Ordinals: the transfinite numbers

I ? is the smallest ordinal: this is 0.

I {?} is the next ordinal: this is 1.

I {?, {?}} is next ordinal: this is 2.

If α is an ordinal then

I α is just the set of all ordinals β such that β is smaller than α,

I α + 1 = α ∪ {α} is the next largest ordinal.

ω denotes the least infinite ordinal, it is the set of all finite ordinals.

V: The Universe of Sets

The power set

Suppose X is a set. The powerset of X is the set

P(X) = {Y Y is a subset of X}.

Cumulative Hierarchy of Sets

The universe V of sets is generated by defining Vα by induction on

the ordinal α:

1. V0 = ?,

2. Vα+1 = P(Vα),

3. if α is a limit ordinal then Vα =

S

β<α Vβ.

I If X is a set then X ∈ Vα for some ordinal α.

I V0 = ?, V1 = {?}, V2 = {?, {?}}.

I These are just the ordinals: 0, 1, and 2.

I V3 has 4 elements (and is clearly not an ordinal).

I V4 has 16 elements.

I V5 has 65, 536 elements.

I V1000 has a lot of elements.

Vω is infinite, it is the set of all (hereditarily) finite sets.

The conception of Vω is mathematically identical to the

conception of the structure (N, +, ·):

I Each structure can be interpreted in the other structure.

Beyond the basic axioms: large cardinal axioms

Shaping the conception of V

I The ZFC axioms of Set Theory formally specify the founding

principles for the conception of V.

I The ZFC axioms are naturally augmented by additional

axioms which assert the existence of “very large” infinite sets.

I Such axioms assert the existence of large cardinals.

These large cardinals include:

I Measurable cardinals

I Strong cardinals

I Woodin cardinals

I Superstrong cardinals

I Supercompact cardinals

I Extendible cardinals

I Huge cardinals

I ω-huge cardinals

Cardinality: measuring the size of sets

Definition: when two sets have the same size

Two sets, X and Y , have the same cardinality if there is a

matching of the elements of X with the elements of Y .

Formally: |X| = |Y | if there is a bijection

f : X → Y

Assuming the Axiom of Choice which is one of the ZFC axioms:

Theorem (Cantor)

For every set X there is an ordinal α such that |X| = |α|.

The Continuum Hypothesis: CH

Theorem (Cantor)

The set N of all natural numbers and the set R of all real numbers

do not have the same cardinality.

I There really are different “sizes” of infinity!

The Continuum Hypothesis

Suppose A ? R is infinite. Then either:

1. A and N have the same cardinality, or

2. A and R have the same cardinality.

I This is Cantor’s Continuum Hypothesis.

Many tried to solve the problem of the Continuum

Hypothesis and failed.

The problem of the Continuum Hypothesis quickly came to be

widely regarded as one of the most important problems in all of

modern Mathematics.

In 1940, G¨odel showed that it is consistent with the axioms of Set

Theory that the Continuum Hypothesis be true.

I One cannot refute the Continuum Hypothesis.

In 1963, on July 4th, Cohen announced in a lecture at Berkeley

that it is consistent with the axioms of Set Theory that the

Continuum Hypothesis be false.

I One cannot prove the Continuum Hypothesis.

Cohen’s method

If M is a model of ZFC then M contains “blueprints” for virtual

models N of ZFC, which enlarge M. These blueprints can be

constructed and analyzed from within M.

I If M is countable then every blueprint constructed within M

can be realized as genuine enlargement of M.

I Cohen proved that every model of ZFC contains a blueprint

for an enlargement in which the Continuum Hypothesis is

false.

I Cohen’s method also shows that every model of ZFC

contains a blueprint for an enlargement in which the

Continuum Hypothesis is true.

I (Levy-Solovay) These enlargements preserve large cardinal

axioms:

I So if large cardinal axioms can help

I it can only be in some unexpected way.

The extent of Cohen’s method: It is not just about CH

A challenging time for the conception of V

I Cohen’s method has been vastly developed in the 5 decades

since Cohen’s original work.

I Many problems have been showed to be unsolvable including

problems outside Set Theory:

I (Group Theory) Whitehead Problem (Shelah)

I (Analysis) Kaplansky’s Conjecture (Solovay)

I (Combinatorics of the real line) Suslin’s Problem

(Solovay-Tennenbaum, Jensen, Jech)

I (Measure Theory) Borel Conjecture (Laver)

I (Operator Algebras) Brown-Douglas-Filmore Automorphism

Problem (Phillips-Weaver, Farah)

I This is a serious challenge to the very conception of

Mathematical Infinity.

I These examples, including the Continuum Hypothesis, are all

statements about just Vω+2.

Ok, maybe it is just time to give up

Claim

I Large cardinal axioms are not provable;

I by G¨odel’s Second Incompleteness Theorem.

I But, large cardinal axioms are falsifiable.

Prediction

No contradiction from the existence of infinitely many Woodin

cardinals will be discovered within the next 1000 years.

I Not by any means whatsoever.

Truth beyond our formal reach

The real claim of course is:

I There is no contradiction from the existence of infinitely

many Woodin cardinals.

Claim

I Such statements cannot be formally proved.

I This suggests there is a component in the evolution of

our understanding of Mathematics which is not formal.

I There is mathematical knowledge which is not

entirely based in proofs.

Claim

The skeptical assessment that the conception of the universe

of sets is incoherent, must be wrong.

I How else can these truths and ensuing predictions be

explained?

I But then either CH must be true or CH must be false.

OK, back to the problem of the Continuum Hypothesis

The skeptic’s challenge

Resolve the problem of CH.

I Perhaps one should begin by trying to more deeply understand

CH.

A natural conjecture

One can more deeply understand CH by looking at special cases.

I But which special cases?

I Does this even make sense?

The simplest uncountable sets

Definition

A set A ? Vω+1 is a projective set if:

I A can be logically defined in the structure

(Vω+1, ∈)

from parameters.

We can easily extend the definition to relations on Vω+1:

Definition

A set A ? Vω+1 × Vω+1 is a projective set if:

I A can be logically defined as a binary relation in the structure

(Vω+1, ∈)

from parameters.

I The countable subsets of Vω+1 and Vω+1 × Vω+1 are

projective sets but so are Vω+1 and Vω+1 × Vω+1 themselves,

and these sets are not countable.

The Continuum Hypothesis and the Projective Sets

The Continuum Hypothesis

Suppose A ? Vω+1 is infinite. Then either:

1. A and Vω have the same cardinality, or

2. A and Vω+1 have the same cardinality.

I This is a statement about all subsets of Vω+1.

The projective Continuum Hypothesis

Suppose A ? Vω+1 is an infinite projective set. Then either:

1. A and Vω have the same cardinality, or

2. There is a bijection

F : Vω+1 → A

such that F is a projective set.

I This is a statement about just the “simple” subsets of Vω+1.

The Axiom of Choice

Definition

Suppose that

A ? X × Y

A function

F : X → Y

is a choice function for A if for all a ∈ X:

I If there exists b ∈ Y such that (a, b) ∈ A then (a, F(a)) ∈ A.

The Axiom of Choice

For every set

A ? X × Y

there exists a choice function for A.

The Axiom of Choice and the Projective Sets

The projective Axiom of Choice

Suppose A ? Vω+1 × Vω+1 is a projective set. Then there is a

function

F : Vω+1 → Vω+1

such that:

I F is a choice function for A.

I F is a projective set.

I There were many attempts in the early 1900s to solve both

the problem of projective Continuum Hypothesis and the

problem of the projective Axiom of Choice:

I Achieving success for the simplest instances.

I However, by 1925 these problems both looked hopeless.

These were both hopeless problems

The actual constructions of G¨odel and Cohen show that both

problems are formally unsolvable.

I In G¨odel’s universe L:

I The projective Axiom of Choice holds.

I The projective Continuum Hypothesis holds.

I In the Cohen enlargement of L (as given by the actual

blueprint which Cohen defined for the failure of CH):

I The projective Axiom of Choice is false.

I The projective Continuum Hypothesis is false.

I This explains why these problems were so difficult.

I But the intuition that these problems are solvable was

correct.

An unexpected entanglement

Theorem (1984)

Suppose there are infinitely many Woodin cardinals. Then:

I The projective Continuum Hypothesis holds.

Theorem (1985: Martin-Steel)

Suppose there are infinitely many Woodin cardinals. Then:

I The projective Axiom of Choice holds.

We now have the correct conception of Vω+1 and the projective

sets.

I This conception yields axioms for the projective sets.

I These (determinacy) axioms in turn are closely related to

(and follow from) large cardinal axioms.

But what about Vω+2? Or even V itself?

Logical definability

The definable power set

For each set X, PDef(X) denotes the set of all Y ? X such that Y

is logically definable in the structure (X, ∈) from parameters in X.

I PDef(X) is the collection of just those subsets of X which are

intrinsic to X itself,

I versus P(X) which is the collection of all subsets of X.

The collection of all the projective subsets of Vω+1 is exactly

given by:

PDef(Vω+1)

The effective cumulative hierarchy: L

Cumulative Hierarchy of Sets

The cumulative hierarchy is defined by induction on α as follows.

1. V0 = ?.

2. Vα+1 = P(Vα).

3. if α is a limit ordinal then Vα =

S

β<α Vβ.

I V is the class of all sets X such that X ∈ Vα for some α.

G¨odel’s constructible universe, L

Define Lα by induction on α as follows.

1. L0 = ?.

2. Lα+1 = PDef(Lα).

3. if α is a limit ordinal then Lα = ∪{Lβ β < α}.

I L is the class of all sets X such that X ∈ Lα for some α.

The missing axiom for V?

The axiom: V = L

Suppose X is a set. Then X ∈ L.

Theorem (G¨odel:1940)

Assume V = L. Then the Continuum Hypothesis holds.

I Suppose there is a Cohen-blueprint for V = L. Then:

I the axiom V = L must hold and the blueprint is trivial.

Claim

Adopting the axiom V = L completely negates the ramifications of

Cohen’s method.

I Could this be the resolution?

I No, there is a serious problem.

The axiom V = L and large cardinals

Theorem (Scott:1961)

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

I In fact there are no ( genuine ) large cardinals.

I Assume V = L. Then there are no Woodin cardinals.

Clearly:

The axiom V = L is false.

A natural conjecture

Perhaps the key is to generalize the construction of L by using the

large cardinals to expand the definable powerset operation.

I But there is an alternative approach which is based on simply

using the large cardinals to directly generalize the projective

sets.

Another way to define the projective sets

Observation

Vω+1 is homeomorphic to the Cantor set, with the topology on

Vω+1 given by the sets

On,a = {X ? Vω X ∩ Vn = a}

as basic open sets where n < ω and a ∈ Vn+1.

I The projective subsets of Vω+1 are exactly the sets generated

from the open sets and closing under the operations:

I Taking images by continuous functions

F : Vω+1 → Vω+1.

I Taking complements.

I This definition generalizes to any topological space.

I In particular, this extends the notion of the projective sets to

the Euclidean spaces R

n

.

Universally Baire sets

Definition (Feng-Magidor-Woodin)

A set A ? R

n

is universally Baire if:

I For all topological spaces ?

I For all continuous functions π : ? → R

n

;

the preimage of A by π has the property of Baire in the space ?.

I Universally Baire sets have the property of Baire

I Simply take ? = R

n

and π to be the identity.

I Universally Baire sets are Lebesgue measurable.

Theorem

Assume V = L. Then every set A ? R is the image of a universally

Baire set by a continuous function

F : R → R.

L(A, R) where A ? R

Relativizing L to A ? R

Suppose A ? R. Define Lα(A, R) by induction on α by:

1. L0(A, R) = Vω+1 ∪ {A},

2. (Successor case) Lα+1(A, R) = PDef(Lα(A, R)),

3. (Limit case) Lα(A, R) = ∪{Lβ(A, R) β < α}.

I L(A, R) is the class of all sets X such that X ∈ Lα(A, R) for

some ordinal α.

I P(R) ∩ Lω1

(A, R) is the smallest σ-algebra containing A and

closed under images by continuous functions f : R → R.

I If B ? R and B ∈ L(A, R) then L(B, R) ? L(A, R). So:

I P(R) ∩ L(A, R) is closed under images by continuous functions

F : R → R.

The universally Baire sets are the ultimate generalization

of the projective sets

Theorem

Suppose that there is a proper class of Woodin cardinals and

suppose A ? R is universally Baire.

I Then every set B ∈ L(A, R) ∩ P(R) is universally Baire.

I Thus every projective set is universally Baire.

I Since clearly there exists a proper class of Woodin cardinals.

Theorem

Suppose that there is a proper class of Woodin cardinals.

(1) (Martin-Steel) Suppose A ? R is universally Baire.

I Then A is determined.

(2) (Steel) Suppose A ? R × R is universally Baire.

I Then A has a choice function which is universally Baire.

I Thus L(A, R) |= AD, where AD is the Axiom of Determinacy.

Measuring the complexity of universally Baire sets

Definition

Suppose A and B are subsets of R.

1. A is weakly Wadge reducible to B, A ≤Wadge B, if there is

a function π : R → R such that:

I π is continuous on R\Q.

I Either A = π

?1

[B] or A = R\π

?1

[B].

2. A and B are weakly Wadge bi-reducible if

I A ≤Wadge B and B ≤Wadge A.

3. The weak Wadge degree of A is the equivalence class of all

sets which are weakly Wadge bi-reducible with A.

I If A is weakly Wadge reducible to B and B is universally Baire

then A is universally Baire.

An indication of deep structure

Theorem (Martin-Steel, Martin, Wadge)

Assume there is a proper class of Woodin cardinals.

Then the weak Wadge degrees of the universally Baire sets are

linearly ordered by weak Wadge reducibility and moreover this is a

wellorder.

Speculation

Perhaps this ultimate generalization of the projective sets can lead

us to the ultimate generalization of the axiom V = L.

I But how?

Defining the axiom V = L without defining L

A sentence ? is a Σ2-sentence if it is of the form:

I There exists an ordinal α such that Vα |= ψ;

for some sentence ψ.

For each ordinal α, let

Nα = ∩{M M is transitive, M |= ZFC\Powerset, and OrdM = α}.

where:

I A set M is transitive if a ? M for each a ∈ M.

Lemma

The following are equivalent.

(1) V = L.

(2) For each Σ2-sentence ?, if V |= ? then there exists a

countable ordinal α such that Nα |= ?.

I We need to somehow use the universally Baire sets in a

reformulation of (2).

G¨odel’s transitive class HOD

Definition

HOD is the class of all sets X such that there exist α ∈ Ord and

M ∈ Vα such that

1. X ∈ M and M is transitive.

2. Every element of M is definable in Vα from ordinal

parameters.

I For every set b there is a minimum transitive set TC(b) which

contains b as an element.

Why HOD?

Suppose N is a model of ZF. Let HODN ? N be HOD as defined

within N. Then for each b ∈ N, the following are equivalent:

1. b ∈ HODN.

2. Every element of (TC(b))N

is definable in N with parameters

from the ordinals of N.

HODL(A,R)

and measurable cardinals

Definition

Suppose that A ? R. Then HODL(A,R)

is the class HOD as

defined within L(A, R).

I The Axiom of Choice must hold in HODL(A,R)

I even if L(A, R) |= AD.

Theorem (Solovay:1967)

Suppose that A ? R and L(A, R) |= AD.

I Then ω

V

1

is a measurable cardinal in HODL(A,R)

.

I Solovay’s theorem gave the first connection between the

Axiom of Determinacy (AD) and large cardinal axioms.

HODL(A,R)

and Woodin cardinals

Theorem

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

A is universally Baire.

I Then ω

V

1

is the least measurable cardinal in HODL(A,R)

.

Definition

Suppose that A ? R is universally Baire.

Then ΘL(A,R)

is the supremum of the ordinals α such that there is

a surjection, π : R → α, such that π ∈ L(A, R).

I ΘL(A,R)

is another measure of the complexity of A.

Theorem

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

A is universally Baire.

I Then ΘL(A,R)

is a Woodin cardinal in HODL(A,R)

.

The axiom V = Ultimate-L

I The existence of a Woodin cardinal is expressible by a

Σ2-sentence.

I Woodin cardinals clearly exist in V;

I If A ? R is universally Baire and there is a proper class of

Woodin cardinals then

HODL(A,R)

|= “There is a Woodin cardinal”.

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)

|= ?.

This is just rank resemblance

Assume there is a proper class of Woodin cardinals. Then the

following are equivalent:

I V = Ultimate-L.

I Suppose ψ is a sentence and there exists an ordinal α such

that

Vα |= ψ.

I Then there exists a universally Baire set A ? R such that

HODL(A,R)

|= “There exists α such that Vα |= ψ”

Some consequences of V = Ultimate-L

Theorem (V = Ultimate-L)

The Continuum Hypothesis holds.

Theorem (V = Ultimate-L)

V = HOD.

Theorem (V = Ultimate-L)

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

Γ

∞ 6= P(R) ∩ L(Γ∞, R)

If V = Ultimate-L then:

I The Axiom of Choice holds in L(?！? R).

I This is the generalization to V = Ultimate-L of the fact that

if V = L then there is a projective wellordering of the reals.

The axiom V = Ultimate-L and Cohen’s method

I Suppose there is a Cohen-blueprint for V = Ultimate-L.

Then:

I The axiom V = Ultimate-L must hold and the blueprint is

trivial.

I The axiom V = Ultimate-L settles (modulo axioms of

infinity) all sentences about “small” sets (like Vω+2) which

have been shown to be independent by Cohen’s method.

Claim

Adopting the axiom V = Ultimate-L completely negates the

ramifications of Cohen’s method.

I But, is the axiom V = Ultimate-L compatible with all large

cardinal axioms?

I Is there a Scott Theorem for V = Ultimate-L?

The language of large cardinals: elementary embeddings

Definition

Suppose X and Y are transitive sets. A function j : X → Y is an

elementary embedding if for all logical formulas ?[x0, . . . , xn]

and all a0, . . . , an ∈ X,

(X, ∈) |= ?[a0, . . . , an] if and only if (Y , ∈) |= ?[j(a0), . . . , j(an)]

I Isomorphisms are elementary embeddings but the only

isomorphisms of (X, ∈) and (Y , ∈) are trivial.

Lemma

Suppose that j : Vα → Vβ is an elementary embedding. Then the

following are equivalent.

(1) j is not the identity.

(2) There is an ordinal η < α such that j(η) 6= η.

I CRT(j) denotes the least ordinal η such that j(η) 6= η.

Extendible cardinals and supercompact cardinals

Definition (Reinhardt:(1974))

Suppose that δ is a cardinal.

I Then δ is an extendible cardinal if for each λ > δ there

exists an elementary embedding

j : Vλ+1 → Vj(λ)+1

such that CRT(j) = δ and j(δ) > λ.

Definition (Solovay, Reinhardt: as reformulated by Magidor(1971))

Suppose that δ is a cardinal.

I Then δ is an supercompact cardinal if for each λ > δ there

exist δ <ˉ λ < δ ˉ and an elementary embedding

j : Vλˉ+1 → Vλ+1

such that CRT(j) = δˉ and j(δˉ) = δ.

Weak extender models

Definition

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

N is a model of ZFC.

I Then N is a weak extender model of δ is supercompact if

for every γ > δ there exist δ <ˉ λ < δ ˉ and an elementary

embedding

π : Vλˉ+1 → Vλ+1

such that CRT(π) = δ

ˉ, π(δ

ˉ) = δ, and such that

I π(N ∩ Vλˉ ) = N ∩ Vλ.

I π|(N ∩ Vλˉ ) ∈ N.

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

that α ≥ δ

+.

I N is uniquely specified by N ∩ Vα.

I N is Σ2-definable from N ∩ Vα.

I The theory of weak extender models is part of the theory of V.

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.

Theorem

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

κ > δ, and that κ is a supercompact cardinal.

I Then κ is a supercompact cardinal in N.

I There are generalizations of this for all large cardinal notions.

The Universality Theorem

Theorem (Universality Theorem)

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

α > δ is a limit ordinal, and that

j : Vα+2 → Vj(α)+2

is an elementary embedding such that δ < CRT(j). Then:

I j(N ∩ Vα) = N ∩ Vj(α)

.

I j|(N ∩ Vα) ∈ 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.

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 axiom V = Ultimate-L.

I By the Universality Theorem.

I The Ultimate-L Conjecture is an existential number theoretic

statement.

I If it is undecidable then it must be false.

Claim

The Ultimate-L Conjecture must be either true or false

I it cannot be meaningless.

Set Theory faces one of two futures

I The Ultimate-L Conjecture reduces the entire post-Cohen

debate on Set Theoretic truth to a single question which

I must have an answer.

Future 1: The Ultimate-L Conjecture is true.

I Then the axiom V = Ultimate-L is very likely the key missing

axiom for V.

I There is no generalization of Scott’s Theorem for the axiom

V = Ultimate-L.

I All the questions which have been shown to be unsolvable by

Cohen’s method are resolved modulo large cardinal axioms.

Future 2: The Ultimate-L Conjecture is false.

I Then the program to understand V by generalizing the

success in understanding Vω+1 and the projective sets fails.

I Which is it?