Introduction to Cardinal Arithmetic (Birkhäuser Advanced Texts Basler Lehrbücher)

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M. Holz K. Steffens E. Weitz

Introduction to Cardinal Arithmetic

Reprint of the 1999 Edition Birkhäuser Verlag Basel · Boston · Berlin

Authors: Michael Holz Unter den Bäumchen 17 30926 Seelze Germany e-mail: [email protected]

Edmund Weitz Bernadottestr. 38 22763 Hamburg Germany

Karsten Steffens Institut für Algebra, Zahlentheorie und Diskrete Mathematik Universität Hannover Welfengarten 1 30167 Hannover Germany e-mail: [email protected]

Originally published under the same title in the Birkhäuser Advanced Texts – Basler Lehrbücher series by Birkhäuser Verlag, Switzerland, ISBN 978-3-7643-6124-7 © 1999 Birkhäuser Verlag, P.O. Box 133, CH-4010 Basel, Switzerland

1991 Mathematics Subject Classification 04-01, 04A10, 03E10 Library of Congress Control Number: 2009937809 Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at .

ISBN 978-3-0346-0327-0 Birkhäuser Verlag AG, Basel · Boston · Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use whatsoever, permission from the copyright owner must be obtained. © 2010 Birkhäuser Verlag AG Basel · Boston · Berlin P.O. Box 133, CH-4010 Basel, Switzerland Part of Springer Science+Business Media Printed on acid-free paper produced of chlorine-free pulp. TCF ∞

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Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1 Foundations 1.1 The Axioms of ZFC . . . . . . . . . . . . . . . . . . . . . . 1.2 Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Transfinite Induction and Recursion . . . . . . . . . . . . . 1.4 Arithmetic of Ordinals . . . . . . . . . . . . . . . . . . . . . 1.5 Cardinal Numbers and their Elementary Properties . . . . . 1.6 Infinite Sums and Products . . . . . . . . . . . . . . . . . . 1.7 Further Properties of κλ – the Singular Cardinal Hypothesis 1.8 Clubs and Stationary Sets . . . . . . . . . . . . . . . . . . . 1.9 The Erd¨ os-Rado Partition Theorem . . . . . . . . . . . . . 2 The 2.1 2.2 2.3

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Galvin-Hajnal Theorem Ideals and the Reduction of Relations . . . . . . . . . . . . . . 103 The Galvin-Hajnal Formula . . . . . . . . . . . . . . . . . . . . 108 Applications of the Galvin-Hajnal Formula . . . . . . . . . . . 121

3 Ordinal Functions 3.1 Suprema and Cofinalities . . . . . . . . . . . . . . . . 3.2 κ-rapid Sequences and the Main Lemma of pcf-Theory 3.3 The Definition and Simple Properties of pcf(a) . . . . 3.4 The Ideal J<λ (a) . . . . . . . . . . . . . . . . . . . . .

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4 Approximation Sequences 4.1 The Sets H(Θ) . . . . . . 4.2 Models and Absoluteness 4.3 Approximation Sequences 4.4 The Skolem Hull in H(Θ) 4.5 3-Club Sequences . . . .

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vi 5 Generators of J<λ+ (a) 5.1 Universal Sequences . . . . . . . . . . . . . . . . . . . 5.2 The Existence of Generators . . . . . . . . . . . . . . . 5.3 Properties of Generators . . . . . . . . . . . . . . . . . 6 The Supremum of pcfμ (a) 6.1 Control Functions . . . . . . . . . . . . . . . . . . . . 6.2 The Supremum of pcfμ (a) . . . . . . . . . . . . . . . . 7 Local Properties p 7.1 The Ideals J∗ (b) and J<λ (a) . . . . . . . . . . . . . . 7.2 Intervals in pcf(a) . . . . . . . . . . . . . . . . . . . . 8 Applications of pcf-Theory 8.1 Cardinal Estimates . . . . . . . . . . . . . . . . . . . . 8.2 J´ onsson Algebras . . . . . . . . . . . . . . . . . . . . . 8.3 A Partition Theorem of Todor˘cevi´c . . . . . . . . . . . 8.4 Cofinalities of Partial Orderings ([λ]≤κ , ⊆) . . . . . . . 9 The Cardinal Function pp(λ) 9.1 pp(λ) and the Theorems of Galvin, Hajnal and Silver 9.2 The Lemma of Galvin and Hajnal for pp(λ) . . . . . .

Contents

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Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

Preface This book is an introduction to modern cardinal arithmetic, developed in the frame of the axioms of Zermelo-Fraenkel set theory together with the axiom of choice. It splits into three parts. Part one, which is contained in Chapter 1, describes the classical cardinal arithmetic due to Bernstein, Cantor, Hausdorff, K¨ onig, and Tarski. The results were found in the years between 1870 and 1930. Part two, which is Chapter 2, characterizes the development of cardinal arithmetic in the seventies, which was led by Galvin, Hajnal, and Silver. The third part, contained in Chapters 3 to 9, presents the fundamental investigations in pcf-theory which has been developed by S. Shelah to answer the questions left open in the seventies. All theorems presented in Chapter 3 and Chapters 5 to 9 are due to Shelah, unless otherwise stated. We are greatly indebted to all those set theorists whose work we have tried to expound. Concerning the literature we owe very much to S. Shelah’s book [Sh5] and to the article by M. R. Burke and M. Magidor [BM] which also initiated our students’ interest for Shelah’s pcf-theory. The enthusiasm of our students for cardinal arithmetic contributed much to the formation of this book. Our thanks are due to S. Shelah for answering several questions and to T. Jech for sending us some papers. Our thanks are also due to S. Neumann for many hints to Chapter 9 and to T. Espley and A. Truong for their contributions concerning the readability of this book.

Introduction If M and N are sets1 and if there exists a bijection2 from M onto N , then we say that M and N are equinumerous, and write M ≈ N . To measure the number of members of a set, we will introduce sets of comparison. With ω we denote the set of natural numbers; 0 is a natural number. If for example M is equinumerous to the set {n ∈ ω : n < 25}, then we say that M has exactly 25 members, and {n ∈ ω : n < 25} is a set of comparison for M . If N is a set and if N and ω are equinumerous, then ω will be a set of comparison for N , and N will be called countably infinite or denumerable. A well known example for such a set is N = {n ∈ ω : n is divisible by 2}. We will assign to each set M a set of comparison |M |, the so-called cardinality of M , such that the following holds. (1) M ≈ N ⇐⇒ |M | = |N |. (2) M ≈ |M |. Chapter 1, § 1.5 shows that such an assignment M → |M | is possible. Any set |M | is called a cardinal number or a cardinal. If |M | and |N | are cardinal numbers, then we write |M | ≤ |N | iff3 there is an injection4 from M into N , and |M | < |N | iff |M | ≤ |N | and ¬(M ≈ N ). If we assume the axiom of choice, then any two cardinal numbers are comparable, i.e., we get |M | ≤ |N | or |N | ≤ |M | for any sets M and N ; conversely it will not prove difficult to show that this assertion implies the axiom of choice. Thus the class CN of cardinal numbers is linearly ordered by ≤ and by <. These linear orderings have no greatest element. G. Cantor proved in 1873 that, for any set M , the so-called power set P(M ) of M , i.e., the set {x : x ⊆ M }, satisfies |M | < |P(M )|. In Section 1.5 1 We assume that the reader has an intuitive knowledge about the mathematical notion of a set. {x : E(x)} will be, as usual, the collection of all mathematical objects satisfying the property E(x). 2 That is a function f from M into N such that, for every y ∈ N , there is exactly one x ∈ M with y = f (x). 3 We use the mathematical word iff as usual as an abbreviation for if and only if. 4 That is a function f from M into N such that, for every y ∈ N , there is at most one x ∈ M with y = f (x).

2

Introduction

we will see that CN is well-ordered by <, that means that every nonempty subset of CN has a <-least element. Consequently, for every cardinal number |M |, there exists a smallest cardinal number greater than |M |; we denote it by |M |+ . Every natural number will be shown to be cardinal, and ω is the set of finite cardinals. The infinite cardinals are enumerated by the so-called aleph function whose values ℵ(α) are usually written as ℵα . The smallest infinite + + cardinal is ω, ℵ0 = ω. So we have ℵ1 = ℵ+ 0 , ℵ2 = ℵ1 , and ℵn+1 = ℵn for every n ∈ ω. A natural  candidate for the next cardinal, the smallest cardinal greater than every ℵn , is {ℵn : n ∈ ω}. The following properties show that this set is the supremum of the set {ℵn : n ∈ ω}, and we denote it by ℵω . Therefore we require further that, for any cardinals κ and μ, (3) κ < μ ⇐⇒ κ ∈ μ.  (4) If K is a set of cardinal numbers, then K is a cardinal number. (5) κ < μ =⇒ κ ⊆ μ. These properties will be realized, and thus CN will be completely ordered by < in the sense that every nonempty set of cardinals has a supremum. The name cardinal number would be inappropriate if one could not calculate with cardinals. We define the addition, multiplication and exponentiation of cardinals by |M | + |N | := |M ∪ N |, if M ∩ N = ∅, |M | · |N | := |M × N |, and |M ||N | := |N M |, where N M is the set of all functions from N into M . Each of these operations is a canonical extension of the corresponding operation for natural numbers. However, addition and multiplication satisfy a simple new rule: if one of the nonempty sets M and N is infinite, then |M | + |N | = |M | · |N | = max{|M |, |N |}. Thus we see that the course of values of addition and multiplication can be easily described. This is not at all the case for the exponentiation (κ, λ) → κλ of cardinals. It is easy to see that |ω 2| = |P(ω)|, where 2 := {0, 1}, and that ω 2 and the set of real numbers are equinumerous. With Cantor’s result we get ℵ0 < 2ℵ0 . Cantor believed that there is no cardinal κ with ℵ0 < κ < 2ℵ0 , meaning that 2ℵ0 = ℵ+ 0 . This assertion is called the continuum hypothesis, the more general assertion that 2κ = κ+ for all infinite cardinals κ is called the generalized continuum hypothesis, written as (GCH). Under the assumption that ZermeloFraenkel set theory with the axiom of choice, denoted by ZFC, is consistent, K. G¨ odel proved that ZFC + (GCH), the theory ZFC with the additional axiom (GCH), is consistent, and P. J. Cohen proved, that ZFC + ¬(GCH)

Introduction

3

is consistent. If the reader wishes to accept (GCH), he need not study this book further, since the exponentiation of cardinals obeys in ZFC + (GCH) simple rules. The difficulties arise when cardinal exponentiation is studied in the axiom system ZFC, and we will describe some of them. Let ICN be the class of infinite cardinal numbers. Clearly the so-called continuum function (2κ : κ ∈ ICN) is increasing, i.e., we have 2κ ≤ 2μ for all κ, μ ∈ ICN with κ < μ. To formulate a further property of the continuum function, we need the notion of the cofinality of an infinite cardinal κ, written as cf (κ). If κ > ℵ0 , then we call κ a successor cardinal iff there exists some cardinal μ with κ = μ+ , and otherwise a limit cardinal. In this introduction let us define cf (ℵ0 ) = ℵ0 , cf (μ+ ) = μ+ for any successor cardinal μ+ , and cf (λ) = min{|K| : K ⊆  {ν ∈ ICN : ν < λ} and K = λ} for limit cardinals λ. This minimum exists  since, for N = {ν ∈ ICN : ν < λ}, we have N = λ. Namely, (5) gives    N ⊆ λ, hence  | N | ≤ |λ|. Furthermore,  N is a cardinal  by (4), and < λ, (1) and (2) yield N ≤ λ. The supposition N < λ leads to ( N )+   + + since λis a limitcardinal, and thus gives ( N ) ∈ N and ( N ) ⊆ N , hence ( N )+ ≤ N , contradicting the definition of a cardinal successor. From N ⊆ λ and thus |N | ≤ λ, together with the other parts of the definition, we get cf (κ) ≤ κ for every infinite cardinal number κ. If κ ∈ ICN, then we call κ regular iff cf (κ) = κ, and singular iff cf (κ) < κ. A well-known theorem of J. K¨ onig says that κ < cf (2κ ). Let Reg denote the class of regular cardinals. W. B. Easton proved the following theorem. Assume that F : Reg −→ ICN is a function satisfying (i) κ < μ =⇒ F (κ) ≤ F (μ) and (ii) κ < cf (F (κ)). Then ZFC + ∀κ ∈ Reg(2κ = F (κ)) is consistent, only if ZFC is consistent. This result shows that the course of values of the continuum function on the class of regular cardinals is only limited by the rules (i) and (ii). However, J. Silver proved for singular cardinals κ with ℵ0 < cf (κ): If 2μ = μ+ for all infinite cardinals μ < κ, then 2κ = κ+ . So we see that the values of the continuum function for singular cardinals κ can in fact depend on its course of values on the set of smaller cardinals. Further investigations on the exponentiation of cardinals showed that the so-called gimel function (κcf(κ) : κ ∈ ICN) is of outstanding importance. Already A. Tarski and K. G¨ odel remarked that one can reduce the study of cardinal arithmetic to the study of this function. If κ is regular, then we have κcf(κ) = 2κ , and for singular κ we have κcf(κ) ≤ 2κ . Galvin and Hajnal found a nontrivial upper bound for κcf(κ) if κ is singular and cf (κ) > ℵ0 . Until recently, one could not give a corresponding estimate for singular cardinals with countable cofinality. For this, Shelah developed a theory of its own, the so-called pcf-theory, which is complete in itself, allows to extend

4

Introduction

the Galvin-Hajnal theorem to singular cardinals with countable cofinality, and has other applications outside of cardinal arithmetic. For cardinal arithmetic, it yields for example the following theorem: If 2ℵn < ℵω for every n ∈ ω, then 2ℵω < ℵℵ4 . Chapters 3 to 9 of this book give an introduction to the foundations of pcf-theory.

Chapter 1 Foundations 1.1 The Axioms of ZFC The year of birth of set theory can be regarded as 1872. In this year, Georg Cantor introduced the notion of a transfinite (infinite) ordinal number. His investigations, which soon made set theory an independent area of mathematics, met with mistrust and sometimes open hostility from other mathematicians. At the start of the last decade of the century, set theory came into vogue and was being applied more and more extensively to analysis and geometry. Just when Cantor had found high recognition, and when he had put the finishing touches to his investigations, he himself discovered the first antinomy in his system, which was also found two years later, in 1897, by Burali-Forti. Neither of them could offer any solution, but they did not take the problem too seriously, since it arose in a pretty technical area, namely, the theory of well-ordered sets. They believed that a revision of proofs and theorems could settle the situation, as was often the case earlier, under similar circumstances. These hopes were radically dashed when, in 1902, Bertrand Russell published his famous antinomy which occurred just at the beginning of set theoretical investigations and demonstrated that something wasn’t right in the foundations of that discipline. The disclosure of that antinomy led to a foundational crisis in mathematics. Several ways out of this crisis have been developed, the most successful one being without any doubt the axiomatic method. And amongst the various axiomatic systems of set theory, the system which was begun by Zermelo and completed by Fraenkel and Skolem is, adding the axiom of choice, perhaps the most known and the most practical one. The results of our book are set forth in this theory, denoted by ZFC, the axioms of which are represented below. We always begin with a naive formulation of the particular axiom of ZFC and will then make precise the notion of a “property” occurring in two axioms.

6

Chapter 1. Foundations

The axioms of ZFC are not independent in the sense that no axiom can be proved with the help of the other ones. But historical and practical reasons speak for choosing these axioms. The system ZF consists of the axioms (A1)– (A9); thus ZFC is the theory ZF together with the axiom of choice, (AC). We write a ∈ A for the assertion “The set a is a member (or element) of the set A”. The formal symbols we are using are explained later on when we will make precise the notion of a property. (A1) Axiom of Extensionality If any two sets have the same members, then they are equal. (A2) Empty Set Axiom There is a set having no members. As usual, we denote this set by ∅; ∅ := {x : x = x}. (A3) Comprehension Axioms or Subset Axioms If an arbitrary property is given, then to each set a there is a set b whose members are exactly those sets in a which satisfy the property. If E(x) is the considered property, then we write b = {x ∈ a : E(x)}. (A4) Pairing Axiom For any sets a and b, there is a set having as its members just a and b. We denote this (uniquely determined) set by {a, b} and call it the unordered pair or pair set of a and b. This means {a, b} := {z : z = a ∨ z = b}. Note that {a, a} = {a} follows from (A1). This set is called the singleton {a} or singleton of a. (A5) Union Axiom For any set a, there exists a set whose members are exactly the members of the members of a.   We donote this set by a. So  we have a := {z : ∃u(u ∈ a ∧ z ∈ u)}. Furthermore we put a∪b := {a, b} and {a1 , . . . , an } := {a1 , . . . , an−1 }∪ {an }. (A6) Power Set Axiom For any set a, there exists a set whose members are exactly the subsets of a. We denote this set by P(a) (i.e., P(a):={z : z ⊆ a}) and call it the power set of a. Hereby we say that a set z is a subset of a iff every member of z is a member of a, formally written as ∀u(u ∈ z =⇒ u ∈ a); of course, we abbreviate this by z ⊆ a. (A7) Infinity Axiom There is a set which has the empty set as a member, and has the member b ∪ {b} whenever b is a member of it. Every set having this property is called an inductive set. Thus we can give a short formulation of the infinity axiom: “There is an inductive set”.

1.1. The Axioms of ZFC

7

(A8) Replacement Axioms If the property E(a, b) is functional, that means that for any sets a, b and c we can conclude b = c from E(a, b) and E(a, c), then for each set A there is a set B whose members are exactly those sets b for which there exists a member a of A satisfying E(a, b). Very roughly speaking: “If the domain of a function is a set, then its range is also a set.”1 (A9) Regularity Axiom or Foundation Axiom Any nonempty set has a member which is minimal amongst its members with respect to the relation ∈; or: Any nonempty set has a member which has no members in common with it. (AC) Axiom of Choice For any set A whose members all are nonempty, there is a function with domain A such that, for any member a of A, the value of this function at a is a member of a; or: For any set A whose members all are nonempty, there is a function with domain A “choosing” one member from any member of A. If we define the cartesian product or the set of choice functions for A by  A :=  {f : f is a function from A into A with f (a) ∈ a for all a ∈ A},2 then the axiom of choice, (AC), says:  If A is a set of nonempty sets, then A = ∅. To make the notion of a property precise, we introduce a formal language LZF of set theory which differs from the so-called metalanguage or informal language, which is English. This formal language is a special case of a language of the first-order predicate calculus with equality, with one non-logical symbol ∈, denoting membership. The basic symbols of LZF , forming its alphabet, are (sentential) connectives ¬, ∧, ∨, =⇒ , and ⇐⇒; quantifiers ∀ and ∃; variables vj for each natural number j; = and ∈; parentheses ( and ) . Intuitively, ¬ means “no”, ∧ means “and”, ∨ means “or”, =⇒ means “implies”3 , ⇐⇒ means “equivalent”, 1 Note that this assertion needs no axiom if a function of the form f : A −→ B from the set A into a set B in the usual sense is given, since in this case its range is a set by a comprehension axiom. We don’t need the replacements axioms before we assign, by a functional property, to each member of a given set a certain set, and do not know at the beginning that these “values” are members of another fixed set.   2 If a and b are sets, then also a × b (see below) is a set. As a subclass of P(A × A), A  is a set by (A5), (A6) and (A3). Note that ∅ = {∅}. 3 This usage of the word “implies” is a conventional part of mathematical jargon, but it is not exactly the way the rest of the world uses the word.

8

Chapter 1. Foundations

∀ means “for all”, and ∃ means “there exists”. A string of symbols or expression is a finite sequence of symbols. The concatenation of strings of symbols is that sequence which results from writing the given strings one behind the other. Certain expressions, corresponding to meaningful assertions, will be called formulas. The formulas of LZF are given by the following inductive definition: (1) For any i and j, (vi ∈ vj ) and (vi = vj ) are formulas (so-called atomic formulas). (2) If ϕ and ψ are formulas, so are (¬ϕ), (ϕ ∧ ψ), (ϕ ∨ ψ), (ϕ =⇒ ψ), (ϕ ⇐⇒ ψ), and (∀vj ϕ) and (∃vj ϕ) for any j. Now a property simply is a formula of LZF . If there is no danger of misunderstanding, we do not use parentheses as restrictively as in the definition of a formula; we will, in a formally incorrect manner, omit them for reasons of readability. The use of bounded quantifiers is very practical: ∀vi ∈ vj ϕ is the formula ∀vi (vi ∈ vj =⇒ ϕ), ∃vi ∈ vj ϕ is the formula ∃vi (vi ∈ vj ∧ ϕ). We also write vi = vj for ¬(vi = vj ) and vi ∈ vj for ¬(vi ∈ vj ). A sentence is a formula with no free variable (see section 4.2 for further details), a theory or axiom system is a set of sentences which are called the axioms of the theory. If a formula ϕ is provable in a theory T , where for our purposes mostly T = ZF or T = ZFC will suffice, we write T  ϕ. As usual, we will introduce new function and relation symbols and socalled operations (as the intersection and union of sets). Furthermore it is very useful to speak about collections which need not be sets. Both aims can be realized by using classes: Every variable is a class; if ϕ(x, p) is a formula, then X := {x : ϕ(x, p)} is a class, where we allow ϕ to have free variables p etc. other than x, which are thought of as parameters upon which the class depends. Members of the class X are exactly those sets a which have the property ϕ(a, p). This means that a ∈ {x : ϕ(x, p)} ⇐⇒ ϕ(a, p). This is the main rule with whose help we can eliminate the new expressions {x : ϕ(a, p)}. We use capitel letters A, B, . . . to denote classes, also with indices, and define the extended formulas as those expressions which may contain the symbols {, } and : besides the symbols of LZF and which are defined in the same way as the formulas with the additional rule that (A = B) and (A ∈ B), for classes A and B, are also atomic formulas. If A = {x : ϕ(x)} and B = {x : ψ(x)}, then y ∈ A is the formula ϕ(y), A = B is the formula ∀x(ϕ ⇐⇒ ψ), A = y is the formula ∀x(ϕ ⇐⇒ x ∈ y), A ∈ B is the formula ∃z(ψ(z) ∧ ∀u(u ∈ z ⇐⇒ ϕ(u))), and A ∈ y is the formula ∃z(∀x(ϕ ⇐⇒ x ∈ z) ∧ z ∈ y), where the variable z can be chosen uniquely (and suitably). Thus any extended formula can be replaced by a formula of LZF . We could go one step further and admit classes of the form {x : Ψ(x)}, where Ψ is allowed to contain further classes.

1.1. The Axioms of ZFC

9

A. Levy proves in the appendix of his book [Le] that these extensions of the language can be eliminated in a formally correct way. In the first chapter of this book, the reader can find an explanation of our motivation and justification for our procedure of using classes. We can now give a precise formulation of the replacement axioms, using formulae or, equivalently, using classes. The formalization of the comprehension axioms is left to the reader. If ϕ(x, y, p1 , . . . , pn ), written as ϕ(x, y), is a formula of LZF , then the following sentence is an axiom of ZF: ∀p1 . . . ∀pn (∀u∀v∀w(ϕ(u, v) ∧ ϕ(u, w) ⇒ v = w) ⇒ ∀x∃y∀v(v ∈ y ⇔ ∃u ∈ x ϕ(u, v))). The class version says that, for any class F , the following formula is an axiom of ZF: Func(F ) =⇒ ∀x∃y(y = ran(F |`x)).4 We assume that the reader is familiar with the basic concepts of set theory as, for example, intersection, union, relation, function etc. Especially he should have seen a development of the elementary properties of the Boolean algebra of sets and should have learned to handle the “language of set theory” in the frame of an undergraduate course in mathematics. However, the basic properties of ordinals and cardinals will be described in detail, and principally no knowledge in set theory is required. We now give the definitions of all basic concepts that will be needed from the beginning. x, y, z, . . . always denote sets, A, B, C, . . . denote classes; since x = {z : z ∈ x} is provable5 , the definitions for classes also apply to sets. The class V := {x : x = x} is the universal class. A is a subclass of B, written as A ⊆ B, iff ∀z(z ∈ A =⇒ z ∈ B) (formally: A ⊆ B is an abbreviation for this formula; we will also state this definition in the form A ⊆ B :⇐⇒ ∀z(z ∈ A =⇒ z ∈ B)). A ⊂ B and A  B stand for the formula A ⊆ B ∧ A = B, to be read as “A is a proper subclass of B”. Any class A is called a set iff ∃y∀z(z ∈ y ⇐⇒ z ∈ A) is provable. We call A a proper class iff ∀y(y = A) is provable. (x, y) := {{x}, {x, y}} is the ordered pair of x and y. The essential property of this set is ∀x∀y∀u∀v((x, y) = (u, v) ⇐⇒ x = u ∧ y = v). The class A × B := {(x, y) : x ∈ A ∧ y ∈ B} is the cartesian product or product of the classes A and B (where {(x, y) : ϕ(x, y)} := {z : ∃x∃y( z = (x, y) ∧ ϕ(x, y))}). The classes A ∪ B := {x : x ∈ A ∨ x ∈ B}, A ∩ B := {x : x ∈ A ∧ x ∈ B} and A \ B := {x : x ∈ A ∧ x ∈ B} are respectively the union, the intersection and the difference of the classes A and B. We say that the classes A and B are 4 For 5 By

the definitions of Func, ran and F |`x, see below. our conventions, this is the formula ∀z(z ∈ x ⇐⇒ z ∈ x).

10

Chapter 1. Foundations

 disjoint iff A ∩ B = ∅. A := {x : ∃a(a ∈ A ∧ x ∈ a)} is the union of A and  A := {x : ∀a(a ∈ A =⇒ x ∈ a)} is the intersection of A. If A and Bare  sets, so are A \ B, A ∪ B, A ∩ B, A × B and A; in the case that A = ∅, A  is also a set. Note that ∅ = V. A relation is a class of ordered pairs: Rel(R) :⇐⇒ R ⊆ V × V. Instead of (x, y) ∈ R we also write xRy. Assume that R is a relation. R is a relation from A into B iff R ⊆ A × B. If R ⊆ A × A, then R is a relation on A or a binary relation on A. Analogously, n-ary relations are defined as subclasses of the n-fold cartesian product of A. The relation R |`A := {(x, y) ∈ R : x ∈ A} is the restriction of the relation R to A, R−1 := {(y, x) : (x, y) ∈ R} is the inverse relation of R. The domain of R is dom(R) := {x : ∃y (x, y) ∈ R}, the range of R is ran(R) := {y : ∃x (x, y) ∈ R}, and fld(R) := dom(R) ∪ ran(R) is the field of the relation R. The image of A under R is R[A] := {y : ∃x ∈ A (x, y) ∈ R}. If S is a further relation, then the composition of R and S is the relation R ◦ S := {(u, w) : ∃v((u, v) ∈ S ∧ (v, w) ∈ R)}. We define F to be a function, written as Func(F ), by Func(F ) :⇐⇒ Rel(F ) ∧ ∀u∀v∀w((u, v) ∈ F ∧ (u, w) ∈ F =⇒ v = w). If F is a function and u ∈ dom(F ), then F (u), the value of F at u, denotes the unique set v with (u, v) ∈ F . Let F : A −→ B

:⇐⇒

Func(F ) ∧ dom(F ) = A ∧ ran(F ) ⊆ B;

then F is a function from A into B. A function F : A −→ V is also called a family and written as F =: (F (a) : a ∈ A). Often one introduces a new symbol, say B, and writes Ba instead of F (a) and (Ba : a ∈ A) instead of F . The cartesian product of a family (F (a) : a ∈ A) is defined by   F (a) := {f : f : A −→ {F (a) : a ∈ A} ∧ ∀a ∈ A(f (a) ∈ F (a))}. a∈A

Any function F : A −→ B is a set, being a subclass of A × ran(F ), if A is a set. If A is a set, A B is the class of functions from A into B. idA is the function {(a, a) : a ∈ A}, the so-called identity function or identity on A. A function F from A into B is called surjective or a surjection or a function from A onto B iff ran(F ) = B. Any function F is called injective or one-to-one or an injection iff ∀u∀v∀w((u, v) ∈ F ∧ (w, v) ∈ F =⇒ u = w); in this case, F −1 is a function, the inverse of the function F . If F is an injective function from A onto B, then F is called bijective or a bijection. Now assume that R and S are binary relations on the classes A and B, respectively. A function F : A −→ B is a homomorphism (with respect to R and S) iff

1.1. The Axioms of ZFC

11

∀x ∀y(xRy =⇒ F (x)SF (y)); it is an embedding (w.r.t. R and S) iff it is injective and ∀x ∀y(xRy ⇐⇒ F (x) S F (y)). Any surjective embedding is an isomorphism. If A and B – and thus also F , R and S – are sets, then we will use for such functions the familiar denotation F : (A, R) −→ (B, S). We say that a relation R is reflexive iff ∀x ∈ fld(R)(xRx); irreflexive iff ∀x ∈ fld(R)(¬xRx); antisymmetric iff ∀x ∀y(xRy ∧ yRx =⇒ x = y); asymmetric iff ∀x ∀y(xRy =⇒ ¬yRx); symmetric iff ∀x ∀y(xRy =⇒ yRx); connected or trichotomic iff ∀x ∈ fld(R) ∀y ∈ fld(R)(xRy ∨ x = y ∨ yRx); transitive iff ∀x ∀y ∀z(xRy ∧ yRz =⇒ xRz); an equivalence relation iff it is reflexive, symmetric, and transitive; a partial ordering iff it is reflexive, antisymmetric, and transitive; a strict partial ordering iff it is irreflexive and transitive6 ; a linear ordering iff R is a connected partial ordering; well-founded, written as wf(R), iff (i) ∀x ∈ fld(R)∃y(x ∈ y ∧ R−1 [y] ⊆ y) and (ii) ∀x ⊆ fld(R)(x = ∅ =⇒ ∃y ∈ x∀z ∈ x(¬zRy)); a well-ordering iff R is a well-founded connected strict partial ordering. If R is a relation, B is a subclass of fld(R) and x ∈ B, then x is a greatest element or largest element or maximum of B (w.r.t. R), denoted by max(B), iff ∀z ∈ B(z = x =⇒ zRx); least element or smallest element or minimum of B (w.r.t. R), denoted by min(B), iff ∀z ∈ B(z = x =⇒ xRz); maximal element (w.r.t. R) or R-maximal element of B iff ¬∃z ∈ B(z = x ∧ xRz); minimal element (w.r.t. R) or R-minimal element of B iff ¬∃z ∈ B(z = x ∧ zRx). If A is a class and R is a relation with fld(R) = A, then we write (A, R). For example, if R is a partial ordering on A with fld(R) = A, then we say that (A, R) is a partial ordering. Assume that R is a relation on H with fld(R) = H and X is a subset of H. X is called a chain iff (X, R ∩ (X × X)) is a linear ordering (in particular, any two elements of X are comparable under R, i.e., for all y, z ∈ X we have yRz ∨y = z ∨zRy). We say that X is an antichain iff any two different elements 6 In most cases we will make no difference between partial orderings and strict partial orderings. For any partial ordering R, we get a strict partial ordering R via xR y :⇐⇒ xRy ∧ x = y, and conversely, any strict partial ordering R yields a partial ordering R by xRy :⇐⇒ xR y ∨ x = y.

12

Chapter 1. Foundations

of X are incomparable under R, i.e., iff for all different y, z ∈ X we have ¬(yRz) and ¬(zRy). Assume further that x ∈ H. x is an upper bound (lower bound) of X iff sRx ∨ s = x for all s ∈ X (xRs ∨ x = s for all s ∈ X). We say that x is a supremum of X and write x = sup(X) iff x is a least upper bound of X; i.e., x is an upper bound of X, and xRy ∨ x = y for any upper bound y of X. Accordingly, x is an infimum of X, written as x = inf(X), iff x is a greatest lower bound of X. If (H, R) is a partial ordering and a supremum or an infimum of X exists, then it is uniquely determined. X is called an initial segment of H iff, for all s ∈ X and all z ∈ H, zRs implies z ∈ X. X is a proper initial segment of H iff X is an initial segment of H and different from H. If (H, ≤) and (K, ) are partial orderings and f : H −→ K is a function, then f is called increasing or monotone or monotonic (decreasing) iff, for all x, y ∈ H, x ≤ y implies f (x)  f (y) (f (y)  f (x)). We say that f is strictly increasing etc. (strictly decreasing) iff, for all x, y ∈ H, x < y implies f (x) ≺ f (y) (f (y) ≺ f (x)). Note: If H is a well-ordering, then every strictly decreasing sequence (for the definition of a sequence, see section 1.4) of elements of H is finite, since its range has a least element. Exercises 1)  Let x = {∅,{∅},{∅,{∅}}}. Write   the following sets in the form {a1 ,...,ak }: x, x, P({x}), P(x) ∩ (x, {x}). 2) Use the axioms of ZF to show for any sets x, y: a) x × y, {P(y) : y ∈ x}, {f : f : x −→ P(x)} are sets; if x = ∅, then x is a set.  b) ∅, {x : y ∈ x}, {x : ¬∃y(x ∈ y ∧ y ∈ x)} are not sets. 3) Prove without using (A9): There is not a set A such that P(A) ⊆ A. Hint: Consider the class {x ∈ A : x ∈ x}. 4) Show with the axiom of regularity, that there is not a finite sequence (xi : i ∈ n) such that x0 ∈ x1 ∈ x2 ∈ . . . ∈ xn−1 ∈ x0 . 5) The following sets [x, y] have, for x = y, exactly two members. Which of them have the property of an ordered pair, i.e., [x, y] = [x , y  ] ⇐⇒ x = x ∧ y = y  ? a) [x, y] := {{x}, {x, y}}. b) [x, y] := {x, {y}}. c) [x, y] := {x, {x, y}}.

1.1. The Axioms of ZFC

13

6) Prove that the pairing axiom follows from the other axioms of ZF. Hint: Observe that {∅, {∅}} is a set by the power set axiom. Now let a and b be sets and apply a replacement axiom to the formula (u = ∅ ∧ v = a) ∨ (u = {∅} ∧ v = b). 7) Assume that a, b, c, d, e are different sets and R = {(a, b), (a, c), (a, d), (b, b), (b, c), (b, d), (c, c)}. Determine the following sets: dom(R),  ran(R), R[{b, e}], R |` {b, e}, R. Is R reflexive, symmetric, R−1 |` {a, b}, R−1 [{a, b}], R ◦ R and antisymmetric, connected, transitive? Has fld(R) a minimal, maximal, greatest, smallest element? 8) In general, a function F : A −→ B needs not be injective and F −1 does not exist. But prove that (F −1 )∗ : P(B) −→ P(A), defined by (F −1 )∗ (x) = F −1 [x] for any x ∈ P(B), is a function, and that (F −1 )∗ is injective if F is surjective. 9) Let f : A −→ B be a function, x ⊆ A and y ⊆ B. Prove: a) b) c) d)

x ⊆ f −1 [f [x]]. f [f −1 [y]] ⊆ y. If f is injective, then in a) equality holds. If f is surjective, then in b) equality holds.

Give examples such that in a) and b) respectively equality does not hold.  10) Let X be a nonempty set. We call a set P a partition of X iff P = X, ∅ ∈ P , and the members of P are pairwise disjoint. If (X, R) is an equivalence relation (we also say that R is an equivalence relation on X) and x ∈ X, then {y ∈ X : (x, y) ∈ R} is the equivalence class of x (w.r.t. R). a) If R is an equivalence relation on X, then the set of equivalence classes is a partition of X. b) If P is a partition of X, and if the relation R ⊆ X × X is defined by (a, b) ∈ R :⇐⇒ ∃Y ∈ P (a ∈ Y ∧ b ∈ Y ), then R is an equivalence relation on X. c) Assume that the set ω of natural numbers and Z, the set of integers, are already known. For any n ∈ ω with n ≥ 2 let Rn be defined by (a, b) ∈ Rn :⇐⇒ n divides b − a. Show that Rn is an equivalence relation on Z, and determine the equivalence classes a, a ∈ Z, for n = 6. (Hint: Use division with remainder.)

14

Chapter 1. Foundations d) Show that, for fixed n ∈ ω, G = {(x, y) : ∃a ∈ Z ∃c ∈ Z(x = (a, c) ∧ y = a · c)} is a function.

11) Assume that X is a nonempty set and Part(X) is the class of partitions of X. Let the relation  on Part(X) be defined by P1  P2 :⇐⇒ ∀Y ∈ P1 ∃Z ∈ P2 (Y ⊆ Z) for P1 , P2 ∈ Part(X). Prove: a) Part(X) is a set. b)  is a partial ordering on Part(X). c) If T is a nonempty subset of Part(X), then T has an infimum under . Hint: inf(T ) is the partition  {Y ⊆ X : Y = ∅ ∧ ∃D(D ⊆ T ∧ Y = D ∧ ∀P ∈ T |P ∩ D| = 1)}. d) If T is a nonempty subset of Part(X), then T has a supremum under . Hint: sup(T ) = inf{P  ∈ Part(X) : ∀P ∈ T P  P  }. 12) Assume that R is a reflexive transitive relation. Put S := {(x, y) : (x, y) ∈ R ∧ (y, x) ∈ R}. Show that S is an equivalence relation on fld(R). Further let [a] := {x ∈ fld(R) : (a, x) ∈ S} be the equivalence class of a ∈ fld(R). Prove that T := {([a], [b]) : (a, b) ∈ R} is a partial ordering on the set {[a] : a ∈ fld(R)}. 13) Assume that A and B are nonempty sets. Further let (X(a,b) : (a, b) ∈ A × B) be a family of sets such that X(a,b) ∩ X(a,c) = ∅ for all a ∈ A and all b, c ∈ B with b = c. Show with the axioms of ZF the existence of the families which are needed in the following equation and prove       X(a,b) = X(a,f (a)) . a∈A

b∈B

f ∈A B

a∈A

14) Any two functions f and g are called compatible iff f (x) = g(x) for all x ∈ dom(f ) ∩ dom(g). Prove: a) f and g are compatible iff f ∪ g is a function iff f |` (dom(f ) ∩ dom(g)) = g |`(dom(f ) ∩ dom(g)).  b) If F is a set of pairwise compatible functions, then G := F is  a function with dom(G) = {dom(f ) : f ∈ F } and ran(G) = {ran(f ) : f ∈ F }.

1.2. Ordinals

15

c) Let A and B be sets and Func(A, B) the class of all functions f with dom(f ) ⊆ A and ran(f ) ⊆ B. Prove with the axioms of ZF that Func(A, B) is a set. We define a relation  on this set by f  g :⇐⇒ f ⊆ g. Show that  is a partial ordering. d) Let X ⊆ Func(A, B). Show that X has a supremum under  in Func(A, B) iff X is a set of pairwise compatible functions.

1.2 Ordinals In this section we will prove that every ordinal is well-ordered by ∈. In Section 1.3 we will show that, for each well-ordered set (A, <), there is a unique ordinal α such that (A, <) is isomorphic to (α, ∈). So it suffices to study the theory of ordinals to know the theory of well-orderings. Furthermore the principles of induction and recursion for natural numbers can be generalized for ordinals. So we get the fundamental tools of transfinite set theory. With the axiom of choice we will prove in Section 1.5 that every set can be well-ordered. This allows us to define for each set its cardinality or cardinal number as a special ordinal number. Definition A class A is transitive iff every member of A is a subset of A; this means formally: ∀y∀z(z ∈ y ∧ y ∈ A =⇒ z ∈ A). An ordinal number or ordinal is a transitive set whose members are all transitive. We denote by ON the class of ordinals. Small greek letters α, β, γ, δ, σ, τ, ξ, ζ are variables for ordinals. The letters κ, λ, μ, ν, ρ are reserved for special ordinals, the cardinals or cardinal numbers. Instead of σ ∈ τ we also write σ < τ , instead of σ < τ ∨ σ = τ we write σ ≤ τ . Example ∅ is an ordinal, and therefore ON = ∅. Further no ordered pair is an ordinal. If we assume that (a, b) = {{a}, {a, b}} is an ordinal, then a ∈ (a, b), since every ordinal is transitive. Thus we get a ∈ a, from which we can infer that the set {a} has no ∈-minimal element, contradicting the axiom of regularity. Lemma 1.2.1 a) Any member of an ordinal is an ordinal. b) Any nonempty class A of ordinals has an ∈-minimal element (i.e., A ⊆ ON ∧ A = ∅ =⇒ ∃y(y ∈ A ∧ ∀z(z ∈ A =⇒ z ∈ y))). c) σ ∈ σ.7 7 By the axiom of foundation, we get x ∈ x for every set x. The proof is exactly the same as that for σ.

16

Chapter 1. Foundations

Proof For a), let x ∈ σ. Since σ is transitive, we get x ⊆ σ. Therefore every member of x is transitive, since this holds for the members of σ. Finally, as a member of σ, x is transitive. For b), let σ ∈ A. If σ ∩ A = ∅, we can choose, by the axiom of foundation, an ∈-minimal element of σ ∩ A, which has no ∈-predecessor in A, since σ is transitive. To prove c), we assume that σ ∈ σ. The axiom of foundation implies that the nonempty subset {σ} of σ has a member y satisfying y ∩ {σ} = ∅. Certainly y = σ, and with σ ∈ σ ∩ {σ} we get a contradiction to σ ∩ {σ} = ∅. Corollary 1.2.2 a) ON is not a set. b) Every class A of ordinals is partially ordered by ∈.8 Proof To prove a), we suppose that there is a set y such that y = ON. By Lemma 1.2.1, y is transitive. Furthermore, by definition, all members of y are transitive, and consequently y is an ordinal. This gives y ∈ ON = y, contradicting Lemma 1.2.1 c). Assertion b) follows immediately from Lemma 1.2.1 c) and the fact that all ordinals are transitive. Lemma 1.2.3 ∈ is connected on ON; that means if σ and τ are ordinals, then σ ∈ τ ∨ σ = τ ∨ τ ∈ σ. Proof To get a contradiction, let us assume that the assumption is false. Then, by Lemma 1.2.1 b), the nonempty class A := {σ : ∃τ (σ ∈ τ ∧ σ = τ ∧ τ ∈ σ)} has an ∈-minimal element, say σ0 . By the same reason, the nonempty class B := {τ : σ0 ∈ τ ∧ σ0 = τ ∧ τ ∈ σ0 } has an ∈-minimal element, say τ0 . Thus σ0 ∈ τ0 and σ0 = τ0 and τ0 ∈ σ0 . The contradiction will have arrived if we show σ0 = τ0 . Every member of an ordinal is an ordinal. So let α ∈ σ0 . Then α ∈ A, which implies α ∈ τ0 or α = τ0 or τ0 ∈ α. From either of the last two assertions we can infer that τ0 ∈ σ0 , which is impossible. So we get α ∈ τ0 , hence σ0 ⊆ τ0 . In the same way we conclude successively from α ∈ τ0 that α ∈ B, α ∈ σ0 , and τ0 ⊆ σ0 . Altogether we have shown that σ0 = τ0 . 8 If

A does not have exactly one member, this means that ∈ ∩ (A×A) is a partial ordering.

1.2. Ordinals

17

Corollary 1.2.4 a) Every class A of ordinals is well-ordered by ∈. b) If A is a transitive proper class and all members of A are transitive, then A = ON. c) σ ∈ τ ⇐⇒ σ  τ . Proof Together with the preceding results, for the proof of a) we only have to verify the first condition in the definition of a well-founded relation. If R :=∈ ∩ (A × A) and σ ∈ fld(R), then y := σ ∪ {σ} is a set satisfying R−1 [y] ⊆ y. Next we prove part b). From the assumption we can easily infer that all members of A are ordinals. Consider, to get a contradiction, the ordinal α := min(ON \ A). Then A, being a proper class, is no subclass of α. So we can choose γ ∈ A \ α. The previous lemma shows that α < γ. But this yields α ∈ A, since A is transitive, contradicting α ∈ A. For the proof of c), we note that one implication of the equivalence holds by the definition of an ordinal and Lemma 1.2.1 c). For the converse, let σ  τ . By Lemma 1.2.3 and τ ∈ τ , we get σ ∈ τ . Definition For any ordinals α and β such that α ≤ β, let (α, β) := (α, β)ON := {x ∈ ON : α < x < β}, (α, β] := (α, β]ON := {x ∈ ON : α < x ≤ β} be the open interval, half open interval of ordinals with endpoints α and β, respectively. The intervals [α, β)ON and [α, β]ON are defined analogously, where [α, β]ON is called the closed interval with endpoints α and β. Lemma 1.2.5

  a) If x is a set of ordinals, then x is an ordinal, and x = sup(x).  b) If A is a nonempty class of ordinals, then A = min(A).  Proof To prove a) we note that every member of x is a member of an ordinal, therefore is itselfan ordinal and in particular transitive. In the same  way one can show that x is transitive. Consequently, x is an ordinal.   It is an upper bound of x, since τ ∈ x implies τ ⊆ x and thus τ ≤ x by  Corollary 1.2.4. If σ < x, then σ is not an upper bound  of x because there is an ordinal τ ∈ x such that σ ∈ τ . So we have proved x = sup(x). For part b), Lemma 1.2.1 implies that A has an ∈-minimal element σ0 .  Obviously, this is the minimum of A. Certainly we have A  ⊆ σ0 . Conversely, if α <  σ0 , then α < β for all β ∈ A, and so we get α ∈ A. This proves σ0 = A.

18

Chapter 1. Foundations

Definition If x is a set, then the set S(x) := x ∪ {x} is called the successor of x. An ordinal α is called a successor ordinal iff there is an ordinal σ such that α = S(σ). We also write σ + 1 instead of S(σ). Let 0 := ∅, called zero. An ordinal α is called a limit ordinal, written Lim(α), iff α = 0 and α is not a successor ordinal. We use Lim to denote the class of limit ordinals 9 . Lemma 1.2.6 a) b) c) d)

S(σ) is an ordinal satisfying σ < S(σ). If σ ≤ τ ≤ S(σ), then τ = σ or τ = S(σ).  Lim(α) ⇐⇒ α = 0 ∧ α = α. Every ordinal is 0 or a successor ordinal or a limit ordinal.

Proof For part b), suppose that σ < τ < S(σ). Then σ ∈ σ, contradicting Lemma 1.2.1. For part c), first let α be a limit  ordinal and β ∈ α. Then α = 0 and, by part b), β < β + 1 < α. So we get β ∈ α and consequently α ⊆ α. Since α   is transitive, we have α ⊆ α. Thus we have shown that α = α. Conversely,  if α = 0 is a successor ordinal β + 1, then α = β = α. Lemma 1.2.7 a) There exists a limit ordinal. b) If α0 is the least limit ordinal  and C := {x : x is inductive} is the class of inductive sets, then α0 = C. Proof First we prove part a). By the axiom of infinity, there is an inductive set y. Let the set x be given by an axiom of comprehension by x := {z ∈ y : z ∈ ON ∧ z ⊆ y}. We want to show that x is an ordinal. First we have 0 ∈ x, hence x = ∅, and every member of x is transitive. If u ∈ z ∈ x, we get u ∈ y since z ⊆ y, and u ∈ ON. Furthermore we have u ⊆ y, since u ⊆ z. This shows u ∈ x; x is transitive. Therefore x is an ordinal. Now suppose x = S(σ). Then S(σ) ⊆ y, and σ ∈ y leads to S(σ) ∈ y. This yields x ∈ x, contradicting Lemma 1.2.1. To prove b), we note that  α0 , as a limit ordinal, is an inductive set by Lemma 1.2.6. This implies C ⊆ α0 . To get a contradiction we assume that   α0 \ C = ∅. Then the ordinal σ := min(α0 \ C) is a member of α0and thus must be a sucessor ordinal S(τ ), since 0 ∈ C. Further we have τ ∈ C.  This implies σ = S(τ ) ∈ C, since any intersection of a nonempty class of inductive sets is inductive again, which contradicts the choice of σ. 9 From the results in Section 1.3 we can easily infer that this is a proper class. If σ ∈ ON and ω is the set of natural numbers defined below, then we can define a function (σn : n ∈ ω) by σ0 := σ and σn+1 := σn + 1; the ordinal τ := sup{σn : n ∈ ω} is a limit ordinal greater than σ.

1.2. Ordinals

19

Definition We define ω := {x ∈ ON : ∀z ≤ x(¬Lim(z))}. ω is a set, since it is a subclass of the smallest limit ordinal. Further it is easy to show that ω is an inductive set. Now Lemma 1.2.7 says that ω is the least limit ordinal and the smallest inductive set. ω is called the set of natural numbers. Each of its members is called a natural number. The first members of ω are 0 := ∅, 1 := {0}, 2 := {0, 1}, . . . . Corollary 1.2.8 (Induction Principle for ω, Complete Induction) inductive subset of ω, then A = ω. Formally:

If A is an

A ⊆ ω ∧ 0 ∈ A ∧ ∀α ∈ A(α + 1 ∈ A) =⇒ A = ω. Exercises 1) Assume that a, b are sets and a is transitive. Prove: a) b is transitive iff b ⊆ P(b). b) If b is transitive, then a ∪ b is transitive. c) If b is transitive, then a ∩ b is transitive. d) If S ⊆ P(a), then a ∪ S is transitive.  e) If every x ∈ b is transitive, then b is transitive.  f) Ist a = ∅, then a is transitive. 2) Let x be a set. Show that x is an ordinal iff x = 1 or R := {(y, z) : y ∈ x ∧ z ∈ x ∧ y ∈ z} is a well-ordering with fld(R) = x such that ∀y ∈ x(y = {z : zRy}). 3) Show that the following assertions are equivalent for any set x: (i) x is an ordinal. (ii) x is transitive ∧ ∀y ∈ x∀z ∈ x(y ∈ z ∨ y = z ∨ z ∈ y). (iii) x is transitive and every transitive proper subset y of x is a member of x.  (iv) (x =0 ∨ 0 ∈ x) ∧ ∀y ∈ x(Sy = x ∨ Sy ∈ x) ∧ ∀y(y ⊆ x =⇒ y = x ∨ y ∈ x). Hint: We have shown above that (i) implies the other assertions. Therefore it is practical to prove that they are all equivalent to (i). For “(iii) =⇒ (i)”, consider the set y := {α ∈ ON : α ∈ x}. For “(iv) =⇒ (i)”, consider α := min(ON \ x).

20

Chapter 1. Foundations 4) Let R be the set of real numbers. Show that any subset of R which is well-ordered by the usual ordering on R is countable. Hint: In a well-ordered set without a maximum, any member has an immediate successor. 5) Prove: If X is a nonempty set of ordinals without a maximum, then sup(X) is a limit ordinal.

1.3 Transfinite Induction and Recursion We remind the reader that a relation R is called well-founded (written as wf(R)) iff (i) ∀x ∈ fld(R)∃y(x ∈ y ∧ R−1 [y] ⊆ y) and (ii) ∀x ⊆ fld(R)(x = ∅ =⇒ ∃y ∈ x∀z ∈ x(¬zRy)). In this definition, condition (i) has technical reasons. We will show that it can be replaced by the condition “the class of R-predecessors of each member x of fld(R) is a set”, which intuitively is more intelligible. From this condition and the axiom of foundation it follows immediately that, for any class A, the relation ∈ ∩ A × A is well-founded. Lemma 1.3.1 If R is a well-founded relation, then R is irreflexive, and any nonempty class A ⊆ fld(R) has an R-minimal element. Proof If x ∈ fld(R), then there is y ∈ {x} such that ¬zRy holds for all z ∈ {x}. Thus we get ¬xRx. To show the second assertion, we take a member x ∈ A and, by the above definition, a set y such that x ∈ y and R−1 [y] ⊆ y. Then the set y ∩ A has an R-minimal element z. If uRz, then u ∈ y, since R−1 [y] ⊆ y, and thus u ∈ A by the choice of z. This shows that z is an R-minimal element of A. Definition If R is a relation and x ∈ fld(R), we say that x

:= {y : yRx∧y = x} is the class of strict R-predecessors of x. If R is well-founded, then x

= {y : yRx}. Theorem 1.3.2 (Transfinite Induction Principle) If R is a well-founded relation, then the following holds: If ∀x ∈ fld(R)(∀y ∈ x

ϕ(y) =⇒ ϕ(x)) , then ∀x ∈ fld(R) ϕ(x). More precisely: If ϕ is a formula of LZF , then ZF  wf(R) ∧ ∀x ∈ fld(R)(∀y ∈ x

ϕ(y) =⇒ ϕ(x)) =⇒ ∀x ∈ fld(R) ϕ(x).

1.3. Transfinite Induction and Recursion

21

Proof We assume, to get a contradiction, that the conclusion is false and the premiss is true. Then Lemma 1.3.1 says that there is an R-minimal element x ∈ fld(R) satisfying ¬ϕ(x). So we have ∀y ∈ x

ϕ(y). From our assumption we get ϕ(x), contradicting ¬ϕ(x). Remark 1.3.3 Important special cases of Theorem 1.3.2 are: a) R = ∈ ∩ (ON × ON). So we get the Transfinite Induction Principle for ON: ZF  ∀α(∀β ∈ α ϕ(β) =⇒ ϕ(α)) =⇒ ∀α ϕ(α). This is also a direct consequence of Lemma 1.2.1. b) R = ∈ ∩ σ × σ for σ ∈ ON. In particular we obtain for σ = ω the induction principle for ω. c) R = ∈: From the axiom of regularity and Lemma 1.3.8 (see below) we can conclude that the relation ∈ is well-founded. Thus we can use in ZF the so-called ∈-Induction Principle: ∀x(∀y ∈ x ϕ(y) =⇒ ϕ(x)) =⇒ ∀xϕ(x). Corollary 1.3.4 (2nd Form of the Transfinite Induction Principle for ON) ZF  ϕ(0) ∧ ∀α(ϕ(α) =⇒ ϕ(α + 1)) ∧ ∀α(Lim(α) ∧ ∀β ∈ α ϕ(β) =⇒ ϕ(α)) =⇒ ∀αϕ(α). Proof This principle follows immediately from the transfinite induction principle for ON if one notices that every ordinal is either 0 or a successor ordinal or a limit ordinal. Theorem 1.3.5 (Transfinite Recursion Theorem Schema) If R, G and F are classes, then let rec(F, R, G) be an abbreviation for the formula Func(F ) ∧ dom(F ) = fld(R) ∧ ∀x ∈ fld(R)(F (x) = G(F |`x

)). If R, F  and G are classes, then we can construct a class F , such that the following formula is provable in ZF: (Rel(R)∧wf(R)∧G : V −→ V) =⇒ rec(F, R, G)∧(rec(F  , R, G) =⇒ F = F  ). We say that the function F is defined by transfinite induction or defined by recursion or defined recursively with respect to the well-founded relation R and the function G.

22

Chapter 1. Foundations

Proof The uniqueness of F can be easily shown, using Theorem 1.3.2.

, then F |` x

= F  |` x

, and Namely, if x ∈ fld(R) and F (z) = F  (z) for all z ∈ x   so we get F (x) = G(F |` x

) = G(F |` x

) = F (x). From Theorem 1.3.2 we can conclude that ∀x ∈ fld(R) F (x) = F  (x). Nowwe define the class F and prove that it has the desired properties. Let F := K, where K := {f : Func(f ) ∧ dom(f ) ⊆ fld(R) ∧ ∀x ∈ dom(f )(

x ⊆ dom(f ) ∧ f (x) = G(f |`x

))}. Claim 1 F is a function. For this we show f (x) = g(x) for any f, g ∈ K and x ∈ dom(f ) ∩ dom(g). We use transfinite induction, i.e., we apply Theorem 1.3.2, and assume that the assertion is true for all z ∈ x

. Since x

⊆ dom(f ) ∩ dom(g), this gives f (z) = g(z) for all z ∈ x

. Thus we get f |` x

= g |` x

and, since G is a function, f (x) = G(f |`x

) = G(g |`x

) = g(x). Claim 2 ∀x ∈ dom(F )(F (x) = G(F |`x

)). Let x ∈ dom(F ), that means x ∈ dom(f ) for some f ∈ K. Then x

⊆ dom(f ) ⊆ dom(F ), hence f |`x

= F |`x

and F (x) = f (x) = G(f |`x

) = G(F |`x

). Claim 3 dom(F ) = fld(R). By the definition of F , we have dom(F ) ⊆ fld(R). Now let x ∈ fld(R) and assume, for an application of Theorem 1.3.2, that z ∈ dom(F ) for all z ∈ x

, i.e., x

⊆ dom(F ). Since R is well-founded, there is a set y such that x ∈ y and R−1 [y] ⊆ y. From {x} ⊆ y we get R−1 [{x}] ⊆ R−1 [y] ⊆ y, hence x

⊆ y. Put f := F |`y. If x ∈ dom(f ), then x ∈ dom(F ), and we are done. So assume that x ∈ dom(f ). Since F is a function, f is also a function, and furthermore f is a set by the axiom of replacement. Obviously we have

))} is dom(f ) = y ∩ dom(F ) and x

⊆ dom(f ). Therefore, f  := f ∪ {(x, G(f |`x  a function such that dom(f ) ⊆ fld(R). Now one easily verifies z ⊆ dom(f  ) and f  (z) = G(f  |` z ) for any z ∈ dom(f  ). Namely, if z = x, then x

⊆ dom(f ) ⊆

= f  |` x

and f  (x) = G(f |` x

) = G(f  |` x

). On the other dom(f  ), hence f |` x hand, if z = x, then z ∈ dom(f ) ⊆ y, which gives z ⊆ R−1 [y] ⊆ y. Furthermore z ⊆ dom(F ), since z ∈ dom(F ). So we can conclude that z ⊆ dom(f ) ⊆ dom(f  ) and f  (z) = f (z) = F (z) = G(F |` z ) = G(f |` z ) = G(f  |` z ). From this, we get f  ∈ K and x ∈ dom(F ). Since x ∈ y, we have x ∈ dom(f ). This contradiction completes the proof of the theorem. Remark 1) Let us call each member of the class K defined in the proof of Theorem 1.3.5 an approximation of F . Then it follows from the preceding arguments that • For each x ∈ fld(R), there is an approximation f of F with F (x) = f (x).

1.3. Transfinite Induction and Recursion

23

• If x ∈ fld(R) and f ∈ V is an arbitrary function satisfying x

⊆ dom(f ) and F (z) = f (z) for all z ∈ x

, then F |` x

= f |` x

, hence F (x) = G(F |` x

) = G(f |` x

). If in addition R is transitive, then f |`x

is an approximation of F . 2) Theorem 1.3.5 is a metatheorem, since it says something about the existence of classes. Given the formulae χ(y, z) and ψ(y, z), defining the classes R and G respectively, we can find a formula ϕ(y, z) defining the class F such that that formula of LZF which results from the formula in Theorem 1.3.5 by the elimination of classes is provable in ZF. For example, under the above premisses, for the formula ∀x ∈ fld(R)(F (x) = G(F |`x

)) (in the extended sense) there is a corresponding formula of ZF: ∀x(∃z(χ(x, z) ∨ χ(z, x)) =⇒ ∃y(ϕ(x, y)∧ ∀f (Func(f ) ∧ dom(f ) = x

∧ ∀z ∈ x

ϕ(z,f (z)) =⇒ (ϕ(x,y) ⇐⇒ ψ(f,y))))), where z ∈ x

is the formula χ(z, x) ∧ z = x and dom(f ) = x

is the formula ∀z(z ∈ dom(f ) ⇐⇒ χ(z, x) ∧ z = x). 3) The proof of Theorem 1.3.5 can be transferred word for word if we formulate Theorem 1.3.5 for functions G whose defining formulae contain parameters. Formally, the above formula ψ(y, z) defining G may contain further free (set) variables in addition to y and z. In this connection we never mention explicitly the parameters. Well known examples are the sum, the product and the exponentiation of ordinals which will be defined in the example after the next corollary. 4) A further important example is the case that we want to define the function F recursively by F (x) = G(x, F |` x

) for all x ∈ fld(R). The proof of Theorem 1.3.5 works for this case if we replace in the definition of the class K the equation f (x) = G(f |`x

) by f (x) = G(x, f |`x

). Corollary 1.3.6 (Transfinite Recursion Principle for ON) a) If G is a class, then we can construct a class F such that ZF  G : V −→ V =⇒ Func(F )∧dom(F ) = ON∧∀σ(F (σ) = G(F |`σ)). b) If a, H1 and H2 are classes, then we can construct a class F such that ZF  H1 : V −→ V ∧ H2 : V −→ V ∧ ∃y(y = a) =⇒ Func(F ) ∧ dom(F ) = ON ∧ F (0) = a ∧ ∀σ(F (σ + 1) = H1 (F (σ)) ∧ (Lim(σ) =⇒ F (σ) = H2 (F |`σ))). Proof Part a) follows from Theorem 1.3.5, if we take into account that the relation ∈ ∩ (ON × ON) is well-founded and satisfies σ

= {z : z ∈ ON ∧ z ∈ σ} = σ. For the proof of b), we note again that ∈ ∩ (ON × ON) is well-founded and has the field ON. To apply Theorem 1.3.5, we define the function G from V into V by G(0) := a; G(x) := H1 (x(σ)), if x is a function such that dom(x) =

24

Chapter 1. Foundations

σ + 1 for some σ ∈ ON; G(x) := H2 (x), if x is a function whose domain is a limit ordinal; G(x) := 0 in all other cases. Let F be the class given by Theorem 1.3.5. Then we get: F (0) = G(F |` 0) = G(0) = a; F (σ + 1) = G(F |` σ + 1) = H1 (F (σ)), since F |` σ + 1 is a function with domain σ + 1; F (σ) = G(F |` σ) = H2 (F |` σ) for limit ordinals σ, since F |` σ is a function whose domain is the limit ordinal σ. This completes the proof of the corollary. Examples 1.3.7 a) From Corollary 1.3.6 and Theorem 1.3.5 we obtain by an obvious choice of H1 , H2 and of G, respectively, the usual recursion principles for ω, taking f := F |`ω : For each function g : ω −→ ω and each a ∈ ω there exists a unique function f : ω −→ ω satisfying f (0) = a and f (n + 1) = g(f (n)) for all n ∈ ω. For each function g with an appropriate domain, there is a unique function f such that f (n) = g(f |`n) for all n ∈ ω. b) Depending of the parameter α ∈ ON, we will define in Section 1.4 for every β ∈ ON the sum α + β, the product α · β and the exponentiation αβ ; that means that we will apply the principle of transfinite recursion. c) von Neumann’s Hierarchy: For a := ∅  and the functions H1 and H2 , given by H1 (x) := P(x) and H2 (x) := ran(x), where ran(x) := {v : ∃u (u, v) ∈ x}, we consider the function F from Corollary 1.3.6 and put Vα := F (α). Then  V0 = ∅; Vα+1 = P(Vα ); Vγ = {Vβ : β < γ} for limit ordinals γ . The sets Vα (α ∈ ON) are called the levels of the von Neumann hierarchy, the function (Vα : α ∈ ON) is called von Neumann’s hierarchy or cumulative hierarchy.  We use the principle of ∈-induction to prove that {Vα : α ∈ ON} is the universal class. First, it is easy to see by transfinite induction that every set Vα is transitive and Vα ⊆ Vβ for any α and  β with α < β. If x ∈ V is an arbitrary set, and if we assume that y ∈ {Vα : α ∈ ON} for every member y of x, then, for each y ∈ x, there is a least ordinal αy such that y ∈ Vαy +1 . By an axiom of replacement and Lemma 1.2.5, there exists the ordinal α := sup{αy : y ∈ x}. Now we can conclude that x ⊆ Vα , hence x ∈ Vα+1 . This implies  V = {Vα : α ∈ ON}.10 10 We

cannot prove this assertion without the axiom of regularity.

1.3. Transfinite Induction and Recursion

25

The least ordinal α with x ∈ Vα+1 is called the rank of the set x, written as rk(x). Sometimes it is practical to use induction on the rank of a set. If y ∈ x ∈ Vα+1 , then y ∈ x ⊆ Vα , and so we get rk(y) < rk(x). Lemma 1.3.8 Let R be a relation. a) The following assertions are equivalent: (i) For every x ∈ fld(R), x

is a set. (ii) For each x ∈ fld(R), there is a set y such that x ∈ y and R−1 [y] ⊆ y. b) R is well-founded iff (1) For every x ∈ fld(R), x

is a set and (2) There is not a family (xn : n ∈ ω) of members of fld(R) satisfying xn+1 Rxn for all n ∈ ω. Proof For the proof of part a), we note first that the implication “(ii) =⇒ (i)” follows from an axiom of subsets, since, for x ∈ y, x

⊆ R−1 [{x}] ⊆ −1 R [y] ⊆ y. Now let us assume (i). Then for every set z the class {

u : u ∈ z∧u ∈ fld(R)} is a set, since it is the range of the restriction of the function F :=  {(u, u

) : u ∈ V} to the set z ∩ fld(R). Consequently, the class {

u : u ∈ z ∩ fld(R)} ∪ z is also a set, and the class R−1 [z] is a subclass of it. Thus R−1 [z] is a set. Take x ∈ fld(R). Then Corollary 1.3.6 b) tells us that there is a function f : ω −→V satisfyingf (0) = {x} and f (n + 1) = R−1 [f (n)] for every n ∈ ω. Put y := ran(f ) = {f (n) : n ∈ ω}. Then we have x ∈ f (0) ⊆ y. Furthermore, if u ∈ R−1 [y], then there is v ∈ y with uRv. Since v ∈ f (m) for some m ∈ ω, our construction gives u ∈ R−1 [f (m)] = f (m + 1) ⊆ y. Next we turn to the proof that every well-founded relation satisfies properties (1) and (2). For (1), this is clear from part a). If there was a family (xn : n ∈ ω) of members of fld(R) satisfying xn+1 Rxn for all n ∈ ω, then the range {xn : n ∈ ω} of this function would not have a minimal element under R. Conversely let R be a relation satisfying (1) and (2). Then a) implies that the first condition in the definition of a well-founded relation is fulfilled. For the second condition, we need (AC). Assume, to get a contradiction, that there is a nonempty subset X of fld(R) without a minimal element. Let g be a choice function for the set P(X) \ {∅}. By the recursion theorem, there is a function (xn : n ∈ ω) such that x0 ∈ X and xn+1 = g({y ∈ X : yRxn }). It is easy to verify that we have obtained a function whose existence contradicts (2). Definition and Example Let x be an arbitrary set. If there exists a transitive superset of x, then we obtain, taking the intersection of all these sets, an ⊆smallest set with this property. The recursion theorem provides us directly with

26

Chapter 1. Foundations

 this set. There is a function  f satisfying f (0) = x and f (n + 1) = f (n) for every n ∈ ω. Put y := ran(f ). Thenclearly x ⊆ y. If u ∈ v ∈ y, then v ∈ f (m) for some m ∈ ω, hence u ∈ f (m) = f (m + 1) ⊆ y. Thus y is transitive. Furthermore we can conclude, using complete induction, that y is a subset of every transitive superset of x. It is called the transitive closure of x and is denoted by tc(x). Lemma 1.3.9 If A1 and A2 are transitive classes and π : A1 −→ A2 is an ∈-isomorphism, then A1 = A2 and π = idA1 = {(x, x) : x ∈ A1 }. Proof Consider x ∈ A1 . We will show that π(x) = {π(y) : y ∈ x}. Then we can use transfinite induction on the well-founded relation ∈ ∩ (A1 × A1 ) and obtain immediately π(x) = x from the inductive hypothesis ∀y ∈ x(π(y) = y), which proves the lemma by Theorem 1.3.2. The inclusion “⊇” holds by assumption. For the reverse inclusion let z ∈ π(x). Then z ∈ A2 , since A2 is transitive. Thus z = π(u) for some u ∈ A1 . This gives u ∈ x, and thus z ∈ {π(y) : y ∈ x}, as desired. Definition and Example A relation R is called extensional iff ∀x, y ∈ fld(R)(

x = y =⇒ x = y). We call a class A extensional iff ∀x, y ∈ A(x ∩ A = y ∩ A =⇒ x = y). The relation ∈ is extensional – as stated by the axiom of extensionality. Any transitive class is extensional. Further every linear ordering < is extensional; if we assume x

= y and x = y, then connectivity implies that, without loss of generality, x < y. Then x ∈ y = x

, and thus we have x < x which is a contradiction. Theorem 1.3.10 (Mostowski’s Collapsing Lemma) If the relation R is well-founded and extensional, then there is a unique transitive class A and a unique isomorphism π : fld(R) −→ A with respect to R and ∈. Proof First we will prove uniqueness. If π1 : fld(R) −→ A1 and π2 : fld(R) −→ A2 are isomorphisms satisfying the properties in the theorem, then π2 ◦ π1−1 : A1 −→ A2 is an ∈-isomorphism. From Lemma 1.3.9 we know that A1 = A2 and π2 ◦ π1−1 = idA1 , hence π1 = π2 . The existence of π follows from the recursion theorem, applied to the wellfounded relation R. Let the function G be defined by G(x) := ran(x) := {z : ∃y (y, z) ∈ x}. Then Theorem 1.3.5 yields a function π : fld(R) −→ V such that π(x) = G(π |`x

) = ran(π |`x

) = {π(y) : yRx}. (π is called the collapsing or Mostowski isomorphism for R.) We put A := π[fld(R)] and show that π and A have the desired properties.

1.3. Transfinite Induction and Recursion

27

Claim 1 π is injective. Assume, to get a contradiction, that there are x, y ∈ fld(R) such that x = y and π(x) = π(y). By Lemma 1.3.1, there is an R-minimal element x0 of the class {x ∈ fld(R) : ∃y ∈ fld(R)(π(x) = π(y) ∧ x = y)}. In the same way we get an R-minimal element y0 of {y ∈ fld(R) : π(x0 ) = π(y)∧x0 = y}. Since R is extensional, we have x 0 = y 0 . So consider, without loss of generality, z ∈ x 0 \ y 0 . Then π(z) ∈ π(x0 ). If we suppose that π(z) ∈ π(y0 ), i.e., π(z) = π(u) for some u ∈ y 0 , then the choice of x0 and z ∈ x 0 yield z = u. This gives z ∈ y 0 , a contradiction. Therefore π(z) ∈ π(x0 ) \ π(y0 ), contradicting π(x0 ) = π(y0 ). Claim 2 π is an isomorphism w.r.t. R and ∈. By Claim 1, π is injective, and by the definition of A, π is a surjective function from fld(R) onto A. If yRx, then π(y) ∈ π(x) by the above properties of π. Conversely, if π(y) ∈ π(x), then there exists u such that uRx and π(y) = π(u). Claim 1 shows that y = u, hence yRx. Claim 3 A is transitive. If z ∈ y ∈ A, then y = π(x) for some x ∈ fld(R). Thus z = π(u) for some u with uRx, and consequently z ∈ A. Corollary 1.3.11 a) Any extensional class B is ∈-isomorphic to a transitive class A. b) If (x, ≺) is a well-ordering, then there is a unique ordinal α and a unique isomorphism π : (x, ≺) −→ (α, ∈). c) If B is a proper class, well-ordered by the relation ≺, then B is isomorphic to ON w.r.t. ≺ and ∈. Proof For the proof of part a), we observe first that for B = ∅ we can take A = ∅ and for B = {x} we can take A = {0}. So we can assume that B has at least two members. By the remark at the beginning of this section, the relation R :=∈ ∩ (B × B) is well-founded, by assumption it is extensional, and clearly fld(R) ⊆ B. To show that fld(R) = B, fix x ∈ B and assume, to get a contradiction, that x ∈ fld(R). The class B \ {x} has an ∈-minimal element y. By the assumption on x, the set of ∈-predecessors of y in B is empty, and the same holds for x. Since B is extensional, we get x = y, contradicting y ∈ B \ {x}. So we have shown that B = fld(R), and Theorem 1.3.10 provides us with an isomorphism π from B onto a transitive class. For b), assume without loss of generality that x has at least two elements and choose A and π according to Theorem 1.3.10. Then A is a transitive set. Furthermore it is easy to see that every member of A is transitive. Namely, if u ∈ v ∈ z and z ∈ A, then u ∈ A and v ∈ A, since A is transitive. Since π is an isomorphism and the relation ≺ is transitive, we get u ∈ z. This shows that A is an ordinal.

28

Chapter 1. Foundations

For c), choose again A and π according to Theorem 1.3.10. As in the proof of b) we can conclude that every member of A is transitive. Furthermore, A is a proper class, since otherwise B = π −1 [A] would be a set. Now Corollary 1.2.4 tells us that A = ON. Definition Let (B, ≺) be a well-ordering. The unique isomorphism given by Corollary 1.3.11 is called the order isomorphism of (B, ≺). If B is a set, then we call the ordinal characterized by Corollary 1.3.11 b) the ordinal number or order type of (B, ≺) and write it as otp(B, ≺). Exercises 1) Prove the following induction principles: a) (Finite Induction) If k ∈ ω, then ϕ(0) ∧ ∀n ∈ k(ϕ(n) =⇒ ϕ(n + 1)) =⇒ ∀n ≤ k ϕ(n). b) If ∀m, n ∈ ω(∀k, l ∈ ω((k ∈ m ∨ (k = m ∧ l ∈ n)) =⇒ ϕ(k, l)) =⇒ ϕ(m, n)), then ∀m, n ∈ ω ϕ(m, n). Hint: Define a suitable well-founded relation on ω × ω. 2)

a) Why are the relations ON × {0} and {(α + 1, α) : α < τ }, for any limit ordinal τ , not well-founded?

b) Let α and β be infinite ordinals and the relation ≺ on α×β be defined by (σ1 , σ2 ) ≺ (τ1 , τ2 ) ⇐⇒ σ1 < τ1 ∧ σ2 < τ2 . Is ≺ well-founded? Is it extensional? 3) Give an example of a linear ordering < on a set A and ofa ⊆-chain K of subsets of A which are well-ordered by <, such that K is not well-ordered by <. 4) For α, β ∈ ON let fin(β, α) be the set of all functions f from β into α for which the set {σ < β : f (σ) = 0} is finite. Now we define the generalized Hebrew lexicographic ordering; for f, g ∈ fin(β, α) we put f ≺ g :⇐⇒ f = g ∧ (τ = max{σ ∈ β : f (σ) = g(σ)} =⇒ f (τ ) < g(τ )). Prove that (fin(β, α), ≺) is a well-ordering. Hint: It is easy to verify that ≺ is a linear ordering. To show that every nonempty subset X of fin(β, α) has a smallest element, use induction on β. Choose f ∈ X, f ≡ 0. Consider τ := max{σ < β : f (σ) = 0} and γ := min{g(τ ) : g ≺ f ∧ g ∈ X}.

1.3. Transfinite Induction and Recursion

29

5) Assume that (A, <) is a well-ordering. Prove: a) If f : (A, <) −→ (A, <) is a homomorphism, then a ≤ f (a) for all a ∈ A. b) (A, <) is not isomorphic to any of its proper initial segments. 6) Assume that (A, <) and (B, ≺) are well-orderings. As usual we write (C, <) instead of (C, < ∩ (C × C)) for substructures C of A. For u ∈ A we put Au := {x ∈ A : x < u}, the initial segment of A determined by u. Prove without using Corollary 1.3.11: Either (A, <) is isomorphic to (B, ≺), or there is v ∈ B such that (A, <) is isomorphic to (Bv , ≺), or there is u ∈ A such that (Au , <) is isomorphic to (B, ≺). Hint: Show that F := {(x, y) ∈ A × B : (Ax , <) is isomorphic to (By , ≺)} is an order preserving function or, alternatively, use the recursion theorem.  7) Assume that (Wα : α ∈ ON) is a function such that Wγ = {Wσ : σ < γ} for limit ordinals γ and Wα ⊆ Wα+1 for all α ∈ ON. Prove that Wα ⊆ Wβ whenever α ∈ β ∈ ON. Hint: Use transfinite induction on β. 8) Prove the following properties of the cumulative hierarchy (Vα : α ∈ ON). a) Vα is transitive. b) α < β =⇒ Vα ⊆ Vβ (Use the previous question.) c) Vα ∩ ON = α. d) rk(x) = sup{rk(y) + 1 : y ∈ x}. e) rk(α) = α. Hint: Use transfinite induction on α.  9) Prove: If x, y ∈ Vα , then P(x), x, x × y, x y ∈ Vα+4 . 10) We call a set x hereditarily finite iff its transitive closure is finite. Prove: Vω = {x : x is hereditarily finite}. 11) Prove: If α is a limit ordinal, then there is a set x such that rk(x) = α, x is transitive, and x ∩ ON = {0, 1}. 12) Assume that A, B and C are classes, C ⊆ A, B and C are transitive, and π : A −→ B is an ∈-isomorphism. Prove that π |`C = idC .

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13) Assume that R is a well-founded relation and fld(R) = P . Prove that there is a function ρR : P −→ ON such that a) xRy =⇒ ρR (x) < ρR (y) for all x, y ∈ P . b) ρR (x) = sup{ρR (y) + 1 : yRx}.

α > ρR (y)}. c) ρR (x) = min{α ∈ ON : ∀y ∈ x The sets Pα := {x ∈ P : ρR (x) = α} are the R-levels of P , ρR (x) is the R-rank of x. Hint: Use the recursion theorem. 14) Prove with the recursion theorem that there is a unique function ρ : V −→ ON such that ρ(x) = sup{ρ(y) + 1 : y ∈ x}. Consequently we have ρ(x) = rk(x) for all sets x. 15) Assume that (A, <) and (B, ≺) are well-orderings. Prove: If g : (A, <) −→ (B, ≺) is a homomorphism, then otp(A, <) ≤ otp(B, ≺).

1.4 Arithmetic of Ordinals Definition Let A be an ordinal or the class ON of ordinals. If F : A −→ V is a function, then we call F a sequence. Notions as upper bound, supremum etc. of a sequence are understood as upper bound, supremum etc. of the range of the sequence. In later chapters we will extend the notion of a sequence to functions whose domain is a club, a stationary set, or a set of regular cardinals. F is called continuous iff for every limit ordinal γ ∈ A we have  F (γ) = {F (σ) : σ < γ}; we also say that F is a continuous sequence of sets. Furthermore every function from A into ON that is continuous and strictly increasing under < is called a normal function or normal sequence. More general, a function from an arbitrary set A into ON is called an ordinal function on A. Remark In most applications of the notion of continuity, the domain A of the function is a limit ordinal or the class ON of ordinals. If ran(F ) ⊆ ON, then the continuity of F is equivalent to ∀γ ∈ dom(F ) ∩ Lim (F (γ) = sup{F (σ) : σ < γ}). If the last property is satisfied and if in addition F (α) < F (α + 1) for every α ∈ A, then F already is a normal function. This can be seen as follows: By

1.4. Arithmetic of Ordinals

31

transfinite induction on β ∈ dom(F ) we show α < β =⇒ F (α) < F (β). For β = 0 there is nothing to prove. If β is a limit ordinal, then α < β implies α < σ for some σ < β. The inductive hypothesis and the continuity of F imply F (α) < F (σ) ≤ sup{F (σ) : σ < β} = F (β). Finally, if β = σ + 1, then we get α ≤ σ, F (α) ≤ F (σ) by the inductive hypothesis, and consequently F (α) ≤ F (σ) < F (σ + 1). Further we note that normal functions always map limit ordinals onto limit ordinals. Lemma 1.4.1 Assume that A ∈ ON or A = ON and F : A −→ ON is a strictly increasing function. Then a) F is continuous iff, for every nonempty subset b of A with sup(b) ∈ A, F (sup(b)) = sup{F (α) : α ∈ b} = sup(F [b]). b) α ≤ F (α) for all α ∈ A. Proof For the proof of a) we note that, by the definition of continuity and since sup(γ) = γ for all limit ordinals γ, we only have to show one direction of the equivalence. So let F be a normal function and b be a nonempty subset of A such that sup(b) ∈ A. If b has a maximum, then sup(b) ∈ b, and the desired equation is obvious. Thus let us assume that b has no maximum. Then sup(b) is a limit ordinal, since otherwise its immediate predecessor would be an upper bound of b, too. This gives F (sup(b)) = sup{F (σ) : σ < sup(b)} ≥ sup{F (α) : α ∈ b}. If γ < F (sup(b)), then there is σ < sup(b) such that γ < F (σ). Furthermore there is α ∈ b with σ < α. By assumption, we have F (σ) < F (α) and consequently γ < F (α). Thus every ordinal γ < F (sup(b)) is not an upper bound of the set {F (α) : α ∈ b}, which was still to be shown. For an indirect proof of b), let α ∈ A be minimal with F (α) < α. Then F (α) ∈ A, and therefore F (F (α)) < F (α) since F is strictly increasing. This contradicts the choice of α, and part b) is shown. The normal functions we are now going to define will play an important role in the sequel. They form a proper class of normal functions, since for every ordinal α the function β → α  β, where β ∈ ON, is normal, when  denotes the addition, multiplication or exponentiation of ordinals. Definition If α is an ordinal, then we define by transfinite recursion on β ∈ ON the ordinal sum α+β, the ordinal product α·β, and the ordinal exponentiation αβ as follows: α + 0 = α; α + (σ + 1) = (α + σ) + 1; Lim(β) =⇒ α + β = sup{α + σ : σ < β}. α · 0 = 0; α · (σ + 1) = (α · σ) + α; Lim(β) =⇒ α · β = sup{α · σ : σ < β}. α0 = 1; ασ+1 = ασ · α; Lim(β) =⇒ αβ = sup{ασ : σ < β}.

32

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Remark 1.4.2 a) In the case of addition note remark 3) after Theorem 1.3.5 and choose, for an application of 1.3.5 (we could apply Corollary 1.3.6 as well – and will do this when defining the multiplication of ordinals), G(0) := α; G(x) := S(x(σ)), ifx is a function such that dom(x) = σ + 1 for some σ ∈ ON; G(x) := ran(x), if x is a function whose domain is a limit ordinal; G(x) := 0 in all other cases. Theorem 1.3.5 says that there is a function F : ON −→ V such that F (β) = G(F |` β). Now define α + β := F (β). In the case of multiplication we apply Corollary 1.3.6: a  := 0; H1 (x) := x + α, if x is an ordinal, H1 (x) := 0 otherwise; H2 (x) := ran(x). b) From the definition of continuity and Lemma 1.4.3 a) (iii) it follows that the function β → α + β is normal for each α ∈ ON. Lemma 1.4.3 b) (iii) implies the corresponding assertion for the multiplication of ordinals if α > 0, and Lemma 1.4.3 c) (iii) for the exponentiation of ordinals if α > 1. c) By complete induction it can be easily seen that the restrictions to the set ω of the functions defined above yield the usual addition, multiplication and exponentiation of natural numbers. (These operations also could have been obtained by applying a recursion theorem for ω.) The well known properties of these operations are proved using complete induction. d) Intuitively speaking, the sum of ordinals corresponds to that order type which results from putting one well-order behind the other. Corollary 1.3.11 tells us that every well-ordered set is isomorphic to exactly one ordinal bearing the relation ∈. We called this ordinal the order type of the well-ordered set. Now, if (A, <1 ) and (B, <2 ) are well-ordered sets of order type α and β respectively, satisfying A ∩ B = ∅, then we obtain a new well-ordering (A ∪ B, <3 ) as follows: <3 := <1 ∪ <2 ∪ {(a, b) : a ∈ A ∧ b ∈ B} (“first (A, <1 ) and then (B, <2 )”). One can show that this is a wellordering of order type α + β. Analogously, α · β is the order type of that well-ordering ≺ that results from lining up β-times well-orderings (Aσ , <σ ), σ < β, of order type α. More precisely: If ((Aσ , <σ ) : σ < β) is a family of well-orderings of order type α satisfying Aσ ∩ Aτ = ∅ for σ = τ  , then we can define a well-ordering ≺ of order type α · β on the set B := {Aσ : σ < β} by a ≺ b :⇐⇒ ∃σ∃τ ((a ∈ Aσ ∧b ∈ Aτ ∧σ < τ ) ∨ (a ∈ Aσ ∧b ∈ Aσ ∧a <σ b)).

1.4. Arithmetic of Ordinals

33

It is simpler to well-order the set α × β by the so-called Hebrew lexicographic ordering: (σ1 , τ1 ) < (σ2 , τ2 ) iff τ1 < τ2 ∨(τ1 = τ2 ∧σ1 < σ2 ). Then (α × β, <) has order type α · β. Shortly speaking, “we take α, β times”. e) The “visual” introduction of addition and multiplication facilitates the understanding of the following special features which immediately follow from the definitions. We have 1 + ω = sup{1 + n : n ∈ ω} = ω. Therefore 1 + ω < ω + 1, the addition of ordinals is not commutative. Further 2 · ω = sup{2 · n : n ∈ ω} = ω, and consequently 2 · ω < ω · 2 = ω + ω, the multiplication of ordinals is also not commutative. In addition, we have shown that (1 + 1) · ω = 1 · ω + 1 · ω – the corresponding distributive law is false. The other distributive law, α · (β + γ) = α · β + α · γ, is true; furthermore the associative laws and certain monotonicity properties hold. We collect all these properties in the following lemma. f) For the sake of completeness, let us mention how to introduce the ordinal exponentiation as an order type of a well-ordered set. For α, β ∈ ON let fin(β, α) be the set of all functions f from β into α for which the set {σ < β : f (σ) = 0} is finite. Now the set fin(β, α) will be well-ordered by a generalized Hebrew lexicographic ordering; for f, g ∈ fin(β, α) we define f ≺ g :⇐⇒ f = g ∧ (τ = max{σ ∈ β : f (σ) = g(σ)} =⇒ f (τ ) < g(τ )). One can show that (fin(β, α), ≺) is a well-ordering of order type αβ . The key for the proof is Theorem 1.4.6. Lemma 1.4.3 If α, β and γ are ordinals, then the following laws hold. a)

b)

c)

(i) 0 + α = α = α + 0 (iii) β < γ =⇒ α + β < α + γ (v) (α + β) + γ = α + (β + γ) (i) 0 · α = 0 = α · 0 (iii) 0 < α ∧ β < γ =⇒ α · β < α · γ (v) (α · β) · γ = α · (β · γ) (i) β = 0 =⇒ 0β = 0 (iii) 1 < α ∧ β < γ =⇒ αβ < αγ (v) 1 < α =⇒ β ≤ αβ (vii) (αβ )γ = αβ·γ

(ii) (iv)

β ≤α+β α ≤ β =⇒ α + γ ≤ β + γ

(ii) (iv) (vi) (ii) (iv) (vi)

α·1=α=1·α α ≤ β =⇒ α · γ ≤ β · γ α · (β + γ) = α · β + α · γ 1β = 1 α ≤ β =⇒ αγ ≤ β γ αβ+γ = αβ · αγ

Proof The assertions are verified by transfinite induction on the “argument on the right”, and their proofs are thought of as exercises for the reader. We indicate some proofs and present all ideas which are necessary for the other ones. For a) (iii), note that there is nothing to prove if γ = 0. If γ = σ + 1 and β < γ, then β ≤ σ, and the inductive hypothesis and the definition of the sum

34

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of ordinals give α + β ≤ α + σ < (α + σ) + 1 = α + (σ + 1) = α + γ. Finally, if γ is a limit ordinal, then γ = sup(γ), and β < γ means β < σ for some σ < γ. By the inductive hypothesis and by definition, we get α + β < α + σ ≤ sup{α + σ : σ < γ} = α + γ. For a) (v): If γ = 0, then (α + β) + 0 = α + β = α + (β + 0). Furthermore, from the inductive hypothesis and the definition of ordinal addition we can conclude that (α + β) + (σ + 1) = ((α + β) + σ) + 1 = (α + (β + σ)) + 1 = α + ((β + σ) + 1) = α + (β + (σ + 1)). For limit ordinals γ we obtain (α + β) + γ = sup{(α + β) + σ : σ < γ} = sup{α + (β + σ) : σ < γ}. Now we apply Lemma 1.4.1 to the normal function Fα : ON −→ ON, given by Fα (τ ) := α + τ , and to the set b := {β + σ : σ < γ}. Since sup(b) = β +γ, it tells us that α+(β +γ) = Fα (sup(b)) = sup{α+(β +σ) : σ < γ}. This completes the proof. Completely analogously, the associative law of multiplication is shown, and in a similar way one obtains the distributive law. For c) (vii): If α = 0, then the assertion follows from the proceding ones by considering the cases γ = 0; β = 0, γ = 0; β = 0, γ = 0. Now let α = 0. We prove (vii) using transfinite induction on γ. If γ = 0, then αβ·0 = α0 = 1 = (αβ )0 . If γ = σ + 1, then we can conclude from b) (vi), c) (vi), the inductive hypothesis and the definition of exponentiation that αβ·(σ+1) = αβ·σ+β = αβ·σ · αβ = (αβ )σ · αβ = (αβ )σ+1 = (αβ )γ . Finally, if γ is a limit ordinal, then again we apply Lemma 1.4.1 to the continuous function Gα , defined by Gα (τ ) := ατ , and obtain with the inductive hypothesis: αβ·γ = αsup{β·σ:σ<γ} = sup{αβ·σ : σ < γ} = sup{(αβ )σ : σ < γ} = (αβ )γ . Lemma 1.4.4 Let α and β be ordinals. a) (Substraction Lemma) If α ≤ β, then there is a unique ordinal γ such that α + γ = β. b) (Division Lemma) If β = 0, then there are unique ordinals ξ and η satisfying α = β · ξ + η and η < β. c) (Logarithm Lemma) If α = 0 and β > 1, then there are unique ordinals σ, τ and γ (the logarithm, the coefficient and the remainder) such that α = βσ · τ + γ ∧ 1 ≤ τ < β ∧ γ < βσ .

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35

Proof For part a) we note, that the uniqueness of γ is obvious if one uses the monotonicity properties of ordinal addition in Lemma 1.4.3. Next we prove existence. Since β ≤ α + β, there is a least ordinal γ with β ≤ α + γ. If equality holds, we are done. Otherwise we have γ = 0. Furthermore γ cannot be a limit ordinal, because β < α + γ yields, by the definition of ordinal addition, for limit ordinals γ the existence of an ordinal σ such that σ < γ and β < α + σ, contradicting the choice of γ. Thus we obtain γ = η + 1 for some η and, again by the choice of γ, α + η < β < α + η + 1, which contradicts Lemma 1.2.6 b). For the proof of part b), we state first that the uniqueness of ξ and η follows again from a) and Lemma 1.4.3. Now we prove existence. Lemma 1.4.3 tells us that α ≤ β · α, and so there is a least ordinal γ such that α ≤ β · γ. If equality holds, we are done taking ξ := γ and η := 0. Suppose that α < β · γ. Then we get in complete analogy to the proof of a), that γ is a successor ordinal ξ + 1. Hence β · ξ < α < β · (ξ + 1) = β · ξ + β. From a) follows the existence of a unique ordinal η satisfying β · ξ + η = α. Since α = β · ξ + η < β · ξ + β, Lemma 1.4.3 a) gives η < β. Finally, we turn to the proof of c). Having studied the previous proofs, the reader will be familiar with the following arguments. To prove existence, we note first that α ≤ β α by Lemma 1.4.3 c). So there is a least ordinal δ with α ≤ β δ . If equality holds, put σ := δ, τ := 1, and γ := 0. In the other case the continuity of exponentiation and the choice of δ ensure again that δ cannot be a limit ordinal. Thus δ = σ + 1 for some σ ∈ ON. Now we have β σ < α, and by part b) there are unique ordinals τ and γ satisfying α = β σ · τ + γ and γ < β σ . Since α < β δ = β σ+1 = β σ · β, Lemma 1.4.3 b) tells us that 1 ≤ τ < β. Uniqueness can be proved again with Lemma 1.4.3; this is left to the reader. Corollary 1.4.5 Let α and β be ordinals. a) α is a limit ordinal iff there is an ordinal ξ = 0 with α = ω · ξ. b) The operation α → ω · (1 + α) defines the unique ∈-isomorphism from ON onto the class of limit ordinals. c) If α is a limit ordinal and 1 ≤ n < ω, then n · α = α. d) If n < ω and ω ≤ α, then n + α = α. e) α · β is a successor ordinal iff α and β are both successor ordinals. Proof First we prove a). By Lemma 1.4.4 we get, for α and ω, unique ordinals ξ and η such that η < ω and α = ω · ξ + η. If α is a limit ordinal, then η = 0 and ξ = 0. Conversely, if η = 0 and ξ = 0 then, by Remark 1.4.2 b), α = ω · ξ is a limit ordinal if this holds for ξ, and also in the case ξ = σ + 1, because ω is a limit ordinal and then α = ω · σ + ω. For b), let the equation G(α) := ω · (1 + α) define the function G. By part a), G(α) is a limit ordinal, and Lemma 1.4.3 says that G is strictly monotonic.

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If α is a limit ordinal and σ < α, then we get with d) 1 + σ ≤ σ + 1 < α and 1 + α = α. Therefore G(α) = ω · α = sup{ω · σ : σ < α} = sup{ω · (1 + σ) : σ < α} = sup{G(σ) : σ < α}, and thus G is continuous. Finally, G is surjective. Consider a fixed limit ordinal τ . By a), we can represent τ as τ = ω·α for some α = 0. If α ≥ ω, then G(α) = τ , since 1 + α = α. On the other hand, if α = σ + 1 is finite, then G(σ) = τ . For c), the assumption and part a) give α = ω · ξ for some ξ = 0. Further, by the definition of ordinal multiplication, n · ω = sup{n · m : m < ω} = ω, since n > 0. The associative law gives n · α = n · (ω · ξ) = ω · ξ = α. Part d) can be proved analogously to c), using Lemma 1.4.4 a). Since α ≥ ω, there is an ordinal β with α = ω + β. For the proof of e), let first α and β be successor ordinals. Then, by the definition of ordinal multiplication, α · β is a successor ordinal. Now let conversely α · β be a successor ordinal γ + 1. Then Lemma 1.4.3 says that both α and β are nonzero. By Lemma 1.4.4, we obtain a unique representation γ = α · ξ + η with η < α. Hence γ + 1 = α · ξ + η + 1 = α · β. The assertion η + 1 < α would contradict the uniqueness property in Lemma 1.4.4 b). Thus η + 1 = α and γ + 1 = α · ξ + α = α · (ξ + 1) = α · β. For the same reasons we get β = ξ + 1, as desired. Theorem 1.4.6 (Cantor Normal Form for the base β) Assume that α and β are ordinals with α = 0 and β > 1. Then there are unique ordinals σ1 , . . . , σn and τ1 , . . . , τn such that α = β σ1 · τ1 + . . . + β σn · τn , σ1 > . . . > σn and 1 ≤ τi < β for i = 1, . . . , n. In particular, for β = ω, there are unique ordinals α1 , . . . , αn satisfying α = ω α1 + . . . + ω αn

and α1 ≥ . . . ≥ αn .

Proof We use transfinite induction on α ∈ ON. Most work has already been done in Lemma 1.4.4 c). This lemma provides us, for α = 0 and β > 1, with unique ordinals σ1 , τ1 and γ such that 1 ≤ τ1 < β, γ < β σ1 and α = β σ1 ·τ1 +γ. Since γ < α, we can represent γ, if γ = 0, by the inductive hypothesis in the desired form with ordinals τ2 , . . . , τn , satisfying 1 ≤ τi < β for i = 1, . . . , n and σ2 > . . . > σn . Now we have β σ2 ≤ γ < β σ1 , from which we can conclude together with Lemma 1.4.3 that σ1 > σ2 . If β = ω, then all ordinals τi are natural numbers, and ωσi ·τi is the τi -fold sum of ω σi . From this, we get the second assertion of the theorem. We have learnt that the operations for ordinals extend the usual operations on ω but satisfy no longer all of its rules. The representation of an ordinal

1.4. Arithmetic of Ordinals

37

in Theorem 1.4.6 permits the definition of a so-called natural sum, satisfying the most important laws of finite arithmetic. Definition The representation of an ordinal α = 0, given by the first part of Theorem 1.4.6, is called the Cantor normal form of α for the base β or the Cantor normal form of α for the base ω, if β = ω. The Cantor normal form of 0 is ω0 · 0. Now assume that α = 0 and β are represented in Cantor normal form for the base ω. If we add powers ω σ of ω, occurring only in one representation, to the other one as the summand ωσ · 0, then we can assume without loss of n σi · ni and β = generality that α and β are uniquely given as α = i=1 ω n σi ω · m such that σ > . . . > σ , 0 ≤ n < ω, 0 ≤ m i 1 n i i < ω, and i=1 mi + ni = 0 for i = 1, . . . , n. Then we can define the natural sum α#β of the ordinals α and β by n

ω σi · (mi + ni ). α#β := i=1

We call an ordinal γ a principal number of addition iff γ = 0 and, for all α < γ, α + γ = γ. γ is called additively decomposable iff there are ordinals α < γ and β < γ such that α + β = γ; otherwise we say that γ is additively indecomposable. Remark The natural sum of ordinals satisfies the associative law and the commutative law. Later on we will need the fact that the equation α#β = γ has, for a fixed ordinal γ, only finitely many solutions (α, β). This is true, since there are only finitely many nonzero coefficients in the Cantor normal form of γ and, for every natural number l, the equation x+y = l has only finitely many solutions (x, y) ∈ ω × ω. In complete analogy to the notion of a principal number of addition one can define principal numbers of multiplication and of exponentiation. It turns σ out that {2}∪{ω(ω ) : σ ∈ ON} is the set of principal numbers of multiplication (see Exercise 1.4.7) and {ω}∪{α ∈ ON : ω α = α} is the set of principal numbers of exponentiation (see Exercise 1.4.8). Clearly 1 is the only principal number of addition less than ω, and Corollary 1.4.5 d) says that ω is the least principal number of addition greater than 1. The next lemma characterizes the principal numbers of addition. Lemma 1.4.7 a) An ordinal γ = 0 is a principal number of addition iff it is additively indecomposable. b) The principal numbers of addition are exactly the ordinals of the form ω α , where α ∈ ON.

38

Chapter 1. Foundations c) (ω α : α ∈ ON) is the unique ∈-isomorphism from ON onto the class of principal numbers of addition.

Proof For a), let γ be a principal number of addition, and let α and β be less than γ. Then, by Lemma 1.4.3, α + β < α + γ = γ, and γ is additively indecomposable. Conversely, if this is the case for a fixed γ = 0, then take α < γ and choose, according to Lemma 1.4.4, an ordinal β ≤ γ with α + β = γ. Then by definition β = γ, and γ is a principal number of addition. For the proof of b), let α ∈ ON. Clearly, ω α = 0. If 0 < β < ω α and β1 ω +. . .+ω βn is the Cantor normal form of β for the base ω, then Lemma 1.4.3 tells us that βi < α for i = 1, . . . , n. If we can show that σ < τ implies ωσ + ω τ = ω τ , then we obtain β + ωα = ω α , and ω α is a principal number of addition. So assume that σ < τ . By Lemma 1.4.4 a), there is an ordinal γ = 0 with τ = σ + γ. Then ω γ ≥ ω, and Corollary 1.4.5 d) says that 1 + ωγ = ω γ . By Lemma 1.4.3, we get ω σ + ω τ = ω σ + ω σ+γ = ω σ + ω σ · ω γ = ωσ · (1 + ω γ ) = ω σ · ωγ = ωτ . To prove the converse, let, for any α, γ = 0 be not of the form ω α . Then in the Cantor normal form ω γ1 + . . . + ω γn of γ occur at least two summands, so we have ω γ1 < γ, and the uniqueness assertion of Theorem 1.4.6 implies that ω γ1 + γ = γ. Thus γ is not a principal number of addition. Part c) is easy to show, since α → ω α defines, by Lemma 1.4.3, a normal function, which is surjective by part b). Exercises 1) If τ is a limit ordinal and (αξ : ξ < τ ) is a strictly increasing sequence of ordinals, then sup{αξ : ξ < τ } is a limit ordinal. 2) If τ is a principal number of addition, (η(ξ) : ξ < τ ) is a normal sequence of ordinals, and σ < τ , then η(ξ) + σ ≤ η(ξ + σ) for all ξ < τ . 3) Use Lemma 1.4.4 to show that the relations <3 on α × {0} ∪ α × {1} and the Hebrew lexicographic ordering on α × β from Remark 1.4.2 d) are well-orderings of order type α + β and α · β, respectively. 4) Prove: If σ1 > . . . > σn and, for all i ∈ {1, . . . , n}, 1 ≤ τi < α, then ασ1 · τ1 + . . . + ασn · τn < ασ1 +1 . 5) Consider the relation defined in Remark 1.4.2 f). It is a well-ordering by Exercise 1.3.4. Use the Cantor normal form theorem to show that this well-ordering has order type αβ .

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Hint: If f ∈ fin(β, α), {σ ∈ β : f (σ) = 0} = {σ1 , . . . , σn } and σ1 > . . . > σn , let Φ(f ) := ασ1 · f (σ1 ) + ασ2 · f (σ2 ) + . . . + ασn f (σn ). Then we know from a previous question that Φ(f ) < ασ1 +1 , hence Φ(f ) ∈ αβ . Prove that Φ is strictly increasing and a function onto αβ . 6) Let ≺ be the canonical well-ordering of ON × ON, defined in Section 1.5, and let ot(α) be the order type of (α × α, ≺). Prove: a) b) c) d) e)

{(σ, τ ) : (σ, τ ) ≺ (0, β)} = β × β. ot(α + 1) = ot(α) + α + α + 1. ot(ω) = ω and ot(ω · 2) = ω · ω. ot(α) ≤ ω α . If α = ω α , then ot(α) = α.

Hint: If A is a set, (A, <) is a well-ordering, and Au = {x ∈ A : x < u} is the initial segment of (A, <) determined by u, then otp(A, <) = sup{otp(Au , <) + 1 : u ∈ A}. 7) An ordinal α is called a principal number of multiplication iff α > 1 and β · α = α for all β with 1 ≤ β < α. Prove: a) 2 and ω are principal numbers of multiplication. If 2 < n < ω, then n is not a principal number of multiplication. τ b) If τ = 0, then ωω is a principal number of multiplication. Hint: Use the Cantor normal form theorem for the base ω and a previous exercise. c) If γ is nonzero and not a principal number of addition, then ωγ is not a principal number of multiplication. d) If α1 = 0, n > 1, and α = ω α1 + · · · + ω αn with α1 ≥ . . . ≥ αn , then α is not a principal number of multiplication. τ e) {α : α is a principal number of multiplication} = {2} ∪ {ω ω : τ ∈ ON}. τ Hint: If α = 2 and α = ω ω for all τ , represent α as α = ω α1 + . . . + ω αn with α1 ≥ α2 ≥ . . . ≥ αn . 8) ε is called a principal number of exponentiation iff ε > 2 and αε = ε for all α with 2 ≤ α < ε. Prove: a) If 2γ = γ, then γ is a principal number of exponentiation. Hint: Prove successively that γ is a limit ordinal, a principal number of addition (using γ = ω · ξ for some ξ > 0), and a principal number of multiplication.

40

Chapter 1. Foundations b) {ω} ∪ {α : ω α = α} is the class of principal numbers of exponentiation. c) If f : ω −→ ON is given by f (0) = α and f (n + 1) = ω f (n) , then γ := sup{f (n) : n ∈ ω} has the property ω γ = γ. So the class of principal numbers of exponentiation is unbounded in ON. For α = 0 we obtain the smallest principal number of exponentiation which is greater than ω.

1.5 Cardinal Numbers and their Elementary Properties Theorem 1.5.1 The following statements are equivalent (in ZF). (i) Axiom of choice. (ii) Well-ordering Principle: For every set x, there exists a relation ≺ on x such that (x, ≺) is a well-ordering. (Shortly speaking: Every set can be well-ordered.) (iii) Zorn’s Lemma11 : If H is a set and ≤ is a partial ordering on H such that every subset of H linearly ordered by ≤, i.e., every chain in H, has a ≤-upper bound in H, then H possesses a ≤-maximal element. Proof For the proof of “(ii)=⇒(i)”, assume that x is a set of nonempty  sets and ≺ is a well-ordering on x, and let the function f be defined by f := {(y, z) : y ∈ x and z is the ≺ -smallest element of y}. Then obviously f is a choice function for x. Next we prove “(i)=⇒(iii)”. By transfinite recursion, we define a function corresponding to the following “visual” proof. Take a0 ∈ H. If a0 is maximal with respect to ≤, then we are ready. Otherwise, there is a member y of H such that a0 ≤ y and a0 = y – choose a1 to be such a member. If we have chosen an ∈ H for each n ∈ ω without finding a maximal element, then our assumption tells us that the chain {an : n ∈ ω} has an upper bound in H. Take one of these upper bounds and call it aω . In this manner we go on and obtain an ascending chain of members of H. The process must stop at some ordinal yielding a ≤maximal element of H, since otherwise, and this is the crucial point of the construction, we would get an injection of ON into H, contradicting the fact that H is a set. Let us make these ideas precise. Take f as a choice function for P(H) \ {∅}, and let for each x ⊆ H the set s(x) be defined as the set of 11 This is only a lemma in the axiom system ZFC. Mostly it is used as an axiom. For historical and habituary reasons, we will keep this name.

1.5. Cardinal Numbers and their Elementary Properties

41

upper bounds of x with respect to ≤ which are not a member of x. Further take b ∈ H, which is possible since H = V . The transfinite recursion theorem provides us with a function F : ON −→ V such that for all σ ∈ ON we have, putting F (σ) =: aσ :  f (s({aτ : τ < σ})) if s({aτ : τ < σ}) = ∅, aσ = b otherwise. It is easy to see that F |` {σ : aσ = b} is an injective function. Thus there is smallest σ0 with aσ0 = b, since, assuming the contrary, we would get F [ON] ⊆ H. Consequently, by the axiom of comprehension, F [ON] would be a set, and the axiom of replacement would imply that this would also hold for F −1 [F [ON]] = ON. This would contradict Corollary 1.2.2. The assumptions of Zorn’s lemma give that σ0 is not a limit ordinal. So let σ0 = τ0 + 1. Since s({aτ : τ < σ0 }) = ∅, aτ0 is a ≤-maximal element of H, as desired. Finally, for the proof of “(iii)=⇒(ii)”, let x be a set. Note that x is wellorderable iff there is a bijection from an ordinal onto x: If this is the case, then each such bijection induces a well-ordering on x by the well-ordering ∈ on this ordinal; the reverse implication follows from Corollary 1.3.11. This motivates the definition of H := {f : ∃σ(f : σ −→ x and f is one-to-one)}. We take the inclusion ⊆ as the partial ordering ≤ on H. If H satisfies the assumptions of Zorn’s Lemma and g is a maximal element of H, say g : α −→ y, then y = x. Otherwise we could choose z ∈ x \ y and define h := g ∪ {(α, z)}. Then h would be a one-to-one function from α + 1 into x, hence h ∈ H, and a proper extension of g, contradicting the fact that g is maximal. Let us check therefore whether H satisfies the assumptions of (iii). First we have H = ∅, since the empty function ∅ is a member of H. If f0 is any union of a chain K of members of H,  then f0 is a function and one-to-one, since K is a chain. The domain of f0 is {dom(g) : g ∈ K}, hence is an ordinal by Lemma 1.2.5, being the union of a set of ordinals. Finally we state that ran(f0 ) ⊆ x, and thus f0 ∈ H. It remains to be shown that H is a set (for proper classes H, Zorn’s Lemma generally fails – take (H, ≤) := (ON, ⊆)). Each f ∈ H induces a well-ordering Rf = {(a, b) : a, b ∈ ran(f ) ∧ f −1 (a) < f −1 (b)} of the subset ran(f ) of x. For each well-ordering (ran(f ), Rf ) there exists, by Corollary 1.3.11, a unique ordinal τf and a unique isomorphism, namely f , from (τf , ∈) onto the given well-ordering. Thus the function F := {(f, (ran(f ), Rf )) :

42

Chapter 1. Foundations

f ∈ H} is one-to-one. Its range is a set as a subset of the set P(x) × P(x × x). Now the axiom of replacement, applied to the function F −1 , tells us that H is a set. Remark There are a lot of assertions that are equivalent in ZF to the axiom of choice. Furthermore there are extensive investigations on weaker versions of this axiom and how they implicate one another. The reader is referred to the literature, especially to Th. Jech, The Axiom of Choice, [Je1], and to H. Rubin and J. E. Rubin, Equivalents of the Axiom of Choice, II, [RR]. We give examples for some further equivalent assertions; other examples can be found in the exercises. (iv) (Cardinal Comparability) ∀x∀y∃f (f : x −→ y is injective ∨f : y −→ x is injective). (v) ∀x∀y(x = ∅ ∧ y = ∅ =⇒ ∃f (f : x −→ y is surjective ∨f : y −→ x is surjective)). (vi) ∀x(¬∃n ∈ ω(|x| = |n|) =⇒ |x × x| = |x|), where |a| = |b| means: There is a bijection from a onto b. (vii) Zorn’s lemma restricted to sets partially ordered by the relation ⊆. (viii) Every vector space has a basis. Definition A cardinal number or cardinal is an ordinal that cannot be mapped one-to-one onto a smaller ordinal. If x is a set, then from (AC), Theorem 1.5.1, and Corollary 1.3.11 we obtain an ordinal that can be mapped one-to-one onto x, since x can be wellordered. The smallest ordinal with this property is called the cardinality or cardinal number of x, written as |x|. Any bijection from |x| onto x is called an enumeration of x. Remark The cardinality of any set is obviously a cardinal number. Furthermore we have, by definition, |α| ≤ α, and for cardinal numbers α holds |α| = α. If α and β are cardinal numbers, and if there is a bijection from α onto β, then α = β. If the cardinal numbers |x| and |y| exist in ZF, then it is easy to obtain the following equivalences. They can be used to formulate in ZF many statements on cardinalities of sets which are of importance for the working mathematician and can be applied without explicitly defining the object |x|. Any two sets with the property in a) are called equinumerous. a) |x| = |y| ⇐⇒ ∃f (f : x −→ y ∧ f is bijective). b) |x| ≤ |y| ⇐⇒ ∃f (f : x −→ y ∧ f is one-to-one). Lemma 1.5.2 (ZF) If z is a set and h : (P(z), ⊆) −→ (P(z), ⊆) is a homomorphism, then h has a fixed point (i.e., there is w ⊆ z such that h(w) = w).

1.5. Cardinal Numbers and their Elementary Properties

43

 Proof Put y := {u ∈ P(z) : u ⊆ h(u)} and  w :=  y. By the definition of y, and since h is a homomorphism, we have y ⊆ {h(u) : u ∈ y} ⊆ h(w). So we get w ⊆ h(w) and thus h(w) ⊆ h(h(w)), which gives h(w) ∈ y. But this implies h(w) ⊆ w, and w is the desired fixed point. Theorem 1.5.3 (Cantor-Schr¨oder-Bernstein) (ZF) If there are injective functions f : x −→ y and g : y −→ x, then there exists a bijection from x onto y. Proof Let the function h∗ : P(x) −→ P(x) be defined by h∗ (u) := x \ g[y \ f [u]]. From u ⊆ v we get h∗ (u) ⊆ h∗ (v). Thus Lemma 1.5.2 says that there is a subset w of x such that w = x \ g[y \ f [w]]. Now the desired bijection h : x −→ y is given by h(z) := f (z), if z ∈ w, and h(z) := g−1 (z), if z ∈ w. The easy verification is left to the reader. Corollary 1.5.4 a) |x| = |y| ⇐⇒ ∃f (f : x −→ y ∧ f is bijective). b) |x| ≤ |y| ⇐⇒ ∃f (f : x −→ y ∧ f is one-to-one). c) x = ∅ =⇒ (|x| ≤ |y| ⇐⇒ ∃f (f : y −→ x ∧ f is onto x)). Lemma 1.5.5 If K is a set of cardinal numbers, then sup(K) (= cardinal number.



K) is a

Proof First we note that K is aset of ordinal numbers. Thus sup(K) isan ordinal, we have sup(K) = K by Lemma1.2.5, and furthermore  and   | K| is an  upper bound of K, then K = sup(K) ≤ | K| | K| ≤ K. If   and K isa cardinal, since K ∈ K.  = | K|. To show that property, let κ  Then κ ⊆ K, and κ = |κ| ≤ | K| follows from Corollary 1.5.4. Thus | K| is an upper bound of K. Lemma 1.5.6 a) (Cantor) If x is a set, then |x| < |P(x)| and |P(x)| = |x 2|. b) (Hartogs’ Theorem)(ZF) If κ is a cardinal number, then there is a cardinal number greater than κ.12 Proof For the proof of a), we note first that {(z, {z}) : z ∈ x} is an injection from x into P(x). We will show that there is no surjection from x onto P(x) and assume, to get a contradiction, that f is such a surjection. Let y := {z ∈ x : z ∈ f (z)}. By assumption, there is u ∈ x such that f (u) = y. Now 12 The smallest cardinal number with this property, constructed in the proof, is called the Hartogs number of κ.

44

Chapter 1. Foundations

we can argue closely to Russell’s antinomy and conclude that u ∈ f (u) ⇐⇒ u ∈ f (u). This contradiction, together with the previous remark, proves |x| < |P(x)|. For the proof of the second assertion we map each subset v of x onto its so-called characteristic function from x into {0, 1}, which has the value 1 at the argument z ∈ x iff z ∈ v. This yields a bijection from P(x) onto x 2. Next we turn to the proof of part b). In ZFC, the power set of each set can be well-ordered. Therefore b) follows at once from a). Since we have carried out the main steps for a proof of this assertion in ZF already in the proof of Theorem 1.5.1, we give Hartogs’ proof of this assertion in ZF. Let A := {α : ∃f (f : α −→ κ ∧ f is one-to-one)}. When proving the well-ordering theorem with the help of Zorn’s lemma, we showed in ZF that the class B := {f : ∃α(f : α −→ x and f is one-to-one)} is a set for each set x. Since, for x = κ, A = {dom(f ) : f ∈ B}, we obtain with an axiom of replacement that A is also a set. Thus there is a smallest ordinal λ such that λ ∈ A. It is easy to show that κ ∈ A and ∀β∀γ(β ∈ A ∧ γ < β =⇒ γ ∈ A). If we could map λ one-to-one onto a smaller ordinal β, then we would get λ ∈ A, since β ∈ A. This contradiction shows that λ is a cardinal greater than κ, as desired. Furthermore, it is the smallest cardinal greater than κ. Corollary 1.5.7 The class CN of cardinal numbers is a proper class. The same holds for CN \ ω, the class of infinite cardinal numbers. Proof Assume that CN is a set y. Then, by Lemma 1.5.5, κ := sup(y) ∈ CN. From Lemma 1.5.6 we obtain a cardinal λ > κ. Together with λ ∈ y, this yields the contradiction λ < λ. Definition Let CN := {κ ∈ ON : κ is a cardinal } be the class of cardinal numbers. By complete induction it is easy to see that every natural number is a cardinal number. The members of ω are called finite cardinal numbers. Thus we call a cardinal number κ infinite or transfinite iff κ ≥ ω. Note that ω is the smallest infinite cardinal. We write ICN for the class of infinite cardinal numbers. Lemma 1.5.6 says that for each cardinal κ there is a smallest cardinal greater than κ. We denote it by κ+ and call it the cardinal successor or successor of κ. Every infinite cardinal of the form κ+ is called a successor cardinal. These cardinal numbers are exactly the values ℵα+1 of the aleph function (see below) at arguments that are successor ordinals. Any infinite cardinal number which is uncountable and not a successor cardinal is called a limit cardinal. A set x is called finite if there is a natural number n and a bijection from n onto x. Otherwise we call x infinite. Pfin (x) denotes the set of finite subsets of the set x. We say that x is denumerable or countably infinite iff there is a bijection from ω onto x. x is called countable iff it is finite or countably infinite.

1.5. Cardinal Numbers and their Elementary Properties

45

By Corollary 1.5.7 and Corollary 1.3.11, there is exactly one ∈-isomorphism from ON onto the class ICN of infinite cardinal numbers. Usually, it is denoted by ℵ and called the aleph function. Instead of ℵ(α) we write ℵα . Especially, ℵ0 , that means ω, is the smallest infinite cardinal, and ℵ1 , that means (ℵ0 )+ , is the smallest uncountable cardinal (ordinal). Often in the literature, part c) of the following lemma is taken as a recursive definition (in ZF – see above) of the aleph function. For each α ∈ ON, the members of the set ℵα+1 \ℵα are exactly the ordinal numbers, and hereby the representatives of well-ordered sets, of cardinality ℵα . The smallest element of this set, ℵα , often is called an initial ordinal. If we want to emphasize this order-theoretical character of ℵα , and also for historical reasons, we use the denotation ωα instead of ℵα .13 If μ is an infinite cardinal, μ = ℵα , and β is an ordinal, then let the cardinal μ+β be defined by μ+β := ℵα+β . In analogy to intervals of ordinals we define, for any cardinals κ and μ such that κ ≤ μ, the interval (κ, μ)CN := (κ, μ)ON ∩ CN of cardinals. The intervals (κ, μ]CN , [κ, μ)CN , and [κ, μ]CN are defined in the same way. Lemma 1.5.8 a) If κ is an infinite cardinal, then there is an ordinal α with κ = ℵα . b) If α < β, then ℵα < ℵβ . c) ℵ0 = ω, ℵα+1 = ℵ+ α and ℵγ = sup{ℵα : α < γ} for limit ordinals γ. Proof Part a) is true by the definition of the aleph function. b) and c) follow from the fact that ℵ : ON −→ ICN is an ∈-isomorphism. Definition (Sum, product and exponentiation of cardinal numbers) If κ and λ are cardinal numbers, then the sets κ × {0} ∪ λ × {1} and κ × λ can be well-ordered (for example by the relation ≺ defined in Theorem 1.5.11). We call the cardinal numbers κ + λ := |κ × {0} ∪ λ × {1}| and κ · λ := |κ × λ| the (cardinal) sum and the (cardinal) product of the cardinal numbers κ and λ (not to be confused with the sum and product of ordinal numbers). Furthermore x y = {f : f : x −→ y} is a set, if x and y are sets. We define κλ := |λ κ| as the exponentiation or power of the cardinal numbers κ and λ. This definition makes sense in ZFC. 13 A

typical example is the denotation ℵωα instead of ℵℵα .

46

Chapter 1. Foundations

By Lemma 1.5.6, we have κ < 2κ for every cardinal κ, hence κ+ ≤ 2κ . The continuum hypothesis is the assertion 2ℵ0 = ℵ1 , the generalized continuum hypothesis, abbreviated by (GCH), is the assertion 2ℵα = ℵα+1 for all α ∈ ON. Remark Usually, one introduces the real numbers as follows: First the set Z of integers is constructed from the set ω of natural numbers, then the set Q of rational numbers, the set of all fractions of integers, is constructed from Z. A Dedekind cut in Q is an ordered pair (A, B) of nonempty subsets of Q satisfying A ∪ B = Q, ∀a ∈ A ∀b ∈ B(a < b) and inf(B) ∈ B, if the infimum exists. Then we can define the set R of real numbers, the so-called continuum, as the set of all Dedekind cuts in Q. Clearly we have |R| ≤ |P(Q)| = 2|Q| = 2ℵ0 by Lemma 1.5.6. If we can specify a subset of R of cardinality 2ℵ0 , then it is shown that |R| = 2ℵ0 . ω We get an injection from into R, if we assign to each sequence f ∈ ω 2 ∞ the set 2−n−1 ∈ [0, 1] := {x ∈ R : 0 ≤ x ≤ 1}. the ternary fraction n=0 2f (n)3 Thus the continuum hypothesis can be reformulated equivalently as ¬∃X ⊆ R(ω < |X| < |R|). Lemma 1.5.9 a) If x and y are sets with |x| = κ and |y| = λ, then κ+λ = |x×{0}∪y×{1}|, κ · λ = |x × y|, and κλ = |y x|. b) For natural numbers, the cardinal sum, product and exponentiation coincide with the corresponding ordinal operations defined in Section 1.4. c) If  is the addition, multiplication, or exponentiation of ordinals, and if α and β are nonzero ordinals, one of them being infinite, then |α  β| ≤ max{|α|, |β|}. In particular, every infinite cardinal number κ is a principal number of addition, i.e., we have α + κ = κ for all α < κ. Proof In a), we immediately obtain the desired bijections with the help of the given bijections from x onto κ and from y onto λ. Part b) is proved by complete induction, and the first assertion in part c) by transfinite induction according to the recursive definitions of the ordinal operations and by Theorem 1.5.14. The second assertion follows from Lemma 1.4.7. Lemma 1.5.10 For cardinal numbers κ, λ, μ a) (i) (κ + λ) + μ = κ + (λ + μ) (ii) b) (i) (κ · λ) · μ = κ · (λ · μ) (ii) (iv) (iii) κ · 0 = 0 (v) κ · (λ + μ) = κ · λ + κ · μ

and ν the following laws hold. κ+λ=λ+κ κ·λ=λ·κ κ·1=κ

1.5. Cardinal Numbers and their Elementary Properties c)

(i) (iii) (v)

d) (i) (iii)

κλ+μ = κλ · κμ (κ · λ)μ = κμ · λμ κ1 = κ, 1κ = 1 If κ ≤ μ and λ ≤ ν, then : κ+λ≤μ+ν λ = 0 =⇒ κλ ≤ μν

(ii) (iv) (vi)

(κλ )μ = κλ·μ κ0 = 1, 0κ = 0 for κ = 0 κ2 = κ · κ

(ii)

κ·λ≤μ·ν

47

Proof We leave it as an exercise for the reader to find suitable bijections and injections. We indicate the proof of (ii) in part c). If f : μ −→ λ κ is a function, then for each β ∈ μ the value f (β) is a function from λ into κ. Let the function φ(f ) : λ × μ −→ κ be defined by φ(f )(α, β) := f (β)(α). Then φ is μ a bijection from (λ κ) onto λ×μ κ. Theorem 1.5.11 There is a well-ordering ≺ on ON × ON such that, for every α ∈ ON, the well-ordering (ℵα × ℵα , ≺) is an initial segment of ≺ which is isomorphic to (ℵα , ∈). In particular we have |ℵα × ℵα | = ℵα , that means ℵα · ℵα = ℵα . Proof We define the relation ≺ on ON × ON by (α, β) ≺ (σ, τ ) :⇐⇒ max{α, β} < max{σ, τ }∨(max{α, β} = max{σ, τ }∧(α < σ ∨(α = σ ∧β < τ ))). That means that we well-order the pairs of ordinal numbers by their maxima and, if their maxima are equal, lexicographically. It is easy to verify that ≺ is a well-ordering on ON × ON, and that (ℵα × ℵα , ≺) is a proper initial segment of this well-ordering for every α ∈ ON. Assume, contrary to our hopes, that α is the smallest ordinal for which (ℵα × ℵα , ≺) is not isomorphic to (ℵα , ∈). By Corollary 1.3.11, there is a unique ordinal γ and an isomorphism f from (γ, ∈) onto (ℵα × ℵα , ≺). Then γ ≥ |γ| = |ℵα × ℵα | ≥ ℵα , and thus, by assumption, γ > ℵα . We conlude that the function f |` ℵα is an isomorphism from (ℵα , ∈) onto a proper initial segment of (ℵα × ℵα , ≺). Since ≺ orders ON × ON first by the maxima of the pairs, the set of first as well as the set of second components of f [ℵα ] must be bounded in ℵα ; take σ < ℵα such that f [ℵα ] ⊆ σ × σ. Now σ is infinite, and so there is β < α with |σ| = ℵβ . Then we get ℵβ < ℵα and, by the choice of α, ℵβ · ℵβ = ℵβ . On the other hand, ℵα = |f [ℵα ]| ≤ |σ × σ| = |σ| · |σ| = ℵβ · ℵβ = ℵβ . This contradiction shows that such an ordinal α does not exist, as desired. Corollary 1.5.12 If κ and λ are cardinal numbers, one of them nonzero and the other infinite, then κ + λ = κ · λ = max{κ, λ}. In particular, ℵα + ℵβ = ℵα · ℵβ = max{ℵα , ℵβ } = ℵmax{α,β} .

48

Chapter 1. Foundations

Proof Without loss of generality let 0 < κ ≤ λ and λ ≥ ω. Then the definitions of the cardinal operations, Corollary 1.5.4, and Theorem 1.5.11 yield λ ≤ κ + λ ≤ κ · λ ≤ λ · λ = λ. This gives the first assertion. The second follows from the monotonicity of the aleph function. Definition If a is a set, then we write [a]λ := {x ⊆ a : |x| = λ} for the set of those subsets of a which have cardinality λ.14 Corollary 1.5.13 Let a be an infinite set of cardinality κ and λ be a cardinal. Then |[a]λ | = |a|λ = κλ , if λ ≤ κ, and |[a]λ | = 0, if λ > κ. Proof The second assertion is obvious. So assume without loss of generality that 0 < λ ≤ κ = |a|. Then |[a]λ | = |[κ]λ |. Using (AC), we can assign uniquely to each subset X of κ of cardinality λ a bijection from λ onto X, that means a member of λ κ. Clearly this defines an injective function, and thus |[κ]λ | ≤ κλ . For the proof of the reverse inequality we observe that every function from λ into κ is a subset of λ × κ with cardinality λ, and that we have |λ × κ| = κ by assumption and by Corollary 1.5.12. Thus we can conclude that κλ = |λ κ| ≤ |[λ × κ]λ | = |[κ]λ | ≤ κλ , as claimed. Theorem 1.5.14 The following assertions hold. They are generally sufficient to get cardinality results in everyday mathematics. a) (ZF) The Cantor-Schr¨ oder-Bernstein theorem. b) If, for any set a and n ∈ ω \ {0}, an is defined by a1 := a and ak+1 := ak × a, then we have for any infinite set a: |a| = |a × a| = |an |

and

|n a| = |a|.

c) If (Ai : i ∈ I) is a family of sets, then  | {Ai : i ∈ I}| ≤ |I| · sup{|Ai | : i ∈ I}. d) If a is an infinite set, then |Pfin (a)| = |a|. Proof The first two statements of b) follow directly from Corollary 1.5.12. Hereby an is the set of all n-tupels of members of a. We will also work with the set n a whose members can be regarded as n-tupels, too.15 If we assume as 14 The

sets [a]<λ , [a]≤λ , . . . are defined analogously in Section 1.6. we have two different definitions of an ordered pair, since the set (c, d) is not equal to the function {(0, c), (1, d)}. 15 Then

1.5. Cardinal Numbers and their Elementary Properties

49

inductive hypothesis that g is a bijection from n a onto a, then the equation h(f ) := (g(f |`n), f (n)) defines a bijection from n+1 a onto a × a, and this gives |n+1 a| = |a × a| = |a|. : i ∈ I}, and let < be a well-ordering of I. The For c),  let κ := sup{|Ai |  function f : {Ai : i ∈ I} −→ {Ai × {i} : i ∈ I}, given by f (a) = (a, i) where i is <-minimal with a ∈ Ai , is injective. So we obtain   | {Ai : i ∈ I}| ≤ | {Ai × {i} : i ∈ I}| = |

  {|Ai | × {i} : i ∈ I}| ≤ | {κ × {i} : i ∈ I}| = |κ × I| = |I| · κ.

Finally, we prove d). With the axiom of choice it is easy to show for nonempty sets x and y that there is an injection  from x into y iff there is a surjection from y onto x.  From c) and b) we get | {n a : n ∈ ω}| = |a|. If we assign to each member of {n a : n ∈ ω} its range, then we obtain a surjective function from this set onto Pfin (a). Example We want to show that any two bases of a vector space W over a field K have the same cardinality. If one of the bases is finite, then the exchange theorem of Steinitz says that the other basis is finite, too, and that the bases have the same number of members. So assume that B1 and B2 are infinite bases of W . For each vector x ∈ B1 , there is a finite subset b x of B2 such that x is a linear combination of the elements of bx . Let C := {bx : x ∈ B1 }, and suppose that C = B2 . Take y ∈ B2 \ C. y is a linear combination of certain members x1 , . . . , xn of B1 , and each xi is a linear combination of bxi . Thus y is a linear combination of {bxi : i ∈ {1, . . . , n}}, and our supposition implies that y is not a member of thisset. This contradicts the fact that B2 is linearly independent, hence B2 = {bx : x ∈ B1 }. Theorem 1.5.14 c) says that |B2 | ≤ |B1 | · ℵ0 = |B1 |. By symmetry, we have |B1 | ≤ |B2 |. Now part a) of the theorem yields |B1 | = |B2 |, as desired. Definition Let R be a relation on A with dom(R) = A. If B ⊆ A, then B is called unbounded or cofinal in A under R iff, for each a ∈ A, there is a member b ∈ B such that aR b. We define the cofinality of (A, R), denoted by cf (A, R), by cf (A, R) := min{|C| : C ⊆ A ∧ C is cofinal in A under R}. Thus cf (A, R) is the smallest cardinality of any subset of A that is cofinal in A under the relation R. Because of dom(R) = A, A itself has this property; so the cardinal number cf (A, R) always exists. If α is an ordinal, then the cofinality of α, written as cf (α), is the smallest ordinal μ with the property that there is a function f : μ −→ α such that

50

Chapter 1. Foundations

ran(f ) is cofinal in α under the relation ≤ on ordinals. Obviously, cf (α) ≤ α. Furthermore cf (α) ≤ cf (α, ≤), since for each cofinal subset C of α satisfying cf (α, ≤) = |C| there is a bijection from |C| onto C whose range is cofinal in α; an analogous argument shows that cf (α) is a cardinal. Lemma 1.5.15 and Lemma 1.5.16 tell us that cf (α) = cf (α, ≤) holds. We call a limit ordinal α regular iff cf (α) = α; we have just shown that regular ordinals are cardinals. A limit ordinal α is called singular iff cf (α) < α. With Reg we denote the class of regular cardinal numbers and let (κ, μ)reg := (κ, μ)CN ∩ Reg etc. Lemma 1.5.15 Let α be an ordinal. Then a) cf (0) = 0 and cf (α + 1) = 1. b) If α is a limit ordinal, then there exists a normal function h : cf (α) −→ α whose range is cofinal in α.  The last condition is equivalent to sup(ran(h)) = α, i.e., to ran(h) = α. Proof The range of the empty function is cofinal in 0, and the range of the function {(0, α)} is cofinal in α + 1. This proves a). Next we turn to the proof of part b). If α ∈ Lim and f : γ −→ α is a function, then we get from the definition of a limit ordinal  ran(f ) is unbounded in α ⇐⇒ sup(ran(f )) = α ⇐⇒ ran(f ) = α. If f : cf (α) −→ α is a function such that sup(ran(f )) = α, then, using transfinite recursion, we obtain a strictly increasing function g : cf (α) −→ α with sup(ran(g)) = α via g(β) := max{f (β), sup{g(σ) + 1 : σ < β}}. Note that, by the definition of cf (α), the ordinal sup{g(σ) + 1 : σ < β} is smaller than α for any β < cf (α). For continuity, we define the function h : cf (α) −→ α by h(0) := g(0), h(γ) := sup{h(β) : β < γ} for limit ordinals γ < α and h(β + 1) := g(β + 1). Then h is continuous. Using transfinite induction, it is easy to verify that h(β) ≤ g(β) for every β < cf (α). From this we get immediately that h is strictly increasing, and clearly we have sup(ran(h)) = α. Lemma 1.5.16 a) If α is a limit ordinal, then cf (cf (α)) = cf (α), hence cf (α) is a regular cardinal. Furthermore we have cf (α, ∈) = cf (α). b) ℵ0 is regular. Furthermore, every infinite successor cardinal κ+ is regular.

1.5. Cardinal Numbers and their Elementary Properties

51

c) If α is a limit ordinal, then cf (α) = cf (ℵα ). d) ℵω is the smallest singular cardinal number. e) The class of singular cardinal numbers is unbounded in ON. Proof For part a), we take strictly increasing functions f : cf (cf (α)) −→ cf (α) and g : cf (α) −→ α such that sup(ran(g)) = α and sup(ran(f )) = cf (α). Then the function g ◦ f is strictly increasing. To show that its range is unbounded in α, consider a fixed β < α. First there is an γ < cf (α) with β < g(γ). Furthermore there is σ < cf (cf (α)) such that γ < f (σ). Thus we have, by the monotonicity of g, β < g(f (σ)). This gives sup(ran(g ◦ f )) = α, and we can conclude that cf (α) ≤ cf (cf (α)). The reverse inequality is trivial. So the first assertion in a) is shown. Since we can find a strictly increasing function f : cf (α) −→ α whose range is cofinal in α, we get cf (α, ∈) ≤ |ran(f )| = |cf (α)| = cf (α). From this together with the remark after the definition of cf (α) we obtain the second assertion of part a). For the proof of b), we note first that obviously ℵ0 is regular, since no finite subset of ω can be cofinal in ω. For the second assertion, consider a function f : μ −→ κ+ . Then, by Theorem 1.5.14 c),   | sup(ran(f ))| = | ran(f )| = | {f (β) : β < μ}| ≤ |μ| · κ, since |f (β)| < κ+ for every β < μ, hence |f (β)| ≤ κ. Now, if μ ≤ κ, then | sup(ran(f ))|< κ+ and thus sup(ran(f )) < κ+ . So we get cf (κ+ ) ≥ κ+ . Since cf (κ+ ) ≤ κ+ , we have cf (κ+ ) = κ+ , and thus κ+ is regular. For the proof of c), we note first that ℵα = sup{ℵβ : β < α} for any limit ordinal α. If h : cf (α) −→ α has the property that sup(ran(h)) = α, then the range of the function h : cf (α) −→ ℵα , given by h (β) = ℵh(β) , satisfies sup(ran(h )) = ℵα , hence cf (ℵα ) ≤ cf (α). Conversely, let g : cf (ℵα ) −→ ℵα be fixed, such that ran(g) is cofinal in ℵα . Let g  : cf (ℵα ) −→ α be defined by g  (γ) = min{σ < α : g(γ) ≤ ℵσ }. If τ < α, then there is γ < cf (ℵα ) such that ℵτ ≤ g(γ). Then τ ≤ g  (γ), hence ran(g  ) is cofinal in α and cf (α) ≤ cf (ℵα ). Next we turn to d). By definition, ℵω = sup{ℵn : n ∈ ω}. Thus cf (ℵω ) = ω = ℵ0 < ℵω , ℵω is singular. By part b), the cardinal numbers ℵn (n ∈ ω) are regular. Consequently, ℵω is the smallest singular cardinal. For e), we remind the reader of the fact that the ordinal sum is continuous in the second argument. Thus each cardinal ℵα+ω is singular, since it is the supremum of the countable set {ℵα+n : n ∈ ω}, where + is the ordinal sum, and furthermore α ≤ ℵα < ℵα+ω . From this we also get that the class of singular cardinals is unbounded in ON; therefore it is a proper class. Remark Further properties of regular and singular cardinal numbers are presented in Lemma 1.6.11, when we have infinite sums of cardinals at our

52

Chapter 1. Foundations

disposal, and of course in the section on clubs and stationary sets. Those readers which are meeting these concepts for the first time, may always imagine ℵω as the prototype for a singular cardinal and ℵ1 as the prototype for a regular cardinal. The regularity of ℵ1 also follows from the well-known fact that any union of countably many countable sets is countable (see Theorem 1.5.14). The corresponding generalization for arbitrary regular cardinal numbers can be found in Lemma 1.6.11. Exercises Assume that a, b, c, . . . , A, B, C, . . . etc. are sets. 1) If n ∈ ω, then there is no injection from n onto a proper subset A of n. Hint: Use complete induction. For the induction step, consider the cases n ∈ A and n ∈ A. 2) Prove that ω and every member of ω are cardinals. 3) Assume that B and C have the same finite cardinality n. Let f : B −→ C be injective (surjective). Prove that f is surjective (injective). Hint: Use complete induction. 4) Prove that the following assertions are equivalent: (i) A is finite. (ii) Every nonempty subset Y of P(A) has a ⊆-maximal element. (iii) ∀f (f : A −→ A =⇒ (f is injective ⇐⇒ f is surjective)). 5) 6) 7) 8)

9) 10)

11)

Hint: For (i) =⇒ (iii), use Exercise 3. Prove that ordinal and cardinal addition (multiplication, exponentiation) coincide on ω × ω. If A is a finite set all of whose elements are nonempty, then A has a choice function.  (ZF) If B is a finite set and each member of B is countable, then B is countable. In ZFC, this assertion is provable for countable sets B. ω {0, 1} is not countable. Hint: Suppose that ω {0, 1} = {fn : n ∈ ω} and construct a new function f : ω −→ {0, 1}. f : ω × ω −→ ω, defined by f (x, y) := 12 (x+ y + 1) · (x+ y) + x, is bijective. Assume that A is a nonempty set and B is a set. Prove: There is an injection from A into B iff there is a surjection from B onto A. For which implication is (AC) not needed? Assume that A has at least three elements. Prove that there is no surjective function from A onto P(A) \ {{a} : a ∈ A}. Hint: The proof is very similar to Cantor’s proof in 1.5.6 a).

1.5. Cardinal Numbers and their Elementary Properties

53

12) Prove that every infinite set has a countably infinite subset. Did you need (AC)? (Yes!) 13) Prove in ZF that (AC) is equivalent to the following assertion: ∀x ∀y ∃f (f is injective ∧ (f : x −→ y ∨ f : y −→ x)). Hint: One implication can be found above. Furthermore we have shown without using (AC) that {σ ∈ ON : ∃f (f : σ −→ A ∧ f is injective)} is a set if A is a set. Use this to show that the above assertion implies that A can be well-ordered. 14) a) If there is a surjection f from an ordinal α onto A, then A can be well-ordered. Hint: For a ∈ A put od(a) := min{σ ∈ α : f (σ) = a}. Define a ≺ b :⇐⇒ od(a) < od(b) for any a, b ∈ A. b) Prove in ZF that (AC) is equivalent to the following assertion: ∀x ∀y(x = ∅ ∧ y = ∅ =⇒ ∃f (f : x −→ y surjective ∨ f : y −→ x surjective)). Hint: Use part a) for one of the implications and show that the class {σ : ∃f (f : A −→ σ ∧ f is surjective)} is a set. Observe that every surjection from A onto σ ∈ ON yields an injection from σ into P(A). 15) (Vaught) Let pd(y) abbreviate the assertion ∀u ∈ y ∀v ∈ y(u = v =⇒ u ∩ v = ∅). Prove in ZF that (AC) is equivalent to the following assertion: ∀x∃y(y ⊆ x ∧ pd(y) ∧ ∀z(z ⊆ x ∧ pd(z) =⇒ ¬ y  z)). 16) The set version of Zorn’s lemma says:

 If X is a set and M ⊆ P(X) is nonempty, such that K ∈ M for every nonempty ⊆-chain K ⊆ M , then M has a ⊆-maximal element.

Show in ZF that this assertion is equivalent to Zorn’s lemma. Hint: If (H, ≤) is a partial ordering that satisfies the assumptions of Zorn’s lemma, then define M := {X ⊆ H : X is a ≤ -chain}. 17) Assume that X is a set and M ⊆ P(X). We say that M has finite character iff any subset Y of X is a member of M if and only if every finite subset of Y is an member of M . The Teichm¨ uller-Tukey Lemma says:

54

Chapter 1. Foundations If M ⊆ P(X) has finite character and M = ∅, then M has a ⊆maximal element.

Show in ZF that this assertion is equivalent to Zorn’s lemma. Hint: For one implication, the proof of the nontrivial direction of the previous question is quite useful. 18) Hausdorff ’s Maximality Principle says: For every nonempty partial ordering (H, ≤), where H is a set, there is a ⊆-maximal ≤-chain in H. a) Prove in ZF that this principle follows from the Teichm¨ uller-Tukey lemma. b) Prove in ZF that Zorn’s lemma follows from this principle. 19) The Principle of Dependent Choices says that, for every nonempty set A and every relation R on A satisfying ∀x ∈ A ∃y ∈ A xRy, there is a sequence f : ω −→ A such that (f (n), f (n + 1)) ∈ R for all n ∈ ω. Prove that this principle follows from (AC), and prove in ZF that it implies that every countable set of nonempty sets has a choice function. The last assertion is called the axiom of countable choice. 20) (Kelley) Let X be a  topological space. An open cover of X is a set U of open sets such that U = X. X is called compact iff every open cover of X has a finite open subcover U  ⊆ U. In lectures on topology, one usually proves with (AC) the following assertion, known as Tychonoff ’s Theorem: a family of compact topological If I is a set and (Xi : i ∈ I) is  spaces, then the topological space i∈I Xi with the standard product topology is compact. Prove in ZF that (AC) follows from this assertion. Hint:Let (Xi : i ∈ I) be a family of nonempty sets. We want to show that i∈I Xi = ∅. Choose a fixed new set a and put Yi := Xi ∪ {a}. Let a topology Ti on Yi be defined by Ti := {∅, {a}} ∪ {Z ⊆ Yi : Yi \ Z is finite}. Now use the fact that X is compact iff every family of closed sets with the finite intersection property has a nonempty intersection, and consider  the family (Ai : i ∈ I), where Ai := {f ∈ k∈I Yk : f (i) ∈ Xi }. 21) Prove without using (AC): a) If B ∩ C = ∅ , then there is a bijection from b) There is a bijection from

B×C

C

A onto (B A).

B∪C

A onto

B

A × C A.

1.5. Cardinal Numbers and their Elementary Properties

55

c) If S is a countable subset of R, then |R \ S| = |R|. Hint: If S is infinite, assume w.l.o.g. that ω \ S is infinite, and map this set bijectively onto ω ∪ S. d) There is a bijection from Rn onto R. e) There is a bijection from R onto the set of all sequences of real numbers. f) There is a bijection from R R onto P(R). 22) (ZF) Show that, for any a, b ∈ R with a < b, |Rn | = |R| = |[a, b]| = |]a, b[| = |]a, b]|. Consequently, the continuum hypothesis is trivially true for open sets: Every nonempty open set is equinumerous to R. 23) Show that, if C = {f ∈ R R : f is continuous} (“continuous” in the usual meaning of calculus), then |C| = |R|. Hint: For any f, g ∈ C, f |`Q = g |`Q implies f = g. −i−1 : (ai : i ∈ ω) ∈ ω {0, 2}} is called the Cantor 24) The set C = { ∞ i=0 ai 3 set. Show: a) The function f : [0, 1] −→ {0, 1}, defined by f (x) = 0 ⇐⇒ x ∈ C, is Riemann integrable. Hint: By the integrability criterion of Riemann, it suffices to specify a sequence of partitions of [0, 1], such that the corresponding lower Riemann sums converge to 1. b) |{f ∈ [0,1] [0, 1] : f is Riemann integrable}| = 2|R| . Hint: Use the hint in a). 25) Prove that the set of open subsets of R (open with respect to the standard topology on R) is equinumerous to R. 26) If A ⊆ R, then a ∈ A is called an isolated point of A if there is ε > 0 such that A ∩ (a − ε, a + ε) = {a}. Show that any subset of R has only countably many isolated points. 27) Assume that (Aα : α < γ) is a sequence of closed subsets of R (with respect to the standard topology on R) which is strictly decreasing under ⊆. Prove that γ is countable. 28) (Cantor-Bendixson Theorem) Prove that every closed subset of R can be represented as A = P ∪ S, where P is perfect (i.e., closed and without isolated points) and S is countable. Hint: Put A0 := A,Aα+1 := Aα := {x ∈ R : x is an accumulation point of Aα } and Aγ := {Aσ : σ < γ} for limit ordinals γ. Show that there is a countable ordinal β such that Aσ = Aβ for all σ ≥ β, and apply the result from a previous question.

56

Chapter 1. Foundations

29) Prove that every perfect subset P of R has the cardinality 2ℵ0 . Hint: Assign to each finite {0, 1}-sequence s a member as of P such that (1) dom(s) = dom(t) =⇒ as = at , (2) f ∈ ω {0, 1} =⇒ af := limn→∞ af |`n exists. Then show that the function (af : f ∈ ω {0, 1}) is injective. 30) Prove that the continuum hypothesis is true for closed sets: For every closed subset A of R, we have |A| ≤ ℵ0 or |A| = 2ℵ0 . Hint: Use the results from the previous questions. 31) Prove: a) |{X ⊆ R : |X| ≤ ℵ0 }| = |R|. b) |{X ⊆ R : X is closed }| = |{X ⊆ R : X is perfect}| = |R|. 32) Prove: If |R| = ℵ1 , then R2 can be partitioned into two sets X and Y such that every straight line parallel to the x-axis contains only countably many points of X, and every straight line parallel to the y-axis contains only countably many points of Y . Hint: The assumption yields a well-ordering < of R of order type ℵ1 . Put X := {(x, y) ∈ R2 : x ≤ y} and Y := {(x, y) ∈ R2 : y < x}. 33) Assume that W is a vector space over the field F . Prove: a) If W has the finite dimension k ∈ ω, then  |F | if |F | ≥ ℵ0 , |W | = mk if |F | = m ∈ ω. b) If B is a basis of W , W = {0}, and if one of the sets F and B is infinite, then |W | = max{|F |, |B|}. n Hint: Every w ∈ W \ {0} is uniquely representable as w = i=1 λi bi for some λ1 , . . . , λn ∈ F and b1 , . . . , bn ∈ B. Show first that the equations Φ(w) := {(λ1 , b1 ), . . . , (λn , bn )} and Φ(0) := ∅ define an injection from W into Pfin (F × B). Now use Theorem 1.5.14. 34) If F is a field, then F [x] := {f ∈ ω F : ∃n0 ∀n ≥ n0 f (n) = 0} is the set of polynomials over F . Prove that |F [x]| = max{ℵ0 , |F |}. 35) Assume that κ is infinite and λ ≤ κ. Let T = {f ∈ κ λ : f is surjective}. Then |T | = λκ . Hint: If λ = 0, consider the set Y := {f ∈ κ×λ λ : f is surjective} and assign to each function f ∈ κ λ one-to-one a member of Y . Observe that |(κ × λ) \ f | = κ, if λ ≥ 2.

1.5. Cardinal Numbers and their Elementary Properties

57

36) Assume that κ is infinite and λ ≤ κ. Let T = {f ∈ λ κ : f is injective}. Then |T | = κλ . Hint: If λ = 0, then λ · κ = κ. Therefore |T | = |T  |, where T  := {g ∈ λ (λ × κ) : g is injective}. Find an injection from λ κ into T  . 37) Assume that κ is infinite. Let T = {f ∈ κ κ : f is bijective}. Then |T | = 2κ . Hint: First κ can be partitioned into two disjoint sets B and C of cardinality κ. Every injection from B into B can be extended to a bijection from A onto A. Now apply the previous exercise. 38) If A is infinite, then there are 2|A| equivalence relations on A. 39) Assume that H is a set and (H, ≤) is a linear ordering. Prove that cf (H, ≤ ) equals 0 or 1 or is a regular cardinal. 40) Give an example of a partial ordering which has singular cofinality. 41) Assume that τ is a limit ordinal, (αξ : ξ < τ ) is a strictly increasing sequence of ordinals, and α = sup{αξ : ξ < τ }. Prove that cf (α) = cf (τ ). 42) Prove that |α| ≤ α < |α|+ for every ordinal α. 43) Assume that α is infinite. Show that α + 1 is not a cardinal. 44) Prove: If a is an infinite set of ordinals, then |a| ≤ sup(a). 45) Let κ be a cardinal. Prove that α < κ+ iff |α| ≤ κ. 46) Prove: If f : B −→ V is a function, then |ran(f )| ≤ |B|. 47) Assume that κ is an infinite cardinal and ≺ is the lexicographical ordering on κ {0, 1}. Prove that there is no strictly increasing or strictly decreasing sequence of length κ+ . Hint: Assume that f¯ = (fα : α < κ+ ) is strictly increasing under ≺. For each α < κ there is an ordinal α∗ < κ such that fα |` α∗ = fα+1 |` α∗ , fα (α∗ ) = 0 and fα+1 (α∗ ) = 1, since fα ≺ fα+1 . Use the regularity of κ+ to obtain γ < κ such that |Aγ | = κ+ for Aγ := {α < κ : α∗ = γ}. Let o(f¯) be the least ordinal γ with this property and take f¯ such that o(f¯) is minimal. If α, β ∈ Ao(f¯) and α < β, then fα+1 (o(f¯)) = 1, fβ (o(f¯)) = 0 and α + 1 < β. So consider the subsequence g¯ of f¯ having the members g ) < o(f¯). fα and fα+1 for all α ∈ Ao(f¯) to obtain the contradiction o(¯ 48) Prove: If 2ℵ0 > ℵω , then ℵℵω0 = 2ℵ0 . 49) Assume that κ is a cardinal, and define κ0 := κ, κn+1 := ℵκn for each n ∈ ω, and μ := sup{κn : n ∈ ω}. Show that μ is the smallest fixed point of the aleph function which is greater or equal than κ. In particular, for κ = ω, we obtain the smallest fixed point of the aleph function.

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Chapter 1. Foundations

1.6 Infinite Sums and Products Definition Let (κi : i ∈ I) be a family of cardinal numbers. Generalizing the definition in Section 1.5, we define the sum and the product of this family as follows:



κi := | {κi × {i} : i ∈ I}|, (κi : i ∈ I) := i∈I

   (κi : i ∈ I) := κi := |{f : f : I −→ {κi : i ∈ I} ∧ ∀i ∈ I f (i) ∈ κi }|. i∈I

The product of a family (κi : i ∈ I) of cardinal numbers is defined as the cardinality of the cartesian product of the family  (κi : i ∈ I) of sets. Thus we obtain a certain ambiguity of the denotation i∈I κi , with which we will put up, since it will not lead to serious confusions. The context  will make plain the intended meaning. So it should be clear that in “f ∈ i∈I κi ” the cartesian product is meant. The following lemma summarizes some laws for the above cardinal operations. It is left as an exercise for the reader to find the suitable bijections and injections. Lemma 1.6.1 Let (κi : i ∈ I) and (λi : i ∈ I) be families of cardinal numbers. Then a) If, for every i ∈ I, xi is a set with |xi | = κi , then

κi = |

   {xi × {i} : i ∈ I}| and κi = | xi |.

i∈I

i∈I

i∈I

  κi = 0, i∈{0,1} κi = κ0 · κ1 , i∈∅ κi = 1.   c) If κi ≤ λi for all i ∈ I, then i∈I κi ≤ i∈I λi and i∈I κi ≤ i∈I λi .  |I| d) i∈I κ = κ · |I| and i∈I κ = κ . b)



i∈{0,1}

κi = κ0 + κ1 ,



i∈∅

Lemma 1.6.2 Let (κi : i ∈ I) be a family of cardinal numbers, and let λ be a cardinal number. Then the following laws hold. a) If (Xj : j ∈ J) is a partition of the index set I, then (i) (General commutative and associative laws)

i∈I

κi =

j∈J i∈Xj

κi

and

 i∈I

κi =

  j∈J i∈Xj

κi .

1.6. Infinite Sums and Products

59

(ii) (Distributive law) 

κi =

j∈J i∈Xj

  ( κf (j) : f ∈ Xj ) . j∈J

j∈J

κi = i∈I λ · κi .  c) λ = i∈I λκi .   d) ( i∈I κi )λ = i∈I κλi . b) λ ·



i∈I i∈I κi

Proof Again it is left to the reader to find the desired bijections.We will only prove the and will specify a bijection  distributive law   φ from j∈J Bj onto the set {Cf × {f } : f ∈ j∈J Xj }, where Bj := {κi × {i} : i ∈ Xj } and  Cf := j∈J κf (j) . Then the assertion follows from Lemma 1.6.1. Fix F ∈ j∈J Bj . Then, for each j ∈ J, there is a unique  i ∈ Xj , say f (j), such that F (j) ∈ κi × {i}. Put φ(F ) := (h, f ) where h ∈ j∈J κf (j) is given by: h(j) is the first component of F (j). Lemma 1.6.3 Let (κi : i ∈ I) be a family of cardinal numbers. Then a) sup{κi : i ∈ I} ≤ i∈I κi . b) If κi > 0 for all i ∈ I, I = ∅ and at least one of the cardinals |I| and κi (i ∈ I) is infinite, then (i) i∈I κi = max{|I|, sup{κi : i ∈ I}} = |I| · sup{κi : i ∈ I}. (ii) If in addition |I| ≤ sup{κi : i ∈ I}, then

κi = sup{κi : i ∈ I}.

i∈I

In particular this holds if the given family is injective. Proof The second equation in b) (i) holds by Corollary 1.5.12. The assertion “≤” in the first equation follows from Theorem 1.5.14. Since κi > 0 for 1 ≤ κ all i ∈ I, we have |I| = i∈I i∈I i . Furthermore it is clear by definition that κj ≤ i∈I κi for every j ∈ I. This gives “≥” and also proves part a). The first equation in b) (ii) follows immediately from (i). Now assume that κi = κj for all i, j ∈ I with i = j. Let f be the order isomorphism from the order type α of the well-ordered set K := {κi : i ∈ I} onto this set. Then |I| ≤ α and f (γ) ≥ γ for every γ < α. From this we can conclude that sup{κi : i ∈ I} = sup(ran(f )) ≥ α ≥ |I|, if α is a limit ordinal. If α = β + 1 and α < ω, then sup{κi : i ∈ I} = f (β) ≥ ω > α = |I|. Otherwise we have α = β + 1, α ≥ ω and sup(ran(f )) = f (β) ≥ β ≥ |α| ≥ |I|.

60

Chapter 1. Foundations

Corollary 1.6.4 Let α and β be ordinals. a) If β is a limit ordinal, (σξ : ξ < β) is a strictly increasing sequence of : ξ < β}, then ordinals and α = sup{σ ξ ξ<β ℵσξ = ℵα . In particular, ℵ = ℵ . ξ β ξ<β b) ξ<α+1 ℵξ = ℵα . Definition If κ and λ are cardinal numbers, then put κ<λ := sup{κν : ν ∈ CN ∧ ν < λ}. If a is a set and λ is a cardinal number, then let [a]<λ := {x ⊆ a : |x| < λ} be the set of all subsets of a whose cardinality is less than λ. In complete analogy the sets [a]≤λ , [a]>λ and [a]≥λ are defined. The set [a]λ has already been defined in Section 1.5. The next corollary establishes the relation between κ<λ and [a]<λ analogously to Corollary 1.5.13. Corollary 1.6.5 a) If κ and λ are cardinal numbers such that κ ≥ 2 and λ ≥ ω, then κ<λ ≥ λ. b) If a is an infinite set, |a| = κ and λ is a cardinal number with 2 ≤ λ ≤ κ, then for all cardinals ν0 < λ

κν = κν .16 κ<λ = |[a]<λ | = ν∈λ∩CN

ν0 ≤ν∈λ∩CN

Proof For the proof of part a), first let λ be a successor cardinal μ+ . Then κ = κμ ≥ 2μ > μ by Lemma 1.5.6. So we get κ<λ ≥ μ+ = λ. Now let λ be a limit cardinal. If ν ∈ λ ∩ CN, then κν ≥ 2ν > ν. From this we can conclude that κ<λ = sup{κν : ν ∈ λ ∩ CN} ≥ sup{ν : ν ∈ λ ∩ CN} = λ. Next we prove b). Corollary 1.5.13 says that the cardinality of [a]ν is κν for cardinal numbers ν ≤ κ and 0 for cardinal numbers ν > κ. Therefore |[a]<λ | = ν ν∈λ∩CN κ . For each cardinal ν0 with ν0 < λ, it follows from Lemma 1.6.3 and 2 ≤ λ ≤ κ that this sum equals sup{κν : ν0 ≤ ν ∈ λ ∩ CN}. In particular, for ν0 = 0, we get κ<λ = sup{κν : ν ∈ λ ∩ CN} = |[a]<λ |. <λ

Corollary 1.6.6 If a is an infinite set and λ is a cardinal number with λ ≤ |a|, then a) |[a]λ | = |a|λ . b) |[a]<λ | = sup{|a|ν : ν ∈ CN ∩ λ}; in particular, |[ω]<ω | = ℵ0 . c) |[a]≤λ | = |a|λ . the literature, also the cardinal number ν∈λ∩CN κν instead of κ<λ is investigated. Further results on κλ can be found in Section 1.7. 16 In

1.6. Infinite Sums and Products

61

Proof Part a) is Corollary 1.5.13, part b) can be found in Corollary 1.6.5 b), and part c) follows at once from b) and c) using [a]≤λ = [a]λ ∪ [a]<λ . Theorem 1.6.7 Assume that (κi : i ∈ I) and (λi : i ∈ I) are families of cardinal numbers. Then the following holds.  a) If λi ≥ 2 and κi ≤ λi for all i ∈ I, then i∈I κi ≤ i∈I λi . b) (J. K¨ onig’s Lemma)  If κi < λi for all i ∈ I, then i∈I κi < i∈I λi . Proof First we prove a) and the relation “≤” of b).  Let I1 := {i ∈ I : λi ≥ ω} and I2 := {i ∈ I : λi < ω}. Since i∈I κi = | {κi × {i} : i ∈ I}|, we  assign uniquely to each pair (α, j) with α ∈ κj and j ∈ I a function fα,j ∈ i∈I (λi + 1) by fα,j (i) := κi , if i = j; fα,j (j) := α . It is easy to verify that the functions (fα,j : α ∈ κj ∧ j ∈ I) and (fα,j |`I1 : α ∈ κj ∧ j ∈ I1 ) are injective. This gives

κi ≤ |

i∈I1



(λi + 1)| =

i∈I1



λi .

i∈I1

Furthermore the same argument, applied to (κi : i ∈ I) using the first function, proves the assertion “≤” in b), since κi < λi for all i ∈ I implies that fα,j ∈  i∈I λi . If I2 is infinite, then, by assumption,

κi ≤ ℵ0 · |I2 | = |I2 | < 2|I2 | ≤ |

i∈I2

 i∈I2

If I2 is finite, then it follows from λi ≥ 2 for all i that Altogether we obtain from Lemma 1.5.10:

i∈I

κi =

i∈I1

κi +

i∈I2

κi ≤

 i∈I1

λi +



λi | =

 i∈I2



λi .

i∈I2

i∈I2

λi ≤

κi ≤



 i∈I2

λi .

λi .

i∈I

For the proof get a contradiction, that there is a  of b) we assume, to  bijection h from {κi × {i} : i ∈ I} onto i∈I λi . Then this cartesian product is the union of the pairwise disjoint sets Ai := h[κi × {i}], i ∈ I. Put Bi := λi ∩ {f (i) : f ∈ Ai }. Since |Bi | ≤ |Ai | = κi < λi , Bi is a proper subset  of λi for every i ∈ I. Therefore the axiom of choice yields a function g ∈ i∈I λi such that g(i) ∈ λi \Bi for every i ∈ I. So we get the contradiction g ∈ {Ai : i ∈ I}. Corollary 1.6.8 Let β be an infinite ordinal and (κξ : ξ < β) be a sequence of nonzero cardinal numbers.

62

Chapter 1. Foundations a) If (κξ : ξ < β) is strictly increasing and β is a limit ordinal, then



κξ < κξ and κξ < ( κξ )|β| . ξ<β

ξ<β

ξ<β

ξ<β

b) If κξ ≥ 2 for every ξ < β, then

 ( κξ )|β| = ( κξ )|β| . ξ<β

ξ<β

Proof The first inequality in a) follows from K¨ onig’s lemma, if we put take into consideration that the factor κ0 ≥ 1 does not make λξ := κξ+1 and  the product 1≤ξ<β κξ smaller. Since κξ ≤ ξ<β κξ for each ξ < β, we get  |β| ξ<β κξ ≤ ( ξ<β κξ ) . Now the second inequality in a) follows from the first one. With the help of Theorem 1.6.7 it can be shown, using the same of part a) argument, that ξ<β κξ ≤ ξ<β κξ ≤ ( ξ<β κξ )|β| . Hereby we obtain 

κξ )|β| ≤ ( κξ )|β| ≤ ( κξ )|β|·|β| , ( ξ<β

ξ<β

ξ<β

and with |β| · |β| = |β| part b) is proved. Theorem 1.6.9 Let κ be a cardinal number. a) If κ ≥ 2, then ℵα < cf (κℵα ). b) If κ is infinite, then κ < κcf(κ) . Proof For a), assume that γ ≤ ℵα and f is a function from γ into κℵα . Then we can infer from Theorem 1.6.7 and Lemma 1.5.10 that

 | sup(ran(f ))| ≤ |f (β)| < κℵα = (κℵα )|γ| = κℵα ·|γ| = κℵα . β<γ

β<γ

From this we get at once cf (κℵα ) > ℵα . For b), assume that f : cf (κ) −→ κ is a function such that sup(ran(f )) = κ. As above, we obtain

 κ≤ |f (β)| < κ = κcf(κ) . β
β
Corollary 1.6.10 a) For any α ∈ ON, cf (2ℵα ) > ℵα . In particular we have 2cf(ℵα ) = ℵα . ℵ

b) If α is a limit ordinal and ℵβ ≥ cf (α), then ℵα = ℵγ β for all ordinals γ; ℵ in particular we have ℵα < ℵαβ .

1.6. Infinite Sums and Products

63

Proof Part a) follows immediately from the previous theorem. So let us prove b). Since cf (ℵα ) = cf (α), it follows from Theorem 1.6.9 that ℵ

α) = ℵcf(α) ≤ ℵαβ . ℵα < ℵcf(ℵ α α

ℵ

ℵ

ℵ

The assumption ℵα = ℵγ β would yield the contradiction ℵαβ = ℵγ β = ℵα . Remark Corollary 1.6.10 tells us that the assertion ℵ0 < cf (2ℵ0 ) is provable in ZFC. Essentially this is the only limitation which is set by the axioms of ZFC for the cardinality of the continuum. The continuum function (2κ : κ ∈ ICN) satisfies two essential laws: (i) κ < μ =⇒ 2κ ≤ 2μ

(monotonicity) and

κ

(ii) κ < cf (2 ). If F : Reg −→ ICN is an arbitrary increasing function satisfying κ < cf (F (κ)) for all κ ∈ Reg, then by the results of W.B. Easton [Ea] it is consistent with the axioms of ZFC that 2κ = F (κ) for all κ ∈ Reg. This means that the axioms of ZFC set no limitations on the continuum function on the regular cardinals except the conditions (i) and (ii). Lemma 1.6.11 Let κ be an infinite cardinal number. a) The following assertions are equivalent. (i) κ is regular. (ii) For any sequence (Sα :  α < γ) of sets, satisfying γ < κ and |Sα | < κ for all α < κ, we have | {Sα : α < κ}| < κ (i.e., κ is not a union of less than κ sets, all of which are of cardinality less than κ). (iii) κ is not a sum of less than κ cardinal numbers, all of which are less than κ. b) Assume that κ is an infinite cardinal. Then κ is a sum of cf (κ) cardinals less than κ. If κ is a limit cardinal, then there is a strictly increasing sequence (λξ : ξ < cf (κ)) of cardinal numbers less than κ such that κ=

λξ = sup{λξ : ξ < cf (κ)} .

ξ
For singular cardinals κ we can arrange that cf (κ) < λξ < κ for all ξ < cf (κ). If in addition cf (κ) > ℵ0 , then there also exists a normal sequence of singular cardinals with these properties.

64

Chapter 1. Foundations

Proof For the proof of a), “(ii)=⇒(i)”, assume that κ is a singular cardinal  and f : cf (κ) −→ κ is a function such that ran(f ) = κ. Put Sα := f (α) for all α < cf (κ). Then (Sα : α < cf (κ)) is a sequence of less than κ sets, all of which are of cardinality < κ. The union of these sets is κ. cardinal numbers less For “(i)=⇒(iii)”, let (κi : i < γ) be a sequence of than κ with domain γ < κ. Then Theorem 1.5.14 yields i<γ κi ≤ |γ| ·sup{κi : i < γ} < κ, since κ is regular.  For “(iii)=⇒(ii)” note that the cardinality of {Sα : α < γ} is at most α<γ |Sα |. Next we turn to the proof of b). Observe that for regular cardinals κ and any ν ∈ CN \ {0} ∩ κ we have κ = ξ<κ ν. Now let κ be a limit cardinal. If, according to Lemma 1.5.15, f : cf (κ) −→ κ is a normal function, i.e., if f is strictly increasing and continuous, such that sup(ran(f )) = κ, then the set c := {|f (β)| : β < cf (κ)} is cofinal in κ. Otherwise the regular cardinal number (sup(c))+ would be smaller than κ and thus be dominated by some value f (σ), and also by |f (σ)|, which would contradict the definition of c. Therefore we obtain by transfinite recursion a strictly increasing continuous sequence (λξ : ξ < cf (κ)) of members of c with supremum κ; from the definition of cofinality we can conclude that it has length cf (κ). Using Lemma 1.6.3, we get κ = ξ cf (κ)}. If in addition we have cf (κ) > ℵ0 , then the cardinals λξ can be chosen as singular. Namely, let λξ+1 for ξ < cf (κ) be the supremum of some strictly increasing sequence in c of length ω having max{λξ , |f (ξ)|} as its first member, let λ0 be defined analogously, and put, for limit ordinals γ < cf (κ), λγ := sup{λξ : ξ < γ}. Then we have, for any limit ordinal γ < cf (κ), cf (λγ ) ≤ γ < cf (κ) < λ0 < λγ , and λγ is singular. Since cf (κ) > ℵ0 , all cardinals λξ are singular. The construction shows that the sequence (λξ : ξ < cf (κ)) is normal and that sup{λξ : ξ < cf (κ)} = κ. Theorem 1.6.12 Let α and β be ordinals. ℵ

a) If α ≤ β, then ℵαβ = 2ℵβ . b) (Hausdorff ’s Formula) ℵ

ℵ

β = ℵα+1 · ℵαβ . ℵα+1

c) (Tarski’s Formula) If |γ| ≤ ℵβ , then ℵ

|γ|

ℵ

β ℵα+γ = ℵα+γ · ℵαβ .

Proof For part a), we can conclude from Lemma 1.5.10 and Corolℵ lary 1.5.12 that 2ℵβ ≤ ℵαβ ≤ (2ℵα )ℵβ = 2ℵα ·ℵβ = 2ℵβ .

1.6. Infinite Sums and Products

65

For b), let first β ≤ α. Then ℵβ ≤ ℵα , and since ℵα+1 is regular, every function from ℵβ into ℵα+1 is already a function from ℵβ into γ for some γ < ℵα+1 . This gives ℵ

β = |ℵβ ℵα+1 | = | ℵα+1

 {ℵβ γ : γ < ℵα+1 }| ≤

ℵ

ℵ

β |γ|ℵβ ≤ ℵα+1 · ℵαβ ≤ ℵα+1 .

γ<ℵα+1

If α < β, then ℵα+1 ≤ ℵβ < 2ℵβ , and the assertion follows from a) and Corollary 1.5.12. Now we prove part c) by transfinite induction on γ. If γ = 0, there is nothing to show. For the successor step, we can infer from part b), the inductive hypothesis, and Corollary 1.5.12 that ℵ

ℵ

ℵ

|γ|

ℵ

|γ+1|

β β ℵα+γ+1 = ℵα+γ · ℵα+γ+1 = ℵαβ · ℵα+γ · ℵα+γ+1

ℵ

|γ+1|

= ℵαβ · ℵα+γ · ℵα+γ+1 = ℵαβ · ℵα+γ+1 . Now let γ be a limit ordinal. Then from Theorem 1.6.7 and the continuity of ordinal addition we can conclude that

 ℵα+σ ≤ ℵα+σ . ℵα+γ = sup{ℵα+σ : σ < γ} ≤ σ<γ

σ<γ

From the inductive hypothesis, Lemma 1.5.10, and Lemma 1.6.1 we can infer that   ℵ  ℵ ℵβ |σ| β ℵα+γ ≤( ℵα+σ )ℵβ = ℵα+σ = (ℵαβ · ℵα+σ ) σ<γ

≤

ℵ (ℵαβ )|γ|



σ<γ |σ| ℵα+σ

≤

σ<γ ℵ ·|γ| ℵαβ

ℵ

|γ|·|γ|

|γ|

ℵ

β · ℵα+γ = ℵαβ · ℵα+γ ≤ ℵα+γ .

σ<γ

The last equation and the last inequality hold because, by assumption, |γ| ≤ ℵβ . Corollary 1.6.13 If n ∈ ω, then the following holds. ℵ

ℵ

β a) (Generalized Hausdorff Formula) ℵα+n = ℵαβ · ℵα+n .

ℵ

b) (Bernstein’s Formula) ℵnβ = 2ℵβ · ℵn . ℵ

|α|

c) If α ≤ ℵβ , then ℵαβ = 2ℵβ · ℵα . Examples 1.6.14 a) ℵℵ1 0 = 2ℵ0 · ℵ1 follows from Corollary 1.6.13, and thus ℵℵ1 0 = 2ℵ0 . b) ℵℵω1 = 2ℵ1 · ℵℵω0 .

66

Chapter 1. Foundations c) If 2ℵ1 = ℵ2 , then ℵℵω0 = ℵω1 . To see this, we conclude with Example b) from 2ℵ1 = ℵ2 that ℵℵω0 = ℵℵω1 . Theorem 1.6.9 gives cf (ℵω1 ) = ℵ1 < cf (ℵℵω1 ). Now the assertion follows. ℵ

d) For any α with ω ≤ α < ℵ1 , we have ℵαβ = 2ℵβ · ℵℵα0 . ℵ

e) If n ∈ ω and ℵωn = ℵγ β , then β ∈ {0, . . . , n − 1}. This follows from Theorem 1.6.9, because, under the given assumptions, we have cf (ℵωn ) = ℵn > ℵβ , hence n > β. In particular ℵℵωm = ℵωn for all m ≥ n and, in view of the surprising role of ℵω4 in Shelah’s Theorem 8.1.4 in Chapter 8, ℵℵωm = ℵω4 for all m ≥ 4. f) 2ℵω = ℵω4 follows from Corollary 1.6.10. Lemma 1.6.15 (The four golden rules of cardinal arithmetic) a) (Tarski) If ν ≥ ℵ0 is a cardinal number and (κξ : ξ < ν) is an increasing sequence of infinite cardinals, then  κξ = (sup{κξ : ξ < ν})ν . ξ<ν

b) If λ and κ are cardinal numbers such that κ ≥ ℵ0 and λ ≥ cf (κ),then κλ = (sup{ν λ : ν ∈ CN ∩ κ})cf(κ) . c) (Tarski) If λ and κ are cardinal numbers such that κ ≥ ℵ0 and 0 < λ < cf (κ), then

κλ = κ · sup{ν λ : ν ∈ CN ∩ κ} = κ · νλ . ν∈κ∩CN

d) If κ is an infinite cardinal number, then 2κ = (2<κ )cf(κ) . Proof First we  prove part a). Since |ν × ν| = ν, we can represent ν as a disjoint union ν = {Bα : α < ν} of subsets Bα of ν such that |Bα | = ν for all α < ν. Then  we get μ := sup{κξ : ξ < ν} = sup{κξ : ξ ∈ Bα } for every α < ν, and μ ≤ ξ∈Bα κξ . Consequently, Lemma 1.6.2 yields      μ≤ κξ = κξ ≤ μ ≤ μν . μν = α<ν

α<ν ξ∈Bα

ξ<ν

ξ<ν

For b), we can choose a sequence (κξ : ξ < cf (κ)) of cardinals κξ with 2 ≤ κξ < κ satisfying κ = ξ
ξ
ξ
1.6. Infinite Sums and Products

67

Next we prove c). If f ∈ λ κ, then there is an α < κ such that f ∈ λ α, since λ < cf (κ). Furthermore we have |α|λ = ν λ for all ν ∈ κ ∩ CN with ω ≤ ν ≤ α < ν + , hence by Lemma 1.6.3 

|α|λ ≤ ν+ · νλ κλ ≤ | {λ α : α < κ}| ≤ ≤κ ·

α<κ

ν∈κ∩CN

ν λ ≤ κ · sup{ν λ : ν ∈ κ ∩ CN} ≤ κλ .

ν∈κ∩CN

For d), we note that the assertion is easy to show for successor cardinals. So let us assume that κ is a limit cardinal. We can apply Lemma 1.6.11 and obtain a strictly increasing sequence (κξ : ξ < cf (κ)) of cardinals less than κ such that κ = ξ
ξ
Corollary 1.6.16 Let κ be an infinite cardinal number and (κξ : ξ < cf (κ)) be a sequence of cardinal numbers such that κ = ξ
ξ
ξ
Part b) follows from part a) together with Corollary 1.5.12 and Lemma 1.6.2. Note that in Lemma 1.6.15 a) we have already proved assertion b) of the following theorem for cardinal numbers β. A slight modification of this proof shows that this assertion is true for any limit ordinal β such that β ≤ ω1 . Tarski believed that it is true for arbitrary limit ordinals. Remark 1.6.18 shows that it is consistent with the axioms of ZFC if ZF is consistent, using the famous result of K. G¨ odel that ZFC + (GCH) is consistent only if ZF is consistent. It was not until the results of P. J. Cohen that this hypothesis was refuted; Magidor [Ma] produced a model with a counterexample for β = ω1 + ω, and Jech and Shelah [JS] showed that if there is any counterexample, then there is one with β = ω1 + ω.

68

Chapter 1. Foundations

Theorem 1.6.17

 |β| a) For any ordinal β > 0, ξ≤β ℵξ = ℵβ . For limit ordinals β, also  |β| ξ<β ℵξ = ℵβ . b) If β is a principal number of addition and (αξ : ξ < β) is a strictly increasing sequence of ordinals with supremum α, then  ℵαξ = ℵ|β| α . ξ<β

Proof We start with the proof of part b). So let β > 1 be a principal number of addition and (αξ : ξ < β) be strictly increasing such that sup{αξ : ξ < β} = α. Then we show first:   ℵαξ )|β| = ℵαξ . (∗) ( ξ<β

ξ<β

For the proof of (∗), we put κξ := ℵαξ and τξ,η := ξ for all η, ξ < β. Since, by assumption and Lemma 1.4.7 a), β is not a sum of smaller ordinals, the natural sum ξ#η is less than β for all ξ, η < β. Further we have ξ ≤ ξ#η and η ≤ ξ#η. This gives {τξ,η : ξ, η < β} = {τξ,η : ζ < β ∧ ξ#η = ζ}. By the definition of the natural sum,the equation ξ#η = ζ has for a given ζ only finitely many solutions, hence ξ#η=ζ κτξ,η = κτζ,0 = κζ by Corollary 1.5.12. From this we can conclude with a general associative law that   |β|   ( κξ )|β| = κξ = ( κξ ) ξ<β

ξ<β

ξ<β η<β

=

     ( κτξ,η ) = ( κτξ,η ) = κζ ,

ξ<β η<β

ζ<β ξ#η=ζ

ζ<β

which proves (∗). Now the sequence (αξ : ξ < β) is strictly increasing and the aleph function is continuous, and thus sup{ℵαξ : ξ < β} = ℵα and β = sup{ξ : ξ < β} ≤ sup{αξ : ξ < β} = α. By Lemma 1.6.3 we can infer that ξ<β ℵαξ = ℵα . From  |β| (∗) and Corollary 1.6.8 it follows that ℵα = ( ξ<β ℵαξ )|β| = ( ξ<β ℵαξ )|β| =  ξ<β ℵαξ . This completes the proof of b). Next we turn to the proof of the second equation in part a). To get a contradiction, we assume that it is false and choose β as the smallest limit ordinal for which it fails. Then part b) tells us that β is not additively indecomposable. If we take the Cantor normal form of β for the base ω, β = i 1, since ω δ is additively indecomposable for every δ > 0, β0 ≥ . . . ≥ βn−1 , and βn−1 > 0. Consequently, γ := ω βn−1 is a principal number of addition,

1.6. Infinite Sums and Products

69

and β has the form β = σ + γ. We have γ ≤ σ < β and |β| = |σ|, since the cardinality of the ordinal sum of two infinite ordinals equals the maximum of the cardinalities of the summands. By the choice of β, we have   |σ| |β| |γ| ξ<σ ℵξ = ℵσ = ℵσ , and by part b), ξ<γ ℵσ+ξ = ℵβ . So we can con   |β| |γ| clude that ξ<β ℵξ = ξ<σ ℵξ · ξ<γ ℵσ+ξ = ℵσ · ℵβ . Furthermore, Tarski’s formula in Lemma 1.6.12 yields  |β| |β| |γ| |γ| |β| ℵβ = ℵσ+γ = ℵ|β| · ℵ = ℵ · ℵ = ℵξ . σ σ σ+γ β ξ<β

This contradiction shows that the second equation in a) is true. Now the first equation in a) follows by transfinite induction. For β = 1 it |β| |β| is true, and for limit ordinals β it follows from the second, since ℵβ · ℵβ = ℵβ . Finally we get, by the inductive hypothesis and Hausdorff’s formula,   |β| |β| |β+1| ℵξ = ( ℵξ ) · ℵβ+1 = ℵβ · ℵβ+1 = ℵβ+1 = ℵβ+1 . ξ≤β+1

ξ≤β

Remark 1.6.18 Assuming the generalized continuum hypothesis, we can conclude for limit ordinals β and strictly increasing sequences (αξ : ξ < β) with supremum α that  ℵαξ = ℵ|β| ℵα+1 = α . ξ<β

To see this we infer from Corollary 1.6.4, K¨ onig’s lemma, and its corollary, that ℵα = ξ<β ℵαξ < ξ<β ℵαξ . From this we get ℵα <



|α| ℵα ℵα ℵαξ ≤ ℵ|β| = ℵα+1 , α ≤ ℵα ≤ ℵα = 2

ξ<β

and this immediately yields the desired claim. Exercises 1) Prove Lemma 1.6.1. 2) Prove Lemma 1.6.2. 3) Assume that κ is an infinite cardinal and n ∈ ω\{0}. Show that ξ<κ n = κ. 4) Assume that κ ∈ ICN, μ ∈ cf (κ) ∩ CN and (κξ : ξ < μ) is a sequence of cardinal numbers less than κ. Prove that ξ<μ κξ < κ.  5) Show that ℵℵω0 = n∈ω ℵn . 6) Assume that κ is an infinite cardinal. Show that (κ+ )κ = 2κ .

70

Chapter 1. Foundations 7) Assume that κ is an infinite cardinal. Show that 2κ = (

2ν )cf(κ) .

ν∈κ∩CN

 8) Prove that α<ℵ1 +ω ℵα = ℵℵℵ11 +ω . Hint: Use Tarski’s formula. 9) Assume that σ is a limit ordinal and (η(ξ) : ξ < σ) is a normal function. a) Prove: sup{ ξ<τ ℵη(ξ) : τ < σ} = ξ<σ ℵη(ξ) .   b) Refute: sup{ ξ<τ ℵη(ξ) : τ < σ} = ξ<σ ℵη(ξ) .  10) Prove that ξ<τ 2ℵξ = 2ℵτ for any limit ordinal τ .  Hint: Assign to X ⊆ ℵτ the function hX ∈ ξ<τ P(ℵξ ) given by hX (ξ) := X ∩ ℵξ . 11) Assume that κ is an infinite cardinal such that 2ν < κ for all ν ∈ κ ∩ CN. Prove that cf (κcf(κ) ) > κ. 12) Prove: If κ is regular and 2μ ≤ κ for all cardinals μ < κ, then μ∈κ∩CN κμ = κ. 13) Prove: If κ = λ+ , then μ∈κ∩CN κμ = 2λ . 14) Refute: If κ and λ are cardinal numbers such that 0 < λ < cf (κ), then κλ = ν∈κ∩CN ν λ . 15) Assume that α is a limit ordinal and γ is an ordinal satisfying ℵγ < cf (α). ℵ ℵγ ℵγ Prove that ℵαγ = ξ<α ℵξ = sup{ℵξ : ξ < α} . Hint: Use a golden rule.

1.7 Further Properties of κλ – the Singular Cardinal Hypothesis In this section we will characterize the cardinal power κλ by its values for smaller bases or exponents17 . Furthermore it will be shown that this power is uniquely determined in the axiom system ZF+(SCH) by the function (2λ : λ ∈ ICN), and that it can be easily computed in the system ZFC+(GCH). (SCH) denotes the so-called “singular cardinal hypothesis”, given by (SCH) ∀κ ∈ ICN(2cf(κ) < κ =⇒ κcf(κ) = κ+ ) . 17 The results in 1.6.12, 1.6.13, and 1.6.15 have certainly already dealt with this subject. We needed them to obtain further findings on products of cardinal numbers.

1.7. Further Properties of κλ – the Singular Cardinal Hypothesis

71

Lemma 1.7.1 (Tarski’s Recursion Formula) If β is a limit ordinal, α is the supremum of the strictly increasing sequence (αξ : ξ < β) and ℵγ < cf (β), then

ℵℵαγ = ℵℵαγξ = sup{ℵℵαγξ : ξ < β} . ξ<β

In particular, for limit ordinals α and ordinals γ satisfying ℵγ < cf (α), we have

ℵ ℵ ℵξ γ = sup{ℵξ γ : ξ < α} . ℵℵαγ = ξ<α

Proof By assumption, we have ℵγ < cf (β) = cf (α) = cf (ℵα ). Therefore we get   ℵγ ℵα = {ℵγ ℵξ : ξ < α} = {ℵγ ℵαξ : ξ < β}. Since obviously the cardinality of the index set is at most the supremum of the summands, Lemma 1.6.3 says that

ℵℵαγξ = sup{ℵℵαγξ : ξ < β} ≤ ℵℵαγ . ℵℵαγ ≤ ξ<β

Definition Assume that κ is an infinite and λ is an uncountable cardinal number. We say that λ is κ-strong iff ρκ < λ for every cardinal number ρ less than λ. λ is called a strong limit cardinal iff λ is ν-strong for every cardinal number ν less than λ, that means iff ρν < λ for all cardinal numbers ρ and ν less than λ. It is easy to see that λ is a strong limit cardinal iff 2ν < λ for all cardinal numbers ν less than λ. We say that κ is a (strongly) inaccessible cardinal number iff κ is regular, uncountable and a strong limit cardinal, and a weakly inaccessible cardinal number iff κ is regular, uncountable and a limit cardinal. Remark and Example Obviously we have κ < λ for any κ-strong cardinal λ. The existence of inaccessible cardinal numbers cannot be proved in ZFC. For strongly inaccessible cardinals κ this results from the fact that the von Neumann level Vκ with the ∈-relation is a model of ZFC. Therefore from the existence of an inaccessible cardinal in ZFC the consistency of ZFC could be proved in ZFC, which would contradict a famous theorem of K. G¨ odel concerning the impossibility of certain consistency proofs. For details, the reader is referred to Drake’s book [Dr], Chapter 4. The strong limit cardinals form a proper class. To show this, consider a fixed infinite cardinal number μ. We define by recursion a strictly increasing sequence (κn : n ∈ ω) of cardinal numbers by κ0 := μ and κn+1 := 2κn . Then κ := sup{κn : n ∈ ω} is a strong limit cardinal. Namely, if ρ and ν are smaller

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than κ, then there is a natural number n such that max{ρ, ν} ≤ κn . We can conclude that ρν ≤ κκnn = 2κn = κn+1 < κ. Thus the class of strong limit cardinals is unbounded in the class of cardinals. Furthermore the supremum of a set of strong limit cardinals is also strong limit. Consequently, the class of strong limit cardinals is unbounded and closed in the class of cardinal numbers, which makes the assertion that these cardinal numbers occur very often precise. The same holds, for any infinite cardinal λ, for the class of λ-strong cardinals. Lemma 1.7.2 If κ is a strong limit cardinal, then 2<κ = κ and 2κ = κcf(κ) . Proof The first assertion is easy to show. For the second we can conclude, using Lemma 1.6.15, that κcf(κ) ≤ κκ = 2κ = (2<κ )cf(κ) = κcf(κ) . Lemma 1.7.3 Let α and β be ordinals with β < α.18 Then ℵ

a) If ℵα is ℵβ -strong, then ℵα = sup{ℵγ β : γ < α} and  ℵα if cf (ℵα ) > ℵβ , ℵβ ℵα = cf(α) cf(ℵα ) = ℵα otherwise. ℵα ℵ

ℵ

ℵ

b) If γ < α and ℵα ≤ ℵγ β , then ℵαβ = ℵγ β . ℵ

c) Assume that ℵαβ > 2ℵβ 19 , ℵα is not ℵβ -strong, and γ < α is the smallest ℵ ordinal such that ℵγ β ≥ ℵα . Then ℵγ is ℵβ -strong and singular, and ℵ cf(ℵ ) satisfies cf (ℵγ ) ≤ ℵβ < ℵγ ; furthermore, ℵαβ = ℵγ γ . Proof First let, for the proof of a), α be a limit ordinal. From the asℵ sumption of a) we can conclude that ℵα = sup{ℵγ : γ < α} ≤ sup{ℵγ β : γ < α} ≤ ℵα , and thus ℵ

ℵα = sup{ℵγ β : γ < α}.  Let first cf (ℵα ) > ℵβ . Then ℵβ ℵα = {ℵβ ℵγ : γ < α}, and Lemma 1.6.3 says that ℵ ℵ ℵαβ ≤ |α| · sup{ℵγ β : γ < α} = ℵα . (∗)

Now assume that cf (ℵα ) ≤ ℵβ . By assumption, we have ℵβ < ℵα , hence ℵα is singular and α is a limit ordinal. From Lemma 1.6.15 and (∗) we can conclude that ℵ ℵ α) = ℵcf(α) . ℵαβ = (sup{ℵγ β : γ < α})cf(ℵα ) = ℵcf(ℵ α α ℵ

ℵ

β If α is the successor ordinal σ + 1, then ℵσ+1 = ℵσ+1 · ℵσ β = ℵσ+1 , since ℵα is ℵβ -strong.

ℵ

α ≤ β it follows from Theorem 1.6.12 that ℵαβ = 2ℵβ . 19 Certainly this is the case if ℵ > 2ℵβ . α 18 For

1.7. Further Properties of κλ – the Singular Cardinal Hypothesis ℵ

ℵ

ℵ

73

ℵ

Part b) follows from ℵγ β ≤ ℵαβ ≤ (ℵγ β )ℵβ = ℵγ β . For the proof of part c), we choose the smallest ordinal γ < α satisfying ℵβ ℵ ℵ ℵ ℵγ ≥ ℵα . Then ℵγ β = ℵαβ , and β < γ, since otherwise we would get ℵαβ = ℵβ ℵ β ℵγ = 2ℵβ , contradicting the assumption 2ℵβ < ℵα . Furthermore we have ℵ ℵ ℵ ℵ ℵ ℵσβ < ℵα and thus ℵσ β < ℵαβ = ℵγ β for all σ < γ. The supposition ℵσ β ≥ ℵγ ℵ ℵ for some σ < γ would give, together with part b), the contradiction ℵσβ = ℵγ β . ℵβ Therefore we get ℵσ < ℵγ for all σ < γ, and ℵγ is ℵβ -strong. The supposition ℵ ℵ cf (ℵγ ) > ℵβ leads with part a) to ℵαβ = ℵγ β = ℵγ < ℵα . So there only remains ℵ ℵ cf(ℵ ) cf (ℵγ ) ≤ ℵβ < ℵγ , and part a) yields ℵαβ = ℵγ β = ℵγ γ . ℵ

Theorem 1.7.4 (Jech) Let α and β be ordinals. If ℵαβ ∈ {2ℵβ , ℵα }, then there is an ordinal γ ≤ α with cf (ℵγ ) ≤ ℵβ < ℵγ , such that ℵγ is ℵβ -strong and ℵ cf(ℵ ) satisfies ℵαβ = ℵγ γ . Therefore ℵ ℵαβ

⎧ ⎨ =

⎩

2ℵβ ℵα

cf(ℵγ )

ℵγ

or or for some γ ≤ α with cf (ℵγ ) ≤ ℵβ < ℵγ .

ℵ

Proof Let ℵαβ ∈ {2ℵβ , ℵα }. Then, by Theorem 1.6.12, β < α. If ℵα is ℵβ -strong, then the assertion follows from Lemma 1.7.3 a), with γ := α, and ℵ from the assumption that ℵαβ = ℵα . Otherwise, it follows from Lemma 1.7.3 c). Now we turn to the discussion of the singular cardinal hypothesis. Let κ be an infinite cardinal number. By Corollary 1.6.10 we have 2cf(κ) = κ. If 2cf(κ) > κ, then κcf(κ) ≤ (2cf(κ) )cf(κ) = 2cf(κ) , and thus 2cf(κ) = κcf(κ) . In other words, in the case that κ < 2cf(κ) , the cardinal κcf(κ) equals the value of the continuum function at the regular argument cf (κ). Since, from Easton’s result, the continuum function for regular arguments is not subject to any limitations except monotonicity and the condition κ < cf (2κ ), this holds in this case also for κcf(κ) ; so there are models of ZFC in which κcf(κ) > κ+ is valid. However, if 2cf(κ) < κ, then 2cf(κ) < κcf(κ) , i.e., the continuum function and the continuum function for κ (s.b.) differ at the regular argument cf (κ). Since κ < κcf(κ) ≤ κκ = 2κ , it clearly follows from 2κ = κ+ that cf(κ) κ = κ+ . Therefore the singular cardinal hypothesis can be inferred from the generalized continuum hypothesis. If we know the values of the continuum function (2κ : κ ∈ ICN), and if we know the values κcf(κ) for any singular κ, then Theorem 1.7.4 shows that we know the set of values of the function (κλ : κ, λ ∈ ICN). The additional axiom (SCH) makes the course of values of the power function for cardinal numbers, which is a subject of actual research and the main topic of this book,

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easily surveyed (see Theorem 1.7.8). This applies all the more if we take the generalized continuum hypothesis as an axiom – see the next lemma. Corollary 1.7.5 Assume (GCH). Then for all cardinals κ ≥ 2 and λ ≥ ω, ⎧ if λ < cf (κ), ⎨ κ κ+ if cf (κ) ≤ λ < κ, κλ = ⎩ + λ if κ ≤ λ. Proof If κ ≤ λ, then κλ = 2λ = λ+ by Theorem 1.6.12. So let λ < κ; then κ is infinite. If λ < cf (κ), then we can infer from ν λ ≤ max{2ν , 2λ } ≤ max{ν + , λ+ }, using Lemma 1.6.15, that κ ≤ κλ = κ · sup{ν λ : ν ∈ CN ∩ κ} ≤ κ · sup{max{ν + , λ+ } : ν ∈ CN ∩ κ} ≤ κ. If cf (κ) ≤ λ < κ, then we get κ < κcf(κ) ≤ κλ ≤ κκ = 2κ = κ+ , hence κλ = κ+ . Definition The function (κcf(κ) : κ ∈ ICN) is called the gimel function; instead of κcf(κ) we also write ‫(ג‬κ). If κ = 2 or if κ is an infinite cardinal number, then we call the function (κρ : ρ ∈ ICN) the continuum function for κ; if κ = 2, we call it the continuum function. We say that the continuum function for κ is eventually constant below λ iff there is a cardinal number ρ0 with ρ0 < λ, such that κρ = κρ0 for all cardinal numbers ρ with ρ0 ≤ ρ < λ. In this case we also say that the continuum function for κ is constant below λ from ρ0 on. Lemma 1.7.6 Assume that κ and λ are cardinal numbers such that κ ≥ 2 and λ ≥ ω. Then a) Either the continuum function for κ is eventually constant below λ – if this happens at ρ0 < λ, then obviously κ<λ = κρ for every cardinal ρ with ρ0 ≤ ρ < λ – or there is a strictly increasing sequence (λξ : ξ < cf (λ)) which is cofinal in λ, such that the sequence (κλξ : ξ < cf (λ)) is strictly increasing and cofinal in κ<λ . b) In a), it holds that cf (κ<λ ) ≥ λ in the first case and cf (κ<λ ) = cf (λ) in the second case. c) If ν > 0 is a cardinal number, then ⎧ <λ if 0 < ν < cf (λ), ⎨ κ <λ ν λ κ if cf (λ) ≤ ν < λ, (κ ) = ⎩ ν if λ ≤ ν. κ d) Assume (GCH). Then κ<λ

⎧ ⎨ κ κ+ = ⎩ λ

if λ ≤ cf (κ), if cf (κ) < λ ≤ κ+ , if κ+ < λ.

1.7. Further Properties of κλ – the Singular Cardinal Hypothesis

75

Proof If λ is a successor cardinal μ+ , then κ<λ = κμ , and clearly the first assertion of the alternative is true. Now let λ be a limit cardinal and assume that the first assertion is not true for λ. Then, for each cardinal ρ < λ, there is a cardinal ν such that ρ < ν < λ and κρ < κν . If ρ runs through the values of a sequence of length cf (λ) which is cofinal in λ, then, by the previous condition, we obtain by recursion a strictly increasing subsequence (λξ : ξ < cf (λ)) of this sequence which is cofinal in λ and therefore has also length cf (λ) and the further property that the sequence (κλξ : ξ < cf (λ)) is strictly increasing. By construction, the last sequence must be cofinal in κ<λ . Now we prove part b). Assume that the first assertion in a) is true. If λ = ℵ0 , then it is easy to see that κ ≥ ω and κ<λ = κ. So we get cf (κ<λ ) = cf (κ) ≥ λ. Now let λ > ℵ0 . Theorem 1.6.9 tells us that cf (κ<λ ) = cf (κρ ) > ρ for every infinite cardinal number ρ with ρ0 ≤ ρ < λ. Thus we get cf (κ<λ ) ≥ λ. If the second assertion holds, the cardinal κ<λ has cofinality cf (λ), since it is the supremum of the strictly increasing sequence (κλξ : ξ < cf (λ)) and since cf (λ) is regular. Next we turn to the proof of part c). Let ν be a cardinal number, and assume first that the continuum function for κ is constant below λ from ρ0 on. If 0 < ν < cf (λ), then we choose a cardinal number ρ1 such that max{ρ0 , ν} ≤ ρ1 < λ. This gives (κ<λ )ν = (κρ1 )ν = κρ1 ·ν = κρ1 = κ<λ . For the case λ ≤ ν we obtain in the same way (κ<λ )ν = κρ0 ·ν = κν . If cf (λ) ≤ ν < λ, then we know that λ is singular. λ can be represented by Lemma 1.6.11 as a sum ξ
ξ
and thus (κ<λ )ν = (κλ )ν = κλ . Now we consider the case that in the alternative in a) the second assertion is true. Let the sequences in this assertion be given. Since cf (λ) = cf (κ<λ ) ≤ <λ λξ <λ κ = sup{κ : ξ < cf (λ)}, Lemma 1.6.3 says that κ = (κλξ : ξ < cf (λ)). Let ν < cf (λ) = cf (κ<λ ) and, without loss of generality, ν ≤ λ0 . Then, by Lemma 1.6.15 c),

(κ<λ )ν = κ<λ · (μν : μ ∈ CN ∩ κ<λ ) = (κλξ )ν = κλξ = κ<λ . ξ
ξ
This proves our claim for the case ν < cf (λ). From Corollary 1.6.8 we can conclude, if ν ≥ ω, that

  (∗) (κ<λ )ν = ( κλξ )ν = ( κλξ )ν = (κλξ )ν = κ ξ
ξ
ξ
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Chapter 1. Foundations

If cf (λ) ≤ ν < λ, we can again assume without loss of generality that λ0 ≥ ν, and we get ξ
cardinal number, then if ∃ν0 ∈ ICN ∩ κ ∀ν ∈ CN(ν0 ≤ ν < κ =⇒ 2ν = 2ν0 ) (the continuum function is eventually constant below κ), otherwise.

b) If κ and λ are infinite cardinal numbers, then ⎧ if 2λ < κ and λ < cf (κ), ⎨ κ λ κ+ if 2λ < κ and λ ≥ cf (κ), κ = ⎩ λ 2 if 2λ ≥ κ. Proof For a), we note that Theorem 1.7.7 yields 2κ = 2ν0 = 2<κ if the continuum function is constant below κ from ν0 on. So let us assume that the continuum function is not constant below κ. Then Lemma 1.7.6 b) tells <κ us that 2<κ has cofinality cf (κ). Our assumption gives 2cf(2 ) = 2cf(κ) < 2<κ . From (SCH) and Lemma 1.6.15 we can conclude that 2κ = (2<κ )cf(κ) = <κ (2<κ )cf(2 ) = (2<κ )+ .

1.7. Further Properties of κλ – the Singular Cardinal Hypothesis

77

The third case in b) is easy to show. If κ ≤ 2λ , then 2λ ≤ κλ ≤ (2λ )λ = 2λ , since λ is infinite. Now we prove part b), using transfinite induction on the cardinal number κ. By the previous remark, we can assume without loss of generality that 2λ < κ, hence λ < κ. First let κ be a successor cardinal ν + . Then 2λ ≤ ν. If 2λ < ν, then the inductive hypothesis gives ν λ ∈ {ν, ν + }. If 2λ = ν, then ν λ = 2λ < κ. In any case we have ν λ ≤ κ, and from Theorem 1.6.12 we can conclude that κλ = (ν + )λ = ν λ · ν + = ν + = κ. Now let κ be a limit cardinal. For ν < κ, the inductive hypothesis gives ν λ ∈ {ν, ν + , 2λ }, in particular ν λ < κ. From Lemma 1.7.3 a) we can conclude that κλ = κ, if cf (κ) > λ, and κλ = κcf(κ) , if cf (κ) ≤ λ. In this case we have 2cf(κ) ≤ 2λ < κ. With (SCH), we obtain κλ = κcf(κ) = κ+ . Lemma 1.7.9 The continuum function (2κ : κ ∈ ICN) can be characterized by the gimel function. More precisely: a) If κ is a regular cardinal number, then 2κ = ‫(ג‬κ). b) If κ is a singular cardinal number and if the continuum function is eventually constant below κ, then 2κ = 2<κ . c) If κ is a singular cardinal number and if the continuum function is not eventually constant below κ, then 2κ = ‫(ג‬2<κ ). Proof Part a) holds, since κ = cf (κ) gives 2κ = 2cf(κ) ≤ κcf(κ) ≤ κκ = 2κ . For b) we note that ‫(ג‬κ) ≤ 2κ , since κcf(κ) ≤ κκ = 2κ . If κ is singular, and if the continuum function is eventually constant below κ, then 2κ = 2<κ follows from the theorem of Bukovsky and Hechler. If the continuum function is not eventually constant below κ, then Lemma 1.7.6 b) says that cf (κ) = cf (2<κ ). <κ Part c) of this lemma tells us that (2<κ )cf(2 ) = 2κ . This proves part c). Remark Lemma 1.7.9 illustrates the importance of the gimel function. An essential aim of this book is to give nontrivial upper bounds for the power κcf(κ) for singular cardinals κ. Exercises 1) Assume that κ is a μ-strong successor cardinal. Show that κ+ is also μ-strong. 2) Prove: If μ and ν are infinite cardinals, then there is a μ-strong regular cardinal greater than ν. 3) Let μ be an infinite cardinal. Find the first ω + 1 μ-strong cardinals. Hint: Consider the cardinal (2μ )+ .

78

Chapter 1. Foundations 4) Prove: If κ is a weakly inaccessible cardinal, then κ = ℵκ . Show that on the other hand the singular fixed points of the aleph function form a proper class. 5) Let κ be an infinite  cardinal. Prove that κ is strongly inaccessible iff κ < κ and α α<λ α<λ κα < κ for all cardinals λ < κ and all sequences (κα : α < λ) of cardinals less than κ. 6) Assume that κ is a strongly inaccessible cardinal number. Prove that |Vα | < κ for all α < κ. Hint: Use transfinite induction on α according to the definition of the sets Vα . 7) The beth function  : ON −→ ICN is defined by (0) = ℵ0 , (α + 1) = 2(α) and (γ) = sup{2(α) : α < γ}, if γ is a limit ordinal. As usual we also write α instead of (α). Prove: a) |Vω+α | = α . b) If κ is an uncountable regular cardinal, then κ is a fixed point of the beth function iff κ is strongly inaccessible.

8) Prove: If 2ℵ1 < ℵω and ‫(ג‬ℵω ) > ℵω1 , then ‫(ג‬ℵω1 ) = ‫(ג‬ℵω ). 9) Prove: If 2ℵ0 ≥ ℵω1 , then ‫(ג‬ℵω ) = 2ℵ0 and ‫(ג‬ℵω1 ) = 2ℵ1 . 10) (Jech) Assume that λ is a cf (λ)-strong singular cardinal. Show that cf (‫(ג‬λ)) ≥ min{ν : ∃μ ∈ λ ∩ CN λ ≤ μν }. 11)

a) Prove: If 2ℵ0 = ℵω+1 and 2ℵ1 = ℵω+2 , then ℵℵω0 < 2ℵω . Hint: Use 1.6.15.

b) Prove that the monotonicity of the gimel function is not provable in ZFC. Hint: By Easton’s result, ZFC is consistent with the assertions 2ℵ0 = ℵω+1 , 2ℵ1 = ℵω+2 and 2ℵ2 = (2ℵω )+ . 12) Assume that λ is a singular cardinal, κ = cf (λ) and Φ ∈ κ κ. Further let (μξ : ξ < κ) be a sequence unbounded in λ such that the set {ξ < κ : +Φ(ξ) } is unbounded in κ. Prove that λ is a strong limit cardinal. 2μξ ≤ μξ 13) Assume that κ and λ are cardinals with κ ≥ 2 and λ ≥ ω, such that cf (κ<λ ) > λ. Prove that the continuum function for κ is eventually constant below λ.

1.8. Clubs and Stationary Sets

79

1.8 Clubs and Stationary Sets Definition If α is a limit ordinal, then we know that a subset C of α is unbounded in α iff sup(C) = α; we call C closed in α iff γ ∈ C for every limit ordinal γ < α in which C ∩ γ is unbounded. Every closed unbounded subset of α is called a club in α. If λ is a singular cardinal of uncountable cofinality, then C is called a special club in λ iff C is a club in λ of order type cf (λ) all of whose members are singular cardinals greater than cf (λ). A club sequence is any sequence of the form (Cα : α ∈ A), where A is a set of limit ordinals and, for every α ∈ A, Cα is a club in α of order type cf (α). For any ordinal α and any sequence (Xξ : ξ < α) of subsets of α, we call the set (Xξ : ξ < α) := ξ<α Xξ := {β ∈ α : ∀ξ < β(β ∈ Xξ )} = {β < α : β ∈



Xξ }

ξ<β

the diagonal intersection and (Xξ : ξ < α) := ξ<α Xξ := {β ∈ α : ∃ξ < β(β ∈ Xξ )} = {β < α : β ∈



Xξ }

ξ<β

the diagonal union of the sequence (Xξ : ξ < α). If C is an arbitraryclass of ordinals, then we say that γ ∈ ON \ {0} is a limit point of C iff γ = (C ∩ γ). Remarks and examples Assume that α is a limit ordinal. a) The order topology on a nonempty subset A of ON is that topology on A having as a base the set {(σ, τ )ON ∩ A : σ, τ ∈ A} ∪ {σ ∩ A : σ ∈ A} ∪ {A \ (σ + 1) : σ ∈ A} ∪ {A}. We remind the reader that a topology on a fixed nonempty set A is a set T of subsets of A which has A and ∅ amongst its members and is closed under finite intersections and arbitrary unions. The members of T are called open sets, their complements in A are called closed sets. We say that a set B ⊆ T is a base of this topology iff every open set is a union of members of B. If A is given without  a topology, then we call a set B ⊆ P(A) a base of a topology on A iff B = A and B1 ∩B2 is a union of members of B for all sets B1 , B2 ∈ B. In this case, the set { B : B ⊆ B} is a topology on A. It is called the topology on A generated by the base B. It is easy to see that any subset C of a limit ordinal α is closed in the above sense iff it is closed relative to the order topology on α.

80

Chapter 1. Foundations b) A subset C of α is closed in α iff the limit, i.e., the supremum, of any sequence of members of C which is bounded in α is an element of C. c) For every ordinal γ < α, the set [γ, α) is a club in α – every final segment of α is a club in α. d) If cf (α) > ω, then α ∩ Lim is a club in α (see the next lemma). Without the assumption on cf (α), this assertion is obviously false, since the set of limit ordinals less than α needs not to be unbounded in α; consider ω or ω · 2 (= ω + ω), where + and · denote the ordinal operations. e) The set of all successor ordinals in α > ω is not a club in α – later on, we will see that it is the standard example of a “thin” set, since its intersection with the club of limit ordinals is empty. f) If α > ℵ0 is regular and thus a regular cardinal, and if f : α −→ α is a function, then the set C := {β < α : f [β] ⊆ β} is a club in α. To prove this, let (βξ : ξ < γ) be a strictly increasing sequence in C with γ < cf (α) = α, and β := sup{βξ : ξ < γ}.Then β ∈ α and, since f [βξ ] ⊆ βξ ⊆ β for every ξ < β, f [β] = f [ {βξ : ξ < γ}] =  {f [βξ ] : ξ < γ} ⊆ β. Consequently, C is closed. For unboundedness, we apply a typical construction. Fix σ < α. Since α is regular, we have β0 := max{σ +1, sup(f [σ])+1} < α. Assume that the monotone sequence (βj : j ≤ n) with βn < α is already defined. Put, analogously to the basis of the induction, βn+1 := max{βn + 1, sup(f [βn ]) + 1}. By assumption  we have β  := sup{βn : n ∈ ω} ∈ α, and we obtain that f [β] = {f [βn ] : n < ω} ⊆ {βn+1 : n < ω} = β. Thus β ∈ C and β > σ, which shows that C is unbounded in α.

Lemma 1.8.1 Let α be a limit ordinal with uncountable cofinality. Then the following holds. a) α ∩ Lim is a club in α. b) α contains a club of order type cf (α) (of course this is also true if cf (α) = ω). Furthermore, if A is a subset of Lim, then there exists, by (AC), a club sequence (Cα : α ∈ A). c) If E is any unbounded subset of α, then the set E  of limit points of E less than α is a club in α; the same holds for E ∪ E  . d) If g : α −→ ON is a normal function, then ran(g) is a club in sup(ran(g)); if in addition g : α −→ α, then ran(g) is a club in α. Conversely, every club C in α is the range of a normal function f with dom(f ) = α if α is regular. If α is singular, then every club C in α is the range of a normal function f : otp(C) −→ α. e) If α is a singular cardinal, then α includes a special club.

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Proof For a), it obviously suffices to prove unboundedness. Fix σ ∈ α∩Lim and put σ0 := σ. If σj < α is already defined for all j ≤ n such that the sequence (σj : j ≤ n) is strictly monotone, then there is an ordinal σn+1 such that σn < σn+1 < α. The ordinal τ := sup{σn : n ∈ ω} is a limit ordinal, has cofinality ω and does not dominate α. The assumption yields that it is a member of α ∩ Lim. For b), observe that we obtain from Lemma 1.5.15 a continuous, strictly monotone function h : cf (α) −→ α such that sup(ran(h)) = α. The last property implies that ran(h) is unbounded in α, and from the continuity of h and Lemma 1.4.1 we can conclude that ran(h) is closed. Since h : cf (α) −→ ran(h) is an ∈-isomorphism, ran(h) is a club in α of order type cf (α), as desired. For c), the unboundedness of E  is proved analogously to the proof of a), using the the unboundedness of E: If σ ∈ α, then choose σ0 ∈ E with σ0 > σ. If σj ∈ E is already defined for all j ≤ n such that the sequence (σj : j ≤ n) is strictly increasing, then there is an ordinal σn+1 ∈ E such that σn < σn+1 < α. The ordinal τ := sup{σn : n ∈ ω} is greater than σ, is a limit point of E, has cofinality ω and does not dominate α. Therefore it is, by assumption, a member of α ∩ E  . If γ is a limit ordinal and E  ∩γ is unbounded in γ, then E∩γ is unbounded in γ. This is easy to see, since for every β < γ there is a limit point σ of E such that β + 1 < σ < γ and, by the definition of a limit point, there exists τ ∈ σ ∩ E with β < τ . Thus we get γ = sup(E ∩ γ) and γ ∈ E  . For d), we observe that C is unbounded in sup(ran(g)), if C is the range of a normal function g from α into ON; in the case that g : α −→ α, C is unbounded in α, since β ≤ g(β) for every β < α. C is closed in sup(ran(g)), since g is continuous. Assume that α is regular, C is a club in α, and, according to Corollary 1.3.11, f : γ −→ C is the order isomorphism from γ := otp(C) onto C. f is strictly monotone and continuous, since C is closed, and ran(f ) = C is unbounded in α. This gives α = cf (α) ≤ γ. Since ξ ≤ f (ξ) for all ξ < γ, we get γ ≤ sup(ran(f )) = α. In the same way one proves the corresponding assertion for singular α. For e), Lemma 1.6.11 says that α is the supremum of a continuous, strictly increasing sequence (λξ : ξ < cf (α)) of singular cardinals greater than cf (α). The range of this sequence is a club in α by part d); by definition, it is a special club. Lemma 1.8.2 Let α be an ordinal with uncountable cofinality. a) The intersection of less than cf (α) clubs in α is a club in α.  b) Assume that (Cγ : γ < α) is a sequence of clubs in α. If the set {Cγ : γ < β} is unbounded in α for all β < α, then the diagonal intersection of

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Chapter 1. Foundations the sequence (Cγ : γ < α) is a club in α. This holds in particular if the sequence (Cγ : γ < α) is decreasing under ⊆. If α > ℵ0 is regular, then (Cγ : γ < α) is a club in α.

Proof For a),let (Cγ : γ < τ ) with 0 < τ < cf (α) be a sequence of clubs in α, and let C := {Cγ : γ < τ }. Certainly C is closed in α. We show that C is unbounded in α. For this we use transfinite induction and assume that every intersection of less than τ clubs in α is unbounded in α. Fix σ < α. If τ = 2, then choose αn ∈ C0 and βn ∈ C1 for each n ∈ ω such that σ < α1 < β1 < α2 < β2 < . . . . We then have σ < β := sup{αn : n ∈ ω} = sup{βn : n ∈ ω} and β ∈ C0 ∩ C1 ∩ α, since cf (α) > ℵ0 . If τ = ξ + 1, then the  same proof, together with the inductive hypothesis, yields that C = Cξ ∩ {Cγ : γ < ξ} is unbounded in α. Finally, if τ is  a limit ordinal, choose for i < τ ordinals γi < α such that γ0 > σ, γi ∈ {Cj : j < i}, and γi > sup{γj : j < i}, which is possible by the inductive hypothesis and because i < cf (α), hence sup{γj : j < i} < α. Put β := sup{γi : i < τ }. We get σ < β < α. Furthermore β = sup{γi : i0 < i < τ } for any i0 < τ , and thus β is a supremum of elements of Ci0 . This gives β ∈ Ci0 for every i0 < τ , and we can conclude that β ∈ C. For the proof of b), we show first that C := Δ(Cγ : γ < α) is closed. So assume that τ < α is a limit ordinal and C ∩ τ is unbounded in τ , and fix σ < τ . Then we have τ = sup{i ∈ C : σ < i < τ }. If σ < i ∈ C, then i ∈ Cσ , and since Cσ is a club, we get τ ∈ Cσ . This gives τ ∈ C.  For unboundedness, fix σ < α. Choose σ0 ∈ C0 \σ and σ {Cγ : γ < n+1 ∈ σn } \ (σn + 1). This is possible, since, by assumption, the set {Cγ : γ < σn } is unbounded in α. Then we have β := sup{σn : n ∈ ω} < α, since cf (α) > ℵ0 . Let i < β. Then there is an n0 ∈ ω with i < σn0 , and further β = sup{σ  n : n > n0 } is a supremum of elements of Ci , which gives β ∈ Ci . Since β ∈ {Ci : i < β}, we can conclude that β ∈ C, and clearly β > σ.  If α > ℵ0 is regular, then, for every β < α, the set {Cγ : γ < β} is a club in α by part a); in particular it is unbounded in α. Using the previous result, we can now conclude that {Cγ : γ < α} is a club in α. Definition If A is a nonempty set, then we say that a set F of subsets of A is a filter on A iff the following conditions are satisfied: (F1) A ∈ F and ∅ ∈ F . (F2) If X, Y ∈ F , then X ∩ Y ∈ F . (F3) If X ∈ F , Y is a subset of A and X ⊆ Y , then Y ∈ F . Every ⊆-maximal filter on A is called an ultrafilter on A. Ifa nonempty set S ⊆ P(A) has the finite intersection property (fip), i.e., if S  = ∅ for all S  ∈ Pfin (S), then (see Lemma 1.8.3)  F := S! := {X ⊆ A : ∃S  ∈ Pfin (S) S  ⊆ X}

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is a filter on A such that S ⊆ F . It is the filter generated by S. If S = {Y }, we also call it the filter generated by Y . If F is an arbitrary filter on A and X is a subset of A such that A \ X ∈ F , we denote, with regard to the corresponding denotations for ideals, by F |`X the filter generated by F ∪ {X}. For any a ∈ A, the set {X ⊆ A : a ∈ X} is an ultrafilter on A; we call it the principal ultrafilter on A generated by {a}. Every ultrafilter which is not principal is called nonprincipal or free. We say that a set I of subsets of A is an ideal on A iff it satisfies the following properties: (I1) A ∈ I and ∅ ∈ I. (I2) If X ∈ I and Y ∈ I, then X ∪ Y ∈ I. (I3) If X ∈ I and Y ⊆ X, then Y ∈ I.  The set dom(I) := I is called the domain of the ideal I. If F is a set of subsets of A, one easily checks that F is a filter on A iff the set I := {A \ X : X ∈ F } is an ideal on A. We call I the dual ideal of F . If F is an ultrafilter on A, then the dual ideal of F is called a prime ideal or a maximal ideal. Conversely, if an ideal I on A is given, we can define a filter F in the same way and call it the dual filter of I. Assume that α is a limit ordinal. We say that a filter F on α is normal iff, for any sequence (Xξ : ξ < cf (α)) of members of F , the diagonal intersection (Xξ : ξ < cf (α)) is a member of F . If cf (α) > ℵ0 , then we call the filter FC = FC (α) := {X ⊆ α : ∃C(C is a club in α ∧ C ⊆ X)} (see Lemma 1.8.2) the clubfilter on α. Any subset S of α is called stationary in α or a stationary subset of α iff C ∩ S = ∅ for every club C in α. Every subset of α which is not stationary in α is called thin or, as usual, nonstationary in α. Lemma 1.8.3 Let A be a nonempty set. a) If a nonempty set S ⊆ P(A) has the finite intersection property, then S! is the smallest filter on A such that S ⊆ S!. b) If U is a filter on A, then the following conditions are equivalent. (i) U is an ultrafilter. (ii) ∀X, Y ∈ P(A)(X ∪ Y ∈ U =⇒ X ∈ U ∨ Y ∈ U ). (iii) ∀X ∈ P(A)(X ∈ / U =⇒ A \ X ∈ U ). c) If I is an ideal and F a filter on A with I ∩ F = ∅20 , then there exists an ultrafilter D on A such that F ⊆ D and D ∩ I = ∅. In particular we get: 20 Such

a filter always exists; take for example the trival filter {A}.

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Chapter 1. Foundations If d ∈ I, then there is an ultrafilter D on A such that D ∩ I = ∅ and d ∈ D. For every filter F on A there is an ultrafilter D on A such that F ⊆ D (“every filter can be extended to an ultrafilter”21 ). There is a free ultrafilter on any infinite set. d) If A is infinite and U an ultrafilter on A, then the following conditions are equivalent: (i) U is nonprincipal. (ii) Fcof := {X ⊆ A : A \ X is finite} ⊆ U . (iii) No member of U is a finite set.

Proof The proof of a) and d) is left to the reader. We give an indirect proof for the implication “(i) =⇒ (iii)” in b) and assume that there is a subset X of A such that X ∈ U and A \ X ∈ U . Then we get Y ∩ X = ∅ for every Y ∈ U , since Y ∩ X = ∅ would imply Y ⊆ A \ X and thus A \ X ∈ U . Put F := {Z ⊆ A : ∃Y ∈ U (Y ∩ X ⊆ Z)}. Obviously, U ⊆ F and X ∈ F . Further it is easy to see with the filter properties of U that F is a filter. This gives U = F , since U is maximal, hence X ∈ U , a contradiction. For the proof of “(iii) =⇒ (ii)” we assume, that there are sets X, Y ∈ P(A) such that X ∪ Y ∈ U , X ∈ / U and Y ∈ / U . From (iii) we get A \ X ∈ U and A \ Y ∈ U . Thus the intersection of these sets, that is Z := A \ (X ∪ Y ), is a member of U . Now we can conclude that ∅ = (X ∪ Y ) ∩ Z ∈ U , contradicting our assumption that U is a filter. To prove “(ii) =⇒ (i)”, assume that F ⊇ U is a filter and X ∈ F . We will show that X ∈ U . Fix B ∈ U . Then B = (B ∩ X) ∪ (B \ X). Assumption (ii) gives X ∈ U or B \ X ∈ U . But from the second condition we could conclude with U ⊆ F that X ∩ (B \ X) = ∅ ∈ F which would contradict the fact that F is a filter. For the proof of c), it is easy to check that the set M := {G ⊆ P(A) : G is a filter on A ∧ F ⊆ G ∧ G ∩ I = ∅} satisfies the assumptions of Zorn’s lemma. So let D be a maximal element of M. We will show that D is an ultrafilter on A and assume, contrary to our hopes, that there is a set X ⊆ A such that X ∈ D and A \ X ∈ D. If Y ∈ D, then X ∩Y = ∅, since otherwise A\X ∈ D would follow from Y ⊆ A\X. In the same way we get Y ∩ (A \ X) = ∅. This shows that the sets D ∪ {X} and D ∪ {A \ X} 21 Of course this assertion follows easily if we apply Zorn’s lemma to the set of all filters on A containing F .

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have the finite intersection property. Let F1 and F2 be the filters generated by them according to part a). Since D is maximal in M, we can choose Z1 ∈ F1 ∩I and Z2 ∈ F2 ∩ I. Part a) gives F1 = {Y ⊆ A : ∃d ∈ D(d ∩ X ⊆ Y )}, since D is a filter, and an analogous representation holds for F2 . Thus we obtain sets d1 , d2 ∈ D, such that d1 ∩ X ∈ I and d2 ∩ (A \ X) ∈ I. Therefore the set (d1 ∩ X) ∪ (d2 ∩ (A \ X)) is a member of I. But d1 ∩ d2 is a subset of it, and so we get d1 ∩ d2 ∈ I ∩ D, contradicting D ∩ I = ∅. This completes the proof of the first assertion in c). If d ⊆ A and d ∈ I, then the filter Fd := {Y ⊆ A : d ⊆ Y } generated by d is a filter on A with I ∩ Fd = ∅, as desired. The second special consequence follows at once, taking I = {∅}. Also the third consequence is easy to show, taking I := {X ⊆ A : X is finite}. Then I is an ideal on A, and from d) we can conclude that any ultrafilter disjoint from I is nonprincipal. Lemma 1.8.4 Assume that α is a limit ordinal with uncountable cofinality. a) Every superset of a club in α is stationary in α. b) Every stationary subset of α is unbounded in α. c) Every union of less than cf (α) sets that are nonstationary in α is nonstationary in α. d) For every regular cardinal λ < cf (α), the set Σα,λ := {β < α : cf (β) = λ} is stationary in α. Furthermore, every club C in α has a member β with cf (β) = λ such that C ∩ β is a club in β. Proof Part a) follows from Lemma 1.8.2 a). The same holds for part c): If sets, then choose for γ < cf (α) and (Tξ : ξ < γ) is a sequence of nonstationary  each ξ < γ a club Cξ inα with Cξ ∩ Tξ = ∅; C := {Cξ : ξ < γ} is a club which does not meet the set {Tξ : ξ < γ}. Part b) follows at once if we choose, for each γ < α, a member of the intersection of the club {β ∈ α : γ ≤ β} with the given stationary set. For the proof of d), consider a fixed club C in α. We can define by recursion, since λ < cf (α) and clubs are unbounded, a strictly increasing continuous sequence g of members of C having length λ. Since C is closed and λ is regular, the supremum β of this sequence is an element of C and has cofinality λ, hence β ∈ Σα,λ . The second assertion follows, too, since ran(g) is a club in β by Lemma 1.8.1, and ran(g) ⊆ C.  Definition Assume that M is a subset of P(A) such that M  = A for every finite subset M  of M . Then  M ! := {X ⊆ A : ∃M  ∈ Pfin (M ) X ⊆ M }

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is an ideal on A called the ideal generated by M . It is the smallest ideal on A including M . If A \ B ∈ I, then there exists the ideal I[B] := I ∪ {B}!. We say that B generates the ideal I[B] over I. If X ∈ I, then the set I |`X := {Z ⊆ A : X ∩ Z ∈ I} is an ideal on A, the restriction of the ideal I to X. Note that I |`X = I[A\X]. Let α be a limit ordinal. We say that an ideal I on α is normal iff, for any sequence (Xξ : ξ < cf (α)) of members of I, the diagonal union (Xξ : ξ < cf (α)) is a member of I. If κ is a cardinal with cf (α) ≥ κ, then an ideal I on α is called κ-complete iff every union of less than κ members of I is a member of I.  I.e., whenever (Xα : α < γ) is a family of members of I and γ < κ, then {Xα : α < γ} ∈ I. Any ℵ1 -complete ideal also is called σ-complete. If cf (α) > ℵ0 , then we call the dual ideal of the clubfilter FC on α the ideal of thin subsets of α, writing Id (α); i.e, Id (α) = {X ⊆ A : ∃C(C is a club in α ∧ C ∩ X = ∅)}. For any nonempty set A of ordinals with sup(A) ∈ A, the ideal Ib (A) := {X ⊆ A : ∃β ∈ A X ⊆ β} on A is called the ideal of bounded subsets of A. An ultrafilter D on A is called unbounded iff D ∩ Ib (A) = ∅. Lemma 1.8.5 If κ is an uncountable regular cardinal, then Id (κ) is a κ-complete normal ideal on κ and Ib (κ) ⊆ Id (κ). Proof By definition, Id (κ) is closed under subsets. If (Xξ : ξ < κ) is a sequence of members of this ideal, then for each ξ < κ there is a club Cξ in κ us that the diagonal intersection D of such that Cξ ∩ Xξ = ∅. Lemma 1.8.2 tells  (Cξ : ξ < κ) and, for every γ < κ, Dγ := {Cξ : ξ < γ} areclubs in κ. An easy calculation shows that D ∩ {Xξ : ξ < κ} = ∅ and Dγ ∩ {Xξ : ξ < γ} = ∅. Thus Id (κ) is κ-complete and normal, as desired. Lemma 1.8.6 Assume that κ is a regular uncountable cardinal and I is a normal ideal on κ. a) If I is κ-complete and dom(I) = κ, then Ib (κ) ⊆ I, and {ξ + 1 : ξ < κ} is a member of I. b) If Ib (κ) ⊆ I, then even Id (κ) ⊆ I. Thus Id (κ) is the smallest κ-complete normal ideal J on κ such that dom(J) = κ.

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Proof For the proof of a) we observe that, for each α < κ, the set Xα := {ξ < κ : ξ ≤ α+1} is a member of I, since I is κ-complete and any singleton {ξ}, ξ < κ, is a member of I on account of dom(I) = κ. Now I is normal, therefore the set α<κ Xα is a member of I, and we have α<κ Xα = {ξ + 1 : ξ < κ}. Next we prove part b). Fix T ∈ Id (κ). Then there is a club C in κ with T ∩ C = ∅. C has order type κ, since κ is regular. So let (δα : α < κ) be the ∈-isomorphism from κ onto C, and let Yα := {ξ < κ : ξ ≤ δα }. By assumption, we have Ib (κ) ⊆ I, and thus Yα ∈ I and Xα ∈ I for every α < κ. From this we can conclude that α<κ Yα ∈ I and α<κ Xα ∈ I, since I is normal. If we can prove the assertion (∗)

T ⊆ α<κ Xα ∪ α<κ Yα ,

then T ∈ I follows, and b) will be shown. We have κ \ (α<κ Xα ∪ α<κ Yα ) = (κ \ α<κ Xα ) ∩ (κ \ α<κ Yα ) = α<κ (κ \ Xα ) ∩ α<κ (κ \ Yα ). If β ∈ κ \ (α<κ Xα ∪ α<κ Yα ), then β is a limit ordinal, since β ∈ α<κ (κ \ Xα ), and δα < β for all α < β, since β ∈ α<κ (κ \ Yα ). Now, firstly the sequence (δα : α < κ) is continuous as a normal function, and we get δβ = sup{δα : α < β} ≤ β, and secondly it is strictly increasing, which gives β ≤ δβ . Altogether we get β = δβ and thus β ∈ C. Since T ∩ C = ∅, we can conclude that β ∈ T , which proves (∗). Lemma 1.8.7 If κ is an uncountable regular cardinal and I is a normal ideal on κ such that Ib (κ) ⊆ I, then I is κ-complete. Proof Consider a cardinal μ < κ and a sequence (Xα : α < μ) of members  of I. We will show that α<μ Xα ∈ I. Let the sequence (Yα : α < κ) be defined by  Xα \ μ if α < μ, Yα := ∅ otherwise. Since I is normal, ∈ I. Furthermore it is easy to see  we know that α<κ Yα  : α < μ} ∈ I. that α<κ Yα = {Yα : α < μ}, and thus {Yα   Now μ ∈ Ib (κ) gives μ ∈  I, and from this we can conclude that α<μ Xα ⊆ α<μ Yα ∪ μ ∈ I and thus α<μ Xα ∈ I, as desired. Theorem 1.8.8 (Fodor) Assume that α is a limit ordinal with cf (α) > ℵ0 and S ⊆ α is stationary in α. Let further f : S −→ α be a regressive function (i.e., f (γ) < γ for all γ ∈ S \ {0}). Then, for some η < α, the set f −1 [η] is stationary in α. If α is regular, then there even exists η < α such that f −1 [{η}] is stationary in α, i.e., f is constant on a stationary subset of S. Proof To get a contradiction we assume that, for each η < α, there is a club Dη in α such that Dη ∩ f −1 [η] = ∅. Choose, according to Lemma 1.8.1, a

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 club E ⊆ α in α of order type cf (α), and put Cη := {Dξ : ξ ∈ E ∩ (τ + 1)}, where τ = τ (η) := min{ξ ∈ E : ξ ≥ η}. Cη is a club in α, being the intersection of less than cf (α) clubs in α. Furthermore we have Cη ∩ f −1 [η] = ∅, since Cη ⊆ Dτ (η) , τ (η) ≥ η and Dτ (η) ∩ f −1 [τ (η)] = ∅ and thus Dτ (η) ∩ f −1 [η] = ∅. Now the sequence (Cη : η < α) is descending, and Lemma 1.8.2 says that C := (Cη : η < α) and thus C ∩Lim are clubs in α. Consequently, there exists a limit ordinal γ such that γ ∈ C ∩ S, since S is stationary in α. Furthermore we have f (γ) < γ, since f is regressive, hence σ := f (γ) + 1 < γ and γ ∈ Cσ . But on the other hand we have γ ∈ f −1 [σ] by the definition of σ, contradicting Cσ ∩ f −1 [σ] = ∅. Now assume in addition that α is regular and f −1 [η] is stationary in α  for some η < α. From f −1 [η] = {f −1 [{σ}] : σ < η} we can conclude from Lemma 1.8.4 that there is an ordinal σ < η such that f −1 [{σ}] is stationary in α. Lemma 1.8.9 Assume that κ is a regular cardinal and I is an ideal on κ. Then I is normal iff there exists, for every set S ⊆ κ with S ∈ I and every function f regressive on S, a set S0 ⊆ S such that S0 ∈ I and f is constant on S0 . Proof First we show that the previous condition is sufficient for the normality of I. So assume that (Xξ : ξ < κ) is a sequence of members of I. We must show that S := ξ<κ Xξ is in I. If σ ∈ S, then the ordinal f (σ) := min{ξ < σ : σ ∈ Xξ } exists. The function f defined by this equation is regressive on S. If S was not a member of I, then our assumption would yield a set S0 ⊆ S with S0 ∈ I and an ordinal γ < κ such that ran(f |` S0 ) = {γ}. We could infer that S0 ⊆ Xγ by the definition of f , and thus S0 ∈ I, a contradiction. Consequently, we get S ∈ I, as desired. Conversely assume that I is normal, S ⊆ κ is a set with S ∈ I, and f : S −→ κ is a regressive function. If 0 ∈ S and S \ {0} ∈ I, then {0} ∈ I, and f is constant on S0 := {0}. Otherwise we have S \ {0} ∈ I or 0 ∈ S, and we can assume without loss of generality that 0 ∈ S. Put Xξ := f −1 [{ξ}] for every ξ < κ. Then there must exist an ordinal ξ < κ with Xξ ∈ I. The supposition Xξ ∈ I for every ξ < κ would lead to ξ<κ Xξ ∈ I, since I is normal. But on the other hand we have S = ξ<κ Xξ , since f is regressive, and S ∈ I. Therefore there exists a set Xξ such that Xξ ∈ I. Clearly we have Xξ ⊆ S, and the function f is constant on Xξ . Corollary 1.8.10 If κ is a regular cardinal, I is a normal ideal on κ and X ∈ P(κ) \ I, then I |`X is a normal ideal on κ. Proof Assume that S ∈ P(κ) \ (I |` X) and f : S −→ κ is a regressive function. Then X ∩ S ∈ I and, by assumption, there is a set S0 ⊆ S ∩ X with S0 ∈ I and an ordinal γ < κ such that ran(f |` S0 ) = {γ}. It is obvious that S0 ∈ I |`X.

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Next we prove a well-known theorem of Solovay which says for regular uncountable cardinals κ: If S ⊆ κ and S ∈ Id (κ), then S can be partitioned into κ classes which are not members of Id (κ). For the proof we need two preparatory lemmas. Lemma 1.8.11 Assume that κ is an uncountable regular cardinal and S is a stationary subset of κ. Then the set T is stationary in κ, where T := {α ∈ S : cf (α) = ω ∨ (cf (α) > ω ∧ S ∩ α is not stationary in α)}. Proof If C is a club in κ, then Lemma 1.8.1 tells us that the set C  := {σ < κ : σ = 0 ∧ sup(C ∩ σ) = σ} of limit points of C is a club in κ. Let α := min(C  ∩ S). We show that α ∈ C ∩ T . If cf (α) = ω, then this holds by definition. So let cf (α) > ω. Then, by Lemma 1.8.1, C  ∩ α is a club in α, since α is a limit point of C. Further we have (C  ∩ α) ∩ (S ∩ α) = ∅ by the choice of α, which means that S ∩ α is not stationary in α. Lemma 1.8.12 Let κ be an uncountable regular cardinal and T be a stationary subset of κ ∩ Lim. If fα is, for all α ∈ T , a normal sequence cofinal in α, then one of the following assertions is true. (1) ∃ξ < κ ∀η < κ(Tη := {α ∈ T : ξ ∈ dom(fα ) ∧ fα (ξ) ≥ η} is stationary in κ). (2) ∃D ⊆ κ(D is a club in κ ∧ ∀γ, α ∈ D ∩ T (γ < α =⇒ γ = fα (γ))). Proof Assume that the first assertion is false, that means that for each ξ < κ there is an ordinal η(ξ) < κ and a club Cξ in κ such that Cξ ∩ Tη(ξ) = ∅. This defines a function η from κ into κ. Put C := ξ<κ Cξ and D := {α ∈ C : η[α] ⊆ α ∧ Lim(α)}. From the example following the definition of a club and Lemma 1.8.2 we can conclude that D is a club in κ. If we consider α, γ ∈ D ∩ T with γ < α, then α ∈ C, and thus α ∈ Cξ for all ξ < α and α ∈ / Tη(ξ) for all ξ < α. From this we can infer for every ξ ∈ γ ∩ dom(fα ) that fα (ξ) < η(ξ), since α ∈ Tη(ξ) , and η(ξ) < γ, since γ ∈ D. Now the supposition dom(fα ) ≤ γ would lead to ran(fα ) ⊆ γ < α, contradicting the assumption on fα . So we get γ ∈ dom(fα ). Since γ is a limit ordinal and fα is continuous, the previous considerations show that fα (γ) = sup{fα (ξ) : ξ < γ} ≤ γ. Furthermore we have γ ≤ fα (γ), since fα is strictly increasing. This gives fα (γ) = γ, and the second assertion is true. Theorem 1.8.13 (Solovay) Assume that κ is an uncountable regular cardinal and S is a stationary subset of κ. Then S can be partitioned into κ stationary subsets of κ.

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Proof Since every superset of a stationary set is also stationary, it obviously suffices to show that the assertion is true for some stationary subset T of S. If f is a function from T into κ and γ1 , γ2 ∈ κ are different, then f −1 [{γ1 }] ∩ f −1 [{γ2 }] = ∅. Thus f yields the desired partition if the set {γ ∈ κ : f −1 [{γ}] is stationary} is unbounded in κ. Now, if f is regressive and Tη := {α ∈ T : f (α) ≥ η} is stationary for all η < κ, then Fodor’s theorem tells us that, for each η < κ, there is an ordinal γη < κ such that f −1 [{γη }] ∩ Tη is stationary in κ. Since γη ≥ η by the definition of Tη , the set {γη : η < κ} is unbounded in κ, and we are ready. We choose the set T ⊆ Lim from Lemma 1.8.11 as the stationary subset T of S which will suit our purposes. For each α ∈ T we can find a sequence fα which is cofinal in α and satisfies ran(fα ) ∩ T = ∅. Namely, if cf (α) > ℵ0 , then S ∩ α is nonstationary in α, and Lemma 1.8.1 d) says that there is a club Cα in α which is the range of a normal function fα and does not meet S ∩ α. If cf (α) = ℵ0 , it is enough to choose a strictly increasing sequence gα : ω −→ α with sup(ran(gα )) = α and to put fα (n) := gα (n) + 1. Now a club D satisfying the conditions in assertion (2) of Lemma 1.8.12 cannot exist. Otherwise D ∩ T would be stationary, in particular there would be ordinals α, γ ∈ D ∩ T such that γ < α and fα (γ) = γ, that means with γ ∈ ran(fα ) ∩ T , contradicting ran(fα ) ∩ T = ∅. Therefore some ξ < κ satisfies the condition in assertion (1) of this lemma, and we define for this ξ and for α ∈ T  if ξ ∈ dom(fα ), fα (ξ) f (α) := 0 otherwise. If η > 0, then {α ∈ T : f (α) ≥ η} = {α ∈ T : ξ ∈ dom(fα ) ∧ fα (ξ) ≥ η}, and thus the set {α ∈ T : f (α) ≥ η} is stationary in κ for every η < κ. Since f is regressive, our introductory remarks show that the proof of the theorem is completed. Definition Let α be a limit ordinal. We say that a stationary subset S of α is nonreflecting iff, for all limit ordinals β < α, the set S ∩ β is not stationary in β. Lemma 1.8.14 If λ is a regular cardinal, then the set Σλ+ ,λ = {σ < λ+ : cf (σ) = λ} is a nonreflecting stationary subset of λ+ . Proof By Lemma 1.8.4, the set Σλ+ ,λ is stationary in λ+ . Now let α < λ+ be a fixed limit ordinal. Then μ := cf (α) ≤ λ. Take f : μ −→ α as a strictly monotone function such that sup(ran(f )) = α. Then the function g with g(γ) := f (γ) + 1 has the same properties. Put C  := ran(g) and C := C  ∪ {γ < α : sup(C  ∩ γ) = γ} .

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For μ = ω, C is a club in α, and the same holds for μ > ω by Lemma 1.8.1. By the definition of C, every limit ordinal γ ∈ C is the supremum of the set {g(σ) : σ < μ ∧ g(σ) < γ}. If we choose αγ as the least ordinal in μ such that γ ≤ g(αγ ), then γ = sup(ran(g |` αγ )), and we get cf (γ) ≤ αγ < μ ≤ λ. So we have shown that Σλ+ ,λ ∩ C = ∅, and Σλ+ ,λ ∩ α is not stationary in α, as desired. Exercises22 A, B, . . . , X, Y, . . . , etc. denote sets.

 1) Prove: If F is a nonempty set of filters on A, then F is a filter on A.  2) Prove: If K is a nonempty ⊆-chain of filters on A, then K is a filter on A. 3) Let X be a subset of A. Prove: a) If F is a filter (ultrafilter) on A and X ∈ F , then F ∩ P(X) is a filter (ultrafilter) on X. b) If G is a filter on X, then G can be extended to an ultrafilter on A.

4) If F is a filter on A, we put F + := {X ⊆ A : A \ X ∈  F }. Prove that + + F ⊆ F and that F is a filter iff F is an ultrafilter. 5) Let A be an infinite set. Prove that the set {X ⊆ A : A \ X is finite} is a filter on A. It is called the Fr´echet-Filter on A. 6) Let A be an infinite set. Prove: a) There is an ultrafilter F on A such that all members of F have the same cardinality as A (a so-called uniform ultrafilter on A). Hint: The set {A \ X : |X| < |A|} can be extended to an ultrafilter. b) The set of all uniform ultrafilters on A and the set of nonprincipal ultrafilters on A coincide iff A is countable (see also Exercise 7 and Exercise 8 below). Hint: If A is not countable, choose X ⊆ A with ℵ0 ≤ |X| < |A|, and apply part a) to X. 7) Prove that there are no normal nonprincipal ultrafilters on ω. Hint: Assume, to get a contradiction, that F is such a filter. Then ω \ {0} is not an element of the dual ideal of F . Define f : ω \ {0} −→ ω by f (n + 1) := n. Now apply Lemma 1.8.9. 8)  Let κ be an infinite cardinal. A filter F on A is called κ-complete iff M ∈ F for any nonempty subset M of F of cardinality less than κ; otherwise it is called κ-incomplete. An ℵ1 -complete filter on A is also called σ-complete, an ℵ1 -incomplete filter on A is also called σ-incomplete. Prove: 22 Exercises

on ideals can be found in Section 2.1.

92

Chapter 1. Foundations a) If A is infinite, then every nonprincipal ultrafilter on A is |A|+ incomplete. In particular, all nonprincipal ultrafilters on countable sets are σ-incomplete. b) If U is a nonprincipal |A|-complete ultrafilter on A, then U is a uniform ultrafilter.

Hint: For any nonprincipal ultrafilter U on A, we have A \ {x} ∈ U for every x ∈ A. 9) We call an uncountable cardinal κ a measurable cardinal iff there is a nonprincipal κ-complete ultrafilter on κ. Prove that every measurable cardinal is regular and strong limit, i.e., it is an inaccessible cardinal. 10) Prove for filters F on A: If μ := sup{κ : F is κ-complete} exists, then μ is regular and F is μ-complete. Hint: Assume that μ is singular and use the fact that F is μ+ -incomplete. 11) Assume that A is infinite and D is an ultrafilter on A. Prove that D is κ-complete iff,  for every τ < κ and every partition (Xσ : σ < τ ) of A (that means {Xσ : σ < τ } = A and Xσ ∩ Xξ = ∅ whenever ξ < σ < τ ), there exists a set Xσ0 ∈ D. 12) (Shelah) If A is infinite, F is a filter on A, and κ is a cardinal, then F is calledκ-regular iff there is a sequence (Xξ : ξ < κ) of members of F such that {Xξ : ξ ∈ N } = ∅ for every denumerable subset N of κ. Prove that reg(F ) := min{κ : F is not κ-regular} exists and is a regular cardinal. 13) We call a filter F on A regular iff F is |A|-regular. Prove: If F is a filter on A, then F is regular iff there is a function f : A −→ [A]<ω , such that {x ∈ A : a ∈ f (x)} ∈ F for all a ∈ A. 14) a) If A is infinite, then there is a regular ultrafilter on A. Hint: Consider a bijection f : A −→ [A]<ω and let Xa := {a ∈ A : a ∈ f (x)} for every a ∈ A. b) If F is a regular ultrafilter, then F is σ-incomplete. 15) If A is denumerable, then every free ultrafilter on A is regular. 16) Assume that A and B are infinite sets and F is an ultrafilter on A. For any functions f, g ∈ A B we define, in analogy to the definition for ideals, f =F g iff {a ∈ A : f (a) = g(a)} ∈ F.

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A B. If [f ] := {g ∈ A B : Show that this gives an equivalence relation on  f =F g} is the equivalence class of F , then let a∈A B/F := {[f ] : f ∈ A B} be the set of equivalence classes. Prove:   If C is a set with |B| = |C|, then | a∈A B/F | = | a∈A C/F |. If U is a  regular ultrafilter on A, then | a∈A B/U | = |B||A| .  Hint: For the last assertion, use κ := |A| instead of A and C := {n B : n ∈ ω} instead of B. 17) (Shelah) Assume that F is a filter on an infinite set A. F is called weakly κ-regular iff there is a sequence (Xξ : ξ < κ) of elements of F + such that  {Xξ : ξ ∈ N } = ∅ for every N ∈ [κ]ℵ0 . Let

reg∗ (F ) := min{κ : F is not weakly κ-regular}. Prove: The cardinal number reg∗ (F ) exists; furthermore, if ν < reg∗ (F ) is a cardinal, then there exists a ν-regular filter D on A with F ⊆ D. Hint: Let ν ∈ reg∗ (F ) ∩ ICN and (Xξ : ξ < ν) be a sequence of members of F + which witnesses that F is weakly ν-regular. Further let (uξ : ξ < ν) be an enumeration of [ν]<ℵ0 . Put

18)

19)

20)

21)

D := {X ⊆ A : ∃u ∈ [ν]<ℵ0 ∀ξ < ν(u ⊆ uξ =⇒ (A \ Xξ ) ∪ X ∈ F )}.  If Yξ := {Xζ : ξ ∈ uζ }, then the sequence (Yξ : ξ < ν) witnesses that D is ν-regular. Let κ be an uncountable regular cardinal and f : κ −→ κ be a normal function. Prove that the set C of all fixed points of f is a club in κ. Hint: It is easy to verify that C is closed. For unboundedness, let σ < κ and define α0 := σ, αn+1 := f (αn ), and α := sup{αn : n ∈ ω}. Assume that α is a limit ordinal and C is a subset of α. Prove that C is closed in α iff, for every sequence of members of C which is bounded in α, its supremum is an element of C. Assume that κ is a cardinal with uncountable cofinality, S ⊆ κ, and (αξ : ξ < cf (κ)) is a normal sequence cofinal in κ. Prove: S is stationary in κ iff {ξ < cf (κ) : αξ ∈ S} is stationary in cf (κ). Assume that κ is an uncountable regular cardinal and S is a subset of κ. Prove that the following assertions are equivalent: (i) S is stationary in κ. (ii) For every regressive function f : S −→ κ there is an ordinal α < κ such that f −1 [α] is unbounded in κ. Hint: Assume that C is a club in κ such that C ∩ S = ∅. Let σ := min(C) and define f : S −→ κ by f (ξ) = 0, if ξ ≤ σ, f (ξ) = sup(C ∩ ξ) otherwise.

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22) Prove: If λ is a κ-strong cardinal with ℵ0 < κ = cf (λ) < λ, then the set {μ < λ : μ is κ-strong} is a club in λ. Hint: For unboundedness, consider a normal sequence (λξ : ξ < cf (κ)) of singular cardinals less than λ such that λ = sup{λξ : ξ < cf (λ)}, according to 1.6.11. 23) a) If κ is an inaccessible cardinal, then the set {μ < κ : μ is strong limit} is a club in κ0 . b) If κ0 is the smallest inaccessible cardinal, then the set {λ < κ0 : λ is strong limit and singular} is a club in κ. 24) Assume that κ is the α-th inaccessible cardinal, where α < κ. Prove that the set T := {λ < κ : λ is regular} is not stationary in κ. Hint: Observe that the set of inaccessible cardinals less than κ is bounded in κ, and use the previous question. 25) A cardinal κ is called a Mahlo cardinal iff κ is inaccessible and the set {λ < κ : λ is regular} is stationary in κ. Assume that κ is a Mahlo cardinal. Prove that {λ < κ : λ is inaccessible} is stationary in κ, and that κ is the κ-th inaccessible cardinal. 26) Assume that κ := min{λ : λ is the λ-th inaccessible cardinal}. Prove that κ is not a Mahlo cardinal. Hint: Take the order isomorphism of the set of inaccessible cardinals less than κ and use Fodor’s theorem to show that this set is not stationary in κ. 27) Assume that κ is a Mahlo cardinal. Prove that the set {λ < κ : λ is the λ-th inaccessible cardinal} is unbounded in κ. Hint: Take the order isomorphism f of the set of inaccessible cardinals less than κ – f (α) is the α-th inaccessible cardinal – and observe, that the set {ν < κ : f −1 (ν) ≤ ν} is stationary and {ν < κ : f −1 (ν) < ν} is not stationary in κ. 28) Assume that κ is a regular uncountable cardinal, and let Xα := κ \ α for  every α < κ. Prove that α<κ Xα = κ and α<κ Xα = ∅.

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95

29) Assume that κ is a regular uncountable cardinal and (Xα : α < κ) is a family of subsets of κ. a) Let Yα := Xα ∩ (κ \ (α + 1)) for every α < κ. Prove that α<κ Xα = α<κ Yα .  b) Prove that α<κ Xα = α<κ (Xα ∪ (α + 1)). 30) Assume that κ is a regular uncountable cardinal, f : κ −→ κ is a normal function and C is the set of fixed points of f which are limit ordinals. Prove that C = α<κ (f (α) + 1, κ)ON . 31) (Jech) Assume that κ and λ are cardinals such that ω < κ ≤ λ and cf (κ) = κ. In analogy to the definitions for cardinals, we call a set C ⊆ [λ]<κ closed in [λ]<κ iff, for every chain K (in the partial order ing ([λ]<κ , ⊆)) of members of C with |K| < κ, we have K ∈ C. C is called unbounded in [λ]<κ iff, for every set X ∈ [λ]<κ , there is a set c ∈ C, so that X ⊆ c. Again, every closed and unbounded subset of [λ]<κ is called a club in [λ]<κ . A set S ⊆ [λ]<κ is called stationary in [λ]<κ iff S ∩ C = ∅ for every club C in [λ]<κ . Prove: a) The intersection of less than κ clubs in [λ]<κ is a club in [λ]<κ . b) If (Cξ : ξ < λ) is a sequence of clubs in [λ]<κ , then the diagonal intersection  ξ<λ Cξ := {X ⊆ [λ]<κ : X ∈ {Cξ : ξ ∈ X} } is a club in [λ]<κ . c) If S ⊆ [λ]<κ is stationary in [λ]<κ and Φ : S −→ λ is a choice function for S, then there is a stationary set S0 ⊆ S and an ordinal α, such that ran(Φ |`S0 ) = {α}. Hint: The proof runs parallel to the proof of Lemma 1.8.2 and Theorem 1.8.8. 32) Assume that λ is a cardinal with uncountable cofinality and S is a stationary subset of λ. Prove that S can be partitioned into cf (λ) stationary subsets of λ. Hint: Take a normal function f : cf (λ) −→ λ with sup(ran(f )) = λ and apply Solovay’s theorem.

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1.9 The Erd¨ os-Rado Partition Theorem There are two versions of the pigeonhole principle, a finite one and an infinite one. Pigeonhole Principle (finite version) If t, k ∈ ω \{0}, and if t·k +1 pigeons fly into k holes, then one hole contains t + 1 pigeons. Pigeonhole Principle (infinite version) If κ is a cardinal and μ an infinite cardinal, and if we partition μ into κ sets, where κ < cf (μ), then one set contains μ members. The finite version of the pigeonhole principle has interesting generalizations. We illustrate this by the following example. At every party of six persons, there are three persons who know each other, or there are three persons who don’t know each other. For a proof let {0, 1, ..., 5} = 6 be the set of participants at the party. We partition the set [6]2 of the subsets of 6 containing two elements into two classes by P0 P1

= {{i, j} ∈ [6]2 : i, j know each other}, = {{i, j} ∈ [6]2 : i, j do not know each other}.

Now we fix some person, without loss of generality the person 0. The set M := {{0, i} : 1 ≤ i < 6} is partitioned by the partition (P0 , P1 ) into two classes. Since |M | = 5, we can infer from the pigeonhole principle that there are three persons i1 , i2 , i3 , so that {0, i1 }, {0, i2 }, {0, i3 } are in the same class, say P0 . If {{i1 , i2 }, {i1 , i3 }, {i2 , i3 }} ∩ P0 = ∅ and, without loss of generality, {i1 , i2 } ∈ P0 , then 0, i1 and i2 know each other. On the other hand, if {{i1 , i2 }, {i1 , i3 }, {i2 , i3 }} ∩ P0 = ∅, then i1 , i2 and i3 do not know each other. Definition Let g : A −→ B be a function. Then we know that the set {g−1 [{b}] : b ∈ ran(g)} is a partition of A; thus we will speak of the partition g. If κ, λ, μ and ν are cardinals such that ν ≤ μ and λ ≤ μ, then we write μ −→ (λ)νκ for the following assertion: If K and M are sets with |M | = μ and |K| = κ, and if f : [M ]ν −→ K is a function, then there is a subset L of M with |L| = λ and an element k0 ∈ K such that f [[L]ν ] = {k0 }. L is called homogeneous with respect to the partition f . In the new notation, our previous example can be written as 6 −→ (3)22 . The pigeonhole principles can be formulated as follows.

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97

Pigeonhole Principle (finite version) If t, k ∈ ω \ {0}, then t · k + 1 −→ (t + 1)1k . Pigeonhole Principle (infinite version) If κ and μ are cardinal numbers with 1 ≤ κ < cf (μ), then μ −→ (μ)1κ . In this section we will prove, for n < ω and μ ∈ ICN, a theorem of the form (∗)

μ −→ (λ)nκ

=⇒

∃ν ∈ ICN(ν −→ (λ)n+1 ). κ

An essential step of the proof is the following lemma. Lemma 1.9.1 Assume that n ∈ ω and μ ∈ ICN satisfy μ −→ (λ)nκ . Further let ν ∈ ICN and f : [ν]n+1 −→ κ be a function. If (τi : i < μ) is an injective function from μ into ν satisfying f ({τi1 , . . . , τin , τj }) = f ({τi1 , . . . , τin , τin +1 }) for all i1 < . . . < in < j < μ, then there is a set M0 ⊆ {τi : i < μ} with |M0 | = λ, and k < κ such that f [[M0 ]n+1 ] = {k}. Proof Let us assume that a function (τi : i < μ) with the above property exists. If i1 < . . . < in < μ, then we put g({τi1 , . . . , τin }) := f ({τi1 , . . . , τin , τin +1 }). From the assumption μ −→ (λ)nκ we can infer that there is a set M0 ⊆ {τi : i < μ} with |M0 | = λ and k ∈ κ, such that g[[M0 ]n ] = {k}. If i1 < . . . < in < in+1 < μ and τi1 , . . . , τin , τin+1 ∈ M0 , then the assumption about f gives us f ({τi1 , . . . , τin+1 }) = g({τi1 , . . . , τin }) = k. Lemma 1.9.1 shows that, for the solution of (∗), we must solve the following problem: If f : [ν]n+1 −→ κ is a partition, find an injective sequence (τi : i < μ) of elements of ν, such that f ({τi1 , . . . , τin , τj }) = f ({τi1 , . . . , τin , τin +1 }) for all i1 < . . . < in < j < μ. Note that we did not yet say anything about the size of ν. Let us solve the problem with a tree argument.

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Definition A partial ordering (T, <) is called a tree iff for every x ∈ T the set x ˆ := {y ∈ T : y < x} is well-ordered by <. A branch in (T, <) is a ⊆-maximal chain in (T, <); thus every branch is well-ordered by <. The order type of this well-ordering is the length of the branch. We define the rank function ρ : T −→ ON of (T, <) by ρ(x) := sup{ρ(y) + 1 : y ∈ T ∧ y < x}. The ordinal α := sup{ρ(x) + 1 : x ∈ T } =: l(T ) is called the length of the tree. An α-tree is a tree of length α. The set Sβ := {x ∈ T : ρ(x) = β} is the β-th level of (T, <). Note that Sβ = ∅ iff β ≥ l(T ). To prove a theorem of the form (∗), let us assume that f : [ν]n+1 −→ κ is a partition and ν ∈ ICN is chosen large enough to carry through the following construction. We define by recursion a tree T of length μ whose elements are nonempty subsets of ν, and define the relation < by A < B :⇐⇒ A  B. It is our aim that, for all α < μ, we can arrange the following for the levels Sα of the tree (T, <).   (i) Sα = ν \ {min(A) : A ∈ {Sβ : β < α} }. (ii) Any two different elements of Sα are disjoint. (iii) If ξ < α and B ∈ Sα , then there is a unique set A ∈ Sξ with A < B. For this, we define by recursion the levels Sα of the tree T . We put S0 = {ν}. For the successor step, let us assume that Sξ is already defined and satisfies properties (i)–(iii) for all ξ with ξ ≤ α. If B ∈ Sα , then, by (i)–(iii), there exists a unique chain of sets such that ν = A0  A1  . . .  Aξ  . . .  Aα = B and Aξ ∈ Sξ for all ξ ≤ α. We put τξ := min(Aξ ) for ξ ≤ α and define on B \ {τα } a relation ∼B by ρ ∼B σ :⇐⇒ f ({τξ1 , . . . , τξn , ρ}) = f ({τξ1 , . . . , τξn , σ}) for all ξ1 < . . . < ξn ≤ α. It is easy to check that ∼B is an equivalence relation. Let EB be the set of its equivalence classes, and let  Sα+1 := {EB : B ∈ Sα } be the α + 1-th level of T . By construction, Sα+1 satisfies properties (i)–(iii). We will give an estimate for |Sα+1 |. Obviously we have |Sα+1 | = (|EB | : B ∈ Sα ). For B ∈ Sα and ρ ∈ B \ {τα } we consider the function gρ : [{τξ : ξ ≤ α}]n −→ κ, defined by the equation gρ ({τξ1 , . . . , τξn }) := f ({τξ1 , . . . , τξn , ρ}). Since ρ ∼B σ iff gρ = gσ , the number of equivalence classes is majorized by

1.9. The Erd¨os-Rado Partition Theorem

99

the number of functions from [{τξ : ξ ≤ α}]n into κ, and thus, using Theorem 1.5.14, by κmax{ℵ0 ,|α|} . So we get (iv) |Sα+1 | ≤ |Sα | · κ|α|+ℵ0 ∧ (κ < ℵ0 ∧ α < ℵ0 =⇒ |Sα+1 | < ℵ0 ). Now let α < μ be a limit ordinal. Take   h(ξ) : h ∈ Sξ } \ {∅}. Sα := { ξ<α

ξ<α

 Putting Tα := {Sξ : ξ < α}, we can infer from property (ii) that Sα is the set of nonempty intersections of branches of length α in the tree Tα . Using transfinite induction, we show that properties (i)–(iii) hold for Sα . So assume that  they are satisfied for every level Sξ , ξ < α. For (i),  let ρ ∈ ν \ {min(A) : : A ∈ A ∈ ξ<α Sξ }. If ξ < α, then ρ ∈ ν \ {min(A) β<ξ Sβ }, and thus the  inductive hypothesis and (i) give ρ ∈ Sξ . By (ii), there  is a unique set Aξ satisfying Aξ ∈ Sξ and ρ ∈ Aξ .Let the function h ∈ ξ<α Sξbe defined by h(ξ) := Aξ for ξ < α. Then ρ ∈ {h(ξ) : ξ < α}, and thus  ρ ∈ Sα . For the  proof of the reverse inclusion, let ρ ∈ Sα . Then there is a  function h ∈ ξ<α Sξ such that ρ ∈ {h(ξ) : ξ < α}. Fix ξ < α. If A ∈ Sξ and A = h(ξ), then ρ ∈ A by (ii), and thus ρ = min(A). If A ∈ Sξ and A = h(ξ), then min(A) = min(h(ξ)) ∈ h(ξ + 1) by (i), and from ρ ∈ h(ξ +1) we can conclude that ρ = min(A). This shows that ρ ∈ ν \ {min(A) : A ∈ {Sξ : ξ < α}}, and (i) is proved for Sα . Properties (ii) and (iii) follow from the inductive hypothesis and the construction. Altogether we have shown that (T, <) is a tree with the desired properties. It is obvious that  (v) Lim(α) =⇒ |Sα | ≤ ξ<α |Sξ |. Let us assume that T is a μ-tree. If we can show that T possesses a branch (Ai : i < μ) of length μ, we can complete the proof of (∗). In this case put τi := min(Ai ) for each i < μ. Assume that i1 < . . . < in < j < μ. Then in + 1 ≤ j, and from Aj ⊆ Ain and Ain +1 ⊆ Ain we can infer that τj , τin +1 ∈ Ain . Consequently, τj and τin +1 are elements of the same equivalence class, that means we have f ({τi1 , . . . , τin , τj }) = f ({τi1 , . . . , τin , τin +1 }). Now, assuming μ −→ (λ)nκ , we can apply Lemma 1.9.1, and (∗) is proved. There remains to show, under an appropriate choice of ν, that T has a branch of length μ. First, we prove two lemmas, due to D. K¨ onig and Ramsey. Lemma 1.9.2 (D. K¨onig) If T is an ω-tree with finite levels only, then T has a branch of length ω.

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Proof As usual let Sn , n ∈ ω, be the levels of T . By induction, we define an ω-branch (xi : i < ω) of T . Since T is infinite and S0 is finite, we can choose x0 ∈ S0 such that |{x ∈ T : x0 < x}| = ℵ0 , using the pigeonhole principle. Assume that xn ∈ Sn with |{x ∈ T : xn < x}| = ℵ0 is already defined. Applying the previous argument to the sets {x ∈ T : xn < x} and Sn+1 , we obtain an element xn+1 of Sn+1 satisfying xn < xn+1 and |{x ∈ T : xn+1 < x}| = ℵ0 . The sequence (xn : n < ω) is a branch of T of length ω. Theorem 1.9.3 (Ramsey) If k, n ∈ ω \ {0}, then ℵ0 −→ (ℵ0 )nk . Proof Consider the above construction of the tree T and take μ = ℵ0 , ν = ℵ0 and κ = k < ω. Property (iv) shows that every level of the tree T is finite, hence T is an ω-tree. Lemma 1.9.2 says that T has an ω-branch, and so we can conclude that ℵ0 −→ (ℵ0 )nk

=⇒

ℵ0 −→ (ℵ0 )n+1 . k

Since ℵ0 −→ (ℵ0 )1k is obviously true, an (n − 1)-times application of this assertion proves the lemma. Theorem 1.9.4 (Erd¨ os and Rado) Assume that μ and λ are infinite cardinals with λ ≤ μ. Let further κ be a cardinal with 0 < κ < μ and n ≥ 1 be a natural number. Then μ −→ (λ)nκ

=⇒

(2<μ )+ −→ (λ)n+1 . κ

Proof If μ = ℵ0 , then λ = ℵ0 , since λ ≤ μ. The proof of Theorem 1.9.3 shows that , ℵ0 −→ (ℵ0 )nκ =⇒ ℵ0 −→ (ℵ0 )n+1 κ and thus all the more ℵ0 −→ (ℵ0 )nκ

=⇒

(2<ℵ0 )+ −→ (ℵ0 )n+1 . κ

So we can assume that μ > ℵ0 . Let T be the μ-tree constructed above. We use induction on α to show that |Sα | ≤ 2|α|+ℵ0 +κ for all α < μ. Clearly, |S0 | = 1 < 2ℵ0 +κ . For the successor step we obtain, using property (iv): |Sα+1 | ≤ |Sα | · κ|α|+ℵ0 ≤ 2|α|+ℵ0 +κ · κ|α|+ℵ0 +κ = 2|α|+ℵ0 +κ = 2|α+1|+ℵ0 +κ . If  α < μ is a limit  ordinal, we canconlude with property (v) that |Sα | ≤ ξ<α |Sξ | ≤ ξ<α 2|ξ|+ℵ0 +κ ≤ ξ<α 2|α|+ℵ0 +κ = 2|α|+ℵ0 +κ . So we get  | {Sα : α < μ}| ≤ μ · sup{|Sα | : α < μ} ≤ μ · sup{2|α|+ℵ0 +κ : α < μ} ≤ μ · 2<μ = 2<μ , where the last equation follows from Corollary 1.6.5.

1.9. The Erd¨os-Rado Partition Theorem

101

Now we put ν := (2<μ )+ and define   Sμ := { h(ξ) : h ∈ Sξ } \ {∅}. ξ<μ

ξ<μ

  As in the proof of (i) it can be shown that Sμ = ν \ {min(A) :  A ∈ {Sξ : ξ < μ} }. Clearly we have |{min(A): A ∈ {Sξ : ξ < μ} }| ≤ | {Sξ : ξ < μ}| ≤ 2<μ . So the choice of ν yields Sμ = ∅, hence Sμ = ∅, and therefore T has a branch (Aξ : ξ < μ) of length μ. The above considerations show that this completes the proof of the theorem. Corollary 1.9.5 (Erd¨ os and Rado) If κ is an infinite cardinal number, then (2κ )+ −→ (κ+ )2κ . Proof Obviously, we have κ+ −→ (κ+ )1κ , and from Theorem 1.9.4 we can + + infer that (2<κ )+ −→ (κ+ )2κ . Since 2<κ = 2κ , the assertion is proved. Exercises 1) Every infinite partial ordering has an infinite chain or an infinite antichain. Hint: Use Ramsey’s theorem. 2) Show that 2κ −→ (3)2κ for all cardinals κ ≥ 1. 3) Assume that κ is an infinite cardinal, and let expn (κ) be defined by exp0 (κ) := κ and expn+1 (κ) := 2expn (κ) for n ∈ ω. Show that (expn (κ))+ −→ (κ+ )n+1 . κ Hint: Use Theorem 1.9.4. 4) Assume that κ ∈ ICN and f : [κ]2 −→ 2 is a partition. Prove: If there exists, for every set A ∈ [κ]κ , a member a of A such that |{x ∈ A : f ({a, x}) = 1}| = κ, then there is a set C ∈ [κ]ℵ0 such that ran(f |`[C]2 ) = {1}. 5) (Dushnik, Miller) Assume that κ ∈ ICN and f : [κ]2 −→ 2 is a partition. Show that there is a subset X of κ such that |X| = κ and ran(f |`[X]2 ) = {0} or |X| = ℵ0 and ran(f |`[X]2 ) = {1}. Hint: Assume that the assertion is true for all μ ∈ κ ∩ ICN and that there no set C ∈ [κ]ℵ0 such that ran(f |` [C]2 ) = {1}. By Exercise 4 there is a set A ∈ [κ]κ such that |{α ∈ A : f ({a, α}) = 1}| < κ for every a ∈ A. Now distinguish the cases that κ is regular and κ is singular and apply the inductive hypothesis in the second case.

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Chapter 1. Foundations

6) Show that there is a partition f : [ω]<ω −→ {0, 1} such that, for every infinite subset x of ω, f −1 [{0}] ∩ [x]<ω = ∅ = f −1 [{1}] ∩ [x]<ω . So we get ω −→ (ω)<ω 2 . Hint: If x ∈ [ω]<ω , consider the property |x| ∈ x. 7) Show that there is a partition f : [ω]ω −→ {0, 1} such that, for every infinite subset x ⊆ ω, f −1 [{0}] ∩ [x]ω = ∅ = f −1 [{1}] ∩ [x]ω . So we get ω −→ (ω)ω 2. Hint: Choose a well-ordering ≺ of [ω]ω and consider the property of a set y ∈ [ω]ω that there is a set z ∈ [y]ω with z ≺ y. 8) Show that there is a partition f : [R]2 −→ {0, 1} such that, for every subset x of R with |x| = ℵ1 , f −1 [{0}] ∩ [x]2 = ∅ = f −1 [{1}] ∩ [x]2 . So we get 2ℵ0 −→ (ℵ1 )22 . In particular, the obvious generalization of Ramsey’s theorem, namely ℵ1 −→ (ℵ1 )22 , is false. Hint: Let < be the usual ordering on R and <∗ be a well-ordering of R. If {x, y} ∈ [R]2 and x < y, then x <∗ y or y <∗ x. 9) (Kurepa, Sierpinski) For every infinite cardinal κ, 2κ −→ (κ+ )22 . Hint: In the exercises of Section 1.5 we have seen that there is no strictly increasing or decreasing sequence of length κ+ in P := κ {0, 1} under the lexicographical ordering. Let 2κ =: λ and {fα : α < λ} be an enumeration of κ {0, 1}, and take ≺ as the linear ordering on λ induced by the lexicographical ordering on P . Now define a partition by F ({α, β}) := 1 iff α ∈ β and α ≺ β, F ({α, β} := 0 otherwise. 10) A cardinal κ is called weakly compact iff it is uncountable and satisfies the partition property κ −→ (κ)22 . Prove that every weakly compact cardinal is inaccessible. Hint: To show that κ is regular, assume that κ is the disjoint union of nonempty sets Aξ , ξ < λ, where λ < κ and |Aξ | < κ for all ξ < λ, and consider the partition of [κ]2 given by this partition of κ. To show that κ is strong limit, assume that κ ≤ 2λ for some λ < κ, and use the previous question. Remark An infinite cardinal κ is called a Ramsey cardinal iff it satisfies the partition property κ −→ (κ)<ω 2 . Exercise 6 shows that ω is not a Ramsey cardinal. One can prove (see for example Jech’s book [Je2]): Every measurable cardinal is a Ramsey cardinal, and every Ramsey cardinal is weakly compact and hence inaccessible, see above.

Chapter 2 The Galvin-Hajnal Theorem The so-called singular cardinal problem consists of the description of the poscf(ℵ ) sible size of the cardinal ℵη η , that is ‫(ג‬ℵη ), the value of the gimel function at the argument ℵη , for singular cardinals ℵη . An estimate for this cardinal power is given by the Galvin-Hajnal theorem if ℵη is an ℵ0 -strong singular cardinal with uncountable cofinality. The centre of our investigations will be the Galvin-Hajnal formula, from which all other results on cardinals in this chapter will follow. For the first time it turns out that a profound cardinal property is a source of cardinal arithmetic.

2.1 Ideals and the Reduction of Relations In the following let A be a nonempty set and I be an ideal on A, that is a nonempty set of subsets of A which does not contain the member A and is closed under finite unions and under inclusions. Any ideal on A provides us with a measure of the size of a set. Elements of I are subsets of A which we can regard as small. We also say that they are of measure zero modulo I. If κ is an uncountable regular cardinal, then Ib (κ) and Id (κ), the set of bounded and the set of thin subsets of κ respectively, are two ideals on κ which will play an important role in this book. We remindthe reader of some denotations. If I is an ideal on the set A, then dom(I) := I is the domain of I and I + := {X ⊆ A : X ∈ I} is the set of subsets of A which are positive modulo I. We say that the sets X ∈ I are null sets or of measure zero modulo I. Any union of two positive sets or null sets is positive or a null set, respectively, any intersection of two null sets is a null set, subsets of null sets are null sets, and supersets of positive sets are positive. If X is positive and Y is a null set, then X \ Y is positive. Since I = P(A), the

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set (A \ X) ∩ (A \ Y ) is positive for any null sets X and Y . Note that X and A \ X can both be positive. Definition We say that a property ϕ(a) holds modulo I for almost all a ∈ A iff the set {a ∈ A : ¬ϕ(a)}, that means the set of exceptions, is a member of I. The elements of the set I + := {X ⊆ A : X ∈ I} are called I-positive sets or sets of positive measure modulo I. Furthermore we say that an ideal J extends an ideal I iff I ⊆ J. Often we will consider ultrafilters instead of maximal ideals. If D is an ultrafilter on A and I is an ideal on A with D ∩ I = ∅, then I is a subset of the maximal ideal dual to D. This justifies the phrase D extends I for ultrafilters D on A with D ∩ I = ∅. To give an example, let us consider the assertion B ⊆I C. It is to be read as “B is a subset of C modulo I”, for B and C subsets of A. B ⊆I C means that modulo I for almost all a ∈ A the implication a ∈ B =⇒ a ∈ C is true. We are mainly interested in such properties later on, when we will investigate products of structures and especially products of ordinals with the standard well-ordering ∈. Definition Let A be a nonempty set. Remember that we have called functions from A into ON ordinal functions on A. We will omit the addition “on A” when the context makes it plain which basic set is meant. Usually, we denote ordinal functions by f, g, h, also with indices. In this chapter, families of almost disjoint functions F, G, . . ., whose members we also denote by f, g, . . ., will play an important role. Therefore we will denote in this chapter, and sometimes in later chapters, ordinal functions also by Φ and Ψ. If f and g are ordinal functions on A and I is an ideal on A, then we say that f is modulo I less than g, and write f
2.1. Ideals and the Reduction of Relations B ⊆I C B =I C f =I g f ≤I g f
:⇐⇒ :⇐⇒ :⇐⇒ :⇐⇒ :⇐⇒ :⇐⇒

105

B \ C ∈ I, (B \ C) ∪ (C \ B) ∈ I, {a ∈ A : f (a) = g(a)} ∈ I, {a ∈ A : g(a) < f (a)} ∈ I, {a ∈ A : g(a) ≤ f (a)} ∈ I, f ≤I g ∧ ¬(f =I g).

Remark The definition of the relation I is a bit out of the ordinary, since it is not a property that holds almost everywhere, but only a property that holds for all members of an I-positive set. We have included it for reasons which will be discussed later on. Lemma 2.1.1 Assume that A is a nonempty set and I is an ideal on A. Then for all subsets B, C and D of A, the following holds. 1. 3. 5. 7. 9.

B =I B. B =I C ∧ C =I D =⇒ B =I D. B ⊆I B. B ⊆ C =⇒ B ⊆I C. B ⊆I C ⇐⇒ A \ C ⊆I A \ B.

2. 4. 6. 8. 10.

B =I C =⇒ C =I B. B = C =⇒ B =I C. B ⊆I C ∧ C ⊆I D =⇒ B ⊆I D. B =I C ⇐⇒ B ⊆I C ∧ C ⊆I B. B ⊆I C ∧ C ∈ I =⇒ B ∈ I.

Lemma 2.1.2 Assume that A is a nonempty set and I is an ideal on A. Then for all ordinal functions f , g and h on A, the following holds. 2. f =I g =⇒ g =I f. 1. f =I f. 3. f =I g ∧ g =I h =⇒ f =I h. 4. f = g =⇒ f =I g. 5. f ≤I f. 6. f ≤I g ∧ g ≤I f ⇐⇒ f =I g. 8. f ≤ g =⇒ f ≤I g. 7. f ≤I g ∧ g ≤I h =⇒ f ≤I h. 10. ¬(f
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Chapter 2. The Galvin-Hajnal Theorem

Lemma 2.1.3 Assume that A is a nonempty set, I is an ideal on A, and J is an ideal on A extending I. Furthermore let ϕ(x) be a formula such that ϕ(a) holds modulo I for almost all a ∈ A. Then ϕ(a) holds modulo J for almost all a ∈ A. In particular this applies to maximal ideals J which are dual to ultrafilters extending I. Thus, if D is an ultrafilter on A extending I, then {a ∈ A : ϕ(a)} ∈ D . Remark By the previous lemma, the relations =I , ⊆I , ≤I and
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107

Lemma 2.1.5 Let A be a nonempty set and D be an ultrafilter on A. If f and g are ordinal functions on A, then a) b) c) d)

f f f f

≤D ≤D
g g g g

∨ g ≤D f . ⇐⇒ f
Exercises 1) Show: If I is an ideal on A and X ∈ I, then X ⊆I Y for every subset Y of A. 2) Show: If I is an ideal on A, X ∈ P(A) \ I, and Z ∈ I, then I |`(X ∪ Z) = I |`X. 3) Assume that I is an ideal on A and X1 , X2 ∈ P(A) \ I. Show that I |`(X1 ∪ X2 ) = I |`X1 ∩ I |`X2 . 4)

a) Give examples of a set A, an ideal I on A, and functions f, g ∈ A ON such that a1) f I g and not f
b) Prove that the relations I and
 7) Assume that I is an ideal on A. Show that I is principal iff I ∈ I. 8) Assume that A is an infinite set and B is a nonempty subset of A. Further let I be an ideal on B and, for each b ∈ B, Ib be an ideal on A. Prove that J := {X ⊆ A : {b ∈ B : X ∈ Ib } ∈ I} is an ideal on A.

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Chapter 2. The Galvin-Hajnal Theorem

9) If I is an ideal on the cardinal κ, then 0 is not an element of any diagonal union. 10) Assume that I is an ideal on A, γ is an ordinal, and (dα : α < γ) is a sequence of I-positive subsets of A such that dβ ⊆I dα for all α, β ∈ γ with α < β. Prove that there is an ultrafilter D on A such that D ∩ I = ∅ and dα ∈ D for all α < γ. Hint: Prove that {dα : α < γ} ∪ {Y ⊆ A : A \ Y ∈ I} has the finite intersection property. 11) If κ is a singular cardinal, then there is no normal ideal on κ which contains all bounded subsets of κ. Hint: As usual, κ is the supremum of a strictly increasing sequence (κα : α < cf (κ)) of cardinals with κ0 ≥ cf (κ). Assume that there is such an ideal. Consider the sequence (Xα : α < κ), where Xα := κα for α < cf (κ), and show that κ \ cf (κ) ⊆ α α : fη (α) = ξ}. The coefficients in the columns of  this matrix are pairwise disjoint subsets of κ. For every α < κ we have {Xα,ξ : ξ < λ} = (α, κ)ON . Since (α, κ)ON ∈ I, there is ξα < λ such that Xα,ξα ∈ I. 13) Assume that κ is a singular cardinal. Show that every κ-complete ideal on κ is principal.  Hint: If I is a κ-complete  ideal on κ, then note that α ∩ I ∈ I for every α < κ and show that I ∈ I.

2.2 The Galvin-Hajnal Formula Definition Every function K : A −→ P(A) is called a set function. We call a subset B of A independent with respect to the set function K iff x ∈ K(y) for all x, y ∈ B with x = y. We can interpret a set function as a function which assigns to each x ∈ A the set of people which are known to x, where this relation needs not be symmetric. An independent set can then be interpreted as a subset of A all of whose members do not know each other. Note that ∅ is independent and that the set of independent subsets of A is closed under union of chains. By Zorn’s lemma, there exists an ⊆-maximal independent subset of A; by the same reasons, we obtain, for any subset Y of A, an independent subset of A which is ⊆-maximal w.r.t. the property of being disjoint to Y .

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109

Lemma 2.2.1 (L´az´ar) Assume that κ is a regular and λ is a – finite or infinite – cardinal with λ < κ, A is a set with |A| = κ, and K : A −→ P(A) is a set function such that |K(x)| < λ for all x ∈ A. Then there is a subset B of A of cardinality κ which is independent w.r.t. K.1 Proof For any subset X of A let Xu := {a ∈ A : K(a) ∩ X = ∅} be the set of members a of A who do not know any member of X. We  call X popular iff |Xu | < κ and unpopular iff |Xu | = κ. Further let kX := {K(x) : x ∈ X} be the set of members of A which are known to members of X. If |X| < κ, then by the assumption on K we have |kX | < κ since κ is regular. First we notice the following fact: If Z ⊆ A is independent and unpopular, |Z| < κ, Y ⊆ A, |Y | < κ, and Z ∩ Y = ∅, then Zu \ (kZ ∪ Y ∪ Z) = ∅. If we take a member z of this set, we have K(z) ∩ Z = ∅ and z ∈ kZ ∪ Y . Thus Z ∗ := Z ∪ {z} is independent and satisfies Z ∗ ∩ Y = ∅. If we can choose Y in such a way that |Y | < κ and Z ∩ Y = ∅ implies that Z is unpopular, then any ⊆-maximal independent set Z disjoint from Y , which exists by Zorn’s lemma, must be of size κ. It is clear that the condition Z ∩ Y = ∅ implies that Z is unpopular if Y is the union of an ⊆-maximal set of pairwise disjoint popular sets which again exists by Zorn’s lemma. If we can show that |Y | < κ for such a set Y , we get the desired result. So assume that  P is a set of pairwise disjoint popular sets such that |P| = λ. The set W := {Xu : X ∈ P} is of size less than κ since κ is regular. Choose y ∈ A\W . Then K(y)∩X = ∅ for all X ∈ P, and consequently K(y) includes a set of representatives for P. So we get |P| ≤ |K(y)| < λ. This contradiction shows that |P| < λ for every ⊆-maximal set Pof pairwise disjoint popular sets. Thus we can choose such a set P, and Y := P has the desired properties. Lemma 2.2.2 If I is a σ-complete ideal on A and α is an ordinal, then the relation
singular cardinals this was proved by A. Hajnal.

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Chapter 2. The Galvin-Hajnal Theorem

Definition Let α be an ordinal and I be a σ-complete ideal on the set A. Lemma 2.2.2 says that α, #Φ#I,α = #Φ#I,β . Proof We prove the assertion by transfinite induction on the well-founded relation
a) b) c) d)

Proof For c), assume that Φ1 ≤I Φ2 . Then, by Lemma 2.1.2, #Φ1 #I = sup{#Ψ#I + 1 : Ψ
2.2. The Galvin-Hajnal Formula

111

By the definition of #Φ#, there is a function Ψ0 ∈ A α such that Ψ0 ω is a regular cardinal, Φ ∈ κ ON, and I is a κ-complete normal ideal on κ with dom(I) = κ. Then, for every ordinal σ < κ, #Φ#I ≤ σ

⇐⇒

Sσ := {ξ < κ : Φ(ξ) ≤ σ} ∈ I + .

Proof Let σ < κ and first #Φ#I ≤ σ. Assume, to get a contradiction, that Sσ ∈ I. We put Ψη := κ × {η} for all η < κ. Then Ψη
112

Chapter 2. The Galvin-Hajnal Theorem

by the inductive hypothesis. These considerations hold for all Ψ with Ψ
In the case that σ ∈ ON and Ψ ∈ κ ON is defined by Ψ(ξ) = σ for all ξ < κ, we write T (κ, ℵσ ) instead of T (ℵ ◦ Ψ). Theorem 2.2.6 (Galvin-Hajnal Formula) Assume that κ > ω is a regular cardinal and I is a κ-complete normal ideal on κ with dom(I) = κ. Further let (η(ξ) : ξ < κ) be an increasing continuous sequence of ordinals and ℵΘ := 2κ · sup{T (κ, ℵη(ξ) ) : ξ < κ}. If Φ ∈ κ ON is a function and Ψ ∈ κ ON is defined by Ψ(ξ) := η(ξ) + Φ(ξ), then T (ℵ ◦ Ψ) ≤ ℵΘ+ Φ I .

 Proof 2 Let F ⊆ ξ<κ ℵη(ξ)+Φ(ξ) be a set of almost disjoint functions. To show that T (ℵ ◦ Ψ) ≤ ℵΘ , we use transfinite induction on the ordinal #Φ#I and will distinguish the cases #Φ#I = 0 and #Φ#I > 0. In the first case, we will represent F, using the fundamental property of normal ideals given by Lemma 1.8.9, as a union of sets FX,β for certain sets X ⊆ κand ordinals  ⊆ ξ∈κ ℵη(β) of β < κ. FX,β can be mapped one-to-one onto a set FX,β almost disjoint functions. This will already motivate the choice of Θ: There are at most 2κ such pairs (X, β), and the definition of T (κ, ℵη(β) ) will yield |FX,β | ≤ T (κ, ℵη(β) ), hence |F| ≤ 2κ · sup{T (κ, ℵη(ξ) ) : ξ < κ} = ℵΘ . The proof of the second case is more complicated and will apply two combinatorial results, Lemma 2.2.1 and the Erd¨ os-Rado theorem. Case 1 #Φ#I = 0. Observe first that Lemma 1.8.6 says that Ib (κ) ⊆ I, hence every set X ∈ I + has cardinality κ. Together with Lemma 2.2.5, Case 1 gives {ξ < κ : Φ(ξ) = 0} ∈ I + . By Lemma 1.8.6, the set {ξ +1 : ξ < κ} is a member of I. Furthermore dom(I) = κ yields {0} ∈ I, and consequently we have X0 := {ξ < κ : Φ(ξ) = 0 ∧ Lim(ξ)} ∈ I + . 2 We

follow the proof in [GH].

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113

 Fix f ∈ F. Since F ⊆ ξ<κ ℵη(ξ)+Φ(ξ) , we get f (ξ) < ℵη(ξ) for any ξ ∈ X0 . By assumption, the function (η(ξ) : ξ < κ) is continuous, and so there is an ordinal αf (ξ) < ξ such that f (ξ) < ℵη(αf (ξ)) . Thus we obtain the regressive function (αf (ξ) : ξ ∈ X0 ), and from Lemma 1.8.9 and the fact that I is normal we can conclude that there is a set Xf ∈ P(X0 ) \ I  and an ordinal βf < κ such that αf (ξ) = {βf } for all ξ ∈ Xf , hence f |`Xf ∈ ξ∈Xf ℵη(βf ) . Fix X ∈ P(X0 ) \ I and β < κ and let  FX,β := {f ∈ F : f |`X ∈ ℵη(β) }. ξ∈X  := Every member of P(X0 ) \ I is unbounded in κ, hence we obtain via FX,β  {f |` X : f  ∈ FX,β } a set of almost disjoint functions with |FX,β | = |FX,β |, included in ℵ . Any bijection from X onto κ induces a bijection from η(β) ξ∈X  ξ∈X ℵη(β) onto ξ<κ ℵη(β) which preserves the property of two functions of being almost disjoint since κ is regular. Therefore we get |FX,β | ≤ T (κ, ℵη(β)) ≤ ℵΘ .  Since F = {FX,β : X ∈ P(X0 ) \ I and β < κ} and there are only 2κ such pairs (X, β), our assertion is proved for the case #Φ#I = 0. Case 2 #Φ#I =: α > 0.  κ Let us assume that the assertion is true for every function Φ  ∈ ON with  #Φ #I < α. Again we want to represent our given set F ⊆ ξ<κ ℵη(ξ)+Φ(ξ) of almost disjoint functions as a union of sets which are not too large; more precisely, as the union of |α| sets, since |α| ≤ α ≤ ℵΘ+ Φ I . Fix f ∈ F. If Φ(ξ) = 0, then there is a least δ < Φ(ξ) such that |f (ξ)| ≤ ℵη(ξ)+δ . But the assumption #Φ#I > 0 implies {ξ < κ : Φ(ξ) = 0} ∈ I by Lemma 2.2.5, and thus Φ(ξ) > 0 holds modulo I for almost all ξ < κ. So we can conclude, if we define functions Φf ∈ κ ON and Ψf ∈ κ ON by

Φf (ξ) := min{δ : |f (ξ)| ≤ ℵη(ξ)+δ } and Ψf (ξ) := η(ξ) + Φf (ξ), that Φf (ξ) < Φ(ξ) holds modulo I for almost all ξ <κ. Thus we get Φf
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Chapter 2. The Galvin-Hajnal Theorem

 f , we have g ∈ ξ<κ (f (ξ)+1). If ξ < κ, then we know that |f (ξ)| ≤ ℵη(ξ)+Φf (ξ) and can choose an injection iξ from  f (ξ)+1 into ℵη(ξ)+Φf (ξ) . For each g ∈ P r(f ) we can define a function g∗ ∈ ξ<κ ℵη(ξ)+Φf (ξ) by g ∗ (ξ) := iξ (g(ξ)). If we put )| = |P r∗ (f )|, and P r∗ (f ) is a set P r∗ (f ) := {g ∗ : g ∈ P r(f )}, then |P r(f of almost disjoint functions included in ξ<κ ℵη(ξ)+Φf (ξ) . From #Φf #I < α and the inductive hypothesis we infer that |P r(f )| = |P r ∗ (f )| ≤ T (ℵ ◦ Ψf ) ≤ ℵΘ+ Φf I = ℵΘ+σ . Now we can apply our combinatorial results. Let us assume, to get a contradiction, that |Fσ | ≥ ℵΘ+σ+2 for some σ < α. Without loss of generality let |Fσ | = ℵΘ+σ+2 , since otherwise we could take a suitable subset of Fσ . Then |Fσ | is a regular cardinal, and we have shown that the assumptions of Lemma 2.2.1 are satisfied if we choose the set function P r : Fσ −→ P(Fσ ) which we have defined by P r(f ) = {g ∈ Fσ : g ≤κ f }, since |P r(f )| ≤ ℵΘ+σ for all f ∈ Fσ . Consequently there exists a subset of Fσ with cardinality ℵΘ+σ+2 which is independent with respect to P r. By the definition of Θ, we have ℵΘ+σ+2 > 2κ , and so there is an injective sequence (fξ : ξ < (2κ )+ ) of elements of Fσ such that fζ ∈ P r(fξ ) for any ζ and ξ with ζ < ξ < (2κ )+ . The last assertion implies that there is a smallest ordinal l(ζ, ξ) < κ such that fζ (l(ζ, ξ)) > fξ (l(ζ, ξ)). By this property, a function l : [(2κ )+ ]2 −→ κ is defined. Now the Erd¨ os-Rado theorem, Corollary 1.9.5, says that (2.1) (2κ )+ −→ (κ+ )2κ . Consequently there is a set A ⊆ (2κ )+ with |A| = κ+ and an ordinal γ < κ satisfying ran(l |`[A]2 ) = {γ}. In particular, we can choose a strictly increasing sequence (ξn : n < ω) of members of A. Then, by construction, (fξn (γ) : n ∈ ω) is a strictly decreasing sequence of ordinals. This contradiction completes the proof of the case #Φ#I > 0 and the proof of the theorem. Lemma 2.2.7 Let κ be an infinite cardinal  and(γ(ξ) : ξ < κ) be a sequence of ordinals. Then there is a set F ⊆ α<κ ( ξ<α ℵγ(ξ) ) of almost disjoint functions such that  |F| = ℵγ(ξ) . ξ<κ

 Proof We put μ := | ξ<κ  ℵγ(ξ) | and choose an enumeration (gξ : ξ < μ) of the cartesian product ξ<κ ℵγ(ξ) . For ξ < μ let the function fξ ∈   ( ℵ ) be defined by fξ (α) = gξ |` α. Then F = {f α<κ ξ<α γ(ξ) ξ : ξ < μ} is a set of almost disjoint functions, and in particular |F| = μ = ξ<κ ℵγ(ξ) .

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115

Theorem 2.2.8 Assume that κ is an uncountable regular cardinal, I is a normal κ-complete ideal on κ with dom(I) = κ, (η(ξ) : ξ < κ) is an increasing continuous sequence of ordinals, and ℵΘ := 2κ · sup{T (κ, ℵη(ξ) ) : ξ < κ}. If Φ ∈ κ ON and (ℵγ(ξ) : ξ < κ) is a sequence of cardinals satisfying 

ℵγ(ξ) ≤ ℵη(α)+Φ(α)

ξ<α

for all α < κ, then



ℵγ(ξ) ≤ ℵΘ+ Φ I .

ξ<κ

Proof By Lemma 2.2.7, there is a set F of almost disjoint functions satisfying    ( ℵγ(ξ) ) and |F| = ℵγ(ξ) . F⊆ 

α<κ ξ<α

ξ<κ

Since ξ<α ℵγ(ξ)≤ ℵη(α)+Φ(α) , there is a set F  of almost disjoint functions such that F  ⊆ α<κ ℵη(α)+Φ(α) and |F| = |F  |. If Ψ ∈ κ ON is defined by Ψ(ξ) := η(ξ) + Φ(ξ), then T (ℵ ◦ Ψ) ≤ ℵΘ+ Φ I by Theorem 2.2.6, and with |F  | ≤ T (ℵ ◦ Ψ) the assertion of the theorem follows. It is obvious that an estimate for #Φ#I would improve the preceding theorem. We will obtain such an estimate in Lemma 2.2.10. For its proof we need the following technical lemma. Lemma 2.2.9 Assume that A is a nonempty set, I is a σ-complete ideal on A, Φ ∈ A ON is a function, and X, Y ∈ I + . Then a) If X ⊆ Y , then #Φ#I |`Y ≤ #Φ#I |`X . b) If Z ∈ I, then #Φ#I |`(X∪Z) = #Φ#I |`X . c) #Φ#I |`(X∪Y ) = min{#Φ#I |`X , #Φ#I |`Y }. Proof For part a), we have I |` Y = I[A \ Y ] ⊆ I[A \ X] = I |` X, and therefore Ψ
116

Chapter 2. The Galvin-Hajnal Theorem For c), we note that part a) implies

(1)

#Φ#I |`(X∪Y ) ≤ min{#Φ#I |`X , #Φ#I |`Y }.

We prove the remaining inequality (2)

min{#Φ#I |`X , #Φ#I |`Y } ≤ #Φ#I |`(X∪Y )

by transfinite induction on the well-founded relation
#Φ#I |`X ≤ #Φ#I |`(X∪Y ) .

On account of #Φ#I |`X = sup{#Ψ#I |`X + 1 : Ψ
Ψ
So assume that Ψ
Ψ
By the inductive hypothesis, we have #Ψ #I |`(X∪Y ) = min{#Ψ #I |`X ,#Ψ #I |`Y }. Furthermore Lemma 2.2.3 implies #Ψ #I |`Y = #Ψ #I |`Y = #Ψ#I |`X = #Ψ #I |`X , since Ψ =I |`Y Ψ and Ψ =I |`X Ψ. So we get #Ψ #I |`(X∪Y ) = #Ψ #I |`X = #Ψ #I |`Y . Now (5) yields #Ψ#I |`X = #Ψ #I |`X = #Ψ #I |`(X∪Y ) < #Φ#I |`(X∪Y ) , hence (4) and hereby (3) is proved. Case 2 X ∩ Y = ∅. If X \ Y ∈ I, then Y ∪ (X \ Y ) = X ∪ Y , and from b) we can conclude that #Φ#I |`(X∪Y ) = #Φ#I |`Y ≥ #Φ#I |`X . If X \ Y ∈ I, we can apply Case 1 to the sets X \ Y and Y . Therefore a) yields the estimate #Φ#I |`X ≤ min{#Φ#I |`(X\Y ) , #Φ#I |`Y } = #Φ#I |`(X∪Y ) . This completes the proof of (3). Lemma 2.2.10 Let κ be an uncountable regular cardinal and I be a κ-complete normal ideal on κ with dom(I) = κ. Then

2.2. The Galvin-Hajnal Formula

117

 a) For every function Φ ∈ κ ON, #Φ#I < ( ξ<κ |Φ(ξ) + 1|)+ . b) For every function Φ ∈ κ ON, I  := I ∪ {Y ⊆ κ : Y ∈ I + and #Φ#I < #Φ#I |`Y } is a κ-complete ideal on κ with dom(I  ) = κ. c) (Shelah) If λ > κ is a regular cardinal with μκ < λ for every cardinal μ < λ with cf (μ) = κ, then there is, for every function Φ ∈ κ λ with #Φ#I ∈ Lim, a set Y ∈ P(κ)\I  and a set F ⊆ κ λ with |F| < λ, such that F is cofinal in Φ modulo I |`Y ; that means that f κ is a regular cardinal satisfying μκ < λ for all cardinals μ < λ with cf (μ) = κ, and if Φ ∈ κ λ, then #Φ#I < λ. Proof For part a), Lemma 2.2.3 says that there is a sequence (Ψα : α < #Φ#I ) of elements of κ ON, so that #Ψα #I = α and Ψα
If Ψ ∈ κ ON, γ ∈ ON and, for all α < β, γ ≤ #Ψ#I |`Yα , then γ ≤ #Ψ#I |`Y .

We assume that this assertion is true for all Ψ ∈ κ ON and all γ  < γ and that, for a given function Ψ, we have γ ≤ #Ψ#I |`Yα for all α < β. Now let γ  < γ be fixed. Using Lemma 2.2.3 we can find, for each α < β, a function Ψα ∈ κ ON, so that Ψα
118

Chapter 2. The Galvin-Hajnal Theorem

Obviously Ψ κ, then let G be our required set F of functions. If ξ ∈ Y , then cf (Φ(ξ)) = μ ≤ Φ(ξ) < λ, and so we get |F| ≤ μ < λ. If μ ≤ κ, then let F be the set of all functions f ∈ κ ON satisfying f (ξ) ∈ {gα (ξ) : α < μ} for all ξ < κ. By assumption, we can conclude that |F| ≤ μκ ≤ κκ < λ. Now in both cases one can easily verify that F is cofinal in Φ modulo I |`Y . Case 2 cf ◦ Φ is not constant on any set Z ∈ P(κ) \ I  . For each cardinal ν < λ, let Xν := {ξ < κ : cf (Φ(ξ)) = ν} (for ν ≥ λ, this set is empty). By Case 2, we have Xν ∈ I  for all ν < λ. Let μ be the smallest cardinal such that  Y := {Xν : ν < μ} \ (X0 ∪ X1 ) ∈ I  .  Since κ\(X0 ∪X1 ) = {Xν : ν ∈ λ}\(X0 ∪X1 ) ∈ I  , such a cardinal exists, and we have μ ≤ λ. Next we observe that μ is a limit cardinal. Namely, if μ = ρ+  and T := {ν < μ : ν∈ {0, 1} ∧ Xν = ∅}, then Y = {Xν : ν ∈ T ∩ ρ} ∪ Xρ , the choice of μ. and Xρ ∈ I  gives {Xν : ν ∈ T ∩ ρ} ∈ I  , contradicting  Now the set T is unbounded in μ, and we have Y = {Xν : ν ∈ T }. From the fact that I  is κ-complete and Y ∈ I  , we can conclude that |T | ≥ κ. Since the sets Xν , ν ∈ T , are pairwise disjoint nonempty subsets of κ, we have |T | = κ, hence cf (μ) ≤ κ. Since T is unbounded in μ, we can easily find a subset T  of T which  is unbounded in μ and has the cardinality cf (μ). For ν ∈ T  , put Zν := {Xρ : ρ ∈ T ∩ ν}; then Zν ∈ I  by the choice of μ, and

2.2. The Galvin-Hajnal Formula

119

 Y = {Zν : ν ∈ T  }, since T  is unbounded in T . So the assumption cf (μ) < κ leads to Y ∈ I  , contradicting again the choice of μ. Therefore we get cf (μ) = κ. Since λ is regular, κ < λ and μ ≤ λ, we get μ < λ and thus, by assumption, μκ < λ. Now we choose, for each ordinal ξ ∈ Y with Φ(ξ) = 0, a cofinal subset Cξ of Φ(ξ) of order  type cf (Φ(ξ)); for the remaining ξ ∈ κ we put Cξ := {0}. We define F := ξ<κ Cξ . Then |F| ≤ μκ , hence |F| < λ. By construction, F is cofinal in Φ modulo I |`Y . For the proof of part d), we assume, to get a contradiction, that there is a function Φ ∈ κ λ such that #Φ#I ≥ λ. If #Φ#I > λ, Lemma 2.2.3 yields a function Ψ ∈ κ λ such that #Ψ#I = λ. So we can assume without loss of generality that #Φ#I = λ. By part c), there is a set F ⊆ κ λ with |F| < λ and a set Y ∈ P(κ) \ I  , such that F is cofinal in Φ modulo I |`Y . Since Y ∈ I  , we have #Φ#I = λ = #Φ#I |`Y . We will show that #Φ#I |`Y = sup{#f #I |`Y : f ∈ F} which will contradict the assumption that λ is a regular cardinal. By definition, we have #Φ#I |`Y = sup{#Ψ#I |`Y + 1 : Ψ
120

Chapter 2. The Galvin-Hajnal Theorem c) If #Φ#I is a limit ordinal, then the set {a ∈ A : Φ(a) is a limit ordinal } is I-positive. Can one show that Φ(a) is a limit ordinal for almost all a ∈ A modulo I? d) If Φ(a) is a limit ordinal for almost all a ∈ A modulo I, then #Φ#I is a limit ordinal.

4) Assume that κ > ℵ0 is a regular cardinal, Φ ∈ κ ON, I = Id (κ), σ < κ is an ordinal, and {ξ < κ : Φ(ξ) = σ} is a club in κ. Prove that #Φ(ξ)#I = σ. 5) Assume that κ > ω is a regular cardinal. Prove that there are functions Φ, Ψ ∈ κ ON such that Φ |` X = Ψ |` X for some X ∈ Id (κ)+ and #Φ#Id (κ) = #Ψ#Id (κ) . Hint: Use Lemma 2.2.5. 6) Assume that κ is an uncountable regular cardinal and Φ ∈ κ ON is a function. Prove: If σ < κ and S := {ξ < κ : Φ(ξ) < σ} is stationary in κ, then #Φ#Id (κ) < σ. 7) Assume that κ > ω is a regular cardinal, I := Id (κ), and α < κ. Further let Φ ∈ κ κ, + be the ordinal sum, and S := {ξ < κ : Φ(ξ) ≤ ξ + α} be stationary in κ. Prove that #Φ#Id (κ) ≤ κ + α. Hint: Use Fodor’s theorem, Lemma 1.8.4 and Lemma 2.2.5. 8) Assume that κ > ω is a regular cardinal, I := Id (κ), and α < κ. Further let #Φ# := #Φ#Id (κ) for any Φ ∈ κ κ and + be the ordinal sum. Prove: If Ψα (ξ) := ξ + α for all ξ < κ, then #Ψα # = κ + α. 9) Assume that κ > ω is a regular cardinal, I := Id (κ), Φ ∈ κ κ, and + is the ordinal sum. Prove: a) If S := {ξ < κ : Φ(ξ) ≤ ξ + ξ} is stationary in κ, then #Φ#Id (κ) ≤ κ + κ. b) If Φ(ξ) := ξ + ξ for all ξ < κ, then #Φ#Id (κ) = κ + κ. Hint: Use transfinite induction, Fodor’s theorem and Lemma 1.8.4. 10) (Galvin, Hajnal) Assume that κ is an uncountable regular cardinal and I is a κ-complete normal ideal on κ with dom(I) = κ. Prove without using Lemma 2.2.10: a) If μ is an infinite cardinal and Φ ∈ κ μ, then #Φ#I < (μκ )+ . b) If λ is an uncountable, regular, κ-strong cardinal and Φ ∈ κ λ, then #Φ#I < λ. 11) Assume that κ is an uncountable regular cardinal, α ≥ 2 is an ordinal, and I is an ideal on κ. a) Prove that =I is an equivalence relation on κ α.

2.3. Applications of the Galvin-Hajnal Formula

121

b) For the case that I = Ib (κ) find two equivalence classes A and B w.r.t. =I such that, for any f ∈ A and g ∈ B, f and g are not almost disjoint. 12) Assume that κ and λ are cardinals satisfying ω ≤ λ ≤ κ. Prove that there is a set F ⊆ λ κ of almost disjoint functions such that |F| > λ. Hint: Every subset of λ κ of almost disjoint functions which is maximal under inclusion has cardinality greater than λ.

2.3 Applications of the Galvin-Hajnal Formula Corollary 2.3.1 Assume that λ and κ are regular cardinals satisfying ℵ0 < κ < λ and ν κ < λ for every cardinal ν < λ with cf (ν) = κ. Further let μ be an infinite cardinal with μκ < μ+λ , and (ℵγ(ξ) : ξ < κ) be a sequence of cardinals satisfying  ℵγ(ξ) < μ+λ ξ<α

for all α < κ. Then



ℵγ(ξ) < μ+λ .

ξ<κ

Proof To apply Theorem 2.2.8, let Φ ∈ κ ON be defined by  ℵγ(ξ) ≤ μ+β }. Φ(α) := min{β ∈ ON : ξ<α

By assumption, we have Φ ∈ κ λ, and Lemma 2.2.10 d) tells us that #Φ#I < λ for every κ-complete normal ideal I on κ with dom(I) = κ, in particular for I := Id (κ)  by Lemma 1.8.5. If μ =: ℵβ and η(ξ) := β for all ξ < κ, then we have ξ<α ℵγ(ξ) ≤ ℵβ+Φ(α) , and the cardinal ℵΘ in Theorem 2.2.8 can be estimated by ℵΘ ≤ 2κ · μκ = μκ < μ+λ = ℵβ+λ . Therefore there is σ < λ such that ℵΘ = ℵβ+σ . From this together with #Φ#I < λ and Lemma 1.5.9 we can conclude that ℵΘ+ Φ I = ℵβ+σ+ Φ I < ℵβ+λ . Moreover, the definitions of the assumptions of Theorem 2.2.8. So we get  the functions η and Φ satisfy +λ ℵ ≤ ℵ < μ , as desired. Θ+ Φ I ξ<κ γ(ξ) Corollary 2.3.2 Assume that ℵη is a κ-strong singular cardinal, where κ := cf (ℵη ) > ω, Φ ∈ κ ON is a function, (η(ξ) : ξ < κ) is a normal function cofinal in η, and  F⊆ ℵη(ξ)+Φ(ξ) ξ<κ

122

Chapter 2. The Galvin-Hajnal Theorem

is a set of almost disjoint functions. Then |F| ≤ ℵη+ Φ Id (κ) . In other words, if Ψ ∈ κ ON is defined by Ψ(ξ) := η(ξ) + Φ(ξ), then T (ℵ ◦ Ψ) ≤ ℵη+ Φ Id (κ) . Proof We get an estimate for the cardinal ℵΘ defined in Theorem 2.2.6 by ℵΘ = 2κ · sup{T (κ, ℵη(ξ) ) : ξ < κ} ≤ 2κ · sup{ℵκη(ξ) : ξ < κ} ≤ 2κ · ℵη = ℵη . From Theorem 2.2.6 we can conclude, putting Ψ := (η(ξ) + Φ(ξ) : ξ < κ), that |F| ≤ T (ℵ ◦ Ψ) ≤ ℵΘ+ Φ Id (κ) ≤ ℵη+ Φ Id (κ) . Lemma 2.3.3 Assume that ℵη is a limit cardinal, κ = cf (ℵη ), (η(ξ) : ξ < κ) is an increasing  sequence cofinal in η, and λ ≥ 1 is a cardinal. Then there exists a set F ⊆ ξ<κ λ ℵη(ξ) of almost disjoint functions such that |F| = ℵλη . Proof For each function Φ ∈ λ ℵη , we define a function fΦ ∈ by

 fΦ (ξ)(α) =

Φ(α) 0



λ ξ<κ

ℵη(ξ)

if Φ(α) < ℵη(ξ) , otherwise.

If Φ, Φ ∈ λ ℵη and Φ = Φ , then there is an ordinal α0 < λ with Φ(α0 ) = Φ (α0 ). Choose an ordinal ξ0 < κ such that ℵη(ξ0 ) is greater than both Φ(α0 ) and Φ (α0 ). Then for every ξ with ξ0 ≤ ξ < κ we can conclude that fΦ (ξ)(α0 ) = Φ(α0 ) = Φ (α0 ) = fΦ (ξ)(α0 ), and thus fΦ (ξ) = fΦ (ξ) for all ξ ∈ [ξ0 , κ)ON . Consequently, the functions fΦ and fΦ are almost disjoint, and the set F := {fΦ : Φ ∈ λ ℵη } has the desired properties. Corollary 2.3.4 (Galvin-Hajnal Lemma) Assume that ℵη is a κ-strong singular cardinal, where κ := cf (ℵη ) > ω, (η(ξ) : ξ < κ) is a normal sequence cofinal in η, and λ ≥ 1 is a cardinal. Further let Φ ∈ κ ON be a function satisfying ℵλη(ξ) = ℵη(ξ)+Φ(ξ) for all ξ < κ. Then ℵλη ≤ ℵη+ Φ Id (κ) .  Proof By Lemma 2.3.3, there is a set F ⊆ ξ<κ λ ℵη(ξ) of almost disjoint functions such that |F| = ℵλη . Corollary 2.3.2 tells us that |F| ≤ ℵη+ Φ Id (κ) .

2.3. Applications of the Galvin-Hajnal Formula

123

Corollary 2.3.5 (Galvin-Hajnal Theorem) Let ℵη be a singular cardinal such that κ := cf (ℵη ) > ω. If ℵη is ℵ0 -strong, then, for every cardinal λ with κ ≤ λ < ℵη , ℵλη < ℵ(|η|λ )+ . If ℵη is even a strong limit cardinal, then 2ℵη < ℵ(2|η| )+ , and if in addition η < ℵη , then 2ℵη < ℵℵη . Proof First assume in addition to our assumption that ℵη is λ-strong. Then clearly ℵη is κ-strong. We take a normal sequence (η(ξ) : ξ < κ) cofinal in η. If Φ ∈ κ ON is defined by the equation ℵλη(ξ) = ℵη(ξ)+Φ(ξ) , then Corollary 2.3.4 tells us that ℵλη ≤ ℵη+ Φ Id (κ) . Since ℵη is λ-strong, we can conclude that ℵλη(ξ) < ℵη for all ξ < κ. Thus Φ ∈ κ η and hereby #Φ#Id (κ) < (|κ η|)+ ≤ (|η|λ )+ . This implies η + #Φ#Id (κ) < (|η|λ )+ , which gives ℵλη < ℵ(|η|λ )+ for this case. Now let, according to our assumption, ℵη be ω-strong but not λ-strong. We distinguish two cases. Case 1 2λ ≥ ℵη . Then we have ℵλη ≤ (2λ )λ = 2λ ≤ ℵ2λ < ℵ(2λ )+ ≤ ℵ(|η|λ )+ . Case 2 2λ < ℵη . Since ℵη is not λ-strong, there is β < η such that ℵλβ ≥ ℵη . Case 2 yields that the assumptions of Theorem 1.7.4 are fulfilled. So there is α ≤ β such that cf(ℵ ) ℵα is singular and λ-strong, has cofinality at most λ and satisfies ℵλβ = ℵα α . Now we can conclude that ℵλη = ℵλβ and cf(ℵα ) . ℵλη = ℵα

(∗)

We want to apply the first part of the proof to ℵα instead of ℵη and assume, contrary to our hopes, that cf (ℵα ) = ω. Since ℵη is ω-strong, we get ℵℵα0 < ℵη , hence ℵℵα0 < ℵη ≤ ℵλη , contradicting (∗). Thus cf (ℵα ) > ω. By the above cf(ℵ )

considerations we get ℵλη = ℵα α < ℵ(|α|cf(ℵα ) )+ ≤ ℵ(|η|λ )+ . If in addition ℵη is a strong limit cardinal, then Lemma 1.7.2 says that cf(ℵ ) 2ℵη = ℵη η . Together with κ = cf (ℵη ) = cf (η) and |η|κ ≤ |η||η| = 2|η| , our claim follows from the inequality ℵκη < ℵ(|η|κ )+ which we have just shown. Corollary 2.3.6 If σ is an ordinal with σ < ω1 such that the set {ξ < ω1 : ℵℵξ 1 ≤ ℵξ+σ } is stationary in ℵ1 , then ℵℵω11 ≤ ℵω1 +σ .

124

Chapter 2. The Galvin-Hajnal Theorem

Proof If β < ω1 , then our assumption yields ξ < ω1 such that β < ξ and ℵℵξ 1 ≤ ℵξ+σ < ℵω1 , hence ℵℵβ 1 ≤ ℵℵξ 1 ≤ ℵξ+σ . Therefore ℵω1 is ℵ1 -strong, and the function Φ, defined by the equation ℵℵξ 1 = ℵξ+Φ(ξ) , is a member of ω1 ω1 . Now the set {ξ < ω1 : Φ(ξ) ≤ σ} is stationary in ℵ1 , i.e., it is an element of Id (ℵ1 )+ . Therefore we know from Lemma 2.2.5 that #Φ#Id (ℵ1 ) ≤ σ. Furthermore we have cf (ℵω1 ) = ℵ1 . Putting η(ξ) := ξ for all ξ < ω1 , we can conclude, using Corollary 2.3.4, that ℵℵω11 ≤ ℵω1 +σ . Corollary 2.3.7 Assume that 2ℵ1 < ℵω1 and σ is an ordinal with σ < ω1 such that the set {ξ < ℵ1 : ℵℵξ 0 ≤ ℵξ+σ } is stationary in ℵ1 . Then ℵℵω11 ≤ ℵω1 +σ . Proof We have 2ℵ1 = ℵρ for some ρ < ω1 . Thus ℵℵξ 0 ≥ ℵℵρ 0 = 2ℵ1 for every ξ with max{ρ, ℵ0 } < ξ < ℵ1 , and from Corollary 1.6.13 we can conclude that ℵℵξ 1 = 2ℵ1 · ℵℵξ 0 = ℵℵξ 0 . It follows that {ξ < ℵ1 : ρ < ξ ∧ω < ξ ∧ℵℵξ 0 ≤ ℵξ+σ } = {ξ < ℵ1 : ρ < ξ ∧ω < ξ ∧ℵℵξ 1 ≤ ℵξ+σ }. Now the desired assertion can be inferred from Corollary 2.3.6. Lemma 2.3.8 If κ > ω is a regular cardinal and Φ ∈ κ ON is a function, then #Φ#Id (κ) < κ iff the set {ξ < κ : Φ(ξ) < ξ} is a member of Id (κ)+ , i.e., is stationary in κ. Proof If the set S := {ξ < κ : Φ(ξ) < ξ} is stationary in κ, then, by Fodor’s Theorem, there is an ordinal α < κ and a stationary set S ∗ ⊆ S such that Φ |` S ∗ = S ∗ × {α}. Together with Lemma 2.2.5 we can conclude that #Φ#Id (κ) ≤ α < κ. If conversely #Φ#Id (κ) = α < κ, then we know from Lemma 2.2.5 that the set Sα := {ξ < κ : Φ(ξ) ≤ α} is stationary in κ. Therefore Sα \ (α + 1) is also stationary in κ; since Sα \ (α + 1) ⊆ S, this also holds for S. Corollary 2.3.9 If the set {ξ < ω1 : ℵℵξ 1 < ℵξ+ξ } is stationary in ℵ1 , then ℵℵω11 < ℵω1 +ω1 . Proof The equation ℵℵξ 1 = ℵξ+Φ(ξ) defines a function Φ ∈ ω1 ON. By assumption, we have {ξ < ω1 : Φ(ξ) < ξ} ∈ Id (ℵ1 )+ , hence #Φ#Id (ω1 ) < ℵ1 by Lemma 2.3.8, and furthermore ℵω1 is ℵ1 -strong. Now the assumptions of Corollary 2.3.4 are fulfilled, taking η(ξ) := ξ for all ξ < κ. So we get ℵℵω11 ≤ ℵω1 + Φ Id (ω1 ) < ℵω1 +ω1 .

2.3. Applications of the Galvin-Hajnal Formula

125

Lemma 2.3.10 Let ℵη be a limit cardinal, κ := cf (ℵη ), and (η(ξ) : ξ < κ) be an increasing sequence that is cofinal in η. Then there is a set F of almost disjoint functions such that F ⊆ ξ<κ ℵη(ξ) 2 and 2ℵη = |F| = |



ℵη(ξ)

2|.

ξ<κ

 Proof For each X ∈ P(ℵη ), we define a function fX ∈ ξ<κ P(ℵη(ξ) ) by fX (ξ) = X ∩ ℵη(ξ) . If X, Y ∈ P(ℵη ) are different, then there is an ordinal ξ0 < κ such that fX (ξ0 ) = fY (ξ0 ). Since (η(ξ) : ξ < κ) is nondecreasing, we get fX (ξ) = fY (ξ) for all ξ ≥ ξ0 , and thus fX and fY are almost disjoint. On  account of 2ℵη(ξ) = |P(ℵη(ξ) )| we can identify fX with a function in ξ<κ ℵη(ξ) 2 and put F := {fX : X ∈ P(ℵη )}.  Then we get 2ℵη = |P(ℵη )| = |F| ≤ | ξ<κ ℵη(ξ) 2| ≤ (2ℵη )κ = 2ℵη . Corollary 2.3.11 Assume that ℵη is a singular cardinal with κ := cf (ℵη ) > ω, (η(ξ) : ξ < κ) is a normal function cofinal in η, and σ < κ is an ordinal, such that {ξ < κ : 2ℵη(ξ) ≤ ℵη(ξ)+σ } ∈ Id (κ)+ . Then 2ℵη ≤ ℵη+σ . Proof As in the proof of Corollary 2.3.6 it can be verified that the equation 2ℵη(ξ) = ℵη(ξ)+Φ(ξ) defines a function Φ ∈ κ κ, and that ℵη is a strong limit cardinal. Since {ξ < κ : Φ(ξ) ≤ σ} ∈ Id (κ)+ , Lemma  2.2.5 says that #Φ#Id (κ) ≤ σ. Lemma 2.3.10 tells us that there is a set F ⊆ ξ<κ ℵη(ξ)+Φ(ξ) of almost disjoint functions such that |F| = 2ℵη . By Corollary 2.3.2, we get |F| ≤ ℵη+ Φ Id (κ) ≤ ℵη+σ . Corollary 2.3.12 (J. Silver) If ℵη is a singular cardinal of uncountable cofinality, and if the set {κ < ℵη : κ ∈ ICN ∧ 2κ = κ+ } is stationary in ℵη , then 2ℵη = ℵη+1 . Corollary 2.3.13 Assume that κ and λ are regular cardinals such that ℵ0 < κ < κ λ and μκ < λ for every cardinal μ <  λ with cf (μ) = κ. Further let Φ ∈ λ be a function. Then, for every set F ⊆ ξ<κ ℵΦ(ξ) of almost disjoint functions, we have |F| < ℵλ . Proof Let the function (η(ξ) : ξ < κ) be defined by η(ξ) = 0 for all ξ < κ. Then we obtain an estimate for the cardinal ℵΘ in Theorem 2.2.6 by ℵΘ ≤ 2κ · ℵκ0 ≤ κκ · κκ < λ ≤ ℵλ , and this gives Θ < λ for the ordinal Θ. Now Lemma 2.2.10 says that #Φ#I < λ, where I := Id (κ), and from Theorem 2.2.6 we can conclude that |F| ≤ ℵΘ+ Φ I < ℵλ .

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Chapter 2. The Galvin-Hajnal Theorem

Corollary 2.3.14 Assume that κ and λ are regular cardinals with ω < κ < λ, with cf (μ) = κ. Further let (μξ : satisfying μκ < λ for all cardinals μ < λ  ξ< κ) be a sequence of cardinals such that ξ<α μξ < ℵλ for all α < κ. Then ξ<κ μξ < ℵλ .  Proof Let the equation ℵΦ(α) = ξ<α μξ , for α < κ, define the function 2.2.7 there Φ on κ. Then Φ ∈ κ λ, andby Lemma   is a set F of almostdisjoint functions such that F ⊆ α<κ ( ξ<α μξ ) = ξ<κ ℵΦ(ξ) and |F| = ξ<κ μξ . Now Corollary 2.3.13 says that |F| < ℵλ . Corollary 2.3.15 Assume that κ and λ are regular cardinals with ω < κ < λ, λ is κ-strong, and μ is an arbitrary cardinal. If μν < ℵλ for all ν ∈ κ ∩ CN, then μκ < ℵλ . Corollary 2.3.16 Assume that κ and λ are regular cardinals, ω < κ, λ is κstrong, and ν is a cardinal with cf (ν) = κ. If 2μ < ℵλ for all μ ∈ CN ∩ ν, then 2ν < ℵλ . Proof We choose a sequence (νξ : ξ < κ) of cardinals satisfing νξ < ν νξ for all ξ < κ and ν = ξ<κ νξ . For each ξ < κ, we put μξ := 2 . Then  ξ<α νξ < ℵ λ for every ordinal α < κ, and this together with ξ<α μξ = 2 νξ ξ<κ Corollary 2.3.14 gives 2 = 2ν < ℵλ . Corollary 2.3.17 Let α be a limit ordinal with cf (α) > ω. If 2ℵξ < ℵ(|α|cf(α) )+ for all ordinals ξ < α, then 2ℵα < ℵ(|α|cf(α) )+ . Proof Let κ := cf (α), λ := (|α|κ )+ , and ν := ℵα . Then κ and λ are regular cardinals with κ < λ, λ is κ-strong, and cf (ν) = cf (α) = κ. Thus all assumptions of Corollary 2.3.16 are satisfied, and we obtain 2ℵα < ℵ(|α|κ )+ . Corollary 2.3.18 If ℵℵξ 0 < ℵ(2ℵ1 )+ for all ξ < ω1 , then ℵℵω11 < ℵ(2ℵ1 )+ . Proof The assumption and Lemma 1.6.15 imply that ℵℵω01 = ℵω1 ·

ℵℵξ 0 < ℵ(2ℵ1 )+ ,

since cf (ℵ(2ℵ1 )+ ) = (2ℵ1 )+ > ℵ1 .

ξ<ω1

If we put κ := ℵ1 , λ := (2ℵ1 )+ , and μ := ℵω1 , then κ and λ are regular cardinals such that κ < λ. Furthermore, λ is κ-strong and μν < ℵλ for all ν ∈ κ ∩ CN. From Corollary 2.3.15 we can conclude that ℵℵω11 < ℵ(2ℵ1 )+ .

2.3. Applications of the Galvin-Hajnal Formula

127

Exercises 1) Assume that ℵη is a κ-strong singular cardinal, where κ := cf (ℵη ) > ℵ0 . Further let (η(ξ) : ξ < κ) be a normal sequence cofinal in η such that the + set S := {ξ < κ : ‫(ג‬ℵη(ξ)) = ℵ+ η(ξ) } is stationary in κ. Prove that ‫(ג‬ℵη ) = ℵη . Hint: Exercise 1.8.22 says that {μ < ℵη : μ is κ-strong} is a club in ℵη , and thus S ∗ := {ξ < κ : ℵη(ξ) is κ-strong and ‫(ג‬ℵη(ξ) ) = ℵ+ η(ξ) } is stationary in κ. If ξ ∈ S ∗ is a limit ordinal, use a golden rule to prove that ℵκη(ξ) = ‫(ג‬ℵη(ξ) ) and apply the Galvin-Hajnal lemma. 2) (Silver) If the singular cardinal hypothesis holds for all singular cardinals with countable cofinality, then it holds for all singular cardinals. Hint: Use transfinite induction on singular cardinals to show that from ω < κ := cf (ℵη ) < ℵη and 2cf(ℵη ) < ℵη it follows that ‫(ג‬ℵη ) = ℵ+ η . Prove first that ℵη is κ-strong, and apply the previous question. 3) Assume that ℵη is a κ-strong singular cardinal, where κ := cf (ℵη ) > ℵ0 . Further let (η(ξ) : ξ < κ) be a normal sequence cofinal in η, and Φ ∈ κ ON be a function, such that 2ℵη(ξ) ≤ ℵη(ξ)+Φ(ξ) for all ξ < κ. Then 2ℵη ≤ ℵη+ Φ Id (κ) . Hint: Apply Lemma 2.3.10 and Corollary 2.3.2. 4) Let β < ℵ1 be an ordinal. If 2ℵη < ℵ+β η holds for all singular cardinals with countable cofinality, then this assertion holds for all singular cardinals. Hint: Apply Exercise 6 from Section 2.2. 5) Assume that 2ℵ1 < ℵω1 and Φ ∈ ω1 ω1 is a function such that the set {ξ < ω1 : ℵℵξ 0 ≤ ℵξ+Φ(ξ) } is a club in ω1 . Prove that ‫(ג‬ℵω1 ) ≤ ℵω1 + Φ Id (ω1 ) . Hint: Remember that ℵℵα1 = max{2ℵ1 , ℵℵα0 } for all α with ℵ0 ≤ α < ℵ1 (Corollary 1.6.13), and apply the Galvin-Hajnal lemma. 6) If the set {ξ < ω1 : 2ℵξ ≤ ℵξ+ξ+2 } is stationary in ω1 , then 2ℵω1 ≤ ℵω1 +ω1 +2 . Hint: Prove that ℵω1 is a strong limit cardinal, and use Lemma 2.3.10 and Corollary 2.3.2. 7) If ℵω1 is a strong limit cardinal, and if {ξ < ω1 : ℵℵξ 0 ≤ ℵξ+ξ } is a club in ω1 , then 2ℵω1 < ℵω1 +ω1 . Hint: Use the previous questions.

Chapter 3 Ordinal Functions In the Galvin-Hajnal theorem, the question of whether it holds for singular cardinals with countable cofinality was left open. Let ℵδ be singular and ν be a cardinal. If δ = ℵδ , then ℵνδ = |δ|ν < ℵ(|δ|ν )+ , and thus the Galvin-Hajnal theorem holds for the case that δ is a fixed point of the aleph function. Let us now turn to the case that δ < ℵδ . Then |δ| < ℵδ , and for every regular cardinal λ with |δ| < λ < ℵδ , the set a := [λ, ℵδ )reg is an interval of regular cardinals satisfying |a| < min(a), since |a| ≤ |δ|. Shelah defines an operator pcf which assigns to each set a of regular cardinals and each cardinal μ a set pcfμ (a) of regular cardinals satisfying the following properties for μ ≥ 1: a) b) c) d)

a ⊆ pcfμ (a). min(a) = min pcfμ (a). If |a| < min(a), then |pcfμ (a)| ≤ |a|μ . If a is an interval of regular cardinals with |a| < min(a), then pcfμ (a) is an interval of regular cardinals.

Now we continue the discussion of the special case that a = [λ, ℵδ )reg . We have λ = ℵα for some ordinal α < δ, and therefore sup pcfμ (a) < ℵα+|pcfμ (a)|+ by a), b) and d). If λ is a μ-strong cardinal, then Theorem 6.2.4, which is of course due to Shelah and represents a main result of this book, tells us that sup pcfμ (a) = ℵμδ . So we obtain ℵμδ < ℵα+|pcfμ (a)|+ . By c) we have |pcfμ (a)| ≤ |a|μ ≤ |δ|μ , and α < δ < (|δ|μ )+ implies (∗)

ℵμδ < ℵ(|δ|μ )+ .

With Lemma 1.7.3 one can show, roughly speaking, that a μ-strong cardinal λ with |δ| < λ < ℵδ can always be found (for details see Section 8.1). Therefore (∗) is a powerful generalization of the Galvin-Hajnal theorem. If in particular μ = |δ|, then |pcfμ (a)| < |a|+4 by Theorem 7.2.6. In the case that ℵδ ≤ 2|δ| we

130

Chapter 3. Ordinal Functions |δ|

have ℵδ ≤ 2|δ| . If 2|δ| < ℵδ , we can conclude, using the fact that λ := (2|δ| )+ |δ| is a |δ|-strong cardinal and the estimate ℵδ < ℵα+|pcfμ (a)|+ from above, that |δ|

ℵδ < ℵ|δ|+4 , since α < |δ|+4 . Altogether we obtain |δ|

ℵδ < max{ℵ|δ|+4 , (2|δ| )+ }.

(∗∗) cf(δ)

|δ|

≤ ℵδ , (∗∗) yields an estimate for the gimel function at singular Since ℵδ arguments. The proofs of the assertions (∗) and (∗∗) are complicated, and the methods and techniques used for it are manifold. The theory developed solely for these proofs, the so-called pcf-theory, originates with Saharon Shelah. It is selfcontained and allows further interesting applications which will be treated in Chapter 8. In this chapter, we define first in Section 3.1 the fundamental notion of the true cofinality of a set S ⊆ A ON of ordinal functions modulo I, written as tcf (S/I), The special case  where I is an ideal on their common domain A. A function f ∈ ON, will play the that S = f := a∈A f (a) for some ordinal  decisive role in our investigations. If tcf ( f /I) exists, then it is a cardinal,   namely 0 in the trivial case that f = ∅, 1 if f has a maximum under ≤I , and a regular cardinal in all other  cases. If I is a maximal ideal on A and  D is the dualultrafilter of I, then tcf  ( f /I) exists, and we also write cf ( f /I) and cf ( f /D) instead of tcf (  f /I). If the cardinal λ = tcf ( f /I) exists, then it turns out  that the partial or every subset S of f whose cardinality dering ( f,
3.1. Suprema and Cofinalities

131

3.1 Suprema and Cofinalities For the rest of this chapter, A denotes a nonempty set and I an ideal on A. The domain of the considered ordinal functions is A if it is not explicitly specified otherwise. We remind the reader of the fact that we often omit the addition “modulo I” when we are dealing with assertions on ordinal functions. Definition Let S be a set (or a family) of ordinal functions and h be an ordinal function. h is called an upper bound of S modulo I iff f ≤I h for all members f of S. We call h a supremum of S modulo I iff h is an upper bound of S modulo I and h ≤I g holds for every upper bound g of S. If T is a subset (subfamily) of S, then T is called unbounded in S modulo I iff no member of S is an upper bound of T modulo I. For ordinal functions f and g, we use the following denotations: B(f, g) := {a ∈ A : f (a) < g(a)}, B[f, g] := {a ∈ A : f (a) ≤ g(a)}. Remark In general, the relation ≤I is not antisymmetric, and therefore suprema are not uniquely determined. On the other hand, modulo I all suprema of S are equal. Further we note: g ≤I f ⇐⇒ B(f, g) ∈ I

and g
A simple and practical necessary and sufficient condition for the existence of suprema is given by the following lemma. Lemma 3.1.1 If S is a set of ordinal functions, then an ordinal function g is a supremum of S iff g is an upper bound of S and there is no upper bound h of S satisfying h I g. Proof Let g be an upper bound of S with the property that there is no upper bound h of S such that h I g. We assume, to get a contradiction, that g is not a supremum of S. Then there is an upper bound h of S such that ¬(g ≤I h). Put p := min{g, h}. Obviously p is an upper bound of S, and we get p ≤ g. Since g ≤I h is not the case and p ≤ h, g ≤I p cannot hold, and therefore we have in particular not p =I g. So we get p I g, contradicting the assumption. Consequently, g is a supremum of S. The proof of the reverse direction of the equivalence is trivial. Definition If S is a set of ordinal functions and T ⊆ S, then T is called cofinal in S modulo I iff T is cofinal in S with respect to the relation ≤I on S, that means iff for all f ∈ S there is g ∈ T such that f ≤I g. The cardinal number

132

Chapter 3. Ordinal Functions

cf (S, ≤I ) from Section 1.5 is also called the cofinality of S modulo I; remember that cf (S, ≤I ) = min{|T | : T is a cofinal subset of S modulo I}. A sequence (fξ : ξ < σ) of ordinal functions is called increasing modulo I (strictly increasing modulo I) iff fα ≤I fβ (fα
 Obviously we have g < f iffg is a member of f . If a sequence of members  of f is cofinal in the set f we also say that it is cofinal in f . The following lemma is a direct consequence of Lemma 2.1.2. It says that the values of members of sequences which are cofinal in f (or more general, which are cofinal in arbitrary sets S of ordinal functions) may be changed appropriately on sets which are members of I. It is a simple standard tool in the following chapters. For those readers who are familiar with reduced products, it is trivial. Lemma 3.1.2 Assume that f is an ordinalfunction on A and (fξ : ξ < σ) and (gξ : ξ < σ) are sequences of members of f satisfying fξ =I gξ for all ξ < σ. Further let (fξ : ξ < σ) be cofinal in f . Then the sequence (gξ : ξ < σ) is also cofinal in f . Consequently, if one changes the values of fξ on a set Xξ ∈ Ifor every ξ < σ in such a way that the resulting function fξ is a member of f ,

3.1. Suprema and Cofinalities

133

then the sequence (fξ : ξ < σ) is also cofinal in f . Corresponding assertions hold for arbitrary sets S of ordinal functions and for other notions such as strict monotonicity, supremum, upper bound, and so on. The next lemma tells us that the true cofinality is 0, 1 or a regular cardinal and that we already know it if we have found one cofinal sequence of regular length. Lemma 3.1.3 Let S be a set of ordinal functions which posesses a sequence (fξ : ξ < σ) which is cofinal in S modulo I. Then we have tcf (S/I) = cf (σ). Furthermore the following assertions hold for λ := tcf (S/I). a) λ ∈ {0, 1} ∪ ICN and cf (λ) = λ. b) λ = 0 iff S = ∅. c) λ = 1 iff S has a maximum under ≤I . Proof By assumption, the ordinal λ := tcf (S/I) exists. b) and c) are obviously true. Now consider an arbitrary sequence (fξ : ξ < σ) which is cofinal in S modulo I. If σ ∈ {0, 1}, it is obvious that λ = cf (σ) = σ. If 1 < σ < ℵ0 or σ is a successor ordinal, then cf (σ) = 1 = tcf (S/I), since clearly the one-element sequence (fσ−1 ) is also cofinal in S modulo I. In particular we get λ ∈ ω \ {0, 1}, and λ is not a successor ordinal. So we can assume that σ is a limit ordinal. Then we can choose a sequence (αi : i < cf (σ)) of ordinal numbers which is cofinal in σ. It is clear that the sequence (fαi : i < cf (σ)) is also cofinal in S, and we obtain λ ≤ cf (σ). Furthermore it is easy to see that λ is infinite. By definition there is a cofinal sequence (gξ : ξ < λ) of members of S. Assume that λ < cf (σ). For each ξ < λ, we can choose an ordinal βξ < cf (σ) such that gξ ≤I fβξ . Since cf (σ) is regular, the ordinal β := sup{βξ + 1 : ξ < λ} is smaller than cf (σ), and so we have by construction gξ
134

Chapter 3. Ordinal Functions

the set X := B(f  , f ) = f  of the sequence such that f  I f . In particular,   {a ∈ A : f (a) < f (a)} is positive. Since f = ∅,  we have f (a) = 0 for all a ∈ A. So we can define the ordinal function f ∗ ∈ f by   f (a) if a ∈ X, ∗ f (a) := 0 if a ∈ A \ X. By assumption, there is an ordinal α < τ such that f ∗ ≤I fα . Since fα |X|, then tcf (X α/{∅}) = cf (α). Proof Since X is nonempty and |X| < cf (α), we have cf (α) ≥ ω. We can choose a normal sequence (αi : i < cf (α)) which is cofinal in α. For each i < cf (α) let fi be the ordinal function with domain X that has the constant value αi . If f ∈ X α is arbitrary, then by assumption γ := sup ran(f ) < α. So there is an i < cf (α) such that γ ≤ αi , hence f ≤X fi . Therefore the sequence (fi : i < cf (α)) is cofinal modulo I := {∅} in X α, and together with Lemma 3.1.3 we are done. A further class of examples is given by the maximal ideals or the ultrafilters on A. Definition Let us assume that S is a set of ordinal functions with the common domain A, D is an ultrafilter on A, and I := P(A) \ D is the maximal ideal dual to D. Then we put cf (S/D) := cf (S, ≤I )

and

tcf (S/D) := tcf (S/I).

When we are working with ultrafilters, we will use, in view of the following lemma, in most cases the denotation “cf ” instead of “tcf ”. Lemma 3.1.6 If D is an ultrafilter on A and S is a set of ordinal functions, then there is a sequence of elements of S which is cofinal in S modulo D, and cf (S/D) = tcf (S/D). Proof Let T be a cofinal subset of S of cardinality cf (S/D). We define by recursion a sequence of elements of T which is strictly increasing modulo D. Assume that σ ∈ ON, and that (fξ : ξ < σ) is already defined. If this sequence is not cofinal in S, we can find a function g ∈ S such that ¬(g ≤D fξ ) for all

3.1. Suprema and Cofinalities

135

ξ < σ. By Lemma 2.1.5, we have fξ
136

Chapter 3. Ordinal Functions

Definition If S is a set of ordinal functions with domain A and λ is a cardinal, then we call the partial ordering (S,
3.1. Suprema and Cofinalities

137

of pj . Therefore the sequence (G(fαi ) : i < λ) is strictly increasing. If h ∈ T , then there is p ∈ S satisfying the property from (3.1) and an ordinal β < λ with p ≤I fβ . Now, by the definition of αβ , we have β ≤ αβ , and this gives h <J G(fαβ ). An essential application of the previous lemma will be the proof of the following assertion. Its importance will be pointed out in the remark after the proof of Lemma 3.1.12. Lemma 3.1.11 If f is an ordinal whose values are limit ordinals,  function all of and if one of the cardinals tcf ( f /I) and tcf ( (cf ◦ f )/I) exists, then so does the other one, and we have   tcf ( (cf ◦ f )/I) = tcf ( f /I).    Proof We will define functions F : (cf ◦ f ) −→ f and G : f −→  (cf ◦ f ) which satisfy the assumption (3.1) in Lemma 3.1.10. For each a ∈ A, we can take a normal sequence (αia: i < cf (f (a)))which is cofinal in the limit ordinal f (a), and put, for any q ∈ (cf ◦ f ), p ∈ f , and a ∈ A a and G(p)(a) := min{i < cf (f (a)) : αia > p(a)}. F (q)(a) := αq(a)   It is clear that F (q) ∈ f and G(p) ∈ (cf ◦ f ).   To show (3.1) for F , I and J = I, let h ∈  f . Define p ∈ (cf ◦ f ) by p(a) := min{i < cf (f (a)) : αia > h(a)}. If q ∈ (cf ◦ f ) and p ≤I q, then a p(a) ≤ q(a) for almost all a ∈ A. By the definition of p we have αp(a) > h(a), a a hence F (q)(a) = αq(a) ≥ αp(a) > h(a) for almost all a ∈ A. This yields h
Lemma 3.1.12 Assume that τ is a limit ordinal, (fξ : ξ < τ ) is a strictly increasing sequence of ordinal functions, and f is a supremum of this sequence. Then fξ
138

Chapter 3. Ordinal Functions

Now we assume, to get a contradiction, that on an I-positive set Z all values of f are successor ordinals. Then we can define an ordinal function f  with f  I f by  f (a) − 1 if a ∈ Z,  f (a) := f (a) if a ∈ A \ Z. By Lemma 3.1.1, f  is not an upper bound of the given sequence, and therefore there is β < τ such that ¬ (fβ ≤I f  ), hence B(f  , fβ ) is positive. Now f  (a) < fβ (a) gives f (a) ≤ fβ (a) for any a ∈ A, hence B(f  , fβ ) ⊆ B[f, fβ ]. Thus B[f, fβ ] is positive, contradicting fβ
3.1. Suprema and Cofinalities

139

Subsequent to a preparatory lemma we now turn to the proof of a theorem which bridges the gap between the notion of almost disjoint functions and Shelah’s idea of the true cofinality of a function. Further it motivates an application of a very important theorem of pcf-theory, namely Theorem 3.3.6, which will play an outstanding role in this book. We already pointed out its importance in the introduction to this chapter. Definition Let I be an ideal on the set A and F ⊆ A ON. We call F I-disjoint iff {a ∈ A : f (a) = g(a)} ∈ I holds for all f, g ∈ F with f = g. A If μ is a cardinal,  then we call f ∈ ON μ-large modulo I iff there is an I-disjoint set F ⊆ f with |F| = μ. Lemma 3.1.14 Assume that λ is a κ-strong singular cardinal, where κ := cf (λ). Then there is an increasing function f ∈ κ λ satisfying ran(f ) ⊆ Reg and sup(ran(f )) = λ which is λκ -large modulo Ib (κ). In particular we have λκ = T(f ), where T(f ) is the cardinal defined in Section 2.2. Proof Let ℵη := λ and choose an increasing sequence (η(ξ): ξ < κ) of ordinals which is cofinal in η. By Lemma 2.3.3, there is a set F ∗ ⊆ ξ<κ κ ℵη(ξ) of almost disjoint functions such that |F ∗ | = λκ . For ξ < κ we put f (ξ) := (ℵκη(ξ) )+ . By assumption, we have f (ξ) < λ for every ξ < κ. Note that ran(f ) ⊆ Reg and sup(ran(f )) = λ. Now we choose, for each ξ < κ, an injective function  τξ : κ ℵη(ξ) −→ f (ξ), and define, for each h∗ ∈ F ∗ , a function h ∈ f by ∗ ∗ ∗ (h (ξ)). Then F := {h : h ∈ F } is an I (κ)-disjoint set of members h(ξ) := τ ξ b  of f satisfying |F| = λκ .  So the function f has the desired properties. Clearly  we have T (f ) = λκ since | f | ≤ | ξ<κ λ| = λκ . Definition For every ordinal function f with dom(f ) = A, we define the limit of f modulo I by limI (f ) := min{μ ∈ CN : {a ∈ A : f (a) ≤ μ} ∈ I + }. Lemma 3.1.15 (Jech, Shelah) Let λ be a κ-strong cardinal and ℵ0 < κ := κ cf (λ) < λ. If μ ∈ (λ, λκ ]CN is regular, then there  is a function f ∈ λ and an ideal I on κ such that limI (f ) = λ and tcf ( f /I) = μ. Proof Let μ ∈ (λ, λκ ]CN be a regular cardinal and I := Ib (κ). I is σcomplete, since κ is uncountable and regular. Therefore, by Lemma 2.2.2, the relation
140

Chapter 3. Ordinal Functions Claim 1 I  is an ideal on κ.

Proof We have ∅ ∈ I ⊆ I  and κ ∈ I  by the minimality of f , since I |`κ = I. Consider X ⊆ Y ∈ I  ; with no loss of generality we can assume that κ X ∈ / I and thus Y ∈ / I. Then  there is a function g ∈ λ with g
3.1. Suprema and Cofinalities

141

Now we can easily construct a sequence (fα : α < μ) of members of F which is strictly increasing modulo I  . If (fβ : β < α) is already defined, then  we infer from Claim  2 that | β<α {h ∈ F : ¬fβ
142

Chapter 3. Ordinal Functions

 7) Assume that I is an ideal on A, f ∈ A ON is a function, and tcf ( f /I) is infinite. Prove that X := {a ∈ A : ∃ξ f (a) = ξ + 1} is a member of I. 8) Assume that A is an infinite set, f ∈ A ON is an ordinal function, I is an ideal on A, and f (a) is a limitordinal for almost all a ∈ A. Further let μ := limI (cf ◦ f ). Prove that ( f, ν for almost all a ∈ A. 10) Assume that I is an ideal on the set  A and f ∈ A ON is an ordinal function. Then the cardinal number tcf ( f /I) exists and is equal to the infinite cardinal λ iff the following conditions are satisfied: (i) λ is regular.  (ii) ( f, κ for all ξ < κ, and μ = tcf ( g/I). 12) (Jech, Shelah) Assume that λ is a cardinal satisfying ℵ0 < κ := cf (λ) < λ and 2κ < λ, and I is a σ-complete ideal on κ. Further let μ be a regular cardinal with λ < μ and f ∈ κ λ be a function which is μ-large modulo I and
3.2. κ-rapid Sequences and the Main Lemma of pcf-Theory

143

3.2 κ-rapid Sequences and the Main Lemma of pcf-Theory For the rest of this chapter, A denotes an infinite set and I an ideal on A. The addition “modulo I” is often omitted. Definition Let κ be a cardinal and f¯ := (fξ : ξ < λ) be a sequence of members of A ON. We say that f¯ is a κ-ub1 sequence modulo I iff f¯ is strictly increasing modulo I and, for every family (Sa : a ∈ A) of subsets of ON satisfying |Sa | < κ for all a ∈ A, there exists an ordinal α < λ such that every function h ∈  ¯ a∈A Sa with fα fαi (a)}. This minimum exists for every a ∈ A by the choice of h0 , and certainly we have fαi
bound.

144

Chapter 3. Ordinal Functions

Now let α := sup{αi : i ∈ Lim ∩ |A|+ }. Then we have α < λ since λ is regular. Further we put for every a ∈ A  Sa := {hi (a) : i < |A|+ } = {Sai : i ∈ Lim ∩ |A|+ }. Analogously to the definition of hi in the limit case, we define for a ∈ A the ordinal h(a) := min{β ∈ Sa : β > fα (a)}.  If i ∈ |A|+ , then fα
3.2. κ-rapid Sequences and the Main Lemma of pcf-Theory

145

Definition Let λ be an infinite cardinal, (fξ : ξ < λ) be a sequence of ordinal functions and κ > ω be a regular cardinal. Then we call this sequence κ-rapid modulo I iff it is strictly increasing modulo I and, for all β < λ with cf (β) = κ, there is a club C in β of limit ordinals such that sup{fξ : ξ ∈ C ∩ γ}
:= B(hi , fαi+1 ) \ (B[fαi , sαi ] ∪ B[hi , fαi ]) = {a ∈ A : sαi (a) < fαi (a) < hi (a) < fαi+1 (a)}

is positive; in particular it has at least one member, say ai . Since κ > |A| is regular, there is a set Z ⊆ κ with |Z| = κ, so that the function (ai : i ∈ κ) has on Z the constant value a. If i, j ∈ Z with i + 1 < j, then αi+1 ∈ C ∩αj , and consequently fαi+1 (a) ≤ sαj (a) by the definition of the function sαj . Furthermore we have a ∈ Bi and a ∈ Bj , which implies hi (a) < fαi+1 (a) and sαj (a) < hj (a). Altogether we get hi (a) < hj (a), and thus the set {hi (a) : i ∈ Z} is a subset of Sa of cardinality κ, contradicting |Sa | < κ.

146

Chapter 3. Ordinal Functions

Definition Let λ be an infinite cardinal and (fξ : ξ < λ) be a sequence of ordinal functions in A ON. We say that an ordinal function g ∈ A ON cuts the sequence (fξ : ξ < λ) diagonally modulo I iff, for each α < λ, there exists β < λ such that the set {a ∈ A : fα (a) ≤ g(a) < fβ (a)} is not a member of I. The sequence (fξ : ξ < λ) is called small modulo I iff there is no ordinal function g ∈ A ON cutting the sequence diagonally modulo I. Next we turn to the proof of the fact that, under certain assumptions, there even exist κ-rapid sequences which are small.  Lemma 3.2.4 Assume that f ∈ A ON  is an ordinal function, f is nonempty, λ is a cardinal, I is an ideal on A, ( f, ℵ1 . Let S := {β < λ : cf (β) ∈ (ℵ0 , μ)reg } and, for each β ∈ S, Bβ := {a ∈ A : cf (f (a)) ≤ cf (β)}

and

Xβ := A \ Bβ

for β ∈ S.

By the definition of μ, we have Bβ ∈ I and consequently Xβ ∈ I + for all β ∈ S. For each β ∈ S, let Cβ be a club in β of order type cf (β), all of whose members are limit ordinals. As well as the requested sequence (fξ : ξ < λ ) we define by  β recursion, for each  β ∈ S, a sequence (gξ : ξ ∈ Cβ ) of members of f . Let f0 ∈ f be arbitrary and ξ < λ be fixed. If the function fξ is already defined, we can choose a function fξ+1 ∈ f such that fξ cf (β) ≥ |Cβ ∩ ξ|. Therefore, if ξ ∈ Cβ , then there  is a function gξβ ∈ f , so that for all a ∈ Xβ sup(

 {{fζ (a), gζβ (a)} : ζ ∈ Cβ ∩ ξ}) < gξβ (a).

  Since ( f,
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147

Now in addition let |A|+ < μ. To get a contradiction we assume that there is a function g ∈ A ON which cuts the sequence (fξ : ξ < λ ) diagonally. By the definition of this property, we can choose by recursion a strictly increasing sequence (αi : i < |A|+ ) of ordinals less than λ , so that for all i < |A|+ {a ∈ A : fαi (a) ≤ g(a) < fαi+1 (a)} ∈ I + . We put β := sup{αi : i < |A|+ }. Then we have cf (β) = |Cβ | = |A|+ , β ∈ S, and Xβ ∈ I + . Fix an ordinal ξ ∈ Cβ . There is i < |A|+ and limit ordinals δ, γ ∈ Cβ such that ξ < αi and αi+1 < γ < δ. We have gξβ ξ such that the set {a ∈ Xβ : gξβ (a) ≤ g(a) < gδβ (a)} is not a member of I and is in particular nonempty. Therefore there exists a strictly increasing sequence (δi : i < |A|+ ) of ordinals in Cβ such that the set Yi := {a ∈ Xβ : gδβi (a) ≤ g(a) < gδβi+1 (a)} is not a member of I for all i ∈ |A|+ . Thus (Yi : i < |A|+ ) is a sequence of nonempty, pairwise disjoint subsets of A. This contradiction shows that in the case |A|+ < μ the sequence (fξ : ξ < λ ) is already small. Definition Let (fξ : ξ < λ) be a sequence of elements of A ON. We call a function f ∈ A ON an exact upper bound of (fξ : ξ < λ) modulo I iff f is an upper bound of the sequence modulo I and, for every function g ∈ A ON with g
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Now g  is also an upper bound of f¯. Namely, if β < λ, then {a ∈ A : g (a) < fβ (a)} = {a ∈ X : g(a) < fβ (a)} ∪ {a ∈ A \ X : f (a) < fβ (a)} ⊆ {a ∈ A : fα (a) ≤ g(a) < fβ (a)} ∪ {a ∈ A : f (a) < fβ (a)} ∈ I, since Yβ ∈ I and f is a supremum of f¯. This implies f ≤I g  , that means {a ∈ A : g  (a) < f (a)} ∈ I. Consequently, with g μ, for which the  partial ordering ( f,
X := {a ∈ A : g(a) ≥ f (a)} = {a ∈ A : g(a) = f (a)} ∈ I.

3.2. κ-rapid Sequences and the Main Lemma of pcf-Theory

149

  Then tcf ( g/I |`X) = λ by Lemma 3.1.7. By assumption,( f, fαγ (a)} for all a ∈ A.

150

5)

6)

7)

8) 9)

Chapter 3. Ordinal Functions Now assume that the sequence (gi : i < σ) is normal and γ < σ is a limit ordinal. Prove that gγ is an upper ≤I -bound of (fξ : ξ < λ) and that gγ I gi for all i < γ. (Jech) Assume that λ is a regular cardinal, the sequence (fξ : ξ < λ) of members of A ON is strictly increasing modulo I, and |A|+ < λ. Further let (gi : i < σ) be a normal sequence of upper ≤I -bounds of (fξ : ξ < λ). Prove that σ < |A|+ . Hint: Assume to get a contradiction that |A|+ ≤ σ. Then also (gi : i < |A|+ ) is a normal sequence of upper bounds of {fξ : ξ < λ} modulo I. If γ < |A|+ is a limit ordinal, let αγ < λ be the ordinal according to property (ii) of a normal sequence. Put α := sup{αγ : Lim(γ) ∧ γ < |A|+ }. For a ∈ A let Sa := {gi (a) : i < |A|+ } and g(a) := min{β ∈ Sa : β > fα (a)}. Prove thatg ≤I gi for all i < |A|+ . Then choose a limit ordinal γ < |A|+ with g ∈ a∈A {gi (a) : i < γ} and show that gγ ≤I g. The assertion g ≤I gγ+1 I gγ ≤I g yields a contradiction. Assume that A is an infinite set and λ is a regular cardinal with 2|A| < λ. Prove that every sequence (fξ : ξ < λ) of members of A ON which is strictly increasing modulo I has a supremum modulo I. Hint: Let (gi : i < σ) be a normal sequence of ≤I -upper bounds of {fξ : ξ < λ}, such that there is no ≤I -upper bound g of {fξ : ξ < λ} with g I gi for all i < σ. If σ = i + 1, then Lemma 3.1.1 tells us that gi is a supremum of {fξ : ξ < λ} modulo I. Now show, using exercises 4 and 5, that σ cannot be a limit ordinal. (Shelah) Assume that κ is an infinite cardinal, I is an ideal on κ, and η is an ordinal greater than 1. Prove that thereis an ordinal α < (|η|κ )+ such that there is no sequence of members of i<κ ℵη of length  ℵα+1 which is strictly increasing modulo I. In particular we have tcf ( i<κ ℵη /I) < ℵ(|η|κ )+ if this cardinal exists. Hint: Suppose to get a contradiction that for each ordinal α < (|η|κ )+  α there is a sequence (fξ : ξ < ℵα+1 ) of members of i<κ ℵη which is strictly increasing modulo I. If 2κ < α < (|η|κ )+ , then 2κ < ℵα+1 , and α α Exercise 6 tells us that (f ξ : ξ < ℵα+1 ) has a supremum g modulo  I. α By Exercise 2 we get tcf ( g /I) = ℵα+1 . Therefore we have tcf ( cf ◦ g α /I) = ℵα+1 for every α ∈ (2κ , (|η|κ )+ )ON . Now consider the function (cf ◦ g α : α ∈ (2κ , (|η|κ )+ )ON ). If ℵη is a singular cardinal and κ = cf (ℵη ), then pp(ℵη ) < ℵ(|η|κ )+ , where pp(λ) = sup{tcf ( f /I) : f ∈ κ λ ∧ ran(f ) ⊆ Reg ∧ limI (f ) = λ}. If ℵη is a κ-strong cardinal, where ℵ0 < κ = cf (ℵη ) < ℵη , then ℵκη < ℵ(|η|κ )+ . Hint: Use Exercise 3.13 and the previous question.

3.3. The Definition and Simple Properties of pcf(a)

151

3.3 The Definition and Simple Properties of pcf(a) Definition If a is an  arbitrary setof ordinal numbers and I is an ideal on a, then we define tcf ( a/I) := tcf ( ida /I) where we know that ida = {(ξ, ξ) : ξ ∈ a} is the identity on a. For any ordinal function f ∈ A ON, we call f progressive iff cf (f (a)) > |A| for all a ∈ A. In this section we investigate the set  pcf(f ) = {tcf ( f /I) : I is an ideal on dom(f )} of true cofinalities of progressive ordinal functions f . Note that cf (f (a)) > |A| for all a ∈ A implies that f (a) is a limit ordinal for all a ∈ A. Thus we can conclude from Lemma 3.1.11 that pcf(f ) = pcf(cf ◦ f ). Clearly also cf ◦ f is progressive. The next lemma says for progressive functions,  that we know pcf(f ) if we can characterize the true cofinalities of the set ran(f ) of ordinal functions. Altogether we will have shown that it suffices for  a characterization of pcf(f ) to investigate the set of true cofinalities of sets a where a is a set of regular cardinals. Definition If I is an ideal on A and f is a function with dom(f ) = A, we put f [I] := {Y ⊆ ran(f ) : f −1 [Y ] ∈ I}. It is easy to verify that this set is an ideal on ran(f ). We call it the image ideal of I under the function f . Lemma 3.3.1 Assume that A = ∅, I is an ideal on A, and f isa progressive  ordinal function on A. If there exists one of the cardinals tcf ( f /I) and tcf ( ran(f )/f [I]), then there exists the other one, and they are equal.  Proof Let tcf (  f /I) exist.  We want to apply Lemma 3.1.10 and hence define a function G  : f −→ ran(f ) which satisfies the condition (3.1) of this lemma. For r ∈ f and α ∈ ran(f ), we put G(r)(α) := sup{r(a) + 1 : a ∈ A ∧ f (a) = α}. Since f is progressive, we have |A| < cf (f (a)) < α for  and hereby G(r)(α)  we get G(r) ∈ ran(f ). If h ∈ ran(f ), then any α ∈ ran(f ). Consequently  h ◦ f is a member of f . This function will play the role of the function  p in Lemma 3.1.10. So consider q ∈ f with h ◦ f ≤I q, i.e. with B(q, h ◦ f ) ∈ I. We will show that h
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If α ∈ ran(f ) and a ∈ A have the properties G(q)(α) ≤ h(α) and f (a) = α, then q(a) < h(α) = h(f (a)) by the definition of G. So we have f −1 [B[G(q), h]] ⊆ B(q, h ◦ f ) ∈ I, and therefore h
cofinalities

3.3. The Definition and Simple Properties of pcf(a)

153

Remark Obviously every subset of a progressive set is also progressive. Further it is clear by our conventionthat in the definition of pcf(a) only those ideals I can appear for which tcf ( a/I) exists. We have pcf(∅) = ∅ = pcf0 (a) since there are no (proper) ideals on the empty  set. If a = ∅, the cardinal tcf ( a/I) is always infinite by Lemma 3.1.3, since ( a, ≤I ) has no maximum. Lemma 3.3.2 Assume that A is a nonempty set and f ∈ A ON is a progressive ordinal function, and let a := ran(cf ◦ f ). Then  pcf(f ) = pcf(cf ◦ f ) = pcf(a) = {cf ( a/D) : D is an ultrafilter on a}. In particular, the last equation is true for any set a of regular cardinals. Proof The first equation holds by Lemma 3.1.11 since cf (f (a)) > |A| ≥ 1 for all a ∈ A and thus all values of f must be limit ordinals. The second equation follows from Lemma 3.3.1. For the last equation, let first D be an  ultrafilter on a. Then cf ( a/D) exists by Lemma 3.1.6, and for the dual ideal   I of D wehave tcf ( a/I) = cf ( a/D). Conversely, if I is an ideal on a such that tcf ( a/I) exists, then there  is an ultrafilter  D on a such that D ∩ I = ∅. Now Lemma 3.1.7 says that cf ( a/D) = tcf ( a/I). Lemma 3.3.3 If a is a nonempty progressive interval of regular cardinals which has no maximum, then all members of the set a \ {min(a)} are successor cardinals. If ℵδ := sup(a), then we have δ < ℵδ , and consequently ℵδ is singular. Proof Let ν be a limit cardinal, ν ∈ (min(a), sup(a))CN . Then obviously ν is the supremum of a ∩ ν. Thus we can conclude by assumption that cf (ν) ≤ |a| < min(a) ≤ ν, and ν is singular. For the proof of the second assertion note first that δ is a limit ordinal and assume, to get a contradiction, that δ = ℵδ = sup(a). For b := [ℵ0 , min(a))reg we have b ∪ a = [ℵ0 , ℵδ )reg , and therefore our supposition gives |b| + |a| = |δ| = ℵδ . Since |b| < ℵδ , we get |a| = ℵδ which contradicts the fact that a is progressive. Now cf (ℵδ ) = cf (δ) ≤ |δ| ≤ δ < ℵδ yields that the cardinal ℵδ is singular. Now we want to list some simple properties of pcf(a) and pcfμ (a), respectively. Lemma 3.3.4 Let μ be a cardinal with μ > 0. Then the following holds. a) pcfμ (a) is a set of regular cardinals. b) sup pcfμ (a) ≤ (sup(a))μ .  c) If D is the principal ultrafilter on a generated by {λ}, then cf ( a/D) = λ and limD (a) = λ. Conversely, if D is an ultrafilter on a, cf ( a/D) = λ and a ∩ λ is finite, then D is principal and generated by {λ}.

154 d) e) f) g)

Chapter 3. Ordinal Functions a ⊆ pcfμ (a). If a is finite, then pcfμ (a) = a. pcfμ (a) ∩ min(a) = ∅. If a has no maximum and D is an unbounded ultrafilter on a, that is an ultrafilter on a which does not contain a bounded subset of a, then  cf ( a/D) ≥ sup(a).

h) If b ⊆ a, then pcfμ (b) ⊆ pcfμ (a) and pcf(b) ⊆ pcf(a). So we get pcfμ (a) ⊆ pcf(a) and pcf|a| (a) = pcf(a). In particular,if λ ∈ pcfμ (a), then there exists an ultrafilter D on a such that λ = cf ( a/D). i) If b ⊆ Reg, then pcfμ (a ∪ b) = pcfμ (a) ∪ pcfμ (b) and pcf(a ∪ b) = pcf(a) ∪ pcf(b). j) If D is an ultrafilter on a such that limD (a) = κ+ or limD (a) = min(a), then D is principal and generated by {κ+ } or {min(a)}, respectively. k) Assume that D is an ultrafilter and I is an ideal on a such that D ∩I = ∅. Then limI (a) ≤ limD (a). If I is the dual ideal of D, then equality holds. Furthermore, if sup(a) ∈ a and Ib (a) ⊆ I, then limI (a) = limD (a) = sup(a). l) Assume that sup(a) ∈ a and D is an ultrafilter on  a such that limD (a) = sup(a). Then D ∩ Ib (a)  = ∅. Furthermore, if cf ( a/D) = κ, μ < sup(a) and a = a \ μ, then cf ( a /D ∩ P(a )) = κ.   Proof If b ⊆ a and f ∈ b, then wehave f + 1 ∈ b and f < f + 1. Consequently, for any ultrafilter D on b, b does not possess a maximum under ≤D . Now the assertion in a) follows from Lemma 3.1.3. To prove b), let λ ∈ pcfμ (a). Then there is a subset b of a with |b| ≤ μ, an ultrafilter D on b and a sequence (fξ : ξ < λ) cofinal in b modulo D. This sequence is certainly injective, and thus we obtain  λ≤| b| ≤ (sup(a))μ . For the proof of part c), consider the principal ultrafilter D on a generated by {λ} for some λ ∈ a. Then we have a ∩ λ+ ∈ D and a ∩ ν + ∈ D for all ν < λ, hence limD (a) = λ. If we put  0 if ν = λ, fξ (ν) := ξ if ν = λ  for ξ < λand ν ∈ a, then the sequence (fξ : ξ < λ) is cofinal in a modulo D. Now cf ( a/D) = λ and part d) follow, and we also get e) since each ultrafilter

3.3. The Definition and Simple Properties of pcf(a)

155

on a finite set is principal. Next we prove the second assertion in c). If a is finite, then D isprincipal, and the first assertion implies that D is generated by {λ} since cf ( a/D) = λ. So we can assume that a is infinite. If a ∩ λ+ ∈ D, then D is principal since it has the finite member a ∩ λ+ , and we can argue as + before. Suppose that D. Then we have a \ λ+ ∈ D, and Lemma 3.1.8  a ∩ λ ∈  says that λ = cf ( a/D) = cf ( (a \ λ+ )/D ∩ P(a \ λ+ )), hence λ ∈ pcf(a \ λ+ ). From f) we can conclude that λ ≥ min(a \ λ+ ) > λ. This contradiction shows that a ∩ λ+ ∈ D, as desired. For the proof of part f) let λ < min(a). Assume that D is an ultrafilter on a nonempty subset  b of a such that |b| ≤ μ, and (fξ : ξ < λ) is a sequence which is cofinal in b modulo D. By the regularity of the members of b,  the equation g(ν) := sup{fξ (ν) + 1 : ξ < λ}, for ν ∈ b, defines a function g ∈ b satisfying  fξ < g for all ξ < λ. Therefore, the sequence (fξ : ξ < λ) is not cofinal in b modulo D, which proves f). For g), assume that the set a has no maximum, and let D be an unbounded ultrafilter on a. Further let λ be a regular cardinal with λ < sup(a) and (fξ :  ξ < λ) be a sequence of members of a. If we define  0 if ν ≤ λ, g(ν) := sup{fξ (ν) + 1 : ξ < λ} if ν > λ for every ν ∈ a, then {ν ∈ a : fξ (ν) < g(ν)} = a ∩ (λ, sup(a))reg ∈ D , i.e., we have fξ
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For j), let D be an ultrafilter on a. If limD (a) = κ+ , then a ∩ κ++ ∈ D and a ∩ κ+ ∈ D, hence (a ∩ κ++ ) \ (a ∩ κ+ ) = {κ+ } ∈ D, and D is the principal ultrafilter generated by {κ+ }. If limD (a) = min(a), then the definition of a limit shows that {min(a)} ∈ D. The proof of the last assertions is left as an exercise. Definition For any set B of cardinals, let sup+ (B) := min{κ ∈ CN : ∀ν ∈ B ν < κ}. Assume that λ is a singular cardinal such that cf (λ) ≤ κ < λ, and let A be the set  {cf ( a/D) : a ⊆ Reg ∧ sup(a) = λ ∧ |a| ≤ κ ∧ D ultrafilter on a ∧ D ∩ Ib (a) = ∅}. We define the cardinals ppκ (λ) and pp+ κ (λ), which will play an important role in the last chapters of this book, as ppκ (λ) := sup(A)

and

+ pp+ κ (λ) := sup (A).

If κ = cf (λ), then we put pp(λ) := ppκ (λ) and pp+ (λ) := pp+ κ (λ). Lemma 3.3.5 Assume that λ and κ are infinite cardinals and cf (λ) ≤ κ < λ. If μ is a regular cardinal such that λ < μ < pp+ κ (λ), then there is a subset a of λ satisfying a ⊆ Reg, sup(a) = λ and |a| ≤ κ, and an ultrafilter D on a such that  D ∩ Ib (a) = ∅ and cf ( a/D) = μ. In particular we have limD (a) = λ, and if κ0 ∈ (κ, λ)CN , we can assume without loss of generality that κ0 < min(a). Proof Let κ0 ∈ (κ, λ)CN and μ be a regular cardinal such that λ < μ < By the definition of pp+ κ (λ), there is a set b ⊆ λ ∩ Reg such that sup(b) = λ and |b| ≤ κ and a maximal ideal I on b such that Ib (b) ⊆ I and  tcf ( b/I) ≥ μ. Without loss of generality we can assume that b ⊆ (λ ∩ Reg) \ κ0 , since Ib (b) ⊆ I.  If μ = ppκ (λ), then tcf ( b/I)  = μ, and we can take a := b and D as the dual ultrafilter of I, and get μ = cf ( a/D). Now we discuss the  case that μ < ppκ (λ). Without loss of generality let μ < tcf ( b/I). Then ( b,
3.3. The Definition and Simple Properties of pcf(a)

157

Theorem 3.3.6 If a is a progressive interval of regular cardinals and μ is a cardinal, then pcfμ (a) is also an interval. Proof If a is a finite set, then pcfμ (a) is the empty set or the set a, hence in each case an interval. So let a be an infinite interval of regular cardinals. We have to show for arbitrary cardinals λ ∈ pcfμ (a) and λ ∈ [min(a), λ)reg that λ is a member of pcfμ (a). By Lemma 3.3.4, this is certainly true for λ < sup(a) since in this case λ ∈ a, and, if sup(a) ∈ a, for λ = sup(a) since then sup(a) = max(a). If sup(a) ∈ a, then cf (sup(a)) ≤ |a| < min(a) ≤ sup(a), and thus sup(a) is a singular cardinal. So let sup(a) < λ and let b be a subset of a satisfying  |b| ≤ μ and λ ∈ pcf(b). Then there is an ultrafilter D on b such that λ = cf ( b/D). D is nonprincipal since λ > λ > sup(a). If μ∗ := limD (b), then Lemma 3.3.4 says that |b| < min(b) <  μ∗ and that μ∗ is a limit cardinal, since D is nonprincipal.  Since λ = cf ( b/D) and (λ )+ ≤ λ, the linear ordering ( b, min(b)}, then d ∈ D, |ran((cf ◦ g) |`d)| ≤ μ, and ran((cf ◦ g) |` d) is a subset of a. The last assertion can be seen as follows. If ν ∈ d, then min(a) ≤ min(b) < cf (g(ν)) ≤ g(ν) < ν; since ν ∈ a, cf (g(ν)) ∈ Reg and a is an interval, we obtain cf (g(ν)) ∈ a. From Theorem 3.1.13 and Lemma 3.1.8 we can conclude that    λ = cf ( g/D) = cf ( (cf ◦ g)/D) = cf ( (cf ◦ g) |`d/D ∩ P(d)), and consequently, by Lemma 3.3.1, we get λ ∈ pcf(ran((cf ◦ g) |`d)) ⊆ pcfμ (a). Therefore we get the desired assertion λ ∈ pcfμ (a). Lemma 3.3.7 Assume that f ∈ A Reg is a function, B is a set with |B| < min(f [A]), I is  b ∈ B there is an ideal Ib on A  an ideal on B, and for each such that tcf ( f /Ib ) exists. Let g := (tcf ( f /Ib ) : b ∈ B) and I ∗ := {X ⊆ A : {b ∈ B : X ∈ Ib } ∈ I}.   Then I ∗ is an ideal on A. If one of the cardinals tcf ( f /I ∗ ) and tcf ( g/I) exists, then so does the other one, and they are equal. Proof The easy proof that I ∗ is an ideal on A is left to the reader.  For each b ∈ B, we can choose a sequence (fξb : ξ < g(b)) which is cofinal in  f Lemma 3.1.10, we can define functions F : g −→ f modulo  Ib . To apply  and G : f −→ g by b F (q)(a) := sup{fq(b) (a) + 1 : b ∈ B} and G(h)(b) := min{ξ < g(b) : h ≤Ib fξb }.

158

Chapter 3. Ordinal Functions

Since |B| < min(f [A]), we have F (q)(a) ∈ f (a) for all a ∈ A. Furthermore b holds for every b ∈ B. h ≤Ib fG(h)(b)  First let tcf ( g/I) exist. According for  to Lemma 3.1.10, we want to find each function h ∈ f a function p ∈ gsuch that h
consequently h
So we get {a ∈ A : p(a) ≤ h(a)} ∈ Ib , and consequently {a ∈ A : h(a) < p(a)} ∈ Ib . Thus we have b ∈ Y , as desired. Corollary 3.3.8 Assume that b is a nonempty subset of pcf(a) such that |b| < min(a)  and I is an ideal on b. For each ν ∈ b, let Iν be an ideal on a satisfying tcf ( a/Iν ) = ν. Further let I ∗ := {X ⊆ a : {ν ∈ b : X ∈ Iν } ∈ I}.  ∗ ∗ Then I is an ideal on a. If there exists one of the cardinals tcf ( a/I ) and tcf ( b/I), then there exists the other one, and they are equal. Exercises 1) Assume that α ∈ ON, A is a nonempty set and g := A×{α} is progressive. Determine pcf(g). 2) Let g ∈ A ON be an ordinal function and a := ran(cf ◦ g). Show that pcf(a) ⊆ pcf(g).

3.4. The Ideal J<λ (a)

159

3) Let g := ω × {ω}. Prove:  a) ( g,
3.4 The Ideal J<λ (a)

  Lemma 3.1.9 presents a necessary condition for λ = tcf ( a/I): ( a,
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Chapter 3. Ordinal Functions

Definition Let λ be a cardinal. We call an ideal I on the set a λ-mighty iff, for each I-positive  set b ⊆ a, there is an ultrafilter D on a satisfying b ∈ D, D ∩ I = ∅, and cf ( a/D) ≥ λ. Lemma 3.4.1 Assume that I is a λ-mighty ideal on a. Then a ∩ λ ⊆ dom(I). Proof Let us assume, to get a contradiction, that there is a cardinal ν ∈ a ∩ λ such that {ν} ∈ I + . Since I isλ-mighty, there is an ultrafilter D on a satisfying {ν}  ∈ D, D ∩ I = ∅ and cf ( a/D) ≥ λ. But now Lemma 3.3.4 c) says that cf ( a/D) = ν < λ. This contradiction proves a ∩ λ ⊆ dom(I). Lemma 3.4.2 (Jech) Assume that f is a progressive ordinal function on the nonempty set A, λ is a regular cardinal with λ > |A|+ , and (fξ : ξ < λ) is a sequence of ordinal functions on A. Then there is an ordinal function g ∈ A ON such that g < f and, for all h ∈ A ON with g ≤ h < f , the set {ξ < λ : B(g, fξ ) = B(h, fξ )} is unbounded in λ and so has cardinality λ. Proof First we consider the case that A is infinite and assume, to get a contradiction, that the assertion of the lemma is false. We will construct by recursion a sequence (gi : i < |A|+ ) of members of f such that gi ≤ gj wheneveri < j < |A|+ , and a sequence (αi : i < |A|+ ) of ordinals less than λ. g0 ∈ f is chosen arbitrarily. If i < |A|+ is a limit ordinal, we define gi := sup{gj : j < i}. Since f is progressive and |i| ≤ |A|, we have gi < f . If i < |A|+ and gi is already defined, then by our supposition there is a function gi+1 with gi ≤ gi+1 < f and an ordinal αi < λ such that B(gi+1 , fξ ) = B(gi , fξ ) for all ξ ∈ [αi , λ)ON . Since gi ≤ gi+1 , we get B(gi+1 , fξ )  B(gi , fξ ) for all ξ ∈ [αi , λ)ON . Since λ is regular and greater than |A|+ , there is an ordinal α < λ such that α ≥ αi for all i < |A|+ . Now, by construction, (B(gi , fα ) : i < |A|+ ) is a strictly decreasing sequence of subsets of A with length |A|+ . This obviously leads to a contradiction. If A is finite, we can proceed in the same way (the limit case does not occur), but will construct a sequence of length |A|++ = |A| + 2. Corollary 3.4.3 Assume that λ is a regular cardinal with λ > |A|+ , f is a progressive ordinal function on A, and f¯ = (fξ : ξ < λ) is a sequence of ordinal   ¯ functions in f which is strictly increasing  modulo I. If f is unbounded in f modulo I, then there is a function g ∈ f and, for each ξ < λ, an I-positive set Bξ ⊆ A with the following properties:

3.4. The Ideal J<λ (a)

161

a) If ζ < ξ < λ, then Bζ ⊆I Bξ . b) For  all ζ < λ, f¯ is cofinal in f modulo I |` Bζ . In particular we get tcf ( f /I |`Bζ ) = λ. c) If J is an ideal on A satisfying I ⊆ J and Bξ ∈ J for all ξ < λ, then g is an upper bound of f¯ modulo J. Proof We can choose an ordinal function g such that g < f and g satisfies the assertion from Lemma 3.4.2. By assumption, g is not an upper bound of the sequence f¯ = (fξ : ξ < λ); hence there is an ordinal α < λ such that B(g, fα ) is positive. We define Bξ := B(g, fα+ξ ) for each ξ < λ. Since f¯ is increasing, we get Bζ \ Bξ ⊆ B(fα+ξ , fα+ζ ) ∈ I for all ζ and ξ with ζ < ξ < λ. This proves part a), and certainly all sets Bξ are positive, since B0 is positive. For the proof of part b), fix ζ < λ. We note first that the given sequence is strictly increasing modulo I |` Bζ , since I ⊆ I |` Bζ . Next we consider an  arbitrary function h ∈ f and will show that h is majorized under
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But in this case we have ν = |S| < λ and therefore {ν} ∈ I by Lemma 3.4.1. So we can change appropriately the value of g at ν. For the inductive proof let μ := |S| > |a|+ and f¯ = (fξ : ξ < μ) be an enumeration of S. Without loss of generality we can assume that this sequence is strictly increasing modulo I, since otherwise we can choose  by recursion, using the inductive hypothesis, for each ξ < μ a function fξ ∈ a such that fξ
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Corollary 3.4.6 If λ is a cardinal, then J<λ (a) = {b ⊆ a : pcf(b) ⊆ λ}. In particular we have pcf(a) ⊆ λ ⇐⇒ J<λ (a) = P(a). Proof Let b be a subset of a. Any ultrafilter on b can be extended to an ultrafilter on a. Conversely, D ∩ P(b) is an ultrafilter on b if D is an ultrafilter on a such that b ∈ D. Using the fact that b is also progressive, we can infer from Lemma 3.3.2 that  pcf(b) = {cf ( a/D) : D is an ultrafilter on a ∧ b ∈ D}. Hereby our assertion follows immediately from the preceding lemma. In particular, pcf(a) ⊆ λ holds iff a ∈ J<λ (a). The last assertion is equivalent to J<λ (a) = P(a). Corollary 3.4.7 If λ is a cardinal and b ⊆ a, then J<λ (b) = J<λ (a) ∩ P(b). Corollary 3.4.8 If κ and λ are cardinals with κ < λ, then J<κ (a) ⊆ J<λ (a). Lemma 3.4.9 If a has no maximum and λ = sup(a), then every member of J<λ (a) is a bounded subset of a, and thus J<λ (a) ⊆ Ib (a). Proof Let b be an unbounded subset of a, i.e., b ∈ Ib (a). Then {b} ∪ {a \ ν : ν < λ} has the finite intersection property and can be extended to an ultrafilter D on a. Clearly D is an unbounded ultrafilter. By Lemma 3.3.4 we have cf ( a/D) ≥ λ, hence b ∈ / J<λ (a). A further characterization of J<λ (a) is the following. Lemma 3.4.10 Let a be progressive, λ be an infinite cardinal, and I be an ideal on a. Then the following assertions are equivalent:  (i) ( a,
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If I is any λ-mighty ideal on a, then J<λ (a) ⊆ I by the definition of J<λ (a). To prove that (iii) implies (i), let us assume that J<λ (a) ⊆ I. Then J<λ (a) = P(a) and thus J <λ (a) is a λ-mighty ideal on a. Theorem 3.4.4 says that  the partial ordering ( a, <J<λ (a) ) is λ-directed and thus the same holds for ( a,
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Corollary 3.4.14 If a is progressive, then |pcf(a)| ≤ 2|a| . Lemma 3.4.15 If a is progressive and λ is a limit cardinal, then  J<λ (a) = {J<μ (a) : μ ∈ CN ∧ μ < λ}. Proof By Corollary 3.4.8, we have  I := {J<μ (a) : μ ∈ CN ∧ μ < λ} ⊆ J<λ (a). Without loss of generality let I = P(a). I is an ideal since it is the union of a chain of ideals. Fix an I-positive set b. There is an ultrafilterD on a extending I which has b as its member. Corollary 3.4.11 says that cf ( a/D) ≥ λ, hence b∈ / J<λ (a) by the same corollary. So we get J<λ (a) ⊆ I. The strength of Corollary 3.4.3 is demonstrated by the following applications.  Lemma 3.4.16 If a is a progressive set of regular cardinals,  ( a, |a| , the existence  of a sequence (bξ : ξ < λ) of subsets of a such that tcf ( a/I |` bξ ) = λ for every ξ < λ and bζ ⊆I bξ whenever ζ < ξ < λ. If at least one bξ is a member of D, then we are done. So let us assume that bξ ∈ D for all ordinals ξ < λ. Then I ∪ {bξ : ξ < λ} generates an  ideal J on a such that D ∩ J = ∅. By Corollary 3.4.3, there is a function g ∈ a such that g is modulo J an upper bound of the set {fξ : ξ < λ}. Then g is also modulo Dan upper bound of {fξ : ξ < λ}, contradicting the fact that f¯ is cofinal in a modulo D. Therefore there is an ordinal α < λ such that bα ∈ D, as desired.

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Lemma 3.4.17 If a is a progressive set of regular cardinals, I is an ideal on a and λ is a regular cardinal, then  J := I ∪ {b ∈ I + : tcf ( a/I |`b) = λ} is an ideal on a iff a ∈ J.

 Proof If a ∈ J, then tcf ( a/I) = λ. For  any subset b of a we have I = I |` a ⊆ I |` b and so can conclude that tcf ( a/I |` b) = λ. Consequently we get J = P(a), hence J is not an ideal on a. Conversely, assume that a ∈ J. We will prove that J is an ideal on a. For this let first b ∈ J and c ⊆ b. We must show that c ∈ J. Without loss of  generality we can assume that c ∈ I and thus b ∈ I, hence tcf ( a/I |` b) =λ. Now c ⊆ b yields I |` b ⊆  I |` c, and thus any sequence which is cofinal in a modulo I |`b is cofinal in a modulo I |`c. So we get c ∈ J. To prove the second property of an ideal, fix b, c ∈ J. If b, c ∈ I, then b ∪ c ∈ I. If b ∈ I + and c ∈ I, then I |` b =  {X ⊆ a : X ∩ b ∈ I} = {X ⊆ a : (X ∩ b) ∪ (X ∩ c) ∈ I} = I |`(b ∪ c), and tcf ( a/I |`b) = λ gives b ∪ c ∈ J. So + let  b, c ∈ I and choose the sequences (fξ : ξ < λ) and (gξ : ξ < λ) cofinal in a modulo I |` b and modulo I |` c, respectively. For all ξ < λ and ν ∈ a, we define  fξ (ν) if ν ∈ b, hξ (ν) = gξ (ν) otherwise.  ¯ = (hξ : ξ < λ) is cofinal in a modulo I |`(b ∪ c), let α < β < λ. To see that h Then Z := {ν ∈ b ∪ c : hβ (ν) ≤ hα (ν)} ⊆ X ∪ Y , where X := {ν ∈ b : fβ (ν) ≤ fα (ν)} ∈ I and Y := {ν ∈ c \ b : gβ (ν) ≤ gα (ν)} ∈ I. So we get Z ∈ I and thus ¯ is strictly increasing modulo I |`(b ∪ c). In the hα
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 Since D ∩ I = ∅, our assumption gives cf ( a/D) = λ. Lemma 3.4.16 says that there is a set b ∈ D such that tcf ( a/I |` b) = λ. This implies that b ∈ J, contradicting D ∩ J = ∅. Lemma 3.4.19 If a is progressive, λ is a cardinal, and b ∈ J<λ+ (a) \ J<λ (a), then  tcf ( a/J<λ (a) |`b) = λ. Proof Let λ be a cardinal and b ∈ J<λ+ (a)\J<λ (a). Then I := J<λ (a) |`b is an ideal on a. For an application of Theorem  3.4.18 let D be an arbitrary ultrafilter on a such that D ∩ I = ∅. Then cf ( a/D) ≥ λ by Corollary  3.4.11. + < λ and consequently cf ( a/D) = Since b ∈ D ∩J<λ+ (a), we get cf ( a/D)  λ. Now Theorem 3.4.18 says that tcf ( a/I) = λ. Lemma 3.4.20 If a is nonempty and progressive, then pcf(a) possesses a maximum, and we have  max pcf(a) ≤ | a| ≤ (sup(a))|a| . In particular, we get with Corollary 3.4.6: J<λ (a) = {b ⊆ a : max pcf(b) < λ} ∪ {∅}. If μ = max pcf(a), then Jμ+ (a) = P(a). Proof To get a contradiction let us assume that λ := sup pcf(a) is not a member of pcf(a). Then λ is a limit cardinal, and we have J<λ (a) = P(a) by Corollary 3.4.6. Lemma 3.4.15 says that a ∈ J<κ (a) for some κ < λ. Using Corollary 3.4.6 we get the contradiction pcf(a) ⊆ κ. The estimate for max pcf(a) is easy to verify since every sequence which  is cofinal in a modulo D, where D is an ultrafilter on a, is injective. The following theorem establishes a connection between the concepts of  the true cofinality modulo I and the cofinality of the partial ordering ( a, ≤a ). Theorem 3.4.21 If a is a nonempty progressive set of regular cardinals, then  max pcf(a) = cf ( a, ≤a ). Proof Let λ := max pcf(a). By Corollary 3.4.6, we have a ∈ J<λ+ (a) \ J<λ (a). Therefore Lemma 3.4.19 says that there is a sequence f¯ = (fξ : ξ < λ)  which is cofinal in a modulo J<λ (a).  First let us prove the relation “≤”. If S ⊆ a and |S| < λ, then every member of S is modulo J<λ (a) smaller than some member of f¯, and since λ

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not is regular we can find an ordinal β < λ satisfying h <J<λ (a) fβ , and thus  fβ ≤ h, for all functions h ∈ S. Consequently, S is not a cofinal subset of a  under ≤. From this we can conclude that λ ≤ cf ( a, ≤a ). Now we will prove the reverse inequality by induction on λ. Since |a|+ ≤ min(a) ≤ λ, we can assume without loss of generality that |a|+ < λ, since otherwise we have a = {λ} and our assertion is true.  By Lemma 3.4.2, applied to ida instead of f , we can choose a function g ∈ a such that g < f and, for  all h ∈ a with g ≤ h < f , the set {ξ < λ : B(g, fξ ) = B(h, fξ )} is unbounded in λ. There exists an ordinal α < λ such that g <J<λ (a) fα . Fix ξ ∈ [α, λ)ON , and let cξ := B[fξ , g]. Then cξ ∈ J<λ (a) and max pcf(c  ξ ) < λ if cξ = ∅. Thus cξ such that |Fcξ | < λ; the inductive hypothesis yields a cofinal subset Fcξ of  if cξ = ∅, let Fcξ = ∅. Now we can define a subset F of a by F := {fξ |`(a \ cξ ) ∪ h : ξ ∈ [α, λ)ON ∧ h ∈ Fcξ }. Obviously we have |F| ≤ λ · sup{|Fcξ | : ξ ∈ [α,  λ)ON } = λ. We will show that F is a cofinal subset of a; this will give cf ( a, ≤a ) ≤ λ. For this, let k ∈ a and, without loss of generality, g ≤ k; otherwise we replace k by the function max{g, k}. By the choice of g, there is an ordinal β ∈ [α, λ)ON satisfying c := cβ = B[fβ , g] = B[fβ , k]. If c = ∅, then we have k ≤ fβ . Otherwise we can choose a function h ∈ Fc such that k |` c ≤ h. Then obviously k ≤ fβ |`(a \ c) ∪ h ∈ F, as desired. Lemma 3.4.22 Assume that a is a progressive set of regular cardinals, I is an ideal on a, and  μ := sup{cf ( a/D) : D is an ultrafilter on a and D ∩ I = ∅}. Then there is a set b ∈ I such that μ = max pcf(a \ b). In particular μ is regular. Proof Consider a fixed member cof I. If D isa ultrafilter on a with D ∩ I = ∅, then a \ c ∈ D. Therefore cf ( a/D) = cf ( (a \ c)/D ∩ P (a \ c)) ≤ max pcf(a \ c). Consequently we get μ ≤ max pcf(a \ c). The following claim will guarantee the existence of the desired set b. Claim There exists b ∈ I such that a \ b ∈ J<μ+ (a). To get a contradiction let us assume that a \ b ∈ J<μ+ (a) for every b ∈ I. If b ∈ I and c ∈ J<μ+ (a) is arbitrary, then b ∪ c = a since otherwise a \ b ⊆ c and hereby a \ b ∈ J<μ+ (a), contradicting our assumption. Consequently, there is an ideal J on a satisfying I ∪ J<μ+ (a) ⊆ J. If D is an ultrafilter on a with

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 D ∩ J = ∅, then Corollary 3.4.11 implies cf ( a/D) ≥ μ+ contradicting the definition of μ. This proves the claim. So choose a set b ∈ I such that  a\b ∈ J<μ+ (a), and let D be an ultrafilter on  a satisfying a \ b ∈ D and cf ( (a \ b)/D ∩ P(a \ b)) = max pcf(a \ b). Then cf ( (a \ b)/D ∩ P(a \ b)) = cf ( a/D) and cf ( a/D) ≤ μ by Lemma 3.4.5. With the first part of the proof we obtain μ = max pcf(a \ b) for this set b ∈ I, as desired. Theorem 3.4.23 Assume that a is a progressive set of regular cardinals and I is an ideal on a. Then   cf ( a, ≤I ) = sup{cf ( a/D) : D is an ultrafilter on a and D ∩ I = ∅}.  Furthermore the cardinal cf ( a, ≤I ) is regular.  Proof Let μ := sup{cf ( a/D) : D is an ultrafilter on a and D ∩ I = ∅}. From Lemma 3.4.22 we know that there is b ∈ I such that μ= max pcf(a \ b). In particular μ is regular. Furthermorewe have μ =cf ( (a \ b), ≤) by Theorem 3.4.21. It is easy to verify that cf ( a, ≤I ) ≤ cf ( (a \ b), ≤), and so arbitrary ultrafilter on we get μ ≥ cf ( a, ≤I ). On the other hand, if D is an   a with D ∩ I = ∅, then  D extends the ideal I, hencecf ( a/D) ≤ cf ( a, ≤I ). This yields μ ≤ cf ( a, ≤I ) and altogether μ = cf ( a, ≤I ), as desired. Exercises 1) Assume that a is a progressive set of regular cardinals and I is an ideal on a. Prove: If tcf ( a/I) > sup(a), then dom(I) = a. 2) Assume that a is a set of regular cardinals and I is an ideal on a. Show that I is ℵ0 -mighty. 3) Assume that a is a  progressive set of regular cardinals and D is an ultrafilter on a with cf ( a/D) = λ. Then there is a set b ∈ D such that  tcf ( a/J<λ (a) |`b) = λ. 4) (S. Neumann) Let a be an infinite progressive set of regular cardinals and  I be an ideal on a. Prove that there is a set b ∈ I + such that tcf ( a/I |`b) exists. 5) Assume that a is an infinite progressive  set of regular cardinals, I is an ideal on a, μ is an infinite cardinal and ( a,
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6) If a is an infinite progressive set of regular cardinals, then for every set b ∈ P(a)\{∅} there is a unique regular cardinal λ such that b ∈ J<λ+ (a)\ J<λ (a). 7) Assume that a is an infinite progressive set of regular cardinals, μ and λ are infinite cardinals with μ < λ, and (bα : α < μ) is a sequence of members of J<λ+ (a) \ J<λ (a). Prove that there exists a modulo J<λ (a) a such strictly increasing sequence f¯ := (fξ: ξ < λ) of members of that, for every α < μ, f¯ is cofinal in a modulo J<λ (a) |`bα . Hint: Use Lemma 3.4.19 and Lemma 3.4.10. 8) Assume that a is an infinite set of regular cardinals such that |a|+ < min(a), μ and λ are infinite cardinals with μ < λ, and (bα : α < μ) is a sequence of members of J<λ+ (a) \ J<λ (a). Prove that there exists ¯ := (Bξ : ξ < λ) of members of P(a) \ J<λ (a) satisfying a sequence B Bζ ⊆J<λ (a) Bξ for all ζ and ξ with ζ < ξ < λ, such that for every γ < μ there is β < λ with bγ ⊆J<λ (a) Bβ . ¯ Hint: If f¯ = (f ξ : ξ < λ) is the sequence from Exercise 7, then f is (a). Corollary 3.4.3 yields a function g∈ unbounded in a modulo J <λ  a and a family (Bξ : ξ < λ) of J<λ (a)-positive subsets of a with the following properties: (1) If ζ < ξ < λ, then Bζ ⊆J<λ (a) Bξ .  (2) For every ζ < λ, f¯ is cofinal in a modulo J<λ (a) |`Bζ . (3) If J is an ideal on a extending J<λ (a) such that Bξ ∈ J for all ξ < λ, then g is an upper bound of f¯ modulo J. Suppose that there is an ordinal α < λ such that, for all β < λ, bα \ Bβ ∈ J<λ (a). For β < λ let dβ := bα \ Bβ . By Exercise 2.10 you can choose an ultrafilter D on a such that D ∩ J<λ (a) = ∅ and dβ ∈ D for all β < λ. Show that D ∩ J<λ (a) |` bα = ∅ = D ∩ J and apply properties (2) and (3). 9) Assume that a is an infinite set of regular cardinals with |a|+ < min(a), μ and λ are infinite cardinals with μ < λ, and (bα : α < μ) is a sequence of members of J<λ+ (a) \ J<λ (a). Prove that there exists a set b ∈ J<λ+ (a) such that bα ∈ J<λ (a)[b] for all α < μ. Hint: Let (Bξ : ξ < λ) be the sequence from Exercise 8. Choose, for every α < μ, an ordinal β(α) < λ such that bα ⊆J<λ (a) Bβ(α) . Show for δ := sup{β(α) : α < μ} that Bδ ∈ J<λ+ (a) and bα ⊆J<λ (a) Bδ for all α < μ. 10) (Shelah) Assume that a is an infinite progressive set of regular cardinals with 2|a| < min(a) and λ is a cardinal. Prove that there exists a set b ∈ P(a) such that J<λ+ (a) = J<λ (a)[b]. Hint: Use the previous question.

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11) Assume that a is an infinite progressive set of regular cardinals and |pcf(a)| < min(a). Then for all subsets d and e of pcf(a) the following holds. (i) pcf(∅) = ∅. (ii) d ⊆ pcf(d). (iii) If d ⊆ e, then pcf(d) ⊆ pcf(e). (iv) pcf(d ∪ e) = pcf(d) ∪ pcf(e). (v) pcf(pcf(d)) = pcf(d). In other words: If |pcf(a)| < min(a), then pcf is the hull operator of a topology on pcf(a). This topology is even compact. 12) (Shelah) Assume that a is a nonempty progressive set of regular cardinals and I is an ideal on a. Prove that there is a set b ∈ I such that   cf ( a, ≤I ) = cf ( (a \ b), ≤(a\b) ). 13) (S. Neumann) Assume that (P, ≤) is a partial ordering and (P, ) is a linear extension of (P, ≤), that means that (P, ) is a linear ordering and ≤ is a subset of . Show that cf (P, ) ≤ cf (P, ≤). 14) (S. Neumann) Prove: If a is a nonempty progressive interval of regular cardinals, then the set    {cf ( a, ≺): ( a, ≺) is a linear extension of ( a,
Chapter 4 Approximation Sequences The methods applied in the previous chapters are classical and, in a certain sense, elementary. In this chapter, we will introduce modern model theoretic methods. Notions such as “model of ZFC” and “absoluteness of a formula” are introduced. For any infinite cardinal number Θ we define the set H(Θ) of those sets which are hereditarily of cardinality less than Θ. We will show that for all regular uncountable cardinals Θ, H(Θ) is a model of all axioms of ZFC except the power set axiom. An approximation sequence (Ni : i < κ) will be any continuous sequence of elementary substructures of H(Θ) satisfying (Nj : j ≤ i) ∈ Ni+1 for all i < κ. It will be shown in the following chapters that these sequences are a technical tool of great importance.

4.1 The Sets H(Θ) Definition For any infinite cardinal number κ let H(κ) := {x : |tc(x)| < κ} be the class of all sets x which are hereditarily of cardinality less than κ. Remark The class H(κ) is a set since Lemma 4.1.2 shows that H(κ) ⊆ Vκ , where Vκ is the κ-th von Neumann level. For any regular cardinal κ, we have H(κ) = Vκ iff κ = ω or κ is a strongly inaccessible cardinal (see Exercise 8). Lemma 4.1.3 represents an alternative definition of H(κ) for regular cardinals κ and is of importance for later applications, since we will consider the sets H(κ) only for such cardinals. Let κ, μ ∈ ICN. Clearly we have H(κ) ⊆ H(μ) if κ ≤ μ. The proof of H(κ) = {H(μ+ ): μ ∈ ICN ∩ κ}, for uncountable κ, is left as an exercise. Furthermore V = {H(κ) : κ ∈ ICN} is true, since, for any set x, the definition of H(κ) yields x ∈ H(λ), where λ := max{ℵ0 , |tc(x)|+ }. Together with the

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fact that H(λ) is transitive (see Lemma 4.1.4), we can conclude in analogy to an assertion about the cumulative hierarchy that, for any set x, there exists λ ∈ ICN such that x ⊆ H(λ).  Lemma 4.1.1 For any set x, we have tc(x) = x ∪ {tc(y) : y ∈ x}.  Proof Let a := x ∪ {tc(y) : y ∈ x}. We will show that a is the smallest transitive superset of x. If z ∈ y ∈ a, then y ∈ x, hence z ∈ y ⊆ tc(y) ⊆ a, or y ∈ tc(u) for some u ∈ x. Therefore we get y ⊆ a, since tc(u) is transitive. Thus a is a transitive superset of x. If b is any transitive superset of x, and if y ∈ x, then y ⊆ b, hence tc(y) ⊆ b. This gives a ⊆ b, as desired. Lemma 4.1.2 For all κ ∈ ICN, H(κ) is transitive. If Vκ is the corresponding von Neumann level, then H(κ) ⊆ Vκ ; in particular H(κ) is a set. Proof Assume that x ∈ H(κ). If y ∈ x, then tc(y) ⊆ tc(x) and thus |tc(y)| ≤ |tc(x)| < κ. So we get y ∈ H(κ) which implies that H(κ) is transitive. As it is already  known – see Example 1.3.7 – we have V0 = ∅, Vα+1 = P(V ), and V = {Vβ : β < τ } for limit ordinals τ . Furthermore, V = τ  α {Vα : α ∈ ON} = {x : x = x}. For any set x, the rank α of x, written as rk(x), is the smallest ordinal β such that x ∈ Vβ+1 . We use transfinite induction on the rank of x to prove that x ∈ H(κ) implies x ∈ Vκ . If κ is a limit cardinal, then it follows from |tc(x)| < κ that |tc(x)| < λ+ for some cardinal λ < κ; thus x ∈ H(λ+ ) ⊆ H(κ). So we can assume without loss of generality that κ is regular. If y ∈ x, then rk(y) < rk(x) and y ∈ H(κ), which yields y ∈ Vκ by the inductive hypothesis. So we get x ⊆ Vκ . Since κ is regular and |x| < κ, there is α < κ such that x ⊆ Vα , and thus x ∈ Vα+1 ⊆ Vκ . Lemma 4.1.3 If Θ is a regular cardinal, then H(Θ) = {x ⊆ H(Θ) : |x| < Θ}. Proof Fix x ∈ H(Θ). Then |tc(x)| < Θ. Furthermore Lemma 4.1.2 says that x ⊆ H(Θ). Now x ⊆ tc(x) yields one inclusion. To verify the reverse inclusion, let x ⊆ H(Θ) and |x| < Θ.For every y ∈ x, we have y ∈ H(Θ) and thus |tc(y)| < Θ, hence |tc(x)| = |x ∪ {tc(y) : y ∈ x}| < Θ, since Θ is regular. Lemma 4.1.4 For any infinite cardinal number κ, we have a) b) c) d) e) f)

H(κ) is transitive. κ ≤ μ =⇒ H(κ) ⊆ H(μ).  κ > ℵ0 =⇒ H(κ) = {H(μ+ ) : μ ∈ ICN ∩ κ}.  If κ is a limit cardinal, then H(κ) = {H(μ) : μ ∈ ICN ∩ κ}. y ⊆ x ∧ x ∈ H(κ) =⇒ y ∈ H(κ). H(κ) ∩ ON = κ.

4.1. The Sets H(Θ) g) x ∈ H(κ) =⇒

175 

x ∈ H(κ).

h) x, y ∈ H(κ) =⇒ {x, y} ∈ H(κ) ∧ (x, y) ∈ H(κ) ∧ x ∪ y ∈ H(κ). i) x, y ∈ H(κ) =⇒ x × y ∈ H(κ) ∧ (f : x −→ y =⇒ f ∈ H(κ)). Proof Part a) is proved in Lemma 4.1.2; the proof of b), c) and d) is left as an exercise. Let x ∈ H(κ). If y ⊆ x, then tc(y) ⊆ tc(x), which implies e). To show f), notice that all ordinals α are transitive, hence satisfy tc(α) = α. So we get |tc(α)| < κ iff α < κ, for all ordinals α, which yields f). For the proof of g) h), let x, y ∈ H(κ), i.e., max{|tc(x)|,|tc(y)|} < κ. It is easy to see that  and  x ⊆ {tc(y) : y ∈ x} ⊆ tc(x), hence tc( x) ⊆ tc(x); furthermore we have tc(x∪y) = tc(x)∪tc(y) and tc({x, y}) = {x, y}∪tc(x)∪tc(y), which yields g) and h). The last assertion can be inferred, for regular cardinals κ, from x×y ⊆ H(κ) and |x × y| < κ, using Lemma 4.1.3. If κ is singular and x, y ∈ H(κ), then we can find a regular cardinal μ < κ such that max{|tc(x)|, |tc(y)|} < μ, hence x, y ∈ H(μ) and thus x × y ∈ H(μ), as we have just shown. With H(μ) ⊆ H(κ) we get the first assertion in i). If f : x −→ y is a function, then we have f ⊆ x × y and thus f ∈ H(κ) by e). Exercises 1) Assume that κ is a regular cardinal. Prove that x ∈ H(κ) iff, for any n ∈ ω and any sequence f : n −→ V satisfying f (n − 1) = x and f (j − 1) ∈ f (j) for all j with 0 < j < n − 1, we have |f (0)| < κ. Show that one direction of the equivalence is false for any singular cardinal κ. Hint: Use Lemma 4.1.1 or the remark in the definition of the transitive closure.  + 2) Prove, for uncountable cardinals κ, that  H(κ) = {H(μ ) : μ ∈ ICN∩κ}. If κ is a limit cardinal, then H(κ) = {H(μ) : μ ∈ ICN ∩ κ}. Why is the last assertion false for successor cardinals? 3) Assume that κ is an infinite cardinal and f, x, y ∈ H(κ). Prove: a) If f is a function, then dom(f ), ran(f ) and, for any x ∈ dom(f ), f (x) are members of H(κ). b) x y ⊆ H(κ), where x y is the set of functions from x into y. 4) Prove that Vω = H(ω). 5) Prove that H(κ+ )  Vκ+ for all κ ∈ ICN. Hint: P(κ) ∈ Vκ+2 . 6) Show that the equation H(κ) = {x ⊆ H(κ) : |x| < κ} is false for κ = ℵω .

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7) Assume that κ is a regular cardinal. Prove x ∈ H(κ) ⇐⇒ |x| < κ ∧ ∀y ∈ tc(x) |y| < κ. Hint: One direction of the equivalence is easy to verify. For the other one, use ∈-induction and Lemma 4.1.1. 8) Assume that κ is a regular cardinal. Prove Vκ = H(κ) ⇐⇒ κ = ω ∨ κ is strongly inaccessible . Hint: If κ > ω is not strongly inaccessible, then there is a cardinal λ < κ such that 2λ ≥ κ. Now use the hint in Exercise 5. Conversely, if κ is strongly inaccessible, use Exercise 1.7.6. 9) Assume that κ is an uncountable regular cardinal. Prove that H(κ) = Vκ iff κ is a fixed point of the beth function. Hint: See the exercise in Section 1.7 in which the beth function is defined. 10) a) Let κ be an infinite cardinal and α ≤ κ+ . We define Hα := H(κ+ ) ∩ Vα . Prove for α < κ+ : a1) a2) a3) a4)

Hα+1 = {X ⊆ Hα : |X| ≤ κ}.  α ∈ Lim =⇒ Hα = {Hβ : β < α}. Hκ+ = H(κ+ ). |Hα | ≤ 2κ .

b) Assume that λ is an infinite cardinal. Prove that |H(λ)| = 2<λ . Hint: If λ = κ+ , use part a). If λ is a limit cardinal, use Exercise 2.

4.2 Models and Absoluteness In this section we will represent some fundamental facts from mathematical logic. The required prereqisites are small, and once the basic Lemmas 4.2.12 and 4.2.13 are available, even a reader who is not familiar with the first order predicate calculus will be able to apply them correctly. The formal language L := LZF of ZF has already been introduced in Section 1.1. A subformula of a formula ϕ of L is a sequence of consecutive symbols in ϕ which is a formula. For example, the formula (∗)

(∃v0 v0 ∈ v1 ∧ ∃v1 v2 ∈ v1 )

has five subformulae, namely v0 ∈ v1 , ∃v0 v0 ∈ v1 , v2 ∈ v1 , ∃v1 v2 ∈ v1 and the formula itself. If there is an occurence of a quantifier ∃vj or ∀vj in a formula,

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then the scope of this quantifier is the unique subformula beginning with this quantifier. An occurence of a variable vj in a formula is called free if it does not lie in the scope of a quantifier ∃vj or ∀vj acting on that variable; otherwise it is called bound. For example, in (∗) the first occurence of v1 is free, but the second and third are bound, whereas v0 is bound at all of its occurrences and v2 only occurs free. Any variable which occurs free in ϕ is called a free variable of ϕ, and a variable which occurs bound in ϕ is called a bound variable of ϕ. Intuitively, a formula expresses a property of its free variables, whereas a bound variable, used to make existential or universal statements, can be replaced in the scope of the quantifier binding it by any other variable which does not not occur in the formula, without changing the “meaning” of the formula. Under the symbols of L are variables vn for every natural number n. To make the text more readable and to avoid double indices, we will also use syntactical variables an , xn , yn etc. for variables. Often we will present a formula in the form ϕ(x0 , . . . , xn−1 ), when its free variables are amongst x0 , . . . , xn−1 , to emphasize its dependence on x0 , . . . , xn−1 . Often we abbreviate the sequence x0 , . . . , xn−1 by x and write ϕ(x) instead of ϕ(x0 , . . . , xn−1 ). If a0 , . . . , an−1 are further variables, then ϕ(a0 , . . . , an−1 ) is that formula which results from substituting an ai for every free occurrence of xi . Such a substitution is called legitimate if no free occurence of an xi is in the scope of a quantifier ∃ai or ∀ai . Intuitively, the formula ϕ(a0 , . . . , an−1 ) says about a0 , . . . , an−1 the same that ϕ(x0 , . . . , xn−1 ) says about x0 , . . . , xn−1 . So let for example ϕ(v1 , v2 ) be the formula in (∗). Then the substitution ϕ(v2 , v3 ) is legitimate, but the substitution ϕ(v0 , v2 ) is not legitimate. Interpreting the symbol ∈ in this formula, namely in (∃v0 v0 ∈ v0 ∧ ∃v1 v2 ∈ v1 ), as the usual naive membership, it says that some set is an element of itself, whereas the formula in (∗) says that v1 has an element. We will always assume without mentioning that all our substitutions are legitimate. Instead of ∃vi (ϕ(vi ) ∧ ∀vj (ϕ(vj ) =⇒ vi = vj )) we also write ∃!vi ϕ(vi ). A closed formula or sentence of LZF is a formula in which no variable occurs free. Intuitively, it states an assertion which is either true or false. If ϕ is a formula, then let a universal closure of ϕ be any sentence ∀x ϕ obtained by universally quantifying all free variables of ϕ. Since such a sentence can be obtained in a unique way and all universal closures of fixed formulae are logically equivalent (see below), we will speak of the universal closure of ϕ. Every formula of LZF is logically equivalent to a formula whose connectives and quantifiers are amongst ¬, ∨, and ∃. Therefore we regard, in definitions and proofs which use induction on the complexity of a formula ϕ, the formulae ϕ∧ψ, ϕ =⇒ ψ, ϕ ⇐⇒ ψ, and ∀xϕ as abbreviations for the formulae ¬(¬ϕ ∨ ¬ψ), ¬ϕ ∨ ψ, (ϕ =⇒ ψ) ∧ (ψ =⇒ ϕ), and ¬∃x(¬ϕ) respectively.

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If T is a set of sentences of LZF , then we also call T a theory and its elements axioms of T . T  ϕ means that ϕ is provable with the axioms of T . T ⊆ ZF means that any formula which is provable with the axioms of T is provable in ZF. A structure for LZF is a pair A = (A, E), where A is a nonempty set, the universe of A, and E = {(x, y) ∈ A × A : x ∈ y}. With L∗ or L∗ZF we denote that language which results from LZF by adding the binary relation symbol <. The notion of a formula is extended correspondingly, adding to the definition of atomic formulae (see Section 1.1) the rule “vi < vj is a formula”. A structure for L∗ZF , shortly, an L∗ -structure, is a triple (A, E, <∗ ), where again A is a nonempty set, E = {(x, y) ∈ A × A : x ∈ y} and <∗ ⊆ A × A is a well-ordering of A. Instead of (A, E) and (A, E, <∗ ) we use the slightly incorrect but more suggestive notation (A, ∈) and (A, ∈, <∗ ), respectively. Assume that A is an L∗ -structure, (a0 , . . . , an−1 ) is a finite sequence of members of the universe A of A, and ϕ is an L∗ -formula with at most the free variables v0 , . . . , vn−1 . By induction on the complexity of ϕ we define the L-formula A |= ϕ[a0 , . . . , an−1 ]. Intuitively, the formula A |= ϕ[a0 , . . . , an−1 ] is provable iff the property of a0 , . . . , an−1 ∈ A formally expressed by ϕ is true, where “for all x” and “there exists x” are to be read as “for all x ∈ A” and “there exists x ∈ A”, and xi is to be read as ai .1 We define all notions for L∗ , and it should be obvious that hereby they are also defined for L. In the same way, most properties which we prove for L∗ -structures also hold for L-structures. To be more precise, let a0 , . . . , an−1 ∈ A. For atomic formulae we define A |= vi ∈ vj [a0 , . . . , an−1 ] ⇐⇒ ai ∈ aj , A |= vi = vj [a0 , . . . , an−1 ] ⇐⇒ ai = aj , A |= vi < vj [a0 , . . . , an−1 ] ⇐⇒ ai <∗ aj . Further we agree that A |= (¬ψ)[a0 , . . . , an−1 ] ⇐⇒ ¬A |= ψ[a0 , . . . , an−1 ], A |= (ψ ∨ χ)[a0 , . . . , an−1 ] ⇐⇒ A |= ψ[a0 , . . . , an−1 ] ∨ A |= χ[a0 , . . . , an−1 ], 1 To be precise, this requires, especially in view of the L¨ owenheim-Skolem theorem 4.2.3, a so-called arithmetization of our formal language. We may identify the symbols of L∗ZF with certain hereditarily finite sets, i.e., with elements of Vω . Then the formulas of L∗ZF are also hereditarily finite sets. A |= ϕ[a0 , . . . , an−1 ] is defined by induction on the complexity of formulae as a relation between structures A, (arithmetized) formulae ϕ, and formal n-tuples (a0 , . . . , an−1 ) ∈ An , that means (if we identify as usual formal n-tuples with functions which have the domain n ∈ ω), elements of n A. The interested reader is referred to the books of F. R. Drake [Dr] (Chapter 3, §5) and K. Kunen [Ku] (Chapter V, §1), in which he can find a detailed formalization of some first order languages.

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A |= ∃vn ψ[a0 , . . . , an−1 ] ⇐⇒ ∃a ∈ A A |= ψ[a0 , . . . , an−1 , a]. With the usual abbreviations we obtain the assertions A |= (ψ ∧ χ)[a0 , . . ., an−1 ], A |= (ψ =⇒ χ)[a0 , . . ., an−1 ], A |= (ψ ⇐⇒ χ)[a0 , . . ., an−1 ], A |= ∀vn ψ[a0 , . . ., an−1 ], where, for example, the last one is the assertion ∀a ∈ A A |= ψ[a0 , . . ., an−1 , a]. If the formula A |= ϕ[a0 . . . an−1 ] is provable in ZFC, then we say that ϕ holds in A under the valuation (a0 , . . . , an−1 ). If the universal closure ∀xϕ of ϕ holds in A, then we say that ϕ holds in A or that A is a model of ϕ, and write A |= ϕ. If T is a set of sentences of L∗ , then we call A a model of T , written as A |= T , if A is a model of ϕ for every sentence ϕ in T . If the context is clear, we identify the structure A with its universe A and write A |= ϕ[a0 , . . . , an−1 ] instead of A |= ϕ[a0 , . . . , an−1 ]. Often we write A |= ϕ(a0 , . . . , an−1 ) instead of A |= ϕ[a0 , . . . , an−1 ]. Definition If A = (A, ∈, <∗1 ) and B = (B, ∈, <∗ ) are structures for L∗ , then A is a substructure of B, written as A ⊆ B iff A ⊆ B and <∗1 =<∗ ∩ A × A. Then we also write (A, ∈, <∗ ) instead of (A, ∈, <∗1 ). If A is a substructure of B, then A is an elementary substructure of B and B is an elementary extension of A, written as A  B, iff, for every formula ϕ of L∗ and every valuation a0 , . . . , an−1 ∈ A of the free variables of ϕ, A |= ϕ[a0 , . . . , an−1 ] ⇐⇒ B |= ϕ[a0 , . . . , an−1 ] . Remark The reader who is not familiar with these notions may visualize A  B as: A is a substructure of B, and any assertion about objects in A holds in A iff it holds in B. Lemma 4.2.1 If A ⊆ B, A  C and B  C, then A  B. The applications of the notion of an elementary substructure are simplified by the following lemma. Lemma 4.2.2 (Tarski, Vaught) Assume that A ⊆ B. Then A  B iff, for any formula ψ of L∗ and any valuation a0 , . . . , an−1 ∈ A, B |= ∃xn ψ[a0 , . . . , an−1 ] =⇒ ∃a ∈ A(B |= ψ[a0 , . . . , an−1 , a]) . Proof If A  B and B |= ∃xn ψ[a], then, by definition, A |= ∃xn ψ[a], where a abbreviates the sequence a0 , . . . , an−1 . The definition of this formula yields a ∈ A such that A |= ψ[a, a], and thus, again using A  B, a ∈ A such that B |= ψ[a, a]. For the proof of the reverse implication we will verify the equivalence (A |= ϕ[a] ⇐⇒ B |= ϕ[a]) by induction on the complexity of the formula ϕ. If

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ϕ is atomic, then this follows from A ⊆ B. The cases ϕ ≡ ¬ψ and ϕ ≡ ψ ∨ χ follow immediately from the inductive hypothesis and the definition of |=. Now assume that ϕ has the form ∃xn ψ. From B |= ∃xn ψ[a] we can infer by assumption, that there is a ∈ A such that B |= ψ[a, a]. The inductive hypothesis gives A |= ψ[a, a], hence A |= ∃xn ψ[a]. Conversely, if the last assertion is true, then there is a ∈ A such that A |= ψ[a, a]. Since A ⊆ B, and since B |= ψ[a, a] holds by inductive hypothesis, we get B |= ∃xn ψ[a]. Theorem 4.2.3 (L¨ owenheim-Skolem Theorem) Assume that A and C are L∗ structures, A ⊆ C, and Y ⊆ C. Then there is an elementary substructure B of C such that A ⊆ B, Y ⊆ B, and |B| ≤ max{ℵ0 , |Y |, |A|}. Proof Put λ := max{ℵ0 , |Y |, |A|} and X0 := Y ∪ A. For each formula ϕ(x0 , . . . , xn ) of L∗ and each valuation (a0 , . . . , an−1 ) ∈ X0n satisfying C |= ∃xn ϕ [a], we can choose, using (AC), some a ∈ C such that C |= ϕ [a, a]. Take ∗ X0∗ as the set of all these  n“satisfiers”. It is easy to verify that X0∗⊆ X0 . Since n |X0 | ≤ λ and thus | {X0 : n ∈ ω}| ≤ ℵ0 · λ = λ, and since L and hereby the set of formulae of L∗ is countable, we can conclude that |X0∗ | ≤ ℵ0 · λ = λ. So put X1 := X0∗ and define, if Xn is already defined, Xn+1 := Xn∗ in complete analogy. If we assume  that |Xn | ≤ λ, then |Xn+1 | ≤ λ follows as above. Now put B := {Xn : n ∈ ω}, and let B be that structure with universe B whose relations are the restrictions of the corresponding relations of C to the subset B of C. Clearly we have B ⊆ C and |B| ≤ λ · ℵ0 = λ. For the proof of B  C we can apply Lemma 4.2.2. If C |= ∃xn ψ [b0 , . . ., bn−1 ] for certain b0 , . . ., bn−1 ∈ B, then there is m ∈ ω such that b0 , . . ., bn−1 ∈ Xm . Our definition of Xm+1 shows that there exists a satisfier b ∈ B such that C |= ψ[b0 , . . ., bn−1 , b], as desired. We will now introduce the notion of absoluteness of a formula. Since the absoluteness of a formula for a set M depends on its complexity, and since we will build up, starting with simple formulae, a list of absolute formulae, it is suitable to introduce the notion of relativization of a formula ϕ to a class M . Definition If M is a transitive class and ϕ is a formula of LZF , then we define the relativization of ϕ to M , denoted by ϕM , by induction on the complexity of ϕ. (x ∈ y)M is the formula x ∈ y, (x = y)M the formula x = y, (¬ψ)M the formula ¬(ψ M ), (ψ ∨ χ)M the formula ψ M ∨ χM , and (∃xψ)M the formula ∃x(x ∈ M ∧ ψ M ). If we take again ψ ∧ χ, ψ =⇒ χ, ψ ⇐⇒ χ and ∀xψ as abbreviations for the formulae ¬(¬ψ ∨ ¬χ), ¬ψ ∨ χ etc., then (ψ ∧ χ)M is the formula ϕM ∧ ψ M , (ϕ =⇒ ψ)M the formula ϕM =⇒ ψ M , (ϕ ⇐⇒ ψ)M the formula ϕM ⇐⇒ ψ M , and (∀xϕ)M the formula ∀x(x ∈ M =⇒ ϕM ).

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Definition Let M be a transitive class. A formula ϕ(x) of LZF is called absolute for M , written as ϕ abs M , iff the following formula is a theorem of ZF:   ∀x0 ∈ M . . . ∀xn−1 ∈ M ϕ ⇐⇒ ϕM . The reader is asked to convince himself that the definitions of validity and relativization immediately yield the following lemma that we will always use without mentioning. In the same way we will use the following theorem. It is intuitively clear, but we omit its proof since it requires an exact listing of the logical axioms and logical rules of first-order logic which are tacitly applied in every-day mathematics. Lemma 4.2.4 If M is a transitive set and ϕ(x0 , . . . , xn−1 ) is a formula of LZF , then a) ∀a0 ∈ M . . . ∀an−1 ∈ M (M |= ϕ[a0 , . . . an−1 ] ⇐⇒ ϕM (a0 , . . . , an−1 )). b) ϕ abs M iff ∀a0 ∈ M . . . ∀an−1 ∈ M (ϕ(a0 , . . . , an−1 ) ⇐⇒ M |= ϕ[a0 , . . . , an−1 ]). Theorem 4.2.5 Assume that T ⊆ ZF is a theory. If M is a model of T and ϕ is a formula of L which is provable with the axioms of T , then the formula ∀a ∈ M ϕM is provable in ZF. Definition A Δ0 -formula is a formula that contains only bounded quantifiers. That means that the Δ0 -formulae are defined by induction in the same way as formulae, with the exception that quantifiers are introduced by the rule “If ϕ is a Δ0 -formula and if vi and vj are variables, then ∃vi ∈ vj ϕ and ∀vi ∈ vj ϕ are Δ0 -formulae”. If T is a set of sentences of LZF , then ϕ is called a ΔT 0 -formula iff there is a Δ0 -formula ψ such that the formula ϕ ⇐⇒ ψ is provable with the axioms of T . For T = ∅ we will denote, in view of Lemma 4.2.6, also ΔT0 -formulae as Δ0 -formulae. Remark a) If one fixes exact rules for the introduction of defined and undefined logical symbols, then most formulae which we are handling are not Δ0 -formulae. For example, x ⊆ y is the formula ∀z ∈ x(z ∈ y) that contains only bounded quantifiers. But if we take ∃, ∨ and ¬ as the logical symbols of our language, then this string of symbols is an abbreviation for the formula ¬∃z¬(¬z ∈ x∨z ∈ y), which is by no means a Δ0 -formula. On the other hand it is logically equivalent to the Δ0 -formula ¬∃z ∈ x(¬z ∈ y). This is a typical example for our later way of proceeding. Δ0 -formulae

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will be absolute, but also formulae which are logically equivalent to Δ0 formulae. Therefore we do not distinguish between the denotations of these two types of formulae. b) Formulae which are in “weak” theories T with M |= T equivalent to Δ0 formulae, will be absolute for M , too. For later purposes we will need only absoluteness results for the structures H(Θ), where Θ > ω is regular, and these sets H(Θ) with the binary relation ∈ are models of all axioms of ZFC except the power set axiom. Therefore we need not take care of such subtle considerations. Lemma 4.2.6 Let M be a transitive class and ϕ and ψ be formulae of L. a) Any formula of the form x = y or x ∈ y is absolute for M . b) If the formulae ϕ and ψ are absolute for M , then so are ¬ϕ and ϕ  ψ, where  is one of the connectives ∨, ∧, =⇒, and ⇐⇒ . c) If the formula ϕ is absolute for M , then so are ∃x ∈ y ϕ and ∀x ∈ y ϕ. d) Every Δ0 -formula is absolute for M . e) If T ⊆ ZF is a theory such that M |= T and T  ∀x(ϕ(x) ⇐⇒ ψ(x)), then ϕ is absolute for M iff ψ is absolute for M . In particular all ΔT0 formulae are absolute for M , if M |= T . f ) Assume that T ⊆ ZF is a theory and M is a model of T . Further let T  ϕ ⇐⇒ ∀xψ and T  ϕ ⇐⇒ ∃xχ,2 where χ and ψ are absolute for M . Then ϕ is absolute for M . Proof Part a) follows immediately from the definition of the relativization of atomic formulae, b) follows with arguments from sentential logic. For c), let y ∈ M . The relativization of the formula ∃x ∈ y ϕ to M is the formula ∃x ∈ M (x ∈ y ∧ ϕM ). Since y ∈ M and M is transitive, x ∈ y implies x ∈ M . Furthermore we can infer from x ∈ M with the absoluteness of ϕ that ϕ ⇐⇒ ϕM . Therefore the discussed relativization is equivalent to ∃x ∈ y ϕ. The proof for the second formula runs similar. Part d) follows at once from a), b) and c). Now we prove e). From Theorem 4.2.5 we get ZF  ∀x ∈ M (ϕM ⇐⇒ M ψ ). Further, by assumption, we have ZF  ∀x(ϕ(x) ⇐⇒ ψ(x)). Thus ∀x ∈ M (ϕ ⇐⇒ ϕM ) is provable in ZF iff ∀x ∈ M (ψ ⇐⇒ ψ M ) is provable in ZF. Therefore ϕ is absolute for M iff ψ is absolute for M . Finally, we turn to the proof of f). Let a be a sequence containing the free variables of the considered formulae, and assume that a ∈ M . From ϕ we get successively ∀xψ, ∀x ∈ M ψ and ∀x ∈ M ψ M , i.e., (∀xψ)M , since ψ is absolute for M . Now Theorem 4.2.5 says that (∀xψ)M =⇒ ϕM , and we have shown 2 If

T in addition ψ and χ are ΔT 0 -formulae, then such formulae ϕ are called Δ1 -formulae.

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that ϕ =⇒ ϕM is provable. In complete analogy one proves, using the second equivalence in the assumption and ∃x ∈ M χ =⇒ ∃xχ, that the implication ϕM =⇒ ϕ is provable. Lemma 4.2.7 Every nonempty transitive set is a model of the axiom of extensionality, of the empty set axiom, and of the axiom of foundation. Proof We must check that M |= ∀a∀b(∀x(x ∈ a ⇐⇒ x ∈ b) =⇒ a = b). So we have to show that (∗)

∀a ∈ M ∀b ∈ M (∀x ∈ M (x ∈ a ⇐⇒ x ∈ b) =⇒ a = b).

But the fact that M is transitive gives, for a ∈ M , x ∈ M ∧ x ∈ a ⇐⇒ x ∈ a. Therefore (∗) follows from the axiom of extensionality. The corresponding proofs for the other axioms is left to the reader. Lemma 4.2.8 Assume that M is a transitive nonempty class. Then the following formulae are absolute for M : (1) (4) (7) (10) (13) (16) (19) (22) (25) (28)

x⊆y x = (a, b) Rel(x) x ∈ ran(u) x= u∩v x= u y = f (x) f : x −→ y x is a successor ordinal x = ω.

(2) (5) (8) (11) (14) (17) (20) (23) (26)

x=∅ ∃u∃v(x = {u, v}) x ∈ dom(u) x = ran(u) x=u∪v Func(x) y = f [x] f is injective Lim(x)

(3) (6) (9) (12) (15) (18) (21) (24) (27)

x = {a, b} ∃u∃v(x = (u, v)) x = dom(u) x=u×v x=u\v x is transitive g = f |`x x ∈ ON x∈ω

Proof To apply Lemma 4.2.6 we show that the formulae (1) to (23) are ΔT0 -formulae, where, as required, T = ∅ or T consists of the extensionality axiom as its unique member. (1) x ⊆ y is the Δ0 -formula ∀z ∈ x(z ∈ y). (2) x = ∅ is the Δ0 -formula ∀z ∈ x(z = z). (3) x = {a, b} is the Δ0 -formula ∀z ∈ x(z = a ∨ z = b) ∧ a ∈ x ∧ b ∈ x. (4) x = (a, b) is the formula x = {{a}, {a, b}}. If we rewrite this formula as in the proof of (3), then the subformulae {a} ∈ x and {a, b} ∈ x still bother us. But the last formula is ∃y ∈ x(y = {a, b}), a Δ0 -formula, and so is the first one. (5) The formula ∃u∃v(x = {u, v}) is equivalent to ∃u ∈ x∃v ∈ x(x = {u, v}). By (3), it is absolute for M .

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(6) The formula ∃u∃v(x = (u, v)) is equivalent in T to the formula ∃w ∈ x∃z ∈ x∃u ∈ w∃v ∈ z(w = {u}∧z = {u, v}∧∀a ∈ x(a = w∨a = z)). (7) (8) (9)

(10) (12) (16) (17) (18) (19) (20) (21) (24) (25) (26) (27) (28)

By (3), it is absolute for M . Rel(x) is the formula ∀z ∈ x∃u∃v(z = (u, v)). (6) says that it is absolute for M . The formula x ∈ dom(u) is equivalent in T to the formula ∃v ∈ u ∃w ∈ v ∃y ∈ w(v = (x, y)). x = dom(u) is equivalent in T to ∀z ∈ x(z ∈ dom(u)) ∧ ∀z ∈ dom(u)(z ∈ x). By (8), we only have to take care of the second formula in the previous conjunction. Now every formula of the form ∀z ∈ dom(u) ψ is equivalent in T to the formula ∀v ∈ u ∀w ∈ v ∀y ∈ w∀z ∈ w(v = (z, y) =⇒ ψ). Thus, if ψ is equivalent in T to a Δ0 -formula, then this also is true for the formula under consideration. and (11) The proof is analogous to that of (8) and (9). The considered formula is equivalent in T to ∀z ∈ x ∃y ∈ u ∃w ∈ v(z = (y, w)) ∧ ∀y ∈ u ∀w ∈ v ∃z ∈ x(z = (y, w)). The formula is equivalent in T to ∀z ∈ x ∃w ∈ u(z ∈ w) ∧ ∀w ∈ u ∀z ∈ w(z ∈ x). Func(x) is an abbreviation of the formula Rel(x) ∧ ∀z ∈ dom(x) ∀u, v ∈ ran(x)((z, u) ∈ x ∧ (z, v) ∈ x =⇒ u = v); now see the proof of (9). “x is transitive” is equivalent to the Δ0 -formula ∀y ∈ x∀z ∈ y(z ∈ x). y = f (x) ⇐⇒ Func(f ) ∧ ∃z ∈ f (z = (x, y)). y = f [x] ⇐⇒ Func(f ) ∧ ∀z ∈ y ∃u ∈ x(z = f (u)) ∧ ∀u ∈ x (u ∈ dom(f ) =⇒ ∃z ∈ y(z = f (u))). g = f |` x ⇐⇒ Func(f ) ∧ g ⊆ f ∧ ∀z ∈ x(z ∈ dom(f ) =⇒ z ∈ dom(g)) ∧ ∀z ∈ dom(g)(z ∈ x). Now see the proof of (9). The formula x ∈ ON is an abbreviation of “x is transitive and every member of x is transitive”. By (18), it is absolute for M . “x is a successor ordinal” is the formula x ∈ ON ∧ ∃v ∈ x(x = v ∪ {v}). Now apply (3) and (14). x is a limit ordinal iff x is an ordinal different from ∅ and x is not a successor ordinal; see (24), (2) and (25). x is a natural number iff ¬Lim(x) ∧ ∀z ∈ x ¬Lim(z). The desired result follows from (26). x = ω is the formula ∀z(z ∈ x ⇐⇒ z ∈ ON ∧ ¬Lim(z) ∧ ∀u ∈ z ¬Lim(u)),

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which is equivalent to the conjunction of the ΔT0 -formula ∀z ∈ x(z ∈ ON ∧ ¬Lim(z) ∧ ∀u ∈ z ¬Lim(u)) and the formula ϕ :≡ ∀z(z ∈ ON ∧ ¬Lim(z) ∧ ∀u ∈ z ¬Lim(u) =⇒ z ∈ x). With the previous results we get ϕ =⇒ ϕM and thus x = ω =⇒ (x = ω)M . Now assume that x ∈ M and (x = ω)M , hence ϕM holds. By (24) and (26), we have ∀z ∈ x(z ∈ ON ∧ ¬Lim(z) ∧ ∀u ∈ z ¬Lim(u)). This gives x ⊆ ω. Next we show that x is transitive. Then x is an ordinal with x ≤ ω, and if x would be a natural number, ϕM and x ∈ M would imply x ∈ x, a contradiction. So assume that v ∈ z and z ∈ x. Then v ∈ M and z ∈ M . We have to show that v ∈ x. We have ¬Lim(z) ∧ ∀u ∈ z ¬Lim(u). The transitivity of z gives ¬Lim(v) ∧ ∀u ∈ v ¬Lim(u). From v ∈ M and ϕM we can conclude that v ∈ x. Definition If ϕ(x, y ) is a formula, then we define the relativization of the class A = A(y ) := {x : ϕ(x, y )} by A(y )M := {x ∈ M : ϕM (x, y )}. A is called absolute for M iff ∀y ∈ M (A(y ) = A(y )M ) is provable in ZF. Remark If A(y ) := {x : ϕ(x, y )}, ϕ is absolute for M and y ∈ M , then A(y )M = {x ∈ M : ϕM (x, y )} = A(y ) ∩ M . An important example is A(y) := P(y). From Lemma 4.2.8 we get (A(y))M = {x ∈ M : (x ⊆ y)M } = {x ∈ M : x ⊆ y} = A(y) ∩ M . Thus the power set is in general not absolute for M . If, for all y ∈ M , A(y ) ⊆ M , and if ϕ is absolute for M , then A is absolute for M . Conversely, if for every y ∈ M we have {x : ϕ(x, y )}M = {x : ϕ(x, y )} ∩ M , then ϕ is absolute for M . Lemma 4.2.9 If M is transitive and closed under unordered pairing (i.e., {x, y} ∈ M for all x, y ∈ M ), then the following classes are absolute for M .   {x, y}, x, x ∪ y,) x (if x = ∅), x ∩ y, x \ y, (x, y), dom(x),

ran(x),

∅,

fld(x),

a × b,

∈ ∩ a × b,

{(u, v) : (v, u) ∈ x} .

If additionally ω ⊆ M , then ω is absolute for M . Proof The proof is easy if one uses Lemma 4.2.8 and the remark after the definition of an absolute class. For example, (x, y)M = {z ∈ M : z = {x} ∨ z = {x, y}} ⊆ (x, y). By assumption we have {x}, {x, y} ∈ M for all x, y ∈ M . Therefore we get (x, y)M = {z : z = {x} ∨ z = {x, y}} = (x, y). Or, to give a further example, dom(x)M = {z ∈ M : ∃u ∈ M (z, u) ∈ x} ⊆ dom(x). If (z, u) ∈ x and x ∈ M , then we infer from the fact that M is transitive that u ∈ M and z ∈ M . This gives z ∈ dom(x)M . The other absoluteness assertions are verified analogously.

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Remark 4.2.10 a) The following formulae and classes are in general not absolute: y = P(x), P(x), |y| = |x|, |x|, x is a cardinal number, cf (x), x is regular. However, by Lemma 4.2.11, the property of being a cardinal number is preserved in M . Intuitively, this is clear, since with fewer sets there are more cardinals, because there is less chance of a one-one correspondence. b) For uncountable cardinals Θ, our models H(Θ) satisfy very strong absoluteness properties, see Lemma 4.2.12. Lemma 4.2.11 If M is transitive, then ∀x ∈ M (x ∈ CN =⇒ M |= x ∈ CN) . Proof Let ϕ(x, f ) be the formula Func(f ) ∧ f is injective ∧ dom(f ) = x ∧ ran(f ) ∈ x. ϕ is absolute for M by Lemma 4.2.8. Further we have by the definition of a cardinal: x ∈ CN ⇐⇒ x ∈ ON ∧ ¬∃f ϕ(x, f ). Thus from x ∈ CN we get x ∈ ON ∧ ¬∃f ∈ M ϕ(x, f ), and from Lemma 4.2.8 we can infer (x ∈ ON)M ∧ ¬∃f ∈ M ϕM (x, f ). But this is the formula (x ∈ CN)M . Lemma 4.2.12 If Θ is an uncountable regular cardinal, then (H(Θ), ∈) is a model of the theory ZFC− , that is ZFC without the power set axiom. Furthermore, for arbitrary uncountable cardinals κ, the following formulae are absolute for H(κ): a) b) c) d)

x ∈ CN. x is a regular cardinal. x ∈ CN ∧ y ∈ CN ∧ y = x+ . x is a limit ordinal and y = cf (x).

Proof We will prove that H(Θ) is a model of all axioms of ZFC− . For this we will use Lemma 4.1.3, Lemma 4.1.4 and the previous results on absoluteness. The validity of (A1), (A2), (A4) and  (A5) is clear. As an exercise, we demonstrate this for (A5). The formula y = x is absolute for H(Θ). Therefore we get H(Θ) |= A5 ⇐⇒ H(Θ) |= ∀x∃y∀z(z ∈ y ⇐⇒ ∃u ∈ x(z  ∈ u)) ⇐⇒ H(Θ) |= ∀x∃y(y = x) ⇐⇒ ∀x ∈ H(Θ)∃y ∈ H(Θ)(H(Θ) |= y = x) ⇐⇒  ∀x ∈ H(Θ)∃y ∈ H(Θ)(y = x) ⇐⇒ ∀x ∈ H(Θ)( x ∈ H(Θ)) , where the last formula holds by Lemma 4.1.4 g).

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(A7) holds in H(Θ), since Θ > ℵ0 and thus, by Lemma 4.1.4, ω is a member of H(Θ). For the comprehension axioms (A3), let ϕ(z, p  ) be a formula of LZF , and let p ∈ H(Θ). We want to show that H(Θ) |= ∀x∃y(y = {z ∈ x : ϕ(z, p )}), that means ∀x ∈ H(Θ) ∃y ∈ H(Θ)(H(Θ) |= y = {z ∈ x : ϕ(z, p )}). Since H(Θ) is transitive, the assertion H(Θ) |= y = {z ∈ x : ϕ(z, p )} is true for x, y ∈ H(Θ) iff y = {z ∈ x : H(Θ) |= ϕ(x, p )}. By Lemma 4.1.4 e), the set {z ∈ x : H(Θ) |= ϕ(x, p )} is a member of H(Θ) since it is a subset of x ∈ H(Θ). For the replacement axioms (A8), let ϕ(x, y) be a formula of LZF . We have to show that H(Θ) |= ∀x∀y∀z(ϕ(x, y) ∧ ϕ(x, z) =⇒ y = z) implies H(Θ) |= ∀x∃y∀z(z ∈ y ⇐⇒ ∃u ∈ x ϕ(u, z)). The last assertion is equivalent to ∀x ∈ H(Θ)∃y ∈ H(Θ)(y = {z ∈ H(Θ) : ∃u ∈ x ϕH(Θ) (u, z)}). Fix x ∈ H(Θ). The premiss says that the formula ϕH(Θ) is functional on H(Θ). So we can apply a replacement axiom and obtain a set y such that y = {z ∈ H(Θ) : ∃u ∈ x ϕH(Θ) (u, z)}. Then y ⊆ H(Θ) and |y| < Θ, since |y| ≤ |x| < Θ. Since Θ is regular, Lemma 4.1.3 yields y ∈ H(Θ). For the axiom of choice, we observe that the formula “f is a choice function for x” is absolute for H(Θ). Let x ∈ H(Θ) be a set of nonempty sets. Using  (AC), we obtain a choice function f for x. Since f ⊆ x × x, Lemma 4.1.4 yields f ⊆ H(Θ). Now we can infer from |f | ≤ |x| together with Lemma 4.1.3 that f ∈ H(Θ). Now we turn to the proof of the desired absoluteness results. For the proof of part a), fix x ∈ H(κ). Note that (x ∈ CN =⇒ H(κ) |= x ∈ CN) by Lemma 4.2.11. For the converse, assume that H(κ) |= x ∈ CN. Then we have x ∈ ON. Assume, to get a contradiction, that x ∈ / CN. If we take the formula ϕ1 (x, f ) as “f is a one-one function ∧ dom(f ) = x ∧ ran(f ) ∈ x”, then we can infer ∃f ϕ1 (x, f ). Since ran(f ) ∈ x, we have ran(f ) ∈ H(κ), which gives f ∈ H(κ) by Lemma 4.1.4. Now we come to the contradiction H(κ) |= x ∈ / CN, since ϕ1 is absolute for H(κ). For part b), observe that the definition of regularity says that x is a cardinal and not regular iff  x ∈ ON ∧ ¬∃f ϕ1 (x, f ) ∧ ∃y ∈ x∃g(g : y −→ x ∧ ran(g) = x). Now we can proceed as in the proof of a). For c), let x, y ∈ H(κ) and observe that x ∈ CN ∧ y ∈ CN ∧ y = x+ is the formula x ∈ CN ∧ y ∈ CN ∧ x ∈ y ∧ ∀z (z ∈ CN ∧ x ∈ z =⇒ y ≤ z), which we abbreviate as ϕ2 (x, y). From Lemma 4.2.11 we know that (ϕ2 (x, y) =⇒ H(κ) |= ϕ2 (x, y)). Assume that H(κ) |= ϕ2 (x, y). Then, by a), x and y are

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cardinal numbers. If z ∈ CN and x ∈ z, then y ≤ z, since z ∈ y would give z ∈ H(κ) and therefore y ≤ z. This proves c). For the proof of d), let x, y ∈ H(κ) and  ϕ3 (y, x) :≡ x, y ∈ ON ∧ y ≤ x ∧ ∃g (g : y −→ x ∧ ran(g) = x). Then “x is a limit ordinal and y = cf (x)” is formalized by ϕ4 (x, y) where ϕ4 (y, x) :≡ Lim(x) ∧ ϕ3 (y, x) ∧ ∀z (ϕ3 (z, x) =⇒ y ≤ z) . Assume that H(κ) |= ϕ4 (y, x). Our absoluteness results yield ϕ3 (y, x) and Lim(x). If z ∈ ON such that ϕ3 (z, x), then z ≤ x gives z ∈ H(κ) and thus y ≤ z. So we have shown that (H(κ) |= ϕ4 (y, x)) =⇒ ϕ4 (y, x).  Conversely assume ϕ4 (y, x). Then there is a function g : y −→ x such that ran(g) = x. Lemma 4.1.4 says that g ∈ H(κ), our absoluteness results give H(κ) |= ϕ4 (y, x), and d) is proved. Lemma 4.2.13 Assume that Θ is a regular cardinal, N is a set and a0 , . . . , an−1 are members of N . Then the following holds. a) If the LZF -formula ϕ(x0 , . . . , xn ) is absolute for H(Θ), N is an elementary substructure of H(Θ) and ϕ(a, a) holds for some a ∈ H(Θ), then there is b ∈ N such that ϕ(a, b). In particular we have: If there is a unique a ∈ H(Θ) such that ϕ(a, a), then a ∈ N . If ϕ(x0 , . . . , xn ) is an arbitrary formula, and if there is a unique member u of H(Θ) such that H(Θ) |= ϕ(a0 , . . . , an−1 , u), then u ∈ N . b) If the LZF -formula ϕ(x0 , . . . , xn ) is absolute for H(Θ), <∗ is a wellordering of H(Θ), (N, ∈, <∗ ) is an elementary substructure of (H(Θ), ∈ , <∗ ), and b is the <∗ -least element of H(Θ) such that ϕ(a, b), then b ∈ N . c) If ϕ(x0 , . . . , xn ) is a formula of L∗ZF , <∗ is a well-ordering of H(Θ), (N, ∈ , <∗ ) is an elementary substructure of (H(Θ), ∈, <∗ ), and b is the <∗ -least element of H(Θ) such that H(Θ) |= ϕ(a, b), then b ∈ N . Proof For the proof of a), we observe first that H(Θ) |= ∃xn ϕ[a] since ϕ(x, xn ) is absolute for H(Θ). From N  H(Θ), Lemma 4.2.2 and the absoluteness of ϕ again we can conclude that the first assertion in a) is true. The second assertion follows from Lemma 4.2.2. Since b) follows immediately from c), let us turn to the proof of part c). Let ψ be the L∗ZF -formula ∃xn (ϕ(x, xn ) ∧ ∀y(y < xn =⇒ ¬ϕ(x, y))). Then we have by assumption (∗)

H(Θ) |= ψ[a].

Furthermore, b is the unique member of H(Θ) with the property H(Θ) |= (ϕ(x, xn ) ∧ ∀y(y < xn =⇒ ¬ϕ(x, y))) [a, b].

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Since (N, ∈, <∗ ) is an elementary substructure of (H(Θ), ∈, <∗ ), we can infer from (∗) and Lemma 4.2.2 that there is c ∈ N such that H(Θ) |= (ϕ(x, xn ) ∧ ∀y(y < xn =⇒ ¬ϕ(x, y))) [a, c]. Obviously this gives b = c ∈ N . Definition If N = (N, ∈) is an elementary substructure of H(Θ), ϕ(x, p ) is a formula of LZF , p0 , . . . , pn−1 are members of N , and the set b is defined by b := {x ∈ H(Θ) : H(Θ) |= ϕ[x, p ]}, then we say that b is definable in H(Θ) by a formula with parameters from N . Lemma 4.2.14 If N = (N, ∈) is an elementary substructure of H(Θ), and b ∈ H(Θ) is definable in H(Θ) by a formula of LZF with parameters p from N , then b ∈ N. Furthermore, there is a formula ψ such that b is the unique member u of H(Θ) which satisfies H(Θ) |= ψ[u, p ]. Proof Let b := {a ∈ H(Θ) : H(Θ) |= ϕ(x, y )[a, p ]} where the parameters of ϕ are members of N , and let b ∈ H(Θ). Obviously we have H(Θ) |= ∃z∀x(x ∈ z ⇐⇒ ϕ(x))[ p ], since b ∈ H(Θ). Thus there is a member c ∈ N such that H(Θ) |= ∀x(x ∈ c ⇐⇒ ϕ(x))[ p ]. Now we get c = {a ∈ H(Θ) : H(Θ) |= ϕ(x, y )[a, p ]}, and b = c gives b ∈ N . ψ can be taken as the formula ∀x(x ∈ z ⇐⇒ ϕ(x, y )). Example 4.2.15 With the help of the following examples, the reader should become familiar with the previous results. At the same time, they are important components of various proofs in later sections and chapters, and we will often refer to them. Θ denotes a regular cardinal. Often we use the absoluteness results from Lemma 4.2.8 without mentioning. a) Assume that N is an elementary substructure of H(Θ). First we will show that the formulae |x| ≤ |y| and |x| = |y| are absolute for H(Θ). In the theory ZFC− , the formula |y| ≤ |x| is equivalent to the formula ψ :≡ y = ∅ ∨ ∃f (f : x −→ y ∧ ran(f ) = y). We know that the formula f : x −→ y ∧ ran(f ) = y is absolute for H(Θ). So, assuming x, y ∈ H(Θ), we get (H(Θ) |= ψ) =⇒ ψ; furthermore we have f ∈ H(Θ) by Lemma 4.1.4 for any function f : x −→ y. Thus we get ψ =⇒ (H(Θ) |= ψ), hence the formula ψ is absolute for H(Θ). Since H(Θ) is a model of ZFC− , Lemma 4.2.6 e) says that the formula |y| ≤ |x|

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Chapter 4. Approximation Sequences is also absolute for H(Θ). An analogous argument works for the formula |x| = |y|. Now fix b ∈ N . We want to show that |b| ∈ N . The previous results show that the formula |y| ≤ |x| ∧ x ∈ ON ∧ ∀z ∈ x ¬(|y| ≤ |z|)

is absolute for H(Θ). Since it is equivalent in the theory ZFC− to x = |y|, this formula is also absolute for H(Θ). Since b ∈ H(Θ) gives |b| ∈ H(Θ), there is a unique x ∈ H(Θ) satisfying x = |b|. Thus Lemma 4.2.13 a) tells us that |b| ∈ N . Consequently we have shown: Every elementary substructure of H(Θ) has as members all cardinalities of its elements. b) Let N be an elementary substructure of H(Θ). Using complete induction it is easy to see that ω ⊆ N and Vω ⊆ N . From the Lemmas 4.2.12 and 4.2.13 we can conclude that the following sets are members of N if they are members of H(Θ). ω; α + 1 for any ordinal α ∈ N ; α for any ordinal α + 1 ∈ N . κ+ for any cardinal κ ∈ N , hence ℵn for n ∈ ω; κ for any cardinal κ+ ∈ N . cf (α) for any ordinal α ∈ N . max(b), min(b), sup(b) for any b ∈ N such that b ⊆ ON. dom(f ), ran(f ), f |`b for any b ∈ N and any relation f ∈ N . f (b) for any function f ∈ N and any b ∈ N ∩ domf . c) Assume that the L∗ -structure (M, ∈, <∗ ) is an elementary substructure of (H(Θ), ∈, <∗ ), and let b ∈ M . By a), we have |b| ∈ M . Any function from b onto |b| is an element of H(Θ) by Lemma 4.1.4, and thus M has a member which is a bijection from b onto |b|. Now Lemma 4.2.8 says that the formula ϕ(x, y, z) :≡ (x is a bijection from y onto z) is absolute for H(Θ). So we can infer, using Lemma 4.2.13 b): b ∈ M ∧ h ∈ H(Θ) is the <∗-least bijection from b onto |b| =⇒ h ∈ M. d) Assume that N is an L-structure and an elementary substructure of H(Θ). First we get, using b) and c): If α ∈ N is a limit ordinal such that cf(α) ⊆ N , then sup(N ∩α) = α. To see this, observe first that cf (α) ∈ N . Since N is an elementary substructure of H(Θ), there exists a function f : cf (α) −→ α such that f ∈ N , sup(ran(f )) = α and ran(f ) ⊆ N .

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Now assume that h, b, c ∈ N have the property that h is a bijection from b onto c. In addition to c) we want to show some further results concerning N . Since H(Θ) |= ϕ[h, b, c], we get N |= ϕ[h, b, c], since the “parameters” are members of N and N is an elementary substructure of H(Θ). We “translate” this according to the definition of |=. First, the assertion N |= ∀u ∈ b∃!v ∈ c(h(u) = v) is equivalent to ∀u ∈ N ∩ b∃!v ∈ N ∩ c(h(u) = v), and similarly N |= ∀v ∈ c∃!u ∈ b(h(u) = v) is equivalent to ∀v ∈ N ∩ c∃!u ∈ N ∩ b(h(u) = v). Thus, for any h, b, c ∈ N , the following holds. h is a bijection from b onto c ⇐⇒ h |`(N ∩ b) is a bijection from b ∩ N onto c ∩ N . e) In part c), let (N, ∈, <∗ ) be a further elementary substructure of (H(Θ), ∈ , <∗ ) and take b = γ ∈ ON ∩ M ∩ N and γ ≥ ℵ0 . Then κ := |γ| ∈ M ∩ N . If we assume that M ∩ κ = N ∩ κ, then we can infer from d) for the <∗ -least bijection h : κ −→ γ that h ∈ M ∩ N and M ∩ γ = h[M ∩ κ] = h[N ∩ κ] = N ∩ γ. Thus we have shown under the given assumptions: If κ is a cardinal such that κ ∈ M ∩ N and M ∩ κ = N ∩ κ, then M ∩ γ = N ∩ γ for all γ ∈ M ∩ N ∩ κ+ . f) The proof of the following assertion is a prototype of our procedure in later chapters. For the set pcf(a) it makes precise the following intuition: Given finitely many sets, for example J<λ (a), max pcf(b), pcf(c) etc., we can choose a regular cardinal Θ “large enough” such that these sets are members of any elementary substructure N of H(Θ) which contains the parameters λ, a, b, c as members. The properties of H(Θ) will guarantee that the considered sets are members of H(Θ), and if we can characterize them uniquely by a formula which is absolute in a restricted sense (see below) and thus allows the proof that the sets have the form {x ∈ H(Θ) : H(Θ) |= ϕ[x, p ]} with p ∈ N , then they will be definable in H(Θ) with parameters from N . So Lemma 4.2.14 will say that they are members of N . Claim If N is an elementary substructure of H(Θ), a ∈ N is a set of regular cardinals, and 2sup(a) < Θ, then pcf(a) ∈ N . If in addition λ ∈ pcf(a) ∩ N , then J<λ (a) ∈ N .  We want to show that Θ is large enough such that P(a), a and pcf(a) are members of H(Θ). Then all objects occurring in the definition of pcf(a) will be members of H(Θ). We repeatedly apply Lemma 4.1.3 and Lemma 4.1.4. Since H(Θ) ∩ ON = Θ, we get x ⊆ H(Θ) for any x ⊆ a, P(a) ⊆ H(Θ) and thus P(a) ∈ H(Θ) since |P(a)| ≤ 2sup(a) < Θ, and also I ∈ H(Θ) for any ideal I on a. The fact that a, sup(a) ∈ H(Θ), hence

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Chapter 4. Approximation Sequences a × sup(a)  ∈ H(Θ), yields in the same way  that P(a × sup(a)) ∈ H(Θ). Since a is a subset of this set, we get a ∈ H(Θ). Next we observe that pcf(a) is a member of H(Θ), since for every λ ∈ pcf(a) we have λ  < | a| + 1 ∈ H(Θ); furthermore this gives f ∈ H(Θ) for any f : λ −→ a by Lemma 4.1.4. Now let ϕ(x, a) be the formula  x ∈ Reg ∧ ∃I∃z∃f (z = a ∧ I is an ideal on a ∧ cof(f, x, z, I)), where the formula cof(f, x, z, I) is given by f : x −→ z ∧ ∀α < x∀β < x(α < β =⇒ fα
Lemma 4.2.16 (Collection Principle) Let ϕ(x, y ) be a formula of LZF . Then ZF  ∀S0 ∃S ⊇ S0 ∀y ∈ S(∃xϕ(x, y ) =⇒ ∃x ∈ S ϕ(x, y )). With the axiom of choice, we have in addition |S| ≤ ℵ0 · |S0 |. Proof If A is a class, then let A∗ := {x ∈ A : ∀y ∈ A(rk(x) ≤ rk(y))} be the set of members of A which have minimal rank in the von Neumann hierarchy. A∗ is set, since there is an ordinal α such that A∗ ⊆ Vα . In particular, for our given formula ϕ and for every y , the class A∗y isalso a set where Ay := {x : ϕ(x, y )}. For any class X we put φ(X) := X ∪ {A∗y : y ∈ X}. If X is a set, then φ(X) is also a set. Now fix a set S0 and let, for each n ∈ ω,

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 Sn+1 := φ(Sn ) and S := {Sn : n ∈ ω}. Clearly we have S0 ⊆ S. Consider y ∈ S, that means y ∈ Sn for some n ∈ ω, such that ∃xϕ(x, y ). Then A∗y = ∅, und furthermore every z ∈ A∗y is a member of φ(Sn ) = Sn+1 ⊆ S. Thus there exists z ∈ S such that ϕ(z, y ), as desired. By (AC), we can take a choice function f for P(S) \ {∅} anddefine  := φ (Sn ) and S  := {Sn : φ (X) := X ∪ {f (A∗y ) : y ∈ X}, S0 := S0 , Sn+1  n ∈ ω}. It is easy to see by induction that |Sn | ≤ ℵ0 · |S0 | for all n ∈ ω. So the set S  instead of S has the desired properties. Remark Let α be an ordinal. We can choose the set S from the collection principle as a von Neumann level Vβ such that β is a limit ordinal with α < β. To see this, take in the construction of S an ordinal β0 such that α < β0 and S0 ⊆ Vβ0 . If βn is defined, choose βn+1 > βn such that φ(Vβn ) ⊆ Vβn+1 and put β = sup{βn : n ∈ ω}. Corollary 4.2.17 If ϕ and ψ are formulae whose free variables occur in the sequence y , then the following sentence is a theorem of ZF: ∀S0 ∃S ⊇ S0 ∀y ∈ S((∃xϕ(x, y ) ⇒ ∃x ∈ S ϕ(x, y )) ∧ (∃xψ(x, y ) ⇒ ∃x ∈ S ψ(x, y ))). An analogous assertion is provable for finitely many formulae ϕ1 , . . . , ϕn . Proof We make the same construction as above with   φ(X) := X ∪ {A∗y : y ∈ X} ∪ {By∗ : y ∈ X}, where By∗ := {x : ψ(x, y )}. Theorem 4.2.18 (Reflection Principle) If ϕ is a formula of LZF , then ZF  ∀M0 ∃M (M0 ⊆ M ∧ ∀y ∈ M (ϕ(y ) ⇐⇒ M |= ϕ(y ))) . Then we say that M reflects ϕ. Proof Let ϕ1 , . . . , ϕn be the subformulae of ϕ such that ϕn is the formula ϕ. Furthermore choose M according to Corollary 4.2.17 such that y ∈ M =⇒ (∃zϕi (z, y ) ⇐⇒ ∃z ∈ M ϕi (z, y )) for i = 1, . . . , n. By induction on the definition of a formula we prove that M reflects every formula ϕi . Firstly, every atomic formula is reflected by every set since it is identical with its relativization. If ϕi and ϕj are reflected by M , then we can conclude at once that M reflects ¬ϕi and (ϕi ∨ ϕj ). Now assume that ϕi has the form ∃zϕj . By the choice of M we have for any y ∈ M : ∃zϕj (z, y ) ⇐⇒ ∃z ∈ M ϕj (z, y ).

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The inductive hypothesis gives ∃z ∈ M ϕj (z,y ) ⇐⇒ ∃z ∈ M (M |= ϕj (z,y )) ⇐⇒ M |= ∃zϕj (z,y ) ⇐⇒ M |= ϕi . Thus we get ϕi ⇐⇒ (M |= ϕi ) for all y ∈ M , as desired. Now the assertion of the theorem follows immediately if we take ϕi as the subformula ϕ of ϕ. Corollary 4.2.19 In the preceding theorem, we can choose the set M as a suitable set Vα for some α ∈ ON. Exercises 1) Prove Lemma 4.2.1. 2) Give an example for isomorphic structures A and B such that A ⊆ B and A  B. Hint: Consider (ω \ {0}, ∈) and (ω, ∈) and take the formula ϕ(v0 ) ≡ ¬∃v1 (v1 ∈ v0 ). 3) If A and B are structures for L, then we say that A is elementary equivalent to B (or that A and B are elementary equivalent) iff, for every sentence ϕ of L, A |= ϕ ⇐⇒ B |= ϕ. If A is an elementary substructure of B, then clearly A and B are elementary equivalent.

4) 5) 6) 7)

8)

a) Show that, for any ordinals α and β such that α is a successor and β is a limit ordinal, (α, ∈) is not elementary equivalent to (β, ∈). b) Assume that β is a limit ordinal with β > ω. Show that (ω, ∈) is not elementary equivalent to (β, ∈). c) Show that (ω · ω, ∈) is not elementary equivalent to (ℵ1 , ∈). d) Show that (ω ω , ∈) is not elementary equivalent to (ℵ1 , ∈).  Prove that, for regular cardinals Θ, the formula z = a is absolute for H(Θ). Prove: If 2ℵ0 = ℵω+1 , then the formula y = pcf(x) is not absolute for H(ℵω+1 ). Prove: If N = (N, ∈) is an elementary substructure of H(Θ), b ∈ N , and Θ is a regular cardinal such that 2|b| < Θ, then P(b) ∈ N . Show that H(ℵω ) is not a model of all replacement axioms. Hint: Let ℵ = {(u, v) : ϕ(u, v)} be the aleph function. Show first, using transfinite induction, that ϕ(u, v) is absolute for H(κ), where κ is an uncountable cardinal. Then consider the function defined by ψ(u, v) :⇐⇒ u ∈ ω ∧ ϕ(u, v), i.e., ψ(u, v) ⇐⇒ u ∈ ω ∧ v = ℵu . If H(ℵω ) was a model for the replacement axiom for ψ, then we would have {ℵn : n ∈ ω} ∈ H(ℵω ) which yields ℵω ∈ H(ℵω ), contradicting Lemma 4.1.4. The beth function  : ON −→ ICN is known from Section 1.7. Prove: If κ > ω is a cardinal, then H(κ) is a model of the power set axiom iff

4.2. Models and Absoluteness

9)

10)

11)

12)

13)

195

κ = (γ) for some limit ordinal γ. Show further that H((ω)) is not a model of all replacement axioms. Hint: For the second assertion, follow the hint in the previous question and take the beth function instead of the aleph function. From the first assertion one can conclude that the formula defining the beth function is absolute for H(ℵω ). Assume that κ is an inaccessible cardinal number. Prove in ZF (in ZFC) that Vκ is a model of all axioms of ZF (of ZFC). Hint: See the exercises in Section 1.3 and Exercise 1.7.6 and prove first that, for all sets x, we have x ∈ Vκ ⇐⇒ x ⊆ Vκ ∧ |x| < κ. Then follow the proof of Lemma 4.2.12. Assume that κ > ω is a regular cardinal. Prove in ZFC that κ is inaccessible iff H(κ) is a model of ZFC. Hint: Use, for example, a previous question and Exercise 1.7.7. Let λ be a singular and κ be an infinite cardinal. Prove that H(λ) is not elementary equivalent to H(κ+ ). Hint: Consider the formula ∃y(y ∈ CN ∧ ∀z(z ∈ CN =⇒ z = y + )) and use the absoluteness results from Lemma 4.2.12. We call a formula ψ of ZF a Σ1 -formula iff it has the form ∃x0 . . . ∃xn−1 ϕ for some Δ0 -formula ϕ. The case n = 0 shows that every Δ0 -formula is a Σ1 -formula. Prove that for each Σ1 -formula ψ there is a Δ0 -formula ϕ such that the formula ∃uϕ ⇐⇒ ψ is provable in ZF without using the power set axiom. Assume that ϕ(u, x1 , . . . , xn ) is a Δ0 -formula. Prove: If κ is an uncountable cardinal, then the formula ∃uϕ is absolute for H(κ). Hint: Observe that it suffices to prove ∀x1 . . . ∀xn ∈ H(λ+ )(∃uϕ =⇒ ∃u ∈ H(λ+ ) ϕ).

For this, use the reflection principle (which yields, for given sets u and a1 , . . . , an ∈ H(λ+ ), some Vα such that Vα |= ϕ(u, a1 , . . . , an )), the L¨owenheim-Skolem theorem (to obtain a sufficiently small elementary substructure A of Vα ) and the Mostowski lemma (to get an ∈-isomorphism from A onto a transitive set B ∈ H(λ+ )). 14) Assume that κ is an uncountable cardinal. Prove that each Σ1 -formula is absolute for H(κ). 15) (Levy) Assume that ψ(x, x1 , . . . , xn ) is a Σ1 -formula. Prove: If κ is an uncountable cardinal, then ∀x1 ∈ H(κ) . . . ∀xn ∈ H(κ)(∃xψ =⇒ ∃x ∈ H(κ) ψ). Hint: Use the previous questions.

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4.3 Approximation Sequences Definition A sequence (Ai : i < γ) of L-structures or L∗ -structures is called a chain iff Ai ⊆ Aj for all i and j with i < j < γ. It is called an elementary chain iff Ai  Aj for all i and j with i < j < γ. If (Ai : i < γ) is a ∗ chain  and Ai = (Ai , ∈,
4.3. Approximation Sequences

197

Nj+1 , Xj ⊆ Nj+1 , |Nj+1 | ≤ max{|Nj |, |Xj |, ℵ0 } ≤ μ, and Nj ⊆ Nj+1 . Since Nj is an elementary substructure of H(Θ), Lemma 4.2.1 says that it is also an elementary substructure of Nj+1 . This completes the definition of the desired sequence. Remark 4.3.2 Assume that Θ is a regular cardinal, κ is an uncountable regular cardinal, and σ is a limit ordinal such that σ < κ and κ ∈ H(Θ). Further let Y ∈ H(Θ), |Y | < κ, and (Xi : i < σ) ∈ H(Θ) be a sequence satisfying |Xi | < κ for all i < σ. Then we can define, similar to the previous construction, an approximation sequence (Ni : i ≤ σ) such that Y ⊆ N0 , Xj ⊆ Nj+1 , |Ni | < κ and Ni ∩ κ ∈ κ for all i ≤ σ. To prove this, we will construct Nj+1 from Nj . N0 is defined analogously, and the limit step is the same as above. Nj+1 will be the union of a countable chain (Mn : n ∈ ω) of elementary substructures of H(Θ) obtained by a standard construction, a so-called ω-iteration, which we already used in the proof of the L¨owenheim-Skolem theorem. Choose M0  H(Θ) according to Theorem 4.2.3 as a superstructure of Nj such that Xj ⊆ M0 and |M0 | < κ, and let Mn with |Mn | < κ be already defined. Since κ is regular, we have αn := sup(Mn ∩κ) < κ, and again Theorem 4.2.3 yields a superstructure Mn+1 of Mn such that α n +1 ⊆ Mn+1 , Mn+1  H(Θ) and |Mn+1 | < κ. Then the structure Nj+1 := {Mn : n ∈ ω} is an elementary substructure of H(Θ) as the union of the elementary chain (Mn : n ∈ ω). Furthermore we get |Nj+1 | ≤ ℵ0 · sup{|Mn | : n ∈ ω} < κ. If α ∈ β ∈ Nj+1 ∩ κ, then there is some n ∈ ω such that β ∈ Mn ∩ κ. By the definition of Mn+1 , we have α ∈ Mn+1 ⊆ Nj+1 . Therefore Nj+1 ∩ κ is an ordinal, since it is a transitive set of ordinals. The following lemma presents some simple properties of approximation sequences. Part d) will be a practical tool for the construction of characterizing functions. Lemma 4.3.3 Assume that M is a set, Θ is a regular cardinal, σ is a limit ordinal with σ < Θ, (Ni : i ≤ σ) is an approximation sequence in H(Θ), and N := Nσ . Then a) If |M | < cf (α), then χM (α) < α. b) For all i < σ, we have Ni ∈ Ni+1 , σ ⊆ N , and {Ni : i < σ} ⊆ N . c) If i < σ and α ∈ Ni , then χNi (α) ∈ Ni+1 ; if in addition χNi (α) < α, then χNi (α) < χNi+1 (α). d) If τ ∈ N0 is a limit ordinal such that |Ni | < cf (τ ) for all i < σ, then (χNi (τ ) : i < σ) is a normal function satisfying sup{χNi (τ ) : i < σ} = χN (τ ), all of whose values are members of N ∩ τ . In particular we have cf (χN (τ )) = cf (σ).

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Proof Part a) is trivial. For b), fix i < σ. Then, by definition, we have f := (Nj : j ≤ i) ∈ Ni+1 , and Example 4.2.15 gives i + 1 = dom(f ) ∈ Ni+1 , i ∈ Ni+1 , and Ni = f (i) ∈ Ni+1 , as desired. For the proof of c), let i < σ and α ∈ Ni . Then we have α ∈ Ni+1 , and Ni ∈ Ni+1 by b). Since Ni+1  N and there is a unique member δ ∈ H(Θ) such that δ = sup(Ni ∩ α), we get χNi (α) = sup(Ni ∩ α) ∈ Ni+1 by Lemma 4.2.13. Furthermore, δ ∈ Ni+1 implies δ + 1 ∈ Ni+1 , hence χNi (α) < χNi+1 (α) if χNi (α) < α. For d), we have just shown that χNi (τ ) is smaller than χNi+1 (τ ). If i ≤ σ is a limit ordinal, we can infer with the continuity of approximation sequences that   sup{χNj (τ ) : j < i} = { (Nj ∩ τ ) : j < i}  = ( {Nj : j < i} ∩ τ ) = (Ni ∩ τ ) = χNi (τ ). Thus (χNi (τ ) : i < σ) is a normal function which is cofinal in χN (τ ) and has all its values in N ∩ τ . A typical application of this lemma is the following important lemma which we need for later purposes. Lemma 4.3.4 Assume that a is an infinite set of regular cardinals satisfying |a|+ < min(a), Θ is a regular cardinal, μ ∈ (|a|, min(a))reg and (Ni : i ≤ μ) is an approximation sequence in H(Θ) such that a ∪ {a} ⊆ N0 and |Ni | ≤ μ for all i < μ. Further let τ be a limit ordinal such that μ < cf (τ ) and τ ∈ N0 , N := Nμ , and (fξ : ξ < τ ) ∈ N0 be a sequence of members of a such that fχN (τ ) (ν) = min{sup{fβ (ν) : β ∈ C} : C is a club in χN (τ )} for every ν ∈ a. Then the following holds. a) There is a strictly increasing continuous sequence (i(β) : β < μ) of limit ordinals in μ such that fχN (τ ) (ν) = sup{fχNi(β) (τ ) (ν) : β < μ} for all ν ∈ a, and we have fχN (τ ) ≤a χN |`a.  b) If 2sup(a) < Θ and (fξ : ξ < τ ) is cofinal in a modulo some ultrafilter D on a with D ∈ N0 , then for any i < μ there is an ordinal ξ0 ∈ Ni+1 ∩ τ such that χNi |`a
4.4. The Skolem Hull in H(Θ)

199

Proof For a), observe that cf (χN (τ )) = cf (sup{χNi (τ ) : i < μ}) = μ by the preceding lemma, and thus χN (τ ) < τ . Let, for each ν ∈ a,Cν be a club in χN (τ ) such that fχN (τ ) (ν) = sup{fβ (ν) : β ∈ Cν }. Then C := {Cν : ν ∈ a} is also a club in χN (τ ). If E is any subclub of C in χN (τ ), then the assumption on fχN (τ ) (ν) yields that fχN (τ ) (ν) = sup{fβ (ν) : β ∈ E} for all ν ∈ a. So we can assume without loss of generality that otp(C) = μ. Lemma 4.3.3 tells us that {χNi (τ ) : i < μ} is a club in χN (τ ). Taking its intersection with C, we obtain a strictly increasing continuous sequence (i(β) : β < μ) of limit ordinals in μ such that for all ν ∈ a (∗)

fχN (τ ) (ν) = sup{fχNi(β) (τ ) (ν) : β < μ} .

Now we have (fξ : ξ < τ ) ∈ N0 , hence dom((fξ : ξ < τ )) = τ ∈ N . Since χNi(β) (τ ) ∈ N , we conclude that fχNi(β) (τ ) ∈ N , hence fχNi(β) (τ ) (ν) ∈ N ∩ ν for all ν ∈ a, since a ⊆ N . This gives fχNi(β) (τ ) (ν) ≤ sup(N ∩ ν) = χN (ν), as desired. and For the proof of b), fix i < μ. Then Ni ∈ Ni+1 , and from a ∈ Ni+1  |Ni | ≤ μ < min(a) it follows, using Lemma 4.2.14, that χNi |` a ∈ Ni+1 ∩ a. Furthermore we have D ∈ Ni+1 . Since Θ > 2sup(a) , Example 4.2.15 says that   a ∈ H(Θ) and that the property of being cofinal in a modulo D is absolute for H(Θ). So we can infer from the assumption, Lemma 4.2.13 and Ni+1  H(Θ) that there exists an ordinal ξ0 ∈ Ni+1 ∩ τ such that χNi |` a
4.4 The Skolem Hull in H(Θ) In this section, Θ denotes a regular cardinal. We associate with each L∗ -formula ϕ(x0 , . . . , xn ) an L∗ -formula ϕ∗ as follows. ϕ∗ (x0 , x1 , . . . , xn ) :≡ ϕ(x0 , . . . , xn ) ∧ ∀xn+1 (xn+1 < x0 =⇒ ¬ϕ(xn+1 , x1 , . . . , xn )). Consequently, if <∗ is a well-ordering of H(Θ), a0 , . . . , an ∈ H(Θ), and if H(Θ) |= ϕ∗ (a0 , . . . , an ), then a0 is the <∗ -least element a ∈ H(Θ) so that H(Θ) |= ϕ(a, a1 , . . . , an ). n n Further let i=1 ϕi be the conjunction ϕ1 ∧ . . . ∧ ϕn and i=1 ϕi be the disjunction ϕ1 ∨ . . . ∨ ϕn of the formulae ϕ1 , . . . , ϕn . The property “elementary substructure” refers in this section always to the language L∗ .

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Lemma 4.4.1 If E is a nonempty set of elementary substructures of H(Θ), then  E is an elementary substructure of H(Θ).  Proof Certainly we have E = ∅, since for example the empty  set is a member of every elementary substructure of H(Θ). Furthermore, E is a substructure of H(Θ).  Consider a1 , . . . , an ∈ E such that ∃b ∈ H(Θ)(H(Θ) |= ϕ(b, a1 , . . . , an )),  for a fixed L∗ -formula ϕ. To show that E is an elementary substructure of H(Θ) it suffices to show, using Lemma 4.2.2, that there is some “satisfier” b in  E. Now there is a unique a ∈ H(Θ) such that H(Θ) |= ϕ∗ (a, a1 , . . . , an ). ∈ N , and thus Lemma 4.2.13 yields a ∈ N , since If N ∈ E, then a1 , . . . , an  N  H(Θ). This gives a ∈ E. This lemma justifies the following definition. Definition If X ⊆ H(Θ), then let  SkolΘ (X) := {N : N  H(Θ) ∧ X ⊆ N } be the Skolem hull of X. Obviously SkolΘ is a hull operator on H(Θ), and SkolΘ (X) is the smallest elementary substructure of H(Θ) whose universe is a superset of X. For each L∗ -formula ϕ(x0 , . . . , xn ) in which xn occurs free, let fϕ be that function from H(Θ)n into H(Θ) which assigns to each tupel (a1 , . . . , an ) the <∗ -least element a ∈ H(Θ) with H(Θ) |= ϕ(a, a1 , . . . , an ), that is the unique a ∈ H(Θ) such that H(Θ) |= ϕ∗ (a, a1 , . . . , an ), if there is such an a; otherwise, we take 0 as the value of fϕ at (a1 , . . . , an ). We say that fϕ is a Skolem function for ϕ. If F is an n-ary function (i.e., F is a function such that dom(F ) ⊆ Vn ) and A is a set, then we put F  A := F [ An ∩ dom(F ) ]. Remark This definition is a bit ambiguous since every function is certainly 1-ary on an appropriate set. However, if there are n, m ∈ ω \ {0} with m = n, such that F is m-ary and n-ary, then m = 1 or n = 1 (if F = ∅). In this case the previous definition refers to the greater one of the numbers n and m. Next we will obtain a simple method to determine Skolem hulls.

4.4. The Skolem Hull in H(Θ)

201

Lemma 4.4.2 If X ⊆ H(Θ), then  SkolΘ (X) = {fϕ X : ϕ is an L∗ -formula}.  Proof Put X0:= X, Xn+1 := {fϕ Xn : ϕ is an L∗ -formula} for all n < ω, and X + := {Xn : n < ω}. Then we know that (Xn : n < ω) is an ascending chain of sets and X + is an elementary substructure of H(Θ); this is just the usual proof of the L¨ owenheim-Skolem theorem if we take as a satisfier always the <∗ -smallest satisfier. By the definition of the Skolem functions and Lemma 4.2.13, we even have X + = SkolΘ (X). Therefore it suffices to show that X2 = X1 , because this yields Xn = X1 for all n ≥ 2 and thus X + = X1 , as desired. Consider a1 , . . . , an ∈ X1 , and let a = fϕ (a1 , . . . , an ) for some L∗ -formula ϕ. Then we can find, for i = 1, . . . , n, members ai1 , . . . , aimi of X0 and an L∗ formula ϕi such that ai = fϕi (ai1 , . . . , aimi ). Now put s0 := 0 and si := ik=1 mk for i = 1, . . . , n, and let ψ be the formula ∃xsn +1 . . . ∃xsn +n (ϕ∗ (x0 , xsn +1 , . . . , xsn +n ) ∧

n 

ϕ∗i (xsn +i , xsi−1 +1 , . . . , xsi )),

i=1

Then we have a = fψ (a11 , . . . , a1m1 , a21 , . . . , a2m2 , . . . , an1 , . . . , anmn ) ∈ X1 . Lemma 4.4.3 Assume that N is an elementary substructure of H(Θ) and that γ, A ∈ N satisfy |A| < cf (γ). If N  is the Skolem hull of N ∪ A, then χN  (γ) = χN (γ). Proof We can assume without loss of generality that γ is a limit ordinal, since otherwise A = ∅, hence N = N  . Clearly we have χN  (γ) ≥ χN (γ). To prove the reverse inequality, let β ∈ N  ∩ γ. By the above lemma, there are a1 , . . . , an ∈ N ∪ A and an L∗ -formula ϕ(x0 , . . . , xn ) such that β = fϕ (a1 , . . . , an ). Let {b1 , . . . , bm } := {a1 , . . . , an } ∩ N ; then β ∈ A+ := fϕ (A ∪ {b1 , . . . , bm }). Note that A+ ⊆ H(Θ) and |A| < Θ, hence A+ ∈ H(Θ). Now let ψ be the formula ∃y1 . . . ∃yn (ϕ∗ (x0 , y1 , . . . , yn ) ∧

n  i=1

(yi ∈ x1 ∨

m 

yi = xj+1 )).

j=1

Then a ∈ A+ iff H(Θ) |= ψ[a, A, b1 , . . . , bm ], i.e., A+ is definable in H(Θ) with parameters from N , hence A+ ∈ N by Lemma 4.2.14. From |A| < cf (γ) we can infer that A+ ∩γ is a bounded subset of γ. Therefore there is a <∗ -minimal ordinal ζ ∈ γ such that A+ ∩ γ ⊆ ζ. Lemma 4.2.13

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shows that ζ ∈ N , hence β < ζ < χN (γ). So we get sup(N  ∩ γ) ≤ χN (γ), as desired. With the previous lemma we now can conclude that we can enlarge elementary substructures of H(Θ) without changing the values of their characteristic functions at sufficiently large arguments. Corollary 4.4.4 Assume that N is an elementary substructure of H(Θ), κ ∈ N is a regular cardinal, and α < χN (κ). If N  is the Skolem hull of N ∪ {α}, then χN  (κ) = χN (κ). Proof Choose some ζ ∈ N ∩κ with α < ζ. Applying the above lemma with A = ζ and γ = κ, we get χSkolΘ (N ∪ζ) (κ) = χN (κ). Since N  is a substructure of SkolΘ (N ∪ ζ), the desired assertion is proved.

4.5 3-Club Sequences For regular cardinals μ and κ with μ < κ, we got acquainted in Lemma 1.8.4 with the stationary subset Σκ, μ of κ: Σκ, μ := {α ∈ κ : cf (α) = μ} .  Definition If A ⊆ ON, then A := { b : b ⊆ A ∧ b = ∅} is called the hull of A with respect to the order topology on ON. Remark If A ⊆ ON is a set whose order type is a limit ordinal, then A has the same order type as the set A∗ := A \ {sup(A)}. To see this, define the function f : A∗ −→ A by f (α) := min{β ∈ A : β > α}. First we notice that every element α of A∗ is smaller than sup(A), hence f (α) ∈ A is defined. If α1 , α2 ∈ A∗ and α1 < α2 , then, by the definition of A (we have α2 = sup(b) for some b ⊆ A), there is β ∈ A such that α1 < β ≤ α2 . Therefore f (α1 ) ≤ β ≤ α2 < f (α2 ), and f turns out to be strictly monotone. Together with A ⊆ A this gives otp(A∗ ) ≤ otp(A) ≤ otp(A) = otp(A∗ )+1. So we can infer by assumption that otp(A∗ ) = otp(A). Definition Assume that μ and κ are uncountable regular cardinals with μ < κ and S is a subset of Σκ+ , μ which is stationary in κ+ . We call a sequence (Cα : α ∈ S) an ecl-sequence3 for S iff, for all α, β ∈ S, (i) Cα is a club in α of order type μ. (ii) If γ is a common limit point of Cα and Cβ , then Cα ∩ γ = Cβ ∩ γ. 3 Equal

at common limit points.

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203

Remark To prove the existence of ecl-sequences, we use for the first time the method of approximation sequences to get a powerful combinatorial result. Lemma 4.5.1 Let κ and Θ be uncountable regular cardinals such that κ+ < Θ and τ < κ a limit ordinal. Further assume that (Ni : i ≤ τ ) is an approximation sequence in H(Θ) such that κ ∈ N0 , |Ni | < κ for all i < τ , and Ni ∩ κ ∈ ON for all i ≤ τ (such a sequence exists by Remark 4.3.2). Then all limit points of Nτ in κ+ whose cofinality differs from cf (τ ) are members of Nτ . Proof Let N := Nτ , and let γ be such a limit point, that means that γ < κ+ , cf (τ ) = cf (γ) and sup(N ∩ γ) = γ. By the definition of an approximation sequence, we have γ = sup{sup(Ni ∩ γ) : i < τ }, hence cf (γ) ≤ cf (τ ) and thus cf (γ) < cf (τ ). Consequently, having chosen a sequence of length cf (γ) cofinal in N ∩ γ, we observe that all its members are already members of a set Nα for some α < τ , and furthermore we get sup(Nα ∩ γ) = γ. Since α, Nα and κ+ are, by Lemma 4.3.3 and Example 4.2.15, members of N , we can conclude from Lemma 4.2.13 that also A := Nα ∩ κ+ ∈ N , since the hull of x is absolute for H(Θ). Again by Example 4.2.15, the cardinal λ := |A| and a bijection h from λ onto A turn out to be members of N . Since |Nα | < κ, we can infer with the remark following the definition of the hull of a set that λ < κ. Thus, if λ ∈ Nβ for some β, our assumption Nβ ∩ κ ∈ ON gives λ ⊆ Nβ ⊆ N . From 4.2.15 we get A = h[λ] ⊆ N , and sup(Nα ∩ γ) = γ yields γ ∈ A, hence γ ∈ N . Lemma 4.5.2 Assume that μ and κ are regular cardinals with max{ℵ1 , μ} < κ. Then there is a set S ⊆ κ+ , μ stationary in κ+ and an ecl-sequence for S. Proof First let us explain why Example 4.2.15 e) motivates us to investigate sets Ni ∩ i for suitable elementary substructures Ni of H(Θ), where If |Ni | < κ and Ni ∩ κ ∈ ON i ∈ Σκ+ ,μ , κ ∈ Ni and Θ is large enough. + ,μ , then Ni ∩ κ ∈ κ, and for all i ∈ Σκ κ+ ,μ is the union of the sets : N ∩ κ = σ}, σ < κ. Since Σκ+ ,μ is stationary in Xσ := {i ∈ i κ+ ,μ + κ , Lemma 1.8.4 says that one of these sets is stationary in κ+ , too; call it S1 := Xσ0 . If α, β ∈ S1 , then the mentioned example shows that Nα ∩γ = Nβ ∩γ for all γ ∈ Nα ∩ Nβ ∩ κ+ . With Example 4.2.15 d) we can infer for α ∈ S1 that Nα ∩ α is unbounded in α, if we can only realize α ∈ Nα and μ = cf (α) ⊆ Nα . However, Nα ∩ α will not be a club in α. On the other hand, if we pass to the hull of this set to obtain a club, we will leave Nα , and our starting idea will be no more applicable. But Lemma 4.5.1 takes care that an appropriate Nα has enough limit points amongst its members. Namely, we can choose Nα as the union of an approximation sequence of length ℵ1 < κ satisfying the assumptions in 4.5.1, such that in addition {α, κ} ⊆ Nα and μ = cf (α) ⊆ Nα (which is possible by Lemma 4.3.1 since μ < κ). Then Nα contains as members all those limit points of Nα in κ+ which have countable cofinality. Furthermore

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it is not difficult to verify that, for the proof of property (ii) in the definition of a ecl-sequence, it suffices to show this property for limit points with countable cofinality. Namely, if γ is a common limit point of the clubs Cα and Cβ with cf (γ) > ℵ0 , then γ is the supremum of a sequence of common limit points of countable cofinality (see the “hand-over-hand” argument in the proof of Lemma 1.8.2 a)), and the validity of (ii) for these points yields at once Cα ∩ γ = Cβ ∩ γ. Let us choose Nα as above and take Dα := Nα ∩ α \ {α} for α ∈ S1 . Certainly Dα is a club in α, but unfortunately Dα has not yet order type μ. However, with the remark following the definition of the hull, we get otp(Dα ) = otp(Nα ∩α) < κ, since |Nα | < κ. Again the set S1 is stationary in κ+ and is the union of the sets Yσ := {α ∈ S1 : otp(Nα ∩ α) = σ}, σ < κ. Therefore one of these sets, say S := Yδ , is stationary in κ+ . Take α ∈ S. Then otp(Nα ∩ α) = δ, α ∈ S1 ⊆ Σκ+ ,μ , hence cf (α) = μ, and Nα ∩ α is unbounded in α. This gives cf (δ) = μ. Consequently there is a club D in δ of order type μ. If we denote by fα , for α ∈ S, the unique order isomorphism from δ onto Dα , then the set Cα := fα [D] is also a club in α of order type μ. Consequently, the club sequence (Cα : α ∈ S) has property (i). If α, β ∈ S, and if γ ≤ min{α, β} is a common limit point of Cα and Cβ with countable cofinality (see above), then γ is also a common limit point of Dα and Dβ , and a limit point of Nα and of Nβ . Now Lemma 4.5.1 says that γ is a member of Nα ∩ Nβ since Nα and Nβ are unions of approximation sequences of length ℵ1 . From Example 4.2.15 e) we can conclude that Nα ∩ γ = Nβ ∩ γ, which gives Dα ∩ γ = Dβ ∩ γ. Since fα and fβ are order isomorphisms, we have τ := fα−1 [Dα ∩ γ] = fβ−1 [Dβ ∩ γ]. Now we can infer that Cα ∩ γ = fα [D ∩ τ ] = fβ [D ∩ τ ] = Cβ ∩ γ, as desired. Thus (Cα : α ∈ S) has property (ii). Remember that a sequence (Cα : α ∈ A) where A is a set of limit ordinals is called a club sequence iff, for every α ∈ A, Cα is a club in α of order type cf (α). If μ and κ are regular cardinals such that μ < κ, and if in particular A ⊆ Σκ,μ , then Cα is a club in α of order type μ for all α ∈ A. Note that then any subset Cα of Cα which is unbounded in α also has the order type μ: We have otp(Cα ) ≤ μ since Cα ⊆ Cα , and otp(Cα ) < μ is impossible since Cα is unbounded in α. The following definition yields an important tool for the next chapter. Definition Assume that μ and κ are regular cardinals, μ < κ, and S ⊆ Σκ, μ is a stationary subset of κ. Any sequence T¯ = (Tα : α ∈ S) is called a 3club sequence for S iff T¯ is a club sequence and (cc) For every club C in κ, {α ∈ S : Tα ⊆ C} is stationary in κ. In general the existence of 3club -sequences for Σκ,μ is not provable in ZFC; for example, this holds for Σω2 ,ω1 .

4.5. 3-Club Sequences

205

The following result, however, is provable in ZFC. Lemma 4.5.3 Let κ and μ be uncountable regular cardinals with μ+ < κ. Further assume that S ⊆ Σκ, μ is a stationary subset of κ and (Fα : α ∈ S) is a fixed club sequence. If T¯ = (Tα : α ∈ S) is a club sequence, then there is a club C in κ such that the sequence T¯ ∗ (C) := (Tα∗ : α ∈ S), given by  Tα ∩ C if Tα ∩ C is a club in α, ∗ Tα = Fα otherwise, is a 3club -sequence for S. Proof 4 Assume that S is a stationary subset of Σκ,μ and T¯ = (Tα : α ∈ S) is a club sequence. To get a contradiction let us assume that, for any club C in κ, the sequence T¯ ∗ (C) = (Tα∗ : α ∈ S) defined in the assertion of the lemma is not a 3club -sequence for S. First we note that any intersection of less than κ clubs in κ is a club in length γ κ. Consequently, given a sequence (Dξ : ξ < γ) of clubs in κ of limit  less than κ, we obtain a sequence (Eξ : ξ < γ) by E0 := κ and Eξ := ζ<ξ Dζ for 0 < ξ < γ which is decreasing under ⊆ and has  the same intersection as (Dξ : ξ < γ). Let E be a club in κ such that E ⊆ {Eξ : ξ < γ}. Assume that we can arrange that Tαξ := Tα ∩ Eξ is a club in α for every member α of the stationary set E ∩ S and every ξ ∈ γ, and denote this property by (∗). Since (Eξ : ξ < γ) is decreasing, the sequence T¯α = (Tαξ : ξ < γ) is decreasing, and |Tα | = μ for all α ∈ S. Thus, if γ := μ+ , T¯α must be constant from some ordinal ξ(α) < γ on, and the function (ξ(α) : α ∈ E ∩ S ∩ (κ \ γ)) is regressive. Since E ∩S ∩(κ\γ) is stationary in κ, we obtain by Fodor’s theorem an ordinal β < γ such that Q = {α < κ : α ∈ E ∩ S ∧ ξ(α) = β} is stationary in κ. If α ∈ Q, then Tαβ = Tαβ+1 = Tα ∩ Eβ+1 = Tα ∩ {Dζ : ζ < β + 1} = Tα ∩ Eβ ∩ Dβ = Tαβ ∩ Dβ , hence Tαβ ⊆ Dβ . So the set {α ∈ S : Tαβ ⊆ Dβ } is stationary in κ. The preceding considerations have lead us to the key of the proof: If we can realize (∗) and can arrange for α ∈ S that T¯α := (Tα ∩ Eξ : ξ < μ+ ) is a club sequence such that for every ξ < μ+ the set {α ∈ S : Tαξ ⊆ Dξ } is not stationary in κ, we will have obtained the desired contradiction. Clearly Dξ will be given by our assumption on T¯, if we will have T¯α = ∗ ¯ T (Eξ ). This fixes our definition of the sequence T¯α :  if Tα ∩ Eξ is a club in α, Tα ∩ E ξ Tαξ := Fα otherwise. There remains the definition of the clubs E and Dξ for any ξ < μ+ . By recursion we will realize the above properties and require 4 We

use ideas from [BM].

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(ns)ξ The set {α ∈ S : Tαξ ⊆ Dξ } is not stationary in κ. (ic)ξ The set {α < κ : α ∈ S or Tα ∩ Eξ is a club in α} includes a club in κ. Then (ic)ξ will guarantee that (∗) holds. For α ∈ S we have Tα0 = Tα since E0 = κ, and clearly (ic)0 is true. By our assumption, T¯∗ (κ) is not a 3club sequence, and there is a club D0 in κ such that (ns)0 is satisfied. + Now fix  ξ < μ and let, for ζ < ξ, clubsζ Dζ in κ be already defined such that Eζ =  {Dδ : δ < ζ} (for ζ = 0) and (Tα : α ∈ S) satisfy (ns)ζ and (ic)ζ . Then Eξ := {Dζ : ζ < ξ} is a club in κ. Let Tαξ be defined as above. Then Tαξ is a club in α of order type μ since Tα ∩ Eξ is unbounded in α, as stated above. Our supposition says that property (cc) does not hold for the club sequence T¯∗ (Eξ ) = (Tαξ : α ∈ S); thus there is a club Dξ in κ, such that (ns)ξ is fulfilled. The set Eξ of limit points of Eξ is also a club in κ. To verify (ic)ξ it suffices to show that Eξ ∩ S ⊆ {α ∈ κ : Tα ∩ Eξ is a club in α}. Fix α ∈ Eξ ∩ S. Then Eξ ∩ α is a club in α, and therefore Tα ∩ (Eξ ∩ α) = Tα ∩ Eξ is also a club in α, since μ = cf (α) > ℵ0 , as desired. Now put   E := {Dξ : ξ < μ+ } ∩ {Eξ : ξ < μ+ }.  Since μ+ < κ, E is a club in κ, E ⊆ {Eξ : ξ < μ+ }, and for every α ∈ E ∩ S we have α ∈ Eξ , hence Tα ∩ Eξ is a club in α. Now all requirements from above are fulfilled, and we obtain the desired contradiction, which proves the lemma. Lemma 4.5.4 Assume that μ and κ are regular cardinals such that ℵ0 < μ < κ. Then there is a set S ⊆ κ+ , μ stationary in κ+ and an ecl-sequence for S, which is a 3club -sequence for S as well. Proof By Lemma 4.5.2, there is a stationary set T ⊆ κ+ , μ and a sequence (Cα : α ∈ T ) such that (i) ∀α ∈ T (Cα is a club in α ∧ otp(Cα ) = μ) . (ii) ∀α, β ∈ T ∀γ ≤ min{α, β} (γ is a limit point of Cα and of Cβ =⇒ Cα ∩ γ = Cβ ∩ γ) . Lemma 4.5.3, applied to κ+ , μ, T , and (Cα : α ∈ T ), yields a club C0 (note that 0 ∈ T ) in κ+ , such that the sequence (Sα∗ : α ∈ T ) defined by  Cα ∩ C0 if Cα ∩ C0 is a club in α, ∗ Sα := otherwise Cα has the following properties: (a) ∀α ∈ T (Sα∗ is a club in α of order type μ). (b) ∀C ⊆ κ+ (C is a club in κ+ =⇒ {α ∈ T : Sα∗ ⊆ C} is stationary in κ+ ).

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207

The set C0 of limit points of the club C0 is also a club in κ+ . If α ∈ T ∩C0 , then C0 ∩ α and Cα are clubs in α, and therefore Cα ∩ (C0 ∩ α) = Cα ∩ C0 is a club in α, since cf (α) = μ > ℵ0 . This gives T ∩ C0 ⊆ {α ∈ T : Cα ∩ C0 is a club in α}, and consequently the set S := {α ∈ T : Cα ∩ C0 is a club in α} is stationary in κ+ as a superset of a stationary set. For each α ∈ S, Eα := Cα ∩ C0 = Sα∗ is a club in α such that otp(Eα ) = μ. If α, β ∈ S and γ ≤ min{α, β}, and if γ is a limit point of Eα and of Eβ , then Eα ∩ γ = Eβ ∩ γ by (ii), hence (Eα : α ∈ S) is a ecl-sequence for S. Finally, if C is a club in κ+ , then C0 ∩ {α ∈ T : Sα∗ ⊆ C} is stationary in + κ by (b), and so the set {α ∈ S : Eα ⊆ C} is also stationary in κ+ , as desired.

Chapter 5 Generators of J<λ+ (a) In this chapter, a denotes an infinite set of regular cardinals. We will prove the following assertion, due to S. Shelah: If |a|+ < min(a) and λ ∈ pcf(a), then there is a set bλ ∈ J<λ+ (a) which generates the ideal J<λ+ (a) over J<λ (a), which means that for each set c ∈ J<λ+ (a) there is a set d ∈ J<λ (a) such that c ⊆ bλ ∪ d. Such a set bλ is called a generator of J<λ+ (a) over J<λ (a). In Section 5.1 we show under the previous assumptions that there exists a universal sequence (fξ : ξ < λ). It has the property  that, for every ultrafilter D on a satisfying cf ( a/D) = λ, it is cofinal in a modulo D. With the results of Section 4.5, we can prove in the second section the existence of the desired sequence (bλ : λ ∈ pcf(a)) of generators.1

5.1 Universal Sequences Definition If λ and μ areregular cardinals with μ < λ, then any sequence (fξ : ξ < λ) of members of a is called μ-continuous iff fξ (ν) = min{sup{fζ (ν) : ζ ∈ C} : C is a club in ξ} for all ξ < λ with cf (ξ) = μ and all ν ∈ a. Lemma 5.1.1 Assume that λ and μ are regular  cardinals with |a| < μ < λ, and let the sequence (fξ : ξ < λ) of elements of a be μ-continuous. a) There is a club C in ξ such that fξ = sup{fζ : ζ ∈ C} for every ξ < λ with cf (ξ) = μ. b) If I is an ideal on a such that fη
this chapter, we owe much to the representation in [BM].

210

Chapter 5. Generators of J<λ+ (a)

Proof For part a), consider a fixed ξ < λ with cf (ξ) = μ. For each ν ∈ a, we choose a club Cν in ξ such that fξ (ν) = sup{fζ (ν) : ζ ∈ Cν }. Then  we can conclude from Lemma 1.8.2, together with |a| < cf (ξ) = μ, that C := {Cν : ν ∈ a} is also a club in ξ. By the definition of μ-continuity, we get fξ = sup{fζ : ζ ∈ C}. For the proof of part b), let ξ < λ and assume, as an inductive hypothesis, that the sequence (fζ : ζ < ξ) is strictly increasing modulo I. We will show that fζ
Lemma 5.1.3 If |a|+ < min(a)  and λ ∈ pcf(a), then there exists a sequence (fξ : ξ < λ) of elements of a which is strictly increasing modulo   J<λ (a) and is cofinal in a modulo D for every ultrafilter D on a with cf ( a/D) = λ. Proof To get a contradiction, let us assume that the assertion of the lemma is false. Then there exists a cardinal λ ∈ pcf(a) such that, for  any (f : ξ < λ) of members of a, modulo J<λ (a) strictly increasing sequence ξ  there is some ultrafilter D on a with cf ( a/D) = λ such that the sequence  (fξ : ξ < λ) is not cofinal in a modulo D. The following proof will again demonstrate the strength of the method of approximation sequences. Assume that Θ is a regular cardinal with 2sup(a) < Θ and (Ni : i ≤ |a|+ ) is an approximation sequence in H(Θ) such that a∪{a} ⊆ N0 and |Ni | = |a|+ for all i < |a|+ . Further let N := N|a|+ . Since |Ni | < cf (λ) = λ for all i < |a|+ , Lemma 4.3.3 says that χN (λ) = sup{χNi (λ) : i < |a|+ } and cf (χN (λ)) = |a|+ < λ. To apply  our assumption, assume that the sequence f¯ = (fξ : ξ < λ) of members of a is strictly increasing modulo J<λ  (a) and is a member of N0 . Thenthere is an ultrafilter D on a such that cf ( a/D) = λ and f¯ is not cofinal in a modulo D. Corollary 3.4.11 says that D ∩ J<λ (a) =

5.1. Universal Sequences

211

∅, and thus we conclude from Lemma 2.1.3 that f¯ is also strictly increasing  modulo D. Consequently, the set {fξ : ξ < λ} has an upper bound g0 in a under the relation
(∗) (∗∗)

+

∀i, j < |a| (i < j =⇒ χNi |`a
Now let b := {ν ∈ a : fχN (λ) (ν) = χN (ν)} and c := {ν ∈ a : gχN (λ) (ν) = χN (ν)}. Since fχN (λ) ≤a gχN (λ) , we have b ⊆ c by (∗). We want to show that b  c, since our indirect proof will allow the definition of |a|+ sequences like f¯ and g¯, and the existence of a strictly increasing sequence of subsets of a of length |a|+ will yield the desired contradiction. Since g¯ is |a|+ -continuous, we obtain from Lemma 4.3.4 a strictly increasing continuous sequence (i(β) : β < |a|+ ) of limit ordinals in |a|+ such that gχN (λ) (ν) = sup{gχNi(β) (λ) (ν) : β < |a|+ } for all ν ∈ a. Fix β < |a|+ . Using (∗) and fχN (λ)
Chapter 5. Generators of J<λ+ (a)

212

The rest of the proof consists of an appropriate recursive definition.  We will define, for each α < |a|+ , sequences f¯α = (fξα : ξ < λ) of members of a  and ultrafilters Dα on a with cf ( a/Dα ) = λ such that, for all α, γ < |a|+ , f¯α is strictly increasing modulo J<λ (a). If γ < α, then fξγ ≤a fξα for all ξ < λ. fξα
5.2. The Existence of Generators

213

cf (ξ) = |a|+ , this is true by definition. Thus assume that cf (ξ) = |a|+ and, according to Lemma 5.1.1 a), fξα = sup{fζα : ζ ∈ C} for some club C in ξ. The inductive hypothesis gives fζγ ≤a fζα for all ζ ∈ C, and thus we have, using the definition of fξγ , fξγ ≤a sup{fζγ : ζ ∈ C} ≤a sup{fζα : ζ ∈ C} = fξα . The case that α is a successor ordinal follows from the definition and the inductive hypothesis. This completes the proof of the lemma.

5.2 The Existence of Generators Lemma 5.2.1 Assume that μ and λ are regular cardinal numbers such that ℵ0 < μ and μ+ < λ. Further let S ⊆ Σμ++ ,μ be a stationary subset of μ++ and (Cα : α ∈ S) be an ecl-sequence for S which is a 3club -sequence for S as well. Then there is a sequence (Rζ : ζ < λ), satisfying Rζ ⊆ [ζ]≤μ and |Rζ | ≤ λ for all ζ < λ, which has the property that, for every regular cardinal Θ > λ and every approximation sequence (Nξ : ξ < μ++ ) in H(Θ) such that (i) |Nξ | < λ for all ξ < μ++ , (ii) {λ, (Cα : α ∈ S)} ⊆ N0 , there exists a normal sequence (ξ(i) : i < μ) of ordinals less than λ such that {χNξ(j) (λ) : j < i} ∈ Nξ(i+1) ∩ RχNξ(i) (λ) for all limit ordinals i < μ. Corollary 5.2.2 If μ and λ are regular cardinals with ℵ0 < μ and μ+ < λ, then there is a set R ⊆ [λ]≤μ with |R| ≤ λ such that, for every regular cardinal Θ > λ, there exists an elementary substructure N of H(Θ) and a set U ⊆ N ∩ λ with the following properties: (i) U is cofinal in N ∩ λ and otp(U ) = μ. (ii) If α ∈ U and otp(U ∩ α) is a limit ordinal, then U ∩ α ∈ N ∩ R. Proof Assume that Θ > λ is regular. Lemma 4.5.4 guarantees the existence of a set S ⊆ Σμ++ , μ which is stationary in μ++ , and of a sequence (Cα : α ∈ S) which is an ecl-sequence and a 3club -sequence for S as well. Let (Rζ : ζ < λ) be a sequence with the properties from Lemma 5.2.1 and R := {Rζ : ζ < λ}. By Lemma 4.3.1 we can choose an approximation sequence (Ni : i < μ++ ) in H(Θ) such that λ, (Cα : α ∈ S) ∈ N0 and |Ni | < μ++ for all i < μ++ . Now we take, according to Lemma 5.2.1, a normal sequence

214

Chapter 5. Generators of J<λ+ (a)

 (ξ(i) : i < μ) of members of μ++ and put N := {Nξ(i) : i < μ}. Then we know from Lemma 4.3.3 that (χNξ(i) (λ) : i < μ) is a normal sequence of members of N ∩ λ, and furthermore U := {χNξ(i) (λ) : i < μ} is a subset of N ∩ λ of order type μ which is cofinal in N ∩ λ. If α ∈ U and otp(U ∩ α) = i for some limit ordinal i < μ, then α = χNξ(i) (λ), and Lemma 5.2.1 says that U ∩ α = {χNξ(j) : j < i} ∈ Nξ(i+1) ∩ RχNξ(i) (λ) ⊆ N ∩ R. Proof of Lemma 5.2.1 Let S ⊆ Σμ++ ,μ be a stationary subset of μ++ and (Cα : α ∈ S) be an ecl-sequence for S which is a 3club -sequence for S as well. Then we have (1) ∀α ∈ S (Cα is a club in α ∧ otp(Cα ) = μ) . (2) ∀α, β ∈ S ∀γ ≤ min{α, β} (γ is a limit point of Cα and of Cβ =⇒ Cα ∩ γ = Cβ ∩ γ) . (3) ∀C ⊆ μ++ (C is a club in μ++ =⇒ {α ∈ S : Cα ⊆ C} is stationary in μ++ ). For each β ≤ λ with cf (β) = μ++ , we choose a club Eβ in β so that otp(Eβ ) = μ++ . If kβ is the ∈-isomorphism from μ++ onto Eβ , we put Eβ, α := kβ [Cα ] for α ∈ S. Then (Eβ, α : α ∈ S) is a copy of (Cα : α ∈ S). If λ = μ++ , we can take Eλ = λ, hence kλ = id. For each ζ < λ let Rζ := {Eβ, α ∩ ζ : β ≤ λ, cf (β) = μ++ , α ∈ S}. From otp(Cα ) = μ we can infer that |Eβ, α | = μ, hence |Eβ, α ∩ ζ| ≤ μ. Since λ · |S| = λ · μ++ = λ, we get |Rζ | ≤ λ. Now let Θ > λ be a regular cardinal and (Nξ : ξ < μ++ ) be an approximation sequence in H(Θ), satisfying |Nξ | < λ for all ξ < μ++ and {λ, (Cα : α ∈ S)} ⊆ N0 . Let δξ := χNξ (λ) for all ξ < μ++ . The sequence (δξ : ξ < μ++ ) is a normal sequence of members of λ, and thus F := {δξ : ξ < μ++ } is a club in β0 := sup(F ). Note that β0 ≤ λ and cf (β0 ) = μ++ . Clearly, also F ∩ Eβ0 is a club in β0 . Furthermore, by definition, kβ0 : μ++ −→ Eβ0 is an isomorphism. We put F ∗ := {σ ∈ β0 : ∃ξ ∈ μ++ (δξ = σ = kβ0 (ξ))}.

5.2. The Existence of Generators

215

Then it is easy to verify that F ∗ is a club in β0 such that F ∗ ⊆ F ∩ Eβ0 . Consequently, F ∗∗ := kβ−1 [F ∗ ] = {ξ ∈ μ++ : δξ = kβ0 (ξ)} is a club in μ++ . 0 By (3), the set {α ∈ S : Cα ⊆ F ∗∗ } is stationary in μ++ ; in particular, it is nonempty. So we can choose α0 ∈ S with Cα0 ⊆ F ∗∗ . Further we take some order preserving enumeration (τi : i < μ) of Cα0 , i.e. Cα0 = {τi : i < μ}. Since Cα0 ⊆ F ∗∗ , we have kβ0 [Cα0 ] = Eβ0 , α0 ⊆ kβ0 [F ∗∗ ] = F ∗ ⊆ F . Thus there is a normal sequence (ξ(i) : i < μ) of members of λ satisfying Eβ0 , α0 = {δξ(i) : i < μ}. For any i < μ, we have {δξ(j) : j < i} = Eβ0 , α0 ∩ δξ(i) ∈ Rδξ(i) . There remains to be shown that {δξ(j) : j < i} ∈ Nξ(i+1) for all limit ordinals i < μ. So let i < μ be a limit ordinal and j ≤ i. Since τj ∈ Cα0 ⊆ F ∗∗ , we get kβ0 (τj ) = δτj . On the other hand, δξ(j) is the j-th element of Eβ0 , α0 , and so we can conclude, using Eβ0 , α0 = kβ0 [Cα0 ], that kβ0 (τj ) = δξ(j) . This gives δξ(j) = δτj , hence ξ(j) = τj

for all j ≤ i,

in particular ξ(i) = τi < α0 . Now we have τi ∈ Nξ(i+1) , and τi is a limit point of Cα0 . Furthermore we have S ∈ Nξ(i+1) , and thus we can infer from H(Θ) |= ∃γ (γ ∈ S ∧ γ > τi ∧ τi is a limit point of Cγ ) and Nξ(i+1)  H(Θ) that there is an ordinal γ ∈ Nξ(i+1) ∩ S with γ > τi such that τi is a limit point of Cγ . From property (2) of the sequence (Cα : α ∈ S) we get {τj : j < i} = Cα0 ∩ τi = Cγ ∩ τi ∈ Nξ(i+1) . From λ ∈ Nξ(i+1) and g := (Nξ : ξ  ≤ ξ(i)) ∈ Nξ(i+1) we can infer for f := (sup(Nξ ∩ λ) : ξ  ≤ ξ(i)), that f ∈ Nξ(i+1) , since f = {(x, y) ∈ H(Θ) : H(Θ) |= x ∈ dom(g)∧y = sup(g(x)∩λ)} is definable in H(Θ) with parameters from Nξ(i+1) . Thus {τj : j < i} = {ξ(j) : j < i} ∈ Nξ(i+1) implies that ran(f |`{ξ(j) : j < i}) = {δξ(j) : j < i} ∈ Nξ(i+1) . So we get {δξ(j) : j < i} ∈ Nξ(i+1) , which completes the proof of the lemma. Theorem 5.2.3 (Main Theorem, S. Shelah) If a is an infinite set of regular cardinals, |a|+ < min(a), and λ is a cardinal, then there is a subset c of a which generates the ideal J<λ+ (a) over J<λ (a).

216

Chapter 5. Generators of J<λ+ (a)

Proof The case that λ ∈ pcf(a) is trivial since then J<λ+ (a) = J<λ (a) and we can choose c = ∅. So let us assume that λ ∈ pcf(a). If λ = max pcf(a), then J<λ+ (a) = P(a), and the assertion of the theorem is true since we can take c := a. If λ = min pcf(a), then J<λ+ (a) = {∅, {min(a)}}, and J<λ+ (a) is generated by {min(a)} over {∅} which proves the theorem for this trivial case. So we can assume that min pcf(a) < λ < max pcf(a). Note that J<λ+ (a) = P(a). Choose a universal sequence f¯ = (fξ : ξ < λ) according to Lemma 5.1.3. Then f¯ is strictly increasing modulo J<λ (a) and, for every ultrafilter D on a  ¯ is cofinal in  a modulo D. Let D be an ultrafilter such that cf ( a/D) = λ, f  on a such that cf ( a/D) = λ. From Corollary 3.4.11 we get D ∩ J<λ (a) = ∅, and thus f¯ is not bounded in a modulo J<λ (a). Now we can apply Corollary 3.4.3 and obtain a sequence (bξ : ξ < λ) of J<λ (a)-positive subsets of a with the following properties: (a) If ζ < ξ < λ, then bζ ⊆J<λ (a) bξ .  (b) For all ξ < λ, f¯ is cofinal in a modulo J<λ (a) |`bξ , and thus tcf ( a/J<λ (a) |`bξ ) = λ.  (c) There is a function g ∈ a such that g is an upper bound of f¯ modulo I if J<λ (a) ∪ {bξ : ξ < λ} generates the ideal I on a. We will show that bξ ∈ J<λ+ (a) for all ξ < λ. For this let us assume that bξ ∈ J<λ+ (a) for some ξ < λ. Then Lemma  3.4.5 says that there is an ultrafilter D on a such that bξ ∈ D and cf ( a/D) ≥ λ+ . Corollary 3.4.11 tells us that D ∩ J<λ+ (a) = ∅, and thus all the more D ∩ J<λ (a) = ∅, hence D ∩J<λ (a) |` bξ = ∅. From (b) we get tcf ( a/J<λ (a) |` bξ ) =  λ. This gives cf ( a/D) = λ, since D extends J<λ (a) |`bξ , contradicting cf ( a/D) ≥ λ+ . So we can conclude that I ⊆ J<λ+ (a). To show that I = J<λ+ (a) assume, to get a contradiction, that b ∈ J<λ+ (a) \ I for some b. Then there exists an ultrafilter D on a such that  b ∈ D and D ∩ I = ∅. From J<λ (a)  ⊆ I we get D ∩ J<λ (a) = ∅, hence cf ( a/D) ≥ λ. Furthermore we have cf ( a/D) < λ+ , since b ∈ J<λ+(a), and altogether we can conclude that cf ( a/D) = λ. Now f¯ is cofinal in a modulo D, but on the other hand we can infer from (c) and the fact that D extends I that the function g is an upper bound of f¯ modulo D. This contradiction and our assumption that J<λ+ (a) = P(a) show that J<λ+ (a) is the ideal generated by J<λ (a) ∪ {bξ : ξ < λ}. Together with (a) we get (∗)

I = {X ⊆ a : ∃Y ∈ J<λ (a)∃ξ < λ(X ⊆ Y ∪ bξ )}.

The inclusion “⊇” is obvious. Conversely, if X ∈ I, then there are ordinals ξ0 < ξ1 < · · · < ξn−1 and a set Y1 ∈ J<λ (a) such that X ⊆ Y1 ∪bξ0 ∪· · ·∪bξn−1 . Now we have X ⊆ Y ∪ bξn−1 where Y := Y1 ∪ (bξ0 \ bξ1 ) ∪ · · · ∪ (bξn−2 \ bξn−1 ) is a member of J<λ (a) by (a).

5.2. The Existence of Generators

217

Claim There is an ordinal ξ < λ such that J<λ+ (a) is generated by J<λ (a) ∪ {bξ }. Let us assume that the claim is false. If ξ < λ, then the assertion bδ ⊆J<λ (a) bξ is not true for all δ with ξ < δ < λ, since otherwise it would follow from (a) and (∗) that J<λ (a) ∪ {bξ } generates the ideal J<λ+ (a). On the other hand, if bδ ⊆J<λ (a) bξ holds for δ with ξ < δ < λ, then (∗) shows that bδ is superfluous as a generating element of the ideal J<λ+ (a) and thus can be omitted in the sequence (bξ : ξ < λ). Therefore we can assume without loss of generality that (∗∗)

∀δ∀ξ(ξ < δ < λ =⇒ ¬(bδ ⊆J<λ (a) bξ )).

Put μ := |a|+ . Since λ > min(a), and, by assumption, min(a) > |a|+ , we have ℵ0 < μ < μ+ < λ and can apply Corollary 5.2.2. So let us choose a set R ⊆ [λ]≤μ with the properties fromthis corollary. For each X ∈ [λ]≤μ let fX ∈ a be defined by fX := sup{fξ : ξ ∈ X} .   Using |X| ≤ μ = |a|+ < min(a) we can infer that fX ∈ a. Since ( a,<J<λ+ (a) )  is λ+ -directed and |R| ≤ λ, the set {fX : X ∈ R} has an upper bound h ∈ a modulo J<λ+ (a). We will show that this leads to a contradiction, using an appropriate elementary substructure N of H(Θ), where Θ is a large enough regular cardinal. The crucial point in the proof is the existence of N  H(Θ) and of a set U which is cofinal in N ∩ λ and has order type μ, such that every initial segment Z of U of limit order type is a member of N ∩ R. We can choose N and U with these properties by Corollary 5.2.2. By starting the construction of the structure N in this corollary with N0 ⊇ {a, h, (bξ : ξ < λ), f¯}, and sup(a) taking , we can arrange, as demonstrated in Example 4.2.15, that  Θ>2 ¯ λ, a, J<λ (a), f and (bξ : ξ < λ) are members of H(Θ) and thus of N . Let (δα : α < μ) be the ∈-isomorphism from μ = otp(U ) onto U , and put Uτ := {δα : α < τ } for every limit ordinal τ < μ and A := {ν ∈ a : fU (ν) > h(ν)}. Since |a| < μ, there is a limit ordinal τ < μ such that A = {ν ∈ a : fZ (ν) > h(ν)}, where Z := Uτ ∈ N ∩ R, and thus fZ ≤J<λ+ (a) h implies A ∈ J<λ+ (a). By (∗), there exists Y ∈ J<λ (a) and an ordinal γ < λ such that A ⊆ Y ∪ bγ . We will show below, as a routine proof, that A ∈ N and that there exists such an ordinal γ which is a member of N . Consequently, we have γ + 1 ∈ N and thus (bξ : ξ < λ)(γ +1) = bγ+1 ∈ N . Now the fact that in (b), for ξ = γ +1,

218

Chapter 5. Generators of J<λ+ (a)

all considered objects are members of N and the formula is absolute for H(Θ), hence holds in H(Θ), yields an ordinal β ∈ N ∩ λ such that h <J<λ (a) |`bγ+1 fβ . Since U is cofinal in N ∩λ and f¯ is also strictly increasing modulo J<λ (a) |`bγ+1 , we can assume that β ∈ U . By the definition of fU , this gives h <J<λ (a) |`bγ+1 fβ ≤a fU . So the set X := {ν ∈ bγ+1 : h(ν) ≥ fU (ν)} is a member of J<λ (a). But now we get bγ+1 \ bγ ⊆ X ∪ {ν ∈ bγ+1 \ bγ : h(ν) < fU (ν)} ⊆ X ∪ (A \ bγ ) and can infer that bγ+1 \bγ ∈ J<λ (a), hence bγ+1 ⊆J<λ (a) bγ , which contradicts (∗∗). This will prove the claim and complete the proof of the theorem if we show that A ∈ N and there is an ordinal γ ∈ N such that A is a member of the ideal J generated by J<λ (a) ∪ {bγ }. First let us note that fZ = {(ν, y) ∈ H(Θ) : H(Θ) |= ν ∈ a ∧ y = sup{f¯(δ)(ν) : δ ∈ Z}} , hence fZ is definable in H(Θ) with parameters from N and thus a member of N . From fZ , h ∈ N we can infer that A = {ν ∈ a : fZ (ν) > h(ν)} ∈ N . Finally, since H(Θ) |= ∃δ∃w ∈ J<λ (a) (A ⊆ w ∪ (bξ : ξ < λ)(δ)) and N  H(Θ), there is some γ ∈ N ∩ λ such that A is a member of the ideal J generated by J<λ (a) ∪ {bγ }, as desired.

5.3 Properties of Generators From Corollary 3.4.12, Lemma 3.4.19 and Corollary 3.4.6 we obtain immediately the following lemma. Lemma 5.3.1 Assume that a is progressive, λ ∈ pcf(a), and bλ is a subset of a generating the ideal J<λ+ (a) over J<λ (a). Then max pcf(bλ ) = λ. Generators yield a further result that illustrates the importance of the ideal J<λ (a). Theorem 5.3.2 Assume that a is progressive, λ ∈ pcf(a) and bλ is a subset of a generating J<λ + (a) over J<λ (a). Then J<λ (a) |`bλ is the smallest ideal I on a such that tcf ( a/I) = λ.

5.3. Properties of Generators

219

Proof First Lemma 3.4.19 tells us that  tcf ( a/J<λ (a) |`bλ ) = λ.  Consider a fixed ideal I on a satisfying tcf ( a/I) = λ. For each I-positive set  b, the ideal I |`b can be extended to an ultrafilter D on a, clearly satisfying cf ( a/D) = λ. From Lemma 3.4.5 we infer that b ∈ / J<λ (a). Thus we have shown that J<λ (a) ⊆ I. It remains to be shown that a \ bλ ∈ I since then J<λ (a) |`bλ = J<λ (a)[a \ bλ ] ⊆ I. We assume that this assertion is false, and choose an ultrafilter D on a such that D ∩ I = ∅ and a \ bλ ∈ D. Then cf ( a/D) = λ, and Corollary 3.4.11 says that there is b ∈ J<λ+ (a) ∩ D. By the choice of bλ , we have b ⊆J<λ (a) bλ . The first part of the proof shows that D, extending I, is also an extension of J<λ (a), and hence b ⊆D bλ . It follows that bλ ∈ D which contradicts a\bλ ∈ D. Corollary 5.3.3 Assume that a is progressive, λ ∈ pcf(a), and bλ is a subset of / pcf(a \ bλ ). a generating J<λ+ (a) over J<λ (a). Then λ ∈ Proof We assume, to get a contradiction, that λ ∈  pcf(a \ bλ ). Then there is an ultrafilter D on a satisfying a \ bλ ∈ D and cf ( a/D) = λ. The previous theorem says that D ∩ J<λ (a) |` bλ = ∅ on account of the minimality of J<λ (a) |` bλ , and a \ bλ ∈ J<λ (a) |` bλ and thus a \ bλ ∈ D leads to a contradiction. Definition If c is a set of cardinal numbers, then we call any sequence (bλ : λ ∈ c) a sequence of generators for a iff, for every λ ∈ c, the ideal J<λ+ (a) is generated by {bλ } ∪ J<λ (a). Corollary 5.3.4 If a is progressive and (bλ : λ ∈ pcf(a)) is a sequence of generators for a, then, for each subset c of a, there is n ∈ ω and λ1 , . . . , λn ∈ pcf(c) such that a) max{λ1 , . . . , λn } = max pcf(c) b) c ⊆ bλ1 ∪ . . . ∪ bλn .

and

Proof Put c0 := c and let, for n < ω,  cn \ bmax pcf(cn ) cn+1 := ∅

if cn = ∅, if cn = ∅.

Corollary 5.3.3 tells us that max pcf(cn+1 ) < max pcf(cn ) if cn+1 = ∅. Consequently, there must be a smallest n < ω such that cn = ∅. Putting λi+1 := max pcf(ci ) for all i < n, we get the assertion of the corollary.

220

Chapter 5. Generators of J<λ+ (a)

Corollary 5.3.5 If a is progressive, (bλ : λ ∈ pcf(a)) is a sequence of generators for a, and D is an ultrafilter on a, then  cf ( a/D) = min{λ ∈ pcf(a) : bλ ∈ D}.  Proof For μ := cf ( a/D) we can infer from Corollary 3.4.11 that D ∩ D ∩ J<μ+ (a), we get J<μ (a) = ∅ and D ∩ J<μ+ (a) = ∅. Taking a member b of  b \ bμ ∈ J<μ (a) and thus bμ ∈ D. If bλ ∈ D, then μ = cf ( a/D) ≤ λ since bλ is a member of J<λ+ (a). This proves the assertion.

Chapter 6 The Supremum of pcfμ(a) In this chapter we introduce first the essential notion of a control function. Its importance will come out when, in Chapter 8, we determine the cofinalities of partial orderings of the form ([λ]≤κ , ⊆). There we will prove the following theorem of Shelah: If κ is an infinite cardinal number and α is an ordinal such that 1 ≤ α < κ, then cf ([κ+α ]≤κ , ⊆) = max pcf({κ+(β+1) : β < α}). Furthermore the notion of a control function will be applied in the proof of a central theorem, due to Shelah: There is no sequence (κi : i < |a|+ ) of elements of pcf(a) such that max pcf({κj : j < i}) < κi for all i < |a|+ . As the main result of this chapter we will get: If a is a nonempty progressive interval of regular cardinals without a maximum, μ is a cardinal, and min(a) is μ-strong, then sup pcfμ (a) = (sup(a))|a| . In the introduction of Chapter 3 we have already described the crucial role this theorem plays in cardinal arithmetic.

6.1 Control Functions In the whole chapter, a denotes an infinite progressive set of regular cardinals. Definition For any set b of regular cardinals with b ⊆ a, we put  max pcf(b) if b =  ∅, λb := 0 if b = ∅.

222

Chapter 6. The Supremum of pcfμ (a)

If μ ∈ (|a|, min(a))reg is a cardinal number,  and, for each subset b of a, f¯b = (fξb : ξ < λb ) is a sequence of members of a such that (a) f¯b is strictly increasing modulo J<λb (a), (b) f¯b is μ-continuous, and  (c) {fξb |`b : ξ < λb } is cofinal in ( b, ≤b ), then we call f¯ := (f¯b : b ⊆ a) a special sequence for a and μ. Lemma 6.1.1 There exists a sequence f¯ = (f¯b : b ⊆ a), which is special for a and μ for all cardinals μ ∈ (|a|, min(a))reg . 3.4.21, we can find a subset G  Proof Fix a subset b of a. Using Theorem  of b with |G| = λb , which is cofinal in b under ≤b . Let (gξ : ξ < λb ) be an enumeration of G. We will define the desired sequence f¯b := (fξb : ξ < λb )  by recursion on ξ. Choose f0b ∈ a and define in the successor step, for every ξ < λb and ν ∈ a,  if ν ∈ b, max{fξb (ν) + 1, gξ (ν)} b fξ+1 (ν) := b fξ (ν) + 1 if ν ∈ / b. Now let ξ < λb be a limit ordinal and (fζb : ζ < ξ) with the desired properties be already defined. If cf (ξ) ∈ (|a|, min(a))reg , then we put for every ν ∈ a fξb (ν) := min{sup{fζb (ν) : ζ ∈ C} : C is a club in ξ}. / (|a|, min(a))reg , then we can  choose, using the fact that  If cf (ξ) ∈ ( a, <J<λ (a) ) is λb -directed, the function fξb ∈ a such that fζb <J<λ (a) fξb b b for all ζ < ξ. This completes the recursive definition. for all μ ∈ (|a|, min(a)) The construction shows that f¯b is μ-continuous reg .   Furthermore {fξb |`b : ξ < λb } is cofinal in b because, for each h ∈ b, there b is α < λb such that h ≤ gα , hence h ≤ fα+1 |` b. Finally Lemma 5.1.1 tells us b ¯ that f is strictly increasing modulo J<λb (a). Lemma 6.1.2 Assume that f¯ = (f¯b : b ⊆ a) is a special sequence for a and sup(a) μ and Θ is a regular cardinal such that 22 < Θ. Then f¯ is a member of H(Θ). Proof Let λ := max pcf(a). Obviously we have   a : b ⊆ a}. (∗) f¯ ⊆ P(a) × {λb  In Example 4.2.15 f) we have shown that P(a) ∈ H(Θ) andλb a ⊆  for H(Θ) every b ⊆ a. From the assumption on Θ we can infer that |λb a| ≤ | a|| a| ≤

6.1. Control Functions

223

    |a| sup(a) < Θ. Since | {λb a :b ⊆ a}| ≤ |P(a)| · |λ a| < 2| a| ≤ 2sup(a) ≤ 22 Θ, we know from Lemma 4.1.4 that P(a) × {λb a : b ⊆ a} is a member of H(Θ). Using (∗), we get f¯ ∈ H(Θ). sup(a)

Lemma 6.1.3 Let Θ be a regular cardinal with Θ > 22 , μ ∈ (|a|, min(a))reg , (Ni : i ≤ μ) be an approximation sequence in H(Θ), and N := Nμ . If f¯ := (f¯b : b ⊆ a) is a special sequence for a and μ, |Ni | < min(a) for all i < μ, and a ∪ {a, f¯} ⊆ N0 , then, for all b ∈ N ∩ P(a): a) If b = ∅, then fχbN (λb ) ≤a χN |`a. b) (i) There is a set c ∈ N ∩ J<λb (b) such that fχbN (λb ) |`(b \ c) = χN |`(b \ c). Moreover, c has the form c = B[fβb , fαb ] ∩ b for some α, β ∈ N ∩ λb with α < β. (ii) There is a club C in μ, such that, for all i, j ∈ C with i < j, the assertion fχbN (λb ) |`(b \ ci,j ) = χN |`(b \ ci,j ) ∩ b. The sets ci,j are definable in H(Θ) by a formula with the parameters f¯, b, Ni , and Nj from N . Furthermore, ci,j ∈ N ∩ J<λb (b). holds for ci,j := B[fχbN

j

(λb ) , χNi ]

Proof Consider b ∈ N ∩ P(a). We want to apply Lemma 4.3.4. Without loss of generality we can assume that b is a member of N0 since otherwise, if b ∈ Nj0 , we can replace the given approximation sequence by its final segment (Ni : j0 ≤ i ≤ μ). Since λb = max pcf(b) ≤ max pcf(a) ≤ sup(a)|a| ≤ 2sup(a) < 22

sup(a)

< Θ,

Lemma 4.1.4 tells us that λb ∈ H(Θ) . b ∈ N0 and N0  H(Θ) yield λb ∈ N0 , see Example 4.2.15 f), and {b, f¯} ⊆ N0 gives f¯b ∈ N0 . Now, together with property (b) from the definition of a special sequence, all assumptions of Lemma 4.3.4 are fulfilled, and we get fχbN (λb ) ≤a χN |` a, which proves a). Futhermore, we obtain a normal sequence (i(ξ) : ξ < μ) of limit ordinals in μ such that, for every ν ∈ a, (∗)

fχbN (λb ) (ν) = sup{fχbN

i(ξ) (λb )

(ν) : ξ < μ}.

Clearly the set C1 := {i(ξ) : ξ < ν} is a club in μ. Together with property (c) from the definition of a special sequence, Lemma 4.3.4 implies that, for each i ∈ C1 , there is an ordinal βi ∈ Ni+1 ∩ λb such that (∗∗)

χNi |`b
Chapter 6. The Supremum of pcfμ (a)

224

For any i ≤ μ let δi := χNi (λb ). If we put ci := {ν ∈ b : χNi (ν) ≥ fδbμ (ν)}, for i < μ, then the sequence (ci : i < μ) of subsets of a is increasing. Since μ > |a| is regular, there must be an ordinal i1 < μ such that ci1 = ci for all i ∈ [i1 , μ)ON . We put C := C1 \ i1 . Let i ∈ C and ν ∈ b \ ci . For every j ∈ [i, μ)ON we have ν ∈ b \ cj , hence χNj (ν) < fδbμ (ν); this gives χN (ν) ≤ fδbμ (ν). Thus we can conclude from a) that fδbμ |`(b \ ci ) = χN |`(b \ ci ). Fix i, j ∈ C such that i < j. Since βi ∈ Ni+1 , we have βi ∈ Nj , hence βi < δj , and property (a) in the definition of a special sequence yields, together with J<λ (b) = J<λ (a) ∩ P(b): c := {ν ∈ b : fβbi (ν) ≥ fδbj (ν)} = B[fδbj , fβbi ] ∩ b ∈ J<λb (b). We put ci,j := {ν ∈ b : χNi (ν) ≥ fδbj (ν)} = B[fδbj , χNi ] ∩ b. Then ci,j ⊆ c, since χNi |` b
:= {fξb |`(b \ c) ∪ g |`c : ξ < λb ∧ g ∈ Fc ∧ ∃α, β < λb (α < β ∧ c = {ν ∈ b :

fαb (ν)

(6.1) ≥

fβb (ν)}) }

for each b ⊆ a. Since f¯b is strictly increasing modulo J<λb (a), we have c ∈ J<λb (a) for the set c in Equation (6.1). Lemma 3.4.20 tell us that λc < λb , hence Fc is already defined.

6.2. The Supremum of pcfμ (a)

225

Any function F given by    P(a) −→ {P( b) : b ⊆ a} F := b −→ Fb is called a control function for a generated by f¯; Fb is called a control set for b. Lemma 6.1.4 If F is a control function for a, and b is a nonempty subset of a, then |Fb | ≤ max pcf(b). Proof The assertion can easily be verified by induction on λb using the recursive definition in Equation (6.1): There are λb choices for ξ, α and β, and, by the inductive hypothesis, for each such choice |Fc | ≤ λc < λb many choices of g. The following theorem allows the construction of an elementary substructure of H(Θ) whose characteristic functions on subsets of a we know in advance since they are members of a given control set. The cardinality of a control set can be estimated with the previous lemma. Thus a very strong assertion is obtained. Theorem 6.1.5 Assume that a is a progressive set of regular cardinals, b ⊆ a, sup(a) be and F is a control function for a generated by f¯. Further let Θ > 22 regular, μ ∈ (|a|, min(a))reg , and (Ni : i ≤ μ) be an approximation sequence in H(Θ) such that {f¯, a, b} ∪ a ⊆ N0 and |Ni | < min(a) for all i < μ. Then χNμ |`b ∈ Fb . Proof We use transfinite induction on λb . Choose the set c as in part b) (i) of Lemma 6.1.3. Without loss of generality we can assume that c ∈ N0 ; otherwise we consider a corresponding final segment of the approximation sequence. So we can apply the inductive hypothesis to c and conclude that χNμ |`c ∈ Fc , hence χNμ |`b = fχbNμ (λb ) |`(b \ c) ∪ χNμ |`c ∈ Fb .

6.2 The Supremum of pcfμ (a) Theorem 6.2.1 There is no sequence (κi : i < |a|+ ) of members of pcf(a) such that max pcf({κj : j < i}) < κi for all i < |a|+ .

Chapter 6. The Supremum of pcfμ (a)

226

Remark Observe that in the previous theorem, for any i < |a|+ , the maximum of pcf({κj : j < i}) exists since the set {κj : j < i} is progressive. Proof 1 Let us assume, to get a contradiction, that there exists a sequence (κi : i < |a|+ ) of members of pcf(a) such that max pcf({κj : j < i}) < κi for all i < |a|+ . Without loss of generality we can assume that |a|++ < min(a). We put a∗ := a ∪ {κi : i < |a|+ } and μ := |a∗ |+ . Then clearly a∗ is progressive. Let f¯ = (f¯b : b ⊆ a∗ ) be a special sequence for a∗ and μ. Further let ¯ := (bν : ν ∈ pcf(a)) b be a sequence of generators for a. sup(a∗ ) regular, and let (Ni : i ≤ μ) be an approximation Choose Θ > 22 sequence in H(Θ), N := Nμ , such that |Ni | < min(a) for all i < μ and ¯ f¯} ⊆ N0 . a∗ ∪ |a|+ ∪ {a, (κi : i < |a|+ ), b, First we claim that for each i < |a|+ there is a sequence (cik : k ≤ ni ) ∈ N such that ∅ = cini ⊆ cini −1 ⊆ . . . ⊆ ci0 = {κj : j < i}, ci

fχkN (λ i ) ≤cik χN |`a∗ c

k

for any k < ni , and ci

fχkN (λ i ) |`(cik \ cik+1 ) = χN |`(cik \ cik+1 ) c

k

for any k < ni . To show its existence, we need only apply Lemma 6.1.3 several times. Since cik+1 ∈ J<λ i (cik ), we have max pcf(cik+1 ) = λcik+1 < λcik = c

k

max pcf(cik ) by Lemma 3.4.20, and thus the descending chain of sets becomes constant after finitely many steps. Put fν,ξ := fξbν |`a for all ν ∈ pcf(a) and ξ < λbν . We define gi := sup{fκj ,χN (κj ) : j < i} for i < |a|+ . Since χN (κj ) < κj by Lemma 4.3.3 and κj = max pcf(bκj ) = λbκj  by Lemma 5.3.1, gi is well-defined, and |i| ≤ |a| < min(a) gives gi ∈ a. It is 1 We

follow the proof in [Sh5], VIII, 3.4.

6.2. The Supremum of pcfμ (a)

227

easy to see that we get a special sequence for a and μ by restricting f¯ to P(a). Consequently, Lemma 6.1.3 yields gi ≤ χN |`a. Put for i < |a|+ Bi := {ν ∈ a : gi (ν) = χN (ν)}. Clearly we have gj ≤a gi for j < i < |a|+ , hence Bj ⊆ Bi . So there must be i0 < |a|+ such that Bi = Bi0 for all i ∈ [i0 , |a|+ )ON . If j < i < |a|+ , then Lemma 6.1.3 says that {ν ∈ bκj : fκj ,χN (κj ) (ν) < χN (ν)} ∈ J<κj (a)

(6.2)

{ν ∈ bκj : gi (ν) < χN (ν)} ∈ J<κj (a).

(6.3)

and thus Putting λi,k := λcik for i < |a|+ and k < ni , we get (χN (κj ) : j < i0 ) = χN |`{κj : j < i0 } = χN |`ci00 i

=

c0 fχ0N (λi ,0 ) 0

i

|`(ci00

\

ci10 )

cn0

−1

∪ . . . ∪ fχNi0(λi

0 ,ni0 −1 )

|`cin0i

0 −1

.

Now we have f¯ ∈ N , i0 ∈ |a|+ ⊆ N , (cik0 : k ≤ ni0 ) ∈ N , and (κi : i < |a|+ ) ∈ N , hence κi0 ∈ N . Therefore the above representation of the sequence (χN (κj ) : j < i0 ) shows that this sequence and thus the function gi0 = sup{fκj ,χN (κj ) : j < i0 } are members of N ∗ := SkolΘ (N ∪ {χN (λi0 ,k ) : k < ni0 }). The assumption on the sequence (κi : i < |a|+ ) yields, for k < ni0 , λi0 ,k = max pcf(cik0 ) ≤ max pcf(ci00 ) < κi0 . Example 4.2.15 tells us that λi0 ,k ∈ N , hence χN (λi0 ,k ) < χN (κi0 ). Since κi0 ∈ N , Corollary 4.4.4 says that χN (κi0 ) = χN ∗ (κi0 ). Now, by the choice of f¯, the ordinal δ := min{ξ < κi0 : gi0 ≤J<κ

i0

(a) |` bκi

0

fκi0 ,ξ }

exists and is a member of N ∗ , since every parameter in the definition of δ is a member of N ∗ , and thus δ < χN (κi0 ). We conclude that e1 := {ν ∈ bκi0 : fκi0 ,δ (ν) ≥ fκi0 ,χN (κi0 ) (ν)} ∈ J<κi0 (a). From 6.2 and 6.3, with i0 , i0 + 1 instead of j, i, together with fκi0 ,χN (κi0 ) ≤a gi0 +1 ≤a χN (ν) |`a, we can infer that e2 := {ν ∈ bκi0 : ¬(fκi0 ,χN (κi0 ) (ν) = χN (ν) = gi0 +1 (ν))} ∈ J<κi0 (a).

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228

If ν is a member of the J<κi0 (a)-positive set bκi0 \ (e1 ∪ e2 ), then fκi0 ,δ (ν) < fκi0 ,χN (κi0 ) (ν) = χN (ν) = gi0 +1 (ν) = gi0 (ν). To justify the last equation we observe first that ν ∈ Bi0 +1 . By the choice of i0 , we have Bi0 +1 = Bi0 , and thus ν ∈ Bi0 , hence gi0 +1 (ν) = gi0 (ν). Altogether we get ¬(gi0 ≤J<κi (a) |`bκi fκi0 ,δ ), 0

0

contradicting the choice of δ. This completes the proof of the theorem. Definition Assume that M is an arbitrary set and b is a set of ordinals. We say that a function D characterizes M over b iff (1) M ∩ b ⊆ dom(D) and (2) For any δ ∈ M ∩ b, D(δ) is a cofinal subset of M ∩ [min(b), δ)ON . The following lemma shows the importance of this definition. Lemma 6.2.2 Assume that Θ is an uncountable regular cardinal, a ⊆ H(Θ) ∩ ICN is nonempty such that a \ {min(a)} is an interval of regular cardinals, and M and N are elementary substructures of H(Θ) with the following properties: (1) M ∩ min(a) ⊆ N . (2) M ∩ a ⊆ N . (3) There is a family of subsets of N characterizing M over a. Then M ∩ sup(a) ⊆ N . Proof Let (Dδ : δ ∈ a ∩ M ) be a characterizing function for M over a such that Dδ ⊆ N for all δ ∈ a ∩ M . By induction on λ we will show that M ∩ λ ⊆ N for all cardinals λ ≤ sup(a). For λ = min(a), this assertion is true by (1), and for limit cardinals λ it follows from the inductive hypothesis. So assume that λ ∈ [min(a), sup(a))CN and M ∩ λ ⊆ N . We will show that M ∩ λ+ ⊆ N . Let α ∈ M ∩ λ+ and, without loss of generality, α ≥ λ. From Example 4.2.15 we can infer that λ = |α| ∈ M and λ+ ∈ M . Clearly λ+ is an element of a. By (2) we have λ+ ∈ N , hence λ ∈ N . From assumption (3) we obtain an ordinal β ∈ Dλ+ such that α < β and β ∈ M ∩ N ∩ λ+ . Since |β| = λ, there is a <∗ -least bijection h from λ onto β, which is a member of both M and N by Example 4.2.15. Again 4.2.15 tells us that h |`(M ∩ λ) is a bijection from λ ∩ M onto β ∩ M . Analogously this holds for N . Using the inductive hypothesis, we can conclude that α ∈ M ∩ β = h[M ∩ λ] ⊆ h[N ∩ λ] = N ∩ β, and thus α ∈ N , as desired.

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229

The set M ∩ sup(a) here is sufficiently characterized by M ∩ min(a), M ∩ a and a corresponding characterizing sequence. This permits an estimate for the number of sets of the form M ∩ sup(a). Lemma 6.2.3 Assume that a ⊆ H(Θ) is a nonempty interval of regular cardinals, μ is an infinite cardinal, and D is a set. We say that an elementary substructure M of H(Θ) is D-nice or nice iff |M | ≤ μ and there is a function which characterizes M over a and is an element of D. Then |{M ∩ sup(a) : M is nice}| ≤ | min(a)|μ · |a|μ · |D|. Proof For every nice M we can choose, by the assumption on a, a member DM of D characterizing M over a. For any nice elementary substructures M1 , M2 of H(Θ) let M1 ∼ M2 iff M1 ∩ sup(a) = M2 ∩ sup(a), and let R be a set of representatives for the equivalence classes of ∼. Define, for M ∈ R, (∗)

g(M ∩ sup(a)) := (M ∩ min(a), M ∩ a, DM ).

If M, N ∈ R and (M ∩ min(a), M ∩ a, DM ) = (N ∩ min(a), N ∩ a, DN ), then Lemma 6.2.2 says that M ∩ sup(a) = N ∩ sup(a). Consequently, equation (∗) defines an injective function g from {M ∩ sup(a) : M is nice} into the set [min(a)]≤μ × [a]≤μ × D. Remark If μ is a cardinal, then sup pcfμ (a) ≤ (sup(a))μ by Lemma 3.3.4. Theorem 6.2.4 Assume that a is a nonempty progressive interval of regular cardinals without a maximum, μ is an infinite cardinal number, and min(a) is μ-strong. Then sup pcfμ (a) = (sup(a))μ . Proof In view of the preceding remark we only have to prove the inequality “≥”. Put λ := min(a). Since λ is μ-strong, we have μ < λ and get λμ = λ. If λ is a successor cardinal, this follows from Hausdorff’s formula, and if λ is a limit cardinal, hence weakly inaccessible, this follows from Lemma 1.6.15. Therefore λ+ is also μ-strong. Since the assertion of the theorem is the same if we consider a \ {min(a)} instead of a, we can assume without loss of generality that λ = κ+ , κ > |a| is regular, and μ < κ. Now a has no members which are limit cardinals, since any limit cardinal ν ∈ [min(a), sup(a))CN obviously is the supremum of a ∩ ν, thus we have cf (ν) ≤ |a| < min(a) ≤ ν, and ν is singular. Since |a| < λ and λ is μ-strong, we further conclude that |[a]≤μ | = |a|μ < λ.

(6.4)

, and let D be a set of functions. If M is Fix a regular cardinal Θ > 22 an elementary substructure of H(Θ) and M is nice according to the definition sup(a)

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230

in Lemma 6.2.3, then |P(M )| ≤ 2μ . Assume that we can choose D in such a way that we can find, for each x ⊆ sup(a) with |x| = μ, a nice M such that x ⊆ M . Then Lemma 6.2.3 says that (sup(a))μ = |[sup(a)]μ | ≤ 2μ · |{M ∩ sup(a) : M is nice}| ≤ | min(a)|μ · |a|μ · |D| ≤ λ · |D|. This will complete the proof of the theorem if we can define in addition the set D in such a way that |D| ≤ sup pcfμ (a). Let F = (Fb : b ⊆ a) be a control function for a generated by a function ¯ h which is special for a and all ν ∈ (|a|, min(a))reg . For each subset b of a we put Fb := {f ∈ Fb : ∀ν ∈ b cf (f (ν)) = κ} and G :=

  {FA : A ∈ [a]≤μ }.

From Lemma 6.1.4 and (6.4) we can infer that |G| ≤ sup pcfμ (a). We will see later on that G is not empty. Now we choose for each limit ordinal α < sup(a) with cf (α) = κ a club Cα in α of order type κ, and define for each f ∈ G a set Df by Df := {(Dν : ν ∈ dom(f )) : ∀ν ∈ dom(f ) Dν ∈ [Cf (ν) ]≤ℵ0 }. Finally we put D :=

 {Df : f ∈ G}.

If f ∈ G, then A := dom(f ) ∈ [a]≤μ , and, for all ν ∈ A, cf (f (ν)) = κ. This means that the club Cf (ν) is defined. Furthermore we have  |Df | ≤ |{φ : φ : dom(f ) −→ [ {Cf (ν) : ν ∈ dom(f )}]≤ℵ0 }| ≤ |μ ([κ]≤ℵ0 )| < λ. Consequently, we get |D| ≤ sup pcfμ (a), and there remains to be shown that D has the desired properties. So fix x ⊆ sup(a) with |x| = μ. We will construct a nice structure M such that x ⊆ M . For this, we choose first an approximation sequence (Ni : i ≤ κ) in H(Θ) such that ¯ a} ⊆ N0 x ∪ a ∪ [a]≤μ ∪ {h, and |Ni | = κ for all i ≤ κ. By Theorem 6.1.5, χNκ |` A is a member of FA and  thus, using Lemma 4.3.3, also a member of FA for any A ∈ [a]≤μ . The same ∗ lemma yields, for each ν ∈ a, a club Cν ⊆ Nκ ∩ ν in χNκ (ν) of order type κ. Without loss of generality we can assume that Cν∗ ⊆ CχNκ (ν) , since otherwise we can take the intersection of the clubs which is possible since κ > ω.

6.2. The Supremum of pcfμ (a)

231

Next we define by recursion an ascending sequence (Mn : n < ω) of elementary substructures of Nκ having cardinality μ. First, applying the L¨ owenheim-Skolem theorem, let M0 be an arbitrary elementary substructure of Nκ such that x ⊆ M0 and |M0 | = μ. Now assume that n < ω, Mn is already defined, and ν ∈ Mn ∩ [min(a), sup(a))CN . Then χMn (ν + ) < χNκ (ν + ), since |Mn | = μ < κ = cf (χNκ (ν + )). Therefore there is an ordinal, say σνn+ , in Cν∗+ such that χMn (ν + ) < σνn+ . Take Mn+1 as an elementary substructure of Nκ satisfying Mn ⊆ Mn+1 , |Mn+1 | = μ and {σνn+ : ν ∈ Mn ∩ [min(a), sup(a))CN } ⊆ Mn+1 . This completes the definition of the structures Mn , and we put  M := {Mn : n < ω}  and f := χNκ |`(M ∩ a). The previous remark shows that f ∈ FM ∩a ⊆ G. Assume that ρ ∈ M ∩ a. If ρ = min(a), we put Σρ := ∅. Otherwise we have shown that ρ = ν + for some ν ∈ [min(a), sup(a))CN , hence ν ∈ M . There is nν < ω with ν ∈ Mnν . Let

Σρ := {σρn : nν ≤ n < ω}. Our construction shows that Σρ is a countable subset of Cρ∗ and thus of Cf (ρ) . Therefore the sequence Σ := (Σρ : ρ ∈ M ∩ a) is a member of Df and thus of D. It remains to be shown that Σ characterizes the set M over a, that means that Σρ is unbounded in M ∩ [min(a), ρ)ON for all ρ ∈ M ∩ a. To see this, let ρ = ν + ∈ M ∩ a and γ ∈ M ∩ ρ. Without loss of generality we can assume that ν ≥ min(a), and, since ν ∈ M ∩ [min(a), ρ)ON and ν < χMnν (ν + ) < σνn+ν , that γ ≥ ν. Then there is n < ω such that n ≥ nν and γ ∈ Mn ∩ [ν, ν + )ON . By construction, σρn is greater than γ and a member of Σρ , as desired. This completes the proof of the theorem. Corollary 6.2.5 If a is a nonempty interval of regular cardinals which has no maximum and satisfies (min(a))|a| < sup(a), then  max pcf(a) = | a|. Proof Put κ := ((min(a))|a| )+ . If μ < κ is a cardinal, then μ|a| ≤ ((min(a))|a| )|a| = (min(a))|a| < κ,  i.e., κ is |a|-strong. We put b := [κ, sup(a))reg . Then ν∈a\b ν ≤ κ|a| = κ.  Since sup(a) ≤ | a| and sup(a) is a limit cardinal, the assumption yields κ <

Chapter 6. The Supremum of pcfμ (a)

232

     ν. From ν∈a ν = ν∈a\b ν · ν∈b ν we conclude that ν∈a ν = ν∈b ν. Theorem 6.2.4, applied to the set b, implies   max pcf(a) ≥ max pcf(b) = sup pcf|b| (b) = (sup(b))|b| ≥ | b| = | a|. 

ν∈a

Now max pcf(a) ≤ |



a| yields the desired assertion.

Remark Under the assumptions of the previous corollary, we can infer that   cf ( a, ≤a ) = | a|, using Theorem 3.4.21.

Chapter 7 Local Properties In his book [Sh5], Shelah introduces the operator pcf∗ which satisfies pcf∗ (pcf∗ (a)) = pcf∗ (a). If b is a set of regular cardinals, which is not necessarily progressive, and if every limit point of b is a singular cardinal, then pcf∗ (b) = pcf(b). Progressive sets a of regular cardinals have only singular limit points, so they satisfy pcf∗ (a) = pcf(a). In this case, we even get pcf(a) = pcf∗ (pcf(a)). If in addition every limit point of pcf(a) is singular, which is for example guaranteed by the condition |pcf(a)| < min(a), then pcf(pcf(a)) = pcf(a). A central tool for the proof of the hull property of pcf∗ is the so-called localisation theorem: If c ⊆ pcf(a) is progressive and λ ∈ pcf(c), then there exists a set d such that d ⊆ c, |d| ≤ |a|, and λ ∈ pcf(d). This theorem will also be applied in the proof of a main result of pcf-theory: If a is a progressive interval of regular cardinals, then |pcf(a)| < |a|+4 . The importance of this result will be demonstrated in Section 8.1. p 7.1 The Ideals J∗ (b) and J<λ (a) In [Sh5], VIII §3, S. Shelah introduces several new concepts which we will investigate and characterize in this chapter1 . Instead of progressive sets a, we will consider the sets pcf(a). The ideal J<λ (pcf(a)) exists, but many results of the preceding chapters cannot be applied to pcf(a), since it is not known whether pcf(a) is progressive. One can extend the ideal J<λ (a) on a to pcf(a) to obtain on this set p (a). Considering not all subsets of pcf(a) but only those being the ideal J<λ members of the set J∗ (pcf(a)), we will obtain a series of analogies to well1 In addition, we use ideas from Shelah’s note “An answer to complaints of E. Weitz on [Sh:371]”, communicated to us in July 1995.

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known results on progressive sets. In the same way this is true if we replace the pcf-operator by the operator pcf∗ . For the rest of this chapter, a denotes a progressive set of regular cardinals. The first lemma contains a definition. Note that this definition makes sense since, by Theorem 5.2.3, a sequence of generators (bμ : μ ∈ pcf(a)) exists. Lemma 7.1.1 Let λ ≤ max pcf(a) be a cardinal. If (bμ : μ ∈ pcf(a)) and (b∗μ : μ ∈ pcf(a)) are sequences of generators for a, then the sets {pcf(bμ ) : μ ∈ λ ∩ pcf(a)} and {pcf(b∗μ ) : μ ∈ λ ∩ pcf(a)} generate the same ideal on pcf(a). p We denote it by J<λ (a). Proof Consider μ1 , . . . , μn ∈ λ ∩ pcf(a) with μ1 < . . . < μn . Then Lemma 5.3.1 says that max(pcf(bμ1 ) ∪ . . . ∪ pcf(bμn )) = max pcf(bμn ) = μn < λ, hence pcf(bμ1 ) ∪ . . . ∪ pcf(bμn ) = pcf(a). Therefore {pcf(bμ ) : μ ∈ λ ∩ pcf(a)} generates an ideal on pcf(a). Now let μ ∈ λ∩pcf(a). By Corollary 5.3.4, there are cardinals ν1 , . . . , νn ∈ pcf(bμ ) such that bμ ⊆ b∗ν1 ∪ . . . ∪ b∗νn , hence pcf(bμ ) ⊆ pcf(b∗ν1 ) ∪ . . . ∪ pcf(b∗νn ). This shows that the ideal generated by {pcf(bμ ) : μ ∈ λ ∩ pcf(a)} is a subset of the ideal generated by {pcf(b∗μ ) : μ ∈ λ ∩ pcf(a)}. So the assertion is proved by symmetry arguments. If λ is a cardinal satisfying λ > max pcf(a), then in addition to the definition of the previous lemma we define p (a) := P(pcf(a)). J<λ

For the rest of this chapter, (bλ : λ ∈ pcf(a)) is a sequence of generators for a such that bmax pcf(a) := a. First we collect some results which easily follow from the definition of p J<λ (a) (see the exercises of this chapter). The first four lemmata are the analogies to 3.4.8, 3.4.12, 3.4.15 and 5.2.3, respectively. p p (a) ⊆ J<λ (a). Lemma 7.1.2 If κ and λ are cardinals with κ < λ, then J<κ

Lemma 7.1.3 For every cardinal λ, p p λ ∈ pcf(a) ⇐⇒ J<λ (a)  J<λ + (a).  p p (a) = {J<μ (a) : μ ∈ CN∧μ < λ}. Lemma 7.1.4 If λ is a limit cardinal, then J<λ

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235

p Lemma 7.1.5 If λ ∈ pcf(a), then the set pcf(bλ ) generates the ideal J<λ + (a) p over J<λ (a). p Lemma 7.1.6 For any cardinal λ, we have J<λ (a) ⊆ J<λ (a). p p Lemma 7.1.7 If b ⊆ a and λ is a cardinal, then J<λ (b) ⊆ J<λ (a).

Proof The sequence (bμ ∩ b : μ ∈ pcf(b)) is a sequence of generators for b.  Definition  We specify a function assigning membersλ of pcf(a) to the members of a. Fix, for each λ ∈ pcf(a), a sequence (fξ : ξ < λ) of members of  a such that (a) (fξλ : ξ < λ) is strictly increasing modulo J<λ (a)  (b) {fξλ |`bλ : ξ < λ} is cofinal in ( bλ , ≤bλ ).

and

Such a sequence can be obtained if we take a special sequence f¯ as in Section 6.12 and put fξλ := fξbλ for each λ ∈ pcf(a) and ξ < λ, since λbλ = max pcf(bλ ) = λ.  a we assign an ordinal function f ∗ ∈  To each ordinal function f ∈ pcf(a) extending f , putting for ν ∈ pcf(a)  f (ν) if ν ∈ a, ∗ f (ν) := if ν ∈ / a. min{γ < ν : f ≤J<ν (a) |`bν fγν } Remark Obviously, we have (fξλ )∗ (λ) = ξ for any ξ < λ and λ ∈ pcf(a) \ a. Now we can state that relations between ordinal functions on a are transferred to the associated functions on pcf(a). The next lemma is immediate from the definition.  Lemma 7.1.8 If f and g are members of a such that f ≤ g, then f ∗ ≤ g ∗ .  Lemma 7.1.9 If f, g ∈ a and λ is a cardinal such that f ≤J<λ (a) g, then ∗ p f ∗ ≤J<λ (a) g .

Proof By assumption, we have c := B(g, f ) = {ν ∈ a : g(ν) < f (ν)} ∈ us that there are elements μ1 , . . . , μn of J<λ (a). Therefore Corollary 5.3.4 tells  λ∩pcf(a) such that c is a subset of b := {bμi : 1 ≤ i ≤ n}. Now, if ν ∈ pcf(a)\a and bν ∩ b ∈ J<ν (a), then bν ∩ c ∈ J<ν (a), which gives f ≤J<ν (a) |`bν g and thus f ∗ (ν) ≤ g ∗ (ν). 2 We need not refer to the cardinal μ since we do not need the fact that the functions f b are μ-continuous.

236

Chapter 7. Local Properties So we have shown that / J<ν (a)}. B(g ∗ , f ∗ ) \ a ⊆ d := {ν ∈ pcf(a) : bν ∩ b ∈

In view of Lemma 7.1.6 and since B(g, f ) ∈ J<λ (a), it suffices to prove that p d ∈ J<λ (a). We have pcf(b) =

 p (a), {pcf(bμi ) : 1 ≤ i ≤ n} ∈ J<λ

so we show that d ⊆ pcf(b). If ν ∈ d, then there is an ultrafilter  D on a satisfying bν ∩ b ∈ D and J<ν(a) ∩ D = ∅. Therefore we get cf ( a/D) ≥ ν by Corollary 3.4.11 and cf ( a/D) ≤ ν by Lemma 5.3.5, since bν ∈ D. Furthermore, b ∈ D yields bμj ∈ D for some j ∈ {1, . . . , n}. This gives  ν = cf ( bμj /D ∩ P(bμj )) ∈ pcf(b). Definition If b is a, not necessarily progressive, set of regular cardinals, we define J∗ (b) := {c ⊆ b : ∀ρ(ρ = sup(ρ ∩ c) =⇒ ρ is not regular)}. It is easy to see that either J∗ (b) is an ideal on b or J∗ (b) = P(b). Especially observe that every progressive subset of b is a member of J∗ (b), and that J∗ (c) = J∗ (b) ∩ P(c) for all c ⊆ b. Lemma 7.1.10 If b is a set of regular cardinals, then J∗ (b) is the smallest subset J of P(b) with the following properties: a) {ν} ∈ J for all ν ∈ b. b) If κ is a cardinal and (ci : i < κ)is a sequence of members of J satisfying κ < min(ci ) for all i < κ, then {ci : i < κ} ∈ J. Proof Certainly J∗ (b) has property a). To verify that it has property b), let (ci : i < κ) bea sequence of members of J∗ (b) such that κ < min(ci ) for all i < κ, and c := {ci : i < κ}. Suppose, contrary to our hopes, that there is a regular cardinal ρ such that  ρ = sup(ρ ∩ c) = sup( {ρ ∩ ci : i < κ}) = sup{sup(ρ ∩ ci ) : i < κ}. By assumption we have sup(ρ ∩ ci ) < ρ for all i < κ, hence ρ = cf (ρ) ≤ κ. This contradicts ρ > min(c) > κ. Now consider a fixed set J ⊆ P(b) with the properties a) and b). We show that J∗ (b) ⊆ J. Note first that from b) it follows that J is closed under

p 7.1. The Ideals J∗ (b) and J<λ (a)

237

finite unions. Let c ∈ J∗ (b). We prove c ∈ J using transfinite induction on the cardinal ν := sup{μ+ : μ ∈ c}. If ν = 0, then c = ∅, and we have c ∈ J, using b) for κ = 0. If ν = μ+ for some μ, then μ = max(c). With the inductive hypothesis we get c \ {μ} ∈ J. Since {μ} ∈ J, this gives c ∈ J. If ν is a limit cardinal, then ν = sup(c). By the definition of J∗ (b), ν is singular. Put κ := cf (ν), and let the sequence (νi : i < κ) be cofinal in ν such that κ < ν0 . From the inductive hypothesis we can conclude that the members c ∩ ν0 and c ∩ [νi , νi+1 )ON of J∗ (b), for i < κ, are members of J. b) yields  c ∩ [ν0 , ν)ON = {c ∩ [νi , νi+1 )ON : i < κ} ∈ J, and thus c ∈ J, as desired. This characterization of J∗ (b) enables us to prove the following important result.  Lemma pcf(a) and b ∈ J∗ (pcf(a)), then there is a function  7.1.11 If g ∈ f ∈ a such that g |`b ≤ f ∗ |`b.  Proof Fix g ∈ pcf(a), and let  J := {b ⊆ pcf(a) : ∃f ∈ a g |`b ≤ f ∗ |`b} We will show, using Lemma 7.1.10, that J∗ (pcf(a)) ⊆ J. First consider ν ∈ pcf(a) and β:= g(ν). To prove property a) of the lemma, we  must find a function f ∈ a such that β ≤ f ∗ (ν). If ν ∈ a, then choose f ∈ a with f (ν) = β. If ν ∈ pcf(a) \ a, put f := fβν . Now we prove property b). Assume that κ is a cardinal and (ci : i < κ) is a sequence of members of Jsatisfying κ < min(ci ) for all i < κ. For each i < κ, choose a function  fi ∈ a according to the definition of J, such that  g |`ci ≤ fi∗ |`ci . Since ( a, <J + (a) ) is κ+ -directed, there is a function f ∈ a <κ such that fi <J + (a) f for all i < κ. Without loss of generality we can assume <κ that g |` a ≤ f , since otherwise  we can replace f by max{f, g |` a}. We verify g |`c ≤ f ∗ |`c for the set c := {ci : i < κ}. Let ν ∈ c. If ν ∈ a, then g(ν) ≤ f (ν) by the choice of f . If ν ∈ / a and i < κ is fixed with ν ∈ ci , then κ < ν and g(ν) ≤ fi∗ (ν). κ < ν gives J<κ+ (a) ⊆ J<ν (a), hence fi <J<ν (a) f , which implies fi ≤J<ν (a) |`bν f . So we get g(ν) ≤ fi∗ (ν) ≤ f ∗ (ν). p Now we will prove a number of results concerning the ideal J<λ (a), and indicate in brackets which results from the preceding chapters are generalized.

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Theorem 7.1.12(3.4.4) If λ is a cardinal, λ ≤ max pcf(a), and b ∈ J∗ (pcf(a))\ p p J<λ (a), then ( pcf(a), <J<λ (a) |` b ) is λ-directed.  Proof Let S be a subset of pcf(a) such that |S|  < λ. Using Lemma 7.1.11, we canchoose for each g ∈ S a function fg ∈ a suchthat g |` b ≤ fg∗ |`b. Since ( a, <J<λ (a) ) is λ-directed, there is a function f ∈ a satisfying ∗ p fg ≤J<λ (a) f for all g ∈ S. Now Lemma 7.1.9 says that fg∗ ≤J<λ (a) f for all g ∈ S, and this together with g |` b ≤ fg∗ |` b yields {ν ∈ b : f ∗ (ν) < g(ν)} ∈ p ∗ p (a). So we get g ≤J<λ J<λ (a) |` b f for all g ∈ S. p Theorem 7.1.13 If λ = max pcf(a) and b ∈ J∗ (pcf(a)) \ J<λ (a), then

 p (a) |`b) = λ. tcf ( pcf(a)/J<λ Proof For each ξ < λ, we put hξ := (fξλ )∗ . (Remember the fixed sequence from page 235. In particular here we have bλ = a.) By Lemma 7.1.9, the ¯ := (hξ : ξ < λ) is increasing modulo J p (a), hence also modulo sequence h <λ p J<λ (a) |`b.  Now considera fixed function g ∈ pcf(a). Lemma 7.1.11 says that there is a function f ∈ a such that (g + 1) |`b ≤ f ∗ |`b. λ (fξλ : ξ < λ) is pointwise There is an  α < λ such that f ≤ fα , since  ordinal ∗ cofinal in bλ = a. Lemma 7.1.8 tells us that f ≤ hα , hence g |`b < hα |`b. p p Together with hα ≤J<λ hα+1 , (a) |` b hα+1 we can conclude that g <J<λ  (a) |`b ¯ which shows that a suitable subsequence of h is cofinal in pcf(a) modulo p J<λ (a) |`b. p Corollary 7.1.14 (3.4.19) Assume that λ ∈ pcf(a) and b ∈ J∗ (pcf(a))∩J<λ + (a)\ p J<λ (a). Then  p tcf ( pcf(a)/J<λ (a) |`b) = λ. p Proof By assumption and Lemma 7.1.5, we have b \ pcf(bλ ) ∈ J<λ (a), p p and thus J<λ (a) |` b = J<λ (a) |` (pcf(bλ ) ∩ b). So we can assume without loss of generality that b ⊆ pcf(bλ ). Applying Theorem 7.1.13 to bλ instead of a, we can conclude that  p (bλ ) |`b) = λ. tcf ( pcf(bλ )/J<λ  Now we choose a sequence of length λ which is cofinal in pcf(bλ ) modulo p J<λ (bλ ) |`b and extend its members arbirarily to pcf(a). By Lemma 7.1.7, the  p resulting sequence is cofinal in pcf(a) modulo J<λ (a) |`b.

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Corollary 7.1.15 (5.3.5) If D is an ultrafilter on pcf(a) such that D ∩ J∗ (pcf(a)) = ∅, then  p cf ( pcf(a)/D) = min{λ ∈ pcf(a) : D ∩ J<λ + (a) = ∅} = min{λ ∈ pcf(a) : pcf(bλ ) ∈ D}. Proof Let D be an ultrafilter on pcf(a) such that D ∩ J∗ (pcf(a)) = ∅. p Clearly there is a least member, say λ, of pcf(a) satisfying D ∩ J<λ + (a) = ∅, p since J<κ (a) = P(pcf(a)) for all κ > max pcf(a). Then for all μ ∈ λ ∩ pcf(a) we have pcf(bμ ) ∈ D p Consider a fixed set d ∈ D ∩ J<λ + (a). Without loss of generality we can assume that d ∈ J∗ (pcf(a)), since otherwise we could take the intersection of p d and some member of D ∩ J∗ (pcf(a)). By the definition of J<λ + (a), there , . . . , μ ∈ λ + 1, which are members of pcf(a) and satisfy d ⊆ are cardinals μ 1 n  {pcf(bμi ) : 1 ≤ i ≤ n}. Now there is k ∈ {1, . . . , n} such that pcf(bμk ) ∈ D. The minimality of λ shows that μk = λ. This yields the second equation of the corollary. Furthermore Corollary 7.1.14 says, together with the choice of λ, that   p (a) |`d) = λ, cf ( pcf(a)/D) = tcf ( pcf(a)/J<λ which proves the first equation of the corollary. Corollary 7.1.16 (3.4.11) Assume that D is an ultrafilter on pcf(a) such that D ∩ J∗ (pcf(a)) = ∅, and λ is a cardinal. Then cf (



p (a) = ∅. pcf(a)/D) < λ ⇐⇒ D ∩ J<λ

p p (a), D ∩ J<λ (a) = ∅ Proof First we observe that, by the definition of J<λ p holds iff D∩J<κ+ (a) =  ∅ for some κ < λ. By Corollary 7.1.15, the last assertion  is equivalent to cf ( pcf(a)/D) ≤ κ.

Corollary 7.1.17 For any cardinal λ, we have p (a) ∩ J∗ (pcf(a)) = {b ∈ J∗ (pcf(a)) : pcf(b) ⊆ λ}. J<λ p (a) and Proof Consider b ∈ J∗ (pcf(a)). First let us assume that b ∈ J<λ (pcf(a)), we D is an  ultrafilter on pcf(a) such that b ∈ D. Since b ∈ D ∩ J ∗  have cf ( pcf(a)/D) = cf ( b/D ∩ P(b)) < λ by Corollary 7.1.16. This gives p (a), then we can choose an ultrafilter D on pcf(b) ⊆ λ. Conversely, if b ∈ J<λ p pcf(a) such that b ∈ D and D ∩ J<λ (a) = ∅. Corollary 7.1.16 implies that  cf ( b/D ∩ P(b)) ≥ λ, and thus pcf(b) is not a subset of λ.

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Corollary 7.1.18 If b ∈ J∗ (pcf(a)) and λ is a cardinal, then p J<λ (b) = J<λ (a) ∩ P(b). p In particular, J<λ (a) = J<λ (a) ∩ P(a).

Proof Use the fact that J<λ (b) = {c ⊆ b : pcf(c) ⊆ λ}. Corollary 7.1.19 (3.4.20) If b ∈ J∗ (pcf(a)), then pcf(b) is a subset of pcf(a) and has a maximum. Proof The first assertion follows immediately from Corollary 7.1.15. To get a contradiction, let us suppose that b ∈ J∗ (pcf(a)) and λ := sup pcf(b) is not a member of pcf(b). Then λ is a limit cardinal, and Corollary 7.1.17 says p p (a), hence b ∈ J<μ (a) for some μ < λ. But this gives pcf(b) ⊆ μ, that b ∈ J<λ contradicting the choice of λ. Lemma 7.1.20 Assume that c ∈ J∗ (pcf(a)) and b ⊆ pcf(c) is progressive. If D is an ultrafilter on b and, for each ν ∈ b, Dν is an ultrafilter on c such that  ν = cf ( c/Dν ), then D∗ := {X ⊆ c : {ν ∈ b : X ∈ Dν } ∈ D}   is an ultrafilter on c satisfying cf ( c/D∗ ) = cf ( b/D). on c. Take, for each Proof One easily verifies that D∗ is an ultrafilter  in c modulo Dν . To apply ν ∈ b, a sequence (gαν : α < ν) which iscofinal  Lemma 3.1.10, we define a function G : c −→ b by G(f )(ν) := min{α < ν : f ≤Dν gαν }  for any f ∈ c and any ν ∈ b. Fix h ∈  Now we show that G satisfies the assumptions of Lemma 3.1.10. p b. Since b is progressive, hence |b| < min(b), we have c ∈ J<|b| + (a), since otherwise Corollary 7.1.17 would yield the contradiction b ⊆ pcf(c) ⊆ |b|+ . From Theorem 7.1.12, applied to |b|+ and  c, we can conclude together with Lemma 2.1.4 that there is a function p ∈ c such that ν gh(ν) <J p

(a)∩P(c) <|b|+

p

 p for all ν ∈ b. If ν ∈ b, then |b|+ ≤ ν = cf ( c/Dν ), hence Dν ∩J<|b| + (a) = ∅ by Corollary 7.1.16; obviously, to apply this corollary, it is no problem to extend ν
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 Assume that q ∈ c and p ≤D∗ q. It remains to be shown that h
fξ ≤J p

<λ+

(a) |` c

f

for all ξ < λ. But then p B(f, fξ ) ∩ c = (B(f, fξ ) ∩ c) \ pcf(bλ ) ∈ J<λ (a) p for all ξ < λ, and thus f is an upper bound of (fξ : ξ < λ) modulo J<λ (a) |` c and, consequently, also modulo D. This contradicts the fact that the sequence  (fξ : ξ < λ) is cofinal in pcf(a) modulo D.

Definition If b is a set of regular cardinals, then we put  pcf∗ (b) := {cf ( b/D) : D is an ultrafilter on b ∧ D ∩ J∗ (b) = ∅}. With the help of Lemma 3.3.4, the following assertions are easy consequences of the definition of the operator pcf∗ . Lemma 7.1.23 (3.3.4) If b and c are sets of regular cardinals, then  a) pcf∗ (b) = {pcf(d) : d ∈ J∗ (b)}. b) If b ∈ J∗ (b), then pcf∗ (b) = pcf(b). In particular, we have pcf∗ (b) = pcf(b) for any progressive set b.

242 c) d) e) f) g)

Chapter 7. Local Properties pcf∗ (b) is a set of regular cardinals. b ⊆ pcf∗ (b). min pcf∗ (b) = min(b). b ⊆ c implies pcf∗ (b) ⊆ pcf∗ (c). pcf∗ (b ∪ c) = pcf∗ (b) ∪ pcf∗ (c).

Theorem 7.1.24 (Localisation Theorem) If a is a progressive set of regular cardinals, c is a subset of pcf(a), and λ ∈ pcf∗ (c), then there exists a subset d of c such that |d| ≤ |a| and λ ∈ pcf(d). Proof We suppose, contrary to our hopes, that the assertion of the theorem is false. By Lemma 7.1.23 a), there is a set c ∈ J∗ (pcf(a)) and a cardinal λ ∈ pcf(c), such that λ ∈ / pcf(d) for all d ∈ [c]≤|a| . We choose λ minimal with this property, and for this λ a set c ∈ J∗ (pcf(a)) satisfying  λ ∈ pcf(c) \ {pcf(d) : d ⊆ c ∧ |d| ≤ |a|}, such that the cardinal κ := sup{μ+ : μ ∈ pcf(c) ∧ μ < max pcf(c)} is minimal. Then Corollary 7.1.19 says that pcf(c) is a subset of pcf(a), hence λ ∈ pcf(a). Lemma 7.1.22 yields λ ∈ pcf(c ∩ pcf(bλ )). Therefore we can assume p without loss of generality that c ⊆ pcf(bλ ). Since c ∈ J<λ + (a), pcf(c) is a subset + of λ by Corollary 7.1.17. From this and λ ∈ pcf(c) we get λ = max pcf(c). Suppose that the set λ ∩ pcf(c) has a maximum, say μ. Then κ = μ+ . From Lemma 7.1.22 we infer that μ ∈ pcf(c \ bμ ). Corollary 7.1.17 together p with c ∩ pcf(bμ ) ∈ J<μ + (a) ∩ J∗ (pcf(a)) tells us that pcf(c ∩ pcf(bμ )) ⊆ μ + 1, hence λ ∈ pcf(c \ pcf(bμ )). This property of c \ pcf(bμ ) contradicts the choice of c and the minimality of κ. Therefore κ = sup(λ ∩ pcf(c)) is a limit cardinal such that κ ∈ / λ ∩ pcf(c). Since pcf(c) ⊆ pcf(a), d ⊆ pcf(c) implies d ⊆ pcf(a). If in addition |d| ≤ |a|, then the set d is progressive, and thus pcf(d) has a maximum. Furthermore pcf(d) is, by Corollary 7.1.21, a subset of pcf(c). If we prove the Claim pcf(d) ⊆ κ for all d ⊆ λ ∩ pcf(c) with |d| ≤ |a|, then the proof of the theorem will be completed, since we can construct by recursion a sequence (κi : i < |a|+ ) of elements of λ ∩ pcf(c), hence of pcf(a), in such a way that its existence contradicts Theorem 6.2.1. To give an indirect proof of the claim, let us assume that there is a set d ⊆ λ ∩ pcf(c) such that |d| ≤ |a| and κ∗ := max pcf(d) ≥ κ.

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Now pcf(d) ⊆ pcf(c) by Corollary 7.1.21, hence max pcf(d) ≤ λ. If max pcf(d) < λ, then (κ∗ )+ < κ, contradicting κ ≤ κ∗ . So we have κ∗ = λ. For each ν ∈ d we obtain, by  the choice of λ, a set eν ⊆ c satisfying ν ∈ pcf(eν ) and |eν | ≤ |a|. With e := {eν : ν ∈ d} we get d ⊆ pcf(e), which gives again pcf(d) ⊆ pcf(e) by Corollary 7.1.21, and in particular λ ∈ pcf(e). But this together with e ⊆ c and |e| ≤ |a| contradicts the choice of c and λ. Lemma 7.1.25 pcf(a) = pcf∗ (pcf(a)). Proof This follows from Lemma 7.1.23 and Corollary 7.1.19. Theorem 7.1.26 If b ⊆ pcf(a), then pcf∗ (b) = pcf∗ (pcf∗ (b)). Proof Fix λ ∈ pcf∗ (pcf∗ (b)). From the assumption, Lemma 7.1.23 and Lemma 7.1.25 we get pcf∗ (b) ⊆ pcf∗ (pcf(a)) = pcf(a), and Theorem 7.1.24 yields λ ∈ pcf(d) for some d ⊆ pcf∗ (b) satisfying |d| ≤ |a|. For each ν ∈ d ⊆ pcf∗ (b) we obtain, again by Theorem 7.1.24, a set dν ⊆ b such that |dν | ≤ |a| and ν ∈ pcf(dν ). The estimate |d| ≤ |a| < min(a) ≤ min pcf(a) ≤ min pcf∗ (b) ≤ min(d) shows that d is progressive. So we can conclude that   λ ∈ pcf(d) = pcf∗ (d) ⊆ pcf∗ ( {pcf(dν ) : ν ∈ d}) ⊆ pcf∗ (pcf( {dν : ν ∈ d}))   = pcf( {dν : ν ∈ d}) = pcf∗ ( {dν : ν ∈ d}) ⊆ pcf∗ (b). The last steps in this chainof inclusions follow from Lemma 7.1.25  and Lemma 7.1.23 respectively, since | {d : ν ∈ d}| ≤ |a| < min(a) ≤ min( {dν : ν ∈ d}) ν  and thus the set {dν : ν ∈ d} is progressive. Now we can state an analogy to Lemma 3.4.20 better than Corollary 7.1.19. Lemma 7.1.27 If b ⊆ pcf(a), then pcf∗ (b) is a subset of pcf(a) and has a maximum. Proof The first assertion follows from Lemma 7.1.23 and Corollary 7.1.19. To get a contradiction, we assume that the set pcf∗ (b) has no maximum and claim that, for any c ⊆ pcf∗ (b) with |c| ≤ |a|, there is λ ∈ pcf∗ (b) such that max pcf(c) < λ. This again gives rise to the construction of a sequence of elements of pcf∗ (b) having length |a|+ whose existence contradicts Theorem 6.2.1. To prove the claim, assume that c ⊆ pcf∗ (b) and |c| ≤ |a|. As in the proof of Theorem 7.1.26 for the set d, we can verify that c is progressive. So we can conclude from Theorem 7.1.26 that pcf(c) = pcf∗ (c) ⊆ pcf∗ (pcf∗ (b)) = pcf∗ (b). Furthermore pcf(c) has a maximum by Lemma 3.4.20. But from our supposition on pcf∗ (b) we can infer that there is λ ∈ pcf∗ (b) greater than this maximum, which proves the claim and completes the proof of the lemma.

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Corollary 7.1.28 (5.3.1) For any λ ∈ pcf(a), max pcf∗ (pcf(bλ )) = λ. Proof Since bλ is progressive as a subset of the progressive set a, we get pcf∗ (pcf(bλ )) = pcf∗ (pcf∗ (bλ )) = pcf∗ (bλ ) = pcf(bλ ), and together with λ = max pcf(bλ ) the corollary is proved. Exercises 1) Prove Lemma 7.1.3. Hint: See the proof of Lemma 7.1.1. 2) Prove Lemma 7.1.4. 3) Prove Lemma 7.1.5. 4) Prove Lemma 7.1.6. 5) If c ⊆ pcf(a), then there are n ∈ ω and λ1 , . . . , λn ∈ pcf∗ (c) such that (a) max{λ1 , . . . , λn } = max pcf∗ (c) and (b) c ⊆ pcf(bλ1 ) ∪ . . . ∪ pcf(bλn ). 6) For any cardinal λ, p J<λ (a) = {c ⊆ pcf(a) : pcf∗ (c) ⊆ λ}.

7) Prove Lemma 7.1.23. 8) If λ is a cardinal and b ⊆ a, then p p (b) = J<λ (a) ∩ P(pcf(b)). J<λ

7.2 Intervals in pcf(a) Definition In analogy to Lemma 1.8.9 regarding normal ideals, we now define the notion of a weakly normal ideal. Let A be a set of ordinals such that sup(A) is not a member of A. We call an ideal I on A weakly normal iff, for every I-positive subset S of A and every regressive function f on S, there is an Ipositive subset S0 of S and an ordinal γ < sup(A) such that f [S0 ] ⊆ γ. Let α be a limit ordinal of uncountable cofinality and C be a subset of α such that sup(C) = α. Then we know that the set Id (C) := {X ⊆ C : X is not stationary in α} of thin subsets of C is an ideal on C. If E is a set of cardinals, then we put E + := {ν + : ν ∈ E}.

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Lemma 7.2.1 If cf (α) > ω and C is a club in α of order type cf (α), then the ideal Id (C) on α is weakly normal. Proof From Section 1.8 we know that Id (C) is an ideal on α. Let Γ be the ∈-isomorphism from κ := cf (α) onto C. Further assume that S is an Id (C)-positive subset of C and f is regressive on S. Then the set S  := Γ−1 [S] ∩ {i < κ : i is a limit ordinal} is a stationary subset of κ. For each i ∈ S  let g(i) be the least j < κ such that f (Γ(i)) < Γ(j). Since all members i of S  are limit ordinals and f is regressive, g is also regressive. Fodor’s theorem says that there is a stationary subset S0 ⊆ S  and an ordinal i0 < κ such that g[S0 ] = {i0 }. Now S0 := Γ[S0 ] is an Id (C)-positive subset of S satisfying f [S0 ] ⊆ Γ(i0 ) < α. Lemma 7.2.2 If I is a weakly normal ideal on a set A and X is an I-positive subset of A, then the ideal I |`X is weakly normal. Proof Let X ⊆ A be I-positive and Y ⊆ A. Further assume that Y is (I |` X)-positive and f is a regressive function on Y . Then f is regressive on the I-positive set Y ∩ X, and thus f [Z] ⊆ γ for some I-positive set Z ⊆ Y ∩ X and some γ < sup(A). This gives Z ∩ X = Z ∈ / I, hence Z is (I |`X)-positive. Theorem 7.2.3 Assume that λ is a singular cardinal with uncountable cofinality. Then there is a progressive club C0 in λ of cardinal numbers such that otp(C0 ) = cf (λ) and max pcf(C0+ ) = λ+ . Proof From Lemma 1.6.11 we obtain a club C in λ such that all members of C are singular cardinals and otp(C) = cf (λ) < min(C). Let a := C + . Obviously a is a progressive set of regular cardinals. Theorem 5.2.3 yields a set bλ+ ⊆ a generating the ideal J<λ++ (a) over J<λ+ (a); if λ+ ∈ pcf(a), then let bλ+ = ∅. It suffices to prove that there is a club C0 ⊆ C in λ such that C0+ ⊆ bλ+ . To see this, note that then we can conclude from Lemma 3.3.4 and Lemma 5.3.1, together with the fact that C0+ is progressive, that λ ≤ max pcf(C0+ ) ≤ max pcf(bλ+ ) = λ+ . Since λ is singular, this will give max pcf(C0+ ) = λ+ . To get a contradiction, let us assume that such a club does not exist. Let b0 be that subset of C with b+ 0 = bλ+ . By our supposition, b0 does not include a club in λ, and thus C \ b0 is stationary in λ. Consequently, by the previous lemmas, J := Id (C) |`(C \ b0 ) = Id (C)[b0 ]

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is a weakly normal ideal on C. We transfer its ideal structure to a, putting I := {X + : X ∈ J}. Clearly I is an ideal on a. Since Id (C) has all bounded subsets of C amongst its members, all bounded subsets of a are members of I. We can infer from Corollary 3.4.12, together with the fact that λ is singular, that J<λ (a) = J<λ+ (a). So we get J<λ+ (a) ⊆ I from Lemma 3.4.9 and λ = sup(a). Furthermore we have λ = limI (a). b0  ∈ J gives bλ+ ∈ I, hence J<λ++ (a) ⊆ I by the choice of bλ+ . Consequently, ( a, κ is an ordinal, and F is a function from P(τ ) into ON, such that for all X, Y ∈ P(τ ) (1) X ⊆ Y =⇒ F (X) ≤ F (Y ). (2) F (X) ≥ sup(X).

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247

(3) If γ < τ and cf (γ) = κ+3 , then there is a club E in γ such that F (E) = γ. (4) If X has order type κ+ , then there is β < sup(X) such that F (X ∩ β) ≥ sup(X). Then we have τ < κ+4 . Proof Consider the cardinal κ ≥ ω. To get a contradiction, let us suppose that there is a function F and an ordinal τ with the properties (1)-(4) such that τ ≥ κ+4 . By Lemma 4.5.3, we can choose a 3Club (κ+3 , κ+ )-sequence C¯ := (Cα : α ∈ Σκ+3 ,κ+ ) for Σκ+3 ,κ+ . Choose the regular cardinal Θ large enough and an approximation sequence (Ni : i ≤ κ+3 ) in H(Θ) such that ¯ F, κ+4 } ⊆ N0 . Let δi := χN (κ+4 ) for |Ni | = κ+3 for all i < κ+3 and κ+3 ∪ {C, i +3 all i ≤ κ . Lemma 4.3.3 says that (δi : i < κ+3 ) is normal sequence which is cofinal in δκ+3 , and that δκ+3 < κ+4 . By property (3), there is a club E in δκ+3 such that F (E) = δκ+3 . Then E ∗ := E ∩ {δi : i < κ+3 } is a club in δκ+3 and C := {j ∈ κ+3 : δj ∈ E ∗ } is a club in κ+3 . Furthermore we have F (E ∗ ) = δκ+3 , since (1) gives F (E ∗ ) ≤ F (E), and (2) yields F (E ∗ ) ≥ sup(E ∗ ) = δκ+3 . So let, without loss of generality, E = {δi : i ∈ C}. By the definition of a 3Club (κ+3 , κ+ )-sequence, we can find an ordinal α ∈ Σκ+3 ,κ+ such that Cα ⊆ C, and furthermore the club Cα has order type κ+ . Therefore this also holds for the set X := {δi : i ∈ Cα }, since (δi : i < κ+3 ) is normal. We have sup(X) = δα , since Cα is unbounded in α. Using (4) and (1), we obtain a limit ordinal σ < sup(X) = δα such that F (X ∩ σ) ≥ δα . We put β := sup{i < κ+3 : δi < σ}. Then we have δβ ≤ σ, hence β < α and ζ := F ({δi : i ∈ Cα ∩ β}) ≥ sup{δi : i ∈ Cα } = δα . From κ+3 ⊆ N0 and C¯ ∈ N0 we can infer that α and thus Cα and β are members of N0 . Furthermore we have (Ni : i < β) ∈ Nβ+1 ⊆ Nα by the definition of an approximation sequence. Altogether we get {δi : i ∈ Cα ∩ β} ∈ Nα , hence ζ ∈ Nα since F ∈ N0 . By (1) we have ζ ≤ F (E) = δκ+3 < κ+4 . This yields the contradiction ζ < χNα (κ+4 ) = δα . Now we can prove one of the most important results of pcf-theory. Theorem 7.2.6 (Shelah) If a is a progressive set of regular cardinals and c is an interval of regular cardinal numbers such that c ⊆ pcf(a), then |c| ≤ |a|+3 . Proof Let us assume that the assertion of the theorem is false. Then there is an interval c ⊆ pcf(a) of regular cardinals of order type |a|+4 . Clearly a is infinite and c has no maximum. Since |a| < min(a) ≤ min(c) and |c| = |a|+4 , we can assume without loss of generality that c is progressive. Hence, by Lemma 3.3.3, all its members are successor cardinals. To get a contradiction, we

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Chapter 7. Local Properties

will construct a function F which satisfies properties (1) to (4) in Lemma 7.2.5 for κ := |a| and τ := κ+4 . Let Γ be the ∈-isomorphism from τ onto c. We define, for every X ⊆ τ ,  −1 Γ (max pcf(Γ[X])) if pcf(Γ[X]) ⊆ c, F (X) := τ otherwise. Property (1) is satisfied since we have pcf(Γ[X]) ⊆ pcf(Γ[Y ]) by Lemma 3.3.4, whenever X ⊆ Y ⊆ τ . Furthermore max pcf(Γ[X]) ≥ sup(Γ[X]) gives (2). For the proof of (3) assume that γ < τ and cf (γ) = κ+3 , and let λ be that cardinal with λ+ = Γ(γ). Then λ is a limit cardinal and λ = sup(Γ[γ]). This gives cf (λ) = cf (γ), and in particular λ is singular. By Theorem 7.2.3, there is a club C0 in λ all of whose elements are cardinals such that otp(C0 ) = cf (λ) = κ+3 and max pcf(C0+ ) = λ+ . Then we have μ ∈ c for every member μ of C0+ which is greater than min(c), and otp(C0+ ) = κ+3 . We put E := Γ−1 [C0+ ∩ c]. By Lemma 3.3.4, the maximum of pcf(C0+ ∩ c) is not less than λ. It is different from λ, since λ is singular, and at most λ+ , since C0+ ∩ c ⊆ C0+ . So we get F (E) = γ. Furthermore, E is a club in γ. To see this, we observe first that Γ−1 [C0+ ∩ c] is unbounded in γ since C0+ ∩ c is unbounded in λ = sup([Γ[γ]). We must further show for an arbitrary limit ordinals σ < γ in which E ∩ σ is unbounded that it is a member of E. We have cf (σ) < κ+3 since otp(E) = κ+3 , and Γ[E ∩ σ] is unbounded in the cardinal μ := sup(Γ[σ]). But μ is also the supremum of the predecessors of members of Γ[σ] and thus an element of C0 . Furthermore we have μ ∈ c since no member of c is of cofinality less than κ+3 . So we get Γ(σ) = μ+ ∈ C0+ ∩ c, and σ is a member of E. Finally we prove (4). Assume that X is a subset of τ of order type κ+ . As above, we have ν := max pcf(Γ[X]) > sup(Γ[X]). Since Γ[X] is progressive, Theorem 7.1.24 tells us that there is a set d ⊆ Γ[X], such that |d| ≤ |a| and ν ∈ pcf(d). With the regularity of κ+ we obtain an ordinal β < sup(X) such that Γ−1 [d] ⊆ β. So we can conclude, using (1) and the definition of F , that F (X ∩ β) ≥ F (Γ−1 [d]) ≥ sup(X), as desired. Now Lemma 7.2.5 implies τ < κ+4 which contradicts the definition of τ . Corollary 7.2.7 If a is a progressive interval of regular cardinals, then |pcf(a)| ≤ |a|+3 . Proof By Theorem 3.3.6, pcf(a) is an interval of regular cardinals. Theorem 7.2.6 says that |pcf(a)| ≤ |a|+3 .

Chapter 8 Applications of pcf-Theory As a first application of pcf-theory, we will prove in this section two estimates for cardinal numbers. Initially, if δ is a limit ordinal and μ is a cardinal with μ < ℵδ , then ℵμδ < ℵ(|δ|μ )+ . This result of Shelah generalizes the Galvin-Hajnal theorem to cardinals ℵδ with countable cofinality. Secondly we will show for limit ordinals δ that |δ|

ℵδ < max{ℵ|δ|+4 , (2|δ| )+ }. A further application of pcf-theory guarantees the existence of a J´ onssonAlgebra on the cardinal ℵω1 . If λ is a cardinal number, then we write p(λ) for the following assertion: There is a function f : [λ]2 −→ λ such that any subset A of λ of cardinality λ satisfies |{f (x) : x ∈ [A]2 }| = λ. A theorem of Todor˘cevi´c will be proved which says: If λ is a singular cardinal and μ < λ is a cardinal satisfying p(κ) for every regular cardinal κ ∈ [μ, λ)reg , then p(λ+ ). Finally, we will get the following theorem concerning the cofinalities of partial orderings of the form ([λ]≤κ , ⊆): If κ is an infinite cardinal number, α is an ordinal such that 1 ≤ α < κ, and λ := κ+α , then cf ([λ]≤κ , ⊆) = max pcf({κ+(β+1) : β < α}).

8.1 Cardinal Estimates Lemma 8.1.1 Let ℵδ be a limit cardinal and μ be an infinite cardinal. Further assume that α is an ordinal with α < δ such that ℵα is μ-strong and a := [ℵα , ℵδ )reg is a progressive interval of regular cardinals. Then ℵμδ < ℵα+|pcfμ (a)|+ .

250

Chapter 8. Applications of pcf-Theory

Proof Theorem 6.2.4 says that sup pcfμ (a) = ℵμδ , and Theorem 3.3.6 tells us that pcfμ (a) is an interval of regular cardinals. Furthermore we have min(a) = min pcfμ (a). It is obvious that any interval b of cardinals with min(b) = ℵα satisfies sup(b) < ℵα+|b|+ . Altogether we get the desired inequality. Theorem 8.1.2 If δ is a limit ordinal and μ is a cardinal with μ < ℵδ , then ℵμδ < ℵ(|δ|μ )+ . Proof Obviously the inequality is true if μ is finite. If δ = ℵδ , then |δ| = ℵδ , hence ℵμδ = |δ|μ < ℵ(|δ|μ )+ . So we can assume that δ < ℵδ and μ is infinite. If ℵδ ≤ 2μ , then ℵμδ ≤ 2μ ≤ |δ|μ < ℵ(|δ|μ )+ . Thus let us assume in addition that 2μ < ℵδ . We distinguish two cases. Case 1 ℵδ is μ-strong. Put ℵα := (|δ|μ )+ . Then we have ℵα < ℵδ , since |δ| ≤ δ < ℵδ , δ is a limit ordinal and ℵδ is μ-strong. Now a := [ℵα , ℵδ )reg is a progressive interval of regular cardinals since |a| ≤ |δ| < (|δ|μ )+ . Furthermore ℵα is μ-strong. With ℵα < ℵδ we get α < δ, and thus ℵμδ < ℵα+|pcfμ (a)|+ by Lemma 8.1.1. Theorem 3.4.13 says that |pcfμ (a)| ≤ |a|μ . So |a| ≤ |δ| gives |pcfμ (a)|+ ≤ (|δ|μ )+ . With α < δ we conclude that α < |δ|+ , hence α < (|δ|μ )+ . On the other hand, Lemma 1.5.9 says that every infinite cardinal is a principal number of addition, hence α+(|δ|μ )+ = (|δ|μ )+ . Now it is easy to check that ℵμδ < ℵ(|δ|μ )+ . Case 2 ℵδ is not μ-strong. Since 2μ < ℵμδ and μ < ℵδ , we can infer from Lemma 1.7.3 that there exists an ordinal γ < δ such that ℵγ is a μ-strong singular cardinal satisfying cf (γ) ≤ μ < ℵγ and ℵμδ = ℵμγ . If γ = ℵγ , then ℵμδ = ℵμγ = |γ|μ < ℵ(|γ|μ )+ ≤ ℵ(|δ|μ )+ . So we can assume that γ < ℵγ . If ℵγ ≤ 2μ , then ℵμγ ≤ 2μ ≤ |γ|μ < ℵ(|γ|μ )+ , and thus ℵμδ < ℵ(|δ|μ )+ . So let in addition 2μ < ℵγ . From Case 1 we get ℵμγ < ℵ(|γ|μ )+ , and thus ℵμδ < ℵ(|δ|μ )+ , as desired. Corollary 8.1.3 If ℵδ is a singular strong limit cardinal, then 2ℵδ < ℵ(2|δ| )+ . cf(ℵδ )

2ℵδ

Proof From Lemma 1.6.15 we get 2ℵδ = ℵδ < ℵ(|δ|cf(δ) )+ ≤ ℵ(2|δ| )+ .

cf(δ)

= ℵδ

. Thus we have

Example ℵℵα0 < ℵ(2ℵ0 )+ holds for all ordinals α < (2ℵ0 )+ . To see this, put δ := α + ω. Then cf (δ) = ω and |δ|ℵ0 ≤ (2ℵ0 )ℵ0 = 2ℵ0 . From Theorem 8.1.2 we can conclude that ℵℵα0 ≤ ℵℵδ 0 < ℵ(|δ|ℵ0 )+ ≤ ℵ(2ℵ0 )+ .

8.2. J´ onsson Algebras

251

Theorem 8.1.4 If δ is a limit ordinal, then |δ|

ℵδ < max{ℵ|δ|+4 , (2|δ| )+ }. In particular, we have 2ℵ0 < ℵω =⇒ ℵℵω0 < ℵω4 . |δ|

Proof If ℵδ ≤ 2|δ| , then ℵδ ≤ 2|δ| . So assume that 2|δ| < ℵδ . Then |δ| + (2 ) < ℵδ , hence a := [ℵα , ℵδ )reg with ℵα := (2|δ| )+ is a progressive interval |δ| of regular cardinals. Lemma 8.1.1 says that ℵδ < ℵα+|pcf|δ| (a)|+ , and from Corollary 7.2.7 we can conclude that |pcf(a)| < |a|+4 . |a| ≤ |δ| gives pcf|δ| (a) = pcf(a). Furthermore we have α < δ and thus α < |δ|+ < |δ|+4 . Every infinite cardinal is a principal number of addition, and therefore α + |δ|+4 = |δ|+4 . |δ| Altogether we get ℵδ < ℵ|δ|+4 . Remark a) If 2ℵ0 < ℵω , then ℵℵω0 is a regular cardinal.  To see this, put a := [ℵ1 , ℵω )CN . Then ℵℵω0 = | a| by Lemma 1.6.15. Hausdorff’s formula yields ℵℵ1 0 = ℵ1 · 2ℵ0 . |a| Now Corollary 6.2.5 says that So our assumption gives min(a) < sup(a). | a| = max pcf(a), and therefore ℵℵω0 is regular since it is the maximum of pcf(a). b) If 2ℵ0 = ℵ1 , then ℵℵω0 < ℵω2 by Theorem 8.1.2 and ℵℵω0 < ℵω4 by Theorem 8.1.4. On the other hand, if 2ℵ0 = ℵ4 , we get ℵℵω0 < ℵω5 from Theorem 8.1.2 and ℵℵω0 < ℵω4 from Theorem 8.1.4.

8.2 J´ onsson Algebras A J´onsson algebra is a structure (A, (fk : k < ω)) with countably many functions fk which has no proper substructure of the same cardinality. We will obtain results on J´ onsson algebras by using suitable elementary substructures of H(Θ) (for suitable cardinals Θ). So it is necessary to generalize the notions language, structure and model from Section 1.1 and Chapter 4. A language L of first order logic is given by its alphabet, its terms and its formulae1 . The alphabet of L consists of the following objects which we call the symbols of L: 1 To be precise, the following objects must actually be defined in ZF as sets, for example as members of Vω . The interested reader is referred to the book of F. R. Drake [Dr], in which he can find a detailed formalization of some first order languages.

252

Chapter 8. Applications of pcf-Theory variables vn for each natural number n; the logical symbols ¬, ∨, ∃ and = (the further symbols =⇒ , ⇐⇒ , ∧ and ∀ are introduced as abbreviations, as in Section 4.2); bracket symbols (, ); function symbols f0n0 , f1n1 , . . ., where n0 , n1 , . . . are natural numbers, and relation symbols Rn0 0 , R1n1 , . . ., where n0 , n1 , . . . are positive natural numbers.

We also use f n and Rn as syntactical variables for function symbols and relation symbols, respectively. n

n

If fj j and Rj j are function symbols and relation symbols, respectively, then nj is called the arity of these symbols. 0-ary function symbols are called constant symbols. An expression is a finite sequence of symbols. The concatenation of expressions is that sequence which results from writing the given expressions one behind the other. The terms of L are defined by recursion on the number of symbols: (T1) Every variable is a term, and every constant symbol is a term. n (T2) If nj > 0, t1 , . . . , tnj are terms, and fj j is a function symbol, then n fj j t1 . . . tnj is a term (resulting from the concatenation of the expresn sions fj j , t1 , . . . , tnj ). The formulae of L are also defined by recursion: n

(F1) If t1 and t2 are terms, then (t1 = t2 ) is a formula; if Rj j is a relation symn bol and t1 , . . . , tn are terms, then (Rj j t1 . . . tnj ) is a formula (resulting, analogously to (T2), from concatenation). (F2) If ϕ and ψ are formulae, then so are (¬ϕ), (ϕ ∨ ψ) and (∃vj ϕ) for any natural number j. A structure A for the language L consists of a) a nonempty set A, called the universe of A; b) a function assigning to each n-ary function symbol f of L, n > 0, a function fA from An into A; if f is a constant symbol, then it assigns to f a member fA of A; n ⊆ An . c) a function assigning to each n-ary relation symbol Rn a relation RA

Usually we write f and R instead of fA and RA . From the context it will be clear whether f denotes a function symbol or a function. For structures A we

8.2. J´ onsson Algebras use the notation

253

(A, (fi : i ∈ α), (Ri : i ∈ β)), 2

where α, β ≤ ω. The cardinality of a structure is the cardinality of its universe. If L has only finitely many function and relation symbols, then we characterize as usual a structure A as an n-tupel, listing its functions and relations. For example, the ordered field of real numbers is the structure (R, 0, 1, +, ·, <). If the language L has ∈ as nonlogical relation symbol, then, for every structure A for L, the binary relation assigned to ∈ is {(x, y) ∈ A × A : x ∈ y}. The free variables of a formula have already been defined in Section 4.2. If ϕ is a formula of the language L whose free variables occur in the sequence (v0 , . . . , vn−1 ), then we define by induction on the complexity of ϕ, for finite sequences (a0 , . . . , an−1 ) of elements of A, the relation A |= ϕ[a0 , . . . , an−1 ] 3 . For this, we need the interpretation of a term under a valuation (a0 , . . . , an−1 ), written as t[a0 , . . . , an−1 ] for terms t. t[a0 , . . . , an−1 ] is defined as a member of A by induction on the complexity of t as follows: a) If t is a variable vi with i < n, then t[a0 , . . . , an−1 ] := ai ; if t is a constant symbol f 0 , then t[a0 , . . . , an−1 ] := fA0 . b) If t is the term f k t1 . . . tk where f k is a function symbol and t1 , . . . , tk are terms, and if all variables occurring in t1 , . . . , tk are members of the sequence (v0 , . . . , vn−1 ), then t[a0 , . . . , an−1 ] := fAk (t1 [a0 , . . . , an−1 ], . . . , tk [a0 , . . . , an−1 ]). Now we can define the relation A |= ϕ[a0 , . . . , an−1 ]: a) If Rk is an k-ary relation symbol, and if t1 , . . . , tk are terms, then let A |= Rk t1 . . . tk [a0 , . . . , an−1 ] k . ⇐⇒ (t1 [a0 , . . . , an−1 ], . . . , tk [a0 , . . . , an−1 ]) ∈ RA

If t1 and t2 are terms, let A |= t1 = t2 [a0 , . . . , an−1 ] ⇐⇒ t1 [a0 , . . . , an−1 ] = t2 [a0 , . . . , an−1 ]. b) If ϕ and ψ are formulae of L, let A |= (¬ϕ)[a0 ,...,an−1 ] ⇐⇒ ¬(A |= ϕ[a0 ,...,an−1 ]), A |= (ϕ ∨ ψ)[a0 ,...,an−1 ] ⇐⇒ A |= ϕ[a0 ,...,an−1 ] ∨ A |= ψ[a0 ,...,an−1 ] and A |= ∃vn ϕ[a0 ,...,an−1 ] ⇐⇒ ∃a ∈ A(A |= ϕ[a0 ,...,an−1 ,a]). 2 or

(A, R, . . .) the footnote on page 178.

3 See

254

Chapter 8. Applications of pcf-Theory

With the usual abbreviations we obtain the assertions A |= (ϕ =⇒ ψ)[a0 , . . . , an−1 ], A |= (ϕ ∧ ψ)[a0 , . . . , an−1 ], A |= (ϕ ⇐⇒ ψ)[a0 , . . . , an−1 ] and A |= ∀vn ϕ[a0 , . . . , an−1 ], where, for example, the last one is the assertion ∀a ∈ A A |= ϕ[a0 , . . . , an−1 , a]. If ϕ is a formula of the language L, whose free variables are occuring in (v0 , . . . , vn−1 ) then we say that the formula ϕ holds in A iff A |= ∀v0 . . . ∀vn−1 ϕ is provable in ZFC. If α, β ≤ ω, and if A = (A, (gk : k < α), (Qk : k < β)) and B = (B, (fk : k < α), (Rk : k < β)) are structures for L, then A is called a substructure of B, written as A ⊆ B iff A is a subset of B and for all k the following holds: If nk is the arity of fk , Rk , then gk = fk |` Ank , Qk = Rk ∩ Ank respectively; in particular, for nk = 0, this means fk = gk . If A is a substructure of B, then A is called an elementary substructure of B, written as A  B, iff, for any formula ϕ of the language L whose free variables occur in the sequence (v0 , . . . , vn−1 ), we have ∀a0 ∈ A . . . ∀an−1 ∈ A(A |= ϕ[a0 , . . . , an−1 ] ⇐⇒ B |= ϕ[a0 , . . . , an−1 ]). Lemma 4.2.2 and the L¨ owenheim-Skolem theorem, Theorem 4.2.3, still hold for languages with countably many function and relation symbols. The proofs run completely analogous to those of Section 4.2. Definition A structure (A, (fk : k < ω)) with countably many functions fk is called a J´ onsson algebra iff it has no proper substructure of the same cardinality. J(κ) is the assertion “There exists a J´ onsson algebra of cardinality κ”. In this section we will deal with the question for which infinite cardinals κ we can prove J(κ). κ, λ and μ are syntactical variables for infinite cardinal numbers. Example Let fn = {(m, n) : m ∈ ω} for each n ∈ ω. Then (ω, (fn : n ∈ ω)) is obviously a J´ onsson algebra. This proves J(ℵ0 ). We prove first a criterion which yields a necessary and sufficient condition for the existence of J´ onsson algebras and will be used frequently. Definition If A is a structure for a language L and ϕ(v0 , . . . , vn ) is a formula of L, then we say that a function f from An into A is a Skolem function for ϕ in A iff, for all a1 , . . . , an ∈ A, A |= ∃v0 ϕ[a1 , . . . , an ] =⇒ A |= ϕ[f (a1 , . . . , an ), a1 , . . . , an ]. If B is a substructure of A such that, for each L-formula ϕ, there is at least one Skolem function for ϕ such that B is closed under this function, it is easy to check that B is an elementary substructure of A.

8.2. J´ onsson Algebras

255

Lemma 8.2.1 Let λ be an infinite cardinal. Then the following propositions are equivalent: (i) J(λ) . (ii) For any regular cardinal κ ≥ λ+ and any elementary substructure (M, ∈) of (H(κ), ∈) satisfying λ ∈ M and |M ∩ λ| = λ, we have λ ⊆ M . (iii) There is a regular cardinal κ ≥ λ+ and a structure H = (H(κ), ∈, . . .), such that, for all M  H, λ ∈ M ∧ |M ∩ λ| = λ =⇒ λ ⊆ M. Proof First we show that (ii) follows from (i). Let κ ≥ λ+ be a regular cardinal and (M, ∈) be an elementary substructure of (H(κ), ∈) such that λ ∈ M and |M ∩ λ| = λ. By Lemma 4.1.4, it is clear that any function from λn into λ and thus any J´ onsson algebra on λ is a member of H(κ). Using the absoluteness for H(κ) of the formula ϕ(x, y) ≡ “x is a J´onsson algebra on y”, we get H(κ) |= “There is a J´ onsson algebra on λ”. Now we obtain, together with Lemma 4.2.13, a J´onsson algebra A = (λ, (fk : k < ω)) on λ which is a member of M . Then (fk : k ∈ ω) ∈ M and thus fk ∈ M for every k ∈ ω since ω ⊆ M. Fix k ∈ ω. Since H(κ) |= fk : λnk −→ λ, M  H(κ) and fk , nk , λ ∈ M , we can conclude that M |= fk : λnk −→ λ. This implies fk ( α) ∈ λ ∩ M for all α  ∈ (λ ∩ M )nk , hence (B, (fk |` B nk : k < ω)) is a substructure of A where B := λ ∩ M . Since A is a J´onsson algebra, we can infer from |B| = |λ ∩ M | = λ that λ = B = λ ∩ M ⊆ M . Clearly (ii) implies (iii). There remains to be shown that (iii) implies (i). So let us assume (iii) and let H be a structure with universe H(κ) for an appropriate language L, for which ∈ is interpreted as usual. By the L¨ owenheimSkolem theorem, there is an elementary substructure M of H such that λ ∈ M , λ ⊆ M and |M | = λ. Choose a bijection g from λ onto M . Clearly g is a binary relation on M , and we identify it with the interpretation of a binary relation symbol of the language Lg which results from the language L by adjoining the symbol g to its alphabet. Claim The structure M := (M, ∈, g, . . .) has no proper elementary substructure of cardinality λ. Proof Consider an elementary substructure N  := (N, ∈, g, . . .) of M such that |N | = λ. Let ϕ(x) be the Lg -formula “x is the least ordinal such that dom(g) ⊆ x”. An easy calculation shows that M |= ϕ(x)[λ], hence M |= ∃x ϕ(x), and from Lemma 4.2.2 and N   M we can infer that there is a ∈ N such that M |= ϕ(a). Now it is easy to see that a = λ, hence λ ∈ N . M |= ∀x∃y g(y) = x gives N  |= ∀x∃y g(y) = x, and thus g[N ∩λ] = N . Therefore we

256

Chapter 8. Applications of pcf-Theory

have |N ∩ λ| = |g[N ∩ λ]| = |N | = λ. If we put N := (N, ∈, . . .), then N  M and consequently N  H. With assumption (iii) we can conclude that λ ⊆ N , and thus we have M = ran(g) = ran(g |`N ∩ λ) = N which proves the claim. Now we choose, for each formula ϕ of Lg , a Skolem function fϕ for ϕ in  M . Then (M, (fϕ : ϕ ∈ Lg )) is a J´onsson algebra. To see this, let (N, (fϕ : ϕ ∈ Lg )) be a substructure of (M, (fϕ : ϕ ∈ Lg )) such that |M | = |N |. Then the structure (N, ∈, g, . . .) is closed under Skolem functions and is thus an elementary substructure of (M, ∈, g, . . .). Now the claim yields N = M . Theorem 8.2.2

J(λ) =⇒ J(λ+ )

Proof We want to apply Lemma 8.2.1. So let us assume that M  H(λ++ ), λ ∈ M , and |M ∩ λ+ | = λ+ . We will show that λ+ is a subset of M and fix β ∈ λ+ . Clearly, there is some set C ∈ [M ∩ λ+ ]λ . Since λ+ is regular, we get α = sup(C ∪ {β}) ∈ λ+ . From |M ∩ λ+ | = λ+ we conclude that there is an ordinal α ∈ M ∩ λ+ with α > α , and hence α satisfies α > β and |M ∩ α| = λ. From λ+ ∈ M and Example 4.2.15 we can infer that λ ∈ M . By Lemma 4.2.13, we obtain a bijection g from λ onto α which is a member of M . Example 4.2.15 implies that g |` λ ∩ M is a bijection from λ ∩ M onto α ∩ M . So we get |λ ∩ M | = |α ∩ M | = λ. Now J(λ) and Lemma 8.2.1 yield λ ⊆ M , hence α = g[λ] = g[λ ∩ M ] = α ∩ M . So we have shown that β ∈ α ⊆ M , which gives λ+ ⊆ M , as desired. +

Theorem 8.2.3 (Tryba, Woodin) If λ is a regular uncountable cardinal number and if there exists a nonreflecting stationary subset of λ, then J(λ). Proof Again we want to apply Lemma 8.2.1. In assertion (iii) we choose κ = λ+ and assume that M  H(λ+ ), λ ∈ M and |M ∩ λ| = λ. We must show that λ ⊆ M . By Lemma 1.8.14, there exists a nonreflecting stationary subset S1 of λ. Then S1 ∈ H(λ+ ), and the considered property of S1 can be expressed by a formula which is absolute for H(λ+ ). Therefore we can find, using Lemma 4.2.13, a nonreflecting stationary subset S of λ which is a member of M . Now Solovay’s theorem, Theorem 1.8.13, says that there is a partition of S into λ stationary sets; it can be regarded as a function from λ into P(S). So let ϕ(x, y, z) be the formula “x is an injection with domain y such that ran(x) is a partition of z and all elements of ran(x) are stationary in y”. The Lemmas 4.1.3 and 4.1.4 say that H(λ+ ) |= ∃x ϕ[λ, S]. Consequently, we can find such a function g = (Sξ : ξ < λ) which is a member of M . Let C := {α < λ : sup(M ∩ α) = α}. We will see that C is a club in λ. If we can prove in addition, for any ξ < λ,

8.2. J´ onsson Algebras

257

Claim 1 C ∩ Sξ ⊆ M , then the rest of the proof is easy: Since C is a club and Sξ is stationary in λ, we can choose an ordinal α ∈ C ∩ Sξ ⊆ M ∩ Sξ . Since g ∈ M and α ∈ M , there is σ ∈ M such that α ∈ Sσ . From S = {Sζ : ζ ∈ λ} and the fact that the sets Sζ are pairwise disjoint, we can infer that σ = ξ ∈ M . So we have shown that λ ⊆ M . Now Lemma 8.2.1 yields the desired assertion. For the proof proof of Claim 1 let us assume that its assertion is false. Then there is α ∈ (C ∩ Sξ ) \ M . Since |M ∩ λ| = λ, the ordinal β := min(M \ α) exists, and we have α < β < λ. Note that β is a limit ordinal, since otherwise the immediate predecessor β  of β would be a member of M and would so be greater than α, contradicting the choice of β. Since S is nonreflecting and S ∈ M , there is a club Cβ in β such that Cβ is a member of M and S and Cβ are disjoint. Now we show that Cβ ∩ α is unbounded in α. Let σ < α. Since α ∈ C and thus sup(M ∩ α) = α, there is δ ∈ M ∩ α such that δ > σ. Since Cβ is a club in β and Cβ ∈ M , we obtain an ordinal  ∈ [δ, β) ∩ Cβ ∩ M . The choice of β gives  < α. Now we can conclude that α ∈ Cβ since Cβ is a club in β, which contradicts α ∈ Sξ ⊆ S and Cβ ∩ S = ∅. This completes the proof of Claim 1. There remains the proof of Claim 2 C is a club in λ. For the proof of Claim 2, consider a limit  ordinal γ < λ such that γ = sup(C ∩ γ). By the definition of C, we have (M ∩ α) = α for all α ∈ C. So we can conclude that      sup(M ∩ γ) = (M ∩ (C ∩ γ)) = (M ∩ α) = (C ∩ γ) = γ, α∈C∩γ

i.e., γ ∈ C. Therefore C is closed in λ. Furthermore, if σ ∈ λ, then we can find, using |M ∩ λ| = λ, a strictly increasing sequence (αi : i < ω) of elements of M ∩ λ such that σ < α0 . Then α := sup{αi : i < ω} ∈ λ, since cf (λ) > ℵ0 , and we can conclude that α = sup(M ∩ {αi : i < ω}) ≤ sup(M ∩ α) ≤ α. This gives α ∈ C and α > σ, and thus C is unbounded in λ. Corollary 8.2.4 For any regular cardinal number κ, we have J(κ+ ). Proof This is an immediate consequence of Theorem 8.2.3, since we obtain from Lemma 1.8.14 a nonreflecting stationary subset of κ+ . In the previous corollary we can omit the assumption that κ is regular if we add the generalized continuum hypothesis to our axioms of ZFC. To prove this, we need a preliminary combinatorial result.

258

Chapter 8. Applications of pcf-Theory

Theorem 8.2.5 (Erd¨ os, Hajnal, Rado) If 2κ = κ+ , then there is a function f : + [κ+ ]2 −→ κ+ such that f [[A]2 ] = κ+ for any A ∈ [κ+ ]κ . Proof By assumption we have |[κ+ ]κ | = (κ+ )κ = (2κ )κ = 2κ = κ+ , and so there exists a bijection g : κ+ −→ [κ+ ]κ . Consider a fixed ordinal α ∈ [κ, κ+ ) such that [α]κ ∩ g[α] = ∅. There α are bijections (αβ : β < κ) and ((Aα β , ηβ ) : β < κ) from κ onto α and onto κ ([α] ∩ g[α]) × α, respectively. We define by recursion for β < κ σβα := min{σ ∈ κ : ασ ∈ Aα β \ {ασγα : γ < β}}. The minimum exists since Aα β is a subset of α of cardinality κ, and it is obvious α that the sequence (σβ : β < κ) is injective. Now we put, for any {α, δ} ∈ [κ+ ]2 with δ < α,  α ηγ if α ≥ κ ∧ [α]κ ∩ g[α] = ∅ ∧ ∃γ ∈ κ δ = ασγα , f ({α, δ}) := 0 otherwise. If A ∈ [κ+ ]κ and η < κ+ , we take a member A of [A]κ and choose an ordinal α > max{sup(A ), g −1 (A ), η} such that α ∈ A. Then we have α ≥ κ, A ∈ [α]κ ∩ g[α], and η ∈ α. Therefore there is γ ∈ κ such that (A , η) = α α  (Aα γ , ηγ ). Putting δ := ασγα , we can conclude that δ ∈ Aγ = A ⊆ A and α 2 η = ηγ = f ({α, δ}) ∈ f [[A] ], as desired. +

Corollary 8.2.6 (GCH) =⇒ ∀κ ∈ ICN J(κ+ ). Proof Fix κ ∈ ICN. Since 2κ = κ+ , we can choose a function f : [κ+ ]2 −→ κ with the properties in Theorem 8.2.5. Let the function g : κ+ × κ+ −→ κ+ be defined by  f ({α, β}) if α = β, g(α, β) = 0 if α = β. +

If (A, g |` A2 ) is a substructure of (κ+ , g) and |A| = κ+ , then A ⊇ g[A2 ] ⊇ f [[A]2 ] = κ+ , hence A = κ+ . Therefore (κ+ , g) is a J´onsson algebra. If one replaces GCH by the stronger assumption V = L, then one can even show that, for any infinite cardinal κ, there is a J´ onsson algebra of cardinality κ. However, we will not present the proof of this theorem since we are mainly interested in results which are provable in ZFC. The preceding theorems say nothing about the provability in ZFC of J(μ+ ) for singular cardinal numbers μ. For such cardinal numbers, an application of the main lemma of pcf-theory will yield a sufficient criterion for the existence of a J´onsson algebra. In particular we will prove that there exists a J´ onsson algebra on ℵω+1 . We need two preliminary technical lemmas.

8.2. J´ onsson Algebras

259

 n Definition For any set A let ξA : n∈ω (A A) −→ ω be that function which assigns to each function f : An −→ A its arity ξA (f ) := n. Further let f  B := f [B n ], if ξA (f ) = n and B ⊆ A.  n Lemma 8.2.7 If A is a set and F is a countable subset  of {A A : n ∈ ω}, n then there is a countable set F  such that F ⊆ F  , F  ⊆ {A A : n ∈ ω}, and F  is closed under the composition of functions (the definition should be clear from the proof ). Proof We define by recursion a sequence (Fi : i < ω) of sets of functions. Put F0 := F . If Fn is already defined, then take Fn+1 as the union of the set Fn and the set of all functions  Ak −→ A (x1 , . . . , xk ) −→ f (f1 (x1 ), . . . , fξA (f ) (xξA (f ) )) such that f, f1 , . . . , fξA (f ) ∈ Fn , k = (ξA (fi ) : 1 ≤ i ≤ ξA (f )), and xi = (x1+ (ξA (fj ):j
where f : A (ξA (fi ):1≤i≤n) −→ A is a composition of the functions g, f1 , . . . , fn . So we get x ∈ B  , hence B  is closed under at least one Skolem function for

260

Chapter 8. Applications of pcf-Theory

each L-formula. This gives B   A. The formula v0 = v1 ensures that idA is a member of G and thus of F , which yields B ⊆ B  . Theorem 8.2.9 Assume that λ is a singular cardinal. If there exists a cardinal μ < λ such that J(κ) holds for all κ ∈ (μ, λ)reg , then there is a J´ onsson algebra on λ+ . λ

Proof Put Θ := (22 )+ . For the structure A := (H(Θ), ∈) we choose a set F of functions as in Lemma 8.2.8. We want to apply Lemma 8.2.1 and consider an elementary substructure (M, ∈, (f : f ∈ F )) of (H(Θ), ∈, (f : f ∈ F )) such that λ+ ∈ M and |M ∩ λ+ | = λ+ . Example 4.2.15 tells us that λ ∈ M . If we can show that λ+ ⊆ M , then Lemma 8.2.1 implies J(λ+ ). First let us note that λ+ ⊆ M if we know that λ ⊆ M : If α ∈ [λ, λ+ ), then |M ∩ λ+ | = λ+ yields an ordinal β ∈ M ∩ λ+ such that α < β. Since |β| = λ and λ, β ∈ M , there is a bijection e from λ onto β which is a member of M , and from λ ⊆ M we can conclude that β = e[λ] = e[λ ∩ M ] ⊆ M and thus α ∈ M , as desired. We will prove the assertion λ ⊆ M with the help of pcf-theory. By Lemma 3.3.5, there is a progressive set a of regular cardinals satisfying |a| = cf (λ) and sup(a) = λ and an ultrafilter D on a such that limD (a) = λ + ¯ and cf ( a/D) = λ+ . Let   h = (hξ : ξ < λ ) be a sequence of members of a which is cofinal in a modulo D. We choose a cardinal μ corresponding to the assumption of the theorem and assume without loss of generality, using ¯ and μ Lemma 3.3.5, that a ⊆ [μ, λ)reg . By the choice of Θ, the sets a, D, h are members of H(Θ). Now let ψ be that formula with the parameter λ ∈ M ¯ and μ with the above which formalizes the assertion that the objects a, D, h properties exist. As in Example 4.2.15 f) one can show, using Θ > 2sup(a) , that ψ is absolute for H(Θ). Consequently, since M is an elementary substructure of H(Θ), we can assume without loss of generality that these sets are members of M . Let A := {κ ∈ M ∩ a : sup(M ∩ κ) = κ}. If κ ∈ A, then we can infer from Lemma 8.2.1 and J(κ) that κ ⊆ M . Now, for the proof of λ ⊆ M , it suffices to show the following Claim A is unbounded in λ. Proof We assume, contrary to our hopes, that the assertion of the claim is false, and let λ := min{ν ∈ λ ∩ CN : A ⊆ ν}. By our assumption, this minimum exists, and we have λ < λ. Subclaim 1 ¬ (M ∩ a ⊆ λ ). Assume that the assertion of the subclaim is false. First we note that (M, ∈) |= ∃x(x ∈ y)[a], and thus we have M ∩ a = ∅, hence λ > 0 by our

8.2. J´ onsson Algebras

261

assumption that M ∩ a ⊆ λ . So there exists κ ∈ A. We can conclude that cf (λ) = |a| < min(a) ≤ κ ⊆ M , hence cf (λ) ⊆ M and cf (λ) ∈ M . If ϕ(x, y, z) is the formula “y, z ∈ ON and x : y −→ z is strictly increasing and ran(x) = z”, then we obtain, using H(Θ) |= ∃xϕ[cf (λ), λ], a strictly increasing function e : cf (λ) −→ λ which is a member of M and whose range is cofinal in λ. Since cf (λ) ⊆ M , we get ran(e) ⊆ M , and thus M ∩ λ is cofinal in λ. Furthermore, (M, ∈) |= “x is cofinal in y”[a, λ] since sup(a) = λ and a, λ ∈ M . Therefore M ∩a is cofinal in M ∩λ and thus cofinal in λ. This contradicts M ∩a ⊆ λ < λ. Now let  M  := {f  (M ∪ a) : f ∈ F }. The choice of F yields M ∪a ⊆ M  and M   H(Θ). We will show that there is a final segment a∩(λ , λ)reg of a such that the restriction of χM  to a∩(λ , λ)reg , extended to a function g on a by taking greater  the value zero for cardinals not  ¯ is cofinal than λ , has the property that g ∈ a. Together with the fact that h  in a, this will lead to a contradiction as follows. Let  sup(M  ∩ κ) if κ ∈ (λ , λ)CN , g(κ) := 0 otherwise,  ¯ yields an for every κ ∈ a. Let us assume that g ∈ a. Then the choice of h + + ordinal α < λ such that g ≤D hα . Since M ∩ λ is unbounded in λ+ , we can find an ordinal β ∈ M  ∩ (α, λ+ ), and so we have g α. By our supposition, we obtain a cardinal κ ∈ a ∩ (λ , λ)CN such that sup(M  ∩ κ) = κ. Since κ ∈ a ⊆ M  , we infer from J(κ) and Lemma 8.2.1 that κ ⊆ M  , hence α ∈ M  . So we get λ ⊆ M  . By Subclaim 1, there exists a cardinal κ ∈ M ∩ a ∩ [λ , λ)CN . Then κ ∈ A, hence sup(M ∩ κ) < κ. We will show that sup(M  ∩ κ) ≤ sup(M ∩ κ). On the other hand, we have κ < λ ⊆ M  , hence κ ⊆ M  . This will yield κ = sup(M  ∩ κ) ≤ sup(M ∩ κ), contradicting sup(M ∩ κ) < κ, and complete the proof of Subclaim 2 and thus of the theorem. So fix α ∈ M  ∩ κ. By the definition of M  , there is a function f : H(Θ)n −→ H(Θ) which is a member of F , and there are s1 , . . . , sn ∈ M ∪ a such that α = f (s1 , . . . , sn ). Put S := {s1 , . . . , sm } := {s1 , . . . , sn } ∩ M . Then

262

Chapter 8. Applications of pcf-Theory

|f  (S ∪ a) ∩ κ| ≤ |a| < min(a) ≤ κ, hence f  (S ∪ a) ∩ κ is bounded in κ. Let ϕ ≡ ϕ(x, y, z, v1 , . . . , vm ) be the formula x ∈ y ∧∀w1 ...∀wn (

n  i=1

(wi ∈ z ∨

m 

wi = vj )∧f (w1 ,...,wn ) ∈ y ⇒ f (w1 ,...,wn ) ∈ x).

j=1

Using (H(Θ), ∈, F ) |= ∃x ϕ[κ, a, s1 , . . . , sm ] and (M, ∈, F )  (H(Θ), ∈, F ), we obtain an ordinal αS,f ∈ M ∩ κ which is a bound for f  (S ∪ a) in M ∩ κ. Now α ∈ f  (S ∪ a) gives α < αS,f ≤ sup(M ∩ κ), and altogether we get sup(M  ∩ κ) ≤ sup(M ∩ κ), as desired. Corollary 8.2.10 There exists a J´ onsson algebra on ℵω+1 . Proof In the example preceding Lemma 8.2.1, we have proved J(ℵ0 ). From Theorem 8.2.2 we get J(ℵn ) for all n ∈ ω. Now the previous theorem, 8.2.9, guarantees for λ = ℵω and μ = ℵ0 the existence of a J´onsson algebra on λ+ = (ℵω )+ = ℵω+1 .

8.3 A Partition Theorem of Todor˘cevi´c Definition If λ is an infinite cardinal, then p(λ), also written as λ → [λ]2λ , is the following assertion: There is a function c : [λ]2 −→ λ such that, for any subset X of λ of cardinality λ, {c(x) : x ∈ [X]2 } = λ. Remark Theorem 8.2.5 says that p(λ) holds if λ = 2κ = κ+ for some cardinal κ. Furthermore, we can conclude from p(λ) that there is a J´onsson algebra on λ, namely, the algebra (λ, g) where, choosing c as in the definition of p(λ), the function g : λ × λ −→ λ is defined by g(α, β) := c({α, β}) for α = β, and g(α, α) := 0. To see that (λ, g) is a J´ onsson algebra, assume that (X, g |`X ×X) is a substructure of (λ, g) such that |X| = λ. Then X ⊇ {c(x) : x ∈ [X]2 } = λ. Lemma 8.3.1 Let λ be an infinite cardinal and c : [λ+ ]2 −→ λ be a function such that, for any subset X of λ+ of cardinality λ+ , we have {c(x) : x ∈ [X]2 } = λ. Then p(λ+ ). Proof To realize λ+ → [λ+ ]2λ+ , we choose for each β < λ+ an injection iβ : β −→ λ, and define the function c∗ : [λ+ ]2 −→ λ+ by c∗ (α, β) = i−1 β (c(α, β)), + ∗ if α < β < λ and c(α, β) ∈ ran(iβ ), c (α, β) = 0 otherwise. Consider X ⊆ λ+ with |X| = λ+ , and δ < λ+ . We search for α, β ∈ X such that α < β and c∗ (α, β) = δ. If Y := {β ∈ X : δ < β}, then |Y | = λ+ and {iβ (δ) : β ∈ Y } ⊆ λ. Since λ+ is regular, there is an ordinal η < λ such that |{β ∈ Y : iβ (δ) = η}| = λ+ . By assumption, we can find α, β ∈ Y ⊆ X such

8.3. A Partition Theorem of Todor˘cevi´c

263

∗ that α < β and c(α, β) = η = iβ (δ). So we get δ = i−1 β (c(α, β)) = c (α, β), as desired.

Theorem 8.3.2 Assume that there is a strictly increasing sequence (κn : n ∈ ω) of regular cardinals such that p(κn ) holds for all n ∈ ω, and let a := {κn : n ∈ ω} and λ := sup(a). If λ+ ∈ pcf(a), then p(λ+ ). Proof Let h := (κn : n ∈ ω) and, without loss of generality, κ0 > ω. Since λ+ ∈ pcf(a), we  know from Lemma 3.3.1 that there is an ultrafilter D on ω such that cf ( h/D) = λ+ . Clearly D is nonprincipal. We choose a  sequence (fξ : ξ < λ+ ) of members of h which is cofinal in h modulo D and assume further that, for each n ∈ ω, cn : [κn ]2 −→ κn is a function satisfying the property from the definition of p(κn ). To apply Lemma 8.3.1, we define a function c : [λ+ ]2 −→ λ and show that, for every subset X of λ+ of cardinality λ+ , c takes on [X]2 each value δ < λ: If α < β < λ+ and n is the least natural number k such that fα (k) = fβ (k), then we put c(α, β) := cn (fα (n), fβ (n)). Now consider a fixed subset Xof λ+ satisfying |X| = λ+ and δ < λ. Then Z = {fξ : ξ ∈ X} is unbounded in h modulo D. Since each cardinal κm is regular, there must be n ∈ ω such that δ < κn and |{fξ (n) : ξ ∈ X}| = κn . Otherwise, we could easily get a bound for Z: Since D is a nonprincipal ultrafilter, we have d := {n ∈ ω : δ <  κn } ∈ D. The supposition sup{fξ (n) : ξ ∈ X} < κn for all n ∈ d leads to g ∈ h, where g is defined by  sup{fξ (n) + 1 : ξ ∈ X} if n ∈ d, g(n) = 0 if n ∈ ω \ d.  g would be an  upper bound of Z modulo D in h which does not exist. Since | {κi : i < n}|  = κn−1 < κn , we obtain, together with the regularity of κn , a function s ∈ {κi : i < n} such that |{fξ (n) : s ⊆ fξ and ξ ∈ X}| = κn . Using p(κn ), we obtain ordinals α, β ∈ X such that α < β, cn (fα (n), fβ (n)) = δ, s ⊆ fα , and s ⊆ fβ . So we have n = min{k ∈ ω : fα (k) = fβ (k)}, hence c(α, β) = δ, which proves the theorem. Theorem 8.3.3 (Todor˘cevi´c) Assume that λ is a singular cardinal. If there is a cardinal μ with μ < λ such that p(κ) holds for every regular cardinal κ ∈ [μ, λ), then p(λ+ ). Proof 4 By Lemma 8.3.1 it suffices to find a function c : [λ+ ]2 −→ λ which maps, for any subset X of λ+ of cardinality λ+ , [X]2 onto λ. Lemma 3.3.5 4 We

follow the proof in [BM].

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Chapter 8. Applications of pcf-Theory

|a| = cf (λ) says that there is a set a ⊆ [μ, λ)reg of regular cardinals with  and |a|+ < min(a) and an ultrafilter D on a such that λ+ = cf ( a/D) and limD (a) = λ. Let I := Ib (a). With Lemma 3.4.9 we get J<λ (a) ⊆ I. Since λ is singular, we have J<λ+ (a) = J<λ (a), hence J<λ+ (a) ⊆ I.  Claim 1 tcf ( a/I |`b) = λ+ for some b ∈ D \ I. limD (a)= λ gives I ∩ D = ∅ and thus J<λ+ (a) ∩ D = ∅. Since ( a, <J<λ+ (a) ) is λ+ -directed, we obtain by Lemma 3.4.16 a set  b ∈ D such that tcf ( a/J<λ+ (a) |` b) = λ+ . Then b ∈ /I, and together with J<λ+ (a) ⊆ I we conclude, using Lemma 3.1.7, that tcf ( a/I |` b) = λ+ . This proves Claim 1.   Now fix b ∈ D \ I such that tcf ( b/I) := tcf ( b/I ∩ P(b)) = λ+ . For technical reasons we need the condition

(∗) (ν : ν ∈ b ∩ κ) < κ for all κ ∈ b . + If κ ∈ b, then κ ∈ a, and κ is regular. Thus |b| ≤ |a| < |a| < min(a) ≤ κ yields (ν : ν ∈ b ∩ κ) < κ. Choose, for each κ ∈ b, a function cκ : [κ]2 −→ κ which  witnesses p(κ). Further let (fξ : ξ < λ+ ) be a sequence which is cofinal in b modulo I. We define the desired function c as follows:

If α < β < λ+ , and if there is a greatest cardinal κ ∈ b such that fα (κ) > fβ (κ), then we put c(α, β) := cκ (fα (κ), fβ (κ)). Otherwise, we let c(α, β) := 0. Consider X ⊆ λ+ with |X| = λ+ , and let δ < λ. We search for α, β ∈ X such that α < β and c(α, β) = δ. Choose, if possible, for each ν ∈ b and each γ < ν a least  ordinal α = α(ν, γ) ∈ X such that fα (ν) = γ, and let the function g ∈ b be defined by g(κ) := sup{fα (κ) : ∃γ∃ν ∈ b (γ < ν < κ ∧ α = α(ν, γ))}. Fix ν ∈ b. Then |{α(ν, γ) : γ < ν}| ≤ ν. Further we conclude from (∗) that (ν : ν ∈ b ∩ κ) < κ. Thus in the definition of g(κ) we take  the supremum of a set of cardinality less than κ, and we get g(κ) ∈ κ and g ∈ b. Claim 2 There is a set Y ⊆ X with |Y | = λ+ , and a cardinal κ0 ∈ b with κ0 > δ, such that g(κ) < fξ (κ) for all ξ ∈ Y and all κ ≥ κ0 . g is majorized under
8.4. Cofinalities of Partial Orderings ([λ]≤κ , ⊆)

265

cardinal μξ ∈ b satisfying μξ > δ. Since |Y1 | = λ+ , we obtain by the pigeonhole principle a subset Y of Y1 with |Y | = λ+ , such that μξ =: κ0 ∈ b for all ξ ∈ Y . Now, if ξ ∈ Y and κ ≥ κ0 , then κ ∈ / Bξ . This gives g(κ) < fξ (κ), and Claim 2 is proved. For each κ ∈ b and each ξ < λ+ , we put Aξ (κ) := {fζ (κ) : ζ ∈ Y ∧ ξ < ζ} . Claim 3 There is a cardinal κ1 ∈ b with κ1 ≥ κ0 such that, for each ξ < λ+ , the set Aξ (κ1 ) is cofinal in κ1 . We give an indirect proof for the claim and assume that, for each κ ≥ κ0 with κ ∈ b, there is an ordinal ξ(κ) < λ+ such that Aξ(κ) (κ) is bounded in κ. Put ξ ∗ = sup{ξ(κ) : κ ≥ κ0 }. Then ξ ∗ < λ+ , since |b| = cf (λ) < λ+ . Therefore Aξ (κ) is bounded in κ for all ξ ≥ ξ ∗ and all κ ≥ κ0 with κ ∈ b. Assume that γ(κ) is a bound for Aξ∗ (κ) for each κ ∈ b \ κ0 , and let h(κ) := γ(κ) + 1 for κ ∈ b \ κ0 and h(κ) := 0 for κ ∈ b ∩ κ0 . If ξ > ξ ∗ and ξ ∈ Y , then {κ ∈ b : fξ (κ) ≥ h(κ)} ⊆ κ0 ∩ b ∈ I, hence fξ ξ ∗ }| = λ+ , and consequently (fξ : ξ ∗ < ξ ∧ ξ ∈ Y ) is a cofinal sequence in b modulo I. This contradiction proves Claim 3. Now we choose κ1 according to Claim 3. Then in particular we get |Aξ (κ1 )| = κ1 < λ for all ξ < λ+ . Furthermore, the sequence (Aξ (κ1 ) : ξ < λ+ ) is decreasing under the relation ⊆. Therefore there is ξ1 < λ+ , such that Aξ (κ1 ) =: A for all ξ ≥ ξ1 . Since δ < κ0 ≤ κ1 , the choice of cκ1 yields γ1 , γ2 ∈ A with γ1 < γ2 < κ1 and cκ1 (γ2 , γ1 ) = δ. From γ2 ∈ A = Aξ1 (κ1 ) we conclude that there is ξ2 ∈ Y satisfying ξ1 < ξ2 and γ2 = fξ2 (κ1 ). So there exists α = α(κ1 , γ2 ) ∈ X, and we have γ2 = fα (κ1 ). Since γ1 ∈ A = Aξ2 (κ1 ), there is β > ξ2 ≥ α with β ∈ Y such that fβ (κ1 ) = γ1 . It remains to be shown that κ1 is the greatest cardinal κ ∈ b with fα (κ) > fβ (κ), since in this case we can conclude that δ = cκ1 (fα (κ1 ), fβ (κ1 )) = c(α, β), as desired. So let κ > κ1 . Then κ0 ≤ κ1 < κ. From the definition of g, β ∈ Y and Claim 2 we infer that fα (κ) ≤ g(κ) < fβ (κ). This completes, together with Lemma 8.3.1, the proof of the theorem.

8.4 Cofinalities of Partial Orderings ([λ]≤κ , ⊆) In this section we will investigate the cofinalities of partial orderings of the form ([λ]≤κ , ⊆). We will prove the following theorem: If κ is an infinite cardinal number, α is an ordinal such that 1 ≤ α < κ, and λ := κ+α , then cf ([λ]≤κ , ⊆) = max pcf({κ+(β+1) : β < α}). First we present some simple properties of these cofinalities.

266

Chapter 8. Applications of pcf-Theory

Lemma 8.4.1 If λ and κ are cardinals such that ω ≤ κ ≤ λ, then max{cf ([λ]≤κ , ⊆), 2κ } = λκ . Proof Obviously it suffices to prove the relation “≥”. Assume  that S is a subset of [λ]≤κ which is cofinal in [λ]≤κ under ⊆. Then [λ]≤κ = {[S]≤κ : S ∈ S}, and Corollary 1.6.6 gives

|[S]≤κ | ≤ |S| · 2κ . λκ = |[λ]≤κ | ≤ S∈S

Hereby we get immediately the desired assertion. Lemma 8.4.2 For any cardinal λ, cf ([λ]≤λ , ⊆) = 1. Lemma 8.4.3 Assume that λ is an infinite cardinal and κ ∈ [1, λ)CN . Then cf ([λ]≤κ , ⊆) ≥ λ. Proof Let us assume, to get a contradiction, that there is a setP which satisfies P ⊆ [λ]≤κ and |P| < λ and is cofinal in [λ]≤κ under ⊆. Then | P| < λ, and so there exists x ∈ λ \ P. However, there is no P ∈ P such that {x} ⊆ P . Lemma 8.4.4 If κ and λ are cardinals such that ω ≤ κ ≤ λ, then cf ([λ+ ]≤κ , ⊆) ≤ max{λ+ , cf ([λ]≤κ , ⊆)}. Proof Choose, for each α ∈ [λ, λ+ )ON , a set Pα cofinal in [α]≤κ under ⊆ such that |Pα | = cf ([λ]≤κ , ⊆), and put  P := {Pα : α ∈ [λ, λ+ )ON }. Obviously we have |P| = max{λ+ , cf ([λ]≤κ , ⊆)}. We show that P is cofinal in [λ+ ]≤κ . If X ⊆ λ+ and |X| ≤ κ, then X ⊆ α for some α ∈ [λ, λ+ )ON , since λ+ is regular. Now there exists a set P ∈ Pα ⊆ P such that X ⊆ P . Corollary 8.4.5 If λ is an infinite cardinal and 1 ≤ n < ω, then cf ([λ+n ]≤λ , ⊆) = λ+n . Lemma 8.4.6 Assume that κ and λ are cardinals such that ω < κ+ < λ. Then cf ([λ]≤κ , ⊆) ≤ cf ([λ]≤κ , ⊆). +

8.4. Cofinalities of Partial Orderings ([λ]≤κ , ⊆)

267

Proof Choose P cofinal in [λ]≤κ under ⊆, and let without loss of general+ ity |P | = κ+ for all P ∈ P. Further assume that, for each P ∈ P, (xP α :α<κ ) + P P is a bijection from κ onto P . We put Mα := {xβ : β < α} for each P ∈ P and each α < κ+ , and +

M := {MαP : P ∈ P ∧ α < κ+ }. Then |M| ≤ max{|P|, κ+ }, and from Lemma 8.4.3 and κ+ < λ we can conclude that |M| ≤ |P|. We show that M is cofinal in [λ]≤κ . Fix X ⊆ λ with |X| ≤ κ. Then X ⊆ P for some P ∈ P. Since κ+ is regular, there exists α < κ+ such that X ⊆ MαP ∈ M. Remark The previous lemma is false for κ+ = λ, as can easily be inferred from Lemmas 8.4.2 and 8.4.3. Theorem 8.4.7 Assume that κ is an infinite cardinal, α is an ordinal such that 1 ≤ α < κ, and λ := κ+α . Then cf ([λ]≤κ , ⊆) = max pcf({κ+(β+1) : β < α}). Proof 5 Put a := {κ+(β+1) : β < α}. Obviously, a is progressive and thus an interval of regular cardinals, since any limit cardinal between κ and λ has a cofinality less than α. We prove first the relation “≥”. Let P be cofinal in [λ]≤κ . Let fP (ν) := sup(P ∩ ν)  for each P ∈ P and any ν ∈ a. |P | ≤ κ < min(a) P. gives fP ∈ a for all P ∈ By Theorem 3.4.21 it suffices to show that cf ( a, ≤a ) ≤ |P|. So fix f ∈ a. Since |a| < κ, we have ran(f ) ∈ [λ]≤κ , and thus ran(f ) ⊆ P for some P ∈ P. So we get f (ν) ≤ sup(P ∩ ν) = fP (ν) for all ν ∈ a, hence f ≤ fP . Now we turn to the proof of the reverse inequality. Without loss of generality we can assume that α ≥ ω, since otherwise max pcf(a) = max(a) = λ and cf ([λ]≤κ , ⊆) = λ by Corollary 8.4.5. If μ := |α|+ , then μ ∈ (|a|, min(a))reg . sup(a) and a control function F = (Fb : Choose a regular cardinal Θ > 22 ¯ where h ¯ is a special sequence for a and μ. We b ⊆ a) for a generated by h, define G := {f ∈ Fa : ∀ν ∈ a cf (f (ν)) = μ}. 5 In

this proof we use ideas of S. Shelah communicated in an e-mail to one of the authors.

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Chapter 8. Applications of pcf-Theory

For each ξ < λ with cf (ξ) = μ, we choose a club Cξ in ξ of order type μ. Furthermore, we choose for each f ∈ G an elementary substructure Nf of H(Θ) such that |Nf | = κ and  κ ∪ {κ} ∪ a ∪ {Cf (ν) : ν ∈ a} ⊆ Nf . Lemma 6.1.4 says that |G| ≤ max pcf(a). So it is enough to show that the set {Nf ∩ λ : f ∈ G} is cofinal in [λ]≤κ . For this, fix X ⊆ λ with |X| ≤ κ. Further let (Mi : i ≤ μ) be an approximation sequence in H(Θ) such that |Mi | = κ for all i < μ and ¯ a} ⊆ M0 . X ∪ κ ∪ {κ} ∪ a ∪ {h, Theorem 6.1.5 tells us that f := χMμ |` a ∈ Fa . By Lemma 4.3.3, there is, for each ν ∈ a, a club Dν ⊆ Mμ ∩ ν of order type μ in f (ν) – in particular we get f ∈ G. Since μ is uncountable, we can assume without loss of generality that Dν ⊆ Cf (ν) ; otherwise we can take the intersection of the clubs. Now we put a∗ := {κ} ∪ a and Dκ := ∅. Then a∗ \ {min(a∗ )} is an interval of regular cardinals and (Dν : ν ∈ Mμ ∩ a∗ ) is a sequence of subsets of Nf characterizing Mμ over a∗ . From Lemma 6.2.2 we can conclude that X ⊆ Mμ ∩ λ ⊆ Nf ∩ λ, as desired. Remark Lemma 8.4.1 says that ℵℵω1 = 2ℵ1 · cf ([ℵω ]≤ℵ1 , ⊆). Therefore we could determine ℵℵω1 if we could “compute” 2ℵ1 and cf ([ℵω ]≤ℵ1 , ⊆). The first cardinal cannot be estimated by the results of P. J. Cohen, but Shelah gives an estimate for the second cardinal: Theorem 8.4.7 tells us that cf ([ℵω ]≤ℵ1 , ⊆) = max pcf({ℵn+2 : n < ω}). By Theorem 3.3.6, pcf({ℵn+2 : n < ω}) is an interval of regular cardinals, and from Corollary 7.2.7 we get |pcf({ℵn+2 : n < ω})| < ℵω4 . So we can conclude that cf ([ℵω ]≤ℵ1 , ⊆) < ℵω4 . Altogether we can infer that ℵℵω1 < max{(2ℵ1 )+ , ℵω4 }. On the other hand, Lemma 8.4.6 says that cf ([ℵω ]≤ℵ0 , ⊆) ≤ cf ([ℵω ]≤ℵ1 , ⊆), hence cf ([ℵω ]≤ℵ0 , ⊆) < ℵω4 . Therefore, if we want to estimate ℵℵω0 , we have to estimate 2ℵ0 which is impossible, and the cardinal cf ([ℵω ]≤ℵ0 , ⊆) which is possible by Shelah’s results.

Chapter 9 The Cardinal Function pp(λ) In this chapter, we introduce Shelah’s cardinal function ppκ (λ) whose properties we now summarize. If λ is a singular cardinal, then the definition of ppκ (λ) gives ppκ (λ) ≤ λκ for any cardinal κ with cf (λ) ≤ κ < λ. For singular cardinals λ with uncountable cofinality and κ = cf (λ) which are in addition κ-strong, we get ppκ (λ) = λκ . In particular ppκ (λ) = 2λ holds for every strong limit cardinal λ with uncountable cofinality κ. When κ = cf (λ), we put pp(λ) := ppκ (λ). So the function pp(λ) takes as values, for certain cardinals, the values of the Gimel function and of the continuum function, respectively. In analogy to the Galvin-Hajnal theorem we will prove that pp(ℵδ ) < ℵ(|δ|cf(δ) )+ for singular cardinals ℵδ . Analogously to Silver’s theorem we get: If λ is a singular cardinal with uncountable cofinality, and if the set {μ < λ : pp(μ) = μ+ } is stationary in λ, then pp(λ) = λ+ . Assume that κ and λ are cardinals and cf (λ) ≤ κ < λ. Then λ will be called ppκ -strong if ppκ (ν) < λ for all singular cardinals ν < λ with cf (ν) ≤ κ < ν. If we rewrite the lemma of Galvin and Hajnal for pp(λ), it reads as follows: If λ is a singular cardinal with uncountable cofinality κ which is ppκ strong, (λξ : ξ < κ) is a normal function cofinal in λ, and Φ ∈ κ ON is a function +Φ(ξ) such that ppκ (λξ ) = λξ for all ξ < κ, then pp(λ) ≤ λ+ Φ Id (κ) .

270

Chapter 9. The Cardinal Function pp(λ)

9.1 pp(λ) and the Theorems of Galvin, Hajnal and Silver As usual, we use κ and λ in this section as syntactical variables for infinite cardinals. We remind the reader on the definition of the cardinals ppκ (λ) and pp+ (λ). Assume that λ is a singular cardinal such that cf (λ) ≤ κ < λ, and let κ A be the set  {cf ( a/D) : a ⊆ Reg ∧ sup(a) = λ ∧ |a| ≤ κ ∧ D ultrafilter on a ∧ D ∩ Ib (a) = ∅}. + + Then ppκ (λ) = sup(A) and pp+ κ (λ) = sup (A), where sup (A) is the cardinal min{κ ∈ CN : ∀ν ∈ A ν < κ}. If κ = cf (λ), then pp(λ) = ppκ (λ) and pp+ (λ) = pp+ κ (λ).  Let κ and λ be given as above. Then we get cf ( a/D) ≤ |κ λ| = λκ for all a and D in question. If μ is a cardinal with μ < λ and a satisfies the properties in the definition of A, then we know from Lemma 3.3.5 that we can assume without loss of generality that μ < min(a); in particular we can always assume that a is progressive. Now we choose a as a set of regular cardinals of cardinality cf (λ) and cofinal in λ and  D as an ultrafilter of final segments on a. Then Lemma 3.3.4 says that cf ( a/D) > λ. So we can conclude that

λ < ppκ (λ) ≤ λκ . The following Lemma yields a characterization of the cardinal ppκ (λ) with the help of ideals. Lemma 9.1.1 If cf (λ) ≤ κ < λ, then  ppκ (λ) = sup{tcf ( f /I) : f ∈ κ λ ∧ ran(f ) ⊆ Reg ∧ limI (f ) = λ}.  Proof Put pp∗κ (λ) := sup{tcf ( f /I) : f ∈ κ λ ∧ ran(f ) ⊆ Reg ∧ limI (f ) = λ}. If, according to the definition of ppκ (λ), a ⊆ λ ∩ Reg, sup(a) = λ, |a| ≤ κ, D is an ultrafilter on a, and D ∩ Ib (a) = ∅, then we can assume without loss of generality that κ < min(a). Let f be a surjection from κ onto a and J be the dual ideal of the ultrafilter D. Note that f is progressive and I := {f −1 [X] : X ∈ J} is an ideal on κ such [I]. Then we  have limI (f ) = λ  that J = f and, using Lemma 3.3.1, cf ( a/D) = tcf ( a/J) = tcf ( f /I). So we get ppκ (λ) ≤ pp∗κ (λ).

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271

Conversely, if f ∈ κ λ is a function satisfying ran(f ) ⊆ Reg and limI (f ) = λ, then c := {ξ < κ : f (ξ) ≤ κ} isa member ofI. Putting g := f |` (κ \ c) and I  := I ∩ P(κ \ c), we obtain tcf ( f /I) = tcf ( g/I  ) from Lemma 3.1.8. For a := ran(g), we can conclude that a ⊆ Reg, sup(a) = λ, and |a| ≤ κ; Ib (a) ⊆ g[I  ] follows from limI (f ) = λ. By construction,  and conse g is progressive, quently we can infer from Lemma 3.3.1 that tcf ( g/I  ) = tcf ( ran(g)/g[I  ]).  Now consider an arbitrary ultrafilter D on a  such that D ∩ g[I we   ] = ∅. Then   (a) = ∅ and tcf ( f /I) = tcf ( g/I ) = tcf ( ran(g)/g[I ]) = have D ∩ I b  cf ( a/D). This yields pp∗κ (λ) ≤ ppκ (λ), as desired. Theorem 9.1.2 Assume that λ is a cardinal, cf (λ) ≤ κ < λ, and λ is κ-strong and not a fixed point of the aleph function. Then ppκ (λ) = λκ . Proof If λ =: ℵδ , then the assumption gives δ < ℵδ . Furthermore, the cardinal μ := (|δ|κ )+ is less than λ and κ-strong. Let a := [μ, λ)reg . Then |a| ≤ |δ| < μ, and thus a is a progressive interval of regular cardinals such that min(a) is κ-strong and sup(a) = λ. From Theorem 6.2.4 we infer that sup pcfκ (a) = λκ . Now we show that sup pcfκ (a) = ppκ (λ), which will prove the theorem. Note that we can infer from the fact that λ is κ-strong that max  pcf(d) < λ for every d ∈ Ib (a) with |d| ≤ κ: If ν < λ satisfies d ⊆ ν, then | d| ≤ ν |d| < λ. For the proof of the relation “≤”, let c ∈ [a]≤κ . If c ∈ Ib (a), then max pcf(c) < λ < ppκ (λ). So we can assume that sup(c) = λ. Consider a fixed ultrafilter D on c. We distinguish two cases. If D ∩ Ib (c) = ∅,choose a 3.1.8, we have cf ( c/D) = set  d ∈ D ∩ Ib (c) and observe that, using Lemma  cf ( d/D ∩ P(d)) < λ. If D ∩ Ib (c) = ∅, then cf ( c/D) is a member of that set in the definition of ppκ (λ) whose supremum is ppκ (λ). So we have shown that sup pcfκ (a) ≤ ppκ (λ). For the proof of the relation “≥” assume that c ⊆ Reg, |c| ≤ κ, sup(c) = λ, and D is an ultrafilter on c such that D ∩ Ib (c) = ∅. Let b := c ∩ μ and ν ∈ pcf(b). Taking pcf(c) = pcf(c\b)∪pcf(b) and max pcf(b) < λ into account, we can assume without loss of generality that c ⊆ [μ, λ)reg = a, which gives ppκ (λ) ≤ sup pcfκ (a). This raises the question of whether Theorem 9.1.2 is true without the assumption that λ is not a fixed point of the aleph function. We can answer it positively if we accept the assumption that λ has uncountable cofinality. Theorem 9.1.3 Let λ be a singular cardinal. If λ has the uncountable cofinality κ and is κ-strong, then ppκ (λ) = λκ .

272

Chapter 9. The Cardinal Function pp(λ)

Proof At the beginning of this section we noted that ppκ (λ) ≤ λκ . For the proof of the reverse inequality, consider a cardinal μ ∈ (λ, λκ ]CN . Without loss of generality we can assume that μ is regular, since any singular cardinal is the supremum of a set of regular cardinals. By Lemma 3.1.15, there is a function  f ∈ κ λ and an ideal I on κsuch that limI (f ) = κ and μ = tcf ( f /I). Theorem 3.1.13 says that tcf ( cf ◦ f /I) = μ. To show that limI (cf ◦ f ) = λ, let us suppose that there is a cardinal ν < λ such that X := {i < κ : cf (f (i)) ≤ ν} ∈ I + . The assumption that λ is κ-strong yields   μ = tcf ( cf ◦ f /I) = tcf ( cf (f (i))/I ∩ P(X)) ≤ |X ν| ≤ ν κ < λ. i∈X

This contradiction shows that limI (cf ◦ f ) = λ. Now we can infer from Lemma 9.1.1 that μ ≤ ppκ (λ), hence λκ ≤ ppκ (λ). In the following we will prove generalizations of the Galvin-Hajnal theorem and of Silver’s theorem. Theorem 9.1.4 (Shelah) If ℵδ is a singular cardinal, then pp(ℵδ ) < ℵ(|δ|cf(δ) )+ . Proof We use Shelah’s interval technique and claim that  [ℵ0 , pp+ (ℵδ ))reg ⊆ {pcf(a) : a ⊆ ℵδ ∩ Reg ∧ |a| ≤ cf (δ) ∧ a is progressive}. So let μ ∈ [ℵ0 , pp+ (ℵδ ))reg . If μ < ℵδ , then there is a finite progressive subset a of ℵδ ∩ Reg, namely a := {μ}, such that |a| ≤ cf (δ) and μ ∈ pcf(a). So let μ ∈ (ℵδ , pp+ (ℵδ ))reg . By Lemma 3.3.5, there is a progressive subset a of ℵδ ∩ Reg such that |a| ≤ cf (ℵδ ) = cf (δ) and μ ∈ pcf(a), as desired. Now |ℵδ ∩Reg| ≤ |δ| gives |[ℵδ ∩Reg]≤cf(δ) | ≤ |δ|cf(δ) . From Corollary 3.4.14 we conclude that |pcf(a)| ≤ 2|a| ≤ 2cf(δ) , and thus we get |[ℵ0 , pp+ (ℵδ ))reg | ≤ |δ|cf(δ) · 2cf(δ) = |δ|cf(δ) . This yields pp(ℵδ ) < ℵ(|δ|cf(δ) )+ . Remark If ℵδ is singular, cf (δ)-strong and not a fixed point of the aleph function, then we can infer from Theorem 9.1.2 and Theorem 9.1.4 that cf(δ)

ℵδ

< ℵ(|δ|cf(δ) )+ .

If ℵδ = δ, this assertion holds trivially. Consequently, Theorem 9.1.4 is a generalization of the Galvin-Hajnal theorem. In the rest of this section we prove Shelah’s generalization of Silver’s theorem.

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273

Lemma 9.1.5 Let λ be a singular cardinal number. If pp(λ) < λ+cf(λ) , then pp(λ) = ppκ (λ) for all κ with cf (λ) ≤ κ < λ. +

+

Proof Consider κ and λ such that cf (λ) ≤ κ < λ and pp(λ) < λ+cf(λ) . By definition we have pp(λ) ≤ ppκ (λ). To get a contradiction, let us assume that there is a cardinal κ ∈ [cf (λ), λ)CN for which equality does not hold. Then we can find a subset a of λ ∩ Reg with sup(a) = λ and  |a| ≤ κ and an ultrafilter D on a, such that D ∩ Ib (a) = ∅ and θ := pp(λ) < cf ( a/D). As usual we can assume without loss of generality that κ < min(a).  Corollary 3.4.11 says that D ∩ J<θ+ (a) = ∅, since cf ( a/D) ≥ θ+ ,  hence J<θ+ (a) = P(a). If ρ < λ is a cardinal, then a \ ρ ∈ D and thus θ < cf ( (a \ ρ)/D), and the same argument yields J<θ+ (a \ ρ) = P(a \ ρ). Therefore, for any b ∈ J<θ+ (a), the set a \ b is unbounded in λ: Since b \ ρ is a member of J<θ+ (a \ ρ), b \ ρ is a proper subset of a \ ρ, and so there is ν ∈ a \ b such that ρ ≤ ν. Choose a sequence (bμ : μ ∈ pcf(a)) of generators for a, let bμ := ∅ if μ ∈ pcf(a), and define B := [ [λ+ , pp+ (λ))reg ]<ℵ0 = {X ⊆ [λ+ , pp+ (λ))reg : X is finite }. + + The assumption pp(λ) < λ+cf(λ) gives  |B| = |[λ , pp (λ))reg | ≤ cf (λ). Consider X ∈ B. Then we have μ∈X bμ ∈ J<θ+ (a), and we canchoose, as shown above, a set cX which is unbounded  in λ and satisfies cX ⊆ a\ μ∈X bμ and |cX | = cf (λ). Then also the set c := X∈B cX satisfies |c| = cf (λ). Clearly c is unbounded in λ. Furthermore the construction gives c \ ρ ∈ / J<θ+ (a) for all cardinals ρ < λ, since otherwise, by Corollary 5.3.4, there would be X0 ∈ B  such that c \ ρ ⊆ μ∈X0 bμ , and thus cX0 would be bounded in λ. Now we can conclude from Exercise 2.1.10 that there exists an ultrafilter  D on a such that ρ < λ, and D ∩ J<θ+ (a) = ∅. But  c \ ρ ∈ D , for all+ cardinals + then we get cf ( c/D ∩P(c)) ≥ θ ≥ pp (λ) and D ∩Ib (c) = ∅, contradicting the definition of pp(λ). +

Theorem 9.1.6 (Shelah) If λ is a singular cardinal with uncountable cofinality, and if the set {ν < λ : pp(ν) = ν + } is stationary in λ, then pp(λ) = λ+ . Proof Let a be a progressive  set of regular cardinals such that sup(a) = λ + ultrafilter D on a with and |a| = cf (λ). To prove that cf ( a/D)  = λ for any + D ∩ Ib (a) = ∅, we will show that tcf ( a/Ib (a)) = λ . Since sup(a) = λ, it is easy to see that the set {ρ < λ : sup(a ∩ ρ) = ρ} is a club of cardinals in λ. By Theorem 7.2.4, there is a club C0 of cardinals in λ such that otp(C0 ) = cf (λ) and  (1) tcf ( C0+ /Ib (C0+ )) = λ+ .

274

Chapter 9. The Cardinal Function pp(λ)

Then C := {ρ < λ : sup(a ∩ ρ) = ρ ∧ ρ = 0} ∩ C0 is also a club in λ with |C| = cf (λ), and from C + ∈ Ib (C0+ ) and (1), together with Lemma 3.1.8, we can conclude that  (2) tcf ( (C + /Ib (C + )) = λ+ . By assumption, the set S := {ν ∈ C : pp(ν) = ν + } is stationary in λ. From S + ∈ Ib (C + ) we infer with (2) that  (3) tcf ( S + /Ib (S + )) = λ+ . Now, if ν ∈ S, then ν = sup(a ∩ ν), and from cf (ν) ≤ |a ∩ ν| ≤ |a| = cf (λ) < + min(a) < ν and pp(ν) = ν + < ν +cf(ν) we conclude, using Lemma 9.1.5, that pp|a∩ν| (ν) = ν + .  For any ultrafilter D on a ∩ ν with D ∩ Ib (a ∩ ν)= ∅, we have cf ( (a ∩ ν)/D) = ν + , and thus Theorem 3.4.18 tells us that tcf ( (a∩ν)/Ib (a∩ν)) = ν + . We put Iν := {X ⊆  a : X ∩ ν ∈ Ib (a ∩ ν)} for each ν ∈ S. Then Lemma 3.1.8 says that tcf ( a/Iν ) = ν + , hence S + ⊆ pcf(a). |S| = |C| = cf (λ) < min(a) gives |S + | < min(a). Taking I ∗ := {X ⊆ a : {ν + ∈ S + : X ∈ Iν } ∈ Ib (S + )}, we can infer from (3) and Corollary 3.3.8 that   tcf ( S + /Ib (S + )) = λ+ = tcf ( a/I ∗ ). It remains to show that I ∗ ⊆ Ib (a). If X ⊆ a and X ∈ Ib (a), then {ρ < λ : sup(X ∩ ρ) = ρ} is a club in λ. Therefore the set {ν ∈ S : sup(X ∩ ν) = ν} = {ν ∈ S : X ∈ Iν } is stationary in λ. Consequently, the set {ν + ∈ S + : X ∈ Iν } is unbounded in S + , and thus X ∈ I ∗ . Remark Let us demonstrate why the preceding theorem “generalizes” Silver’s theorem. If λ is a singular cardinal with uncountable cofinality and is not a limit point of fixed points of the aleph function, then there is some final segment E of λ containing no fixed points of the aleph function. Assume that S := {ν < λ : 2ν = ν + } is stationary in λ. Then λ is a strong limit cardinal. Furthermore the set L of all singular strong limit cardinals in E is a club in λ, and thus L ∩ S is stationary in λ. For any cardinal ν ∈ L ∩ S we get pp(ν) = 2ν , using Theorem 9.1.2 and Lemma 1.6.15 d). Now Theorem 9.1.6 says that pp(λ) = λ+ . Since λ is a strong limit cardinal, we get pp(λ) = λcf(λ) from Theorem 9.1.3, and λcf(λ) = 2λ from Theorem 1.6.15 d). In particular we have 2λ = λ+ . In this sense, Theorem 9.1.6 is a generalization of Silver’s theorem.

9.2. The Lemma of Galvin and Hajnal for pp(λ)

275

9.2 The Lemma of Galvin and Hajnal for pp(λ) In this section we will prove a theorem that is analogous to Corollary 2.3.4: If λ is a singular cardinal with uncountable cofinality κ and ppκ -strong, (λξ : ξ < κ) is a normal sequence cofinal in λ, and Φ ∈ κ ON is an ordinal function such +Φ(ξ) that pp(λξ ) = λξ for all ξ < κ, then pp(λ) ≤ λ+ Φ Id (κ) . The proof of this 1 theorem of Shelah needs several preparatory considerations. Lemma 9.2.1 Assume that μ and λ are singular cardinals of cofinality ≤ κ, where κ is a regular cardinal and κ < μ < λ. Further let λ ≤ ppκ (μ). Then ppκ (λ) ≤ ppκ (μ). Proof Fix ν ∈ Reg such that λ < ν < pp+ κ (λ). From Lemma 3.3.5 we obtain a progressive set b ⊆ λ of regular cardinals satisfying |b| ≤ κ, sup(b) = λ, and μ < min(b), and an ultrafilter D on b such that cf ( b/D) = ν. In the same way we can infer from λ ≤ ppκ (μ) that, for each ρ ∈ b, there exists a progressive set aρ ⊆ μ ∩ Reg with|aρ | ≤ κ < min(aρ ) and sup(aρ ) = μ, and an ultrafilter Dρ on aρ , such that cf ( aρ /Dρ ) = ρ and Dρ ∩ Ib (aρ ) = ∅. Since also Dρ ∩ Ib (μ) = ∅, we can extend each of the ultrafilters Dρ to an ultrafilter Dρ on c := {aρ : ρ ∈ b}, and we can easily verify that Dρ ∩ Ib (c) = ∅. Furthermore, we have b ⊆ pcf(c), |b| ≤ min(c), and c is a progressive set. Now Corollary 3.3.8 saysthat D ∗ := {X⊆ c : {ρ ∈ b : X ∈ Dρ } ∈ D} is an ultrafilter on c such that cf ( c/D∗ ) = cf ( b/D) = ν. No member of D∗ is bounded in c, since this holds for all Dρ and ∅ ∈ D. Furthermore we have c ⊆ μ ∩ Reg, |c| ≤ κ · κ = κ, and sup(c) = μ. So we can conclude that ν ≤ ppκ (μ). Lemma 9.2.2 Assume that λ is a singular cardinal with uncountable cofinality κ, (λξ : ξ < κ) is a normal sequence of cardinals which is cofinal in λ, and I ⊇ Ib (κ) is a normal ideal on κ. If f ∈ κ Reg is a function with limI (f ) = λ, then the set {ξ < κ : f (ξ) ≤ λξ } is a member of I. Proof Lemma 1.8.6 says that we can infer from I ⊇ Ib (κ) that I ⊇ Id (κ). Let us assume that the set X := {ξ < κ : κ < f (ξ) ≤ λξ } is not a member of I. If C is a club in κ, then κ \ C is not stationary and is thus a member of I. This gives X ∩ C ∈ / I since X ⊆ (X ∩ C) ∪ (κ \ C), hence in particular S := X ∩ Lim ∈ I. Fix ξ ∈ S. With the continuity of the sequence (λξ : ξ < κ), we can conclude that cf (λξ ) ≤ ξ < κ < λξ , hence λξ is singular. Since f (ξ) is regular, we get f (ξ) < λξ , and since ξ is a limit ordinal, there exists an ordinal j(ξ) < ξ such that f (ξ) < λj(ξ) . Now the function j : S −→ κ is regressive. Therefore we obtain, by Lemma 1.8.9, an ordinal α < κ and a set S0 ⊆ S such that S0 = {ξ ∈ S : j(ξ) = α} ∈ I. Consequently, the set {ξ < κ : f (ξ) ≤ λα } is not a member of I. This contradicts the assumption that limI (f ) = λ. 1 Our

proof elaborates Shelah’s representation in [Sh5], VIII, §1.

276

Chapter 9. The Cardinal Function pp(λ) Remember that, for any ordinal function g ∈ A ON,  pcf(g) = {λ : ∃ I(I is an ideal on A ∧ λ = tcf ( g/I))}.

If in addition g is progressive, then Lemma 3.3.1 says that pcf(g) = pcf(ran(g)). Lemma 9.2.3 Assume that g ∈ A ON is a progressive function with ran(g) ⊆ Reg and I is an ideal on  A. Further assume that λ = max pcf(g |` B) for all B ∈ I + . Then λ = tcf ( g/I).  Proof It suffices to show that λ = tcf ( ran(g)/g[I]). From A ∈ I and the assumption we can infer that λ = max pcf(g), hence λ = max pcf(ran(g)) since g is progressive. We put a := ran(g) and want to apply Theorem 3.4.18. For this we suppose, contrary to our hopes, that there is an ultrafilter D on  a such that D ∩ g[I] = ∅ and μ := cf ( a/D) = λ. Then we get μ < λ and D ∩ J<μ+ (a) = ∅. So we choose a set b ∈ D ∩ J<μ+ (a) and can conclude that max pcf(b) < μ+ ≤ λ. The set B := g−1 [b] is not a member of I, since b ∈ g[I]. This yields pcf(g |` B) = pcf(g[B]) = pcf(b), and thus max pcf(b) = max pcf(g |`B) < λ, contradicting the assumption. Theorem 9.2.4 Assume that a is a set of regular cardinals with |a| ≤  min(a), κ > ℵ0 is regular, and (bξ : ξ < κ) is a sequence of subsets of a witha = ξ<κ bξ which is increasing under inclusion. Further let λ ∈ pcf(a) \ ξ<κ pcf(bξ ). Then there is an ordinal ξ0 < κ and a sequence (λξ : ξ0 ≤ ξ < κ) satisfying λξ ∈ pcf(bξ ) and λξ > κ+ for all ξ with  ξ0 ≤ ξ < κ, and a subset A of κ which is unbounded in κ, such that λ = tcf ( ξ∈A λξ /Ib (A)).  If κ = ℵ0 , then there is a countable set b ⊆ {pcf(bξ ) : ξ < κ} such that λ ∈ pcf(b). Proof Let a, κ, (bξ : ξ < κ), and λ satisfy the assumptions of the theorem. Clearly a is infinite since λ ∈ pcf(a) \ a. Claim 1 We can assume without loss of generality that a) b) c) d)

max{|a|+ , ℵ2 } < min(a). λ = max pcf(a). a has no maximum. bξ = ∅ for all ξ < κ.

Proof Put a := a \ {|a|, |a|+ , ℵ2 }. Then a is progressive, and it is obvious that λ ∈ pcf(a ) and thus J<λ+ (a ) \ J<λ (a ) = ∅. We choose a set a ⊆ a such that a ∈ J<λ+ (a ) \ J<λ (a ). From Lemma 3.4.19 and max pcf(a ) < λ+ we get λ = max pcf(a ). Let ρ := min{ν ∈ ICN : |a \ ν| < ℵ0 }. Then we have sup(a ∩ ρ) = ρ. Since pcf(a ) = pcf(a ∩ ρ) ∪ pcf(a \ ρ) = pcf(a ∩ ρ) ∪ (a \ ρ)

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277

and λ ∈ pcf(a ) \ a , we get λ = max pcf(a ∩ ρ), and a ∩ ρ has no maximum. It is easy to see that it suffices to prove the theorem for a ∩ ρ and (bξ : ξ < κ), where bξ := bξ ∩ a ∩ ρ. Since the sequence (bξ : ξ < κ) is increasing, this holds also for the sequence (bξ : ξ < κ), and by assumption there is ξ0 < κ such that bξ = ∅ for all ξ with ξ0 ≤ ξ < κ. This proves Claim 1. Next we put μ := max{|a|+ , ℵ2 }. Then μ ∈ (|a|, min(a))reg . For S := μ,ℵ0 we further assume that (Eβ : β ∈ S) is a 3Club -sequence for S according to Lemma 4.5.3, and, according to Lemma 6.1.1, f¯ = (f¯b : b ⊆ a) is a special sequence for a and all cardinals ρ ∈ (|a|, min(a))reg . We choose a regular sup(a) and an approximation sequence (Ni : i ≤ μ) in H(Θ) such cardinal Θ > 22 that ¯ ⊆ N0 a ∪ {a, λ, (bξ : ξ < κ), (Eβ : β ∈ S), f}

and |Ni | ≤ μ for all i < μ. Let N := Nμ . From Lemma 4.3.3 we get μ ⊆ N , and thus, using |a| < μ, we can assume without loss of generality that |a| ⊆ N0 . Example 4.2.15 yields |a|+ ∈ N . Clearly ℵ2 ∈ H(Θ), hence ℵ2 ∈ N and thus μ ∈ N . The  supposition |a| < κ leads, together with the fact that κ is regular, to a ⊆ {bξ : ξ < σ} for some σ < κ. Then a ⊆ bσ , which contradicts λ ∈ pcf(a) \ pcf(bσ ). Therefore κ ≤ |a|, and we also get κ + 1 ⊆ N0 . Claim 2 There is a sequence h := (((cξ,k , λξ,k ) : k ≤ n(ξ)) : ξ < κ) such that (i) For all ξ < κ (a) n(ξ) < ω and bξ = cξ,0 ⊇ . . . ⊇ cξ,n(ξ) = ∅. (b) If k < n(ξ), then cξ,k+1 ∈ J<λξ,k (a), where λξ,k = max pcf(cξ,k ). c

(c) If k < n(ξ) and θ ∈ cξ,k \ cξ,k+1 , then χN (θ) = fχξ,k (θ). N (λξ,k ) (ii) h ∈ N . Proof By an iterated application of Lemma 6.1.3 it is easy to find finite sequences with the properties in (i). The crucial point is the proof of (ii). First we define by recursion for each n < ω sets Tn ⊆ 2n μ of strictly increasing sequences, and, for each η ∈ Tn and ξ < κ, sets cηξ ⊆ bξ and clubs Cξη and C η in μ as follows. Put T0 := {∅} and c∅ξ := bξ for all ξ < κ. Note that c∅ξ ∈ N for all ξ < κ. Fix n < ω, and assume that Tn and cηξ ∈ N are defined for all η ∈ Tn and all ξ < κ. Now consider ξ < κ. If η ∈ Tn , we can apply Lemma 6.1.3 to cηξ and (Ni : i ≤ μ) instead of b and (Ni : i ≤ μ) and obtain clubs Cξη in μ which have that properties with

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Chapter 9. The Cardinal Function pp(λ)

respect to cηξ which are satisfied by the club C in the lemma with respect to b.  Then the set C η := ξ<κ Cξη is also a club in μ, since κ ≤ |a| < μ. Now we put Tn+1 := {η  (i, j) : η ∈ Tn ∧ i, j ∈ C η ∧ η(2n − 1) < i < j} , and get from the lemma, for any η  (i, j) ∈ Tn+1 and ξ < κ, that η  (i,j)

cξ

cη

:= B[fχξN

j

(λξ,η ) , χNi ]

∩ cηξ ∈ N ∩ J<λξ,η (cηξ ), η  (i,j)

where λξ,η is the maximum of pcf(cηξ ). Furthermore, for any θ ∈ cηξ \ cξ we have χN (θ) =

cη fχξN (λξ,η ) (θ).

,

In addition we obtain from Lemma 6.1.3 and η  (i,j)

is the unique member u of H(Θ) Lemma 4.2.14 a formula ϕ such that cξ satisfying H(Θ) |= ϕ(u, cηξ , f¯, Ni , Nj ). This completes the definition of the desired sets. Next we want to find a club F in μ such that η |` 2n ∈ Tn for all strictly increasing functions η : ω → F and all n < ω. 2n 2n  Considerη n < ω and α < μ. Then |Tn ∩ α| ≤ |α| < μ, hence Fn,α := η∈Tn ∩2n α C is a club in μ. For each n ∈ ω we put Fn := α<μ (Fn,α ∩ {δ < μ : δ ∈ Lim}). Now, if η ∈ Tn and i, j ∈ Fn satisfy β := η(2n − 1) < i < j, then η  (i, j) ∈ Tn+1 . To see this, we first observe that i, j ∈ Lim, η ∈ Tn ∩ 2n (β + 1), and β + 1 < i < j. So the definition of Fn yields i, j ∈ Fn,β+1 ; the definition of Fn,β+1 , together with η ∈ Tn ∩ 2n (β + 1), gives i, j ∈ C η . Thus, by definition, η  (i, j) ∈ Tn+1 . By induction it is easy to see that F := n<ω Fn satisfies our claims. (Eβ : β ∈ S) is a 3Club -sequence for S, and thus the set {β ∈ S : Eβ ⊆ F } is stationary in μ. Consequently, there is α ∈ S such that Eα ⊆ F . Since μ ⊆ N , we have α ∈ N and thus Eα ∈ N . Furthermore Eα has order type ω, and the unique order isomorphism q from ω onto Eα is a member of H(Θ), hence of N . From ran(q) = Eα ⊆ F and the fact that q is strictly increasing we can now infer that q |` 2n ∈ Tn for all n ∈ ω. This enables us to define the q |` 2k desired set cξ,k . We put cξ,k := cξ and λξ,k := max pcf(cξ,k ). Then by construction, for ξ fixed, cξ,k+1 is the unique member u of H(Θ) such that H(Θ) |= ϕ(u, cξ,k , f¯, Nq(2k) , Nq(2k+1) ). Furthermore, Lemma 3.4.20 together q |` 2(k+1) q |` 2k with cξ,k = ∅ gives λξ,k+1 < λξ,k , using cξ ∈ J<λξ,q |`2k (cξ ). Consequently, there must be k ∈ ω such that cξ,k = ∅, and we can define the natural number n(ξ) := min{k ∈ ω : cξ,k = ∅}. The construction shows that, for each ξ < κ, the sequence (cξ,k : k ≤ n(ξ)) satisfies (i).

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279

Since (Ni : i ≤ α) ∈ N and Eα ⊆ α, we get g := (Nq(n) : n < ω) ∈ N . With the above property of the formula ϕ we can conclude that the sequence h1 := ((cξ,n : n < ω) : ξ < κ) is a member of N . Namely, h1 is the unique member of H(Θ) such that H(Θ) |= Func(h1 ) ∧ dom(h1 ) = κ ∧ ∀ξ < κ(Func(h1 (ξ)) ∧ dom(h1 (ξ)) = ω ¯ g(2n), g(2n + 1)))), ∧h1 (ξ)(0) = bξ ∧ ∀n ∈ ω(ϕ(h1 (ξ)(n + 1), h1 (ξ)(n), f, and thus h1 ∈ N by Lemma 4.2.13 since the parameters of the defining formula are members of N . It is easy to see that the function h2 := ((cξ,k : k ≤ nξ ) : ξ < κ) is definable in H(Θ) with the parameter h1 , hence h2 ∈ N . From Example 4.2.15, together with the assumption on Θ, we know that x ⊆ a =⇒ pcf(x) ∈ H(Θ) ∧ ∀y ∈ H(Θ)(y = pcf(x) ⇐⇒ H(Θ) |= y = pcf(x)). So we can conclude for h3 := ((λξ,k : k ≤ nξ ) : ξ < κ), where max pcf(∅) := 0, that h3 = {(ξ, x) : H(Θ) |= ξ < κ ∧ Func(x) ∧ dom(x) = dom(h2 (ξ))∧ ∀k ∈ dom(x)(x(k) = max pcf(h2 (ξ)(k)))}. This gives h3 ∈ N . Now it is easy to define the function h in H(Θ) with the parameters h2 , h3 ∈ N , which proves (ii) and completes the proof of Claim 2. h3 ∈ N yields immediately that (dξ : ξ < κ) ∈ N , where dξ := {λξ,k : k < n(ξ)} for all ξ < κ.  Claim 3 If A ⊆ κ is unbounded in κ, then λ = max pcf( ξ∈A dξ ). Proof To get a contradiction,  we assume that thereis an unbounded subset A of κ such that max pcf( ξ∈A dξ ) < λ. Let d := ξ∈A dξ . We have λξ,k ∈ pcf(bξ ) ⊆ pcf(a), hence κ+ < λξ,k for all ξ and k. So we get d ⊆ pcf(a) and |d |+ < min(d ). Now Corollary 7.1.19 says that pcf(d ) ⊆ pcf(a). Since (dξ : ξ < κ) ∈ N , we can assume in addition that A is a member of N , hence d ∈ N . We want to choose a special sequence g¯ = (¯ g e : e ⊆ d) for appropriate d and μ. Of course we want to ensure the existence of such a sequence which in addition is a member of N . A natural candidate for d is d , but unfortunately we do not know that d is infinite. So let d := d if d is infinite. If |d | < ℵ0 and κ > ℵ0 , then there exists ξ ∈ A such that ℵ0 ≤ |bξ |. Let d := d ∪ bξ . Then we have |d| = |bξ | < |a|+ < min(a) ≤ min(d), hence d is an infinite, progressive set satisfying max pcf(d) < λ. Furthermore, bξ ∈ N yields d ∈ N . If |d | < ℵ0 , κ = ℵ0 , and at least one bζ is infinite for some ζ ∈ A, then the same procedure works. Finally we consider the case that |d | < ℵ0 , κ = ℵ0 , and bξ is finite for all ξ ∈ A. Fix ζ ∈ A. Then pcf(bζ ) = bζ and thus

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Chapter 9. The Cardinal Function pp(λ)

 λζ,0 = max(bζ ). Since a = ξ∈A bξ and a has no maximum, there exists ξ ∈ A such that ζ < ξ and max(bζ ) < max(bξ ). From this we get |d | ≥ ℵ0 , and we can can infer that this case cannot occur. Next we choose a special sequence g¯ = (¯ g e : e ⊆ d) for d and μ. Lemma sup(a) 6.1.2 says that all such sequences are members of H(Θ) if only Θ > 22 . So we know why we have chosen Θ so large. The previous considerations show, together with d ∈ N and μ ∈ N , that we can assume without loss of generality that g¯ ∈ N . Again we obtain with Lemma 6.1.3 a sequence ((el , ρl ) : l ≤ m) ∈ N , such that d = e0 ⊇ . . . ⊇ em = ∅, ρl = max pcf(el ), and χN (ρ) = gχelN (ρl ) (ρ) for any ρ ∈ d and l < m with ρ ∈ el \ el+1 . Consider θ ∈ a ⊆ N . Then ξ := min{ζ ∈ A : θ ∈ bζ } ∈ N . Furthermore we have k := min{k < n(ξ) : θ ∈ / cξ,k+1 } ∈ N , and we can conclude from Claim 2 that c χN (θ) = fχξ,k (θ). N (λξ,k ) / ej+1 } ∈ N , we infer that Since λξ,k ∈ d ∩ N and thus l := min{j < m : λξ,k ∈ χN (λξ,k ) = gχelN (ρl ) (λξ,k ). These considerations show that the function χN |` a is definable in H(Θ) with the parameters a, f¯, A, (((cξ,k , λξ,k ) : k ≤ n(ξ)) : ξ < κ), g¯, (el : l ≤ m), and (χN (ρl ) : l ≤ m). Therefore, putting N + := SkolΘ (N ∪ {χN (ρl ) : l ≤ m}), we can conclude from Lemma 4.2.14 and Lemma 4.4.2 that χN |`a ∈ N + . Now we have ρl ∈ N , and our supposition gives ρl = max pcf(el ) ≤ max pcf(d) < λ for all l ≤ m. So we get χN (ρl ) < χN (λ), and Corollary 4.4.4 tells us that + χN (λ) = χN + (λ). On  the other hand, we have χN |` a ∈ N and, using +|N | < a. Therefore there is α < λ such that α ∈ N and min(a), χN |` a ∈ a χN |` a
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281

∗ Claim  4 There is n < ω and an unbounded subset A of κ such that λ = tcf ( ξ∈A λξ,n∗ /Ib (A)).

Proof Since κ+ < λξ,n for all ξ < κ and n ∈ ω, the function (λξ,n : ξ < κ) is progressive. Therefore it suffices to prove, by Claim 3 and Lemma 9.2.3, that there is n∗ ∈ ω and an unbounded subset A of κ such that λ ∈ pcf{λξ,n∗ : ξ ∈ B} for all B ∈ Ib (A)+ . Let us assume, contrary to our hopes, that this assertion is false. First we obtain with κ > ℵ0 an n ∈ ω \ {0} such that A0 := {ξ < κ : n(ξ) = n } is unbounded in κ. If the set Ak is already defined and unbounded in κ, we can choose, by the assumption of the indirect proof, a set Ak+1 ⊆ Ak which is unbounded in κ and satisfies λ ∈ / pcf{λξ,k : ξ ∈ Ak+1 }. With An ⊆ A0  and by the choice of n we can conclude that    dξ = {λξ,k : ξ ∈ An } ⊆ {λξ,k : ξ ∈ Ak+1 }. ξ∈An

k
k
This inclusion also holds for  the sets of possible cofinalities of these sets. But Claim 3 says that λ ∈ pcf( ξ∈An dξ ), and so we can infer from pcf(a1 ∪ a2 ) = pcf(a1 ) ∪ pcf(a2 ) that there is k < n such that λ ∈ pcf({λξ,k : ξ ∈ Ak+1 }). This contradicts the choice of Ak+1 . We can slightly improve the previous result. Lemma 9.2.5 Assume that a is a progressive set of regular cardinals, κ > ℵ0 is a regular cardinal, and (bξ : ξ < κ) isa sequence of subsets of a which  is increasing under inclusion, such that a = {bξ : ξ < κ} and λ ∈ pcf(a)\ ξ<κ pcf(bξ ). in κ and a sequence (μξ :  ξ ∈ C ∗ ) of regular cardinals Then there is a club C ∗  such that μξ ∈ pcf(bξ ) \ ζ<ξ pcf(bζ ) and λ = tcf ( ξ∈C ∗ μξ /Ib (C ∗ )). Proof With the previous theorem we obtain an unbounded subset A of κ and, without loss of generality, a progressive sequence (λξ : ξ < κ) of regular cardinals such that λξ ∈ pcf(bξ ) and λξ > κ+ for all ξ < κ and  λ = tcf ( ξ∈A λξ /Ib (A)). We will use Lemma 3.3.1 without mentioning. Let C := {ξ < κ : sup(A ∩ ξ) = ξ} be the union of {0} and the set of limit points of A. We know that C is a club in κ. Fix ξ ∈ C. Extending the ultrafilter on A ∩ ξ which is generated by {ξ \ α : α < ξ}, we obtain an ultrafilter Dξ on A such that A ∩ ξ ∈ Dξ and sup(X ∩ ξ) = ξ for all X ∈ Dξ . In particular we have Dξ ∩ Ib (A ∩ ξ) = ∅. The cardinal   μξ := cf ( λζ /Dξ ) = cf ( λζ /Dξ ∩ P(A ∩ ξ)) ζ∈A

ζ∈A∩ξ

282

Chapter 9. The Cardinal Function pp(λ)

is a member of pcf{λζ : ζ ∈ A ∩ ξ}. Since {λζ : ζ ∈ A ∩ ξ} is progressive, we can infer from Corollary 7.1.21 that pcf{λζ : ζ ∈ A ∩ ξ} ⊆ pcf(bξ ). So we get μξ ∈ pcf(bξ ).  Now we show that tcf ( ξ∈C μξ /Ib (C)) = λ. It is natural to apply Lemma 3.3.7. For this, put f := (λξ : ξ ∈ A), g := (μξ : ξ ∈ C), and let, for ξ ∈ C, Iξ be the dual ideal of Dξ and I ∗ := {X ⊆ A : {ξ ∈ C : X ∈ Iξ } ∈ Ib (C)}.   Then |C| < min(f [A]), and we know that tcf ( f /Iξ ) = cf ( f /Dξ ) = μξ = g(ξ) for every ξ ∈ C. From Lemma 3.3.7 we can conclude that tcf ( g/Ib (C)) = tcf ( f /I ∗ ). If we can prove that Ib (A) ⊆ I ∗ , then we get     λ = tcf ( f /Ib (A)) = tcf ( f /I ∗ ) = tcf ( g/Ib (C)) = tcf ( μξ /Ib (C)). ξ∈C

To show that Ib (A) ⊆ I ∗ , fix X ⊆ A bounded in κ. Then X ⊆ A ∩ ζ for some ζ ∈ C. By the choice of the sequence (Dξ : ξ ∈ C), we have X ∈ Dξ for all ξ ∈ C with ξ > ζ. Consequently, the set {ξ ∈ C : X∈ Dξ } is a member of Ib (C), hence X ∈ I ∗ . This completes the proof of tcf ( ξ∈C μξ /Ib (C)) = λ. For all ξ < κ, the set {ζ ∈ C : μζ ∈ pcf(bξ )} is bounded in C, since otherwise λ ∈ pcf{μζ : ζ ∈ C ∧ μζ ∈ pcf(bξ )} and thus λ ∈ pcf(bξ ). So we can choose, for each ζ < κ, a club Cζ ⊆ {ξ ∈ C : μξ ∈ / pcf(bζ )}, for example a final segment of C, and put C ∗ := ζ<κ Cζ . Now C ∗ has the desired properties: If ξ ∈ C ∗ , then μξ ∈ / pcf(bζ ) for all ζ < ξ and μξ ∈ pcf(bξ ), as demonstrated above. Furthermore we have C ∗ ⊆ C, Ib (C) ∩ P(C ∗ ) = Ib (C ∗ ), and C ∗ ∈ Ib (C), hence Ib (C) |` C ∗ is an ideal on C. Now we can infer from Lemma 3.1.8 that    tcf ( μξ /Ib (C ∗ )) = tcf ( μξ /(Ib (C) |`C ∗ )) = tcf ( μξ /Ib (C)) = λ, ξ∈C ∗

ξ∈C

ξ∈C

which proves the lemma. Lemma 9.2.6 (Jech) Assume that λ is a singular cardinal of uncountable cofinality κ and (λξ : ξ < κ) is a normal sequence of infinite cardinals which is cofinal in λ. Further let I ⊇ Ib (κ) be a normal ideal on κ. If f ∈ κ Reg satisfies +Φ(ξ) for all ξ < κ, and limI (f ) = λ, Φ ∈ κ ON is a function such that f (ξ) ≤ λξ   + Φ I tcf ( f /I) exists, then tcf ( f /I) ≤ λ . Proof Let us assume that the assertion is false. The Lemmas 1.8.7 and 2.2.2 show that
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283

+Φ(ξ)

for all ξ < κ, limI (f ) = λ, there is a function f ∈ κ Reg satisfying f (ξ) ≤ λξ  + Φ I and tcf ( f /I) > λ . Furthermore we can assume without loss of gener+Φ(ξ) and thus in particular Φ(ξ) = 0 for all ξ < κ. ality that λξ < f (ξ) = λξ Namely, Lemma 9.2.2 says that the set B := {ξ < κ : f (ξ) ≤ λξ } is a member  of I. Putting f  (ξ) := λ+ ξ for ξ ∈ B and f (ξ) := f (ξ) for ξ ∈ κ\B, and defining +Φ (ξ)

, we get Φ ≤I Φ. Consequently, Φ the function Φ ∈ κ ON by f  (ξ) =: λξ   is also minimal under
) ≤ cf (Φ(ξ)) ≤ Φ(ξ). Any modulo I |`X cofinal sequence of members cf (λ ξ of f is strictly says that  increasing under   λ+ Φ I we can conclude that #Φ#I |`X ≥ tcf ( f /I |` X) = tcf ( f /I) > λ+ Φ I ≥ #Φ#I . Lemma 2.2.9 b) tells us that κ \ X ∈ I + , and so we can infer from Lemma 2.2.9 c) that #Φ#I |`(κ\X) = #Φ#I , since #Φ#I = #Φ#I |`(X∪κ\X) = min{#Φ#I |`X , #Φ#I |`(κ\X) }. However, the function Ψ ∈ κ ON defined above satisfies #Φ#I |`(κ\X) = #Ψ#I |`(κ\X) + 1, hence #Φ#I = #Ψ#I |`(κ\X) + 1. This contradicts our assumption, and we have shown that #Φ#I is a successor ordinal. By the main lemma of pcf-theory, Theorem 3.2.6, and by Theorem  3.1.13 we obtain a function g ∈ κ Reg such that g λξ , h(ξ) := λ+ if g(ξ) ≤ λξ , ξ we get h =I g and thus h λ+ Ψ I . Definition Assume that λ is a singular cardinal and κ is a cardinal with cf (λ) ≤ κ < λ. We say that λ is ppκ -strong iff ppκ (ν) < λ for all ν < λ with cf (ν) ≤ κ < ν.

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Chapter 9. The Cardinal Function pp(λ)

Lemma 9.2.7 Assume that λ is a singular cardinal and κ is a cardinal such sup(a) = λ that cf (λ) ≤ κ < λ, and let λ be ppκ -strong. If a ⊆ Reg satisfies  and |a| ≤ κ < min(a) and D is an ultrafilter on a with cf ( a/D) > λ, then  D ∩ Ib (a) = ∅ and, in particular, cf ( a/D) ≤ ppκ (λ). Therefore we have ppκ (λ) = sup{max pcf(a) : a ⊆ Reg ∧ |a| ≤ κ < min(a) ∧ sup(a) = λ}. Proof Let λ, κ, a and D satisfy the assumptions of the lemma. Assume, to get a contradiction, that there is ν < λ such that a ∩ ν + ∈ D, and put μ := min{ν < λ : a ∩ ν + ∈ D}. Then we have μ = limD (a) and κ < μ, and from Lemma 3.3.4 we can infer that μ is a limit cardinal and a ∩ μ ∈ D. The choice of μ gives μ = sup(a ∩ μ) and thus cf (μ) ≤ |a ∩ μ| ≤ |a| ≤ κ < μ. Furthermore, D := {X ⊆ a∩μ : X ∈ D} is an ultrafilter on a∩μ which satisfies D ∩ Ib (a ∩ μ) = ∅ bythe choice of μ. So we conclude from our supposition that cf ( a/D) = cf ( (a ∩ μ)/D ) ≤ ppκ (μ) < λ, and this contradicts the assumption. This yields D ∩ Ib (a) = ∅. Lemma 9.2.8 Assume that λ is a ppcf(λ) -strong singular cardinal and κ := cf (λ) > ℵ0 . Then there is a normal sequence (λξ : ξ < κ) of singular cardinals which is cofinal in λ such that λ0 > κ and ppκ (ν) < λξ for all ξ with 0 < ξ < κ and all cardinals ν < λξ satisfying cf (ν) ≤ κ < ν. Proof First we show that, for any cardinal λ < λ, p(λ ) := sup{ppκ (ν) : ν ∈ ICN ∧ cf (ν) ≤ κ < ν < λ } < λ. Without loss of generality we can assume that p(λ ) > λ . If the above assertion is false, then there is a cardinal ν < λ such that cf (ν) ≤ κ < ν and λ < ppκ (ν). Choose ν minimal with respect to this property. Then we can conclude that ppκ (ν  ) ≤ ppκ (ν) for all ν  < λ with cf (ν  ) ≤ κ < ν  : If ν  < ν, then ppκ (ν  ) ≤ λ by the choice of ν. If ν < ν  , then ν < ν  < λ < ppκ (ν), and Lemma 9.2.1 gives ppκ (ν  ) ≤ ppκ (ν). So we get p(λ ) = ppκ (ν), and since λ is ppκ -strong, we conclude that p(λ ) = ppκ (ν) < λ, contradicting our supposition. Now we can define by recursion the desired normal sequence (λξ : ξ < κ). Let h : cf (λ) −→ λ be a normal function such that sup(ran(h)) = λ. Choose λ0 singular such that κ < λ0 < λ and put, inevitably, λξ := sup{λζ : ζ < ξ} for limit ordinals ξ. For the successor step, define λξ,0 := max{λξ , h(ξ)}, λξ,n+1 := max{λ+ ξ,n , p(λξ,n )} for each n ∈ ω, and λξ+1 := supn<ω λξ,n . Then λξ+1 < λ, since cf (λ) > ℵ0 . If cf (ν) ≤ κ < ν and ν < λξ+1 , that means ν < λξ,n for some n ∈ ω, then, by definition, ppκ (ν) ≤ p(λξ,n ) ≤ λξ,n+1 < λξ+1 .

9.2. The Lemma of Galvin and Hajnal for pp(λ)

285

Lemma 9.2.9 Assume that λ is a singular cardinal of uncountable cofinality κ which is ppκ -strong. Further let (λξ : ξ < κ) be a normal sequence of singular cardinals which is cofinal in λ. If μ ∈ (λ, pp+ (λ))CN is regular, then there is a  club C in κ and a function f ∈ C (Reg ∩ λ), such that tcf ( f /Ib (C)) = μ and λξ < f (ξ) < pp+ κ (λξ ) for all ξ ∈ C. Proof Without loss of generality let λξ > κ for all ξ < κ. By Lemma 3.3.5, there is a set a of regular cardinals satisfying sup(a) = λ, |a| ≤ κ < min(a), and μ ∈ pcf(a). The assumption on λ and Lemma 9.2.7 imply that we can  assume without loss of generality that μ = max pcf(a). To see this, let μ = cf ( a/D) for some ultrafilter D on a. We can replace, if necessary, a by a generator bμ of J<μ+ (a) over J<μ (a). Corollary 5.3.5 says that bμ ∈ D, hence sup(bμ ) = λ since, by Lemma 9.2.7, D ∩ Ib (a) = ∅, and from Lemma 5.3.1 we infer that max pcf(bμ ) = μ. Furthermore we can assume without loss of generality that sup(a ∩ λξ ) = λξ , hence cf (λξ ) ≤ κ < λξ , and ppκ (ν) < λξ for all ξ < κ and all ν < λξ such that cf (ν) ≤ κ < ν, that means that λξ is ppκ -strong. Otherwise we could thin out the given sequence h := (λξ : ξ < κ) with the help of Lemma 9.2.8. If (νξ : ξ < κ) is the sequence of singular cardinals yielded by this lemma, the set E := {ξ < κ : λξ = νξ ∧ sup(a ∩ λξ ) = λξ } is a club in κ, and we can replace h by the sequence (λξ : ξ ∈ E). Now fix ξ < κ and put aξ := a ∩ λξ . Note that λξ < max pcf(aξ ). From Lemma 9.2.7 we infer that max pcf(aξ ) ≤ ppκ (λξ ) < λ for all ξ < κ. Consequently we can apply Lemma 9.2.5 to obtain  a club C1 in κ and a function {pcf(aζ ) : ζ < ξ} for all ξ ∈ C1 f1 := (μ ξ : ξ ∈ C1 ) satisfying μξ ∈ pcf(aξ ) \ and tcf ( f1 /Ib (C1 )) = μ. Let  C := C1 ∩ Lim, f := f1 |` C and fix ξ ∈ C. First we note that μξ ∈ {aζ : ζ < ξ} = aξ . We claim that μξ ≥ λξ . Otherwise we could infer from sup(aξ ) = λξ that μξ < sup(aζ ) = λζ for some ζ < ξ. From μξ ∈ pcf(aζ ) we can conclude that μξ ∈ pcf(aξ \aζ ), hence λζ ≤ min(aξ \aζ ) ≤ μξ , contradicting μξ < λζ . Furthermore we have μξ < λ, since μξ ∈ pcf(aξ ) and max pcf(aξ ) < λ. To apply Lemma 9.2.2, we extend the function f to κ, putting f (ξ) := λ+ ξ for all ξ ∈ κ \ C. From λξ ≤ f (ξ) for all ξ < κ and Ib (κ) ⊆ Id (κ) we get limId (κ) (f ) = λ. Now Lemma 9.2.2 tells us that {ξ < κ : f (ξ) ≤ λξ } ∈ Id (κ). So we can assume without loss of generality that λξ < f (ξ) = μξ for all ξ ∈ C. Next we claim that μξ < pp+ κ (λξ ). Namely, there is an ultrafilter D on aξ such that μξ = cf ( aξ /D) > λξ . Now Lemma 9.2.7, together with the fact that + we can conclude λξ is ppκ -strong, implies that  μξ = f (ξ) < ppκ (λξ ). Finally  from C ∈ Ib (C1 ) and tcf ( f1 /Ib (C1 )) = μ that tcf ( f /Ib (C)) = μ. This completes the proof of the lemma. Theorem 9.2.10 (Shelah) Assume that ℵ0 < κ = cf (λ) < λ, λ is a ppκ -strong cardinal, (λξ : ξ < κ) is a normal sequence of singular cardinals which is

286

Chapter 9. The Cardinal Function pp(λ) +Φ(ξ)

for all cofinal in λ, and Φ ∈ κ ON is a function such that ppκ (λξ ) = λξ ξ < κ. Further let the norm of Φ be determined by the ideal Id (κ), i.e., let #Φ# := #Φ#Id (κ) . Then pp(λ) ≤ λ+ Φ . Proof Consider μ ∈ (λ, pp+ (λ))reg . It suffices to show that μ ≤ λ+ Φ . By Lemma a function f ∈ C Reg such  9.2.9, we obtain a club C in κ and that tcf ( f /Ib (C)) = μ and λξ < f (ξ) < pp+ κ (λξ ) for all ξ ∈ C. This gives +Φ(ξ) f (ξ) ≤ λξ for all ξ ∈ C. To apply Lemma 9.2.6, we extend the function f to κ, putting f (ξ) := λ+ ξ for every ξ ∈ κ \ C. From Corollary 1.8.10 we can conclude that I := Id (κ) |`C is a normal ideal on κ satisfying Ib (κ) ⊆ I d (κ) ⊆ I. Obviously we  have limI (f ) = λ, and thus Lemma 9.2.6 yields tcf ( f /I) ≤ λ+ Φ I if tcf ( f /I) exists. But    tcf ( f /I) = tcf ( f |`C/Id (C)) = tcf ( f |`C/Ib (C)) = μ  shows that tcf ( f /I) exists, and we can infer that μ ≤ λ+ Φ I . / Now the supposition #Φ# < #Φ#I gives Id (κ) = Id (κ) |` C, hence κ \ C ∈ Id (κ), contradicting the fact that C is a club in κ. Therefore we get #Φ# = #Φ#I and thus μ ≤ λ+ Φ , as desired.

Bibliography [Ba]

H. Bachmann, Transfinite Zahlen, Springer-Verlag, Berlin, Heidelberg, New York 1967. [BM] M. R. Burke, M. Magidor, Shelah’s pcf theory and its applications, Ann. Pure Appl. Logic 50 (1990), 207–254. [Dr] F. R. Drake, Set Theory. An Introduction to Large Cardinals, Studies in Logic and the Foundations of Mathematics 76, North-Holland, Amsterdam 1974. [Ea] W. B. Easton, Powers of regular cardinals, Ann. Math. Logic 1 (1970), 139–178. [En] H. B. Enderton, Elements of Set Theory, Academic Press, New York 1977. [EHM] P. Erd˝ os, A. Hajnal, A. M´ at´e, R. Rado, Combinatorial Set Theory: Partition Relations for Cardinals, Studies in Logic and the Foundations of Mathematics 76, North-Holland, Amsterdam 1984. [GH] F. Galvin, A. Hajnal, Inequalities for cardinal powers, Ann. of Math. 101 (1975), 491–498. [Je1] T. Jech, The Axiom of Choice, Studies in Logic and the Foundations of Mathematics 75, North-Holland, Amsterdam 1973. [Je2] T. Jech, Set Theory, 2nd edition, Springer-Verlag, Berlin 1997. [Je3] T. Jech, Shelah’s theorem on 2ℵω , Bull. London Math. Soc. 24 (1992), 127–139. [Je4] T. Jech, A variation on a theorem of Galvin and Hajnal, Bull. London Math. Soc. 25 (1993), 97–103. [Je5] T. Jech, Singular cardinals and the pcf theory, Bull. Symbolic Logic 1 (1995), 408–424. [JS] T. Jech, S. Shelah, On a conjecture of Tarski on products of cardinals, Proc. Amer. Math. Soc. 112 (1991), 1117–1124. [JW1] W. Just, M. Weese, Discovering Modern Set Theory. I, The Basics, Graduate Studies in Mathematics, vol. 8, American Mathematical Society, 1996.

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[JW2] W. Just, M. Weese, Discovering Modern Set Theory. II, Set-Theoretic Tools for Every Mathematician, Graduate Studies in Mathematics, vol. 18, American Mathematical Society, 1997. [Ku] K. Kunen, Set Theory. An Introduction to Independence Proofs, Studies in Logic and the Foundations of Mathematics 102, North-Holland, Amsterdam 1980. [Le] A. Levy, Basic Set Theory, Springer-Verlag, Berlin, Heidelberg, New York 1979. [Ma] M. Magidor, On the singular cardinals problem I, Israel J. Math. 28 (1977), 1–31. [RR] H. Rubin, J. E. Rubin, Equivalents of the Axiom of Choice, II, Studies in Logic and the Foundations of Mathematics 102, North-Holland, Amsterdam 1985. [Sh1] S. Shelah, Proper Forcing, Lecture Notes in Mathematics 940, SpringerVerlag, Berlin 1982. [Sh2] S. Shelah, Successors of singulars, cofinalities of reduced products of cardinals and productivity of chain conditions, Israel J. Math. 62 (1988), 213–256. [Sh3] S. Shelah, Products of regular cardinals and cardinal invariants of products of Boolean algebras, Israel J. Math. 70 (1990), 129–187. [Sh4] S. Shelah, Cardinal arithmetic for sceptics, Bull. Am. Math. Soc. 26 (1992), 197–210. [Sh5] S. Shelah, Cardinal Arithmetic, Oxford Logic Guides 29, Clarendon Press, Oxford 1994. [Sh6] S. Shelah, The pcf theorem revisited, The mathematics of Paul Erd¨ os, II, 420–459, Algorithms Combin. 14, Springer-Verlag, Berlin 1997. [Si] J. Silver, On the singular cardinals problem, in Proc. Internat. Congress of Math., Vancouver 1974, vol. I, 265–268. [We] E. Weitz, Untersuchungen u ¨ber die Grundlagen der pcf-Theorie von Saharon Shelah, Dissertation, Hannover 1996.

List of Symbols ≈ .............................. P(M ) . . . . . . . . . . . . . . . . . . . . . . . 1, ZF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ZFC . . . . . . . . . . . . . . . . . . . . . . . . . . . a ∈ A .......................... ∅ ............................... {a, b} . . . . . . . . . . . . . . . . . . . . . . . . . . {a}  ............................ a ............................ a ∪ b ........................... {a1 , . . . , an } . . . . . . . . . . . . . . . . . . . . z⊆ a . . . . . . . . . . . . . . . . . . . . . . . . . . A ........................... (AC) . . . . . . . . . . . . . . . . . . . . . . . . . . . vj . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ¬ .............................. ∧ .............................. ∨ .............................. =⇒ . . . . . . . . . . . . . . . . . . . . . . . . . . . ⇐⇒ . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∀ .............................. ∃ .............................. = .............................. ∈ .............................. ( ............................... ) ............................... LZF . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∀vi ∈ vj . . . . . . . . . . . . . . . . . . . . . . . . ∃vi ∈ vj . . . . . . . . . . . . . . . . . . . . . . . . = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∈ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  .............................. ϕ(x, p) . . . . . . . . . . . . . . . . . . . . . . . . .

1 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8

{x : ϕ(x, p)} . . . . . . . . . . . . . . . . . . . 8 A = B ......................... 8 A ∈ B ......................... 8 y ∈ A .......................... 8 A = y .......................... 8 V .............................. 9 A ⊆ B ......................... 9 A ⊂ B ......................... 9 A  B ......................... 9 (x, y) . . . . . . . . . . . . . . . . . . . . . . . . . . 9 A × B ......................... 9 {(x, y) : ϕ(x, y)} . . . . . . . . . . . . . . . 9 A ∪ B .......................... 9 A ∩ B .......................... 9 A \ B .......................... 9  A . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 A . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Rel(R) . . . . . . . . . . . . . . . . . . . . . . . . 10 xRy . . . . . . . . . . . . . . . . . . . . . . . . . . 10 R |`A . . . . . . . . . . . . . . . . . . . . . . . . . 10 R−1 . . . . . . . . . . . . . . . . . . . . . . . . . . 10 dom(R) . . . . . . . . . . . . . . . . . . . . . . . 10 ran(R) . . . . . . . . . . . . . . . . . . . . . . . . 10 fld(R) . . . . . . . . . . . . . . . . . . . . . . . . 10 R[A] . . . . . . . . . . . . . . . . . . . . . . . . . . 10 R ◦ S . . . . . . . . . . . . . . . . . . . . . . . . . 10 Func(F ) . . . . . . . . . . . . . . . . . . . . . . 10 F (u) . . . . . . . . . . . . . . . . . . . . . . . . . . 10 F : A −→ B . . . . . . . . . . . . . . . . . . 10 (F (a) : a ∈ A) . . . . . . . . . . . . . . . . 10 Ba . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 (B  a : a ∈ A) . . . . . . . . . . . . . . . . . . 10 a∈A F (a) . . . . . . . . . . . . . . . . . . . . 10

290 A

B ........................... idA . . . . . . . . . . . . . . . . . . . . . . . . . . . F −1 . . . . . . . . . . . . . . . . . . . . . . . . . . F : (A, R) −→ (B, S) . . . . . . . . . . wf(R) . . . . . . . . . . . . . . . . . . . . . . . . . max(B) . . . . . . . . . . . . . . . . . . . . . . . min(B) . . . . . . . . . . . . . . . . . . . . . . . (A, R) . . . . . . . . . . . . . . . . . . . . . . . . sup(X) . . . . . . . . . . . . . . . . . . . . . . . inf(X) . . . . . . . . . . . . . . . . . . . . . . . . Z ............................. ON . . . . . . . . . . . . . . . . . . . . . . . . . . . α, β, γ, σ, τ, ξ, ζ . . . . . . . . . . . . . . . . κ, λ, μ, ν, ρ . . . . . . . . . . . . . . . . . . . . σ < τ ......................... σ ≤ τ ......................... (α, β) . . . . . . . . . . . . . . . . . . . . . . . . . (α, β)ON . . . . . . . . . . . . . . . . . . . . . . (α, β] . . . . . . . . . . . . . . . . . . . . . . . . . (α, β]ON . . . . . . . . . . . . . . . . . . . . . . [α, β)ON . . . . . . . . . . . . . . . . . . . . . . [α, β]ON . . . . . . . . . . . . . . . . . . . . . . . S(x) . . . . . . . . . . . . . . . . . . . . . . . . . . 0 ............................. Lim(α) . . . . . . . . . . . . . . . . . . . . . . . Lim . . . . . . . . . . . . . . . . . . . . . . . . . . . ω ............................. 1 ............................. 2 ............................. R ............................. x

............................. rec(F, R, G) . . . . . . . . . . . . . . . . . . . Vα . . . . . . . . . . . . . . . . . . . . . . . . . . . . rk(x) . . . . . . . . . . . . . . . . . . . . . . . . . tc(x) . . . . . . . . . . . . . . . . . . . . . . . . . . otp(B, ≺) . . . . . . . . . . . . . . . . . . . . . fin(α, β) . . . . . . . . . . . . . . . . . . . . . . α + β ......................... α · β .......................... αβ . . . . . . . . . . . . . . . . . . . . . . . . . . . . α#β . . . . . . . . . . . . . . . . . . . . . . . . . .

List of Symbols 10 10 10 11 11 11 11 11 12 12 13 15 15 15 15 15 17 17 17 17 17 17 18 18 18 18 19 19 19 20 20 21 24 25 26 28 28 31 31 31 37

|x| . . . . . . . . . . . . . . . . . . . . . . . . . . . . |x| ≤ |y| . . . . . . . . . . . . . . . . . . . . . . |x| = |y| . . . . . . . . . . . . . . . . . . . . . . κ+ . . . . . . . . . . . . . . . . . . . . . . . . . . . . CN . . . . . . . . . . . . . . . . . . . . . . . . . . . ICN . . . . . . . . . . . . . . . . . . . . . . . . . . Pfin . . . . . . . . . . . . . . . . . . . . . . . . . . . ℵ ............................. ℵ0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . ℵα . . . . . . . . . . . . . . . . . . . . . . . . . . . . ωα . . . . . . . . . . . . . . . . . . . . . . . . . . . . μ+β . . . . . . . . . . . . . . . . . . . . . . . . . . . (κ, μ)CN . . . . . . . . . . . . . . . . . . . . . . (κ, μ]CN . . . . . . . . . . . . . . . . . . . . . . . [κ, μ)CN . . . . . . . . . . . . . . . . . . . . . . . [κ, μ]CN . . . . . . . . . . . . . . . . . . . . . . . κ + λ ......................... κ · λ .......................... κλ . . . . . . . . . . . . . . . . . . . . . . . . . . . . (GCH) . . . . . . . . . . . . . . . . . . . . . . . . Q ............................. [a]λ . . . . . . . . . . . . . . . . . . . . . . . . . . . an . . . . . . . . . . . . . . . . . . . . . . . . . . . . cf (A, R) . . . . . . . . . . . . . . . . . . . . . . cf (α) . . . . . . . . . . . . . . . . . . . . . . . . . Reg . . . . . . . . . . . . . . . . . . . . . . . . . . . (κ, μ)reg . . . . . . . . . . . . . . . . . . . . . . (κi : i ∈ I) . . . . . . . . . . . . . . . . . .  i∈I κi . . . . . . . . . . . . . . . . . . . . . . . (κi : i ∈ I) . . . . . . . . . . . . . . . . . . i∈I κi . . . . . . . . . . . . . . . . . . . . . . . κ<λ . . . . . . . . . . . . . . . . . . . . . . . . . . . [a]<λ . . . . . . . . . . . . . . . . . . . . . . . . . [a]≤λ . . . . . . . . . . . . . . . . . . . . . . . . . [a]>λ . . . . . . . . . . . . . . . . . . . . . . . . . [a]≥λ . . . . . . . . . . . . . . . . . . . . . . . . . (SCH) . . . . . . . . . . . . . . . . . . . . . . . . ‫(ג‬κ) . . . . . . . . . . . . . . . . . . . . . . . . . .  ............................. (Xξ : ξ < α) . . . . . . . . . . . . . . . . ξ<α Xξ . . . . . . . . . . . . . . . . . . . . . .

42 42 42 44 44 44 44 45 45 45 45 45 45 45 45 45 45 45 45 46 46 48 48 49 49 50 50 58 58 58 58 60 60 60 60 60 70 74 74 79 79

List of Symbols (Xξ : ξ < α) . . . . . . . . . . . . . . . . 79 ξ<α Xξ . . . . . . . . . . . . . . . . . . . . . . 79 (fip) . . . . . . . . . . . . . . . . . . . . . . . . . . 82 S! . . . . . . . . . . . . . . . . . . . . . . . . 82, 85 F |`X . . . . . . . . . . . . . . . . . . . . . . . . . 83 dom(I) . . . . . . . . . . . . . . . . . . . . . . . 83 Fcof . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Σα,λ . . . . . . . . . . . . . . . . . . . . . . . . . . 85 I[B] . . . . . . . . . . . . . . . . . . . . . . . . . . 86 I |`X . . . . . . . . . . . . . . . . . . . . . . . . . 86 Id (α) . . . . . . . . . . . . . . . . . . . . . . . . . 86 Ib (A) . . . . . . . . . . . . . . . . . . . . . . . . . 86 F + . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 reg∗ (F ) . . . . . . . . . . . . . . . . . . . . . . . 93 μ → (λ)νκ . . . . . . . . . . . . . . . . . . . . . 96 ρ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . 98 l(T ) . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Sβ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 I + . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 f
291 tcf  (S/I) . . . . . . . . . . . . . . . . . . . . . f .......................... cf (S/D) . . . . . . . . . . . . . . . . . . . . . tcf (S/D) . . . . . . . . . . . . . . . . . . . . limI (f ) . . . . . . . . . . . . . . . . . . . . . . f¯ . . . . . . . . . . . . . . . . . . . . . . . . . . . . f [I] . . . . . . . . . . . . . . . . . . . . . . . . . . pcf(g) . . . . . . . . . . . . . . . . . . . . . . . pcf(a) . . . . . . . . . . . . . . . . . . . . . . . . pcfμ (a) . . . . . . . . . . . . . . . . . . . . . . limI (a) . . . . . . . . . . . . . . . . . . . . . . limD (a) . . . . . . . . . . . . . . . . . . . . . . sup+ (B) . . . . . . . . . . . . . . . . . . . . . ppκ (λ) . . . . . . . . . . . . . . . . . . . . . . . pp+ κ (λ) . . . . . . . . . . . . . . . . . . . . . . pp(λ) . . . . . . . . . . . . . . . . . . . . . . . . pp+ (λ) . . . . . . . . . . . . . . . . . . . . . . J<λ (a) . . . . . . . . . . . . . . . . . . . . . . H(κ) . . . . . . . . . . . . . . . . . . . . . . . . . L ............................ ϕ(x0 , . . . , xn−1 ) . . . . . . . . . . . . . . ∃!vj . . . . . . . . . . . . . . . . . . . . . . . . . T ⊆ ZF . . . . . . . . . . . . . . . . . . . . . . A = (A, E) . . . . . . . . . . . . . . . . . . L∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . L∗ZF . . . . . . . . . . . . . . . . . . . . . . . . . (A, E, <∗ ) . . . . . . . . . . . . . . . . . . . A |= ϕ[a0 , . . . , an−1 ] . . . . . . . . . . A |= ϕ(a0 , . . . , an−1 ) . . . . . . . . . A |= ϕ . . . . . . . . . . . . . . . . . . . . . . . A |= T . . . . . . . . . . . . . . . . . . . . . . . A ⊆ B ....................... A  B ....................... ϕM . . . . . . . . . . . . . . . . . . . . . . . . . . ϕ abs M . . . . . . . . . . . . . . . . . . . . . Δ0 . . . . . . . . . . . . . . . . . . . . . . . . . . ΔT0 . . . . . . . . . . . . . . . . . . . . . . . . . . ΔT1 . . . . . . . . . . . . . . . . . . . . . . . . . . {x : ϕ(x, y )}M . . . . . . . . . . . . . . . ZFC− . . . . . . . . . . . . . . . . . . . . . . .  {Ai : i < γ} . . . . . . . . . . . . . . . .

132 132 134 134 139 143 151 152 152 152 152 152 156 156 156 156 156 162 173 176 177 177 178 178 178 178 178 178 179 179 179 179 179 180 181 181 181 182 185 186 196

292  {Ai : i < γ} . . . . . . . . . . . . . . . . χM (α) . . . . . . . . . . . . . . . . . . . . . . . ϕ∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . SkolΘ (X) . . . . . . . . . . . . . . . . . . . . fϕ . . . . . . . . . . . . . . . . . . . . . . . . . . . F  A . . . . . . . . . . . . . . . . . . . . . . . . . A ............................ ecl-sequence . . . . . . . . . . . . . . . . . 3club . . . . . . . . . . . . . . . . . . . . . . . . bλ . . . . . . . . . . . . . . . . . . . . . . . . . . . (bλ : λ ∈ c) . . . . . . . . . . . . . . . . . . λb . . . . . . . . . . . . . . . . . . . . . . . . . . . (f¯b : b ⊆ a) . . . . . . . . . . . . . . . . . . Fb . . . . . . . . . . . . . . . . . . . . . . . . . . .

List of Symbols 196 196 199 200 200 200 202 202 204 218 219 221 222 224

p (a) . . . . . . . . . . . . . . . . . . . . . . J<λ ∗ f ........................... J∗ (b) . . . . . . . . . . . . . . . . . . . . . . . . pcf∗ (b) . . . . . . . . . . . . . . . . . . . . . . E+ . . . . . . . . . . . . . . . . . . . . . . . . . . Id (C) . . . . . . . . . . . . . . . . . . . . . . . . fini . . . . . . . . . . . . . . . . . . . . . . . . . . Rini . . . . . . . . . . . . . . . . . . . . . . . . . . t[a0 , . . . , an−1 ] . . . . . . . . . . . . . . . (A, (fk : k < ω)) . . . . . . . . . . . . . J(κ) . . . . . . . . . . . . . . . . . . . . . . . . . ξA (f ) . . . . . . . . . . . . . . . . . . . . . . . . p(λ) . . . . . . . . . . . . . . . . . . . . . . . . . λ −→ [λ]2λ . . . . . . . . . . . . . . . . . . .

234 235 236 241 244 244 252 252 253 254 254 259 262 262

Index absolute class, 185 formula for M , 181 addition , cardinal, 45 , ordinal, 31 , principal number of, 37 additively decomposable, 37 indecomposable, 37 aleph function, 45 almost all, 104 almost disjoint functions, 112 alphabet, 7, 251 antichain, 11 antisymmetric, 11 approximation sequence in H(Θ), 196 arity, 10, 252 asymmetric, 11 atomic formula, 8 axiom of choice (AC), 7 , comprehension (A3), 6 , empty set (A2), 6 of extensionality (A1), 6 , foundation (A9), 7 , infinity (A7), 6 , pairing (A4), 6 , power set (A6), 6 , regularity (A9), 7 , replacement (A8), 7 , subset (A3), 6

of a theory, 8, 178 , union (A5), 6 axiom schema of comprehension (A3), 6 of replacement (A8), 7 axiom system, 8 of Zermelo-Fraenkel, 5 ZF, 6 ZFC, 6 axiom of a theory, 8, 178 Bachmann, H., 287 base in a Cantor normal form, 37 ω, 37 of a topology, 79 Bernstein, F., vii, 43, 65 Bernstein’s formula, 65 Cantor-Schr¨ oder-Bernstein theorem, 43 Bendixson, 55 Cantor-Bendixson theorem, 55 beth function, 78 bijection, 10 bijective, 10 binary, 10 bound , greatest lower, 12 , least upper, 12 , lower, 12 , upper, 12 modulo I, 131 variable, 177 bounded quantifier, 8

294 branch α-branch, 98 , length of a, 98 of a tree, 98 Bukovsky, L., 76 Bukovsky-Hechler theorem, 76 Burke, M. R., 287 Cantor, G., vii, 1, 2, 5, 36, 37, 43, 55 Cantor-Bendixson theorem, 55 Cantor normal form, 37 theorem, 36 Cantor-Schr¨ oder-Bernstein theorem, 43 set, 55 cardinal, 42 addition, 45 comparability, 42 exponentiation, 45 , finite, 44 , inaccessible, 71 , infinite, 44 , κ-strong, 71 , limit, 44 , Mahlo, 94 , measurable, 92 multiplication, 45 number, 42 power, 45 product, 45, 58 , Ramsey, 102 , regular, 50 , singular, 50 , strong limit, 71 , strongly inaccessible, 71 , successor, 44 sum, 45, 58 , transfinite, 44 , variable for a, 15 , weakly compact, 102 , weakly inaccessible, 71

Index cardinal arithmetic , the four golden rules of, 66 cardinality, 42 of a basis of a vector space, 56 of the set of bijections from κ onto κ, 57 of a closed set of reals, 56 of the set of continuous real functions, 55 of the set of injections from λ ≥ ℵ0 into κ, 57 of the set of isolated points of a set of reals, 55 of the set of open sets of reals, 55 of a perfect set of reals, 56 of the set of polynomials over a field F , 55 of the set of Riemann integrable real functions, 55 of a structure, 253 of the set of surjections from κ ≥ ℵ0 onto λ, 56 of a vector space, 56 cardinal number of a set, 42 cartesian product, 7, 9 of a family, 10 chain, 11 , elementary, 196 of structures, 196 character, finite, 53 characteristic function, 196 characterizes D characterizes M over b, 228 choice , axiom of, 7 , countable, 54 , dependent, 54 function, 7 class, 8 , absolute, 185

Index , equivalence, 13 , extensional, 26 , proper, 9 of R-predecessors, 20 subclass, 9 , transitive, 15 , universal, 9 variables, 9 closed formula, 177 interval, 17 set in a topology, 79 subset of [λ]<κ , 95 subset of an ordinal, 79 closure, transitive, 26 club, 79 filter, 83 in [λ]<κ , 95 sequence, 79 , special, 79 cofinal, 49 in f , 132 modulo I, 131 sequence modulo I, 132 subset, 49 cofinality of (A, R), 49 modulo I, 132 of an ordinal, 49 , true, 132 Cohen, P. J., 2, 67 collapsing isomorphism, 26 lemma, 26 collection principle, 192 compact topological space, 54 comparable, 11 compatible, 14 complete, 86 induction, 19 κ-complete, 86

295 composition, 10 comprehension axiom schema, 6 concatenation, 8, 252 connected, 11 connectives , sentential, 7 constant below, 74 constant symbol, 252 continuous, 30 chain of structures, 196 function, 30 μ-continuous, 209 sequence, 30 continuum, 46 function, 74 hypothesis, 46 , generalized, 46 control function, 225 set, 225 countable, 44 choice, 54 set, 44 countably infinite, 44 cover, open, 54 cumulative hierarchy, 24 cuts diagonally modulo I, 146 decomposable, 37 decreasing, 12 Dedekind cut, 46 definable with parameters from N , 189 denumerable, 44 dependent choices, 54 diagonal intersection, 79 union, 79 diamond club sequence, 204 difference, 9 directed λ-directed, 136

296 disjoint , almost, 112 classes, 10 I-disjoint, 139 division lemma, 34 domain of a function, 10 of an ideal, 83 of a relation, 10 Drake, F. R., 71, 178, 287 dual filter, 83 ideal, 83 Easton, W. B., 3, 73, 76, 287 ecl-sequence, 202 element , greatest, 11 , least, 11 , maximal, 11 modulo I, 132 , minimal, 11 , R-maximal, 11 , R-minimal, 11 , smallest, 11 elementary chain, 196 equivalent, 194 extension, 179 substructure, 179, 254 embedding, 11 empty set, 6 axiom, 6 enumeration, 42 equinumerous, 1, 42 equivalence class, 13 relation, 11 Erd¨ os, P., 100, 101, 258, 287 Erd¨ os-Hajnal-Rado theorem, 258

Index Erd¨ os-Rado partition theorem, 100, 101 eventually constant, 74 from ρ0 on, 74 exact upper bound modulo I, 147 exponentiation , cardinal, 45 , ordinal, 31 , principal number of, 39 expression, 8, 252 extended formula, 8 extends D extends I, 104 J extends I, 104 extension, elementary, 179 extensionality, axiom of, 6 family, 10 field of a relation, 10 filter, 82 clubfilter, 83 , dual, 83 , Fr´echet, 91 generated by S, 83 , κ-complete, 91 , κ-incomplete, 91 , κ-regular, 92 , normal, 83 , regular, 92 , σ-complete, 91 , σ-incomplete, 91 , weakly κ-regular, 93 finite cardinal, 44 character, 53 , hereditarily, 29 set, 44 finite intersection property, 82 first order predicate calculus, 7, 251 Fodor, G., 87 Fodor’s theorem, 87

Index formal language, 7, 251 formula, 8, 252 absolute for M , 181 , atomic, 8 , closed, 177 Δ0 -formula, 181 ΔT0 -formula, 181 ΔT1 -formula, 182 , extended, 8 , relativized, 180 Σ1 -formula, 193 foundation, axiom of, 7 Fraenkel, A., vii, 2, 5 Zermelo-Fraenkel axiom system, 5 Fr´echet filter, 91 free ultrafilter, 83 variable, 177 function, 10 , aleph, 45 almost disjoint functions, 112 , beth, 78 , bijective, 10 , characteristic, 196 , choice, 7 , continuous, 30 , continuum, 74 , control, 225 , decreasing, 12 from A into B, 10 , gimel, 74 , identity, 10 , increasing, 12 , injective, 10 , inverse, 10 , monotone, 12 , normal, 30 , one-to-one, 10 onto, 10 , ordinal, 30

297 , progressive, 151 , regressive, 87 , set, 108 , Skolem, 200 , strictly increasing, 12 , surjective, 10 functional property, 7 Galvin, F., vii, 3, 112, 122, 123, 287 Galvin-Hajnal formula, 112 Galvin-Hajnal lemma, 122 Galvin-Hajnal theorem, 123 generalized continuum hypothesis (GCH), 46 Hausdorff formula, 65 generated, 83, 86 generator, 209 sequence of generators, 219 gimel function, 74 G¨ odel, K., 2, 3, 67, 71 greatest element, 11 lower bound, 12 Hajnal, A., vii, 3, 112, 122, 123, 258, 287 Galvin-Hajnal formula, 112 Galvin-Hajnal lemma, 122 Galvin-Hajnal theorem, 123 Hartogs, F., 43 number, 43 theorem, 43 Hausdorff, F., vii, 54, 64, 65 generalized Hausdorff formula, 65 Hausdorff’s formula, 64 Hausdorff’s maximality principle, 54 Hebrew lexicographic ordering, 28, 33 Hechler, S., 76

298 Bukovsky-Hechler theorem, 76 hereditarily of cardinality less than κ, 173 finite, 29 hierarchy , cumulative, 24 , von Neumann, 24 holds in A, 179, 254 under a valuation, 179 modulo I, 104 homogeneous set, 96 homomorphism, 10 hull of a set, 202 , Skolem, 200 hypothesis , continuum, 46 , generalized continuum, 46 , singular cardinal (SCH), 70 ideal, 83 of bounded subsets, 86 , domain of an, 83 , dual, 83 generated by M , 86 generated over an ideal, 86 , image, 151 , κ-complete, 86 , λ-mighty, 160 , maximal, 83 , normal, 86 restriction of an ideal to a set, 86 , σ-complete, 86 of thin subsets, 86 , weakly normal, 244 identity function, 10 on a set, 10 image of a class under a relation, 10 ideal, 151

Index inaccessible cardinal, 71 , strongly, 71 , weakly, 71 incomparable, 12 increasing, 12 modulo I, 132 indecomposable, additively, 37 independent of a theory, 6 set, 108 induction , complete, 19 ∈-induction, 21 , finite, 28 principle, 19 , transfinite, 20 inductive set, 6 infimum, 12 infinite cardinal, 44 product, 58 set, 44 sum, 58 infinity, axiom of, 6 initial ordinal, 45 segment, 12 determined by u, 29 injection, 10 injective function, 10 interpretation of a term, 253 intersection of classes, 9 , diagonal, 79 of structures, 196 interval, 17 of cardinals, 45 , closed, 17 , half-open, 17 , open, 17 of regular cardinals, 50

Index inverse of a function, 10 of a relation, 10 isolated point, 55 isomorphism, 11 , collapsing, 26 , Mostowski, 26 , order, 28 Jech, T. J., vii, 42, 67, 73, 102, 139, 160, 282, 287 Jech’s theorem, 282 J´ onsson algebra, 254 Just, W., 287 K¨ onig, D., 99 D. K¨ onig’s theorem, 99 K¨ onig, J., vii, 3, 61 J. K¨ onig’s lemma, 61 κ-complete, 86 κ-rapid modulo I, 145 κ-regular filter, 92 , weakly, 93 κ-ub sequence modulo I, 143 Kunen, K., 178, 288 language of first order logic, 251 , formal, 7 of ZF, 7 large μ-large modulo I, 139 L´az´ar’s lemma, 109 least element, 11 least upper bound, 12 legitimate, 177 length of a branch, 98 of a tree, 98 less modulo I, 104 , pointwise, 104 level

299 , von Neumann, 24 of a tree, 98 Levy, A., 9, 195, 288 limit cardinal, 44 limit of a function modulo D, 139 modulo I, 139 limit ordinal, 18 limit point, 79 limit of a sequence, 80 limit of a set modulo D, 152 modulo I, 152 linear ordering, 11 localisation theorem, 242 logarithm lemma, 34 L¨owenheim-Skolem theorem, 180, 254 lower bound, 12 Magidor, M., vii, 67, 288 Mahlo cardinal, 94 M´ at´e, A., 287 main lemma, 148 main theorem, 215 maximal element, 11 modulo I, 132 maximality principle, Hausdorff’s, 54 maximum, 11 measurable cardinal, 92 mighty λ-mighty, 160 minimal element, 11 minimum, 11 model of a formula, 179 of a theory, 179 modulo I, 104 monotone, 12 monotonic, 12

300 Mostowski, A., 26 Mostowski’s collapsing lemma, 26 Mostowski isomorphism, 26 multiplication , cardinal, 45 , ordinal, 31 , principal number of, 39 natural number, 19 sum, 37 von Neumann hierarchy, 24 , level of the, 24 nice D-nice, 229 nonprincipal ultrafilter, 83 nonreflecting stationary set, 90 nonstationary, 83 norm, 110 normal filter, 83 form, 37 function, 30 ideal, 86 , weakly, 244 sequence, 30 null set, 103 number , cardinal, 42 , Hartogs, 43 , natural, 19 , ordinal, 15 principal number of addition, 37 principal number of exponentiation, 39 principal number of multiplication, 39 one-to-one, 10 open cover, 54 interval, 17

Index set in a topology, 79 order isomorphism, 28 topology, 79 type, 28 ordered pair, 9 ordering , generalized Hebrew lexicographic, 28 , Hebrew lexicographic, 33 , linear, 11 , partial, 11 , strict partial, 11 well-ordering, 11 ordinal, 15 addition, 31 , Cantor normal form of an, 37 exponentiation, 31 function, 30 , initial, 45 , limit, 18 multiplication, 31 number, 15 product, 31 , regular, 50 , successor, 18 sum, 31 , variable for an, 15 pair , ordered, 9 set, 6 , unordered, 6 parameter, 8 partial ordering, 11 partition, 13, 96 pcf(a), 152 pcf(g), 152 perfect set of real numbers, 55 pigeonhole principle, 96, 97 polynomial over a field, 56 pointwise

Index less than, 104 supremum, 106 positive set, 104 possible cofinality, 152 power , cardinal, 45 set, 6 ppκ -strong, 283 predecessor R-predecessor, 20 predicate calculus, 7 prime ideal, 83 principal ideal, 107 principal number of addition, 37 of exponentiation, 39 of multiplication, 39 principal ultrafilter, 83 principle , collection, 192 of dependent choices, 54 , ∈-induction, 21 , induction, 19 , reflection, 193 , transfinite induction, 20 , well-ordering, 40 product , cardinal, 45 , cartesian, 7, 9 of a family of cardinals, 58 , ordinal, 31 progressive function, 151 set, 152 proper class, 9 initial segment, 12 property, 8 , functional, 7 quantifier, 7 , bounded, 8

301 Rado, R., 100, 101, 258, 287 Erd¨ os-Hajnal-Rado theorem, 258 Erd¨ os-Rado partition theorem, 100, 101 Ramsey, F. P., 100, 102 cardinal, 102 Ramsey’s theorem, 100 range, 10 rank in the cumulative hierarchy, 25 function of a tree, 98 of a function, 110 rapid κ-rapid modulo I, 145 recursion, 21 theorem, 21 , transfinite, 21 transfinite recursion principle for ON, 23 transfinite recursion theorem schema for well-founded relations, 21 reflection principle, 193 M reflects ϕ, 193 regressive function, 87 regular cardinal, 50 filter, 92 ordinal, 50 relation, 10 , antisymmetric, 11 , asymmetric, 11 , binary, 10 , connected, 11 , equivalence, 11 , extensional, 26 , inverse, 10 , irreflexive, 11 , n-ary, 10 , reflexive, 11

302 , symmetric, 11 , transitive, 11 , trichotomic, 11 , well-founded, 11 relativization of a formula, 180 replacement axiom, 7 restriction, 10 of an ideal, 86 of a relation, 10 Rubin, H., 42, 288 Rubin, J. E., 42, 288 Russell, B., 5 Schr¨ oder, E., 43 Cantor-Schr¨ oder-Bernstein theorem, 43 scope of a quantifier, 177 segment, initial, 12 sentence, 8, 177 sequence, 30 , approximation, 196 , club, 79 , continuous, 30 club -sequence, 204 ecl-sequence, 202 of generators, 219 , κ-rapid, 145 , κ-ub, 143 , limit of a, 80 , normal, 30 , special, 222 set, 9 of almost disjoint functions, 112 , Cantor, 55 closed in [λ]<κ , 95 in an ordinal, 79 in a topology, 79 unbounded, 79 , control, 225 , countable, 44 , countably infinite, 44

Index , definable with parameters from N , 189 , denumerable, 44 , empty, 6 , finite, 44 function, 105 , homogeneous, 96 , independent, 108 , inductive, 6 , infinite, 44 of measure zero, 103 of natural numbers, 19 , nonreflecting, 90 , nonstationary, 83 , null, 103 , open, 79 , pair, 6 , perfect, 55 , positive, 104 of possible cofinalities, 152 , power, 6 , progressive, 152 , stationary, 83 , thin, 83 , transitive, 15 , unbounded, 49 in an ordinal, 79 variables, 7 Shelah, S., vii, 3, 117, 129, 130, 139, 148, 215, 233, 247, 272, 273, 285, 288 Shelah’s main lemma, 148 Shelah’s main theorem, 215 Silver, J., vii, 3, 125, 288 Silver’s theorem, 125 singleton, 6 singular cardinal, 50 singular cardinal hypothesis (SCH), 70 singular cardinal problem, 103 Skolem, Th., 5, 180, 200, 254

Index function, 200, 254 hull, 200 L¨owenheim-Skolem theorem, 180, 254 small modulo I, 146 Solovay, R. M., 89 Solovay’s theorem, 89 special sequence, 222 stationary in [λ]<κ , 95 set, 83 , nonreflecting, 90 strictly decreasing, 12 strictly increasing, 12 modulo I, 132 strict partial ordering, 11 string of symbols, 8 strong κ-strong, 71 limit cardinal, 71 ppκ -strong, 283 structure, 178, 252 , cardinality of a, 253 for LZF , 178 for L∗ZF , 178 , universe of a, 178, 252 subclass, 9 subformula, 176 subset axiom, 6 substraction lemma, 34 substructure, 179, 254 , elementary, 179, 254 successor, 18 , cardinal, 44 , ordinal, 18 of a set, 18 sum , cardinal, 45 of a family of cardinals, 58 , ordinal, 31 supremum, 12

303 modulo I, 131 , pointwise, 106 surjection, 10 surjective function, 10 symbol, 7, 251 , constant, 252 , function, 252 , relation, 252 symmetric relation, 11 Tarski, A., vii, 3, 64, 67, 71 Tarski’s formula, 64 Tarski’s recursion formula, 71 Teichm¨ uller, O., 53 Teichm¨ uller-Tukey lemma, 53 term, 252 theory, 8, 178 thin, 83 topology, 79 , base of a, 79 , compact, 54 generated by B, 79 , order, 79 transfinite induction principle, 20 recursion, 21 transitive class, 15 closure, 26 Todorcevic, 263 Todorcevic’s partition theorem, 263 tree, 98 α-tree, 98 , branch of a, 98 , level of, 98 true cofinality modulo I, 132 Tryba-Woodin theorem, 256 Tukey, J. W., 53 Teichm¨ uller-Tukey lemma, 53 Tychonoff’s theorem, 54

304 ultrafilter, 82 , free, 83 generated by {a}, 83 , nonprincipal, 83 , principal, 83 , unbounded, 86 , uniform, 91 unbounded in A, 49 modulo I, 131 subset of an ordinal, 79 ultrafilter, 86 union , diagonal, 79 of classes, 10 of structures, 196 universal closure, 177 universal class, 9 universe, 178, 252 unordered pair, 6 upper bound, 12 modulo I, 131 , exact, 147 valuation, 179 , ϕ holds in A under a, 179 value, 10 variable, 7 , bound, 177

Index for cardinals, 15 for classes, 9 , free, 177 for ordinal functions, 104 for ordinals, 15 for sets, 9 von Neumann, see Neumann, von weakly compact, 102 inaccessible, 71 κ-regular filter, 93 normal, 244 Weese, M., 287 Weitz, E., 288 well-founded relation, 11 well-ordering, 11 principle, 40 Woodin, W. H., 256 Tryba-Woodin theorem, 256 Zermelo, E., vii, 2, 5 Zermelo-Fraenkel set theory, 5 zero, 18 ZF, 6 ZFC, 6 Zorn, M., 40 Zorn’s lemma, 40 , set version of, 53