Measurable Cardinals and Scott's Theorem

Measurable Cardinals and Scott’s Theorem

Chris Mierzewski

Mathematical Logic Seminar

1/29 Measurable Cardinals and Scott’s Theorem 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.

2/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals

Measurable Cardinals

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())?

3/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals

Measurable Cardinals

Ulam: If is the smallest cardinal with a non- trivial ﬁnite measure over , then 2ℵ0 or admits a non-trivial measure that takes only values 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}-valued measure over it?

4/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals

Measurable cardinals

Relation between -additive measures and non-principal, -complete ultraﬁlters over : U = X | (X) = 1

Deﬁnition (Measurable Cardinal) An uncountable cardinal is mesurable if there exists a non-principal, -complete ultraﬁlter over .

5/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals

Some facts about measurable cardinals Suppose 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 the existence of measurable cardinals. They are indeed large. In fact: I (Hanf–Tarski (1960)) Least inaccessible cardinal < least measurable cardinal. J.Bell: “GARGANTUAN proportions...”

6/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals

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 statement V=L] is just not true.”

7/29 Measurable Cardinals and Scott’s Theorem From measures to cardinals

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 statement V=L] is just not true.”

8/29 Measurable Cardinals and Scott’s Theorem 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 class containing all ordinals, then L M.

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

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

9/29 Measurable Cardinals and Scott’s Theorem 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) := S 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

10/29 Measurable Cardinals and Scott’s Theorem Background about L

Alternative characterisation of L: Gödel operations We can deﬁne:

L0 := ;

L +1 := cl(L [ {L }) \P(L ) [ L := L 2On

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 some Y 2 M.

11/29 Measurable Cardinals and Scott’s Theorem 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 a transitive 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 embedding → of V . (V , 2) (M, 2)

(V /U, 2U )

12/29 Measurable Cardinals and Scott’s Theorem 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.

13/29 The ultrapower V /U is the class of minimal-rank representatives of equivalence 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 →

Measurable Cardinals and Scott’s Theorem Scott’s Proof

Ultrapowers of proper classes: Scott’s trick

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

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

14/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

Ultrapowers of proper classes: Scott’s trick

I We want to build ultrapowers of proper class models. Let a cardinal and 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 of equivalence 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 → 14/29 Measurable Cardinals and Scott’s Theorem 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 any ﬁrst order formula '(x1, ..., xn), write

'(f1, ..., fn) := < '(f1( ), ..., fn( )) .

15/29 Measurable Cardinals and Scott’s Theorem 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 'f , ..., f 2 U ( ) 1( ) n( )

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

16/29 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 )

Measurable Cardinals and Scott’s Theorem Scott’s Proof

I Assume V = L, and assume there is a measurable cardinal. Let be the least measurable cardinal. I Take the ultrapower (V /U, 2U ), where U is a non-principal, -complete ultraﬁlter on .

17/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

I Assume V = L, and assume there is a measurable cardinal. Let be the 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 )

17/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

I Assume V = L, and assume there is a measurable cardinal. Let be the 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, 2), giving an elementary embedding of (V , 2) into (M, 2).

(V , 2) (M, 2)

(V /U, 2U )

18/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

I Assume V = L, and assume there is a measurable cardinal. Let be the 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, 2), giving an elementary embedding of (V , 2) into (M, 2).

(V , 2) (M, 2)

(V /U, 2U )

19/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

Embedding

(V , 2) (M, 2)

(V /U, 2U )

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] → → ' 2 L By Łoś, for any →2 we have :

(V , 2) ' a1, ..., an iﬀ (V /U, 2U ) ' [ca1 ], ..., [can ]

so that id is an elementary embedding of V into V /U. 20/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

Mostowski collapse

Let K a class (possibly proper) and E a binary relation on K. Suppose the structure (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.

21/29 → Measurable Cardinals and Scott’s Theorem 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 required properties hold.

I Well-founded( !1-completeness of U) I Set-like S [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 proper class M and an isomorphism :(V /U, 2U ) (M, 2). ⇒ →

22/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

Embedding V in M

(V , 2) (M, 2)

(V /U, 2U ) I The collapsing map : V /U M is deﬁned (by induction on 2U -rank) as

([f]) := →([g]) [g] 2U [f] . Let :(V , 2) (M, 2) be the map deﬁned as := id, so that for any set x 2 V , → (x) := ([cx]).

23/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

Embedding V in M (V , 2) (M, 2)

(V /U, 2U ) I The collapsing map : V /U M is deﬁned (by induction on 2U -rank) as

([f]) := →([g]) [g] 2U [f] . Deﬁnition Let :(V , 2) (M, 2) be the map deﬁned as := id, so that for any set x 2 V , → (x) := ([cx]).

24/29 Crucially, the embedding is non-trivial. In particular, it does not map the least measurable cardinal to itself.

Measurable Cardinals and Scott’s Theorem 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 . By the above, ( ) 2 M. M is transitive, so is in M.

25/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

The embedding is non-trivial

Some immediate consequences of Łoś’ Theorem:

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

25/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

The embedding is non-trivial

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 we have [c] < [d] < [c], which yields

for any < , ([c]) < ([d]) < ([c]) () < ([d]) < () so that = sup () ([d]) < (). < 2

26/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

The embedding is non-trivial

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: it is closed under all Gödel operations, and almost universal). This means 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 .

27/29 Measurable Cardinals and Scott’s Theorem Scott’s Proof

Non-trivial embeddings of V

Assuming V = L, we have constructed a non-trivial elementary embedding 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 that j( ) =6 . → I Close relationship between measurable cardinals and critical points of non-trivial embeddings on V .

Theorem (Keisler) If j : V M is a non-trivial elementary embedding of the universe, then cp(j) is a measurable cardinal. → (Kunen: there are no non-trivial elementary embeddings of V into V .) 28/29 Measurable Cardinals and Scott’s Theorem 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-trivial elementary embeddings I Measurable cardinals are Large large cardinals: assuming the existence of measurable cardinals takes us beyond L.

thanks.

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