PCF: THE ADVANCED PCF THEOREMS E69 SAHARON SHELAH

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1 PCF: THE ADVANCED PCF THEOREMS E69 SAHARON SHELAH Abstract. This is a revised version of [Sh:430, 6]. Date: January 21, The author thanks Alice Leonhardt for the beautiful typing. I thank Peter Komjath for some comments. 1

2 2 SAHARON SHELAH 0. Introduction

3 PCF: THE ADVANCED PCF THEOREMS E On pcf This is a revised version of [Sh:430, 6] more self-contained, large part done according to lectures in the Hebrew University Fall 2003 Recall Definition 1.1. Let f = f α : α < δ,f α κ Ord,I an ideal on κ. 1) We say that f κ Ord is a I -l.u.b. of f when: (a) α < δ f α I f (b) if f κ Ord and ( α < δ)(f α I f ) then f I f. 2) We say that f is a I -e.u.b. of f when (a) α < δ f α I f (b) if f κ Ordand f < I Max{f,1 κ }then f < I Max{f α,1 κ } forsomeα < δ. 3) f is I -increasing if α < β f α I f β, similarly < I -increasing. We say f is eventually < I -increasing: it is I -increasing and ( α < δ)( β < δ)(f α < I f β ). 4) We may replace I by the dual ideal on κ. {1.1} Remark 1.2. For κ,i, f as in Definition 1.1, if f is a I -e.u.b. of f then f is a {1.1} I -l.u.b. of f. {1.2} Definition ) We say that s witness or exemplifies f is (< σ)-chaotic for D when, for some κ (a) f = f α : α < δ is a sequence of members of κ Ord (b) D is a filter on κ (or an ideal on κ) (c) f is < D -increasing (d) s = s i : i < κ,s i a non-empty set of < σ ordinals (e) for every α < δ for some β (α,δ) and g s i we have f α D g D f β. 2) Instead (< σ + )-chaotic we may say σ-chaotic. Claim 1.4. Assume (a) I an ideal on κ (b) f = f α : α < δ is < I -increasing, f α κ Ord (c) J I is an ideal on κ and s witnesses f is (< σ)-chaotic for J. {1.3} Then f has no I -e.u.b. f such that {i < κ : cf(f(i)) σ} J. {1.4} Discussion 1.5. What is the aim of clause (c) of 1.4? For I -increasing sequence {1.3} f, f α : α < δ in κ Ordwe areinterested whether it has an appropriate I -e.u.b. Of course, I may be a maximal ideal on κ and f t : t cf((ω,<) κ /D)) is < I -increasing cofinalin(ω,<) κ /D, soithasan< I -e.u.b. thesequenceω κ = ω : i < κ, butthisis notwhatinterestsusnow; weliketohavea I -e.u.b. g suchthat ( i)(cf(g(i)) > κ). Proof. Toward contradiction assume that f κ Ord is a I -e.u.b. of f and A 1 := {i < κ : cf(f(i)) σ} / I hence A / I. We define a function f κ Ord as follows: (a) if i A then f (i) = sup(s i f(i))+1

4 4 SAHARON SHELAH (b) if i κ\a then f (i) = 0. {1.5} Now that i A cf(g(i)) σ > s i f (i) < f(i) Max{g(i),1} and i κ\a f (i) = 0 f (i) < Max{f(i),1}. So by clause (b) of Definition {1.1} 1.1(2) we know that for some α < δ we have f < I Max{f α,1}. But s witness that f is (< σ)-chaotic hence we can find g s i and β (α,δ) such that f α I g I f β and as f is < I -increasing without loss of generality g < I f β. So A 2 := {i < κ : f α (i) g(i) < f β (i) f(i) and f (i) < Max{f α (i),1} = κ} mod I hence A := A 1 A 2 mod I hence A. So for any i A we have f α (i) g(i) < f β (i) f(i) and f(i) s i hence g(i) < f (i) := sup(s i f(i)) +1 and so f (i) 1. Also f (i) < Max{f α (i,1)} hence f (i) < f α (i). Together f (i) < f α (i) g(i) < f (i), contradiction. 1.4 Lemma 1.6. Suppose cf(δ) > κ +,I an ideal on κ and f α κ Ord for α < δ is I -increasing. Then there are J, s, f satisfying: (A) s = s i : i < κ, each s i a set of κ ordinals, (B) sup{f α (i) : α < δ} s i ; moreover is max(s i ) (C) f = f α : α < δ where f α s i is defined by f α (i) = Min{s i\f α (i)}, (similar to rounding!) (D) cf[f α(i)] κ (e.g. f α(i) is a successor ordinal) implies f α(i) = f α (i) (E) J = J α : α < δ,j α is an ideal on κ extending I (for α < δ), decreasing with α (in fact for some a α,β κ (for α < β < κ) we have a α,β /I decreases with β, increases with α and J α is the ideal generated by I {a α,β : β belongs to (α,)}) so possibly J α = P(κ) and possibly J α = I such that: (F) if D is an ultrafilter on κ disjoint to J α then f α /D is a < D-l.u.b and even < D -e.u.b. of f β /D : β < α which is eventually < D -increasing and {i < κ : cf[f α(i))] > κ} D. Moreover (F) + if κ / J α then f α is an < Jα -e.u.b (= exact upper bound) of f β : β < δ and β (α,δ) f β = J α f α (G) if D is an ultrafilter on κ disjoint to I but for every α not disjoint to J α then s exemplifies f α : α < δ is κ chaotic for D as exemplified by s (see Definition 1.3), i.e., for some club E of δ,β < γ E f β D f β < {1.2} D f γ (H) if cf(δ) > 2 κ then f α : α < δ has a I -l.u.b. and even I -e.u.b. and for every large enough α we have I α = I (I) if b α =: {i : f α(i) has cofinality κ (e.g., is a successor)} / J α then: for every β (α,δ) we have f α b α = f β b α mod J α. Remark 1.7. Compare with [Sh:506]. Proof. Let α = {f α (i) + 1 : α < δ,i < κ} and S = {j < α : j has cofinality κ},ē = e j : j S be such that

5 PCF: THE ADVANCED PCF THEOREMS E69 5 (a) e j j, e j κ for every j S (b) if j = i+1 then e j = {i} (c) if j is limit, then j = sup(e j ) and j S e j e j e j. For a set a α let clē(a) = a e j hence by clause (c) clearly clē(clē(a)) = j a S clē(a) and [a b clē(a) clē(b)] and clē(a) a + κ. We try to choose by induction on ζ < κ +, the following objects: α ζ, D ζ, g ζ, s ζ = s ζ,i : i < κ, f ζ,α : α < δ such that: (a) g ζ κ Ord and g ζ (i) {f α (i) : α < δ} (b) s ζ,i = clē[{g ǫ (i) : ǫ < ζ} {sup α<δ f α (i)}] so it is a set of κ ordinals increasing with ζ and sup α<δ f α (i) s ζ,i, moreover sup α<δ f α (i) = max(s ζ,i ) (c) f ζ,α κ Ord is defined by f ζ,α (i) = Min{s ζ,i \f α (i)}, (d) D ζ is an ultrafilter on κ disjoint to I (e) f α Dζ g ζ for α < δ (f) α ζ is an ordinal < δ (g) α ζ α < δ g ζ < Dζ f ζ,α. If we succeed, let α( ) = sup{α ζ : ζ < κ + }, so as cf(δ) > κ + clearly α( ) < δ. Now let i < κ and look at f ζ,α( ) (i) : ζ < κ + ; by its definition (see clause (c)), f ζ,α( ) (i) is the minimal member of the set s ζ,i \f α( ) (i). This set increases with ζ, so f ζ,α( ) (i) decreases with ζ (though not necessarily strictly), hence is eventually constant; so for some ξ i < κ + we have ζ [ξ i,κ + ) f ζ,α( ) (i) = f ξi,α( )(i). Let ξ( ) = sup ξ i, so ξ( ) < κ +, hence 1 ζ [ξ( ),κ+ )andi < κ f ζ,α( ) (i) = f ξ( ),α( ) (i). By clauses (e) + (g) of we know that f α( ) Dξ( ) g ξ( ) < Dξ( ) f ξ( ),α( ) hence for some i < κ we have f α( ) (i) g ξ( ) (i) < f ξ( ),α( ) (i). But g ξ( ) (i) s ξ( )+1,i by clause (b) of hence recalling the definition of f ξ( )+1,α( ) (i) in clause (c) of and the previous sentence f ξ( )+1,α( ) (i) g ξ( ) (i) < f ξ( ),α( ) (i), contradicting the statement 1. So necessarily we are stuck in the induction process. Let ζ < κ + be the first ordinal that breaks the induction. Clearly s ζ,i (i < κ),f ζ,α (α < δ) are well defined. Let s i =: s ζ,i (for i < κ) and f α = f ζ,α (for α < δ), as defined in, clearly they are well defined. Clearly s i is a set of κ ordinals and: ( ) 1 f α f α ( ) 2 α < β f α I f β ( ) 3 if b = {i : f α (i) < f β (i)} / I and α < β < δ then f α b < I f β b. We let for α < δ 2 J α = { b κ : b I or b / I and for every β (α,δ) we have: f α (κ\b) = I f β (κ\b)} 3 for α < β < δ we let a α,β =: {i < κ : f α(i) < f β (i)}. Then as f α : α < δ is I-increasing (i.e., ( ) 2 ):

6 6 SAHARON SHELAH {1.2} ( ) 4 a α,β /I increases with β, decreases with α, J α increases with α ( ) 5 J α is an ideal on κ extending I, in fact is the ideal generated by I {a α,β : β (α,δ)} ( ) 6 if D is an ultrafilter on κ disjoint to J α, then f α /D is a < D-lub of {f β /D : β < δ}. [Why? We know that β (α,δ) a α,β = mod D, so f β f β = D f α for β (α,δ), so f α /D is an D-upper bound. If it is not a least upper bound then for some g κ Ord, for every β < δ we have f β D g < D f α and we can get a contradiction to the choice of ζ, s,f β because: (D,g,α) could serve as D ζ,g ζ,α ζ.] ( ) 7 If D is an ultrafilter on κ disjoint to I but not to J α for every α < δ then s exemplifies that f α : α < δ is κ + -chaotic for D, see Definition 1.3. [Why? For every α < δ for some β (α,δ) we have a α,β D, i.e., {i < κ : f α(i) < f β (i)} D, so f α /D : α < δ is not eventually constant, so if α < β,f α < D f β then f α < D f β (by ( ) 3 ) and f α D f α (by (c)). So f α D f α < D f β as required.] ( ) 8 if κ / J α then f α is an J α -e.u.b. of f β : β < δ. [Why? By ( ) 6, f α is a J α -upper bound of f β : β < δ ; so assume that it is not a Jα -e.u.b. of f β : β < δ, hence there is a function g with domain κ, such that g < Jα Max{1,f α}, but for no β < δ do we have c β =: {i < κ : g(i) < Max{1,f β (i)}} = κ mod J α. Clearly c β : β < δ is increasingmodulo J α so there is an ultrafilter D on κ disjoint to J α {c β : β < δ}. So β < δ f β D g D f α, so we get a contradiction to ( ) 6 except when g = D f α and then f α = D 0 κ (as g(i) < 1 g(i) < f α(i)). If we can demand c = {i : f α (i) = 0} / D we are done, but easily c \c β J α so we finish.] ( ) 9 If cf[f α (i)] κ then f α (i) = f α(i) so clause (D) of the lemma holds. [Why? By the definition of s ζ = clē[...] and the choice of ē, and of f α(i).] ( ) 10 Clause (I) of the conclusion holds. [Why? As f α Jα f β Jα f α and f α b α = Jα f α b α by ( ) 9.] ( ) 11 if α < β < δ then f α = f β mod J α, so clause (F) + holds. [Why? First, f is I -increasing hence it is Jα -increasing. Second, β α f β I f α f α f β Jα f α. Third, if β (α,δ) then a α,β = {i < κ : f α(i) < f β (i)} J α, hence f β J α f α but as f α I f β clearly f α I f β hence f α J α f β, so together f α = Jα f β.] ( ) 12 if cf(δ) > 2 κ then for some α( ),J α( ) = I (hence f has a I -e.u.b.) [Why? As J α : α < δ isa -decreasingsequenceofsubsetsof P(κ)itiseventually constant, say, i.e., there is α( ) < δ such that α( ) α < δ J α = J α( ). Also I J α( ), but if I J α( ) then there is an ultrafilter D of κ disjoint to I but not to J α( ) hence s i : i < κ witness being κ-chaotic. But this implies cf(δ) s i κ κ = 2 κ, contradiction.] The reader can check the rest. 1.6

7 PCF: THE ADVANCED PCF THEOREMS E69 7 Example ) We show that l.u.b and e.u.b are not the same. Let I be an ideal on κ,κ + < = cf(),ā = a α : α < be a sequence of subsets of κ, (strictly) increasing modulo I, κ\a α / I but there is no b P(κ)\I such that b a α I. [Does this occur? E.g., for I = [κ] <κ, the existence of such ā is α known to be consistent; e.g., MA andκ = ℵ 0 and = 2 ℵ0. Moreover, for any κ and κ + < = cf() 2 κ we can find a α κ for α < such that, e.g., any Boolean combination of the a α s has cardinality κ (less needed). Let I 0 be the ideal on κ generated by [κ] <κ {a α \a β : α < β < }, and let I be maximal in {J : J an ideal on κ,i 0 J and [α < β < a β \a α / J]}. So if G.C.H. fails, we have examples.] For α <, we let f α : κ Ord be: f α (i) = { α if i κ\a α, +α if i a α. {1.6} Now the constant function f κ Ord,f(i) = + is a l.u.b of f α : α < but not an e.u.b. (both mod I) (no e.u.b. is exemplified by g κ Ord which is constantly ). 2) Why do we require cf(δ) > κ + rather than cf(δ) > κ? As we have to, by Kojman-Shelah [KjSh:673]. Recall (see [Sh:506, 2.3(2)]) Definition 1.9. We say that f = f α : α < δ obeys u α : α S when {1.7} (a) f α : w Ord for some fixed set w (b) S a set of ordinals (c) u α α (d) if α S δ and β u α then t w f β (t) f α (t). Claim Assume I is an ideal on κ, f = f α : α < δ is I -increasing and obeys ū = u α : α S. The sequence f has a I -e.u.b. when for some S + we have 1 or 2 where 1 (a) S + {α < δ : cf(α) > κ} (b) S + is a stationary subset of δ (c) for each α S + there are unbounded subsets u,v of α for which β v u β u β. 2 S + = {δ} and for δ clause (c) of 1 holds. Proof. By [Sh:506] {1.8} Remark ) Connected to Ǐ[], see [Sh:506]. {1.10} Claim Suppose J a σ-complete ideal on δ,µ > κ = cf(µ),µ = tlim J i : i < δ,δ < µ, i = cf( i ) > δ for i < δ and = tcf( i<δ i /J), and f α : α < exemplifies this. Then we have

8 8 SAHARON SHELAH ( ) if u β : β < is a sequence of pairwise disjoint non-empty subsets of, each of cardinality σ (not < σ!) and α < µ +, then we can find B such that: (a) otp(b) = α, (b) if β B,γ B and β < γ then sup(u β ) < min(u γ ), (c) we can find s ζ J for ζ u i such that: if ζ u β,ξ i B β B u β,ζ < ξ and i δ\(s ζ s ξ ), then f ζ (i) < f ξ (i). β B Proof. First assume α < µ. For each regular θ < µ, as θ + < = cf() there is a stationary S θ {δ < : cf(δ) = θ < δ} which is in Ǐ[] (see [Sh:420, 1.5]) which is equivalent (see [Sh:420, 1.2(1)]) to: ( ) there is C θ = C θ α : α < (α) C θ α a subset of α, with no accumulation points (in C θ α), (β) [α nacc(c θ β ) Cθ α = Cθ β α], (γ) for some club E 0 θ of, [δ S θ Eθ 0 cf(δ) = θ < δ δ = sup(cδ) otp(c θ δ) θ = θ]. Without loss of generality S θ Eθ 0, and otp(cα θ ) θ. By [Sh:365, 2.3,Def.1.3] α<δ for some club E θ of, gl(cα θ,e θ) : α S θ guess clubs (i.e., for every club E E θ of, for stationarily many ζ S θ, gl(cζ θ,e θ) E) (remember gl(cδ θ,e θ) = {sup(γ E θ ) : γ Cδ θ;γ > Min(E θ)}). Let Cα θ, = {γ Cα θ : γ = Min(Cα\sup(γ θ E θ ))}, they have all the properties of the Cα θ s and guess clubs in a weak sense: for every club E of for some α S θ E, if γ 1 < γ 2 are successive members of E then (γ 1,γ 2 ] Cα θ, 1; moreover, the function γ sup(e γ) is one to one on Cθ,. Now we define by induction on ζ <, an ordinal α ζ and functions g ζ θ α (for each θ Θ =: {θ : θ < µ,θ regular uncountable}). For given ζ, let α ζ < be minimal such that: ξ < ζ α ξ < α ζ ξ < ζ θ Θ g ξ θ < f α ζ mod J. Now α ζ exists as f α : α < is < J -increasing cofinal in θ Θ we define g ζ θ as follows: i<δ i i<δ i /J. Now for each for i < δ,g ζ θ (i) is sup[{gξ θ (i)+1 : ξ Cθ ζ } {f α ζ (i)+1}] if this number is < i, and f αζ (i)+1 otherwise. Having made the definition we prove the assertion. We are given u β : β <, a sequence of pairwise disjoint non-empty subsets of, each of cardinality σ and α < µ. We should find B as promised; let θ =: ( α + δ ) + so θ < µ is regular > δ. Let E = {δ E θ : ( ζ)[ζ < δ sup(u ζ ) < δ u ζ δ α ζ < δ]}. Choose α S θ acc(e) such that gl(cζ θ,e θ) E; hence letting Cα θ, = {γ i : i < θ} (increasing), γ(i) = γ i, we know that i < δ (γ i,γ i+1 ) E. Now let

9 PCF: THE ADVANCED PCF THEOREMS E69 9 B =: {γ 5i+3 : i < α } we shall prove that B is as required. For α u γ(5ζ+3),ζ < α, let s o α = {i < δ : g γ(5ζ+1) θ (i) < f α (i) < g γ(5ζ+4) θ (i)}, for each ζ < α let α ζ,ε : ε < u γ(5ζ+3) enumerate u γ(5ζ+3) and let s 1 α ζ,ε = {i : for every ξ < ǫ,f αζ,ξ (i) < f αζ,ǫ (i) α ζ,ξ < α ζ,ǫ f αζ,ξ (i) f αζ,ǫ (i)}. Lastly,forα ζ<α u 5ζ+3 lets α = s o α s1 α anditisenoughtocheckthat ζ α : α B witness that B is as required. Also we have to consider α [µ,µ + ), we prove this by induction on α and in the induction step we use θ = (cf(α ) + δ ) + using a similar proof {1.11} Remark In 1.12: {1.10} 1) We can avoid guessing clubs. 2) Assume σ < θ 1 < θ 2 < µ are regular and there is S {δ < : cf(δ) = θ 1 } from I[] such that for every ζ < (or at least a club) of cofinality θ 2, S ζ is stationary and f α : α < obey suitable C θ (see [Sh:345a, 2]). Then for some A unbounded, for every u β : β < θ 2 sequence of pairwise disjoint non-empty subsets ofa, each ofcardinality < σ with [minu β,supu β ] pairwisedisjoint we have: for every B 0 A of order type θ 2, for some B B 0, B = θ 1, (c) of ( ) of 1.12 {1.10} holds. 3) In ( ) of 1.12, α < µ can be replaced by α < µ + (prove by induction on {1.10} α ). {1.12} Observation Assume < <,µ = Min{τ : 2 τ > }. Then there are δ,χ and T, satisfying the condition ( ) below for χ = 2 µ or at least arbitrarily large regular χ < 2 µ ( ) T a tree with δ levels, (where δ µ) with a set X of χ δ-branches, and for α < δ, T β <. β<α Proof. So let χ 2 µ be regular, χ >. Case 1: 2 α <. Then T = µ> 2, T α = α 2 are O.K. (the set of branches µ 2 α<µ has cardinality 2 µ ). Case 2: Not Case 1. So for some θ < µ, 2 θ, but by the choice of µ,2 θ, so 2 θ =,θ < µ and so θ α < µ 2 α = 2 θ. Note µ> 2 = as µ. Note also that µ = cf(µ) in this case (by the Bukovsky-Hechler theorem). Subcase 2A: cf() µ = cf(µ). Let µ> 2 = B j,b j increasing with j, B j <. For each η µ 2, (as cf() j< cf(µ)) for some j η <, µ = sup{ζ < µ : η ζ B jη }. So as cf(χ) µ, for some ordinal j < we have {η µ 2 : j η j } has cardinality χ.

10 10 SAHARON SHELAH As cf() cf(µ) and µ (by its definition) clearly µ <, hence B j µ <. Let It is as required. T = {η ǫ : ǫ < lg(η) and η B j }. {6.4} {1.21} Subcase 2B: Not 2A so cf() = µ = cf(µ). If = µ we get = < contradicting an assumption. So > µ, so singular. Now if α < µ,µ < σ i = cf(σ i ) < for i < α then (see [Sh:g,?, 1.3(10)]) maxpcf{σ i : i < α} σ i α (2 θ ) α 2 <µ =, but as is singular and maxpcf{σ i : i < α} is regular (see [Sh:345a, 1.9]), clearly the inequality is strict, i.e., maxpcf{σ i : i < α} <. So let σ i : i < µ be a strictly increasing sequence of regulars in (µ,) with limit, and by [Sh:355, 3.4] there is T σ i satisfying {ν i : ν T } maxpcf{σ j : j < i} <, and number of i<µ µ-branches >. In fact we can get any regular cardinal in (,pp + ()) in the same way. Let = min{ : µ <,cf( ) = µ and pp( ) > }, so (by [Sh:355, 2.3]), also has those properties and pp( ) pp(). So if pp + ( ) = (2 µ ) + or pp( ) = 2 µ is singular, we are done. So assume this fails. If µ > ℵ 0, then (as in [Sh:430, 3.4]) α < 2 µ cov(α,µ +,µ +,µ) < 2 µ and we can finish as in subcase 2A (actually cov(2 <µ,µ +,µ +,µ) < 2 µ suffices which holds by the previous sentence and [Sh:355, 5.4]). If µ = ℵ 0 all is easy Claim Assume b 0... b k b k+1 for k < ω,a = b k (and k<ω a + < Min(a)) and pcf(a)\ pcf(b k ). k<ω 1) We can find finite d k pcf(b k \b k 1 ) (stipulating b 1 = ) such that pcf( {d k : k < ω}). 2) Moreover, we can demand d k pcf(b k )\(pcf(b k 1 )). Proof. We start to repeat the proof of [Sh:371, 1.5] for κ = ω. But there we apply [Sh:371, 1.4] to b ζ : ζ < κ and get c ζ,l : l n(ζ) : ζ < κ and let ζ,l = maxpcf(c ζ,l ). Here we apply the same claim ([Sh:371, 1.4]) to b k \b k 1 : k < ω to get part (1). As for part (2), in the proof of [Sh:371, 1.5] we let δ = a + +ℵ 2 choose N i : i < δ, but now we have to adapt the proof of [Sh:371, 1.4] (applied to a, b k : k < ω, N i : i < δ ); we have gotten there, toward the end, α < δ such that E α E. Let E α = {i k : k < ω},i k < i k+1. But now instead of applying [Sh:371, 1.3] to each b l separately, we try to choose c ζ,l : l n(ζ) by induction on ζ < ω. Forζ = 0weapply[?, 1.3]. Forζ > 0, weapply[sh:371, 1.3]tob ζ buttheredefining by induction on l,c l = c ζ,l a such that max(pcf(a\c ζ,0 \ \c ζ,l 1 ) pcf(b ζ ) is strictly decreasing with l. We use: i<α Observation If a i < Min(a i ) for i < i, then c = element or is empty. i<i pcf(a i ) has a last Proof. By renaming without loss of generality a i : i < i is non-decreasing. By [Sh:345b, 1.12]

11 PCF: THE ADVANCED PCF THEOREMS E69 11 ( ) 1 d cand d < Min(d) pcf(d) c. By [Sh:371, 2.6] or 2.7(2) ( ) 2 if pcf(d),d c, d < Min(d) then for some d we have Min(a 0 ), pcf( ). {6.7C.1} Now choose by induction on ζ < a 0 +,θ ζ c, satisfying θ ζ > maxpcf{θ ǫ : ǫ < ζ}. If we are stuck in ζ, maxpcf{θ ǫ : ǫ < ζ} is the desired maximum by ( ) 1. If we succeed the cardinal θ = maxpcf{θ ǫ : ǫ < a 0 + } is in pcf{θ ǫ : ǫ < ζ} for some ζ < a 0 + by ( ) 2 ; easy contradiction {1.22} Conclusion Assume ℵ 0 = cf(µ) κ µ 0 < µ,[µ (µ 0,µ)andcf(µ ) κ pp κ (µ ) < ] and pp + κ (µ) > = cf() > µ. Then we can find n for n < ω,µ 0 < n < n+1 < µ, µ = n and = tcf( n /J) for some ideal J on ω (extending J bd ω ). n<ω Proof. Let a (µ 0,µ) Reg, a κ, pcf(a). Without loss of generality = maxpcf(a), let µ = µ 0 n, µ 0 µ 0 n < µ 0 n+1 < µ, let µ 1 n = µ 0 n+sup{pp κ (µ ) : µ 0 < n<ω µ µ 0 n and cf(µ ) κ}, by [Sh:355, 2.3] µ 1 n < µ,µ 1 n = µ 0 n + sup{pp κ (µ ) : µ 0 < µ < µ 1 n and cf(µ ) κ} and obviously µ 1 n µ1 n+1 ; by replacing by a subsequence without loss of generality µ 1 n < µ 1 n+1. Now let b n = a µ 1 n and apply the previous claim 1.15: to b k =: a (µ 1 n )+, note: {6.4} n<ω maxpcf(b k ) µ 1 k < Min(b k+1\b k ) {1.23} Claim ) Assume ℵ 0 < cf(µ) = κ < µ 0 < µ,2 κ < µ and [µ 0 µ < µandcf(µ ) κ pp κ (µ ) < µ]. If µ < = cf() < pp + (µ) then there is a tree T with κ levels, each level of cardinality < µ, T has exactly κ-branches. 2) Suppose i : i < κ is a strictly increasing sequence of regular cardinals, 2 κ < 0,a =: { i : i < κ}, = maxpcf(a), j > maxpcf{ i : i < j} for each j < κ (or at least i<j i > maxpcf{ i : i < j}) and a / J where J = {b a : b is the union of countably many members of J < [a]} (so J Ja bd and cf(κ) > ℵ 0 ). Then the conclusion of (1) holds with µ = i. Proof. 1) By (2) and [Sh:371, 1] (or can use the conclusion of [Sh:g, AG,5.7]). 2) For each b a define the function g b : κ Reg by g b (i) = maxpcf[b { j : j < i}]. Clearly[b 1 b 2 g b1 g b2 ]. Ascf(κ) > ℵ 0,J isℵ 1 -complete,thereisb a,b / J such that: c bandc / J g c < J g b. Let i = maxpcf(b { j : j < i}). For each i let b i = b { j : j < i} and f,α b : α < : pcf(b) be as in [Sh:371, 1]. Let

12 12 SAHARON SHELAH {1.24} {1.25} {1.25} T 0 i = { Max 0<l<n fb l,α l b i : l pcf(b i ),α l < l,n < ω}. Let T i = {f T 0 i : for every j < i,f b j T 0 j moreover for some f for every j, f b j T 0 j and f f }, and T = j<κ j, T i, clearly it is a tree, T i its ith level (or empty), T i i. By [Sh:371, 1.3,1.4] for every g b for some f b, f b i Ti 0 hence f b i T i. So T i = i, and T has κ-branches. By the observation below we can finish (apply it essentially to F = {η: for some f b for i < κ we have η(i) = f b i and for every i < κ,f b i Ti 0}), then find A κ,κ \ A J and g ( i +1) such that Y =: {f F : f A < g A} has cardinality and then the tree will be T where T i =: {f b i : f Y } and T = T i. (So actually this proves that if we have such a tree with θ(cf(θ) > 2 κ ) κ-branches then there is one with exactly θ κ-branches.) 1.18 Observation If F i, J an ℵ 1 -complete ideal on κ, and [f g F f J g] and F θ,cf(θ) > 2 κ, then for some g ( i +1) we have: (a) Y = {f F : f < J g } has cardinality θ, (b) for f < J g, we have {f F : f J f } < θ, (c) there 1 are f α Y for α < θ such that: f α < J g,[α < β < θ f β < J f α ]. (Also in [Sh:829, 1]). Proof. Let Z =: {g : g ( i +1) and Y g =: {f F : f J g} has cardinality θ}. Clearly i : i < κ Z so there is g Z such that: [g Z g < J g ]; so clause (b) holds. Let Y = {f F : f < J g }, easily Y Y g and Y g \Y 2 κ hence Y θ, also clearly [f 1 f 2 Fandf 1 J f 2 f 1 < J f 2 ]. If (a) fails, necessarily by the previous sentence Y > θ. For each f Y let Y f = {h Y : h J f}, so by clause (b) we have Y f < θ hence by the Hajnal free subset theorem for some Z Z, Z = +, and f 1 f 2 Z f 1 / Y f2 so [f 1 f 2 Z f 1 < J f 2 ]. But there is no such Z of cardinality > 2 κ ([Sh:111, 2.2,p.264]) so clause (a) holds. As for clause (c): choose f α F by induction on α, suchthat f α Y \ β<α Y f β ; it existsby cardinalityconsiderationsand f α : α < θ is as required (in (c)) Observation Let κ < be regular uncountable, 2 κ < µ i < (for i < κ), µ i increasing in i. The following are equivalent: (A) there is F κ such that: (i) F =, (ii) {f i : f F} µ i, (iii) [f g F f J bd g]; κ 1 Or straightening clause (i) see the proof of 1.20

13 PCF: THE ADVANCED PCF THEOREMS E69 13 (B) there be a sequence i : i < κ such that: (i) 2 κ < i = cf( i ) µ i, (ii) maxpcf{ i : i < κ} =, (iii) for j < κ,µ j maxpcf{ i : i < j}; (C) there is an increasing sequence a i : i < κ suchthat pcf( a i ),pcf(a i ) µ i (so Min( a i ) > a i ). Proof. (B) (A): By [Sh:355, 3.4]. (A) (B): If ( θ)[θ 2 κ θ κ θ + ] we can directly prove (B) if for a club of i < κ,µ i > µ j, and contradict (A) if this fails. Otherwise every normal filter D j<i on κ is nice (see [Sh:386, 1]). Let F exemplify (A). Let K = {(D,g) : D a normal filter on κ, g κ (+1), = {f F : f < D g} }. ClearlyK is not empty (let g be constantly ) so by [Sh:386] we can find (D,g) K such that: ( ) 1 if A κ,a mod D,g 1 < D+A g then > {f F : f < D+A g 1 }. Let F = {f F : f < D g}, so (as in the proof of 1.18) F =. {1.23} We claim: ( ) 2 if h F then {f F : h D f} has cardinality <. [Why? Otherwise for some h F, F =: {f F : h D f} has cardinality, for A κ let F A = {f F : f A h A} so F = {F A : A κ,a mod D}, hence (recall that 2 κ < ) for some A κ,a mod D and F A = ; now (D+A,h) contradicts ( ) 1 ]. By ( ) 2 we can choose by induction on α <, a function f α F such that f β < D f α. By [Sh:355, 1.2A(3)] f α : α < has an e.u.b. f. Let i = β<α cf(f (i)), clearly {i < κ : i 2 κ } = mod D, so without loss of generality cf(f (i)) > 2 κ so i is regular (2 κ,], and = tcf( i /D). Let J i = {A i : maxpcf{ j : j A} µ i }; so (remembering (ii) of (A)) we can find h i j<if (i) such that: ( ) 3 if {j : j < i} / J i, then for every f F,f i < Ji h i. Let h f (i) be defined by: h(i) = sup{h j (i) : j (i,κ) and {j : j < i} / J i }. As cf[f (i)] > 2 κ, clearly i h < f hence by the choice of f for some α( ) < we have: h < D f α( ) and let A =: {i < κ : h(i) < f α( ) (i)}, so A D. Define i as follows: i is i if i A, and is (2 κ ) + if i κ\a. Now i : i < κ is as required in (B). (B) (C): Straightforward. (C) (B): By [Sh:371, 1] {1.26} Claim If F κ Ord,2 κ < θ = cf(θ) F then we can find g κ Ord and a proper ideal I on κ and A κ,a I such that:

14 {1.27} {1.26} {1.28} {1.28a} {1.28} {1.29} {1.27} 14 SAHARON SHELAH (a) g (i)/i has true cofinality θ, and for each i κ\a we have cf[g (i)] > 2 κ, (b) for every g κ Ord satisfying g A = g A, g (κ\a) < g (κ\a) we can find f F such that: f A = g A,g (κ\a) < f (κ\a) < g (κ\a). Proof. As in [Sh:410, 3.7], proof of (A) (B). (In short let f α F for α < θ be distinct, χ large enough, N i : i < (2 κ ) + as there, δ i =: sup(θ N i ),g i κ Ord,g i (ζ) =: Min[N Ord\f δi (ζ)],a κ and S {i < (2 κ ) + : cf(i) = κ + } stationary, [i S g i = g ], [ζ < αandi S [f δi (ζ) = g (ζ) ζ A] and for some i( ) < (2 κ ) +, g N i( ), so [ζ κ\a cf(g (ζ)) > 2 κ ] Claim Suppose D is a σ-complete filter on θ = cf(θ),κ an infinite cardinal, θ > α κ for α < σ, and for each α < θ, β = βǫ α : ǫ < κ is a sequence of ordinals. Then for every X θ,x mod D there is βǫ : ǫ < κ (a sequence of ordinals) and w κ such that: (a) ǫ κ\w σ cf(βǫ) θ, (b) if β ǫ β ǫ and [ǫ w β ǫ = β ǫ ], then {α X: for every ǫ < κ we have β ǫ βǫ α βǫ and [ǫ w βǫ α = βǫ]} mod D. Proof. Essentially by the same proof as 1.21 (replacing δ i by Min{α X: for every Y N i D we have α Y}). See more [Sh:513, 6]. (See [Sh:620, 7]) Remark We can rephrase the conclusion as: (a) B =: {α X: if ǫ w then β α ǫ = β ǫ, and: if ǫ κ\w then βα ǫ is < β ǫ but > sup{β ζ : ζ < ǫ,βα ζ < β ǫ}} is mod D (b) If β ǫ < β ǫ for ǫ κ \ w then {α B: if ǫ κ \ w then βα ǫ > β ǫ } mod D (c) ǫ κ\w cf(β ǫ) is θ but σ. Remark If a < min(a), F Πa, F = θ = cf(θ) / pcf(a) and even θ > σ = sup(θ + pcf(a)) then for some g Πa, the set {f F : f < g} is unbounded in θ (or use a σ-complete D as in 1.23). (This is as Πa/J <θ [a] is min(pcf(a) \ θ)-directed as the ideal J <θ [a] is generated by σ sets; this is discussed in [Sh:513, 6].) Remark It is useful to note that 1.22 is useful to use [Sh:462, 4,5.14]: e.g., for if n < ω, θ 0 < θ 1 < < θ n, satisfying ( ) below, for any β ǫ βǫ satisfying [ǫ w β ǫ < βǫ] we can find α < γ in X such that: where ǫ w β α ǫ = β ǫ, {ǫ,ζ} κ\wand{cf(β ε),cf(β ζ)} [θ l,θ l+1 ))andl even β α ǫ < β γ ζ, {ǫ,ζ} κ\wand{cf(β ε ),cf(β ζ )} [θ l,θ l+1 )andl odd β γ ǫ < βα ζ ( ) (a) ǫ κ\w cf(β ǫ) [θ 0,θ n ), and

15 PCF: THE ADVANCED PCF THEOREMS E69 15 (b) maxpcf[{cf(β ǫ ) : ǫ κ\w} θ l] θ l (which holds ifθ l = σ + l, σκ l = σ l for l {l,...,n}).

16 16 SAHARON SHELAH {1.31} 2. Nice generating sequences Claim 2.1. For any a, a < Min(a), we can find b = b : a such that: (α) b is a generating sequence, i.e. a J [a] = J < [a]+b, (β) b is smooth, i.e., for θ < in a, θ b b θ b, {1.32} {1.33} {1.33} (γ) b is closed, i.e., for a we have b = a pcf(b ). Definition ) Foraset a and set a ofregularcardinalslet Ch a a be the function with domain a a defined by Ch a a (θ) = sup(a θ). 2) We may write N instead of N, where N is a model (usually an elementary submodel of (H (χ),,< χ) for some reasonable χ. Observation 2.3. If a a and a < Min(a) then ch a a Πa. Proof. Let b θ [a] : θ pcf(a) be as in [Sh:371, 2.6] or Definition [Sh:506, 2.12]. For a, let f a, = fα a, : α < be a < J< [a]-increasing cofinal sequence of members of a, satisfying: ( ) 1 if δ <, a < cf(δ) < Min(a) and θ a then: f a, δ (θ) = Min{ fα a, (θ) : C a club of δ} α C [exists by [Sh:345a, Def.3.3,(2) b + Fact 3.4(1)]]. Let χ = ℶ ω (sup(a)) + and κ satisfies a < κ = cf(κ) < Min(a) (without loss of generality there is such κ) and let N = N i : i < κ be an increasing continuous sequenceof elementarysubmodels of(h (χ),,< χ), N i κ an ordinal, N (i+1) N i+1, N i < κ, and a, f a, : a and κ belong to N 0. Let N κ = N i. Clearly by 2.3 ( ) 2 Ch a N i Πa for i κ. Now for every a the sequence Ch a N i () : i κ is increasing continuous (note that N 0 N i N i+1 and N i, N i+1 hence sup(n i ) N i+1 hence Ch a N i () is < sup(n i+1 )). Hence {Ch a N i () : i < κ} is a club of Ch a N κ (); moreover, for every club E of κ the set {Ch a N i () : i E} is a club of Ch a N κ (). Hence by ( ) 1, for every a, for some club E of κ, ( ) 3 (α) ifθ aande E isaclubofκthenf a, sup(n κ ) (θ) = α E f a, sup(n α ) (θ) (β) f a, sup(n (θ) cl(θ N κ ) κ), (i.e., the closure as a set of ordinals). Let E = E, so E is a club of κ. For any i < j < κ let a b i,j = {θ a : Cha N i (θ) < f a, sup(n j ) (θ)}.

17 PCF: THE ADVANCED PCF THEOREMS E69 17 ( ) 4 for i < j < κ and a, we have: (α) J [a] = J < [a]+b i,j [Why? (hence bi,j = b [ā] mod J < [a]), (β) b i,j + a, (γ) b i,j : a N j+1, (δ) f a, sup(n κ ) Cha N κ = sup(n κ θ) : θ a. Clause (α): First as Ch a N i Πa (by 2.3) there is γ < such that Ch a N i < J= [a] {1.33} fγ a, and as a {a,n i } Ch a N i+1 clearly Ch a N i N i+1 hence without loss of generality γ N i+1 but i + 1 j hence N i+1 N j hence γ N j hence γ < sup(n j ) hence fγ a, < J= [a] f a, sup(n. Together j ) Cha N i < J= [a] f a, sup(n j ) hence by the definition of b i,j we have a\bi,j J =[a] hence / pcf(a\b i,j ) so J [a] J < [a]+b i,j. Second, (Πa,< J [a]) is + -directed hence there is g Πa such that α < f a, α < J [a] g. As f a, N 0 without loss of generality g N 0 hence g N i so g < Ch a N i. By the choice of g,f a, sup(n < j ) J [a] g so together f a, sup(n < j ) J [a] Ch a N i hence b i,j J [a]. As J < [a] J [a] clearly J < [a]+b i,j J [a]. Together we are done. Clause (β): Because Π(a\ + ) is + -directed we have θ a\ + {θ} / J [a]. Clause (γ): As Ch a N i,f,a sup(n j ), f belongs to N j+1. Clause (δ): For θ a( N 0 ) we have f a, sup(n κ ) (θ) = {fa, sup(n (θ) : ε E ε ) } sup(n κ θ). So we have proved ( ) 4.] ( ) 5 ε( ) < κ when ε( ) = {ε,θ : θ < are from a} where ε,θ = Min{ε < κ: if f a, sup(n (θ) < sup(n κ ) κ θ) then f a, sup(n (θ) < sup(n κ ) ε θ)}. [Why? Obvious.] ( ) 6 f a, sup(n κ ) bi,j = Cha N κ b i,j when i < j are from E\ε( ). [Why? Let θ b i,j, so by ( ) 3(β) we know that f a, sup(n (θ) κ ) Cha N κ (θ). If the inequality is strict then there is β N κ θ such that f a, sup(n (θ) β < κ ) Ch a N κ (θ) hence for some ε < κ,β N ε hence ζ (ε,κ) f a, sup(n (θ) < κ ) Cha N ζ (θ) hence (as i ε,θ holds) we have f a, sup(n (θ) < κ ) Cha N i (θ) so f a, sup(n (θ) j ) f a, sup(n (θ) < κ ) Cha N, (thefirstinequalityholdsasj E i(θ) ). Butbythedefinition of b i,j this contradicts θ bi,j.] We now define by induction on ǫ < a +, for a (and i < j < κ), the set b i,j,ǫ ( ) 7 (α) b i,j,0 (β) b i,j,ǫ+1 j (γ) b i,j,ǫ = b i,j = b i,j,ǫ = ζ<ǫ b i,j,ζ {b i,j,ǫ θ : θ b i,j,ǫ } {θ a : θ pcf(b i,j,ǫ )}, for ǫ < a + limit. :

18 18 SAHARON SHELAH Clearlyfor a, b i,j,ǫ : ǫ < a + belongston j+1 andisanon-decreasingsequence of subsets of a, hence for some ǫ(i,j,) < a +, we have {1.31} [ǫ (ǫ(i,j,), a + ) b i,j,ǫ So letting ǫ(i,j) = sup a ǫ(i,j,) < a + we have: ( ) 8 ǫ(i,j) ǫ < a + b i,j,ǫ(i,j) = b i,j,ǫ. a = b i,j,ǫ(i,j,) ]. We restrict ourselves to the case i < j are from E\ε( ). Which of the properties required from b : a are satisfied by b i,j,ǫ(i,j) : a? In the conclusion of 2.1 properties (β), (γ) hold by the inductive definition of b i,j,ǫ (and the choice of ǫ(i,j)). As for property (α), one half, J [a] J < [a]+b i,j,ǫ(i,j) hold by ( ) 4 (α) (and b i,j = bi,j,0 b i,j,ǫ(i,j) ), so it is enough to prove (for a): ( ) 9 b i,j,ǫ(i,j) J [a]. For this end we define by induction on ǫ < a + functions f a,,ǫ α for every pair (α,) satisfying α < a, such that ζ < ǫ fα a,,ζ domain increases with ǫ. We let f a,,0 α = f a, α b i,j,fa,,ε α defined by defining each f a,,ǫ+1 α Case 1: If θ b i,j,ǫ then f a,,ε+1 α = ζ<ǫ f a,,ζ α (θ) as follows: (θ) = f a,,ǫ (θ). α with domain b i,j,ǫ fα a,,ǫ, so the for limit ǫ < a + and f a,,ǫ+1 α Case 2: If µ b i,j,ǫ,θ b i,j,ǫ µ and not Case 1 and µ minimal under those conditions, then fα a,,ε+1 (θ) = f a,µ,ǫ (θ where we choose β = fα a,,ǫ (µ). β Case 3: If θ a pcf(b i,j,ǫ ) and neither Case 1 nor Case 2, then fα a,,ǫ+1 (θ) = Min{γ < θ : fα a,,ǫ b θ [a] J<θ [a] fγ a,θ,ǫ }. Now b i,j,ǫ : a : ǫ < a + can be computed from a and b i,j : a. But the latter belongs to N j+1 by ( ) 4 (γ), so the former belongs to N j+1 and as b i,j,ǫ : a : ǫ < a + is eventually constant, also each member of the sequence belongs to N j+1. As also fα a, : α < : pcf(a) belongs to N j+1 we clearly get that f a,,ǫ α : ǫ < a + : α < : a belongs to N j+1. Next we prove by induction on ǫ that, for a, we have: 1 θ b i,j,ǫ and a f a,,ǫ sup(n κ ) (θ) = sup(n κ θ). For ǫ = 0 this holds by ( ) 6. For ǫ limit this holds by the induction hypothesis and the definition of fα a,,ǫ (as union of earlier ones). For ǫ+1, we check f a,,ǫ+1 sup(n (θ) κ ) according to the case in its definition; for Case 1 use the induction hypothesis applied to f a,,ǫ sup(n κ ). For Case 2 (with µ), by the induction hypothesis applied to f a,µ,ǫ sup(n κ µ). Lastly, for Case 3 (with θ) we should note: is

19 (i) b i,j,ǫ PCF: THE ADVANCED PCF THEOREMS E69 19 b θ [a] / J <θ [a]. [Why? By the case s assumption b i,j,ε (ii) f a,,ǫ sup(n κ ) (bi,j,ǫ b i,j,ǫ θ ) f a,θ,ǫ sup(n. κ θ) (J θ [a]) + and ( ) 4 (α) above.] [Why? By the induction hypothesis for ǫ, used concerning and θ.] Hence (by the definition in case 3 and (i) + (ii)), (iii) f a,,ǫ+1 sup(n κ ) (θ) sup(n κ θ). Now if γ < sup(n κ θ) then for some γ(1) we have γ < γ(1) N κ θ, so letting b =: b i,j,ǫ b θ [a] b i,j,ǫ θ, it belongs to J θ [a]\j <θ [a] and we have f a,θ γ b < J<θ [a] f a,θ γ(1) b fa,θ,ǫ sup(n κ θ) hence f a,,ǫ+1 sup(n κ ) (θ) > γ; as this holds for every γ < sup(n κ θ) we have obtained (iv) f a,,ǫ+1 sup(n κ ) (θ) sup(n κ θ); together we have finished proving the inductive step for ǫ+1, hence we have proved 1. This is enough for proving b i,j,ǫ J [a]. Why? If it fails, as b i,j,ǫ N j+1 and fα a,,ǫ : α < belongs to N j+1, there is g b i,j,ǫ such that ( ) α < f a,,ǫ α b i,j,ǫ < g mod J [a]. Without loss of generality g N j+1 ; by ( ), f a,,ǫ sup(n < g mod J κ ) [a]. But g < sup(n κ θ) : θ b i,j,ǫ. Together this contradicts 1! This ends the proof of {1.31} If pcf(a) < Min(a) then 2.1 is fine and helpful. But as we do not know this, we {1.31} shall use the following substitute. {6.7A} Claim 2.4. Assume a < κ = cf(κ) < Min(a) and σ is an infinite ordinal satisfying σ + < κ. Let f, N = Ni : i < κ, N κ be as in the proof of 2.1. Then we can find {1.31} ī = i α : α σ, ā = a α : α < σ and b β [ā] : a β : β < σ such that: (a) ī is a strictly increasing continuous sequence of ordinals < κ, (b) for β < σ we have i α : α β N iβ+1 hence N iα : α β N iβ+1 and b γ [ā] : a γ and γ β N iβ+1, we can get ī (β + 1) N iβ +1 if κ succesor of regular (we just need a suitable partial square) (c) a β = N iβ pcf(a), so a β is increasing continuous with β,a a β pcf(a) and a β < κ, (d) b β [ā] a β (for a β ), (e) J [a β ] = J < [a β ]+b β [ā] (so bβ [a] and bβ [ā] + ), (f) if µ < are from a β and µ b β [ā] then bβ µ [ā] bβ [ā] (i.e., smoothness), (g) b β [ā] = a β pcf(b β [ā]) (i.e., closedness),

20 20 SAHARON SHELAH {6.7B} {6.7A} {6.7A} {6.7B} {6.7C} {6.7C} {6.7B} {6.7B} {6.7A} {6.7C.1} {6.7A} {6.7A} {6.7A} {6.7A} {6.7A} (h) if c a β,β < σ and c N iβ+1 then for some finite d a β+1 pcf(c), we have c b β+1 µ [ā]; µ d more generally (note that in (h) + if θ = ℵ 0 then we get (h)). (h) + if c a β,β < σ,c N iβ+1,θ = cf(θ) N iβ+1, then for some d N iβ+1,d a β+1 pcf θ complete (c) we have c b β+1 µ [ā] and d < θ, (i) b β [ā] increases with β. This will be proved below. Claim 2.5. In 2.4 we can also have: (1) if we let b [ā] = b σ [a] = b β [ā], a σ = a β then also for β = σ we have β<σ (b) (use N iβ +1), (c), (d), (f), (i) (2) If σ = cf(σ) > a then for β = σ also (e), (g) (3) If cf(σ) > a,c N iσ, c a σ (hence c < Min(c) and c a σ ), then for some finite d (pcf(c)) a σ we have c b µ [ā]. Similarly for θ-complete, µ d β<σ θ < cf(σ) (i.e., we have clauses (h), (h) + for β = σ). (4) We can have continuity in δ σ when cf(δ) > a, i.e., b δ [ā] = We shall prove 2.5 after proving 2.4. µ d β<δ b β [ā]. Remark ) If we would like to use length κ, use N as produced in [Sh:420, L2.6] so σ = κ. 2) Concerning 2.5, in 2.6(1) for a club E of σ = κ, we have α E b α [ā] = b [ā] a α. 3) We can also use 2.4,2.5 to give an alternative proof of part of the localization theorems similar to the one given in the Spring 89 lectures. For example: Claim ) If a < θ = cf(θ) < Min(a), for no sequence i : i < θ of members of pcf(a), do we have [ α > maxpcf{ i : i < α}]. α<θ 2) If a < Min(a), b < Min(b),b pcf(a) and pcf(a), then for some c b we have c a and pcf(c). Proof. Relying on 2.4: 1) Without loss of generality Min(a) > θ +3, let κ = θ +2, let N, N κ, ā, b (as a function), i α : α σ =: a + be as in 2.4 but we in addition assume that i : i < θ N 0. So for j < θ, c j =: { i : i < j} N 0 (so c j pcf(a) N 0 = a 0 ) hence (by clause (h) of 2.4), for some finite d j a 1 pcf(c j ) = N i1 pcf(a) pcf(c j ) we have c j d j b 1 [ā]. Assume j(1) < j(2) < θ. Now if µ a [ā] then d j(1) b 1 for some µ 0 d j(1) we have µ b 1 µ 0 [ā]; now µ 0 d j(1) pcf(c j(1) ) pcf(c j(2) ) pcf( b 1 [ā]) = (pcf(b 1 [ā]) hence (by clause (g) of 2.4 as µ 0 d j(0) N 1 ) d j(2) d j(2) forsomeµ 1 d j(2),µ 0 b 1 µ 1 [ā]. Sobyclause(f) of2.4wehaveb 1 µ 0 [ā] b 1 µ 1 [ā]hence remembering µ b 1 µ 0 [ā], we have µ b 1 µ 1 [ā]. Remembering µ was any member of

21 PCF: THE ADVANCED PCF THEOREMS E69 21 a b 1 [ā], we have a b 1 [ā] a b 1 [ā] (holds also without a d j(1) dj(1) dj(2) but not used). So a [ā] : j < θ is a -increasing sequence of subsets of a, d j b 1 but cf(θ) > a, so the sequence is eventually constant, say for j j( ). But maxpcf(a d j b 1 [ā]) maxpcf( b 1 [ā]) d j = max(maxpcf(b 1 [ā])) d j = max maxpcf{ i : i < j} < j d j = maxpcf(a b 1 [ā]) d j+1 (last equality as b j [ā] b 1 [ā] mod J <[a 1 ]). Contradiction. 2) (Like [Sh:371, 3]): If this fails choose a counterexample b with b minimal, and amongthosewithmaxpcf(b)minimalandamongthosewith {µ + : µ pcf(b)} minimal. So by the pcf theorem ( ) 1 pcf(b) has no last member ( ) 2 µ = sup[ pcf(b)] is not in pcf(b) or µ =. ( ) 3 maxpcf(b) =. Try to choose by induction on i < a +, i pcf(b), i > maxpcf{ j : j < i}. Clearly by part (1), we will be stuck at some i. Now pcf{ j : j < i} has a last member and is included in pcf(b), hence by ( ) 3 and being stuck at necessarily pcf({ j : j < i}) but it is pcf(b) +, so = maxpcf{ j : j < i}. For each j, by the choice of minimal counterexample for some b j b, we have b j a, j pcf(b j ). So pcf{ j : j < i} pcf( b j ) but frb j is a subset of b of j<i cardinality i a = a, so we are done. 2.7 {6.7D} Proof. Without loss of generality σ = ωσ (as we can use ω ω σ so ω ω σ = σ ). Let f a = f a, = fα a, : α < : pcf(a) and N i : i κ be chosen as in the proof of 2.1 and without loss of generality f a belongs to N 0. For ζ < κ we {1.31} define a ζ =: N ζ pcf(a); we also define ζ f as f aζ, α : α < : pcf(a) where a ζ is defined as follows: f aζ, α (a) if θ a,f aζ, α (θ) = fα a,(θ), (b) if θ a ζ \a and cf(α) / ( a ζ,min(a)), then f aζ, α (θ) = Min{γ < θ : fα a, b θ[a] J<θ [b θ [a]] fγ a,θ b θ [a]}, (c) if θ a ζ \a and cf(α) ( a ζ,min(a)), define f aζ, α (θ) so as to satisfy ( ) 1 in the proof of 2.1. {1.31} Now ζ f is legitimate except that we have only j<i β < γ < pcf(a) f aζ, β f aζ, γ mod J < [a ζ ]

22 22 SAHARON SHELAH {1.31} {1.31} {1.31} (insteadofstrictinequality)howeverwestillhave [f aζ, β<γ< β < f aζ, γ mod J < [a ζ ]], but this suffices. (The first statement is actually proved in [Sh:371, 3.2A], the second in [Sh:371, 3.2B]; by it also ζ f is cofinal in the required sense.) For every ζ < κ we can apply the proof of 2.1 with (N ζ pcf(a)), ζ f and N ζ+1+i : i < κ here standing for a, f, N there. In the proofof 2.1 get a club E ζ of κ (corresponding to E there and without loss of generality ζ+min(e ζ ) = Min(E ζ ) so any i < j from E ζ are O.K.). Now we can define for ζ < κ and i < j from E ζ, ζ b i,j and ζ b i,j,ǫ : ǫ < a ζ +, ǫ ζ (i,j,) : a ζ,ǫ ζ (i,j), as well as in the proof of 2.1. Let: E = {i < κ : i is a limit ordinal ( j < i)(j +j < iandj j < i) and i E j }. j<i So by [Sh:420, 1] we can find C = C δ : δ S,S {δ < κ : cf(δ) = cf(σ)} stationary, C δ a club of δ,otp(c δ ) = σ such that: (1) foreachα <,{C δ α : α nacc(c δ )}hascardinality< κ. Ifκissuccessor of regular, then we can get [γ C α C β C α γ = C β γ] and (2) for every club E of κ for stationarily many δ S,C δ E. {1.31} {1.31} {6.7A} {1.31} Without loss of generality C N 0. For some δ,c δ E, and let {j ζ : ζ ω 2 σ} enumerate C δ {δ }. So j ζ : ζ ω 2 σ is a strictly increasing continuous sequence of ordinals from E κ such that j ǫ : ǫ ζ N jζ+1 and if, e.g., κ is a successor of regulars then j ε : ε ζ N jζ +1. Let j(ζ) = j ζ and for l {0,2} let i l (ζ) = i l ζ =: j ω (1+ζ),a l ζ = Ni l ζ pcf(a), and ā l =: a l ζ : ζ < σ,l b ζ [ā] =: i l (ζ) b j(ωl ζ+1),j(ω l ζ+2),ǫ ζ (j(ω l ζ+1),j(ω l ζ+2)). Recall that σ = ωσ so σ = ω 2 σ; if the value of l does not matter we omit it. Most of the requirements follow immediately by the proof of 2.1, as for each ζ < σ, we have b ζ, b ζ [ā] : a ζ are as in the proof (hence conclusion of 2.1) and belongs to N iβ +3 N iβ+1. We are left (for proving 2.4) with proving clauses (h) + and (i) (remember that (h) is a special case of (h) + choosing θ = ℵ 0 ). For proving clause (i) note that for ζ < ξ < κ, f aζ, α f aξ, α hence ζ b i,j ξ b i,j. Now we can prove by induction on ǫ that ζ b i,j,ǫ ξ b i,j,ǫ for every a ζ (check the definition in ( ) 7 in the proof of 2.1) and the conclusion follows. Instead of proving (h) + we prove an apparently weaker version (h) below, but having (h) for the case l = 0 gives (h) + for l = 2 so this is enough [[then note that ī = i ω 2 ζ : ζ < σ, ā = a ω 2 ζ : ζ < σ, N i(ω 2 ζ) : ζ < σ, b ω2 ζ [ā ] : ζ < σ, a ζ = a ω 2 ζ will exemplify the conclusion]] where: (h) if c a β, β < σ, c N iβ+1,θ = cf(θ) N iβ+1 then for some frd N iβ+ω+1 +1 satisfying d a β+ω pcf θ complete (c) we have c b β+ω µ [ā] and d < θ. µ d

23 PCF: THE ADVANCED PCF THEOREMS E69 23 Proof. Proof of (h) So let θ,β,c be given; let b µ [ā] : µ pcf(c) ( N iβ+1 ) be a generating sequence. We define by induction on n < ω, A n, (c η, η ) : η A n such that: (a) A 0 = { },c = c, = maxpcf(c), (b) A n n θ, A n < θ, (c) if η A n+1 then η n A n, c η c η n, η < η n and η = maxpcf(c η ), (d) A n, (c η, η ) : η A n belongs to N iβ+1+n hence η N iβ+1+n, (e) if η A n and η pcf θ-complete (c η ) and c η b β+1+n η [ā] then ( ν)[ν A n+1 andη ν ν = ηˆ 0 ] and c ηˆ 0 = c η \b β+1+n η [ā] (so ηˆ 0 = maxpcf(c ηˆ 0 ) < η = maxpcf(c η ), (f) if η A n and η / pcf θ-complete (c η ) then c η = {b γˆ i [c] : i < i n < θ,ηˆ i A n+1 }, and if ν = ηˆ i A n+1 then c ν = b ν [c], (g) if η A n, and η pcf θ-complete (c η ) but c η b β+1 n n [ā], then ( ν)[η ν A n+1 ]. There is no problem to carry the definition (we use 2.8(1), the point is that c {6.7F} N iβ+1+n implies b (c) : pcf θ [c] N iβ+1+n and as there is d as in 2.8(1), there {6.7F} is one in N iβ+1+n+1 so d a β+1+n+1 ). Now let d n =: { η : η A n and η pcf θ-complete (c η )} and d =: d n ; we shall show that it is as required. n<ω The main point is c b β+ω [ā]; note that d henceitsufficestoshowc [ η d,η A n b β+1+n η n<ω b β+1+n d n [ā] b β+ω [ā]] η [ā],soassumeθ c\ n<ω b β+1+n d n and we choose by induction on n, η n A n such that η 0 =<>, η n+1 n = η n and θ c η ; by clauses (e) + (f) above this is possible and maxpcf(c ηn ) : n < ω is (strictly) decreasing, contradiction. The minor point is d < θ; if θ > ℵ 0 note that A n < θ and θ = cf(θ) clearly n d n A n < θ +ℵ 1 = θ. If θ = ℵ 0 (i.e. clause (h)) we should show that n A n finite; the proof is as above noting that the clause (f) is vacuous now. So n < ω A n = 1 and for some n n A n =, so n A n is finite. Another minor point is d N iβ+ω+1 ; this holds as the construction is unique from c, b µ [c] : µ pcf(c), N j : j < i β+ω, i j : j β +ω, (a i(ζ), b ζ [ā] : a i(ζ) ) : ζ β + ω ; no outside information is used so (A n, (c η, η ) : η A n ) : n < ω N iβ+ω+1, so (using a choice function) really d N iβ+ω [ā],

24 {6.7B} {6.7} {6.7F} {6.7G} {6.7C.1} {6.7A} {6.7A} {6.7A} {6.7F} {6.7A} {6.7F} {6.7A} 24 SAHARON SHELAH Proof. Let b [ā] = b σ = b β [a β] and a σ = a ζ. Part (1) is straightforward. β<σ ζ<σ For part (2), for clause (g), for β = σ, the inclusion is straightforward; so assume µ a β pcf(b β [ā]). Then by 2.4(c) for some β 0 < β, we have µ a β0, {6.7A} and by 2.7 (which depends on 2.4 only) for some β 1 < β, µ pcf(b β1 [ā]); by {6.7C.1} {6.7A} monotonicity without loss of generality β 0 = β 1, by clause (g) of 2.4 applied to β 0, {6.7A} µ b β0 [ā]. Hence by clause (i) of 2.4, µ bβ [ā], thus proving the other inclusion. {6.7A} The proof of clause (e) (for 2.5(2)) is similar, and also 2.5(3). For??(B)(4) for δ < σ,cf(δ) > a redefine b δ [ā] as i<α β<δ Claim 2.8. Let θ be regular. 0) If α < θ,pcf θ-complete ( a i ) = pcf θ-complete (a i ). i<α b β+1 [ā] ) If b [a] : pcf(a) is a generating sequence for a, c a, then for some d pcf θ-complete (c) we have: d < θ and c b θ [a]. 2) If a c < Min(a),c pcf θ-complete (a), pcf θ-complete (c) then pcf θ-complete (a). 3) In (2) we can weaken a c < Min(a) to a < Min(a), c < Min(c). Proof. (0) and (1): Left to the reader. 2) See [Sh:345b, ]. 3) Similarly. 2.8 Claim ) Let θ be regular a. We cannot find α pcf θ-complete (a) for α < a + such that i > suppcf θ-complete ({ j : j < i}). 2) Assume θ a,c pcf θ-complete (a) (and c < Min(c); of course a < Min(a)). If pcf θ-complete (c) then for some d c we have d a and pcf θ-complete (d). Proof. 1) If θ = ℵ 0 we already know it (see 2.7), so assume θ > ℵ 0. We use 2.4 with {θ, i : i < a + } N 0, σ = a +, κ = a +3 where, without loss of generality, κ < Min(a). For each α < a + by (h) + of 2.4 there is a α N i1,d α pcf θ-complete ({ i : i < α}), d α < θ such that { i : i < α} b 1 θ [ā]; hence by clause (g) of 2.4 θ d α and part (0) Claim 2.8 we have a 1 pcf θ-complete ({ i : i < α}) [ā]. So θ a θ d α b 1 θ for α < β < a +, d α a 1 pcf θ-complete { i : i < α} a 1 pcf θ-complete { i : i < β} b 1 θ [ā]. As the sequence is smooth (i.e., clause (f) of 2.4) clearly θ d β α < β b 1 µ [ā] b 1 µ [ā]. µ d α µ d β So b 1 µ [ā] a : α < a + is a non-decreasing sequence of subsets of a of µ d α length a +, hence for some α( ) < a + we have: ( ) 1 α( ) α < a + b 1 µ[ā] a = b 1 µ[ā] a. µ d α µ d α( ) If τ a 1 pcf θ-complete ({ i : i < α}) then τ pcf θ-complete (a) (by parts (2),(3) of Claim 2.8), and τ b 1 µ τ [ā] for some µ τ d α so b 1 τ[ā] b 1 µ τ [ā], also τ pcf θ-complete (b 1 τ [ā] a) (by clause (e) of 2.4), hence τ pcf θ-complete (b 1 τ[ā] a) pcf θ-complete (b 1 µ τ [ā] a) pcf θ-complete ( µ d α b 1 µ [ā] a).

25 PCF: THE ADVANCED PCF THEOREMS E69 25 So a 1 pcf θ-complete ({ i : i < α}) pcf θ-complete ( µ d α b 1 µ [ā] a). But for each α < a + we have α > suppcf θ-complete ({ i : i < α}), whereas d α pcf σ-complete { i : i < α}, hence α > supd α hence ( ) 2 α > sup µ dα maxpcf(b 1 µ[ā]) suppcf θ-complete ( b 1 µ[ā] a). µ d α On the other hand, ( ) 3 α pcf θ-complete { i : i < α+1} pcf θ-complete ( b 1 µ [ā] a). µ d α+1 For α = α( ) we get contradiction by ( ) 1 +( ) 2 +( ) 3. 2) Assume a,c, form a counterexample with minimal. Without lossof generality a +3 < Min(a) and = maxpcf(a) and = maxpcf(c) (just let a =: b [a],c =: c pcf θ [a ]; if / pcf θ-complete (c ) then necessarily pcf(c\c ) (by 2.8(0)) and {6.7F} similarly c\c pcf θ-complete (a\a ) hence by parts (2),(3) of Claim 2.8 we have {6.7F} pcf θ-complete (a\a ), contradiction). Also without loss of generality / c. Let κ,σ, N, i α = i(α) : α σ,ā = a i : i σ be as in 2.4 with a N 0,c N 0, N 0,σ = a +,κ = a +3 < Min(a). We {6.7A} choose by induction on ǫ < a +, ǫ,d ǫ such that: (a) ǫ a ω2 ǫ+ω+1,d ǫ N i(ω2 ǫ+ω+1), (b) ǫ c, (c) d ǫ a ω 2 ǫ+ω+1 pcf θ-complete ({ ζ : ζ < ǫ}), (d) d ǫ < θ, (e) { ζ : ζ < ǫ} (f) ǫ / pcf θ-complete ( b ω 2 ǫ+ω+1 θ θ d ǫ [ā], b ω 2 ǫ+ω+1 θ θ d ǫ [ā]). Foreveryǫ < a + wefirstchoosed ǫ asthe< χ-firstelementsatisfying(c)+(d)+(e) andthenifpossible ǫ asthe< χ-firstelementsatisfying(b)+(f). Itiseasytocheck the requirements and in fact ζ : ζ < ǫ N ω 2 ǫ+1, d ζ : ζ < ǫ N ω 2 ǫ+1 (so clause (a) will hold). But why can we choose at all? Now / pcf θ-complete { ζ : ζ < ǫ} as a,c, form a counterexample with minimal and ǫ < a + (by 2.8(3)). As {6.7F} = maxpcf(a) necessarily pcf θ-complete ({ ζ : ζ < ǫ}) hence d ǫ (by clause (c)). By part (0) of Claim 2.8 (and clause (a)) we know: {6.7F} pcf θ-complete [ b ω 2 ǫ+ω+1 µ [ā]] = pcf θ-complete [b ω 2 +ω+1 µ [ā]] µ d ǫ µ d ǫ (µ+1) µ d ǫ (note µ = maxpcf(b β µ[ā])). So / pcf θ-complete ( b ω 2 ǫ+ω+1 µ [ā]) hence by part (0) µ d ǫ of Claim 2.8 c b ω 2 ǫ+ω+1 µ [ā] so ǫ exists. Now d ǫ exists by 2.4 clause (h) +. µ d ǫ Now clearly a 2 ǫ+ω+1 µ [ā] : ǫ < a + is non-decreasing(as in the earlier µ d ǫ b ω proof) hence eventually constant, say for ǫ ǫ( ) (where ǫ( ) < a + ). But {6.7A} {6.7F}

26 {6.7A} {6.7A} 26 SAHARON SHELAH (α) ǫ b ω 2 ǫ+ω+1 µ d ǫ+1 µ [ā] [clause (e) in the choice of ǫ,d ǫ ], (β) b ω2 ǫ+ω+1 ǫ [ā] µ d ǫ+1 b ω2 ǫ+ω+1 µ [ā] [by clause (f) of 2.4 and (α) alone], (γ) ǫ pcf θ-complete (a) [as ǫ c and a hypothesis], (δ) ǫ pcf θ-complete (b ω2 ǫ+ω+1 ǫ [ā]) [by (γ) above and clause (e) of 2.4], (ǫ) ǫ / pcf(a\b ω2 ǫ+ω+1 ǫ ), (ζ) ǫ pcf θ-complete (a b ω 2 ǫ+ω+1 µ d ǫ+1 µ [ā]) [by (δ)+(ǫ)+(β)]. But for ǫ = ǫ( ), the statement (ζ) contradicts the choice of ǫ( ) and clause (f) above. 2.9