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Extremely Large Cardinals in the Absence of Choice

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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 σ, whether or not σ is in T ), T has sufficient expressive power, and • T is consistent. • Then: 1 There is a sentence σ such that T σ and • 0 T 0 σ • ¬ (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 σ 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 −! σ–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<! is a sequence of pairwise disjoint subsets of X, then µ( n Yn)= n µ(Yn). S P 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 σ–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: ↵<βiff ↵ β. 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 : β<δ if δ = 0 is a limit ordinal (i.e., not of • δ { β } 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 ↵<β, V V . ↵ ✓ β 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 λ>. Some pieces of notation and definitions The least cardinal λ>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 λ, there is a cardinal such that (λ) = µ (i.e., such that there is a bijection f : µ (λ)). |P | −!P This cardinal is denoted by 2λ. A cardinal is a strong limit iff 2λ <for all λ<. Fact (ZFC) For every cardinal , + is regular. • For every cardinal λ there is a strong limit cardinal >λ. • The least one is the supremum of λ λ 2λ 22 λ, 2 = i1(λ), 2 = i2(λ), 2 = i3(λ),... { } 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 σ (that is, a formula without free variables), we write M = σ | to indicate that M = σ[~a] for some (equivalently, for every) | assignment a : Var M. −! Fact (ZFC) If is an inaccessible cardinal, then V = σ for every | axiom σ 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 σ (that is, a formula without free variables), we write M = σ | to indicate that M = σ[~a] for some (equivalently, for every) | assignment a : Var M. −! Fact (ZFC) If is an inaccessible cardinal, then V = σ for every | axiom σ 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.
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