1 The closed unbounded filter Closed unbounded sets Stationary sets in generic extensions... 9


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1 Contents I Stationary Sets 3 by Thomas Jech 1 The closed unbounded filter Closed unbounded sets Splitting stationary sets Generic ultrapowers Stationary sets in generic extensions Some combinatorial principles Reflection Reflecting stationary sets A hierarchy of stationary sets Canonical stationary sets Full reflection Saturation κ + saturation Precipitousness The closed unbounded filter on P κ λ Closed unbounded sets in P κ A Splitting stationary sets
2 2 CONTENTS 4.3 Saturation Proper forcing and other applications Proper forcing Projective and Cohen Boolean algebras Reflection Reflection principles Nonreflecting stationary sets Stationary tower forcing
3 I. Stationary Sets Thomas Jech 1. The closed unbounded filter 1.1. Closed unbounded sets Stationary sets play a fundamental role in modern set theory. This chapter attempts to explain this role and to describe the structure of stationary sets of ordinals and their generalization. The concept of stationary sets first appeared in the 1950 s; the definition is due to G. Bloch [16], and the fundamental theorem on stationary sets was proved by G. Fodor in [24]. However, the concept of a stationary set is implicit in the work of P. Mahlo [71]. The precursor of Fodor s Theorem is the 1929 result of P. Alexandroff and P. Urysohn [2]: if f(α) <αfor all α such that 0 <α<ω 1,thenf is constant on an uncountable set. Let us call an ordinal function f regressive if f(α) <αwhenever α>0. Fodor s Theorem (Theorem 1.5) states that every regressive function on a stationary set is constant on a stationary set. As a consequence, a set S ω 1 is stationary if and only if every regressive function on S is constant on an uncountable set. In this section we develop the theory of closed unbounded and stationary subsets of a regular uncountable cardinal. If X is a set of ordinals, then α is a limit point of X if α > 0and sup(x α) =α. AsetX κ is closed (in the order topology on κ) ifand only if X includes Lim(X), the set of all limit points of X less than κ. 3
4 4 I. Stationary Sets 1.1 Definition. Let κ be a regular uncountable cardinal. A set C κ is closed unbounded (or club for short) if it is closed and also an unbounded subset of κ. AsetS κ is stationary if S C for every closed unbounded C κ. It is easily seen that the intersection of any number of closed sets is closed. The basic observation is that if C 1 and C 2 are both closed unbounded, then C 1 C 2 is also closed unbounded. This leads to the following basic property. 1.2 Proposition. The intersection of less than κ closed unbounded subsets of κ is closed unbounded. Consequently, the closed unbounded sets generate a κcomplete filter on κ called the closed unbounded filter. The dual ideal (which is κcomplete and contains all singletons) consists of all sets that are disjoint from some closed unbounded sets the nonstationary sets, and is thus called the nonstationary ideal, denoted I NS. If I is any nontrivial ideal on κ, theni + denotes the set P (κ) I of all Ipositive sets. Thus stationary subsets of κ are exactly those that are I NS positive. 1.3 Definition. Let X α : α<κ be a κsequence of subsets of κ. Its diagonal intersection is the set α<κ X α = {ξ <κ: ξ α<ξ X α} ; its diagonal union is Σ α<κ X α = {ξ <κ: ξ α<ξ X α}. The following lemma states that the closed unbounded filter is closed under diagonal intersections (or dually, that the nonstationary ideal is closed under diagonal unions): 1.4 Lemma. If C α : α<κ is a sequence of closed unbounded subsets of κ, then its diagonal intersection is closed unbounded. This immediately implies Fodor s Theorem: 1.5 Theorem (Fodor [24]). If S is a stationary subset of κ and if f is a regressive function on S, then there exists some γ<κsuch that f(α) =γ on a stationary subset of S. Proof. Let us assume that for each γ<κthere exists a closed unbounded set C γ such that f(α) γ for each α S C γ.letc = γ<κ C γ.asc is closed unbounded, there exists an α>0ins C. By the definition of C it follows that f(α) α, a contradiction.
5 1. The closed unbounded filter 5 A nontrivial κcomplete ideal I on κ is called normal (and so is its dual filter) if I is closed under diagonal unions; equivalently, if for every A I +, every regressive function on A is constant on some Ipositive set. Thus Fodor s Theorem (or Lemma 1.4) states that the nonstationary ideal (and the club filter) is normal. In fact, the nonstationary ideal is the smallest normal κcomplete ideal on κ: 1.6 Proposition. If F is a normal κcomplete filter on κ, thenf contains all closed unbounded sets. Proof. If C is a club subset of κ, let a α : α<κ be the increasing enumeration of C. Then C α<κ {ξ : a α+1 <ξ<κ} F, because F contains all final segments (being nontrivial and κcomplete). In other words, if I is normal, then every Ipositive set is stationary. The quotient algebra B = P (κ)/i NS is a κcomplete Boolean algebra, where the Boolean operations α<γ and α<γ for γ<κare induced by α<γ and α<γ. Fodor s Theorem implies that B is in fact κ+ complete: if {X α : α<κ} is a collection of subsets of κ, then α<κ X α and Σ α<κ X α are, respectively, the greatest lower bound and the least upper bound of the equivalence classes X α /I NS B. This observation also shows that if X α : α<κ and Y α : α<κ are two enumerations of the same collection, then α X α and α Y α differ only by a nonstationary set. The following characterization of the club filter is often useful, in particular when used in its generalized form (see Section 6). Let F :[κ] <ω κ; an ordinal γ < κ is a closure point of F if F (α 1,...,α n ) <γwhenever α 1,...,α n <γ. It is easy to see that the set Cl F of all closure points of F is a club. Conversely, if C is a club, define F :[κ] <ω κ by letting F (e) be the least element of C greater than max(e). It is clear that Cl F =Lim(C). Thus every club contains Cl F for some F, and we have this characterization of the club filter: 1.7 Proposition. The club filter is generated by the sets Cl F, for all F : [κ] <ω κ. AsetS κ is stationary if and only if for every F :[κ] <ω κ, S contains a closure point of F Splitting stationary sets It is not immediately obvious that the club filter is not an ultrafilter, that is that there exist stationary sets that are costationary, i.e. whose comple
6 6 I. Stationary Sets ment is stationary. The basic result is the following theorem of Solovay: 1.8 Theorem (Solovay [85]). Let κ be a regular uncountable cardinal. Then every stationary subset of κ can be partitioned into κ disjoint stationary sets. Solovay s proof of this basic result of combinatorial set theory uses methods of forcing and large cardinals, and we shall describe it later in this section. For an elementary proof, see e.g. [49], p To illustrate the combinatorics involved, let us prove a special case of Solovay s theorem. 1.9 Proposition. There exist ℵ 1 pairwise disjoint stationary subsets of ω 1. Proof. For every limit ordinal α<ω 1, choose an increasing sequence {a α n } n=0 with limit α. We claim that there is an n such that for all η<ω 1,there are stationary many α such that a α n η: Otherwise there exists, for each n, someη n such that a α n η n for only a nonstationary set of α s. By ω 1  completeness, for all but a nonstationary set of α s the sequences {a α n} n are bounded by sup n η n. A contradiction. Thus let n be such that for all η, thesets η = {α : a α n η} is stationary. The function f(α) =a α n is regressive and so by Fodor s Theorem, there is some γ η η such that T η = {α : a α n = γ η} is stationary. Clearly, there are ℵ 1 distinct values of γ η and therefore ℵ 1 mutually disjoint sets T η. Let κ be a regular uncountable cardinal, and let λ<κbe regular. Let E κ λ = {α <κ:cf α = λ}. For each λ, E κ λ is a stationary set. An easy modification of the proof of 1.9 above shows that for every regular λ<κ, every stationary subset of E κ λ can be split into κ disjoint stationary sets. The union λ Eκ λ is the set of all singular limit ordinals. Its complement is the set Reg of all regular cardinals α<κ. The set Reg is stationary just in case κ is a Mahlo cardinal Generic ultrapowers Let M be a transitive model of ZFC, and let κ be a cardinal in M. LetU be an Multrafilter, i.e. an ultrafilter on the set algebra P (κ) M. Using functions f M on κ, onecanformanultrapowern = Ult U (M), which is
7 1. The closed unbounded filter 7 a model of ZFC but not necessarily wellfounded: f = g {α : f(α) =g(α)} U, f g {α : f(α) g(α)} U. The (equivalence classes of) constant functions c x (α) =x provide an elementary embedding j :(M,ɛ) (N, ), where j(x) =c x, for all x M. An Multrafilter U is Mκcomplete if it is closed under intersections of families {X α : α<γ} M, for all γ<κ; U is normal if every regressive f M is constant on a set in U Proposition. Let U be a nonprincipal Mκcomplete, normal M ultrafilter on κ. Then the ordinals of N have a wellordered initial segment of order type at least κ +1, j(γ) =γ for all γ<κ,andκ is represented in N by the diagonal function d(α) =α. Now let κ be a regular uncountable cardinal and consider the forcing notion (P, <) wherep is the collection of all stationary subsets of κ, andthe ordering is by inclusion. Let B be the complete Boolean algebra B = B(P ), the completion of (P, <). Equivalently, B is the completion of the Boolean algebra P (κ)/i NS. Let us consider the generic extension V [G] givenby a generic G P. It is rather clear that G is a nonprincipal V κcomplete normal ultrafilter on κ. Thus Proposition 1.10 applies, where N = Ult G (V ). The model Ult G (V ) is called a generic ultrapower. There is more on generic ultrapowers in Foreman s chapter in this volume; here we use them to present the original argument of Solovay s [85]. First we prove a lemma (that will be generalized in Section 2): 1.11 Lemma. Let κ be a regular uncountable cardinal, and let S be a stationary set. Then the set T = {α S : either α/ Reg or S α is not a stationary subset of α} is stationary. Proof. Let C be a club and let us show that T C is nonempty. Let α be the least element of the nonempty set S C where C =Lim(C ω). If α is not regular, then α T C and we are done, so assume that α Reg. Now C α is a club subset of α disjoint from S α, andsoα T. We shall now outline the proof of Solovay s Theorem: Proof. (Theorem 1.8.) Let S be a stationary subset of κ that cannot be partitioned into κ disjoint stationary sets. By 1.9 and the remarks following
8 8 I. Stationary Sets its proof, we have S Reg. Let I = I NS S, i.e. I = {X κ : X S I NS }.TheidealI is κsaturated, i.e. every disjoint family W I + has size less than κ; equivalently,b = P (κ)/i has the κchain condition. I is also κcomplete and normal. Let G I + be generic, and let N = Ult G (V ) be the generic ultrapower. As I is κsaturated, N is wellfounded (this is proved by showing that every name f for a function in V on κ can be replaced by an actual function on κ). Thus we have (in V [G]) an elementary embedding j : V N where N is a transitive class, j(γ) =γ for all γ<κ,andκ is represented in N by the diagonal function d(α) =α. Note that if A κ is any set (in V ), then A N: this is because A = j(a) κ; infacta is represented by the function f(α) =A α. Now we use the fact that κc.c. forcing preserves stationarity (cf. Theorem 1.13 below). Thus S is stationary in V [G], and because N V [G], S is a stationary set in the model N. By the ultrapower theorem we have V [G] S α is stationary for Galmost all α. This, translated into forcing, gives {α S : S α is not stationary} I but that contradicts Lemma Another major application of generic ultrapowers is Silver s Theorem: 1.12 Theorem (Silver [84]). Let λ be a singular cardinal of uncountable cofinality. If 2 α = α + for all cardinals α<λ,then2 λ = λ +. Silver s Theorem is actually stronger than this. It assumes only that 2 α = α + for a stationary set of α s (see Section 2 for the definition of stationary when λ is not regular). The proof uses a generic ultrapower. Even though Ult G (V ) is not necessarily well founded, the method of generic ultrapowers enables one to conclude that 2 λ = λ + when 2 α = α + holds almost everywhere. Silver s Theorem can be proved by purely combinatorial methods [10, 11]. In [30], Galvin and Hajnal used combinatorial properties of stationary sets to prove a substantial generalization of Silver s Theorem (superseded only by Shelah s powerful pcf theory). For further generalizations using stationary sets and generic ultrapowers, see [51] and [52]. One of the concepts introduced in [30] is the GalvinHajnal norm of an ordinal function. If f and g are ordinal functions on a regular uncountable
9 1. The closed unbounded filter 9 cardinal κ, letf<gif {α <κ: f(α) <g(α)} contains a club. The relation < is a wellfounded partial order, and the norm f is the rank of f in the relation <. We remark that if f<g, then in the generic ultrapower (by I NS ), the ordinal represented by f is smaller than the ordinal represented by g. By induction on η one can easily show that for each η<κ + there exists a canonical function f η : κ κ of norm η, i.e. f η = η and whenever h = η, then{α : f η (α) h(α)} contains a club. (Proof: Let f 0 (α) =0, f η+1 (α) =f η (α) +1. If η<κ + is a limit ordinal, let λ =cfη and let η = lim ξ λ η ξ. If λ < κ, let f η (α) = sup ξ<λ f ηξ (α) andifλ = κ, let f η (α) =sup ξ<α f ηξ (α).) A canonical function of norm κ + may or may not exist, but is consistent with ZFC (cf. [53]). The existence of canonical function f η for all η is equiconsistent with a measurable cardinal [50] Stationary sets in generic extensions Let M and N be transitive models and let M N. Let κ be a regular uncountable cardinal and let S M be a subset of κ. Clearly, if S is stationary in the model N, thens is stationary in M; the converse is not necessarily true, and κ may even not be regular or uncountable in N. It is important to know which forcing extensions preserve stationarity and we shall return to the general case in Section 5. For now, we state two important special cases: 1.13 Theorem. Let κ be a regular uncountable cardinal and let P be a notion of forcing. (a) If P satisfies the κchain condition, then every club C V [G] has aclubsubsetd in the ground model. Hence every stationary S remains stationary in V [G]. (b) If P is λclosed for every λ < κ, then every stationary S remains stationary in V [G]. Proof. (outline) (a) This follows from this basic fact on forcing: if P is κc.c., then every unbounded A κ in V [G] has an unbounded subset in V. (b) Let p Ċ is a club; we find a γ S and a q p such that q γ Ċ as follows: we construct an increasing continuous ordinal sequence {γ α } α<κ and a decreasing sequence {p α } of conditions such that p α+1 γ α+1 Ċ,
10 10 I. Stationary Sets and if α is a limit ordinal, then γ α = lim ξ<α γ ξ and p α is a lower bound of {p ξ } ξ<α. There is some limit ordinal α such that γ α S. It follows that p α γ α Ċ. We shall now describe the standard way of controlling stationary sets in generic extensions, so called shooting a club. First we deal with the simplest case when κ = ℵ 1. Let S be a stationary subset of ω 1, and consider the following forcing P S (cf. [9]): The forcing conditions are all bounded closed sets p of countable ordinals such that p S. A condition q is stronger than p if q endextends p, i.e. p = q α for some α. It is clear that this forcing produces ( shoots ) a closed unbounded subset of S in the generic extension, thus the complement of S becomes nonstationary. The main point of [9] is that ω 1 is preserved and in fact V [G] adds no new countable sets. Also, every stationary subset of S remains stationary. The forcing P S has the obvious generalization to κ>ℵ 1, but more care is required to guarantee that no new small sets of ordinals are added. For instance, this is the case when S contains the set Sing of all singular ordinals <κ. For a more detailed discussion of this problem see [1] Some combinatorial principles There has been a proliferation of combinatorial principles involving closed unbounded and stationary sets. Most can be traced back to Jensen s investigation of the fine structure of L [59] and generalize either Jensen s diamond ( ) or square ( ). These principles are discussed elsewhere in this volume; we conclude this section by briefly mentioning diamond and clubguessing, and only their typical special cases Theorem ( (ℵ 1 ), Jensen [59]). Assume V = L. There exists a sequence a α : α<ω 1 with each a α α, such that for every A ω 1,theset {α <ω 1 : A α = a α } is stationary. (Note that every A ω is equal to some a α,andso (ℵ 1 ) implies 2 ℵ0 = ℵ 1.) 1.15 Theorem ( (E ℵ2 ℵ 0 ), Gregory [40]). Assume GCH. There exists a sequence a α : α E ℵ2 ℵ 0 with each a α α, such that for every A ω 2,the set {α <ω 2 : A α = a α } is stationary Theorem (Clubguessing, Shelah [82]). There exists a sequence c α : α E ℵ3 ℵ 1,whereeachc α isaclosedunboundedsubsetofα, such that for every club C ω 3,theset{α : c α C} is stationary.
11 2. Reflection 11 Unlike most generalizations of square and diamond, Theorem 1.16 is a theorem of ZFC but we note that the gap (between ℵ 1 and ℵ 3 )isessential. 2. Reflection 2.1. Reflecting stationary sets An important property of stationary sets is reflection. Itisusedinseveral applications, and provides a structure among stationary sets it induces a well founded hierarchy. Natural questions about reflection and the hierarchy are closely related to large cardinal properties. We start with a generalization of stationary sets. Let α be a limit ordinal of uncountable cofinality, say cf α = κ>ℵ 0.AsetS α is stationary if it meets every closed unbounded subset of α. The closed unbounded subsets of α generate a κcomplete filter, and Fodor s Theorem 1.5 yields this: 2.1 Lemma. If f is a regressive function on a stationary set S α, then there exists a γ<αsuch that f(ξ) <γ on a stationary subset of S. If S is a set of ordinals and α is a limit ordinal such that cf α>ω,we say that S is stationary in α if S α is a stationary subset of α. 2.2 Definition. Let κ be a regular uncountable cardinal and let S be a stationary subset of κ. If α < κ and cf α > ω, S reflects at α if S is stationary in α. S reflects if it reflects at some α<κ. It is implicit in the definition that κ>ℵ 1. For our first observation, let α<κbe such that cf α>ω. There is a club C α of order type cf α such that every element of C has cofinality < cf α. Thus if S κ is such that every β S has cofinality cf α, thens does not reflect at α. In particular, if κ = λ + where λ is regular, then the stationary set E κ λ does not reflect. On the other hand, if λ<κis regular and λ + <κ,thene κ λ reflects at every α<κsuch that cf α>λ. To investigate reflection systematically, let us first look at the simplest case, when κ = ℵ 2. Let E 0 = E ℵ2 ℵ 0 and E 1 = E ℵ2 ℵ 1. The set E 1 does not reflect; can every stationary S E 0 reflect? Let us recall Jensen s Square Principle [59]: ( κ ) There exists a sequence C α : α Lim (κ + ) such that
12 12 I. Stationary Sets (i) C α is club in α, (ii) if β Lim(C α ), then C β = C α β, (iii) if cf α<κ,then C α <κ. Now assume that ω1 holds and let C α : α Lim (ω 2 ) be a square sequence. Note that for each α E 1, the order type of C α is ω 1. It follows that there exists a countable limit ordinal η such that the set S = {γ E 0 : γ is the η th element of some C α } is stationary. But for every α E 1, S has at most one element in common with C α,andsos does not reflect. Thus ω1 implies that there is a nonreflecting stationary subset of E ℵ2 ℵ 0. Since ω1 holds unless ℵ 2 is Mahlo in L, the consistency strength of every S E ℵ2 ℵ 0 reflects is at least a Mahlo cardinal. This is in fact the exact strength: 2.3 Theorem (HarringtonShelah [41]). The following are equiconsistent: (i) the existence of a Mahlo cardinal. (ii) every stationary set S E ℵ2 ℵ 0 reflects. Theorem 2.3 improves a previous result of Baumgartner [6] who proved the consistency of (ii) from a weakly compact cardinal. Note that (ii) implies that every stationary set S E ℵ2 ℵ 0 reflects at stationary many α E ℵ2 ℵ 1. A related result of Magidor (to which we return later in this section) gives this equiconsistency: 2.4 Theorem (Magidor [70]). The following are equiconsistent: (i) the existence of a weakly compact cardinal, (ii) every stationary set S E ℵ2 ℵ 0 reflectsatalmostallα E ℵ2 ℵ 1. Here, almost all means all but a nonstationary set. Let us now address the question whether it is possible that every stationary subset of κ reflects. We have seen that this is not the case when κ is the successor of a regular cardinal. Thus κ must be either inaccessible or κ = λ + where λ is singular. Note that because a weakly compact cardinal is Π 1 1 indescribable, every stationary subset of it reflects. In [68], Kunen showed that it is consistent that every stationary S κ reflects while κ is not weakly compact. In [76] it is shown that the consistency strength of every stationary subset of κ reflects is strictly between greatly Mahlo and weakly compact. (For definition of greatly Mahlo, see Section 2.2.)
13 2. Reflection 13 If, in addition, we require that κ be a successor cardinal, then much stronger assumptions are necessary. The argument we gave above using ω1 works for any κ: 2.5 Proposition (Jensen). If λ holds, then there is a nonreflecting stationary subset of E λ+ ℵ 0. As the consistency strength of λ for singular λ is at least a strong cardinal (as shown by Jensen), one needs at least that for the consistency of every stationary S λ + reflects. In [70], Magidor proved the consistency of every stationary subset of ℵ ω+1 reflects from the existence of infinitely many supercompact cardinals. We mention the following applications of nonreflecting stationary sets: 2.6 Theorem (MeklerShelah [76]). The following are equiconsistent: (i) every stationary S κ reflects, (ii) every κfree abelian group is κ + free. 2.7 Theorem (Tryba [90]). If a regular cardinal κ is Jónsson, then every stationary S κ reflects. 2.8 Theorem (Todorčević [88]).If Rado s Conjecture holds, then for every regular κ>ℵ 1, every stationary S E κ ℵ 0 reflects A hierarchy of stationary sets Consider the following operation (the Mahlo operation) on stationary sets. For a stationary set S κ, thetrace of S is the set of all α at which S reflects: Tr(S) ={α <κ:cfα>ωand S α is stationary}. The following basic properties of trace are easily verified. 2.9 Lemma. (a) If S T,thenTr(S) Tr(T ), (b) Tr(S T )=Tr(S) Tr(T ), (c) Tr(Tr(S)) Tr(S), (d) if S T mod I NS,thenTr(S) Tr(T ) mod I NS. Property (d) shows that the Mahlo operation may be considered as an operation on the Boolean algebra P (κ)/i NS.
14 14 I. Stationary Sets If λ<κis regular, let M κ λ = {α <κ:cfα λ}, and note that Tr(Eκ λ ) = Tr(M κ λ )=Mκ λ +. The Mahlo operation on P (κ)/i NS can be iterated α times, for α<κ +. Let M 0 = κ M α+1 = Tr(M α ) M α = ξ<κ M αξ (α limit,α= {α ξ : ξ<κ}). The sets M α are defined mod I NS (the limit stages depend on the enumeration of α). The sequence {M α } α<κ + is decreasing mod I NS,andwhen α<κ,thenm α = M κ λ where λ is the αth regular cardinal. Note that κ is (weakly) Mahlo just in case M κ = Reg is stationary, and that by Lemma 1.11, {M α } α is strictly decreasing (mod I NS,aslongasM α is stationary). Following [13], κ is called greatly Mahlo if M α is stationary for every α<κ +. We shall now consider the following relation between stationary subsets of κ Definition (Jech [47]). S<T iff S α is stationary for almost all α T. In other words, S < T iff Tr(S) T mod I NS. As an example, if λ<µare regular, then E κ λ < Eκ µ. Note also that the language of generic ultrapowers gives this description of <: 2.11 Proposition. S<T iff T S is stationary in Ult G (V ). The following lemma states the basic properties of < Lemma. (a) A<Tr(A), (b) if A<Band B<Cthan A<C, (c) if A A and B B mod I NS,andifA<B,thenA <B. By (c), < can be considered a relation on P (κ)/i NS. By Proposition 1.11, < is irreflexive and so it is a partial ordering. The next theorem shows that the partial ordering < is well founded Theorem (Jech [47]). The relation < is well founded. Proof. Assume to the contrary that there are stationary sets such that A 1 > A 2 >A 3 >. Therefore there are clubs C n such that A n C n Tr(A n+1 )
15 2. Reflection 15 for n =1, 2,...Foreachn, let B n = A n C n Lim(C n+1 ) Lim(Lim(C n+2 )) Each B n is stationary, and for every n, B n Tr(B n+1 ). Let α n =min(b n ). Since B n+1 α n is stationary, we have α n+1 <α n, and therefore, a decreasing sequence α 1 >α 2 >α 3 >. A contradiction. As < is well founded, we can define the order of stationary sets A κ, and of the cardinal κ: o(a) =sup{o(x)+1:x<a}, o(κ) =sup{o(a)+1:a κ stationary}. We also define o(ℵ 0 ) = 0, and o(α) =o(cf (α)) for every limit ordinal α. Note that o(e κ ℵ 0 ) = 0, and in general o(e κ λ )=o(mκ λ )=α, ifλ is the α th regular cardinal. Also, o(ℵ n )=n, o(κ) κ +1 iff κ is Mahlo, and o(κ) κ + iff κ is greatly Mahlo Canonical stationary sets If λ is the α th regular cardinal, then E κ λ has order α; moreover, the set is canonical, in the sense explained below. In fact, canonical stationary sets exist for all orders α<κ +. Let E be a stationary set of order α. IfX E is stationary, then o(x) o(e). We call E canonical of order α if (i) every stationary X E has order α, and (ii) E meets every set of order α. Clearly, a canonical set of order α is unique (mod I NS ), and two canonical sets of different orders are disjoint (mod I NS ). In the following proposition, maximal and is meant mod I NS Proposition (Jech [47]). A canonical set E of order α exists iff there exists a maximal set M of order α. Then(modI NS ) E M Tr(M),M E Tr(E), and Tr(E) Tr(M). One can show that the sets M α obtained by iterating the Mahlo operation are maximal (as long as they are stationary). Thus when we let E α = M α Tr(M α ), we get canonical stationary sets, of all orders α<κ + (for α<o(κ)). The canonical stationary sets E α and the canonical function f α (of Galvin Hajnal norm α) are closely related:
16 16 I. Stationary Sets 2.15 Proposition (Jech [47]). For every α<κ +, α<o(κ), E α {ξ<κ: f α (ξ) =o(ξ)} Full reflection Let us address the question of what is the largest possible amount of reflection, for stationary subsets of a given κ. As A<Bmeans that A reflects at almost all points of B, we would like to maximize the relation <. But A<Bimplies that o(a) <o(b), so we might ask whether it is possible that A<Bfor any two stationary sets such that o(a) <o(b). By Magidor s Theorem 2.4 it is consistent that S<E ℵ2 ℵ 1, and therefore S<Tfor every S of order 0 and every T of order 1. However, this does not generalize, as the following lemma shows that when κ ℵ 3, then there exist S and T with o(s) =0ando(T ) = 1 such that S T Lemma (JechShelah [54]). If κ ℵ 3, then there exist stationary sets S E κ ℵ 0 and T E κ ℵ 1 such that S does not reflect at any α T. Proof. Let S γ, γ<ω 2, be pairwise disjoint stationary subsets of E κ ℵ 0,and let C α, α E κ ℵ 1, be such that for every α, C α is a club subset of α, oforder type ω 1. Because at most ℵ 1 of the sets S γ meet each C α,thereexistsfor each α some γ(α) such that C α S γ(α) =. There exists some γ such that the set T = {α : γ(α) =γ} is stationary; let S = S γ. For every α T, S C α = and so S does not reflect at α. This lemma illustrates some of the difficulties involved when dealing with reflection at singular ordinals. This problem is investigated in detail in [54], where the best possible consistency result is proved for stationary subsets of the ℵ n, n<ω. Let us say that a stationary set S κ reflects fully at regular cardinals if for any stationary set T of regular cardinals o(s) <o(t ) implies S<T, and let us call Full Reflection the statement that every stationary subset of κ reflects fully at regular cardinals. Full Reflection is of course nontrivial only if κ is a Mahlo cardinal. A modification of Theorem 2.4 shows that Full Reflection for a Mahlo cardinal is equiconsistent with weak compactness. The following theorem establishes the consistency strength of Full Reflection for cardinals in the Mahlo hierarchy:
17 3. Saturation Theorem (JechShelah [55]). The following are equiconsistent, for every α κ + : (i) κ is Π 1 αindescribable, (ii) κ is αmahlo and Full Reflection holds. (A regular cardinal κ is αmahlo if o(κ) κ + α; κ is Π 1 1 indescribable iff it is weakly compact.) Full Reflection is also consistent with large cardinals. The paper [57] proves the consistency of Full Reflection with the existence of a measurable cardinal. This has been improved and further generalized in [38]. Finally, the paper [91] shows that any wellfounded partial order of size κ + can be realized by the reflection ordering < on stationary subsets of κ, in some generic extension (using P 2 κstrong κ in the ground model). 3. Saturation 3.1. κ + saturation By Solovay s 1.8 every stationary subset of κ can be split into κ disjoint stationary sets. In other words, for every stationary S κ, theideali NS S is not κsaturated. A natural question is if the nonstationary ideal can be κ + saturated. An ideal I on κ is κ + saturated if the Boolean algebra P (κ)/i has the κ + chain condition. Thus I NS S is κ + saturated when there do not exist κ + stationary subsets of S such that the intersection of any two of them is nonstationary. The existence and properties of κ + saturated ideals have been thoroughly studied since their introduction in [85], and involve large cardinals. The reader will find more details in Foreman s chapter in this volume. We shall concentrate on the special case when I is the nonstationary ideal. The main question, whether the nonstationary ideal can be κ + saturated, has been answered. But a number of related questions are still open. 3.1 Theorem (GitikShelah [37]). The nonstationary ideal on κ is not κ +  saturated, for every regular cardinal κ ℵ Theorem (Shelah). It is consistent, relative to the existence of a Woodin cardinal, that the nonstationary ideal on ℵ 1 is ℵ 2 saturated.
18 18 I. Stationary Sets The consistency result in Theorem 3.2 was first proved in [87] using a strong determinacy assumption. That hypothesis was reduced in [92] to AD, while in [27], the assumption was the existence of a supercompact cardinal. Shelah s result (announced in [81]) is close to optimal: by Steel [86], the saturation of I NS plus the existence of a measurable cardinal imply the existence of an inner model with a Woodin cardinal. All the models mentioned in the preceding paragraph satisfy 2 ℵ0 > ℵ 1. This may not be accidental, and it has been conjectured that the saturation of I NS on ℵ 1 implies that 2 ℵ0 > ℵ 1. In fact, Woodin proved this [94] under the addional assumption that there exists a measurable cardinal. We note in passing that by [27], 2 ℵ0 = ℵ 1 is consistent with I NS S being saturated for some stationary S. Woodin s construction [94] yields a model (starting from AD) in which the ideal I NS is ℵ 1 dense, i.e. the algebra P (ω 1 )/I NS has a dense set of size ℵ 1. This, and Woodin s more recent work using Steel s inner model theory, gives the following equiconsistency. 3.3 Theorem (Woodin). The following are equiconsistent: (i) ZF + AD, (ii) there are infinitely many Woodin cardinals, (iii) the nonstationary ideal on ℵ 1 is ℵ 1 dense. As for the continuum hypothesis, Shelah proved in [80] that if I NS is ℵ 1 dense, then 2 ℵ0 =2 ℵ1. We remark that the mere existence of a saturated ideal affects cardinal arithmetic, cf. [63] and [52]. Let us now return to Theorem 3.1. The general result proved in [37] is this: 3.4 Theorem (GitikShelah [37]). If ν is a regular cardinal and ν + <κ, then I NS E κ ν is not κ+ saturated. The proof of 3.4 combines an earlier result of Shelah (Theorem 3.7 below) with an application of the method of guessing clubs (as in 1.16). The earlier result uses generic ultrapowers and states that if κ = λ + and ν cfλ is regular, then no ideal concentrating on E κ ν is κ + saturated. The method of generic ultrapowers is well suited for κ + saturated ideals. Forcing with P (κ)/i where I is a normal κcomplete κ + saturated ideal makes the generic ultrapower N = Ult G (V ) well founded, preserves the cardinal κ +, and satisfies P N (κ) =P V [G] (κ). It follows that all cardinals
19 3. Saturation 19 <κare preserved in V [G], and it is obvious that if E κ ν G, thenn (and therefore V [G] as well) satisfies cf κ = ν. Shelah s Theorem 3.7 below follows from a simple combinatorial lemma. Let λ be a cardinal and let α<λ + be a limit ordinal. Let us call a family {X ξ : ξ<λ + } a strongly almost disjoint (s.a.d.) family of subsets of α if every X ξ α is unbounded, and if for every ϑ<λ + there exist ordinals δ ξ <α,forξ<ϑ, such that the sets X ξ δ ξ, ξ<ϑ, are pairwise disjoint. Note that if κ is a regular cardinal than there is a s.a.d. family {X ξ : ξ<κ + } of subsets of κ. 3.5 Lemma. If α < λ + and cf α cf λ, then there exists no strongly almost disjoint family of subsets of α. Proof. Assume to the contrary that {X ξ : ξ<λ + } is a s.a.d. family of subsets of α. We may assume that each X ξ has order type cf α. Let f be a function mapping λ onto α. Since cf λ cfα there exists for each ξ some γ ξ <λsuch that X ξ f γ ξ is cofinal in α. There is some γ and a set W λ + of size λ such that γ ξ = γ for all ξ W. Let ϑ>sup W. By the assumption on the X ξ there exist ordinals δ ξ <α, ξ<ϑ, such that the X ξ δ ξ are pairwise disjoint. Thus f 1 (X ξ δ ξ ), ξ W,areλ pairwise disjoint subsets of γ. A contradiction. 3.6 Corollary (Shelah [79]). If κ is a regular cardinal and if a forcing P makes cf κ cf κ, thenp collapses κ +. Proof. Assume that κ + is not collapsed; thus in V [G], (κ + ) V = λ + where λ = κ. In V there is a s.a.d. family {X ξ : ξ<(κ + ) V }, and it remains a s.a.d. family in V [G], of size λ +. Since cf κ cfλ, inv [G], this contradicts Lemma Theorem (Shelah). If κ = λ +,ifν cf λ is regular and if I is a normal κcomplete κ + saturated ideal on κ, thene κ ν I. Proof. If not, then forcing with Ipositive subsets of E κ ν well as cf λ, andmakescfκ = ν; a contradicton. preserves κ + as Theorem 3.4 leaves open the following problem: If λ is a regular cardinal, can I NS E λ+ λ be λ ++ saturated? (For instance can I NS E ℵ2 ℵ 1 be ℵ 3  saturated?) Let us also mention that for all regular ν and κ not excluded by Corollary 3.7, it is consistent that I NS S is κ + saturated for some S E κ ν (see [33]).
20 20 I. Stationary Sets If κ is a large cardinal, then I NS Reg can be κ + saturated, as the following theorem shows. Of course, κ cannot be too large: if κ is greatly Mahlo, then the canonical stationary sets E α κ α<κ + witness nonsaturation. 3.8 Theorem (JechWoodin [58]). For any α < κ +, the following are equiconsistent: (i) κ is measurable of order α, (ii) κ is αmahlo and the ideal I NS Reg on κ is κ + saturated Precipitousness An important property of saturated ideals is that the generic ultrapower is wellfounded. It has been recognized that this property is important enough to single out and study the class of ideals that have it. The ideals for which the generic ultrapower is well founded are called precipitous. They are described in detail in Foreman s chapter in this volume; here we address the question of when the nonstationary ideal is precipitous. Precipitous ideals were introduced by Jech and Prikry in [51]. There are several equivalent formulations of precipitousness. Let I be an ideal on some set E. AnIpartition is a maximal family of Ipositive sets such that the intersection of any two of them is in I. LetG I denote the infinite game of two players who alternately pick Ipositive sets S n such that S 1 S 2 S 3. The first player wins if n=1 S n =. 3.9 Theorem (JechPrikry [51, 45, 46, 29]). Let I be an ideal on a set E. The following are equivalent: (i) forcing with P (E)/I makes the generic ultrapower wellfounded, (ii) for every sequence {W n } n=1 of Ipartitions there exists a sequence {X n } n=1 such that X n W n for each n, and n=1 X n, (iii) the first player does not have a winning strategy in the game G I. The problem of whether the nonstationary ideal on κ can be precipitous involves large cardinals. For κ = ℵ 1 the exact consistency strength is the existence of a measurable cardinal: 3.10 Theorem (JechMagidorMitchellPrikry [50]). The following are equiconsistent: (i) there exists a measurable cardinal, (ii) the nonstationary ideal on ℵ 1 is precipitous.
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