Large Cardinals from Determinacy - Logic at Harvard

Large Cardinals from DeterminacyPeter Koellner and W. Hugh WoodinContents1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 21.1 Determinacy and Large Cardinals . . . . . . . . . . . 21.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . 162 Basic Results . . . . . . . . . . . . . . . . . . . . . . . 182.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 182.2 Boundedness and Basic Coding . . . . . . . . . . . . 222.3 Measurability . . . . . . . . . . . . . . . . . . . . . . 262.4 The Least Stable . . . . . . . . . . . . . . . . . . . . 322.5 Measurability of the Least Stable . . . . . . . . . . . 403 Coding . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.1 Coding Lemma . . . . . . . . . . . . . . . . . . . . . 443.2 Uniform Coding Lemma . . . . . . . . . . . . . . . . 493.3 Applications . . . . . . . . . . . . . . . . . . . . . . . 554 A Woodin Cardinal in HOD L(R) . . . . . . . . . . . . 584.1 Reflection . . . . . . . . . . . . . . . . . . . . . . . . 594.2 Strong Normality . . . . . . . . . . . . . . . . . . . . 724.3 A Woodin Cardinal . . . . . . . . . . . . . . . . . . . 915 Woodin Cardinals in General Settings . . . . . . . . 955.1 First Abstraction . . . . . . . . . . . . . . . . . . . . 975.2 Strategic Determinacy . . . . . . . . . . . . . . . . . 1001

1. Introduction 25.3 Generation Theorem . . . . . . . . . . . . . . . . . . 1095.4 Special Cases . . . . . . . . . . . . . . . . . . . . . . 1376 Definable Determinacy . . . . . . . . . . . . . . . . . 1456.1 Lightface Definable Determinacy . . . . . . . . . . . . 1466.2 Boldface Definable Determinacy . . . . . . . . . . . . 1697 Second-Order Arithmetic . . . . . . . . . . . . . . . 1777.1 First Localization . . . . . . . . . . . . . . . . . . . . 1797.2 Second Localization . . . . . . . . . . . . . . . . . . . 1868 Further Results . . . . . . . . . . . . . . . . . . . . . 1878.1 Large Cardinals and Determinacy . . . . . . . . . . . 1888.2 HOD-Analysis . . . . . . . . . . . . . . . . . . . . . . 192Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . 1991. IntroductionIn this chapter we give an account of Woodin’s technique for deriving largecardinal strength from determinacy hypotheses. These results appear herefor the first time and for this reason we have gone into somewhat more detailthan is customary in a handbook. All unattributed results that follow areeither folklore or due to Woodin.1.1. Determinacy and Large CardinalsIn the era of set theory following the discovery of independence a majorconcern has been the discovery of new axioms that settle the statements leftundecided by the standard axioms (ZFC). One interesting feature that hasemerged is that there are often deep connections between axioms that springfrom entirely different sources. In this chapter we will be concerned withone instance of this phenomenon, namely, the connection between axioms ofdefinable determinacy and large cardinal axioms.In this introduction we will give a brief overview of axioms of definabledeterminacy and large cardinal axioms (in sections A and B), discuss their

1. Introduction 3interconnections (in sections C and D), and give an overview of the chapter(in section E). At some points we will draw on notation and basic notionsthat are explained in fuller detail in sections 1.2 and 2.1.A. DeterminacyFor a set of reals A ⊆ ω ω consider the game where two players take turnsplaying natural numbers:I x(0) x(2) x(4) . . .II x(1) x(3) . . .At the end of a round of this game the two players will have produced areal x, obtained through “interleaving” their plays. We say that Player Iwins the round if x ∈ A; otherwise Player II wins the round. The set A issaid to be determined if one of the players has a “winning strategy” in theassociated game, that is, a strategy which ensures that the player wins around regardless of how the other player plays. The Axiom of Determinacy(AD) is the statement that every set of reals is determined.It is straightforward to see that very simple sets are determined. Forexample, if A is the set of all reals then clearly I has a winning strategy; ifA is empty then clearly II has a winning strategy; and if A is countable thenII has a winning strategy (by “diagonalizing”). A more substantive resultis that if A is closed then one player must have a winning strategy. Thismight lead one to expect that all sets of reals are determined. However,it is straightforward to use the Axiom of Choice (AC) to construct a nondeterminedset (by listing all winning strategies and “diagonalizing” acrossthem). For this reason AD was never really considered as a serious candidatefor a new axiom. However, there is an interesting class of related axioms thatare consistent with AC, namely, the axioms of definable determinacy. Theseaxioms extend the above pattern by asserting that all sets of reals at a givenlevel of complexity are determined, notable examples being, ∆ ∼11 -determinacy(all Borel sets of reals are determined), PD (all projective sets of reals aredetermined) and AD L(R) (all sets of reals in L(R) are determined).One issue is whether these are really new axioms or whether they followfrom ZFC. In the early development of the subject the result on the determinacyof closed sets was extended to higher levels of definability. Thesedevelopments culminated in Martin’s proof of ∆ ∼11 -determinacy in ZFC. Itturns out that this result is close to optimal—as one climbs the hierarchy

1. Introduction 4of definability, shortly after ∆ ∼11 one arrives at axioms that fall outside theprovenance of ZFC. For example, this is true of PD and AD L(R) . Thus,we have here a hierarchy of axioms (including PD and AD L(R) ) which aregenuine candidates for new axioms.There are actually two hierarchies of axioms of definable determinacy,one involving lightface notions of definability (by which we mean notions(such as ∆ 1 2) that do not involve real numbers as parameters) and the otherinvolving boldface notions of definability (by which we mean notions (suchas ∆ ∼12 ) that do involve real numbers as parameters). (See Jackson’s chapterin this Handbook for details concerning the various grades of definabilityand the relevant notation.) Each hierarchy is, of course, ordered in termsof increasing complexity. Moreover, each hierarchy has a natural limit: thenatural limit of the lightface hierarchy is OD-determinacy (all OD sets of realsare determined) and the natural limit of the boldface hierarchy is OD(R)-determinacy (all OD(R) sets of reals are determined). The reason these arenatural limits is that the notions of lightface and boldface ordinal definabilityare candidates for the richest lightface and boldface notions of definability.To see this (for the lightface case) notice first that any notion of definabilitywhich does not render all of the ordinals definable can be transcended (ascan be seen by considering the least ordinal which is not definable accordingto the notion) and second that the notion of ordinal definability cannot be sotranscended (since by reflection OD is ordinal definable). It is for this reasonthat Gödel proposed the notion of ordinal definability as a candidate for an“absolute” notion of definability. Our limiting cases may thus be regardedas two forms of absolute definable determinacy.So we have two hierarchies of increasingly strong candidates for new axiomsand each has a natural limit. There are two fundamental questionsconcerning such new axioms. First, are they consistent? Second, are theytrue? In the most straightforward sense these questions are asked in an absolutesense and not relative to a particular theory such as ZFC. But sincewe are dealing with new axioms, the traditional means of answering suchquestions—namely, by establishing their consistency or provability relativeto the standard axioms (ZFC)—is not available. Nevertheless, one can hopeto establish results—such as relative consistency and logical connections withrespect to other plausible axioms—that collectively shed light on the original,absolute question. Indeed, there are a number of results that one canbring to bear in favour of PD and AD L(R) . For example, these axioms lift thestructure theory that can be established in ZFC to their respective domains,

1. Introduction 5namely, second-order arithmetic and L(R). Moreover, they do so in a fashionwhich settles a remarkable number of statements that are independentof ZFC. In fact, there is no “natural” statement concerning their respectivedomains that is known to be independent of these axioms. (For more on thestructure theory provided by determinacy and the traditional considerationsin their favour see [10] and for more recent work see Jackson’s chapter inthis Handbook.) The results of this chapter figure in the case for PD andAD L(R) . However, our concern will be with the question of relative consistency;more precisely, we wish to calibrate the consistency strength of axiomsof definable determinacy—in particular, the ultimate axioms of lightface andboldface determinacy—in terms of the large cardinal hierarchy.There are some reductions that we can state at the outset. In terms ofconsistency strength the two hierarchies collapse at a certain stage: Kechrisand Solovay showed that ZF+DC implies that in the context of L[x] for x ∈ω ω , OD-determinacy and ∆ 1 2-determinacy are equivalent (see Theorem 6.6).And it is a folklore result that ZFC+OD(R)-determinacy and ZFC+AD L(R)are equiconsistent. Thus, in terms of consistency strength, the lightfacehierarchy collapses at ∆ 1 2-determinacy and the boldface hierarchy collapsesat AD L(R) . So if one wishes to gauge the consistency strength of lightfaceand boldface determinacy it suffices to concentrate on ∆ 1 2-determinacy andAD L(R) .Now, it is straightforward to see that if ∆ 1 2-determinacy holds then itholds in L[x] for some real x and likewise if AD L(R) (or AD) holds then itholds in L(R). Thus, the natural place to study the consistency strength oflightface definable determinacy is L[x] for some real x and the natural placeto study the consistency strength of boldface definable determinacy (or fulldeterminacy) is L(R). For this reason these two models will be central inwhat follows.To summarize: We shall be investigating the consistency strength of lightfaceand boldface determinacy. This reduces to ∆ 1 2 -determinacy and ADL(R) .The settings L[x] and L(R) will play a central role. Consistency strengthwill be measured in terms of the large cardinal hierarchy. Before turning toa discussion of the large cardinal hierarchy let us first briefly discuss strongerforms of determinacy.Our concern in this chapter is with axioms of determinacy of the aboveform, where the games have length ω and the moves are natural numbers.However, it is worthwhile to note that there are two directions in which onecan generalize these axioms.

1. Introduction 6First, one can consider games of length greater than ω (where the movesare still natural numbers). A simple argument shows that one cannot havethe determinacy of all games of length ω 1 but there is a great deal of roombelow this upper bound and much work has been done in this area. For moreon this subject see [11].Second, one can consider games where the moves are more complex thannatural numbers (and where the length of the game is still ω). One alternativeis to consider games where the moves are real numbers. The axiom AD Rstates that all such games are determined. One might try to continue inthis direction and consider the axiom AD P(R) asserting the determinacy ofall games where the moves are sets of real numbers. It is straightforwardto see that this axiom is inconsistent. In fact, even the definable versionasserting that all OD subsets of P(R) ω is inconsistent. Another alternativeis to consider games where the moves are ordinal numbers. Again, a simpleargument shows that one cannot have the determinacy of all subsets of ω 1 ω .However, a result of Harrington and Kechris shows that in this case if oneadds a definability constraint then one can have determinacy at this level.In fact, OD-determinacy implies that every OD set A ⊆ ω 1 ω is determined.It is natural then to extend this to large ordinals. The ultimate axiom inthis direction would simply assert that every OD set A ⊆ On ω is determined.Perhaps surprisingly, at this stage a certain tension arises since recent work ininner model theory provides evidence that this axiom is in fact inconsistent.See [13] for more on this subject.B. Large CardinalsOur aim is to calibrate the consistency strength of lightface and boldfacedeterminacy in terms of the large cardinal hierarchy. The importance ofthe large cardinal hierarchy in this connection is that it provides a canonicalmeans of climbing the hierarchy of consistency strength. To show, for a givenhypothesis ϕ and a given large cardinal axiom L, that the theories ZFC + ϕand ZFC+L are equiconsistent one typically uses the dual methods of innermodel theory and outer model theory (also known as forcing). Very roughly,given a model of ZFC + L one forces to obtain a model of ZFC + ϕ andgiven a model of ZFC + ϕ one uses the method of inner model theory toconstruct a model of ZFC + L. The large cardinal hierarchy is (for the mostpart) naturally well-ordered and it is a remarkable phenomenon that givenany two “natural” theories extending ZFC one can compare them in terms of

1. Introduction 7consistency strength (equivalently, interpretability) by lining them up withthe large cardinal hierarchy.In a very rough sense large cardinal axioms assert that there are “large”levels of the universe. A template for formulating a broad class of largecardinal axioms is in terms of elementary embeddings. The basic format ofthe template is as follows: There is a transitive class M and a non-trivialelementaryj : V → M.To say that the embedding is non-trivial is simply to say that it is not theidentity, in which case one can show that there is a least ordinal moved. Thisordinal is denoted crit(j) and called the critical point of j. A cardinal is saidto be a measurable cardinal if and only if it is the critical point of such anembedding.It is easy to see that for any such elementary embedding there is necessarilya certain degree of agreement between V and M. In particular, it followsthat V κ+1 ⊆ M, where κ = crit(j). This degree of agreement in conjunctionwith the elementarity of j can be used to show that κ has strong reflectionproperties, in particular, κ is strongly inaccessible, Mahlo, weakly compact,etc.One way to strengthen a large cardinal axiom of the above form is todemand greater agreement between M and V . For example, if one demandsthat V κ+2 ⊆ M then the fact that κ is measurable is recognized within M andhence it follows that M satisfies that there is a measurable cardinal belowj(κ), namely, κ. Thus, by the elementarity of the embedding, V satisfiesthat there is a measurable cardinal below κ. The same argument shows thatthere are arbitrarily large measurable cardinals below κ.This leads to a natural progression of increasingly strong large cardinalaxioms. It will be useful to discuss some of the major axioms in this hierarchy:If κ is a cardinal and η > κ is an ordinal then κ is η-strong if there is atransitive class M and a non-trivial elementary embedding j : V → M suchthat crit(j) = κ, j(κ) > η and V η ⊆ M. A cardinal κ is strong iff it isη-strong for all η. As we saw above if κ is (κ+2)-strong then κ is measurableand there are arbitrarily large measurable cardinals below κ. Next, one candemand that the embedding preserve certain classes: If A is a class, κ isa cardinal, and η > κ is an ordinal then κ is η-A-strong if there exists aj : V → M which witnesses that κ is η-strong and which has the additionalfeature that j(A ∩ V κ ) ∩ V η = A ∩ V η . The following large cardinal notion

1. Introduction 8will play a central role in this chapter.1.1 Definition. A cardinal κ is a Woodin cardinal if κ is strongly inaccessibleand for all A ⊆ V κ there is a cardinal κ A < κ such thatfor each η such that κ A < η < κ.κ A is η-A-strong,It should be noted that in contrast to measurable and strong cardinals,Woodin cardinals are not characterized as the critical point of an embeddingor collection of embeddings. In fact, a Woodin cardinal need not be measurable.However, if κ is a Woodin cardinal, then V κ is a model of ZFC andfrom the point of view of V κ there is a proper class of strong cardinals.Going further, a cardinal κ is superstrong if there is a transitive class Mand a non-trivial elementary embedding j : V → M such that crit(j) = κand V j(κ) ⊆ M. If κ is superstrong then κ is a Woodin cardinal and thereare arbitrarily large Woodin cardinals below κ.One can continue in this vein, demanding greater agreement betweenM and V . The ultimate axiom in this direction would, of course, demandthat M = V . This axiom was proposed by Reinhardt. But shortly afterits introduction Kunen showed that it is inconsistent with ZFC. In fact,Kunen showed that assuming ZFC, there can be no non-trivial elementaryembedding j : V λ+2 → V λ+2 . (An interesting open question is whether theseaxioms are inconsistent with ZF or whether there is a hierarchy of “choiceless”large cardinal axioms that climb the hierarchy of consistency strength farbeyond what can be reached with ZFC.)There is a lot of room below the above upper bound. For example, avery strong axiom is the statement that there is a non-trivial elementaryembedding j : V λ+1 → V λ+1 . The strongest large cardinal axiom in the currentliterature is the axiom asserting that there is a non-trivial elementaryembedding j : L(V λ+1 ) → L(V λ+1 ) such that crit(j) < λ. Surprisingly, thisaxiom yields a structure theory of L(V λ+1 ) which is closely analogous to thestructure theory of L(R) under the axiom AD L(R) . This parallel betweenaxioms of determinacy and large cardinal axioms suggests seeking strongerlarge cardinal axioms by following the guide of the strong axioms of determinacydiscussed at the close of the previous section. In fact, there is evidencethat the parallel extends. For example, there is a new large cardinal axiomthat is the analogue of AD R . See [13] for more on these recent developments.

1. Introduction 9C. Determinacy from Large CardinalsLet us return to the questions of the truth and the consistency of axiomsof definable determinacy, granting that of large cardinal axioms. In the late1960s Solovay conjectured that AD L(R) is provable from large cardinal axioms(and hence that ZF+AD is consistent relative to large cardinal axioms). Thisconjecture was realized in stages.In 1970 Martin showed that if there is a measurable cardinal then all1Σ ∼1 sets of reals are determined. Later, in 1978, he showed that under themuch stronger assumption of a non-trivial iterable elementary embedding1j : V λ → V λ all Σ ∼ 2 sets of reals are determined. It appeared that therewould be a long march up the hierarchy of axioms of definable determinacy.However, in 1984 Woodin showed that if there is a non-trivial elementaryembedding j : L(V λ+1 ) → L(V λ+1 ) with crit(j) < λ, then AD L(R) holds.The next major advances concerned reducing the large cardinal hypothesisused to obtain AD L(R) . The first step in this direction was made shortlyafter, in 1985, when Martin and Steel proved the following remarkable result,using a completely different technique:Theorem 1.2 (Martin and Steel). Assume ZFC. Suppose that there are nWoodin cardinals with a measurable cardinal above them all. Then Σ ∼1n+1 -determinacy holds.It follows that if there is a Woodin cardinal with a measurable cardinalabove, then ∆ 1 2-determinacy holds and if there are infinitely many Woodincardinals then PD holds. Finally, the combination of Martin and Steel’swork and Woodin’s work on the stationary tower (see [9]) led to a significantreduction in the hypothesis required to obtain AD L(R) .Theorem 1.3. Assume ZFC. Suppose there are infinitely many Woodincardinals with a measurable cardinal above them all. Then AD L(R) .A more recent development is that, in addition to being implied by largecardinal axioms, AD L(R) is implied by a broad array of other strong axioms,which have nothing to do with one another—in fact, there is reason to believethat AD L(R) is implied by all sufficiently strong “natural” theories. For furtherdiscussion of this subject and other more recent results that contributeto the case for certain axioms of definable determinacy see [8], [12], and [14].Each of the above results concerns the truth of axioms of definable determinacy,granting large cardinal axioms. A closely related question concerns

1. Introduction 10the consistency of axioms of definable determinacy, granting that of largecardinal axioms. For this one can get by with slightly weaker large cardinalassumptions.Theorem 1.4. Assume ZFC. Suppose δ is a Woodin cardinal. SupposeG ⊆ Col(ω, δ) is V -generic. Then V [G] |= ∆ 1 2 -determinacy.Theorem 1.5. Assume ZFC. Suppose that λ is a limit of Woodin cardinals.Suppose G ⊆ Col(ω,

1. Introduction 11axioms that have been investigated. But it has a complementary defect inthat one cannot carry out a full analysis of this structure within ZFC.A major program in set theory—the inner model program—aims to combinethe advantages of the two approaches by building inner models thatresemble L in having a highly ordered inner structure but which resembleHOD in that they can accommodate strong large cardinal axioms.“L-like” inner models at the level of Woodin cardinals were developed instages beginning with work of Martin and Steel, and continuing with workof Mitchell and Steel. The Mitchell-Steel inner models are true analogs of L.Martin and Steel used their models to show that the large cardinal hypothesesin their proofs of determinacy were essentially optimal. For example, theyshowed that if there is a Woodin cardinal then there is a canonical innermodel M that contains a Woodin cardinal and has a ∆ 1 3 well-ordering of the1reals. It follows that one cannot prove Σ ∼2 -determinacy from the assumptionof a Woodin cardinal alone.However, this still left open a number of questions. First, does the consistencyof ZFC + “There is a Woodin cardinal” follow from that of ZFC+∆ 1 2 -determinacy? Second, can one build an inner model of a Woodin cardinaldirectly from ZFC + ∆ 1 2-determinacy? Third, what is the strength ofZFC + AD L(R) ? To approach these questions it would seem that one wouldneed fine-structural inner model theory. However, at the time when the centralresults of this chapter were proved, fine-structural inner model theoryhad not yet reached the level of Woodin cardinals. One option was to proceedwith HOD.In contrast to L the structure of HOD is closely tied to the universe inwhich it is constructed. In the general setting, where one works in ZF andconstructs HOD in V , the structure theory of HOD is almost as intractableas that of V . Surprisingly if one strengthens the background theory then thestructure theory of HOD becomes tractable. For example, Solovay showedthat under ZF + AD, HOD satisfies that ω1V is a measurable cardinal. Itturns out that both lightface and boldface definable determinacy are ableto illuminate the structure of HOD (when constructed in the natural innermodels of these axioms—L[x] and L(R)) to the point where one can recoverthe large cardinals that are necessary to establish their consistency.In the case of lightface definable determinacy the result is the following:Theorem 1.6. Assume ZF+DC+∆ 1 2-determinacy. Then for a Turing cone

1. Introduction 12of x,HOD L[x] |= ZFC + ω L[x]2 is a Woodin cardinal.Thus, the consistency strength of ZFC+OD-determinacy is precisely that ofZFC + “There is a Woodin cardinal”. For the case of boldface determinacylet us first state a preliminary result of which the above result is a localization.First we need a definition. LetΘ = sup{α | there is a surjection π : R → α}.Theorem 1.7. Assume ZF + AD. ThenHOD L(R) |= Θ L(R) is a Woodin cardinal.In fact, both of these results are special instances of a general theorem onthe generation of Woodin cardinals—the Generation Theorem. In additionto giving the above results, the Generation Theorem will also be used toestablish the optimal large cardinal lower bound for boldface determinacy:Theorem 1.8. Assume ZF+AD. Suppose Y is a set. There is a generalizedPrikry forcing P Y through the Y -degrees such that if G ⊆ P Y is V -genericand 〈[x i ] Y | i < ω〉 is the associated sequence, thenHOD V [G]Y,〈[x i ] Y |i

1. Introduction 13E. OverviewThe results on the strength of lightface and boldface determinacy were establishedin the late 1980s. However, the current presentation and many ofthe results that follow are quite recent. One of the key new ingredients is thefollowing abstract theorem on the generation of Woodin cardinals, which liesat the heart of this chapter:Theorem 1.9 (Generation Theorem). Assume ZF. Supposeis such that(1) M |= T 0 ,(2) Θ M is a regular cardinal,(3) T ⊆ Θ M ,M = L ΘM (R)[T, A, B](4) A = 〈A α | α < Θ M 〉 is such that A α is a prewellordering of the reals oflength greater than or equal to α,(5) B ⊆ ω ω is nonempty, and(6) M |= Strategic determinacy with respect to B.ThenHOD M T,A,B |= ZFC + There is a T-strong cardinal.Here T 0 is the theory ZF + AC ω (R) − Power Set + “P(ω) exists ′′ and thenotion of “strategic determinacy” is a technical notion that we will stateprecisely later.The Generation Theorem provides a template for generating models containingWoodin cardinals. One simply has to show that in a particular settingthe various conditions can be met, though this is often a non-trivial task. Thetheorem is also quite flexible in that it is a result of ZF that does not presupposeDC and has applications in both lightface and boldface settings. Itwill play a central role in the calibration of the strength of both lightface andboldface determinacy.We shall approach the proof of the Generation Theorem by proving aseries of increasingly complex approximations.

1. Introduction 14In Section 2 we take the initial step by proving Solovay’s theorem thatunder ZF + AD, ω1V is a measurable cardinal in HOD and we show thatthe associated measure is normal. The proof that we give is slightly morecomplicated than the standard proof but has the virtue of illustrating in asimple setting some of the key components that appear in the more complexvariations. We illustrate this at the end of the section by showing that theproof of Solovay’s theorem generalizes to show that under ZF + AD, the2ordinal (δ ∼1 ) L(R) is a measurable cardinal in HOD L(R) . Our main aim in thissection is to illustrate the manner in which “boundedness” and “coding”combine to yield normal ultrafilters. In subsequent sections stronger formsof boundedness (more precisely, “reflection”) and stronger forms of codingwill be used to establish stronger forms of normality.In Section 3 we prove the strong forms of coding that will be centralthroughout.In Section 4, as a precursor to the proof of the Generation Theorem, weprove the following theorem:Theorem 1.10. Assume ZF + DC + AD. ThenHOD L(R) |= ZFC + Θ L(R) is a Woodin cardinal.The assumption of DC is merely provisional—it will ultimately be eliminatedwhen we prove the Generation Theorem. Toward the proof of the abovetheorem, we begin in §4.1 by establishing the reflection phenomenon that willplay the role played by boundedness in §2. We will then use this reflectionphenomenon in L(R) to define for cofinally many λ < Θ, an ultrafilter µ λ on2δ∼ 1 that is intended to witness that δ 2∼1 is λ-strong. In §4.2 we shall introduceand motivate the notion of strong normality by showing that the strong2normality of µ λ ensures that ∼δ 1 is λ-strong. We will then show how reflectionand uniform coding combine to secure strong normality. In §4.3 we will provethe above theorem by relativizing the construction of §4.2 to subsets of Θ L(R) .In Section 5 we extract the essential components of the above constructionand prove two abstract theorems on Woodin cardinals in a general setting,one that involves DC and one that does not. The first theorem is proved in§5.1. The importance of this theorem is that it can be used to show that incertain strong determinacy settings HOD can contain many Woodin cardinals.The second theorem is the Generation Theorem, the proof of which willoccupy the remainder of the section. The aim of the Generation Theorem

1. Introduction 15is to show that the construction of Section 4 can be driven by lightface determinacyalone. The difficulty is that the construction of Section 4 involvesgames that are defined in terms of real parameters. To handle this we introducethe notion of “strategic determinacy”, a notion that resembles boldfacedeterminacy in that it involves real parameters but which can nonethelesshold in settings where one has AC. To motivate the notion of “strategicdeterminacy” we shall begin in §5.2 by examining one such setting, namely,L[S, x] where S is a class of ordinals and x is a real. Once we show that“strategic determinacy” can hold in this setting we shall return in §5.3 tothe general setting and prove the Generation Theorem. In the final subsection,we prove a number of special cases, many of which are new. Althoughsome of these applications involve lightface settings, they all either involveassuming full AD or explicitly involve “strategic determinacy”.In Section 6 we use two of the special cases of the Generation Theoremto calibrate the consistency strength of lightface and boldface definable determinacyin terms of the large cardinal hierarchy. In the case of the firstresult the main task is to show that ∆ 1 2-determinacy suffices to establish that“strategic determinacy” can hold. In the case of the second result the maintask is to show that the Generation Theorem can be iteratively applied togenerate ω-many Woodin cardinals.In Section 7 we show that the Generation Theorem can itself be localizedin two respects. In the first localization we show that ∆ 1 2-determinacy impliesthat for a Turing cone of x, ω L[x]1 is a Woodin cardinal in an inner model ofL[x]. In the second localization we show that the proof can in fact be carriedout in second-order arithmetic.In Section 8 we survey some further results. First, we discuss results concerningthe actual equivalence of axioms of definable determinacy and axiomsasserting the existence of inner models with Woodin cardinals. Second, werevisit the analysis of HOD L(R) and HOD L[x][g] , for certain generic extensionsL[x][g], in light of the advances that have been made in fine-structuralinner model theory. Remarkably, it turns out that not only are these modelswell-behaved in the context of definable determinacy—they are actuallyfine-structural inner models, but of a kind that falls outside of the traditionalhierarchy.We have tried to keep the account self-contained, presupposing only acquaintancewith the constructible universe, the basics of forcing, and thebasics of large cardinal theory. In particular, we have tried to minimize ap-

1. Introduction 16peal to fine-structure and descriptive set theory. Fine-structure enters onlyin Section 8 where we survey more recent developments, but even there oneshould be able to get a sense of the lay of the land without following the details.For the relevant background and historical development of the subjectsee [1], [2] and [10].Acknowledgements. We would like to thank Scott Cramer, Moti Gitik,Aki Kanamori, Richard Ketchersid, Paul Larson, and Nam Trang for sendingus helpful comments at various stages. We are especially grateful to AndrésCaicedo and an anonymous referee for sending us extensive comments onan earlier draft of this chapter. The first author would like to acknowledgethe support of the Clarke/Cooke fund and the second author would like toacknowledge the support of the National Science Foundation under grantsDMS-9970255 and DMS-0355334.1.2. NotationFor the most part our notational conventions are standard. Nevertheless,some comments are in order.(1) We use µα ϕ(α) to indicate the least ordinal α such that ϕ(α).(2) In writing OD X and HOD X we always mean that X itself (as opposedto its elements) is allowed as a parameter. The notation OD {X} issometimes used for this, for example, in contexts where one would liketo speak of both OD {X} and OD X . However, in this chapter we willhave no occasion to speak of the latter and so we have dropped the curlybrackets on the ground that they would only serve to clutter the text.We also use OD X as both a name (for the class of sets which are ordinaldefinable from X) and as an adjective (for example when we say thata particular class is OD X .) We use < ODX for a fixed canonical OD X -well-ordering of OD X sets. The notation OD(R) is used in analogywith L(R).(3) A strategy for Player I is a function σ : ⋃ i

1. Introduction 17are defined similarly. We typically reserve σ for strategies for PlayerI and τ for strategies for Player II. The play that results from havingII play y against σ is denoted σ∗y and likewise the play that resultsfrom having I play x against τ is denoted x∗τ. We write x∗y for thereal that results from having Player I play x and Player II play y andin this case we let (x∗y) I = x and (x∗y) II = y. For example, (σ∗y) I isthe real that Player I plays when following the strategy σ against II’splay of y. If σ is a strategy for Player I and τ is a strategy for PlayerII we write σ∗τ for the real produced by playing the strategies againstone another. Occasionally, when z = x∗y we write z even to indicate xand z odd to indicate y.(4) If X is a subset of the plane ω ω ×ω ω we use proj 1 (X) for the “projectionto the first coordinate” and proj 2 (X) for the “projection to the secondcoordinate”.(5) For n 0 , . . .,n k−1 ∈ ω, we use 〈n 0 , . . .,n k−1 〉 to denote the natural numberencoding (n 0 , . . .,n k−1 ) via a recursive bijection between ω k and ω(which we fix throughout) and we let (n) i be the associated projectionfunctions. For x ∈ ω ω and i ∈ ω we also use (x) i for the projection functionassociated to a recursive bijection between (ω ω ) ω and ω ω . See [10,Chapter 3] for further details on such recursive coding and decodingfunctions.There is a slight conflict in notation in that angle brackets are alsotraditionally used for sequences and n-tuples. We have lapsed into thisusage at points but the context resolves the ambiguity; for example,when we write 〈x α | α < λ〉 it is clear that we are referring to asequence.(6) In this chapter by the “reals” we mean ω ω , which, under the standardtopology, is homeomorphic to the irrationals as normally construed.However, we continue to use the symbol ‘R’ in contexts where it istraditional, for example, in L(R).(7) We use tc(x) for the transitive closure of x.(8) A base theory that will play a central role throughout isT 0 = ZF + AC ω (R) − Power Set + “P(ω) exists ′′ .

2. Basic Results 182. Basic ResultsThe central result of this section is Solovay’s theorem to the effect that underZF+AD, ω 1 is a measurable cardinal. The proof that we will give is slightlymore involved than the standard proof but it has the advantage of illustratingsome of the key components in the more general theorems to be provedin later sections. One thing we would like to illustrate is the manner inwhich “boundedness” and “coding” combine to yield normal ultrafilters. Insubsequent sections stronger forms of boundedness (more precisely, “reflection”)and stronger forms of coding will be used to establish stronger formsof normality. This will culminate in the production of Woodin cardinals.In §2.1 we review some basic consequences of ZF + AD. In §2.2 we prove-boundedness and use it to prove the Basic Coding Lemma, a simple caseof the more general coding lemmas to be proved in Section 3. In §2.3 we1use Σ ∼1 -boundedness to show that the club filter on ω 1 witnesses that ω 1 is1a measurable cardinal and we use Σ ∼ 1-boundedness and the Basic Coding2Lemma to show that this ultrafilter is normal. In §2.4 we introduce δ ∼1 andestablish its basic properties. Finally, in §2.5 we draw on the Coding Lemmaof Section 3 to show that the proof of Solovay’s theorem generalizes to showthat assuming ZF+DC+AD then in the restricted setting of L(R) the ordinal2(δ ∼1 ) L(R) is a measurable cardinal. Later, in Section 4, we will dispense withDC and reprove this theorem in ZF + AD.Σ ∼112.1. PreliminariesIn order to keep this account self-contained, in this subsection we shall collecttogether some of the basic features of the theory of determinacy. Theseconcern (1) the connection between determinacy and choice, (2) the implicationsof determinacy for regularity properties, and (3) the implications ofdeterminacy for the Turing degrees. See [2], [10], and Jackson’s chapter inthis Handbook for further details and references.Let us begin with the axiom of choice. A straightforward diagonalizationargument shows that AD contradicts the full axiom of choice, AC. However,certain fragments of AC are consistent with AD and, in fact, certainfragments of AC follow from AD.

2. Basic Results 192.1 Definition. The Countable Axiom of Choice, AC ω , is the statement thatevery countable set consisting of non-empty sets has a choice function. TheCountable Axiom of Choice for Sets of Reals, AC ω (R), is the statement thatevery countable set consisting of non-empty sets of reals has a choice function.Theorem 2.2. Assume ZF + AD. Then AC ω (R).Proof. Let {X n | n < ω} be a countable collection of non-empty sets of reals.Consider the gameI x(0) x(1) x(2) . . .II y(0) y(1) . . .where I wins if and only if y ∉ X x(0) . (Notice that we are leaving the definitionof the payoff set of reals A implicit. In this case the payoff set is {x ∈ ω ω |x odd ∉ X x(0) }. In the sequel we shall leave such routine transformations tothe reader.) Thus, Player I is to be thought of as playing an element X nof the countable collection and Player II must play a real which is not inthis element. Of course, Player I cannot win. So there must be a winningstrategy τ for Player II. The functionf : ω → ω ωis a choice function for {X n | n < ω}.n ↦→ (〈n, 0, 0, . . .〉∗τ) IICorollary 2.3. Assume ZF + AD. Then ω 1 is regular.2.4 Definition. The Principle of Dependent Choices, DC, is the statementthat for every non-empty set X and for every relation R ⊆ X × X such thatfor all x ∈ X there is a y ∈ X such that (x, y) ∈ R, there is a functionf : ω → X such that for all n < ω, (f(n), f(n + 1)) ∈ R. The Principle ofDependent Choices for Sets of Reals, DC R , is simply the restricted version ofDC where X is R.It is straightforward to show that DC implies AC ω and Jensen showed thatthis implication cannot be reversed. Solovay showed that Con(ZF + AD R )implies Con(AD + ¬DC) and this was improved by Woodin.Theorem 2.5. Assume ZF + AD + V =L(R). Then in a forcing extensionthere is an inner model of AD + ¬AC ω .

2. Basic Results 20Theorem 2.6 (Kechris). Assume ZF + AD. Then L(R) |= DC.1 Open Question. Does AD imply DC R ?Thus, of the above fragments of AC, AC ω (R) is known to be withinthe reach of AD, DC R could be within the reach of AD, and the strongerprinciples AC ω and DC are known to be consistent with but independentof AD (assuming consistency of course). For this reason, to minimize ourassumptions, in what follows we shall work with AC ω (R) as far as this ispossible. There are two places where we invoke DC, namely, in Kunen’stheorem (Theorem 3.11) and in Lemma 4.8 concerning the well-foundednessof certain ultrapowers. However, in our applications, DC will reduce to DC Rand so if the above open question has a positive answer then these appealsto DC can also be avoided.We now turn to regularity properties. The axiom of determinacy hasprofound consequences for the structure theory of sets of real numbers. See[10] and Jackson’s chapter in this Handbook for more on this. Here wemention only one central consequence that we shall need below.Theorem 2.7 (Mycielski-Swierczkowski; Mazur, Banach; Davis). AssumeZF+AD. Then all sets are Lebesgue measurable, have the property of Baire,and have the perfect set property.Proof. See [2, Section 27].Another important consequence we shall need is the following:Theorem 2.8. Assume ZF + AD. Then every ultrafilter is ω 1 -complete.Proof. Suppose U ⊆ P(X) is an ultrafilter. If U is not ω 1 -complete thenthere exists {X i | i < ω} such that(1) for all i < ω, X i ∈ U and(2) ⋂ i

2. Basic Results 21Finally, we turn to the implications of determinacy for the Turing degrees.For x, y ∈ ω ω , we say that x is Turing reducible to y, x T y, if x is recursivein y and we say that x is Turing equivalent to y, x ≡ T y, if x T y andy T x. The Turing degrees are the corresponding equivalence classes [x] T ={y ∈ ω ω | y ≡ T x}. LettingD T = { [x] T | x ∈ ω ω}the relation T lifts to a partial ordering on D T . A cone of Turing degreesis a set of the form {[y]T | y T x 0}for some x 0 ∈ ω ω . A Turing cone of reals is a set of the form{y ∈ ω ω | y T x 0}for some x 0 ∈ ω ω . In each case x 0 is said to be the base of the cone. In latersections we will discuss different degree notions. However, when we speak ofa “cone of x” without qualification we always mean a “Turing cone of x”.The cone filter on D T is the filter consisting of sets of Turing degrees thatcontain a cone of Turing degrees.Theorem 2.9 (Cone Theorem) (Martin). Assume ZF + AD. The conefilter on D T is an ultrafilter.Proof. For A ⊆ D T consider the gameI x(0) x(1) x(2) . . .II y(0) y(1) . . .where I wins iff [x∗y] T ∈ A. If I has a winning strategy σ 0 then σ 0 witnessesthat A is in the cone filter since for y T σ 0 , [y] T = [σ 0 ∗y] T ∈ A. If II has awinning strategy τ 0 then τ 0 witnesses that D T A is in the cone filter sincefor x T τ 0 , [x] T = [x∗τ 0 ] T ∈ D T A.It follows that under ZF + AD each statement ϕ(x) either holds for aTuring cone or reals x or fails for a Turing cone of reals x.The proof of the Cone Theorem easily relativizes to fragments of definabledeterminacy. For example, assuming Σ 1 2-determinacy every Σ 1 2 set which isinvariant under Turing equivalence either contains or is disjoint from a Turingcone of reals.

2. Basic Results 22It is of interest to note that when Martin proved the Cone Theorem hethought that he would be able to refute AD by finding a property that “toggles”.He started with Borel sets and, when no counterexample arose, movedon to more complicated sets. We now know (assuming there are infinitelymany Woodin cardinals with a measurable above) that no counterexamplesare to be found in L(R). Moreover, the statement that there are no counterexamplesin L(R) (i.e. the statement that Turing determinacy holds inL(R)) actually implies AD L(R) (over ZF + DC). Thus, the basic intuitionthat the Cone Theorem is strong is correct—it is just not as strong as 0=1.2.2. Boundedness and Basic CodingWe begin with some definitions. For x ∈ ω ω , let E x be the binary relationon ω such that mE x n iff x(〈m, n〉) = 0, where recall that 〈·, ·〉 : ω × ω → ωis a recursive bijection. The real x is said to code the relation E x . LetWO = {x ∈ ω ω | E x is a well-ordering}. For x ∈ WO, let α x be the ordertypeof E x and, for α < β < ω 1 let WO α = {x ∈ WO | α x = α}, WO

2. Basic Results 23weaker than T 0 yield an equivalent definition. For example, one can use thefinite theory ZF N for some sufficiently large N.As an illustration of the utility of this model-theoretic characterization ofΣ ∼11 we shall use it to show that for each x ∈ WO, WO

2. Basic Results 24a contradiction (making free use of AC) as follows: Let z ∈ ω ω be such thatboth X and WO are Σ 1 1(z). Let α be such that V α |= T 0 and choose Y ≺ V αsuch that Y is countable and z ∈ Y . Let N be the transitive collapse of Y .By correctness, X ∩ N = X N . Choose a uniform ultrafilter U ⊆ P(ω 1 ) Nsuch that ifj : N → Ult(N, U)is the associated embedding then crit(j) = ω1 N and j(ωN 1 ) is not well-founded.(To obtain such an ultrafilter build a generic for (P(ω 1 )/countable) N . SeeLemma 22.20 of [1].) Since Ult(N, U) is an ω-model of T 0 it correctly computesWO. It follows that (WO) Ult(N,U) ⊆ WO, which in turn contradicts thefact that ω Ult(N,U)1 is not well-founded.)Lemma 2.11 (Basic Coding) (Solovay). Assume ZF + AD. Suppose Z ⊆WO × ω ω . Then there exists a Σ ∼12 set Z ∗ such that(1) Z ∗ ⊆ Z and(2) for all α < ω 1 , Z ∗ ∩ (WO α × ω ω ) ≠ ∅ iff Z ∩ (WO α × ω ω ) ≠ ∅.Moreover, there is such a Z ∗ which is of the form X ∩ (WO × ω ω ) whereX ⊆ ω ω × ω ω is Σ ∼11 .Proof. Here is the picture:ω ωZThe space WO × ω ω is sliced into sections WO α × ω ω for α < ω 1 . Z isrepresented by the unshaded ellipse and Z ∗ is represented by the shadedregion. Basic Coding says that whenever Z meets one of the sections so doesZ ∗ . In such a situation we say that Z ∗ is a selector for Z.To see that Z ∗ exists, consider the gameWOI x(0) x(1) x(2) . . .II y(0) y(1) . . .where II wins iff whenever x ∈ WO then y codes a countable set Y such that

2. Basic Results 25(1) Y ⊆ Z and(2) for all α α x , Y ∩ (WO α × ω ω ) ≠ ∅ iff Z ∩ (WO α × ω ω ) ≠ ∅.The idea is that Player I challenges by playing a countable ordinal α x andPlayer II meets this challenge provided he can play (a code for a) a selectorY for Z ∩ (WO αx × ω ω ).Claim. There can be no winning strategy for Player I in this game.Proof. Suppose σ is a winning strategy for I. As the play unfolds, Player Ican attempt to increase α x as Player II’s play is revealed. However, PlayerII can anticipate all such attempts as follows: The setX = {(σ∗y) I | y ∈ ω ω }is Σ 1 1 (σ) and, since σ is winning for I, X ⊆ WO. So, by Σ 1∼1 -boundedness,there is an β < ω 1 such that X ⊆ WO

2. Basic Results 262.3. MeasurabilityTheorem 2.12 (Solovay). Assume ZF + AD. Then the club filter is anω 1 -complete ultrafilter on ω 1 .Proof. The ultrafilter on ω 1 will be extracted from a game. As motivation,for the moment work in ZFC. For S ⊆ ω 1 , consider the gameI α 0 α 1 α 2 . . .II β 0 β 1 . . .where we demand that α 0 < β 0 < α 1 < · · · < ω 1 and where the first playerthat fails to meet this demand loses and if both players meet the demandthen I wins provided sup i

2. Basic Results 27with the following rules: Rule 1: For all i < ω, (x) i , (y) i ∈ WO. If Rule 1 isviolated then, letting i be least such that either (x) i ∉ WO or (y) i ∉ WO, Iwins if (x) i ∈ WO; otherwise II wins. Now suppose Rule 1 is satisfied. Rule2: α (x)0 < α (y)0 < α (x)1 < α (y)1 · · ·. The first failure defines who wins asabove. If both rules are satisfied then I wins if and only if sup i

2. Basic Results 28Proof. For S ⊆ ω 1 let G(S) be the game from the previous proof and letµ be as defined there. We know that µ is the club filter. To motivate theproof of normality we give a proof of ω 1 -completeness that will generalizeto produce normal ultrafilters on ordinals larger than ω 1 . This is merely forillustration—the proof uses DC but this will be eliminated in Claim 2.Claim 1. µ is ω 1 -complete.Proof. Suppose S j ∈ µ for j < ω. We have to show that S = ⋂ j

2. Basic Results 29(1.2) a sequence of collections of strategies 〈X i | i < ω〉 where X i containswinning strategies for I in games G(S α ) for α ∈ [η i−1 , η i ), whereη −1 = 0, and(1.3) a sequence 〈y i | i < ω〉 of plays such that y i is legal for II againstany σ ∈ X i and sup j

2. Basic Results 30For the first step letη 0 = some ordinal η such that η < ω 1X 0 = proj 2(X ∩ (WO[0,η0 ) × ω ω ) )Y 0 = { ((σ∗y) I ) 0 | σ ∈ X 0 ∧ y ∈ ω ω}z 0 = some real z such that Y 0 ⊆ WO

2. Basic Results 31This completes the proof of the theorem.It should be noted that using DC normality can be proved without usingBasic Coding since in place of the sequence 〈X i | i ∈ ω〉 one can use DCto construct a sequence 〈σ α | α < η〉 of strategies. This, however, relies onthe fact that η is countable. Our reason for giving the proof in terms ofBasic Coding is that it illustrates in miniature how we will obtain normalultrafilters on ordinals much larger than ω 1 .Corollary 2.14 (Solovay). Assume ZF + AD. ThenProof. We have thatHOD |= ω V 1 is a measurable cardinal.HOD |= µ ∩ HOD is a normal ultrafilter on ω V 1 ,since µ ∩ HOD ∈ HOD (as µ is OD and OD is OD).Thus, if ZF + AD is consistent, then ZFC + “There is a measurablecardinal” is consistent.There is also an effective version of Solovay’s theorem, which we shallneed.Theorem 2.15. Assume ZFC + OD-determinacy. ThenHOD |= ω V 1is a measurable cardinal.Proof. If S is OD then the game G(S) is OD and hence determined. Itfollows (by the above proof) that if I has a winning strategy in G(S) then Scontains a club and if II has a winning strategy in G(S) then ω 1 S containsa club. Thus,V |= µ ∩ HOD is an ultrafilter on HODand soHOD |= µ ∩ HOD is an ultrafilter.Similarly the proof of Claim 1 in Theorem 2.13 shows thatand sowhich completes the proof.V |= µ ∩ HOD is ω 1 -completeHOD |= µ ∩ HOD is ω 1 -complete,

2. Basic Results 322.4. The Least StableWe now take the next step in generalizing the above result. For this purposeit is useful to think of ω 1 in slightly different terms: Recall the followingdefinition:11δ∼ 1 = sup{α | there is a ∆ ∼ 1 -surjection π : ω ω → α}.It is a classical result that ω 1 = δ ∼11 . Now consider the following higher-orderanalogue of δ ∼11 :22δ∼ 1 = sup{α | there is a ∆ ∼ 1 -surjection π : ω ω → α}.In this section we will work without determinacy and establish the basicfeatures of this ordinal in the context of L(R). In the next section we willsolve for U in the equation1δ∼ 1WO = δ ∼Uin such a way that U is accompanied by the appropriate boundedness andcoding theorems required to generalize Solovay’s proof to show that ZF +DC + AD L(R) 2implies that (δ ∼1 ) L(R) is a measurable cardinal in HOD L(R) .2The following model-theoretic characterization of the pointclass Σ ∼ 1 willbe useful in complexity calculations: A ⊆ ω ω 2is Σ ∼1 iff for some formula ϕand some real z ∈ ω ω ,A = {y ∈ ω ω | there is a transitive set M such that(a) ω ω ⊆ M,21(b) there is a surjection π : ω ω → M, and(c) M |= T 0 + ϕ[y, z]}As before, theories much weaker than T 0 yield an equivalent definition andour choice of T 0 is simply one of convenience. The lightface version Σ 2 1 isdefined similarly by omitting the parameter z.2We wish to study δ ∼1 in the context of L(R). In the interest of keepingour account self-contained and free of fine-structure we will give a brief introductionto the basic features of L(R) under the stratification L α (R) forα ∈ On. For credits and references see [2].Definability issues will be central. Officially our language is the languageof set theory with an additional constant Ṙ which is always to be interpreted

2. Basic Results 33as R. For a set M such that X ∪ {R} ⊆ M, let Σ n (M, X) be the collectionof sets definable over M via a Σ n -formula with parameters in X ∪ {R}. Forexample, x is Σ 1 (L(R), X) iff x is Σ 1 -definable over L(R) with parametersfrom X ∪{R}. It is important to note that the parameter R is always allowedin our definability calculations. To emphasize this we will usually make itexplicit.The basic features of L carry over to L(R), one minor difference beingthat R is allowed as a parameter in all definability calculations. For example,for each limit ordinal λ, the sequence 〈L α (R) | α < λ〉 is Σ 1 (L λ (R), {R}).For X ∪ {R} ⊆ M ⊆ N, let M ≺ X n N mean that for all parametersequences ⃗a ∈ (X ∪ {R})

2. Basic Results 34Proof. It suffices (by the Tarski-Vaught criterion) to show that if A is a nonemptyset which is definable over L λ (R) from parameters in R ∪ {R}, thenA ∩ X ≠ ∅. Let A be such a non-empty set and choose x 0 ∈ A. Since everyset in L λ (R) is definable over L λ (R) from a real and an ordinal parameter,{x 0 } = {x ∈ L λ (R) | L λ (R) |= ϕ 0 [x, c 0 , α 0 , R]}for some formula ϕ 0 , and parameters c 0 ∈ ω ω and α 0 ∈ On. Let α 1 be leastsuch that there is exactly one element x such that L λ (R) |= ϕ 0 [x, c 0 , α 1 , R]and x ∈ A. Notice that α 1 is definable in L λ (R) from c 0 and the real parameterused in the definition of A. Thus, letting x 1 be the unique element suchthat L λ (R) |= ϕ 0 [x 1 , c 0 , α 1 , R] we have a set which is in A (by the definitionof x 1 ) and in X (since it is definable in L λ (R) from c 0 and the real parameterused in the definition of A.)Lemma 2.18. Assume ZF + AC ω (R) + V =L(R). For each α < Θ, there isan OD surjection π : ω ω → α.Proof. Fix α < Θ. Since every set in L(R) is OD x for some x ∈ ω ω there isan OD x surjection π : ω ω → α. For each x ∈ ω ω , let π x be the < ODx -leastsuch surjection if one exists and let it be undefined otherwise. We can now“average over the reals” to eliminate the dependence on real parameters,lettingThis is an OD surjection.π : ω ω → α{π (x)0 ((x) 1 ) if π (x)0 is definedx ↦→0 otherwise.Lemma 2.19 (Solovay). Assume ZF + AC ω (R) + V = L(R). Then Θ isregular in L(R).Proof. By the proof of the previous lemma, there is an OD sequence〈π α | α < Θ〉such that each π α : ω ω → α is an OD surjection. Assume for contradictionthat Θ is singular. Letf : α → Θ

2. Basic Results 35be a cofinal map witnessing the singularity of Θ. Let g : ω ω → α be asurjection. It follows that the mapπ : ω ω → Θx ↦→ π f◦g((x)0 )((x) 1 )is a surjection, which contradicts the definition of Θ.Lemma 2.20. Assume ZF + AC ω (R) + V = L(R). ThenL Θ (R) = {x ∈ L(R) | there is a surjection π : ω ω → tc(x)}.Thus, P(R) ⊆ L Θ (R).Proof. For the first direction suppose x ∈ L Θ (R). Let λ < Θ be a limitordinal such that x ∈ L λ (R). Thus tc(x) ⊆ L λ (R). Moreover, there is asurjection π : ω ω → L λ (R), since every element of L λ (R) is definable froman ordinal and real parameters.For the second direction suppose x ∈ L(R) and that there is a surjectionπ : ω ω → tc(x). We wish to show that x ∈ L Θ (R). Let λ be a limit ordinalsuch that x ∈ L λ (R). Thus tc(x) ⊆ L λ (R). LetX = { y ∈ L λ (R) | y is definable over L λ (R)from parameters in tc(x) ∪ R ∪ {R} } ,By the proof of Lemma 2.17, X ≺ L λ (R) and tc(x) ⊆ X. By Condensation,the transitive collapse of X is L¯λ(R) for some ¯λ. Since there is a surjectionπ : ω ω → tc(x) and since all members of L¯λ(R) are definable from parametersin tc(x) ∪ R ∪ {R}, there is a surjection ρ : ω ω → L¯λ(R). So ¯λ < Θ and sincex ∈ L¯λ(R) this completes the proof.Lemma 2.21. Assume ZF + AC ω (R) + V = L(R). ThenL Θ (R) |= T 0 .Proof. It is straightforward to see that L Θ (R) satisfies T 0 − Separation −Replacement.To see that L Θ (R) |= Separation note that if S ⊆ x ∈ L Θ (R) thenS ∈ L Θ (R), by Lemma 2.20. To see that L Θ (R) |= Replacement we verifyCollection, which is equivalent to Replacement, over the other axioms.SupposeL Θ (R) |= ∀x ∈ a ∃y ϕ(x, y),

2. Basic Results 36where a ∈ L Θ (R). Letf : a ↦→ Θx → µα (∃y ∈ L α (R) such that L Θ (R) |= ϕ(x, y)).The ordertype of ran(f) is less that Θ since otherwise there would be asurjection π : ω ω → Θ (since there is a surjection π : ω ω → a). Moreover,since Θ is regular, it follows that ran(f) is bounded by some λ < Θ. Thus,which completes the proof.L Θ (R) |= ∀x ∈ a ∃y ∈ L λ (R) ϕ(x, y),Lemma 2.22. Assume ZF+AC ω (R)+V = L(R). There are arbitrarily largeα such that L α (R) |= T 0 .Proof. The proof is similar to the previous proof. Let us say that α is anR-cardinal if for every γ < α there does not exist a surjection π : R ×γ → α.For each limit ordinal γ ∈ On, lettingΘ(γ) = sup{α | there is a surjection π : R × γ → α}we have that Θ(γ) is an R-cardinal. For each γ which is closed under theGödel pairing function, the argument of Lemma 2.19 shows that Θ(γ) isregular. The proof of the previous lemma generalizes to show that for everyregular Θ(γ), L Θ(γ) (R) |= T 0 .Lemma 2.23 (Solovay). Assume ZF+AC ω (R)+V = L(R). L Θ (R) ≺ 1 L(R).Proof. SupposeL(R) |= ϕ[a],where a ∈ L Θ (R) and ϕ is Σ 1 . By Reflection, let λ be a limit ordinal suchthatL λ (R) |= ϕ[a].LetX = { y ∈ L λ (R) | y is definable over L λ (R)from parameters in tc(a) ∪ R ∪ {R} } ,By Condensation and Lemma 2.20, the transitive collapse of X is L¯λ(R) forsome ¯λ < Θ. Thus, by upward absoluteness,L Θ (R) |= ϕ[a].

2. Basic Results 37Lemma 2.24. Assume ZF+AC ω (R)+V = L(R). There are arbitrarily largeα < δ F such that L α (R) |= T 0Proof. Suppose ξ < δ F . Since L Θ (R) |= T 0 ,L(R) |= ∃α > ξ (L α (R) |= T 0 ).The formula is readily seen to be Σ 1 with parameters in {R, ξ} by our modeltheoreticcharacterization. Thus, by the definition of δ F ,which completes the proof.L δF (R) |= ∃α > ξ (L α (R) |= T 0 ),Lemma 2.25. Assume ZF + AC ω (R) + V = L(R). Suppose ϕ is a formulaand a ∈ ω ω . Suppose λ is least such that L λ (R) |= T 0 + ϕ[a]. LetX = { x ∈ L λ (R) | y is definable over L λ (R)from parameters in R ∪ {R} } ,Then X = L λ (R). Moreover, there is a surjection π : ω ω → L λ (R) such thatπ is definable over L λ+1 (R) from R and a.Proof. By Lemma 2.17 we have that X ≺ L λ (R). By condensation thetransitive collapse of X is some L¯λ(R). So L¯λ(R) |= T 0 + ϕ[a] and thus bythe minimality of λ we have ¯λ = λ. Since every x ∈ X is definable froma real parameter and since L λ (R) ∼ = X, we have that every x ∈ L λ (R) isdefinable from a real parameter, in other words, X = L λ (R). The desiredmap π : ω ω → L λ (R) is the map which takes a real coding the Gödel numberof ϕ and a real parameter a to the set {x ∈ L λ (R) | L λ (R) |= ϕ[x, a]}. Thismap is definable over L λ+1 (R).Lemma 2.26. Assume ZF + AC ω (R) + V = L(R). Suppose 0 < α < δ R .Then there is a surjection π : ω ω → L α (R) such that {(x, y) | π(x) ∈ π(y)}is ∆ ∼21 . Thus, δ R δ ∼21 .Proof. Fix α such that 0 < α < δ R . By the minimality of δ R ,L α (R) ⊀ R∪{R}1 L(R).So there is an a ∈ ω ω and a Σ 1 -formula ϕ such that if β is the least ordinalsuch that L β (R) |= ϕ[a] then β > α. Let γ be least such that γ > β > α

2. Basic Results 38and L γ (R) |= T 0 (which exists by Lemma 2.22). So γ is least such thatL γ (R) |= T 0 + ϕ[a] and, by Lemma 2.25, there is a surjection π : ω ω →L γ (R) which is definable over L γ+1 (R) with the parameters R and a. LetA = {(x, y) | π(x) ∈ π(y)}. Let ψ 1 and ψ 2 be the formulas defining A and(ω ω ) 2 A over L γ+1 (R), respectively. By absoluteness,(x, y) ∈ A iff there is a transitive set M such thatω ω ⊆ M,there is a surjection π : ω ω → M, andM |= T 0 + ∃γ (L γ (R) |= T 0 + ϕ[a] andL γ+1 (R) |= ψ 1 [x, y]).2 2This shows, by our model-theoretic characterization of Σ ∼1 that A is Σ ∼1 . Asimilar argument shows that (ω ω ) 2 2 A is Σ ∼ 1 . Finally, the desired map canbe extracted from π.We now use a universal Σ ∼21 set to knit together all of these “∆ ∼21 projectionmaps”.Lemma 2.27. Assume ZF + AC ω (R) + V = L(R). Then there is a partialsurjection ρ : ω ω → L δR (R) such that dom(ρ) and ρ are both Σ 1 -definable overL δR (R) with the parameter R. Thus, L δR (R) ≺ 1 L(R) and hence δ F δ R .Proof. Let U be a Σ 2 1 subset of ωω ×ω ω ×ω ω that is universal for Σ ∼21 subsetsof ω ω × ω ω , that is, such that for each Σ ∼21 subset A ⊆ ω ω × ω ω there is anx ∈ ω ω such that A = U x where by definitionU x = {(y, z) ∈ ω ω × ω ω | (x, y, z) ∈ U}.We define ρ using U. For the domain of ρ we takedom(ρ) = {x ∈ ω ω | ∃α ∈ On (L α (R) |= T 0 andL α (R) |= U (x)0 = (ω ω × ω ω ) U (x)1 )}.Notice that dom(ρ) is Σ 1 (L(R), {R}) and hence Σ 1 (L δR (R), {R}). Notice alsothat in general if L α (R) |= T 0 thenand thus, if in addition,(U (x)0 ) Lα(R) ⊆ U (x)0(U (x)0 ) Lα(R) = (ω ω × ω ω ) (U (x)1 ) Lα(R) ,

2. Basic Results 39then,(U (x)0 ) Lα(R) = U (x)0 .We can now define ρ as follows: Suppose x ∈ dom(ρ). Let α(x) be theleast α as in the definition of dom(ρ). If there is an ordinal η and a surjectionπ : ω ω → L η (R) such that{(t 1 , t 2 ) | π(t 1 ) ∈ π(t 2 )} = (U (x)0 ) L α(x)(R)then let ρ(x) = π((x) 2 ); otherwise let ρ(x) = ∅. Notice that the map ρ isΣ 1 (L(R), {R}) and hence Σ 1 (L δR (R), {R}). By Lemma 2.26, ρ : dom(ρ) →L δR (R) is a surjection.For the last part of the proof recall that by definition L δR (R) ≺ R∪{R}1L(R). The partial surjection ρ : ω ω → L δR (R) allows us to reduce arbitraryparameters in L δR (R) to parameters in ω ω .Theorem 2.28. Assume ZF + AC ω (R) + V = L(R). δ ∼21 = δ R = δ F .Proof. We have δ R δ ∼21 (by Lemma 2.26), δ R δ F (by definition), andδ F δ R (by Lemma 2.27). It remains to show δ ∼21 δ R .Suppose γ < δ ∼21 . We wish to show that γ < δ R . Let π : ω ω → α be asurjection such that A = {(x, y) | π(x) < π(y)} is ∆ ∼21 . Using the notationfrom the previous proof let x be such thatU (x)0 = A and U (x)1 = (ω ω × ω ω ) A.There is an ordinal α such that L α (R) |= T 0 andSince(U (x)0 ) Lα(R) = (ω ω × ω ω ) (U (x)1 ) Lα(R) .L δR (R) ≺ R∪{R}1 L(R),the least such ordinal, α(x), is less than δ R . Thus,(U (x)0 ) L α(x)(R) = A.Finally, since L α(x) (R) |= T 0 , this model can compute the ordertype, γ, of A.Thus, γ < α(x) < δ R .2.29 Remark. Although we will not need these facts it is worthwhile to notethat the above proofs show

2. Basic Results 40(1) (Σ ∼21 ) L(R) = Σ ∼ 1(L δ˜21(R)) ∩ P(ω ω ),(2) (∆ ∼21 ) L(R) = L δ˜21(R) ∩ P(ω ω ), and(3) (Solovay’s Basis Theorem) if L(R) |= ∃X ϕ(X) where ϕ is Σ ∼21 thenL(R) |= ∃X ∈ ∆ ∼21 ϕ(X).2.5. Measurability of the Least StableWe are now in a position to show that under ZF + DC + AD,HOD L(R) |= (δ ∼21 ) L(R) is a measurable cardinal.This serves as a warm-up to Section 4, where we will show that under ZF +DC + AD,HOD L(R) |= (δ ∼21 ) L(R) is λ-strong,for each λ < Θ L(R) , and, in fact, thatHOD L(R) |= Θ L(R) is a Woodin cardinal.The proof that we give in Section 4 will show that DC can be eliminatedfrom the result of the present section.First we need an analogue U of WO that enables us to encode (unboundedlymany) ordinals below δ ∼21 and is accompanied by the boundedness andcoding theorems required to push the above proof through for δ ∼21 . The followingworks: Let U be a Σ 2 1 subset of ω ω × ω ω that is universal for Σ ∼21 subsetsof ω ω . For y ∈ ω ω we let U y = {z ∈ ω ω | (y, z) ∈ U}. For (y, z) ∈ U, letΘ (y,z) be least such thatL Θ(y,z) (R) |= T 0 and (y, z) ∈ U L Θ (y,z) (R) .Let δ (y,z) = (δ ∼21 ) L Θ (y,z) (R) . These ordinals are the analogues of α x from theproof that ω 1 is measurable. For notational convenience we will routinelyuse our recursive bijection from ω ω ×ω ω to ω ω to identify pairs of reals (y, z)with single reals x = 〈y, z〉. Thus we will write Θ x and δ x instead of Θ (y,z)and δ (y,z) .Lemma 2.30. Assume ZF+AC ω (R)+V = L(R). {δ x | x ∈ U} is unboundedin δ ∼21 .

2. Basic Results 41Proof. Let α < δ ∼21 . Let A be (the set of reals coding) a ∆ ∼21 prewellordering oflength greater than α. Let y, z ∈ ω ω be such that U y = A and U z = ω ω A.So L(R) |= “U y = ω ω U z ”. Since δ ∼21 is the least stable, there is a β < δ ∼21 suchthat L β (R) |= “U y = ω ω U z ” and since (U y ) L β(R) ⊆ A and (U z ) L β(R) ⊆ ω ω Awe have that A = (U y ) L β(R) . Now, letting x ∈ U U L β(R) and γ < δ ∼21 besuch that L γ (R) |= “T 0 + x ∈ U”, we have that α < δ x since A ∈ L γ (R) andL γ (R) can compute the ordertype of A.In analogy with WO, for x ∈ U let U δx = {y ∈ U | δ y = δ x }, U

2. Basic Results 42with the following rules: Rule 1: For all i < ω, (x) i , (y) i ∈ U. If Rule 1is violated then, letting i be least such that either (x) i ∉ U or (y) i ∉ U, Iwins if (x) i ∈ U; otherwise II wins. Now suppose Rule 1 is satisfied. Rule 2:δ (x)0 < δ (y)0 < δ (x)1 < δ (y)1 · · ·. The first failure defines who wins as above.If both rules are satisfied then I wins iff sup i∈ω δ (x)i ∈ S.Now letµ = {S ⊆ δ ∼21 | I wins G(S)}.Notice that as before (using ∆ ∼21 -boundedness) if I has a winning strategyin G(S) then S contains a set C which is unbounded and closed under ω-sequences. The proof that U is an ultrafilter is exactly as before. To see thatit is δ ∼21 -complete and normal one uses the new versions of Boundedness andCoding. We note the minor changes in the proof of normality.Assume for contradiction that f : δ ∼21 → δ ∼21 and that there is no α < δ ∼21such that {ξ | f(ξ) = α} ∈ µ or, equivalently (by AD) that for all α < δ ∼21 ,S α = {ξ | f(ξ) ≠ α} ∈ µ.Let 〈δ α | α < δ ∼21 〉 enumerate 〈δ x | x ∈ U〉. Here we are appealing to the factthat δ ∼21 is regular, which can be shown using the Coding Lemma (See [10,p.433]). In analogy with WO, for α < β < δ ∼21 , let U α = {x ∈ U | δ x = δ α },U (α,β] = {x ∈ U | δ α < δ x δ β } and likewise for other intervals. Let U bethe associated prewellordering.As before, our strategy is to inductively define(1.1) an increasing sequence 〈η i | i < ω〉 of ordinals with supremum η,(1.2) a sequence of collections of strategies 〈X i | i < ω〉 where X i containswinning strategies for I in games G(S α ) for α ∈ [η i−1 , η i ), whereη −1 = 0, and(1.3) a sequence 〈y i | i < ω〉 of plays such that y i is legal for II againstany σ ∈ X i and sup j

2. Basic Results 43(2.1) Z ∗ ∩ (U α × ω ω ) ≠ ∅ iff Z ∩ (U α × ω ω ) ≠ ∅(2.2) Z ∗ ∩ (U α × ω ω ) is ∆ ∼21 .This puts us in a position to apply ∆ ∼21 -boundedness.For the first step letη 0 = some ordinal η such that η < δ ∼21X 0 = proj 2(Z ∗ ∩ (U [0,η0 ) × ω ω ) )Y 0 = { ((σ∗y) I ) 0 | σ ∈ X 0 ∧ y ∈ ω ω}z 0 = some real z such that Y 0 ⊆ U

3. Coding 44The above proof uses DC. However, as we shall see in Section 4.1 thetheorem can be proved in ZF + AD. See Lemma 4.7.The coding lemma was used to enable II to “collect together” the relevantstrategies and then the ∆ ∼21 -boundedness lemma was used to enable II to “takecontrol of the ordinal played” in all such games by devising a play that is legalagainst all of the relevant strategies and (in each case) has the same fixedordinal as output. This technique is central in what follows. It is importantto note, however, that the above ultrafilter (and, more generally, ultrafiltersobtained by such a “sup” game) concentrates on points of cofinality ω. Laterwe will use a slightly different game, where the role of the ∆ ∼21 -boundednesslemma will be played by a certain reflection phenomenon. Before turning tothis we prove the coding lemmas we shall need.3. CodingIn the Basic Coding Lemma we constructed selectors relative to WO; we nowdo so relative to more general prewellorderings.3.1. Coding LemmaWe begin by fixing some notation. For P ⊆ ω ω 1, the notion of a Σ ∼1 (P) set is1defined exactly like that of a Σ ∼1 set only now we allow reference to P and toω ω P. In model-theoretic terms, X ⊆ ω ω 1is Σ ∼ 1 (P) iff there is a formula ϕand a real z such thatX = { y ∈ ω ω | there is an ω-model M such thaty, z, P ∩ M ∈ M and M |= T 0 + ϕ[y, z, P ∩ M] } .1The notion of a Σ ∼ 1 (P, P ′ ) set is defined in the same way, only now referenceto both P and P ′ and their complements is allowed. The lightface versionsof these notions and the versions involving P ⊆ (ω ω ) n are all defined in theobvious way.Let U (n) (P) be a Σ 1 1(P) subset of (ω ω ) n+1 1that is universal for Σ ∼1 (P)subsets of (ω ω ) n 1, that is, such that for each Σ ∼ 1 (P) set A ⊆ (ωω ) n there is ane ∈ ω ω such that A = U e(n) (P) = {y ∈ (ω ω ) n | (e, y) ∈ U (n) (P)}. We do thisin such a way that the same formula is used, so that the definition is uniformin P. Likewise, for U (n) (P, P ′ ) etc. (The existence of such a universal set

3. Coding 45U (n) (P) is guaranteed by the fact that the pointclass in question, namely,Σ ∼11 (P), is ω-parameterized and closed under recursive substitution. See [10],3E.4 on p. 160 and especially 3H.1 on p. 183. We further assume that theuniversal sets are “good” in the sense of [10], p. 185 and we are justified indoing so by [10], 3H.1. A particular component of this assumption is thatour universal sets satisfy the s-m-n-theorem (uniformly in P (or P and P ′ )).See Jackson’s chapter in this Handbook for further details.)Theorem 3.1 (Recursion Theorem) (Kleene). Suppose f : ω ω → ω ω isΣ ∼11 (P). Then there is an e ∈ ω ω such thatProof. For a ∈ ω ω , letU (2)e (P) = U (2)f(e) (P).T a = {(b, c) | (a, a, b, c) ∈ U (3) (P)}.Let d : ω ω → ω ω be Σ 1 1 such that T a = U (2)d(a)(P). (The function d comesfrom the s-m-n-theorem. In fact, d(a) = s(a, a) (in the notation of Jackson’schapter) and d is continuous.) LetY = {(a, b, c) | (b, c) ∈ U (2)f(d(a)) (P)}and let a 0 be such that Y = U a (3)10 (P). Notice that Y is Σ ∼ 1 (P) using theparameter for Y (as can readily be checked using the model-theoretic characterizationof Σ ∼1 (P)). We1haveand so d(a 0 ) is as desired.(b, c) ∈ U (2)d(a 0 ) (P) iff (a 0, a 0 , b, c) ∈ U (3) (P)iff (a 0 , b, c) ∈ U (3)a 0(P) = Yiff (b, c) ∈ U (2)f(d(a 0 )) (P)Theorem 3.2 (Coding lemma) (Moschovakis). Assume ZF+AD. SupposeX ⊆ ω ω and π : X → On. Suppose Z ⊆ X × ω ω . Then there is an e ∈ ω ωsuch that(1) U e(2) (Q) ⊆ Z and(2) for all a ∈ X, U (2)e (Q) ∩ (Q a × ω ω ) ≠ ∅ iff Z ∩ (Q a × ω ω ) ≠ ∅,

3. Coding 46where Q = {〈a, b〉 | π(a) π(b)}.Proof. Assume toward a contradiction that there is no such e. Consider theset G of reals e for which (1) in the statement of the theorem is satisfied:G = { e ∈ ω ω | U (2)e (Q) ⊆ Z } .So, for each e ∈ G, (2) in the statement of the theorem fails for some a ∈ X.Let α e be the least α such that (2) fails at the α th -section:α e = min { α | ∃a ∈ X (π(a) = α ∧ U e(2) (Q) ∩ (Q a × ω ω ) = ∅ ∧Z ∩ (Q a × ω ω ) ≠ ∅) } .Now play the gameI x(0) x(1) x(2) . . .II y(0) y(1) . . .where I wins if x ∈ G and either y ∉ G or α x α y . Notice that by ourassumption that there is no index e as in the statement of the theorem,neither I nor II can win a round of this game by playing a selector. Thebest they can do is play “partial” selectors. For a play e ∈ G, let us callU (2)e(Q) ∩(Q

3. Coding 47By assumption, U (2)e σ (Q) is not a selector. So α eσ exists. Since for all y ∈ ω ω ,α eσ α (σ∗y)I , we can take α eσ as our bound. Choose a ∈ X such thatπ(a) = α eσ . Pick (x 1 , x 2 ) ∈ Z ∩ (Q a × ω ω ). Let e ∗ be such thatU (2)e ∗(Q) = U(2) e σ∪ {(x 1 , x 2 )}.So e ∗ ∈ G. But α eσ < α e ∗. In summary, we have that for all y ∈ ω ω ,α (σ∗y)I α eσ < α e ∗. Thus, by playing e ∗ , II defeats σ.Claim 2. Player II does not have a winning strategy.Proof. Assume toward a contradiction that τ is a winning strategy for II. Weshall show that τ yields a selector for Z; in other words, it yields an e ∗ suchthat(1) U (2)e (Q) ⊆ Z and∗(2) for all a ∈ X, U (2)e ∗ (Q) ∩ (Q a × ω ω ) ≠ ∅ iff Z ∩ (Q a × ω ω ) ≠ ∅.Choose h 0 : ω ω × X → ω ω such that h 0 is Σ 1 1 (Q) and for all e, a ∈ ωω ,U (2)h 0 (e,a)(Q) = U(2) e (Q) ∩ (Q

3. Coding 48By the recursion theorem there is a fixed point for h 1 ; that is, there is ane ∗ such thatU (2)e (Q) = ∗ U(2)h 1 (e ∗ ) (Q).This set has the following closure property: if I plays an initial segment of itthen II responds with a subset of it. We shall see that e ∗ ∈ G. Moreover, ifU (2)e (Q) is not a selector then having I play the largest initial segment which is∗a partial selector, II responds with a larger selector, which is a contradiction.Thus, e ∗ codes a selector. Here are the details.Subclaim 1. e ∗ ∈ G.Proof. Suppose for contradiction that U (2)e ∗(Q) Z with π(x 1) minimal. SoU (2)e ∗(x 1 , x 2 ) ∈ U (2)e ∗Fix a ∈ X such that(Q) = U(2)h 1 (e ∗ ) (Q)= ⋃ a∈X(Q) Z ≠ ∅. Choose (x 1, x 2 ) ∈(U(2)(h 0 (e ∗ ,a)∗τ) II(Q) ∩ (Q a × ω ω ) ) .(x 1 , x 2 ) ∈ U (2)(h 0 (e ∗ ,a)∗τ) II(Q) ∩ (Q a × ω ω ).The key point is that h 0 (e ∗ , a) ∈ G since we chose (x 1 , x 2 ) with π(x 1 ) = π(a)minimal. Thus, since τ is a winning strategy, (h 0 (e ∗ , a)∗τ) II ∈ G, and so(x 1 , x 2 ) ∈ Z, which is a contradiction.Subclaim 2. α e ∗does not exist.Proof. Suppose for contradiction that α e ∗ exists. Let a ∈ X be such thatπ(a) = α e ∗. Thus h 0 (e ∗ , a) ∈ G and α h0 (e ∗ ,a) = α e ∗. Since τ is a winningstrategy for II,α (h0 (e ∗ ,a)∗τ) II> α h0 (e ∗ ,a) = α e ∗,which is impossible sinceThus α e ∗does not exist.U (2)(h 0 (e ∗ ,a)∗τ) II(Q) ⊆ U (2)e (Q).∗Hence e ∗ is the code for a selector.This completes the proof of the Coding Lemma.

3. Coding 493.2. Uniform Coding LemmaWe shall need a uniform version of the above theorem. The version we proveis different than that which appears in the literature ([5]). We shall need thefollowing uniform version of the recursion theorem.Theorem 3.3 (Uniform Recursion Theorem) (Kleene). Suppose f :ω ω → ω ω is Σ ∼11 . Then there is an e ∈ ω ω such that for all P, P ′ ⊆ ω ω ,U (2)e (P, P ′ ) = U (2)f(e) (P, P ′ ).Proof. The proof is the same as before. The key point is that the definitionof the fixed point d(a 0 ) depends only on f and, of course, d, which is uniformin P, P ′ .Theorem 3.4 (Uniform Coding Lemma)). Assume ZF + AD. SupposeX ⊆ ω ω and π : X → On. Suppose Z ⊆ X × ω ω . Then there exists ane ∈ ω ω such that for all a ∈ X,(1) U e(2) (Q

3. Coding 50in the parameters Q

3. Coding 51andChoose a ∈ X such that π(a) = α eσ . SoU (2)e σ(Q

3. Coding 52Notice that for e ∈ ω ω and z ∈ X, the set U (2)h 0 (e,z)(·, ·) is such that it agrees(·, ·) for parameters Q

3. Coding 53SoFix z ∗ ∈ Q a ∗(x 1 , x 2 ) ∈ U (2)e (Q ∗

3. Coding 54So, α h0 (e ∗ ,z ∗ ) = α e ∗. Since τ is a winning strategy for II,which is impossible sincefor all a ∈ X.α (h0 (e ∗ ,z ∗ )∗τ) II> α h0 (e ∗ ,z ∗ ) = α e ∗,U (2)(h 0 (e ∗ ,z ∗ )∗τ) II(Q

3. Coding 553.3. ApplicationsIn this section we will bring together some basic results and key applicationsof the above coding lemmas that will be of use later. It will be useful to dothings in a slightly more general fashion than is customary.For a set X, letΘ X = sup{α | there is an OD X surjection π : ω ω → α}.Lemma 3.7. Assume ZF and suppose X is a set. Then there is an OD Xsequence A = 〈A α | α < Θ X 〉 such that A α is a prewellordering of the realsof length α.Proof. Let A α be the < ODX -least prewellordering of the reals of length α,where < ODX is the canonical OD X well-ordering of the OD X sets.Lemma 3.8. Assume ZF and suppose X is a set. Suppose that every set isOD X,y for some real y. Then Θ = Θ X .Proof. Fix α < Θ. We have to show that there is an OD X surjection π : ω ω →α. There is certainly an OD X,y surjection for some y. For each y ∈ ω ω , letπ y be the < ODX,y -least such surjection if one exists and let it be undefinedotherwise. We can now “average over the reals” to eliminate the dependenceon real parameters, lettingThis is an OD X surjection.π : ω ω → α{π (y)0 ((y) 1 ) if π (y)0 is definedy ↦→0 otherwise.The following theorem is essentially due to Moschovakis. We are justreplacing AD with OD X -determinacy and the changes are straightforward.Theorem 3.9. Assume ZF + OD X -determinacy, where X is a set. ThenHOD X |= Θ X is strongly inaccessible.Proof. First we show that Θ X is regular in HOD X . By Lemma 3.7 there isan OD X sequence〈π α | α < Θ X 〉

3. Coding 56where each π α : ω ω → α is a surjection. Assume for contradiction that Θ Xis singular in HOD X and letf : η → Θ Xbe an OD X cofinal map. Let g be an OD X surjection from ω ω onto η. Thenthe mapπ : ω ω → Θ Xx ↦→ π f◦g((x)0 )((x) 1 )is an OD X surjection, which contradicts the definition of Θ X .We now show that Θ X is a strong limit in HOD X . For each η < Θ X , wehave to show that |P(η)| HOD X< Θ X . For this it suffices to show that thereis an OD X surjectionπ : ω ω → P(η) HOD X,since if |P(η)| HOD X Θ X then there would be an OD X surjection ρ :P(η) → Θ X and so ρ ◦ π : ω ω → Θ X would be an OD X surjection, whichcontradicts the definition of Θ X .Let π η : ω ω → η be an OD X surjection and, for α < η, let Q

3. Coding 57(1) HOD L(R) |= Θ is strongly inaccessible and(2) HOD L(R) ∩ V Θ = HOD L Θ(R) .Proof. (1) This follows immediately from Theorem 3.9 and Lemma 3.8.(2) Since HOD L(R) is Σ 1 -definable over L(R) (with the parameter R) andsince L Θ (R) ≺ 1 L(R) (by Lemma 2.23),Thus, it suffices to showHOD L Θ(R) = HOD L(R) ∩ L Θ (R).HOD L(R) ∩ V Θ = HOD L(R) ∩ L Θ (R).The right-to-left inclusion is immediate. For the left-to-right inclusion supposex ∈ HOD L(R) ∩ V Θ . We have to show that x ∈ L Θ (R). Since Θ isstrongly inaccessible in HOD L(R) , x is coded by a set of ordinals A ⊆ αwhere α < Θ. However, by the proof of Theorem 3.9, P(α) ∈ L Θ (R), foreach α < Θ. Thus, x ∈ L Θ (R), which completes the proof.3 Open Question. Assume ZF + DC + V =L(R).(1) Suppose that for every α < Θ there is a surjection π : ω ω → P(α).Must AD hold in L(R)?(2) Suppose Θ is inaccessible. Must AD hold in L(R)?Theorem 3.11 (Kunen). Assume ZF + DC + AD. Suppose λ < Θ and µ isan ultrafilter on λ. Then µ is OD.Proof. Let be a prewellordering of ω ω of length λ. Let π : ω ω → P(λ) bethe surjection derived from as in the above proof. For x ∈ ω ω , letA x = ⋂ {π(y) | π(y) ∈ µ ∧ y T x}.Since there are only countably many such y and AD implies that all ultrafiltersare countably complete (Theorem 2.8), A x is non-empty. Letf(x) = ⋂ A x .Notice that A x and f(x) depend only on the Turing degree of x. In particular,we can regard f as a function from the Turing degrees D T into the ordinals.Notice also thatA ∈ µ iff for a cone of x, f(x) ∈ A

4. A Woodin Cardinal in HOD L(R) 58since if B ∈ µ then, for any x T x 0 we have f(x) ∈ B, where x 0 is such thatπ(x 0 ) = B. We can now “erase” reference to the prewellordering by takingthe ultrapower. Let µ T be the cone ultrafilter on the Turing degrees (seeTheorem 2.9) and consider the ultrapower V D T/µ T . By DC the ultrapoweris well-founded. So we can let M be the transitive collapse of V D T/µ T andletj : V → Mbe the canonical map. Letting γ be the ordinal represented by f, we haveand so µ is OD.B ∈ µ iff γ ∈ j(B)4. A Woodin Cardinal in HOD L(R)Our main aim in this section is to prove the following theorem:Theorem 4.1. Assume ZF + DC + AD. ThenHOD L(R) |= ZFC + Θ L(R) is a Woodin cardinal.This will serve as a warm-up for the proof of the Generation Theorem inthe next section. The proof that we give appeals to DC at only one point(Lemma 4.8) and as we shall see in the next section one can avoid this appealand prove the result in ZF + AD. See Theorem 5.36.In §4.1 we will establish the reflection phenomenon that will play the roleplayed by boundedness in §2 and we will define for cofinally many λ < Θ,2 2an ultrafilter µ λ on δ ∼1 that is intended to witness that ∼δ 1 is λ-strong. In §4.2we shall introduce and motivate the notion of strong normality by showing2that the strong normality of µ λ ensures that δ ∼1 is λ-strong. We will thenshow how reflection and uniform coding combine to secure strong normality.In §4.3 we will prove the main theorem by relativizing the construction tosubsets of Θ. Throughout this section we work in L(R) and so when we write2δ∼ 1 and Θ we will always be referring to these notions as interpreted in L(R).

4. A Woodin Cardinal in HOD L(R) 594.1. ReflectionWe have seen that ZF+AD implies that Θ is strongly inaccessible in HOD L(R) .Our next task is to show thatHOD L(R) |= δ ∼21 is λ-strong,for all λ < Θ. The proof will then relativize to subsets of Θ that are inHOD L(R) and thereby establish the main theorem.The ultrafilters that witness strength cannot come from the “sup” gameof Section 2 since the ultrafilters produced by this game concentrate on ω-club sets, whereas to witness strength we will need ultrafilters according towhich there are measure-one many measurable cardinals below δ ∼21 . For thisreason we will have to use a variant of the “sup” game. In this variant therole of boundedness will be played by a certain reflection phenomenon.The reflection phenomenon we have in mind does not presuppose anydeterminacy assumptions. For the time being work in ZF + AC ω (R). Themain claim is that there is a function F : δ ∼21 → L δ˜21(R) which is ∆ 1 -definableover L δ˜21(R) and for which the following reflection phenomenon holds:For all X ∈ L(R) ∩OD L(R) , for all Σ 1 -formulas ϕ, and for all z ∈ ω ω , ifL(R) |= ϕ[z, X, δ ∼21 , R]then there exists a δ < δ ∼21 such thatL(R) |= ϕ[z, F(δ), δ, R].One should think of F as a sequence that contains “proxies” or “genericwitnesses” for each OD L(R) set X: Given any Σ 1 -fact (with a real parameter)about any OD L(R) set X there is a “proxy” F(δ) in our fixed sequence thatwitnesses the same fact.The function F is defined (much like ♦) in terms of the least counterexample.To describe this in more detail let us first recall some basic facts fromSection 2.4 concerning L(R) and the theory T 0 : There are arbitrarily largeα such that L α (R) |= T 0 . In particular,L Θ (R) |= T 0 .Moreover, sinceL Θ (R) ≺ 1 L(R),

4. A Woodin Cardinal in HOD L(R) 60there are arbitrarily large α < Θ such that L α (R) |= T 0 . Similarly, there arearbitrarily large α < δ ∼21 such that L α (R) |= T 0 . However, notice that it isnot the case L δ˜21(R) |= T 0 (by Lemma 2.27).Because of the greater manoeuvring room provided by levels L α (R) thatsatisfy T 0 we will concentrate (for example, in reflection arguments) on suchlevels. For example, we can use these levels to give a first-order definition ofOD L(R) and the natural well-ordering < OD L(R) on the OD L(R) sets. For thelatter, given X ∈ OD L(R) , letα X = the least α such that(1) L α (R) |= T 0 ,(2) X ∈ L α (R), and(3) X is definable in L α (R) from ordinal parameters;let ϕ X be the least formula that defines X from ordinal parameters in L α (R);and let ⃗ ξ X be the lexicographically least sequence of ordinal parameters usedto define X in L α (R) via ϕ X . Given X and Y in OD L(R) , working in L(R)setX < OD Y iff α X < α Y orα X = α Y and ϕ X < ϕ Y orα X = α Y and ϕ X = ϕ Y and ⃗ ξ X < lex⃗ ξY .Since the L α (R) hierarchy is Σ 1 -definable in L(R), it follows that OD L(R)and (< OD ) L(R) are Σ 1 -definable in L(R). (This is in contrast to the usualdefinitions of these notions, which are Σ 2 since they involve existential quantificationover the V α hierarchy, which is Π 1 .) Notice that if L α (R) |= T 0 ,then(< OD ) Lα(R) (< OD ) L(R) .Furthermore, if L α (R) ≺ 1 L(R), thenOD Lα(R) = OD L(R) ∩ L α (R) and(< OD ) Lα(R) = (< OD ) L(R) ↾L α (R).For example,HOD L Θ(R) = HOD L(R) ∩ L Θ (R).(For this it is crucial that we use the Σ 1 -definition given above since theΣ 2 -definition involves quantification over the V α hierarchy and yet in L δ˜21(R)

4. A Woodin Cardinal in HOD L(R) 61even the level V ω+2 does not exist.) Our goal can thus be rephrased as thatof showingHOD L Θ(R) |= δ ∼21 is a strong cardinal.We are now in a position to define the reflection function F. If the reflectionphenomenon fails in L(R) with respect to F↾δ ∼21 then (by Replacement)there is some level L α (R) which satisfies T 0 over which the reflection phenomenonfails with respect to F↾δ ∼21 . This motivates the following definition:4.2 Definition. Assume T 0 . Suppose that F↾δ is defined. Let ϑ(δ) be leastsuch thatL ϑ(δ) (R) |= T 0 and there is an X ∈ L ϑ(δ) (R) ∩ OD L ϑ(δ)(R) such that(⋆) there is a Σ 1 -formula ϕ and a real z such thatand for all ¯δ < δ,L ϑ(δ) (R) |= ϕ[z, X, δ, R]L ϑ(δ) (R) ̸|= ϕ[z, F(¯δ), ¯δ, R](if such an ordinal exists) and then set F(δ) = X where X is (< OD ) L ϑ(δ) -leastsuch that (⋆) holds.We have to establish two things: First, F(δ) is defined for all δ < δ ∼21 .Second, F(δ ∼21 ) is not defined. This implies that the reflection phenomenonholds with respect to F.Lemma 4.3. Assume ZF + AC ω (R). Then(1) if L α (R) |= T 0 , then (F) Lα(R) = F↾γ for some γ,(2) F L (R)δ˜21=2 F↾δ∼ 1 , and(3) F(δ) is defined for all δ < δ ∼21 .Proof. For (1) suppose that (F↾δ) Lα(R) = F↾δ with the aim of showing that(F(δ)) Lα(R) = (F(δ)) L(R) . The point is thatL α (R) |= ϑ(δ) exists

4. A Woodin Cardinal in HOD L(R) 62if and only ifin which case(ϑ(δ)) L(R) < α,(ϑ(δ)) Lα(R) = (ϑ(δ)) L(R) and (F(δ)) Lα(R) = (F(δ)) L(R) ,by the locality of the definition of F and the assumption that (F↾δ) Lα(R) =F↾δ.For (2) first notice that we can make sense of F as defined over levels(such as L δ˜21(R)) that do not satisfy T 0 by letting, for an arbitrary ordinal ξ,F L ξ(R) = ⋃ {F Lα(R) | α < ξ and L α (R) |= T 0 }.Thus, F L (R)δ˜21= F↾γ for some γ, by (1). Assume for contradiction that2(2) fails, that is, for some γ < ∼δ 1 , F(γ) is defined and yet F L (R)δ˜21(γ) is notdefined. Since in L(R), ϑ(γ) and F(γ) are defined, the following is a trueΣ 1 -statement about γ:∃α > γ (L α (R) |= T 0 + ϑ(γ) exists.)Since L δ˜21(R) ≺ 1 L(R), this statement holds in L δ˜21(R) and so F Lα(R) (γ) isdefined and hence F L δ˜21(R)(γ) is defined, which is a contradiction.For (3) assume for contradiction that γ < δ ∼21 , where γ = dom(F). By (2)(and the definition of F L δ˜21(R)) there is an α < δ∼21 such that L α (R) |= T 0 andF Lα(R) = F↾γ = F. We claim that this implies thatL α (R) ≺ R∪{R}1 L(R),which is a contradiction (by Theorem 2.28). SupposeL(R) |= ψ[z, R]where ψ is a Σ 1 -formula and z ∈ ω ω . We have to show that L α (R) |= ψ[z, R].By Replacement there is an ordinal β such thatL β (R) |= ψ[z, R].Consider the Σ 1 -statement ϕ[z, X, R] expressing “There exists ξ such thatX = L ξ (R) and X |= ψ[z, R]”. Letting ϑ > β be such that L ϑ (R) |= T 0 wehave: there exists an X ∈ L ϑ (R) ∩ OD L ϑ(R) (namely, X = L β (R)) such thatL ϑ (R) |= T 0 + ϕ[z, X, R].

4. A Woodin Cardinal in HOD L(R) 63Moreover, since ϑ(γ) does not exist, it follows (by the definition of ϑ(γ)) thatthere exists a ¯δ < γ such thatL ϑ (R) |= ϕ[z, F(¯δ), R].Thus (unpacking ϕ[z, X, R]) there exists a ξ such that F(¯δ) = L ξ (R) andL ξ (R) |= ψ[z, R]. Since F ⊆ L α (R), ξ < α and so, by upward absoluteness,which completes the proof.L α (R) |= ψ[z, R],It follows that F↾δ ∼21 : δ ∼21 → L δ˜21(R) is total and ∆ 1 -definable over L δ˜21(R).It remains to see that F(δ ∼21 ) is not defined.Theorem 4.4. Assume ZF + AC ω (R). For all X ∈ OD L(R) , for all Σ 1 -formulas ϕ, and for all z ∈ ω ω ifthen there exists a δ < δ ∼21 such thatL(R) |= ϕ[z, X, δ ∼21 , R]L(R) |= ϕ[z, F(δ), δ, R].Proof. The idea of the proof is straightforward but the details are somewhatinvolved.Assume for contradiction that there is an X ∈ OD L(R) , a Σ 1 -formula ϕ,and z ∈ ω ω such thatL(R) |= ϕ[z, X, δ ∼21 , R]and for all δ < δ ∼21 ,L(R) ̸|= ϕ[z, F(δ), δ, R].Step 1. By Replacement, let ϑ 0 > δ ∼21 be least such that(1.1) L ϑ0 (R) |= T 0 and there is an X ∈ L ϑ0 (R) ∩ OD L ϑ 0 (R) and(⋆) there is a Σ 1 -formula ϕ and a real z such thatL ϑ0 (R) |= ϕ[z, X, δ ∼21 , R]and for all δ < δ ∼21L ϑ0 (R) ̸|= ϕ[z, F(δ), δ, R].

4. A Woodin Cardinal in HOD L(R) 64Let X 0 be least (in the order of definability) such that (1.1) and for this choicepick ϕ 0 and z 0 such that (⋆). (Thus we have let ϑ 0 = ϑ(δ ∼21 ), X 0 = F(δ ∼21 ), andwe have picked witnesses ϕ 0 and z 0 to the failure of reflection with respectto F(δ ∼21 ).)Notice that L ϑ0 (R) |= δ ∼21 exists + F(δ) is defined for all δ < δ ∼21 . SinceF L ϑ 0 (R) ↾δ ∼21 = F↾δ ∼21 ,by Lemma 4.3, (1.1) is equivalent to the internal statement L ϑ0 (R) |= T 0 +“reflection fails with respect to F↾δ ∼21 ”. It is this internal statement that wewill reflect to get a contradiction. We have that for all δ < δ ∼21 ,(1.2) L ϑ0 (R) ̸|= ϕ 0 [z 0 , F(δ), δ, R].Our strategy is to reflect to get ¯ϑ < δ ∼21 such thatL¯ϑ(R) |= ϕ 0 [z 0 , F((δ ∼21 ) L¯ϑ (R) ), (δ ∼21 ) L¯ϑ (R) , R].By upward absoluteness, this will contradict (1.2). To implement this strategywe need the appropriate Σ 1 -fact (in a real) to reflect.Step 2. The following is a true Σ 1 -statement about ϕ 0 and z 0 (as witnessedby taking α to be ϑ 0 from Step 1): There is an α such that(2.1) L α (R) |= δ ∼21 exists + F(δ) is defined for all δ < δ ∼21 ,(2.2) L α (R) |= T 0 and there is an X ∈ L α (R) ∩ OD Lα(R) and(⋆) there is a Σ 1 -formula ϕ and a real z such thatand for all δ < (δ ∼21 ) Lα(R)L α (R) |= ϕ[z, X, (δ ∼21 ) Lα(R) , R]L α (R) ̸|= ϕ[z, F Lα(R) (δ), δ, R],(2.3) if β < α then it is not the case that L β (R) |= T 0 and there is anX ∈ L β (R) ∩ OD L β(R) and(⋆) there is a Σ 1 -formula ϕ and a real z such thatand for all δ < (δ ∼21 ) Lα(R)L β (R) |= ϕ[z, X, (δ ∼21 ) Lα(R) , R]L β (R) ̸|= ϕ[z, F Lα(R) (δ), δ, R],

4. A Woodin Cardinal in HOD L(R) 65and(2.4) if ¯X is least (in the order of definability) such that (2.2) thenand for all δ < (δ ∼21 ) Lα(R)L α (R) |= ϕ 0 [z 0 , ¯X, (δ ∼21 ) Lα(R) , R]L α (R) ̸|= ϕ 0 [z 0 , F Lα(R) (δ), δ, R].(Notice that in (2.3) the ordinal δ ∼21 and the function F are computed inL α (R) while the formulas are evaluated in L β (R).) Thus (2.1) ensures (byLemma 4.3) that F Lα(R) ↾(δ ∼21 ) Lα(R) = F↾(δ ∼21 ) Lα(R) , (2.2) says that L α (R) satisfies“reflection is failing with respect to F Lα(R) ↾(δ ∼21 ) Lα(R) ” and, because of(2.1), this ensures that ϑ((δ ∼21 ) Lα(R) ) exists, (2.3) ensures in addition thatα = ϑ((δ ∼21 ) Lα(R) ), and (2.4) says that ϕ 0 and z 0 (as chosen in Step 1) witnessthe existence of ϑ((δ ∼21 ) Lα(R) ).Since L δ˜21(R) ≺ R 1 L(R) and ϕ 0 and z 0 can be coded by a single real, theleast ordinal α witnessing the existential of the above statement must be lessthan δ ∼21 . Let ¯ϑ be this ordinal.Step 3. We claim thatL¯ϑ(R) |= ϕ 0 [z 0 , F((δ ∼21 ) L¯ϑ (R) ), (δ ∼21 ) L¯ϑ (R) , R],which finishes the proof since by upward absoluteness this contradicts (1.2).The ordinal ¯ϑ has the Σ1 -properties listed in (2.1)–(2.4) for α. So wehave: (4.1) L¯ϑ(R) |= “δ ∼21 exists”+“F(δ) is defined for all δ < δ ∼21 ” and so (byLemma 4.3) F L¯ϑ (R) ↾(δ ∼21 ) L¯ϑ (R) = F↾(δ ∼21 ) L¯ϑ (R) , (4.2) L¯ϑ(R) satisfies “reflectionis failing with respect to F L¯ϑ (R) ↾(δ ∼21 ) L¯ϑ (R) ” and, because of (4.1), this ensuresthat ϑ((δ ∼21 ) L¯ϑ (R) ) exists, (4.3) ¯ϑ = ϑ((δ ∼21 ) L¯ϑ (R) ), and (4.4) ϕ 0 and z 0 (as chosenin Step 1) witness the existence of ϑ((δ ∼21 ) L¯ϑ (R) ). Therefore, by the definitionof F, (4.4) implies thatwhich contradicts (1.2).L¯ϑ(R) |= ϕ 0 [z 0 , F((δ ∼21 ) L¯ϑ (R) ), (δ ∼21 ) L¯ϑ (R) , R],We will need a slight strengthening of the above theorem. This involvesthe notion of the reflection filter, which in turn involves various universalsets.

4. A Woodin Cardinal in HOD L(R) 662Let U X be a good universal Σ 1 (L(R), {X, ∼δ 1 , R}) set of reals. So U X is a2Σ 1 (L(R), {X, ∼δ 1 , R}) subset of ω ω × ω ω 2such that each Σ 1 (L(R), {X, ∼δ 1 , R ∪{R}}) subset of ω ω is of the form (U X ) t for some t ∈ ω ω 2. For each δ < ∼δ 1 ,let U δ be the universal Σ 1 (L(R), {F(δ), δ, R}) set obtained using the same2definition used for U X except with X and δ ∼1 replaced by the reflected proxiesF(δ) and δ. As before, we shall identify each of U X and U δ with a set of realsusing our recursive bijection between ω ω × ω ω and ω ω .For each Σ 1 -formula ϕ and for each real y, there exists a z ϕ,y ∈ ω ω suchthat2z ϕ,y ∈ U X iff L(R) |= ϕ[y, X, ∼δ 1 , R].In such a situation we say that z ϕ,y certifies the Σ 1 -fact ϕ about y. The keyproperty is, of course, that if z ϕ,y ∈ U δ then L(R) |= ϕ[y, F(δ), δ, R]. Noticethat the real z ϕ,y is recursive in y (uniformly).In what follows we will drop reference to ϕ and y and simply write z ∈ U X ,it being understood that the formula and parameter are encoded in z. Inthese terms Theorem 4.4 can be recast as stating that if z ∈ U X then thereis an ordinal δ < δ ∼21 such that z ∈ U δ , in other words, U X ⊆ ⋃ δ

4. A Woodin Cardinal in HOD L(R) 67and consider the following Σ 1 -statement:There is an α such thatL α (R) |= T 0 ,δ = (δ ∼21 ) Lα(R) , F↾δ = (F) Lα(R) , and F(δ) ∈ OD Lα(R) , andL α (R) |= ψ[y, F(δ), δ, R].If we replace δ by δ ∼21 and F(δ) by X then this statement is true. It followsthat the statement holds for F X -almost all δ. The second clause ensures thateach such δ is a “local δ ∼21 ” and that the “local computation of F up to δ”coincides with F. By altering ψ and y we can increase the degree to whichthe proxies F(δ), δ resemble X, δ ∼21 .Lemma 4.5. Assume ZF + AC ω (R). Then L(R) |= F X is a countablycomplete filter.Proof. Upward closure and the non-triviality condition are immediate. Itremains to prove countable completeness. Suppose {S n | n < ω} ⊆ F X . Forn < ω, let z n ∈ U X be such that S zn ⊆ S n . Let z ∈ ω ω be such that (z) n = z nfor all n < ω. The following is a true Σ 1 -statement about z, X, δ ∼21 , and R:There is an α such that(1) δ ∼21 < α,(2) L α (R) |= T 0 ,(3) X ∈ OD Lα(R) and(4) for all n < ω, (z) n ∈ (U X ) Lα(R) .Let z ∗ ∈ U X certify this statement. It follows that for each δ < δ ∼21 such thatz ∗ ∈ U δ the following holds:There is an α such that(1) δ < α,(2) L α (R) |= T 0 ,(3) F(δ) ∈ OD Lα(R) and(4) for all n < ω, (z) n ∈ (U δ ) Lα(R) .But then, by upward absoluteness, δ ∈ ⋂ {S zn | n < ω} and so S z ∗ ⊆ ⋂ {S zn |n < ω} ⊆ ⋂ {S n | n < ω}.

4. A Woodin Cardinal in HOD L(R) 68We shall call F X the reflection filter since, by definition, there are F X -many reflecting points in the Reflection Theorem.We wish now to extend the Reflection Theorem by allowing various parametersS ⊆ δ ∼21 and their “reflections” S ∩ δ. For this we bring in AD.Theorem 4.6 (Reflection Theorem). Assume ZF + AD. Suppose f :2 2 2δ∼ 1 → δ ∼1 and S ⊆ ∼δ 1 are in L(R). For all X ∈ OD L(R) , for all Σ 1 -formulasϕ, and for all z ∈ ω ω , ifthen for F X -many δ < δ ∼21 ,L(R) |= ϕ[z, X, f, S, δ ∼21 , R]L(R) |= ϕ[z, F(δ), f↾δ, S ∩ δ, δ, R],where here f and S occur as predicates.Proof. First we show that the theorem holds for S ⊆ δ ∼21 . For each δ < δ ∼21 , letThe sequenceQ δ = U δ ⋃ {U γ | γ < δ}.〈Q δ | δ < δ ∼21 〉gives rise to a prewellordering of length δ ∼21 . By the Uniform Coding Lemma,there is an e(S) ∈ ω ω such thatU (2)e(S) (Q

4. A Woodin Cardinal in HOD L(R) 69It follows that for δ ∈ S z , e(S) codes S ∩ δ.This enables us to associate with each Σ 1 -sentence ϕ involving the predicateS, a Σ 1 -sentence ϕ ∗ involving instead the real e(S) in such a way thatL(R) |= ϕ[z, X, δ ∼21 , S, R]if and only ifand, for δ ∈ S z ∈ F X ,L(R) |= ϕ ∗ [z, X, δ ∼21 , e(S), R]if and only ifL(R) |= ϕ[z, F(δ), δ, S ∩ δ, R]L(R) |= ϕ ∗ [z, F(δ), δ, e(S), R].In this fashion, the predicate S can be eliminated in favour of the real e(S),thereby reducing the present version of the reflection theorem to the originalversion (Theorem 4.4).To see that we can also include parameters of the form f : δ ∼21 → δ ∼21 simplynote that F X -almost all δ are closed under the Gödel pairing function andso we can include functions f : δ ∼21 → δ ∼21 by coding them as subsets of δ ∼21 .We are now in a position to define, for cofinally many λ < Θ, an ultrafilterµ λ on δ ∼21 . For the remainder of this section fix an ordinal λ < Θ and (bythe results of Section 3.3) an OD-prewellordering λ of ω ω of length λ. Ourinterest is in applying the Reflection Theorem toX = ( λ , λ).For each S ⊆ δ ∼21 , let G X (S) be the gameI x(0) x(1) x(2) . . .II y(0) y(1) . . .with the following winning conditions: Main Rule: For all i < ω, (x) i , (y) i ∈U X . If the rule is violated then, letting i be the least such that either (x) i ∉U X or (y) i ∉ U X , I wins if (x) i ∈ U X ; otherwise II wins. If the rule is satisfiedthen, letting δ be least such that for all i < ω, (x) i , (y) i ∈ U δ , (which existsby reflection since (as in Lemma 4.5) we can regard this as a Σ 1 -statementabout a single real) I wins iff δ ∈ S. Thus, I is picking δ by steering into theδ th -approximation U δ . (Note that the winning condition is not Σ 1 .)

4. A Woodin Cardinal in HOD L(R) 70Now setµ X = {S ⊆ δ ∼21 | I wins G X (S)}.We let µ λ = µ X but shall typically write µ X to emphasize the dependenceon the prewellorder. For z ∈ U X , Player I can win G X (S z ) by playing x suchthat (x) i ∈ U X for all i < ω and, for some i < ω, (x) i = z. Thus,F X ⊆ µ X .It is easy to see that µ X is upward closed and contains either S or δ ∼21 Sfor each S ⊆ δ ∼21 .Lemma 4.7. Assume ZF+AD. Then L(R) |= µ X is a δ ∼21 -complete ultrafilter.Proof. The proof is similar to the proof of Theorem 2.33 (which traces backto the proof of Theorem 2.13). Consider {S α | α < γ} where S α ∈ µ X andγ < δ ∼21 . Let S = ⋂ α

4. A Woodin Cardinal in HOD L(R) 71(1.2) ∀y ∈ ω ω ∀σ ∈ Σ ((σ∗y) I ) 0 ∈ U Xsince σ ′ and σ (as in (1.2)) are winning strategies for I. Since Σ is ∆ ∼21 this is aΣ 1 (L(R), {X, δ ∼21 , R} ∪ R) fact about σ ′ and hence certified by a real z 0 ∈ U Xsuch that z 0 T σ ′ ; more precisely, z 0 T σ ′ is such that for all δ if z 0 ∈ U δthen(1.3) ∀y ∈ ω ω ((σ ′ ∗y) I ) 0 ∈ U δ and(1.4) ∀y ∈ ω ω ∀σ ∈ Σ ((σ∗y) I ) 0 ∈ U δ .(n + 1) st Step. Assume we have defined z 0 , . . .,z n in such a way thatz n T · · · T z 0 and(2.1) ∀y ∈ ω ω ( ∀i n (y) i = z i → ((σ ′ ∗y) I ) n+1 ∈ U X)and(2.2) ∀y ∈ ω ω ∀σ ∈ Σ ( ∀i n (y) i = z i → ((σ∗y) I ) n+1 ∈ U X).Let z n+1 ∈ U X be such that z n+1 T z n and for all δ, if z n+1 ∈ U δ then(2.3) ∀y ∈ ω ω ( ∀i n (y) i = z i → ((σ ′ ∗y) I ) n+1 ∈ U δ)and(2.4) ∀y ∈ ω ω ∀σ ∈ Σ ( ∀i n (y) i = z i → ((σ∗y) I ) n+1 ∈ U δ).Finally, let z ∈ ω ω be such that (z) i = z i for all i < ω and let δ 0 be leastsuch that (z) i ∈ U δ0 for all i ∈ ω. Notice that by our choice of z n no DC isrequired to define z. Then, for all i < ω,So(3.1) ((σ ′ ∗z) I ) i ∈ U δ0 by (1.3) and (2.3) and(3.2) ((σ∗z) I ) i ∈ U δ0 for all σ ∈ Σ by (1.4) and (2.4).(4.1) δ 0 is the ordinal produced by σ ′ ∗z, i.e. δ 0 ∈ δ ∼21 S and(4.2) δ 0 is the ordinal produced by σ α ∗z where σ α ∈ Σ is a winning strategyfor I in G X (S α ), i.e. δ 0 ∈ S α for all α < γ.This is a contradiction.

4. A Woodin Cardinal in HOD L(R) 724.2. Strong NormalityAssuming ZF + AD, in L(R) we have defined, for cofinally many λ < Θ, anOD L(R) ultrafilter on δ ∼21 and shown that these ultrafilters are δ ∼21 -complete. Wenow wish to take the ultrapower of HOD L(R) with these ultrafilters and showthat collectively they witness that for each λ < Θ, δ ∼21 is λ-strong in HOD L(R) .This will be achieved by showing that reflection and uniform coding combineto show that µ λ is strongly normal.We begin with the following basic lemma on the ultrapower construction,which we shall prove in greater generality than we need at the moment.Lemma 4.8. Assume ZF+DC. Suppose µ is a countably complete ultrafilteron δ and that µ is OD. Suppose T is a set. Let (HOD T ) δ be the class ofall functions f : δ → HOD T . Then the transitive collapse M of (HOD T ) δ /µexists, the associated embeddingj : HOD T → Mis OD T , andM ⊆ HOD T .Proof. For f, g : δ → HOD T , let f ∼ µ g iff {α < δ | f(α) = g(α)} ∈ µ andlet [f] µ be the set consisting of the members of the equivalence class of fwhich have minimal rank. The structure (HOD T ) δ /µ is the class consistingof all such equivalence classes. Let E be the associated membership relation.So [f] µ E [g] µ if and only if {α < δ | f(α) ∈ g(α)} ∈ µ. Notice that both(HOD T ) δ /µ and E are OD T .The mapj µ : HOD T → (HOD T ) δ /µa ↦→ [c a ] µ ,where c a ∈ (HOD T ) δ is the constant function with value a, is an elementaryembedding, since ̷Loś’s theorem holds, as HOD T can be well-ordered. Noticethat j µ is OD T .Claim 1. ((HOD T ) δ /µ, E) is well-founded.Proof. Suppose for contradiction that ((HOD T ) δ /µ, E) is not well-founded.Then, by DC, there is a sequence〈[fn ] µ∣ ∣ n < ω 〉

4. A Woodin Cardinal in HOD L(R) 73such that [f n+1 ] µ E [f n ] µ for all n < ω. For each n < ω, letA n = {α < δ | f n+1 (α) ∈ f n (α)}.For all n < ω, A n ∈ µ and since µ is countable complete,⋂ {An | n < ω} ∈ µ.This is a contradiction since for each α in this intersection, f n+1 (α) ∈ f n (α)for all n < ω.Claim 2. ((HOD T ) δ /µ, E) is isomorphic to a transitive class (M, ∈).Proof. We have established well-foundedness and extensionality is immediate.It remains to show that for each a ∈ (HOD T ) δ /µ,{b ∈ (HOD T ) δ /µ | b E a}is a set. Fix a ∈ (HOD T ) δ /µ and choose f ∈ (HOD T ) δ such that a = [f] µ .Let α be such that f ∈ V α . Then for each b ∈ (HOD T ) δ /µ such that bEa,letting g ∈ (HOD T ) δ be such that b = [g] µ ,Thus,{β < δ | g(β) ∈ V α } ∈ µ.{b ∈ (HOD T ) δ /µ | b E a} = {[g] µ | [g] µ E [f] µ and g ∈ V α },which completes the proof.Letπ : ((HOD T ) δ /µ, E) → (M, ∈)be the transitive collapse map and letj : HOD T → Mbe the composition map π ◦ j µ . Since π and j µ are OD T , j and M are OD T .It remains to see that M ⊆ HOD T . For this it suffices to show that forall α, M ∩ V j(α) ⊆ HOD T . We haveM ∩ V j(α) = j(HOD T ∩ V α ).

4. A Woodin Cardinal in HOD L(R) 74Let A ∈ HOD T ∩ P(γ) be such thatfor some γ. We haveHOD T ∩ V α ⊆ L[A]M ∩ V j(α) = j(HOD T ∩ V α ) ⊆ L[j(A)].But j and A are OD T . Thus, j(A) ∈ HOD T and hence L[j(A)] ⊆ HOD T ,which completes the proof.4.9 Remark. The use of DC in this lemma is essential in that assumingmild large cardinal axioms (such as the existence of a strong cardinal) thereare models of ZF + AC ω in which the lemma is false. In these models theclub filter on ω 1 is an ultrafilter and the ultrapower of On by the club filteris not well-founded.The ultrafilter µ X defined in Section 4.1 is OD L(R) . Thus, by Lemma 4.8(with T = ∅), lettingπ : (HOD L(R) ) δ˜21 /µX → M Xbe the transitive collapse map and lettingj X : HOD L(R) → M Xbe the induced elementary embedding we have that M X ⊆ HOD L(R) and thefragments of j X are in HOD L(R) (in other words, j X is amenable to HOD L(R) ).Moreover, since µ X is δ ∼21 -complete, the critical point of j X is δ ∼21 .Our next aim is to show thatand for this it remains to show thatHOD L(R) |= δ ∼21 is λ-strongj X (δ ∼21 ) > λ and HOD L(R) ∩ V λ ⊆ M X .From now on we will also assume that λ is such that L λ (R) ≺ L Θ (R) and2δ∼ 1 < λ. There are arbitrarily large λ < Θ with this feature (by the proof ofLemma 2.20). SinceHOD L(R) ∩ V Θ = HOD L Θ(R)

4. A Woodin Cardinal in HOD L(R) 75(by Theorem 3.10), it follows thatHOD L(R) ∩ V λ = HOD L λ(R) .Thus, letting A ⊆ λ be an OD L(R) set coding HOD L λ(R) , we haveHOD L(R) ∩ V λ = L λ [A].Thus, it remains to show that A ∈ M X . In fact, we will show thatLetP(λ) ∩ HOD L(R) ⊆ M X .2S 0 = {δ < ∼δ 1 |F(δ) = ( δ , λ δ ) where δ is aprewellordering of length λ δ and L λδ (R) |= T 0 }.Note that S 0 ∈ F X . For α < λ, let Q δ˜21α be the α th -component of λ and,for δ ∈ S 0 and α < λ δ , let Q δ α be the αth -component of δ . Each t ∈ ω ωdetermines a canonical function f t as follows: For δ ∈ S 0 , let αt δ be the uniqueordinal α such that t ∈ Q δ α and then setf t : S 0 → δ ∼21δ ↦→ α δ t .f t δ1 δ2 λα δ1tα δ2tα tS 0For t ∈ ω ω , let α t = |t| λα t = |t| λ = µα (t ∈ Q δ˜21α ).be the rank of t according to λ , that is,Lemma 4.10. Assume ZF + AD. j X (δ ∼21 ) > λ.

4. A Woodin Cardinal in HOD L(R) 76Proof. Suppose t 1 , t 2 ∈ ω ω and |t 1 | λ = |t 2 | λ . This is a true Σ 1 -statement inL(R) about t 1 , t 2 , X and R. Thus, by reflection (Theorem 4.6), it follows thatfor F X -almost all δ < δ ∼21 , |t 1 | δ = |t 2 | δ and so the ordinal [f t ] µX representedby f t only depends on |t| λ . Likewise, if |t 1 | λ < |t 2 | λ then [f t1 ] µX < [f t2 ] µX .Therefore, the mapρ : λ → ∏ λ δ /µ X|t| λ ↦→ [f t ] µXis well-defined and order-preserving and it follows that λ ∏ λ δ /µ X

4. A Woodin Cardinal in HOD L(R) 77µ X -almost all δ, |t| δ ∈ A δ , that is, {δ < δ ∼21 | f t (δ) ∈ A δ } ∈ µ X . Likewise,if |t| λ ∉ A then since this is a true Σ 1 -statement about e(A), t and X, forµ X -almost all δ, |t| δ ∉ A δ .So it suffices to show that the mapρ : λ → ∏ λ δ /µ X|t| λ ↦→ [f t ] µXis an isomorphism since then π([h A ] µX ) = A ∈ M X , where recall that π :(HOD L(R) ) δ˜21 /µX ∼ = MX is the transitive collapse map. We already knowthat ρ is well-defined and order-preserving (by Lemma 4.10). It remains toshow that ρ is onto, that is, that every function f ∈ ∏ λ δ /µ X is equivalent(modulo µ X ) to a canonical function f t . To say that this is true is to saythat µ X is strongly normal:4.11 Definition (Strong normality). µ X is strongly normal iff wheneverf : S 0 → δ ∼21 is such thatthen there exists a t ∈ ω ω such that{δ ∈ S 0 | f(δ) < λ δ } ∈ µ X{δ ∈ S 0 | f(δ) = f t (δ)} ∈ µ X .Notice that normality is a special case of strong normality since if{δ < δ ∼21 | f(δ) < δ} ∈ µ Xthen (since for F X -almost all δ, λ δ > δ), by strong normality there is at ∈ ω ω such that{δ < δ ∼21 | f t (δ) = f(δ)} ∈ µ X .So if β is such that t ∈ Q δ˜21βthen β < δ 2∼1, since otherwise by reflection thiswould contradict the assumption that{δ < δ ∼21 | f(δ) < δ} ∈ µ X .Thus,{δ < δ ∼21 | f(δ) = β} ∈ µ X .Theorem 4.12. Assume ZF + AD. L(R) |= µ X is strongly normal.

4. A Woodin Cardinal in HOD L(R) 78Proof. Assume toward a contradiction that f is a counterexample to strongnormality. So, for each t ∈ ω ω ,{δ ∈ S 0 | f(δ) ≠ f t (δ)} ∈ µ X .Letη = min { β < λ | ∀t ∈ Q δ˜21β {δ ∈ S 0 | f(δ) < f t (δ)} ∈ µ X}if such β exist; otherwise, let η = λ. Fix y η ∈ Q δ˜21η (unless η = λ, in which2case we ignore this parameter) and, for δ ∈ S 0 , let η δ = f yη (δ) and for δ = δ ∼1 ,let η δ = η. Note that f yη (δ) > f(δ) for µ X -almost all δ. In the proof we willbe working on this set and so we modify S 0 by intersecting it with this set ifnecessary. For convenience letS(t) = {δ ∈ S 0 | f t (δ) < f(δ)}.Notice that by the definition of η and our assumption that f is a counterexampleto strong normality, we have thatfor all t ∈ Q δ˜21

4. A Woodin Cardinal in HOD L(R) 79e 0 is the index of the universal set that selects the Zα δ ’s (represented in thediagrams as ellipses) from Z ′ (represented in the diagrams as chimneys).e 1 is the index of the universal set that selects subsets of the Zα δ ’s (representedin the diagrams as black dots inside the ellipses).y η is the real in Q δ˜21η that determines η δ for δ ∈ S 0 .y f is the real in Q δ˜21“f(δ˜2 1 )” that determines f(δ) for δ ∈ S 3.We will successively shrink S 0 to S 1 , S 2 , and finally S 3 . All four of these setswill be members of µ X . We now proceed with the proof.LetZ ′ = { (t, σ) | t ∈ Q δ˜21

4. A Woodin Cardinal in HOD L(R) 80Claim A (Disjointness Property). There is an S 2 ⊆ S 1 , S 2 ∈ µ X suchthat for δ 1 , δ 2 ∈ S 2 ∪ {δ ∼21 } with δ 1 < δ 2 δ ∼21 ,for all α ∈ [f(δ 1 ), η δ1 ) and β ∈ [0, η δ2 ).Proof. Here is the picture:Z δ 1α ∩ Zδ 2β = ∅η δ1. δ1η δ2. δ2η λf(δ 1 )Z δ1αZ δ2βWe begin by establishing a special case.Subclaim. For µ X -almost all δ,S 2Z δ α ∩ Zδ˜21β= ∅for all α ∈ [f(δ), η δ ) and β ∈ [0, η).Proof. The picture is similar:η δf(δ). δZ δ α λ.Z δ˜21βηS 2LetT = { δ ∈ S 1 | Z δ α ∩ Z δ˜21β = ∅ for all α ∈ [f(δ), η δ) and β ∈ [0, η) }

4. A Woodin Cardinal in HOD L(R) 812and assume, toward a contradiction, that T ∉ µ X . So (δ ∼1 T) ∩ S 1 ∈ µ X .Let σ ′ be a winning strategy for I in G X 2((δ ∼1 T) ∩ S 1 ).Let us first motivate the main idea: Suppose z is a legal play for II againstσ ′ (by which we mean a play for II that satisfies the Main Rule) and suppose2that the ordinal associated with this play is δ 0 . So δ 0 ∈ (δ ∼1 T) ∩ S 1 and(by the definition of T) there exists an α 0 ∈ [f(δ 0 ), η δ ) and β 0 ∈ [0, η) suchthat Z δ 0α 0∩ Z δ˜21β 0 ≠ ∅. Pick (t 0 , σ 0 ) ∈ Z δ 0α 0∩ Z δ˜21β 0. In virtue of the fact that(t 0 , σ 0 ) ∈ Z δ 0α 0we have(3.1) f t0 (δ 0 ) = α 0 f(δ 0 )and in virtue of the fact that (t 0 , σ 0 ) ∈ Z δ˜21β 0we have(3.2) σ 0 is a winning strategy for I in G X (S(t 0 )), whereS(t 0 ) = {δ ∈ S 0 | f t0 (δ) < f(δ)}.So we get a contradiction if δ 0 happens to be in S(t 0 ) (since then f t0 (δ 0 )

4. A Woodin Cardinal in HOD L(R) 82(5.1) ∀y ∈ ω ( ) ω ∀i n (y) i = z i → ((σ ′ ∗y) I ) n+1 ∈ U X and( )(5.2) ∀y ∈ ω ω ∀(t, σ) ∈ Z δ˜21∀i n (y)i = z i → ((σ∗y) I ) n+1 ∈ U X .Again, all of this is a Σ 1 (L(R), {X, δ ∼21 , R})-fact about σ ′ , e 0 , z 0 , . . ., z n and soit is certified by a real z n+1 ∈ U X such that z n+1 T z n ; so z n+1 is such thatif z n+1 ∈ U δ then(5.3) ∀y ∈ ω ω ( ∀i n (y) i = z i → ((σ ′ ∗y) I ) n+1 ∈ U δ)and(5.4) ∀y ∈ ω ω ∀(t, σ) ∈ Z δ ( ∀i n (y) i = z i → ((σ∗y) I ) n+1 ∈ U δ).Finally, let z ∈ ω ω be such that (z) i = z i for all i < ω and let δ 0 be least suchthat (z) i ∈ U δ0 for all i ∈ ω. Notice that since we chose z n+1 to be recursivein z n no DC is required to form z. Since (z) i ∈ U X for all i ∈ ω, z is a legalplay for II in any of the games G X (S) relevant to the argument. Moreover,for all i ∈ ω,(6.1) ((σ ′ ∗z) I ) i ∈ U δ0 by (4.3) and (5.3) and(6.2) ((σ∗z) I ) i ∈ U δ0 for all σ ∈ proj 2 (Z δ 0) by (4.4) and (5.4)and so(7.1) δ 0 is the ordinal produced by σ ′ ∗z, i.e. δ 0 ∈ (δ ∼21 T) ∩ S 1 and(7.2) δ 0 is the ordinal produced by σ∗z for any σ ∈ proj 2 (Z δ 0).2Since δ 0 ∈ (δ ∼1 T)∩S 1 , by the definition of T there exists an α 0 ∈ [f(δ 0 ), η δ0 )and β 0 ∈ [0, η) such that Z δ 0α 0∩ Z δ˜21β 0 ≠ ∅. Pick (t 0 , σ 0 ) ∈ Z δ 0α 0∩ Z δ˜21β 0. In virtueof the fact that (t 0 , σ 0 ) ∈ Z δ 0α 0we have(8.1) f t0 (δ 0 ) = α 0 f(δ 0 )and in virtue of the fact that (t 0 , σ 0 ) ∈ Z δ˜21β 0we have(8.2) σ 0 is a winning strategy for I in G X (S(t 0 )), whereS(t 0 ) = {δ ∈ S 0 | f t0 (δ) < f(δ)}.Combined with (7.2) this implies δ 0 ∈ S(t 0 ), in other words, f t0 (δ 0 ) < f(δ 0 ),which contradicts (8.1).

4. A Woodin Cardinal in HOD L(R) 83Thus, T ∈ µ X and we have(9.1) ∀δ ∈ T ∀β ∈ [0, η δ˜21) ∀α ∈ [f(δ), η δ ) (Z δ α ∩ Z δ˜21β= ∅).This is a true Σ 1 -statement in L(R) about e 0 , y η , f, X, R, δ ∼21 , and T. SinceT is Σ 1 (L(R), {e 0 , y η , f, X, δ ∼21 , R}), the above statement is Σ 1 (L(R), {e 0 , y η ,f, X, δ ∼21 , R}). Thus by the Reflection Theorem (Theorem 4.6) there existsS 2 ⊆ S 1 , S 2 ∈ µ X such that for all δ 2 ∈ S 2 ,(9.2) ∀δ 1 ∈ T ∩ δ 2 ∀β ∈ [0, η δ2 ) ∀α ∈ [f(δ 1 ), η δ1 ) (Z δ 1α ∩ Zδ 2β = ∅).Notice that S 2 is Σ 1 (L(R), {X, δ ∼21 , R}) in e 0 , y η and the parameters forcoding. This completes the proof of Claim A.Claim B (Tail Computation). There exists an index e 1 ∈ ω ω such thatfor all δ ∈ S 2 ,(1) U (2)e 1 (P δ , Z δ β ) ⊆ Zδ β for all β < η δ,(2) U e (2)1 (P δ , Zf(δ) δ ) = ∅, and(3) U (2)e 1(P δ , Z δ β ) ≠ ∅ for β such that f(δ) < β < η δ,where P δ = ⋃ {Z¯δα | ¯δ ∈ S 2 ∩ δ and α ∈ [f(¯δ), η¯δ)} and S 2 is from the end ofthe proof of Claim A.Proof. Here is the picture of the “tail parameter” P δ : δf(δ)η¯δ.¯δη δ.f(¯δ)Here is the picture of the statement of Claim B:S 2

4. A Woodin Cardinal in HOD L(R) 84P δη δS 2.. δU(2) e 1(P δ , Z δ β )f(δ)Assume toward a contradiction that there is no such e 1 . We follow theproof of the Uniform Coding Lemma. To begin with, notice that it sufficesto find e 1 ∈ ω ω satisfying (2) and(3 ′ ) U (2)e 1(P δ , Z δ β ) ∩ Zδ β ≠ ∅ for β such that f(δ) < β < η δsince given the parameter Zβ δ we can easily ensure (1).Consider the set of reals such that (2) of the (revised) claim holds, thatis,G = { (e ∈ ω ω | ∀δ ∈ S 2 U(2)e (P δ , Zf(δ) δ ) = ∅)} .So, for each e ∈ G, (3 ′ ) in the claim fails for some δ ∈ S 2 and β ∈ (f(δ), η δ ).For each e ∈ G, letNow play the gameα e = lexicographically least pair (δ, β) such that(1) δ ∈ S 2 ,(2) f(δ) < β < η δ , and(3) U (2)e (P δ , Z δ β) ∩ Z δ β = ∅.I x(0) x(1) x(2) . . .II y(0) y(1) . . .where II wins iff (x ∈ G → (y ∈ G ∧ α y > lex α x ))Claim 1. Player I does not have a winning strategy.Proof. Suppose toward a contradiction that σ is a winning strategy for I. Asin the proof of the Uniform Coding Lemma, we aim to “bound” all of I’s

4. A Woodin Cardinal in HOD L(R) 85plays and then use this bound to construct a play e ∗ for II which defeats σ.We will make key use of the Disjointness Property.Choose e σ ∈ ω ω such that for all P, P ′ ⊆ ω ω ,U (2)e σ(P, P ′ ) = ⋃ y∈ω ω U (2)(σ∗y) I(P, P ′ ).In particular, for all δ ∈ S 2 and β < η δ ,U (2)e σ(P δ , Z δ β ) = ⋃ y∈ω ω U (2)(σ∗y) I(P δ , Z δ β ).Note two things: First, since σ is a winning strategy for I, (σ∗y) I ∈ G for ally ∈ ω ω ; so e σ ∈ G. Second, for all y ∈ ω ω , α (σ∗y)I lex α eσ . So e σ is “at leastas good” as any (σ∗y) I . We have to do “better”.Pick x 0 ∈ Z δ 0β 0where (δ 0 , β 0 ) = α eσ . Choose e ∗ such that for all P, P ′ ⊆ ω ω ,U (2)e ∗ (P, P ′ ) ={U e (2)σ (P, P ′ ) if x 0 ∉ P ′U e (2)σ (P, P ′ ) ∪ {x 0 } if x 0 ∈ P ′ .In particular, for all δ ∈ S 2 and β < η δ ,{U (2)e (P δ , Z δ ∗ β ) = U e (2)σ (P δ , Zβ δ)if x 0 ∉ ZβδU e (2)σ (P δ , Zβ δ) ∪ {x 0} if x 0 ∈ Zβ δ.Since we chose x 0 ∈ Z δ 0β 0, by the Disjointness Property (and the fact that forfixed δ, Z δ α ∩ Zδ β = ∅ for α < β < η δ) we have(10.1) x 0 ∉ Z δ β for δ ∈ S 2 ∩ [0, δ 0 ) and β ∈ [f(δ), η δ ),(10.2) x 0 ∉ Z δ 0βfor β ∈ [0, η δ 0) {β 0 }, and(10.3) x 0 ∉ Z δ β for δ ∈ S 2 ∩ (δ 0 , δ ∼21 ) and β ∈ [0, η δ ).Thus, by the definition of e ∗ , we have, by (10.1–3),U (2)e ∗ (P δ , Z δ f(δ) ) = U(2) e σ(P δ , Z δ f(δ) )for all δ ∈ S 2 . Since e σ ∈ G, this means e ∗ ∈ G. So α e ∗the definition of e ∗ , we have, by (10.1) and (10.2),exists. Similarly, byU (2)e ∗ (P δ , Z δ β ) = U(2) e σ(P δ , Z δ β )

4. A Woodin Cardinal in HOD L(R) 86for all δ ∈ S 2 ∩ [0, δ 0 ) and β ∈ [f(δ), η δ ) and for δ = δ 0 and β ∈ [f(δ 0 ), β 0 ).So e ∗ is “at least as good” as e σ . But since x 0 ∈ Z δ 0β 0, we have that x 0 ∈U (2)e (P δ 0, Z δ ∗ 0β 0), by the definition of e ∗ ; that is, e ∗ is “better” than e σ . In otherwords, α e ∗ > lex α eσ lex α (σ∗y)I for all y ∈ ω ω and so, by playing e ∗ , II defeatsσ.Claim 2. Player II does not have a winning strategy.Proof. Suppose toward a contradiction that τ is a winning strategy for II.We shall find an e ∗ such that e ∗ ∈ G (Subclaim 1) and α e ∗ does not exist(Subclaim 2), which is a contradiction.Choose h 0 : ω ω × (ω ω × ω ω ) → ω ω such that h 0 is Σ 1 1 and for all (e, x) ∈ω ω × (ω ω × ω ω ) and for all P, P ′ ⊆ ω ω ,{U (2)h 0 (e,x) (P, P ′ U e (2) (P, P ′ ) if x ∉ P ∪ P ′) =∅ if x ∈ P ∪ P ′ .In particular, for δ ∈ S 2 and β < η δ ,{U (2)h 0 (e,x) (P δ , Zβ δ ) = U e(2) (P δ , Zβ δ) if x ∉ P δ ∪ Zβδ∅ if x ∈ P δ ∪ Zβ δ.Choose h 1 : ω ω → ω ω such that h 1 is Σ 1 1 and for all P, P ′ ⊆ ω ω ,U (2)h 1 (e) (P, P ′ ) = ⋃ x∈P ′ U (2)(h 0 (e,x)∗τ) II(P, P ′ ).In particular, for δ ∈ S 2 and β < η δ ,U (2)h 1 (e) (P δ , Zβ) δ = ⋃ x∈ZU (2)βδ (h 0 (e,x)∗τ) II(P δ , Zβ δ).By the recursion theorem, there is an e ∗ ∈ ω ω such that for all δ ∈ S 2 andβ < η δ ,U (2)e (P δ , Z δ ∗ β ) = U(2) h 1 (e ∗ ) (P δ , Zβ δ ).Subclaim 1. e ∗ ∈ G.Proof. Suppose for contradiction that e ∗ ∉ G. Let δ 0 ∈ S 2 be least such thatU (2)e ∗ (P δ 0, Z δ 0f(δ 0 ) ) ≠ ∅.

4. A Woodin Cardinal in HOD L(R) 87NowSo choose x 0 ∈ Z δ 0f(δ 0 )U (2)e ∗ (P δ 0, Z δ 0f(δ 0 ) ) = U(2) h 1 (e ∗ ) (P δ 0, Z δ 0f(δ 0 ) )such that= ⋃ x∈Z δ 0 U (2)(hf(δ 0 ) 0 (e ∗ ,x)∗τ) II(P δ 0, Z δ 0f(δ 0 ) ).U (2)(h 0 (e ∗ ,x 0 )∗τ) II(P δ 0, Z δ 0f(δ 0 ) ) ≠ ∅.If we can show h 0 (e ∗ , x 0 ) ∈ G then we are done since this implies that(h 0 (e ∗ , x 0 )∗τ) II ∈ G (as τ is a winning strategy for II), which contradicts theprevious statement.Subsubclaim. h 0 (e ∗ , x 0 ) ∈ G, that is, for all δ ∈ S 2 ,U (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ f(δ) ) = ∅.Proof. By the definition of h 0 , for all δ ∈ S 2 ,{U (2)h 0 (e ∗ ,x 0 ) (P δ , Zf(δ)) δ U (2)e=(P δ , Z δ ∗ f(δ) ) if x 0 ∉ P δ ∪ Zf(δ)δ∅ if x 0 ∈ P δ ∪ Zf(δ) δ .Since x 0 ∈ Z δ 0f(δ 0 ), by the Disjointness Property, this definition yields thefollowing: For δ ∈ S 2 ∩ [0, δ 0 ) we have x 0 ∉ P δ ∪ Zf(δ) δ and so,U (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ f(δ) ) = U(2) e ∗ (P δ , Z δ f(δ) ) = ∅,where the latter holds since we chose δ 0 to be least such thatU (2)e ∗ (P δ 0, Z δ 0f(δ 0 ) ) ≠ ∅;for δ = δ 0 we have x 0 ∈ Z δ f(δ)and soU (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ f(δ)) = ∅;and for δ ∈ S 2 ∩ (δ 0 , δ ∼21 ) we have x 0 ∈ P δ and soThus, h 0 (e ∗ , x 0 ) ∈ G.U (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ f(δ) ) = ∅.

4. A Woodin Cardinal in HOD L(R) 88This completes the proof of Subclaim 1.Subclaim 2. α e ∗does not exist.Proof. Suppose for contradiction that α e ∗exists. Recall thatα e ∗ = lexicographically least pair (δ, β) such that(1) δ ∈ S 2 ,(2) f(δ) < β < η δ , and(3) U (2)e ∗ (P δ , Z δ β ) ∩ Zδ β = ∅.Let (δ 0 , β 0 ) = α e ∗. We shall show U (2)e ∗ (P δ 0, Z δ 0β 0) ∩ Z δ 0β 0 ≠ ∅, which is acontradiction. By the definition of h 1 ,U (2)e ∗ (P δ 0, Z δ 0β 0) = U (2)h 1 (e ∗ ) (P δ 0, Z δ 0β 0)= ⋃ x∈Z δ 0 U (2)(hβ 0 (e ∗ ,x)∗τ) II(P δ 0, Z δ 0β 0).0Fix x 0 ∈ Z δ 0β 0. Since e ∗ ∈ G, h 0 (e ∗ , x 0 ) ∈ G, by the Disjointness Property.(This is because for δ ∈ S 2 ∩ [0, δ 0 ) we have x 0 ∉ P δ ∪ Zf(δ) δ and soU (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ f(δ)) = U (2)e ∗ (P δ , Z δ f(δ)) = ∅,where the latter holds since e ∗ ∈ G; for δ = δ 0 we have x 0 ∉ P δ ∪ Z δ f(δ) andsince e ∗ ∈ G this impliesU (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ f(δ) ) = ∅;and for δ ∈ S 2 ∩ (δ 0 , δ ∼21 ) we have x 0 ∈ P δ and soSo α h0 (e ∗ ,x 0 ) exists.Subsubclaim. α h0 (e ∗ ,x 0 ) = α e ∗.Proof. By the definition of h 0 ,U (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ β) =U (2)h 0 (e ∗ ,x 0 ) (P δ , Z δ f(δ) ) = ∅.){U (2)e ∗ (P δ , Z δ β ) if x 0 ∉ P δ ∪ Z δ β∅ if x 0 ∈ P δ ∪ Z δ β

4. A Woodin Cardinal in HOD L(R) 89for δ ∈ S 2 and β < η δ . SoU (2)h 0 (e ∗ ,x 0 ) (P δ 0, Z δ 0β 0) ∩ Z δ 0β 0= ∅,since x 0 ∈ P δ 0∪ Z δ 0β 0. And, when either δ = δ 0 and β ∈ (f(δ 0 ), β 0 ) orδ ∈ S 2 ∩ [0, δ 0 ) and β ∈ [f(δ), η δ ), we have, by the Disjointness Property,x 0 ∉ P δ ∪ Z δ β , hence U (2)Thus, α h0 (e ∗ ,x 0 ) = α e ∗.h 0 (e ∗ ,x 0 ) (P δ , Z δ β ) = U(2) e ∗ (P δ , Z δ β ),Since τ is winning for II,andSoSincewe havewhich is a contradiction.(h 0 (e ∗ , x 0 )∗τ) II ∈ Gα (h0 (e ∗ ,x 0 )∗τ) II> lex α h0 (e ∗ ,x 0 ) = α e ∗.U (2)(h 0 (e ∗ ,x 0 )∗τ) II(P δ 0, Z δ 0β 0) ∩ Z δ 0β 0 ≠ ∅.U (2)e (P δ 0, Z δ ∗ 0β 0) = ⋃ x∈Z δ 0 U (2)(hβ 0 (e ∗ ,x)∗τ) II(P δ 0, Z δ 0β 0)0This completes the proof of Claim 2.U (2)e ∗ (P δ 0, Z δ 0β 0) ∩ Z δ 0β 0 ≠ ∅,We have a contradiction and therefore there is an e 1 is as desired.Notice that U e (2)1 (P δ , Zα δ ), for variable α, allows us to pick out f(δ).2Now we can consider the ordinal “f(δ ∼1 )” picked out in this fashion.Claim C. There exists a β 0 < η such that(1) U (2)e 1 (P δ˜21 , Zδ˜21β 0) = ∅ and(2) U (2)e 1 (P δ˜21 , Zδ˜21β ) ≠ ∅ for all β ∈ (β 0, η), whereP δ˜21 =⋃ {Zδα | δ ∈ S 2 and α ∈ [f(δ), η δ )}.

4. A Woodin Cardinal in HOD L(R) 90Proof. Suppose for contradiction that the claim is false. The statement thatthe claim fails is a true Σ 1 -statement about e 0 , e 1 , y η , X, R, f and S 2 . Butthen by the Reflection Theorem (Theorem 4.6) this fact reflects to F X -almostall δ, which contradicts Claim B.Pick y f ∈ Q δ˜21β 0. Now the statement that y f ∈ Q δ˜21β 0where β 0 is such that(1) and (2) of Claim C hold is a true Σ 1 -statement about e 0 , e 1 , y η , y f , f, X,2 2R, and δ ∼1 . Thus, by Theorem 4.6, for F X -almost every δ < ∼δ 1 this statementreflects. Let S 3 ⊆ S 2 be in µ X and such that the above statement reflectsto each point in S 3 . Now by Claim B, for δ ∈ S 3 , the least β 0 such thaty f ∈ Q δ β 0is f(δ). Thus,and hence µ X is strongly normal.{δ ∈ S 0 | f yf (δ) = f(δ)} ∈ µ XThis completes the proof of the following:Theorem 4.13. Assume ZF + DC + AD. Then, for each λ < Θ L(R) ,HOD L(R) |= ZFC + (δ ∼21 ) L(R) is λ-strong.4.14 Remark. For simplicity we proved Lemma 4.7 and Claim A of Theorem4.12 using a proof by contradiction. This involves an appeal to determinacy.However, one can prove each result more directly, without appealing to determinacy.Call a real y suitable if (y) i ∈ U X for all i < ω. Call a strategy σ aproto-winning strategy if σ is a winning strategy for I in G X (δ ∼21 ). Thus if yis suitable and σ is a proto-winning strategy thenand so we can let{((σ∗y) I ) i , (y) i | i < ω} ⊆ U Xδ (σ,y) = the least δ such that {((σ∗y) I ) i | i < ω} ∪ {(y) i | i < ω} ⊆ U δ .Let κ be least such that X ∈ L κ (R) and L κ (R) ≺ 1 L(R). This is the “leaststable over X”. It is easy to see thatP(R) ∩ L κ (R) = ∆ 1 (L(R), R ∪ {X, δ ∼21 , R})

4. A Woodin Cardinal in HOD L(R) 91and so if Σ ∈ P(R) ∩ L κ (R) then for F X -almost all δ there is a reflectedversion Σ δ of Σ. We can now state the relevant result:Suppose Σ ∈ P(R) ∩ L κ (R) is a set of proto-winning strategies for I.Then there is a proto-winning strategy σ such that for all suitable reals y,for all τ ∈ Σ δ(σ,y) ∩ Σ, there is a suitable real y τ such thatδ (σ,y) = δ (τ,yτ).The proof of this is a variant of the above proofs and it provides a moredirect proof of completeness and strong normality.4.3. A Woodin CardinalWe now wish to show that Θ L(R) is a Woodin cardinal in HOD L(R) . Ingeneral, in inner model theory there is a long march up from strong cardinalsto Woodin cardinals. However, in our present context, where we have thepower of AD and are working with the special inner model HOD L(R) , thisnext step comes almost for free.Theorem 4.15. Assume ZF + DC + AD. ThenHOD L(R) |= Θ L(R) is a Woodin cardinal.Proof. For notational convenience let Θ = Θ L(R) . To show thatHOD L(R) |= ZFC + Θ is a Woodin cardinalit suffices to show that for each T ∈ P(Θ) ∩ OD L(R) , there is an ordinal δ Tsuch that|= ZFC + δ T is λ-T-strong,HOD L Θ(R)[T]Tfor each λ < Θ. Since Θ is strongly inaccessible in HOD L(R) , HOD L(R)satisfies that VΘHODL(R) is a model of second-order ZFC. Thus, since T ∈P(Θ) ∩ HOD L(R) and VΘ HODL(R) = HOD LΘ(R) ,HOD L Θ(R)[T]T|= ZFC.It remains to establish strength. Since this is almost exactly as before wewill just note the basic changes.

4. A Woodin Cardinal in HOD L(R) 92The model L Θ (R)[T] comes with a natural Σ 1 -stratification, namely,〈Lα (R)[T ∩ α] ∣ ∣ α < Θ〉.Since Θ is regular in L(R) and L Θ (R) |= T 0 , the set{α < Θ | Lα (R)[T ∩ α] ≺ L Θ (R)[T] }contains a club in Θ. To see this is note that for each n < ω,C n = { α < Θ | L α (R)[T ∩ α] ≺ n L Θ (R)[T] }is club (by Replacement) and, since Θ is regular, ⋂ {C n | n < ω} is club.Thus, there are arbitrarily large α < Θ such thatL α (R)[T ∩ α] |= T 0 .For this reason OD T , < ODT and HOD T are Σ 1 -definable in L Θ (R)[T] exactlyas before. (Here, as usual, we are working in the language of set theorysupplemented with a predicate for T, which is assumed to be allowed in allof our definability calculations.)Letδ T = the least λ such that L λ (R)[T ∩ λ] ≺ 1 L Θ (R)[T].As will be evident, the relevant facts concerning δ ∼21 carry over to the presentcontext. For example, δ T is the least ordinal λ such thatL λ (R)[T ∩ λ] ≺ R∪{R}1 L Θ (R)[T].The function F T : δ T → L δT (R)[T ∩ δ T ] is defined as before as follows:Work in T 0 . Suppose that F T ↾δ is defined. Let ϑ(δ) be least such thatL ϑ(δ) (R)[T ∩ ϑ(δ)] |= T 0 and there is an X ∈ L ϑ(δ) (R)[T ∩ ϑ(δ)] ∩OD L ϑ(δ)(R)[T ∩ϑ(δ)]Tand(⋆) there is a Σ 1 -formula ϕ and a real z such thatand for all ¯δ < δ,L ϑ(δ) (R)[T ∩ ϑ(δ)] |= ϕ[z, X, δ T , T ∩ ϑ(δ), R]L ϑ(δ) (R)[T ∩ ϑ(δ)] ̸|= ϕ[z, F(¯δ), ¯δ, T ∩ ϑ(δ), R]

4. A Woodin Cardinal in HOD L(R) 93(if such an ordinal exists) and then let F T (δ) be the (< ODT ) L ϑ(δ)(R)[T ∩ϑ(δ)] -leastX such that (⋆).The proof of the Reflection Theorem carries over exactly as before toestablish the following: For all X ∈ OD L Θ(R)[T]T, for all Σ 1 -formulas ϕ, andfor all z ∈ ω ω ifL Θ (R)[T] |= ϕ[z, X, δ T , T, R]then there exists a δ < δ T such thatL Θ (R)[T] |= ϕ[z, F T (δ), δ, T ∩ δ, R].Let U T X be a universal Σ 1(L Θ (R)[T], {X, δ T , T, R}) set of reals and, forδ < δ T , let U T δ be the universal Σ 1(L Θ (R)[T], {F T (δ), δ, T ∩δ, R}) set obtainedby using the same definition. For z ∈ U T X , let ST z = {δ < δ T | z ∈ U T δ } andsetF T X = {S ⊆ δ T | ∃z ∈ U T X (ST z ⊆ S)}.As before, FX T is a countably complete filter and in the Reflection Theoremwe can reflect to FX T-many points δ < δ T and allow parameters A ⊆ δ T andf : δ T → δ T .Fix an ordinal λ < Θ. By the results of Section 3.3, there is an OD L Θ(R)[T]T-prewellordering λ of ω ω of length λ. Our interest is in applying the ReflectionTheorem toX = ( λ , λ).Working in L Θ (R)[T], for each S ⊆ δ T , let G X T (S) be the gameI x(0) x(1) x(2) . . .II y(0) y(1) . . .with the following winning conditions: Main Rule: For all i < ω, (x) i , (y) i ∈UX T. If the rule is violated then, letting i be the least such that either (x) i ∉UX T or (y) i ∉ UX T I wins if (x) i ∈ UX T ; otherwise II wins. If the rule is satisfiedthen, letting δ be least such that for all i < ω, (x) i , (y) i ∈ Uδ T , I wins iffδ ∈ S.Now setµ T X = {S ⊆ δ T | I wins G X T (S)}.Notice that µ T X ∈ ODL Θ(R)[T]T. As before F T X ⊆ µT X and µT X is a δ T-completeultrafilter.

4. A Woodin Cardinal in HOD L(R) 94LetS 0 = {δ < δ T | F T (δ) = ( δ , λ δ ) where δ is aprewellordering of length λ δ and L λδ (R)[T ∩ λ δ ] |= T 0 }.By reflection, S 0 ∈ F T X .As before we say that µ T X is strongly normal iff whenever f : S 0 → δ T issuch that{δ ∈ S 0 | f(δ) < λ δ } ∈ µ T Xthen there exists a t ∈ ω ω such that{δ ∈ S 0 | f(δ) = f t (δ)} ∈ µ T X .The proof that µ T X is strongly normal is exactly as before. As in theproof of Lemma 4.8 we can use µ T X to take the ultrapower of HODL Θ(R)[T] .In L Θ (R)[T] form(L HODΘ (R)[T]) δT/T µTX .As before we get an elementary embeddingj λ : HOD L Θ(R)[T]T→ M,where M is the transitive collapse of the ultrapower. By completeness, thisembedding has critical point δ T and as in Lemma 4.10 the canonical functionswitness that j λ (δ T ) > λ. Assuming further that λ is such thatwe have thatL λ (R)[T ∩ λ] ≺ 1 L Θ (R)[T]HOD L Θ(R)[T]T⊆ M λ .As before, strong normality implies thatρ : λ → ∏ λ δ /µ T X|t| λ ↦→ [f t ] µ TXis an isomorphism. It remains to establish T-strength, that is,|t| λ ∈ T ∩ λ iff {δ < δ T | f t (δ) ∈ T ∩ λ δ } ∈ µ T X.The point is that both|t| λ ∈ T ∩ λ

5. Woodin Cardinals in General Settings 95and|t| λ ∉ T ∩ λare Σ 1 (L Θ (R)[T], {X, δ T , T, R}) and so the result follows by the ReflectionTheorem (Theorem 4.6) and the fact that FX T ⊆ µT X .Thus,HOD L Θ(R)[T]Twhich completes the proof.|= ZFC + δ T is λ-T-strong,In the above proof DC was only used in one place—to show that the ultrapowerswere well-founded (Lemma 4.8). This was necessary since althoughthe ultrapowers were ultrapowers of HOD and HOD satisfies AC, the ultrapowerswere “external” (in that the associated ultrafilters were not in HOD)and so we had to assume DC in V to establish well-foundedness. However,this use of DC can be eliminated by using the extender formulation of beinga Woodin cardinal. In this way one obtains strength through a network of“internal” ultrapowers (that is, via ultrafilters that live in HOD) and thisenables one to bypass the need to assume DC in V . We will take this routein the next section.5. Woodin Cardinals in General SettingsOur aim in this section is to abstract the essential ingredients from the previousconstruction and prove two abstract theorems on Woodin cardinals ingeneral settings, one that requires DC and one that does not.The first abstract theorem will be the subject of Section 5.1:Theorem 5.1. Assume ZF + DC + AD. Suppose X and Y are sets. LetΘ X,Y = sup{α | there is an OD X,Y surjection π : ω ω → α}.ThenHOD X |= ZFC + Θ X,Y is a Woodin cardinal.There is a variant of this theorem (which we will prove in Section 5.4) whereone can drop DC and assume less determinacy, the result being that Θ X isa Woodin cardinal in HOD X . The importance of the version involving Θ X,Yis that it enables one to show that in certain settings HOD X can have manyWoodin cardinals. To describe one such key application we introduce the

5. Woodin Cardinals in General Settings 96following notion due to Solovay. Assume ZF + DC R + AD + V = L(P(R))and work in V = L(P(R)). The sequence 〈Θ α | α Ω〉 is defined to be theshortest sequence such that Θ 0 is the supremum of all ordinals γ for whichthere is an OD surjection of ω ω onto γ, Θ α+1 is the supremum of all ordinalsγ for which there is an OD surjection of P(Θ α ) onto γ, Θ λ = sup α

5. Woodin Cardinals in General Settings 97ordinals. Since (OD S,x ) L[S,x] = L[S, x] and L[S, x] satisfies AC one cannothave boldface determinacy in L[S, x]. However, by assuming full determinacyin the background universe, strong forms of lightface determinacy holdin L[S, x], for an S-cone of x. (The notion of an S-cone will be defined inSection 5.2). We will extract stronger and stronger forms of lightface determinacyuntil ultimately we reach the notion of “strategic determinacy”,which is sufficiently rich to simulate boldface determinacy and drive the construction.With this motivation in place we will return to the general settingin Section 5.3 and prove the Generation Theorem. Finally, in Section 5.4 wewill use the Generation Theorem as a template reprove the theorem of theprevious section in ZF +AD and to deduce a number of special cases, two ofwhich are worth mentioning here:Theorem 5.4. Assume ZF + AD. Then for an S-cone of x,HOD L[S,x]S|= ZFC + ω L[S,x]2 is a Woodin cardinal.Theorem 5.5. Assume ZF + AD. Suppose Y is a set and a ∈ H(ω 1 ). Thenfor a Y -cone of x,HOD Y,a,[x]Y |= ZFC + ω HOD Y,a,x2 is a Woodin cardinal,where [x] Y = {z ∈ ω ω | HOD Y,z = HOD Y,x }.(The notion of a Y -cone will be defined in Section 5.4). In Section 6 these tworesults will be used as the basis of a calibration of the consistency strengthof lightface and boldface definable determinacy in terms of the large cardinalhierarchy. The second result will also be used to reprove and generalizeKechris’ classical result that ZF + AD implies that DC holds in L(R). Forthis reason it is important to note that the theorem does not presuppose DC.5.1. First AbstractionTheorem 5.6. Assume ZF + DC + AD. Suppose X and Y are sets. ThenProof. By Theorem 3.9,HOD X |= ZFC + Θ X,Y is a Woodin cardinal.HOD X,Y |= Θ X,Y is strongly inaccessible

5. Woodin Cardinals in General Settings 98and soHOD X |= Θ X,Y is strongly inaccessible.A direct approach to showing that in additionHOD X |= Θ X,Y is a Woodin cardinalwould be to follow Section 4.3 by showing that for each T ∈ P(Θ X,Y )∩OD Xthere is an ordinal δ T such thatHOD X ∩ V ΘX,Y |= δ T is λ-T-strongfor each λ < Θ X,Y . However, such an approach requires that for eachλ < Θ X,Y , there is a prewellordering of ω ω of length λ which is OD inL ΘX,Y (R)[T] and in our present, more general setting we have no guaranteethat this is true. So our strategy is to work with a larger model (where suchprewellorderings exist), get the ultrafilters we need, and then pull them backdown to L ΘX,Y (R)[T] by Kunen’s theorem (Theorem 3.11).We will actually first show thatHOD X,Y |= Θ X,Y is a Woodin cardinal.Let T be an element of P(Θ X,Y ) ∩ OD X,Y and let (by Lemma 3.7)A = 〈A α | α < Θ X,Y 〉be an OD X,Y sequence such that each A α is a prewellordering of ω ω of lengthα. We will work with the structureL ΘX,Y (R)[T, A]and the natural hierarchy of structures that it provides.To begin with we note some basic facts. First, notice that(For the first equivalence we havebecause of A and we haveΘ X,Y = (Θ T,A ) L(R)[T,A] = Θ L(R)[T][A] .(Θ T,A ) L(R)[T,A] Θ X,Y(Θ T,A ) L(R)[T,A] Θ X,Y

5. Woodin Cardinals in General Settings 99because L(R)[T, A] is OD X,Y . The second equivalence holds since every elementin L(R)[T, A] is OD L(R)[T][A]T,A,yfor some y ∈ ω ω . So the “averaging overreals” argument of Lemma 3.8 applies.) It follows that our earlier argumentsgeneralize. For example, by the proof of Theorem 3.10,andΘ X,Y is strongly inaccessible in HOD L(R)[T,A]Θ X,Y is regular in L(R)[T, A].(Note that Θ X,Y need not be regular in V . For example, assuming ZF +DC + AD R , Θ 0 has cofinality ω in V .) Moreover, the proof of Lemma 2.21shows thatL ΘX,Y (R)[T, A] |= T 0and the proof of Lemma 2.23 shows thatL ΘX,Y (R)[T, A] ≺ 1 L(R)[T, A].This implies (in conjunction with the fact that Θ X,Y is regular in L(R)[T, A])that{α < Θ X,Y | L α (R)[T↾α, A↾α] ≺ L ΘX,Y (R)[T, A]}is club in Θ X,Y and hence that each such level satisfies T 0 .So we are in exactly the situation of Section 4.3 except that now theprewellorderings are explicitly part of the structure. The proof of Theorem4.15 thus shows that: For each T ∈ P(Θ X,Y ) ∩ OD X,Y there is an ordinalδ T,A such thatHOD L Θ X,Y(R)[T,A]T,A|= δ T,A is λ-T-strongfor each λ < Θ X,Y , as witnessed by an ultrafilter µ T λ on δ T,A. These ultrafiltersare OD L Θ X,Y(R)[T,A]T,A.The key point is that all of these ultrafilters µ T λ are actually OD byKunen’s theorem (Theorem 3.11). This is where DC is used.Now we return to the smaller model L ΘX,Y (R)[T]. Since Θ X,Y is stronglyinaccessible in HOD X there is a set H ∈ P(Θ X,Y ) L Θ (R)[T] X,Ysuch thatHOD X ∩ V ΘX,Y = L ΘX,Y [H].We may assume without loss of generality that H is folded into T. ThusHOD L Θ X,Y(R)[T]T= HOD X ∩ V ΘX,Y

5. Woodin Cardinals in General Settings 100and this structure contains all of the ultrafilters µ T λ . These ultrafilters cannow be used (as in the proof of Lemma 4.8) to take the ultrapower and sowe haveHOD L Θ X,Y(R)[T]T|= δ T,A is λ-T-strong,which completes the proof.5.2. Strategic DeterminacyLet us now turn to the Generation Theorem. We shall begin by motivatingthe notion of “strategic determinacy” by examining the special case of L[S, x]where S is a class of ordinals.For x ∈ ω ω , the S-degree of x is [x] S = {y ∈ ω ω | L[S, y] = L[S, x]}. TheS-degrees are the sets of the form [x] S for some x ∈ ω ω . Let D S = {[x] S |x ∈ ω ω }. Define x S y to hold iff x ∈ L[S, y] and define the notions x ≡ S y,x < S y, x S y, [x] S S [y] S in the obvious way. A cone of S-degrees is a setof the form {[y] S | y S x 0 } for some x 0 ∈ ω ω . An S-cone of reals is a setof form {y ∈ ω ω | y S x 0 } for some x 0 ∈ ω ω . The cone filter on D S is thefilter consisting of sets of S-degrees that contain a cone of S-degrees. Givena formula ϕ(x) we say that ϕ holds for an S-cone of x if there is a real x 0such that for all y S x 0 , L[S, y] |= ϕ(y). The proof of the Cone Theorem(Theorem 2.9) generalizes.Theorem 5.7 (Martin). Assume ZF + AD. The cone filter on D S is anultrafilter.Proof. For A ⊆ D S consider the gameI x(0) x(1) x(2) . . .II y(0) y(1) . . .where I wins iff [x∗y] S ∈ A. If I has a winning strategy σ 0 then σ 0 witnessesthat A is in the S-cone filter since for y S σ 0 , [y] S = [σ 0 ∗y] S ∈ A. If II hasa winning strategy τ 0 then τ 0 witnesses that D S A is in the S-cone filtersince for x S τ 0 , [x] S = [x∗τ 0 ] S ∈ D S A.It follows that each statement ϕ either holds on an S-cone or fails on an S-cone. In fact, the entire theory stabilizes. However, in order to fully articulatethis fact one needs to invoke second-order assumptions (like the existence ofa satisfaction relation). Without invoking second-order assumptions one hasthe following:

5. Woodin Cardinals in General Settings 101Corollary 5.8. Assume ZF +AD. For each n < ω, there is an x n such thatfor all x S x n ,L[S, x] |= ϕ iff L[S, x n ] |= ϕ,for all Σ 1 n-sentences ϕ.Proof. Let 〈ϕ i | i < ω〉 enumerate the Σ 1 n-sentences of the language of settheory and, for each i, let y i be the base of an S-cone settling ϕ i . Now usingAC ω (R) (which is provable in ZF + AD) let x n encode 〈y i | i < ω〉.A natural question then is: “What is the stable theory?”Theorem 5.9. Assume ZF + AD. Then for an S-cone of x,L[S, x] |= CH.Proof. Suppose for contradiction (by Theorem 5.7) that ¬CH holds for anS-cone of x. Let x 0 be the base of this cone.We will arrive at a contradiction by producing an x S x 0 such thatL[S, x] |= CH. This will be done by forcing over L[S, x 0 ] in two stages, firstto get CH and then to get a real coding this generic (while preserving CH).It will be crucial that the generics actually exist.Claim. ω V 1is strongly inaccessible in any inner model M of AC.Proof. We first claim that there is no ω1 V -sequence of distinct reals: Letµ be the club filter on ω1 V . By Solovay’s theorem (Theorem 2.12, whichdoesn’t require DC) µ is a countably complete ultrafilter on ω1 V . Suppose〈a α | α < ω1 V 〉 is a sequence of characteristic functions for distinct reals. Bycountable completeness there is a µ-measure one set X n of elements of thissequence that agree on their n th -coordinate. Thus, ⋂ n

5. Woodin Cardinals in General Settings 102The trouble is that L[S, x 0 ][G] is not of the form L[S, x] for x ∈ R. (Wecould code G via a real by brute force but doing so might destroy CH. Amore delicate approach is needed.)Step 2. Code G using almost disjoint forcing: First, view G as a subset ofω L[S,x 0]1 by letting A ⊆ ω L[S,x 0]1 be such thatNow letL[S, x 0 ][G] = L[S, x 0 , A].〈σ α | α < ω L[S,x 0]1 〉 ∈ L[S, x 0 ]be a sequence of infinite almost disjoint subsets of ω (that is, such that ifα ≠ β then σ α ∩σ β is finite). By almost disjoint forcing, in L[S, x 0 , A] there isa c.c.c. forcing P A of size ω L[S,x 0,A]1 such that if H ⊆ P A is L[S, x 0 , A]-genericthen there is a c(A) ⊆ ω such thatα ∈ A iff c(A) ∩ σ α is infinite.(See [1, pp.267-8] for details concerning this forcing notion.) AlsoFinally,L[S, x 0 , A][H] = L[S, x 0 , A][c(A)] = L[S, x 0 , c(A)].L[S, x 0 , c(A)] |= CHas P A is c.c.c., |P A | = ω L[S,x 0,A]1 , and L[S, x 0 , A] |= CH, and so there are, upto equivalence, only ω L[S,x 0,A]1 -many names for reals.Corollary 5.10. Assume ZF + AD. For an S-cone of x,L[S, x] |= GCH below ω V 1 .Proof. Let x 0 be such that for all x S x 0 ,L[S, x] |= CH.Fix x S x 0 . We claim that L[S, x] |= GCH below ω V 1 : Suppose for contradictionthat there is a λ < ω V 1 such that L[S, x] |= 2 λ > λ + . LetG ⊆ Col(ω, λ) be L[S, x]-generic. Thus L[S, x][G] |= ¬CH. But L[S, x][G] =L[S, y] for some real y and so L[S, x][G] |= CH.

5. Woodin Cardinals in General Settings 103A similar proof shows that ♦ holds for an S-cone of x, the point beingthat adding a Cohen subset of ω 1 forces ♦ and this forcing is c.c.c. and ofsize ω 1 . See [1], Exercises 15.23 and 15.24.5.11 Conjecture. Assume ZF + AD. For an S-cone of x,L[S, x] ∩ V ω V1is an “L-like” model in that it satisfies Condensation, □, Morasses, etc.Corollary 5.10 tells us that for an S-cone of x,Θ L[S,x] = (c + ) L[S,x] = ω L[S,x]2 .Thus, to prove that for an S-cone of x,L[S, x] |= ω 2 is a Woodin cardinal in HOD S ,we can apply our previous construction concerning Θ provided we haveenough determinacy in L[S, x].Theorem 5.12 (Kechris and Solovay). Assume ZF + AD. For an S-cone ofx,L[S, x] |= OD S -determinacy.Proof. Play the following gameIIIa, bc, dwhere, letting p = 〈a, b, c, d〉, I wins if L[S, p] ̸|= OD S -determinacy andL[S, p] |= “a∗d ∈ A p ”, where A p is the least (in the canonical ordering)undetermined OD L[S,p]Sset in L[S, p]. In such a game the reals are playedso as to be “interleaved” in the pattern (a(0), c(0), b(0), d(0), . . .). Here thetwo players are to be thought of as cooperating to determine the playingfield L[S, p] in which they will simultaneously play (via a and d) an auxiliaryround of the game on the least undetermined OD S set A p (assuming, ofcourse, that such a set exists, as I is trying to ensure.)Case 1: I has a winning strategy σ 0 .We claim that for all x S σ 0 , L[S, x] |= OD S -determinacy, which contradictsthe assumption that σ 0 is a winning strategy for I. For consider such

5. Woodin Cardinals in General Settings 104a real x and suppose for contradiction that L[S, x] ̸|= OD S -determinacy. Asabove let A x ∈ OD L[S,x]Sbe least such that A x is not determined. We willarrive at a contradiction by deriving a winning strategy σ for I in A x fromthe strategy σ 0 . Run the game according to σ 0 while having Player II feedin x for c and playing some auxiliary play d ∈ L[S, x]. This ensures that theresulting model L[S, p] that the two players jointly determine is just L[S, x]and so A p = A x . We can now derive a winning strategy σ for I in A x fromσ 0 as follows: For d ∈ L[S, x], let σ be the strategy such that (σ∗d) I is the asuch that (σ 0 ∗〈x, d〉) I = 〈a, b〉.(It is crucial that we have II play c = x and d ∈ L[S, x] since otherwise wewould get a∗d ∈ A p for varying p. By having II play c = x and d ∈ L[S, x], IIhas “steered into the right model”, namely L[S, x], and we have “fixed” theset A x . This issue will become central later on when we refine this proof.)Case 2: II has a winning strategy τ 0 .We claim that for x S τ 0 , L[S, x] |= OD S -determinacy. This is as aboveexcept that now we run the game according to τ 0 , having I steer into L[S, x]by playing x for b and some a ∈ L[S, x]. Then, as above, we derive a winningstrategy for I in ω ω A x and hence a winning strategy τ for II in A x .To drive the construction of a model containing a Woodin cardinal weneed more than OD S -determinacy since some of the games in the constructionare definable in a real parameter. Unfortunately, we cannot hope to getL[S, x] |= OD S,y -determinacyfor each y since (OD S,x ) L[S,x] = L[S, x] and L[S, x] is a model of AC. Nevertheless,it is possible to have OD S,y -determinacy in L[S, x] for certain speciallychosen reals y. There is therefore hope of approximating a sufficientamount of boldface definable determinacy to drive the construction. To makeprecise the approximation we need, we introduce the notion of a “prestrategy”.Let A and B be sets of reals. A prestrategy for I (respectively II ) in Ais a continuous function f such that for all x ∈ ω ω , f(x) is a strategy for I(respectively II) in A. A prestrategy f in A (for either I or II) is winningwith respect to the basis B if, in addition, for all x ∈ B, f(x) is a winningstrategy in A. The strategic game with respect to the predicates P 1 , . . .,P k

5. Woodin Cardinals in General Settings 105and the basis B is the game SG B P 0 ,...,P kwhere we requireI A 0 · · · A n · · ·II f 0 · · · f n · · ·(1) A 0 ∈ P(ω ω ) ∩ OD P0 ,...,P k, A n+1 ∈ P(ω ω ) ∩ OD P0 ,...,P k ,f 0 ,...,f nand(2) f n is a prestrategy for A n that is winning with respect to B,and II wins iff II can play all ω rounds. We say that strategic definabledeterminacy holds with respect to the predicates P 0 , . . ., P k and the basis B(ST B P 0 ,...,P k-determinacy) if II wins SG B P 0 ,...,P kand we say that strategic definabledeterminacy for n moves holds with respect to the predicates P 0 , . . .,P kand the basis B (ST B P 0 ,...,P k-determinacy for n moves) if II can play n roundsof SG B P 0 ,...,P k. When these parameters are clear from context we shall oftensimply refer to SG and ST-determinacy.In the context of L[S, x] the predicate will be S and the basis B will bethe S-degree of x. Thus to say that L[S, x] satisfies ST B S -determinacy (orST-determinacy for short) is to say that II can play all rounds of the gamewhere we requireI A 0 · · · A n · · ·II f 0 · · · f n · · ·(1) A 0 ∈ P(ω ω ) ∩ OD L[S,x]S, A n+1 ∈ P(ω ω ) ∩ OD L[S,x]S,f 0 ,...,f n, and(2) f n ∈ L[S, x] is a prestrategy for A n that is winning with respect to [x] S .The ability to survive a single round of this game implies that L[S, x] satisfiesOD S -determinacy. So this notion is indeed a generalization of OD S -determinacy.Before turning to the main theorems, some remarks are in order. First,notice that the games ST B P 0 ,...,P kare closed for Player II, hence determined.The only issue is whether II wins.Second, notice also that if I wins then I has a canonical strategy. Thiscan be seen as follows: Player I can rank partial plays, assigning rank 0 topartial plays in which he wins; Player I can then play by reducing rank. Theresult is a quasi-strategy that is definable in terms of the tree of partial plays

5. Woodin Cardinals in General Settings 106which in turn is ordinal definable. Since I is essentially playing ordinals thisquasi-strategy can be converted into a strategy in a definable fashion. Wetake this to be I’s canonical strategy.Third, notice that each prestrategy can be coded by a real number in acanonical manner. We assume that such a coding has been fixed and, fornotational convenience, we will identify a prestrategy with its code.Fourth, it is important to note that if II is to have a hope of winning thenwe must allow II to play prestrategies and not strategies. To see this, work inL[S, x] and consider the variant of SG B S where we have II play strategies τ 0,τ 1 , . . . instead of prestrategies. The set A 0 = {y ∈ ω ω | L[S, y even ] = L[S, x]}is OD L[S,x]Sand hence a legitimate first move for I. But then II’s response mustbe a winning strategy for I in A 0 since I can win a play of A 0 by playing x.However, OD L[S,x]S,τ 0= L[S, x] and so in the next round I is allowed to playany A 1 ∈ L[S, x]. But then II cannot hope to always respond with a winningstrategy since L[S, x] ̸|= AD. The upshot is that if II is to have a hope ofwinning a game of this form then we must allow II to be less committal.Fifth, although one can use a base B which is slightly larger than [x] S ,the previous example motivates the choice of B = [x] S . Let A 0 be as in theprevious paragraph and let f 0 be II’s response. By the above argument, itfollows that for all z ∈ B, 〈f 0 , z〉 ∈ [x] S and so in a sense we are “one stepaway” from showing that one must have B ⊆ [x] S .Finally, as we shall show in the next section, for every OD basis B ⊆ ω ωthere is an OD set A ⊆ ω ω such that there is no OD prestrategy which iswinning for A with respect to B (Theorem 6.11). Thus, for each basis B,ST B -determinacy does not trivially reduce to OD-determinacy.Theorem 5.13. Assume ZF + AD. Then for an S-cone of x, for each n,where B = [x] S .L[S, x] |= ST B S-determinacy for n moves,Theorem 5.14. Assume ZF + DC R + AD. Then for an S-cone of x,where B = [x] S .L[S, x] |= ST B S -determinacy,Proofs of theorems 5.13 and 5.14. Assume toward a contradiction that thestatement of Theorem 5.14 is false. By Theorem 5.7, there is a real x 0 suchthat if x T x 0 ,L[S, x] |= I wins SG,

5. Woodin Cardinals in General Settings 107(where here and below we drop reference to S and B since these are fixedthroughout). For x T x 0 , let σ x be I’s canonical winning strategy in SG L[S,x] .Note that the strategy depends only on the model, that is, if y ≡ S x thenσ y = σ x .Our aim is to construct a sequence of games G 0 , G 1 , . . .,G n , . . . such thatthe winning strategies (for whichever player wins) enable us to define, foran S-cone of x, prestrategies f x 0 , f x 1 , . . .,f x n, . . . which constitute a non-losingplay against σ x in SG L[S,x] .Step 0. Consider (in V ) the game G 0IIIa, bc, dwhere, letting p = 〈a, b, c, d, x 0 〉 and A p 0 = σ p (∅), I wins iff a∗d ∈ A p 0. Noticethat by including x 0 in p we have ensured that σ p is defined and hence thatthe winning condition makes sense. In this game I and II are cooperating tosteer into the model L[S, p] and they are simultaneously playing (via a andd) an auxiliary round of the game A p 0 , where Ap 0 is I’s first move accordingto the canonical strategy in the strategic game SG L[S,p] . I wins a round iff Iwins the auxiliary round of this auxiliary game.Claim 1. There is a real x 1 such that for all x S x 1 there is a prestrategyf x 0 that is a non-losing first move for II against σ x in SG L[S,x] .Proof. Case 1: I has a winning strategy σ 0 in G 0 .For x T σ 0 , let f0 x be the prestrategy derived from σ 0 by extractingthe response in the auxiliary game where we have II feed in y for c, thatis, for y ∈ (ω ω ) L[S,x] let f0 x (y) be such that f0 x (y)∗d = a∗d where a is suchthat (σ 0 ∗〈y, d〉) I = 〈a, b〉. Note that f0 x ∈ L[S, x] as it is definable fromσ 0 . Let x 1 = 〈σ 0 , x 0 〉 and for x S x 1 let A x 0 = σ x (∅). We claim thatfor x S x 1 , f0 x is a prestrategy for I in Ax 0 that is winning with respect to{y ∈ ω ω | L[S, y] = L[S, x]}, that is, f0 x is a non-losing first move for II againstσ x in SG L[S,x] . To see this fix x S x 1 and y such that L[S, y] = L[S, x] andconsider d ∈ L[S, x]. The value f0 x (y) of the prestrategy was defined byrunning G 0 , having II feed in y for c:IIIa, by, d

5. Woodin Cardinals in General Settings 108By our choice of y and d, we have solved the “steering problem”, that is, wehave L[S, p] = L[S, x] and A p 0 = A x 0 where p = 〈a, b, y, d, x 0 〉. Now, f x 0 is suchthat f x 0 (y)∗d = a∗d where a is such that (σ 0∗〈y, d〉) I = 〈a, b〉. Since σ 0 iswinning for I, we have f x 0 (y)∗d = a∗d ∈ Ap 0 = Ax 0 .Case 2: II has a winning strategy τ 0 in G 0 .Let f0 x be the prestrategy derived from τ 0 by extracting the response inthe auxiliary game where we have I feed in y for b, that is, for y ∈ (ω ω ) L[S,x] letf0 x(y) be such that a∗fx 0 (y) = a∗d where d is such that (〈a, y〉∗τ 0) II = 〈c, d〉.Let x 1 = 〈τ 0 , x 0 〉 and for x S x 1 let A x 0 = σ x (∅). As before, we have thatfor x S x 1 , f0 x is a prestrategy for II in Ax 0 that is winning with respectto {y ∈ ω ω | L[S, y] = L[S, x]}, that is, f0 x is a non-losing first move for IIagainst σ x in SG L[S,x] .Let x 1 be as described in whichever case holds.Step n+1. Assume that we have defined games G 0 , . . .,G n , reals x 0 , . . .,x n+1 such that x 0 S x 1 S · · · S x n+1 , and prestrategies f x 0 , . . .,fx n whichdepend only on the degree of x and such that for all x S x n+1 ,f x 0 , . . .,fx nis a non-losing partial play for II against σ x in SG L[S,x] .Consider (in V ) the game G n+1IIIa, bc, dwhere, letting p = 〈a, b, c, d, x n+1 〉 and A p n+1 be I’s response via σ p to thepartial play f p 0 , . . .,fp n , I wins iff a∗d ∈ Ap n+1 . Notice that we have includedx n+1 in p to ensure that σ p , f0, p . . ., fn p are defined and hence that the winningcondition makes sense. In this game I and II are cooperating to steer into themodel L[S, p] and they are simultaneously playing an auxiliary round (via aand d) on A p n+1, where A p n+1 is I’s response via σ p to II’s non-losing partialplay f p 0 , . . .,fp n in the strategic game SGL[S,p] . I wins a round iff he wins theauxiliary round of this auxiliary game.Claim 2. There is a real x n+2 such that for all x S x n+2 there is a prestrategyfn+1 x such that fx 0 , . . ., fx n , fx n+1 is a non-losing partial play for II againstσ x in SG L[S,x] .

5. Woodin Cardinals in General Settings 109Proof. Case 1: I has a winning strategy σ n+1 in G n+1 .Let fn+1 x be the prestrategy derived from σ n+1 by extracting the responsein the auxiliary game, that is, for y ∈ (ω ω ) L[S,x] let fn+1 x (y) be such thatfn+1 x (y)∗d = a∗d where a is such that (σ n+1∗〈y, d〉) I = 〈a, b〉. Let x n+2 =〈σ n+1 , x n+1 〉 and for x S x n+2 let A x n+1 = σ x (〈f0 x , . . .,fn〉), x i.e. A x n+1 is the(n+2) nd move of I in SG L[S,x] following σ x against II’s play of f0 x, . . .,fx n . Asin Claim 1, fn+1 x is a prestrategy for I in A x n+1 that is winning with respectto {y ∈ ω ω | L[S, y] = L[S, x]}, that is, fn+1 x is a non-losing (n + 2)nd movefor II against σ x in SG L[S,x] .Case 2: II has a winning strategy τ n+1 in G n+1 .Let fn+1 x be the prestrategy derived from τ n+1 by extracting the responsein the auxiliary game, that is, for y ∈ (ω ω ) L[S,x] let fn+1 x (y) be such thata∗fn+1 x (y) = a∗d where d is such that (〈a, y〉∗τ n+1) II = 〈c, d〉. Let x n+2 =〈τ n+1 , x n+1 〉 and for x S x n+2 let A x n+1 = σx (〈f0 x, . . .,fx n 〉), as above. Asbefore, we have that for x S x n+2 , fn+1 x is a prestrategy for II in Ax n+1 thatis winning with respect to {y ∈ ω ω | L[S, y] = L[S, x]}, that is, fn+1 x is anon-losing (n + 2) nd move for II against σ x in SG L[S,x] .Let x n+2 be as described in whichever case holds.Finally, using DC R , we get a sequence of reals x 0 , . . .,x n , . . . and prestrategiesf0 x, . . .,fx n , . . . as in each of the steps. Letting x∞ S x n , for alln, we have that for all x S x ∞ , f0 x, . . .,fx n , . . . is a non-losing play for IIagainst σ x in SG L[S,x] , which is a contradiction. This completes the proof ofTheorem 5.14.For Theorem 5.13 simply note that DC R is not needed to define the finitesequences x 0 , . . .,x n , x n+1 and f0 x , . . ., fn x for x S x n+1 (as these prestrategiesare definable from x 0 , . . ., x n , x n+1 ).5.3. Generation TheoremIn the previous section we showed (assuming ZF + AD) that for an S-coneof x,L[S, x] |= OD S -determinacy,and (even more) that for each n,L[S, x] |= ST B S -determinacy for n moves,

5. Woodin Cardinals in General Settings 110where B = [x] S . It turns out that for a sufficiently large choice of n thisdegree of determinacy is sufficient to implement the previous arguments andshow thatL[S, x] |= ω 2 is a Woodin cardinal in HOD S .At this stage we could proceed directly to this result but instead, with thismotivation behind us, we return to the more general setting. The maintheorem to be proved is the Generation Theorem:Theorem 5.15 (Generation Theorem). Assume ZF. Supposeis such that(1) M |= T 0 ,(2) Θ M is a regular cardinal,(3) T ⊆ Θ M ,M = L ΘM (R)[T, A, B](4) A = 〈A α | α < Θ M 〉 is such that A α is a prewellordering of the reals oflength greater than or equal to α,(5) B ⊆ ω ω is nonempty, and(6) M |= ST B T,A,B-determinacy for four moves.ThenHOD M T,A,B|= ZFC + There is a T-strong cardinal.The importance of the restriction to strategic determinacy for four movesis that in a number of applications of this theorem strategic determinacy forn moves (for each n) can be established without any appeal to DC (as forexample in Theorem 5.13) in contrast to full strategic determinacy which(just as in Theorem 5.14) uses DC R .The external assumption that Θ M is a regular cardinal is merely forconvenience—it ensures that there are cofinally many stages in the stratificationof M where T 0 holds. The dedicated reader can verify that this assumptioncan be dropped by working instead with the theory ZF N +AC ω (R)for some sufficiently large N.The remainder of this section is devoted to a proof of the GenerationTheorem.

5. Woodin Cardinals in General Settings 111Proof. Let us start by showing that HOD M T,A,B satisfies ZFC. When workingwith structures of the form L ΘM (R)[T, A, B] it is to be understood that weare working in the language of ZFC augmented with constant symbols for T,A, B, and R. The first step is to show that HOD M T,A,B is first-order over M.For γ < Θ M , letM γ = L γ (R)[T↾γ, A↾γ, B],it being understood that the displayed predicates are part of the structure.Since Θ M is regular and M |= T 0 there are cofinally many γ < Θ M such thatM γ |= T 0 . So a set x ∈ M is OD M T,A,B if and only if there is a γ < Θ M suchthat M γ |= T 0 and x is definable in M γ from ordinal parameters (and theconstant symbols for the parameters). It follows that OD M T,A,B and HODM T,A,Bare Σ 1 -definable over M (in the expanded language).With this first-order characterization of HOD M T,A,B all of the standardresults carry over to our present setting. For example, since M |= ZF −Power Set we have that HOD M T,A,B |= ZFC − Power Set. (The proofs thatAC holds in HOD M T,A,B and that for all α < Θ M , V α ∩HOD M T,A,B ∈ HOD M T,A,Brequire that OD M T,A,B be ordinal definable.)Lemma 5.16. HOD M T,A,B|= ZFC.Proof. We have seen that HOD M T,A,B |= ZFC−Power Set. Since HODM T,A,B |=AC it remains to show that for all λ < Θ M ,P(λ) HODM T,A,B ∈ HODMT,A,B .The point is that since M |= OD T,A,B -determinacy, for each S ∈ OD M T,A,B ∩P(λ) the game for coding S relative to the prewellordering A λ is determined:Without loss of generality, we may assume A λ has length λ. For α < λ, letQ κ

5. Woodin Cardinals in General Settings 112is an OD M T,A,B surjection. Thus, P(λ)HODM T,A,B ∈ M and so, by our first-ordercharacterization, P(λ) HODM T,A,B ∈ HODMT,A,B .The ordinal κ that we will show to be T-strong in HOD M T,A,Bstable in M”:is “the least5.17 Definition. Let κ be least such thatM κ ≺ 1 M.As before the ♦-like function F : κ → M κ is defined inductively in termsof the least counterexample: Given F↾δ let ϑ(δ) be least such thatM ϑ(δ) |= T 0 and there is an X ∈ M ϑ(δ) ∩ OD M ϑ(δ)T↾ϑ(δ),A↾ϑ(δ),B and(⋆) there is a Σ 1 -formula ϕ and a t ∈ ω ω such thatand for all ¯δ < δM ϑ(δ) |= ϕ[t, X, δ, R]M ϑ(δ) ̸|= ϕ[t, F(¯δ), ¯δ, R](if such an ordinal exists) and then let F(δ) be the < M ϑ(δ)OD T↾ ϑ(δ),A↾ ϑ(δ),B-least Xsuch that (⋆) holds.Theorem 5.18. For all X ∈ OD M T,A,B , for all Σ 1-formulas ϕ, and for allt ∈ ω ω , ifM |= ϕ[t, X, κ, R]then there exists a δ < κ such thatM |= ϕ[t, F(δ), δ, R].Proof. Same as the proof of Theorem 4.6.Our interest is in applying Theorem 5.18 toX = ( λ , λ)where λ = A λ is the prewellordering of length λ, for λ < Θ M . Clearly X isOD M T,A,B .

5. Woodin Cardinals in General Settings 113Let U X be a universal Σ 1 (M, {X, κ, R}) set of reals and, for δ < κ, letU δ be the reflected version (using the same definition used for U except withF(δ) and δ in place of X and κ). For z ∈ U X , let S z = {δ < κ | z ∈ U δ } andsetF X = {S ⊆ κ | ∃z ∈ U X (S z ⊆ S)}.As before, F X is a countably complete filter and in Theorem 5.18 we canreflect to F X -many points δ < κ. LetS 0 = {δ < κ | F(δ) = (A λδ , λ δ ) for some λ δ > δ}.Notice that S 0 ∈ F X . For notational conformity let δ be A λδ . For α < λ,let Q κ α be the αth -component of λ and, for δ ∈ S 0 and α < λ δ , let Q δ α bethe α th -component of δ (where without loss of generality we may assumethat each A α has length exactly α).In our previous settings we went on to do two things. First, using theUniform Coding Lemma we showed that one can allow parameters of theform A ⊆ κ and f : κ → κ in the Reflection Theorem. Second, for S ⊆ κ,we defined the games G X (S) that gave rise to the ultrafilter extending thereflection filter, an ultrafilter that was either explicitly OD in the backgrounduniverse (as in Section 4.3) or shown to be OD by appeal to Kunen’s theorem(as in Section 5.1). In our present setting (where we have a limited amountof determinacy at our disposal) we will have to manage our resources morecarefully. The following notion will play a central role.5.19 Definition. A set x ∈ M is n-good if and only if II can play n roundsof (SG B T,A,B,x )M . For y ∈ M, a set x ∈ M is n-y-good if and only if (x, y) isn-good.Notice that if M satisfies ST B T,A,B,y-determinacy for n + 1 moves thenII’s first move f 0 is n-y-good. Notice also that if x is 1-y-good then everyOD M T,A,B,x,y set of reals is determined. For example, if S ⊆ κ is 1-good thenthe game for coding S relative to A κ using the Uniform Coding Lemma isdetermined. Thus we have the following version of the Reflection Theorem.Theorem 5.20. Suppose f : κ → κ, G ⊆ κ, S ⊆ κ and (f, G, S) is 1-good.For all X ∈ M ∩ OD M T,A,B, for all Σ 1 -formulas ϕ, and for all t ∈ ω ω , ifthen for F X -many δ < κ,M |= ϕ[t, X, κ, R, f, G, S]M |= ϕ[t, F(δ), δ, R, f↾δ, G ∩ δ, S ∩ δ].

5. Woodin Cardinals in General Settings 114For each S ⊆ κ, let G X (S) be the gameI x(0) x(1) x(2) . . .II y(0) y(1) . . .with the following winning conditions: Main Rule: For all i < ω, (x) i , (y) i ∈U X . If the rule is violated, then, letting i be the least such that either(x) i ∉ U X or (y) i ∉ U X , I wins if (x) i ∈ U X ; otherwise II wins. If the ruleis satisfied, then, letting δ be least such that for all i < ω, (x) i , (y) i ∈ U δ , Iwins iff δ ∈ S.As before, if S ∈ F X then I wins G X (S) by playing any x such that forall i < ω, (x) i ∈ U X and for some i < ω, (x) i = z, where z is such thatS z ⊆ S. But we cannot setµ X = {S ⊆ κ | I wins G X (S)}since we have no guarantee that G X (S) is determined for an arbitrary S ⊆ κ.However, if S is 1-good then G X (S) is determined. In particular, G X (S)is determined for each S ∈ P(κ) ∩ HOD M T,A,B . Thus, settingµ = {S ∈ P(κ) ∩ HOD M T,A,B | I wins GX (S)}we have directly shown that κ is measurable in HOD M T,A,B.It is useful at this point to stand back and contrast the present approachwith the two earlier approaches. In both of the earlier approaches (namely,that of Section 4.3 and that of Section 5.1) the ultrafilters were ultrafilters inV and seen to be complete and normal in V and the ultrafilters were OD V ,the only difference being that in the first case the ultrafilters were directlyseen to be OD V , while in the second case they were indirectly seen to beOD V by appeal to Kunen’s theorem (Theorem 3.11). Now, in our presentsetting, there is no hope of getting such ultrafilters in V since we do not haveenough determinacy. Instead we will get ultrafilters in HOD M T,A,B . However,the construction will still be “external” in some sense since we will be definingthe ultrafilters in V .We also have to take care to ensure that the ultrafilters “fit together” insuch a way that they witness that κ is T-strong. In short, we will define a(κ, λ)-pre-extender E X ∈ HOD M T,A,B , a notion we now introduce.For n ∈ ω and z ∈ [On] n , we write z = {z 1 , . . .,z n }, where z 1 < · · · < z n .Suppose b ∈ [On] n and a ⊆ b is such that a = {b i1 , . . .b ik }, where i 1 < · · ·

5. Woodin Cardinals in General Settings 115Thus the elements of z a,b sit in z in the same manner in which the elementsof a sit in b. For α ∈ On and X ⊆ [α] k , letX a,b = {z ∈ [α] n | z a,b ∈ X}.For α ∈ On and f : [α] k → V , let f a,b : [α] n → V be such thatf a,b (z) = f(z a,b ).Thus we use ‘a, b’ as a subscript to indicate that z a,b is the “drop of z fromb to a” and we use ‘a, b’ as a superscript to indicate that X a,b is the “lift ofX from a to b”.5.21 Definition. Let κ be an uncountable cardinal and let λ > κ be anordinal. The sequenceE = 〈E a | a ∈ [λ]

5. Woodin Cardinals in General Settings 116If conditions (1) and (2) alone are satisfied then we say that E is a (κ, λ)-pre-extender.We need to ensure that the ultrafilter E a on [κ] |a| depends on a in such away that guarantees coherence and the other properties. The most naturalway to define an ultrafilter E a on [κ] |a| that depends on a is as follows:(1) For F X -almost all δ define a “reflected version” a δ ∈ [λ δ ]

5. Woodin Cardinals in General Settings 117Lemma 5.22. Assume z is (n + 1)-good. Then for each a ∈ [λ]

5. Woodin Cardinals in General Settings 118that a has. To see this, consider a statement such as the following: For allx, x ′ ∈ B, if α 1 , . . .,α k are such thatthenandf a (x) ∈ Q κ α 1× · · · × Q κ α kf a (x ′ ) ∈ Q κ α 1× · · · × Q κ α kα 1 < · · · < α k .This is a true Σ 1 (M, {X, R, κ}) statement. Thus, for F X -almost all δ, thestatement reflects.5.24 Definition. Suppose a ∈ [λ]

5. Woodin Cardinals in General Settings 119The only trouble with this definition is that there is no guarantee that E X (f a )is in HOD M T,A,B because there is no guarantee that E a (f a ) is in HOD M T,A,B.We have to “erase” the dependence on the choice of f a in the definition ofE a .Lemma 5.27. Suppose a ∈ [λ]

5. Woodin Cardinals in General Settings 120Note that E a ∈ OD M T,A,B and E a ⊆ HOD M T,A,B . Thus, E a ∈ HOD M T,A,B andE X ∈ HOD M T,A,B .Our definition of the extender E X presupposes that for each a ∈ [λ]

5. Woodin Cardinals in General Settings 121The key point is thatS(a, f a , Y ) = ⋂ {S(a, f a , Y α ) | α < γ}and so we are in almost exactly the situation as Lemma 4.7, only now wehave to carry along the parameter f a .Since Y ∈ HOD M T,A,B , S(a, f a, Y ) ∈ OD M T,A,B,f a. Since f a is 1-good itfollows that G X (S(a, f a , Y )) is determined. Assume for contradiction that Idoes not win G X (S(a, f a , Y )) and let σ ′ be a winning strategy for I in G X (κS(a, f a , Y )). We will derive a contradiction by finding a play that is legalagainst σ ′ and against winning strategies for I in each game G X (S(a, f a , Y α )),for α < γ.As in the case of Lemma 4.7, for the purposes of coding the winningstrategies (in the games G X (S(a, f a , Y α )) for α < γ) we need a prewellorderingof length γ which is such that in a reflection argument we can ensurethat it reflects to itself. For this purpose, for δ < κ, letThe sequenceQ δ = U δ ⋃ {U ξ | ξ < δ}.〈Q ξ | ξ < κ〉gives rise to an OD M T,A,B prewellordering with the feature that for F X-almostall δ,〈Q ξ | ξ < δ〉 = 〈Q ξ | ξ < δ〉 M ϑ(δ)and, by choosing a real, we can ensure that we always reflect to some suchpoint δ > γ.Now setZ = {(x, σ) | for some α < γ, x ∈ Q α andσ is a winning strategy for I in G X (S(a, f a , Y α ))}.This set is OD M T,A,B,f a, hence determined (as f a is 1-good). So the game inthe Uniform Coding Lemma is determined. The rest of the proof is exactlyas before.(b) By κ-completeness, E a is principal if and only if there exists b ∈ [κ] |a|such thatE a = {Y ∈ P([κ] |a| ) ∩ HOD M T,A,B | b ∈ Y }.

5. Woodin Cardinals in General Settings 122Suppose that a ∈ [κ] |a| . We claim that b = a witnesses that E a is principal.Let f a be a 1-good code of a. For F X -almost all δ, a δ f a= a. So, for Y ∈P([κ] |a| ) ∩ HOD M T,A,B ,Y ∈ E a ↔ I wins G X (S(a, f a , Y ))↔ I wins G X ({δ ∈ S 0 (f a ) | a δ f a= a ∈ Y })↔ a ∈ Y.Suppose that a ∉ [κ] |a| . We claim that no β ∈ [κ] |a| witnesses that E a isprincipal. Consider b ∈ [κ] |a| and let f b be a 1-good code for b and let f a bea 1-good code for a. For F X -almost all δ, a δ f a ≠ b δ f b= b. Let S be the set ofsuch δ and let Y = {a δ f a| δ ∈ S}. Then Y ∈ E a and b ∉ Y . Hence E a is notprincipal.(2) Suppose a ⊆ b ∈ [λ]

5. Woodin Cardinals in General Settings 123the proof of Lemma 5.22 shows that there are f ai such that 〈f ai | i < ω〉 is1-good. So G X (S) is determined. As in the proof of the completeness of E awe have that I wins G X (S).As in the proof of coherence there is a set S 0 (f a1 , . . .,f an , . . .) ∈ F X suchthat for all δ ∈ S 0 (f a1 , . . .,f an , . . .),Fix δ ∈ S ∩ S 0 (f a1 , . . .,f an , . . .). Set〈a δ 1, . . .,a δ i, . . .〉 ∼ = 〈a 1 , . . .,a i , . . .〉.h i : a i → κ(a i ) j ↦→ ((a i ) δ f ai) j .Since δ ∈ S 0 (f a1 , . . .,f an , . . .), the functionh = ⋃ i

5. Woodin Cardinals in General Settings 124Notice that the manner in which the ordinal αf δ αis determined is differentthan in Section 4. In Section 4 we just chose t ∈ Q α and let αt δ be uniquesuch that t ∈ Q δ . Notice also that gα δ fα is 1-good since it is OD M T,A,B,fta.The statement and proof of strong normality are similar to before, onlynow we have to ensure that the objects are sufficiently good to guaranteethe determinacy of the games defined in terms of them. The real parametersthat arise in the proof of strong normality will now have to be generatedusing the technique of Lemma 5.22 and every time we use this techniquewe will sacrifice one degree of goodness. There will be finitely many suchsacrifices and so it suffices to assume that M satisfies ST B T,A,B -determinacyfor n moves for some sufficiently large n. Furthermore, there is no loss ingenerality in making this assumption since in all of the applications of theGeneration Theorem, one will be able to show without DC that M satisfiesST B T,A,B-determinacy for n < ω. As we shall see there will in fact only betwo sacrifices of goodness. Thus, since we want our final object to be 1-good(to ensure that the games defined in terms of it are determined) it sufficesto start with an object which is 3-good.Theorem 5.32 (Strong Normality). Suppose g : κ → κ is such that(1) g is 3-good and(2) I wins G X ({δ ∈ S 0 | g(δ) < λ δ }).Then there exists an α < λ such thatwhere f α is any 1-g-good code of α.I wins G X ({δ ∈ S 0 (f α ) | g(δ) = g fα (δ)}),Proof. We begin with a few comments. First, note that since g is 3-good, byLemma 5.22 we have that for each α < λ there is a 1-g-good code f α of α (infact, there is a 2-g-good code) and hence each game G X ({δ ∈ S 0 (f α ) | g(δ) =g fα (δ)}) is determined. The only issue is whether I wins some such game.Second, notice that α is uniquely specified. For suppose fˆα is a 1-g-goodcode of ˆα such that I wins the corresponding game. If α < ˆα, then{δ ∈ S 0 (f α ) ∩ S 0 (fˆα ) | g fα (δ) < g fˆα(δ)} ∈ F Xand I wins G X (S) where S is this set. But then I cannot win bothG X ({δ ∈ S 0 (f α ) | g(δ) = g fα (δ)})

5. Woodin Cardinals in General Settings 125andG X ({δ ∈ S 0 (fˆα ) | g(δ) = g fˆα(δ)}).Third, it will be useful at this point to both list the parameters thatwill arise in the proof and motivate the need for assuming that g is 3-good.In outline the proof will follow that of Theorem 4.12. The final game inthe present proof (the one involving e 1 ) will be defined in terms of threeparameters: g, f η and e 0 , corresponding respectively to f, y η , and e 0 fromTheorem 4.12. To ensure the determinacy of the final game we will need totake steps to ensure that (g, f η , e 0 ) is 1-good. Now, the parameter e 0 will beobtained by applying the technique of Lemma 5.22 to the parameter (g, f η )and so we will need to take steps to ensure that this parameter is 2-good.And the parameter f η will in turn be obtained by applying the technique ofLemma 5.22 to the parameter g and so we have had to assume from the startthat g is 3-good.We now turn to the proof proper. Suppose g : κ → κ is such that(1.1) g is 3-good and(1.2) I wins G X ({δ ∈ S 0 | g(δ) < λ δ }).Assume for contradiction that for each α < λ and for each 1-g-good code f αof α,(2.1) I does not win G X ({δ ∈ S 0 (f α ) | g(δ) = g fα (δ)}),and hence (since each such game is determined, as f α is 1-g-good)(2.2) I wins G X ({δ ∈ S 0 (f α ) | g(δ) ≠ g fα (δ)}).Step 1. Letη = min ({ β < λ | I wins G X ({δ ∈ S 0 (f β ) | g(δ) < g fβ (δ)})for each 1-g-good code f β of β })if such β exist; otherwise let η = λ. (So if there are such β then η is a limitordinal.) Notice that(3.1) whenever α < η and f α is a 1-g-good code of α,I wins G X ({δ ∈ S 0 (f α ) | g(δ) > g fα (δ)}),

5. Woodin Cardinals in General Settings 126which is the desired situation. By Lemma 5.22, letf η be a 2-g-good code of η.For notational convenience, for δ ∈ S 0 (f η ), let η δ be η δ f η. By the definition ofη, I wins G X ({δ ∈ S 0 (f η ) | g(δ) < g fη (δ)}). Now update S 0 to be S 0 ∩ {δ ∈S 0 (f η ) | g(δ) < g fη (δ)}. We will work on this “large” set. Notice that S 0 isOD M T,A,B,g,f η. If η = λ then η δ = δ and we may omit mention of f η in whatfollows.For convenience let us write “S ∈ µ X ” as shorthand for “I wins G X (S)”.To summarize:(4.1) g is 3-good,(4.2) (g, f η ) is 2-good (First Drop in Goodness),(4.3) S 0 is OD M T,A,B,g,f η,(4.4) S 0 ∈ µ X and for all δ ∈ S 0 , g(δ) < g fη (δ), and(4.5) for all α < η and for all 1-g-good codes f α of α,{δ ∈ S 0 (f α ) | g(δ) > g fα (δ)} ∈ µ X .Step 2. We now establish the “disjointness property”.LetWe haveZ ′ = { (x, 〈y, σ〉) | x ∈ Q κ α for some α < η,y codes a 1-g-good code f α of αsuch that x ∈ ran(f α ↾B), andσ is a winning strategy for I in(5.1) Z ′ is OD M T,A,B,g,f ηand Z ′ ⊆ Q κ g fα (δ)}) } .(5.2) for all α < η,Z ′ ∩ (Q κ α × ωω ) ≠ ∅,by (3.1).

5. Woodin Cardinals in General Settings 127Since (g, f η ) is 2-good the game in the proof of the Uniform Coding Lemma(Theorem 3.4) is determined. So there is an index e ∈ ω ω such that for allα < η,(6.1) U e(2) (Q κ

5. Woodin Cardinals in General Settings 128Omitting Z ′ , (8.2) is Σ 1 (M, {X, κ, R, f η , e 0 }). So, for F X -almost all δ,(8.3) for all α < η δ ,(1) U (2)e 0 (B, Q δ

5. Woodin Cardinals in General Settings 129(9.1) z is a legal play for II against σ ′ and δ 0 is the associated ordinal and(9.2) z is a legal play for II against each σ ∈ (proj 2 (Z δ 0)) 1 and in eachcase δ 0 is the associated ordinal.This will finish the proof: By (9.1) and the definition of G, there is an α 0 ∈[g(δ 0 ), η δ0 ) and a β 0 ∈ [0, η) such that Z δ 0α 0∩Zβ κ 0 ≠ ∅. Fix (x 0 , 〈y 0 , σ 0 〉) ∈ Z δ 0α 0∩Zβ κ 0. Since (x 0 , 〈y 0 , σ 0 〉) ∈ Zβ κ 0⊆ Z ′ ∩ (Q κ β 0× ω ω ) we have, by the definitionof Z ′ , x 0 ∈ Q κ β 0, y 0 codes a 1-g-good code f β0 of β 0 , x 0 ∈ ran(f β0 ↾B), andσ 0 is a winning strategy for I in S({δ ∈ S 0 (f β0 ) | g(δ) > g fβ0 (δ)}). Since(x 0 , 〈y 0 , σ 0 〉) ∈ Z δ 0α 0, σ 0 ∈ (proj 2 (Z δ 0)) 1 . Now, by (9.2), z is a legal play for IIagainst σ 0 with associated ordinal δ 0 , and since σ 0 is a winning strategy forI in S({δ ∈ S 0 (f β0 ) | g(δ) > g fβ0 (δ)}), this implies(10.1) δ 0 ∈ {δ ∈ S 0 (f β0 ) | g(δ) > g fβ0 (δ)},that is, g(δ 0 ) > g fβ0 (δ 0 ). We now argue(10.2) g fβ0 (δ 0 ) = α 0 ,which is a contradiction since α 0 g(δ 0 ). Recall that by definition g fβ0 (δ 0 ) =|f β0 (x)| δ0 , where x is any element of B. Since we arranged x 0 ∈ ran(f β0 ↾B)and since (x 0 , 〈y 0 , σ 0 〉) ∈ Z δ 0α 0, this implies that g fβ0 (δ 0 ) = |f β0 (x)| δ0 = α 0 ,where x is any element of B. Thus, a play z as in (9.1) and (9.2) will finishthe proof.The play z is constructed as before:Base Case. We have(11.1) ∀y ∈ ω ω ((σ ′ ∗y) I ) 0 ∈ U X and(11.2) ∀y ∈ ω ω ∀σ ∈ (proj 2 (Z κ )) 1 ((σ∗y) I ) 0 ∈ U X .This is a true Σ 1 (M, {X, κ, R, σ ′ , e 0 , f η })-statement. So there is a z 0 ∈ U Xsuch that z 0 T 〈σ ′ , e 0 , f η 〉 and for all δ if z 0 ∈ U δ then(11.3) ∀y ∈ ω ω ((σ ′ ∗y) I ) 0 ∈ U δ and(11.4) ∀y ∈ ω ω ∀σ ∈ (proj 2 (Z δ )) 1 ((σ∗y) I ) 0 ∈ U δ .(n + 1) st Step. Assume we have defined z 0 , . . .,z n in such a way that(12.1) ∀y ∈ ω ω ( ∀i n (y) i = z i → ((σ ′ ∗y) I ) n+1 ∈ U X)and

5. Woodin Cardinals in General Settings 130(12.2) ∀y ∈ ω ω σ ∈ (proj 2 (Z κ )) 1 , ( ∀i n (y) i = z i → ((σ∗y) I ) n+1 ∈ U X).This is a true Σ 1 (M, {X, κ, R, σ ′ , e 0 , f η , z 0 , . . .,z n })-statement. So there is az n+1 ∈ U X such that z n+1 T z n and for all if z n+1 ∈ U δ then(12.3) ∀y ∈ ω ω ( ∀i n (y) i = z i → ((σ ′ ∗y) I ) n+1 ∈ U δ)and(12.4) ∀y ∈ ω ω ∀σ ∈ (proj 2 (Z δ )) 1(∀i n (y)i = z i → ((σ∗y) I ) n+1 ∈ U δ).Finally, let z ∈ ω ω be such that (z) i = z i for all i ∈ ω and let δ 0 be least suchthat (z) i ∈ U δ0 for all i ∈ ω. Notice that by our choice of z n , for n < ω, noDC is required to construct z. We have that for all i ∈ ω,(13.1) ((σ ′ ∗z) I ) i ∈ U δ0 by (11.3) and (12.3) and(13.2) ((σ∗z) I ) i ∈ U δ0 for all σ ∈ (proj 2 (Z δ 0)) 1 by (11.4) and (12.4).So we have (9.1) and (9.2), which is a contradiction.By the Subclaim,(14.1) ∀δ ∈ G ∀α ∈ [g(δ), η δ ) ∀β ∈ [0, η) (Z δ α ∩ Zκ β = ∅).This is a true Σ 1 (M, {X, κ, R, 〈f η , e 0 〉, g, G})-statement ϕ. Notice that sinceG is OD M T,A,B,g,f η,e 0and (g, f η , e 0 ) is 1-good, it follows that (G, g, f η , e 0 ) is1-good. In particular, (G, g) is 1-good, and so Theorem 5.20 applies (taking〈f η , e 0 〉 for the real t in the statement of that theorem) and we have that(14.3) for F X -almost all δ 2 ,(1) M |= ϕ[〈f η , e 0 〉, F(δ 2 ), δ 2 , g↾δ 2 , G ∩ δ 2 ], that is,(2) ∀δ 1 ∈ G ∩ δ 2 ∀α ∈ [g(δ 1 ), η δ1 ) ∀β ∈ [0, η δ2 ) (Z δ 1α ∩ Zδ 2β = ∅).Let S ′ 2 be the set of such δ 2 in (14.3) and let S 2 = S ′ 2∩G. Since S ′ 2 ∈ F X ⊆ µ Xand G ∈ µ X , we have that S 2 ∈ µ X . Hence S 2 is as desired in Claim A. Also,S 2 is OD M T,A,B,g,f η,e 0.Notice that two additional parameters have emerged, namely, G and S 2 ,but these do not lead to a drop in goodness since(15.1) OD M T,A,B,g,f η,e 0 ,G,S 2= OD M T,A,B,g,f η,e 0 ,G = ODM T,A,B,g,f η,e 0, and so(15.2) (g, f η , e 0 , G, S 2 ) is 1-good.

5. Woodin Cardinals in General Settings 131Step 3. We are now in a position to “compute g”.For δ ∈ S 2 , letBy (15.1), P δ ∈ OD M T,A,B,g,f η,e 0.P δ = ⋃{ Z¯δα | ¯δ ∈ S 2 ∩ δ ∧ α ∈ [g(¯δ), η¯δ) } .Claim B (Tail Computation). There exists an index e 1 ∈ ω ω such thatfor all δ ∈ S 2 ,(1) U (2)e 1 (P δ , Z δ α ) ⊆ Zδ α for all α ∈ [0, η δ),(2) U e (2)1(P δ , Zg(δ) δ ) = ∅, and(3) U (2)e 1 (P δ , Z δ α ) ≠ ∅ for α ∈ (g(δ), η δ).Proof. As before it suffices to show (2) and (3 ′ ) U e (2)1(P δ , Zα δ) ∩ Zδ α ≠ ∅ forα ∈ (g(δ), η δ ).LetG = { (e ∈ ω ω | ∀δ ∈ S 2 U(2)e (P δ , Zg(δ) δ ) = ∅)} .Toward a contradiction assume that for each e ∈ G, (3 ′ ) in the claim fails forsome δ and α. For each e ∈ G, letNow play the gameα e = lexicographically least pair (δ, α) such that(1) δ ∈ S 2 ,(2) g(δ) < α < η δ , and(3) U (2)e (P δ , Z δ α ) ∩ Zδ α = ∅.I x(0) x(1) x(2) . . .II y(0) y(1) . . .where II wins iff (x ∈ G → (y ∈ G ∧ α y > lex α x )).The key point is that this payoff condition is OD M T,A,B,g,f η,e 0, by (15.1),and hence, the game is determined, since (g, f η , e 0 ) is 1-good.The rest of the proof is exactly as before.From this point on there are no uses of determinacy that require further“joint goodness”.

5. Woodin Cardinals in General Settings 132Claim C. There exists an α 0 < η such that(1) U (2)e 1 (P κ , Z κ α 0) = ∅ and(2) U (2)e 1 (P κ , Z κ α) ≠ ∅ for all α ∈ (α 0 , η), whereP κ = ⋃ {Z δ α | δ ∈ S 2 ∧ α ∈ [g(δ), η δ )}.Proof. The statement that there is not a largest ordinal α 0 which is “empty”is Σ 1 (M, {X, κ, R, 〈f η , e 0 , e 1 〉, g, G, S 2 }). Since (g, f η , e 0 ) is 1-good and G andS 2 are OD M T,A,B,g,f η,e 0, it follows that (g, G, S 2 ) is 1-good. Thus, the ReflectionTheorem (Theorem 5.20) applies and we have that for F X -many δ, thestatement reflects, which contradicts Claim B.Let α 0 be the unique ordinal as above and let f α0 be a 1-g-good code ofα 0 (which exists by Lemma 5.22). The statement(16.2) ∀x ∈ B f α0 (x) ∈ Q κ α where α is such that(1) U (2)e 1 (P κ , Z κ α) = ∅ and(2) U (2)e 1 (P κ , Z κ β ) ≠ ∅ for β ∈ (α, η δ).is Σ 1 (M, {X, κ, R, 〈f η , f α0 , e 0 , e 1 〉, g, G, S 2 }). Since (g, G, S 2 ) is 1-good, theReflection Theorem (Theorem 5.20) applies and hence for F X -almost all δthe statement reflects. Let S ′ 3 ∈ F X be this set. Let S 3 = S ′ 3 ∩ S 2. SoS 3 ∈ µ X . By Claim B and Claim C, for δ ∈ S 3 the ordinal α in question isg(δ). So I wins G X ({δ ∈ S 0 (f α0 ) ∩ S 3 | g(δ) = g fα0 (δ)}) and hence I winsG X ({δ ∈ S 0 (f α0 ) | g(δ) = g fα0 (δ)}). This game is determined since f α0 is1-g-good.To summarize:(17.1) f α0 is 1-g-good and(17.2) I wins G X ({δ ∈ S 0 (f α0 ) | g(δ) = g fα0 (δ)}),which completes the proof of strong normality.Since every g : κ → κ in HOD M T,A,B is 3-good and since in the context ofthe main theorem we assume that M satisfies ST B T,A,B-determinacy for fourmoves, we have shown:

5. Woodin Cardinals in General Settings 133Corollary 5.33. Suppose g : κ → κ is in HOD M T,A,B and such that I winsG X ({δ ∈ S 0 | g(δ) < λ δ }). Then there exists an α < λ and a 1-g-good codef α of α such thatI wins G X ({δ ∈ S 0 (f α ) | g(δ) = g fα (δ)}).Lemma 5.34 (Normality). In HOD M T,A,B : If a ∈ [λ]

5. Woodin Cardinals in General Settings 134Note that β < a i since if β a i then for F X -almost all δ, g fβ (δ) (a δ f a) iand we get that I wins G X ({δ ∈ S 0 (f β ) ∩ S 0 (f a ) | f(a δ f a) (a δ f a) i }), whichcontradicts (1.1).Let k be such that β = (a∪{β}) k . Let f a∪{β} be a 1-good code of a∪{β}.Note that(2.1) for F X -almost all δ,((a ∪ {β})δfa∪{β})and(2.2) for F X -almost all δ,((a ∪ {β})δfa∪{β})k = g f β(δ)a,a∪{β} = aδ f aand, moreover, I wins on these sets (since the parameters in the definitionsare 1-good). So (1.2)(3) yields(3.1) I wins G X({ δ ∈ S 0 (f a∪{β} ) | f (( )(a ∪ {β}) δ f a∪{β})a,a∪{β}= ( })(a ∪ {β}) δ f a∪{β})kthat is,(3.2) {z ∈ [κ] |a∪{β}| | f(z a,a∪{β} ) = z k } ∈ E a∪{β} ,as desired.We are now in a position to take the “ultrapower” of HOD M T,A,B by E X.It will be useful to recall this construction and record some basic facts concerningit. For further details see Steel’s chapter in this Handbook.LetD = {〈a, f〉 ∈ HOD M T,A,B | a ∈ [λ]

5. Woodin Cardinals in General Settings 135and membership relation defined by[a, f] ∈ EX [b, g] ↔ {z ∈ [κ] |a∪b| | f(z a,a∪b ) ∈ g(z b,a∪b )} ∈ E a∪b .Since HOD M T,A,B satisfies AC, ̷Loś’s theorem holds in the following form: Forall formulas ϕ(x 1 , . . .,x n ) and all elements [a 1 , f 1 ], . . .,[a n , f n ] ∈ Ult,Ult |= ϕ [ [a 1 , f 1 ], . . .,[a n , f n ] ] ↔{z ∈ [κ] |b| | HOD M T,A,B |= ϕ[f 1(z a1 ,b), . . .,f n (z an,b)]} ∈ E b ,where b = ⋃ 1in a i. It follows thatj ′ E : HODM T,A,B → Ultx ↦→ [∅, c x ],where c x is the constant function with value x, is an elementary embedding.The countable completeness of E X ensures that Ult is well-founded and it isstraightforward to see that it is extensional and set-like. So we can take thetransitive collapse. Letπ : Ult → M Xbe the transitive collapse map and letj E : HOD M T,A,B → M Xbe the elementary embedding obtained by letting j E = π ◦ j E ′ . The κ-completeness of each E a , for a ∈ [λ]

5. Woodin Cardinals in General Settings 136where A ⊆ λ and A ∈ HOD M T,A,B . Let λ be such an ordinal and let κ, j E,etc. be as above. We have that j E (κ) λ and it remains to show thatV HODM T,A,Bλ⊆ M Xandj E (T ∩ κ) ∩ λ = T ∩ λ.The proof of each is the same. Let us start with the latter. We have to showthat for all α < λ,α ∈ j E (T ∩ κ) ↔ α ∈ T.We haveSo we have to show thatα ∈ j E (T ∩ κ) ↔ π([{α}, z ↦→ ∪z]) ∈ π([∅, c T ∩κ ])↔ [{α}, z ↦→ ∪z] ∈ EX [∅, c T ∩κ ]↔ {z ∈ [κ] 1 | ∪z ∈ T ∩ κ} ∈ E {α} .α ∈ T ↔ {{z} | z ∈ T ∩ κ} ∈ E {α} .Let f {α} be a 1-good code of {α}.Assume α ∈ T. We have to show thatI wins G X( S({α}, f {α} , {{z} | z ∈ T ∩ κ}) ) .The key point is that the statement “for all x ∈ B, |f {α} (x)| λ ∈ T” is a trueΣ 1 (M, {X, κ, R, f {α} })-statement. So the set S of δ to which this statementreflects is in F X . Since S ∈ OD M T,A,B,f {α}and f {α} is 1-good, G X (S) isdetermined and I wins. ButS({α}, f {α} , {{z} | z ∈ T ∩ κ}) = S 0 (f {α} ) ∩ Sand so I wins this game as well.Assume α ∉ T. We have to show thatI does not win G X( S({α}, f {α} , {{z} | z ∈ T ∩ κ}) ) .Again, the point is that the statement “for all x ∈ B, |f {α} (x)| λ ∉ T” isa true Σ 1 (M, {X, κ, R, f {α} })-statement. So this statement reflects to F X -almost all δ, which implies that I cannot win the above game.

5. Woodin Cardinals in General Settings 137Exactly the same argument with ‘A’ in place of ‘T’ shows thatand hence thatwhich completes the proof.j E (A ∩ κ) ∩ λ = A ∩ λ,V HODM T,A,Bλ= L λ [A] ⊆ M X ,This completes the proof of the Generation Theorem5.4. Special CasesWe now consider a number of special instances of the Generation Theorem.In each case all we have to do is find appropriate values for the parametersΘ M , T, A, and B. We begin by recovering the main result of Section 4.Theorem 5.36. Assume ZF + AD. ThenHOD L(R) |= Θ L(R) is a Woodin cardinal.Proof. For notational convenience let Θ = Θ L(R) . Our strategy is to meetthe conditions of the Generation Theorem while at the same time arrangingthat M = L Θ (R)[T, A, B] is such thatHOD M T,A,B = HOD L(R) ∩ V Θ .We will do this by taking care to ensure that the ingredients T, A, and Bare in HOD L(R) while at the same time packaging HOD L(R) ∩ V Θ as part ofT. It will then follow from the Generation Theorem thatHOD L(R) ∩ V Θ |= ZFC + There is a T-strong cardinal,and by varying T the result follows.To begin with let Θ M = Θ L(R) and, for notational convenience, we continueto abbreviate this as Θ. By Theorem 3.9, Θ is strongly inaccessible inHOD L(R) . Also,HOD L(R) ∩ V Θ = HOD L Θ(R) ,by Theorem 3.10. So we can let H ∈ P(Θ) ∩ HOD L(R) codeHOD L(R) ∩ V Θ .

5. Woodin Cardinals in General Settings 138Fix T ′ ∈ P(Θ) ∩HOD L(R) and let T ∈ P(Θ) ∩HOD L(R) code T ′ and H. ByLemmas 3.7 and 3.8, there is an OD L(R) sequence A = 〈A α | α < Θ〉 suchthat each A α is a prewellordering of reals of length α. Let B = R.LetM = L Θ (R)[T, A, B]where Θ, T, A, and B are as above. Conditions (1)–(5) of the GenerationTheorem are clearly met and condition (6) follows since L(R) satisfies ADand M contains all reals. Moreover, since we have arranged that all of theingredients T, A, and B are in OD L(R) and also that T codes HOD L(R) ∩ V Θ ,we haveHOD M T,A,B = HODL(R) ∩ V Θand, since T ′ was arbitrary, the result follows as noted above.We can also recover the following approximation to Theorem 5.6.Theorem 5.37. Assume ZF + AC ω (R). Suppose ST B X-determinacy holds,where X is a set and B is non-empty and OD X . ThenHOD X |= Θ X is a Woodin cardinal.Proof. Let Θ M = Θ X . Let A = 〈A α | α < Θ X 〉 be such that A α codesthe OD X -least prewellordering of reals of length α. By Theorem 3.9, Θ X isstrongly inaccessible in HOD X and so there exists an H ∈ P(Θ X ) ∩ HOD Xcoding HOD X ∩ V ΘX . Let T ∈ P(Θ X ) ∩ HOD X code H and some arbitraryT ′ ∈ P(Θ X ) ∩ HOD X .LetM = L ΘM (R)[T, A, B]where Θ M , T, A, and B are as above. Work in HOD {X}∪R . Conditions (3)–(5) of the Generation Theorem are clearly met. For condition (2) note thatby Lemma 3.7, Θ X = Θ HOD {X}∪R and that by the arguments of Lemma 3.8and Lemma 3.9, Θ HOD {X}∪R is regular in HOD{X}∪R . Thus, Θ M is regularin HOD {X}∪R . Condition (6) follows from the fact that M is OD X and Mcontains all of the reals. It remains to see that condition (1) can be met.For this we just have to see that Replacement holds in M. If Replacementfailed in M then there would be a cofinal map π : ω ω → Θ X that is definablefrom parameters in M, which in conjunction with A would lead to an OD Xsurjection from ω ω onto Θ X , which is a contradiction.

5. Woodin Cardinals in General Settings 1395.38 Remark. Work in ZF+DC. For µ a δ-complete ultrafilter on δ let E µ bethe (δ, λ)-extender derived from µ where λ = j(δ) (or λ = δ δ /µ) and j is theultrapower map. We have the following corollary: Assume ZF + AD + DC.Then Θ X is Woodin in HOD X and this is witnessed by the collection ofE µ ∩ HOD X where µ is a normal ultrafilter on some δ < Θ X .5.39 Remark. Theorem 5.6 cannot be directly recovered from the GenerationTheorem and this is why we have singled it out for special treatment.However, it follows from the proof of the Generation Theorem, as can beseen by noting that in the case where one has full boldface determinacy theultrafilters are actually in HOD X by Kunen’s theorem (Theorem 3.11).4 Open Question. There are some interesting questions related to Theorem5.37.(1) Suppose Θ X = Θ 0 . Suppose ST B X-determinacy, where B is non-emptyand OD X . Is Θ 0 a Woodin cardinal in HOD?(2) Suppose ST B X-determinacy, where X is a set and B is non-empty andOD X . Is Θ X a Woodin cardinal in HOD?(3) In the AD + setting, every Θ X is of the form Θ α and there are constraintson this sequence. For example, each Θ X must be of the form Θ α+1 .Does this constraint apply in the lightface setting?Theorem 5.40. Assume ZF + AD. Let S be a class of ordinals. Then foran S-cone of x,HOD L[S,x]SProof. For an S-cone of x,by Corollary 5.10, and, for all n < ω,|= ω L[S,x]2 is a Woodin cardinal.L[S, x] |= ZFC + GCH below ω V 1 ,L[S, x] |= ST B S -determinacy for n moves,where B = [x] S , by Theorem 5.13. Let x be in this S-cone.Let Θ M = ω L[S,x]2 . Since L[S, x] satisfies GCH below ω1 V, by Lemma 3.8 we have thatOD L[S,x]S,xand L[S, x] =ω L[S,x]2 = sup{α | there is an OD L[S,x]Sprewellordering of length α},

5. Woodin Cardinals in General Settings 140in other words, ω L[S,x]2 = (Θ S ) L[S,x] . Let A = 〈A α | α < ω L[S,x]2 〉 be such thatA α is the OD L[S,x]S-least prewellordering of length α. Since L[S, x] |= OD S -determinacy, it follows (by Theorem 3.9) that ω L[S,x]in HOD L[S,x]S2 is strongly inaccessible. So there is a set H ⊆ ω L[S,x]2 coding HOD L[S,x]S∩ V L[S,x] ω. Let2T ′ be in P(ω L[S,x]2 ) ∩ OD L[S,x]SH. Let B = [x] S .Letand let T ∈ P(ω L[S,x]2 ) ∩ OD L[S,x]SM = L ΘM (R L[S,x] )[T, A, B],code T andwhere Θ M , T, A, and B are as above. Conditions (1)–(5) of the GenerationTheorem are clearly met and condition (6) follows since L[S, x] satisfies ST B S -determinacy for four moves, M is OD S in L[S, x] and M contains the realsof L[S, x]. Thus,HOD M T,A,B |= ZFC + There is a T-strong cardinal.Since we have arranged that all of the ingredients T, A, and B are in OD L[S,x]and also that T codes HOD L[S,x] ∩ V L[S,x] ω, we have2HOD M T,A,B = HODL[S,x] ∩ V L[S,x] ω.2Since T ′ was arbitrary, the result follows.Theorem 5.41. Assume ZF + AD. Then for an S-cone of x,HOD S,HODS,x |= ω HOD S,x2 is a Woodin cardinal.Proof. This will follow from the next theorem which is more general.The next two theorems require some notation. Suppose Y is a set anda ∈ H(ω 1 ). For x ∈ ω ω , the (Y, a)-degree of x is the set[x] Y,a = {z ∈ ω ω | HOD Y,a,z = HOD Y,a,x }.The (Y, a)-degrees are the sets of the form [x] Y,a for some x ∈ ω ω . Definex Y,a y to hold iff x ∈ HOD Y,a,y . A cone of (Y, a)-degrees is a set of the form{[y] Y,a | y Y,a x 0 } for some x 0 ∈ ω ω and a (Y, a)-cone of reals is a set of theform {y ∈ ω ω | y Y,a x 0 } for some x 0 ∈ ω ω . The proof of the Cone Theorem(Theorem 2.9) generalizes to the present context. In the case where a = ∅we speak of Y -degrees, etc.

5. Woodin Cardinals in General Settings 141Theorem 5.42. Assume ZF+AD. Suppose Y is a set and a ∈ H(ω 1 ). Thenfor a (Y, a)-cone of x,HOD Y,a,[x]Y,a |= ω HOD Y,a,x2 is a Woodin cardinal,where [x] Y,a = {z ∈ ω ω | HOD Y,a,z = HOD Y,a,x }.Proof. By determinacy it suffices to show that the above statement holdsfor a Turing cone of x, which is what we shall do. The key issues in thiscase are getting a sufficient amount of GCH and strategic determinacy. Toestablish the first we need two preliminary claims. Recall that a set A ⊆ ω ωis comeager if and only if ω ω A is meager.Claim 1. Assume ZF+AD. Suppose that 〈A α | α < γ〉 is a sequence of setswhich are comeager in the space ω ω , where either γ ∈ On or γ = On, in whichcase the sequence is a definable proper class. Then ⋂ α

5. Woodin Cardinals in General Settings 142Claim 2. Assume ZF + AD. Suppose Y is a set, a ∈ H(ω 1 ) and P ∈HOD Y,a ∩H(ω 1 ) is a partial order. Then for comeager many HOD Y,a -genericG ⊆ P,HOD Y,a,G = HOD Y,a [G].Proof. For each G we clearly have HOD Y,a [G] ⊆ HOD Y,a,G . We seek a setA that is comeager in the Stone space of P and such that for all G ∈ A,HOD Y,a,G = HOD Y,a [G]. We will do this by showing that for each G ∈ A thelatter model can compute the “ordinal theory” of the former model.For every Σ 2 -statement ϕ and finite sequence of ordinals ⃗ ξ consider thestatement ϕ[ ⃗ ξ, Y, a, G] about a generic G. Let B ϕ,⃗ ξ,Y,a be the associatedcollection of filters on P and letP ϕ,⃗ξ = {p ∈ P | B ϕ,⃗ ξ,Y,a is comeager in O p } andN ϕ,⃗ξ = {p ∈ P | B ¬ϕ,⃗ ξ,Y,a is comeager in O p },where O p is the open set of generics containing p. These are the sets of conditionswhich “positively” and “negatively” decide ϕ[ ⃗ ξ, Y, a, G], respectively.So P ϕ,⃗ξ ∪ N ϕ,⃗ξ is predense. Now letA ϕ, ⃗ ξ={G ⊆ P | ϕ[ ⃗ ξ, Y, a, G] ↔ G ∩ P ϕ,⃗ξ ≠ ∅}∪ {G ⊆ P | ¬ϕ[ ⃗ ξ, Y, a, G] ↔ G ∩ N ϕ,⃗ξ ≠ ∅}.Each such set is comeager. We thus have a class size well-order of comeagersets and so, by the previous lemma,A = ⋂ {A ϕ, ⃗ ξ| ϕ is a Σ 2 -formula and ⃗ ξ ∈ On

5. Woodin Cardinals in General Settings 143Claim 3. Assume ZF + AD. Suppose Y is a set and a ∈ H(ω 1 ). Then for aTuring cone of x,HOD Y,a,x |= GCH below ω V 1 .Proof. It suffices to show that CH holds on a cone since given this the proofthat GCH below ω V 1 holds on a cone goes through exactly as before.Suppose for contradiction (by the Cone Theorem (Theorem 2.9)) thatthere is a real x 0 such that for all x T x 0 ,HOD Y,a,x |= ¬CH.We will arrive at a contradiction by producing an x T x 0 with the featurethat HOD Y,a,x |= CH. As before x is obtained by forcing (in two steps) overHOD Y,a,x0 . First, we get a HOD Y,a,x0 -genericG ⊆ Col(ω HOD Y,a,x 01 , R HOD Y,a,x 0)and then we use almost disjoint forcing to code G with a real. Viewing thegeneric g as a real, by the previous claim we have that for comeager many g,HOD Y,a,x0 ,g = HOD Y,a,x0 [g] |= CH,and hencewhich is a contradiction.HOD Y,a,〈x0 ,g〉 |= CH,Claim 4. Suppose Y is a set and a ∈ H(ω 1 ). Then for a Turing cone of x,for each n < ω, II can play n moves of SG B Y,a,[x] Y,a, where B = [x] Y,a , and wedemand in addition that II’s moves belong to HOD Y,a,x , in other words, IIcan play n moves of the gamewhere we require, for i + 1 < n,I A 0 · · · A n−1II f 0 · · · f n−1(1) A 0 ∈ P(ω ω ) ∩ OD V Y,a,[x] Y,a, A i+1 ∈ P(ω ω ) ∩ OD V Y,a,[x] Y,a ,f 0 ,...,f iand(2) f i+1 is prestrategy for A i+1 that belongs to HOD Y,a,x and is winningwith respect to [x] Y,a .

5. Woodin Cardinals in General Settings 144Proof. The proof of Theorem 5.13 actually establishes this stronger result.We are now in a position to meet the conditions of the Generation Theorem.For a Turing cone of x,by Claim 3, and for all n < ω,HOD Y,a,x |= ZFC + GCH below ω V 1 ,ST B Y,a,[x] Y,a-determinacy for n movesholds in V where B = [x] Y,a , by Claim 4. Let x be in this cone.Let Θ M = ω HOD Y,a,x2 . Since HOD Y,a,x |= ZFC + GCH below ω V 1 ,Θ M = Θ.Since every set is OD Y,a,x , and hence OD Y,a,[x]Y ,x,by Lemma 3.8. Thus,Θ = Θ Y,a,[x]Y ,ω HOD Y,a,x2 = Θ HOD Y,a,xY,a,[x] Y,a.Letting A = 〈A α | α < ω HOD Y,a,x2 〉 be such that A α is the OD Y,a,[x]Y,a -leastprewellordering of length α we have that A is OD Y,a,[x]Y,a . We also havethat ω HOD Y,a,x2 is strongly inaccessible in HOD Y,a,[x]Y,a , by Theorem 3.9. Sothere is a set H ⊆ ω HOD Y,a,x2 coding HOD Y,a,[x]Y,a ∩ V HOD ω Y,a,x . Let T ′ be inP(ω HOD Y,a,x2 ) ∩ OD Y,a,[x]Y,a and let T ∈ P(ω HOD Y,a,x2 ) ∩ OD Y,a,[x]Y,a code T ′and H. Let B = [x] Y,a .LetM = L ΘM (R HOD Y,a,x)[T, A, B],where Θ M , T, A, and B are as above. Conditions (1)–(5) of the GenerationTheorem are clearly met. Condition (6) follows from the fact that Mis OD Y,a,[x]Y,a and we have arranged (in Claim 4) that all of II’s moves inSG B Y,a,[x] Y,aare in M.Thus,HOD M T,A,B|= ZFC + There is a T-strong cardinal,2

6. Definable Determinacy 145and since we have arranged thatHOD M T,A,B = HOD Y,a,[x]Y,a ∩ V HOD ω Y,a,x ,2and T was arbitrary, the result follows.In the above theorem the degree notion [x] Y,a depends on the initial choiceof a. However, later (in Section 6.2) we will want to construct models withmany Woodin cardinals. A natural approach to doing this is to iterativelyapply the previous theorem, starting off with a = ∅, increasing the degreeof x to get that ω HOD Y,x2 is a Woodin cardinal in HOD Y,[x]Y , and then takinga = [x] Y , increasing the degree of x yet again to get that ω HOD Y,[x] Y ,x2 is aWoodin cardinal in HOD Y,[x]Y ,[x] Y,[x]Y, etc. This leads to serious difficultiessince the degree notion is changing. We would like to keep the degree notionfixed as we supplement a and for this reason we need the following variantof the previous theorem.Theorem 5.43. Assume ZF+AD. Suppose Y is a set and a ∈ H(ω 1 ). Thenfor a Y -cone of x,HOD Y,a,[x]Y |= ω HOD Y,a,x2 is a Woodin cardinal,where [x] Y = {z ∈ ω ω | HOD Y,z = HOD Y,x }.Proof. The proof is essentially the same as that of the previous theorem.Claims 1 to 3 are exactly as before. The only difference is that now in Claim4 we have [x] Y in place of [x] Y,a . The proof of this version of the claim is thesame, as is that of the rest of the theorem.6. Definable DeterminacyWe now use the Generation Theorem to derive the optimal amount of largecardinal strength from both lightface and boldface definable determinacy.The main result concerning lightface definable determinacy is the following:Theorem 6.1. Assume ZF+DC+∆ 1 2-determinacy. Then for a Turing coneof x,HOD L[x] |= ZFC + ω L[x]2 is a Woodin cardinal.

6. Definable Determinacy 146When combined with the results mentioned in the introduction this has theconsequence that the theories ZFC + OD-determinacy and ZFC + “There isa Woodin cardinal” are equiconsistent. In order to prove this theorem wewill have to get into the situation of the Generation Theorem. The issuehere is that ∆ 1 2-determinacy does not imply that for a cone of x strategicdeterminacy holds in L[x] with respect to the constructibility degree of x.Instead we will use a different basis set B, one for which we can establishST B -determinacy for four moves, using ∆ 1 2-determinacy alone.The main result concerning boldface definable determinacy is the following:Theorem 6.2. Assume ZF+AD. Suppose Y is a set. There is a generalizedPrikry forcing P Y through the Y -degrees such that if G ⊆ P Y is V -genericand 〈[x i ] Y | i < ω〉 is the associated sequence, thenHOD V [G]Y,〈[x i ] Y |i

6. Definable Determinacy 147for an appropriate basis B. In the boldface setting we took our basis B tobe the constructibility degree of x. But as we shall see (in Theorem 6.12) inour present setting one cannot secure this version of strategic determinacy.Nevertheless, it turns out that strategic determinacy holds for a different,smaller basis. This leads to the notion of restricted strategic determinacy.We shall successively extract stronger and stronger forms of determinacyuntil we ultimately reach the version we need. The subsection closes witha series of limitative results, including results that motivate the need forstrategic and restricted strategic determinacy.Theorem 6.3 (Martin). Assume ZF + DC + ∆ 1 2-determinacy. Then Σ 1 2-determinacy.Proof. Consider A = {x ∈ ω ω | ϕ(x)} where ϕ is Σ 1 2. We have to show thatA is determined. Our strategy is to show that if II (the Π 1 2 player) does nothave a winning strategy for A then I (the Σ 1 2 player) has a winning strategyfor A.Assume that II does not have a winning strategy for A. First, we haveto shift to a “local” setting where we can apply ∆ 1 2-determinacy. For eachx ∈ ω ω ,L[x] |= II does not have a winning strategy in {y ∈ ω ω | ϕ(y)}(since otherwise, by Σ 1 3 upward absoluteness, II would have a winning strategyin V , contradicting our initial assumption) and so, by the Löwenheim-Skolem theorem, there is a countable ordinal λ such thatL λ [x] |= T + II does not have a winning strategy in {y ∈ ω ω | ϕ(y)},where T is some fixed sufficiently strong fragment of ZFC (such as ZFC Nwhere N is large or ZFC − Replacement + Σ 2 -Replacement). For x ∈ ω ω , letλ(x) = µλ (L λ [x] |= T +II does not have a winningFor convenience let A x = {y ∈ ω ω | ϕ(y)} Lλ(x)[x] .Consider the game GI a, xII bstrategy in {y ∈ ω ω | ϕ(y)}).

6. Definable Determinacy 148where I wins iff a∗b ∈ L λ(x) [x] and L λ(x) [x] |= ϕ(a∗b). Here Player I is tobe thought of as choosing the playing field L λ(x) [x] in which the two playersare to play an auxiliary round (via a and b) of the localized game A x . Thekey point is that since λ(x) is always defined this game is ∆ 1 2 and hencedetermined.We claim that I has a winning strategy in G (and so I wins each roundof the localized games A x ) and, furthermore, that (by ranging over theserounds and applying upward Σ 1 2-absoluteness) this winning strategy yields awinning strategy for I in A.Assume for contradiction (by ∆ 1 2-determinacy) that II has a winning strategyτ 0 in G. For each x T τ 0 , in L λ(x) [x] we can derive a winning strategyτ x for II in A x as follows: For a ∈ (ω ω ) Lλ(x)[x] , let (a∗τ x ) II = b where b issuch that (〈a, x〉∗τ 0 ) II = b. Since τ 0 is a winning strategy for II in G and wehave arranged that a∗b ∈ L λ(x) [x], II must win in virtue of the second clause,which means that a∗b /∈ A x . Thus, L λ(x) [x] |= “τ x is a winning strategy forII in A x ”, which is a contradiction.Thus I has a winning strategy σ 0 in G. Consider the derived strategy σsuch that for b ∈ ω ω , (σ∗b) I = a where a is such that (σ 0 ∗b) I = 〈a, x〉. Sinceσ 0 is a winning strategy for I in G, σ∗b ∈ L λ(x) [x] and L λ(x) [x] |= ϕ(σ∗b) andso, by upward Σ 1 2-absoluteness, V |= ϕ(σ∗b). Thus, σ is a winning strategyfor I in A.6.4 Remark.(1) The above proof relativizes to a real parameter to show that ∆ 1 2 (x)-determinacy implies Σ 1 2 (x)-determinacy.(2) A similar but more elaborate argument shows that if ∆ 1 2-determinacyholds and for every real x, x # exists, then Th(L[x]) is constant for aTuring cone of x. See [7].Theorem 6.5 (Martin). Assume ZF + DC + ∆ 1 2-determinacy. If I has awinning strategy in a Σ 1 2-game then I has a ∆ 1 3 strategy.Proof. Consider A = {x ∈ ω ω | ϕ(x)} where ϕ is Σ 1 2. Our strategy is to showthat if II (the Π 1 2 player) does not win A then I (the Σ1 2 player) wins A viaa ∆ 1 3 strategy.Assume that II does not have a winning strategy in A. For x ∈ ω ω ,let λ(x), A x , G, and σ 0 be as in the previous proof. Since σ 0 is a winningstrategy for I in G, for x T σ 0 , in L λ(x) [x] we can derive a winning strategy

6. Definable Determinacy 149σ x for I in A x as follows: For x T σ 0 and b ∈ (ω ω ) L λ(x)[x]let (σ∗b) I = awhere a is such that (σ 0 ∗b) I = 〈a, x〉.Next we show that there is an x 0 T σ 0 such that for all x T x 0 ,L λ(x) [x] |= ∆ 1 2 -determinacy. Let 〈ϕ n, ψ n 〉 enumerate the pairs of Σ 1 2 -formulasand let A ϕ = {x ∈ ω ω | ϕ(x)}. Using DC let z n be such that z n codes awinning strategy for A ϕn if A ϕn = ω ω A ψn (i.e. A ϕn is ∆ 1 2 ); otherwise letz n = 〈0, 0, . . .〉. Finally, let x 0 code 〈z n | n < ω〉. Thus, for x T x 0 ,L λ(x) [x] |= I has a Σ 1 4 strategy in A xby the Third Periodicity Theorem of Moschovakis.(For a proof of Third Periodicity see Jackson’s chapter in this Handbook.The statement of Third Periodicity typically involves boldface determinacy.However, the proof shows that lightface ∆ 1 2 determinacy suffices toget Σ 1 4 winning strategies for Σ1 2 games that I wins. To see this note thatScale(Σ 1 2) holds in ZF + DC. Furthermore, we also have the determinacy ofthe Σ 1 2 games (denoted Gn s,t in Jackson’s chapter) that are used to define theprewellorderings and ultimately the definable strategies. It follows that theseprewellorderings and strategies are Σ 1 2 ⊆ Σ 1 14. (Notice that if we had ∆ ∼ 2 -determinacy then we could flip the quantifiers and conclude that Σ 1 2 = Π1 3and hence get ∆ 1 3 strategies. However, in our present lightface setting somemore work is required.))For x T x 0 , let ˆσ x be the Σ 1 4 -strategy for I in Ax . For a Turing cone ofx the formula ϕ(y, z) defining this strategy is constant. We can now “freezeout” the value of ˆσ x on a Turing cone of x. The key point is that the functionx ↦→ L λ(x) [x] is ∆ 1 2 . So, for each s ∈ ω2n and m ∈ ω the statementL λ(x) [x] |= ϕ(s, m)is ∆ 1 2 . Thus, for each s ∈ ω2n , the m such that L λ(x) [x] |= ϕ(s, m) is fixed fora Turing cone of x. Since there are only countably many s ∈ ω 2n this meansthat the value of ˆσ x is fixed on a Turing cone of x. Finally, lettingσ(s) = m ↔ ∃x 0 ∀x T x 0 (L λ(x) [x] |= ϕ(s, m))↔ ∀x 0 ∃x T x 0 (L λ(x) [x] |= ϕ(s, m))(where we have used ∆ 1 2 -determinacy to flip the quantifiers) we have that σis a ∆ 1 3 winning strategy for I in A.

6. Definable Determinacy 150Kechris and Solovay showed (in [6]) that under ZF+DC+∆ 1 2 -determinacythere is a real x 0 that “enforces” OD-determinacy in the following sense: Forall x T x 0 , L[x] |= OD-determinacy. We will need the following strengtheningof this result, which involves a stronger notion of “enforcement”. Weneed the following definition: An ordinal λ is additively closed (a.c.) iff forall α, β < λ, α + β < λ.Theorem 6.6. Assume ZF + DC + ∆ 1 2-determinacy. Then there is a real x 0such that for all additively closed λ > ω, and for all reals x, if x 0 ∈ L λ [x],then L λ [x] |= OD-determinacy.Proof. Some preliminary remarks are in order. First, for λ additively closed,L λ [x] might satisfy only a very weak fragment of ZFC; so the statement“L λ [x] |= OD-determinacy” is to be taken in the following external sense:For each ξ < λ and for each formula ϕ, L λ [x] |=“Either I or II has a winningstrategy for {z ∈ ω ω | ϕ(z, ξ)}”. The point is that this statement makessense even when {z ∈ ω ω | ϕ(z, ξ)} is a proper class from the point of view ofL λ [x]. Second, the key feature of additively closed λ > ω, is that if y ∈ L λ [x]then L λ [y] ⊆ L λ [x]. This is true since additively closed ordinals λ > ω aresuch that α + λ = λ for all α < λ and so if y is constructed at stage α, thenL λ [x] still has λ-many remaining stages in which to “catch up” and constructeverything in L λ [y]. Third, the proof of the theorem is a “localization” ofthe proof of Theorem 5.12.Assume for contradiction that for every real x 0 there is an additivelyclosed λ > ω and a real x such that x 0 ∈ L λ [x] and L λ [x] ̸|= OD-determinacy.So, for every real x 0 there is an additively closed λ > ω and a real x ′ T x 0such that L λ [x ′ ] ̸|= OD-determinacy (since we can take x ′ = 〈x, x 0 〉 where xand x 0 are as in the first statement) and hence, by the Löwenheim-Skolemtheorem,∀x 0 ∈ ω ω ∃x T x 0 ∃λ λ is a.c. ∧ ω < λ < ω 1 ∧L λ [x] ̸|= OD-determinacy,where ‘a.c.’ abbreviates ‘additively closed’. Since the condition on x in thisstatement is Σ 1 2 and since we have Σ 1 2-determinacy (by Theorem 6.3)∃x 0 ∈ ω ω ∀x T x 0 ∃λ λ is a.c. ∧ ω < λ < ω 1 ∧L λ [x] ̸|= OD-determinacy

6. Definable Determinacy 151by the Cone Theorem (Theorem 2.9). Let{µλ (ω < λ < ω 1 ∧ λ is a.c. ∧ L λ [x] ̸|= OD-det) if such a λ existsλ(x) =undefinedotherwise.Notice that for a Turing cone of xλ(x) is definedand that there are x 0 of arbitrarily large Turing degree such that for allx T x 0λ(x) λ(x 0 ).To see this last point it suffices to observe that otherwise (by Σ 1 2-determinacyand the Cone Theorem (Theorem 2.9)) there would be an infinite descendingsequence of ordinals. This point will be instrumental below in ensuring thatPlayer II can “steer into the right model”.For each x such that λ(x) is defined let (ϕ x , ξ x ) be lexicographically leastsuch thatL λ(x) [x] |= {z ∈ ω ω | ϕ x (z, ξ x )} is not determinedand let A x = {z ∈ ω ω | ϕ x (z, ξ x )}. (However, note that since A x might bea proper class from the point of view of L λ(x) [x], when we write ‘L λ(x) [x] |=a ∈ A x ’ we really mean ‘L λ(x) [x] |= ϕ x (a, ξ x )’.)Consider the gameI a, bII c, dwhere, letting p = 〈a, b, c, d〉, I wins iff λ(p) is defined and L λ(p) [p] |= “a∗d ∈A p ”. This game is Σ 1 2 , hence determined.(Notice that in contrast to the proof of Theorem 5.12 we cannot includex 0 in p since we need our game to be lightface definable. However, in theplays of interest we will have one player fold in x 0 . This will ensure that thefirst clause of the winning condition is satisfied and so the players are to bethought of as cooperating to determine the model L λ(p) [p] and simultaneouslyplaying an auxiliary game (via a and d) on the least non-determined OD setof this model, namely, A x .)We will arrive at a contradiction by showing that neither player can win.Case 1: I has a winning strategy σ 0 .

6. Definable Determinacy 152Let x 0 T σ 0 be such that for all x T x 0 , λ(x) is defined and λ(x) λ(x 0 ). We claim that L λ(x0 )[x 0 ] |= “I has a winning strategy σ in A x 0”, whichis a contradiction. The strategy σ is the strategy derived by playing themain game according to σ 0 while having II feed in x 0 for c and playingsome auxiliary play d ∈ L λ(x0 )[x 0 ]; that is, (σ∗d) I = a where a is such that(σ 0 ∗〈x 0 , d〉) I = 〈a, b〉:I a, bII x 0 , d.Let p = 〈a, b, x 0 , d〉. Since we have ensured that p T x 0 we know that λ(p)is defined and, since σ 0 is winning for I, I must win in virtue of the firstclause and so L λ(p) [p] |= “a∗d ∈ A p ”. It remains to see that II has managedto “steer into the right model”, that is, thatand henceL λ(p) [p] = L λ(x0 )[x 0 ]A p = A x 0.Since x 0 T σ 0 and d ∈ L λ(x0 )[x 0 ] we have that p ∈ L λ(x0 )[x 0 ] andL λ(x0 )[p] = L λ(x0 )[x 0 ](where for the left to right inclusion we have used that λ(x 0 ) is additivelyclosed). Furthermore, by arrangement, λ(p) λ(x 0 ) since p T x 0 . Butλ(p) is the least additively closed λ such that ω < λ < ω 1 and L λ [p] ̸|= ODdeterminacy.Thus, λ(p) = λ(x 0 ) andL λ(p) [p] = L λ(x0 )[x 0 ].So L λ(x0 )[x 0 ] |= “σ∗d ∈ A x 0”. Since this is true for any d ∈ L λ(x0 )[x 0 ], thismeans that L λ(x0 )[x 0 ] |= “σ is a winning strategy for I in A x 0”, which is acontradiction.Case 2: II has a winning strategy τ 0 .Let x 0 T τ 0 be such that for all x T x 0 , λ(x) is defined and λ(x) λ(x 0 ). For a ∈ L λ(x0 )[x 0 ] let (a∗τ) II where d is such that (〈a, x 0 〉∗τ 0 ) II =〈c, d〉. Since p T x 0 , II must win in virtue of the second clause. The restof the argument is exactly as above. So we have that L λ(x0 )[x 0 ] |= “τ is awinning strategy for II in A x 0”, which is a contradiction.

6. Definable Determinacy 1536.7 Remark. The proof relativizes to a real parameter to show ZF + DC +Σ 1 2(x)-determinacy implies that there is a real enforcing (in the strong senseof Theorem 6.6) OD x -determinacy.Corollary 6.8 (Kechris and Solovay). Assume ZF. Suppose L[x] |= ∆ 1 2 -determinacy, where x ∈ ω ω . Then L[x] |= OD-determinacy.Proof. This follows by reflection.We will now extract an even stronger form of determinacy from ∆ 1 2 -determinacy. We begin by recalling some definitions. The strategic gamewith respect to the basis B is the game SG Bwhere we requireI A 0 · · · A n · · ·II f 0 · · · f n · · ·(1) A 0 ∈ P(ω ω ) ∩ OD, A n+1 ∈ P(ω ω ) ∩ OD f0 ,...,f nand(2) f n is a prestrategy for A n that is winning with respect to B,and II wins iff he can play all ω rounds. We say that strategic determinacyholds with respect to the basis B (ST B -determinacy) if II wins SG B .In the context of L[S, x] we dropped reference to the basis B since it wasalways understood to be {y ∈ ω ω | L[S, y] = L[S, x]}. In our present lightfacesetting we will have to pay more careful attention to B since (as we will seein Theorem 6.12) ∆ 1 2-determinacy is insufficient to ensure that for a Turingcone of x, L[x] |= ST B -determinacy, where B = {y ∈ ω ω | L[y] = L[x]}. Wewill now be interpreting strategic determinacy in the local setting of modelsL λ [x] where x ∈ ω ω and λ is a countable ordinal and the relevant basis willbe of the form C ∩ {y ∈ ω ω | L λ [y] = L λ [x]} where C is a Π 1 2 set of L λ[x].It is in the attempt to “localize” the proof of Theorem 5.14 that the needfor the Π 1 2 set becomes manifest. The issue is one of “steering into the rightmodel” and can be seen to first arise in the proof of Claim 3 below.Let RST-determinacy abbreviate the statement “There is a Π 1 2 set Csuch that C contains a Turing cone and ST B -determinacy holds where B =C ∩ {y ∈ ω ω | L[y] = V }”. Here ‘R’ stands for ‘restrictive’. We will beinterpreting this notion over models L λ [x] that do not satisfy full Replacement.In such a case it is to be understood that the statement involves theΣ 1 definition of ordinal definability.

6. Definable Determinacy 154Theorem 6.9. Assume ZF+DC+∆ 1 2-determinacy. Then for a Turing coneof x,L[x] |= RST-determinacy.Proof. We will actually prove something stronger: Assume ZF + V = L[x] +∆ 1 2 -determinacy for some x ∈ ωω . Let T be the theory ZFC−Replacement+Σ 2 -Replacement. Then there is a real z 0 such that if L λ [z] is such thatz 0 ∈ L λ [z] and L λ [z] |= T then L λ [z] |= RST-determinacy. The theoremfollows by reflection.Assume for contradiction that for every real z 0 there is a real z T z 0and an ordinal λ such that L λ [z] |= T+¬RST-determinacy. The preliminarystep is to reduce to a local setting where we can apply ∆ 1 2-determinacy. Bythe Löwenheim-Skolem theorem∀z 0 ∈ ω ω ∃z T z 0 ∃λ < ω 1 (L λ [z] |= T + ¬RST-determinacy).Since the condition on z in this statement is Σ 1 2 and since we have ODdeterminacy(by Corollary 6.8) it follows (by the Cone Theorem (Theorem2.9)) that∃z 0 ∈ ω ω ∀z T z 0 ∃λ < ω 1 (L λ [z] |= T + ¬RST-determinacy).For z ∈ ω ω , let{µλ (L λ [z] |= T + ¬RST-determinacy) if such a λ existsλ(z) =undefinedotherwise.Thus, if λ(z) is defined, then for every (Π 1 2 )L λ(z)[z] set C that contains a Turingcone, I wins the gamewhere we requireI A 0 · · · A n · · ·II f 0 · · · f n · · ·(1) A 0 ∈ OD L λ(z)[z] , A n+1 ∈ OD L λ(z)[z]f 0 ,...,f n, and(2) f n is a prestrategy for A n that is winning with respect to C ∩ {y ∈ ω ω |L λ(z) [y] = L λ(z) [z]}.

6. Definable Determinacy 155We now need to specify a particular (Π 1 2 )L λ(z)[z] set since (i) we want to get ourhands on a canonical winning strategy σ z for I and (ii) we need to solve the“steering problem”. The naïve approach would be to forget about the Π 1 2 setsand just work with {y ∈ ω ω | L λ(z) [y] = L λ(z) [z]}. The trouble is that for anelement y of this set we might have λ(y) < λ(z) and yet (when we implementthe proof of Theorem 5.13) we will need to ensure that L λ(y) [y] = L λ(z) [z] andthus A y = A z and for this we require that λ(y) = λ(z). So we will need tointersect with a set C that “holds up the value of λ(y)”. A good candidateis the following: For each z such that λ(z) is defined letC z = {y ∈ ω ω | λ(y) is undefined} L λ(z)[z] .This is a (Π 1 2 )L λ(z)[z] -set. It contains z (since in L λ(z) [z] the ordinal λ(z) iscertainly undefined). We would like to ensure that it contains the cone abovez.Claim 1. For a Turing cone of z,(1) λ(z) is defined,(2) for all reals y ∈ L λ(z) [z], if y T z then λ(y) = λ(z).Proof. We have already proved (1). Assume for contradiction that (2) doesnot hold on a Turing cone. Then (by OD-determinacy) for every real z thereis a real z ′ T z such that λ(z ′ ) is defined and in L λ(z ′ )[z ′ ] there is a realz ′′ such that z ′′ T x ′ and λ(z ′′ ) < λ(z). But then, for each n < ω, wecan successively choose z n+1 T z n such that λ(z n+1 ) < λ(z n ), which is acontradiction.For each z as in Claim 1 we now have that C z contains the Turing coneabove z (since, by (2) of Claim 1, λ(y) = λ(z) and so L λ(y) [y] = L λ(z) [z] andagain in L λ(z) [z] the ordinal λ(y) = λ(z) is undefined). Lettingwe have thatB z = {y ∈ C z | L λ(z) [y] = L λ(z) [z]}I wins (SG Bz ) L λ(z)[z] .Moreover, since we have arranged that L λ(z) [z] |= T, Player I has a canonicalstrategy σ z ∈ HOD L λ(z)[z] . (This is because, since L λ(z) [z] |= T, the OD L λ(z)[z]sets of reals are sets (and not proper classes) in L λ(z) [z]. So the tree on which

6. Definable Determinacy 156(SG Bz ) L λ(z)[z]is played is an element of HOD L λ(z)[z] .) Notice also that σ zdepends only on the model L λ(z) [z], in the sense that if L λ(y) [y] = L λ(z) [z]then σ y = σ z .Our aim is to obtain a contradiction by defeating σ z for some z in theTuring cone of Claim 1. We will do this by constructing a sequence ofgames G 0 , G 1 , . . .,G n , . . . such that I must win via σ 0 , σ 1 , . . .,σ n , . . . and,for a Turing cone of z, the winning strategies give rise to prestrategiesf z 0 , fz 1 , . . .,fz n , . . . that constitute a non-losing play against σz in the game(SG Bz ) L λ(z)[z] .Step 0. Consider (in L[x]) the game G 0IIIǫ a, bc, dwhere ǫ is either 1 or 2 and, letting p = 〈a, b, c, d〉, I wins iff(1) p satisfies the condition on z in Claim 1 (so σ p makes sense) and(2) ǫ = 1 iff L λ(p) [p] |= “a∗d ∈ A p 0”, where A p 0 = σ p (∅).In the plays of interest we will ensure that p is in the cone of Claim 1. Soclause (1) of the winning condition will be automatically satisfied and thedecisive factor will be whether in L λ(p) [p] Player ǫ wins the auxiliary round(via a and d) of A p 0 . This game is Σ1 2 (for Player I), hence determined.Claim 2. I has a winning strategy σ 0 in G 0 .Proof. Assume for contradiction that I does not have a winning strategyin G 0 . Then, by Σ 1 2 -determinacy, II has a winning strategy τ 0 in G 0 . Letz 0 T τ 0 be such that for all z T z 0 ,(1) z satisfies the conditions of Claim 1 and(2) if λ and z are such that z 0 ∈ L λ [z] and L λ [z] |= T then L λ [z] |=OD-determinacy (by Theorem 6.6).Consider A z 00 = σ z 0(∅). Since L λ(z0 )[z 0 ] |= OD-determinacy, L λ(z0 )[z 0 ] |=“A z 00 is determined”. We will use τ 0 to show that neither player can win thisgame. Suppose for contradiction that L λ(z0 )[z 0 ] |= “σ is a winning strategyfor I in A z 00 ”. Run G 0 according to τ 0 , having Player I (falsely) predict that

6. Definable Determinacy 157Player I wins the auxiliary game, while steering into L λ(z0 )[z 0 ] by playingb = z 0 and using σ to respond to τ 0 on the auxiliary play:I 1 (σ∗d) I , z 0II c, dWe have to see that Player I has indeed managed to steer into L λ(z0 )[z 0 ],that is, we have to see that L λ(p) [p] = L λ(z0 )[z 0 ], where p = 〈(σ∗d) I , z 0 , c, d〉.Since σ, z 0 , τ 0 ∈ L λ(z0 )[z 0 ] and λ(z 0 ) is additively closed, we have L λ(z0 )[p] =L λ(z0 )[z 0 ]. But λ(p) = λ(z 0 ) since z 0 satisfies Claim 1. Thus, L λ(p) [p] =L λ(z0 )[z 0 ] and hence A p 0 = A z 00 . Finally, since τ 0 is a winning strategy for IIin G 0 and ǫ = 1, we have that L λ(p) [p] |= “σ∗d ∉ A p 0 ”, and hence L λ(z 0 )[z 0 ] |=“σ∗d ∉ A z 00 ”, which contradicts the assumption that σ is a winning strategyfor I. Similarly, we can use τ 0 to defeat any strategy τ for II in A z 0Since the game is Σ 1 2 for Player I, Player I has a ∆1 3 -strategy σ 0, byTheorem 6.5.Claim 3. For every real z T σ 0 in the Turing cone of Claim 1, there isa prestrategy f z 0 such that fz 0 is definable in L λ(z)[z] from σ 0 and f z 0 is anon-losing first move for II against σ z in (SG Bz ) L λ(z)[z] .Proof. Fix z T σ 0 as in Claim 1. Consider A z 0 = σz (∅). Let f0 z be theprestrategy derived from σ 0 in L λ(z) [z] by extracting the response in theauxiliary game, that is, for y ∈ (ω ω ) L λ(z)[z]let f0 z (y) be such that for d ∈(ω ω ) Lλ(z)[z] , f0 z(y)∗d = a∗d where a is such that (σ 0∗〈y, d〉) I = 〈ǫ, a, b〉. f0 z isclearly definable in L λ(z) [z] from σ 0 . We claim that f0 z is a non-losing firstmove for II against σ z in (SG Bz ) Lλ(z)[z] .To motivate the need for the Π 1 2 set, let us first see why fz 0 need not be aprestrategy for II in A z 0 that is winning with respect to {y ∈ (ω ω ) L λ(z)[z]|L λ(z) [y] = L λ(z) [z]}. Consider such a real y and an auxiliary play d ∈(ω ω ) Lλ(z)[z] . By definition f0(y) z is such that f0(y)∗d z = a∗d where a is suchthat (σ 0 ∗〈y, d〉) I = 〈ǫ, a, b〉. Assume first that ǫ = 1. Since σ 0 is a winningstrategy for I in G 0 , z(y)∗d = a∗d ∈ Ap 0 where p = 〈a, b, y, d〉. What weneed, however, is that f0(y)∗d z = a∗d ∈ A z 0. The trouble is that we may haveL λ(p) [p] = L λ(y) [y] L λ(z) [z] because although L λ(z) [y] = L λ(z) [z] we mighthave λ(y) < λ(z). And if this is indeed the case then we cannot concludethat A p 0 = A z 0 . If ǫ = 0 then fz 0 (y)∗d = a∗d ∉ Ap 0 but again what we need isthat f0 z(y)∗d = a∗d ∉ Az 0 and the same problem arises.0 .

6. Definable Determinacy 158The above problem is solved by demanding in addition that y ∈ C z ,since then λ(y) = λ(z) and so ǫ = 1 iff L λ(z) [z] |= “f z 0(y)∗d = a∗d ∈A p 0 = Az 0 ” as desired. Thus fz 0 is a non-losing first move for II against σ z in(SG Bz ) L λ(z)[z] .Step n+1. Assume that we have defined (in L[x]) games G 0 , . . .,G nwith winning strategies σ 0 , . . .,σ n ∈ HOD such that for all z T 〈σ 0 , . . .,σ n 〉in the Turing cone of Claim 1 there are prestrategies f0 z, . . .,fz n such thatfi z is definable in L λ(z) [z] from σ 0 , . . ., σ i (for all i n) and f0, z . . .,fn z is anon-losing partial play for II in (SG Bz ) Lλ(z)[z] .Consider (in L[x]) the game G n+1IIIǫ a, bc, dwhere ǫ is 1 or 2 and, letting p = 〈a, b, c, d, σ 0 , . . .,σ n 〉, I wins iff(1) p satisfies the condition on z in Claim 1 (so σ p makes sense) and(2) ǫ = 1 iff L λ(p) [p] |= “a∗d ∈ A p n+1 ”, where Ap n+1 is I’s response via σp toII’s partial play f p 0,...,f p nIf p satisfies condition (1) then, since p T 〈σ 0 , . . .,σ n 〉, we have, by theinduction hypothesis, prestrategies f0, p . . .,fn p such that f p i is definable inL λ(p) [p] from σ 0 , . . ., σ i (for all i n) and f0, p . . .,fn p is a non-losing partialplay for II in (SG Bp ) Lλ(p)[p] . Thus, condition (2) in the definition of the gamemakes sense.This game is Σ 1 2 (σ 0, . . .,σ n ) (for Player I) and hence determined (sinceσ 0 , . . ., σ n ∈ HOD and we have OD-determinacy, by Theorem 6.6).Claim 4. I has a winning strategy σ n+1 in G n+1 .Proof. Assume for contradiction that I does not have a winning strategy.Then, by OD-determinacy, II has a winning strategy τ n+1 . Let z n+1 T〈τ n+1 , σ 0 , . . ., σ n 〉 be such that for all z T z n+1 ,(1) z satisfies the conditions of Claim 1 and(2) if λ and z are such that z n+1 ∈ L λ [z] and L λ [z] |= T then L λ [z] |=OD σ0 ,...,σ n-determinacy (by the relativized version of Theorem 6.6).

6. Definable Determinacy 159It follows thatL λ(zn+1 )[z n+1 ] |= A z n+1n+1 is determined,where A z n+1n+1 = σ z n+1(〈f z n+10 , . . .,f z n+1n 〉). This is because A z n+1n+1 is an elementof HOD L λ(z n+1 )[z n+1 ] (σ 0 , . . .,σ n ) (as all of the ingredients σ z n+1, f z n+10 ,. . ., f z n+1n used to define A z n+1n+1 are in this model) and we arranged thatL λ(zn+1 )[z n+1 ] satisfies OD σ0 ,...,σ n-determinacy.[The enforcement of the parameterized version of OD-determinacy in (2)appears to be necessary. The point is that even though, in Step 1 for example,σ 0 is ∆ 1 3 and σ 0 ∈ L λ(z) [z] we have no guarantee that in L λ(z1 )[z 1 ], σ 0 satisfiesthis definition. If we did then we would have that A z 1 is in HOD L λ(z 1 )[z 1 ] andhence just enforce OD-determinacy.]We will use τ n+1 to show that neither player can win this game. The argumentis exactly as in Step 0 except with the subscripts ‘0’ replaced by ‘n+1’:Suppose for contradiction that L λ(zn+1 )[z n+1 ] |= “σ is a winning strategy forI in A z n+1n+1 ”. Run G n+1 according to τ n+1 , having Player I (falsely) predictthat Player I wins the auxiliary game, while steering into L λ(zn+1 )[z n+1 ] byplaying b = z n+1 and using σ to respond to τ n+1 on the auxiliary play:I 1 (σ∗d) I , z n+1II c, dWe have to see that Player I has indeed managed to steer into L λ(zn+1 )[z n+1 ],that is, we have to see that L λ(p) [p] = L λ(zn+1 )[z n+1 ], where p is the set〈(σ∗d) I , z n+1 , c, d〉. Since σ, z n+1 , τ n+1 ∈ L λ(zn+1 )[z n+1 ] and λ(z n+1 ) is additivelyclosed, we have L λ(p) [p] = L λ(zn+1 )[z n+1 ]. Since p T z n+1 and z n+1 satisfiesthe condition of Claim 1, λ(p) = λ(z), and so L λ(p) [p] = L λ(zn+1 )[z n+1 ]and hence A p n+1 = Az n+1n+1 . Finally, since τ n+1 is a winning strategy for IIin G n+1 and ǫ = 1, we have that L λ(p) [p] |= “σ∗d ∉ A p n+1”, and henceL λ(zn+1 )[z n+1 ] |= “σ∗d ∉ A z n+1n+1 ”, which contradicts the assumption that σ isa winning strategy for I. Similarly, we can use τ n+1 to defeat any strategy τfor II in A z n+1n+1 .Since the game is Σ 1 2 (σ 0, . . .,σ n ) for Player I, Player I has a ∆ 1 3 (σ 0, . . ., σ n )strategy σ n+1 , by the relativized version of Theorem 6.5.Claim 5. For every real z T 〈σ 0 , . . .,σ n 〉 as in Claim 1, there is a prestrategyf z n+1 that is definable in L λ(z)[z] from σ 0 , . . .,σ n+1 and such thatf z 0 , . . .,fz n+1 is a non-losing first move for II against σz in (SG Bz ) L λ(z)[z] .

6. Definable Determinacy 160Proof. The proof is just like the proof of Claim 3. Fix z T 〈σ 0 , . . .,σ n 〉 as inClaim 1 and consider A z n+1 = σ z (〈f0, z . . .,fn〉). z Let fn+1 z be the prestrategyderived from σ n+1 in L λ(z) [z] by extracting the response in the auxiliarygame, that is, for y ∈ (ω ω ) Lλ(z)[z] let fn+1 z (y) be such that for d ∈ (ωω ) Lλ(z)[z] ,fn+1(y)∗d z = a∗d where a is such that (σ n+1 ∗〈y, d〉) I = 〈ǫ, a, b〉. Clearly, fn+1zis definable in L λ(z) [z] from σ 0 , . . .,σ n+1 . We claim that fn+1 z is a non-losingfirst move for II against σ z in (SG Bz ) Lλ(z)[z] . Again the point is that fory ∈ B z , L λ(y) [y] = L λ(z) [z], hence A y n+1 = A z n+1 . Thus, ǫ = 1 iff L λ(z)[z] |=“fn+1 z (y)∗d = a∗d ∈ Ap n+1 = Az n+1 ” as desired. Hence 〈fz 0 , . . .,fz n+1 〉 is anon-losing play for II against σ z in (SG Bz ) Lλ(z)[z] .Finally, letting z ∞ be such that z ∞ T z n for all n and z ∞ is as in Claim1, we have that f0 z∞ , . . ., fn z∞,. . . defeats in (SG B σz∞ z ∞ ) L λ(z ∞ )[z ∞] , whichis a contradiction.Theorem 6.10. Assume ZF + DC + ∆ 1 2-determinacy. Then for a Turingcone of x,HOD L[x] |= ZFC + ω L[x]2 is a Woodin cardinal.Proof. For a Turing cone of x, L[x] |= RST-determinacy, by Theorem 6.9.Let x be in this cone. We have to meet the conditions of the GenerationTheorem. Let Θ M = ω L[x]2 . Since L[x] satisfies GCH and L[x] = OD L[x]ω L[x]2 = sup{α | there is an OD L[x] prewellordering of length α},in other words, ω L[x]2 = (Θ 0 ) L[x] . Let A = 〈A α | α < ω L[x]2 〉 be such that A α isthe OD L[x] -least prewellordering of length α. Since L[x] |= OD-determinacy,it follows (by Theorem 3.9) that ω L[x]2 is strongly inaccessible in HOD L[x] . Sothere is a set H ⊆ ω L[x]2 coding HOD L[x] ∩V L[x] ω. Let T ′ be in P(ω L[x]2 )∩OD L[x]and let T ∈ P(ω L[x]2 ) ∩ OD L[x] code T ′ and H. Let B be as in the statementof RST-determinacy.LetM = (L ΘM (R)[T, A, B]) L[x] ,where Θ M , T, A, B are as above. Conditions (1)–(5) of the GenerationTheorem are clearly met and condition (6) follows since L[x] satisfies RSTdeterminacy,M is OD in L[x] and M contains the reals of L[x]. Thus,HOD M T,A,B2|= ZFC + There is a T-strong cardinal.x ,

6. Definable Determinacy 161Since, by arrangement, HOD M T,A,B = HODL[x] ∩ V L[x] ω, it follows that2HOD L[x] |= ZFC + There is a T-strong cardinal.Since T ′ was arbitrary, the theorem follows.We close with four limitative results. The first result motivates the needfor the notion of strategic determinacy by showing that strategic determinacydoes not follow trivially from OD-determinacy in the sense that for some ODbasis there are OD prestrategies.Theorem 6.11. Assume ZF. Then for each non-empty OD set B ⊆ ω ω ,there is an OD set A ⊆ ω ω such that there is no OD prestrategy in A whichis winning with respect to the basis B.Proof. Assume for contradiction that there is a set B ⊆ ω ω which is ODand such that for all OD sets A ⊆ ω ω there is an OD prestrategy f A in Awhich is winning with respect to B. We may assume OD-determinacy since ifOD-determinacy fails then the theorem trivially holds (as clearly one cannothave a prestrategy which is winning with respect to a non-empty basis for anon-determined game).We shall need to establish three claims.Claim 1. Assume ZF. Then⋂{A ⊆ ω ω | A ∈ OD, A is Turing invariant,Proof. For each α < ω 1 , letand A contains a Turing cone } = ∅.A α = {z ∈ ω ω | ∃x, y ∈ ω ω such thatx ≡ T y T z and x codes α}.Notice that A α is OD, Turing invariant, and contains a Turing cone. Butclearly⋂α

6. Definable Determinacy 162Claim 2. Assume ZF + OD-determinacy. ThenHOD |= There is a countably complete ultrafilter on ω V 1 .Proof. Since we are not assuming AC ω (R), the proof of Theorem 2.12 doesnot directly apply. To see this note that ω1 V may not be regular in V —infact, we do not even know whether ω1 V is regular in HOD. Nevertheless, wewill be able to implement some of the previous arguments by dropping intoan appropriate model of AC ω (R). In the case of countable completeness anadditional change will be required since without AC ω (R) we cannot choosecountably many strategies as we did in the earlier proof. Letµ = {S ⊆ ω V 1 | S ∈ HOD and I has a winning strategy in G(S)},where G(S) is the game from Theorem 2.12.Subclaim 1. HOD |= µ ∩ HOD is an ultrafilter.Proof. It is clear that ω1 V ∈ µ and ∅ /∈ µ. It is also clear that if S ∈ µ andS ′ ∈ HOD ∩ P(ω1 V ) and S ⊆ S ′ then S ′ ∈ µ.Suppose that S ∈ HOD ∩ P(ω1 V ) and that II has a winning σ strategyin G(S). We claim that I has a winning strategy in G(ω1 V S). Supposefor contradiction that I does not have a winning strategy. Then, by ODdeterminacy,II has a winning strategy σ ′ . Now work in L[σ, σ ′ ]. Using1Σ ∼1 -boundedness, by the usual arguments, one can construct a play x for Iwhich is legal against both σ and σ ′ and in each case has the same associatedordinal α < ω L[σ,σ′ ]1 . This is a contradiction.We now show that if S 1 , S 2 ∈ µ then S 1 ∩ S 2 ∈ µ. Let σ 1 be a winningstrategy for I in G(S 1 ) and let σ 2 be a winning strategy for I in G(S 2 ).Suppose for contradiction that S 1 ∩ S 2 /∈ µ. Since S 1 ∩ S 2 is OD, G(S 1 ∩ S 2 )is determined and so II has a winning strategy in G(S 1 ∩ S 2 ), which impliesthat I has a winning strategy σ in G(ω1 V (S 1 ∩ S 2 )). Work in L[σ 1 , σ 2 , σ].1The strategy σ 1 witnesses (by the usual argument using Σ ∼ 1 -boundedness)that S 1 ∩ω L[σ 1,σ 2 ,σ]1 contains a club. Likewise, σ 2 witnesses that S 2 ∩ω L[σ 1,σ 2 ,σ]1contains a club and σ witnesses that (ω1 V (S 1 ∩ S 2 )) ∩ ω L[σ 1,σ 2 ,σ]1 contains aclub. This contradiction completes the proof of Subclaim 1.Subclaim 2. HOD |= µ ∩ HOD is countably complete.

6. Definable Determinacy 163Proof. Suppose for contradiction that the subclaim fails. Let 〈S i | i < ω〉 ∈HOD be such that for each i < ω, S i ∈ µ and ⋂ i

6. Definable Determinacy 164Proof. By our original supposition for contradiction recall that we let f A bean OD prestrategy which is winning with respect to B. Since A contains aTuring cone f A must be winning for Player I. This means that for all z ∈ B,for all y ∈ ω ω , f A (z)∗y ∈ A. Now let y T 〈y ∗ , z〉. We wish to show thaty ∈ A. The point is thaty ≡ T f A (z)∗y ∈ Aand since A is Turing invariant, this implies that y ∈ A.Claim 3 contradicts Claim 1, which completes the proof.The second result motivates the need for restricted strategic determinacyby showing that V =L[x]+∆ 1 2 -determinacy does not imply STB -determinacy,where B is the constructibility degree of x. Thus, in Theorem 6.9 it was necessaryto drop down to a restricted form of strategic determinacy. It alsofollows from the theorem that something close to ∆ ∼12 -determinacy is requiredto establish that ST B -determinacy holds with respect to the constructibilitydegree of x since the statement “∆ ∼12 -determinacy” is equivalent to the statement“for every y ∈ ω ω there is an inner model M such that y ∈ M andM |= ZFC + There is a Woodin cardinal.”Theorem 6.12. Assume ZF + V = L[x] for some x ∈ ω ω . Suppose ST B -determinacy, where B = {y ∈ ω ω | L[y] = L[x]}. Suppose there exists anα > ω L[x]1 such that L α [x] |= ZFC. Then for every y ∈ ω ω there is a transitivemodel M such that y ∈ M and M |= ZFC + There is a Woodin cardinal.Proof. LetA 0 = {y ∈ ω ω | ¬∃M (M is transitive ∧ y ∈ M ∧M |= ZFC + There is a Woodin cardinal)}.Suppose for contradiction that A 0 ≠ ∅. Let t ∈ A 0 . It follows that A 0contains a Turing cone of reals. Let Player I play A 0 in SG B and let f 0 beII’s response. Since Player I can win a round of A 0 by playing t, f 0 is winningfor I with respect to B, that is, for all y ∈ B, f 0 (y) ∈ A 0 . We will arrive ata contradiction by constructing a real y ∈ B such that f 0 (y) /∈ A 0 .We claim thatHOD Lα[x]f 0|= ZFC + There is a Woodin cardinal.

6. Definable Determinacy 165First note thatL[x] |= ST B f 0-determinacy.Since L α [x] is ordinal definable in L[x] (as α > ω L[x]1 and so L α [x] = L α [x ′ ]for any real x ′ such that V = L[x ′ ]) it follows thatL α [x] |= ST B f 0-determinacy.Thus, by the relativized version of Theorem 6.10,HOD Lα[x]f 0|= ZFC + There is a Woodin cardinal.Therefore f 0 ∉ A 0 .By Σ 1 2(f 0 )-absoluteness, L[f 0 ] satisfies that there is a countable transitivemodel M such that f 0 ∈ M andM |= ZFC + There is a Woodin cardinal.Since L α [x] |= “f # 0 exists” (by the effective version of Solovay’s Theorem(Theorem 2.15) there is a countable ordinal λ such that L λ [f 0 ] satisfies ZFC+“M is countable”. In L λ [f 0 ] let P be a perfect set of reals that are Cohengeneric over M. Since P is perfect in L λ [f 0 ] there is a path c ∈ [P] whichcodes x in the sense that c T x.Our desired real y is 〈f 0 , c〉. To see that 〈f 0 , c〉 ∈ B note that L λ (〈f 0 , c〉)can compute x and hence L ω1 [〈f 0 , c〉] = L ω1 [x]. To see that f 0 (〈f 0 , c〉) ∉ A 0note that since c is Cohen generic over M, the model M[c] is a transitivemodel containing f 0 (〈f 0 , c〉) satisfying ZFC + “There is a Woodin cardinal”.This is a contradiction.The third result shows that Martin’s “lightface form” of Third Periodicity(Theorem 6.3) does not generalize to higher levels. In fact, the result showsthat even ZFC + OD-determinacy (assuming consistency of course) does notimply that for every Σ 1 4 game which Player I wins, Player I has a ∆ 1 5 strategy(or even an OD strategy). The reason that the “lightface form” of ThirdPeriodicity holds at the level of Σ 1 2 but not beyond is that in Third Periodicity1boldface determinacy is used to get scales but in ZF + DC we get Scale(Σ ∼2 )for free.Theorem 6.13. Assume ZF+V = L[x]+OD-determinacy for some x ∈ ω ω .There is a Π 1 2 set of reals which contains a Turing cone but which does notcontain a member in HOD.

6. Definable Determinacy 166Proof. Consider the setA = {y ∈ ω ω | for all additively closed λ < ω 1 ,for all z T y, if x ∈ OD L λ[z] then x T y}.This is a Π 1 2 set. Notice that each y ∈ A witnesses that R ∩ HODL[z] iscountable for each z T y.Claim 1. A contains a Turing cone.Proof. For y ∈ ω ω and α such that ω < α < ω 1 , let R α,y be the set of realswhich are ordinal definable in L α [y] and let < α,y the canonical well-orderingof R α,y , where we arrange that < α,y is an initial segment of < α ′ ,y when α < α ′ .For y ∈ ω ω , let R y = ⋃ {R α,y | ω < α < ω 1 } and let < y be the induced orderon R y (where we order first by α and then by < α,y ). Let z y α be the αth realin < y and let ϑ y be the ordertype of < y . Notice that R α,y , R y , < α,y , < y , z y αand ϑ y depend only on the Turing degree of y.Our strategy is to “freeze out” the values of R y and < y on a Turing coneof y. For α < ω 1 , the setA α = {y ∈ ω ω | ϑ y > α}is OD and hence, by OD-determinacy, either it or its complement containsa Turing cone. Moreover, if A α contains a Turing cone and ᾱ < α then Aᾱcontains a Turing cone. Thus,A ′ = {α < ω 1 | A α contains a Turing cone}is an initial segment of ω 1 . For each α ∈ A ′ , and for each y ∈ ω ω , thestatement “ϑ y > α and zα y (n) = m” is an OD-statement about y. So, byOD-determinacy, the value of zα y is fixed for a Turing cone of y. We writez α for this stable value. It follows that 〈z α | α ∈ A ′ 〉 is a definable wellorderingof reals and hence, by OD-determinacy, A ′ must be countable (bythe effective version of Solovay’s theorem (Theorem 2.15) and the argumentin the Claim in Theorem 5.9). Let ϑ = sup{α + 1 | α ∈ A ′ }. Finally, letR ∞ = {z α | α < ϑ} and < ∞ = {(z α , z β ) | α < β < ϑ}. We claim that for aTuring cone of y, R y = R ∞ . To see this let y ∈ L[x] be such that x T y (so,in particular L[x] = L[y]) and y belongs to all of the (countably many) conesfixing z α for α ∈ A ′ . Then ϑ y = ϑ and R y = R ∞ . (In fact, R y = R HOD .)Let z 0 be such that for all z T z 0 , R z = R z0 = R ∞ . Since R ∞ iscountable, we can choose y 0 T z 0 such that R ∞ T y 0 . Then for all z T y 0 ,R z = R ∞ T y 0 , that is, y 0 ∈ A. Likewise, if y T y 0 , then y ∈ A.

6. Definable Determinacy 167Claim 2. A ∩ HOD L[x] = ∅.Proof. Suppose for contradiction that y ∈ A ∩ HOD L[x] . Since y ∈ A, ywitnesses that R z is countable for all z T y. Let z be such thatThen since y ∈ HOD L[x] ,which is impossible.This completes the proof.R x = R z .HOD L[x] |= R is countable,The final result is a refinement of a theorem of Martin ([4, Theorem 13.1]).It shows that ZF + DC + ∆ 1 2-determinacy implies that for a Turing cone ofx, HOD L[x] has a ∆ 1 3 well-ordering of reals and hence that for a Turing coneof x, ∆ 1 2-determinacy fails in HOD L[x] .Theorem 6.14. Assume ZF+V = L[x]+∆ 1 2 -determinacy, for some x ∈ ωω .Then in HOD there is a ∆ 1 3-well-ordering of the reals.Proof. For y ∈ ω ω and α such that ω < α < ω 1 , let R α,y , < α,y , z y α , R y, < y ,and ϑ y be as in the proof of Theorem 6.13. Let A ′ and R ∞ be as in theproof of Theorem 6.13. The argument of Claim 1 of Theorem 6.13 showsthat R ∞ = R HOD : To see this let x ′ ∈ L[x] be such that x T x ′ (so, inparticular L[x] = L[x ′ ]) and x ′ belongs to all of the (countably many) conesfixing z α for α ∈ A ′ . Then ϑ x ′ = ϑ and R ∞ = R x ′ = R ∩ HOD L[x′] = R HOD .Notice that∃y 0 ∀y T y 0 ∀ω < α < ω 1 (R α,y ⊆ R ∞ ∧ < α,y < ∞ ),where denotes ordering by initial segment; x ′ as above is such a y 0 . SinceR ∞ and < ∞ are countable they can be coded by a real. Let y 0 be the baseof the above cone and let a be a real coding 〈y 0 , R ∞ , < ∞ 〉. The statement“a codes 〈y 0 , R ∞ , < ∞ 〉 and for all y T y 0 , for all α < ω 1 , R α,y ⊆ R ∞ and< α,y < ∞ ” is a Π 1 2 truth about a. Writing ψ(a) for this statement we havethe following Π 1 3-definitions (in L[x]) of ω ω ∩ HOD and < ∞ :z ∈ R ∞ ↔ ∀a [ a codes (z, R,

6. Definable Determinacy 168andz 0 < ∞ z 1 ↔ ∀a [ a codes (z, R,

6. Definable Determinacy 169cardinal and a ∆ 1 3 well-ordering of reals. It follows that ∆1 2-determinacy failsin HOD L[x] for a Turing cone of x.Some interesting questions remain. For example: Does HOD L[x] satisfyGCH, for a Turing cone of x? Is HOD L[x] a fine-structural model, for a Turingcone of x? We will return to this topic in Section 8.6.2. Boldface Definable DeterminacyIn this section we will work in ZF + AD. Our aim is to extract the optimalamount of large cardinal strength from boldface determinacy by constructinga model of ZFC that contains ω-many Woodin cardinals.We shall prove a very general theorem along these lines. Our strategyis to iteratively apply Theorem 5.43. Recall that this theorem states thatunder ZF + AD, for a Y -cone of x,whereHOD Y,a,[x]Y |= ω HOD Y,a,x2 is a Woodin cardinal,[x] Y = {z ∈ ω ω | HOD Y,z = HOD Y,x }.We start by taking a to be the empty set. By Theorem 5.43, there exists anx 0 such that for all x Y x 0 ,HOD Y,[x]Y |= ω HOD Y,x2 is a Woodin cardinal.To generate a model with two Woodin cardinals we would like to applyTheorem 5.43 again, this time taking a to be [x 0 ] Y . This gives us an x 1 Y x 0such that for all x Y x 1 ,HOD Y,[x0 ] Y ,[x] Y|= ω HOD Y,[x 0 ] Y ,x2 is a Woodin cardinaland we would like to argue thatHOD Y,[x0 ] Y ,[x] Y|= ω HOD Y,x 02 < ω HOD Y,[x 0 ] Y ,x2 are Woodin cardinals.But there are two difficulties in doing this. First, in the very least, we needto ensure thatω HOD Y,x 02 < ω HOD Y,[x 0 ] Y ,x2

6. Definable Determinacy 170and this is not immediate. Second, in moving to the larger model we needto ensure that we have not collapsed the first Woodin cardinal; a sufficientcondition for this is thatP(ω HOD Y,x 02 ) ∩ HOD Y,[x0 ] Y ,[x] Y= P(ω HOD Y,x 02 ) ∩ HOD Y,[x0 ] Y,but again this is not immediate. It turns out that both difficulties can beovercome by taking x to be of sufficiently high “Y -degree”. This will bethe content of an elementary observation and a “preservation” lemma. Oncethese two hurdles are overcome we will be able to generate models with nWoodin cardinals for each n < ω. We shall then have to take extra steps toensure that we can preserve ω-many Woodin cardinals. This will be achievedby shooting a Prikry sequence through the Y -degrees and proving an associated“generic preservation” lemma.6.15 Remark. It is important to note that in contrast to Theorem 5.42here the degree notion in Theorem 5.43 does not depend on a and this isinstrumental in iteratively applying the theorem to generate several Woodincardinals. In contexts such as L(R) where HOD “relativizes” in the sensethat HOD a = HOD[a], one could also appeal to Theorem 5.42, since in sucha case HOD Y,a,[x]Y,a = HOD Y,a,[x]Y . Our reason for not taking this approachis twofold. First, it would take us too far afield to give the argument thatHOD a = HOD[a] in, for example, L(R). Second, it is of independent interestto work in a more general setting.We shall be working with the “Y -degrees”D Y = {[x] Y | x ∈ ω ω }.Let µ Y be the cone filter over D Y . As noted earlier, the argument of Theorem2.9 shows that µ Y is an ultrafilter. Also, by Theorem 2.8 we know that µ Yis countably complete.Lemma 6.16 (Preservation Lemma). Assume ZF + AD. Suppose Y isa set, a ∈ H(ω 1 ), and α < ω 1 . Then for a Y -cone of x,P(α) ∩ HOD Y,a,[x]Y = P(α) ∩ HOD Y,a .Proof. The right-to-left direction is immediate. Suppose for contradictionthat the left-to-right direction fails. For sufficiently large [x] Y , letf([x] Y ) = least Z ∈ P(α) ∩ HOD Y,a,[x]Y HOD Y,a ,

6. Definable Determinacy 171where the ordering is the canonical ordering of OD Y,a,[x]Y . This function isdefined for a Y -cone of x and it is OD Y,a . Let Z 0 ∈ P(α) be such thatξ ∈ Z 0 iff ξ ∈ f([x] Y ) for a Y -cone of x.Since α is countable and since µ Y is countably complete Z 0 = f([x] Y ) forsufficiently large x. Thus, Z 0 ∈ HOD Y,a , which is a contradiction.We are now in a position to iteratively apply Theorem 5.43 to generatea model with n Woodin cardinals.Step 0. By Theorem 5.43, let x 0 be such that for all x Y x 0 ,HOD Y,[x]Y |= ω HOD Y,x2 is a Woodin cardinal.Step 1. Recall that ω1V is strongly inaccessible in any inner model ofZFC, by the Claim of Theorem 5.9. It follows that ω HOD Y,x 02 < ω1 V and sowhen we choose x 1 Y x 0 we may assume that x 1 codes ω HOD Y,x 02 . Thus,there exists an x 1 Y x 0 such that for all x Y x 1 ,ω HOD Y,x 02 < ω HOD Y,[x 0 ] Y ,x2 ,and, by the Preservation Lemma (taking a to be [x 0 ] Y ),P(ω HOD Y,x 02 ) ∩ HOD Y,[x0 ] Y ,[x] Y= P(ω HOD Y,x 02 ) ∩ HOD Y,[x0 ] Y,and, by Theorem 5.9 (taking a to be [x 0 ] Y ),It follows thatHOD Y,[x0 ] Y ,[x] Y|= ω HOD Y,[x 0 ] Y ,x2 is a Woodin cardinal.HOD Y,[x0 ] Y ,[x 1 ] Y|= ω HOD Y,x 02 < ω HOD Y,[x 0 ] Y ,x 12 are Woodin cardinals.Step n+1. It is useful at this stage to introduce a piece of notation: Forx 0 Y · · · Y x n+1 , letandδ 0 (x 0 ) = ω HOD Y,x 02δ n+1 (x 0 , . . .,x n+1 ) = ω HOD Y,〈[x 0 ] Y ,...,[xn] Y 〉,x n+12 .

6. Definable Determinacy 172Suppose that we have chosen x 0 Y x 1 Y · · · Y x n such thatHOD Y,〈[x0 ] Y ,...,[x n] Y 〉 |= δ 0 (x 0 ) < · · · < δ n (x 0 , . . .,x n )are Woodin cardinals.Again, since ω1 V is strongly inaccessible in any inner model of ZFC, it followsthat each of these ordinals is countable in V and so when we choose x n+1 Yx n we may assume that x n+1 collapses these ordinals. Thus, there exists anx n+1 Y x n such that for all x Y x n+1 ,δ n (x 0 , . . .,x n ) < δ n+1 (x 0 , . . .,x n , x)and, by the Preservation Lemma (taking a to be 〈[x 0 ] Y , . . .,[x n ] Y 〉),P(δ n (x 0 , . . .,x n )) ∩HOD Y,〈[x0 ] Y ,...,[x n] Y 〉,[x] Y= P(δ n (x 0 , . . .,x n )) ∩ HOD Y,〈[x0 ] Y ,...,[x n] Y 〉and, by Theorem 5.43 (taking a to be 〈[x 0 ] Y , . . .,[x n ] Y 〉),HOD Y,〈[x0 ] Y ,...,[x n] Y 〉,[x] Y|= δ n+1 (x 0 , . . ., x n , x) is a Woodin cardinal.It follows thatHOD Y,〈[x0 ] Y ,...,[x n] Y 〉,[x] Y|= δ 0 (x 0 ) < · · · < δ n+1 (x 0 , . . .,x n , x)are Woodin cardinals.We now need to ensure that when we do the above stacking for ω-manystages, the Woodin cardinals δ n (x 0 , . . ., x n ) are preserved in the final model.This is not immediate since, for example, if we are not careful then thereals x 0 , x 1 , . . . might code up a real that collapses sup n

6. Definable Determinacy 173(1) [x i+1 ] Y ∈ F(〈[x 0 ] Y , . . .,[x i ] Y 〉) for all i n and(2) F ∗ (p) ⊆ F(p) for all p ∈ D

6. Definable Determinacy 174We begin by noting the following three persistence properties which will aidus in shrinking F so as to decide ϕ. First, if ρ(q) = ∞ then setA q = { a | ρ(q ⌢ a) = ∞ }and notice that A q ∈ µ Y since otherwise we would have { a | ρ(q ⌢ a) ∈ On } ∈µ Y (as µ Y is an ultrafilter) and letting β = sup{ ρ(q ⌢ a) | ρ(q ⌢ a) ∈ On } wewould have q ∈ Z β+1 , a contradiction. Second, if ρ(q) ∈ On { 0 } then setB q = { a | ρ(q ⌢ a) < ρ(q) }and notice that B q ∈ µ Y since ρ(q) is clearly a successor, say α + 1, andq ∈ Z α and so (by our claim) D α q ∈ µ Y ; but D α q ⊆ B q. Third, if ρ(q) = 0then setC q = { a | ρ(q ⌢ a) = 0 }and notice that C q ∈ µ Y since clearly q ∈ Z 0 and so, by the claim, D 0 q ∈ µ Y ;but D 0 q ⊆ C q.Now either ρ(p) = ∞ or ρ(p) ∈ On.Claim 1. If ρ(p) = ∞ then there is an F ∗ such that 〈p, F ∗ 〉 PY 〈p, F 〉 and〈p, F ∗ 〉 ⊩ ¬ϕ.Proof. Define F ∗ as follows:F ∗ (q) ={F(q) ∩ A qF(q)if ρ(q) = ∞otherwise.Suppose that it is not the case that 〈p, F ∗ 〉 ⊩ ¬ϕ. Then ∃〈q, G〉 PY 〈p, F ∗ 〉such that 〈q, G〉 ⊩ ϕ. But then q is such that ρ(q) = 0. However, F ∗ witnessesthat in fact ρ(q) = ∞: Suppose q = p ⌢ a 0⌢ · · · ⌢a k . Since ρ(p) = ∞ anda 0 ∈ F ∗ (p), we have that a 0 ∈ A p and so ρ(p ⌢ a 0 ) = ∞. Continuing in thismanner, we get that ρ(q) = ∞. This is a contradiction.Claim 2. If ρ(p) ∈ On then there is an F ∗ such that 〈p, F ∗ 〉 PY 〈p, F 〉 and〈p, F ∗ 〉 ⊩ ϕ.Proof. Define F ∗ as follows:⎧⎪⎨ F(q) ∩ B q if ρ(q) ∈ On {0}F ∗ (q) = F(q) ∩ C q if ρ(q) = 0⎪⎩F(q) otherwise.

6. Definable Determinacy 175We claim that 〈p, F ∗ 〉 ⊩ ϕ. Assume not. Then ∃〈q, G〉 PY 〈p, F ∗ 〉 such that〈q, G〉 ⊩ ¬ϕ. Since 〈q, G〉 PY 〈p, F ∗ 〉 and ρ(p) ∈ On we have that ρ(q) ∈ On(by an easy induction using the definition of F ∗ ). We may assume that 〈q, G〉is chosen so that ρ(q) is as small as possible. But then ρ(q) = 0 as otherwisethere is an a such 〈q ⌢ a, G〉 PY 〈q, G〉, 〈q ⌢ a, G〉 ⊩ ¬ϕ and ρ(q ⌢ a) < ρ(q),contradicting the minimality of ρ(q). Now since ρ(q) = 0, q ∈ Z 0 and so∃G ′ 〈q, G ′ 〉 ⊩ ϕ. But this is a contradiction since 〈q, G ′ 〉 is compatible with〈q, G〉 and 〈q, G〉 ⊩ ¬ϕ.This completes the proof of the lemma.We can now obtain the following “generic preservation” lemma.Lemma 6.19 (Generic Preservation Lemma). There exists an F suchthat if G ⊆ P Y is V -generic and 〈∅, F 〉 ∈ G and 〈[x i ] Y | i < ω〉 is the genericsequence associated to G, then, for all i < ω,P(δ i (x 0 , . . ., x i )) V [G] ∩ HOD V [G]Y,〈[x j ] Y |j

6. Definable Determinacy 176that the left-to-right inclusion fails. Let A ⊆ α be

7. Second-Order Arithmetic 177By the Generic Preservation Lemma it suffices to show that for each n < ωHOD Y,〈[x0 ] Y ,...,[x n] Y 〉 |= δ 0 (x 0 ) < · · · < δ n (x 0 , . . .,x n )are Woodin cardinals,which follows by genericity and the argument for the finite case.As an interesting application of this theorem in conjunction with theDerived Model Theorem (Theorem 8.12), we obtain Kechris’ theorem thatunder ZF + AD, DC holds in L(R). This alternate proof is of interest sinceit is entirely free of fine structure and it easily generalizes.Theorem 6.21 (Kechris). Assume ZF + AD. Then L(R) |= DC.Proof Sketch. Work in ZF + AD + V = L(R). Let Y = ∅ and let N =HOD V [G]〈[x i ] Y |i

7. Second-Order Arithmetic 178Theorem 7.1. Assume ZF + V = L[x] + ∆ 1 2 -determinacy, for some x ∈ ωω .Suppose N is such that(1) On ⊆ N ⊆ HOD L[x] and(2) N |= δ is a Woodin cardinal.Then δ ω L[x]2 .However, it turns out that ω L[x]1 can be a Woodin cardinal in an inner modelthat overspills HOD L[x] .Theorem 7.2. Assume ZF + V = L[x] + ∆ 1 2 -determinacy, for some x ∈ ωω .Then there exists an N ⊆ L[x] such thatMoreover, this result is optimal.N |= ZFC + ω L[x]1 is a Woodin cardinal.Theorem 7.3. Assume ZF + ∆ 1 2-determinacy. Then there is a real x suchthat(1) L[x] |= ∆ 1 2-determinacy, and(2) for all α < ω L[x]1 , α is not a Woodin cardinal in any inner model Nsuch that On ⊆ N.In Section 7.1 we prove Theorem 7.2. More precisely, we prove the following:Theorem 7.4. Assume ZF+DC+∆ 1 2-determinacy. Then for a Turing coneof x,HOD L[x][x] T|= ZFC + ω L[x]1 is a Woodin cardinal.This involves relativizing the previous construction to the Turing degree of x,replacing the notions that concerned reals (for example, winning strategies)with relativized analogues that concern only those reals in the Turing degreeof x.In Section 7.2 we show that the relativized construction goes through inthe setting of second-order arithmetic.Theorem 7.5. Assume that PA 2 + ∆ 1 2 -determinacy is consistent. ThenZFC + “On is Woodin” is consistent.

7. Second-Order Arithmetic 179Here PA 2 is the standard axiomatization of second-order arithmetic (withoutAC). The statement that On is Woodin is to be understood schematically.Alternatively, one could work with the conservative extension GBC of ZFCand the analogous conservative extension of PA 2 . This would enable one tofuse the schema expressing that On is Woodin into a single statement.7.1. First LocalizationTo prove Theorem 7.4 we have to prove an analogue of the Generation Theoremwhere ω 2 is replaced by ω 1 . The two main steps are (1) getting a suitablenotion of strategic determinacy and (2) getting definable prewellorderings forall ordinals less than ω 1 .For x ∈ ω ω we “relativize” our previous notions to the Turing degree of x.The relativized reals are R x = {y ∈ ω ω | y T x}. Fix A ⊆ R x . A relativizedstrategy for I is a function σ : ⋃ n

7. Second-Order Arithmetic 180We caution the reader that in the context of the relativized notions weare dealing only with definable versions of relativized determinacy such asOD-[x] T -determinacy and SG-[x] T -determinacy. In fact, full relativized determinacycan never hold. But as we shall see both OD-[x] T -determinacyand SG-[x] T -determinacy can hold.Theorem 7.6. Assume ZF + DC + ∆ 1 2-determinacy. Let T be the theoryZFC − Replacement + Σ 2 -Replacement. There is a real x 0 such that for allreals x and for all ordinals λ if x 0 ∈ L λ [x] and L λ [x] |= T, then L λ [x] |=OD-[x] T -determinacy.Proof. The proof is similar to that of Theorem 6.6. Assume for contradictionthat for every real x 0 there is an ordinal λ and a real x such that x 0 ∈ L λ [x]and L λ [x] |= T + ¬OD-[x] T -determinacy, where T = ZFC − Replacement +Σ 2 -Replacement. As before, by the Löwenheim-Skolem theorem and Σ 1 2 -determinacy the ordinal{µλ (L λ [x] |= T + ¬OD-[x] T -determinacy) if such a λ existsλ(x) =undefinedotherwiseis defined for a Turing cone of x. For each x such that λ(x) is defined, letA x ⊆ R x be the (OD [x]T ) Lλ(x)[x] -least counterexample.Consider the gameI a, bII c, dwhere, letting p = 〈a, b, c, d〉, I wins iff λ(p) is defined and L λ(p) [p] |= “a∗d ∈A p ”, where a and d can be thought of as strategies. This game is Σ 1 2, hencedetermined.We arrive at a contradiction by showing that neither player can win.Case 1: I has a winning strategy σ 0 .Let x 0 T σ 0 be such that for all x T x 0 , λ(x) is defined. We claimthat L λ(x0 )[x 0 ] |= “I has a relativized winning strategy σ in A x 0”, which is acontradiction. The relativized strategy σ is derived as follows: Given d↾n ∈ω n have II play x 0 ↾n, d↾n in the main game. Let a↾n, b↾n be σ 0 ’s responsealong the way and let a(n) be σ 0 ’s next move. Then set σ(d↾n) = a(n).(Clearly, σ is continuous, and the real a = σ(d) it defines is to be thought ofas coding a strategy for Player I.) This strategy σ is clearly recursive in σ 0 ,hence it is a relativized strategy.

7. Second-Order Arithmetic 181It remains to show that for every d ∈ R x0 , σ∗d ∈ A x 0. The point isthat for d ∈ R x0 , p ≡ T x 0 , where p = 〈a, b, x 0 , d〉 is the play obtainedby letting 〈a, b〉 = (σ 0 ∗〈x 0 , d〉) I . It follows that λ(p) = λ(x 0 ) and henceL λ(p) [p] = L λ(x0 )[x 0 ] and A p = A x 0. Thus, L λ(x0 )[x 0 ] |= “σ(d)∗d ∈ A x 0”. SoL λ(x0 )[x 0 ] |= “σ is a relativized winning strategy for I in A x 0”.Case 2: II has a winning strategy τ 0 .Let x 0 T τ 0 be such that for all x T x 0 , λ(x) is defined and λ(x) λ(x 0 ). Given a↾(n + 1) ∈ ω n+1 have I play a↾(n + 1), x 0 ↾(n + 1) in the maingame. Let c↾n, d↾n be τ 0 ’s response along the way. Then set τ(a↾n) = d(n).This strategy is clearly recursive in τ 0 , hence it is a relativized strategy, and,as above, L λ(x0 )[x 0 ] |= “τ is a relativized winning strategy for II in A x 0”.Theorem 7.7. Assume ZF+DC+∆ 1 2-determinacy. Then for a Turing coneof x,L[x] |= ST-[x] T -determinacy.Proof. The proof is a straightforward variant of the proof of Theorem 6.9.In fact it is simpler. We note the major changes.As before we assume that V =L[x] and show that there is a real z 0 with thefeature that if z 0 ∈ L λ [z] and L λ [z] |= T, then L λ [z] |= ST-[x] T -determinacy.Assume for contradiction that this fails. For z ∈ ω ω , let{µλ (L λ [z] |= T + ¬ST-[x] T -determinacy) if such a λ existsλ(z) =undefinedotherwise.The following is immediate.Claim 1. For a Turing cone of z, λ(z) is defined.For each z in the cone of Claim 1 Player I has a canonical strategy σ zthat depends only on the Turing degree of z, the point being that if y ≡ T zthen L λ(y) [y] = L λ(z) [z].As before our aim is to obtain a contradiction by defeating σ z for somez in the Turing cone of Claim 1. We do this by constructing a sequence ofgames G 0 , G 1 , . . .,G n , . . . such that I must win via σ 0 , σ 1 , . . .,σ n , . . . and, fora cone of z, the winning strategies give rise to prestrategies f z 0 , fz 1 , . . .,fz n , . . .that constitute a non-losing play against σ z in (SG-[x] T ) L λ(z)[z] .

7. Second-Order Arithmetic 182Step 0. Consider (in L[x]) the game G 0IIIǫ a, bc, dwhere ǫ is either 1 or 2 and, letting p = 〈a, b, c, d〉, I wins iff(1) p satisfies the condition on z in Claim 1 (so σ p makes sense) and(2) ǫ = 1 iff L λ(p) [p] |= “a∗d ∈ A p 0 ”, where Ap 0 = σp (∅).Claim 2. I has a winning strategy σ 0 in G 0 .Proof. Assume for contradiction that I does not have a winning strategyin G 0 . Then, by Σ 1 2-determinacy, II has a winning strategy τ 0 in G 0 . Letz 0 T τ 0 be such that for all z T z 0 ,(1) z satisfies the conditions of Claim 1 and(2) if λ and z are such that z 0 ∈ L λ [z] and L λ [z] |= T then L λ [z] |=OD-[x] T -determinacy (by Theorem 7.6).Consider A z 00 = σ z 0(∅). Since L λ(z0 )[z 0 ] |= OD-[x] T -determinacy, assumewithout loss of generality that L λ(z0 )[z 0 ] |= “σ is a relativized winning strategyfor I in A z 00 ”. We use τ 0 to defeat this relativized strategy. Run G 0 accordingto τ 0 , having Player I (falsely) predict that Player I wins the auxiliary game,while steering into L λ(z0 )[z 0 ] by playing b = z 0 and using σ to respond to τ 0on the auxiliary play:I 1 (σ∗d) I , z 0II c, dThe point is that p ≡ T z 0 (since σ, τ 0 ∈ R z0 ) and so λ(p) = λ(z 0 ). Thusthe “steering problem” is immediately solved and we have a contradiction asbefore.Since the game is Σ 1 2 for Player I, Player I has a ∆ 1 3 strategy σ 0 , byTheorem 6.5.Claim 3. For every real z T σ 0 there is a prestrategy f0 z such that f0 z isrecursive in σ 0 as in Claim 1 and f0 z is a non-losing first move for II againstσ z in (SG-[x] T ) Lλ(z)[z] .

7. Second-Order Arithmetic 183Proof. Fix z T σ 0 as in Claim 1 and consider A z 0 = σz (∅). Let f0 z bethe prestrategy derived from σ 0 as follows: Given y↾n and d↾n have II playy↾n, d↾n in G 0 . Let ǫ, a↾n, b↾n be σ 0 ’s response along the way and let a(n)be σ 0 ’s next move. Then let f0 z(y↾n) = a(n). We have that fz 0 is recursive inσ 0 T z and for y ∈ [z] T , f0(y) z ∈ R z . It remains to see that for y ∈ [z] T , f0(y)zis a relativized winning strategy for I in A z 0 . The point is that since y ∈ [z] T,λ(y) = λ(z) and so L λ(y) [y] = L λ(z) [z] and A y 0 = A z 0. For d ∈ R z , by definitionf0 z(y)∗d = a∗d where a is such that (σ 0∗〈y, d〉) I = 〈ǫ, a, b〉. So, lettingp = 〈a, b, y, d〉 we have p ≡ T y. Thus, ǫ = 1 iff L λ(z) [z] |= “f0 z(y)∗d ∈ Az 0 ”.Step n+1. Assume that we have defined (in L[x]) games G 0 , . . .,G nwith winning strategies σ 0 , . . .,σ n ∈ HOD such that for all z T 〈σ 0 , . . .,σ n 〉as in Claim 1 there are prestrategies f z 0 , . . .,fz n such that fz i is recursive in〈σ 0 , . . .,σ i 〉 (for all i n) and f z 0, . . .f z n is a non-losing partial play for II in(SG-[x] T ) L λ(z)[z] .Consider (in L[x]) the game G n+1IIIǫ a, bc, dwhere ǫ is 1 or 2 and, letting p = 〈a, b, c, d, σ 0 , . . .,σ n 〉, I wins iff(1) p satisfies the condition on z in Claim 1 (so σ p makes sense) and(2) ǫ = 1 iff L λ(p) [p] |= “a∗d ∈ A p n+1”, where A p n+1 is I’s response via σ p toII’s partial play f p 0,...,f p nThis game is Σ 1 2 (σ 0, . . .,σ n ) (for Player I) and hence determined (sinceσ 0 , . . ., σ n ∈ HOD and we have OD-determinacy).Claim 4. I has a winning strategy σ n+1 in G n+1 .Proof. The proof is as before, only now we use the relativized version ofTheorem 7.6 to enforce OD σ0 ,...,σ n-[x] T -determinacy.Since the game is Σ 1 2 (σ 0, . . .,σ n ) for Player I, Player I has a ∆ 1 3 (σ 0, . . ., σ n )strategy σ n+1 , by the relativized version of Theorem 6.5.Claim 5. For every real z T 〈σ 0 , . . .,σ n 〉 there is a prestrategy fn+1 z suchthat fn+1 z is recursive in 〈σ 0, . . .,σ n+1 〉 and f0 z, . . .,fz n+1 is a non-losing firstmove for II against σ z in (SG-[x] T ) Lλ(z)[z] .

7. Second-Order Arithmetic 184Proof. The proof is just like the proof of Claim 3.Finally, letting z ∞ as in Claim 1 be such that z ∞ T z n for all n wehave that f0 z∞ , . . .,fn z∞ , . . . defeats σ z∞ in (SG-[x] T ) L λ(z ∞ )[z ∞] , which is acontradiction.Theorem 7.8. Assume ZF + DC. Then for every x ∈ ω ω and for everyα < ω L[x]1 there is an OD [x]T surjection ρ : [x] T → α.Proof. First we need to review Vopěnka’s theorem. Work in L[x] and letd = [x] T . LetB ′ d = {A ⊆ d | A ∈ OD d},ordered under ⊆. There is an OD d isomorphism π between (B ′ d , ⊆) and apartial ordering (B d , ) in HOD d .Claim 1. (B d , ) is complete in HOD d and every real in d is HOD d -genericfor B d .∨Proof. For completeness consider S ⊆ B d in HOD d . We have to show thatS exists. Let S′= π −1 [S]. Then ∨ S ′ = ⋃ S ′ ∈ B ′ d as this set is clearlyOD d . So ∨ S = π( ∨ S ′ ).Now consider z ∈ d. Let G ′ z = {A ∈ B ′ d | z ∈ A} and let G z = π[G ′ z]. Weclaim that G z is HOD d -generic for B d . Let S ⊆ B d be a maximal antichain.So ∨ S = 1. Let S ′ = π −1 [S]. Note ∨ S ′ = d. Thus there exists a b ∈ S suchthat z ∈ π(b). So G z is HOD d -generic for B d . Now the map f : ω → B d ,defined by f(n) = π({x ∈ d | n ∈ x}), is in HOD d . Moreover, n ∈ z ifff(n) ∈ G z . Thus z ∈ HOD d [G z ].Notice that HOD d [G] = L[x] for every G = G z that is HOD d -generic(where z ∈ d) since such a generic adds a real in [x] T . Thus, if HOD d ≠ L[x]then B d is non-trivial. This is a key difference between our present settingand that of Vopěnka’s—in general our partial order does not have atoms.If L[x] = HOD d then clearly for each α < ω L[x]1 there exists a surjectionρ : d → α such that ρ ∈ OD d . So we may assume that L[x] ≠ HOD d . Thus,for every z ∈ dω L[x]1 = ω HOD d[G z]1 .Claim 2. Assume ZFC. Suppose λ is an uncountable regular cardinal, B isa complete Boolean algebra, and V B |= λ = ω 1 . Then for every α < λ thereis an antichain in B of size |α|.

7. Second-Order Arithmetic 185Proof. If λ is a limit cardinal then since B collapses all uncountable cardinalsbelow λ it cannot be ¯λ-c.c. for any uncountable cardinal ¯λ < λ.Suppose λ = ¯λ + . We need to show that there is an antichain of size ¯λ. If¯λ > ω then this is immediate since B collapses ¯λ and so it cannot be ¯λ-c.c.So assume ¯λ = ω. There must be an antichain of size ω since not everycondition in B is above an atom.Letting λ = ω L[x]1 , we are in the situation of the claim. So, for every α < λthere is an antichain S α in B of size |α|. Letting S ′ α = π −1 [S α ] we have thatS ′ α is an OD d subset of B ′ d consisting of pairwise disjoint OD d subsets of d.Picking an element from each set we get an OD d -surjection ρ : d → α.Theorem 7.9. Assume ZF+DC+∆ 1 2-determinacy. Then for a Turing coneof x,HOD L[x][x] T|= ω L[x]1 is a Woodin cardinal.Proof. For a Turing cone of x, L[x] |= OD [x]T -determinacy (by the relativizedversion of Theorem 6.6) and L[x] |= ST-[x] T -determinacy (by Theorem 7.7).Let x be in this cone and work in L[x]. Let d = [x] T . Since L[x] |= OD [x]T -determinacy, ω L[x]1 is strongly inaccessible in HOD d . Let H ⊆ ω L[x]HOD d ∩ V L[x] ω 1Let A = 〈A α | α < ω L[x]. Fix T ∈ P(ω L[x]1 ) ∩ OD d and let T 0 ⊆ ω L[x]1 code1 code T and H.1 〉 be such that A α is an OD d prewellordering oflength greater than or equal to α (by Theorem 7.8). Let B = d. Considerthe structureM = ( L L[x] ω(R)[T 0 , A, B] ) L[x].1We claim thatHOD M |= There is a T 0 -strong cardinal,which completes the proof as before. The reason is that we are in the situationof the Generation Theorem, except with ω L[x]1 replacing ω L[x]2 andST-[x] T -determinacy replacing ST B -determinacy. The proof of the GenerationTheorem goes through unchanged. One just has to check that all ofthe operations we performed before (which involved definability in variousparameters) are in fact recursive in the relevant parameters.

7. Second-Order Arithmetic 1867.2. Second LocalizationWe now wish to show that the above construction goes through when wereplace ZF + DC with PA 2 . Notice that if we had ∆ ∼12 -determinacy then thiswould be routine.Theorem 7.10. Assume PA 2 + ∆ ∼12 -determinacy. Then for all reals x, thereis a model N such that x ∈ N andN |= ZFC + There is a Woodin cardinal.Proof. Working in PA 2 if one has ∆ ∼12 -determinacy then for every x ∈ ω ω , x #exists. It follows that for all x ∈ ω ω , there is an ordinal α < ω 1 such thatL α [x] |= ZFC. Using ∆ ∼12 -determinacy one can find a real x 0 enforcing ODdeterminacy.Thus we have a model L α0 [x 0 ] satisfying ZFC+V = L[x 0 ]+ODdeterminacyand this puts us in the situation of Theorem 6.10.The situation where one only has ∆ 1 2-determinacy is bit more involved.Theorem 7.11. Assume that PA 2 + ∆ 1 2-determinacy is consistent. ThenZFC + “On is Woodin” is consistent.Proof Sketch. First we pass to a theory that more closely resembles the theoryused to prove Theorem 7.9. In PA 2 one can simulate the constructionof L ω1 [x]. Given a model M of PA 2 and a real x ∈ M, there is a definableset of reals A coding the elements of L ω1 [x]. One can then show that the“inner model” L ω1 [x] satisfies ZFC−Power Set+V =L[x] (using, for exampleComprehension to get Replacement). Thus, ZFC − Power Set + V =L[x] isa conservative extension of PA 2 .Next we need to arrange a sufficient amount of definable determinacy.The most natural way to secure ∆ 1 2-determinacy is to let x encode winningstrategies for all ∆ 1 2 games. However, this approach is unavailable to us sincewe have not included AC in PA 2 and, in any case, we wish to work with ODdeterminacy(understood schematically). For this we simultaneously run (anelaboration of) the proof of Theorem 6.6 while defining L ω1 [x]. In this way, forany model M of PA 2 , there is a real x and an associated definable set of realsA which codes a model L ω1 [x] satisfying ZFC − Power Set + V = L[x] + ODdeterminacy.Working in ZFC − Power Set + V =L[x] + OD-determinacy we wish nowto show thatHOD [x]T |= ZFC + On is Woodin.

8. Further Results 187So we have to localize the construction of the previous section to the structure〈L ω1 [x], [x] T 〉. The first step is to show that〈L ω1 [x], [x] T 〉 |= ST-[x] T -determinacy for n moves,for each n. Here by ST-[x] T -determinacy we mean what we meant in theprevious section. However, there is a slight metamathematical issue thatarises when we work without Powerset, namely, at each stage of the gamethe potential moves for Player I are a proper class from the point of view of〈L ω1 [x], [x] T 〉. So in quantifying over these moves we have to use the firstorderdefinition of OD in 〈L ω1 [x], [x] T 〉. The winning condition for the n-moveversion of the game is first-order over 〈L ω1 [x], [x] T 〉 but since the complexityof the definition increases as n increases the full game is not first-order over〈L ω1 [x], [x] T 〉. This is why we have had to restrict to the n-move version.The proof of this version of the theorem is just like that of Theorem 7.7,only now one has to keep track of definability and verify that there is noessential use of Powerset (for example, in the proof of Third Periodicity).The proof of Theorem 7.8 goes through as before. Finally, as in the proof ofTheorem 7.9, the proof of the Generation Theorem gives a structure M suchthatHOD M [x] T|= ZFC + On is T-strong,for an arbitrary OD 〈Lω 1 [x],[x] T 〉[x] Tresult.class T of ordinals, which implies the finalThis raises the following question: Are the theories PA 2 +∆ 1 2-determinacyand ZFC + “On is Woodin” equiconsistent? We turn to this and other moregeneral issues in the next section.8. Further ResultsIn this section we place the above results in a broader setting by discussingsome results that draw on techniques that are outside the scope of this chapter.The first topic concerns the intimate connection between axioms ofdefinable determinacy and large cardinal axioms (as mediated through innermodels). The second topic concerns the surprising convergence betweentwo very different approaches to inner model theory—the approach basedon generalizations of L and the approach based on HOD. In both cases the

8. Further Results 188relevant material on inner model theory can be found in Steel’s chapter inthis Handbook.8.1. Large Cardinals and DeterminacyThe connection between axioms of definable determinacy and inner models oflarge cardinals is even more intimate than indicated by the above results. Wehave seen that certain axioms of definable determinacy imply the existenceof inner models of large cardinal axioms. For example, assuming ZFC + ∆ ∼12 -determinacy, for each x ∈ ω ω , there is an inner model M such that x ∈ MandM |= ZFC + There is a Woodin cardinal.And, assuming ZFC + AD L(R) , in L(R) there is an inner model M such thatM |= ZFC + There is a Woodin cardinal.In many cases these implications can be reversed—axioms of definable determinacyare actually equivalent to axioms asserting the existence of innermodels of large cardinals. We discuss what is known about this connection,starting with a low level of boldface definable determinacy and proceedingupward. We then turn to lightface determinacy, where the situation is moresubtle. It should be emphasized that our concern here is not merely withconsistency strength but rather with outright equivalence (over ZFC).Theorem 8.1. The following are equivalent:(1) ∆ ∼12 -determinacy.(2) For all x ∈ ω ω , there is an inner model M such that x ∈ M andM |= There is a Woodin cardinal.Theorem 8.2. The following are equivalent:(1) PD (Schematic).(2) For every n < ω, there is a fine-structural, countably iterable innermodel M such that M |= There are n Woodin cardinals.Theorem 8.3. The following are equivalent:(1) AD L(R) .

8. Further Results 189(2) In L(R), for every set S of ordinals, there is an inner model M and anα < ω L(R)1 such that S ∈ M and M |= α is a Woodin cardinal.Theorem 8.4. The following are equivalent:(1) AD L(R) and R # exists.(2) M # ωexists and is countably iterable.Theorem 8.5. The following are equivalent:(1) For all B, V B |= AD L(R) .(2) M # ω exists and is fully iterable.The above examples concern boldface definable determinacy. The situationwith lightface definable determinacy is more subtle. For example,assuming ZFC + ∆ 1 2-determinacy, must there exist an α < ω 1 and an innermodel M such that α is a Woodin cardinal in M? In light of Theorem 8.1 onewould expect that this is indeed the case. However, since Theorem 8.1 alsoholds in the context of PA 2 one would then expect that the theories PA 2 +∆ 1 2 -determinacy and PA 2 + “There is an α < ω 1 and an inner model M suchthat M |= α is a Woodin cardinal” are equivalent, and yet this expectation isin conflict with the expectation that the theories PA 2 + ∆ 1 2-determinacy andZFC + “On is Woodin” are equiconsistent. In fact, this seems likely, but thedetails have not been fully checked. We state a version for third-order Peanoarithmetic, PA 3 , and second-order ZFC. But first we need a definition andsome preliminary results.8.6 Definition. A partial order P is δ-productive if for all δ-c.c. partial ordersQ, the product P × Q is δ-c.c.Theorem 8.7. In the fully iterable, 1-small, 1-Woodin Mitchell-Steel modelthe extender algebra built using all extenders on the sequence which are strongto their length is δ-productive.This is a warm-up since in the case of interest we do not have iterability. Itis unknown if iterability is necessary.Theorem 8.8. Suppose δ is a Woodin cardinal. Then there is a proper classinner model N ⊆ V such that

8. Further Results 190(1) N |= δ is a Woodin cardinal and(2) N |= There is a complete δ-c.c. Boolean algebra B such thatLet ZFC 2 be second-order ZFC.N B |= ∆ 1 2-determinacy.Theorem 8.9. The following are equiconsistent:(1) PA 3 + ∆ 1 2 -determinacy.(2) ZFC 2 + On is Woodin.We now turn from theories to models and discuss the manner in whichone can pass back and forth between models of infinitely many Woodin cardinalsand models of definable determinacy at the level of AD L(R) and beyond.We have already dealt in detail with one direction of this—the transferfrom models of determinacy to models with Woodin cardinals—and theother direction—the transfer from models with Woodin cardinals to modelsof determinacy—was briefly discussed in the introduction, but the situationis much more general. To proceed at the appropriate level of generality weneed to introduce a potential strengthening of AD.A set A ⊆ ω ω is ∞ -borel if there is a set S ⊆ On, an ordinal α, and aformula ϕ such thatA = {y ∈ ω ω | L α [S, y] |= ϕ[S, y]}.It is fairly straightforward to show that to say that A is ∞ -borel is equivalentto saying that it has a “transfinite borel code”. Notice that under AC everyset of reals is ∞ -borel.8.10 Definition. Assume ZF+DC R . The theory AD + consists of the axioms:(1) Every set A ⊆ ω ω is ∞ -borel.(2) Suppose λ < Θ and π : λ ω → ω ω is a continuous surjection. Then foreach A ⊆ ω ω the set π −1 [A] is determined.8.11 Conjecture. AD implies AD + .

8. Further Results 191It is known that the failure of this implication has strong consistency strength.For example, AD + ¬AD + proves Con(AD R ).The following theorem—the Derived Model Theorem—is a generalizationof Theorem 1.5, mentioned in the introduction.Theorem 8.12. Suppose that δ is a limit of Woodin cardinals. Suppose thatG ⊆ Col(ω,

8. Further Results 1928.2. HOD-AnalysisThere is also an intimate connection between the two approaches to innermodel theory mentioned in the introduction—the approach based on generalizationsof L and the approach based on HOD.As mentioned in the introduction, the two approaches have opposingadvantages and disadvantages. The disadvantage of the first approach isthat the problem of actually defining the models that can accommodate largecardinals—the inner model problem—is quite a difficult problem. However,the advantage is that once the inner model problem is solved at a givenlevel of large cardinals the inner structure of the models is quite transparentand so these models are suitable for extracting the large cardinal contentinherent in a given statement. The advantage of the approach based onHOD is that this model is trivial to define and it can accommodate virtuallyevery large cardinal. The disadvantage—the tractability problem—is that ingeneral the inner structure of HOD is about as tractable as that of V andso it is not generally suitable for extracting the large cardinal content froma given statement.Nevertheless, we have taken the approach based on HOD and we havefound that AD L(R) and ∆ 1 2-determinacy are able to overcome (to some extent)the tractability problem for their natural models, L(R) and L[x] for a Turing(or constructibility) cone of x. For example, we have seen that under AD L(R) ,HOD L(R) |= Θ L(R) is a Woodin cardinal,and that under ∆ 1 2-determinacy, for a Turing cone of reals x,HOD L[x] |= ω L[x]2 is a Woodin cardinal.Despite this progress, much of the structure of HOD in these contexts is farfrom clear. For example, it is unclear whether under ∆ 1 2-determinacy, fora Turing cone of reals x, HOD L[x] satisfies GCH, something that would beimmediate in the case of “L-like” inner models.Since the above results were first proved, Mitchell and Steel developed thefine-structural version of the “L-like” inner models at the level of Woodincardinals. These models have the form L[ E] ⃗ where E ⃗ is a sequence of(partial) extenders and (as noted above) their inner structure is very wellunderstood—for example, they satisfy GCH and many of the other combinatorialproperties that hold in L. A natural question, then, is whether

8. Further Results 193there is any connection between these radically different approaches, that is,whether HOD as computed in L(R) under AD L(R) or in L[x], for a Turingcone of x, under ∆ 1 2 -determinacy, bears any resemblance to the L[⃗ E] models.The remainder of this section is devoted to this question. We begin withHOD L(R) and its generalizations (where a good deal is known) and then turnto HOD L[x] (where the central question is open). Again, the situation withlightface determinacy is more subtle.The theorems concerning HOD L(R) only require AD L(R) but they are simplerto state under the stronger assumption that AD L(R) holds in all genericextensions of V . By Theorem 8.5, this assumption is equivalent to the statementthat M ω # exists and is fully iterable.The first hint that HOD L(R) is a fine-structural model is the remarkablefact thatHOD L(R) ∩ R = M ω ∩ R.The agreement between HOD L(R) and M ω fails higher up but HOD L(R) agreeswith an iterate of M ω at slightly higher levels. More precisely, letting N bethe result of iterating M ω by taking the ultrapower ω1 V -many times using the(unique) normal ultrafilter on the least measurable cardinal, we have thatHOD L(R) ∩ P(ω V 1 ) = N ∩ P(ω V 1 ).Steel improved this dramatically by showing thatHOD L(R) ∩ V (δ˜2 1 ) L(R)is the direct limit of a directed system of iterable fine-structural inner models.Theorem 8.14 (Steel). HOD L(R) ∩ V δ is a Mitchell-Steel model, where δ =(δ ∼21 ) L(R) .For a proof of this result see Steel’s chapter in this Handbook. As a corollaryone has that HOD L(R) satisfies GCH along with the combinatorial principles(such as ♦ and □) that are characteristic of fine-structural models.The above results suggest that all of HOD L(R) might be a Mitchell-Steelinner model of the form L[ ⃗ E]. This is not the case.Theorem 8.15. HOD L(R) is not a Mitchell-Steel inner model.

8. Further Results 194Nevertheless, HOD L(R) is a fine-structural inner model, one that belongs toa new, quite different, hierarchy of models. LetD = { L[ ⃗ E] ∣ ∣ L[ ⃗ E] is an iterate of Mω by a countable treewhich is based on the first Woodin cardinaland has a non-dropping cofinal branch } .Any two structures in D can be compared and the iteration halts in countablymany steps (since we have full iterability) with iterates lying in D. So D is adirected system under the elementary embeddings given by iteration maps.By the Dodd-Jensen lemma the embeddings commute and hence there is adirect limit. Let L[ E ⃗ ∞ ] be the direct limit of D. Let 〈δi∞ | i < ω〉 be theWoodin cardinals of L[ E ⃗ ∞ ].Theorem 8.16. Let L[ ⃗ E ∞ ] be as above. Then(1) L[ ⃗ E ∞ ] ⊆ HOD L(R) ,(2) L[ ⃗ E ∞ ] ∩ V δ = HOD L(R) ∩ V δ , where δ = δ ∞ 0 ,(3) Θ L(R) = δ ∞ 0 , and(4) (δ ∼21 ) L(R) is the least cardinal in L[ ⃗ E ∞ ] which is λ-strong for all λ < δ ∞ 0 .To reach HOD L(R) we need to supplement L[ ⃗ E ∞ ] with additional innermodel-theoreticinformation. A natural candidate is the iteration strategy.It turns out that by folding in the right fragment of the iteration strategyone can capture HOD L(R) . LetandT ∞ = { T ∣ ∣ T is a maximal iteration tree on L[ ⃗ E ∞ ] based on δ ∞ 0 ,T ∈ L[ ⃗ E ∞ ], and length(T) < sup{δ ∞ n | n < ω}}P = { 〈b, T 〉 ∣ ∣ T ∈ T ∞ and b is the true branch through T } .Theorem 8.17. Let L[ ⃗ E ∞ ] and P be as above. ThenHOD L(R) = L[ ⃗ E ∞ , P].In fact, there is a single iteration tree T ∈ T ∞ such that if b is the branchthrough T chosen by P thenHOD L(R) = L[ ⃗ E ∞ , b].

8. Further Results 195This analysis has an interesting consequence. Notice that the modelL[ E ⃗ ∞ ] is of the form L[A] for A ⊆ δ0 ∞ . Thus, although the addition ofP does not add any new bounded subsets of Θ L(R) it does a lot of damage tothe model above Θ L(R) , for example, it collapses ω-many Woodin cardinals.One might think that this is an artifact of L[ E ⃗ ∞ ] but in fact the situation ismuch more general: Suppose L[ E] ⃗ is ω-small, fully iterable, and has ω-manyWoodin cardinals. Let P be defined as above except using the Woodin cardinalsof L[ E]. ⃗ Then L[ E, ⃗ P] ∩ V δ = L[ E] ⃗ ∩ V δ , where δ is the first Woodincardinal of L[ E], ⃗ and L[ E] ⃗ L[ E, ⃗ P] L[ E ⃗ # ]. For example, applying thisresult to L[ E] ⃗ = M ω , one obtains a canonical inner-model-theoretic objectbetween M ω and M ω # . In this way, what appeared to be a coarse approachto inner model theory has actually resulted in a hierarchy that supplementsand refines the standard fine-structural hierarchy.The above results generalize. We need a definition.8.18 Definition (Mouse Capturing). MC is the statement: For all x, y ∈ω ω , x ∈ OD y iff there is an iterable Mitchell-Steel model M of the form L[ ⃗ E, y]such that x ∈ M.The Mouse Set Conjecture, MSC, is the conjecture that it is a theorem ofAD + that MC holds if there is no iterable model with a superstrong cardinal.There should be a more general version of MC, one that holds for extensionsof the Mitchell-Steel models that can accommodate long extenders. And thisversion of MC should follow from AD + . However, the details are still beingworked out. See [13].Theorem 8.19. Assume AD + + V = L(P(R)) + Θ 0 = Θ + MSC. Then theinner model HOD L(P(R)) is of the form L[ ⃗ E ∞ , P], with the key differencebeing that L[ ⃗ E ∞ ] need not be ω-small.Theorem 8.20. Assume AD + + V =L(P(R)) + Θ 0 < Θ + MSC. Then(1) Θ 0 is the least Woodin cardinal in HOD,(2) HOD ∩ V Θ0 is a Mitchell-Steel model,(3) HOD ∩ V Θ0 +1 is not a Mitchell-Steel model, and(4) HOD ∩V Θ1 is a model of the form L[ ⃗ E ∞ , P] (assuming the appropriateform of the Mouse Set Theorem).

8. Further Results 196One can move on to stronger hypotheses. For example, assuming AD +and V = L(P(R)), AD R is equivalent to the statement that Ω (defined atthe beginning of Section 5) is a non-zero limit ordinal. There is a minimalinner model N of ZF+AD R that contains all of the reals. The model HOD Nhas ω-many Woodin cardinals and these are exactly the members of the Θ-sequence. This model belongs to the above hierarchy and has been usedto calibrate the consistency strength of AD R in terms of the large cardinalhierarchy. This hierarchy extends and a good deal is known about it.We now turn to the case of lightface determinacy and the setting L[x] fora Turing cone of x. Here the situation is less clear. In fact, the basic questionis open.5 Open Question. Assume ∆ 1 2-determinacy. For a Turing cone of x, whatis HOD L[x] from a fine-structural point of view?We close with partial results in this direction and with a conjecture. Tosimplify the discussion we state these results under a stronger assumptionthan is necessary: Assume ∆ 1 2 -determinacy and that for all x ∈ ωω , x # exists.It follows that M 1 and M # 1 exist. Let x 0 ∈ ω ω be such that M # 1 ∈ L[x 0].Let κ x0 be the least inaccessible of L[x 0 ] and let G ⊆ Col(ω, < κ x0 ) be L[x 0 ]-generic. The Kechris-Solovay result carries over to show thatFurthermore,L[x 0 ][G] |= OD-determinacy.HOD L[x 0][G] = HOD L(R)L[x 0 ][G]and ω L[x 0][G]2 = Θ L(R)L[x 0 ][G] .Thus, the model L(R) L[x 0][G]is a “lightface” analogue of L(R). In fact theconditions of the Generation Theorem hold in L[x 0 ][G] and as a consequenceone has thatHOD L[x 0][G] |= ω L[x 0][G]2 is a Woodin cardinal.For a model L[ E] ⃗ Econtaining at least one Woodin cardinal let δ ⃗ 0 be the leastWoodin cardinal. LetD = { L[ E] ⃗ ⊆ L[x 0 ][G] ∣ L[ E] ⃗ is an iterate ofE M1 and δ ⃗ 0 < ω L[x }0][G]1 .

8. Further Results 197Let L[ ⃗ E ∞ ] be the direct limit of D. Let δ ∞ be the least Woodin of L[ ⃗ E ∞ ]and let κ ∞ be the least inaccessible above δ ∞ . LetandT ∞ = { T ∣ ∣ T is a maximal iteration tree on L[ ⃗ E ∞ ],T ∈ L[ ⃗ E ∞ ], and length(T) < κ ∞}P = { 〈b, T 〉 ∣ ∣ T ∈ T ∞ and b is the true branch through T } .Theorem 8.21. Let L[ ⃗ E ∞ , P] be as above. Then(1) HOD L[x 0][G] ∩ V δ ∞ = L[ ⃗ E ∞ ] ∩ V δ ∞,(2) HOD L[x 0][G] = L[ ⃗ E ∞ , P], and(3) ω L[x 0][G]2 = δ ∞ .A similar analysis can be carried out for other hypotheses that place onein an “L(R)-like” setting. For example, suppose again that x 0 is such thatM # 1 ∈ L[x 0 ]. One can “generically force” MA as follows: In L[x 0 ] let P bethe partial order where the conditions 〈B α | α < γ〉 are such that (i) for eachα < γ, B α is c.c.c., (ii) |B α | = ω 1 , (iii) if α β < γ then B α is a completesubalgebra of B β , and (iv) γ < ω 2 , and the ordering is by extension. Theforcing is

8. Further Results 1981would suggest that δ is ∼δ 2 as computed in L[x 0 ][G]. It turns out this analogyfails: the least cardinal δ that is λ-strong in HOD L[x0][G] for all λ < ω L[x 0][G]2is in fact strictly less δ2 1 as computed in L[x 0 ][G].But there is another analogy that does hold. First we need some definitions.A set A ⊆ ω ω is γ-Suslin if there is an ordinal γ and a tree T on ω ×γsuch that A = p[T] = {x ∈ ω ω | ∃y ∈ γ ω ∀n (x↾n, y↾n) ∈ T }. A cardinal κis a Suslin cardinal if there exists a set A ⊆ ω ω such that A is κ-Suslin butnot γ-Suslin for any γ < κ. A set A ⊆ ω ω is effectively γ-Suslin if there is anordinal γ and an OD tree T ⊆ ω × γ such that A = p[T]. A cardinal κ is aneffective Suslin cardinal if there exists a set A ⊆ ω ω such that A is effectivelyκ-Suslin but not effectively γ-Suslin for any γ < κ.2In L(R), ∼δ 1 is the largest Suslin cardinal. Since L[x 0 ][G] is a lightfaceanalogue of L(R) one might expect that in L[x 0 ][G], δ2 1 is the largest effectiveSuslin cardinal in L[x 0 ][G]. This is indeed the case.There is one more advance on the HOD-analysis for L[x] that is worthmentioning.Theorem 8.22. Assume ∆ ∼12 -determinacy. For a Turing cone of x there isa predicate A such that(1) HOD L[x]Astrategy,(2) HOD L[x]A(3) HOD L[x]Ahas the form L[ ⃗ E, P] where P is a fragment of the iteration|= ω L[x]2 is a Woodin cardinal,is of the form L[ ⃗ E] below ω L[x]2 ,(4) L[x] |= ST A -determinacy, and(5) HOD L[x] ∩ V δ = HOD L[x]A∩ V δ where δ is the least cardinal of HOD L[x]Athat is λ-strong for all λ < ω L[x]2 .Moreover, there exists a definable collection of such A and the collection hassize ω L[x]1 .This provides some evidence that HOD L[x] is of the form L[ ⃗ E] below ω L[x]2and that HOD L[x] is not equal to a model of the form L[ ⃗ E].8.23 Conjecture. HOD L[x] is of the form L[ ⃗ E, P] where P selects branchesthrough all trees in L[ ⃗ E] based on the Woodin cardinal and with length lessthan the successor of the Woodin cardinal.

8. Bibliography 199Bibliography[1] Thomas J. Jech. Set Theory, Third Millenium Edition, Revised andExpanded. Springer-Verlag, Berlin, 2003.[2] Akihiro Kanamori. The Higher Infinite: Large Cardinals in Set Theoryfrom their Beginnings. Springer Monographs in Mathematics. Springer,Berlin, second edition, 2003.[3] Alexander S. Kechris. The axiom of determinacy implies dependentchoices in L(R). The Journal of Symbolic Logic, 49(1):161–173, 1984.[4] Alexander S. Kechris, Donald A. Martin, and Robert M. Solovay. Introductionto Q-theory. In Cabal Seminar 79–81, volume 1019 of LectureNotes in Mathematics, pages 199–282. Springer, 1983.[5] Alexander S. Kechris, Yiannis N. Moschovakis, W. Hugh Woodin, andE. M. Kleinberg. The axiom of determinacy, partition properties, andnon-singular measures. In Cabal Seminar 77–79, volume 839 of LectureNotes in Mathematics, pages 75–99. Springer, 1981.[6] Alexander S. Kechris and Robert M. Solovay. On the relative strengthof determinacy hypotheses. Transactions of the American MathematicalSociety, 290(1):179–211, 1985.[7] Alexander S. Kechris and W. Hugh Woodin. Equivalence of partitionproperties and determinacy. Proceedings of the National Academy ofScience U.S.A, 80:1783–1786, 1983.[8] Peter Koellner. On the question of absolute undecidability. PhilosophiaMathematica, 14(2):153–188, 2006. Revised and reprinted in Kurt Gödel:Essays for his Centennial, edited by Solomon Feferman, Charles Parsons,and Stephen G. Simpson. Lecture Notes in Logic, 33. Associationof Symbolic Logic, 2009.[9] Paul B. Larson. The Stationary Tower: Notes on a Course by W. HughWoodin, volume 32 of University Lecture Series. American MathematicalSociety, 2004.[10] Yiannis N. Moschovakis. Descriptive Set Theory. North-Holland, Amsterdam,1980.

Bibliography 200[11] Itay Neeman. The Determinacy of Long Games, volume 7 of de GruyterSeries in Logic and its Applications. de Gruyter, Berlin, 2004.[12] John Steel. Mathematics needs new axioms. The Bulletin of SymbolicLogic, 6(4):422–433, December 2000.[13] W. Hugh Woodin. Suitable extender sequences.[14] W. Hugh Woodin. The continuum hypothesis. In Rene Cori, AlexanderRazborov, Stevo Todorcevic, and Carol Wood, editors, Logic Colloquium2000, volume 19 of Lecture Notes in Logic, pages 143–197. Associationof Symbolic Logic, 2005.

IndexL(V λ+1 ), 9Θ, 12Θ α , 962δ∼ 1 , 32AC ω , 19AC ω (R), 19AD, 3AD + , 177, 190–191, 195–196AD L(R) , 3AD R , 6DC, 19DC R , 19PD, 3T 0 , 18absolute definability, 4additively closed ordinal, 150boundednessΣ∼ 1 1-boundedness, 23∆∼ 2 1-boundedness, 41canonical function (for Chapter 23),75, 123certifies a fact, 66coding lemmasapplications, 55–58Basic Coding Lemma, 24Coding Lemma, 45∆∼ 1 2-Coding Lemma, 41Strong Coding Lemma, 54Uniform Coding Lemma, 49critical point (ordinal), 7Dependent Choice, 19for Sets of Reals, 19Derived Model Theorem, 10, 190determinacyAD + , 177, 190–191, 195–196AD L(R) , 3AD R , 6Axiom of Determinacy (AD), 3definable determinacy, 3, 145–177absolute, 4boldface, 4, 169–177lightface, 4, 147–169∆ 1 2-determinacy, 5OD-determinacy, 4OD(R)-determinacy, 4Projective Determinacy (PD), 3Σ∼ 1 n+1 -determinacy, 9strategic determinacyST B P 0 ,...,P k-determinacy, 105RST-determinacy, 153ST-[x] T -determinacy, 179Turing determinacy, 22disjointness property, 80, 128extender, 115pre-extender, 116Generation Theorem, 12, 13, 96, 109–137201

INDEX 202applications, 137–177good code, 117good set, 113HOD-analysis, 192–198HOD L(R) , 193–195HOD L[x] , 196–198∞ -borel, 190least stable ordinal, 33long game, 6measurable cardinal, 7Mouse Capturing, 195Mouse Set Conjecture, 195α-strong, 7strong normality, 72–91, 124superstrong cardinal, 8Suslin cardinal, 198effective, 198tail computation, 83, 131Third Periodicity Theorem, 149, 165Turingcone, 21cone filter, 21degrees, 21determinacy, 22Woodin cardinal, 8ω-model, 22pre-extender, 116prestrategy, 104Prikry forcingthrough degrees, 172Recursion Theorem, 45Uniform, 49reflected generator, 118reflection filter, 66, 68, 113Reflection Theorem (for Chapter 23),63, 68, 93, 113relativized strategic game, 179relativized prestrategy, 179relativized reals, 179relativized strategy, 179S-cone, 100S-degree, 100second-order arithmetic, 177–187strategic game, 104strong cardinal, 7