Generalizing L

Relativizing L to an arbitrary predicate P

Suppose P is a set. Define Lα[P] by induction on α by:

1. L0[P] = ?,

2. (Successor case) Lα+1[P] = PDef(Lα[P]) ∪ {P ∩ Lα[P]},

3. (Limit case) Lα[P] = S

β<α Lβ[P].

I L[P] is the class of all sets X such that X ∈ Lα[P] for some

ordinal α.

I If P ∩ L ∈ L then L[P] = L.

I L[R] = L versus L(R) which is not L unless R ? L.

Lemma

For every set X, there exists a set P such that X ∈ L[P].

I This is equivalent to the Axiom of Choice.

Normal ultrafilters and L[U]

Definition

Suppose that U is a uniform ultrafilter on δ. Then U is a normal

ultrafilter if for all functions, f : δ → δ, if

I {α < δ f (α) < α} ∈ U,

then for some β < δ,

I {α < δ f (α) = β} ∈ U.

I A normal ultrafilter on δ is necessarily δ-complete.

Theorem (Kunen)

Suppose that δ1 ≤ δ2, U1 is a normal ultrafilter on δ1, and U2 is a

normal ultrafilter on δ2. Then:

I L[U2] ? L[U1]

I If δ1 = δ2 then

I L[U1] = L[U2] and U1 ∩ L[U1] = U2 ∩ L[U2].

I If δ1 < δ2 there is an elementary embedding j : L[U1] → L[U2].

L[U] is a generalization of L

Theorem (Silver)

Suppose that U is a normal ultrafilter on δ. Then in L[U]:

I 2

λ = λ

+ for infinite cardinals λ.

I There is a projective wellordering of the reals.

Theorem (Kunen)

Suppose that U is a normal ultrafilter on δ.

I Then δ is the only measurable cardinal in L[U].

I This generalizes Scott’s Theorem to L[U] and so:

I V 6= L[U].

Weak Extender Models

Theorem

Suppose N is a transitive class, N contains the ordinals, and that

N is a model of ZFC. Then for each cardinal δ the following are

equivalent.

I N is a weak extender model of δ is supercompact.

I For every γ > δ there exists a δ-complete normal fine

ultrafilter U on Pδ(γ) such that

I N ∩ Pδ(γ) ∈ U,

I U ∩ N ∈ N.

I If δ is a supercompact cardinal then V is a weak extender

model of δ is supercompact.

Why weak extender models?

The Basic Thesis

If there is a generalization of L at the level of a supercompact

cardinal then it should exist in a version which is a weak extender

model of δ is supercompact for some δ.

I Suppose U is δ-complete normal fine ultrafilter on Pδ(γ), such

that δ

+ ≤ γ, and such that γ is a regular cardinal. Then:

I L[U] = L.

I Let W be the induced uniform ultrafilter on γ by restricting U

to a set Z on which the “sup function” is 1-to-1. Then:

I L[W ] is a Kunen inner model for 1 measurable cardinal.

Theorem

Suppose N is a weak extender model of δ is supercompact.

I Then:

I N has the δ-approximation property.

I N has the δ-covering property.

Corollary

Suppose N is a weak extender model of δ is supercompact and let

A = N ∩ H(δ

+). Then:

I N ∩ H(γ) is (uniformly) definable in H(γ) from A, for all

strong limit cardinals γ > δ.

I N is Σ2-definable from A.

I The theory of weak extender models for supercompactness is

part of the first order theory of V.

I There is no need to work in a theory with classes.

Weak extender models of δ is supercompact are close to V

above δ

Theorem

Suppose N is a weak extender model of δ is supercompact and

that γ > δ is a singular cardinal. Then:

I γ is a singular cardinal in N.

I γ

+ = (γ

+)

N.

This theorem strongly suggests:

I There can be no generalization of Scott’s Theorem to any

axiom which holds in some weak extender model of δ is

supercompact, for any δ.

I Since a weak extender model of δ is supercompact cannot be

far from V.

The Universality Theorem

I The following theorem is a special case of the Universality

Theorem for weak extender models.

Theorem

Suppose that N is a weak extender model of δ is supercompact,

α > δ is an ordinal, and that

j : N ∩ Vα+1 → N ∩ Vj(α)+1

is an elementary embedding such that δ ≤ CRT(j).

I Then j ∈ N.

I Conclusion: There can be no generalization of Scott’s

Theorem to any axiom which holds in some weak extender

model of δ is supercompact, for any δ.

Large cardinals above δ are downward absolute to weak

extender models of δ is supercompact

Theorem

Suppose that N is a weak extender model of δ is supercompact.

κ > δ,

and that κ is an extendible cardinal.

I Then κ is an extendible cardinal in N.

(sketch) Let A = N ∩ H(δ

+) and fix an elementary embedding

j : Vα+ω → Vj(α)+ω

such that κ < α and such that CRT(j) = κ > δ.

I N ∩ H(γ) is uniformly definable in H(γ) from A for all strong

limit cardinals γ > δ+.

I This implies that j(N ∩ Vα+ω) = N ∩ Vj(α)+ω since j(A) = A.

I Therefore by the Universality Theorem, j|(N ∩ Vα+1) ∈ N.

Magidor’s characterization of supercompactness

Lemma (Magidor)

Suppose that δ is strongly inaccessible. Then the following are

equivalent.

(1) δ is supercompact.

(2) For all λ > δ there exist δ <ˉ λ < δ ˉ and an elementary

embedding

π : Vλˉ+1 → Vλ+1

such that CRT(π) = δˉ and such that π(δˉ) = δ.

Theorem

Suppose that N is a weak extender model of δ is supercompact,

κ > δ, and that κ is supercompact.

I Then N is a weak extender model of κ is supercompact.

Too close to be useful?

I Are weak extender models for supercompactness simply too

close to V to be of any use in the search for generalizations of

L?

Theorem (Kunen)

There is no nontrivial elementary embedding

π : Vλ+2 → Vλ+2.

Theorem

Suppose that N is a weak extender model of δ is supercompact

and λ > δ.

I Then there is no nontrivial elementary embedding

π : N ∩ Vλ+2 → N ∩ Vλ+2

such that CRT(π) ≥ δ.

Perhaps not

I Weak extender models for supercompactness can be

nontrivially far from V in one key sense.

Theorem (Kunen)

The following are equivalent.

1. L is far from V (as in the Jensen Dichotomy Theorem).

2. There is a nontrivial elementary embedding j : L → L.

Theorem

Suppose that δ is a supercompact cardinal.

I Then there exists a weak extender model N for δ is

supercompact such that

I N

ω ? N.

I There is a nontrivial elementary embedding j : N → N.

I This theorem shows that the restriction in the Universality

Theorem on CRT(j) is necessary.

The HOD Dichotomy (full version)

Theorem (HOD Dichotomy Theorem)

Suppose that δ is an extendible cardinal. Then one of the following

holds.

(1) No regular cardinal κ ≥ δ is ω-strongly measurable in HOD.

Further:

I HOD is a weak extender model of δ is supercompact.

(2) Every regular cardinal κ ≥ δ is ω-strongly measurable in HOD.

Further:

I HOD is not a weak extender model of λ is supercompact, for

any λ.

I There is no weak extender model N of λ is supercompact such

that N ? HOD, for any λ.

A unconditional corollary

Theorem

Suppose that δ is an extendible cardinal, κ ≥ δ, and that κ is a

measurable cardinal.

I Then κ is a measurable cardinal in HOD.

Two cases by appealing to the HOD Dichotomy Theorem:

I Case 1: HOD is close to V. Then HOD is a weak extender

model of δ is supercompact.

I Apply (a simpler variation of) the Universality Theorem.

I Case 2: HOD is far from V. Then every regular cardinal

κ ≥ δ is a measurable cardinal in HOD;

I since κ is ω-strongly measurable in HOD.

The axiom V = Ultimate-L

The axiom for V = Ultimate-L

I There is a proper class of Woodin cardinals.

I For each Σ2-sentence ?, if ? holds in V then there is a

universally Baire set A ? R such that

HODL(A,R)

|= ?.

Scott’s Theorem and the rejection of V = L

Theorem (Scott)

Assume V = L. Then there are no measurable cardinals.

The key question

Is there a generalization of Scott’s theorem to the axiom

V = Ultimate-L?

I If so then we must reject the axiom V = Ultimate-L.

V = Ultimate-L and the structure of ?！?/p>

Theorem (V = Ultimate-L)

For each x ∈ R, there exists a universally Baire set A ? R such

that

x ∈ HODL(A,R)

.

I Assume there is a proper class of Woodin cardinals and that

for each x ∈ R there exists a universally Baire set A ? R such

that x ∈ HODL(A,R)

.

I This is in general yields the simplest possible wellordering of

the reals.

I It implies R ? HOD.

Question

Does some large cardinal hypothesis imply that there must exist

x ∈ R such that

x ∈/ HODL(A,R)

for any universally Baire set?

V = Ultimate-L and the structure of ?！?/p>

Lemma

Suppose that there is a proper class of Woodin cardinals and that

A, B ∈ P(R) are each universally Baire. Then the following are

equivalent.

(1) L(A, R) ? L(B, R).

(2) ΘL(A,R) ≤ ΘL(B,R)

.

Corollary

Suppose that there is a proper class of Woodin cardinals and that

A ? R is universally Baire. Then

HODL(A,R) ? HOD.

Corollary (V = Ultimate-L)

Let ?！?be the set of all universally Baire sets A ? R.

I Then ?！?6= P(R) ∩ L(?！? R).

Projective Sealing Theorems

Theorem (Unconditional Projective Sealing)

Suppose that there is a proper class of Woodin cardinals and that

V[G] is a generic extension of V.

I Then Vω+1 ? V[G]ω+1.

I Suppose Vω+1 ? V[G]ω+1 for generic extensions of V. Then

there is no projective wellordering of the reals.

Theorem (Martin-Steel)

Suppose there are infinitely many Woodin cardinals. Then for each

n < ω there exists a model M such that:

(1) M |= ZFC + “There exist n-many Woodin cardinals”.

(2) M |= ZFC + “There is a projective wellordering of the reals”.

Strong cardinals and conditional projective sealing

Suppose δ is a Woodin cardinal. Then:

I Vδ |= ZFC + “There is a proper class of strong cardinals”

Thus:

I ZFC + “There is a proper class of strong cardinals” cannot

prove projective sealing.

Theorem (Conditional Projective Sealing)

Suppose that δ is a limit of strong cardinals and V[G] is a generic

extension of V in which δ is countable.

Suppose V[H] is a generic extension of V[G].

I Then V[G]ω+1 ? V[H]ω+1.

I Thus after collapsing a limit of strong cardinals to be

countable, one obtains projective sealing.

I Can ?！?be sealed?

A Sealing Theorem for Γ∞

Notation

Suppose V[H] is a generic extension of V. Then

I ?！?/p>

H = (Γ∞)

V [H]

I RH = (R)

V [H]

.

Theorem (Conditional ?！?Sealing)

Suppose that δ is a supercompact cardinal and that there is a

proper class of Woodin cardinals.

Suppose that V[G] is a generic extension of V in which (2δ

)

V is

countable.

Suppose that V[H] is a generic extension of V[G].

I Then:

I ?！?/p>

G = P(RG ) ∩ L(?！?/p>

G

, RG ).

I There is an elementary embedding

j : L(?！?/p>

G

, RG ) → L(Γ∞

H

, RH ).

What about an Unconditional ?！?Sealing Theorem?

A natural conjecture

By analogy with the Projective Sealing Theorems, there should be

some large cardinal hypothesis which suffices to prove:

I Unconditional Γ∞ Sealing.

But:

If some large cardinal hypothesis proves that

I ?！?= P(R) ∩ L(Γ∞, R)

then the axiom V = Ultimate-L is false.

I So there are potential paths to generalizing Scott’s Theorem

to the axiom V = Ultimate-L.

I Is there a potential path to showing that there is no

generalization of Scott’s Theorem to the axiom

V = Ultimate-L?

The Ultimate-L Conjecture

Ultimate-L Conjecture

(ZFC) Suppose that δ is an extendible cardinal. Then (provably)

there is a transitive class N such that:

1. N is a weak extender model of δ is supercompact.

2. N |= “V = Ultimate-L”.

I The Ultimate-L Conjecture implies there is no generalization

of Scott’s Theorem to the case of V = Ultimate-L.

I By the Universality Theorem.

I The Ultimate-L Conjecture is a number theoretic statement

I It is an existential statement, so if it is undecidable it must be

false. Therefore:

I It must be either true or false (it cannot be meaningless).

I Just like the HOD Conjecture.

I The Ultimate-L Conjecture implies a slightly weaker version

of the HOD Conjecture.

The summary from Tuesday’s lecture

There is a progression of theorems from large cardinal hypotheses

that suggest:

I Some version of V = L is true.

Further:

I The theorems become much stronger as the large cardinal

hypothesis is increased.

Large cardinals are amplifiers of the structure of V.

A natural conjecture building on this theme

One should be able to augment large cardinal axioms with some

simple consequences of V = Ultimate-L and actually

I recover that V = Ultimate-L,

I laying the foundation for an argument that the axiom

V = Ultimate-L is true.

Close embeddings and finitely generated models

Definition

Suppose that M, N are transitive sets, M |= ZFC, and that

π : M → N

is an elementary embedding. Then π is close to M if for each

X ∈ M and each a ∈ π(X),

{Z ∈ P(X) ∩ M a ∈ π(Z)} ∈ M.

Definition

Suppose that N is a transitive set such that

N |= ZFC + “V = HOD”.

Then N is finitely generated if there exists a ∈ N such that every

element of N is definable from a.

Why close embeddings?

Lemma

Suppose that M, N are transitive sets,

M |= ZFC + “V = HOD”,

and that M is finitely generated.

I Suppose that

I π0 : M → N

I π1 : M → N

are elementary embeddings each of which is close to M.

I Then π0 = π1.

I Without the requirement of closeness, the conclusion that

π0 = π1 can fail.

Weak Comparison

Definition

Suppose that V = HOD. Then Weak Comparison holds if for all

X, Y ?Σ2 V the following hold where MX is the transitive collapse

of X and MY is the transitive collapse of Y .

I Suppose that MX and MY are finitely generated models of

ZFC, MX 6= MY , and

I MX ∩ R = MY ∩ R.

I Then there exist a transitive set M?

, and elementary

embeddings

I πX : MX → M?

I πY : MY → M?

such that πX is close to MX and πY is close to MY .

Why weak comparison?

I By Shoenfield’s Absoluteness Theorem, the conclusion of

Weak Comparison is absolute.

I Weak Comparison holds in the current generation of

generalizations of L.

I Weak Comparison looks difficult to force.

Summary:

I Weak Comparison provides a good test question for

generalizing L to levels of the large cardinal hierarchy.

Question

Assume there is a supercompact cardinal and that V = HOD.

I Can Weak Comparison hold?

I (conjecture) V = Ultimate-L implies Weak Comparison.

Goldberg’s Ultrapower Axiom

Notation

Suppose that N |= ZFC is an inner model of ZFC, U ∈ N and

N |= “U is a countably complete ultrafilter”

I NU denotes the transitive collapse of Ult0(N,U)

I j

N

U

: N → NU denotes the associated ultrapower embedding.

Definition (The Ultrapower Axiom)

Suppose that U and W are countably complete ultrafilters. Then

there exist W ? ∈ VU and U

? ∈ VW such that the following hold.

(1) VU |= “W ?

is a countably complete ultrafilter”.

(2) VW |= “U

?

is a countably complete ultrafilter”.

(3) (VU)W? = (VW )U? .

(4) j

VU

W? ? j

V

U = j

VW

U? ? j

V

W .

I If V = HOD then (3) implies (4).

Weak Comparison and the Ultrapower Axiom

I The Ultrapower Axiom simply asserts that amalgamation

holds for the ultrapowers of V by countably complete

ultrafilters.

I If there are no measurable cardinals then the Ultrapower

Axiom holds trivially

I since every countably complete ultrafilter is principal.

Theorem (Goldberg)

Suppose that V = HOD and that there exists

X ?Σ2 V

such that MX |= ZFC where MX is the transitive collapse of X.

Suppose that Weak Comparison holds.

I Then the Ultrapower Axiom holds.

I If X does not exist then Weak Comparison holds vacuously.

I If there is a supercompact cardinal, or even just a strong

cardinal, then X must exist.

Strongly compact cardinals

Definition

Suppose that κ is an uncountable regular cardinal. Then κ is a

strongly compact cardinal if for each λ > κ there exists an

ultrafilter U on Pκ(λ) such that:

1. U is a κ-complete ultrafilter,

2. U is a fine ultrafilter.

I Every supercompact cardinal is a strongly compact cardinal.

A natural question immediately arises:

Question

Suppose κ is a strongly compact cardinal. Must κ be a

supercompact cardinal?

Menas’ Theorem

Theorem (Menas)

Suppose κ is a measurable cardinal and that κ is a limit of strongly

compact cardinals.

I Then κ is a strongly compact cardinal.

Lemma

Suppose κ is a supercompact cardinal and let S be the set of

γ < κ such that γ is a measurable cardinal.

I Then S is a stationary subset of κ.

Corollary (Menas)

Suppose that κ is the least measurable cardinal which is a limit of

supercompact cardinals.

I Then κ is a strongly compact cardinal and κ is not a

supercompact cardinal.

The Ultrapower Axiom and strongly compact cardinals

I The Identity Crisis Theorem of Magidor:

Theorem (Magidor)

Suppose κ is a supercompact cardinal. Then there is a (class)

generic extension of V in which:

I κ is a strongly compact cardinal.

I κ is the only measurable cardinal.

Theorem (Goldberg)

Assume the Ultrapower Axiom and that for some κ:

I κ is a strongly compact cardinal.

I κ is not a supercompact cardinal.

Then κ is a limit of supercompact cardinals.

I The Ultrapower Axiom resolves the “identity crisis”.

I By Menas’ Theorem, this is best possible.

The Ultrapower Axiom and the GCH

Theorem (Goldberg)

Asume the Ultrapower Axiom and that κ is a supercompact

cardinal.

I Then 2

λ = λ

+ for all λ ≥ κ.

I The Ultrapower Axiom is absolute between V and V[G] for

all generic extensions whose associated Boolean algebra is of

cardinality below the least strongly inaccessible cardinal of V.

I Therefore the Ultrapower Axiom even augmented by large

cardinal assumptions cannot imply either of:

I The Continuum Hypothesis.

I V = HOD.

Supercompact cardinals and HOD

Lemma

Suppose κ is a supercompact cardinal and that V = HOD. Then

Vκ |= “V = HOD”

I The converse is not true: if κ is supercompact and

Vκ |= “V = HOD”

then V 6= HOD can hold.

I However, if in addition κ is an extendible cardinal then

necessarily

V = HOD.

The Ultrapower Axiom and HOD

Theorem (Goldberg)

Assume the Ultrapower Axiom , κ is a supercompact cardinal, and

Vκ |= “V = HOD”.

Then:

I For all regular cardinals γ ≥ κ,

H(γ

++) = HODH(γ

++)

More precisely,

I Every set x ∈ H(γ

++) is definable in H(γ

++) from some

α < γ++.

I V = HOD.

I Thus in the context of the Ultrapower Axiom, the existence of

a supercompact cardinal greatly amplifies the assumption that

V = HOD by giving:

I A uniform local version which must hold above the

supercompact cardinal.

I Just like with GCH, this is best possible.

HODA and Vopˇenka’s Theorem

Definition

Suppose A is a set. HODA is the class of all sets X such that

there exist α ∈ Ord and M ? Vα such that

1. A ∈ Vα.

2. X ∈ M and M is transitive.

3. Every element of M is definable in Vα from ordinal parameters

and A.

Theorem (Vopˇenka)

For each set A, HODA is a set-generic extension of HOD.

I From the perspective of Set Theoretic Geology:

I For each set A, HOD is a ground of HODA.

The Ultrapower Axiom and the grounds of V

Theorem (Goldberg)

Asume the Ultrapower Axiom and that κ is a supercompact

cardinal. Suppose A is a wellordering of Vκ.

I Then V = HODA.

Corollary (Goldberg)

Asume the Ultrapower Axiom and that there is a supercompact

cardinal.

I Then HOD is a ground of V.

The HOD of the mantle of V

Putting everything together:

Theorem

Asume the Ultrapower Axiom and that there is an extendible

cardinal. Let M be the mantle of V.

I Then M |= “V = HOD”.

(sketch)

I By Goldberg’s Theorem, V = HODA for some set A.

I Therefore by Vopˇenka’s Theorem:

I If N is a ground of V then HODN

is a ground of N and so:

I HODN

is a ground of V.

I By Usuba’s Mantle Theorem, M is a ground of V.

I Thus HODM is a ground of V.

I Therefore M ? HODM and so M = HODM.

The mantle, V, HOD, and large cardinals

Theorem (after Hamkins et al)

Suppose V[G] is the Easton extension of V where for each limit

cardinal γ, if Vγ ?Σ2 V then G adds a fast club at γ

+. Then:

I V is not a ground of V[G].

I V is the mantle of V[G] and HODV = HODV [G]

.

I Many large cardinals are preserved, but:

I There are no extendible cardinals in V[G].

Theorem (after Hamkins et al)

Suppose V[G] is the Backward Easton extension of V where for

each strong limit cardinal γ, G adds a fast club at γ

+. Then:

I V[G] is the mantle of V[G].

I HODV [G] ? HODV

.

I Every extendible cardinal of V is extendible in V[G].

I By changing G slightly one can arrange HODV [G] = V.

The mantle of V and HOD when V = Ultimate-L

Theorem

Assume V = Ultimate-L. Then:

I V has no nontrivial grounds.

I Suppose V[G] is a set-generic extension of V. Then

I V is the mantle of V[G].

Theorem

Assume V = Ultimate-L. Then:

I V = HOD.

I An obvious conjecture emerges.

The Mantle Conjecture

Mantle Conjecture

Asume the Ultrapower Axiom and that there is an extendible

cardinal. Let M be the mantle of V.

I Then M |= “V = Ultimate-L”.

I The conjunction of the Ultimate-L Conjecture and the

Mantle Conjecture would provide the basis for a powerful

argument that the axiom, V = Ultimate-L, is true, by citing

as reasons:

I convergence (of different approaches to the same axiom).

I recovery (of axioms from their basic consequences).