ON NICE EQUIVALENCE RELATIONS ON λ 2 SH724

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1 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH724 Saharon Shelah The Hebrew University of Jerusalem Einstein Institute of Mathematics Edmond J. Safra Campus, Givat Ram Jerusalem 91904, Israel Department of Mathematics Hill Center-Busch Campus Rutgers, The State University of New Jersey Piscataway, NJ USA Abstract. Let E be an equivalence relation on the powerset of an uncountable set, which is reasonably definable. We assume that any two subsets with symmetric difference of size exactly 1 are not equivalent. We investigate whether for E there are many pairwise non-equivalent sets Mathematics Subject Classification. (2000); FILL!! 03E47,03E35; 20K94. Key words and phrases. set theory; definable equivalence relations, generalizing descriptive set theory to uncountable cardinals; perfect sets of pairwise non-equivalent; abelian groups, Ent. I would like to thank Alice Leonhardt for the beautiful typing Written 99/11 First Typed /Feb/12; (revised with proofreading for the journal and little more) Latest Revision - 03/Jan/29 1 Typeset by AMS-TEX

2 2 SAHARON SHELAH Annotated content 0 Introduction 1 Dichotomical results on nice equivalence relations [Assume E is a Π 1 1 [λ]-equivalence relation on λ 2 such that η,ν are not E- equivalent whenever they differ in exactly one place. Assume further that this holds even after adding a λ-cohen subset of λ. If λ = λ <λ ℶ ω (alternatively, E is more nicely defined or other requirement on λ) then E has a perfect set (so 2 λ elements) of pairwise non E-equivalent members of λ 2. There are related results.] 2 Singular of uncountable cofinality [Assume λ = λ <κ > cf(λ) = κ > ℵ 0. We find on κ λ quite nice equivalence relations for which the parallel of the results of 1 fail badly. If λ is strong limit, we can use λ 2.] 3 Countable cofinality: positive results [Assume that λ > cf(λ) = ℵ 0 and λ is the limit of measurables, or just a related property (which consistently holds for ℵ ω = ℶ ω ) is satisfied. We prove the parallel of the result in 1 on ω λ.] 4 The countable cofinality case: negative results [We show that if our universe is far enough from large cardinals (and close to L) then we can build counterexamples as in 2.] 5 On r p (Ext(G,Z)) [We return to the p-rank of the abelian group Ext(G,Z) where G is torsion free abelian group (ℵ 1 -free, without loss of generalitysee [Fu](xxx)). We show that if κ is compact, λ strong limit (singular) cardinal > κ and r p (Ext(G,Z)) λ then r p (Ext(G,Z)) 2 λ. This is preserved by adding κ Cohens, κ super-compact. If GCH holds above κ we have a complete characterization of {Ext(G, Z) : G}.]

3 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Introduction The main topic here is the possible generalizations of the following theorem from [Sh 273] on simple equivalence relation on ω 2 to higher cardinals. 0.1 Theorem. 1) Assume that (a) E is a Borel 2-place relation on ω 2 (b) E is an equivalence relation (c) if η,ν ω 2 and (!n)(η(n) ν(n)), then η,ν are not E-equivalent. Then there is a perfect subset of ω 2 of pairwise non E-equivalent members. 2) Instead of E is Borel, E is analytic (or even a Borel combination of analytic relations) is enough. 3) If E is a Π 1 2 relation which is an equivalence relation satisfying clauses (b) + (c) also in V Cohen, then the conclusion of (1) holds. In [Sh 273], Theorem 0.1 was used to prove a result on the homotopy group: if X is a Hausdorff metric topological space which is compact, separable, arc-connected, and locally arc-connected, and the homotopy group is not finitely generated then it has the cardinality of the continuum; the proof of 0.1 used forcing in [Sh 273], see [PaSr98] without the forcing. We may restrict E to be like the natural equivalence relation in presenting r p (Ext(G,Z)) or just closer to group theory as in Grossberg Shelah [GrSh 302], [GrSh 302a], Mekler-Roslanowski-Shelah [MRSh 314], [Sh 664]. In 5 we say somewhat more. We here continue [Sh 664] but do not rely on it. Turning to λ 2 the problem split according to the character of λ and the simplicity ofe. If E isπ 1 1 andλ = λ<λ andλ ℶ ω (or just (Dl) λ holds), a generalization holds. If E is Σ 1 1 and λ = λ<λ, the generalizationin general fails; all this in 1. Now if λ is singular, strong limit for simplicity, it is natural to consider cf(λ) λ instead of λ 2. If λ has uncountable cofinality we get strong negative results in 2. If λ has countable cofinality, and is the limit of somewhat large cardinals, e.g. measurable cardinals, (but λ = ℵ ω may be O.K., i.e., consistently) the generalization holds (in 3), but if the universe is close to L (e.g. in L there is no weakly compact) then we get negative results (see 4). Note that theorems of the form if E has many equivalence classes it has continuum many equivalence classes do not generalize well, see [ShVs 719] even for λ weakly compact. We thank Alex Usuyatsov for many helpful comments and corrections.

4 4 SAHARON SHELAH 0.2 Definition. For a cardinal λ let B λ be λ 2 (or λ λ or cf(λ) λ); we write B for such set. 1) For a logic L we say that E is a L-nice, (say 2-place for simplicity), relation on B if there is a model M with universe λ and finite vocabulary τ, and unary function symbols F 1,F 2 / τ (denoting possibly partial unary functions), such that letting τ + = τ {F 1,F 2 }, for some sentence ψ = ψ(f 1,F 2 ) in L(τ + ) we have for any η 1,η 2 B letting M η1,η 2 = (M,η 1,η 2 ) be the τ + -model expanding M with F M η 1,η 2 l = η l for l = 1,2 we have η 1 Eη 2 (M,η 1,η 2 ) = ψ. We may write M = ψ[η 1,η 2 ] and ψ[η 1,η 2,M] or ψ(x,y,m) or write a λ coding M instead of M. 2)E isaπ 1 1-relationon B meansthataboveweallowψ tobeoftheform( X λ)ϕ where ϕ is first order or even in inductive logic (i.e., we have variables on sets and are allowed to form the first fix point for formula ϕ(x,x) such that ϕ(x,x 1 ) & X 1 X 2 ϕ(x,x 2 )); if we allow just first order ϕ we say strictly if we allow formulas ϕ from L we say L-strictly. Similarly Σ 1 1,Π1 2, projective; writing nice means L is L(induction) i.e. first order + definition by induction. We may write E nice(b λ ),Σ 1 1[B] etc, and may replace B by λ if this holds for every B = λ 2. We write very nice for L-nice when L is L first order logic. 0.3 Notation: ( i < δ) means for every large enough i < δ. Jδ bd is the ideal of bounded subsets of δ. L denotes a logic, L(τ) denotes the language (i.e, a set of formulas, for the logic L in the vocabulary τ), L denotes first order logic, L λ,κ denotes the extension of L by allowing ϕ α (when α( ) < λ) and ( x 0,...,x i ) i<α( ) ϕ for α( ) < κ. α<α( ) We note the obvious (by now) relation 0.4 Fact. 1) If λ = λ <κ and R is a [strict] L λ +,κ-nice relation on B λ then R is a [strict] Σ 1 1 -relation and also a [strict] Π1 1-relation (with parameter a relation of λ, of course). If κ > ℵ 0,L λ +,κ = L λ +,κ (induction). 2) If R is a L λ +,κ(induction)-nice relation on B λ and κ > ℵ 0, then R is a L λ +,κ- [strict] Π 1 1 -relation on B λ and Σ 1 1-relation on B. 3) If cf(λ) > ℵ 0 then if R is L(induction)-nice relation on B λ then R is strictly Σ 1 1 -nice (hence being Σ1 1 is equivalent to being strictly Σ1 1 ).

5 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH724 5 Proof. 1) The quantification on X λ can code the satisfaction relation for any subformula. 2) Easy. 3) It is well known that a linear order < on such λ is a well ordering iff for every α < λ,< {β : β < α} is isomorphic to (γ,<) for some γ < λ (e.g. [Na85]) Definition. Let (Dl) λ means that λ is regular, uncountable and there is a sequence P = P α : α < λ such that P α is a family of < λ subsets of α and for every X λ the set {δ < λ : X δ P δ } is stationary; hence λ = λ <λ. (By [Sh 460], λ = λ <λ ℶ ω (Dl) λ and (by Kunen) λ = µ + (Dl) λ λ ). 0.6 Definition. Q λ 2 is called perfect or λ-perfect if: (a) Q (b) if η Q then {lg(η ν) : ν Q\{η}} λ is an unbounded subset of λ (c) the set {η ζ : η Q and ζ λ} is closed under the union of -increasing sequences. Equivalently, Q = {ρ η : η λ 2} such that (a) ρ η λ 2 (b) η 1 η 2 λ 2 ρ η2 ρ η2 (c) if η 0,η 1,η 2 λ 2 are distinct and (η 1 η 2 ) (η 1 η 0 ) (so η 1 η 2 η 1 η 0 ) then (ρ η1 ρ η2 ) (ρ η1 ρ η0 ) and ρ η1 (lg(ρ η1 ρ η2 )) = η 1 (lg(η 1 η 2 )).

6 6 SAHARON SHELAH 1 Dichotomical results on nice equivalence relations on λ 2 We here continue [Sh 664, 2], the theorem and most proofs can be read without it. The claims below generalize [Sh 273]. 1.1 Claim. Assume 1 (a) λ = λ <λ and λ ℶ ω or just (Dl) λ (see 0.4) (b) E is a nice 2-place relation on λ 2 (c)(α) E is an equivalence relation on λ 2 (β) if η,ν λ 2 and (!α < λ)(η(α) ν(α)) then (ηeν). Then E has 2 λ equivalence classes, moreover a perfect set of pairwise non E- equivalent members of λ 2. Proof. Note that If P is a λ-complete forcing (or just λ-strategically complete) then P clauses (c), (α),(β) are still true. So we can apply 1.2 below. 1.1 A relative is 1.2 Claim. Assume 2 (a),(c) as in 1 (b) E is a Π 1 1[λ] 2-place relation on λ 2, say defined by ( Z)ϕ(x,y,Z,ā) see Definition 0.2 (c) + = (c) + Cohen if P = ( λ> 2, ), i.e. λ-cohen, then in V P clauses (c) from 1.1 still hold. Then the conclusion of 1.1 holds. Proof. Stage A: Let (η 0,η 1 ) λ 2 λ 2 be generic over V for the forcing Q = ( λ> 2) ( λ> 2) ordered naturally, i.e., (η 0,η 1 ) (ν 0,ν 1 ) iff η 0 ν 0 & η 1 ν 1. Now do we have V[η 0,η 1 ] = η 0 Eη 1? If so, then for some (p 0,p 1 ) ( λ> 2) ( λ> 2) we have (p 0,p 1 ) Q η let α < λ be > lg(p 0 ),lg(p 1 ) and by clause (c) 0Eη 1, + (β) in V[η 0,η 1 ] we can find η 1 λ 2 such that η 1 α = η 1 α, and for some β (α,λ),η 1 [β,λ) = η 1 [β,λ), (here β = α+1 is O.K. but not so in some generalizations) and V[η 0,η 1 ] = (η 1 Eη 1).

7 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH724 7 So V[η 0,η 1 ] = (η 0 Eη 1 ) (again as in V[η 0,η 1 ],E is an equivalence relation by clause (c) + and we are assuming for the time being that V[η 0,η 1 ] = η 0 Eη 1 ). But also (η 0,η 1 ) is generic over V for (λ> 2) ( λ> 2) with (p 0,p 1 ) in the generic set and V[η 0,η 1 ] = V[η 0,η 1] so we get a contradiction to (p 0,p 1 ) (η Hence 0Eη 1). 1 ( λ> 2) ( λ> 2) (η 0Eη 1). Stage B: Let χ be large enough and let N (H (χ), ) be such that N = λ,n <λ N and the definition of E belongs to N. Note that 2 if (η 0,η 1 ) ( λ 2) ( λ 2) (and is in V) and N[η 0,η 1 ] = (η 0 Eη 1 ), then (η 0 Eη 1 ). [Why? As E is Π 1 1, in N[η 0,η 1 ], there is a witness λ 2 for failure, and it also witnesses in V that (η 0 Eη 1 ).] Clearly to finish proving 1.1, it suffices to prove 1.3 Subclaim. 1) Assume λ = λ <λ and (Dl) λ. If H (λ) N,N <λ N, N = λ and N = ZFC, then there is a perfect Q λ 2 such that for any η 0 η 1 from Q the pair (η 0,η 1 ) is generic over N for [( λ> 2) ( λ> 2)] N. 2) Assume that λ is regular and (a) T is a tree with λ levels each of cardinality < λ and 2 λ λ-branches (or just µ) and (b) N = Nα : α < λ is -increasing, N (α + 1) Nα+1, T N 0 and α N α, N α < λ and N = α<λn α and T α N α+1 (if λ is regular it is enough that N (α+1) N, T α N) (c) < is a well ordering of N such that < N α N α+1. Then for some X λ 2, X = 2 λ (or just X = µ) and η 0 η 1 X the pair (η 0,η 1 ) is generic over N for ( λ> 2) ( λ> 2). 3) Like part (2) but we weaken clause (a) to (a) T is a tree with λ levels each of cardinality λ and Ȳ = Y α : α λ,y α is a set of < λ nodes of T of level α if α < λ and a set of λ-branches of T if α = λ and Y λ µ and η ν Y λ ( α < λ)(η α,ν α Y α ).

8 8 SAHARON SHELAH 1.4 Remark. A tree T as in clause (a) of 1.3(2) is called a λ-kurepa tree and much is known on its existence (and non existence). E.g. if λ is strong limit then such T exists. Proof. 1) Let P α : α < λ be such that P α P(α), P α < λ, and for every X λ the set {α : X α P α } is stationary. So by coding we can find P α {(η 0,η 1 ) : η 0,η 1 α 2} of cardinality < λ such that for every η 0,η 1 λ 2 the set {α < λ : (η 0 α,η 1 α) P α } is stationary. Lastly, let I α : α < λ list the dense open subsets of ( λ> 2) ( λ> 2) which belong to N. Now we define by induction on α < λ, ρ η : η α 2 such that: (a) ρ η λ> 2 (b) β < lg(η) ρ η β ρ η (c) ρ ηˆ l ρ ηˆ l (d) ifαisalimitordinal and(η 0,η 1 ) P α,l 0 < 2,l 1 < 2 andη 0ˆ l 0 η 1ˆ l 1 then (ρ η0ˆ l 0,ρ η1ˆ l 1 ) I β. β α Thereisnoproblemtocarrythedefinition(using P α < λ = cf(λ))and{ α<λρ η α : η λ 2} is a perfect set as required. 2) Similar. We choose by induction on α, ρ η : η T α such that (a),(b),(c) above hold and (d) if η 0 η 1 are in T α+1 then (ρ η0,ρ η1 ) {I : I is a dense open subset of P and belongs to N α } (e) if ρ η : η T α is the < χ-sequence satisfying (a)-(d). So ρ η : η T α can be defined from N β : β < α. The proof in part (2) is easier as we can assume that such a tree belongs to N. 3) Left to the reader. So we have finished proving claim 1.2 hence claim , Claim. 1) In claims 1.1, 1.2 we can weaken clause (β) (in (c),(c) +, call it (c),(c) ± respectively) to: (β) if η λ 2 and α < λ then for some β (α,λ) and ρ [α,β) 2 the sequences η,((η α)ˆρˆη [β,λ)) are not E-equivalent.

9 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH ) In claims 1.1, 1.2 and in 1.5(1), for any ε λ we can replace E by E ε : ε < ε, each E ε satisfying clauses (b) and (c),(c) +,(c),(c) ± there respectively and we strengthen the conclusion: ( ) there is a λ-perfect set Q such that (α) Q = {ρ η : η λ 2} and (β) if η 1 η 2 are from λ 2 then ρ η1 ρ η2 and ε < ε (ρ η1 E ε ρ η2 ) (γ) for η λ 2 the set {lg(ρ η ρ ν ) : ν λ 2\{η}} is a closed unbounded subset of λ. 3) In 1.2, 1.5(1),(2) we can weaken (c) + or (c) ± to ( ) for a stationary set of N [H (λ + )] λ there is (in V) η λ 2 which is Cohen over N such that Π 1 1[λ] sentences are absolute from N[η] to V (for Σ 1 1 [λ]-sentences this is necessarily true) and clause (c) (or (c) ) holds. Proof. 1), 2) The same as the proof of ) The only place it makes a difference is in Stage A of the proof of Claim 1.1. We choose N,η as in ( ) of 1.5(3), and let η l = η(2α+l) : α < λ in N[η] = N[η 0,η 1 ] instead of working with V[η 1.5 0,η 1]. Now we would like not to restrict ourselves to Π 1 1 [λ]-equivalence relations. 1.6 Claim. 1) Assume (a) λ = λ <λ,µ 2 λ (b) E isaπ 1 2 [λ] 2-place relation on λ 2, saydefinable by ( Z 1 )( Z 2 )ϕ(x,y,z 1,Z 2,a) (c)(α) E is an equivalence relation on λ 2 (β) if η,ν λ 2 and (!α < λ)(η(α)) ν(α)) then (ηeν) (c) + if η λ 2 is generic over V for ( λ> 2, ), i.e. is a Cohen sequence over V then in V[η], clause (c) still holds (note that for ρ 1,ρ 2 ( λ 2) V anyhow V = ρ 1 Eρ 2 V[η] = ρ 1 Eρ 2 ) (d) for every A λ and χ > 2 λ there are N, ρ ε : ε < µ such that (i) N (H (χ), ),N <λ N, N = λ,a N (ii) ρ ε λ 2 and [ε < ζ ρ ε ρ ε ] (iii) for ε ζ the pair (ρ ε,ρ ζ ) is generic over N for the forcing notion ( λ> 2 λ> 2) (iv) Π 1 1 [λ] formulas are preserved from N[ρ ε,ρ ζ ] to V for ε < ζ < µ.

10 10 SAHARON SHELAH Then E has µ equivalence classes. 2) We can replace µ by perfect in the conclusion if in (d), {ρ ε : ε < µ} λ 2 is perfect [see 0.6]. 3) We can replace λ> 2 by a subtree T λ> 2 such that forcing with T adds no bounded subset to λ. Proof. By [Sh 664, 2.2t]. 1.7 Definition. Clause (d) of 1.6 is called λ is [λ, µ)-weakly Cohen-absolute: [λ, µ)-w.c.a., in short (as in [Sh 664, 2.1t] s notation). 1.8 Claim. We can strengthen 1.6 just as 1.5 strenghthens 1.1. We may wonder when does clause (d) of 1.6 hold. 1.9 Claim. 1) Assume (i) λ = λ <λ in V (ii) P is a forcing notion (iii) η : ε < µ is a sequence of P-names, ε (iv) P η η ζ ε λ 2 for ε < ζ < µ (v) if A λ,p P,χ large enough then there are N (H (χ), ), N = λ,n <λ N,{A,p} N and q such that p q P,q is (N,P)-generic, q ( λ> 2) VP P] and P N[G P such that q P for some u [µ] µ, for every ε ζ fromu, the pair (η is generic over P ε,η ζ) N[G ] for (λ> 2 λ> 2) V P and the forcing P/(P + η + η is λ-complete (or at least λ-strategically ε ζ) complete). Then λ is (λ,µ)-w.c.a. (see 1.7) in the universe V P. Proof. Straightforward.

11 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Singulars of uncountable cofinality In this section we show that the natural generalization of 0.1 usually provably fails badly for cf(λ) λ,λ singular of uncountable cofinality. 2.1 Claim. Assume (a) λ > κ = cf(λ) > ℵ 0 (b) 2 κ +λ <κ = λ. Then there is E such that (α) E is an equivalence relation on κ λ (β) E is very nice 1 (see Definition 0.2) (γ) if η 1,η 2 κ λ and ( i < κ)(η 1 (i) = η 2 (i)) then η 1 Eη 2 η 1 = η 2 (δ) E has exactly λ equivalence classes. 2.2 Observation. In 2.1, and in the rest of this section: (of course, we have to translate the results; we leave it as an exercise to the reader). 1) We can restrict ourselves to λ i where i < κ λ i < λ = j, see the proof; i<κ j<κλ similarly in ) We can consider κ λ as a subset of λ 2, in fact a very nice one: we identify η κ λ with ν η λ 2 when ν η (i) = 1 i {pr(ζ,η(ζ)) : ζ < κ} for any choice of a pairing function pr, that is, any one to one function pr from κ λ onto λ is O.K. 3)Ifλisstronglimitwecanidentify i<κλ i with λ 2asfollows: without loss of generalityλ i = 2 µ i with µ i increasing, let g i ε : ε < µ i 2 list the functions from [ j<iµ j,µ i ) to {0,1} and we identify η λ i with η(i) i<κ i<κg i λ 2. 4) We can translate our results to any λ i when λ i λ = i = lim sup λ i : i<κ i<κλ i < κ. 5) Even without assuming 2 κ +λ <κ = λ, the union of λ Tichonov closed subsets 1 In fact we have a closed division of κ λ to κ 2 sets such that E refines this division and on each part E is closed, see 2.2(5)

12 12 SAHARON SHELAH of ( κ λ) ( κ λ) is very nice where A κ λ κ λ is closed when: if (η,ν) κ λ κ λ and for every finite u κ for some (η,ν ) A we have η u = η u & ν u = ν u then (η,ν) A. 6) If λ = λ <κ (5) holds even for tree closed subsets (the topology we normally use). Proof of 2.2. (1),(2),(3) left to the reader. 4) Define the function F from ζ λ to λ i by defining F(η) by induction on lg(η) as follows: (a) F(<>) =<> ζ κ ζ κ i<ζ (b) F(ηˆ α ) is F(η)ˆρ η,α when: ε η,α = Min{ε : α < λ lg(f(η))+ε },ρ η,α = 0 εη,αˆ 1+α (c) for η of limit length, F(η) = F(η ε). ε<lg(η) Clearly lg(η) lg(f(η)) and η,ν are -incomparable implies F(η),F(ν) are - incomparable, so F is one to one. Also F maps κ λ into i<κλ i continuously so Range(F) is a closed set. Also,whencf(κ) > ℵ 0 foranyη,ν κ λwehave( ε)(η(ε) = ν(ε)) ( ε)((f(η))(ε) = F(ν))(ε)). This is enough to translate 2.1 to i<κλ i instead of κ λ. Alternatively, we can repeat the proof. 5) Why is it very nice? Assume E = {E i : i < i( )},i( ) λ and each E i is a closed subset of ( κ λ) ( κ λ). Let {ν α : α < λ} list κ> λ with no repetitions, and we define a model M: its universe is λ F 0 is unary, F 0 (α) = lg(ν α ) F 1 is binary, F 1 (α,ε) = β iff ν β = ν α (min{ε,lg(ν α )) R is a three-place relation, R M (α,β,i) iff for some (η 0,η 1 ) E i we have ν α η 0,ν β η 1 P is unary predicate P M = i( ) < is binary relation, the order on λ,κ an individual constant. Now for f,g : κ λ we have feg iff (M,f,g) = ( i)[p(i) & ( ε < κ)( α,β)(f 0 (α) = ε & F 0 (β) = ε & R(α,β,i) & ( ζ < ε)[f(ζ) = F 1 (α,ζ) & g(ζ) = F 1 (β,ζ)]). Normally we do not elaborate such things. 2.2

13 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Proof of 2.1. We choose λ = λ i : i < κ, nondecreasing, i.e. i < j λ i λ j with limit λ, (e.g. λ i = λ which is the case stated in the claim) let µ j = i<jλ i so µ j λ and let f i = f i α : α < µ i list j<iλ j or be just a set of representatives of j<i λ j /J bd i. For every η i<κλ i let (a) for limit i < κ let α i (η) = Min{α : η i = fα i mod Jbd i } (b) for ε < κ let B ε (η) = {i : i < κ is a limit ordinal, ε < i and fα i i (η)(ε) = η(ε)} and lastly (c) A(η) = {ε < κ : B ε (η) is not stationary}. Now we define two binary relations E 0,E 1 on i<κλ i : (d) η 1 E 0 η 2 iff for every ε < κ we have B ε (η 1 ) = B ε (η 2 ) (e) η 1 E 1 η 2 iff η 1 E 0 η 2 & η 1 A(η 1 ) = η 2 A(η 2 ). Clearly (α) E 0 is an equivalence relation on i<κλ i with 2 κ λ classes (β) E 1 is an equivalence relation on κ λ, refining E 0 (γ) E 0,E 1 are very nice; in details: (a) E 0 is a closed subset of ( i<κ λ i ) ( i<κλ i ) (under the initial segment topology, that is, for (η 0,η 1 ) ( i<κ {u ε (η 0 ε,η 1 ε) : ε < κ} where uε ρ = {{(ν 0,ν 1 ) ( i<κ ε,ν 1 ε) = ρ} is a neighborhood basis of (η 0,η 1 )) λ i ) ( i<κλ i ) the family λ i ) ( i<κλ i ) : (ν 0 (b) E 1 is the union of 2 κ closed subsets of ( κ λ) ( κ λ) under the initial segment topology [Why? (a) as if (η 0,η 1 ) λ i i \E 0, then for some ε < κ i<κ i<κλ and i < κ, we have (i B ε (η 0 )) (i / B ε (η 1 )) so ε < i < κ

14 14 SAHARON SHELAH (b) and so u = u i (η 0 ε,η 1 ε) is a neighbor of (η 0,η 1 ) and by the definition of B ε ( ) we have u E 0 = hence u E 1 = for B = B ε : ε < κ,b ε κ let Γ B = {η i<κλ i : B ε (η) = B ε for every ε < κ}. Now Γ B : B κ P(κ) list the E 0 -equivalence classes (and ) and each E 1 Γ B is closed.] (δ) if η 1,η 2 κ λ and η 1 E 0 η 2 then A(η 1 ) = A(η 2 ) [Why? Check the definitions] (ε) for η κ λ,a(η) is a bounded subset of κ [why? otherwise let C = {δ < κ : δ = sup(a(η) δ)}, it is a club of κ, and for each i C there is j i < i such that η [j i,i) = f i α i (η) [j i,i), clearly j i exists by the definition of α i (η). By Fodor lemma, for some j( ) < κ the set S j( ) = {i C : j i = j( )} is stationary, now choose ε A(η)\j( ), so clearly B ε (η) includes S j( ) \ε hence is a stationary subset of κ hence by the definition of A(η) clearly ε does not belong to A(η), contradiction.] So clearly (ζ) E 1 has ( i<κ λ i /E 0 )+Σ{ j<iλ j : i < κ} λ equivalence classes. Now (η) if η 1,η 2 κ λ and η 1 = η 2 mod Jκ bd then for every limit i < κ large enough we have α i (η 1 ) = α i (η 2 ) [why? let i = sup{j+1 : η 1 (j) η 2 (j)} so by the assumption, if i is a limit ordinal and i (i,κ) then η 1 i = η 1 i mod Ji bd hence α i (η 1 ) = α i (η 2 ) by the definition of α i ( ), which is the desired conclusion of clause (η).] (θ) if η 1,η 2 κ λ and η 1 = η 2 mod J bd κ then η 1E 1 η 2 η 1 = η 2 [why? if η 1 = η 2 clearly η 1 E 1 η 2 ; so assume η 1 E 1 η 2 and we shall show that η 1 = η 2, i.e. ε < κ η 1 (ε) = η 2 (ε). By the definition of E 1 we have η 1 E 0 η 2 hence by clause (δ) we have A(η 1 ) = A(η 2 ), call it A. If ε A, by the definition of E 1 we have η 1 A = η 2 A hence η 1 (ε) = η 2 (ε). So assume ε κ\a, first we can find j < κ such that for every limit i (j,κ) we have α i (η 1 ) = α i (η 2 ), it exists by clause (η). Second, the sets B ε (η 1 ),B ε (η 2 ) are stationary (as ε / A(η l )) and equal (as η 1 E 0 η 2 ); so we can find i B ε (η 1 ) B ε (η 2 ) which satisfy i > j. Now η 1 (ε) = f i α i (η 1 ) (ε) by the definition of B ε (η 1 ) as i B ε (η 1 ) and α i (η 1 ) = α i (η 2 ) as i > j and f i α i (η 2 ) (ε) = η 2(ε) by the definition of B ε (η 2 ) as i B ε (η 2 ); together η 1 (ε) = η 2 (ε). So we have completed the proof that ε < κ η 1 (ε) = η 2 (ε) thus proving η 1 = η 2 as required.]

15 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH (ι) E 1 has λ j equivalence classes for any j < κ [why? let η i<κλ i and for α < λ j let ηα κ λ be defined by ηα (ε) is α if ε = j andisη (ε)otherwise. Byclause(θ)wehaveα < β < λ j η α E 1η β, hence i<κλ i /E 1 λ i.] (κ) E 1 has exactly λ equivalence classes [why? by clause (ι),e 1 has sup{λ i : i < κ} = λ equivalence classes and by clause (ζ),e has λ equivalence classes.] We could have defined E 0 as ( ) η 1 E 0 η 2 iff for every ε < κ we have B ε (η 1 ) = B ε (η 2 ) mod D κ where D κ is the club filter on κ. This causes no change except that E 0 is not a closed subset of ( κ λ) ( κ λ), but a union of 2 κ ones Claim. Assume (a) λ > κ = cf(λ) > ℵ 0 (b) 2 κ +λ <κ = λ (c) λ θ λ κ. Then there is E such that (α) E is an equivalence relation on κ λ (β) E is very nice 2 (γ) if η 1,η 2 κ λ and η 1 = η 2 mod J bd κ then η 1 Eη 2 η 1 = η 2 (δ) E has exactly θ equivalence classes. Proof. Let λ be as in the proof of 2.1 except that we add i < κ j<iλ j λ i, (this holds if e.g. if i < κ λ i = λ). We can find a tree T κ> λ with λ nodes and exactly θ κ-branches ([Sh 262]); we can easily manage that η ν lim κ (T ) ( κ i < κ)(η(i)) ν(i)). We proceed as in the proof of 2.1, but in the definition of E 1 we add 2 in fact, again union of 2 κ closed sets of pairs

16 16 SAHARON SHELAH η 1 lim κ (T) η 2 lim κ (T ) & (η 1 lim κ (T ) η 1 = η 2 ) Claim. In Claim 2.1 we can replace clauses (β),(γ) by (β) 1 E is very nice, moreover is the union of λ closed sets minus the union of λ closed sets (γ) 1 for every η κ λ, the set {η κ λ : η = η modjκ bd } is a set of representatives for the family of E-equivalence classes. Proof. Let λ be as there but ε < κ κ + λ i. Let K i be a group, with universe λ i and unit 0 Ki. Let < be a well ordering of κ (P(κ)). For every η i<κλ i let Ξ η = { B ε (ν) : ε < κ : ν i<κ λ i and ν = η mod J bd κ }. So Ξ η is a non-empty subset of κ (P(κ)) and let B η = B η,ε : ε < κ be its < -first member. Note that for η 1,η 2 i<κλ i if η 1 = η 2 mod J bd κ then B η 1 = B η 2 and Ξ η1 = Ξ η2. Let Θ η = {ν i<κλ i : B ε (ν) = B η,ε Now note ( ) 0 Θ η. [Why? By the definition of Ξ η, B η and Θ η.] for every ε < κ and ν = η mod Jbd κ }. ( ) 1 if ν Θ η then for every limit i < κ large enough we have α i (ν) = α i (η). [Why? As ν = η mod J bd κ.] ( ) 2 if ν 1,ν 2 Θ η and ε < κ, then for every limit i large enough we have: α i (ν 1 ) = α i (ν 2 ) hence f i α i (ν 1 ) (ε) = fi α i (ν 2 ) (ε). Now for η i<κ λ i we define ρ η i<κλ i by ρ η (ε) is :f i α i (η) (ε) for every i B η,ε large enough if B η,ε is stationary 0 Ki if Bη,ε is not stationary.

17 It is easy to see that ON NICE EQUIVALENCE RELATIONS ON λ 2 SH ( ) 3 if η i<κλ i then ρ η (ε) = η(ε) for every ε < κ large enough. [Why? We can find ν i<κλ i such that ν = η mod J bd κ and B ε (ν) : ε < κ = B η. Now apply (ε) inside the proof of 2.1.] hence ( ) 4 ρ η = η mod J bd κ ( ) 5 if η 1,η 2 i<κλ i and η 1 = η 2 mod J bd κ then ρ η 1 = ρ η2. Lastly, we define the equivalence relation E: for η 1,η 2 i<κλ i we define: η 1 Eη 2 iff (for every i < κ we have K i = η 1 (i)(ρ η1 (i)) 1 = η 2 (i)(ρ η2 (i)) 1 ). Now clearly ( ) 6 if η 1,η 2 i<κλ i and η 1 = η 2 mod J bd κ then η 1 Eη 2 η 1 = η 2. [Why? By ( ) 5 we have ρ η1 = ρ η2, call it ρ; we are done by and the properties of groups (i.e. x 1 y 1 = x 2 y 1 x 1 = x 2.] ( ) 7 if η i<κ λ i then {η : η i<κλ i and η = η mod Jκ bd } is a set of representatives of the E-equivalence classes. [Why? Let η,ν λ i and we shall define η i such that η i<κ i<κλ ν/e and η = η mod J bd κ. For i < κ we choose η (i) K i, i.e. < λ i such that K i = η (i)(ρ η (i)) 1 = ν(i)(ρ ν (i)) 1. [Why this is solvable? As K i is a group and ρ η (i),ν(i),ρ ν (i) are well defined members of K i.] Also we know that ν = ρ ν mod Jκ bd by ( ) 4 hence for some i 1 < κ we have i [i 1,κ) ν(i) = ρ ν (i); this implies that i [i 1,κ) η (i) = ρ η (i), so η = ρ η mod Jκ bd ; however ρ η = η mod Jκ bd hence η = η mod Jκ bd, as required. Hence ρ η = ρ η so by the definition of η we have K i = η (i)(ρ η (i)) 1 = ν(i)(ρ ν (i)) 1 which means that η Eν, so we have proved ( ) 7.] Lastly, how complicated is E? Define a two-place relation E on i<κλ i :

18 18 SAHARON SHELAH 2 η 1 E η 2 iff (a) B η1 = B η 2. Clearly ( ) 8 E is an equivalence relation on κ λ and is the union of λ closed minus the union of λ closed subsets of ( λ i ) ( i ) with 2 i<κ i<κλ κ equivalence classes ( ) 9 on each E -equivalence class the function η ρ η is continuous (even under the Tichonov topology, even more) ( ) 10 if Y 1,Y 2 are E -equivalence classes, then E (Y 1 Y 2 ) is closed (even under the Tichonov topology). Now check. 2.4 We may like to weaken the cardinal arithmetic assumptions. 2.5 Remark. Assume that κ = θ + and instead the ideal J bd κ we use the ideal [κ] <θ. Then we can define α j (η) for η i<κλ i and j < κ if cf(j) = cf(θ). Let α j (η) be Min{α : fα j = η j mod J j} where J j = {A j: for some i < j we have A\i < θ} so J j replaces Jj bd in the earlier proof. So η = ν mod [κ] <θ implies that α j (η) = α j (ν) for all suitable j. There are no marked changes. Now ( ) if η 1 E η 2 then B ε (η 1 ) = B ε (η 2 ),Ξ η1 = Ξ η2 and B η 1 = B η 2 0 E can serve as well and it is an equivalence relation with 2 κ equivalence classes, each closed even under the Tichonov topology. We can use λ > κ θ,j = [κ] <θ but in general the number of ideals necessary is κ θ. Most interesting is the case θ = ℵ 0 dealt with in the next claim. 2.6 Claim. 1) Assume (a) λ > κ = cf(λ) > ℵ 0 (b) κ ℵ 0 < λ = λ ℵ 0.

19 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Then the results 2.1, 2.4 and 2.3 holds if we replace the ideal J bd κ by the ideal [κ]<ℵ 0. 2) This applies also to 2.3 if (c) λ θ λ κ and there is a tree T with λ nodes and κ-branches. 3) The natural topology for (1) + (2) is the ℵ 1 -box product. Proof. Without loss of generality λ i > κ ℵ 0, λ i : i < κ as in the proof of 2.1. Let D i : i < κ ℵ 0 list the subsets of κ of order type ω and let f i = fα i : α < λ j j D i list λ j (or just a set of representatives modulo JD bd i ). For η ε let j D i ε<κλ (a) α i (η) = Min{α : η D i = f i α mod J bd D i } for i < κ ℵ 0 (b) for ε < κ let B ε (η) = {i < κ ℵ 0 : ε D i and η(ε) = f αi (η)(ε)} (c) A(η) = {ε < κ : B ε (η) is finite} (d) B ε (η) = {i B ε (η) : i B ε(η) is finite}. With those choices the proofs are similar Claim. 1) If 2 ℵ 0 < λ = λ <κ,ℵ 0 < κ = cf(λ) < λ, then we can find E as in 2.1(α),(β),(δ) (but not necessarily (γ)) and (γ) if η κ λ and i < κ then X η,i = {ν κ λ : ( j)(j < κ & j i ν(j) = η(j)} is a set of representatives for E. 2) If 2 κℵ 0 λ = λ ℵ 0,ℵ 0 < κ = cf(λ) < λ,1 θ λ and ( µ < λ)[(µ+θ) <κ λ), then we can find E as in 2.1(α),(β) and (γ) if η κ λ and i < κ then X η,i contains a set of representatives (δ) E has θ equivalence classes. Proof. 1) First the proof in short. Wechooseλ i = λfori < κ. WeletK beagroupwithuniverseλandlet D j : j < κ ℵ 0 be as in the proof of 2.6 and define E by: ηeν iff K = (η(i)(ρ η (i)) 1 ) = i A(ν) (ν(i)(ρ ν (i)) 1 ). We give a more detailed proof below. i A(η) 2)First, theproofinshort. Wechooseλ i but θ λ i ; without loss of generalityeach

20 20 SAHARON SHELAH λ i is a subgroup of K but we use equality in cosets of xk 1 = yk 1,K 1 a subgroup of K such that [K : K 1 ] = θ and a,c K {abck 1 : b {ε : ε < λ 0 }} = {bk 1 ;b < λ}. Now in detail (for (2) so including a proof of (1)). We repeat the proof of , so for η κ λ we let Ξ η = { B ε (ν) : ε < κ : ν κ λ and ν = η mod[κ] <ℵ 0 } where B ε (ν) = {j < κ ℵ 0 : f αj (ν)(ε) = ν(ε)} and let B ν be the < -first member of Ξ ν and let Θ η = {ν κ λ : B ε (ν) = Bη,ε for every ε < κ and ν = η mod [κ] <ℵ 0 } and for η κ λ let ρ η κ λ be defined by (a) ρ η (ε) = f αj (η)(ε) if ( ν)(ν Θ η & ν(ε) = f αj (η)(ε)) & j B η,ε (b) ρ η (ε) = 0 if there are no j,ν as in (a). Easily ρ η κ λ is well defined and ρ η = η mod [κ] <ℵ 0. Lastly, let a η = {ε < κ : η(ε) ρ η (ε)} and we define the two-place relation E on λ i by η 1 Eη 2 iff a η1 = a η2 & ( η 1 (ε)ρ η (i) 1 K 1 = ( η 2 (ε)η 2 (ε) 1 )K 1. i<κ i a η1 ε a η2 Is this well defined? The product η l (ε) is a finite product in the group K, so ε a ηl in general we have to choose an order of η l (ε) : ε a ηl, i.e., of a ηi. We use the most natural choice: the order on κ (if K is abelian clearer). Obviously E is an equivalence relation on ε<κλ ε and it has {xk 1 : λ K} = [K : K 1 ] equivalence classes. Now suppose that η κ λ and ε < κ and we shall prove that X η,ε is the set of representatives for E, recall X η,ε is defined in (γ) of 2.7(1). Let a = a η ε,a + = a η \(ε+1), let g = (η(i)(ρ η (i)) 1 ) and g + = (η(i)(ρ η (i)) 1 ), i a i a so: + ( ) g,g + K again well defined as a,a + are finite ( ) if ν X η,ε then a ν a η {ε} and the product in K, of course. i a ν (ν(i)(ρ ν (i)) 1 ) = g ν(ε)g + K, Now for part (1), g ν the sequence ν(ε) : ν X η,ε lists K without repetition (as the universe of K is λ) hence (by basic group theory), g 1 ν(ε)g + : ν X η,ε lists K without repetitions hence i a ν (ν(i)(ρ ν (i)) 1 ) : ν X α,ε lists K without repetitions, so if we use the trivial K 1,X η,i is a set of representatives of E, as required. For (2) the sequence g ν K 2 : ν X η,ε lists {xk 1 : x K} possibly withrepetition. 3.5

21 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Concluding Remark. 1) Instead of JD bd i : i < κ ℵ 0 we can use (D i,j i ) : i < i λ,d i κ,j i an ideal on D i such that λ ε /J i λ,i = {D κ: for every ε D i i < i we have D D i J i } is included in Jκ bd. The author has not pursued this. 2) Assume K is a group of cardinality λ,k 1 a subgroup and [K : K 1 ] = θ λ. Then we can find B K, B = θ such that if K is a subgroup of K including B such that K,K1,K if a,c K then {ack 1 : b K } = {bk 1 : b K}. [Why? Let {b i : i < θ} be such that {b i K 1 : i < θ} = {bk 1 : b K} and let B = {b i : i < θ}. If B K K and ac K and i < θ there is b K such that ab c = b i so ab ck 1 = b i K 1.]

22 22 SAHARON SHELAH 3 Countable cofinality: positive results We first phrase sufficient conditions which relate to large cardinals. Then we prove that they suffice. The proof of 3.1 is presented later in this section. 3.1 Lemma. Assume (a) λ is strong limit of cofinality ℵ 0 (b) λ is a limit of measurables, or just (b) for every θ < λ for some µ,χ satisfying θ µ χ < λ, there is a (χ,µ,θ)- witness (see Definition 3.2 below) (c) E is a nice equivalence relation on ω λ (or has enough absoluteness, as proved in 3.12), i.e., fact 3.13, so being 1 1 (λ) over zc is enough (d) if η,ν ω λ and (!n)(η(n) ν(n)) then (ηeν). Then E has 2 λ equivalence classes, moreover if λ n < λ n+1 < λ = Σ n<ω λ n then there is a subtree of ω> λ isomorphic to λ n, whose ω-branches are pairwise m n<m non E-equivalent (even somewhat more, see 3.17). Remark. For the simplest example of witness defined below see 3.4(2) so a witness is a weak form of λ being measurable. 3.2 Definition. 1) We say (Q,s 1,s 2 ) is a (λ,µ,θ)-witness if (λ µ θ and): (a) Q is a θ-complete forcing notion (b) s 1 is a function from Q to P(λ)\{ } (c) s 2 is a function from Q to {A : A {(α,β) : α < β < λ}} (d) if Q = p q then s l (q) s l (p) for l = 1,2 (e) (α,β) s 2 (p) {α,β} s 1 (p) for p Q (f) for every p Q there is q such that p q Q and ( β)( α,γ)[β s 1 (q) (α,β) s 2 (p) & (β,γ) s 2 (p)] (g) if p Q and A λ λ, then for some q we have p q Q and (s 2 (q) A) (s 2 (q) A = ) (h) if p Q then for some Y [λ] µ for every α < β from Y we have (α,β) s 2 (p) (hence Y s 1 (p)).

23 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH ) We say (Q,s 1,s 2 ) is a (λ,µ,θ, )-witness if is a cardinal λ and we can strengthen clause (g) to 3 (g) + if f : 2 λ and p Q then for some q we have p q Q and f s 2 (q) is constant. 3) We call (Q,s 1,s 2 ) a uniform (λ,µ,θ)-witness if λ = {s 1 (p) : p Q} and for every p Q and α < λ for some q we have p q Q and s 1 (q) α =. Similarly a uniform (λ, µ, θ, )-witness. 4) We replace by < if we demand only (g) + < which means that Rang(f) is a subset of of cardinality <. We write < µ instead of µ if in clause (h) of Definition 3.2(1) we demand just that for each α < µ there is Y λ of order type α and as there (so µ can be an ordinal). 3.3 Definition. 1) We say that (Q, s) is a (λ,µ,θ, ;n)-witnessif λ µ θ,λ and s = s m : m = 1,...,n and (a) Q is a θ-complete forcing (b) s m is a function from Q to P({ᾱ : ᾱ = α l : l < m m λ and α l < α l+1 < λ for l < m 1}) (c) if Q = p q and m {1,...,n} then s m (q) s m (p) (d) if α l : l < m+1 s m+1 (p) and k < m+1 then α l : l < k ˆ α l : l = k +1,...,m s m (p) (e) for every m {1,...,n 1},k < mandp Q thereisq satisfyingp q Q and ( ᾱ s m (q))( β s m+1 (p))[ᾱ = ( β k)ˆ( β [k +1,m))] (f) + if m {1,...,n} and f : m λ and p Q then for some q we have p q Q and f s m (q) is constant (g) if p Q then for some Y [λ] µ every increasing ᾱ n Y belongs to s n (p). 2) (Q, s) is a (λ,µ,θ, ;ω)-witness is defined similarly (i.e., s = s m : m [1,ω) ) and in clause (g) the same Y works for all n. 2A) (Q, s) is a (λ,µ,θ, ;ω)-witness is defined similarly, except that in clause (h), for each n < ω there is Y [λ] µ such that every increasing ᾱ n Y belongs to s n (p). 3) If = 2 we may omit it, as in Definition 3.2. Also uniform and < and < µ mean as in Definition 3.2. We first give some basic facts on witnesses, including cases of existence. 3 note that (g) + is equal to (g) if = 2

24 24 SAHARON SHELAH 3.4 Claim. 1) If (Q, s) is a (λ,µ,θ;n)-witness and < θ,n < ω, then (Q, s) is a (λ,µ,θ,2 ;n)-witness. 2) If D is a normal ultrafilter on λ so λ is a measurable cardinal and we choose, Q = (D, ),s 1 (A) = A,s 2 (A) = {(α,β) : α < β are from A}, then (Q,s 1,s 2 ) is a uniform (λ, λ, λ, < λ)-witness. 3) If in (2), s m (A) = {ᾱ : ᾱ = α l : l < m is increasing, α l A}, s = s m+1 : 1+m n and n ω then (Q, s) is a (λ,λ,λ,< λ;n)-witness. 4) If there is a (λ,µ,θ, ;n)-witness and 2 <θ λ, then there is such (Q, s) with Q 2 λ. 5) Definition 3.2(1) is the case n = 2 of Definition 3.3(1) that is, (Q,s 1,s 2 ) is a (λ,µ,θ, )-witness iff (Q,(s 1,s 2 )) is a (λ,µ,θ, ;2)-witness. 6) If (Q, s) is a (λ,µ,θ)-witness and p Q, then we can find q such that p q Q and for every β s 1 (q) there are α 1 < α 2 < β such that (α 1,β),(α 2,β) s 2 (p) (this strengthens clause (f) of 3.2). Proof. Easy. 1) Checking Definition 3.3 the least easy clause is (f) +, so assume m {1,...,n} and p Q and f is a function from m λ to 2 and we should find q satisfying p q Q and f s m (q) is constant. Let h be a one to one function from 2 2 into and define f ε : m λ {0,1} for ε < by f ε ( s) = (h(f( s)))(ε). Now we choose p ε Q, increasing (by Q ) by induction on ε such that p 0 = p,f ε s m (p ε+1 ) is constant, say is l ε. For ε = 0 this is trivial, for ε successor use (Q, s) is (λ,µ,θ;n)- witness, i.e. clause (f) + in Definition 3.3. For ε a limit ordinal we use Q is θ-complete, i.e., clause (a) in Definition 3.3 for (Q, s) is a (λ,µ,θ;n)-witness, recall < θ. Lastly, let q = p so we are done. 2), 3) Note that Q is λ-complete as D is λ-complete as D is a λ-complete ultrafilter (being normal) and clause (f) + holds because if f n : [λ] n µ and µ < λ then for some A n D we have f [A] n is constant (see, e.g., [J]) and as D is closed under intersection of < λ (hence of ℵ 0 ) we are done (if p Q, let q = p A n ). 4) Let (Q, s) be a (λ,µ,θ, ;n)-witness and let χ be large enough. Choose an elementary submodel N of (H (χ), ) to which (Q, s) satisfying N = 2 λ,[n] λ N so 2 λ N. Lastly, choose Q = Q N and s m = s m Q. Now check that (Q, s m+1 : m < n ) is a (λ,µ,θ, ;n)-witness recalling µ,θ, λ. 5) Read the definitions. 6) For l {0,1,2} let A l = {α s 1 (p): the number {α < α : (α,α) s 2 (p)} is equal to l or l = 2 and the number is l}. So A 0,A 1,A 2 is a partition of s 1 (p). n<ω

25 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Define a function f from A 1 to λ: for α A 1,f(α) is the unique α < α s 2 (p). It is known (and easy) that we can find a partition B 1,B 2,B 3 of A 1 such that l {1,2,3} & α B l f(α) / B l. Let B 0 = A 0,B 4 = A 2, so B 0,...,B 4 2 is a partition of A l that is of s 1 (p). By clause (g) of Definition 3.2 (applied l=0 three times, see 3.4(1)) we can find l( ) < 5 and q Q such that p q Q and s 1 (q) B l( ). s s 2 (q) necessarily l( ) = 4 and so we are done. 3.4 Something of the largeness remains if we collapse a large cardinal, see, e.g., [JMMP]. We shall need 3.5 Claim. 1) Assume (a) 2 n < ω and λ = ℶ n 1 (θ) + (b) θ is a compact cardinal or just a λ-compact cardinal (c) µ = µ <µ < θ (d) P = Levy(µ,< θ). Then in V P (and of course in V), there is a (λ,µ,θ;n)-witness (Q, s) which is even a (λ,µ,θ,<µ;n)-witness. 2) If there are λ n for n < ω,λ n < λ n+1 and λ n is 2 (2λn ) + -compact and λ = Σ{λ n : n < ω}, then for some set forcing P, in V P the cardinal λ = ℶ ω = ℵ ω is dichotomically good (see Definition 3.8 below). Proof. By [Sh 124]. 3.6 Remark. 1) In fact we can weaken the consistency strength. Assume that (G.C.H. holds for simplicity) and: (a) µ n : n < ω is strictly increasing sequence of cardinals (b) µ n λ n < µ n+1 (c) D n+1 is a µ n+1 -complete ultrafilter on I n+1 = {a [λ n+1 ] < µ n+1 : min(a) < µ n+1 } (d) let ι n+1 : I n+1 λ n+1 is ι n+1 (a) = min(µ n+1 a) and if A D n+1,f : A µ n+1 is regressive, i.e., f(a) < ι n+1 (a) then f is constant on some B D n+1,b A (e) if g : [λ n+1 ] n µ n then {a I n+1 : g [a\ι n+1 (a)] n is constant} D n+1 (f) Q 0 = Levy(ℵ 0,µ 0 ),Q n+1 = Levy(λ ++ n,< µ n+1),q = n<ωq n.

26 26 SAHARON SHELAH Then V Q is as required in ) If µ n is µ n +(n+2) -hyper-measurable and we let λ n = µ n +(n+1) and µ n < µ n+1, then there is j n : V M n,µ n is the critical cardinal of j n,m µ n n M n,j n (µ n ) > µ n +(n+2) and H (λ n ) M n. So in V we can find b [j n (µ n )\µ n ] λ n 1 such that ( ) if f : [µ n ] n λ n 1 then j n (f) [b] n is constant. Let a = {µ n } b so a M n and D n = {A [µ n ] λ n 1 : a j n (A)}. Those D n are as required for λ n = µ n. Toward proving Lemma 3.1 assume (from 3.10 till the end of this section) that 3.7 Hypothesis. m = λ n,µ n,θ n,p n,s n,1,s n,2 n<ω = λ m n,µm n,θm n,pm n,sm n,1,sm n,2 n<ω satisfies λ = Σ{λ n : n < ω} and ℵ 1 +{2 λ l : l < n} < θ n λ n and (P n,s n,1,s n,2 ) is a (λ n,< µ + n,θ n)-witness and for simplicity µ n < µ n+1 and λ = {µ n : n < ω}. n 3.8 Definition. We call λ dichotomically good if there is m, i.e., there are λ n,µ n,θ n,p n,s n,1,s n,2 as in 3.7. The hypothesis 3.7 is justified because 3.9 Observation. 1) If λ satisfies (a) + (b) or at least (a) + (b) of Lemma 3.1 then λ is dichotomically good. 2) It is consistent that G.C.H. and ℵ ω is dichotomically good (if CON(ZFC + there is a supercompact cardinal). 3) For proving 3.1 without loss of generalitye is a nice equivalence relation on λ n satisfying clause (d) of 3.1. n<ω Proof. 1) By 3.4(2) we know (b) (b) in 3.1, now read the definitions. 2) By )?? 3.10 Definition. 1) We define the forcing notionq 1 (reallyq 1 = Q[m]) as follows: (a) Q 1 = { p : p = (η,ā) = (ηp,āp ) such that letting n p = n(p) = lg(η) we have n p < ω,η p λ l and l<n[p] Ā p = A p l : l [n(p),ω) and Ap l P } l

27 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH (b) p Q1 q iff η p η q (so n(p) n(q) and [l [n(q),ω) P l = A p l Aq l ] and [n(p) l < n(q) η q (l) s 1 (A p l )] (c) We define the Q 1 -name η (d) We define by: η[g] = {η p : p Q 1 } G (α) p Q 1 pr q iff p Q 1 q & n(p) = n(q) (β) p Q 1 apr q iff p Q1 q & (A q l = Ap l ) l n(q) (γ) p Q 1 pr,n q iff p Q 1 pr q and n > n(p) Āp [n(p),n) = Āq [n(p),n). 2) We define the forcing notion Q 2 (really Q 2 [m]) by: (a) Q 2 = { p : p = (η 0,η 1,Ā) = (ηp 0,ηp 1,Āp ) where for some n(p) < ω we have: η p 0,ηp 1 λ l and Āp = A p l : l [n(p),ω) and Ap l P } l (b) p Q2 q iff l<n(p) (i) n(p) n(q) (ii) η p l ηq l for l = 0,1 (iii) A q l Ap l for l [n(q),ω) (iv) the pair (η q 0 (l),ηq 1 (l)) is from s 2(A p l ) for l [n(p),n(q)) (c) we define the Q 2 -name η (for l = 0,1) by η = {η l l[g] p l (d) we define (α) p Q 2 pr q iff p Q 1 q & n(p) = n(q) and (β) p Q 2 apr q iff p Q2 q & A q l = Ap l and l n(q) : p G Q2 } (γ) p Q 2 pr,n q iff p Q 2 pr q and n > n(p) Āp [n(p),n) = Āq [n(p),n). 3) If for a fixed k < ω, we have (P n, s n ) is a (λ n,µ n,θ n ;k)-witness for n < ω then we can define Q k naturally. 4) If (P n, s n ) is a (λ n,µ n,θ n ;n)-witness for n < ω then we can define Q = {(η,ā) : n < ω,η(l) l (λ l ) and Ā = A l : l [n,ω),a l P l }} with the natural order.

28 28 SAHARON SHELAH Remark. 1) We shall not pursue here parts (3) and (4) of Definition 3.10 because we deal with equivalence relations which are binary. We can prove parallel theorems for relations with higher arity using 3.10(3),(4). 2) In the definition of the set of elements p of Q 2, why don t we ask ( l < n p )(η p 0 (l) < ηp 1 (l))? To be able to construct the perfect set, but, of course, p Q2 η < η for l [n(p),ω). 0(l) 1(l) 3) Those forcing notions are in the (large) family of relatives of Prikry forcing Fact. Let l {1,2}. 0) For p,q Q l we have: (i) p Q l pr q p Ql q (ii) p Q l apr q p q (iii) p Q l pr,n+1 q p Q l pr,n q p Q l pr q. 1) If p Ql r then for some q we have p Q l pr,n(q) q Q l apr r. 2) If p = p i : i < α is Q l pr -increasing and α < θ n(p 0 )(= θn(p m 0 )), then p has a Q l pr-upper bound; similarly for Q l pr,n and α < θ n. 3) If is a Q l -name of an ordinal and p Q l, then for some q and n we have: τ (a) p pr q (b) if q apr r and n(r) n, then r forces a value to τ. 4) In (3), if Ql τ < ω or just < α < θ n(p) then without loss of generalityn = n(p). Proof. Easy Claim. Recall that by 3.9(3) without loss of generalitye is a nice definition of a two-place relation on n<ωλ n. Then forcing by Q 2 preserves E is an equivalence relation on n<ωλ n satisfying clause (d) of 3.1 or more exactly the definition E defines in V Q 2 an equivalence relation on n<ωλ n satisfying clause (d) of 3.1 (and, of course, E (VQ 2) ( n<ωλ n ) V = E V ).

29 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Proof. Assume toward contradiction that p Q2 ν 0,ν 1,ν 2 l<ωλ l form a counterexample, that is: ν 0Eν 1 ν 1Eν 2 ν 0Eν 2 or ν 0Eν 0 or ν 0Eν 1 ν 1Eν 0 or ν 0Eν 1 (!n)(ν 0(n) ν 1(n)). Choose χ large enough and N = N n : n < ω,n such that: χ N(i) N n Lλ + n,λ + n (H (χ), )and N n = 2 λ n and{p,e,ν 0,ν 1,ν 2,N 0,...,N n 1 } belong to N n (ii) N n N n+1 hence N n Lλ + n,λ + n N n+1 and N = n<ω N n so N (H (χ), ). Now we choose p n by induction on n < ω such that: ( )(i) p 0 = p, (ii) p n N n Q 2 and n(p n ) = max{n,n(p )} (iii) p n p n+1 (iv) if N n is a Q 2 -name of an ordinal then for some k n (τ ) > n+1 we have: τ if p n+1 q and n(q) k n (τ ) then q forces a value to τ. This is possible by 3.11(2),(3). Now let G = {q : q N Q 2 and q p n or just p n q G for some n}; it is a subset of Q N 2 generic over N. (Why? If N = I Q 2 is dense then I Q 2 is dense and there is I I, a maximal antichain of Q 2 which belongs to N hence to some N n ; there is g N n, a one to one function from I onto I, so it defines a Q 2 -name τ by τ [G] = γ ( q)(q I G f(q) = γ) ( q)(q I G & f(q) = γ), so k n (τ ) < ω is well defined (see clause (iv) above) and so p kn (τ ) forces a value to τ hence forces q G for some q I I, hence q G so G I as required). Now by straightforward absoluteness argument, ν 0[G],ν 1[G],ν 2[G] l<ωλ l give contradiction to an assumption. In details let ν l = ν l[g]. Let M be the Mostowski collapse of N, so there is an isomorphism g from N onto M. Clearly λ n N n hence λ N hence λ+1 N so g(x) = x if x λ + 1 or x λ + 1 or x H (λ). Clearly G = g (G) is a generic subset of Q 2 = (g(q 2)) M and M = M[G] is a generic extension of M (for g(q 2 ) M ) and so 1 M is a transitive model of enough set theory (i.e. of ZC if χ is strong limit) which includes H (λ) {λ, H (λ)} { λ n : n < ω }.

30 30 SAHARON SHELAH Also easily in M,ν l[g ] = ν l, so as g(p ) G, clearly (E stands for the formula defining it, its parameter is a subset of λ so it is mapped by g to itself): M = ν 0,ν 1,ν 2 Π{λ n : n < ω} and ν 0 Eν 1 & ν 1 Eν 2 & ν 0 Eν 1 or ν 0 Eν 0 or ν 0 Eν 1 & ν 1 Eν 0 or ν 0 Eν 1 & (!n)(ν 0 (n) ν 1 (n)). So it is enough to prove (see Lemma 3.1, clause (c)) Fact. Assume M satisfies 1 above, E is a nice two-place relation on Π{λ n : n < ω} so a definition with parameter which is a subset of λ (equivalently: a model on λ) as in Definition 0.2(1). Then 2 if M satisfies η 1 Eη 2 & η 3 Eη 4 and η 0,η 1,η 2,η 3 Π{λ n : n < ω} then so does V. Proof. Immediate. In fact 3.14 Observation. Assume (a)(i) λ is strong limit of cofinality ℵ 0, (ii) λ = n<ωλ n (iii) λ n < λ n+1 for n < ω, for simplicity 2λ n < λ n+1 (b)(i) Q is a forcing notion (ii) pr is included in Q (iii) n : Q ω is a function satisfying for each n the set I n = {p Q : n(p) n} is a dense subset of Q (iv) for p Q,{q Q : p pr q} is λ n(p) -complete (v) Q has pure decidability for Q-names of truth values (vi) if p Q and τ is a Q-name of an ordinal, then there are m < ω and q satisfying: p pr q and (q r & m n(r)) (r forces a value to τ ) (c) N, N n : n < ω as in the proof of 3.12 for λ n : n < ω, {Q,, pr } N 0.

31 ON NICE EQUIVALENCE RELATIONS ON λ 2 SH Then there is G Q N generic over N hence H (λ) N[G] = H (λ) = H (λ) N. Proof. Should be clear Claim. Assume that F is a permutation of ( l<n( ) λ l ) ( l<n( ) λ l ) and let Q n( ) 2 = {p Q 2 : n(p) n( )}. We let ˆF be the following function from Q n( ) 2 to Q n( ) 2 ˆF(p) = q iff n(q) = n(p) (η q 0 n( ),ηq 1 n( )) = F((ηp 0 n( ),ηp 1 n( ))) η q 0 [n( ),n(p)) = ηp 0 [n( ),n(p)) η q 1 [n( ),n(p)) = ηp 1 [n( ),n(p)) Ā q = Āp. Then the following holds: 1) For p Q n( ) 2, ˆF(p) is well defined Q n( ) 2. 2) ˆF is a permutation of Q n( ) 2 preserving, pr, pr,n, apr and their negations, and F ˆF is a group homomorphism (hence embedding). 3) If G Q 2 is generic over V then (a) ˆF(G) =: {r Q 2 : for some q G Q n( ) 2 we have r ˆF(q)} is a subset of Q 2 generic over V (b) G = {p Q 2 : there is q Q n( ) 2 such that p Q2 q and ˆF(q) ˆF(G)} (c) and V[ˆF(G)] = V[G] and even N[ˆF(G)] = N[G] if, e.g., N (H (χ), ),Q 2 N,F N,λ N. Proof. Easy Claim. Q2 η 0Eη 1. Proof. If not, let p Q 2 be such that p Q2 η Now by clause (f) of 0Eη 1. Definition 3.2(1), we can find p 1 such that:

32 32 SAHARON SHELAH (i) Q 2 = p pr p 1 (ii) if n(p) n < ω and β s 1 (A p 1 n ) then for some α,γ we have (α,β),(β,γ) s 2 (A p n ). Let G 1 Q 2 be generic over V such that p 1 G 1 and let η l = η 1 ] for l = 1,2 l[g so V[G 1 ] = η 0 Eη 1. By 3.12 in V[G 1 ],E is still an equivalence relation satisfying clause (d) of 3.1 and trivially n [n(p),ω) η 1 (n) s 1 (A p 1 n ). Let n =: n(p), by 3.4(6)we can find α < λ n such that α < η 1 (n ),α η 0 (n ) and (α,η 1 (n )) s 2 (A p n ). Let us define η 0 n<ωλ n by η 0(n) is α if n = n and η 0 (n) otherwise; as α < η 1 (n ) < η 0 (n ) necessarily η 0 η 0. Nowthepairs(η 0 (n( )+1),η 1 (n( )+1))and(η 0 (n( ))+1),η 1 (n( )+1))are from ( λ n ) ( λ n ), so there is a permutation F of this set interchanging n n( ) n n( ) those two pairs and is the identity otherwise. Let ˆF be the automorphism of Q (n +1) 2 from Claim Let G 2 = ˆF(G 1 ). Now by 3.15: ( ) 1 G 2 is a generic subset of Q 2 over V ( ) 2 V[G 2 ] = V[G 1 ] ( ) 3 η 0[G 2 ] = η 0,η 1[G 2 ] = η 1. By 3.12 (and the choice of η 0 ) we have ( ) 4 V[G 1 ] = η 0 Eη 0. As p p 1 G 1, by the choice of p clearly ( ) 5 V[G 1 ] = η 0 Eη 1. By the choice of p 1 and (α,η 1 (n )) clearly p (η 0 [ n( )+1)),η 1 (n( )+1),Ā [n( )+1,ω) G 2 so (using ( ) 1 ) ( ) 6 V[G 2 ] = (η 0[G 2 ])E(η 1[G 2 ]) hence by ( ) 2 +( ) 3 we have ( ) 7 V[G 2 ] = η 0 Eη 1. Now ( ) 4 +( ) 5 +( ) 7 contradict