Extremely large cardinals in the absence of Choice

Transcription

1 Extremely large cardinals in the absence of Choice David Asperó University of East Anglia UEA pure math seminar, 8 Dec 2014

2 The language First order language of set theory. Only non logical symbol: 2

3 The axioms Axiom of Extensionality: 8x, y(x = y! (8w)(w 2 x $ w 2 y)) Axiom of Unordered Pairs: 8x, y9z8w(w 2 z! w = x _ w = y) Union Axiom: 8x9y8z(z 2 y! 9w(w 2 x ^ z 2 w)) Power Set Axiom: 8x9y8z(z 2 y! 8w(w 2 z! w 2 x)) Axiom Scheme of Replacement: For all x, v 0,...,v n, if '(v 0,...v n, x, u, v) is functional, then there is y such that for all v, v 2 y if and only if there is some u 2 x such that '(v 0,...v n, x, u, v), for every formula '(v 0,...v n, x, u, v) such that y does not occur as bound variable, and where x, y, u, v, v 0,...,v n are distinct variables. Axiom of Infinity: There is some X whose members are exactly all natural numbers. Axiom of Choice (AC): For every X, if all members of X are nonempty, then there is a choice function for X.

4 The background theories Zermelo Fraenkel set theory, ZF, is the first order theory with all the above axioms, except for the Axiom of Choice. Zermelo Fraenkel set theory with the Axiom of Choice, ZFC, is the first order theory with all the above axioms (including the Axiom of Choice).

5 Most mathematical constructions (the usual number systems, functions, spaces of functions, your favourite algebraic structures, etc.) can be done using sets only. Therefore, most mathematical assertions are (or can be translated to) assertions about sets. ZFC answers many questions arising naturally in mathematics. Sociological fact: ZFC has become the standard foundation (i.e., background theory) in mathematics.

6 Gödel s Incompleteness Theorems Suppose T is a theory in the language of set theory such that T is computable (i.e., there is an algorithm deciding, for any given sentence, whether or not is in T ), T has sufficient expressive power, and T is consistent. Then: 1 There is a sentence such that T 0 and T 0 (First Incompleteness Theorem) 2 T does not prove that T is consistent (T 0 Con(T ), where Con(T ) is an arithmetical sentence expressing T is consistent. ) (Second Incompleteness Theorem)

7 The sentence in the First Incompleteness Theorem is not a natural mathematical statement. However, there are infinitely many natural questions in mathematics that ZFC does not decide (if ZFC is consistent). For example: (1) 1 the cardinality of R? Is 24567? (2) Is the real line the only (up to order isomorphism) complete linear order without end points and without an uncountable collection of pairwise disjoint nonempty intervals? (3) Is there a list of 5 uncountable linear orders such that every uncountable linear order contains a suborder order isomorphic to some of the members of the list?

8 (4) Let X be a closed subset of R 3. Let Y be the projection of X on R 2. Let Z be the projection of R 2 \ Y on R. Is Z necessarily Lebesgue measurable? (5) Let X be a set. µ : P(X)! R is a (probabilistic) additive measure on X iff: (a) µ(;) =0 and µ(x) =1. (b) If µ({a}) =0 for all a 2 X. (c) If (Y n ) n<! is a sequence of pairwise disjoint subsets of X, then µ( S n Y n)= P n µ(y n). Lebesgue measure (restricted to subsets of the interval [0, 1]) satisfies (a) (c) but not every subset of [0, 1] is Lebesgue measurable. Is there any additive measure on [0, 1]?

9 All these results are proved by the complementary methods of building forcing extensions, or building inner models or by a combination of these methods.

10 For this talk, our background theory will be sometimes ZFC and sometimes ZF. I will always make this precise (when not clear from the context).

11 The set theoretical universe Ordinals: An ordinal is a set well ordered by 2. Ordinals are well ordered by: < iff 2. They provide the natural labels for counting in any length we want: The first ordinal is 0 = ;, the second is 1 = {0} = {;}, the third is 2 = {0, 1} = {;, {;}},... the first ordinal after all natural numbers is! = {n : n is a natural number}, the next one is! + 1 =! [{!},... Let Ord = { : an ordinal}.

12 We define (V : 2 Ord) by recursion on Ord: V 0 = ; V +1 = P(V )={X : X V } V = S {V : < } if 6= 0 is a limit ordinal (i.e., not of the form + 1;! is the first nonzero limit ordinal,! +! the second, etc.). (V : 2 Ord) is called the cumulative hierarchy.

13 V 0 = ; V 1 = {;} = 1 V 2 = {;, {;}} = 2 V 3 = {;, {;}, {{;}}, {{;, {;}}} V 4 = 2 4 = 16 V 5 = 2 16 = V 6 = (which, according to Wikipedia, is much bigger than the number of atoms of the observable universe!) V 7 = 2 ( )... V! 0 V!+1 = 0 = i 1 V!+2 = 2 i 1 = i 2 For every ordinal, V!+ = i.

14 Fact (ZF) For all <, V V. Let V denote the set theoretical universe. Fact (ZF) V = S 2Ord V. In other words, every set belongs to some V. Hence, (1) the cumulative hierarchy provides a very appealing picture of the set theoretical universe (every set is generated at some stage as a collection of objects generated at earlier stages, and every collection generated in this way is a set), and (2) ZF says that the universe is described by precisely this picture.

15 Cardinals and large cardinals Definition An ordinal apple is a cardinal iff there is no <applefor for which there is a bijection f :! apple. Examples: Every natural number is a cardinal.! is a cardinal.! + 1 is not a cardinal.! + 2 is not a cardinal.! +! is not a cardinal. Fact (ZF) For every cardinal apple there is a cardinal >apple.

16 Some pieces of notation and definitions The least cardinal >appleis called the successor of apple and is denoted by apple +.! is denoted 0. Given an ordinal, both! denote the th infinite cardinal. A cardinal which is not of the form apple + is called a limit cardinal. An ordinal apple is regular if there is no <applefor which there is a function f :! apple with range cofinal in apple (i.e., such that for every <applethere is some < with apple f ( )).

17 (ZFC) Given a cardinal, there is a cardinal apple such that P( ) = µ (i.e., such that there is a bijection f : µ!p( )). This cardinal is denoted by 2. A cardinal apple is a strong limit iff 2 <applefor all <apple. Fact (ZFC) For every cardinal apple, apple + is regular. For every cardinal there is a strong limit cardinal apple>. The least one is the supremum of {, 2 = i 1 ( ), 2 2 = i 2 ( ), 2 22 = i 3 ( ),...}. Definition A cardinal is inaccessible if it is regular and a strong limit. Question: Is there an inaccessible cardinal?

18 Satisfaction Let L be the language of set theory and let M =(M, R) be an L structure (i.e., R M M). Let Var be the set of variables. Given an assignment ~a : Var! M: M = (v i 2 v j )[~a] if and only if (~a(v i ),~a(v j )) 2 R. M = (v i = v j )[~a] if and only if ~a(v i )=~a(v j ). M = ( ')[~a] if and only if M = '[~a] does not hold. M = (' 0 _ ' 1 )[~a] if and only if M = ' 0 [~a] or =' 1 [~a]; and similarly for the other connectives. M = (9v')[~a] if and only if there is some b 2 M such that M = '[~a(v/b)], where ~a(v/b) is the assignment ~ b such that ~ b(v i )=~a(v i ) if v i 6= v and ~ b(v) =b. M = (8v')[~a] if and only if for every b 2 M, M = '[~a(v/b)].

19 Given a sentence (that is, a formula without free variables), we write M = to indicate that M = [~a] for some (equivalently, for every) assignment a : Var! M. Fact (ZFC) If apple is an inaccessible cardinal, then V apple = for every axiom of ZFC. In other words, V apple is a model of ZFC. By the Completeness Theorem of first order logic, this implies that if there is an inaccessible cardinal, then ZFC is consistent. Corollary If ZFC is consistent, then ZFC does not prove the existence of inaccessible cardinals. Even more: We cannot prove Con(ZFC + There is an inaccessible cardinal ) starting from just Con(ZFC). To sum up: ZFC is strictly stronger (more daring ) than ZFC.

20 Given a sentence (that is, a formula without free variables), we write M = to indicate that M = [~a] for some (equivalently, for every) assignment a : Var! M. Fact (ZFC) If apple is an inaccessible cardinal, then V apple = for every axiom of ZFC. In other words, V apple is a model of ZFC. By the Completeness Theorem of first order logic, this implies that if there is an inaccessible cardinal, then ZFC is consistent. Corollary If ZFC is consistent, then ZFC does not prove the existence of inaccessible cardinals. Even more: We cannot prove Con(ZFC + There is an inaccessible cardinal ) starting from just Con(ZFC). To sum up: ZFC is strictly stronger (more daring ) than ZFC.

21 Inaccessible cardinals are the first (natural) step in the large cardinal hierarchy. There is no official definition of large cardinal, but you recognise a large cardinal notion when you see it.

22 The skeptical response: Maybe the corollary indicates that ZFC + There is an inaccessible cardinal is inconsistent. Reply of the average set theorist: Well, if you feel uncomfortable with ZFC + There is an inaccessible cardinal, then why would you feel confident in ZFC? ZFC \Axiom of Infinity turns out to be essentially the same theory as Peano Artihmetic (PA) and ZFC certainly proves Con(PA), so in a way! is already a large cardinal (taking PA as our base theory). And then: Why would you even feel confident with PA? Conclusion: Stronger theories give you more tools to prove theorems (see below). The large cardinals you are willing to accept (and this includes even!) are a measure of how much risk you are willing to take.

23 Some classical theories in the large cardinal hierarchy ZFC is equiconsistent with ZF and is strictly stronger than ZFC \{Infinity}. ZFC + There is an inaccessible cardinal is strictly stronger than ZFC. ZFC + There is a weakly compact cardinal is strictly stronger than ZFC + There is an inaccessible cardinal. ZFC + There is a measurable cardinal is strictly stronger than ZFC + There is a weakly compact cardinal. ZFC + There is a Woodin cardinal is strictly stronger than ZFC + There is a measurable cardinal. ZFC + There is a supercompact cardinal is strictly stronger than ZFC + There is a Woodin cardinal. ZFC + There is a huge cardinal is strictly stronger than ZFC + There is a supercompact cardinal....

24 Empirical fact: Natural large cardinal notions seem to be linearly ordered under consistency strength. If P 0 and P 1 are two large cardinal notions, one can usually prove Con(ZFC + There is some apple such that P 0 (apple) )! Con(ZFC + There is some apple such that P 1 (apple) ), or Con(ZFC + There is some apple such that P 1 (apple) )! Con(ZFC + There is some apple such that P 0 (apple) ).

25 A word or two on applications of large cardinals Remember: ZFC, if consistent, doesn t settle any of the following questions: (1) 1 the cardinality of R? Is 24567? (2) Is the real line the only (up to order isomorphism) complete linear order without end points and without an uncountable collection of pairwise disjoint nonempty intervals? (3) Is there a list of 5 uncountable linear orders such that every uncountable linear order contains a suborder order isomorphic to some of the members of the list?

26 (4) Let X be a closed subset of R 3. Let Y be the projection of X on R 2. Let Z be the projection of R 2 \ Y on R. Is Z necessarily Lebesgue measurable? (5) Let X be a set. µ : P(X)! R is a (probabilistic) additive measure on X iff: (a) µ(;) =0 and µ(x) =1. (b) If µ({a}) =0 for all a 2 X. (c) If (Y n ) n<! is a sequence of pairwise disjoint subsets of X, then µ( S n Y n)= P n µ(y n). Lebesgue measure (restricted to subsets of the interval [0, 1]) satisfies (a) (c) but not every subset of [0, 1] is Lebesgue measurable. Is there any additive measure on [0, 1]?

27 It turns out that: (a) The theories ZF, ZFC + The answer to question (n) is YES, for n = 1, 2, and ZFC + The answer to question (n) is NO for n = 1, 2, 3, 4, 5 are equiconsistent. (b) Con(ZFC + There is a weakly compact cardinal ) is more than enough to prove Con(ZFC + The answer to question (3) is YES ), and it is not known if, for example, Con(ZFC) suffices.

28 (c) The theories ZFC + There is an inaccessible cardinal, and ZFC + The answer to Question (4) is YES are equiconsistent. (d) The theories ZFC + There is a measurable cardinal and ZFC + The answer to Question (5) is YES are equiconsistent.

29 Astrological properties of large cardinals A lot of mathematics takes place in rather small initial segments of the cumulative hierarchy. V!+n, for some small n <!, is where the situations described in all of the questions (1) (5) live, and where most of ordinary analysis, etc., lives. It is therefore a remarkable fact that the mere existence of large certain large cardinals, very high up, has a direct influence on the properties of down to earth things like sets of reals. Example: Theorem (ZFC) Suppose there are infinitely many Woodin cardinals. Then every projective set of reals (i.e., every set obtained from a closed subset of R n, for some n, in finitely many steps by projecting and taking complements) is Lebesgue measurable.

30 In fact this theorem is a corollary of the following. Theorem (ZFC) Woodin, Martin Steel, mid 1980 s. The following are equivalent. 1 Projective Determinacy. 2 For every n and every real x there is an iterable inner model with n Woodin cardinals and containing x.

31 Measurable cardinals and beyond Let X be a set. F P(X) is a filter on F iff: (1) X 2 F and ; /2 X. (2) For all Y Z X, if Y 2 F, then Z 2 F (F is closed under supersets). (3) For all Y 0, Y 1 2 F, Y 0 \ Y 1 2 F (F is closed under finite intersections). A filter F is an ultrafilter iff: (4) For all Y X, either Y 2 F or else X \ Y 2 F. Examples: If a 2 X, U a = {Y X : a 2 Y } is an ultrafilter. U a is the principal ultrafilter generated by a. F = {Y! :! \ Y <!} is a filter on! (F is called the Fréchet filter).

32 An ultrafilter on X wich is not of the form {Y X : a 2 Y } for any a 2 X is a non principal ultrafilter on X. Given a filter F on X and a cardinal apple, F is apple complete iff T i< Y i 2 F whenever <appleand {Y i : i < } F. Definition A cardinal apple is measurable iff there is a apple complete non principal ultrafilter on apple. Measurable cardinals are a very central large cardinal notion. One reason is:

33 Theorem (Keisler Tarski, Scott, early 1960 s) The following are equivalent. 1 apple is a measurable cardinal. 2 There is a proper class M and an elementary embedding j :(V, 2)! (M, 2), j 6= id, such that crit(j) =apple, and M is closed under apple sequences (i.e., (x ) <apple 2 M for every apple sequence (x ) <apple such that x 2 M for all ).

34 Here, j :(V, 2)! (M, 2) is an elementary embedding iff j preserves satisfaction, i.e., for every formula '(v 0,...,v n ) in the language of set theory and for all a 0,...,a n 2 V, (V, 2) = '(a 0,...,a n ) iff (M, 2) = '(j(a 0 ),...,j(a n )) Also: If j :(V, 2)! (M, 2) is an elementary embedding and is not the identity, then <j( ) for some ordinal. The least such is the critical point of j, denoted crit(j).

35 Proof sketch of the theorem: Suppose apple is measurable. Let U be a apple complete non-principal ultrafilter on apple. Given functions f, g with domain apple, set f = U g iff { <apple: f ( ) =g( )} 2U and f 2 U g iff { <apple: f ( ) 2 g( )} 2U Let M be the Mostowski collapse of the ultraproduct Ult(V, U) =({[f ] =U f : apple! V}, 2 U ) and for every set a let j(a) be the image, under the Mostowski collapsing function, of [c a ] =U, where c a is the function sending every 2 apple to a. Then j :(V, 2)! (M, 2) is an elementary embedding with critical point apple and M is closed under apple sequences.

36 Conversely, suppose j :(V, 2)! (M, 2) is an elementary embedding with crit(j) =apple, and let U = {X apple : apple 2 j(x)} Then U is a apple complete non-principal ultrafilter on apple.

37 The characterisation of measurable cardinal in the theorem provides a very useful blueprint for generating large cardinal notions. Look at: apple is the critical point of an elementary embedding such that M is close to V. j :(V, 2)! (M, 2) The closer to V that M is required to be in this definition, the stronger the large cardinal notion is. It is then usually a routine verification to show that if apple is such a large cardinal, then apple is a limit of cardinals such that is the critical point of an elementary embedding j :(V, 2)! (M, 2) such that M is close 0 to V, where close 0 to V is now any reasonable weaker notion of being close to V.

38 It is then typically the case that V apple, satisfies ZFC since apple is necessarily inaccessible, and thinks that there are many cardinals such that is the critical point of an elementary embedding j :(V, 2)! (M, 2) such that M is close 0 to V. This shows that ZFC + There is an elementary embedding j :(V, 2)! (M, 2) such that M is close to V implies the consistency of ZFC + There is an elementary embedding j :(V, 2)! (M, 2) such that M is close 0 to V. Caveat: Strictly speaking what I am saying doesn t make sense as There is an elementary embedding j :(V, 2)! (M, 2) such that... is a second order statement about the universe (!). However there are typically first order characterisations of precisely these situations, exactly like in the above characterisation of measurable cardinal. In the end everything makes sense.

39 A prominent example: Given an ordinal, a cardinal apple is supercompact iff there is an elementary embedding j :(V, 2)! (M, 2) such that crit(j) =apple, and M is closed under sequences (i.e., (x ) < 2 M for every sequence (x ) < such that x 2 M for all ). A cardinal apple is supercompact if it is supercompact for all. If apple is apple + supercompact, then apple is measurable (trivially) and is also a limit of measurable cardinals. In fact, there is a apple complete non principal ultrafilter U on apple and some X 2U such that every 2 X is a measurable cardinal.

40 Reinhardt cardinals The following is a natural limit for this type of large cardinal notions. A Reinhardt cardinal is the critical point of an elementary j :(V, 2)! (V, 2) This very natural large cardinal axiom was proposed by William Reinhardt in his 1967 PhD thesis.

41 However: Theorem (Kunen, 1971) (ZFC) There are no Reinhardt cardinals. Kunen s proof actually shows that there is no elementary embedding j : V +2! V +2 different from the identity for any ordinal. This is a better result. This remarkable result of Kunen puts an absolute unattainable upper bound to all large cardinal notions, at least in the presence of the Axiom of Choice.

42 However, Kunen s proof doesn t rule out the existence of an elementary embedding j : V +1! V +1 for some ordinal. In fact it doesn t even rule out the existence of an elementary embedding j : L(V +1 )! L(V +1 ) for some ordinal. These axioms have been extensively studied for several decades and no contradiction has arisen yet! They are among the strongest hypotheses not known to be inconsistent (with ZFC as background theory).

43 What about dropping AC? The Axiom of Choice figures prominently in Kunen s inconsistency proof and in all alternative proofs found since then. The following question remains open. Question Is the existence of a Reinhardt cardinal consistent with ZF?

44 Literally speaking the above question doesn t make sense as it again involves second order quantification over the universe. However there are many ways to make good sense of it. For example, one may ask: Question: Let L j be the first order language {2, j} and let T be the first order theory in L comprising the ZF axioms, the axiom 8x 0,...x n,'(x 0,...,x n ) $ '(j(x 0 ),...,j(x n )) for every formula ' in the language of set theory, (9x)(x 6= j(x)), and the axiom Scheme of Replacement for L j formulas. Is T a consistent theory? This question is open.

45 The possibility that Reinhardt cardinals are consistent with ZF opens exciting prospects: Perhaps, if we drop the Axiom of Choice, there is room for extending the large cardinal hierarchy beyond what has been considered so far, and perhaps these new axioms provide an interesting theory. If you believe in large cardinals, you would be inclined to thinking that as we climb up the cumulative hierarchy we would eventually have to abandon the Axiom of Choice. In Koellner s words, perhaps we need to break the AC barrier as we climb up the large cardinal hierarchy, in very much the same way that we need to break the V = L barrier a little before we reach the level of a measurable cardinal.

46 Study of the universe under the existence of a Reinhardt cardinal, or similar large cardinal notions, is difficult since giving up the Axiom of Choice seems to put many limitations to our methods. On the other hand, there are surprising results: In ZF, the existence of certain large cardinals actually implies certain non trivial instances of AC. Example: Theorem (Woodin, 2000 s) (ZF) Suppose is a singular (i.e., not regular) limit of supercompact cardinals. Then + is regular and the club filter on + is + complete.

47 It is conceivable that something like the existence of a Reinhardt cardinal might imply enough of AC to actually recover Kunen s proof (or some other proof of his result) and obtain a contradiction. In any event it is worth developing the theory of Reinhardt cardinals and other large cardinal notions in this region. A sample result:

48 Theorem (A. 2010) (ZF) Suppose there is a Reinhardt cardinal.then there is an ordinal such that for every ordinal there is some > and some elementary embedding i :(V, 2)! (V, 2) such that i( ) >. Proof sketch: Let j :(V, 2)! (V, 2) be an elementary embedding, j 6= id, and let apple = crit(j). Given an ordinal, let apple be the least cardinal apple for which there is an ordinal and an elementary embedding i : V! V with crit(i) =apple and such that { 2 : apple, i( ) = } has order type (if there is such a cardinal).

49 Note that suitable fragments of j witness that apple is defined for every. Moreover, < implies apple apple apple apple crit(j) (apple apple apple is immediate, and apple apple crit(j) is, again, witnessed by a suitable restriction of j). Hence, there is some limit ordinal and some apple apple apple such that apple = apple for every ordinal. Note that in fact apple<apple since apple is definable in (V, 2). Now pick any limit ordinal > and any embedding i : V! V witnessing apple = apple. All we need to do is check that i( ).

50 Claim For every 2 [, ), V = apple = apple. Proof. Let be the -th member of the strictly increasing enumeration of { <µ: apple, i( ) = }. Then i V +1 : V +1! V +1 can be naturally coded as a set belonging to V +2 V (V +2 V is true since { < : apple, i( ) = } has order type, which is a limit ordinal above ). It follows that V thinks that apple exists and that apple apple apple (as witnessed by i V +1 ). On the other hand, V = apple < apple is clearly impossible since. From the above claim it follows that if i( ) <, then V = apple i( ) = apple. But also V = apple i( ) = i(apple )=i(apple) > apple by elementarity of i and since apple = crit(i). This contradiction shows i( ).