Extremely Large Cardinals in the Absence of Choice

Extremely large cardinals in the absence of Choice

David Aspero´

University of East Anglia

UEA pure math seminar, 8 Dec 2014 The language

First order language of set theory. Only non–logical symbol: 2 The axioms Axiom of Extensionality: • x, y(x = y ( w)(w x w y)) 8 ! 8 2 $ 2 Axiom of Unordered Pairs: • x, y z w(w z w = x w = y) 8 9 8 2 ! _ Union Axiom: x y z(z y w(w x z w)) • 8 9 8 2 ! 9 2 ^ 2 Power Set Axiom: • x y z(z y w(w z w x)) 8 9 8 2 ! 8 2 ! 2 Axiom Scheme of Replacement: “For all x, v ,...,v , if • 0 n '(v0,...vn, x, u, v) is functional, then there is y such that for all v, v y if and only if there is some u x such that 2 2 '(v0,...vn, x, u, v),” for every formula '(v0,...vn, x, u, v) such that y does not occur as bound variable, and where x, y, u, v, v0,...,vn 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. 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). 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. Godel’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 and • 0 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) 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) Is 1 the cardinality of R? Is it 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? (4) Let X be a closed subset of R3. Let Y be the projection of X on R2. Let Z be the projection of R2 Y on R. Is Z \ necessarily Lebesgue measurable? (5) Let X be a set. µ : (X) R is a (probabilistic) P ! –additive measure on X iff: (a) µ( )=0 and µ(X)=1. ; (b) If µ( a )=0 for all a X. { } 2 (c) If (Yn)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]? All these results are proved by the complementary methods of building forcing extensions, or • building inner models • or by a combination of these methods. 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). The set–theoretical universe Ordinals: An ordinal is a set well–ordered by . 2

Ordinals are well–ordered by: ↵<iff ↵ . They provide the 2 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 . { } We deﬁne (V : ↵ Ord) by recursion on Ord: ↵ 2

V = • 0 ; V = (V )= X : X V • ↵+1 P ↵ { ✓ ↵} V = V : < if = 0 is a limit ordinal (i.e., not of • { } 6 the form ↵ + 1; ! is the ﬁrst nonzero limit ordinal, ! + ! the S second, etc.).

(V : ↵ Ord) is called the cumulative hierarchy. ↵ 2 V = • 0 ; V = = 1 • 1 {;} V = , = 2 • 2 {; {;}} V = , , , , • 3 {; {;} {{;}} {{; {;}}} V = 24 = 16 • | 4| V = 216 = 65536 • | 5| V = 265536 (which, according to Wikipedia, is much • | 6| bigger than the number of atoms of the observable universe!) 65536 V = 2(2 ) • | 7| ... • V = • | !| @0 V!+1 = 2@0 = i1 • | | i1 V!+2 = 2 = i2 • | | For every ordinal ↵, V!+↵ = i↵. • | | Fact (ZF) For all ↵<, V V . ↵ ✓ Let V denote the set–theoretical universe. Fact (ZF) V = ↵ Ord V↵. In other words, every set belongs to some 2 V↵. S

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. Cardinals and large cardinals

Deﬁnition An ordinal  is a cardinal iff there is no ↵<for for which there is a bijection f : ↵ . ! 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  there is a cardinal >. Some pieces of notation and deﬁnitions

The least cardinal >is called the successor of  and is denoted by +.

! is denoted by . Given an ordinal ↵, both ! and denote @0 ↵ @↵ the ↵–th inﬁnite cardinal.

A cardinal which is not of the form + is called a limit cardinal.

An ordinal  is regular if there is no ↵<for which there is a function f : ↵  with range coﬁnal in  (i.e., such that for ! every ⇠<there is some ⌘<↵with ⇠ f (⌘)).  (ZFC) Given a cardinal , there is a cardinal  such that () = µ (i.e., such that there is a bijection f : µ ()). |P | !P This cardinal is denoted by 2.

A cardinal  is a strong limit iff 2 <for all <.

Fact (ZFC) For every cardinal , + is regular. • For every cardinal there is a strong limit cardinal >. • The least one is the supremum of 2 22 , 2 = i1(), 2 = i2(), 2 = i3(),... . { } Deﬁnition A cardinal is inaccessible if it is regular and a strong limit.

Question: Is there an inaccessible cardinal? Satisfaction

Let 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. L ✓ ⇥ Given an assignment ~a : Var M: ! M =(v v )[~a] if and only if (~a(v ),~a(v )) R. • | i 2 j i j 2 M =(v = v )[~a] if and only if ~a(v )=~a(v ). • | i j i j M =( ')[~a] if and only if M = '[~a] does not hold. • | ¬ | M =(' ' )[~a] if and only if M = ' [~a] or =' [~a]; and • | 0 _ 1 | 0 | 1 similarly for the other connectives. M =( v')[~a] if and only if there is some b M such that • | 9 2 M = '[~a(v/b)], where ~a(v/b) is the assignment ~b such | that ~b(v )=~a(v ) if v = v and ~b(v)=b. i i i 6 M =( v')[~a] if and only if for every b M, • | 8 2 M = '[~a(v/b)]. | 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  is an inaccessible cardinal, then V = for every  | axiom of ZFC. In other words, V 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. 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  is an inaccessible cardinal, then V = for every  | axiom of ZFC. In other words, V 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. 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. 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. 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”. ... • Empirical fact: Natural large cardinal notions seem to be linearly ordered under consistency strength. If P0 and P1 are two large cardinal notions, one can usually prove Con(ZFC +“There is some  such that P ()”) • 0 ! Con(ZFC +“There is some  such that P1()”), or Con(ZFC +“There is some  such that P ()”) • 1 ! Con(ZFC +“There is some  such that P0()”). A word or two on applications of large cardinals

Remember: ZFC, if consistent, doesn’t settle any of the following questions:

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]? 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. (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. 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

Examples: If a X, U = Y X : a Y is an ultraﬁlter. U is the • 2 a { ✓ 2 } a principal ultraﬁlter generated by a. F = Y ! : ! Y

Deﬁnition A cardinal  is measurable iff there is a –complete non–principal ultraﬁlter on .

Measurable cardinals are a very central large cardinal notion. One reason is: Theorem (Keisler–Tarski, Scott, early 1960’s) The following are equivalent. 1  is a measurable cardinal. 2 There is a proper class M and an elementary embedding

j :(V, ) (M, ), 2 ! 2 j = id, such that 6 crit(j)=, and • M is closed under –sequences (i.e., (x ) M for every • ⇠ ⇠< 2 –sequence (x ) such that x M for all ⇠). ⇠ ⇠< ⇠ 2 Here, j :(V, ) (M, ) is an elementary embedding iff j 2 ! 2 preserves satisfaction, i.e., for every formula '(v0,...,vn) in the language of set theory and for all a ,...,a V, 0 n 2 (V, ) = '(a ,...,a ) 2 | 0 n iff (M, ) = '(j(a ),...,j(a )) 2 | 0 n

Also: If j :(V, ) (M, ) is an elementary embedding and is 2 ! 2 not the identity, then ↵

Given functions f , g with domain , set f = g iff • U ⇠<: f (⇠)=g(⇠) { }2U and f g iff • 2U ⇠<: f (⇠) g(⇠) { 2 }2U

Let M be the “Mostowski collapse” of the ultraproduct

Ult(V, )=( [f ]= f :  V , ) U { U | ! } 2U and for every set a let j(a) be the image, under the Mostowski collapsing function, of [ca]= , where ca is the function sending U every ⇠  to a. Then j :(V, ) (M, ) is an elementary 2 2 ! 2 embedding with critical point  and M is closed under –sequences. Conversely, suppose

j :(V, ) (M, ) 2 ! 2 is an elementary embedding with crit(j)=, and let

= X  :  j(X) U { ✓ 2 } Then is a –complete non-principal ultraﬁlter on . U ⇤ The characterisation of measurable cardinal in the theorem provides a very useful blueprint for generating large cardinal notions. Look at:

“ is the critical point of an elementary embedding

j :(V, ) (M, ) 2 ! 2 such that M is ‘close to V’.”

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  is such a large cardinal, then  is a limit of cardinals such that is the critical point of an elementary embedding

j :(V, ) (M, ) 2 ! 2 such that M is close0 to V, where ‘close0 to V’ is now any reasonable weaker notion of being close to V. It is then typically the case that V, satisﬁes ZFC since  is necessarily inaccessible, and • thinks that there are many cardinals such that is the • critical point of an elementary embedding j :(V, ) (M, ) such that M is close to V. 2 ! 2 0 This shows that ZFC + “There is an elementary embedding j :(V, ) (M, ) such that M is close to V” implies the 2 ! 2 consistency of ZFC + “There is an elementary embedding j :(V, ) (M, ) such that M is close to V”. 2 ! 2 0 Caveat: Strictly speaking what I am saying doesn’t make sense as “There is an elementary embedding j :(V, ) (M, ) 2 ! 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. A prominent example:

Given an ordinal , a cardinal  is –supercompact iff there is an elementary embedding j :(V, ) (M, ) such that 2 ! 2 crit(j)=, and • M is closed under –sequences (i.e., (x ) M for every • ⇠ ⇠< 2 –sequence (x ) such that x M for all ⇠). ⇠ ⇠< ⇠ 2 A cardinal  is supercompact if it is –supercompact for all .

If  is +–supercompact, then  is measurable (trivially) and is also a limit of measurable cardinals. In fact, there is a –complete non–principal ultraﬁlter on  and some X U 2U such that every ↵ X is a measurable cardinal. 2 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, ) (V, ) 2 ! 2

This very natural large cardinal axiom was proposed by William Reinhardt in his 1967 PhD thesis. However:

Theorem (Kunen, 1971) (ZFC) There are no Reinhardt cardinals.

Kunen’s proof actually shows that there is no elementary embedding j : V V different from the identity for any +2 ! +2 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. However, Kunen’s proof doesn’t rule out the existence of an elementary embedding

j : V V +1 ! +1 for some ordinal . In fact it doesn’t even rule out the existence of an elementary embedding

j : L(V ) L(V ) +1 ! +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). 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? 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 j be the ﬁrst order language , j and let T be L {2 } the ﬁrst order theory in comprising L the ZF axioms, • the axiom x ,...x ,'(x ,...,x ) '(j(x ),...,j(x )) for • 8 0 n 0 n $ 0 n every formula ' in the language of set theory, ( x)(x = j(x)), and • 9 6 the axiom Scheme of Replacement for j –formulas. • L Is T a consistent theory?

This question is open. 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. 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. • 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: 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 , ) (V , ) 2 ! 2 such that i(↵) >.

Proof sketch: Let j :(V, ) (V, ) be an elementary 2 ! 2 embedding, j = id, and let  = crit(j). Given an ordinal , let 6 ⇤  be the least cardinal  for which there is an ordinal and an elementary embedding i : V V with crit(i)= and such ! that ⇠ : ⇠, i(⇠)=⇠ has order type (if there is such { 2  } a cardinal). Note that suitable fragments of j witness that  is deﬁned for every . Moreover, ↵<implies ↵  crit(j) (↵  is    immediate, and  crit(j) is, again, witnessed by a suitable  restriction of j). Hence, there is some limit ordinal ↵ and some   such that  =  for every or-  ⇤ dinal ↵. Note that in fact < since  is deﬁnable in (V, ). ⇤ 2

Now pick any limit ordinal >↵and any embedding i : V V witnessing  = . All we need to do is check that ! i(↵) . Claim For every [↵,), V =  = . 2 | Proof. Let be the -th member of the strictly increasing enumeration of ⇠<µ: ⇠, i(⇠)=⇠ . Then i V : V V can {  } +1 +1 ! +1 be naturally coded as a set belonging to V V (V V +2 ✓ +2 ✓ is true since ⇠<: ⇠, i(⇠)=⇠ has order type , which {  } is a limit ordinal above ). It follows that V thinks that  exists and that   (as witnessed by i V ). On the other hand,  +1 V =  <  is clearly impossible since ↵. |

From the above claim it follows that if i(↵) <, then V = i(↵) = . But also V = i(↵) = i(↵)=i() >  by | | elementarity of i and since  = crit(i). This contradiction shows i(↵) . ⇤