Speculations on nature of set theoretic truth: A thesis in 4 parts followed by 4 conjectures W. Hugh Woodin University of California, Berkeley C September 21, 2011 A thesis in four parts The axioms of ZFC Axiom 1 (Extensionality) Two sets A and B are equal if and only if they have the same elements. Axiom 2 (Pairing) If A and B are sets then there exists a set C = fA; Bg whose only elements are A and B. Axiom 3 (Union) If A is a set then there exists a set C whose elements are the elements of the elements of A. Axiom 4 (Regularity or Foundation) If A is a set then either A is empty (i. e. A has no elements) or there exists an element C of A which is disjoint from A. Axiom 5 (Comprehension) If A is a set and '(x) formalizes a property of sets then there exists a set C whose elements are the elements of A with this property. Axiom 6 (Powerset) If A is a set then there exists a set C whose elements are the subsets of A. Axiom 7 (Axiom of Choice) If A is a set whose elements are pairwise disjoint and each nonempty then there exists a set C which contains exactly one element from each element of A. Axiom 8 (Replacement) If A is a set and '(x) formalizes a property which defines a function of sets then there exists a set C which contains as elements all the values of this function acting on the elements of A. Axiom 9 (Infinity) There exists a set W which is nonempty and such that for each element A of W there exists an element B of W such that A is an element of B. The axioms of ZFC0: Finite set theory Axiom 1 (Extensionality) Two sets A and B are equal if and only if they have the same elements. Axiom 2 (Bounding) There exists a set C such that every set is a subset of C. Axiom 3 (Union) If A is a set then there exists a set C whose elements are the elements of the elements of A. Axiom 4 (Regularity) If A is a set then either A is empty or there exists an element C of A which is disjoint from A. Axiom 5 (Comprehension) If A is a set and '(x) formalizes a property of sets then there exists a set C whose elements are the elements of A with this property. Axiom (6a) (Powerset) For all sets A one of the following holds. I P(A) exists I there exists a set C such that V = P(C) and A is not an element of C Axiom (6b) (Powerset) For all sets A one of the following holds. I P(A) exists I there exist B 2 A and C ⊆ B such that C 2= A I V = P(A) Axiom 7 (Axiom of Finiteness) If A is a nonempty set then there is an element B of A such that for all sets C, if C is an element of A then B is not an element of C. The cumulative hierarchy Definition Define for each ordinal α a set Vα by induction on α. 1. V0 = ;. 2. Vα+1 = P(Vα) = fX X ⊆ Vαg. 3. If β is a limit ordinal then Vα = [fVβ β < αg. I It is a consequence of the ZFC axioms that for each set A there exists an ordinal α such that A 2 Vα. I The ZFC0 axioms (1)-(6) imply that V = Vα+1 for some ordinal α, adding Axiom (7), α is finite. For each finite ordinal n > 0,Vn j= ZFC0. I ZFC0 is a very weak theory. Theorem ZFC0 proves it own consistency. I But ZFC0 does not prove there is a model of ZFC0. The axioms of the form \Vn exists" for specific n are \large cardinal" axioms for ZFC0. G¨odelsentences and ZFC0 Theorem (after G¨odel) There is a sentence Φ such that for all models (M; E) j= ZFC0 the following are equivalent: 1.( M; E) j= Φ. 2.( M; E) j= \ ZFC0 ` (:Φ) " The sentence Φ0 Φ0 asserts: 1. Vn exists where n = jV1000j. 24 2. There is a proof of (:Φ0) with length at most 10 from the theory: ZFC0 + \ Vn exists where n = jV1000j ": Φ0 if true is physically verifiable from the witness. I Φ0 is a meaningful statement about the actual physical universe. Question Is Φ0 true? I Of course not. But we have no evidence (physical or mathematical) that Φ0 is false. Claim Any coherent basis (at present) for the assertion that Φ0 is false must also yield that the conception of Vn where n = jV1000j is meaningful. Thesis: Part 1 Any coherent basis for the mathematical claim of the consistency of a formal (recursive) theory T must be paired ultimately with a conception of mathematical objects with structure, whose existence implies the consistency of T . Consistency and independence in ZFC I !1 is the least uncountable ordinal I it is the set of all countable ordinals. Definition 1. A set C ⊆ !1 is closed if for all α < !1 if C \ α is cofinal in α then α 2 C. 2. A set S ⊆ !1 is stationary if S \ C 6= ; for all closed, cofinal, sets C ⊆ !1. I The sets, S ⊂ !1, which are stationary and co-stationary are the simplest manifestation of the Axiom of Choice. I How complicated is the structure of the stationary, co-stationary, subsets of !1? I Can exist a small generating family for these sets? The combinatorics of stationary subsets of !1 A precise question along these lines is the following: Question Can there exist !1 many stationary sets, hSα : α < !1i, such that for every stationary set S ⊆ !1, there exists α < !1 such that Sα ⊆ S modulo a non-stationary set? The assertion that Sα ⊆ S modulo a non-stationary set is simply the assertion that the set, SαnS = fβ < !1 β 2 Sα and β2 = Sg; is not stationary. Observation Such a sequence, hSα : α < !1i, of stationary subsets of !1 would give in a natural sense, a basis for the stationary subsets of !1 which is of cardinality !1. Infinite games on ! and Determinacy Axioms I Associated to a set A ⊆ R is an infinite game involving two players, Player I and Player II. The players construct a function, f : ! ! f0; 1g, in stages, (Stage 0) : Player I specifies f (0), (Stage 1) : Player II specifies f (1), (Stage 2) : Player I specifies f (2), ············ After infinitely many stages a function f : ! ! f0; 1g is constructed. I Player I wins this run of the game if 1 X f (k)2−(k+1) 2 A; k=0 otherwise Player II wins. Strategies A strategy is a function τ : fs : k ! f0; 1g k 2 !g ! f0; 1g and a player follows τ in a run of the game yielding f if at each stage k for that player, f (k) = τ(f jk). Definition (Mycielski, Steinhaus: 1961) The Axiom of Determinacy, AD, is the axiom which asserts that for all sets A ⊆ R there is a winning strategy for either Player I or Player II in the game given by A. I AD contradicts the Axiom of Choice. Question Is the Axiom of Choice necessary to construct a set A ⊆ R for which the corresponding game is not determined? Large Cardinal Axioms Basic template for (modern) large cardinal axioms A cardinal κ is a large cardinal if there exist an ordinal α, a transitive set M, and an elementary embedding, j : Vα ! M such that κ is the least ordinal such that j(β) 6= β. I CRT(j) denotes the least ordinal β such that j(β) 6= β. I If j is the identity on α then j is the identity on Vα. I One can require more sets to belong to M, possibly in a way that depends on the action of j on the ordinals. I A hierarchy of notions. I (Axiom of Choice) If M = Vα then either α = λ or α = λ + 1 where λ is the supremum of hκi : i < !i, κ0 = CRT(j) and for all i < !, κi+1 = j(κi ). Three theories Theory 1 ZFC + \ There exist !1 many stationary sets, hSα : α < !1i, such that for every stationary set S ⊆ !1, there exists α < !1 such that Sα ⊆ S modulo a non-stationary set". Theory 2 ZF + AD Theory 3 ZFC + \There exist infinitely many Woodin cardinals". Theorem These three theories are equiconsistent. Thesis: Part 2 The conception of the universe of sets with the structure from large cardinals can account for all possible consistency claims. Two large cardinal axioms in ZF Definition: κ is an Enormous Cardinal There exist κ < λ < γ and an elementary embedding j : Vλ+1 ! Vλ+1 such that 1. κ = CRT(j) and λ > κ is least such that j(λ) = λ, 2. Vλ ≺ Vγ. Definition: κ is a Weak Reinhardt Cardinal There exist κ < λ < γ and an elementary embedding j : Vλ+2 ! Vλ+2 such that 1. κ = CRT(j) and λ > κ is least such that j(λ) = λ, 2. Vλ ≺ Vγ. Theorem (Kunen) Assume the Axiom of Choice. Suppose λ is an ordinal and j : Vλ+2 ! Vλ+2 is an elementary embedding. Then j is the identity. Corollary (ZFC) There are no Weak Reinhardt Cardinals. Theorem (ZF) Assume there is a Weak Reinhardt Cardinal.