A Sharp for the Chang Model - Logic at Harvard - Harvard University

2011-02-21A Sharp for the Chang ModelOutlineOutline1 What is the Chang Model C?2 Introducing the Sharp3 A Sketch of the ProofThese are the slides which I used for my talk at the MAMLS meeting atHarvard University, “Inner Model Theory & Large Cardinals—A 50 YearCelebration” on February 19, 2011. I have corrected (but not, I believe,all) of the typos and have added some notes.The forcing used in this paper is based on that of Gitik in his proof of theindependence of the SCH. The argument that a generic set for theforcing can be constructed using the indiscernibles coming from theiteration is based on work of Carmi Merimovich.

DefinitionDefinitionThe Chang model C is the smallest model of ZF containing the ordinalsand containing all of its countable subsets.It can be obtained by modifying the definition of L at ordinals α ofcofinality ω:C α+1 = Def Cα∪[α]

Simple ObservationsR ⊆ C, and hence L(R) ⊆ C.C satisfies DC (Dependent choice), and any integer game decided inV is decided in C.Any member of C is definable in C by using a countable sequence ofordinals as a parameter.[C]

ExamplesC L = L.¬0 ♯ =⇒ C = L([ω 2 ] ω ).In L[U], let i U ω : L[U] → Ult ω (L, U) = L[U ω ].ThenC L[U] = L[U ω ][〈 i U n (κ) | n < ω 〉]= L[〈 i U n (κ) | n < ω 〉]andHOD CL[U] = L[U ω ].The point is that U ω can be reconstructed from its Prikry sequence〈 i U n (κ) | n ∈ ω 〉, which is countable.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 5 / 22

Examples, continuedWith more measures, we have the theorem of Kunen:Theorem (Kunen)If there are uncountably many measurable cardinals, then the Axiom ofChoice is false in C.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 6 / 22

Examples, continuedNevertheless, if V = L[U] for a sequence U of measures thenC = L[an iterate of U]({all countable sequences of critical points})= L[({all countable sequences of critical points})HOD C is an iterate of L[U].and if K = L[U] then, by the covering lemma,C = L([ω 2 ] ω )({countable sequences of critical points}).Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 7 / 22

Where does this fail?Woodin gave an upper bound:Theorem (Woodin)If there is a measurable Woodin cardinal, then there is a sharp for theChang model.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 8 / 22

What this meansThere is a closed, unbounded class I of ordinals such that the followingholds: Say B ⊆ I is suitable ifThenB is closed and countable, andevery successor member of B either is a successor point of I or hasuncountable cofinality.If B and B ′ are suitable and have the same length, thenC |= (φ(B) ⇐⇒ φ(B ′ )) for all formulas φ.The theory of C (with parameters for suitable sets) is fixed.There are Skolem functions (not in C) such that C is the hull of Rand [I ]

QuestionQuestion (Woodin)With only measurable cardinals, K HODC is an iterate of K.With a measurable Woodin cardinal, we have a sharp for C.What happens inside this large gap? Does HOD C continue to be aniterate of K C ?When Woodin asked this in a conversation at the Mittag-Leffler Institute,James Cummings and Ralf Schindler stated that arguments of Gitiksuggested that this would break down at o(κ) = κ +ω .I demurred, thinking that this situation was different.I was wrong.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 10 / 22

2011-02-21A Sharp for the Chang ModelIntroducing the SharpQuestionQuestionQuestion (Woodin)With only measurable cardinals, K HODC is an iterate of K.With a measurable Woodin cardinal, we have a sharp for C.What happens inside this large gap? Does HOD C continue to be aniterate of K C ?When Woodin asked this in a conversation at the Mittag-Leffler Institute,James Cummings and Ralf Schindler stated that arguments of Gitiksuggested that this would break down at o(κ) = κ +ω .I demurred, thinking that this situation was different.I was wrong.Note: Woodin observes that with some large cardinal strength we haveHOD C = K C , so it is not necessary to specify K HODC in the next slide. Ido not think that this is true at the level of measurable cardinals.However the main point is that C has all the large cardinal strength of V .

First, Pushing the Lower BoundsAn argument due to Gitik allows extenders of length less than κ +ω tobe reconstructed from their indiscernibles.Hence K HODC is an iterate of K if there is no extender of length κ +ω .This can easily be extended to “no extender of length κ +ω+1 .”This breaks down at an extender of length κ +(ω+1) (calculated in amouse before the extender is added)..Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 11 / 22

A Conjecture. . .ConjectureSuppose that M = L α (R)[E] is the least mouse over the reals such that Ehas a last extender E of length κ +(ω+1) .Then M is a sharp for C.Note that M projects onto R by a Σ 1 function — In particular, if Msatisfies the CH then |M| = ω 1 .Hence all the cardinal calculations will be made in M (0r iterates ofM) below the final extender.It may well be that the extender E should survive for a few levels ofconstruction after E is added.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 12 / 22

A Bit of Explaination . . .0 ♯ , the sharp for L, can be viewed in (at least) three ways:A closed proper class I = { ι n | ν ∈ Ω } of indiscernibles for L, suchthat every member of L is definable with parameters from [I ]

A Bit of Explaination . . .That E is an extender of length (κ +(ω+1) ) M means thatP ω+1 (κ) ⊆ M andUlt(M, E) = { i E (f )(ν) | f ∈ M & ν ∈ [κ, κ +(ω+1) } .Thus, if M Ω = Ult Ω (M, E) then every member of M can be writteni Ω (f )(a) for some f ∈ M anda ∈ ⋃ { [iν (κ), i ν (κ +(ω+1) ) ) }| ν ∈ Ω .Thus, since ω M ⊆ M, the Chang model C can be obtained by addingevery countable subset of ⋃ { [ i ν (κ), i ν (κ +(ω+1) ) ) | ν ∈ Ω } .Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 14 / 22

An Upper boundTheoremSuppose that M = L α (R)[E] is a mouse over the the reals with a lastextender E of length κ +ω 1.Then there is a sharp for the Chang model C.I will try to outline some ideas of the proof.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 15 / 22

For simplicity, we will assume that M |= CH. If not, then we couldbegin with a generic collapse of R onto ω 1 .Let M Ω = Ult Ω (M, E) be the class iteration of M (with the topextender stripped off). Let I be the set of critical points ofi E Ω : M → M Ω.If B is suitable, then let M B ≺ M Ω be the Skolem hull in M Ω ofmembers of (and indiscernibles belonging to members of) B.If δ + 1 = otp(B) and B ′ is the set containing the first δ + 1 membersof I , then M B ′ = Ult δ (M, E) ∼ = M B (with the top extender i δ (E) cutoff).Let C B be the Chang model in the model obtained by closing M Bunder countable sequences.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 16 / 22

Some ObservationsIf B = { ι ξ : ξ ≤ δ } is the first δ + 1 members of I thenM B is transitive.C B = C α where α = Ord(M B ).and C B can be obtained from M B byStart with M B .Add { all further countable threads, } i.e., the sets of the formi −1ξ,δ (ν) | ν < δ & β ∈ a for a ∈ [ [ι, ι +ω 1) ]

i(E)νgeneratorsofi(E)νMitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 18 / 22

More ObservationsProposition[Ω] ω ⊆ ⋃ B C B.C B |= ZF + V = C.If B and B ′ have the same order type, then C B∼ = CB ′.Lemma (Main Lemma)If B ′ ⊂ B then C B ′ ≺ C B .CorollaryC = ⋃ B C B — and all the rest.Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 19 / 22

2011-02-21A Sharp for the Chang ModelA Sketch of the ProofMore ObservationsMore ObservationsProposition[Ω] ω ⊆ ⋃ B CB.CB |= ZF + V = C.If B and B ′ have the same order type, then CB ∼ = CB ′.Lemma (Main Lemma)If B ′ ⊂ B then CB ′ ≺ CB.CorollaryC = ⋃ BCB — and all the rest.The crucial point is that this consideration allows us to deal directly onlywith countable iterations. This is important because the forcing Idescribe only works for countably many indiscernibles.

Proof of Main LemmaWe can assume B is the first δ members of I for some countable δ:B = { ι ν | ν < δ }B ′ = { ι σ(ν) | ν < δ ′ } .Where σ : δ ′ → δ is continuous, strictly increasing, and maps successorordinals to sucessor ordinals.We want to define a forcing P in M B , and a M B -generic set G ∈ V , sothatC M B[G] = C B .Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 20 / 22

Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 21 / 22

2011-02-21A Sharp for the Chang ModelA Sketch of the ProofPerhaps some more explaination of this diagram should be given. Thepoint is that we want to take threads from the actual iteration (labeledat β ′ in the picture and introduce them into the forcing. We can’t do sodirectly—for one thing, E is not in the model M B and so can’t be used inthe forcing. We simply assign the thread to an arbitrary ordinal β lessthan κ + . However M B does require some guidance for its forcing. Wefollow Gitik in having M B regard the νth element of the thread asassigned to an ordinal β ′′ in a model N ν ≺ M B , using the extenderE ν = E ↾ N ν (which is a member of M B ). However (still following Gitik)it is necessary to introduce an ambiguity since β ′′ can’t be fully like β ′ .This is done by using an equivalence relation ↔ on the conditions.Note that 〈 N ν | ν < δ 〉 is a member of M B , but 〈 N ν | ν < ω 1 〉 isnot—indeed its union is probably all of M B .

We start by defining a chain 〈 N ν | ν ∈ ω 1 〉 in V withN ν ′ ≺ ? N ν ≺ ? (M, E)κ +ν ⊆ N νN ν ′ ⊆ N ν⃗N ↾ γ ∈ N γ for each γ < ω 1 .⃗N ↾ γ ∈ M for each γ < ω 1 .Mitchell (University of Florida) A Sharp for the Chang Model Harvard MAMLS 22 / 22