Measurable Cardinals and Scott's Theorem - Stanford University · 2015-12-01 · 4/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals MeasurableCardinals Ulam:If

Measurable Cardinals and Scott’s Theorem

Chris Mierzewski

Mathematical Logic Seminar

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From measures to cardinals

Measurable Cardinals

Lebesgue’s Measure Problem

Is there a measure � : P(R)→ R such that

I � is not the constant zero function

I � is �-additive

I For any sets X,Y � R, if X = {y + r | y 2 Y } for some fixed r 2 R, then�(X) = �(Y )?

Vitali (1905): No such thing.

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Banach:

Is there any set S admitting a measure � : P(S)→ [0, 1] such that

I �(S) = 1

I � is �-additive

I �({s}) = 0 for all s 2 S?

For which cardinals � is there a non-trivial finite measure defined over �(i.e. on P(�))?

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Ulam: If � is the smallest cardinal with a non-trivial finite measure over �, then � � 2ℵ0 or� admits a non-trivial measure that takes onlyvalues in {0, 1}.

I If � is the smallest cardinal with a non-trivial finite measure �, then �

is �-additive.

I So we can ask: does there exist any uncountable � with a {0, 1}-valuedmeasure over it?

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Measurable cardinals

Relation between �-additive measures and non-principal, �-completeultrafilters over �:

U ={X � � |�(X) = 1

}

Definition (Measurable Cardinal)

An uncountable cardinal � is mesurable if there exists a non-principal,�-complete ultrafilter over �.

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Some facts about measurable cardinalsSuppose � is measurable, and U a non-principal, �-complete ultrafilter over�. Then the following hold:

I If X 2 U , then |X | = �.

I � is regular.

I (Tarski–Ulam) � is inaccessible.

So if � is measurable, then (V�,2) � ZFC; thus ZFC cannot prove theexistence of measurable cardinals. They are indeed large. In fact:

I (Hanf–Tarski (1960)) Least inaccessible cardinal < least measurablecardinal.

J.Bell: “GARGANTUAN proportions...”

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Scott’s Theorem

Theorem (Scott)

If there exists a measurable cardinal, then V 6= L.

Woodin’s "Meta-Corollary"

“The whole point of set theory is to study infinity.You can’t deny large cardinals. So [the statementV=L] is just not true.”

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Scott’s Theorem

Theorem (Scott)

If there exists a measurable cardinal, then V 6= L.

Woodin’s "Meta-Corollary"

“The whole point of set theory is to study infinity.You can’t deny large cardinals. So [the statementV=L] is just not true.”

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Background about L

The constructible universe: Göd-L

Recall:

I L is the smallest inner model (in V ). That is, if M is a transitive classcontaining all ordinals, then L �M .

Aternative characterisation of L: via Gödel operations and the associatedclosure operator.

Def(A) = cl(A[ {A})\P(A)

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Background about L

Alternative characterisation of L: Gödel operations

I G1(X,Y ) := {X,Y }

I G2(X,Y ) := X � Y

I G3(X,Y ) :={(u, v) |u 2 X ∧ v 2 Y ∧ u 2 v

}I G4(X,Y ) := X \ Y

I G5(X,Y ) := X \ Y

I G6(X) :=⋃X

I G7(X) := dom(X)

I G8(X) :={(u, v) | (v,u) 2 X

}I G9(X) :=

{(u, v,w) | (v,w, v) 2 X

}I G10(X) :=

{(u, v,w) | (v,w,u) 2 X

}

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Background about L

Alternative characterisation of L: Gödel operationsWe can define:

L0 := ;

L�+1 := cl(L� [ {L�})\P(L�)

L :=⋃

�2On

L�

Theorem (Gödel)

A transitive class M is an inner model of ZF if and only if

I M is closed under Gödel operations

I M is almost universal (whenever X �M , then X � Y for someY 2M .

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Scott’s Proof

Scott’s Proof

I Take �, the least measurable cardinal, and U a non-principal,�-complete ultrafilter over �.

I Build an ultrapower (V �/U ,2U ) of V , and collapse it to form atransitive class model (M ,2).

I Using a variant of Łoś’ Theorem, there is an elementary embedding� : V →M .

I Using V = L, show that M = V , so that � is an elementary embeddingof V .

(V ,2) (M ,2)

(V �/U ,2U )

id

�

�

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Scott’s Proof

Scott’s Proof

I Using the �-completeness of U , show that

(i) for any � < �, �(�) < �, and

(ii) we have �(�) > �.

I By Łoś’ Theorem, we have that

(M ,2) � ‘�(�) is the least measurable cardinal’,

and since M = V , �(�) is the least measurable cardinal in V .

I But we have shown that �(�) > �, which contradicts the minimality of�(�).

I Contradiction. If V = L, there can be no measurable cardinals.

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Scott’s Proof

Ultrapowers of proper classes: Scott’s trick

I We want to build ultrapowers of proper class models. Let � a cardinaland U an ultrafilter over �. For any f , g 2 V �, we let

f∼U g if and only if {� < � | f(�) = g(�)} 2 U

The ultrapower V �/U is the class of minimal-rank representatives ofequivalence classes (Scott’s trick):

[f]:={g2 V �

∣∣ f ∼U g and 8h 2 V �, if f ∼U h then rk(g) � rk(h)

}Then each [f ] is a set.

I Thus we get a proper class

V�/U :={[f ]∣∣ f : �→ V

}

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Scott’s Proof

Ultrapowers of proper classes: Scott’s trick

I We want to build ultrapowers of proper class models. Let � a cardinaland U an ultrafilter over �. For any f , g 2 V �, we let

f∼U g if and only if {� < � | f(�) = g(�)} 2 U

The ultrapower V �/U is the class of minimal-rank representatives ofequivalence classes (Scott’s trick):

[f]:={g2 V �

∣∣ f ∼U g and 8h 2 V �, if f ∼U h then rk(g) � rk(h)

}Then each [f ] is a set.

I Thus we get a proper class

V�/U :={[f ]∣∣ f : �→ V

}

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Scott’s Proof

Ultrapowers of proper classes

We have

V �/U ={[f ]∣∣ f 2 V �

}The ultrapower structure comes with a membership relation 2U , defined as

[f ] 2U [g] if and only if{� < �

∣∣ f(�) 2 g(�)}2 U

We obtain the ultrapower structure

(V �/U ,2U ).

Useful notation: write ||f 2 g|| := {� < � | f(�) 2 g(�)}. In general, for anyfirst order formula '(x1, ...,xn), write∣∣∣∣'(f1, ..., fn)∣∣∣∣ := {� < �

∣∣'(f1(�), ..., fn(�))}.

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Scott’s Proof

Łoś’ Theorem

For any L2-formula ', and any f1, ..., fn 2 V �, we have:

(V �,2U ) � '

([f1], ..., [fn]

)if and only if{

� < �∣∣∣ (V ,2) � '

(f1(�), ..., fn(�)

)}2 U

(Equivalently, iff ||'(f1, ..., fn)|| 2 U .)

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Scott’s Proof

I Assume V = L, and assume there is a measurable cardinal. Let � bethe least measurable cardinal.

I Take the ultrapower (V �/U ,2U ), where U is a non-principal,�-complete ultrafilter on �.

I We will show that V can be elementarily embedded in (V �/U ,2U ),and (V �/U ,2U ) can be collapsed to a transitive class model(M in),giving an elementary embedding of (V ,2) into (M ,2).

(V ,2)

(V �/U ,2U )

id

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Scott’s Proof



I We will show that V can be elementarily embedded in (V �/U ,2U ),and (V �/U ,2U ) can be collapsed to a transitive class model(M in),giving an elementary embedding of (V ,2) into (M ,2).

(V ,2)

(V �/U ,2U )

id

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Scott’s Proof



I We will show that V can be elementarily embedded in (V �/U ,2U ),and (V �/U ,2U ) can be collapsed to a transitive class model (M ,2),giving an elementary embedding of (V ,2) into (M ,2).

(V ,2) (M ,2)

(V �/U ,2U )

id �

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Scott’s Proof



I We will show that V can be elementarily embedded in (V �/U ,2U ),and (V �/U ,2U ) can be collapsed to a transitive class model (M ,2),giving an elementary embedding of (V ,2) into (M ,2).

(V ,2) (M ,2)

(V �/U ,2U )

id

�

�

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Scott’s Proof

Embedding

(V ,2) (M ,2)

(V �/U ,2U )

id

�

�

I For any set x, let cx : �→ V be the constant map cx : � 7→ x.

I The map id : V → V �/U is defined as id(x) = [cx]

By Łoś, for any ' 2 L2 we have :

(V ,2) � '(a1, ..., an

)iff (V �/U ,2U ) � '

([ca1 ], ..., [can ]

)so that id is an elementary embedding of V into V �/U .

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Scott’s Proof

Mostowski collapse

Let K a class (possibly proper) and E a binary relation on K. Suppose thestructure (K,E) satisfies the following:

I (K,E) is extensional – 8a, b 2 K, a = b iff 8c 2 K, cEa↔ cEb.

I E is set-like – 8a 2 K, {b 2 K | b 2 a} is a set.

I E is well-founded (as seen from V ).

Then there is a unique transitive class (M ,2) isomorphic to (K,E).

(M ,2) is the Mostowski collapse of (K,E).The (unique) isomorphism � : (K,E)→ (M ,2) is the collapsing function.

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Scott’s Proof

Collapsing the ultrapower

The ultrapower is not necessarily transitive. In order to collapse(V �/U ,2U ) to a transitive structure, we need to ensure that the requiredproperties hold.

I Well-founded (!1-completeness of U)

I Set-like[g] 2U [f ] iff 9h 2 (

⋃�<� f(�)[ {;})� such that h ∼U g and [h] 2U [f ]

I Extensional(Closure properties of U)

⇒ By Mostowski’s Collapse Lemma, there is a transitive properclass M and an isomorphism � : (V �/U ,2U )→ (M ,2).

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Scott’s Proof

Embedding V in M

(V ,2) (M ,2)

(V �/U ,2U )

id

�

�

I The collapsing map � : V �/U →M is defined (by induction on2U -rank) as

�([f ]) :={�([g])

∣∣ [g] 2U [f ]}.

Let � : (V ,2)→ (M ,2) be the map defined as � := � � id, so that for anyset x 2 V ,

�(x) := �([cx]).

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Scott’s Proof

Embedding V in M

(V ,2) (M ,2)

(V �/U ,2U )

id

�

�

I The collapsing map � : V �/U →M is defined (by induction on2U -rank) as

�([f ]) :={�([g])

∣∣ [g] 2U [f ]}.

DefinitionLet � : (V ,2)→ (M ,2) be the map defined as � := � � id, so that for anyset x 2 V ,

�(x) := �([cx]).

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Scott’s Proof

The embedding is non-trivial

Some immediate consequences of Łoś’ Theorem:

I � sends On to On.

I For �, � 2 On, if � < �, then �(�) < �(�). So for any � 2 On,�(�) � �.

I M contains all ordinals.By Łoś, we have (M ,2) |= ‘�(�) is an ordinal’, for any ordinal �. Bythe above, � � �(�) 2M . M is transitive, so � is in M.

Crucially, the embedding is non-trivial. In particular, it does not map theleast measurable cardinal to itself.

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Scott’s Proof


Some immediate consequences of Łoś’ Theorem:

I � sends On to On.

I For �, � 2 On, if � < �, then �(�) < �(�). So for any � 2 On,�(�) � �.

I M contains all ordinals.By Łoś, we have (M ,2) |= ‘�(�) is an ordinal’, for any ordinal �. Bythe above, � � �(�) 2M . M is transitive, so � is in M.

Crucially, the embedding is non-trivial. In particular, it does not map theleast measurable cardinal to itself.

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Scott’s Proof


I �(�) > �.

Let d : �→ V defined as d(�) = � for all � < �. Then||d < �|| = {� < � | d(�) < �} = � 2 U , so [d] < [c�] holds in the ultrapower.Further, let � < � and consider the function c�. We have||c� < d|| = {� < � | c�(�) < d(�)} = {� < � |� < �}. The complement{� < � |� � �} is bounded, and thus not in U . So ||c� < d|| 2 U , and so wehave [c�] < [d] < [c�], which yields

for any � < �, �([c�]) < �([d]) <�([c�])

�(�) < �([d]) <�(�)

so that � = sup

�<�

�(�) � �([d])<�(�).

2

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Scott’s Proof


I �(�) > �.

Now for the final contradiction.

I By Łoś’ Theorem, we have that

(M ,2) � ‘�(�) is the least measurable cardinal’.

I M is a transitive proper class containing all ordinals (alternatively: itis closed under all Gödel operations, and almost universal). Thismeans L �M . But we assumed V = L: so M = V !

I So �(�) is the least measurable cardinal in V .

I But �(�) > �, contradicting the choice of �.

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Scott’s Proof

Non-trivial embeddings of V

→ Assuming V = L, we have constructed a non-trivial elementaryembedding of V into itself.

I � is the least ordinal that is not fixed by �.

I We say that � is a critical point of �. If j : V →M is an embedding(with M an inner model), let cp(j) := least ordinal � such thatj(�) 6= �.

I Close relationship between measurable cardinals and critical points ofnon-trivial embeddings on V .

Theorem (Keisler)

If j : V →M is a non-trivial elementary embedding of the universe, thencp(j) is a measurable cardinal.

(Kunen: there are no non-trivial elementary embeddings of V into V .)

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Conclusion

Conclusion

I The notion of measurable cardinals can be traced back to relatively‘natural’ questions from measure theory

I Scott’s proof, relying on a generalised ultrapower construction,highlights the connection between large cardinals and non-trivialelementary embeddings

I Measurable cardinals areLarge large cardinals: assuming theexistence of measurable cardinals takes us beyond L.

thanks.

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Measurable Cardinals and Scott's Theorem - Stanford University · 2015-12-01 · 4/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals MeasurableCardinals Ulam:If