Mathematical Research Letters 9, 1 7 (2002) GENERIC ABSOLUTENESS AND THE CONTINUUM. Stevo Todorcevic

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1 Mathematical Research Letters 9, 1 7 (2002) GENERIC ABSOLUTENESS AND THE CONTINUUM Stevo Todorcevic Let H ω2 denote the collection of all sets whose transitive closure has size at most ℵ 1. Thus, (H ω2, ) is a natural model of ZFC minus the ower-set axiom which correctlyestimates manyof the roblems left oen bythe smaller and better understood structure (H ω1, ) of hereditarilycountable sets. One of such roblems is, for examle, the Continuum Hyothesis. It is largely for this reason that the structure (H ω2, ) has recentlyreceived a considerable amount of study (see e.g. [15] and [16]). Recall the well-known Levy-Schoenfield absoluteness theorem ([10, 2]) which states that for everyσ 0 sentence ϕ(x, a) with one free variable x and arameter a from H ω2, if there is an x such that ϕ(x, a) holds then there is such an x in H ω2, or in other words, (1) (H ω2, ) 1 (V, ). Strictlyseaking, what is usuallycalled the Levy-Schoenfield absoluteness theorem is a bit stronger result than this, but this is the form of their absoluteness theorem that allows a variation of interest to us here. The generic absoluteness considered in this aer is a natural strengthening of (1) where the universe V is relaced byone of its boolean-valued extensions V B, i.e. the statement of the form (2) (H ω2, ) 1 (V B, ) for a suitablychosen boolean-valued extension V B. This sort of generic absoluteness has aarentlybeen first considered byj. Stavi (see [11]) and then by J. Bagaria [4] who has also observed that anyof the Bounded Forcing Axioms introduced bym.goldstern and S.Shelah [9] is equivalent to the corresonding generic absoluteness statement for the structure (H ω2, ). The most rominent such statement (besides of course Martin s axiom; see [6]) is the Bounded Martin s Maximum which asserts (2) for anyboolean-valued extension V B which reserves stationarysubsets of ω 1 (see [1]-[4], [6]-[9], [11]-[12], [14]-[16]). The urose of this note is to answer the natural question (aearing exlicitelyor imlicitlyin some of the listed aers) which asks whether anyof the standard forms of the generic absoluteness (2) discussed above decides the size of the continuum. Theorem 1. Assume generic absoluteness (2) for boolean-valued extensions which reserve stationary subsets of ω 1. Then there is a well ordering of the continuum of length ω 2 which is definable in the structure (H ω2, ). Received May 6,

2 2 STEVO TODORCEVIC Remark 2. It should be noted that rior to our work, substantial and quite insiring advances towards Theorem 1 had alreadybeen made. For examle, W. H. Woodin [16, 10.3] roved Theorem 1 under the additional assumtion of the existence of a measurable cardinal and D. Asero [2] roved the weaker conclusion in Theorem 1 to the effect that there is F ω ω of size ℵ 2 which is cofinal in the ordering of eventual dominance of ω ω. We start roving Theorem 1 byfixing the onlyarameter of our definition, a one-to-one sequence r ξ (ξ<ω 1 ) of elements of the Cantor set 2 ω. This allows us to associate to everycountable set of ordinals X, the real r X = r ot(x). (See [13] for a more recise functor X r ot(x) which can also be used in the definitions that follow.) For a air x and y of distinct members of 2 ω, set (x, y) = min{n <ω: x(n) y(n)}. Note that for three distinct members x, y and z of 2 ω, the set (x, y, z) ={ (x, y), (y, z), (x, z)} has exactlytwo elements. Definition 3. Let θ AC denote the statement that for every S ω 1 there exist ordinals γ>β>α ω 1 and an increasing continuous decomosition γ = ν<ω 1 N ν of the ordinal γ into countable sets such that for all ν<ω 1, (3) N ν ω 1 S iff (r Nν α,r Nν β) = max (r Nν α,r Nν β,r Nν ). For a given S ω 1 we let θ 1 AC (S, {r ξ : ξ<ω 1 }) denote the Σ 1 sentence of (H ω2, ) asserting the existence of γ > β > α ω 1 and the decomosition N ν (ν<ω 1 ) of γ satisfying the equivalence (3) for all ν<ω 1. Theorem 1 follows from the following more informative result. Theorem 4. For every S ω 1 there is a boolean-valued extension which reserves stationary subsets of ω 1 and satisfies θ 1 AC (S, {r ξ : ξ<ω 1 }). Corollary 5. The generic absoluteness (2) for boolean extensions which reserve stationary subsets of ω 1 imlies θ AC. We need a few definitions, so let λ =2 2ℵ 1. For ω 1 α<β<γ λ + and P {min, max}, set Sαβγ P = {N [γ] ω : (r N α,r N β )=P (r N α,r N β,r N )}. For A ω 1 and α, β, γ and P as above, set Sαβγ(A) P ={N Sαβγ P : N ω 1 A}.

3 GENERIC ABSOLUTENESS AND THE CONTINUUM 3 For ω 1 α<β<λ +,A ω 1 and P {min, max}, set Γ P αβ(a) ={γ <λ + : S P αβγ(a) is stationary}. Lemma 6. There exist ω 1 α<β<λ +, such that Γ P αβ (A) is stationary for all stationary A ω 1 and P {min, max}. Proof. Otherwise, for each ω 1 α < β < λ +, we can choose a stationary A(α, β) ω 1 and P (α, β) {min, max} such that D(α, β) ={γ <λ + : S P (α,β) αβγ (A(α, β)) is nonstationary} contains a closed and unbounded subset of λ +. It follows that there is a single closed and unbounded set D λ + such that for all α<βin D, D \ (β +1) D(α, β). Alying λ + (ω +2) 2 to the coloring of (α, β) (A(α, β),p(α, β)), we 2 ℵ 1 can find E D of order-tye ω+2, a stationaryset A ω 1, and P {min, max} such that A(α, β) =A and P (α, β) =P for all α<βin E. It follows that (4) S P αβγ(a) is nonstationaryfor all α<β<γin E. Let κ be a large enough regular cardinal and M ν (ν<ω 1 ) a continuous - chain of countable elementarysubmodels of (H κ, ) containing all the objects accumulated so far. Let N ν = M ν λ +, (ν <ω 1 ), and A 0 = {ν A : N ν ω 1 = ν}. Then A 0 is stationaryas A \ A 0 is not. Note that for ν<ω 1 and α<β<γ from E, the set Sαβγ P (A) belongs to M ν,soby(4) N ν γ = M ν γ/ Sαβγ(A). P It follows that: (5) (r Nν α,r Nν β) = P (r Nν α,r Nν β,r Nν γ) for all ν A 0 and α<β<γfrom E, where P = min if P = max and P = max if P = min. Case 1. P = min. For α<βin E and k ω, set A αβ [ k] ={ν A 0 : (r Nν α,r Nν β) k}. For α<βin E, set k αβ = max{k ω : A \ A αβ [ k] is nonstationary}. Fix α<β<γ<δin E. Consider ν A αβ [ k αβ ]. Alying (5) to triles α<β<γand α<β<δ,we get that

4 4 STEVO TODORCEVIC (r Nν α,r Nν β) < (r Nν β,r Nν γ ), and (r Nν α,r Nν β) < (r Nν β,r Nν δ). Note that this gives the equalities r Nν γ n = r Nν β n = r Nν δ n for n = (r Nν α,r Nν β) + 1. It follows that, k αβ (r Nν α,r Nν β) < (r Nν γ,r Nν δ). Since ν was chosen to be an arbitrarymember of A αβ [ k αβ ], we conclude that the set A \ A γδ [ k αβ +1] is nonstationary. Hence k γδ k αβ + 1. This shows that k αβ <k γδ for all α<β<γ<δfrom E, and this is in direct contradiction with the fact that E has order-tye ω +2. Case 2. P = max. Fix α<β<γ<δin E and consider an arbitrary ν Aγδ [ k γδ ]. Alying (5) to triles α<γ<δand β<γ<δ,we get that (r Nν α,r Nν γ) > (r Nν γ,r Nν δ ), and (r Nν β,r Nν γ) > (r Nν γ,r Nν δ). This gives that r Nν α n = r Nν γ n = r Nν β n for n = (r Nν γ,r Nν δ) + 1. It follows that, k γδ (r Nν γ,r Nν δ) < (r Nν α,r Nν β). Since ν is an arbitrarymember of A γδ [ k γδ ], we conclude that the set A \ A αβ [ k γδ +1] is nonstationary. Hence k αβ k γδ + 1. This shows that k αβ >k γδ for all α<β<γ<δfrom E, and this can haen onlyif the set E is finite, a contradiction. The case-analysis shows that our initial assumtion that the conclusion of Lemma 6 is false, leads to contradictions finishing thus the roof. Fix ω 1 α<β<λ + satisfying the conclusion of Lemma 6. Lemma 7. For every stationary set A ω 1 and P {min, max}, the set Sαβ(A) P ={N [λ + ] ω : N ω 1 A & (r Nν α,r Nν β) = = P (r Nν α,r Nν β,r N )} is stationary.

5 GENERIC ABSOLUTENESS AND THE CONTINUUM 5 Proof. Consider an f :(λ + ) <ω λ +. We need to find N Sαβ P (A) such that f (N) <ω N. Since α and β satisfythe conclusion of Lemma 6, the set Γ P αβ (A) is a stationarysubset of λ +, so there is γ Γ P αβ (A) such that f γ <ω γ. Then Sαβγ P (A) is stationaryso alying this to the restriction g = f γ<ω we find N Sαβγ P (A) such that g N <ω N. Note that N Sαβ P (A) and that g N <ω = f N <ω. This finishes the roof. We are now readyto start the roof of Theorem 4. Proof of Theorem 4. Fix a subset S of ω 1. Let P be the oset of all countable increasing continuous transfinite sequences of the form = Nν : ν ν such that for all ν ν, (6) Nν ω 1 S iff (r N ν α,r N ν β) = max (r N ν α,r N ν β,r N ν ). Clearly, it suffices to show that P reserves stationarysubsets of ω 1. So, consider a stationarysubset A of ω 1, P, and a P-term τ for a closed and unbounded subset of ω 1. Let κ be a large enough regular cardinal so that the structure (H κ, ) contains all the objects accumulated so far. Suose first that A S is stationary. By Lemma 7, the set Sαβ max (A S) is stationary, so we can find a countable elementarysubmodel M of (H κ, ) containing all the relevant objects such that M λ + Sαβ max (A S). Working in M we build a decreasing sequence = 0 1 n of elements of P M and a sequence ξ n (n<ω) of ordinals from M ω 1 such that: (a) M λ + = n<ω ν ν n Nν n, (b) n forces that ξ n belongs to τ, (c) su n<ω ξ n = M ω 1. Let ν q = su n<ω ν n and q =( n ) { ν q,m λ + }. n<ω Then q P,q and q forces that M ω 1 belongs to the intersection of τ and A. The case when A \ S is stationaryis considered similarlyusing Lemma 7 for A \ S and P = min, i.e., the fact that Sαβ min (A \ S) is stationary. This finishes the roof of Theorem 4. We finish this note with the following observation which shows that θ AC not onlygives an uer bound on the size of the continuum but also the exact value. Theorem 8. θ AC imlies 2 ℵ0 =2 ℵ1 = ℵ 2. Proof. It remains to show that θ AC imlies 2 ℵ0 =2 ℵ1. This will be done by showing that θ AC violates the weak-diamond rincile of Devlin and Shelah [5]. Define F :2 <ω1 2 byletting F (f) =0ifff codes (in the usual way) a well

6 6 STEVO TODORCEVIC ordering < f of its domain ν which has to be a countable limit ordinal bigger than 0 such that if we let then α f <β f <γ f and α f = ot ({ξ <ν: ξ< f 0},< f ), β f = ot ({ξ <ν: ξ< f 1},< f ), γ f = ot (ν, < f ), (r αf,r βf ) = max (r αf,r βf,r γf ). If weak-diamond holds it would alyto F giving us g 2 ω1 such that for every f 2 ω1, the set {ν <ω 1 : F (f ν) =g(ν)} is a stationarysubset of ω 1. Let S be equal to g 1 (1). Alying θ AC to S we get ω 1 α<β<γ<ω 2 and an increasing continuous decomosition γ = N ν ν<ω 1 of the ordinal γ into countable subsets such that for all ν<ω 1, (7) N ν ω 1 S iff (r Nν α,r Nν β) = max (r Nν α,r Nν β,r Nν ). Then we can find f 2 ω1 which codes a well-ordering < f of ω 1 of order tye γ so that for almost all countable limit ordinals ν, It follows that for almost all ν<ω 1, α f ν = ot(n ν α), β f ν = ot(n ν β), γ f ν = ot(n ν ). (8) ν S iff (r αf ν,r βf ν ) = max (r αf ν,r βf ν,r γf ν ). Let A = {ν <ω 1 : F (f ν) =g(ν)}. Then bythe choice of g, the set A is stationary. If A S is stationary, we can find ν A S for which the equivalence (8) is true, i.e., such that (r αf ν,r βf ν ) = max (r αf ν,r βf ν,r γf ν ). Going back to the definition of F we see that F (f ν) =0, and therefore, g(ν) = 0 which means that ν / S, a contradiction. If A \ S is stationarywe chose ν A \ S satisfying the equivalence (8), i.e., such that (r αf ν,r βf ν ) = min (r αf ν,r βf ν,r γf ν ). Alying the definition of F, we see that F (f ν) =1, and therefore g(ν) =1 which means ν S, a contradiction. This finishes the roof of Theorem 8.

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