Andr es Eduardo Caicedo Joint work with Martin Zeman · Downward transference of mice Andr es...

Downward transference of mice

Andres Eduardo CaicedoJoint work with Martin Zeman

Department of MathematicsBoise State University

BEST XIXBoise, March 27–29, 2010

Caicedo Downward transference of mice

Introduction

The results in this talk are part of a program whose goal is tostudy the structure of (not necessarily fine structural) inner modelsM of the set theoretic universe V , perhaps in the presence ofadditional axioms, assuming that there is agreement between(some of) the cardinals of V and of M . The idea is that thisagreement measures closeness of M to V .


Introduction

Although our results do not concern forcing axioms, they weremotivated by a problem in the theory of forcing axioms.

Theorem (Velickovic)

Assume MM and that M is an inner model that computes ω2

correctly. Then R ⊂M .

The arguments I know actually use the weak reflection principle, aconsequence of both MM and PFA+, but not of PFA.


Introduction

Although our results do not concern forcing axioms, they weremotivated by a problem in the theory of forcing axioms.

Theorem (Velickovic)

Assume MM and that M is an inner model that computes ω2

correctly. Then R ⊂M .

The arguments I know actually use the weak reflection principle, aconsequence of both MM and PFA+, but not of PFA.


Introduction

It is natural to wonder whether MM can be replaced with theweaker PFA in the theorem above. There is a partial result in thisdirection:

Theorem (C.-Velickovic)

If M is an inner model that computes ω2 correctly, and both Vand M satisfy BPFA, then R ⊂M.

Question

Assume PFA and let M be an inner model with the same ω2. DoesR ⊂M?


Introduction




Question



Introduction




Question



Introduction

It is perhaps too optimistic to expect that this is the case ingeneral. However, it does not seem completely unreasonable thatin the setting of the question, the reals of V \M would be genericin some sense.

A possible way of formalizing this intuition consists in trying toshow that the large cardinal strength coded by reals in V is alsopresent in M .

When approaching this problem, we realized that there may besome ZFC results that the presence of forcing axioms could behiding.


Examples

Here are some examples illustrating how strength transfersdownward to inner models:

Assume that M is an inner model such that CARM = CAR,and that PFA holds. Then ADL(R) holds in M and in all itsset generic extensions, as shown by Steel.The point here is that ADL(R) holds in any generic extensionof the universe by a forcing of size at most κ, whenever �κ

fails for κ a strong limit singular cardinal.But if PFA holds, �κ fails for all uncountable κ. Since �κ

relativizes up to any outer model where κ and κ+ are stillcardinals, it follows that it fails in M as well.


Examples






Examples






Examples






Examples

Suppose that 0 |•

exists, and that M is an inner model withωM3 = ω3. Then 0 |

• ∈M.To see this, recall the following key property of K:

Theorem (Schindler)

Assume that 0 |•

does not exist. Let β ≥ ω2. Then cf(β+K) ≥ |β|.

If 0 |•/∈M , then M |= 0 |

•does not exist, and weak covering

holds in M with respect to KM . Thus

cfM (ω+KM

2 ) ≥ ω2.

However, cf(ω+KM

2 ) = ω, since 0 |•

exists. Since

cfM (ω+KM

2 ) ≤ ωM3 , this is a contradiction.


Examples

Suppose that 0 |•



Theorem (Schindler)

Assume that 0 |•


If 0 |•/∈M , then M |= 0 |



cfM (ω+KM

2 ) ≥ ω2.

However, cf(ω+KM

2 ) = ω, since 0 |•

exists. Since

cfM (ω+KM



Examples

Suppose that 0 |•



Theorem (Schindler)

Assume that 0 |•


If 0 |•/∈M , then M |= 0 |



cfM (ω+KM

2 ) ≥ ω2.

However, cf(ω+KM

2 ) = ω, since 0 |•

exists. Since

cfM (ω+KM



Examples

Suppose that 0 |•



Theorem (Schindler)

Assume that 0 |•


If 0 |•/∈M , then M |= 0 |



cfM (ω+KM

2 ) ≥ ω2.

However, cf(ω+KM

2 ) = ω, since 0 |•

exists. Since

cfM (ω+KM



Examples

If PFA holds and M is an inner model that satisfies BPFA andcomputes ω2 correctly, then H(ω2) ⊂M so, by recent resultsof Schindler, M is closed under any mouse operator that doesnot go beyond M ]

1.

The moral here is that it seems that we only need to assume alocal version of agreement of cardinals rather than a global one toconclude that an inner model M must absorb a significant amountof the large cardinal strength present in V.


Mice

However, for anything like the argument with 0 |•

to hold, we needto be able to invoke weak covering, which seems to require at leastagreement of cardinals up to ω3.

It is then interesting to note that if M computes ω2 correctly and0] exists, then 0] ∈M . This fact is probably due to Friedman.

We have started a systematic approach to the question of how farthis fact can be generalized. Using completely different techniquesfrom those of Friedman, we reprove the fact, and our methodallows us to show that the same conclusion can be obtained with0] replaced by stronger mice.

Conjecture

Let r be a 1-small sound (iterable) mouse that projects to ω.Assume that M is an inner model and that ωM2 = ω2. Then r∈M .


Mice





Conjecture



Mice





Conjecture



Mice

Our result is that this is indeed the case under some anti-largecardinal assumptions. For example:

Theorem

Let M is an inner model such that ωM2 = ω2. Then either there isan inner model with a Woodin cardinal, or else KM |ωM1 = K|ω1.


Mice

For countable mice not projecting to ω, one cannot expect that thetheorem will go through.

This is because, using a construction due to Jensen, one canproduce models M ⊂ V with the same ω2 and the same K, and acountable mouse in V \M that iterates to K‖ω1 .

However, the argument for our theorem shows:

Theorem

Let M is an inner model such that ωM2 = ω2. Then KM iteratespast every countable sound mouse not past 0P in the mouse order.


Mice




Theorem



Mice




Theorem



Mice

Although this does not literally give us that every countable soundmouse below 0P is in M , which is actually impossible as Jensen’sexample illustrates, in fact KM captures all the large cardinalstrength coded by reals.

I was partially supported by the NSF through grant DMS-0801189.Martin Zeman was partially supported by the NSF through grantDMS-0500799. We want to thank Matt Foreman and Uri Abrahamfor personal communications. I also want to thank Martin’s familyfor being such wonderful hosts during my visit to UCI in Summer2009, when several of these results were obtained.


Mice

Although this does not literally give us that every countable soundmouse below 0P is in M , which is actually impossible as Jensen’sexample illustrates, in fact KM captures all the large cardinalstrength coded by reals.

I was partially supported by the NSF through grant DMS-0801189.Martin Zeman was partially supported by the NSF through grantDMS-0500799. We want to thank Matt Foreman and Uri Abrahamfor personal communications. I also want to thank Martin’s familyfor being such wonderful hosts during my visit to UCI in Summer2009, when several of these results were obtained.


Forcing the same ω1

It is a folklore result that traces back at least to Hjorth’s thesisthat not much can be concluded about the reals of an inner modelM if only ωM1 = ω1 is assumed.

Fact

If 0] exists, there exists an inner model N which is a set forcingextension of L and computes ω1 correctly. Thus 0] /∈ N.



It is a folklore result that traces back at least to Hjorth’s thesisthat not much can be concluded about the reals of an inner modelM if only ωM1 = ω1 is assumed.

Fact

If 0] exists, there exists an inner model N which is a set forcingextension of L and computes ω1 correctly. Thus 0] /∈ N.



Proof.

Use the first ω1 L-indiscernibles (ια : α ≤ ωV1 ), to guide aninductive construction of a Coll(ω,< ωV1 )-generic over L asfollows:

Only ω many dense sets need to be met to extend aColl(ω,< ια)-generic Gα to a Coll(ω,< ια+1)-generic Gα+1

whenever α is countable.

Since Coll(ω,< κ) is κ-cc for κ regular and theL-indiscernibles form a club, it follows that

⋃α<κGα is

Coll(ω,< ικ)-generic for κ ≤ ω1 limit.

This gives the result. q



Proof.





⋃α<κGα is





Proof.





⋃α<κGα is





Proof.





⋃α<κGα is




0]

First, we show that if 0] exists, and M is an inner model withωM2 = ω2, then it belongs to M . We actually prove somethingmore general.

Lemma (Shelah)

Let M be an inner model, and suppose that ωM2 = ω2. ThenPω1(ω2) ∩M is stationary.

Proof.

Let F : [ω2]<ω → ω2, and pick γ < ω2 of size ω1 and closed underF. Since M computes ω2 correctly, it sees a bijection between ωV1and γ. Then club many countable subsets of γ are both in M andclosed under F. q


0]


Lemma (Shelah)


Proof.



0]


Lemma (Shelah)


Proof.



0]

Theorem

Let M be an inner model, and suppose that Pω1(ω2) ∩M isstationary. Then 0] ∈M.

We sketch the proof.

Assume P is the (unique, sound) mouse representing 0], so

P = (Jτ ,∈,U)

where U is an amenable measure on the largest cardinal of Jτ , and%1P = ω. Let θ be regular and large. By stationarity ofPω1(ω2) ∩M, we can find some countable X ≺ Hθ such that:

1 P ∈X, and

2 X ∩ ω2 ∈M .


0]

Theorem




P = (Jτ ,∈,U)


1 P ∈X, and

2 X ∩ ω2 ∈M .


0]

Theorem




P = (Jτ ,∈,U)


1 P ∈X, and

2 X ∩ ω2 ∈M .


0]

Theorem




P = (Jτ ,∈,U)


1 P ∈X, and

2 X ∩ ω2 ∈M .


0]

Theorem




P = (Jτ ,∈,U)


1 P ∈X, and

2 X ∩ ω2 ∈M .


Proof (continued)

Let κ = X ∩ ω1. Let H be the transitive collapse of X and

σ : H → Hθ

be the inverse of the collapsing map. Then the critical point of σ isκ, σ(κ) = ω1, and P ∈H.

Let P ′ be the κ-th iterate of P, formed by applying the ultrapowerconstruction using U and its images under the correspondingembeddings. Then P ′ ∈H and also

P(κ) ∩ L ∈H ∩ L;

note that P ′ = (Jτ ′ ,U ′), where τ ′ is the cardinal successor ofcp(U ′) in L.


Proof (continued)


σ : H → Hθ



P(κ) ∩ L ∈H ∩ L;



Proof (continued)


σ : H → Hθ



P(κ) ∩ L ∈H ∩ L;



Proof (continued)


σ : H → Hθ



P(κ) ∩ L ∈H ∩ L;



Proof (continued)


σ : H → Hθ



P(κ) ∩ L ∈H ∩ L;



Proof (continued)

One can then check that σ � Jτ ′ ∈M and

σ � P ′ = πκ,ω1 ,

where the map on the right side is the iteration map.

Hence U ′ is the L-measure on κ derived from σ � Jτ ′ . It followsthat U ′ ∈M. But then P ∈M. q

Corollary

Assume 0] exists and M is an inner model such that ωM2 = ω2.Then 0] ∈M . q


Proof (continued)


σ � P ′ = πκ,ω1 ,



Corollary



Proof (continued)


σ � P ′ = πκ,ω1 ,



Corollary



0†

We briefly sketch how to show that if 0† exists, and M is an innermodel with ωM2 = ω2, then 0† ∈M . This case indicates in somedetail our general approach, while avoiding many of the finestructural issues that are present in the general situation.

Theorem

Assume that 0† exists. Let M be an inner model such thatωM2 = ω2. Then 0† ∈M .


Proof

The idea is to coiterate KM against 0†. If 0† /∈M , then it winsthe coiteration. Notice that we can carry out the iteration, sinceKM is iterable in M , but iterability is Π1

2 and therefore absolute.

First, one shows that KM has a measurable cardinal κ < ωV1 .

Then, a simplified version of this argument allows us now toconclude that in fact 0† is indeed in M .


Proof






Proof






Proof (continued)

For our purposes, 0† is a sound mouse of the form

N = (JEτ ,∈,U)

with %1N = ω, where, as before, U is an amenable measure on the

largest cardinal of JEτ . Here, E is a coherent sequence of (partial)measures and

JEτ |= There is exactly one measurable cardinal.

Assume 0† /∈M and there are no measurable cardinals in KM . Wecoiterate KM vs. N = 0† for ω1 many stages. Since KM = L[EM ]has only partial measures, it does not move in the coiteration.


Proof (continued)

For our purposes, 0† is a sound mouse of the form

N = (JEτ ,∈,U)

with %1N = ω, where, as before, U is an amenable measure on the

largest cardinal of JEτ . Here, E is a coherent sequence of (partial)measures and

JEτ |= There is exactly one measurable cardinal.

Assume 0† /∈M and there are no measurable cardinals in KM . Wecoiterate KM vs. N = 0† for ω1 many stages. Since KM = L[EM ]has only partial measures, it does not move in the coiteration.


Proof (continued)

By Fodor’s lemma and normality, the same critical point (in fact,the same measure) is iterated on a club. Let

Nα = (JEα

τα ,∈,Uα)

be the αth model on the N -side, let κα be the αth critical point ofthe iteration, and let να = (κ+

α )Nα , so

Eα � να = EM � να.

We try to reconstruct some πα,β : Nα → Nβ inside M . We haveνω1 < ωV2 = ωM2 , so we can fix in M a surjection f : ω1 → νω1 .Note that

C = {α < ω1 : f ′′να = ran(πα,ω1) ∩ νω1}

is a club.


Proof (continued)


Nα = (JEα

τα ,∈,Uα)


α )Nα , so



C = {α < ω1 : f ′′να = ran(πα,ω1) ∩ νω1}

is a club.


Proof (continued)


Nα = (JEα

τα ,∈,Uα)


α )Nα , so



C = {α < ω1 : f ′′να = ran(πα,ω1) ∩ νω1}

is a club.


Proof (continued)


Nα = (JEα

τα ,∈,Uα)


α )Nα , so



C = {α < ω1 : f ′′να = ran(πα,ω1) ∩ νω1}

is a club.


Proof (continued)


Nα = (JEα

τα ,∈,Uα)


α )Nα , so



C = {α < ω1 : f ′′να = ran(πα,ω1) ∩ νω1}

is a club.


Proof (continued)

If α < β are in C, then f ′′να and f ′′νβ are in M , and therefore sois

πα,β : JEα

να → JEβ

νβ,

since πα,β � να : να → νβ is the collapse of the inclusion map, andcompletely determines πα,β.

Now notice that

Eανα = {x ∈ JEανα ∩ P(κα) : κα ∈ πα,β(x)}

so Eανα ∈M . We also have that

(JEM

να ,∈, Eανα) = Nα‖να

is a premouse.


Proof (continued)


πα,β : JEα

να → JEβ

νβ,


Now notice that



(JEM


is a premouse.


Proof (continued)


πα,β : JEα

να → JEβ

νβ,


Now notice that



(JEM


is a premouse.


Proof (continued)

Since KM does not move during the coiteration, it is easy to seethat

ult(KM , Eανα)

is well-founded.

This is a contradiction: It follows from standard arguments thatEανα = EK

M

να . But then, by definition, Eανα would not have beenthe measure used on the N -side at the αth stage.

We conclude that KM has a measurable cardinal, and the sameargument gives us that it is below ωV1 . A similar argument allowsus to conclude that 0† ∈M . q


Proof (continued)


ult(KM , Eανα)

is well-founded.


M




Proof (continued)


ult(KM , Eανα)

is well-founded.


M




Proof (continued)


ult(KM , Eανα)

is well-founded.


M




Pω1(ω2)

We now sketch how to replace the assumption that ωM2 = ω2 withthe weaker assumption that M ∩ Pω1(ω2) is stationary.

This follows an unpublished argument of Abraham (pointed out tous by Matt Foreman).

Theorem

Let M be an inner model such that M ∩ Pω1(ω2) is stationary.Then there is an ω2-preserving forcing extension of V by a forcing

in M that adds a generic set G such that ωM [G]2 = ω

V [G]2 = ωV2 .


Pω1(ω2)

We now sketch how to replace the assumption that ωM2 = ω2 withthe weaker assumption that M ∩ Pω1(ω2) is stationary.

This follows an unpublished argument of Abraham (pointed out tous by Matt Foreman).

Theorem

Let M be an inner model such that M ∩ Pω1(ω2) is stationary.Then there is an ω2-preserving forcing extension of V by a forcing

in M that adds a generic set G such that ωM [G]2 = ω

V [G]2 = ωV2 .


Proof sketch

Let g be V -generic over Coll(ω,< ω1).

This forcing is equivalent to Add(ω, ω1), so we can identify g witha sequence

〈rα : α < ω1〉

of mutually generic Cohen reals.

Let P = Coll(ω,< ω1) ∗ Coll(ω1, < ω2)M [g].

We claim that P is the desired forcing notion. The key issue is thatω1 is preserved, from this the result follows easily.

To prove the key issue, let θ be regular and sufficiently large, andpick a countable N ≺ H(θ) that contains all the relevant sets andsuch that N ∩ ω2 ∈M ; this is possible by our assumption on M .Then N [g] ≺ H(θ)V [g], and its transitive collapse has the formN [〈rα : α < δ〉] where N is the collapse of N and δ = N ∩ ω1.


Proof sketch



〈rα : α < ω1〉






Proof sketch



〈rα : α < ω1〉






Proof sketch



〈rα : α < ω1〉






Proof sketch



〈rα : α < ω1〉






Proof sketch



〈rα : α < ω1〉






Proof sketch (continued)

Let τ = ωN2 . The collapse P of P inside N is contained in

Coll(ω,< δ) ∗ Coll(δ,< τ)M [〈rα : α<δ〉].

The point is that rδ is a genuine generic over V [〈rα : α < δ〉].Therefore we can easily use it to guide the construction of aP-generic over N [〈rα : α < δ〉], and so (via the inverse of thecollapse), of an N -generic P-condition, from which preservation ofω1 follows. q












Questions

It is natural to ask how far our result can be generalized. There iscurrently a scenario that would allow us to show that M ]

1 transfersdownward and, in fact, PD does.

Does ADL(R) transfer downward, perhaps under the additionalassumption of CH?


Questions





Questions





Questions

In a different direction, we mentioned earlier that if PFA holds andM is an inner model with the same cardinals, then ADL(R) holds inall set generic extensions of M . One should actually get muchstronger results from significantly weaker hypothesis. For example,Jensen, Schimmerling, Schindler, and Steel showed that if2ℵ0 ≤ ℵ2 and both �(ℵ3) and �ℵ3 fail (these are consequences ofPFA), then there is a sharp for a proper class model with a properclass of strong cardinals and a proper class of Woodin cardinals.

Assume PFA and let M be an inner model with the same ω4.Does the conclusion of the Jensen et al. theorem hold in M?

The issue is that preservation of cardinals is not enough to ensurethat �(ℵ3) fails in M .


Questions





Questions





Questions





Questions





(The Mouse’s Tale, illustration by John Tenniel for Alice’sAdventures in Wonderland.)

The end.


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