start with an arbitrary A ⊆ ωω, which we will show to be Suslin. Since DC and Uniformization hold, there exists a suitable sequence h≤i: i ∈ ωi of prewellorderings, as witnessed by a good sequence hPi : i ∈ ωi, and a γ < Θ such that

• A ≤W P0;

• γ is equal to both the supremum of the lengths of the prewellorderings ≤i 6 and the Wadge ranks of the sets Pi.

For each i ∈ ω, let γi be the length of ≤i and let πi be the corresponding rank function. We assume also that A is γ0-Borel. S Q Say that a tree on γ is good if it is a subset of n∈ω i

• x¯ = hxn : n ∈ ωi ∈ Tr and • y ∈ A if and only if Tr(¯x) is illfounded. If II has a winning strategy in this game, then A is Suslin. A theorem of Kechris7 shows that this game is determined in our context, but the proof of Theorem 12.3.1 does not use this fact. Let Λ = S Σ1(P ), and say that a strategy σ for Player I is simple if, n∈ω ∼1 n for each n, the fragment of σ for the ﬁrst n moves of the game is in Λ. The main result of Section 12.2 gives the following claim.

Claim 1. Player I does not have a simple winning strategy in G1.

Proof. Suppose that Σ is a simple winning strategy for Player I in G1. Consider the tree T of sequences ht, x,¯ yni, where ht, x¯i is a run of the integer/real game according to Σ and yn is a ﬁnite partial path through the tree coded by II’s moves. Order T by equivalence on the ﬁrst two coordinates and extension on the third. Fixing any pair (t, x¯), Tt,x¯ has some ordinal rank. For each wellfounded tree (with ordinals from the right levels), if II plays it then I plays something + (in A0), so the union of the ranks of these trees (and thus the rank of T ) is γ .

6and maybe one more thing about the Turing ultrapower 7citation 144 CHAPTER 12. SCALES FROM UNIFORMIZATION

Any wellfounded tree can be spaced out with 0’s to make a tree of the same rank with ordinals from the right levels.

0 We now code G1 with a game G2 in the same way that G (Q, A) coded G(Q, A) ω in Section 12.1. To do this, we fix a Borel surjection π : 2 → ω and va (a ∈ ω) such that π(va) = a. We again use a fixed suppressed a recursive bijection between 2ω and 2ω × ω×2ω ×ωω, assigning a 4-tuple to each element of 2ω in such a way that the first three coordinates of the output are continuous in the input. For each i ∈ ω, we ω ω let Fi be the function on ω×2 ×ω which on input (n, w, x) returns the 4-tuple ω coded by x as a path through Ti(n, w)n (i.e., assigned to the element of 2 which ω is the image of x under the natural bijection between [Ti(n, w)n] and 2 ). For ω all i, n ∈ ω and w ∈ 2 , then, the function sending x to Fi(n, w, x) is a bijection between 2ω and 2ω×ω×2ω×ωω. Given (n, w, a, m, y, z) ∈ ω×2ω×ω×ω×2ω×ωω, ∗∗ we let Fi (n, w, a, m, y, z) be the unique x ∈ [Ti(n, w)n] such that Fi(n, w, x) = hva, m, y, zi. We let G2 be the integer game in which Player I plays (n0, w0, x0) ∈ ω × ω ω ω ω 2 × ω and Player II plays (m0, y0, z0) ∈ ω × 2 × ω , both one coordinate per move. To determine the winner of the game, we then let, for each i ∈ ω,

(u2i, ni+1, wi+1, xi+1) be F2i(ni, wi, xi) and (u2i+1, mi+1, yi+1, zi+1) 0 be F2i+1(mi, yi, zi). Player I wins this run of G (h≤i: i ∈ ωi,A) if and only if either

• there is an i ∈ ω such that (mi, yi, zi) is not (2i + 1)-good, and such that, for all j ≤ i,(nj, wj, xj) is (2j)-good, or

• hπ(u0), u1, π(u2), u2,...i is a winning run of G1 for Player I.

Since G2 is an integer game, AD implies that it is determined. The proof8 in Section 12.1 gives the following.

Claim 2. If Player I has a winning strategy in G2, then he has a simple winning strategy in G1. Finally, a modiﬁcation of the proof in Section 12.1 gives the following claim, which completes the proof of Theorem 12.3.1.

Claim 3. If Player II has a winning strategy in G2, then she has one in G1. The rest of this section is a proof of Claim 3. Suppose that σ is a winning <ω strategy for Player II. Givena ¯ = ha0, . . . , aki ∈ ω ,s ¯ = h(n0, w0),..., (nk, wk)i ∈ ω <ω ∗ (ω × 2 ) and xk ∈ [T2k(nk, wk)nk ], let B (¯a, s,¯ xk) be u2k+1, where ∗∗ • for all i < k, xi = F2i (ni, wi, ai, ni+1, wi+1, xi+1); 8of? 12.3. BECKER’S ARGUMENT 145

• (m0, y0, z0) = σ(n0, w0, x0);

• for all i ≤ k,(u2i+1, mi+1, yi+1, zi+1) = F2i+1(mi, yi, zi); Lemma 12.3.2. There is a function E0 : ω<ω \ {∅} → (ω × 2ω)<ω such that

1. for all t, t0 ∈ (2ω)<ω, if t ⊆ t0 then E0(t) ⊆ E0(t0);

2. for all nonempty t ∈ ω<ω, length(E0(t)) = length(t);

3. for all a¯ = ha0, . . . , aki ∈ ω, if

0 E (a0, . . . , ak) =s ¯ = h(n0, w0),..., (nk, wk)i,

then ¯ Q (a) for each xk ∈ [T2k(nk, wk)nk ] and each ξ ∈ i

(b) there exist ρ1, . . . , ρk such that for each xk ∈ [T2k(nk, wk)nk ], if for ∗∗ all i < k, xi = F2i (ni, wi, ai, ni+1, wi+1, xi+1), then xik = ρk for all i ≤ k.

Given a function E0 as in Lemma 12.3.2, we define a strategy for player 0 II in G1 : in response toa ¯ = ha0, . . . , aki, lettings ¯ = E (¯a), II plays a u ω (fixing these u’s requires only countably many choices from 2 ) such that Sk(u) ∗ 0 is the union of all trees of the form Sk(B (¯a, E (¯a), x)), where u2k+1 for some x ∈ [T2k(nk, wk)nk ]. To see this that is a winning strategy for player II, fix ∗ hai : i ∈ ωi and let hui : i ∈ ωi be the sequence of moves played by Player II ∗ S ∗ in response, according to this strategy. Let S = i∈ω Si(ui ). Let ni and wi 0 (i ∈ ω) be as given by hai : i ∈ ωi and E . k k k As in the proof of the Theorem 12.1.1, for each k ∈ ω define C0 ,C1 ,...,Ck as follows :

k • Ck is the set of x ∈ [T2k(nk, wk)nk ] such that

0 F2k(nk, wk, z) = hvak , nk+1, wk+1, x i

for some z0 ∈ ωω;

k • for each i < k, Ci is the set of x ∈ [T2i(ni, wi)ni ] such that

0 F2i(ni, wi, x) = hvai , ni+1, wi+1, x i

0 k for some x ∈ Ci+1.

k Then hC0 : k ∈ ωi is a ⊆-decreasing sequence of nonempty closed sets, so we may pick a x0 in their intersection and let Player I play (n0, w0, x0). Then the induced values (ni, wi, xi) according to the rules of the game agree with the 146 CHAPTER 12. SCALES FROM UNIFORMIZATION

0 values (ni, wi) given by E , are each respectively (2i)-good, and induce the plays vai which map by π to the corresponding values ai. Let (m0, y0, z0) be σ(n0, w0, x0), and let

(u2i+1, mi+1, yi+1, zi+1) = F2i+1(mi, yi, zi) S for each i ∈ ω. Let S = i∈ω Si(u2i+1). Since σ is a winning strategy for Player II, hai : i ∈ ωi is in A if and only if S is wellfounded. It suﬃces then to see that S is wellfounded if and only if S∗ is. Since S is contained in S∗, one direction of this is clear. ∗ For the other direction, consider an inﬁnite path ξi through S and some initial segment ξ¯of some length k which is not in S. Let ` ≥ k be as in conclusion ¯ (3a) of Lemma 12.3.2, with respect to ξ and u2k+1. Since S`(u2`+1)k is a subset ¯ ¯ of Sk(u2k+1), ξ is not in S`(u2k+1). Since x` ∈ [T2`(n`, x`)n` ], ξ is not in

∗ 0 S`(B (ha0, . . . , a`i,E (ha0, . . . , a`i), x)

for any x ∈ [T2`(n`, x`)n` ], by conclusion (3b) of Lemma 12.3.2. It follows then ¯ ∗ ∗ that ξ is not in S`(u`+1). This completes the proof that S is wellfounded, and the proof that the strategy deﬁned above is winning for Player II.

It remains to prove Lemma 12.3.2.

Proof of Lemma 12.3.2. Let a0, . . . , ak ∈ ω be given and suppose, in the case 0 where k > 1, that E (a0, . . . , ak−1) has been chosen. Fix a Borel coding w 7→ ω k−1 ω ew of elements of (ω × 2 ) by elements of 2 , and let wk be such that 0 E (a0, . . . , ak−1) = ewk . Let ew,i refer to the ith pair from ew, for each i < k. We need to choose nk in such a way that

(nk, wk,T2k(nk, wk)nk ) ∈ D, where D is the set of (n, w, T ) such that (n, w) ∈ ω ×2ω, T ⊆ ω<ω is a 2-perfect tree and ¯ Q • for all x ∈ [T ] and all ξ ∈ i

• there exist ρ1, . . . , ρk such that for each xk ∈ [T ], if for all i < k, xi = ∗∗ F2i (ew,i, ai, ew,i+1, xi+1), then xik = ρk for all i ≤ k (letting ew,k denote the pair (n, w)).

Note that D is projective in P2k−1. For each (n, w) there is a T such that (n, w, T ) ∈ D. To see this, note ﬁrst ω ¯ Q that (since every subset of ω has the Baire Property) for each ξ ∈ i

¯ ∗ 0 _ 0 ξ ∈ Sk(B (ha0, . . . , aki,E (ha0, . . . , ak−1i (n, w), x )).

By the Kuratowski-Ulam theorem, the intersection of these sets over all xi¯ (ﬁxing n and w) is also comeager. Again applying the Baire Property, we can shrink this set to a somewhere comeager set of x for which the values xik are ∗∗ all the same, where, for each i < k, xi = F2i (ni, wi, ai, ni+1, wi+1, xi+1). Any T such that [T ] is contained in this somewhere comeager set suﬃces. The rest of the proof is the same as the end of the proof of Lemma 12.1.2. 1 ω Since hPi : i ∈ ωi is good, there is a Σ1(P2k) function g on ω × 2 such that, for all (n, w) ∈ ω × ωω,(n, w, g(n, w)) ∈ D. Moreover, the function sending each ω 1 w ∈ 2 to the sequence hg(n, w): n ∈ ωi is Σ1(P2k), so for some nk ∈ ω we ω have that g(n, w) = T2k(nk, w)n for all (n, w) ∈ ω × 2 . Then T2k(nk, wk)m = g(nk, wk), so (nk, wk,T2k(nk, wk)nk ) ∈ D. 148 CHAPTER 12. SCALES FROM UNIFORMIZATION Chapter 13

ADR

In this chapter we will prove the following theorem, which is due to Martin and Woodin independently.

Theorem 13.0.1 (Martin, Woodin). If AD holds and every subset of ωω is

Suslin, then ADR holds. We will give two versions of one of the key steps in our proof of this theorem. One is a theorem of Martin (Theorem 13.0.2 below) which we will not prove. The other is a lemma which we will prove in Section 13.1, which can be used in place of Martin’s theorem. Our proof of this lemma will not be entirely self- contained, either. We prove a key lemma in Section 13.2 and finish the proof of Theorem 13.0.1 in Section 13.3. We will be using material on towers of measures. We briefly review some of this material here. More thorough treatments can be found in [24, 22, 32], among other places. Given a set X and an i ∈ ω, an ultrafilter µ on Xi+1 projects to the ultrafilter µ0 on Xi whose members are the set of A ⊆ Xi for which {s_hxi : s ∈ A, x ∈ X} is in µ (we sometimes refer to ultrafilters as measures). Each function on Xi has a natural reinterpretation as function on Xi+1 (ignoring the last coordinate of the input) and this reinterpretation 0 induces a factor map jµ0,µ from a µ -ultrapower to a µ-ultrapower. A tower of ultrafilters on X is a sequence hµi :∈ ωi such that each µi is an ultrafilter on i X , and each µi+1 projects to the corresponding µi. There is a natural direct limit embedding associated to a tower, and the tower is said to be wellfounded if the corresponding image model is wellfounded. Given a tree T on a product X × Y , and an s ∈ X<ω, T (s) denotes the set {t ∈ Y |s| :(s, t) ∈ T }. A tree T on ω × Y (for some set Y ) is homogeneous if <ω there exists a collection of measures {µs : s ∈ ω } such that,

<ω • for each s ∈ ω , T (s) ∈ µs, and

ω • for each x ∈ ω , x ∈ p[T ] if and only if hµxn : n ∈ ωi is a wellfounded tower.

149 150 CHAPTER 13. ADR

A tree T on ω × Y (for some set Y ) is weakly homogeneous if there exists a countable set σ such that, for each x ∈ ωω, x ∈ p[T ] if and only if there exists a wellfounded tower hµn : n ∈ ωi such that, for each n ∈ ω, µn ∈ σ and T (xn) ∈ µn. We refer the reader to [24, 22, 32] for more on homogeneous trees and weakly homogeneous trees. A subset of ωω is homogeneously Suslin if it is the projection of a homogeneous tree, and weakly homgeneously Suslin if it is the projection of a weakly homogeneous tree (equivalently, if it is the projection, in a diﬀerent sense,1 of a homogeneously Suslin subset of ωω × ωω2). The following theorem of Martin appears as Theorem 1.1 of [26]. ω Theorem 13.0.2 (Martin). Assuming ZF + AD + DCR, for any A ⊆ ω , A is homogeneously Suslin if and only if A and its complement are Suslin.3 13.0.3 Remark. The construction of the Martin-Solovay tree (in [23]; see also Section 1.3 of [22]) implies that the complement of a weakly homogeneously Suslin set is Suslin. It is the other direction of Theorem 13.0.2 which is used in the proof of Theorem 13.0.1.

13.1 Weakly homogeneous trees

We start by proving a theorem of Kunen on definability of measures. Given a set X, we let meas(X) denote the set of countable complete ultrafilters on X. Recall (from part (3) of Remark 1.0.2) that AD implies that every set of reals has the property of Baire, which implies that every nonprincipal ultrafilter (on any set) is countably complete. The following theorem shows that, assuming AD, every nonprincipal ultrafilter on an ordinal below Θ is ordinal definable. Our proof is taken from [32]. We use the ordered equivalence relation (≡M, ≤M) from Chapter 8 (although any ordered equivalence relation as thick as (≡M, ≤M), such as the Turing degrees or the constructibility degrees) would work just as well). By Theorem 1.1.2, AD implies that the ≡M-cone measure µM is an ultrafilter on the set CM of ≡M equivalence classes.

Theorem 13.1.1 (Kunen). If AD + DCR holds and κ < Θ, then every element of meas(κ) is ordinal definable. Proof. Fix κ < Θ, µ ∈ meas(κ) and A ⊆ ωω of Wadge rank κ. By the Moschovakis Coding Lemma4, there is a surjection F : ωω → P(κ) in L(A, R). Let γ be the supremum of κ and the Wadge rank of F −1[µ]. The model L(B, R) is the same for all B ⊆ ωω of Wadge rank γ, which means that this model is ordinal definable. It suffices then to fix such a B and prove that µ is ordinal definable in L(B, R). Note that F and µ are elements of L(B, R). CM L(B,R) Since DCR holds, DC holds in L(B, R), which means that (Ord /µM ) is wellfounded. Let 1used somewhere in this book? 2reference? 3Look at [26]. Maybe we just need, and are proving, a weak version of this? 4and something else 13.1. WEAKLY HOMOGENEOUS TREES 151

ω • Fµ be the function on ω deﬁned by setting Fµ(x) to be F (x) if F (x) ∈ µ, and κ \ F (x) otherwise;

• fµ be the function on CM deﬁned by setting fµ(d) to be the least member T of {Fµ(x): x ≤M y}, for any y ∈ d (this does not depend on the choice of y).

CM L(B, ) • γµ be the ordinal represented by fµ in the ultrapower (Ord /µM) R .

Then µ is is deﬁnable from γµ in L(B, R), as it is the set of C ⊆ κ such that, for any function g representing γµ in

CM L(B, ) (Ord /µM ) R ,

{d ∈ CM : g(d) ∈ C} ∈ µM. 13.1.2 Remark. Let κ be an ordinal below Θ, and let F : ωω → P(κ) be a surjection. If the Wadge ranks of the members of {F −1[µ]: µ ∈ meas(κ)} are not cofinal in Θ, then there is a surjection from ωω onto meas(κ). By Theorem 13.1.1, each set F −1[µ] is ordinal definable. By Theorem 6.1.4, if the Wadge ranks of the members of {F −1[µ]: µ ∈ meas(κ)} are cofinal in Θ, then there is a counterexample to Uniformization. The following theorem is our alternative to Theorem 13.0.2. 5 6 Lemma 13.1.3. If AD holds, κ is an ordinal below Θ and there is a surjection from ωω to meas(κ<ω), then every tree on ω × κ is weakly homogeneous. Proof. Let F : ωω → meas(λ<ω) be a surjection, and define

<ω FM : CM → Pℵ1 (meas(κ ))

7 <ω by setting FM(d) to be F [d]. Let U be the set of C ⊆ Pℵ1 (meas(κ )) −1 such that FM [C] ∈ µM. Then U is a ﬁne, countably complete measure on <ω Pℵ1 (meas(κ )). ω Fix a tree T on ω × κ. As usual, given x ∈ ω , we let Tx be the set <ω <ω {s ∈ κ :(x|s|, s) ∈ T }. We will show that the set of σ ∈ Pℵ1 (meas(κ )) witnessing the weak homogeneity of T is in U. To do this, we deﬁne for each such σ a game Gσ. This game will be closed for player I, and player I will have a winning strategy if σ does not witness the weak homogeneity of T . For each i ∈ ω, in round i of the game Gσ, player I plays ki ∈ ω, αi ∈ κ and βi ∈ Ord, and player II plays a measure µi ∈ σ concentrating

5??? We’re probably not using any of this (why not???). Now suppose that we can map ωω onto the measures on λ. By Kunen, all measures are ordinal deﬁnable. Let B be a prewellordering of length λ, inducing a surjection ρ from R into P(λ). Then every measure on HOD λ induces an ODA set of reals. Since Θ A < Θ, we can map the reals onto the measures. 6Or this. Under AD all measures on ordinals are countably complete and ordinal-deﬁnable. By the coding of measures theorem of Kechris (Kunen??) assuming AD and that there is a Suslin cardinal above κ, there are fewer than Θ many measures on κ (or κ<ω for that matter.) 7This proof is copied from Trevor’s MO post 152 CHAPTER 13. ADR

i on κ . Player II is required to play so that for each i ∈ ω, µi+1 projects to µi. Player I is required to play so that for each i ∈ ω,(hk0, . . . , kii, hα0, . . . , αii) is in T and jµi,µi+1 (βi) > βi+1, where jµi,µi+1 is the factor map corresponding to the projection of µi+1 to µi. The ﬁrst player to break these rules loses; if there is no such player then player I wins.

I x0, α0, β0 x1, α1, β1 x2, α2, β2 ... II µ0 µ1 ...

The game Gσ

If σ does not witness that T is weakly homogeneous then there is an x ∈ p[T ] for which there is no wellfounded tower hµi : i < ωi consisting of measures from i σ, with {s ∈ κ :(xi, s) ∈ T } ∈ µi for each i. Lemma 1.3.8 of [22] shows that in this case there is a continuous witness to the illfoundedness of towers of measures in σ concentrating on Tx. This gives a winning strategy for player I in Gσ, letting x = hki : i < ωi, hαi : i < ωi be any fixed witness to x ∈ p[T ], and the values βi come from this continuous witness. It suffices now to show that for U-many σ, I does not have a winning strategy for Gσ. Assume toward a contradiction that for U-almost every σ, player I does has a winning strategy in Gσ. Since each game Gσ is closed, and since the set of possible moves is for player I wellorderable (this is also true for player II, since all measures on κ<ω are ordinal-definable, but not needed for the argument that follows), we can let, for each σ for which I has a winning strategy in in Gσ, H(σ) be the winning strategy for I where I plays the least move leading to a subgame where she still has a winning strategy. Now:

• let k0 be the element of ω played by H(σ) in round 0 for U-many σ;

σ • let µ0 be the set of A ⊆ κ such that, for U-many σ, the ordinal α0 played by H(σ) in round 0 is in A (then µ0 is a countably complete measure on κ);

• let k1 be the element of ω played by H(σ) in round 1 in response to µ0, for U-many σ;

2 σ σ • let µ1 be the set of A ⊆ κ such that, for U-many σ, letting hα0 , α1 i be as σ σ played by H(σ) in rounds 0 and 1 in response to µ0, hα0 , α1 i ∈ A (then 2 µ1 is a countably complete measure on κ projecting to µ0).

ω Continuing in this way, we get x = hki : i ∈ ωi ∈ ω and a tower of measures σ σ ~µ. Each µi concentrates on Tx because (α0 , . . . , αi ) ∈ Tx. It is a wellfounded tower because if Ai ∈ µi for all i < ω then by the countable completeness of σ σ U there is a σ such that (α0 , . . . , αi ) ∈ Ai for all i ∈ ω. However, by the countable completeness of U there is a σ such that ~µ is a legal play by player 13.2. NORMAL MEASURES 153

II against player I’s winning strategy H(σ), so player I’s moves βi induced by playing H(σ) against ~µ continuously witness the illfoundedness of ~µ, giving a contradiction.

13.1.4 Remark. If all sets of reals are weakly homogeneously Suslin, then all set of reals are homogeneously Suslin. Otherwise, the homogeneously Suslin sets would form a proper initial segment of the Wadge hierarchy, and every subset of ωω would be the continuous image of a member of this initial segment. This would induce a surjection from ωω onto P(ωω).

13.1.5 Remark. Suppose that AD holds and that κ is the largest Suslin cardinal ω below some member θ∗ of the Solovay sequence. Let Γ be the set of subsets of ω HODΓ of Wadge rank less than θ∗. Then Θ = θ∗. By Theorem 6.0.7 there exists in HODΓ a tree T on ω × κ projecting to a set whose complement is not Suslin ω in HODΓ. There is no surjection in HODΓ from ω onto the measures on κ or even onto any countable set of measures witnessing the weak homogeneity of T , since otherwise, the Martin-Solovay construction (see Remark 13.0.3) would produce Suslin representations for the complement of the projection of T . By ω Theorem 13.1.1, each measure on κ is coded by a subset of ω in HODΓ. It follows that if θ∗ < Θ, then there is a countable set of measures witnessing the T is weakly homogeneously Suslin by Lemma 13.1.3, and the least Wadge ranks ω of subsets of ω in HODΓ coding these measures must be coﬁnal in θ∗. The Martin-Solovay construction also shows in this case that the complement of the projection of T is θ∗-Suslin, which implies that θ∗ is a Suslin cardinal.

Theorem 13.1.6. If AD+ holds then every successor member of the Solovay sequence below Θ has coﬁnality ω.

Proof. Let θ∗ be a successor member of the Solovay sequence. By Theorems 6.0.13 and 6.1.4, and Corollary 11.4.6, there is largest Suslin cardinal below θ∗. If θ∗ < Θ, then Remark ?? shows that θ∗ has countable coﬁnality.

13.2 Normal measures

In this section we show (assuming DCR) that if λ is a Suslin cardinal and λ-

Determinacy holds, then there is a normal ﬁne ultraﬁlter on Pℵ1 (λ).

13.2.1 Deﬁnition. We say that an ultraﬁlter µ on P(Pℵ1 (λ)) is

• ﬁne if for all α ∈ λ, {σ ∈ Pℵ1 (λ): α ∈ σ} is in µ;

• normal if whenever f : Pℵ1 (λ) → Pℵ1 (λ) is such that p(σ) is a nonempty

subset of σ for each σ ∈ Pℵ1 (λ), there exists an α ∈ λ such that

{σ ∈ Pℵ1 (λ): α ∈ f(σ)}

is in µ. 154 CHAPTER 13. ADR

Recall (from Deﬁnition 6.0.4) that a Suslin cardinal is an ordinal λ (necessarily a cardinal) such that some subset of ωω is λ-Suslin but not γ-Suslin for any γ < λ.

Theorem 13.2.2. If λ is a Suslin cardinal, and λ-Determinacy + DCR holds, 8 then there is a normal ﬁne measure on Pℵ1 (λ).

Proof. Fix a tree T as given by Lemma 6.0.11. For each σ ⊆ λ, let T σ be the set of nodes (s, t) of T for which the range of t is contained in σ. We say that σ is full if, for each node (s, t) of T σ, there exists an x ∈ p[T σ] whose leftmost branch in T contains (s, t). Given an injection g : ω → λ, we let Tg be the tree on ω × ω consisting of those nodes (s, t) ∈ ω<ω × ω<ω such that (s, g ◦ t) is in T . Let us say that a function g : ω → λ is full if it is injective and its range is is full. We call the function (s, t) 7→ (s, g ◦ t) embedding Tg into T the g-induced map.

0 Claim. If g and g are distinct full functions from ω to λ, then Tg 6= Tg0 . Proof of Claim. We want to see that, given a full function g : ω → λ, the g- induced map can be recovered from Tg without using g. This suﬃces, as the g-induced map is uniquely determined by g (by the second condition in the conclusion of Lemma 6.0.11; this is the reason for that condition). We work recursively through Tg (using some wellordering of Tg in ordertype ω which lists each node before its successors), building a ⊆-increasing sequence of ﬁnite partial length-preserving embeddings ek (k ∈ ω) of Tg into T (and possibly halting if our construction breaks down, although we will eventually see that it doesn’t). We let e0 be the map which sends the empty node to itself. Now suppose that we have built ek (for some k ∈ ω), and that (s, t) is the least node of Tg (in our wellordering) outside the domain of ek. We may suppose that (s, t) is of length n + 1, for some n ∈ ω, and that (sn, tn) is in the domain of ek. Let Ek be the set of γ < λ for which there exists a length-preserving _ embedding of Tg into T extending ek and mapping (s, t) to (s, (ek ◦(tn)) hγi). If Ek is empty, then we stop the construction. Otherwise, letting γ∗ be the least element of Ek, we let

_ ek+1 = ek ∪ {(s, t), (s, (ek ◦ (tn)) hγ∗i)}. This completes the construction. We now verify inductively that our maps ek agree with the g-induced map. For e0 this is clear. Supposing that is is true for a given k ∈ ω, and letting (s, t) be the least node of Tg outside domain of ek, we need to see that g(t(n)) is the least element of Ek (where (s, t) has length n + 1). Our induction hypothesis implies that g(t(n)) is in Ek, which means that ek+1 was deﬁned. Since g is full, there is an x ∈ ωω whose T -leftmost branch goes through (s, g ◦ t). This shows that g(t(n)) is the least element of Ek, as desired, since a length-preserving embedding of Tg into T witnessing the contrary would produce a path though

8This is due to Woodin? Compare the result here with Kechris-Harrington. 13.2. NORMAL MEASURES 155

[T ] contradicting our assumption that the leftmost branch of x in T contains (s, g ◦ t). The completes the proof of the claim.

We deﬁne a function F whose domain is the set of trees on ω × ω:

• for trees of the form Tg, for g a full function from ω to λ, F (Tg) is the range of g;

• for all other trees on ω × ω, F takes the value ∅. By the claim, this function is well deﬁned. We now apply λ-Determinacy.

Given a set A ⊆ Pℵ1 (λ), let GA be the game where

• players I and II build a ⊆-increasing sequence hpn : n ∈ ωi of ﬁnite partial injections from ω to λ, with n contained in the domain of pn for each n ∈ ω; S • letting g = n∈ω pn, player II wins if g is full and F (Tg) is in A. This game is determined by λ-Determinacy.

Let µ be the set of A ⊆ Pℵ1 (λ) for which II has a winning strategy. We wish to see that µ is a normal fine ultrafilter. To check that µ is fine, fix an α < λ and a strategy Σ for player I in the game GA, where A = {x ∈ Pℵ1 (λ): α ∈ x}. Using DCR and the assumption that every node of T is part of the leftmost branch of some element of p[T ], we can find a winning play for II against Σ.

To see that µ is an ultrafilter, fix A ⊆ Pℵ1 (λ). We show first that at least one of A and P (λ) \ A is in µ. Fix strategies Σ and Σ for player I in G ℵ1 A Pℵ1 (λ)\A A and G , respectively. Using DC we can find a countable full set σ ⊆ λ Pℵ1 (λ)\A R which is closed under both Σ and Σ (in the sense that a response by A Pℵ1 (λ)\A either strategy to a sequence of moves whose ranges are contained in σ will a function with range contained in σ). Then, playing as II, we can produce runs of G and G where I plays with Σ and Σ respectively, and A Pℵ1 (λ)\A A Pℵ1 (λ)\A the range of the resulting function g is σ in each case. This shows that ΣA and Σ cannot both be winning strategies for I. The same argument, with I Pℵ1 (λ)\A and II reversed, shows that A and Pℵ1 (λ) \ A cannot both be in µ. To see that µ is normal, fix a function f which picks a nonempty subset of each σ ∈ Pℵ1 (λ). For each α ∈ λ, let Aα be the set of σ ∈ Pℵ1 (λ) for which α is in f(σ). We want to see that that player II has a winning strategy in some Aα. Let G be the game in which player II goes first, picking α < λ, after which the players play GAα , with I going first as usual. We want to see that player II has a winning strategy in this game. If she doesn’t, then I does, by λ-Determinacy. Player I’s winning strategy in the game G induces a sequence hΣα : α < λi, where each Σα is a winning strategy for I in the corresponding game GAα . Using DCR again we can find a countable full set σ ⊆ λ which is closed under

Σα for each α in σ. Then we can ﬁnd runs of the games GAγ (γ ∈ σ), where in each case I plays according to Σγ and the range of the resulting function g is σ. This gives a contradiction. 156 CHAPTER 13. ADR

We note the following theorem, whose proof is essentially the same as the proof of the claim in the proof of Theorem 13.2.2.9 Theorem 13.2.3. Let M be a transitive model of ZF, let λ be an ordinal of M and let T be a tree on ω × λ such that for each node (s, f) of T there exists an x ∈ p[T ] ∩ M such that (s, f) is part of the leftmost branch of x through T . Let j : M → N be an elementary embedding, with N transitive. Then j[T ] is in L[T, j(T )]. Theorem 13.2.4. Let λ be an ordinal, and let T be a tree on ω × λ. Suppose that M is a transitive set such that T ∈ M and such that for each node (s, f) of T there exists an x ∈ p[T ] ∩ M such that (s, f) is part of the leftmost branch of x through T . Let N be a transitive set and let j : M → N be an elementary embedding. Then j[T ] is in L[T, j(T )].

Proof. The map jT embeds T into j(T ). We want to ﬁnd this map in L[T, j(T )]. We use the fact that if

• P is a wellfounded model of ZF (in particular, a forcing extension of L[T, j(T )]),

• S and R are trees in P , • S is countable in P , • g is, in P , a length-preserving partial embedding of S into R, and • there is a length-preserving embedding of S into R extending g (in V ), then there is, in P , a length-preserving embedding of S into R extending g. We work recursively through T , using some wellordering h(sα, tα): α < δi of T which lists each node before its successors, building a continuous ⊆-increasing sequence of partial length-preserving embeddings eα (α < δ) of T into j(T ) (and possibly halting if our construction breaks down, although we will eventually see that it doesn’t). Each eα will have {(sβ, tβ): β < α} as its domain. Then e0 is the empty function. We let e1 be the map which sends the empty node to itself. Now suppose that we have built eα, for some α ∈ (1, δ). Then (sα, tα) is of length n + 1, for some n ∈ ω, and (sαn, tαn) is in the domain of eα. Let Eα be the set of γ < λ for which, in any generic extension of L[T, j(T )] in which T is countable, there exists a length-preserving embedding of T into j(T ) extending _ eα and mapping (sα, tα) to (sα, (eα(tαn)) hγi). If Eα is empty, then we stop the construction. Otherwise, letting γ∗ be the least element of Eα, we let

_ eα+1 = eα ∪ {(sα, tα), (sα, (eα(tαn)) hγ∗i)}. This completes the construction. We now verify inductively that our maps eα agree with jT . For e0 and e1 this is clear. The limit case of the induction also follows immediately. Supposing

9Choose a version, or remove the theorem. 13.3. AD IMPLIES ADR IF ALL SETS ARE SUSLIN 157 that it is true for a given α < δ, we need to see that j(tα(n)) is the least element of Eα (where (sα, tα) has length n + 1). Our induction hypothesis implies that ω j(tα(n)) is in Eα, which means that ek+1 was deﬁned. Fix an x ∈ ω ∩ M whose T -leftmost branch goes through (sα, tα). Then the leftmost branch for x in j(T ) (as computed in any wellfounded model of ZF) goes through (sα, j(tα)). This shows that j(tα(n)) is the least element of Eα, as desired, since a length- preserving embedding of T into j(T ) witnessing the contrary would produce a path though [T ] contradicting the fact that the leftmost branch for x in j(T ) (as computed in any wellfounded model of ZF) goes through (sα, j(tα)).

13.3 AD implies ADR if all sets are Suslin In this section we complete our proofs of Theorem 13.0.1. Our proofs will show that quasi-ADR holds. Then we will be done by Remark 6.1.3. Recall that if every subset of ωω is Suslin, then Uniformization holds, and that Uniformization implies DCR. Fix a bijection π : ω × ω → ω, and let π∗ be the induced bijection ω ω ω from (ω ) → ω . We fix a real game G, and let A be the π∗ image of its payoff set. By Theorem 13.0.2, we may fix homogeneous trees S and T (on ω × κ, for some cardinal κ) such that p[S] = A and p[T ] = ωω \ A. The key remaining step is given in Lemma 13.3.1 below. Note that in the statement of the lemma the trees S and T are homogeneous in V , but not necessarily in M.

Lemma 13.3.1. Assume that AD holds. Let κ be an ordinal. Let G be a real game with payoﬀ set A ⊆ (ωω)ω, and let S and T be homogeneous trees on ω ×κ ω such that p[S] = π∗[A] and p[T ] = ω \ π∗[A]. Let M be an inner model of ZF containing S and T , such that ωω ∩ M is countable. Then, in M, the real game −1 with payoﬀ set π∗ [p[S]] is quasi-determined.

−1 Proof. Let GM be the real game with payoﬀ set π∗ [p[S]], as computed in M. ∗ ∗ ∗ We deﬁne games GI and GII , both in M. In GI , players I and II alternate ω playing xi ∈ ω ∩ M. In each turn, I also plays αi/2 ∈ κ. Player I wins if and only if (π∗(hxi : i < ωi), hαi : i < ωi) is in [S].

I x0, α0 x2, α1 x4, α2 ... II x1 x3 ...

∗ The game GI .

∗ ω In the game GII , players I and II alternate playing xi ∈ ω ∩ M. In each turn, II also plays α(i−1)/2 ∈ κ. Player II wins if and only if

(π∗(hxi : i < ωi), hαi : i < ωi) 158 CHAPTER 13. ADR is in [T ].

I x0 x2x x4 ... II x1, α0 x3, α1 ...

∗ The game GII .

The first of these games is closed, and the second is open, so they are both quasi-determined. Moreover, for each game there is a quasi-strategy in M which is a winning quasi-strategy in V , since M and V compute the same ranking function when applying the proof of open determinacy to these games. Since ωω ∩ M is countable, such a winning quasi-strategy can be converted in V into a winning strategy. ∗ We want to see that, in M, either I has a winning quasi-strategy in GI or II ∗ has a winning quasi-strategy in GII , since the corresponding player could use his strategy in the game GM . It suffices then to derive a contradiction from the ∗ assumption that, in V , player II has a winning strategy ΣII in GI and player ∗ I has a winning strategy ΣI in GII . This contradiction follows from the fact ∗ that if player II has a winning strategy in GI , then player II has a winning ∗ strategy in GM and, similarly that if if player I has a winning strategy in GII , then player II has a winning strategy in GM . These facts are essentially the same, and essentially the same as Martin’s theorem that homogeneously Suslin sets are determined.10 We give a proof for the case where player I has a winning ∗ strategy in GII . <ω Let ΣI be such a strategy. Let {νs : s ∈ ω } be a set of measures witnessing <ω that T is homogeneously Suslin. Then for each s ∈ ω , νs is a countably |s| |s| complete ultrafilter on κ , and {t ∈ κ :(s, t) ∈ T } is in νs. Define a strategy Σ∗ for player I in GM as follows. Let Σ∗(hi) = ΣI (hi). Now fix a sequence ω <ω (n+1)/2 x¯ = hx0, . . . , xni ∈ (ω ∩ M) with n odd, and let sx¯ ∈ ω be the common initial segment of length (n + 1)/2 of the π∗-values of the elements of ω ω ω (ω ) extendingx ¯. Let Σ∗(¯x) be the unique y ∈ ω ∩ M such that the set

y (n+1)/2 Rx¯ = {t ∈ κ :ΣI (hx0, (x1, t(0)),..., (xn, t((n − 1)/2))i = y}

is in νsx¯ . Now suppose that hxi : i < ωi is a run of GM where I has played according to Σ∗ and lost. Since the measures νs witness the homogeneity of T , there is an f ∈ κω such that f ((n + 1)/2) is in Rxn+1 for each odd n ∈ ω. hxi:i

Proof of Theorem 13.0.1. Fix A ⊆ (ωω)ω and let S and T be homogeneous trees projecting to π∗[A] and its complement respectively. By Lemma 10.0.4,

10citation 13.3. AD IMPLIES ADR IF ALL SETS ARE SUSLIN 159 we may fix a partial order P on ΘHOD{S,T } and a set K ⊆ ΘHOD{S,T } , both in ∗ HOD{S,T }, such that, in any forcing extension of V by Col (ω, R), there is a V HOD{S,T }-generic filter G ⊆ P such that R is contained in L[S, T, P, K][G]. Let λ be a Suslin cardinal greater than ΘHOD{S,T } (which exists by Remark 6.0.9 ω if all subsets of ω are Suslin) and let µλ be a normal fine measure on Pℵ1 (λ), as given by Theorem 13.2.2. For each countable σ ⊆ λ, let Pσ and Kσ be the restrictions of P to σ ∩ κ. By Lemma 13.3.1, for any such σ, in any inner model of any forcing extension of L[S, T, Pσ,Kσ] by Pσ, the real game with payoff set −1 π∗ [p[S]] is determined. Since µλ is a normal fine measure, the ultraproduct Y L[S, T, Pσ,Kσ]/µλ

σ∈Pℵ1 (λ) is a wellfounded model L[S0,T 0,P,K], and by elementarity, in any inner model of any forcing extension of L[S0,T 0,P,K] by P , the real game with payoff set −1 0 0 0 0 π∗ [p[S ]] is determined. However, p[S ] = p[S] = π∗[A], and L[S ,T ,P,K](R) −1 is such an inner model, which shows that the real game with payoff set π∗ [p[S]] is determined. 13.3.2 Remark. Our proofs of Theorem 13.0.1 do not appear to easily induce local versions. Each proof uses Theorem 13.2.2, which in turn uses λ- Determinacy. Theorem 7.0.3 (along with Theorem 5.0.2) gives that AD implies the restriction of λ-Determinacy to games with Suslin, co-Suslin payoff sets. Our proofs of Theorem 13.0.1 seem to require then that every subsets of ωω is Suslin. 160 CHAPTER 13. ADR Chapter 14

Questions

1. Does ZF + AD imply DCR?

2. Does ZF+AD imply that the Wadge hierarchy is wellfounded? (yes if DCR is also assumed) 3. Does ZF + AD imply that all subsets of ωω are ∞-Borel?

ω 4. Does ZF + AD + DCR imply that all subsets of ω are ∞-Borel? 5. Does ZF + AD imply <Θ-Determinacy?

6. Does ZF + AD + DCR imply <Θ-Determinacy?

7. Does ZF + AD imply WFT?

8. Does ZF + AD imply that χB = κB? (yes if DCR is also assumed)

9. Does ZF + AD imply that κB = λB? (yes if DCR is also assumed)

10. Does Turing Determinacy + CCR imply AD?

11. Does Turing Determinacy imply CCR?

12. Does ZF + AD imply ADR for Suslin-co-Suslin sets?

13. Does ZF + ADR imply <Θ-Determinacy?

161 162 CHAPTER 14. QUESTIONS Bibliography

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(µ, κ)-club directedness, 62 ODX , 107

2-perfect tree, 138 ODt1,...,tn , 69 < γ-Determinacy, 75 Ord, 7

166 INDEX 167

∞-Borel set, 97 hereditarily countable set, 8 ∞-Borel subset of ωδ, 101 hereditarily finite set, 8 ∞-Borel∗ code, 101 homogeneous tree, 149 κ-Borel set, 97 homogeneously Suslin set, 150 λ-Determinacy, 75 λ-reasonable ordinal, 59 induced map, 154 lb≤(T ), 63 Kunen-Martin property, 69 ≤L, 19 ≤Wa, 19 leftmost branch, 63 ≤Gö, 8 length of a prewellordering, 48 ≤Tu, 8 <κ-Borel set, 97 lightface pointclass, 26 X , 26 Lipschitz class, 19 ≺α, β , 8 Lipschitz function, 19 meas(X), 150 Lipschitz game, 19 pos-Σ1(A), 32 Lipschitz rank, 21 ∼1 Lipschitz reducibility, 19 rank≤, 8 Ma, 15 Lusin-Sierpińskiorder, 76 p[T ], 63 measure, 149 σ ∗ x, 14 minimal tree, 64 σ ◦ x, 37 2 minimization of a tree, 64 ∼∆1(A), 50 1 ∼Πn(A), 33 nonselfdual class, 19 2 ∼Π1(A), 50 nonselfdual pointclass, 26 1 ∼Σn(A), 33 norm, 48 2 normal measure, 153 Σ1(A), 49 ∼rmc fx , 33 o(Γ), 26 ordered equivalence relation, 16 s_x, 22 Ordinal Determinacy, 75 x ∗ σ, 14 x ◦ σ, 37 pointclass, 26 preorder, 8 almost-disjoint coding, 10 prewellordering, 48 prewellordering property, 49 base of a cone, 16 projection of a tree, 63 basic open interval of a space in X , 26 projection of an ultrafilter, 149 boldface pointclass, 26 projective algebra, 54 projective in A, 33 computing a set of reals on a cone, 94 projectively closed pointclass, 33 determined set, 13 quasi-AD , 68 diffuse function, 88 X quasi-determined set, 68 downward cone, 16 quasi-strategy, 68 extension-preserving function, 75 rank function, 8 fine ultrafilter, 153 real, 7 168 INDEX recursion property, 29 Recursion Theorem, 32 recursive function, 26 recursive subset of HF, 8 reduction property, 44 regular norm, 48 s-m-n property, 29 selfdual class, 19 selfdual pointclass, 26 semi-recursive subset of HF, 8 separation property, 43 strategy, 13 strict supremum, 8 strong partition cardinal, 57 strongly Lipschitz function, 43 strongly maximal set of ordinals, 95 Suslin cardinal, 64 Suslin set, 63 tower of ultrafilters, 149 Turing cone, 16 Turing Determinacy, 17 Turing equivalence, 8 Turing measure, 16 Turing reducibility, 8

Uniform Coding Lemma, 39 uniformization, 113 uniformizing, 68 upward cone, 16

Vopˇenka algebra, 107

Wadge class, 19 Wadge rank, 21 Wadge reducibility, 19 weakly ∞-Borel set, 102 weakly κ-Borel set, 102 weakly homogeneous tree, 150 weakly homogeneously Suslin set, 150 wellfounded preorder, 8 wellfounded tower of ultraﬁlters, 149 winning quasi strategy, 68 winning strategy, 13 2 witness for a Σ∼1 statement, 49