Large Cardinals from Determinacy - Logic at Harvard
Large Cardinals from Determinacy - Logic at Harvard
Large Cardinals from Determinacy - Logic at Harvard
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<strong>Large</strong> <strong>Cardinals</strong> <strong>from</strong> <strong>Determinacy</strong>Peter Koellner and W. Hugh WoodinContents1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 21.1 <strong>Determinacy</strong> and <strong>Large</strong> <strong>Cardinals</strong> . . . . . . . . . . . 21.2 Not<strong>at</strong>ion . . . . . . . . . . . . . . . . . . . . . . . . . 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 Applic<strong>at</strong>ions . . . . . . . . . . . . . . . . . . . . . . . 554 A Woodin Cardinal in HOD L(R) . . . . . . . . . . . . 584.1 Reflection . . . . . . . . . . . . . . . . . . . . . . . . 594.2 Strong Normality . . . . . . . . . . . . . . . . . . . . 724.3 A Woodin Cardinal . . . . . . . . . . . . . . . . . . . 915 Woodin <strong>Cardinals</strong> in General Settings . . . . . . . . 955.1 First Abstraction . . . . . . . . . . . . . . . . . . . . 975.2 Str<strong>at</strong>egic <strong>Determinacy</strong> . . . . . . . . . . . . . . . . . 1001
1. Introduction 25.3 Gener<strong>at</strong>ion Theorem . . . . . . . . . . . . . . . . . . 1095.4 Special Cases . . . . . . . . . . . . . . . . . . . . . . 1376 Definable <strong>Determinacy</strong> . . . . . . . . . . . . . . . . . 1456.1 Lightface Definable <strong>Determinacy</strong> . . . . . . . . . . . . 1466.2 Boldface Definable <strong>Determinacy</strong> . . . . . . . . . . . . 1697 Second-Order Arithmetic . . . . . . . . . . . . . . . 1777.1 First Localiz<strong>at</strong>ion . . . . . . . . . . . . . . . . . . . . 1797.2 Second Localiz<strong>at</strong>ion . . . . . . . . . . . . . . . . . . . 1868 Further Results . . . . . . . . . . . . . . . . . . . . . 1878.1 <strong>Large</strong> <strong>Cardinals</strong> and <strong>Determinacy</strong> . . . . . . . . . . . 1888.2 HOD-Analysis . . . . . . . . . . . . . . . . . . . . . . 192Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . 1991. IntroductionIn this chapter we give an account of Woodin’s technique for deriving largecardinal strength <strong>from</strong> determinacy hypotheses. These results appear herefor the first time and for this reason we have gone into somewh<strong>at</strong> more detailthan is customary in a handbook. All un<strong>at</strong>tributed results th<strong>at</strong> follow areeither folklore or due to Woodin.1.1. <strong>Determinacy</strong> and <strong>Large</strong> <strong>Cardinals</strong>In the era of set theory following the discovery of independence a majorconcern has been the discovery of new axioms th<strong>at</strong> settle the st<strong>at</strong>ements leftundecided by the standard axioms (ZFC). One interesting fe<strong>at</strong>ure th<strong>at</strong> hasemerged is th<strong>at</strong> there are often deep connections between axioms th<strong>at</strong> spring<strong>from</strong> 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 not<strong>at</strong>ion and basic notionsth<strong>at</strong> are explained in fuller detail in sections 1.2 and 2.1.A. <strong>Determinacy</strong>For a set of reals A ⊆ ω ω consider the game where two players take turnsplaying n<strong>at</strong>ural 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 th<strong>at</strong> 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 str<strong>at</strong>egy” in theassoci<strong>at</strong>ed game, th<strong>at</strong> is, a str<strong>at</strong>egy which ensures th<strong>at</strong> the player wins around regardless of how the other player plays. The Axiom of <strong>Determinacy</strong>(AD) is the st<strong>at</strong>ement th<strong>at</strong> every set of reals is determined.It is straightforward to see th<strong>at</strong> very simple sets are determined. Forexample, if A is the set of all reals then clearly I has a winning str<strong>at</strong>egy; ifA is empty then clearly II has a winning str<strong>at</strong>egy; and if A is countable thenII has a winning str<strong>at</strong>egy (by “diagonalizing”). A more substantive resultis th<strong>at</strong> if A is closed then one player must have a winning str<strong>at</strong>egy. Thismight lead one to expect th<strong>at</strong> 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 str<strong>at</strong>egies and “diagonalizing” acrossthem). For this reason AD was never really considered as a serious candid<strong>at</strong>efor a new axiom. However, there is an interesting class of rel<strong>at</strong>ed axioms th<strong>at</strong>are consistent with AC, namely, the axioms of definable determinacy. Theseaxioms extend the above p<strong>at</strong>tern by asserting th<strong>at</strong> all sets of reals <strong>at</strong> 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 follow<strong>from</strong> ZFC. In the early development of the subject the result on the determinacyof closed sets was extended to higher levels of definability. Thesedevelopments culmin<strong>at</strong>ed in Martin’s proof of ∆ ∼11 -determinacy in ZFC. Itturns out th<strong>at</strong> this result is close to optimal—as one climbs the hierarchy
1. Introduction 4of definability, shortly after ∆ ∼11 one arrives <strong>at</strong> axioms th<strong>at</strong> 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 candid<strong>at</strong>es 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) th<strong>at</strong> do not involve real numbers as parameters) and the otherinvolving boldface notions of definability (by which we mean notions (suchas ∆ ∼12 ) th<strong>at</strong> do involve real numbers as parameters). (See Jackson’s chapterin this Handbook for details concerning the various grades of definabilityand the relevant not<strong>at</strong>ion.) Each hierarchy is, of course, ordered in termsof increasing complexity. Moreover, each hierarchy has a n<strong>at</strong>ural limit: then<strong>at</strong>ural limit of the lightface hierarchy is OD-determinacy (all OD sets of realsare determined) and the n<strong>at</strong>ural limit of the boldface hierarchy is OD(R)-determinacy (all OD(R) sets of reals are determined). The reason these aren<strong>at</strong>ural limits is th<strong>at</strong> the notions of lightface and boldface ordinal definabilityare candid<strong>at</strong>es for the richest lightface and boldface notions of definability.To see this (for the lightface case) notice first th<strong>at</strong> 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 th<strong>at</strong> the notion of ordinal definability cannot be sotranscended (since by reflection OD is ordinal definable). It is for this reasonth<strong>at</strong> Gödel proposed the notion of ordinal definability as a candid<strong>at</strong>e 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 candid<strong>at</strong>es for new axiomsand each has a n<strong>at</strong>ural 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 rel<strong>at</strong>ive 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 rel<strong>at</strong>iveto the standard axioms (ZFC)—is not available. Nevertheless, one can hopeto establish results—such as rel<strong>at</strong>ive consistency and logical connections withrespect to other plausible axioms—th<strong>at</strong> collectively shed light on the original,absolute question. Indeed, there are a number of results th<strong>at</strong> one canbring to bear in favour of PD and AD L(R) . For example, these axioms lift thestructure theory th<strong>at</strong> 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 st<strong>at</strong>ements th<strong>at</strong> are independentof ZFC. In fact, there is no “n<strong>at</strong>ural” st<strong>at</strong>ement concerning their respectivedomains th<strong>at</strong> is known to be independent of these axioms. (For more on thestructure theory provided by determinacy and the traditional consider<strong>at</strong>ionsin 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 rel<strong>at</strong>ive consistency;more precisely, we wish to calibr<strong>at</strong>e the consistency strength of axiomsof definable determinacy—in particular, the ultim<strong>at</strong>e axioms of lightface andboldface determinacy—in terms of the large cardinal hierarchy.There are some reductions th<strong>at</strong> we can st<strong>at</strong>e <strong>at</strong> the outset. In terms ofconsistency strength the two hierarchies collapse <strong>at</strong> a certain stage: Kechrisand Solovay showed th<strong>at</strong> ZF+DC implies th<strong>at</strong> 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 th<strong>at</strong> ZFC+OD(R)-determinacy and ZFC+AD L(R)are equiconsistent. Thus, in terms of consistency strength, the lightfacehierarchy collapses <strong>at</strong> ∆ 1 2-determinacy and the boldface hierarchy collapses<strong>at</strong> AD L(R) . So if one wishes to gauge the consistency strength of lightfaceand boldface determinacy it suffices to concentr<strong>at</strong>e on ∆ 1 2-determinacy andAD L(R) .Now, it is straightforward to see th<strong>at</strong> 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 n<strong>at</strong>ural place to study the consistency strength oflightface definable determinacy is L[x] for some real x and the n<strong>at</strong>ural placeto study the consistency strength of boldface definable determinacy (or fulldeterminacy) is L(R). For this reason these two models will be central inwh<strong>at</strong> follows.To summarize: We shall be investig<strong>at</strong>ing 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 n<strong>at</strong>ural numbers.However, it is worthwhile to note th<strong>at</strong> there are two directions in which onecan generalize these axioms.
1. Introduction 6First, one can consider games of length gre<strong>at</strong>er than ω (where the movesare still n<strong>at</strong>ural numbers). A simple argument shows th<strong>at</strong> one cannot havethe determinacy of all games of length ω 1 but there is a gre<strong>at</strong> 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 thann<strong>at</strong>ural numbers (and where the length of the game is still ω). One altern<strong>at</strong>iveis to consider games where the moves are real numbers. The axiom AD Rst<strong>at</strong>es th<strong>at</strong> 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 th<strong>at</strong> this axiom is inconsistent. In fact, even the definable versionasserting th<strong>at</strong> all OD subsets of P(R) ω is inconsistent. Another altern<strong>at</strong>iveis to consider games where the moves are ordinal numbers. Again, a simpleargument shows th<strong>at</strong> one cannot have the determinacy of all subsets of ω 1 ω .However, a result of Harrington and Kechris shows th<strong>at</strong> in this case if oneadds a definability constraint then one can have determinacy <strong>at</strong> this level.In fact, OD-determinacy implies th<strong>at</strong> every OD set A ⊆ ω 1 ω is determined.It is n<strong>at</strong>ural then to extend this to large ordinals. The ultim<strong>at</strong>e axiom inthis direction would simply assert th<strong>at</strong> every OD set A ⊆ On ω is determined.Perhaps surprisingly, <strong>at</strong> this stage a certain tension arises since recent work ininner model theory provides evidence th<strong>at</strong> this axiom is in fact inconsistent.See [13] for more on this subject.B. <strong>Large</strong> <strong>Cardinals</strong>Our aim is to calibr<strong>at</strong>e the consistency strength of lightface and boldfacedeterminacy in terms of the large cardinal hierarchy. The importance ofthe large cardinal hierarchy in this connection is th<strong>at</strong> it provides a canonicalmeans of climbing the hierarchy of consistency strength. To show, for a givenhypothesis ϕ and a given large cardinal axiom L, th<strong>at</strong> 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) n<strong>at</strong>urally well-ordered and it is a remarkable phenomenon th<strong>at</strong> givenany two “n<strong>at</strong>ural” 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 th<strong>at</strong> there are “large”levels of the universe. A templ<strong>at</strong>e for formul<strong>at</strong>ing a broad class of largecardinal axioms is in terms of elementary embeddings. The basic form<strong>at</strong> ofthe templ<strong>at</strong>e is as follows: There is a transitive class M and a non-trivialelementaryj : V → M.To say th<strong>at</strong> the embedding is non-trivial is simply to say th<strong>at</strong> it is not theidentity, in which case one can show th<strong>at</strong> 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 th<strong>at</strong> for any such elementary embedding there is necessarilya certain degree of agreement between V and M. In particular, it followsth<strong>at</strong> V κ+1 ⊆ M, where κ = crit(j). This degree of agreement in conjunctionwith the elementarity of j can be used to show th<strong>at</strong> 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 gre<strong>at</strong>er agreement between M and V . For example, if one demandsth<strong>at</strong> V κ+2 ⊆ M then the fact th<strong>at</strong> κ is measurable is recognized within M andhence it follows th<strong>at</strong> M s<strong>at</strong>isfies th<strong>at</strong> there is a measurable cardinal belowj(κ), namely, κ. Thus, by the elementarity of the embedding, V s<strong>at</strong>isfiesth<strong>at</strong> there is a measurable cardinal below κ. The same argument shows th<strong>at</strong>there are arbitrarily large measurable cardinals below κ.This leads to a n<strong>at</strong>ural 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 <strong>at</strong>ransitive class M and a non-trivial elementary embedding j : V → M suchth<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> κ is η-strong and which has the additionalfe<strong>at</strong>ure th<strong>at</strong> 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 th<strong>at</strong>for each η such th<strong>at</strong> κ A < η < κ.κ A is η-A-strong,It should be noted th<strong>at</strong> 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 and<strong>from</strong> 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 th<strong>at</strong> 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 gre<strong>at</strong>er agreement betweenM and V . The ultim<strong>at</strong>e axiom in this direction would, of course, demandth<strong>at</strong> M = V . This axiom was proposed by Reinhardt. But shortly afterits introduction Kunen showed th<strong>at</strong> it is inconsistent with ZFC. In fact,Kunen showed th<strong>at</strong> 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 th<strong>at</strong> climb the hierarchy of consistency strength farbeyond wh<strong>at</strong> can be reached with ZFC.)There is a lot of room below the above upper bound. For example, avery strong axiom is the st<strong>at</strong>ement th<strong>at</strong> there is a non-trivial elementaryembedding j : V λ+1 → V λ+1 . The strongest large cardinal axiom in the currentliter<strong>at</strong>ure is the axiom asserting th<strong>at</strong> there is a non-trivial elementaryembedding j : L(V λ+1 ) → L(V λ+1 ) such th<strong>at</strong> 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 <strong>at</strong> the close of the previous section. In fact, there is evidenceth<strong>at</strong> the parallel extends. For example, there is a new large cardinal axiomth<strong>at</strong> is the analogue of AD R . See [13] for more on these recent developments.
1. Introduction 9C. <strong>Determinacy</strong> <strong>from</strong> <strong>Large</strong> <strong>Cardinals</strong>Let us return to the questions of the truth and the consistency of axiomsof definable determinacy, granting th<strong>at</strong> of large cardinal axioms. In the l<strong>at</strong>e1960s Solovay conjectured th<strong>at</strong> AD L(R) is provable <strong>from</strong> large cardinal axioms(and hence th<strong>at</strong> ZF+AD is consistent rel<strong>at</strong>ive to large cardinal axioms). Thisconjecture was realized in stages.In 1970 Martin showed th<strong>at</strong> if there is a measurable cardinal then all1Σ ∼1 sets of reals are determined. L<strong>at</strong>er, in 1978, he showed th<strong>at</strong> 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 th<strong>at</strong> therewould be a long march up the hierarchy of axioms of definable determinacy.However, in 1984 Woodin showed th<strong>at</strong> 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 th<strong>at</strong> there are nWoodin cardinals with a measurable cardinal above them all. Then Σ ∼1n+1 -determinacy holds.It follows th<strong>at</strong> 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 combin<strong>at</strong>ion of Martin and Steel’swork and Woodin’s work on the st<strong>at</strong>ionary 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 th<strong>at</strong>, 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 believeth<strong>at</strong> AD L(R) is implied by all sufficiently strong “n<strong>at</strong>ural” theories. For furtherdiscussion of this subject and other more recent results th<strong>at</strong> 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 rel<strong>at</strong>ed question concerns
1. Introduction 10the consistency of axioms of definable determinacy, granting th<strong>at</strong> 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 th<strong>at</strong> λ is a limit of Woodin cardinals.Suppose G ⊆ Col(ω,
1. Introduction 11axioms th<strong>at</strong> have been investig<strong>at</strong>ed. But it has a complementary defect inth<strong>at</strong> 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 th<strong>at</strong>resemble L in having a highly ordered inner structure but which resembleHOD in th<strong>at</strong> they can accommod<strong>at</strong>e strong large cardinal axioms.“L-like” inner models <strong>at</strong> 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 th<strong>at</strong> the large cardinal hypothesesin their proofs of determinacy were essentially optimal. For example, theyshowed th<strong>at</strong> if there is a Woodin cardinal then there is a canonical innermodel M th<strong>at</strong> contains a Woodin cardinal and has a ∆ 1 3 well-ordering of the1reals. It follows th<strong>at</strong> one cannot prove Σ ∼2 -determinacy <strong>from</strong> 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 <strong>from</strong> th<strong>at</strong> of ZFC+∆ 1 2 -determinacy? Second, can one build an inner model of a Woodin cardinaldirectly <strong>from</strong> ZFC + ∆ 1 2-determinacy? Third, wh<strong>at</strong> is the strength ofZFC + AD L(R) ? To approach these questions it would seem th<strong>at</strong> one wouldneed fine-structural inner model theory. However, <strong>at</strong> 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 th<strong>at</strong> of V . Surprisingly if one strengthens the background theory then thestructure theory of HOD becomes tractable. For example, Solovay showedth<strong>at</strong> under ZF + AD, HOD s<strong>at</strong>isfies th<strong>at</strong> ω1V is a measurable cardinal. Itturns out th<strong>at</strong> both lightface and boldface definable determinacy are ableto illumin<strong>at</strong>e the structure of HOD (when constructed in the n<strong>at</strong>ural innermodels of these axioms—L[x] and L(R)) to the point where one can recoverthe large cardinals th<strong>at</strong> 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 th<strong>at</strong> ofZFC + “There is a Woodin cardinal”. For the case of boldface determinacylet us first st<strong>at</strong>e a preliminary result of which the above result is a localiz<strong>at</strong>ion.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 gener<strong>at</strong>ion of Woodin cardinals—the Gener<strong>at</strong>ion Theorem. In additionto giving the above results, the Gener<strong>at</strong>ion 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 th<strong>at</strong> if G ⊆ P Y is V -genericand 〈[x i ] Y | i < ω〉 is the associ<strong>at</strong>ed 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 l<strong>at</strong>e 1980s. However, the current present<strong>at</strong>ion and many ofthe results th<strong>at</strong> follow are quite recent. One of the key new ingredients is thefollowing abstract theorem on the gener<strong>at</strong>ion of Woodin cardinals, which lies<strong>at</strong> the heart of this chapter:Theorem 1.9 (Gener<strong>at</strong>ion Theorem). Assume ZF. Supposeis such th<strong>at</strong>(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 th<strong>at</strong> A α is a prewellordering of the reals oflength gre<strong>at</strong>er than or equal to α,(5) B ⊆ ω ω is nonempty, and(6) M |= Str<strong>at</strong>egic 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 “str<strong>at</strong>egic determinacy” is a technical notion th<strong>at</strong> we will st<strong>at</strong>eprecisely l<strong>at</strong>er.The Gener<strong>at</strong>ion Theorem provides a templ<strong>at</strong>e for gener<strong>at</strong>ing models containingWoodin cardinals. One simply has to show th<strong>at</strong> in a particular settingthe various conditions can be met, though this is often a non-trivial task. Thetheorem is also quite flexible in th<strong>at</strong> it is a result of ZF th<strong>at</strong> does not presupposeDC and has applic<strong>at</strong>ions in both lightface and boldface settings. Itwill play a central role in the calibr<strong>at</strong>ion of the strength of both lightface andboldface determinacy.We shall approach the proof of the Gener<strong>at</strong>ion Theorem by proving aseries of increasingly complex approxim<strong>at</strong>ions.
1. Introduction 14In Section 2 we take the initial step by proving Solovay’s theorem th<strong>at</strong>under ZF + AD, ω1V is a measurable cardinal in HOD and we show th<strong>at</strong>the associ<strong>at</strong>ed measure is normal. The proof th<strong>at</strong> we give is slightly morecomplic<strong>at</strong>ed than the standard proof but has the virtue of illustr<strong>at</strong>ing in asimple setting some of the key components th<strong>at</strong> appear in the more complexvari<strong>at</strong>ions. We illustr<strong>at</strong>e this <strong>at</strong> the end of the section by showing th<strong>at</strong> theproof of Solovay’s theorem generalizes to show th<strong>at</strong> under ZF + AD, the2ordinal (δ ∼1 ) L(R) is a measurable cardinal in HOD L(R) . Our main aim in thissection is to illustr<strong>at</strong>e 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 th<strong>at</strong> will be centralthroughout.In Section 4, as a precursor to the proof of the Gener<strong>at</strong>ion 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 ultim<strong>at</strong>ely be elimin<strong>at</strong>edwhen we prove the Gener<strong>at</strong>ion Theorem. Toward the proof of the abovetheorem, we begin in §4.1 by establishing the reflection phenomenon th<strong>at</strong> 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 th<strong>at</strong> is intended to witness th<strong>at</strong> δ 2∼1 is λ-strong. In §4.2 we shall introduceand motiv<strong>at</strong>e the notion of strong normality by showing th<strong>at</strong> the strong2normality of µ λ ensures th<strong>at</strong> ∼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 rel<strong>at</strong>ivizing 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 th<strong>at</strong> involves DC and one th<strong>at</strong> does not. The first theorem is proved in§5.1. The importance of this theorem is th<strong>at</strong> it can be used to show th<strong>at</strong> incertain strong determinacy settings HOD can contain many Woodin cardinals.The second theorem is the Gener<strong>at</strong>ion Theorem, the proof of which willoccupy the remainder of the section. The aim of the Gener<strong>at</strong>ion Theorem
1. Introduction 15is to show th<strong>at</strong> the construction of Section 4 can be driven by lightface determinacyalone. The difficulty is th<strong>at</strong> the construction of Section 4 involvesgames th<strong>at</strong> are defined in terms of real parameters. To handle this we introducethe notion of “str<strong>at</strong>egic determinacy”, a notion th<strong>at</strong> resembles boldfacedeterminacy in th<strong>at</strong> it involves real parameters but which can nonethelesshold in settings where one has AC. To motiv<strong>at</strong>e the notion of “str<strong>at</strong>egicdeterminacy” 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 th<strong>at</strong>“str<strong>at</strong>egic determinacy” can hold in this setting we shall return in §5.3 tothe general setting and prove the Gener<strong>at</strong>ion Theorem. In the final subsection,we prove a number of special cases, many of which are new. Althoughsome of these applic<strong>at</strong>ions involve lightface settings, they all either involveassuming full AD or explicitly involve “str<strong>at</strong>egic determinacy”.In Section 6 we use two of the special cases of the Gener<strong>at</strong>ion Theoremto calibr<strong>at</strong>e 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 th<strong>at</strong> ∆ 1 2-determinacy suffices to establish th<strong>at</strong>“str<strong>at</strong>egic determinacy” can hold. In the case of the second result the maintask is to show th<strong>at</strong> the Gener<strong>at</strong>ion Theorem can be iter<strong>at</strong>ively applied togener<strong>at</strong>e ω-many Woodin cardinals.In Section 7 we show th<strong>at</strong> the Gener<strong>at</strong>ion Theorem can itself be localizedin two respects. In the first localiz<strong>at</strong>ion we show th<strong>at</strong> ∆ 1 2-determinacy impliesth<strong>at</strong> for a Turing cone of x, ω L[x]1 is a Woodin cardinal in an inner model ofL[x]. In the second localiz<strong>at</strong>ion we show th<strong>at</strong> 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 th<strong>at</strong> have been made in fine-structuralinner model theory. Remarkably, it turns out th<strong>at</strong> not only are these modelswell-behaved in the context of definable determinacy—they are actuallyfine-structural inner models, but of a kind th<strong>at</strong> 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 <strong>at</strong> various stages. We are especially gr<strong>at</strong>eful 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 N<strong>at</strong>ional Science Found<strong>at</strong>ion under grantsDMS-9970255 and DMS-0355334.1.2. Not<strong>at</strong>ionFor the most part our not<strong>at</strong>ional conventions are standard. Nevertheless,some comments are in order.(1) We use µα ϕ(α) to indic<strong>at</strong>e the least ordinal α such th<strong>at</strong> ϕ(α).(2) In writing OD X and HOD X we always mean th<strong>at</strong> X itself (as opposedto its elements) is allowed as a parameter. The not<strong>at</strong>ion 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 l<strong>at</strong>ter and so we have dropped the curlybrackets on the ground th<strong>at</strong> 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 <strong>from</strong> X) and as an adjective (for example when we say th<strong>at</strong>a particular class is OD X .) We use < ODX for a fixed canonical OD X -well-ordering of OD X sets. The not<strong>at</strong>ion OD(R) is used in analogywith L(R).(3) A str<strong>at</strong>egy for Player I is a function σ : ⋃ i
1. Introduction 17are defined similarly. We typically reserve σ for str<strong>at</strong>egies for PlayerI and τ for str<strong>at</strong>egies for Player II. The play th<strong>at</strong> results <strong>from</strong> havingII play y against σ is denoted σ∗y and likewise the play th<strong>at</strong> results<strong>from</strong> having I play x against τ is denoted x∗τ. We write x∗y for thereal th<strong>at</strong> results <strong>from</strong> 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 th<strong>at</strong> Player I plays when following the str<strong>at</strong>egy σ against II’splay of y. If σ is a str<strong>at</strong>egy for Player I and τ is a str<strong>at</strong>egy for PlayerII we write σ∗τ for the real produced by playing the str<strong>at</strong>egies againstone another. Occasionally, when z = x∗y we write z even to indic<strong>at</strong>e xand z odd to indic<strong>at</strong>e y.(4) If X is a subset of the plane ω ω ×ω ω we use proj 1 (X) for the “projectionto the first coordin<strong>at</strong>e” and proj 2 (X) for the “projection to the secondcoordin<strong>at</strong>e”.(5) For n 0 , . . .,n k−1 ∈ ω, we use 〈n 0 , . . .,n k−1 〉 to denote the n<strong>at</strong>ural 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 associ<strong>at</strong>ed projectionfunctions. For x ∈ ω ω and i ∈ ω we also use (x) i for the projection functionassoci<strong>at</strong>ed 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 not<strong>at</strong>ion in th<strong>at</strong> angle brackets are alsotraditionally used for sequences and n-tuples. We have lapsed into thisusage <strong>at</strong> points but the context resolves the ambiguity; for example,when we write 〈x α | α < λ〉 it is clear th<strong>at</strong> we are referring to asequence.(6) In this chapter by the “reals” we mean ω ω , which, under the standardtopology, is homeomorphic to the irr<strong>at</strong>ionals 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 th<strong>at</strong> 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 th<strong>at</strong> underZF+AD, ω 1 is a measurable cardinal. The proof th<strong>at</strong> we will give is slightlymore involved than the standard proof but it has the advantage of illustr<strong>at</strong>ingsome of the key components in the more general theorems to be provedin l<strong>at</strong>er sections. One thing we would like to illustr<strong>at</strong>e 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 culmin<strong>at</strong>e 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 th<strong>at</strong> the club filter on ω 1 witnesses th<strong>at</strong> ω 1 is1a measurable cardinal and we use Σ ∼ 1-boundedness and the Basic Coding2Lemma to show th<strong>at</strong> 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 th<strong>at</strong> the proof of Solovay’s theorem generalizes to showth<strong>at</strong> assuming ZF+DC+AD then in the restricted setting of L(R) the ordinal2(δ ∼1 ) L(R) is a measurable cardinal. L<strong>at</strong>er, 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 fe<strong>at</strong>ures of the theory of determinacy. Theseconcern (1) the connection between determinacy and choice, (2) the implic<strong>at</strong>ionsof determinacy for regularity properties, and (3) the implic<strong>at</strong>ions 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 diagonaliz<strong>at</strong>ionargument shows th<strong>at</strong> AD contradicts the full axiom of choice, AC. However,certain fragments of AC are consistent with AD and, in fact, certainfragments of AC follow <strong>from</strong> AD.
2. Basic Results 192.1 Definition. The Countable Axiom of Choice, AC ω , is the st<strong>at</strong>ement th<strong>at</strong>every countable set consisting of non-empty sets has a choice function. TheCountable Axiom of Choice for Sets of Reals, AC ω (R), is the st<strong>at</strong>ement th<strong>at</strong>every 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 th<strong>at</strong> 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 transform<strong>at</strong>ions 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 winningstr<strong>at</strong>egy 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 st<strong>at</strong>ementth<strong>at</strong> for every non-empty set X and for every rel<strong>at</strong>ion R ⊆ X × X such th<strong>at</strong>for all x ∈ X there is a y ∈ X such th<strong>at</strong> (x, y) ∈ R, there is a functionf : ω → X such th<strong>at</strong> 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 th<strong>at</strong> DC implies AC ω and Jensen showed th<strong>at</strong>this implic<strong>at</strong>ion cannot be reversed. Solovay showed th<strong>at</strong> 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 wh<strong>at</strong> 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 applic<strong>at</strong>ions, 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 th<strong>at</strong> 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 th<strong>at</strong>(1) for all i < ω, X i ∈ U and(2) ⋂ i
2. Basic Results 21Finally, we turn to the implic<strong>at</strong>ions of determinacy for the Turing degrees.For x, y ∈ ω ω , we say th<strong>at</strong> x is Turing reducible to y, x T y, if x is recursivein y and we say th<strong>at</strong> 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 rel<strong>at</strong>ion 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 l<strong>at</strong>ersections we will discuss different degree notions. However, when we speak ofa “cone of x” without qualific<strong>at</strong>ion we always mean a “Turing cone of x”.The cone filter on D T is the filter consisting of sets of Turing degrees th<strong>at</strong>contain 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 str<strong>at</strong>egy σ 0 then σ 0 witnessesth<strong>at</strong> A is in the cone filter since for y T σ 0 , [y] T = [σ 0 ∗y] T ∈ A. If II has awinning str<strong>at</strong>egy τ 0 then τ 0 witnesses th<strong>at</strong> D T A is in the cone filter sincefor x T τ 0 , [x] T = [x∗τ 0 ] T ∈ D T A.It follows th<strong>at</strong> under ZF + AD each st<strong>at</strong>ement 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 rel<strong>at</strong>ivizes 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 <strong>from</strong> a Turingcone of reals.
2. Basic Results 22It is of interest to note th<strong>at</strong> when Martin proved the Cone Theorem hethought th<strong>at</strong> he would be able to refute AD by finding a property th<strong>at</strong> “toggles”.He started with Borel sets and, when no counterexample arose, movedon to more complic<strong>at</strong>ed sets. We now know (assuming there are infinitelymany Woodin cardinals with a measurable above) th<strong>at</strong> no counterexamplesare to be found in L(R). Moreover, the st<strong>at</strong>ement th<strong>at</strong> there are no counterexamplesin L(R) (i.e. the st<strong>at</strong>ement th<strong>at</strong> Turing determinacy holds inL(R)) actually implies AD L(R) (over ZF + DC). Thus, the basic intuitionth<strong>at</strong> 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 rel<strong>at</strong>ionon ω such th<strong>at</strong> mE x n iff x(〈m, n〉) = 0, where recall th<strong>at</strong> 〈·, ·〉 : ω × ω → ωis a recursive bijection. The real x is said to code the rel<strong>at</strong>ion 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 illustr<strong>at</strong>ion of the utility of this model-theoretic characteriz<strong>at</strong>ion ofΣ ∼11 we shall use it to show th<strong>at</strong> for each x ∈ WO, WO
2. Basic Results 24a contradiction (making free use of AC) as follows: Let z ∈ ω ω be such th<strong>at</strong>both X and WO are Σ 1 1(z). Let α be such th<strong>at</strong> V α |= T 0 and choose Y ≺ V αsuch th<strong>at</strong> 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 th<strong>at</strong> ifj : N → Ult(N, U)is the associ<strong>at</strong>ed 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 th<strong>at</strong> (WO) Ult(N,U) ⊆ WO, which in turn contradicts thefact th<strong>at</strong> 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 th<strong>at</strong>(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 th<strong>at</strong> whenever Z meets one of the sections so doesZ ∗ . In such a situ<strong>at</strong>ion we say th<strong>at</strong> Z ∗ is a selector for Z.To see th<strong>at</strong> 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 th<strong>at</strong>
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 th<strong>at</strong> 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 str<strong>at</strong>egy for Player I in this game.Proof. Suppose σ is a winning str<strong>at</strong>egy for I. As the play unfolds, Player Ican <strong>at</strong>tempt to increase α x as Player II’s play is revealed. However, PlayerII can anticip<strong>at</strong>e all such <strong>at</strong>tempts 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 th<strong>at</strong> 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 <strong>from</strong> a game. As motiv<strong>at</strong>ion,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 th<strong>at</strong> α 0 < β 0 < α 1 < · · · < ω 1 and where the first playerth<strong>at</strong> 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 isviol<strong>at</strong>ed then, letting i be least such th<strong>at</strong> either (x) i ∉ WO or (y) i ∉ WO, Iwins if (x) i ∈ WO; otherwise II wins. Now suppose Rule 1 is s<strong>at</strong>isfied. Rule2: α (x)0 < α (y)0 < α (x)1 < α (y)1 · · ·. The first failure defines who wins asabove. If both rules are s<strong>at</strong>isfied then I wins if and only if sup i
2. Basic Results 28Proof. For S ⊆ ω 1 let G(S) be the game <strong>from</strong> the previous proof and letµ be as defined there. We know th<strong>at</strong> µ is the club filter. To motiv<strong>at</strong>e theproof of normality we give a proof of ω 1 -completeness th<strong>at</strong> will generalizeto produce normal ultrafilters on ordinals larger than ω 1 . This is merely forillustr<strong>at</strong>ion—the proof uses DC but this will be elimin<strong>at</strong>ed in Claim 2.Claim 1. µ is ω 1 -complete.Proof. Suppose S j ∈ µ for j < ω. We have to show th<strong>at</strong> S = ⋂ j
2. Basic Results 29(1.2) a sequence of collections of str<strong>at</strong>egies 〈X i | i < ω〉 where X i containswinning str<strong>at</strong>egies 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> Y 0 ⊆ WO
2. Basic Results 31This completes the proof of the theorem.It should be noted th<strong>at</strong> 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 str<strong>at</strong>egies. This, however, relies onthe fact th<strong>at</strong> η is countable. Our reason for giving the proof in terms ofBasic Coding is th<strong>at</strong> it illustr<strong>at</strong>es in mini<strong>at</strong>ure how we will obtain normalultrafilters on ordinals much larger than ω 1 .Corollary 2.14 (Solovay). Assume ZF + AD. ThenProof. We have th<strong>at</strong>HOD |= 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) th<strong>at</strong> if I has a winning str<strong>at</strong>egy in G(S) then Scontains a club and if II has a winning str<strong>at</strong>egy 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 th<strong>at</strong>and 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 th<strong>at</strong> 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 basicfe<strong>at</strong>ures of this ordinal in the context of L(R). In the next section we willsolve for U in the equ<strong>at</strong>ion1δ∼ 1WO = δ ∼Uin such a way th<strong>at</strong> U is accompanied by the appropri<strong>at</strong>e boundedness andcoding theorems required to generalize Solovay’s proof to show th<strong>at</strong> ZF +DC + AD L(R) 2implies th<strong>at</strong> (δ ∼1 ) L(R) is a measurable cardinal in HOD L(R) .2The following model-theoretic characteriz<strong>at</strong>ion of the pointclass Σ ∼ 1 willbe useful in complexity calcul<strong>at</strong>ions: 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 th<strong>at</strong>(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 fe<strong>at</strong>ures of L(R) under the str<strong>at</strong>ific<strong>at</strong>ion 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 th<strong>at</strong> 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 parameters<strong>from</strong> X ∪{R}. It is important to note th<strong>at</strong> the parameter R is always allowedin our definability calcul<strong>at</strong>ions. To emphasize this we will usually make itexplicit.The basic fe<strong>at</strong>ures of L carry over to L(R), one minor difference beingth<strong>at</strong> R is allowed as a parameter in all definability calcul<strong>at</strong>ions. 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 th<strong>at</strong> for all parametersequences ⃗a ∈ (X ∪ {R})
2. Basic Results 34Proof. It suffices (by the Tarski-Vaught criterion) to show th<strong>at</strong> if A is a nonemptyset which is definable over L λ (R) <strong>from</strong> 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) <strong>from</strong> 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 th<strong>at</strong> there is exactly one element x such th<strong>at</strong> L λ (R) |= ϕ 0 [x, c 0 , α 1 , R]and x ∈ A. Notice th<strong>at</strong> α 1 is definable in L λ (R) <strong>from</strong> c 0 and the real parameterused in the definition of A. Thus, letting x 1 be the unique element suchth<strong>at</strong> 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) <strong>from</strong> 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 elimin<strong>at</strong>e 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 th<strong>at</strong> each π α : ω ω → α is an OD surjection. Assume for contradictionth<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 <strong>from</strong>an ordinal and real parameters.For the second direction suppose x ∈ L(R) and th<strong>at</strong> there is a surjectionπ : ω ω → tc(x). We wish to show th<strong>at</strong> x ∈ L Θ (R). Let λ be a limit ordinalsuch th<strong>at</strong> x ∈ L λ (R). Thus tc(x) ⊆ L λ (R). LetX = { y ∈ L λ (R) | y is definable over L λ (R)<strong>from</strong> parameters in tc(x) ∪ R ∪ {R} } ,By the proof of Lemma 2.17, X ≺ L λ (R) and tc(x) ⊆ X. By Condens<strong>at</strong>ion,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 <strong>from</strong> 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 th<strong>at</strong> L Θ (R) s<strong>at</strong>isfies T 0 − Separ<strong>at</strong>ion −Replacement.To see th<strong>at</strong> L Θ (R) |= Separ<strong>at</strong>ion note th<strong>at</strong> if S ⊆ x ∈ L Θ (R) thenS ∈ L Θ (R), by Lemma 2.20. To see th<strong>at</strong> 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 th<strong>at</strong> L Θ (R) |= ϕ(x, y)).The ordertype of ran(f) is less th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> L α (R) |= T 0 .Proof. The proof is similar to the previous proof. Let us say th<strong>at</strong> 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 th<strong>at</strong> Θ(γ) is an R-cardinal. For each γ which is closed under theGödel pairing function, the argument of Lemma 2.19 shows th<strong>at</strong> Θ(γ) isregular. The proof of the previous lemma generalizes to show th<strong>at</strong> 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 suchth<strong>at</strong>L λ (R) |= ϕ[a].LetX = { y ∈ L λ (R) | y is definable over L λ (R)<strong>from</strong> parameters in tc(a) ∪ R ∪ {R} } ,By Condens<strong>at</strong>ion 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 th<strong>at</strong> 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 modeltheoreticcharacteriz<strong>at</strong>ion. 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 th<strong>at</strong> L λ (R) |= T 0 + ϕ[a]. LetX = { x ∈ L λ (R) | y is definable over L λ (R)<strong>from</strong> parameters in R ∪ {R} } ,Then X = L λ (R). Moreover, there is a surjection π : ω ω → L λ (R) such th<strong>at</strong>π is definable over L λ+1 (R) <strong>from</strong> R and a.Proof. By Lemma 2.17 we have th<strong>at</strong> X ≺ L λ (R). By condens<strong>at</strong>ion 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 <strong>from</strong>a real parameter and since L λ (R) ∼ = X, we have th<strong>at</strong> every x ∈ L λ (R) isdefinable <strong>from</strong> 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 th<strong>at</strong> {(x, y) | π(x) ∈ π(y)}is ∆ ∼21 . Thus, δ R δ ∼21 .Proof. Fix α such th<strong>at</strong> 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 th<strong>at</strong> if β is the least ordinalsuch th<strong>at</strong> L β (R) |= ϕ[a] then β > α. Let γ be least such th<strong>at</strong> γ > β > α
2. Basic Results 38and L γ (R) |= T 0 (which exists by Lemma 2.22). So γ is least such th<strong>at</strong>L 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 th<strong>at</strong>ω 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 characteriz<strong>at</strong>ion of Σ ∼1 th<strong>at</strong> A is Σ ∼1 . Asimilar argument shows th<strong>at</strong> (ω ω ) 2 2 A is Σ ∼ 1 . Finally, the desired map canbe extracted <strong>from</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> is universal for Σ ∼21 subsetsof ω ω × ω ω , th<strong>at</strong> is, such th<strong>at</strong> for each Σ ∼21 subset A ⊆ ω ω × ω ω there is anx ∈ ω ω such th<strong>at</strong> 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 th<strong>at</strong> dom(ρ) is Σ 1 (L(R), {R}) and hence Σ 1 (L δR (R), {R}). Notice alsoth<strong>at</strong> 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 th<strong>at</strong>{(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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> γ < δ R . Let π : ω ω → α be asurjection such th<strong>at</strong> A = {(x, y) | π(x) < π(y)} is ∆ ∼21 . Using the not<strong>at</strong>ion<strong>from</strong> the previous proof let x be such th<strong>at</strong>U (x)0 = A and U (x)1 = (ω ω × ω ω ) A.There is an ordinal α such th<strong>at</strong> 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 noteth<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> under ZF +DC + AD,HOD L(R) |= (δ ∼21 ) L(R) is λ-strong,for each λ < Θ L(R) , and, in fact, th<strong>at</strong>HOD L(R) |= Θ L(R) is a Woodin cardinal.The proof th<strong>at</strong> we give in Section 4 will show th<strong>at</strong> DC can be elimin<strong>at</strong>ed<strong>from</strong> the result of the present section.First we need an analogue U of WO th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong>L 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 <strong>from</strong> theproof th<strong>at</strong> ω 1 is measurable. For not<strong>at</strong>ional convenience we will routinelyuse our recursive bijection <strong>from</strong> 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 gre<strong>at</strong>er than α. Let y, z ∈ ω ω be such th<strong>at</strong> 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 suchth<strong>at</strong> 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 th<strong>at</strong> A = (U y ) L β(R) . Now, letting x ∈ U U L β(R) and γ < δ ∼21 besuch th<strong>at</strong> L γ (R) |= “T 0 + x ∈ U”, we have th<strong>at</strong> 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 viol<strong>at</strong>ed then, letting i be least such th<strong>at</strong> either (x) i ∉ U or (y) i ∉ U, Iwins if (x) i ∈ U; otherwise II wins. Now suppose Rule 1 is s<strong>at</strong>isfied. 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 s<strong>at</strong>isfied then I wins iff sup i∈ω δ (x)i ∈ S.Now letµ = {S ⊆ δ ∼21 | I wins G(S)}.Notice th<strong>at</strong> as before (using ∆ ∼21 -boundedness) if I has a winning str<strong>at</strong>egyin G(S) then S contains a set C which is unbounded and closed under ω-sequences. The proof th<strong>at</strong> U is an ultrafilter is exactly as before. To see th<strong>at</strong>it 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 th<strong>at</strong> f : δ ∼21 → δ ∼21 and th<strong>at</strong> there is no α < δ ∼21such th<strong>at</strong> {ξ | f(ξ) = α} ∈ µ or, equivalently (by AD) th<strong>at</strong> for all α < δ ∼21 ,S α = {ξ | f(ξ) ≠ α} ∈ µ.Let 〈δ α | α < δ ∼21 〉 enumer<strong>at</strong>e 〈δ x | x ∈ U〉. Here we are appealing to the factth<strong>at</strong> 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 associ<strong>at</strong>ed prewellordering.As before, our str<strong>at</strong>egy 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 str<strong>at</strong>egies 〈X i | i < ω〉 where X i containswinning str<strong>at</strong>egies 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 relevantstr<strong>at</strong>egies 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 th<strong>at</strong> is legalagainst all of the relevant str<strong>at</strong>egies and (in each case) has the same fixedordinal as output. This technique is central in wh<strong>at</strong> follows. It is importantto note, however, th<strong>at</strong> the above ultrafilter (and, more generally, ultrafiltersobtained by such a “sup” game) concentr<strong>at</strong>es on points of cofinality ω. L<strong>at</strong>erwe 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 rel<strong>at</strong>ive to WO; we nowdo so rel<strong>at</strong>ive to more general prewellorderings.3.1. Coding LemmaWe begin by fixing some not<strong>at</strong>ion. For P ⊆ ω ω 1, the notion of a Σ ∼1 (P) set is1defined exactly like th<strong>at</strong> 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 th<strong>at</strong>X = { y ∈ ω ω | there is an ω-model M such th<strong>at</strong>y, 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 1th<strong>at</strong> is universal for Σ ∼1 (P)subsets of (ω ω ) n 1, th<strong>at</strong> is, such th<strong>at</strong> for each Σ ∼ 1 (P) set A ⊆ (ωω ) n there is ane ∈ ω ω such th<strong>at</strong> A = U e(n) (P) = {y ∈ (ω ω ) n | (e, y) ∈ U (n) (P)}. We do thisin such a way th<strong>at</strong> the same formula is used, so th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong>our universal sets s<strong>at</strong>isfy 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 th<strong>at</strong>Proof. 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 th<strong>at</strong> T a = U (2)d(a)(P). (The function d comes<strong>from</strong> the s-m-n-theorem. In fact, d(a) = s(a, a) (in the not<strong>at</strong>ion 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 th<strong>at</strong> Y = U a (3)10 (P). Notice th<strong>at</strong> Y is Σ ∼ 1 (P) using theparameter for Y (as can readily be checked using the model-theoretic characteriz<strong>at</strong>ionof 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 th<strong>at</strong>(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 th<strong>at</strong> there is no such e. Consider theset G of reals e for which (1) in the st<strong>at</strong>ement of the theorem is s<strong>at</strong>isfied:G = { e ∈ ω ω | U (2)e (Q) ⊆ Z } .So, for each e ∈ G, (2) in the st<strong>at</strong>ement of the theorem fails for some a ∈ X.Let α e be the least α such th<strong>at</strong> (2) fails <strong>at</strong> 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 th<strong>at</strong> by ourassumption th<strong>at</strong> there is no index e as in the st<strong>at</strong>ement 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 th<strong>at</strong>π(a) = α eσ . Pick (x 1 , x 2 ) ∈ Z ∩ (Q a × ω ω ). Let e ∗ be such th<strong>at</strong>U (2)e ∗(Q) = U(2) e σ∪ {(x 1 , x 2 )}.So e ∗ ∈ G. But α eσ < α e ∗. In summary, we have th<strong>at</strong> for all y ∈ ω ω ,α (σ∗y)I α eσ < α e ∗. Thus, by playing e ∗ , II defe<strong>at</strong>s σ.Claim 2. Player II does not have a winning str<strong>at</strong>egy.Proof. Assume toward a contradiction th<strong>at</strong> τ is a winning str<strong>at</strong>egy for II. Weshall show th<strong>at</strong> τ yields a selector for Z; in other words, it yields an e ∗ suchth<strong>at</strong>(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 th<strong>at</strong> 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 ; th<strong>at</strong> is, there is ane ∗ such th<strong>at</strong>U (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 th<strong>at</strong> 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 th<strong>at</strong> U (2)e ∗(Q) Z with π(x 1) minimal. SoU (2)e ∗(x 1 , x 2 ) ∈ U (2)e ∗Fix a ∈ X such th<strong>at</strong>(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 th<strong>at</strong> 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 str<strong>at</strong>egy, (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 th<strong>at</strong> α e ∗ exists. Let a ∈ X be such th<strong>at</strong>π(a) = α e ∗. Thus h 0 (e ∗ , a) ∈ G and α h0 (e ∗ ,a) = α e ∗. Since τ is a winningstr<strong>at</strong>egy 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 th<strong>at</strong> which appears in the liter<strong>at</strong>ure ([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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> for all a ∈ X,(1) U e(2) (Q
3. Coding 50in the parameters Q
3. Coding 51andChoose a ∈ X such th<strong>at</strong> π(a) = α eσ . SoU (2)e σ(Q
3. Coding 52Notice th<strong>at</strong> for e ∈ ω ω and z ∈ X, the set U (2)h 0 (e,z)(·, ·) is such th<strong>at</strong> 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 str<strong>at</strong>egy 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. Applic<strong>at</strong>ionsIn this section we will bring together some basic results and key applic<strong>at</strong>ionsof the above coding lemmas th<strong>at</strong> will be of use l<strong>at</strong>er. 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 th<strong>at</strong> 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 th<strong>at</strong> every set isOD X,y for some real y. Then Θ = Θ X .Proof. Fix α < Θ. We have to show th<strong>at</strong> 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 elimin<strong>at</strong>e 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 th<strong>at</strong> 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 th<strong>at</strong> Θ Xis singular in HOD X and letf : η → Θ Xbe an OD X cofinal map. Let g be an OD X surjection <strong>from</strong> 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 th<strong>at</strong> Θ X is a strong limit in HOD X . For each η < Θ X , wehave to show th<strong>at</strong> |P(η)| HOD X< Θ X . For this it suffices to show th<strong>at</strong> 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 immedi<strong>at</strong>ely <strong>from</strong> 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 immedi<strong>at</strong>e. For the left-to-right inclusion supposex ∈ HOD L(R) ∩ V Θ . We have to show th<strong>at</strong> 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 th<strong>at</strong> 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 <strong>from</strong> 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 th<strong>at</strong> all ultrafiltersare countably complete (Theorem 2.8), A x is non-empty. Letf(x) = ⋂ A x .Notice th<strong>at</strong> A x and f(x) depend only on the Turing degree of x. In particular,we can regard f as a function <strong>from</strong> the Turing degrees D T into the ordinals.Notice also th<strong>at</strong>A ∈ 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 th<strong>at</strong>π(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 Gener<strong>at</strong>ion Theorem inthe next section. The proof th<strong>at</strong> we give appeals to DC <strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> is intended to witness th<strong>at</strong> ∼δ 1 is λ-strong. In §4.2we shall introduce and motiv<strong>at</strong>e the notion of strong normality by showing2th<strong>at</strong> the strong normality of µ λ ensures th<strong>at</strong> 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 rel<strong>at</strong>ivizing 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 th<strong>at</strong> ZF+AD implies th<strong>at</strong> Θ is strongly inaccessible in HOD L(R) .Our next task is to show th<strong>at</strong>HOD L(R) |= δ ∼21 is λ-strong,for all λ < Θ. The proof will then rel<strong>at</strong>ivize to subsets of Θ th<strong>at</strong> are inHOD L(R) and thereby establish the main theorem.The ultrafilters th<strong>at</strong> witness strength cannot come <strong>from</strong> the “sup” gameof Section 2 since the ultrafilters produced by this game concentr<strong>at</strong>e 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 th<strong>at</strong> 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 th<strong>at</strong>L(R) |= ϕ[z, F(δ), δ, R].One should think of F as a sequence th<strong>at</strong> 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 th<strong>at</strong>witnesses 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 <strong>from</strong>Section 2.4 concerning L(R) and the theory T 0 : There are arbitrarily largeα such th<strong>at</strong> 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 th<strong>at</strong> L α (R) |= T 0 . Similarly, there arearbitrarily large α < δ ∼21 such th<strong>at</strong> L α (R) |= T 0 . However, notice th<strong>at</strong> it isnot the case L δ˜21(R) |= T 0 (by Lemma 2.27).Because of the gre<strong>at</strong>er manoeuvring room provided by levels L α (R) th<strong>at</strong>s<strong>at</strong>isfy T 0 we will concentr<strong>at</strong>e (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 n<strong>at</strong>ural well-ordering < OD L(R) on the OD L(R) sets. For thel<strong>at</strong>ter, given X ∈ OD L(R) , letα X = the least α such th<strong>at</strong>(1) L α (R) |= T 0 ,(2) X ∈ L α (R), and(3) X is definable in L α (R) <strong>from</strong> ordinal parameters;let ϕ X be the least formula th<strong>at</strong> defines X <strong>from</strong> 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 th<strong>at</strong> 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 quantific<strong>at</strong>ionover the V α hierarchy, which is Π 1 .) Notice th<strong>at</strong> 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 th<strong>at</strong> we use the Σ 1 -definition given above since theΣ 2 -definition involves quantific<strong>at</strong>ion 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 th<strong>at</strong>of 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 s<strong>at</strong>isfies T 0 over which the reflection phenomenonfails with respect to F↾δ ∼21 . This motiv<strong>at</strong>es the following definition:4.2 Definition. Assume T 0 . Suppose th<strong>at</strong> F↾δ is defined. Let ϑ(δ) be leastsuch th<strong>at</strong>L ϑ(δ) (R) |= T 0 and there is an X ∈ L ϑ(δ) (R) ∩ OD L ϑ(δ)(R) such th<strong>at</strong>(⋆) there is a Σ 1 -formula ϕ and a real z such th<strong>at</strong>and 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 th<strong>at</strong> (⋆) 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 th<strong>at</strong> 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 th<strong>at</strong> (F↾δ) Lα(R) = F↾δ with the aim of showing th<strong>at</strong>(F(δ)) Lα(R) = (F(δ)) L(R) . The point is th<strong>at</strong>L 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 th<strong>at</strong> (F↾δ) Lα(R) =F↾δ.For (2) first notice th<strong>at</strong> we can make sense of F as defined over levels(such as L δ˜21(R)) th<strong>at</strong> do not s<strong>at</strong>isfy 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 th<strong>at</strong>2(2) fails, th<strong>at</strong> 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 -st<strong>at</strong>ement about γ:∃α > γ (L α (R) |= T 0 + ϑ(γ) exists.)Since L δ˜21(R) ≺ 1 L(R), this st<strong>at</strong>ement 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 th<strong>at</strong> 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 th<strong>at</strong> L α (R) |= T 0 andF Lα(R) = F↾γ = F. We claim th<strong>at</strong> this implies th<strong>at</strong>L 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 th<strong>at</strong> L α (R) |= ψ[z, R].By Replacement there is an ordinal β such th<strong>at</strong>L β (R) |= ψ[z, R].Consider the Σ 1 -st<strong>at</strong>ement ϕ[z, X, R] expressing “There exists ξ such th<strong>at</strong>X = L ξ (R) and X |= ψ[z, R]”. Letting ϑ > β be such th<strong>at</strong> L ϑ (R) |= T 0 wehave: there exists an X ∈ L ϑ (R) ∩ OD L ϑ(R) (namely, X = L β (R)) such th<strong>at</strong>L 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)) th<strong>at</strong>there exists a ¯δ < γ such th<strong>at</strong>L ϑ (R) |= ϕ[z, F(¯δ), R].Thus (unpacking ϕ[z, X, R]) there exists a ξ such th<strong>at</strong> 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 th<strong>at</strong> F↾δ ∼21 : δ ∼21 → L δ˜21(R) is total and ∆ 1 -definable over L δ˜21(R).It remains to see th<strong>at</strong> 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 th<strong>at</strong>L(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 somewh<strong>at</strong>involved.Assume for contradiction th<strong>at</strong> there is an X ∈ OD L(R) , a Σ 1 -formula ϕ,and z ∈ ω ω such th<strong>at</strong>L(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 th<strong>at</strong>(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 th<strong>at</strong>L 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 th<strong>at</strong> (1.1) and for this choicepick ϕ 0 and z 0 such th<strong>at</strong> (⋆). (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 th<strong>at</strong> 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 st<strong>at</strong>ement L ϑ0 (R) |= T 0 +“reflection fails with respect to F↾δ ∼21 ”. It is this internal st<strong>at</strong>ement th<strong>at</strong> wewill reflect to get a contradiction. We have th<strong>at</strong> for all δ < δ ∼21 ,(1.2) L ϑ0 (R) ̸|= ϕ 0 [z 0 , F(δ), δ, R].Our str<strong>at</strong>egy is to reflect to get ¯ϑ < δ ∼21 such th<strong>at</strong>L¯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 str<strong>at</strong>egywe need the appropri<strong>at</strong>e Σ 1 -fact (in a real) to reflect.Step 2. The following is a true Σ 1 -st<strong>at</strong>ement about ϕ 0 and z 0 (as witnessedby taking α to be ϑ 0 <strong>from</strong> Step 1): There is an α such th<strong>at</strong>(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 th<strong>at</strong>and 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 th<strong>at</strong> 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 th<strong>at</strong>and 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 th<strong>at</strong> (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 th<strong>at</strong> in (2.3) the ordinal δ ∼21 and the function F are computed inL α (R) while the formulas are evalu<strong>at</strong>ed in L β (R).) Thus (2.1) ensures (byLemma 4.3) th<strong>at</strong> F Lα(R) ↾(δ ∼21 ) Lα(R) = F↾(δ ∼21 ) Lα(R) , (2.2) says th<strong>at</strong> L α (R) s<strong>at</strong>isfies“reflection is failing with respect to F Lα(R) ↾(δ ∼21 ) Lα(R) ” and, because of(2.1), this ensures th<strong>at</strong> ϑ((δ ∼21 ) Lα(R) ) exists, (2.3) ensures in addition th<strong>at</strong>α = ϑ((δ ∼21 ) Lα(R) ), and (2.4) says th<strong>at</strong> 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 st<strong>at</strong>ement must be lessthan δ ∼21 . Let ¯ϑ be this ordinal.Step 3. We claim th<strong>at</strong>L¯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) s<strong>at</strong>isfies “reflectionis failing with respect to F L¯ϑ (R) ↾(δ ∼21 ) L¯ϑ (R) ” and, because of (4.1), this ensuresth<strong>at</strong> 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 th<strong>at</strong>which 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 th<strong>at</strong> 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 suchth<strong>at</strong>2z ϕ,y ∈ U X iff L(R) |= ϕ[y, X, ∼δ 1 , R].In such a situ<strong>at</strong>ion we say th<strong>at</strong> z ϕ,y certifies the Σ 1 -fact ϕ about y. The keyproperty is, of course, th<strong>at</strong> if z ϕ,y ∈ U δ then L(R) |= ϕ[y, F(δ), δ, R]. Noticeth<strong>at</strong> the real z ϕ,y is recursive in y (uniformly).In wh<strong>at</strong> follows we will drop reference to ϕ and y and simply write z ∈ U X ,it being understood th<strong>at</strong> the formula and parameter are encoded in z. Inthese terms Theorem 4.4 can be recast as st<strong>at</strong>ing th<strong>at</strong> if z ∈ U X then thereis an ordinal δ < δ ∼21 such th<strong>at</strong> z ∈ U δ , in other words, U X ⊆ ⋃ δ
4. A Woodin Cardinal in HOD L(R) 67and consider the following Σ 1 -st<strong>at</strong>ement:There is an α such th<strong>at</strong>L 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 st<strong>at</strong>ement is true. It followsth<strong>at</strong> the st<strong>at</strong>ement holds for F X -almost all δ. The second clause ensures th<strong>at</strong>each such δ is a “local δ ∼21 ” and th<strong>at</strong> the “local comput<strong>at</strong>ion 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 immedi<strong>at</strong>e. Itremains to prove countable completeness. Suppose {S n | n < ω} ⊆ F X . Forn < ω, let z n ∈ U X be such th<strong>at</strong> S zn ⊆ S n . Let z ∈ ω ω be such th<strong>at</strong> (z) n = z nfor all n < ω. The following is a true Σ 1 -st<strong>at</strong>ement about z, X, δ ∼21 , and R:There is an α such th<strong>at</strong>(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 st<strong>at</strong>ement. It follows th<strong>at</strong> for each δ < δ ∼21 such th<strong>at</strong>z ∗ ∈ U δ the following holds:There is an α such th<strong>at</strong>(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 predic<strong>at</strong>es.Proof. First we show th<strong>at</strong> 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 th<strong>at</strong>U (2)e(S) (Q
4. A Woodin Cardinal in HOD L(R) 69It follows th<strong>at</strong> for δ ∈ S z , e(S) codes S ∩ δ.This enables us to associ<strong>at</strong>e with each Σ 1 -sentence ϕ involving the predic<strong>at</strong>eS, a Σ 1 -sentence ϕ ∗ involving instead the real e(S) in such a way th<strong>at</strong>L(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 predic<strong>at</strong>e S can be elimin<strong>at</strong>ed in favour of the real e(S),thereby reducing the present version of the reflection theorem to the originalversion (Theorem 4.4).To see th<strong>at</strong> we can also include parameters of the form f : δ ∼21 → δ ∼21 simplynote th<strong>at</strong> 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 viol<strong>at</strong>ed then, letting i be the least such th<strong>at</strong> either (x) i ∉U X or (y) i ∉ U X , I wins if (x) i ∈ U X ; otherwise II wins. If the rule is s<strong>at</strong>isfiedthen, letting δ be least such th<strong>at</strong> 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 -st<strong>at</strong>ementabout a single real) I wins iff δ ∈ S. Thus, I is picking δ by steering into theδ th -approxim<strong>at</strong>ion U δ . (Note th<strong>at</strong> 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 suchth<strong>at</strong> (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 th<strong>at</strong> 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 str<strong>at</strong>egies 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 th<strong>at</strong> z 0 T σ ′ ; more precisely, z 0 T σ ′ is such th<strong>at</strong> 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 th<strong>at</strong>z 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 th<strong>at</strong> 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 th<strong>at</strong> (z) i = z i for all i < ω and let δ 0 be leastsuch th<strong>at</strong> (z) i ∈ U δ0 for all i ∈ ω. Notice th<strong>at</strong> 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 str<strong>at</strong>egyfor 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 th<strong>at</strong> these ultrafilters are δ ∼21 -complete. Wenow wish to take the ultrapower of HOD L(R) with these ultrafilters and showth<strong>at</strong> collectively they witness th<strong>at</strong> for each λ < Θ, δ ∼21 is λ-strong in HOD L(R) .This will be achieved by showing th<strong>at</strong> reflection and uniform coding combineto show th<strong>at</strong> µ λ is strongly normal.We begin with the following basic lemma on the ultrapower construction,which we shall prove in gre<strong>at</strong>er generality than we need <strong>at</strong> the moment.Lemma 4.8. Assume ZF+DC. Suppose µ is a countably complete ultrafilteron δ and th<strong>at</strong> 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 associ<strong>at</strong>ed 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 associ<strong>at</strong>ed membership rel<strong>at</strong>ion.So [f] µ E [g] µ if and only if {α < δ | f(α) ∈ g(α)} ∈ µ. Notice th<strong>at</strong> 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. Noticeth<strong>at</strong> j µ is OD T .Claim 1. ((HOD T ) δ /µ, E) is well-founded.Proof. Suppose for contradiction th<strong>at</strong> ((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 th<strong>at</strong> [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 immedi<strong>at</strong>e.It remains to show th<strong>at</strong> 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 th<strong>at</strong> a = [f] µ .Let α be such th<strong>at</strong> f ∈ V α . Then for each b ∈ (HOD T ) δ /µ such th<strong>at</strong> bEa,letting g ∈ (HOD T ) δ be such th<strong>at</strong> 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 th<strong>at</strong> M ⊆ HOD T . For this it suffices to show th<strong>at</strong> 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 th<strong>at</strong>for 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong>and for this it remains to show th<strong>at</strong>HOD 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 th<strong>at</strong> λ is such th<strong>at</strong> L λ (R) ≺ L Θ (R) and2δ∼ 1 < λ. There are arbitrarily large λ < Θ with this fe<strong>at</strong>ure (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 th<strong>at</strong>HOD 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 th<strong>at</strong> A ∈ M X . In fact, we will show th<strong>at</strong>LetP(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 th<strong>at</strong> 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 th<strong>at</strong> 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 , th<strong>at</strong> 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 -st<strong>at</strong>ement inL(R) about t 1 , t 2 , X and R. Thus, by reflection (Theorem 4.6), it follows th<strong>at</strong>for 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 th<strong>at</strong> λ ∏ λ δ /µ X
4. A Woodin Cardinal in HOD L(R) 77µ X -almost all δ, |t| δ ∈ A δ , th<strong>at</strong> is, {δ < δ ∼21 | f t (δ) ∈ A δ } ∈ µ X . Likewise,if |t| λ ∉ A then since this is a true Σ 1 -st<strong>at</strong>ement about e(A), t and X, forµ X -almost all δ, |t| δ ∉ A δ .So it suffices to show th<strong>at</strong> 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 th<strong>at</strong> π :(HOD L(R) ) δ˜21 /µX ∼ = MX is the transitive collapse map. We already knowth<strong>at</strong> ρ is well-defined and order-preserving (by Lemma 4.10). It remains toshow th<strong>at</strong> ρ is onto, th<strong>at</strong> is, th<strong>at</strong> every function f ∈ ∏ λ δ /µ X is equivalent(modulo µ X ) to a canonical function f t . To say th<strong>at</strong> this is true is to sayth<strong>at</strong> µ X is strongly normal:4.11 Definition (Strong normality). µ X is strongly normal iff wheneverf : S 0 → δ ∼21 is such th<strong>at</strong>then there exists a t ∈ ω ω such th<strong>at</strong>{δ ∈ S 0 | f(δ) < λ δ } ∈ µ X{δ ∈ S 0 | f(δ) = f t (δ)} ∈ µ X .Notice th<strong>at</strong> 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 <strong>at</strong> ∈ ω ω such th<strong>at</strong>{δ < δ ∼21 | f t (δ) = f(δ)} ∈ µ X .So if β is such th<strong>at</strong> t ∈ Q δ˜21βthen β < δ 2∼1, since otherwise by reflection thiswould contradict the assumption th<strong>at</strong>{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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> by the definition of η and our assumption th<strong>at</strong> f is a counterexampleto strong normality, we have th<strong>at</strong>for all t ∈ Q δ˜21
4. A Woodin Cardinal in HOD L(R) 79e 0 is the index of the universal set th<strong>at</strong> selects the Zα δ ’s (represented in thediagrams as ellipses) <strong>from</strong> Z ′ (represented in the diagrams as chimneys).e 1 is the index of the universal set th<strong>at</strong> 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 th<strong>at</strong> determines η δ for δ ∈ S 0 .y f is the real in Q δ˜21“f(δ˜2 1 )” th<strong>at</strong> 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 suchth<strong>at</strong> 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, th<strong>at</strong> T ∉ µ X . So (δ ∼1 T) ∩ S 1 ∈ µ X .Let σ ′ be a winning str<strong>at</strong>egy for I in G X 2((δ ∼1 T) ∩ S 1 ).Let us first motiv<strong>at</strong>e the main idea: Suppose z is a legal play for II againstσ ′ (by which we mean a play for II th<strong>at</strong> s<strong>at</strong>isfies the Main Rule) and suppose2th<strong>at</strong> the ordinal associ<strong>at</strong>ed 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) suchth<strong>at</strong> 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 th<strong>at</strong>(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 th<strong>at</strong> (t 0 , σ 0 ) ∈ Z δ˜21β 0we have(3.2) σ 0 is a winning str<strong>at</strong>egy 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 th<strong>at</strong> z n+1 T z n ; so z n+1 is such th<strong>at</strong>if 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 th<strong>at</strong> (z) i = z i for all i < ω and let δ 0 be least suchth<strong>at</strong> (z) i ∈ U δ0 for all i ∈ ω. Notice th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> (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 th<strong>at</strong> (t 0 , σ 0 ) ∈ Z δ˜21β 0we have(8.2) σ 0 is a winning str<strong>at</strong>egy 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 -st<strong>at</strong>ement 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 st<strong>at</strong>ement 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 th<strong>at</strong> 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 th<strong>at</strong> 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 Comput<strong>at</strong>ion). There exists an index e 1 ∈ ω ω such th<strong>at</strong>for 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 th<strong>at</strong> f(δ) < β < η δ,where P δ = ⋃ {Z¯δα | ¯δ ∈ S 2 ∩ δ and α ∈ [f(¯δ), η¯δ)} and S 2 is <strong>from</strong> 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 st<strong>at</strong>ement 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 th<strong>at</strong> there is no such e 1 . We follow theproof of the Uniform Coding Lemma. To begin with, notice th<strong>at</strong> it sufficesto find e 1 ∈ ω ω s<strong>at</strong>isfying (2) and(3 ′ ) U (2)e 1(P δ , Z δ β ) ∩ Zδ β ≠ ∅ for β such th<strong>at</strong> f(δ) < β < η δsince given the parameter Zβ δ we can easily ensure (1).Consider the set of reals such th<strong>at</strong> (2) of the (revised) claim holds, th<strong>at</strong>is,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 th<strong>at</strong>(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 str<strong>at</strong>egy.Proof. Suppose toward a contradiction th<strong>at</strong> σ is a winning str<strong>at</strong>egy 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 defe<strong>at</strong>s σ.We will make key use of the Disjointness Property.Choose e σ ∈ ω ω such th<strong>at</strong> 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 str<strong>at</strong>egy 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 “<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 “<strong>at</strong> least as good” as e σ . But since x 0 ∈ Z δ 0β 0, we have th<strong>at</strong> x 0 ∈U (2)e (P δ 0, Z δ ∗ 0β 0), by the definition of e ∗ ; th<strong>at</strong> 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 defe<strong>at</strong>sσ.Claim 2. Player II does not have a winning str<strong>at</strong>egy.Proof. Suppose toward a contradiction th<strong>at</strong> τ is a winning str<strong>at</strong>egy for II.We shall find an e ∗ such th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> e ∗ ∉ G. Let δ 0 ∈ S 2 be least such th<strong>at</strong>U (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 th<strong>at</strong>= ⋃ 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 th<strong>at</strong>(h 0 (e ∗ , x 0 )∗τ) II ∈ G (as τ is a winning str<strong>at</strong>egy for II), which contradicts theprevious st<strong>at</strong>ement.Subsubclaim. h 0 (e ∗ , x 0 ) ∈ G, th<strong>at</strong> 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 l<strong>at</strong>ter holds since we chose δ 0 to be least such th<strong>at</strong>U (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 th<strong>at</strong> α e ∗exists. Recall th<strong>at</strong>α e ∗ = lexicographically least pair (δ, β) such th<strong>at</strong>(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 l<strong>at</strong>ter 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 th<strong>at</strong> 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 th<strong>at</strong>(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 th<strong>at</strong> the claim is false. The st<strong>at</strong>ement th<strong>at</strong>the claim fails is a true Σ 1 -st<strong>at</strong>ement 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 st<strong>at</strong>ement th<strong>at</strong> y f ∈ Q δ˜21β 0where β 0 is such th<strong>at</strong>(1) and (2) of Claim C hold is a true Σ 1 -st<strong>at</strong>ement 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 st<strong>at</strong>ementreflects. Let S 3 ⊆ S 2 be in µ X and such th<strong>at</strong> the above st<strong>at</strong>ement reflectsto each point in S 3 . Now by Claim B, for δ ∈ S 3 , the least β 0 such th<strong>at</strong>y 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 str<strong>at</strong>egy σ aproto-winning str<strong>at</strong>egy if σ is a winning str<strong>at</strong>egy for I in G X (δ ∼21 ). Thus if yis suitable and σ is a proto-winning str<strong>at</strong>egy thenand so we can let{((σ∗y) I ) i , (y) i | i < ω} ⊆ U Xδ (σ,y) = the least δ such th<strong>at</strong> {((σ∗y) I ) i | i < ω} ∪ {(y) i | i < ω} ⊆ U δ .Let κ be least such th<strong>at</strong> X ∈ L κ (R) and L κ (R) ≺ 1 L(R). This is the “leaststable over X”. It is easy to see th<strong>at</strong>P(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 st<strong>at</strong>e the relevant result:Suppose Σ ∈ P(R) ∩ L κ (R) is a set of proto-winning str<strong>at</strong>egies for I.Then there is a proto-winning str<strong>at</strong>egy σ such th<strong>at</strong> for all suitable reals y,for all τ ∈ Σ δ(σ,y) ∩ Σ, there is a suitable real y τ such th<strong>at</strong>δ (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 th<strong>at</strong> Θ L(R) is a Woodin cardinal in HOD L(R) . Ingeneral, in inner model theory there is a long march up <strong>from</strong> 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 not<strong>at</strong>ional convenience let Θ = Θ L(R) . To show th<strong>at</strong>HOD L(R) |= ZFC + Θ is a Woodin cardinalit suffices to show th<strong>at</strong> for each T ∈ P(Θ) ∩ OD L(R) , there is an ordinal δ Tsuch th<strong>at</strong>|= 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)s<strong>at</strong>isfies th<strong>at</strong> 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 n<strong>at</strong>ural Σ 1 -str<strong>at</strong>ific<strong>at</strong>ion, 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 th<strong>at</strong> 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 th<strong>at</strong>L 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 predic<strong>at</strong>e for T, which is assumed to be allowed in allof our definability calcul<strong>at</strong>ions.)Letδ T = the least λ such th<strong>at</strong> 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 th<strong>at</strong>L 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 th<strong>at</strong> F T ↾δ is defined. Let ϑ(δ) be least such th<strong>at</strong>L 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 th<strong>at</strong>and 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 th<strong>at</strong> (⋆).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 th<strong>at</strong>L 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 viol<strong>at</strong>ed then, letting i be the least such th<strong>at</strong> either (x) i ∉UX T or (y) i ∉ UX T I wins if (x) i ∈ UX T ; otherwise II wins. If the rule is s<strong>at</strong>isfiedthen, letting δ be least such th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> µ T X is strongly normal iff whenever f : S 0 → δ T issuch th<strong>at</strong>{δ ∈ S 0 | f(δ) < λ δ } ∈ µ T Xthen there exists a t ∈ ω ω such th<strong>at</strong>{δ ∈ S 0 | f(δ) = f t (δ)} ∈ µ T X .The proof th<strong>at</strong> 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 th<strong>at</strong> j λ (δ T ) > λ. Assuming further th<strong>at</strong> λ is such th<strong>at</strong>we have th<strong>at</strong>L λ (R)[T ∩ λ] ≺ 1 L Θ (R)[T]HOD L Θ(R)[T]T⊆ M λ .As before, strong normality implies th<strong>at</strong>ρ : λ → ∏ λ δ /µ T X|t| λ ↦→ [f t ] µ TXis an isomorphism. It remains to establish T-strength, th<strong>at</strong> is,|t| λ ∈ T ∩ λ iff {δ < δ T | f t (δ) ∈ T ∩ λ δ } ∈ µ T X.The point is th<strong>at</strong> both|t| λ ∈ T ∩ λ
5. Woodin <strong>Cardinals</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> the ultrapowerswere well-founded (Lemma 4.8). This was necessary since althoughthe ultrapowers were ultrapowers of HOD and HOD s<strong>at</strong>isfies AC, the ultrapowerswere “external” (in th<strong>at</strong> the associ<strong>at</strong>ed 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 elimin<strong>at</strong>ed by using the extender formul<strong>at</strong>ion of beinga Woodin cardinal. In this way one obtains strength through a network of“internal” ultrapowers (th<strong>at</strong> is, via ultrafilters th<strong>at</strong> 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 <strong>Cardinals</strong> in General SettingsOur aim in this section is to abstract the essential ingredients <strong>from</strong> the previousconstruction and prove two abstract theorems on Woodin cardinals ingeneral settings, one th<strong>at</strong> requires DC and one th<strong>at</strong> 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 th<strong>at</strong> Θ X isa Woodin cardinal in HOD X . The importance of the version involving Θ X,Yis th<strong>at</strong> it enables one to show th<strong>at</strong> in certain settings HOD X can have manyWoodin cardinals. To describe one such key applic<strong>at</strong>ion we introduce the
5. Woodin <strong>Cardinals</strong> 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 th<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 97ordinals. Since (OD S,x ) L[S,x] = L[S, x] and L[S, x] s<strong>at</strong>isfies 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 ultim<strong>at</strong>ely we reach the notion of “str<strong>at</strong>egic determinacy”,which is sufficiently rich to simul<strong>at</strong>e boldface determinacy and drive the construction.With this motiv<strong>at</strong>ion in place we will return to the general settingin Section 5.3 and prove the Gener<strong>at</strong>ion Theorem. Finally, in Section 5.4 wewill use the Gener<strong>at</strong>ion Theorem as a templ<strong>at</strong>e 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 calibr<strong>at</strong>ion 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 th<strong>at</strong> ZF + AD implies th<strong>at</strong> DC holds in L(R). Forthis reason it is important to note th<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 98and soHOD X |= Θ X,Y is strongly inaccessible.A direct approach to showing th<strong>at</strong> in additionHOD X |= Θ X,Y is a Woodin cardinalwould be to follow Section 4.3 by showing th<strong>at</strong> for each T ∈ P(Θ X,Y )∩OD Xthere is an ordinal δ T such th<strong>at</strong>HOD X ∩ V ΘX,Y |= δ T is λ-T-strongfor each λ < Θ X,Y . However, such an approach requires th<strong>at</strong> 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 guaranteeth<strong>at</strong> this is true. So our str<strong>at</strong>egy 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 th<strong>at</strong>HOD 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 th<strong>at</strong> each A α is a prewellordering of ω ω of lengthα. We will work with the structureL ΘX,Y (R)[T, A]and the n<strong>at</strong>ural hierarchy of structures th<strong>at</strong> it provides.To begin with we note some basic facts. First, notice th<strong>at</strong>(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 <strong>Cardinals</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong>L ΘX,Y (R)[T, A] |= T 0and the proof of Lemma 2.23 shows th<strong>at</strong>L ΘX,Y (R)[T, A] ≺ 1 L(R)[T, A].This implies (in conjunction with the fact th<strong>at</strong> Θ X,Y is regular in L(R)[T, A])th<strong>at</strong>{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 th<strong>at</strong> each such level s<strong>at</strong>isfies T 0 .So we are in exactly the situ<strong>at</strong>ion of Section 4.3 except th<strong>at</strong> now theprewellorderings are explicitly part of the structure. The proof of Theorem4.15 thus shows th<strong>at</strong>: For each T ∈ P(Θ X,Y ) ∩ OD X,Y there is an ordinalδ T,A such th<strong>at</strong>HOD 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 th<strong>at</strong> 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 th<strong>at</strong>HOD X ∩ V ΘX,Y = L ΘX,Y [H].We may assume without loss of generality th<strong>at</strong> H is folded into T. ThusHOD L Θ X,Y(R)[T]T= HOD X ∩ V ΘX,Y
5. Woodin <strong>Cardinals</strong> 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. Str<strong>at</strong>egic <strong>Determinacy</strong>Let us now turn to the Gener<strong>at</strong>ion Theorem. We shall begin by motiv<strong>at</strong>ingthe notion of “str<strong>at</strong>egic 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 th<strong>at</strong> contain a cone of S-degrees. Givena formula ϕ(x) we say th<strong>at</strong> ϕ holds for an S-cone of x if there is a real x 0such th<strong>at</strong> 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 str<strong>at</strong>egy σ 0 then σ 0 witnessesth<strong>at</strong> A is in the S-cone filter since for y S σ 0 , [y] S = [σ 0 ∗y] S ∈ A. If II hasa winning str<strong>at</strong>egy τ 0 then τ 0 witnesses th<strong>at</strong> 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 th<strong>at</strong> each st<strong>at</strong>ement ϕ either holds on an S-cone or fails on an S-cone. In fact, the entire theory stabilizes. However, in order to fully articul<strong>at</strong>ethis fact one needs to invoke second-order assumptions (like the existence ofa s<strong>at</strong>isfaction rel<strong>at</strong>ion). Without invoking second-order assumptions one hasthe following:
5. Woodin <strong>Cardinals</strong> in General Settings 101Corollary 5.8. Assume ZF +AD. For each n < ω, there is an x n such th<strong>at</strong>for 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〉 enumer<strong>at</strong>e 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 n<strong>at</strong>ural question then is: “Wh<strong>at</strong> 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) th<strong>at</strong> ¬CH holds for anS-cone of x. Let x 0 be the base of this cone.We will arrive <strong>at</strong> a contradiction by producing an x S x 0 such th<strong>at</strong>L[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 th<strong>at</strong> the generics actually exist.Claim. ω V 1is strongly inaccessible in any inner model M of AC.Proof. We first claim th<strong>at</strong> 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 th<strong>at</strong> agree on their n th -coordin<strong>at</strong>e. Thus, ⋂ n
5. Woodin <strong>Cardinals</strong> in General Settings 102The trouble is th<strong>at</strong> 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 delic<strong>at</strong>e 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 th<strong>at</strong>Now 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 (th<strong>at</strong> is, such th<strong>at</strong> 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 th<strong>at</strong> if H ⊆ P A is L[S, x 0 , A]-genericthen there is a c(A) ⊆ ω such th<strong>at</strong>α ∈ 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 th<strong>at</strong> for all x S x 0 ,L[S, x] |= CH.Fix x S x 0 . We claim th<strong>at</strong> L[S, x] |= GCH below ω V 1 : Suppose for contradictionth<strong>at</strong> there is a λ < ω V 1 such th<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 103A similar proof shows th<strong>at</strong> ♦ holds for an S-cone of x, the point beingth<strong>at</strong> 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 th<strong>at</strong> it s<strong>at</strong>isfies Condens<strong>at</strong>ion, □, Morasses, etc.Corollary 5.10 tells us th<strong>at</strong> for an S-cone of x,Θ L[S,x] = (c + ) L[S,x] = ω L[S,x]2 .Thus, to prove th<strong>at</strong> 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 p<strong>at</strong>tern (a(0), c(0), b(0), d(0), . . .). Here thetwo players are to be thought of as cooper<strong>at</strong>ing 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, th<strong>at</strong> such a set exists, as I is trying to ensure.)Case 1: I has a winning str<strong>at</strong>egy σ 0 .We claim th<strong>at</strong> for all x S σ 0 , L[S, x] |= OD S -determinacy, which contradictsthe assumption th<strong>at</strong> σ 0 is a winning str<strong>at</strong>egy for I. For consider such
5. Woodin <strong>Cardinals</strong> in General Settings 104a real x and suppose for contradiction th<strong>at</strong> L[S, x] ̸|= OD S -determinacy. Asabove let A x ∈ OD L[S,x]Sbe least such th<strong>at</strong> A x is not determined. We willarrive <strong>at</strong> a contradiction by deriving a winning str<strong>at</strong>egy σ for I in A x <strong>from</strong>the str<strong>at</strong>egy 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 th<strong>at</strong> theresulting model L[S, p] th<strong>at</strong> the two players jointly determine is just L[S, x]and so A p = A x . We can now derive a winning str<strong>at</strong>egy σ for I in A x <strong>from</strong>σ 0 as follows: For d ∈ L[S, x], let σ be the str<strong>at</strong>egy such th<strong>at</strong> (σ∗d) I is the asuch th<strong>at</strong> (σ 0 ∗〈x, d〉) I = 〈a, b〉.(It is crucial th<strong>at</strong> 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 l<strong>at</strong>er on when we refine this proof.)Case 2: II has a winning str<strong>at</strong>egy τ 0 .We claim th<strong>at</strong> for x S τ 0 , L[S, x] |= OD S -determinacy. This is as aboveexcept th<strong>at</strong> 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 winningstr<strong>at</strong>egy for I in ω ω A x and hence a winning str<strong>at</strong>egy 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. Unfortun<strong>at</strong>ely, 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 approxim<strong>at</strong>ing a sufficientamount of boldface definable determinacy to drive the construction. To makeprecise the approxim<strong>at</strong>ion we need, we introduce the notion of a “prestr<strong>at</strong>egy”.Let A and B be sets of reals. A prestr<strong>at</strong>egy for I (respectively II ) in Ais a continuous function f such th<strong>at</strong> for all x ∈ ω ω , f(x) is a str<strong>at</strong>egy for I(respectively II) in A. A prestr<strong>at</strong>egy 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 winningstr<strong>at</strong>egy in A. The str<strong>at</strong>egic game with respect to the predic<strong>at</strong>es P 1 , . . .,P k
5. Woodin <strong>Cardinals</strong> 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 prestr<strong>at</strong>egy for A n th<strong>at</strong> is winning with respect to B,and II wins iff II can play all ω rounds. We say th<strong>at</strong> str<strong>at</strong>egic definabledeterminacy holds with respect to the predic<strong>at</strong>es 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 th<strong>at</strong> str<strong>at</strong>egic definabledeterminacy for n moves holds with respect to the predic<strong>at</strong>es 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 <strong>from</strong> context we shall oftensimply refer to SG and ST-determinacy.In the context of L[S, x] the predic<strong>at</strong>e will be S and the basis B will bethe S-degree of x. Thus to say th<strong>at</strong> L[S, x] s<strong>at</strong>isfies ST B S -determinacy (orST-determinacy for short) is to say th<strong>at</strong> 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 prestr<strong>at</strong>egy for A n th<strong>at</strong> is winning with respect to [x] S .The ability to survive a single round of this game implies th<strong>at</strong> L[S, x] s<strong>at</strong>isfiesOD S -determinacy. So this notion is indeed a generaliz<strong>at</strong>ion of OD S -determinacy.Before turning to the main theorems, some remarks are in order. First,notice th<strong>at</strong> the games ST B P 0 ,...,P kare closed for Player II, hence determined.The only issue is whether II wins.Second, notice also th<strong>at</strong> if I wins then I has a canonical str<strong>at</strong>egy. 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-str<strong>at</strong>egy th<strong>at</strong> is definable in terms of the tree of partial plays
5. Woodin <strong>Cardinals</strong> in General Settings 106which in turn is ordinal definable. Since I is essentially playing ordinals thisquasi-str<strong>at</strong>egy can be converted into a str<strong>at</strong>egy in a definable fashion. Wetake this to be I’s canonical str<strong>at</strong>egy.Third, notice th<strong>at</strong> each prestr<strong>at</strong>egy can be coded by a real number in acanonical manner. We assume th<strong>at</strong> such a coding has been fixed and, fornot<strong>at</strong>ional convenience, we will identify a prestr<strong>at</strong>egy with its code.Fourth, it is important to note th<strong>at</strong> if II is to have a hope of winning thenwe must allow II to play prestr<strong>at</strong>egies and not str<strong>at</strong>egies. To see this, work inL[S, x] and consider the variant of SG B S where we have II play str<strong>at</strong>egies τ 0,τ 1 , . . . instead of prestr<strong>at</strong>egies. The set A 0 = {y ∈ ω ω | L[S, y even ] = L[S, x]}is OD L[S,x]Sand hence a legitim<strong>at</strong>e first move for I. But then II’s response mustbe a winning str<strong>at</strong>egy 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 winningstr<strong>at</strong>egy since L[S, x] ̸|= AD. The upshot is th<strong>at</strong> 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 motiv<strong>at</strong>es 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 th<strong>at</strong> for all z ∈ B, 〈f 0 , z〉 ∈ [x] S and so in a sense we are “one stepaway” <strong>from</strong> showing th<strong>at</strong> 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 th<strong>at</strong> there is no OD prestr<strong>at</strong>egy 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 th<strong>at</strong> thest<strong>at</strong>ement of Theorem 5.14 is false. By Theorem 5.7, there is a real x 0 suchth<strong>at</strong> if x T x 0 ,L[S, x] |= I wins SG,
5. Woodin <strong>Cardinals</strong> 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 str<strong>at</strong>egy in SG L[S,x] .Note th<strong>at</strong> the str<strong>at</strong>egy depends only on the model, th<strong>at</strong> 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 th<strong>at</strong>the winning str<strong>at</strong>egies (for whichever player wins) enable us to define, foran S-cone of x, prestr<strong>at</strong>egies 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. Noticeth<strong>at</strong> by including x 0 in p we have ensured th<strong>at</strong> σ p is defined and hence th<strong>at</strong>the winning condition makes sense. In this game I and II are cooper<strong>at</strong>ing 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 str<strong>at</strong>egy in the str<strong>at</strong>egic 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 th<strong>at</strong> for all x S x 1 there is a prestr<strong>at</strong>egyf x 0 th<strong>at</strong> is a non-losing first move for II against σ x in SG L[S,x] .Proof. Case 1: I has a winning str<strong>at</strong>egy σ 0 in G 0 .For x T σ 0 , let f0 x be the prestr<strong>at</strong>egy derived <strong>from</strong> σ 0 by extractingthe response in the auxiliary game where we have II feed in y for c, th<strong>at</strong>is, for y ∈ (ω ω ) L[S,x] let f0 x (y) be such th<strong>at</strong> f0 x (y)∗d = a∗d where a is suchth<strong>at</strong> (σ 0 ∗〈y, d〉) I = 〈a, b〉. Note th<strong>at</strong> f0 x ∈ L[S, x] as it is definable <strong>from</strong>σ 0 . Let x 1 = 〈σ 0 , x 0 〉 and for x S x 1 let A x 0 = σ x (∅). We claim th<strong>at</strong>for x S x 1 , f0 x is a prestr<strong>at</strong>egy for I in Ax 0 th<strong>at</strong> is winning with respect to{y ∈ ω ω | L[S, y] = L[S, x]}, th<strong>at</strong> 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 th<strong>at</strong> L[S, y] = L[S, x] andconsider d ∈ L[S, x]. The value f0 x (y) of the prestr<strong>at</strong>egy was defined byrunning G 0 , having II feed in y for c:IIIa, by, d
5. Woodin <strong>Cardinals</strong> in General Settings 108By our choice of y and d, we have solved the “steering problem”, th<strong>at</strong> 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 suchth<strong>at</strong> f x 0 (y)∗d = a∗d where a is such th<strong>at</strong> (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 str<strong>at</strong>egy τ 0 in G 0 .Let f0 x be the prestr<strong>at</strong>egy derived <strong>from</strong> τ 0 by extracting the response inthe auxiliary game where we have I feed in y for b, th<strong>at</strong> is, for y ∈ (ω ω ) L[S,x] letf0 x(y) be such th<strong>at</strong> a∗fx 0 (y) = a∗d where d is such th<strong>at</strong> (〈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 th<strong>at</strong>for x S x 1 , f0 x is a prestr<strong>at</strong>egy for II in Ax 0 th<strong>at</strong> is winning with respectto {y ∈ ω ω | L[S, y] = L[S, x]}, th<strong>at</strong> 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 th<strong>at</strong> we have defined games G 0 , . . .,G n , reals x 0 , . . .,x n+1 such th<strong>at</strong> x 0 S x 1 S · · · S x n+1 , and prestr<strong>at</strong>egies f x 0 , . . .,fx n whichdepend only on the degree of x and such th<strong>at</strong> 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 th<strong>at</strong> we have includedx n+1 in p to ensure th<strong>at</strong> σ p , f0, p . . ., fn p are defined and hence th<strong>at</strong> the winningcondition makes sense. In this game I and II are cooper<strong>at</strong>ing 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 str<strong>at</strong>egic 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 th<strong>at</strong> for all x S x n+2 there is a prestr<strong>at</strong>egyfn+1 x such th<strong>at</strong> fx 0 , . . ., fx n , fx n+1 is a non-losing partial play for II againstσ x in SG L[S,x] .
5. Woodin <strong>Cardinals</strong> in General Settings 109Proof. Case 1: I has a winning str<strong>at</strong>egy σ n+1 in G n+1 .Let fn+1 x be the prestr<strong>at</strong>egy derived <strong>from</strong> σ n+1 by extracting the responsein the auxiliary game, th<strong>at</strong> is, for y ∈ (ω ω ) L[S,x] let fn+1 x (y) be such th<strong>at</strong>fn+1 x (y)∗d = a∗d where a is such th<strong>at</strong> (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 prestr<strong>at</strong>egy for I in A x n+1 th<strong>at</strong> is winning with respectto {y ∈ ω ω | L[S, y] = L[S, x]}, th<strong>at</strong> 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 str<strong>at</strong>egy τ n+1 in G n+1 .Let fn+1 x be the prestr<strong>at</strong>egy derived <strong>from</strong> τ n+1 by extracting the responsein the auxiliary game, th<strong>at</strong> is, for y ∈ (ω ω ) L[S,x] let fn+1 x (y) be such th<strong>at</strong>a∗fn+1 x (y) = a∗d where d is such th<strong>at</strong> (〈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 th<strong>at</strong> for x S x n+2 , fn+1 x is a prestr<strong>at</strong>egy for II in Ax n+1 th<strong>at</strong>is winning with respect to {y ∈ ω ω | L[S, y] = L[S, x]}, th<strong>at</strong> 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 prestr<strong>at</strong>egiesf0 x, . . .,fx n , . . . as in each of the steps. Letting x∞ S x n , for alln, we have th<strong>at</strong> 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 th<strong>at</strong> 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 prestr<strong>at</strong>egiesare definable <strong>from</strong> x 0 , . . ., x n , x n+1 ).5.3. Gener<strong>at</strong>ion TheoremIn the previous section we showed (assuming ZF + AD) th<strong>at</strong> for an S-coneof x,L[S, x] |= OD S -determinacy,and (even more) th<strong>at</strong> for each n,L[S, x] |= ST B S -determinacy for n moves,
5. Woodin <strong>Cardinals</strong> in General Settings 110where B = [x] S . It turns out th<strong>at</strong> for a sufficiently large choice of n thisdegree of determinacy is sufficient to implement the previous arguments andshow th<strong>at</strong>L[S, x] |= ω 2 is a Woodin cardinal in HOD S .At this stage we could proceed directly to this result but instead, with thismotiv<strong>at</strong>ion behind us, we return to the more general setting. The maintheorem to be proved is the Gener<strong>at</strong>ion Theorem:Theorem 5.15 (Gener<strong>at</strong>ion Theorem). Assume ZF. Supposeis such th<strong>at</strong>(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 th<strong>at</strong> A α is a prewellordering of the reals oflength gre<strong>at</strong>er 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 str<strong>at</strong>egic determinacy for four movesis th<strong>at</strong> in a number of applic<strong>at</strong>ions of this theorem str<strong>at</strong>egic determinacy forn moves (for each n) can be established without any appeal to DC (as forexample in Theorem 5.13) in contrast to full str<strong>at</strong>egic determinacy which(just as in Theorem 5.14) uses DC R .The external assumption th<strong>at</strong> Θ M is a regular cardinal is merely forconvenience—it ensures th<strong>at</strong> there are cofinally many stages in the str<strong>at</strong>ific<strong>at</strong>ionof M where T 0 holds. The dedic<strong>at</strong>ed reader can verify th<strong>at</strong> 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 Gener<strong>at</strong>ionTheorem.
5. Woodin <strong>Cardinals</strong> in General Settings 111Proof. Let us start by showing th<strong>at</strong> HOD M T,A,B s<strong>at</strong>isfies ZFC. When workingwith structures of the form L ΘM (R)[T, A, B] it is to be understood th<strong>at</strong> weare working in the language of ZFC augmented with constant symbols for T,A, B, and R. The first step is to show th<strong>at</strong> 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 th<strong>at</strong> the displayed predic<strong>at</strong>es are part of the structure.Since Θ M is regular and M |= T 0 there are cofinally many γ < Θ M such th<strong>at</strong>M γ |= T 0 . So a set x ∈ M is OD M T,A,B if and only if there is a γ < Θ M suchth<strong>at</strong> M γ |= T 0 and x is definable in M γ <strong>from</strong> ordinal parameters (and theconstant symbols for the parameters). It follows th<strong>at</strong> OD M T,A,B and HODM T,A,Bare Σ 1 -definable over M (in the expanded language).With this first-order characteriz<strong>at</strong>ion 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 th<strong>at</strong> HOD M T,A,B |= ZFC − Power Set. (The proofs th<strong>at</strong>AC holds in HOD M T,A,B and th<strong>at</strong> for all α < Θ M , V α ∩HOD M T,A,B ∈ HOD M T,A,Brequire th<strong>at</strong> OD M T,A,B be ordinal definable.)Lemma 5.16. HOD M T,A,B|= ZFC.Proof. We have seen th<strong>at</strong> HOD M T,A,B |= ZFC−Power Set. Since HODM T,A,B |=AC it remains to show th<strong>at</strong> for all λ < Θ M ,P(λ) HODM T,A,B ∈ HODMT,A,B .The point is th<strong>at</strong> since M |= OD T,A,B -determinacy, for each S ∈ OD M T,A,B ∩P(λ) the game for coding S rel<strong>at</strong>ive 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 <strong>Cardinals</strong> in General Settings 112is an OD M T,A,B surjection. Thus, P(λ)HODM T,A,B ∈ M and so, by our first-ordercharacteriz<strong>at</strong>ion, P(λ) HODM T,A,B ∈ HODMT,A,B .The ordinal κ th<strong>at</strong> 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 th<strong>at</strong>M 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 th<strong>at</strong>M 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 th<strong>at</strong>and 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 th<strong>at</strong> (⋆) 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 th<strong>at</strong>M |= 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 <strong>Cardinals</strong> 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 th<strong>at</strong> S 0 ∈ F X . For not<strong>at</strong>ional 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 assumeth<strong>at</strong> 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 th<strong>at</strong> 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) th<strong>at</strong> gave rise to the ultrafilter extending thereflection filter, an ultrafilter th<strong>at</strong> 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 <strong>at</strong> 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 th<strong>at</strong> if M s<strong>at</strong>isfies ST B T,A,B,y-determinacy for n + 1 moves thenII’s first move f 0 is n-y-good. Notice also th<strong>at</strong> 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 rel<strong>at</strong>ive 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 <strong>Cardinals</strong> 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 viol<strong>at</strong>ed, then, letting i be the least such th<strong>at</strong> either(x) i ∉ U X or (y) i ∉ U X , I wins if (x) i ∈ U X ; otherwise II wins. If the ruleis s<strong>at</strong>isfied, then, letting δ be least such th<strong>at</strong> 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 th<strong>at</strong> forall i < ω, (x) i ∈ U X and for some i < ω, (x) i = z, where z is such th<strong>at</strong>S z ⊆ S. But we cannot setµ X = {S ⊆ κ | I wins G X (S)}since we have no guarantee th<strong>at</strong> 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 th<strong>at</strong> κ is measurable in HOD M T,A,B.It is useful <strong>at</strong> this point to stand back and contrast the present approachwith the two earlier approaches. In both of the earlier approaches (namely,th<strong>at</strong> of Section 4.3 and th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> the ultrafilters “fit together” insuch a way th<strong>at</strong> they witness th<strong>at</strong> 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 th<strong>at</strong> a = {b i1 , . . .b ik }, where i 1 < · · ·
5. Woodin <strong>Cardinals</strong> 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 th<strong>at</strong>f a,b (z) = f(z a,b ).Thus we use ‘a, b’ as a subscript to indic<strong>at</strong>e th<strong>at</strong> z a,b is the “drop of z <strong>from</strong>b to a” and we use ‘a, b’ as a superscript to indic<strong>at</strong>e th<strong>at</strong> X a,b is the “lift ofX <strong>from</strong> 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 <strong>Cardinals</strong> in General Settings 116If conditions (1) and (2) alone are s<strong>at</strong>isfied then we say th<strong>at</strong> E is a (κ, λ)-pre-extender.We need to ensure th<strong>at</strong> the ultrafilter E a on [κ] |a| depends on a in such away th<strong>at</strong> guarantees coherence and the other properties. The most n<strong>at</strong>uralway to define an ultrafilter E a on [κ] |a| th<strong>at</strong> depends on a is as follows:(1) For F X -almost all δ define a “reflected version” a δ ∈ [λ δ ]
5. Woodin <strong>Cardinals</strong> in General Settings 117Lemma 5.22. Assume z is (n + 1)-good. Then for each a ∈ [λ]
5. Woodin <strong>Cardinals</strong> in General Settings 118th<strong>at</strong> a has. To see this, consider a st<strong>at</strong>ement such as the following: For allx, x ′ ∈ B, if α 1 , . . .,α k are such th<strong>at</strong>thenandf 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}) st<strong>at</strong>ement. Thus, for F X -almost all δ, thest<strong>at</strong>ement reflects.5.24 Definition. Suppose a ∈ [λ]
5. Woodin <strong>Cardinals</strong> in General Settings 119The only trouble with this definition is th<strong>at</strong> there is no guarantee th<strong>at</strong> E X (f a )is in HOD M T,A,B because there is no guarantee th<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 120Note th<strong>at</strong> 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 th<strong>at</strong> for each a ∈ [λ]
5. Woodin <strong>Cardinals</strong> in General Settings 121The key point is th<strong>at</strong>S(a, f a , Y ) = ⋂ {S(a, f a , Y α ) | α < γ}and so we are in almost exactly the situ<strong>at</strong>ion 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 th<strong>at</strong> G X (S(a, f a , Y )) is determined. Assume for contradiction th<strong>at</strong> Idoes not win G X (S(a, f a , Y )) and let σ ′ be a winning str<strong>at</strong>egy for I in G X (κS(a, f a , Y )). We will derive a contradiction by finding a play th<strong>at</strong> is legalagainst σ ′ and against winning str<strong>at</strong>egies 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 winningstr<strong>at</strong>egies (in the games G X (S(a, f a , Y α )) for α < γ) we need a prewellorderingof length γ which is such th<strong>at</strong> in a reflection argument we can ensureth<strong>at</strong> 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 fe<strong>at</strong>ure th<strong>at</strong> for F X-almostall δ,〈Q ξ | ξ < δ〉 = 〈Q ξ | ξ < δ〉 M ϑ(δ)and, by choosing a real, we can ensure th<strong>at</strong> we always reflect to some suchpoint δ > γ.Now setZ = {(x, σ) | for some α < γ, x ∈ Q α andσ is a winning str<strong>at</strong>egy 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 th<strong>at</strong>E a = {Y ∈ P([κ] |a| ) ∩ HOD M T,A,B | b ∈ Y }.
5. Woodin <strong>Cardinals</strong> in General Settings 122Suppose th<strong>at</strong> a ∈ [κ] |a| . We claim th<strong>at</strong> b = a witnesses th<strong>at</strong> 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 th<strong>at</strong> a ∉ [κ] |a| . We claim th<strong>at</strong> no β ∈ [κ] |a| witnesses th<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 123the proof of Lemma 5.22 shows th<strong>at</strong> there are f ai such th<strong>at</strong> 〈f ai | i < ω〉 is1-good. So G X (S) is determined. As in the proof of the completeness of E awe have th<strong>at</strong> I wins G X (S).As in the proof of coherence there is a set S 0 (f a1 , . . .,f an , . . .) ∈ F X suchth<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 124Notice th<strong>at</strong> 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 th<strong>at</strong> t ∈ Q δ . Notice also th<strong>at</strong> gα δ fα is 1-good since it is OD M T,A,B,fta.The st<strong>at</strong>ement and proof of strong normality are similar to before, onlynow we have to ensure th<strong>at</strong> the objects are sufficiently good to guaranteethe determinacy of the games defined in terms of them. The real parametersth<strong>at</strong> arise in the proof of strong normality will now have to be gener<strong>at</strong>edusing 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 th<strong>at</strong> M s<strong>at</strong>isfies 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 applic<strong>at</strong>ions of theGener<strong>at</strong>ion Theorem, one will be able to show without DC th<strong>at</strong> M s<strong>at</strong>isfiesST 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 th<strong>at</strong> 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 th<strong>at</strong>(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 th<strong>at</strong>where 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 th<strong>at</strong> since g is 3-good, byLemma 5.22 we have th<strong>at</strong> 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 th<strong>at</strong> α is uniquely specified. For suppose fˆα is a 1-g-goodcode of ˆα such th<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 125andG X ({δ ∈ S 0 (fˆα ) | g(δ) = g fˆα(δ)}).Third, it will be useful <strong>at</strong> this point to both list the parameters th<strong>at</strong>will arise in the proof and motiv<strong>at</strong>e the need for assuming th<strong>at</strong> g is 3-good.In outline the proof will follow th<strong>at</strong> 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 <strong>from</strong>Theorem 4.12. To ensure the determinacy of the final game we will need totake steps to ensure th<strong>at</strong> (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 th<strong>at</strong> 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 <strong>from</strong> the startth<strong>at</strong> g is 3-good.We now turn to the proof proper. Suppose g : κ → κ is such th<strong>at</strong>(1.1) g is 3-good and(1.2) I wins G X ({δ ∈ S 0 | g(δ) < λ δ }).Assume for contradiction th<strong>at</strong> 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 th<strong>at</strong>(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 <strong>Cardinals</strong> in General Settings 126which is the desired situ<strong>at</strong>ion. By Lemma 5.22, letf η be a 2-g-good code of η.For not<strong>at</strong>ional 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 upd<strong>at</strong>e S 0 to be S 0 ∩ {δ ∈S 0 (f η ) | g(δ) < g fη (δ)}. We will work on this “large” set. Notice th<strong>at</strong> 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 wh<strong>at</strong>follows.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 th<strong>at</strong> x ∈ ran(f α ↾B), andσ is a winning str<strong>at</strong>egy 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 <strong>Cardinals</strong> 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 th<strong>at</strong> for allα < η,(6.1) U e(2) (Q κ
5. Woodin <strong>Cardinals</strong> 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 <strong>Cardinals</strong> in General Settings 129(9.1) z is a legal play for II against σ ′ and δ 0 is the associ<strong>at</strong>ed 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 associ<strong>at</strong>ed 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 th<strong>at</strong> 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 str<strong>at</strong>egy 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 associ<strong>at</strong>ed ordinal δ 0 , and since σ 0 is a winning str<strong>at</strong>egy 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)},th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 })-st<strong>at</strong>ement. So there is a z 0 ∈ U Xsuch th<strong>at</strong> 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 th<strong>at</strong>(12.1) ∀y ∈ ω ω ( ∀i n (y) i = z i → ((σ ′ ∗y) I ) n+1 ∈ U X)and
5. Woodin <strong>Cardinals</strong> 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 })-st<strong>at</strong>ement. So there is az n+1 ∈ U X such th<strong>at</strong> 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 th<strong>at</strong> (z) i = z i for all i ∈ ω and let δ 0 be least suchth<strong>at</strong> (z) i ∈ U δ0 for all i ∈ ω. Notice th<strong>at</strong> by our choice of z n , for n < ω, noDC is required to construct z. We have th<strong>at</strong> 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})-st<strong>at</strong>ement ϕ. Notice th<strong>at</strong> sinceG is OD M T,A,B,g,f η,e 0and (g, f η , e 0 ) is 1-good, it follows th<strong>at</strong> (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 st<strong>at</strong>ement of th<strong>at</strong> theorem) and we have th<strong>at</strong>(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 ], th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 <strong>Cardinals</strong> 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 Comput<strong>at</strong>ion). There exists an index e 1 ∈ ω ω such th<strong>at</strong>for 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 th<strong>at</strong> 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 th<strong>at</strong>(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 th<strong>at</strong> 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 th<strong>at</strong> require further“joint goodness”.
5. Woodin <strong>Cardinals</strong> in General Settings 132Claim C. There exists an α 0 < η such th<strong>at</strong>(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 st<strong>at</strong>ement th<strong>at</strong> 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 th<strong>at</strong> (g, G, S 2 ) is 1-good. Thus, the ReflectionTheorem (Theorem 5.20) applies and we have th<strong>at</strong> for F X -many δ, thest<strong>at</strong>ement 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 st<strong>at</strong>ement(16.2) ∀x ∈ B f α0 (x) ∈ Q κ α where α is such th<strong>at</strong>(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 st<strong>at</strong>ement 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 th<strong>at</strong> M s<strong>at</strong>isfies ST B T,A,B-determinacy for fourmoves, we have shown:
5. Woodin <strong>Cardinals</strong> in General Settings 133Corollary 5.33. Suppose g : κ → κ is in HOD M T,A,B and such th<strong>at</strong> 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 th<strong>at</strong>I 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 <strong>Cardinals</strong> in General Settings 134Note th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> β = (a∪{β}) k . Let f a∪{β} be a 1-good code of a∪{β}.Note th<strong>at</strong>(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})kth<strong>at</strong> 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 <strong>Cardinals</strong> in General Settings 135and membership rel<strong>at</strong>ion 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 s<strong>at</strong>isfies 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 th<strong>at</strong>j ′ 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 th<strong>at</strong> Ult is well-founded and it isstraightforward to see th<strong>at</strong> 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 <strong>Cardinals</strong> 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 th<strong>at</strong> j E (κ) λ and it remains to show th<strong>at</strong>V HODM T,A,Bλ⊆ M Xandj E (T ∩ κ) ∩ λ = T ∩ λ.The proof of each is the same. Let us start with the l<strong>at</strong>ter. We have to showth<strong>at</strong> for all α < λ,α ∈ j E (T ∩ κ) ↔ α ∈ T.We haveSo we have to show th<strong>at</strong>α ∈ 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 th<strong>at</strong>I wins G X( S({α}, f {α} , {{z} | z ∈ T ∩ κ}) ) .The key point is th<strong>at</strong> the st<strong>at</strong>ement “for all x ∈ B, |f {α} (x)| λ ∈ T” is a trueΣ 1 (M, {X, κ, R, f {α} })-st<strong>at</strong>ement. So the set S of δ to which this st<strong>at</strong>ementreflects 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 th<strong>at</strong>I does not win G X( S({α}, f {α} , {{z} | z ∈ T ∩ κ}) ) .Again, the point is th<strong>at</strong> the st<strong>at</strong>ement “for all x ∈ B, |f {α} (x)| λ ∉ T” isa true Σ 1 (M, {X, κ, R, f {α} })-st<strong>at</strong>ement. So this st<strong>at</strong>ement reflects to F X -almost all δ, which implies th<strong>at</strong> I cannot win the above game.
5. Woodin <strong>Cardinals</strong> in General Settings 137Exactly the same argument with ‘A’ in place of ‘T’ shows th<strong>at</strong>and hence th<strong>at</strong>which 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 Gener<strong>at</strong>ion Theorem5.4. Special CasesWe now consider a number of special instances of the Gener<strong>at</strong>ion Theorem.In each case all we have to do is find appropri<strong>at</strong>e 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 not<strong>at</strong>ional convenience let Θ = Θ L(R) . Our str<strong>at</strong>egy is to meetthe conditions of the Gener<strong>at</strong>ion Theorem while <strong>at</strong> the same time arrangingth<strong>at</strong> M = L Θ (R)[T, A, B] is such th<strong>at</strong>HOD M T,A,B = HOD L(R) ∩ V Θ .We will do this by taking care to ensure th<strong>at</strong> the ingredients T, A, and Bare in HOD L(R) while <strong>at</strong> the same time packaging HOD L(R) ∩ V Θ as part ofT. It will then follow <strong>from</strong> the Gener<strong>at</strong>ion Theorem th<strong>at</strong>HOD 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 not<strong>at</strong>ional convenience, we continueto abbrevi<strong>at</strong>e 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 <strong>Cardinals</strong> 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〉 suchth<strong>at</strong> 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 Gener<strong>at</strong>ionTheorem are clearly met and condition (6) follows since L(R) s<strong>at</strong>isfies ADand M contains all reals. Moreover, since we have arranged th<strong>at</strong> all of theingredients T, A, and B are in OD L(R) and also th<strong>at</strong> 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 approxim<strong>at</strong>ion 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 th<strong>at</strong> 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 Gener<strong>at</strong>ion Theorem are clearly met. For condition (2) note th<strong>at</strong>by Lemma 3.7, Θ X = Θ HOD {X}∪R and th<strong>at</strong> 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 <strong>from</strong> the fact th<strong>at</strong> M is OD X and Mcontains all of the reals. It remains to see th<strong>at</strong> condition (1) can be met.For this we just have to see th<strong>at</strong> Replacement holds in M. If Replacementfailed in M then there would be a cofinal map π : ω ω → Θ X th<strong>at</strong> is definable<strong>from</strong> parameters in M, which in conjunction with A would lead to an OD Xsurjection <strong>from</strong> ω ω onto Θ X , which is a contradiction.
5. Woodin <strong>Cardinals</strong> 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 <strong>from</strong> 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 <strong>from</strong> the Gener<strong>at</strong>ionTheorem and this is why we have singled it out for special tre<strong>at</strong>ment.However, it follows <strong>from</strong> the proof of the Gener<strong>at</strong>ion Theorem, as can beseen by noting th<strong>at</strong> 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 rel<strong>at</strong>ed 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] s<strong>at</strong>isfies GCH below ω1 V, by Lemma 3.8 we have th<strong>at</strong>OD 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 <strong>Cardinals</strong> 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 th<strong>at</strong>A 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) th<strong>at</strong> 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 Gener<strong>at</strong>ionTheorem are clearly met and condition (6) follows since L[S, x] s<strong>at</strong>isfies 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 th<strong>at</strong> all of the ingredients T, A, and B are in OD L[S,x]and also th<strong>at</strong> 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 <strong>from</strong> the next theorem which is more general.The next two theorems require some not<strong>at</strong>ion. 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 <strong>Cardinals</strong> 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 th<strong>at</strong> the above st<strong>at</strong>ement holdsfor a Turing cone of x, which is wh<strong>at</strong> we shall do. The key issues in thiscase are getting a sufficient amount of GCH and str<strong>at</strong>egic determinacy. Toestablish the first we need two preliminary claims. Recall th<strong>at</strong> a set A ⊆ ω ωis comeager if and only if ω ω A is meager.Claim 1. Assume ZF+AD. Suppose th<strong>at</strong> 〈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 <strong>Cardinals</strong> 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 th<strong>at</strong> is comeager in the Stone space of P and such th<strong>at</strong> for all G ∈ A,HOD Y,a,G = HOD Y,a [G]. We will do this by showing th<strong>at</strong> for each G ∈ A thel<strong>at</strong>ter model can compute the “ordinal theory” of the former model.For every Σ 2 -st<strong>at</strong>ement ϕ and finite sequence of ordinals ⃗ ξ consider thest<strong>at</strong>ement ϕ[ ⃗ ξ, Y, a, G] about a generic G. Let B ϕ,⃗ ξ,Y,a be the associ<strong>at</strong>edcollection 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 “neg<strong>at</strong>ively” 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 <strong>Cardinals</strong> 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 th<strong>at</strong> CH holds on a cone since given this the proofth<strong>at</strong> GCH below ω V 1 holds on a cone goes through exactly as before.Suppose for contradiction (by the Cone Theorem (Theorem 2.9)) th<strong>at</strong>there is a real x 0 such th<strong>at</strong> for all x T x 0 ,HOD Y,a,x |= ¬CH.We will arrive <strong>at</strong> a contradiction by producing an x T x 0 with the fe<strong>at</strong>ureth<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 prestr<strong>at</strong>egy for A i+1 th<strong>at</strong> belongs to HOD Y,a,x and is winningwith respect to [x] Y,a .
5. Woodin <strong>Cardinals</strong> 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 Gener<strong>at</strong>ion 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 th<strong>at</strong> A α is the OD Y,a,[x]Y,a -leastprewellordering of length α we have th<strong>at</strong> A is OD Y,a,[x]Y,a . We also haveth<strong>at</strong> 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 Gener<strong>at</strong>ionTheorem are clearly met. Condition (6) follows <strong>from</strong> the fact th<strong>at</strong> Mis OD Y,a,[x]Y,a and we have arranged (in Claim 4) th<strong>at</strong> 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 <strong>Determinacy</strong> 145and since we have arranged th<strong>at</strong>HOD 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, l<strong>at</strong>er (in Section 6.2) we will want to construct models withmany Woodin cardinals. A n<strong>at</strong>ural approach to doing this is to iter<strong>at</strong>ivelyapply the previous theorem, starting off with a = ∅, increasing the degreeof x to get th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> of the previous theorem.Claims 1 to 3 are exactly as before. The only difference is th<strong>at</strong> 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 th<strong>at</strong> of the rest of the theorem.6. Definable <strong>Determinacy</strong>We now use the Gener<strong>at</strong>ion Theorem to derive the optimal amount of largecardinal strength <strong>from</strong> 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 <strong>Determinacy</strong> 146When combined with the results mentioned in the introduction this has theconsequence th<strong>at</strong> 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 situ<strong>at</strong>ion of the Gener<strong>at</strong>ion Theorem. The issuehere is th<strong>at</strong> ∆ 1 2-determinacy does not imply th<strong>at</strong> for a cone of x str<strong>at</strong>egicdeterminacy 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 th<strong>at</strong> if G ⊆ P Y is V -genericand 〈[x i ] Y | i < ω〉 is the associ<strong>at</strong>ed sequence, thenHOD V [G]Y,〈[x i ] Y |i
6. Definable <strong>Determinacy</strong> 147for an appropri<strong>at</strong>e 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 str<strong>at</strong>egic determinacy.Nevertheless, it turns out th<strong>at</strong> str<strong>at</strong>egic determinacy holds for a different,smaller basis. This leads to the notion of restricted str<strong>at</strong>egic determinacy.We shall successively extract stronger and stronger forms of determinacyuntil we ultim<strong>at</strong>ely reach the version we need. The subsection closes witha series of limit<strong>at</strong>ive results, including results th<strong>at</strong> motiv<strong>at</strong>e the need forstr<strong>at</strong>egic and restricted str<strong>at</strong>egic 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 th<strong>at</strong>A is determined. Our str<strong>at</strong>egy is to show th<strong>at</strong> if II (the Π 1 2 player) does nothave a winning str<strong>at</strong>egy for A then I (the Σ 1 2 player) has a winning str<strong>at</strong>egyfor A.Assume th<strong>at</strong> II does not have a winning str<strong>at</strong>egy 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 str<strong>at</strong>egy in {y ∈ ω ω | ϕ(y)}(since otherwise, by Σ 1 3 upward absoluteness, II would have a winning str<strong>at</strong>egyin V , contradicting our initial assumption) and so, by the Löwenheim-Skolem theorem, there is a countable ordinal λ such th<strong>at</strong>L λ [x] |= T + II does not have a winning str<strong>at</strong>egy 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 bstr<strong>at</strong>egy in {y ∈ ω ω | ϕ(y)}).
6. Definable <strong>Determinacy</strong> 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 th<strong>at</strong> since λ(x) is always defined this game is ∆ 1 2 and hencedetermined.We claim th<strong>at</strong> I has a winning str<strong>at</strong>egy in G (and so I wins each roundof the localized games A x ) and, furthermore, th<strong>at</strong> (by ranging over theserounds and applying upward Σ 1 2-absoluteness) this winning str<strong>at</strong>egy yields awinning str<strong>at</strong>egy for I in A.Assume for contradiction (by ∆ 1 2-determinacy) th<strong>at</strong> II has a winning str<strong>at</strong>egyτ 0 in G. For each x T τ 0 , in L λ(x) [x] we can derive a winning str<strong>at</strong>egyτ x for II in A x as follows: For a ∈ (ω ω ) Lλ(x)[x] , let (a∗τ x ) II = b where b issuch th<strong>at</strong> (〈a, x〉∗τ 0 ) II = b. Since τ 0 is a winning str<strong>at</strong>egy for II in G and wehave arranged th<strong>at</strong> a∗b ∈ L λ(x) [x], II must win in virtue of the second clause,which means th<strong>at</strong> a∗b /∈ A x . Thus, L λ(x) [x] |= “τ x is a winning str<strong>at</strong>egy forII in A x ”, which is a contradiction.Thus I has a winning str<strong>at</strong>egy σ 0 in G. Consider the derived str<strong>at</strong>egy σsuch th<strong>at</strong> for b ∈ ω ω , (σ∗b) I = a where a is such th<strong>at</strong> (σ 0 ∗b) I = 〈a, x〉. Sinceσ 0 is a winning str<strong>at</strong>egy 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 str<strong>at</strong>egyfor I in A.6.4 Remark.(1) The above proof rel<strong>at</strong>ivizes to a real parameter to show th<strong>at</strong> ∆ 1 2 (x)-determinacy implies Σ 1 2 (x)-determinacy.(2) A similar but more elabor<strong>at</strong>e argument shows th<strong>at</strong> 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 str<strong>at</strong>egy in a Σ 1 2-game then I has a ∆ 1 3 str<strong>at</strong>egy.Proof. Consider A = {x ∈ ω ω | ϕ(x)} where ϕ is Σ 1 2. Our str<strong>at</strong>egy is to showth<strong>at</strong> if II (the Π 1 2 player) does not win A then I (the Σ1 2 player) wins A viaa ∆ 1 3 str<strong>at</strong>egy.Assume th<strong>at</strong> II does not have a winning str<strong>at</strong>egy 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 winningstr<strong>at</strong>egy for I in G, for x T σ 0 , in L λ(x) [x] we can derive a winning str<strong>at</strong>egy
6. Definable <strong>Determinacy</strong> 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 th<strong>at</strong> (σ 0 ∗b) I = 〈a, x〉.Next we show th<strong>at</strong> there is an x 0 T σ 0 such th<strong>at</strong> for all x T x 0 ,L λ(x) [x] |= ∆ 1 2 -determinacy. Let 〈ϕ n, ψ n 〉 enumer<strong>at</strong>e the pairs of Σ 1 2 -formulasand let A ϕ = {x ∈ ω ω | ϕ(x)}. Using DC let z n be such th<strong>at</strong> z n codes awinning str<strong>at</strong>egy 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 str<strong>at</strong>egy in A xby the Third Periodicity Theorem of Moschovakis.(For a proof of Third Periodicity see Jackson’s chapter in this Handbook.The st<strong>at</strong>ement of Third Periodicity typically involves boldface determinacy.However, the proof shows th<strong>at</strong> lightface ∆ 1 2 determinacy suffices toget Σ 1 4 winning str<strong>at</strong>egies for Σ1 2 games th<strong>at</strong> I wins. To see this note th<strong>at</strong>Scale(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) th<strong>at</strong> are used to define theprewellorderings and ultim<strong>at</strong>ely the definable str<strong>at</strong>egies. It follows th<strong>at</strong> theseprewellorderings and str<strong>at</strong>egies are Σ 1 2 ⊆ Σ 1 14. (Notice th<strong>at</strong> if we had ∆ ∼ 2 -determinacy then we could flip the quantifiers and conclude th<strong>at</strong> Σ 1 2 = Π1 3and hence get ∆ 1 3 str<strong>at</strong>egies. However, in our present lightface setting somemore work is required.))For x T x 0 , let ˆσ x be the Σ 1 4 -str<strong>at</strong>egy for I in Ax . For a Turing cone ofx the formula ϕ(y, z) defining this str<strong>at</strong>egy is constant. We can now “freezeout” the value of ˆσ x on a Turing cone of x. The key point is th<strong>at</strong> the functionx ↦→ L λ(x) [x] is ∆ 1 2 . So, for each s ∈ ω2n and m ∈ ω the st<strong>at</strong>ementL λ(x) [x] |= ϕ(s, m)is ∆ 1 2 . Thus, for each s ∈ ω2n , the m such th<strong>at</strong> L λ(x) [x] |= ϕ(s, m) is fixed fora Turing cone of x. Since there are only countably many s ∈ ω 2n this meansth<strong>at</strong> 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 th<strong>at</strong> σis a ∆ 1 3 winning str<strong>at</strong>egy for I in A.
6. Definable <strong>Determinacy</strong> 150Kechris and Solovay showed (in [6]) th<strong>at</strong> under ZF+DC+∆ 1 2 -determinacythere is a real x 0 th<strong>at</strong> “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 th<strong>at</strong> 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 s<strong>at</strong>isfy only a very weak fragment of ZFC; so the st<strong>at</strong>ement“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 winningstr<strong>at</strong>egy for {z ∈ ω ω | ϕ(z, ξ)}”. The point is th<strong>at</strong> this st<strong>at</strong>ement makessense even when {z ∈ ω ω | ϕ(z, ξ)} is a proper class <strong>from</strong> the point of view ofL λ [x]. Second, the key fe<strong>at</strong>ure of additively closed λ > ω, is th<strong>at</strong> if y ∈ L λ [x]then L λ [y] ⊆ L λ [x]. This is true since additively closed ordinals λ > ω aresuch th<strong>at</strong> α + λ = λ for all α < λ and so if y is constructed <strong>at</strong> stage α, thenL λ [x] still has λ-many remaining stages in which to “c<strong>at</strong>ch up” and constructeverything in L λ [y]. Third, the proof of the theorem is a “localiz<strong>at</strong>ion” ofthe proof of Theorem 5.12.Assume for contradiction th<strong>at</strong> for every real x 0 there is an additivelyclosed λ > ω and a real x such th<strong>at</strong> 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 th<strong>at</strong> L λ [x ′ ] ̸|= OD-determinacy (since we can take x ′ = 〈x, x 0 〉 where xand x 0 are as in the first st<strong>at</strong>ement) 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.’ abbrevi<strong>at</strong>es ‘additively closed’. Since the condition on x in thisst<strong>at</strong>ement 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 <strong>Determinacy</strong> 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 th<strong>at</strong> for a Turing cone of xλ(x) is definedand th<strong>at</strong> there are x 0 of arbitrarily large Turing degree such th<strong>at</strong> for allx T x 0λ(x) λ(x 0 ).To see this last point it suffices to observe th<strong>at</strong> 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 th<strong>at</strong>Player II can “steer into the right model”.For each x such th<strong>at</strong> λ(x) is defined let (ϕ x , ξ x ) be lexicographically leastsuch th<strong>at</strong>L 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 th<strong>at</strong> since A x might bea proper class <strong>from</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> thefirst clause of the winning condition is s<strong>at</strong>isfied and so the players are to bethought of as cooper<strong>at</strong>ing 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 <strong>at</strong> a contradiction by showing th<strong>at</strong> neither player can win.Case 1: I has a winning str<strong>at</strong>egy σ 0 .
6. Definable <strong>Determinacy</strong> 152Let x 0 T σ 0 be such th<strong>at</strong> for all x T x 0 , λ(x) is defined and λ(x) λ(x 0 ). We claim th<strong>at</strong> L λ(x0 )[x 0 ] |= “I has a winning str<strong>at</strong>egy σ in A x 0”, whichis a contradiction. The str<strong>at</strong>egy σ is the str<strong>at</strong>egy 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 ]; th<strong>at</strong> is, (σ∗d) I = a where a is such th<strong>at</strong>(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 th<strong>at</strong> p T x 0 we know th<strong>at</strong> 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 th<strong>at</strong> II has managedto “steer into the right model”, th<strong>at</strong> is, th<strong>at</strong>and 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> L λ(x0 )[x 0 ] |= “σ is a winning str<strong>at</strong>egy for I in A x 0”, which is acontradiction.Case 2: II has a winning str<strong>at</strong>egy τ 0 .Let x 0 T τ 0 be such th<strong>at</strong> 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 th<strong>at</strong> (〈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 th<strong>at</strong> L λ(x0 )[x 0 ] |= “τ is awinning str<strong>at</strong>egy for II in A x 0”, which is a contradiction.
6. Definable <strong>Determinacy</strong> 1536.7 Remark. The proof rel<strong>at</strong>ivizes to a real parameter to show ZF + DC +Σ 1 2(x)-determinacy implies th<strong>at</strong> 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 <strong>from</strong> ∆ 1 2 -determinacy. We begin by recalling some definitions. The str<strong>at</strong>egic 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 prestr<strong>at</strong>egy for A n th<strong>at</strong> is winning with respect to B,and II wins iff he can play all ω rounds. We say th<strong>at</strong> str<strong>at</strong>egic 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 <strong>at</strong>tention to B since (as we will seein Theorem 6.12) ∆ 1 2-determinacy is insufficient to ensure th<strong>at</strong> for a Turingcone of x, L[x] |= ST B -determinacy, where B = {y ∈ ω ω | L[y] = L[x]}. Wewill now be interpreting str<strong>at</strong>egic 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 <strong>at</strong>tempt to “localize” the proof of Theorem 5.14 th<strong>at</strong> 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 abbrevi<strong>at</strong>e the st<strong>at</strong>ement “There is a Π 1 2 set Csuch th<strong>at</strong> 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] th<strong>at</strong> do not s<strong>at</strong>isfy full Replacement.In such a case it is to be understood th<strong>at</strong> the st<strong>at</strong>ement involves theΣ 1 definition of ordinal definability.
6. Definable <strong>Determinacy</strong> 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 th<strong>at</strong> if L λ [z] is such th<strong>at</strong>z 0 ∈ L λ [z] and L λ [z] |= T then L λ [z] |= RST-determinacy. The theoremfollows by reflection.Assume for contradiction th<strong>at</strong> for every real z 0 there is a real z T z 0and an ordinal λ such th<strong>at</strong> 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 st<strong>at</strong>ement is Σ 1 2 and since we have ODdeterminacy(by Corollary 6.8) it follows (by the Cone Theorem (Theorem2.9)) th<strong>at</strong>∃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 th<strong>at</strong> 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 prestr<strong>at</strong>egy for A n th<strong>at</strong> is winning with respect to C ∩ {y ∈ ω ω |L λ(z) [y] = L λ(z) [z]}.
6. Definable <strong>Determinacy</strong> 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 str<strong>at</strong>egy 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 th<strong>at</strong> 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 th<strong>at</strong> L λ(y) [y] = L λ(z) [z] andthus A y = A z and for this we require th<strong>at</strong> λ(y) = λ(z). So we will need tointersect with a set C th<strong>at</strong> “holds up the value of λ(y)”. A good candid<strong>at</strong>eis the following: For each z such th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> (2) doesnot hold on a Turing cone. Then (by OD-determinacy) for every real z thereis a real z ′ T z such th<strong>at</strong> λ(z ′ ) is defined and in L λ(z ′ )[z ′ ] there is a realz ′′ such th<strong>at</strong> z ′′ T x ′ and λ(z ′′ ) < λ(z). But then, for each n < ω, wecan successively choose z n+1 T z n such th<strong>at</strong> λ(z n+1 ) < λ(z n ), which is acontradiction.For each z as in Claim 1 we now have th<strong>at</strong> 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 th<strong>at</strong>B 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 th<strong>at</strong> L λ(z) [z] |= T, Player I has a canonicalstr<strong>at</strong>egy 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 <strong>Determinacy</strong> 156(SG Bz ) L λ(z)[z]is played is an element of HOD L λ(z)[z] .) Notice also th<strong>at</strong> σ zdepends only on the model L λ(z) [z], in the sense th<strong>at</strong> if L λ(y) [y] = L λ(z) [z]then σ y = σ z .Our aim is to obtain a contradiction by defe<strong>at</strong>ing 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 th<strong>at</strong> I must win via σ 0 , σ 1 , . . .,σ n , . . . and,for a Turing cone of z, the winning str<strong>at</strong>egies give rise to prestr<strong>at</strong>egiesf z 0 , fz 1 , . . .,fz n , . . . th<strong>at</strong> 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 s<strong>at</strong>isfies 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 th<strong>at</strong> p is in the cone of Claim 1. Soclause (1) of the winning condition will be autom<strong>at</strong>ically s<strong>at</strong>isfied 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 str<strong>at</strong>egy σ 0 in G 0 .Proof. Assume for contradiction th<strong>at</strong> I does not have a winning str<strong>at</strong>egyin G 0 . Then, by Σ 1 2 -determinacy, II has a winning str<strong>at</strong>egy τ 0 in G 0 . Letz 0 T τ 0 be such th<strong>at</strong> for all z T z 0 ,(1) z s<strong>at</strong>isfies the conditions of Claim 1 and(2) if λ and z are such th<strong>at</strong> 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 th<strong>at</strong> neither player can win thisgame. Suppose for contradiction th<strong>at</strong> L λ(z0 )[z 0 ] |= “σ is a winning str<strong>at</strong>egyfor I in A z 00 ”. Run G 0 according to τ 0 , having Player I (falsely) predict th<strong>at</strong>
6. Definable <strong>Determinacy</strong> 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 th<strong>at</strong> Player I has indeed managed to steer into L λ(z0 )[z 0 ],th<strong>at</strong> is, we have to see th<strong>at</strong> 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 s<strong>at</strong>isfies 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 str<strong>at</strong>egy for IIin G 0 and ǫ = 1, we have th<strong>at</strong> 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 th<strong>at</strong> σ is a winning str<strong>at</strong>egyfor I. Similarly, we can use τ 0 to defe<strong>at</strong> any str<strong>at</strong>egy τ for II in A z 0Since the game is Σ 1 2 for Player I, Player I has a ∆1 3 -str<strong>at</strong>egy σ 0, byTheorem 6.5.Claim 3. For every real z T σ 0 in the Turing cone of Claim 1, there isa prestr<strong>at</strong>egy f z 0 such th<strong>at</strong> fz 0 is definable in L λ(z)[z] <strong>from</strong> 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 theprestr<strong>at</strong>egy derived <strong>from</strong> σ 0 in L λ(z) [z] by extracting the response in theauxiliary game, th<strong>at</strong> is, for y ∈ (ω ω ) L λ(z)[z]let f0 z (y) be such th<strong>at</strong> for d ∈(ω ω ) Lλ(z)[z] , f0 z(y)∗d = a∗d where a is such th<strong>at</strong> (σ 0∗〈y, d〉) I = 〈ǫ, a, b〉. f0 z isclearly definable in L λ(z) [z] <strong>from</strong> σ 0 . We claim th<strong>at</strong> f0 z is a non-losing firstmove for II against σ z in (SG Bz ) Lλ(z)[z] .To motiv<strong>at</strong>e the need for the Π 1 2 set, let us first see why fz 0 need not be aprestr<strong>at</strong>egy for II in A z 0 th<strong>at</strong> 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 th<strong>at</strong> f0(y)∗d z = a∗d where a is suchth<strong>at</strong> (σ 0 ∗〈y, d〉) I = 〈ǫ, a, b〉. Assume first th<strong>at</strong> ǫ = 1. Since σ 0 is a winningstr<strong>at</strong>egy for I in G 0 , z(y)∗d = a∗d ∈ Ap 0 where p = 〈a, b, y, d〉. Wh<strong>at</strong> weneed, however, is th<strong>at</strong> f0(y)∗d z = a∗d ∈ A z 0. The trouble is th<strong>at</strong> 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 concludeth<strong>at</strong> A p 0 = A z 0 . If ǫ = 0 then fz 0 (y)∗d = a∗d ∉ Ap 0 but again wh<strong>at</strong> we need isth<strong>at</strong> f0 z(y)∗d = a∗d ∉ Az 0 and the same problem arises.0 .
6. Definable <strong>Determinacy</strong> 158The above problem is solved by demanding in addition th<strong>at</strong> 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 th<strong>at</strong> we have defined (in L[x]) games G 0 , . . .,G nwith winning str<strong>at</strong>egies σ 0 , . . .,σ n ∈ HOD such th<strong>at</strong> for all z T 〈σ 0 , . . .,σ n 〉in the Turing cone of Claim 1 there are prestr<strong>at</strong>egies f0 z, . . .,fz n such th<strong>at</strong>fi z is definable in L λ(z) [z] <strong>from</strong> 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 s<strong>at</strong>isfies 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 s<strong>at</strong>isfies condition (1) then, since p T 〈σ 0 , . . .,σ n 〉, we have, by theinduction hypothesis, prestr<strong>at</strong>egies f0, p . . .,fn p such th<strong>at</strong> f p i is definable inL λ(p) [p] <strong>from</strong> 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 str<strong>at</strong>egy σ n+1 in G n+1 .Proof. Assume for contradiction th<strong>at</strong> I does not have a winning str<strong>at</strong>egy.Then, by OD-determinacy, II has a winning str<strong>at</strong>egy τ n+1 . Let z n+1 T〈τ n+1 , σ 0 , . . ., σ n 〉 be such th<strong>at</strong> for all z T z n+1 ,(1) z s<strong>at</strong>isfies the conditions of Claim 1 and(2) if λ and z are such th<strong>at</strong> z n+1 ∈ L λ [z] and L λ [z] |= T then L λ [z] |=OD σ0 ,...,σ n-determinacy (by the rel<strong>at</strong>ivized version of Theorem 6.6).
6. Definable <strong>Determinacy</strong> 159It follows th<strong>at</strong>L 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 th<strong>at</strong>L λ(zn+1 )[z n+1 ] s<strong>at</strong>isfies OD σ0 ,...,σ n-determinacy.[The enforcement of the parameterized version of OD-determinacy in (2)appears to be necessary. The point is th<strong>at</strong> even though, in Step 1 for example,σ 0 is ∆ 1 3 and σ 0 ∈ L λ(z) [z] we have no guarantee th<strong>at</strong> in L λ(z1 )[z 1 ], σ 0 s<strong>at</strong>isfiesthis definition. If we did then we would have th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> L λ(zn+1 )[z n+1 ] |= “σ is a winning str<strong>at</strong>egy forI in A z n+1n+1 ”. Run G n+1 according to τ n+1 , having Player I (falsely) predictth<strong>at</strong> 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 th<strong>at</strong> Player I has indeed managed to steer into L λ(zn+1 )[z n+1 ],th<strong>at</strong> is, we have to see th<strong>at</strong> 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 s<strong>at</strong>isfiesthe 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 str<strong>at</strong>egy for IIin G n+1 and ǫ = 1, we have th<strong>at</strong> 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 th<strong>at</strong> σ isa winning str<strong>at</strong>egy for I. Similarly, we can use τ n+1 to defe<strong>at</strong> any str<strong>at</strong>egy 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 )str<strong>at</strong>egy σ n+1 , by the rel<strong>at</strong>ivized version of Theorem 6.5.Claim 5. For every real z T 〈σ 0 , . . .,σ n 〉 as in Claim 1, there is a prestr<strong>at</strong>egyf z n+1 th<strong>at</strong> is definable in L λ(z)[z] <strong>from</strong> σ 0 , . . .,σ n+1 and such th<strong>at</strong>f z 0 , . . .,fz n+1 is a non-losing first move for II against σz in (SG Bz ) L λ(z)[z] .
6. Definable <strong>Determinacy</strong> 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 prestr<strong>at</strong>egyderived <strong>from</strong> σ n+1 in L λ(z) [z] by extracting the response in the auxiliarygame, th<strong>at</strong> is, for y ∈ (ω ω ) Lλ(z)[z] let fn+1 z (y) be such th<strong>at</strong> for d ∈ (ωω ) Lλ(z)[z] ,fn+1(y)∗d z = a∗d where a is such th<strong>at</strong> (σ n+1 ∗〈y, d〉) I = 〈ǫ, a, b〉. Clearly, fn+1zis definable in L λ(z) [z] <strong>from</strong> σ 0 , . . .,σ n+1 . We claim th<strong>at</strong> fn+1 z is a non-losingfirst move for II against σ z in (SG Bz ) Lλ(z)[z] . Again the point is th<strong>at</strong> 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 th<strong>at</strong> z ∞ T z n for all n and z ∞ is as in Claim1, we have th<strong>at</strong> f0 z∞ , . . ., fn z∞,. . . defe<strong>at</strong>s 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 Gener<strong>at</strong>ionTheorem. Let Θ M = ω L[x]2 . Since L[x] s<strong>at</strong>isfies 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 th<strong>at</strong> A α isthe OD L[x] -least prewellordering of length α. Since L[x] |= OD-determinacy,it follows (by Theorem 3.9) th<strong>at</strong> 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 st<strong>at</strong>ementof 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 Gener<strong>at</strong>ionTheorem are clearly met and condition (6) follows since L[x] s<strong>at</strong>isfies 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 <strong>Determinacy</strong> 161Since, by arrangement, HOD M T,A,B = HODL[x] ∩ V L[x] ω, it follows th<strong>at</strong>2HOD L[x] |= ZFC + There is a T-strong cardinal.Since T ′ was arbitrary, the theorem follows.We close with four limit<strong>at</strong>ive results. The first result motiv<strong>at</strong>es the needfor the notion of str<strong>at</strong>egic determinacy by showing th<strong>at</strong> str<strong>at</strong>egic determinacydoes not follow trivially <strong>from</strong> OD-determinacy in the sense th<strong>at</strong> for some ODbasis there are OD prestr<strong>at</strong>egies.Theorem 6.11. Assume ZF. Then for each non-empty OD set B ⊆ ω ω ,there is an OD set A ⊆ ω ω such th<strong>at</strong> there is no OD prestr<strong>at</strong>egy in A whichis winning with respect to the basis B.Proof. Assume for contradiction th<strong>at</strong> there is a set B ⊆ ω ω which is ODand such th<strong>at</strong> for all OD sets A ⊆ ω ω there is an OD prestr<strong>at</strong>egy 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 prestr<strong>at</strong>egy 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 th<strong>at</strong>x ≡ T y T z and x codes α}.Notice th<strong>at</strong> A α is OD, Turing invariant, and contains a Turing cone. Butclearly⋂α
6. Definable <strong>Determinacy</strong> 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 th<strong>at</strong> 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 appropri<strong>at</strong>e 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 str<strong>at</strong>egies as we did in the earlier proof. Letµ = {S ⊆ ω V 1 | S ∈ HOD and I has a winning str<strong>at</strong>egy in G(S)},where G(S) is the game <strong>from</strong> Theorem 2.12.Subclaim 1. HOD |= µ ∩ HOD is an ultrafilter.Proof. It is clear th<strong>at</strong> ω1 V ∈ µ and ∅ /∈ µ. It is also clear th<strong>at</strong> if S ∈ µ andS ′ ∈ HOD ∩ P(ω1 V ) and S ⊆ S ′ then S ′ ∈ µ.Suppose th<strong>at</strong> S ∈ HOD ∩ P(ω1 V ) and th<strong>at</strong> II has a winning σ str<strong>at</strong>egyin G(S). We claim th<strong>at</strong> I has a winning str<strong>at</strong>egy in G(ω1 V S). Supposefor contradiction th<strong>at</strong> I does not have a winning str<strong>at</strong>egy. Then, by ODdeterminacy,II has a winning str<strong>at</strong>egy 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 associ<strong>at</strong>edordinal α < ω L[σ,σ′ ]1 . This is a contradiction.We now show th<strong>at</strong> if S 1 , S 2 ∈ µ then S 1 ∩ S 2 ∈ µ. Let σ 1 be a winningstr<strong>at</strong>egy for I in G(S 1 ) and let σ 2 be a winning str<strong>at</strong>egy for I in G(S 2 ).Suppose for contradiction th<strong>at</strong> 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 str<strong>at</strong>egy in G(S 1 ∩ S 2 ), which impliesth<strong>at</strong> I has a winning str<strong>at</strong>egy σ in G(ω1 V (S 1 ∩ S 2 )). Work in L[σ 1 , σ 2 , σ].1The str<strong>at</strong>egy σ 1 witnesses (by the usual argument using Σ ∼ 1 -boundedness)th<strong>at</strong> S 1 ∩ω L[σ 1,σ 2 ,σ]1 contains a club. Likewise, σ 2 witnesses th<strong>at</strong> S 2 ∩ω L[σ 1,σ 2 ,σ]1contains a club and σ witnesses th<strong>at</strong> (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 <strong>Determinacy</strong> 163Proof. Suppose for contradiction th<strong>at</strong> the subclaim fails. Let 〈S i | i < ω〉 ∈HOD be such th<strong>at</strong> for each i < ω, S i ∈ µ and ⋂ i
6. Definable <strong>Determinacy</strong> 164Proof. By our original supposition for contradiction recall th<strong>at</strong> we let f A bean OD prestr<strong>at</strong>egy which is winning with respect to B. Since A contains aTuring cone f A must be winning for Player I. This means th<strong>at</strong> for all z ∈ B,for all y ∈ ω ω , f A (z)∗y ∈ A. Now let y T 〈y ∗ , z〉. We wish to show th<strong>at</strong>y ∈ A. The point is th<strong>at</strong>y ≡ T f A (z)∗y ∈ Aand since A is Turing invariant, this implies th<strong>at</strong> y ∈ A.Claim 3 contradicts Claim 1, which completes the proof.The second result motiv<strong>at</strong>es the need for restricted str<strong>at</strong>egic determinacyby showing th<strong>at</strong> 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 str<strong>at</strong>egic determinacy. It alsofollows <strong>from</strong> the theorem th<strong>at</strong> something close to ∆ ∼12 -determinacy is requiredto establish th<strong>at</strong> ST B -determinacy holds with respect to the constructibilitydegree of x since the st<strong>at</strong>ement “∆ ∼12 -determinacy” is equivalent to the st<strong>at</strong>ement“for every y ∈ ω ω there is an inner model M such th<strong>at</strong> 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 th<strong>at</strong> L α [x] |= ZFC. Then for every y ∈ ω ω there is a transitivemodel M such th<strong>at</strong> 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 th<strong>at</strong> A 0 ≠ ∅. Let t ∈ A 0 . It follows th<strong>at</strong> 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, th<strong>at</strong> is, for all y ∈ B, f 0 (y) ∈ A 0 . We will arrive <strong>at</strong>a contradiction by constructing a real y ∈ B such th<strong>at</strong> f 0 (y) /∈ A 0 .We claim th<strong>at</strong>HOD Lα[x]f 0|= ZFC + There is a Woodin cardinal.
6. Definable <strong>Determinacy</strong> 165First note th<strong>at</strong>L[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 th<strong>at</strong> V = L[x ′ ]) it follows th<strong>at</strong>L α [x] |= ST B f 0-determinacy.Thus, by the rel<strong>at</strong>ivized 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 ] s<strong>at</strong>isfies th<strong>at</strong> there is a countable transitivemodel M such th<strong>at</strong> 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 th<strong>at</strong> L λ [f 0 ] s<strong>at</strong>isfies ZFC+“M is countable”. In L λ [f 0 ] let P be a perfect set of reals th<strong>at</strong> are Cohengeneric over M. Since P is perfect in L λ [f 0 ] there is a p<strong>at</strong>h c ∈ [P] whichcodes x in the sense th<strong>at</strong> c T x.Our desired real y is 〈f 0 , c〉. To see th<strong>at</strong> 〈f 0 , c〉 ∈ B note th<strong>at</strong> L λ (〈f 0 , c〉)can compute x and hence L ω1 [〈f 0 , c〉] = L ω1 [x]. To see th<strong>at</strong> f 0 (〈f 0 , c〉) ∉ A 0note th<strong>at</strong> since c is Cohen generic over M, the model M[c] is a transitivemodel containing f 0 (〈f 0 , c〉) s<strong>at</strong>isfying ZFC + “There is a Woodin cardinal”.This is a contradiction.The third result shows th<strong>at</strong> Martin’s “lightface form” of Third Periodicity(Theorem 6.3) does not generalize to higher levels. In fact, the result showsth<strong>at</strong> even ZFC + OD-determinacy (assuming consistency of course) does notimply th<strong>at</strong> for every Σ 1 4 game which Player I wins, Player I has a ∆ 1 5 str<strong>at</strong>egy(or even an OD str<strong>at</strong>egy). The reason th<strong>at</strong> the “lightface form” of ThirdPeriodicity holds <strong>at</strong> the level of Σ 1 2 but not beyond is th<strong>at</strong> 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 <strong>Determinacy</strong> 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 th<strong>at</strong> each y ∈ A witnesses th<strong>at</strong> R ∩ HODL[z] iscountable for each z T y.Claim 1. A contains a Turing cone.Proof. For y ∈ ω ω and α such th<strong>at</strong> 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 th<strong>at</strong> < 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 th<strong>at</strong> R α,y , R y , < α,y , < y , z y αand ϑ y depend only on the Turing degree of y.Our str<strong>at</strong>egy 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 , thest<strong>at</strong>ement “ϑ y > α and zα y (n) = m” is an OD-st<strong>at</strong>ement 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 th<strong>at</strong> 〈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 th<strong>at</strong> for aTuring cone of y, R y = R ∞ . To see this let y ∈ L[x] be such th<strong>at</strong> 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 th<strong>at</strong> for all z T z 0 , R z = R z0 = R ∞ . Since R ∞ iscountable, we can choose y 0 T z 0 such th<strong>at</strong> R ∞ T y 0 . Then for all z T y 0 ,R z = R ∞ T y 0 , th<strong>at</strong> is, y 0 ∈ A. Likewise, if y T y 0 , then y ∈ A.
6. Definable <strong>Determinacy</strong> 167Claim 2. A ∩ HOD L[x] = ∅.Proof. Suppose for contradiction th<strong>at</strong> y ∈ A ∩ HOD L[x] . Since y ∈ A, ywitnesses th<strong>at</strong> R z is countable for all z T y. Let z be such th<strong>at</strong>Then 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 th<strong>at</strong> ZF + DC + ∆ 1 2-determinacy implies th<strong>at</strong> for a Turing cone ofx, HOD L[x] has a ∆ 1 3 well-ordering of reals and hence th<strong>at</strong> 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 th<strong>at</strong> 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 showsth<strong>at</strong> R ∞ = R HOD : To see this let x ′ ∈ L[x] be such th<strong>at</strong> 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 th<strong>at</strong>∃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 st<strong>at</strong>ement“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 st<strong>at</strong>ement we havethe following Π 1 3-definitions (in L[x]) of ω ω ∩ HOD and < ∞ :z ∈ R ∞ ↔ ∀a [ a codes (z, R,
6. Definable <strong>Determinacy</strong> 168andz 0 < ∞ z 1 ↔ ∀a [ a codes (z, R,
6. Definable <strong>Determinacy</strong> 169cardinal and a ∆ 1 3 well-ordering of reals. It follows th<strong>at</strong> ∆1 2-determinacy failsin HOD L[x] for a Turing cone of x.Some interesting questions remain. For example: Does HOD L[x] s<strong>at</strong>isfyGCH, 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 <strong>Determinacy</strong>In this section we will work in ZF + AD. Our aim is to extract the optimalamount of large cardinal strength <strong>from</strong> boldface determinacy by constructinga model of ZFC th<strong>at</strong> contains ω-many Woodin cardinals.We shall prove a very general theorem along these lines. Our str<strong>at</strong>egyis to iter<strong>at</strong>ively apply Theorem 5.43. Recall th<strong>at</strong> this theorem st<strong>at</strong>es th<strong>at</strong>under 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 th<strong>at</strong> for all x Y x 0 ,HOD Y,[x]Y |= ω HOD Y,x2 is a Woodin cardinal.To gener<strong>at</strong>e 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 th<strong>at</strong> 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 th<strong>at</strong>HOD 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 th<strong>at</strong>ω HOD Y,x 02 < ω HOD Y,[x 0 ] Y ,x2
6. Definable <strong>Determinacy</strong> 170and this is not immedi<strong>at</strong>e. Second, in moving to the larger model we needto ensure th<strong>at</strong> we have not collapsed the first Woodin cardinal; a sufficientcondition for this is th<strong>at</strong>P(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 immedi<strong>at</strong>e. It turns out th<strong>at</strong> both difficulties can beovercome by taking x to be of sufficiently high “Y -degree”. This will bethe content of an elementary observ<strong>at</strong>ion and a “preserv<strong>at</strong>ion” lemma. Oncethese two hurdles are overcome we will be able to gener<strong>at</strong>e models with nWoodin cardinals for each n < ω. We shall then have to take extra steps toensure th<strong>at</strong> we can preserve ω-many Woodin cardinals. This will be achievedby shooting a Prikry sequence through the Y -degrees and proving an associ<strong>at</strong>ed“generic preserv<strong>at</strong>ion” lemma.6.15 Remark. It is important to note th<strong>at</strong> in contrast to Theorem 5.42here the degree notion in Theorem 5.43 does not depend on a and this isinstrumental in iter<strong>at</strong>ively applying the theorem to gener<strong>at</strong>e several Woodincardinals. In contexts such as L(R) where HOD “rel<strong>at</strong>ivizes” in the senseth<strong>at</strong> 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 th<strong>at</strong>HOD 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 th<strong>at</strong> µ Y is an ultrafilter. Also, by Theorem 2.8 we know th<strong>at</strong> µ Yis countably complete.Lemma 6.16 (Preserv<strong>at</strong>ion 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 immedi<strong>at</strong>e. Suppose for contradictionth<strong>at</strong> 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 <strong>Determinacy</strong> 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 th<strong>at</strong>ξ ∈ 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 iter<strong>at</strong>ively apply Theorem 5.43 to gener<strong>at</strong>ea model with n Woodin cardinals.Step 0. By Theorem 5.43, let x 0 be such th<strong>at</strong> for all x Y x 0 ,HOD Y,[x]Y |= ω HOD Y,x2 is a Woodin cardinal.Step 1. Recall th<strong>at</strong> ω1V is strongly inaccessible in any inner model ofZFC, by the Claim of Theorem 5.9. It follows th<strong>at</strong> ω HOD Y,x 02 < ω1 V and sowhen we choose x 1 Y x 0 we may assume th<strong>at</strong> x 1 codes ω HOD Y,x 02 . Thus,there exists an x 1 Y x 0 such th<strong>at</strong> for all x Y x 1 ,ω HOD Y,x 02 < ω HOD Y,[x 0 ] Y ,x2 ,and, by the Preserv<strong>at</strong>ion 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 th<strong>at</strong>HOD 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 <strong>at</strong> this stage to introduce a piece of not<strong>at</strong>ion: 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 <strong>Determinacy</strong> 172Suppose th<strong>at</strong> we have chosen x 0 Y x 1 Y · · · Y x n such th<strong>at</strong>HOD 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 followsth<strong>at</strong> each of these ordinals is countable in V and so when we choose x n+1 Yx n we may assume th<strong>at</strong> x n+1 collapses these ordinals. Thus, there exists anx n+1 Y x n such th<strong>at</strong> for all x Y x n+1 ,δ n (x 0 , . . .,x n ) < δ n+1 (x 0 , . . .,x n , x)and, by the Preserv<strong>at</strong>ion 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 th<strong>at</strong>HOD 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 th<strong>at</strong> 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 immedi<strong>at</strong>e since, for example, if we are not careful then thereals x 0 , x 1 , . . . might code up a real th<strong>at</strong> collapses sup n
6. Definable <strong>Determinacy</strong> 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 <strong>Determinacy</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 〈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 th<strong>at</strong> it is not the case th<strong>at</strong> 〈p, F ∗ 〉 ⊩ ¬ϕ. Then ∃〈q, G〉 PY 〈p, F ∗ 〉such th<strong>at</strong> 〈q, G〉 ⊩ ϕ. But then q is such th<strong>at</strong> ρ(q) = 0. However, F ∗ witnessesth<strong>at</strong> in fact ρ(q) = ∞: Suppose q = p ⌢ a 0⌢ · · · ⌢a k . Since ρ(p) = ∞ anda 0 ∈ F ∗ (p), we have th<strong>at</strong> a 0 ∈ A p and so ρ(p ⌢ a 0 ) = ∞. Continuing in thismanner, we get th<strong>at</strong> ρ(q) = ∞. This is a contradiction.Claim 2. If ρ(p) ∈ On then there is an F ∗ such th<strong>at</strong> 〈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 <strong>Determinacy</strong> 175We claim th<strong>at</strong> 〈p, F ∗ 〉 ⊩ ϕ. Assume not. Then ∃〈q, G〉 PY 〈p, F ∗ 〉 such th<strong>at</strong>〈q, G〉 ⊩ ¬ϕ. Since 〈q, G〉 PY 〈p, F ∗ 〉 and ρ(p) ∈ On we have th<strong>at</strong> ρ(q) ∈ On(by an easy induction using the definition of F ∗ ). We may assume th<strong>at</strong> 〈q, G〉is chosen so th<strong>at</strong> 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 comp<strong>at</strong>ible with〈q, G〉 and 〈q, G〉 ⊩ ¬ϕ.This completes the proof of the lemma.We can now obtain the following “generic preserv<strong>at</strong>ion” lemma.Lemma 6.19 (Generic Preserv<strong>at</strong>ion Lemma). There exists an F suchth<strong>at</strong> if G ⊆ P Y is V -generic and 〈∅, F 〉 ∈ G and 〈[x i ] Y | i < ω〉 is the genericsequence associ<strong>at</strong>ed 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 <strong>Determinacy</strong> 176th<strong>at</strong> the left-to-right inclusion fails. Let A ⊆ α be
7. Second-Order Arithmetic 177By the Generic Preserv<strong>at</strong>ion Lemma it suffices to show th<strong>at</strong> 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 applic<strong>at</strong>ion of this theorem in conjunction with theDerived Model Theorem (Theorem 8.12), we obtain Kechris’ theorem th<strong>at</strong>under ZF + AD, DC holds in L(R). This altern<strong>at</strong>e 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 th<strong>at</strong>(1) On ⊆ N ⊆ HOD L[x] and(2) N |= δ is a Woodin cardinal.Then δ ω L[x]2 .However, it turns out th<strong>at</strong> ω L[x]1 can be a Woodin cardinal in an inner modelth<strong>at</strong> 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 th<strong>at</strong>Moreover, 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 suchth<strong>at</strong>(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 th<strong>at</strong> 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 rel<strong>at</strong>ivizing the previous construction to the Turing degree of x,replacing the notions th<strong>at</strong> concerned reals (for example, winning str<strong>at</strong>egies)with rel<strong>at</strong>ivized analogues th<strong>at</strong> concern only those reals in the Turing degreeof x.In Section 7.2 we show th<strong>at</strong> the rel<strong>at</strong>ivized construction goes through inthe setting of second-order arithmetic.Theorem 7.5. Assume th<strong>at</strong> PA 2 + ∆ 1 2 -determinacy is consistent. ThenZFC + “On is Woodin” is consistent.
7. Second-Order Arithmetic 179Here PA 2 is the standard axiom<strong>at</strong>iz<strong>at</strong>ion of second-order arithmetic (withoutAC). The st<strong>at</strong>ement th<strong>at</strong> On is Woodin is to be understood schem<strong>at</strong>ically.Altern<strong>at</strong>ively, one could work with the conserv<strong>at</strong>ive extension GBC of ZFCand the analogous conserv<strong>at</strong>ive extension of PA 2 . This would enable one tofuse the schema expressing th<strong>at</strong> On is Woodin into a single st<strong>at</strong>ement.7.1. First Localiz<strong>at</strong>ionTo prove Theorem 7.4 we have to prove an analogue of the Gener<strong>at</strong>ion Theoremwhere ω 2 is replaced by ω 1 . The two main steps are (1) getting a suitablenotion of str<strong>at</strong>egic determinacy and (2) getting definable prewellorderings forall ordinals less than ω 1 .For x ∈ ω ω we “rel<strong>at</strong>ivize” our previous notions to the Turing degree of x.The rel<strong>at</strong>ivized reals are R x = {y ∈ ω ω | y T x}. Fix A ⊆ R x . A rel<strong>at</strong>ivizedstr<strong>at</strong>egy for I is a function σ : ⋃ n
7. Second-Order Arithmetic 180We caution the reader th<strong>at</strong> in the context of the rel<strong>at</strong>ivized notions weare dealing only with definable versions of rel<strong>at</strong>ivized determinacy such asOD-[x] T -determinacy and SG-[x] T -determinacy. In fact, full rel<strong>at</strong>ivized 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 th<strong>at</strong> 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 th<strong>at</strong> of Theorem 6.6. Assume for contradictionth<strong>at</strong> for every real x 0 there is an ordinal λ and a real x such th<strong>at</strong> 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 th<strong>at</strong> 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 str<strong>at</strong>egies. This game is Σ 1 2, hencedetermined.We arrive <strong>at</strong> a contradiction by showing th<strong>at</strong> neither player can win.Case 1: I has a winning str<strong>at</strong>egy σ 0 .Let x 0 T σ 0 be such th<strong>at</strong> for all x T x 0 , λ(x) is defined. We claimth<strong>at</strong> L λ(x0 )[x 0 ] |= “I has a rel<strong>at</strong>ivized winning str<strong>at</strong>egy σ in A x 0”, which is acontradiction. The rel<strong>at</strong>ivized str<strong>at</strong>egy 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 str<strong>at</strong>egy for Player I.) This str<strong>at</strong>egy σ is clearly recursive in σ 0 ,hence it is a rel<strong>at</strong>ivized str<strong>at</strong>egy.
7. Second-Order Arithmetic 181It remains to show th<strong>at</strong> for every d ∈ R x0 , σ∗d ∈ A x 0. The point isth<strong>at</strong> 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 th<strong>at</strong> 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 rel<strong>at</strong>ivized winning str<strong>at</strong>egy for I in A x 0”.Case 2: II has a winning str<strong>at</strong>egy τ 0 .Let x 0 T τ 0 be such th<strong>at</strong> 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 str<strong>at</strong>egy is clearly recursive in τ 0 , hence it is a rel<strong>at</strong>ivized str<strong>at</strong>egy, and,as above, L λ(x0 )[x 0 ] |= “τ is a rel<strong>at</strong>ivized winning str<strong>at</strong>egy 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 th<strong>at</strong> V =L[x] and show th<strong>at</strong> there is a real z 0 with thefe<strong>at</strong>ure th<strong>at</strong> if z 0 ∈ L λ [z] and L λ [z] |= T, then L λ [z] |= ST-[x] T -determinacy.Assume for contradiction th<strong>at</strong> 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 immedi<strong>at</strong>e.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 str<strong>at</strong>egy σ zth<strong>at</strong> depends only on the Turing degree of z, the point being th<strong>at</strong> if y ≡ T zthen L λ(y) [y] = L λ(z) [z].As before our aim is to obtain a contradiction by defe<strong>at</strong>ing 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 th<strong>at</strong> I must win via σ 0 , σ 1 , . . .,σ n , . . . and, fora cone of z, the winning str<strong>at</strong>egies give rise to prestr<strong>at</strong>egies f z 0 , fz 1 , . . .,fz n , . . .th<strong>at</strong> 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 s<strong>at</strong>isfies 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 str<strong>at</strong>egy σ 0 in G 0 .Proof. Assume for contradiction th<strong>at</strong> I does not have a winning str<strong>at</strong>egyin G 0 . Then, by Σ 1 2-determinacy, II has a winning str<strong>at</strong>egy τ 0 in G 0 . Letz 0 T τ 0 be such th<strong>at</strong> for all z T z 0 ,(1) z s<strong>at</strong>isfies the conditions of Claim 1 and(2) if λ and z are such th<strong>at</strong> 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 th<strong>at</strong> L λ(z0 )[z 0 ] |= “σ is a rel<strong>at</strong>ivized winning str<strong>at</strong>egyfor I in A z 00 ”. We use τ 0 to defe<strong>at</strong> this rel<strong>at</strong>ivized str<strong>at</strong>egy. Run G 0 accordingto τ 0 , having Player I (falsely) predict th<strong>at</strong> 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 th<strong>at</strong> p ≡ T z 0 (since σ, τ 0 ∈ R z0 ) and so λ(p) = λ(z 0 ). Thusthe “steering problem” is immedi<strong>at</strong>ely solved and we have a contradiction asbefore.Since the game is Σ 1 2 for Player I, Player I has a ∆ 1 3 str<strong>at</strong>egy σ 0 , byTheorem 6.5.Claim 3. For every real z T σ 0 there is a prestr<strong>at</strong>egy f0 z such th<strong>at</strong> 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 prestr<strong>at</strong>egy derived <strong>from</strong> 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 th<strong>at</strong> fz 0 is recursive inσ 0 T z and for y ∈ [z] T , f0(y) z ∈ R z . It remains to see th<strong>at</strong> for y ∈ [z] T , f0(y)zis a rel<strong>at</strong>ivized winning str<strong>at</strong>egy for I in A z 0 . The point is th<strong>at</strong> 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 th<strong>at</strong> (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 th<strong>at</strong> we have defined (in L[x]) games G 0 , . . .,G nwith winning str<strong>at</strong>egies σ 0 , . . .,σ n ∈ HOD such th<strong>at</strong> for all z T 〈σ 0 , . . .,σ n 〉as in Claim 1 there are prestr<strong>at</strong>egies f z 0 , . . .,fz n such th<strong>at</strong> 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 s<strong>at</strong>isfies 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 str<strong>at</strong>egy σ n+1 in G n+1 .Proof. The proof is as before, only now we use the rel<strong>at</strong>ivized 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 )str<strong>at</strong>egy σ n+1 , by the rel<strong>at</strong>ivized version of Theorem 6.5.Claim 5. For every real z T 〈σ 0 , . . .,σ n 〉 there is a prestr<strong>at</strong>egy fn+1 z suchth<strong>at</strong> 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 th<strong>at</strong> z ∞ T z n for all n wehave th<strong>at</strong> f0 z∞ , . . .,fn z∞ , . . . defe<strong>at</strong>s 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 th<strong>at</strong>S 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 th<strong>at</strong> 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 suchth<strong>at</strong> 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 th<strong>at</strong> HOD d [G] = L[x] for every G = G z th<strong>at</strong> 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 th<strong>at</strong> of Vopěnka’s—in general our partial order does not have <strong>at</strong>oms.If L[x] = HOD d then clearly for each α < ω L[x]1 there exists a surjectionρ : d → α such th<strong>at</strong> ρ ∈ OD d . So we may assume th<strong>at</strong> 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 th<strong>at</strong> there is an antichain of size ¯λ. If¯λ > ω then this is immedi<strong>at</strong>e 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 <strong>at</strong>om.Letting λ = ω L[x]1 , we are in the situ<strong>at</strong>ion 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 th<strong>at</strong>S ′ α is an OD d subset of B ′ d consisting of pairwise disjoint OD d subsets of d.Picking an element <strong>from</strong> 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 rel<strong>at</strong>ivizedversion 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 th<strong>at</strong> A α is an OD d prewellordering oflength gre<strong>at</strong>er 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 th<strong>at</strong>HOD M |= There is a T 0 -strong cardinal,which completes the proof as before. The reason is th<strong>at</strong> we are in the situ<strong>at</strong>ionof the Gener<strong>at</strong>ion Theorem, except with ω L[x]1 replacing ω L[x]2 andST-[x] T -determinacy replacing ST B -determinacy. The proof of the Gener<strong>at</strong>ionTheorem goes through unchanged. One just has to check th<strong>at</strong> all ofthe oper<strong>at</strong>ions we performed before (which involved definability in variousparameters) are in fact recursive in the relevant parameters.
7. Second-Order Arithmetic 1867.2. Second Localiz<strong>at</strong>ionWe now wish to show th<strong>at</strong> the above construction goes through when wereplace ZF + DC with PA 2 . Notice th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> for all x ∈ ω ω , there is an ordinal α < ω 1 such th<strong>at</strong>L α [x] |= ZFC. Using ∆ ∼12 -determinacy one can find a real x 0 enforcing ODdeterminacy.Thus we have a model L α0 [x 0 ] s<strong>at</strong>isfying ZFC+V = L[x 0 ]+ODdeterminacyand this puts us in the situ<strong>at</strong>ion of Theorem 6.10.The situ<strong>at</strong>ion where one only has ∆ 1 2-determinacy is bit more involved.Theorem 7.11. Assume th<strong>at</strong> PA 2 + ∆ 1 2-determinacy is consistent. ThenZFC + “On is Woodin” is consistent.Proof Sketch. First we pass to a theory th<strong>at</strong> more closely resembles the theoryused to prove Theorem 7.9. In PA 2 one can simul<strong>at</strong>e 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 th<strong>at</strong> the“inner model” L ω1 [x] s<strong>at</strong>isfies ZFC−Power Set+V =L[x] (using, for exampleComprehension to get Replacement). Thus, ZFC − Power Set + V =L[x] isa conserv<strong>at</strong>ive extension of PA 2 .Next we need to arrange a sufficient amount of definable determinacy.The most n<strong>at</strong>ural way to secure ∆ 1 2-determinacy is to let x encode winningstr<strong>at</strong>egies 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 schem<strong>at</strong>ically). For this we simultaneously run (anelabor<strong>at</strong>ion 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 associ<strong>at</strong>ed definable set of realsA which codes a model L ω1 [x] s<strong>at</strong>isfying ZFC − Power Set + V = L[x] + ODdeterminacy.Working in ZFC − Power Set + V =L[x] + OD-determinacy we wish nowto show th<strong>at</strong>HOD [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 th<strong>at</strong>〈L ω1 [x], [x] T 〉 |= ST-[x] T -determinacy for n moves,for each n. Here by ST-[x] T -determinacy we mean wh<strong>at</strong> we meant in theprevious section. However, there is a slight metam<strong>at</strong>hem<strong>at</strong>ical issue th<strong>at</strong>arises when we work without Powerset, namely, <strong>at</strong> each stage of the gamethe potential moves for Player I are a proper class <strong>from</strong> 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 th<strong>at</strong> of Theorem 7.7,only now one has to keep track of definability and verify th<strong>at</strong> 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 Gener<strong>at</strong>ion Theorem gives a structure M suchth<strong>at</strong>HOD 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 th<strong>at</strong> draw on techniques th<strong>at</strong> are outside the scope of this chapter.The first topic concerns the intim<strong>at</strong>e connection between axioms ofdefinable determinacy and large cardinal axioms (as medi<strong>at</strong>ed through innermodels). The second topic concerns the surprising convergence betweentwo very different approaches to inner model theory—the approach basedon generaliz<strong>at</strong>ions of L and the approach based on HOD. In both cases the
8. Further Results 188relevant m<strong>at</strong>erial on inner model theory can be found in Steel’s chapter inthis Handbook.8.1. <strong>Large</strong> <strong>Cardinals</strong> and <strong>Determinacy</strong>The connection between axioms of definable determinacy and inner models oflarge cardinals is even more intim<strong>at</strong>e than indic<strong>at</strong>ed by the above results. Wehave seen th<strong>at</strong> 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 th<strong>at</strong> x ∈ MandM |= ZFC + There is a Woodin cardinal.And, assuming ZFC + AD L(R) , in L(R) there is an inner model M such th<strong>at</strong>M |= ZFC + There is a Woodin cardinal.In many cases these implic<strong>at</strong>ions can be reversed—axioms of definable determinacyare actually equivalent to axioms asserting the existence of innermodels of large cardinals. We discuss wh<strong>at</strong> is known about this connection,starting with a low level of boldface definable determinacy and proceedingupward. We then turn to lightface determinacy, where the situ<strong>at</strong>ion is moresubtle. It should be emphasized th<strong>at</strong> our concern here is not merely withconsistency strength but r<strong>at</strong>her 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 th<strong>at</strong> x ∈ M andM |= There is a Woodin cardinal.Theorem 8.2. The following are equivalent:(1) PD (Schem<strong>at</strong>ic).(2) For every n < ω, there is a fine-structural, countably iterable innermodel M such th<strong>at</strong> 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 th<strong>at</strong> 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 situ<strong>at</strong>ionwith 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 th<strong>at</strong> α is a Woodin cardinal in M? In light of Theorem 8.1 onewould expect th<strong>at</strong> this is indeed the case. However, since Theorem 8.1 alsoholds in the context of PA 2 one would then expect th<strong>at</strong> the theories PA 2 +∆ 1 2 -determinacy and PA 2 + “There is an α < ω 1 and an inner model M suchth<strong>at</strong> M |= α is a Woodin cardinal” are equivalent, and yet this expect<strong>at</strong>ion isin conflict with the expect<strong>at</strong>ion th<strong>at</strong> 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 st<strong>at</strong>e 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 th<strong>at</strong>
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 th<strong>at</strong>Let 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 <strong>from</strong> 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 <strong>at</strong> the level of AD L(R) and beyond.We have already dealt in detail with one direction of this—the transfer<strong>from</strong> models of determinacy to models with Woodin cardinals—and theother direction—the transfer <strong>from</strong> models with Woodin cardinals to modelsof determinacy—was briefly discussed in the introduction, but the situ<strong>at</strong>ionis much more general. To proceed <strong>at</strong> the appropri<strong>at</strong>e 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 th<strong>at</strong>A = {y ∈ ω ω | L α [S, y] |= ϕ[S, y]}.It is fairly straightforward to show th<strong>at</strong> to say th<strong>at</strong> A is ∞ -borel is equivalentto saying th<strong>at</strong> it has a “transfinite borel code”. Notice th<strong>at</strong> 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 th<strong>at</strong> the failure of this implic<strong>at</strong>ion has strong consistency strength.For example, AD + ¬AD + proves Con(AD R ).The following theorem—the Derived Model Theorem—is a generaliz<strong>at</strong>ionof Theorem 1.5, mentioned in the introduction.Theorem 8.12. Suppose th<strong>at</strong> δ is a limit of Woodin cardinals. Suppose th<strong>at</strong>G ⊆ Col(ω,
8. Further Results 1928.2. HOD-AnalysisThere is also an intim<strong>at</strong>e connection between the two approaches to innermodel theory mentioned in the introduction—the approach based on generaliz<strong>at</strong>ionsof 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 isth<strong>at</strong> the problem of actually defining the models th<strong>at</strong> can accommod<strong>at</strong>e largecardinals—the inner model problem—is quite a difficult problem. However,the advantage is th<strong>at</strong> once the inner model problem is solved <strong>at</strong> 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 st<strong>at</strong>ement. The advantage of the approach based onHOD is th<strong>at</strong> this model is trivial to define and it can accommod<strong>at</strong>e virtuallyevery large cardinal. The disadvantage—the tractability problem—is th<strong>at</strong> ingeneral the inner structure of HOD is about as tractable as th<strong>at</strong> of V andso it is not generally suitable for extracting the large cardinal content <strong>from</strong>a given st<strong>at</strong>ement.Nevertheless, we have taken the approach based on HOD and we havefound th<strong>at</strong> AD L(R) and ∆ 1 2-determinacy are able to overcome (to some extent)the tractability problem for their n<strong>at</strong>ural models, L(R) and L[x] for a Turing(or constructibility) cone of x. For example, we have seen th<strong>at</strong> under AD L(R) ,HOD L(R) |= Θ L(R) is a Woodin cardinal,and th<strong>at</strong> 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 far<strong>from</strong> clear. For example, it is unclear whether under ∆ 1 2-determinacy, fora Turing cone of reals x, HOD L[x] s<strong>at</strong>isfies GCH, something th<strong>at</strong> would beimmedi<strong>at</strong>e 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 <strong>at</strong> 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 s<strong>at</strong>isfy GCH and many of the other combin<strong>at</strong>orialproperties th<strong>at</strong> hold in L. A n<strong>at</strong>ural question, then, is whether
8. Further Results 193there is any connection between these radically different approaches, th<strong>at</strong> 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 generaliz<strong>at</strong>ions (where a good deal is known) and then turnto HOD L[x] (where the central question is open). Again, the situ<strong>at</strong>ion withlightface determinacy is more subtle.The theorems concerning HOD L(R) only require AD L(R) but they are simplerto st<strong>at</strong>e under the stronger assumption th<strong>at</strong> AD L(R) holds in all genericextensions of V . By Theorem 8.5, this assumption is equivalent to the st<strong>at</strong>ementth<strong>at</strong> M ω # exists and is fully iterable.The first hint th<strong>at</strong> HOD L(R) is a fine-structural model is the remarkablefact th<strong>at</strong>HOD L(R) ∩ R = M ω ∩ R.The agreement between HOD L(R) and M ω fails higher up but HOD L(R) agreeswith an iter<strong>at</strong>e of M ω <strong>at</strong> slightly higher levels. More precisely, letting N bethe result of iter<strong>at</strong>ing M ω by taking the ultrapower ω1 V -many times using the(unique) normal ultrafilter on the least measurable cardinal, we have th<strong>at</strong>HOD L(R) ∩ P(ω V 1 ) = N ∩ P(ω V 1 ).Steel improved this dram<strong>at</strong>ically by showing th<strong>at</strong>HOD 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 th<strong>at</strong> HOD L(R) s<strong>at</strong>isfies GCH along with the combin<strong>at</strong>orial principles(such as ♦ and □) th<strong>at</strong> are characteristic of fine-structural models.The above results suggest th<strong>at</strong> 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 th<strong>at</strong> belongs toa new, quite different, hierarchy of models. LetD = { L[ ⃗ E] ∣ ∣ L[ ⃗ E] is an iter<strong>at</strong>e 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 iter<strong>at</strong>ion halts in countablymany steps (since we have full iterability) with iter<strong>at</strong>es lying in D. So D is adirected system under the elementary embeddings given by iter<strong>at</strong>ion 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-theoreticinform<strong>at</strong>ion. A n<strong>at</strong>ural candid<strong>at</strong>e is the iter<strong>at</strong>ion str<strong>at</strong>egy.It turns out th<strong>at</strong> by folding in the right fragment of the iter<strong>at</strong>ion str<strong>at</strong>egyone can capture HOD L(R) . LetandT ∞ = { T ∣ ∣ T is a maximal iter<strong>at</strong>ion 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 iter<strong>at</strong>ion tree T ∈ T ∞ such th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> this is an artifact of L[ E ⃗ ∞ ] but in fact the situ<strong>at</strong>ion 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, wh<strong>at</strong> appeared to be a coarse approachto inner model theory has actually resulted in a hierarchy th<strong>at</strong> supplementsand refines the standard fine-structural hierarchy.The above results generalize. We need a definition.8.18 Definition (Mouse Capturing). MC is the st<strong>at</strong>ement: 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 th<strong>at</strong> x ∈ M.The Mouse Set Conjecture, MSC, is the conjecture th<strong>at</strong> it is a theorem ofAD + th<strong>at</strong> MC holds if there is no iterable model with a superstrong cardinal.There should be a more general version of MC, one th<strong>at</strong> holds for extensionsof the Mitchell-Steel models th<strong>at</strong> can accommod<strong>at</strong>e long extenders. And thisversion of MC should follow <strong>from</strong> 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 th<strong>at</strong> 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 appropri<strong>at</strong>eform 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 st<strong>at</strong>ement th<strong>at</strong> Ω (defined <strong>at</strong>the beginning of Section 5) is a non-zero limit ordinal. There is a minimalinner model N of ZF+AD R th<strong>at</strong> 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 calibr<strong>at</strong>e 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 situ<strong>at</strong>ion is less clear. In fact, the basic questionis open.5 Open Question. Assume ∆ 1 2-determinacy. For a Turing cone of x, wh<strong>at</strong>is HOD L[x] <strong>from</strong> a fine-structural point of view?We close with partial results in this direction and with a conjecture. Tosimplify the discussion we st<strong>at</strong>e these results under a stronger assumptionthan is necessary: Assume ∆ 1 2 -determinacy and th<strong>at</strong> for all x ∈ ωω , x # exists.It follows th<strong>at</strong> M 1 and M # 1 exist. Let x 0 ∈ ω ω be such th<strong>at</strong> 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 th<strong>at</strong>Furthermore,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 Gener<strong>at</strong>ion Theorem hold in L[x 0 ][G] and as a consequenceone has th<strong>at</strong>HOD L[x 0][G] |= ω L[x 0][G]2 is a Woodin cardinal.For a model L[ E] ⃗ Econtaining <strong>at</strong> least one Woodin cardinal let δ ⃗ 0 be the leastWoodin cardinal. LetD = { L[ E] ⃗ ⊆ L[x 0 ][G] ∣ L[ E] ⃗ is an iter<strong>at</strong>e 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 iter<strong>at</strong>ion 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 th<strong>at</strong> place onein an “L(R)-like” setting. For example, suppose again th<strong>at</strong> x 0 is such th<strong>at</strong>M # 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 th<strong>at</strong> (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 th<strong>at</strong> δ is ∼δ 2 as computed in L[x 0 ][G]. It turns out this analogyfails: the least cardinal δ th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> 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 th<strong>at</strong> A = p[T]. A cardinal κ is aneffective Suslin cardinal if there exists a set A ⊆ ω ω such th<strong>at</strong> 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 th<strong>at</strong> 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] th<strong>at</strong> is worthmentioning.Theorem 8.22. Assume ∆ ∼12 -determinacy. For a Turing cone of x there isa predic<strong>at</strong>e A such th<strong>at</strong>(1) HOD L[x]Astr<strong>at</strong>egy,(2) HOD L[x]A(3) HOD L[x]Ahas the form L[ ⃗ E, P] where P is a fragment of the iter<strong>at</strong>ion|= 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]Ath<strong>at</strong> 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 th<strong>at</strong> HOD L[x] is of the form L[ ⃗ E] below ω L[x]2and th<strong>at</strong> 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: <strong>Large</strong> <strong>Cardinals</strong> in Set Theory<strong>from</strong> their Beginnings. Springer Monographs in M<strong>at</strong>hem<strong>at</strong>ics. Springer,Berlin, second edition, 2003.[3] Alexander S. Kechris. The axiom of determinacy implies dependentchoices in L(R). The Journal of Symbolic <strong>Logic</strong>, 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 M<strong>at</strong>hem<strong>at</strong>ics, 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 M<strong>at</strong>hem<strong>at</strong>ics, pages 75–99. Springer, 1981.[6] Alexander S. Kechris and Robert M. Solovay. On the rel<strong>at</strong>ive strengthof determinacy hypotheses. Transactions of the American M<strong>at</strong>hem<strong>at</strong>icalSociety, 290(1):179–211, 1985.[7] Alexander S. Kechris and W. Hugh Woodin. Equivalence of partitionproperties and determinacy. Proceedings of the N<strong>at</strong>ional Academy ofScience U.S.A, 80:1783–1786, 1983.[8] Peter Koellner. On the question of absolute undecidability. PhilosophiaM<strong>at</strong>hem<strong>at</strong>ica, 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 <strong>Logic</strong>, 33. Associ<strong>at</strong>ionof Symbolic <strong>Logic</strong>, 2009.[9] Paul B. Larson. The St<strong>at</strong>ionary Tower: Notes on a Course by W. HughWoodin, volume 32 of University Lecture Series. American M<strong>at</strong>hem<strong>at</strong>icalSociety, 2004.[10] Yiannis N. Moschovakis. Descriptive Set Theory. North-Holland, Amsterdam,1980.
Bibliography 200[11] Itay Neeman. The <strong>Determinacy</strong> of Long Games, volume 7 of de GruyterSeries in <strong>Logic</strong> and its Applic<strong>at</strong>ions. de Gruyter, Berlin, 2004.[12] John Steel. M<strong>at</strong>hem<strong>at</strong>ics needs new axioms. The Bulletin of Symbolic<strong>Logic</strong>, 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, <strong>Logic</strong> Colloquium2000, volume 19 of Lecture Notes in <strong>Logic</strong>, pages 143–197. Associ<strong>at</strong>ionof Symbolic <strong>Logic</strong>, 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 lemmasapplic<strong>at</strong>ions, 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 <strong>Determinacy</strong> (AD), 3definable determinacy, 3, 145–177absolute, 4boldface, 4, 169–177lightface, 4, 147–169∆ 1 2-determinacy, 5OD-determinacy, 4OD(R)-determinacy, 4Projective <strong>Determinacy</strong> (PD), 3Σ∼ 1 n+1 -determinacy, 9str<strong>at</strong>egic determinacyST B P 0 ,...,P k-determinacy, 105RST-determinacy, 153ST-[x] T -determinacy, 179Turing determinacy, 22disjointness property, 80, 128extender, 115pre-extender, 116Gener<strong>at</strong>ion Theorem, 12, 13, 96, 109–137201
INDEX 202applic<strong>at</strong>ions, 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 comput<strong>at</strong>ion, 83, 131Third Periodicity Theorem, 149, 165Turingcone, 21cone filter, 21degrees, 21determinacy, 22Woodin cardinal, 8ω-model, 22pre-extender, 116prestr<strong>at</strong>egy, 104Prikry forcingthrough degrees, 172Recursion Theorem, 45Uniform, 49reflected gener<strong>at</strong>or, 118reflection filter, 66, 68, 113Reflection Theorem (for Chapter 23),63, 68, 93, 113rel<strong>at</strong>ivized str<strong>at</strong>egic game, 179rel<strong>at</strong>ivized prestr<strong>at</strong>egy, 179rel<strong>at</strong>ivized reals, 179rel<strong>at</strong>ivized str<strong>at</strong>egy, 179S-cone, 100S-degree, 100second-order arithmetic, 177–187str<strong>at</strong>egic game, 104strong cardinal, 7