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The Continuum Hypothesis, Part II W fea-woodin.qxp 6/29/01 1:11 PM Page 681 The Continuum Hypothesis, Part II W. Hugh Woodin Introduction So the resolution of the theory of the structure In the first part of this article, I identified the cor- H(ω2), ∈ could well be a far more difficult rect axioms for the structure P(N), N, +, ·, ∈ , challenge than was the resolution of the theory of which is the standard structure for Second Order the structure H(ω1), ∈. Number Theory. The axioms, collectively “Projec- One example of the potential subtle aspects of tive Determinacy”, solve many of the otherwise un- the structure H(ω2), ∈ is given in the following solvable, classical problems of this structure. theorem from 1991, the conclusion of which is in Actually working from the axioms of set theory, essence a property of the structure H(ω2), ∈. ZFC, I identified a natural progression of structures Theorem (Woodin). Suppose that the axiom Mar- ∈ ∈ increasing in complexity: H(ω), , H(ω1), , tin’s Maximum holds. Then there exists a surjection and H(ω2), ∈, where for each cardinal κ, H(κ) ρ : R → ω2 such that {(x, y) | ρ(x) <ρ(y)} is a pro- denotes the set of all sets whose transitive closure jective set. has cardinality less than κ. The first of these struc- tures is logically equivalent to N, +, ·, the stan- As we saw in Part I, assuming the forcing axiom, dard structure for number theory; the second is Martin’s Maximum, CH holds projectively in that logically equivalent to the standard structure for if X ⊆ R is an uncountable projective set, then Second Order Number Theory; and the third struc- |X| = |R|. This is because Projective Determinacy ture is where the answer to the Continuum Hy- must hold. However, the preceding theorem shows pothesis, CH, lies. The main topic of Part I was the that assuming Martin’s Maximum, CH fails pro- structure H(ω1), ∈. jectively in that there exists a surjection Are there analogs of these axioms, say, some ρ : R → ω2 generalization of Projective Determinacy, for the such that {(x, y) | ρ(x) <ρ(y)} is a projective set. structure H(ω2), ∈? Any reasonable generaliza- tion should settle the Continuum Hypothesis. Such a function ρ is naturally viewed as a “pro- An immediate consequence of Cohen’s method jective counterexample” to CH, for it is a coun- of forcing is that large cardinal axioms are not terexample to the following reformulation of CH: R → R terribly useful in providing such a generalization. Suppose that π : α is a surjection of onto Indeed it was realized fairly soon after the dis- the ordinal α; then α<ω2. covery of forcing that essentially no large cardinal There is a curious asymmetry which follows hypothesis can settle the Continuum Hypothesis. from (the proofs of) these results. Assume there This was noted independently by Cohen and by exist infinitely many Woodin cardinals. Then: Levy-Solovay. Claim (1) There can be no projective “proof” of CH (there can be no projec- W. Hugh Woodin is professor of mathematics at the Uni- tive well-ordering of R of length ω1). versity of California, Berkeley. His e-mail address is [email protected]. Part I of this article appeared in the June/July 2001, Claim (2) There can be a projective issue, pp. 567-576. “proof” of ¬CH (there can be, in The research reported was supported in part by NSF the sense just defined, a projective grant number 9322442. counterexample to CH). AUGUST 2001 NOTICES OF THE AMS 681 fea-woodin.qxp 6/29/01 1:11 PM Page 682 Therefore, if there exist infinitely many Woodin car- Can the theory of the structure dinals and if the Continuum Hypothesis is to be H(ω2), ∈ be finitely axiomatized decided on the basis of “simple” evidence (i.e., (over ZFC) in a (reasonable) logic which projective evidence), then the Continuum Hy- extends first order logic? pothesis must be false. This is the point of the first The logics arising naturally in this analysis sat- claim. But is this an argument against CH? If so, the isfy two important conditions, Generic Soundness playful adversary might suggest that a similar line and Generic Invariance. As a consequence, any ax- ∈ of argument indicates that ZFC is inconsistent, for ioms we find will yield theories for H(ω2), , while we can have a finite proof that ZFC is in- whose “completeness” is immune to attack by ap- consistent, we, by Gödel’s Second Incompleteness plications of Cohen’s method of forcing, just as is Theorem, cannot have a finite proof that ZFC is con- the case for number theory. sistent (unless ZFC is inconsistent). There is a key How shall we define the relevant strong logics? difference here, though, which is the point of the There is a natural strategy motivated by the Gödel second claim. If there is more than one Woodin car- Completeness Theorem. If φ is a sentence in the L ∈ dinal, then a projective “proof” that CH is false can language (ˆ=, ˆ) for set theory, then “ZFC φ” in- always be created by passing to a Cohen extension. dicates that there is a formal proof of φ from ZFC. More precisely, if M,E is a model of ZFC to- This is an arithmetic statement. gether with the statement “There exist 2 Woodin The Gödel Completeness Theorem shows that cardinals”, that is, if if φ is a sentence, then ZFC φ if and only if M,E φ for every structure M,E such that M,E ZFC+“There exist 2 Woodin cardinals,” M,E ZFC. ∗ ∗ then there is a Cohen extension, M ,E , of Therefore a strong logic 0 can naturally be M,E such that M∗,E∗ ZFC + φ, where φ is defined by first specifying a collection of test struc- the sentence which asserts that there exists a sur- tures—these are structures of the form M = M,E, where E ⊂ M × M—and then defining “ZFC φ” jection ρ : R → ω2 such that {(x, y) | ρ(x) <ρ(y)} 0 is a projective set. This theorem, which is the the- if for every test structure M, if M ZFC, then orem behind the second claim above, shows that M φ. what might be called the Effective Continuum Hy- Of course, we shall only be interested in the case pothesis is as intractable as the Continuum Hy- that there actually exists a test structure M such pothesis itself. that M ZFC. In other words, we require that ZFC These claims are weak evidence that CH is false, be consistent in our logic. so perhaps large cardinal axioms are not quite so The smaller the collection of test structures, useless for resolving CH after all. the stronger the logic, i.e., the larger the set of sen- Of course there is no a priori reason that CH tences φ which are proved by ZFC. Note that if should be decided solely on the basis of projective there were only one test structure, then for each evidence. Nevertheless, in the 1970s Martin con- sentence φ either ZFC 0 φ or ZFC 0 ¬φ. So in jectured that the existence of projective evidence the logic 0 defined by this collection of test against CH will eventually be seen to follow from structures, no propositions are independent of reasonable axioms. the axioms ZFC. By the Gödel Completeness Theorem, first order Axioms for H(ω2) logic is the weakest (nontrivial) logic. Encouraged by the success in Part I in finding the To formulate the notion of Generic Soundness, correct axioms for H(ω1), and refusing to be dis- I first define the cumulative hierarchy of sets: couraged by the observation that large cardinal ax- this is a class of sets indexed by the ordinals. The ioms cannot settle CH, we turn our attention to set with index α is denoted Vα, and the definition H(ω2). Here we have a problem if we regard large is by induction on α as follows: V0 = ∅ ; cardinal axioms as our sole source of inspiration: Vα+1 = P(Vα) ; and if β is a limit ordinal, then Even if there is an analog of Projective Determinacy Vβ = ∪{Vα | α<β}. It is easily verified that the for H(ω2), how can we find it or even recognize it sets Vα are increasing, and it is a consequence of if we do find it? the axioms that every set is a member of Vα for My point is simply that the axiom(s) we seek can- large enough α. not possibly be implied by any (consistent) large It follows from the definitions that Vω = H(ω) cardinal hypothesis remotely related to those cur- and that Vω+1 ⊆ H(ω1). However, Vω+1 = H(ω1). rently accepted as large cardinal hypotheses. Nevertheless, Vω+1 and H(ω1) are logically equiv- Strong Logics alent in that each can be analyzed within the other. The solution is to take an abstract approach. This The relationship between Vω+2 and H(ω2) is far we shall do by considering strengthenings of first more subtle. If the Continuum Hypothesis holds, order logic and analyzing the following question, then these structures are logically equivalent, but which I shall make precise. the assertion that these structures are logically 682 NOTICES OF THE AMS VOLUME 48, NUMBER 7 fea-woodin.qxp 6/29/01 1:11 PM Page 683 equivalent does not imply the Continuum Hy- Definition. For a given strong logic 0, the theory pothesis. of the structure H(ω2), ∈ is “finitely axiomatized Suppose that M is a transitive set such that over ZFC” if there exists a sentence Ψ such that for M,∈ ZFC.
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