the evolution of hod mice - harvard universitylogic.harvard.edu/mamls 2011 - grigor sargsyan.pdf ·...

The evolution of hod mice


Grigor Sargsyan

UCLA

Harvard MamlsFebruary 20, 2011

Cambridge, Massachusetts

The evolution of hod mice Grigor Sargsyan


The beginnings

CH in HOD

Theorem (Harrington-Kechris)Assume V = L(R) + AD. Then HOD � CH.

Question: Assume AD+ + V = L(P(R)). Does HOD � GCH?



The beginnings

Mice and HOD

LetMω(y) be the minimal proper class y -mouse with ω Woodincardinals.

Theorem (Woodin)IfMω exists then AD holds in L(R).

Theorem (Steel-Woodin)AssumeMω exists. Then L(R) � x ∈ OD(y) iff x ∈Mω(y).



The beginnings

Mice and HOD cont.

CorollaryAssumeMω-exists. Let κ be the least measurable ofMω andlet µ be the measure on κ inMω. Let N be the result ofiterating µ through the ordinals. Then

HOD|ω1 = N|ω1.

Question: Assume AD+ + V = L(P(R)). Is HOD a mouse?



The beginnings

Capturing by mice

Recall,

Theorem (Steel-Woodin)AssumeMω exists. Then L(R) � x ∈ OD(y) iff x ∈Mω(y).

Question: Assume AD+ + V = L(P(R)). Is it true thatwhenever x , y ∈ R, then x ∈ OD(y) iff x is in a y -mouse?

Conjecture (The Mouse Set Conjecture, MSC)Assume AD+ and that there is no inner model with asuperstrong cardinal. Then for x , y ∈ R,

x ∈ OD(y) iff x is in a y-mouse.



The beginnings

Summary

Assume AD+ + V = L(P(R)).1 Does HOD � GCH?2 Is HOD a mouse?3 Is MSC true?



The beginnings

An answer to 2

TheoremAssume V = L(R) + AD.

1 (Steel) V HODΘ is a mouse.

2 (Woodin) HOD isn’t a mouse, it is a hybrid mouse. Moreprecisely, it has a fragment of the strategy of V HODΘ on itssequence.



The beginnings

An answer to 2

TheoremAssume V = L(R) + AD.

1 (Steel) V HODΘ is a mouse.2 (Woodin) HOD isn’t a mouse, it is a hybrid mouse. More

precisely, it has a fragment of the strategy of V HODΘ on itssequence.



Hod mice

Hod mice

1 Hod mice were invented to answer Question 2 but they areuseful in answering all three questions and have otherapplications.

2 Hod mice besides having extender sequence are alsoclosed under the iteration strategies of their initialsegments.

3 η is a layer of a hod mouse P if P is closed under theiteration strategy of P|η.

4 All hod mice have a largest layer for which no strategy isadded.

5 For “simple” hod mice, the only layers are its Woodincardinals and the limit of Woodin cardinals.



Hod mice

P, δ+ωω = µω

δω,Σ


Hod mice

The Solovay hierarchy

The hierarchy of hod mice grows according to the Solovayhierarchy.

First, recall that assuming AD,

Θ = sup{α : there is a surjection f : R→ α}.

Then, assuming AD, the Solovay sequence is a closedsequence of ordinals 〈θα : α ≤ Ω〉 defined by:

1 θ0 = sup{α : there is an OD surjection f : P(ω)→ α},2 If θα < Θ then θα+1 = sup{β : there is an OD surjection

f : P(θα)→ β},3 θλ = supα


Hod mice


The hierarchy of hod mice grows according to the Solovayhierarchy.First, recall that assuming AD,






Hod mice





1 θ0 = sup{α : there is an OD surjection f : P(ω)→ α},

2 If θα < Θ then θα+1 = sup{β : there is an OD surjectionf : P(θα)→ β},

3 θλ = supα


Hod mice






f : P(θα)→ β},

3 θλ = supα


Hod mice








Hod mice


AD+ + Θ = θ0


Hod mice

The structure of HOD

The motivation behind hod mice is the following theorem.

Theorem (Woodin)Assume AD+ and suppose θα+1 ≤ Θ. Then HOD � “θα+1 isWoodin”.



Hod mice

Important limit points of the Solovay hierarchy

1 ADR + “Θ is regular”.

2 The largest Suslin cardinal is a theta.



Hod mice

Important limit points of the Solovay hierarchy

1 ADR + “Θ is regular”.2 The largest Suslin cardinal is a theta.



Hod mice below ADR + “Θ is regular”

ADR + “Θ is regular” and the Solovay hierarchy

Theorem (Woodin)Assume AD+ + V = L(P(R)). Then ADR is equivalent toAD+ + “Θ = θα for some limit α”.

Hence, ADR + “Θ is regular”is equivalent to AD+ + “Θ = θΘ and Θ is regular”




ADR + “Θ is regular” and the Solovay hierarchy

Theorem (Woodin)Assume AD+ + V = L(P(R)). Then ADR is equivalent toAD+ + “Θ = θα for some limit α”. Hence, ADR + “Θ is regular”is equivalent to AD+ + “Θ = θΘ and Θ is regular”





Q is a shortening of a hod premouse if there is a hod premouseP such that Q = P|η where η is the largest layer of P.

Theorem (S.)Assume AD+ + V = L(P(R)) and that there is no inner modelof ADR + “Θ is regular”. Then

1 V HODΘ is a shortening of a hod premouse.2 HOD � GCH.3 MSC holds.







1 V HODΘ is a shortening of a hod premouse.

2 HOD � GCH.3 MSC holds.







1 V HODΘ is a shortening of a hod premouse.2 HOD � GCH.

3 MSC holds.







1 V HODΘ is a shortening of a hod premouse.2 HOD � GCH.3 MSC holds.




Large cardinals and ADR + “Θ is regular”

Theorem (S.)Suppose there is a Woodin limit of Woodins with a measurableabove. Then there is a proper class inner model containing thereals and satisfying ADR + “Θ is regular.

Theorem (Steel, building on a prior work of Woodin)Assume ADR. Then there is an inner model in which there is λwhich is a limit of Woodin cardinals and < λ-strong cardinals.




Large cardinals and ADR + “Θ is regular” cont.

Remark: One should be able to get many more Woodins.

However, the exact large cardinal corresponding to ADR + “Θ isregular” isn’t known and has been rather mysterious. Perhapsthat isn’t even the right question to ask.Question: Is every model of AD+ a derived model of a mouse?





Remark: One should be able to get many more Woodins.However, the exact large cardinal corresponding to ADR + “Θ isregular” isn’t known and has been rather mysterious.

Perhapsthat isn’t even the right question to ask.Question: Is every model of AD+ a derived model of a mouse?





Remark: One should be able to get many more Woodins.However, the exact large cardinal corresponding to ADR + “Θ isregular” isn’t known and has been rather mysterious. Perhapsthat isn’t even the right question to ask.

Question: Is every model of AD+ a derived model of a mouse?





Remark: One should be able to get many more Woodins.However, the exact large cardinal corresponding to ADR + “Θ isregular” isn’t known and has been rather mysterious. Perhapsthat isn’t even the right question to ask.Question: Is every model of AD+ a derived model of a mouse?




One application

Theorem (S.)Suppose ¬�κ for some singular strong limit κ. Then there is aproper class model containing R and satisfying ADR + “Θ isregular”.




HOD below ADR + “Θ is regular”

Theorem (S.)Assume AD+ + V = L(P(R)) and that there is no inner modelof ADR + “Θ is regular”. Then V HODΘ is a shortening of a hodpremouse.




The proof

Comparison of hod mice involves comparing strategies.

We say (P,Σ) is a hod pair if P is a hod premouse and Σ is itsstrategy.

DefinitionSuppose (P,Σ) and (Q,Λ) are two hod pairs. Then comparisonholds for (P,Σ) and (Q,Λ) if there are (R,Ψ) and (S,Φ) suchthat

1 R and S are respectively Σ-iterate of P and a Λ-iterate ofQ,

2 Ψ and Φ are the corresponding tails of respectively Σ andΛ.

3 R Ehod S and ΦR = Ψ or S Ehod R and ΨS = Φ.




The proof

Comparison of hod mice involves comparing strategies.We say (P,Σ) is a hod pair if P is a hod premouse and Σ is itsstrategy.

DefinitionSuppose (P,Σ) and (Q,Λ) are two hod pairs. Then comparisonholds for (P,Σ) and (Q,Λ) if there are (R,Ψ) and (S,Φ) suchthat

1 R and S are respectively Σ-iterate of P and a Λ-iterate ofQ,

2 Ψ and Φ are the corresponding tails of respectively Σ andΛ.

3 R Ehod S and ΦR = Ψ or S Ehod R and ΨS = Φ.




Directed system associated to hod pairs

Suppose (P,Σ) is a hod pair. Let

F = {Q : Q is a Σ-iterate of P}.

Define ≤ on F by Q ≤ R iff R is a ΣQ-iterate of Q. LetiQ,R : Q → R be the iteration embedding.Let

M∞(P,Σ) = dirlim(F ,≤)

under iQ,R’s.







Define ≤ on F by Q ≤ R iff R is a ΣQ-iterate of Q. LetiQ,R : Q → R be the iteration embedding.

Let


under iQ,R’s.







Define ≤ on F by Q ≤ R iff R is a ΣQ-iterate of Q. LetiQ,R : Q → R be the iteration embedding.Let


under iQ,R’s.




Proof cont.

One then shows that for every α such that θα < Θ, there is ahod pair (P,Σ) such that

V HODθα =M∞(P,Σ)|θα.

Remark: It follows from comparison thatM∞(P,Σ) isindependent of (P,Σ).




Largest Suslin is a theta

A ⊆ R is κ-Suslin if for some tree T on κ, A = p[T ].

κ is a Suslin cardinal if there is A ⊆ R which is κ-Suslin but notλ-Suslin for all λ < κ.LST: AD+ + Θ = θα+1 + “θα is the largest Suslin cardinal belowΘ”.

FactLST → Con(ADR + “Θ is regular”)





A ⊆ R is κ-Suslin if for some tree T on κ, A = p[T ].κ is a Suslin cardinal if there is A ⊆ R which is κ-Suslin but notλ-Suslin for all λ < κ.

LST: AD+ + Θ = θα+1 + “θα is the largest Suslin cardinal belowΘ”.






A ⊆ R is κ-Suslin if for some tree T on κ, A = p[T ].κ is a Suslin cardinal if there is A ⊆ R which is κ-Suslin but notλ-Suslin for all λ < κ.LST: AD+ + Θ = θα+1 + “θα is the largest Suslin cardinal belowΘ”.




Hod mice below LST

Hod mice below LST

Theorem (S-Steel)Assume AD+ + V = L(P(R)) and that there is no proper classinner model containing the reals and satisfying LST . Then

1 V HODΘ is a shortening of a hod premouse.2 HOD � GCH3 MSC holds.



Hod mice below LST

Large cardinals and LST

Theorem (S.-Steel)Suppose there is a Woodin limit of Woodins and a measurablecardinal above. Then there is a hod mouse satisfying LST .

ConjectureLST is equiconsistent with Woodin limit of Woodins.

The conjecture is important as it will probably lead to showingthat Con(¬�κ for singular strong limit κ) is stronger thenCon(Woodin limit of Woodins).



Beyond LST

Two limitations

Theorem (Woodin)Assume AD+ + V = L(P(R)).

1 If Θ = θα for some limit α then Θ isn’t Woodin in HOD.

2 If θα+2 ≤ Θ and θα is the largest Suslin cardinal below θα+1then, in HOD, θα isn’t κ-strong where κ is the least Suslinabove θα+1.



Beyond LST

Two limitations

Theorem (Woodin)Assume AD+ + V = L(P(R)).

1 If Θ = θα for some limit α then Θ isn’t Woodin in HOD.2 If θα+2 ≤ Θ and θα is the largest Suslin cardinal below θα+1

then, in HOD, θα isn’t κ-strong where κ is the least Suslinabove θα+1.



Beyond LST

First simple case

P is 1-overlapped LST if1 P has a largest Woodin cardinal δ,

2 if κ is the least cardinal which is < δ-strong then κ is a limitof Woodins,

3 δ is only overlapped by extenders with critical point κ,4 For every η < δ if η isn’t a cutpoint then there is Q E P

such that Q � “η isn’t Woodin” and in Q, all the extendersoverlapping η have the same critical point.



Beyond LST

First simple case

P is 1-overlapped LST if1 P has a largest Woodin cardinal δ,2 if κ is the least cardinal which is < δ-strong then κ is a limit

of Woodins,

3 δ is only overlapped by extenders with critical point κ,4 For every η < δ if η isn’t a cutpoint then there is Q E P




Beyond LST

First simple case


of Woodins,3 δ is only overlapped by extenders with critical point κ,

4 For every η < δ if η isn’t a cutpoint then there is Q E Psuch that Q � “η isn’t Woodin” and in Q, all the extendersoverlapping η have the same critical point.



Beyond LST

First simple case


of Woodins,3 δ is only overlapped by extenders with critical point κ,4 For every η < δ if η isn’t a cutpoint then there is Q E P




Beyond LST

Comparison theory for minimally overlapped LST ’s

Theorem (S.)Assume MSC. Then comparison holds for 1-overlapped LSTpairs.

Remark: This isn’t the theorem we want to prove as it assumesMSC. There are also arguments for proving the theoremwithout assuming MSC but again they are not fully worked out.



Beyond LST

How far will these arguments go?

1 They seem to generalize up to a Woodin limit of Woodinsand somewhat beyond.

2 There seem to be much more serious problems afterWoodin limit of Woodins.

3 However, maybe as far as superstrongs are concerned allwe need is the theory up to Woodin limit of Woodins.



Beyond LST

The superstrong question

Question: Assume AD+ and that there is some α such thatHOD � “θα is Woodin limit of Woodins”. Is there an inner modelwith a superstrong?

Remark: In general, a Woodin in a hod mouse is much strongerthan a Woodin in a mouse. Recall ADR-equiconsistency result.



Beyond LST

The superstrong question

Question: Assume AD+ and that there is some α such thatHOD � “θα is Woodin limit of Woodins”. Is there an inner modelwith a superstrong?Remark: In general, a Woodin in a hod mouse is much strongerthan a Woodin in a mouse. Recall ADR-equiconsistency result.



The inner model problem


Inner model problem: Construct mice with large cardinals.

Remark: Neeman solved it for a Woodin limit of Woodins, butthe solution cannot be used for lower bound evaluations.Neeman’s result is the best partial result.






Remark: Neeman solved it for a Woodin limit of Woodins, butthe solution cannot be used for lower bound evaluations.

Neeman’s result is the best partial result.






Remark: Neeman solved it for a Woodin limit of Woodins, butthe solution cannot be used for lower bound evaluations.Neeman’s result is the best partial result.




Large cardinals and the Solovay hierarchy

One approach to IMP has been via descriptive set theory. Oneapproach using DST has been to consider HOD of models ofAD+ + V = L(P(R)) and translate the strategies into extendersto get an equivalent model with large cardinals.

If P is a hodmouse then let PS be this translated mouse.

Theorem (Steel)Assume V � “I am the minimal model of ADR” and letP = HOD. Then PS � ADR-hypo.

For this to work in general, we need that the Solovay hierarchycatches up with the large cardinal hierarchy.





One approach to IMP has been via descriptive set theory. Oneapproach using DST has been to consider HOD of models ofAD+ + V = L(P(R)) and translate the strategies into extendersto get an equivalent model with large cardinals. If P is a hodmouse then let PS be this translated mouse.

Theorem (Steel)Assume V � “I am the minimal model of ADR” and letP = HOD. Then PS � ADR-hypo.

For this to work in general, we need that the Solovay hierarchycatches up with the large cardinal hierarchy.





Is it possible that Large Cardinal hierarchy and the Solovayhierarchy catch up?

Let φ be a large cardinal axiom and let

Sφ =def AD+ + V HODΘ � φ.

ConjectureFor every large cardinal φ, Sφ is consistent relative to somelarge cardinal.

Remark: Even the case of φ =def “δ is Woodin and there is aδ-strong cardinal” is open. This isn’t a superficial way ofstrengthening the Solovay hierarchy. It seems that anyreasonable approach to the problems outlines in this talk willinevitably lead to the study of axiom like Sφ.





Is it possible that Large Cardinal hierarchy and the Solovayhierarchy catch up?Let φ be a large cardinal axiom and let











Remark: Even the case of φ =def “δ is Woodin and there is aδ-strong cardinal” is open.

This isn’t a superficial way ofstrengthening the Solovay hierarchy. It seems that anyreasonable approach to the problems outlines in this talk willinevitably lead to the study of axiom like Sφ.










The beginningsHod miceHod mice below ADR+`` is regular"Hod mice below LSTBeyond LSTThe inner model problem

the evolution of hod mice - harvard universitylogic.harvard.edu/mamls 2011 - grigor sargsyan.pdf ·...

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