A Course on Set Theory

www.cambridge.org/9781107008175

A Course on Set Theory

Set theory is the mathematics of infinity and part of the core curriculum formathematics majors. This book blends theory and connections with other parts ofmathematics so that readers can understand the place of set theory within the widercontext. Beginning with the theoretical fundamentals, the author proceeds to illustrateapplications to topology, analysis and combinatorics, as well as to pure set theory.Concepts such as Boolean algebras, trees, games, dense linear orderings, ideals, filtersand club and stationary sets are also developed.

Pitched specifically at undergraduate students, the approach is neither esoteric norencyclopedic. The author, an experienced instructor, includes motivating examples andover 100 exercises designed for homework assignments, reviews and exams. It isappropriate for undergraduates as a course textbook or for self-study. Graduatestudents and researchers will also find it useful as a refresher or to solidify theirunderstanding of basic set theory.

ernest schimmerling is a Professor of Mathematical Sciences at CarnegieMellon University, Pennsylvania.

A Course on Set Theory

ERNEST SCHIMMERLINGCarnegie Mellon University, Pennsylvania

cambridge university pressCambridge, New York, Melbourne, Madrid, Cape Town,

Singapore, Sao Paulo, Delhi, Tokyo, Mexico City

Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK

Published in the United States of America by Cambridge University Press, New York

www.cambridge.orgInformation on this title: www.cambridge.org/9781107008175

C© E. Schimmerling 2011

This publication is in copyright. Subject to statutory exceptionand to the provisions of relevant collective licensing agreements,no reproduction of any part may take place without the written

permission of Cambridge University Press.

First published 2011

Printed in the United Kingdom at the University Press, Cambridge

A catalogue record for this publication is available from the British Library

Library of Congress Cataloguing in Publication data

ISBN 978-1-107-00817-5 HardbackISBN 978-1-107-40048-1 Paperback

Cambridge University Press has no responsibility for the persistence oraccuracy of URLs for external or third-party internet websites referred to inthis publication, and does not guarantee that any content on such websites is,

or will remain, accurate or appropriate.

Contents

Note to the instructor page viiAcknowledgements x

1 Preliminaries 12 ZFC 73 Order 22

3.1 Wellorderings 233.2 Ordinal numbers 283.3 Ordinal arithmetic 41

4 Cardinality 534.1 Cardinal numbers 534.2 Cardinal arithmetic 604.3 Cofinality 67

5 Trees 825.1 Topology fundamentals 825.2 The Baire space 855.3 Illfounded and wellfounded trees 965.4 Infinite games 1055.5 Ramsey theory 1155.6 Trees of uncountable height 119

6 Dense linear orderings 1256.1 Definitions and examples 1256.2 Rational numbers 1286.3 Real numbers 131

7 Filters and ideals 1417.1 Motivation and definitions 1417.2 Club and stationary sets 153

vi Contents

Appendix Summary of exercises on Boolean algebra 163Selected further reading 164Bibliography 166Index 167

Note to the instructor

This book was written for an undergraduate set theory course,which is taught at Carnegie Mellon University every spring. Itis aimed at serious students who have taken at least one proof-based mathematics course in any area. Most are mathematics orcomputer science majors, or both, but life and physical science,engineering, economics and philosophy students have also donewell in the course. Other students have used this book to learnthe material on their own or as a refresher. Mastering this bookand learning a bit of mathematical logic, which is not included,would prepare the student for a first-year graduate level set theorycourse in the future. The book also contains the minimum amountof set theory that everyone planning to go on in math should know.

I have included slightly more than the maximum amount ofmaterial that I have covered in a fifteen-week semester. But I donot reach the maximum every time; in fact, only once. For a slowerpace or shorter academic term, one of several options would be toskip Sections 5.6 and 7.2, which are more advanced.

There are over one hundred exercises, more than enough foreight homework assignments, two midterm exams, a final examand review problems before each exam. Exercises are located atthe ends of Chapters 1, 2, 3, 4 and 6. They are also dispersedthroughout Chapters 5 and 7. This slight lack of uniformity istied to the presentation and ultimately makes sense.

In roughly the first half of the book, through Chapter 4, I de-velop ordinal and cardinal arithmetic starting from the axioms ofZermelo–Fraenkel Set Theory with the Axiom of Choice (ZFC). Inother words, this is not a book on what some call naive set theory.There is one minor way in which the presentation is not entirely

viii Note to the instructor

rigorous. Namely, in listing the axioms of ZFC, I use the impreciseword property instead of the formal expression first-order formulabecause mathematical logic is not a prerequisite for the course.

Some other textbooks develop the theory of cardinality for aslong as possible without using the Axiom of Choice (AC). I donot take this approach because it adds technicalities, which arenot used later in the course, and gives students the misleadingimpression that AC is controversial. By assuming AC from thestart, I am able to streamline the theory of cardinality. I may notehow AC has been used in a proof but I do not belabor the point.Once, when an alternate proof without AC exists, it is outlined inan exercise.

The second half of the book is designed to give students a senseof the place of set theory within mathematics. Where I draw con-nections to other fields, I include all the necessary backgroundmaterial. Some of the other areas that come up in Chapter 5 aretopology, metric spaces, trees, games and Ramsey theory. Thereal numbers are constructed using Dedekind cuts in Chapter 6.Chapter 7 introduces the student to filters and ideals, and takes upthe combinatorics of uncountable sets. There is no section specif-ically on Boolean algebra but it is one of the recurring themes inthe exercises throughout the book. For the reader’s convenience,I have briefly summarized the results on Boolean algebra in theAppendix. All of this material is self-contained.

As I mentioned, before starting this book, students should haveat least one semester’s worth of experience reading and writingproofs in any area of mathematics; it does not matter which area.They should be comfortable with sets, relations and functions,having seen and used them at a basic level earlier. They shouldknow the difference between integers, rational numbers and realnumbers, even if they have not seen them explicitly constructed.And they should have experience with recursive definitions alongthe integers and proofs by induction on the integers. These no-tions come up again here but in more sophisticated ways than ina first theoretical mathematics course. There are no other pre-requisites. However, because of the emphasis on connections toother fields, students who have taken courses on logic, analysis,algebra, or discrete mathematics will enjoy seeing how set theoryand these other subjects fit together. The unifying perspective of

Note to the instructor ix

set theory will give students significant advantages in their futuremathematics courses.

Acknowledgements

As an undergraduate, I studied from Elements of set theory byHerbert Enderton and Set theory: an introduction to independenceproofs by Kenneth Kunen. When I started teaching undergraduateset theory, I recommended Introduction to set theory by KarelHrbacek and Thomas Jech to my students. The reader who knowsthese other textbooks will be aware of their positive influence.

This book began as a series of handouts for undergraduate stu-dents at Carnegie Mellon University. Over the years, they foundtypographical errors and indicated what needed more explana-tion, for which I am grateful. I also thank Michael Klipper forproofreading a draft of the book in Spring 2008, when he was agraduate student in the CMU Doctor of Philosophy program.

During the writing of this book, I was partially supported byNational Science Foundation Grant DMS-0700047.

1

Preliminaries

In one sense, set theory is the study of mathematics using thetools of mathematics. After millennia of doing mathematics, math-ematicians started trying to write down the rules of the game.Since mathematics had already fanned out into many subareas,each with its own terminology and concerns, the first task wasto find a reasonable common language. It turns out that every-thing mathematicians do can be reduced to statements about sets,equality and membership. These three concepts are so fundamen-tal that we cannot define them; we can only describe them. Aboutequality alone, there is little to say other than “two things areequal if and only if they are the same thing.” Describing sets andmembership has been trickier. After several decades and somefalse starts, mathematicians came up with a system of laws thatreflected their intuition about sets, equality and membership, atleast the intuition that they had built up so far. Most importantly,all of the theorems of mathematics that were known at the timecould be derived from just these laws. In this context, it is com-mon to refer to laws as axioms, and to this particular system asZermelo–Fraenkel Set Theory with the Axiom of Choice, or ZFC.In the first unit of the course, through Chapter 4, we examine thissystem and get some practice using it to build up the theory ofinfinite numbers.

In another sense, set theory is a part of mathematics like anyother, rich in ideas, techniques and connections to other areas.This perspective is emphasized more than the foundational aspectsof set theory throughout the course but especially in the secondhalf, Chapters 5–7. There, our choice of topics within set theory is

2 Preliminaries

designed to give the reader an impression of the depth and breadthof the subject and where it fits within the whole of mathematics.

To get started, we review some basic notation and terminology.We expect that the reader is familiar with the following notionsbut perhaps has not seen them expressed in exactly the same way.

Ordered pairs are used everywhere in mathematics, for example,to refer to points on the plane in geometry. The precise meaningof (x, y) is left to the imagination in most other courses but weneed to be more specific.

Definition 1.1 (x, y) = x, x, y is the ordered pair withfirst coordinate x and second coordinate y.

It is convenient that (x, y) is defined in terms of sets. After all,this is set theory, so everything should be a set! The main pointof the definition is that from looking at x, x, y we can tellwhich is the first coordinate and which is the second coordinate.Namely, if x, x, y has exactly two elements, then the firstcoordinate is

x = the unique z such that z ∈ x, x, yand the second coordinate is

y = the unique z = x such that x, z ∈ x, x, y.And, if x, x, y has just one element, which can only happenif x = y, then the first and second coordinates are both

x = the unique z such that z ∈ x.To understand this formula, keep in mind that

x, y = y, xand

x, x = x.In particular,

x, x, x = x, x = xand x is the only element of x.Definition 1.2 A × B = (x, y) | x ∈ A and y ∈ B is theCartesian product of A and B.

Preliminaries 3

Definition 1.3 R is a relation from A to B iff R is a subset ofA×B, that is

R ⊆ A×B.

Sometimes, if we know that R is a relation, then we write xRyinstead of (x, y) ∈ R. For example, we write

√2 < π

not

(√

2, π) ∈<

because the latter is confusing.

Definition 1.4 Let R be a relation from A to B and S ⊆ A.

1. The domain of R is

dom(R) = x ∈ A | there exists y such that xRy.

2. The image of S under R is

R[S] = y ∈ B | there exists x ∈ S such that xRy.

3. The range of R is

ran(R) = y ∈ B | there exists x such that xRy.

Notice that ran(R) = R[dom(R)].

Definition 1.5 f is a function from A to B iff f is a relationfrom A to B and, for every x ∈ A, there exists a unique y suchthat (x, y) ∈ f .

If we happen to know that f is a function, then we write

f(x) = y

instead of (x, y) ∈ f . When we write f : A→ B, it is implicit thatf is a function from A to B. In certain situations, we refer to afunction f by writing x → f(x) or 〈f(x) | x ∈ A〉. There are timeswhen we write fx instead of f(x); this is when we are thinking ofelements x of A as indices and 〈fx | x ∈ A〉 as an indexed family.If the domain of f consists of ordered pairs, then it is commonto write f(x, x′) instead of f((x, x′)). Functions are also called

4 Preliminaries

operations and maps. Some people distinguish between a functionf : A → B and its graph,

graph(f) = (x, f(x)) | x ∈ A,

but we do not. To us they are the same, that is, f = graph(f), aswe see from Definition 1.5.

Definition 1.6 If f : A → B is a function and S ⊆ A, then therestriction of f to S is

f S = (x, f(x)) | x ∈ S.

Definition 1.7 Let f : A → B be a function.

1. f is an injection iff for all x, x′ ∈ A, if x = x′, then f(x) = f(x′).2. f is a surjection iff for every y ∈ B, there exists x ∈ A such

that f(x) = y.3. f is a bijection iff f is both an injection and a surjection.

Injections are also called one-to-one functions. Surjections fromA to B are also called functions from A onto B. Bijections arealso called one-to-one correspondences.

Definition 1.8 If f is an injection from A to B, then we writef−1 for the unique injection g : f [A] → A with the property thatg(f(x)) = x for every x ∈ A. In other words,

f−1 = (f(x), x) | x ∈ A.

Finally, we assume that the reader has good intuition about theset of integers,

Z = . . . ,−2,−1, 0, 1, 2, . . . ,

the set of rational numbers,

Q = m/n | m, n ∈ Z and n = 0

and the set of real numbers, R. One thing we will do in this courseis define all these kinds of numbers, starting from the naturalnumbers 0, 1, 2, 3, 4, etc. Each natural number will be the set ofnatural numbers that precedes it. Thus 0 = ∅, where ∅ is the setwith no members. After that, 1 = 0, 2 = 0, 1, 3 = 0, 1, 2,

Preliminaries 5

4 = 0, 1, 2, 3, etc. This happens to be very convenient becausethen

m < n ⇐⇒ m ∈ n.

In other words, the usual ordering on the natural numbers coin-cides with membership.

We use natural numbers to denote cardinality, for example,when we say, “Lance Armstrong won the Tour de France seventimes.” And we use natural numbers to denote order, for example,when we say, “the attorney general is seventh in the presidentialline of succession.” Another thing we will do in this course is ex-tend the notions of cardinality and order into the infinite. Finitecardinal and ordinal numbers are basically the same thing; onecould say that the difference between “seven” and “seventh” isjust grammatical. However, the difference between infinite cardi-nal and ordinal numbers is more profound, as we will explain inChapters 3 and 4.

Exercises

Exercise 1.1 If R is a relation, then we define

R−1 = (y, x) | xRy.

Give an example where R is a function but R−1 is not.

Exercise 1.2 How many functions whose domain is the emptyset are there? In other words, given a set B, how many functionsf : ∅ → B are there?

Exercise 1.3 Explain why (x, y, z) = (x, (y, z)) is a reasonabledefinition of an ordered triple.

Exercise 1.4 Equivalence relations play an important role inthis book. We assume that the reader has studied them beforebut this exercise reviews all the necessary definitions and facts.Let A be a set and R be a relation on A, that is, R ⊆ A × A.Then:

• R is a reflexive relation on A iff for every x ∈ A, xRx.• R is a symmetric relation on A iff for all x, y ∈ A, if xRy, then

yRx.

6 Preliminaries

• R is a transitive relation on A iff for all x, y, z ∈ A, if xRy andyRz, then xRz.1

• R is an equivalence relation on A iff R is a reflexive, symmetricand transitive relation on A.

Assuming that R is an equivalence relation on A, for every x ∈ A,we define the equivalence class of x to be

[x]R = y ∈ A | xRy.It is also standard to write

A/R = [x]R | x ∈ A.A partition of A is a family F of non-empty subsets of A such that

• A is the union of F , that is,

A =⋃F = x | there exists X ∈ F such that x ∈ X

and• the elements of F are pairwise disjoint, that is, for all X, Y ∈ F ,

if X = Y , then X ∩ Y = ∅.Now here are the exercises:

1. Let R be an equivalence relation on A. Prove that A/R is apartition of A.

2. Let F be a partition of A. Prove that there exists a uniqueequivalence relation R such that F = A/R.

1 Later in the book we will define transitive set, which is different from transitiverelation. Unfortunately, it will be important to pay attention to this subtledifference in terminology.

2

ZFC

In the most general terms, when we talk about a mathematical the-ory, we have in mind a collection of axioms in a certain language.The language of set theory has two symbols, = and ∈, althoughsometimes we add symbols that are defined in terms of these twoto make things easier to read. For example, we write A ⊆ B whenwe mean that, for every x, if x ∈ A, then x ∈ B.

Zermelo–Fraenkel Set Theory with the Axiom of Choice, or ZFCfor short, is a certain theory in the language of set theory that wewill describe in this chapter. There are infinitely many axioms ofZFC, each of which says something rather intuitive about sets,equality and membership. In our list below, some axioms of ZFCare presented individually whereas others are presented as schemesfor generating infinitely many axioms. One last comment aboutterminology before we begin: throughout the course,

set = collection = family

and

member = element.

Also, the three phrases,

• x belongs to A,• x is an element of A and• x is a member of A,

all mean the same thing, namely x ∈ A.

8 ZFC

Empty Set Axiom

This axiom says that there is a unique set without members. For-mally, it is written

∃!A ∀x (x ∈ A) .

In plain English, this says:

There exists a unique A such that, for every x,x is not an element of A.

The unique set without elements is written ∅.

Extensionality Axiom

This axiom says that two sets are equal if they have the samemembers. Formally, it is written

∀A ∀B [ ∀x (x ∈ A ⇐⇒ x ∈ B) =⇒ A = B ] .

Because we defined

A ⊆ B ⇐⇒ ∀x (x ∈ A =⇒ x ∈ B) ,

another way to write the Extensionality Axiom is

∀A ∀B [ (A ⊆ B and B ⊆ A) =⇒ A = B ] .

In other words, two sets are equal if each is a subset of the other.By logic alone, if A = B, then A and B have the same members.

Combining this fact with the Extensionality Axiom, we have that

∀A ∀B [ ∀x (x ∈ A ⇐⇒ x ∈ B) ⇐⇒ A = B ] .

Equivalently,

∀A ∀B [ (A ⊆ B and B ⊆ A) ⇐⇒ A = B ] .

Pairing Axiom

This axiom allows us to form singletons and unordered pairs. Itsformal statement is

∀x ∀y ∃!A ∀z [z ∈ A ⇐⇒ (z = x or z = y)] .

If x = y, then we write x, y for the unique set whose onlymembers are x and y and call it an unordered pair. We always

ZFC 9

write x instead of x, x and call it a singleton. At this point,it makes sense to define the first three natural numbers 0 = ∅,1 = 0 and 2 = 0, 1. We can also justify defining ordered pairsby setting

(x, y) = x , x, y

whenever we are given x and y as we did in Definition 1.1. As areminder, when x = y, what we really have is

(x, x) = x .

Notice that, based on this definition, when we write (x, y), we cantell that x is the first coordinate and y is the second coordinate.Formally, this means we can prove that for all x, y, x′ and y′,

(x, y) = (x′, y′) ⇐⇒ (x = x′ and y = y′).

Union Axiom

This axiom allows us to form unions. Its formal statement is

∀F ∃!A ∀x [x ∈ A ⇐⇒ ∃Y ∈ F (x ∈ Y )] .

We write⋃F for the unique set whose members are exactly the

members of the members of F . In other words,⋃F = x | there exists Y ∈ F such that x ∈ Y .

It is important to note that, in the Union Axiom, the family F isallowed to be infinite. We often use different notation when F isfinite. For example, we define

A ∪B =⋃A, B

and

A ∪B ∪ C =⋃A, B, C.

At this point, we can define the remaining natural numbers

3 = 2 ∪ 2 = 0, 1, 2,

4 = 3 ∪ 3 = 0, 1, 2, 3,

5 = 4 ∪ 4 = 0, 1, 2, 3, 4

10 ZFC

and, in general,

n + 1 = n ∪ n = 0, . . . , n.

Power Set Axiom

This axiom allows us to form the set of all subsets of a given set.Its formal statement is

∀A ∃!F ∀X (X ∈ F ⇐⇒ X ⊆ A).

We write P(A) for the unique set of subsets of A. In other words,

P(A) = X | X ⊆ A.We call P(A) the power set of A. As an example, let us see whathappens when we start with the empty set and take power setsover and over. Define

V0 = ∅,

V1 = P(V0) = ∅,

V2 = P(V1) = ∅, ∅,

V3 = P(V2) = ∅, ∅, ∅, ∅, ∅and, in general,

Vn+1 = P(Vn).

The sets Vn will come up again later.

Comprehension Scheme

This axiom scheme gives us a way to form specific subsets of agiven set. It says the following.

For each “property” P (x), the following is an axiom:

∀A ∃!B ∀x [x ∈ B ⇐⇒ (x ∈ A and P (x))] .

Notice that the word “property” appears in quotes. There areinfinitely many properties, which is why ZFC has infinitely manyaxioms. We will not give a formal definition of “property” becauseit involves first-order logic, which is not a prerequisite. It is enough

ZFC 11

for students in this course to depend on their intuition about themeaning of “property”. Given a property P (x), we write

x ∈ A | P (x)

for the set of elements x of A for which P (x) is true. For example,

x ∈ 10 | x is even = 0, 2, 4, 6, 8

and

x ∈ 10 | x is odd = 1, 3, 5, 7, 9.

It is important to note that the Comprehension Scheme does not,in general, permit us to define sets by writing x | P (x). In fact,for some P (x), what we think of as x | P (x) is not a set. Forexample x | x = x is not a set by the result in Exercise 2.6.

At this point, we are justified in making many familiar defini-tions. For example, Definition 1.2 says that, given sets A and B,we have the Cartesian product

A×B = (x, y) | x ∈ A and y ∈ B .

In order to see that this is a legitimate definition, note that ifx ∈ A and y ∈ B, then

(x, y) = x , x, y ∈ P (P (A ∪B)) .

Thus A×B =

z ∈ P (P (A ∪B)) | z = (x, y) for some x ∈ A and y ∈ B ,

which means the definition of A×B is justified by a combinationof the axioms we have listed so far. See Exercise 2.4.

Infinity Axiom

In an indirect way, this axiom tells us that the set of naturalnumbers exists. Its formal statement is

∃I [ ∅ ∈ I and ∀x (x ∈ I =⇒ x ∪ x ∈ I) ].

We say that a set I is inductive if I is a witness to the InfinityAxiom. In other words,

I is inductive ⇐⇒ [ ∅ ∈ I and ∀x (x ∈ I =⇒ x ∪ x ∈ I) ].

12 ZFC

Proposition 2.1 There is a unique inductive set I such thatI ⊆ J for every inductive set J .

The proof of Proposition 2.1 is broken up into smaller steps inExercise 2.5. The set of natural numbers is defined to be the uniqueinductive set that is a subset of every other inductive set. We writeω (lower case Greek omega) for the set of natural numbers. Inother words,

ω = 0, 1, 2, 3, . . . .It is also common for mathematicians to write N instead of ωalthough we will not. Not only does the Infinity Axiom allow usto define ω, it implies that we can prove statements by inductionon n ∈ ω and make recursive definitions for n ∈ ω just as you havedone in your other mathematics courses.

To see the relationship with induction, suppose we wish to provethat a given property P (n) holds for every natural number n. ByProposition 2.1 and the definition of ω, it would be enough toshow that n ∈ ω | P (n) is an inductive set. In other words,show that P (0) holds and if P (n) holds, then so does P (n + 1).

Replacement Scheme

This is a scheme for generating infinitely more axioms.

For each “property” P (x, y), the following is an axiom:

∀A [(∀x ∈ A ∃y P (x, y)) =⇒ (∃B ∀x ∈ A ∃y ∈ B P (x, y))] .

The same comments we made about meaning of “property” whenwe discussed the Comprehension Scheme apply here too.

Because we will not emphasize how the Replacement Scheme isused later in the book, let us give a concrete example and someintuition here. Suppose we want to define

Vω =⋃Vn | n ∈ ω .

That is, we want to let Vω be the union of the infinite family

Vn | n ∈ ω.This family is supposed to be the range of the infinite sequence

〈Vn | n ∈ ω〉.

ZFC 13

But why does this infinite sequence exist? In other words, why isit a set? Given a particular natural number, say 5, we can provefrom the other axioms that the finite sequence

〈Vm | m < 5〉 = 〈V0, V1, V2, V3, V4〉

exists. The Replacement Scheme can be used to make the leapfrom infinitely many finite sequences to one infinite sequence. Tosee what we mean, let P (x, y) be the following property: x is anatural number and there exists a function f with domain

dom(f) = x + 1 = 0, . . . , x

such that

• f(0) = ∅,• for every n < x, f(n + 1) is the power set of f(n), and• f(x) = y.

Then P (n, Vn) holds for every n ∈ ω. By the Replacement Scheme,there is a set B such that, for every n ∈ ω, Vn ∈ B. Now use theComprehension Scheme to define

〈Vn | n < ω〉 = (x, y) ∈ ω ×B | P (x, y).

Finally, use the Union Axiom to define Vω as we originally wantedto do.

A slogan that captures the intuition behind the ReplacementScheme is: If an assignment looks like a function and its domainis a set, then its range is also a set, so it really is a function. Inthe example above, we used the Replacement Scheme to see thatthe assignment n → Vn really is a function with domain ω andrange Vn | n ∈ ω.

Foundation Axiom

This axiom says that if S is a non-empty set, then there existsx ∈ S such that, for every y ∈ S, y ∈ x. In symbols,

∀S (S = ∅ =⇒ ∃x ∈ S ∀y ∈ S (y ∈ x)).

For example, should S = 2, the only witness to the FoundationAxiom would be x = 0 because 2 = 0, 1 and 0 ∈ 0 = 1 but1 ∈ ∅ = 0. On the other hand, when S = 0, 1, both elements

14 ZFC

of S satisfy the requirement of the Foundation Axiom because0 ∈ 1 and 1 ∈ 0.

The Foundation Axiom turns out to be equivalent to the state-ment that there is no sequence 〈xn | n ∈ ω〉 such that

· · ·xn+1 ∈ xn ∈ · · · ∈ x1 ∈ x0.

In particular, it implies that no set is an element of itself. Forotherwise, if x ∈ x, then

· · ·x ∈ x ∈ x ∈ x,

which contradicts the Foundation Axiom. We will make additionalcomments about this axiom later but it is not a focus of the book.

Axiom of Choice

This axiom says that every family of sets has a choice function.

∀F ∃ function c ∀A ∈ F (A = ∅ =⇒ c(A) ∈ A) .

The function c is called a choice function for F because it choosesan element c(A) out of every non-empty A that belongs to F . Fromexperience, given finitely many non-empty sets, you can pick oneelement from each. The Axiom of Choice says this is possible evenif you start with infinitely many non-empty sets.

The first time the Axiom of Choice is used is in Chapter 4.There, it is essential for being able to assign a numerical cardinal-ity (size) to each set, which is one of the most important thingswe do in this book. As we go along, we will point out where andhow the Axiom of Choice is used.

Exercises

In the following exercises, you only need to be attentive to theaxioms of ZFC when so instructed, namely, in Exercises 2.1, 2.4,2.6, the second part of 2.8 and the first part of 2.11. Otherwise,use the same style of mathematical argumentation as in your otherproof-based mathematics courses without reference to ZFC.

ZFC 15

Exercise 2.1 Prove that the following theories are equivalent.

1. Empty Set Axiom + Extensionality Axiom.2. ∃A ∀x (x ∈ A) + Extensionality Axiom.

Hint: Obviously 1 implies 2. In the other direction, uniqueness isthe issue.

Exercise 2.2 Recall that 0 = ∅ and n+1 = 0, . . . , n for everyn ∈ ω. Recall also that V0 = ∅ and Vn+1 = P(Vn) for every n ∈ ω.

1. Prove by induction that if n ∈ ω, then P(n) has 2n elements.2. List the elements of V4.3. Make a conjecture regarding the size of Vn. Then prove your

conjecture by induction on n ∈ ω.

Exercise 2.3 Define z to be a transitive set iff for all x and y,if x ∈ y and y ∈ z, then x ∈ z. This is equivalent to saying that,for every y, if y ∈ z, then y ⊆ z.

1. Prove that Vn is a transitive set for every n ∈ ω.2. Prove that Vn ⊆ Vn+1 for every n ∈ ω.3. Prove that Vn ∩ ω = n for every n ∈ ω.

Exercise 2.4 Give a detailed explanation of how the definitionsof A × B, dom(R) and R[S] given in Definitions 1.2 and 1.4 arejustified by the axioms of ZFC.

Exercise 2.5 Prove Proposition 2.1 by showing the following.

1. If I and J are inductive sets, then I ∩ J is an inductive set.2. Suppose that K is an inductive set. Let

F = J ∈ P(K) | J is an inductive setand

I = x ∈ K | ∀J ∈ F ( x ∈ J ) .

Prove the following statements.(a) I is an inductive set.(b) For every J , if J is an inductive set, then I ⊆ J .(c) Suppose that I ′ is an inductive set and, for every inductive

set J ,I ′ ⊆ J.

Then I ′ = I.

16 ZFC

Exercise 2.6 Prove that there is no set V such that x ∈ V forevery set x. (This shows that x | x = x is not a set.) If youuse the Foundation Axiom in your proof, then find a second proofthat does not use the Foundation Axiom.

Exercise 2.7 In general, define the intersection of a non-emptyfamily G to be ⋂

G = a | ∀X ∈ G ( a ∈ X ).

Let S be a set and F ⊆ P(S). Assume that F = ∅. Prove that

S −⋂F =

⋃S −X | X ∈ F

andS −

⋃F =

⋂S −X | X ∈ F.

Exercise 2.8 In general, defineAB = f | f is a function from A to B.

Consider a function of the form (a, b) → S(a,b) with domain A×B.In other words, consider an indexed family

〈S(a,b) | (a, b) ∈ A×B〉.1. Prove that ⋂

a∈A

⋃b∈B

S(a,b) =⋃

f∈A B

⋂a∈A

S(a,f (a)).

2. Assume in addition that

S(a,b) ∩ S(a,b′) = ∅whenever a ∈ A and b, b′ ∈ B but b = b′. Now prove the sameequation as in part 1 but without using the Axiom of Choice.

Exercise 2.9 By definition, the symmetric difference of A andB is

AB = (A−B) ∪ (B −A).

Verify the distributive law

A ∩ (B C) = (A ∩B) (A ∩ C).

Remark: This is one step in showing that if S is a set, then P(S)

ZFC 17

is a ring if addition is taken to be symmetric difference and mul-tiplication is taken to be intersection. We only mention this forreaders who happen to know some abstract algebra.

Exercise 2.10 Define a relation E on P(ω) according to theformula

(x, y) ∈ E ⇐⇒ x y is finite.

(See Exercise 2.9 for the definition of x y.)

1. Prove that E is an equivalence relation on P (ω).

2. We write [x]E for the equivalence class of x, in other words,

[x]E = y ∈ P(ω) | xEy.

Prove that for every E-equivalence class [x]E is infinite.

3. By P (ω) /E we mean the family of E-equivalence classes, thatis,

P (ω) /E = [x]E | x ∈ P(ω).

Prove that P (ω) /E is infinite.

Exercise 2.11 Let A be a set. By recursion on n ∈ ω, define

B0 = A

and

Bn+1 =⋃

Bn.

Let

C =⋃Bn | n < ω.

1. Which axioms of ZFC are used to see that C is a set?2. Prove that C is a transitive set. (See Exercise 2.3 for the defi-

nition of transitive set.)3. Prove that if D is a transitive set and A ⊆ D, then C ⊆ D.

We call C the transitive closure of A.

18 ZFC

Beginning exercises on Boolean algebra

To understand Exercises 2.12 and 2.13 below, and many exerciseslater in the book,1 you must first read the following definitionsand example.

A Boolean algebra is a 6-tuple of the form

B = (B,∨,∧,¬,⊥,)

where B is a set, ∨ and ∧ are binary operations on B, ¬ is a unaryoperation on B, ⊥ and are distinct elements of B, and for allX, Y, Z ∈ B, the following ten laws hold.

Associativity

X ∨ (Y ∨ Z) = (X ∨ Y ) ∨ Z

X ∧ (Y ∧ Z) = (X ∧ Y ) ∧ Z

Commutativity

X ∨ Y = Y ∨X

X ∧ Y = Y ∧X

Distributivity

X ∨ (Y ∧ Z) = (X ∨ Y ) ∧ (X ∨ Z)X ∧ (Y ∨ Z) = (X ∧ Y ) ∨ (X ∧ Z)

Identity

X ∨ ⊥ = X

X ∧ = X

Complementation

X ∨ ¬X = X ∧ ¬X = ⊥

Each Boolean algebra, B = (B,∨,∧,¬,⊥,), has an associatedBoolean algebra relation, , which is defined by

X Y ⇐⇒ X = X ∧ Y

for all X, Y ∈ B. Here is some special terminology for Booleanalgebras that will be used in the exercises. If A ∈ B, then we say1 The Appendix lists the exercises to which we are referring.

ZFC 19

that A is an atom iff A = ⊥ and, for every X ∈ B, if X A, thenX = ⊥ or X = A. We say that B is finite iff B is finite.

Given Boolean algebras

B = (B,∨B,∧B,¬B,⊥B,B)

andC = (C,∨C,∧C,¬C,⊥C,C),

an isomorphism from B to C is defined to be a function f suchthat

• f is a bijection from B to C and• for all X, Y ∈ B,

f(X ∨B Y ) = f(X) ∨C f(Y ),

f(X ∧B Y ) = f(X) ∧C f(Y ),

f(¬BX) = ¬Cf(X),

f(⊥B) = ⊥C

andf(B) = C.

Example You should work out the details of the following asser-tions before attempting the exercises that follow. Let S = ∅ andput

B(S) = (P(S),∪,∩,−, ∅, S)

where −X means S −X in this context. Then B(S) is a Booleanalgebra. It is called the Boolean algebra of subsets of S. If S isfinite and has n elements, then P(S) has 2n elements hence B(S)is finite. For B(S), the Boolean algebra relation amounts to

X Y ⇐⇒ X = X ∩ Y ⇐⇒ X ⊆ Y.

The atoms of B(S) are exactly the singletons a for a ∈ S. Thefunction X → S−X is an isomorphism from B(S) to the Booleanalgebra

(P(S),∩,∪,−, S, ∅).Notice that union and intersection are exchanged, as are ∅ and S.

20 ZFC

Exercise 2.12 Consider an arbitrary finite Boolean algebra

B = (B,∨,∧,¬,⊥,).

Let S be the set of atoms of B.

1. Prove that if X ∈ B and X = ⊥, then there exists A ∈ S suchthat A X.

2. Let X ∈ B. Suppose that X = ⊥ and

A ∈ S | A X = A1, . . . , Am.Let

Y = A1 ∨ · · · ∨Am.

ProveX = Y.

Hint: Certain basic facts about sets generalize to Boolean al-gebras. For example, for sets we know that

if X ⊆ Y and Y ⊆ X, then X = Y ,

and for Boolean algebras we have that

if X Y and Y X, then X = Y .

The reason is that if X Y and Y X, then

X = X ∧ Y = Y ∧X = Y

by the commutativity law for ∧ and the definition of . Themoral is that you should base your intuition about Booleanalgebras on what you already know about Boolean algebras ofsets. Of course, ultimately, you need to prove your intuitionis correct using just the laws of Boolean algebras. It shouldalso be said that the solution to this exercise is relatively longso organizing your answer into a well-chosen series of lemmaswould be very helpful.

3. Let f : B → P(S) be defined by

f(X) = A ∈ S | A X.Prove that f is an isomorphism from B to B(S). Remark: Thisexplains why intuition coming from Boolean algebras of setsreally is valuable intuition about finite Boolean algebras.

ZFC 21

Exercise 2.13 As in Exercise 2.10, let E be the equivalencerelation on P(ω) defined by

x E y ⇐⇒ x y is finite.

1. Prove that the following table of equations determines a Booleanalgebra B = (B,∨,∧,¬,⊥,).

B = P(ω)/E

[x]E ∨ [y]E = [x ∪ y]E[x]E ∧ [y]E = [x ∩ y]E

¬[x]E = [ω − x]E⊥ = [∅]E = [ω]E

Before proving the laws of Boolean algebras, you must showthat the operations ∨, ∧ and ¬ are well-defined by the equationslisted above. So part of what you must show is that if x E x′

and y E y′, then(x ∪ y)E (x′ ∪ y′),

(x ∩ y)E (x′ ∩ y′)

and(ω − x)E (ω − x′).

Remark: This is an example of a quotient Boolean algebra. Inthe literature, it is referred to as P(ω)/Finite.

2. Prove that the Boolean algebra B that was defined in part 1has no atoms.

3

Order

At the end of Chapter 1, we gave examples of sentences in Englishthat illustrated the difference between ordinal and cardinal num-bers. Let us expand on our example of ordinal numbers, whichinvolved the presidential line of succession:

1st Vice President

2nd Speaker of the House

3rd President pro tempore of the Senate

4th Secretary of State

5th Secretary of the Treasury

6th Secretary of Defense

7th Attorney General

8th Secretary of the Interior

9th Secretary of Agriculture

10th Secretary of Commerce

11th Secretary of Labor

12th Secretary of Health and Human Services

13th Secretary of Housing and Urban Development

14th Secretary of Transportation

15th Secretary of Energy

3.1 Wellorderings 23

For fun, here are a few more:

16th Secretary of Education

17th Secretary of Veterans Affairs

18th Secretary of Homeland Security

What is the point? In English, we usually count ordinal numbers1st, 2nd, 3rd, etc. However, sometimes it makes sense to count 0th,1st, 2nd, 3rd, etc. For example, at the top of the presidential lineof succession, we really have:

0th President

In set theory, we start counting ordinals from 0. For example, inthe sequence of length 6,

〈x0, x1, x2, x3, x4, x5〉 = 〈11, 6, 18, 9, 72, 31〉

the 0th number is x0 = 11, the 1st number is x1 = 6, etc., andthe 5th and final number is x5 = 31. Keep in mind that, in plainEnglish, it would be very strange to say that the fifth item is thelast in a list of six items!

In set theory, we also continue counting ordinal numbers pastall the finite ordinal numbers. For example, ω is the first infiniteordinal number, followed by ω + 1, ω + 2, etc. It takes a bit oftheoretical work to make concrete sense of this idea, so this iswhat we do first.

3.1 Wellorderings

Definition 3.1 Let A be a set and ≺ be a relation on A.

1. (A,≺) is transitive iff for all x, y, z ∈ A, if x ≺ y and y ≺ z,then x ≺ z.

2. (A,≺) is irreflexive iff for every x ∈ A, x ≺ x.3. (A,≺) is total iff for all x, y ∈ A, either x ≺ y or x = y or

y ≺ x.

24 Order

4. (A,≺) is a strict linear ordering iff it is transitive, irreflexiveand total.

Definition 3.2 Let ≺ and be two relations on A. Supposethat

x y ⇐⇒ (x ≺ y or x = y)

for all x, y ∈ A. Then (A, ) is a linear ordering iff (A,≺) is astrict linear ordering.

The definition tells us how to pass from a strict linear orderingto its associated linear ordering. In the other direction, we havethe fact that if (A, ) is a linear ordering and

x ≺ y ⇐⇒ (x y and x = y)

for all x, y ∈ A, then (A,≺) is a strict linear ordering and (A, )is the linear ordering associated to (A,≺). Occasionally, we mightdrop the word strict when it is clear from the choice of symbolsor context which we mean.

Here is another remark on notation. If we are given that (A,≺)is a strict linear ordering, then we might write without botheringto explain that it is the linear ordering associated to (A,≺) eventhough officially we should explain. Suppose, instead, that we aretold that (A, R) is a strict linear ordering. It is unlikely that wewould write R for the associated linear ordering because it looksso strange. If, for some reason, we wrote R, then we could notassume that the reader knows what we mean, so we would have toexplain. Officially, ≺ and are two completely different symbolseven though looks like a combination of ≺ and =.

Definition 3.3 Let (A,≺) be a strict linear ordering, S ⊆ Aand x ∈ S. Then x is the ≺-least element of S iff for every y ∈ S,x y.

Definition 3.4 (A,≺) is a wellordering iff it is a strict linearordering and, for every non-empty S ⊆ A, S has a ≺-least element.

The extra property that turns a strict linear ordering into awellordering is called wellfoundedness. Every finite linear orderingis a wellordering. Also, the usual linear ordering of the natural


numbers is a wellordering. On the other hand, the usual linearordering on the set of integers,

Z = · · · − 2,−1, 0, 1, 2, . . . ,

is obviously not a wellordering because there is no least integer.However, if we define a brand new relation ≺ on Z by

0 ≺ 1 ≺ −1 ≺ 2 ≺ −2 ≺ 3 ≺ −3 ≺ · · · ,

then (Z,≺) is a wellordering.

Lemma 3.5 Let (A,≺) be a strict linear ordering. Then (A,≺)is a wellordering iff there is no sequence 〈xn | n ∈ ω〉 such that,for every n ∈ ω, xn+1 ≺ xn.

Proof First suppose that (A,≺) is a wellordering. Consider anarbitrary sequence 〈xn | n ∈ ω〉. Let S = xn | n ∈ ω. Let y bethe ≺-least element of S. Say y = xn. Then xn xn+1. Hencexn+1 ≺ xn.

Second, suppose that (A,≺) is not a wellordering. Therefore,there exists S ⊆ A such that S = ∅ and S does not have a ≺-least element. Define a sequence xn by recursion on n ∈ ω asfollows. For the base case, let x0 ∈ S if possible. Otherwise, leavex0 undefined. For the successor case, let xn+1 ∈ S with xn+1 ≺ xn

if possible. Otherwise, leave xn+1 undefined.We claim that, for every n ∈ ω, xn is defined, xn ∈ S and,

if n ≥ 1, then xn ≺ xn−1. We prove this claim by induction onn ∈ ω. The base case is n = 0. Since S = ∅, it is possible tolet x0 ∈ S. So that is what we did. The successor case has theinduction hypothesis that xn is defined and xn ∈ S. Since S doesnot have a ≺-least element, it is possible to let xn+1 ∈ S withxn+1 ≺ xn. So that is what we did. Note that x(n+1)−1 = xn.

In the second part of the previous proof, we made a definition byrecursion on n ∈ ω and proved a statement by induction on n ∈ ω.Our ability to do this is tied to the fact that the usual ordering ofω is a wellordering. The following fundamental theorems put thephenomena of induction and recursion in the general context ofwellorderings.

Theorem 3.6 (Proofs by induction) Let (A,≺) be a wellorder-ing. Let P (x) be a statement about a variable x. Suppose that, for

26 Order

every y ∈ A,

(∀x ≺ y P (x) holds) =⇒ P (y) holds.

Then, for every y ∈ A, P (y) holds.

Proof Let S = x ∈ A | P (x) does not hold. For contradiction,suppose that S = ∅. Let y be the ≺-least element of S. Then,y ∈ S and, for every x ≺ y, x ∈ S. Hence, for every x ≺ y, P (x)holds. From the hypothesis of the theorem, P (y) holds. That is,y ∈ S. This is a contradiction.

Definition 3.7 f is a partial function from A to B iff thereexists A′ ⊆ A such that f : A′ → B.

Theorem 3.8 (Recursive definitions) Let (A,≺) be a wellorder-ing. Let

F : A× P → B

be a function where B is a set and P is the set of partial functionsfrom A to B. Then there is a unique function G : A → B suchthat

G(y) = F (y, G x | x ≺ y)for every y ∈ A.

The point is that the equation in Theorem 3.8 determines thefunction G. We describe this as a recursive definition of G becausethe equation is a recipe for finding G(y) based on the earlier valuesG(x) for x ≺ y. Specifically, the function F outputs G(y) whenwe input y and the restriction of G to x | x ≺ y.

The set P in the Theorem 3.8 might seem a bit mysterious atfirst. Not all partial functions are relevant to the statement andproof of the theorem. Really, we only use that, for every y ∈ A,

G x | x ≺ y ∈ P.

But there would be no advantage in shrinking down P to just thepartial functions we need.

Proof of Theorem 3.8 Let us say that g is a z-approximation iff

• z ∈ A,• g is a partial function from A to B,


• dom(g) = y | y z and• for every y z,

g(y) = F (y, g x | x ≺ y) .

The reason we call it a z-approximation is that g approximates Gup to and including z.

Claim 3.8.1 For every z ∈ A, there exists at most one z-approximation.

Proof of claim Fix z ∈ A. Let g and h be z-approximations. Weprove that g(y) = h(y) by induction on y z using Theorem 3.6.Let y z. Our induction hypothesis is that, for every x ≺ y,

g(x) = h(x).

In other words,

g x | x ≺ y = h x | x ≺ y.Therefore,

g(y) = F (y, g x | x ≺ y) = F (y, h x | x ≺ y) = h(y),

which completes the proof of the claim.

Claim 3.8.2 For every z ∈ A, there exists a z-approximation.

Proof of claim We argue by induction using Theorem 3.6. Letz ∈ A. Our induction hypothesis is that, for every y ≺ z, thereexists a y-approximation; call it gy . By Claim 3.8.1, gy is the onlyy-approximation. Observe that if x ≺ y ≺ z, then

gy w | w xis an x-approximation. Since there is only one x-approximation,

gx = gy w | w xwhenever x ≺ y ≺ z. Let

h =⋃y≺z

gy.

Then h is a partial function from A to B with

dom(h) =⋃y≺z

x | x y = x | x ≺ z

28 Order

and, for every y ≺ z,

gy = h x | x y.

Now let

gz = h ∪ (z, F (z, h).

In other words,

gz(y) =

h(y) if y ≺ z

F (z, h) if y = z.

It is easy to see that gz is a z-approximation.

By the two claims, for each z ∈ A, there is a unique z-approximation,which we call gz . As we already calculated, if y ≺ z, then

gy = gz x | x y.

Define

G =⋃z∈A

gz.

Then G is a function with

dom(G) =⋃z∈A

y | y z = A

and, for every z ∈ A,

G(z) = gz(z) = F (z, gz y | y ≺ z) = F (z, G y | y ≺ z) .

Thus G witnesses the conclusion of Theorem 3.8. To see that G isthe unique witness to the theorem, argue just like in the proof ofClaim 3.8.1.

3.2 Ordinal numbers

We start by repeating a definition from Exercise 2.3.

Definition 3.9 A is a transitive set iff for all x and y, if x ∈ yand y ∈ A, then x ∈ A.

3.2 Ordinal numbers 29

Equivalently, A is a transitive set iff for every y ∈ A, y ⊆ A.The importance of transitivity will not really begin to be apparentuntil Definition 3.15. First, here are three basic facts about howcertain operations preserve transitivity.

Lemma 3.10 Let A be a transitive set. Then A ∪ A is also atransitive set.

Proof Let y ∈ A∪A and x ∈ y. We must show that x ∈ A∪A.In fact, we will show that x ∈ A. Either y ∈ A or y ∈ A. Ify ∈ A, then x ∈ A since A is transitive. If y ∈ A, then y = A,so x ∈ A.

Lemma 3.11 If A is a transitive set, then so is P(A).

Proof Assume that A is transitive, z ∈ P(A) and y ∈ z. Wemust show that y ∈ P(A). By our assumption, z ⊆ A and y ∈ z, soy ∈ A. Since A is transitive, y ⊆ A. Thus, y ∈ P(A) as desired.

Lemma 3.12 If F is a family of transitive sets, then⋃F is a

transitive set too.

Proof Let y ∈⋃F . Pick A ∈ F such that y ∈ A. Since A is

transitive, y ⊆ A. Hence y ⊆⋃F as desired.

As motivation for the next definition, observe that membershipis not a relation. This is because (x, y) | x ∈ y is not a set. Forif it were a set, then it would be a relation and its domain wouldalso be a set. But its domain would be x | x = x, which is nota set by Exercise 2.6. On the other hand, if A is a set, then so is

(x, y) ∈ A×A | x ∈ yby the fact that A×A is a set and the Comprehension Scheme.

Definition 3.13 When we write (A,∈) we really mean

(A, (x, y) ∈ A×A | x ∈ y) .

Similarly, when we say that (A,∈) is transitive, we really meanthat the restriction of ∈ to A is a transitive relation on A.

Do not get confused between A being transitive and (A,∈) beingtransitive. The meaning is different. For example, if A = x,then ∈ is a transitive relation when restricted to A because A onlyhas one element, but A is not a transitive set because x ∈ Aand x ⊆ A. The two notions of transitivity are related though.

30 Order

Lemma 3.14 Suppose that A is a transitive set. Then (A,∈) istransitive iff for every z ∈ A, z is a transitive set.

Proof Given that A is a transitive set, the following six state-ments are equivalent.

1. (A,∈) is transitive.2. ∀ x, y, z ∈ A ((x ∈ y and y ∈ z) =⇒ x ∈ z).3. ∀ z ∈ A ∀ y ∈ z ∩A ∀ x ∈ y ∩A (x ∈ z).4. ∀ z ∈ A ∀ y ∈ z ∀ x ∈ y ∩A (x ∈ z).5. ∀ z ∈ A ∀ y ∈ z ∀ x ∈ y (x ∈ z).6. ∀ z ∈ A (z is a transitive set).

The first equivalence (1 ⇐⇒ 2) is by definition. The second(2 ⇐⇒ 3) is just logic. The third (3 ⇐⇒ 4) is because if z ∈ A,then z ⊆ A since A is a transitive set, so z ∩ A = z. The fourth(4 ⇐⇒ 5) is similar: if z ∈ A and y ∈ z, then y ∈ A (since A is atransitive set), hence y ⊆ A (again since A is a transitive set), soy ∩A = y. The fifth (5 ⇐⇒ 6) is by definition.

Now we come to one of the most important definitions of thecourse.

Definition 3.15 A set α is an ordinal iff α is a transitive setand (α,∈) is a wellordering.

By the Foundation Axiom, α is an ordinal iff α is a transitiveset and (α,∈) is a strict linear ordering. Also, by the FoundationAxiom, (α,∈) is always irreflexive. Combining these observationswith Lemma 3.14 we get the following useful characterization ofwhen a set is an ordinal.

Lemma 3.16 A set α is an ordinal iff α is a transitive set, everyelement of α is a transitive set and (α,∈) is total.

In fact, we have seen some ordinals. If n ∈ ω, then n is anordinal. Also, ω is an ordinal. We have also seen transitive setsthat are not ordinals. For example, for every n ∈ ω, if n > 2, thenVn is a transitive set that is not an ordinal.

It is important to have a reasonably good picture of where weare headed before plunging into technical facts about ordinals. Asyou read this paragraph, beware of the significant work requiredto justify this description of the ordinals, work that is capturedby Lemmas 3.18, 3.19, 3.20, 3.21 and 3.22, and results on ordinal


addition in the next section. Here is the picture you should havein mind. Starting from the empty set, we use the operation

α → α ∪ α

at successor stages and take unions at limit stages to generate allthe ordinals beginning with the natural numbers 0 = ∅, 1 = 0,2 = 0, 1, 3 = 0, 1, 2, etc. The next ordinal after all the naturalnumbers is the set of natural numbers,

ω = 0, 1, 2, . . . .

After ω comes the infinite sequence of ordinals

ω + 1 = 0, 1, 2, . . . , ωω + 2 = 0, 1, 2, . . . , ω, ω + 1ω + 3 = 0, 1, 2, . . . , ω, ω + 1, ω + 2

...

followed by infinitely more ordinals

ω + ω = 0, 1, 2, . . . , ω, ω + 1, ω + 2, . . . ω + ω + 1 = 0, 1, 2, . . . , ω, ω + 1, ω + 2, . . . , ω + ωω + ω + 2 = 0, 1, 2, . . . , ω, ω + 1, ω + 2, . . . , ω + ω, ω + ω + 1

...

after which comes the ordinal

ω + ω + ω.

Skipping ahead, we eventually get to

ω + ω + ω + ω

and, somewhat later, to

ω · ω = ω + · · ·+ ω + · · · .

The list of ordinals never ends. Notice that, for natural numbers,the membership relation ∈ coincides with the usual strict linearordering < on the natural numbers, and that this pattern contin-ues through the ordinals we have listed above. Namely,

0 ∈ 1 ∈ 2 ∈ 3 ∈ · · · ∈ ω ∈ ω + 1 ∈ ω + 2 ∈ ω + 3 ∈ · · · .

32 Order

Returning now to a rigorous exposition, we make the followinggeneral notational rules.

Definition 3.17 If α and β are ordinals, then we may write

α < β ⇐⇒ α ∈ β

and

α ≤ β ⇐⇒ (α < β or α = β).

This convention will allow us to write

0 < 1 < 2 < 3 < · · · < ω < ω + 1 < ω + 2 < ω + 3 < · · ·

with the same meaning as

0 ∈ 1 ∈ 2 ∈ 3 ∈ · · · ∈ ω ∈ ω + 1 ∈ ω + 2 ∈ ω + 3 ∈ · · ·

once we finish developing the theory of ordinals.Next are some important basic principles about ordinals to

which we have alluded above. The proofs might seem confusingthe first time through because they involve all three notions: ∈, <and ⊆. Read slowly and attentively. Remember, in general, oncewe establish that α and β are ordinals (but not before) we are freeto write α < β instead of α ∈ β.

Lemma 3.18 Let β be an ordinal. Then every element of β isan ordinal. Thus

β = α | α is an ordinal and α < β.

Proof Let α ∈ β. By the forward direction of Lemma 3.14, everymember of β is a transitive set. Thus α is a transitive set. Becauseβ is a transitive set, α ⊆ β. Since (β,∈) is a wellordering, sois (α,∈). (Every subset of a wellordered set is also wellordered.)Therefore, α is an ordinal by Definition 3.15.

Lemma 3.19 Let γ and δ be ordinals. Then

γ ≤ δ ⇐⇒ γ ⊆ δ.

Proof First we prove the forward (left to right) direction. Assumeγ ≤ δ. If γ = δ, we are done. So assume γ < δ. In other words,γ ∈ δ. Since δ is a transitive set, γ ⊆ δ. So, again, we are done.

Now, for the proof of the reverse (right to left) direction, assumethat γ ⊆ δ. If γ = δ, then we are done, so assume that γ = δ.


Then δ − γ (set difference) is a non-empty subset of δ and (δ, <)is a wellordering, so we may let β be the <-least element of δ− γ.

Claim 3.19.1 β = γ.

Proof of claim By the Extensionality Axiom, Lemma 3.18 andthe fact that both β and γ are subsets of δ, the claim is equivalentto saying that, for every α < δ,

α < β ⇐⇒ α < γ.

Let α < δ be given. If α < β, then α < γ by the definition of β. Inorder to finish the proof of the claim, we assume that α < γ butα < β and work towards a contradiction. Since α and β are bothelements of δ and (δ, <) is a total ordering, saying that α < β isequivalent to saying that α ≥ β. Combining facts, we have that

β ≤ α < γ.

This is the same as saying that either

β ∈ α and α ∈ γ

or else

β = α and α ∈ γ.

Either way, since γ is a transitive set, we conclude that β ∈ γ.But β ∈ γ by the definition of β. This contradiction proves theclaim.

Now we are done proving the reverse direction of Lemma 3.19because γ = β < δ.

Lemma 3.20 If β and γ are ordinals, then either β < γ orβ = γ or β > γ.

Proof Assume that γ ≤ β. By Lemma 3.19, γ ⊆ β. Because(γ, <) is a wellordering, we may let α be the <-least element ofγ − β (set difference). Then α ⊆ β but α ∈ β. Hence, α ≤ β byLemma 3.19 but α < β. Thus β = α < γ.

Lemma 3.21 Let α be an ordinal and β = α ∪ α. Then β isan ordinal and α < β. Moreover, if γ is an ordinal and α < γ,then β ≤ γ.

34 Order

Proof By Lemma 3.10, β is a transitive set. Obviously, every ele-ment of β is also a transitive set and (β,∈) is total. By Lemma 3.16,β is an ordinal. Clearly, α < β. Suppose that γ is an ordinal andα < γ. Then α ⊆ γ since γ is transitive. Hence β = α ∪ α ⊆ γ.By Lemma 3.19, β ≤ γ.

Because of Lemma 3.21, it is natural to write α + 1 = α ∪ αfor ordinals α. We call α+1 a successor ordinal. Non-zero ordinalsthat are not successor ordinals are called limit ordinals.

Lemma 3.22 Let A be a set of ordinals and β =⋃

A. Then βis an ordinal and, for every α ∈ A, α ≤ β. Moreover, if γ is anordinal and α ≤ γ for every α ∈ A, then β ≤ γ.

The proof of Lemma 3.22 is left as a practice problem; it buildson the chain of lemmas starting from Lemma 3.12. Recall thatsupremum is another way to say least upper bound. Lemma 3.22says that if A is a set of ordinals, then sup(A) =

⋃A. If A is a set

of ordinals and sup(A) ∈ A, then sup(A) is the maximum elementof A, in which case we may write max(A) = sup(A) =

⋃A. But

not every set of ordinals has a maximum element. For example,

sup (5, 6, 7, . . . ) =⋃5, 6, 7, . . . = ω

butω ∈ 5, 6, 7, . . . ,

so 5, 6, 7, . . . does not have a maximum element. On the otherhand,

sup (5, 6, 7, . . . , ω) =⋃5, 6, 7, . . . , ω = ω

andω ∈ 5, 6, 7, . . . , ω ,

somax (5, 6, 7, . . . , ω) = ω.

If A is a set of ordinals and A = ∅, then A has a <-least ele-ment called the minimum of A and denoted min(A). To justifythe definition of min(A) use the fact that A ⊆ sup(A) + 1 and(sup(A) + 1, <) is a wellordering.

Since they are important, let us state versions of Theorems 3.6and 3.8 that relate to ordinals.


Theorem 3.23 (Proofs by induction) Let P (α) be a statementabout a variable α. Assume that, for every ordinal β,

(∀α < β P (α) holds) =⇒ P (β) holds.

Then, for every ordinal γ, P (γ) holds.

Proof Consider an arbitrary ordinal γ and let θ = γ + 1. Weprove that P (β) holds for every β < θ by induction along thewellordering (θ, <) using Theorem 3.6. But there is really nothingto do because the assumption here is at least as strong as theassumption of Theorem 3.6. To see the connection more clearly,substitute (A,≺) = (θ, <), y = β and x = α in the statement ofTheorem 3.6.

Notice that Theorem 3.23 gives us a method for showing thata property holds for every ordinal, not just every ordinal up to agiven ordinal. Often, the verification of the hypothesis of Theo-rem 3.23 is broken up into three cases: β = 0, β = α + 1 and βa limit ordinal. Similarly, in applications of Theorem 3.24 below,sometimes the definition of F (β, ·) is broken up into three cases:β = 0, β = α + 1 and β a limit ordinal.

Theorem 3.24 (Recursive definitions) Suppose that θ is an or-dinal. Let

F : θ × P → B

be a function where B is a set and P is the set of partial functionsfrom θ to B. Then there is a unique function G : θ → B such that

G(β) = F (β, G β)

for every β < θ.

Notice that Theorem 3.24 has a different flavor than Theo-rem 3.23 in that it only tells us how to make recursive definitionsup to a given ordinal θ. Although it is possible to state a theoremon recursive definitions through all the ordinals, the statementwould be overly technical, so we prefer to show how such defini-tions can be made and justified by giving an example.

Here is an illustration of a recursive definition followed by aproof by induction on the ordinals. If θ is an ordinal, then usingTheorem 3.24, we may define a function α → Vα with domain θ

36 Order

by saying thatV0 = ∅,

Vα+1 = P (Vα)

andVβ =

⋃Vα | α < β

if β is a limit ordinal. If we really want to match up this definitionof Vα with Theorem 3.24, then we end up with G : α → Vα startingwith

F (β, g) =

⎧⎪⎨⎪⎩∅ if β = 0P(g(α)) if β = α + 1⋃g(α) | α < β if β is a limit ordinal.

Notice that the definition Vα does not depend on θ so really whatwe have is a way of assigning a set Vα to each ordinal α. Theassignment α → Vα is not a function because what should be itsdomain is not a set by the following lemma.

Lemma 3.25 There is no set of all ordinals.

Proof Suppose for contradiction that

Ω = α | α is an ordinalis a set. It is easy to see that Ω is transitive and (Ω,∈) is awellordering. But then Ω is an ordinal. In other words, Ω ∈ Ω.But then

· · · ∈ Ω ∈ Ω ∈ Ω

is an infinite descending sequence of members of Ω. This contra-dicts that (Ω,∈) is a wellordering.

Notice that, once we realized that Ω ∈ Ω, we could have used theFoundation Axiom to finish the proof of Lemma 3.25 slightly morequickly. The Foundation Axiom automatically holds for ordinals,which is essentially how we got away without using it above.

We have been discussing recursion; now let us see an exampleof induction.

Lemma 3.26 Let δ be an ordinal and β < δ. Then the followinghold.


(1)δ Vδ is a transitive set.(2)δ For every β < δ, Vβ ⊆ Vδ .

Proof We prove the lemma by induction on δ. The inductionhypothesis is that the lemma holds for every γ < δ.

Base case δ = 0.

In this case, (1)0 holds because V0 = ∅ is a transitive set,whereas (2)0 holds because there is nothing to check.

Successor case δ = γ + 1.

In this case, Vδ = P (Vγ). By Lemma 3.11, since (1)γ holds,(1)δ holds too. Also, since (1)γ holds, if y ∈ Vγ , then y ⊆ Vγ , soy ∈ Vδ . This shows that Vγ ⊆ Vδ . This conclusion together with(2)γ implies that (2)δ holds.

Limit case δ is a limit ordinal.

In this case, Vδ =⋃Vγ | γ < δ. The fact that (1)δ holds is

immediate from Lemma 3.12 and the assumption that (1)γ forγ < δ. Also, (2)δ follows from (2)γ for γ < δ.

Here is an interesting fact that plays no role in the rest of thebook. It turns out that the Foundation Axiom is equivalent to thestatement

∀x ∃α (α is an ordinal and x ∈ Vα).

That is, every set belongs to some Vα. The proof can be found invarious graduate level textbooks on set theory.

Now we continue developing the theory of ordinal numbers.

Definition 3.27 Let R ⊆ A×A and S ⊆ B ×B. Suppose thatπ : A → B. Then π is an isomorphism from (A, R) to (B, S) iffπ is a bijection from A to B and, for every x, y ∈ A,

xRy ⇐⇒ π(x)Sπ(y).

We write π : (A, R) (B, S) in this case.

The relationship between ordinals and arbitrary wellorderingsis summarized by the following theorem, which is a special case ofa more general theorem by Mostowski.

38 Order

Theorem 3.28 (Mostowski collapse) Let (A,≺) be a wellorder-ing. Then there exists an ordinal α and an isomorphism

π : (A,≺) (α, <).

Moreover, α and π are unique in the sense that if α′ is an ordinaland

π′ : (A,≺) (α′, <)

is an isomorphism, then α′ = α and π′ = π.

The isomorphism π in the conclusion of Theorem 3.28 is calledthe Mostowski collapse of (A,≺). Here is a specific example. Let

A = ω − 0, 1, 4

and

m ≺ n ⇐⇒ (m, n ∈ A and m < n) .

Then the Mostowski collapse π : (A,≺) (ω, <) is determined bythe following table.

m π(m)2 03 15 26 37 4...

...

In general, if x is the ≺-least element of A, then π(x) = 0. If,in addition, y is the ≺-least element of A − x, then π(y) = 1.And so on. Notice that the list on the left has gaps (0, 1 and 4are missing) but the list on the right has no gaps. Intuitively, thefunction π is called the collapse because it gets rid of the gaps.

Another way to think about the Mostowski collapse is as follows.Suppose we list A as a0, a1, . . . in increasing order according to≺ using ordinals as indices. In the example above, we would havea0 = 2, a1 = 3, a2 = 5, a3 = 6, etc. What we would end up withis, for every b ∈ A,

b = aπ(b).

In other words, the Mostowski collapse tells us the index.


Proof of Theorem 3.28 We are given a wellordering (A,≺) andwe are looking for an ordinal α and an isomorphism

π : (A,≺) (α, <).

Let us work backwards to see what the definitions of α and πmust be. Suppose we already have α and π as above. Then, for allx, y ∈ A,

x ≺ y ⇐⇒ π(x) < π(y) ⇐⇒ π(x) ∈ π(y).

Therefore, for every y ∈ A,

π(y) = π(x) | x ≺ y.Notice that this is a recursive equation because the value of π(y) isdetermined from π x | x ≺ y. By Theorem 3.8, there is at mostone function π that satisfies this recursive equation. Moreover,once we know π, we also know α because α = π[A]. This explainswhy α and π are unique. In other words, it proves the moreoverpart of Theorem 3.28. It also gives us the clue we need to provethe existence part of Theorem 3.28, which is what we do next.

Without assuming that α is an ordinal and π is an isomorphism,recursively define π : A → B by setting

π(y) = π(x) | x ≺ y.This is not entirely legitimate, at least not if we wish to implementTheorem 3.8, because we are required to name the set B before wedefine π by recursion. We will leave out the argument that we canchoose B in advance since it involves reasoning from the axiomsof ZFC that is overly technical from the perspective of this book.

Claim 3.28.1 π is an isomorphism from (A,≺) to (π[A],∈).

Proof First we show that π is an injection. Consider arbitraryx, y ∈ A such that x = y. Then either x ≺ y or y ≺ x. Supposethat x ≺ y. Then π(x) ∈ π(y) by the definition of π. On the otherhand, π(x) ∈ π(x) by the Foundation Axiom. Hence π(x) = π(y).The proof is similar if y ≺ x.

Every injection is a bijection with its range. It remains to seethat π is order preserving. We already noted that if x ≺ y, thenπ(x) ∈ π(y). For the converse, suppose that π(x) ∈ π(y). Thenthere exists x′ ≺ y such that π(x′) = π(x). Since π is an injection,x′ = x. Hence x ≺ y as required.

40 Order

Claim 3.28.2 π[A] is an ordinal.

Proof If u ∈ v and v ∈ π[A], then there are x, y ∈ A such thatu = π(x) and v = π(y), so u ∈ π[A]. This shows that π[A] is atransitive set. By the previous claim, (π[A],∈) is an isomorphiccopy of (A,≺). Since (A,≺) is a wellordering, so is (π[A],∈).

Setting α = π[A] concludes the proof of Theorem 3.28.

The previous theorem justifies the following definition.

Definition 3.29 Let (A,≺A) be a wellordering. Then

type(A,≺A)

is the unique ordinal isomorphic to (A,≺A).

Suppose that A is a set of ordinals. Say A ⊆ β. Then (β, <) is awellordering and hence so is (A, <). This motivates the followingnotation.

Definition 3.30 Let A be a set of ordinals. Then

type(A) = type(A, <).

We end this section with a technical lemma about the Mostowskicollapse of a set of ordinals. It will be used in Chapter 4.

Lemma 3.31 Let A be a set of ordinals and suppose that A ⊆ β.Let

α = type(A)

andπ : (A, <) (α, <)

be the Mostowski collapse of (A, <). Then, for every η ∈ A,

π(η) ≤ η.

Moreover, α ≤ β.

Proof First we prove that π(η) ≤ η by induction on η ∈ A. Recallthat if ζ and η are ordinals, then

ζ ≤ η ⇐⇒ ζ ⊆ η.

This is by Lemma 3.19. Therefore, what we must show is equiva-lent to

π(η) ⊆ η

3.3 Ordinal arithmetic 41

for every η ∈ A. By the definition of the Mostowski collapse,

π(η) = π(ζ) | ζ ∈ A ∩ η.

By the induction hypothesis, for every ζ ∈ A ∩ η,

π(ζ) ≤ ζ

so

π(ζ) < ζ + 1 ≤ η.

Therefore,

π(η) ⊆⋃ζ + 1 | ζ ∈ A ∩ η ⊆ η.

This completes the proof by induction. In particular, we have seenthat if η ∈ A, then

π(η) ≤ η < β.

Since α = π[A], this implies that α ⊆ β, hence α ≤ β.

3.3 Ordinal arithmetic

In this section, we extend the usual notions of addition, multipli-cation and exponentiation of natural numbers to all ordinals.

There are various ways to join a pair of sets. For example, givenA and B, we can form their union A∪B. But sometimes we wantto take a disjoint union instead. This means that we take theunion of disjoint copies of A and B. The advantage is that, givena point in the disjoint union, it comes either from A or from Bbut not both. A convenient way to define the disjoint union is(0 × A) ∪ (1 × B). Building on this idea, sometimes we aregiven two wellorderings and we want to put one after the other toform a new wellordering. The following definition tells us how.

Definition 3.32 Let (A,≺A) and (B,≺B) be wellorderings. Thentheir concatenation is

(A,≺A)(B,≺B) = (C,≺C )

where

C = (0 ×A) ∪ (1 ×B)

42 Order

and, for all (i, x), (j, y) ∈ C,

(i, x) ≺C (j, y) ⇐⇒

⎛⎜⎜⎜⎜⎝

(i = j = 0 and x ≺A y)or

(i = 0 and j = 1)or

(i = j = 1 and x ≺B y)

⎞⎟⎟⎟⎟⎠ .

Lemma 3.33 Let (A,≺A) and (B,≺B) be wellorderings, and

(C,≺C ) = (A,≺A)(B,≺B).

Then (C,≺C ) is a wellordering.

Proof It is straightforward to verify that (C,≺C ) is a strict linearordering. For contradiction, suppose that 〈(in, xn) | n < ω〉 is aninfinite descending sequence from C. Since (B,≺B) is a wellorder-ing, there is some m < ω such that, for every n < ω, if m < n,then in = 0. But then 〈xn | n < ω −m〉 is an infinite descendingsequence from A, which is a contradiction.

Definition 3.34 Let α and β be ordinals. Then their sum is

α + β = type ((α, <)(β, <)) .

In other words, α + β is the unique ordinal isomorphic to

(α, <)(β, <).

It is the combination of Lemma 3.33 and Theorem 3.28 that justi-fies Definition 3.34. We also remark that, earlier, we defined α+1to mean α ∪ α whereas here we used a different definition ofα + 1. The two definitions coincide because

(α, <)(1, <) (α ∪ α, <).

Make sure you understand what is being asserted here and why itis true!

Example 3 + 2 = 5. This is because if (C,≺C ) = (3, <)(2, <),then

(0, 0) ≺C (0, 1) ≺C (0, 2) ≺C (1, 0) ≺C (1, 1)

and so we see that (C,≺C ) (5, <).


Example 2 + 3 = 5. This is because if (C,≺C ) = (2, <)(3, <),then

(0, 0) ≺C (0, 1) ≺C (1, 0) ≺C (1, 1) ≺C (1, 2)

and so we see that (C,≺C ) (5, <).

Example 3 + ω = ω. This is because (C,≺C ) = (3, <)(ω, <)consists of the initial segment

(0, 0) ≺C (0, 1) ≺C (0, 2)

followed by the infinite tail

(1, 0) ≺C (1, 1) ≺C (1, 2) ≺C (1, 3) ≺C (1, 4) ≺C (1, 5) ≺C · · ·

from which we see that

(C,≺C ) (ω, <)

according to the isomorphism

(i, n) →

n if i = 03 + n if i = 1.

Notice that 3 + ω = ω = ω + 3, so ordinal addition is notcommutative! However, ordinal addition is associative.

Lemma 3.35 For all ordinals α, β and γ,

(α + β) + γ = α + (β + γ).

Proof Let

(C,≺C ) = (α, <)(β, <)

and

D = (C,≺C )(γ, <).

Let

(E,≺E) = (β, <)(γ, <)

and

F = (α, <)(E,≺E).

It is enough to see that there is an isomorphism

π : (D,≺D) (F,≺F ).

44 Order

Define π by cases according to the following list of equations.

π((0, (0, ξ))) = (0, ξ)π((0, (1, ξ))) = (1, (0, ξ))

π((1, ξ)) = (1, (1, ξ))

It is clear that this works.

Addition can also be defined recursively in terms of the assign-ment that takes an ordinal α to its successor α + 1. The followinglemma shows this.

Lemma 3.36 Let α and θ be ordinals. Then there is a uniquefunction f with domain θ such that, for every γ < θ,

f(γ) =

⎧⎪⎨⎪⎩

α if γ = 0f(β) + 1 if γ = β + 1sup (f(β) | β < γ) if γ is a limit ordinal,

namely, the function given by

f(γ) = α + γ.

Sketch of proof Use induction on γ < θ to see that that ordinaladdition satisfies the three conditions we specified for f . Namely,

• α + 0 = α,• α + (β + 1) = (α + β) + 1 and• if γ is a limit ordinal, then α + γ = sup (α + β | β < γ).

Then apply Theorem 3.24. The details comprise Exercise 3.5.

Here is an entertaining false argument. We saw that

3 + ω = ω = 0 + ω.

Subtracting ω from both sides of this equation, we see that 3 = 0.Do you see why this is nonsense? There is no inverse operation forordinal addition. The following lemma is as close as we come toordinal subtraction.

Lemma 3.37 Let α ≤ β be ordinals. Then there is a uniqueordinal δ such that α + δ = β.


Sketch of proof Let D = η | α ≤ η < β. It is clear that

(α, <)(D,<) (β, <).

By Theorem 3.28, there is a unique ordinal δ such that

(D,<) (δ, <),

namely,δ = type(D).

It is clear that(α, <)(δ, <) (β, <).

This is equivalent to saying that

α + δ = β.

It remains to see that δ is the unique solution to this equation.Suppose that

α + δ′ = β.

Then(α, <)(δ′, <) (β, <),

from which one can argue that

(D,<) (δ′, <).

By the uniqueness clause of Theorem 3.28, we see that δ′ = δ. Theremaining details form Exercise 3.12.

Just before Lemma 3.37, we saw that δ + ω = 3 + ω does notimply δ = 3. In other words, we cannot cancel on the right. It is aconsequence of the uniqueness clause in Lemma 3.37 that we cancancel on the left. For example, if ω + δ = ω + 3, then δ = 3.

Now we work towards defining multiplication. For this, we useanother way of putting together two wellorderings.

Definition 3.38 Let (A,≺A) and (B,≺B) be wellorderings. Thenthe lexicographic ordering on A × B is the relation ≺ such that,for all (x, y), (x′, y′) ∈ A×B,

(x, y) ≺ (x′, y′) ⇐⇒

⎛⎝ (x ≺A x′)

or(x = x′ and y ≺B y′)

⎞⎠ .

46 Order

Lemma 3.39 Let (A,≺A) and (B,≺B) be wellorderings. Then(C,≺C ) is a wellordering where C = A×B and ≺C is the lexico-graphic ordering on C.

The proof of Lemma 3.39 is Exercise 3.7.

Definition 3.40 Let α and β be ordinals. Let γ be the uniqueordinal isomorphic to the lexicographic ordering on α × β. Thenthe product is

β · α = γ.

This is not a misprint; by tradition, ordinal multiplication isread from right to left. By β · α, we mean α many copies of β,not the other way around, and sometimes it matters for infiniteordinals.

Example 3 · 2 = 6. This is because, if C = 2 × 3 and ≺C is thelexicographic order on C, then

(0, 0) ≺C (0, 1) ≺C (0, 2) ≺C (1, 0) ≺C (1, 1) ≺ (1, 2)

and so we see that (C,≺C ) (6,∈).

Example 2 · 3 = 6. This is because, if C = 3 × 2 and ≺C is thelexicographic order on C, then

(0, 0) ≺C (0, 1) ≺C (1, 0) ≺C (1, 1) ≺C (2, 0) ≺ (2, 1)

and so we see that (C,≺C ) (6,∈).

Example 3 · ω = ω. This is because, if C = ω × 3 and ≺C is thelexicographic order on C, then (C,≺C ) looks like

(0, 0) ≺C (0, 1) ≺C (0, 2)

followed by(1, 0) ≺C (1, 1) ≺C (1, 2)

then(2, 0) ≺C (2, 1) ≺C (2, 2)

then(3, 0) ≺C (3, 1) ≺C (3, 2)

then(4, 0) ≺C (4, 1) ≺C (4, 2)


and so on. From this, we see that

(C,≺C ) (ω, <)


(i, n) → 3 · i + n.

Example ω · 3 = ω +ω +ω. This is because, if C = 3×ω and ≺C

is the lexicographic order on C, then (C,≺C ) looks like

(0, 0) ≺C (0, 1) ≺C (0, 2) ≺C (0, 3) ≺C (0, 4) ≺C · · ·

followed by

(1, 0) ≺C (1, 1) ≺C (1, 2) ≺C (1, 3) ≺C (1, 4) ≺C · · ·

then

(2, 0) ≺C (2, 1) ≺C (2, 2) ≺C (2, 3) ≺C (2, 4) ≺C · · · .

From this we see that

(C,≺C ) (ω + ω + ω, <)


(i, n) →

⎧⎪⎨⎪⎩

n if i = 0ω + n if i = 1ω + ω + n if i = 2,

which we could also write as

(i, n) → ω · i + n.

Notice that

3 · ω = ω = ω · 3,

so ordinal multiplication is not commutative! However, ordinalmultiplication is associative as the following lemma shows. Theproof is Exercise 3.6(5).


γ · (β · α) = (γ · β) · α.

Ordinal multiplication distributes over ordinal addition in thefollowing way. The proof is the first part of Exercise 3.8.

48 Order


α · (β + γ) = (α · β) + (α · γ).

Ordinal multiplication can also be defined recursively in termsof ordinal addition as the following lemma shows. The proof isExercise 3.9.

Lemma 3.43 Let α and θ be ordinals. Then there is a uniquefunction f with domain θ such that, for every γ < θ,

f(γ) =

⎧⎪⎨⎪⎩

0 if γ = 0f(β) + α if γ = β + 1sup (f(β) | β < γ) if γ is a limit ordinal,

namely, the function given by

f(γ) = α · γ.

For ordinal addition and ordinal multiplication, we started withdefinitions in terms of wellorderings, then stated lemmas givingequivalent recursive definitions. For variety and because it is eas-ier in this case, we give the recursive definition of ordinal expo-nentiation and leave the interpretation in terms of wellorderingsto the reader as a rather tricky exercise. (Exercise 3.15 will getthe reader started.)

Definition 3.44 For every ordinal α,

• α0 = 1,• for every ordinal β,

αβ+1 = αβ · α

and• for every limit ordinal γ,

αγ = sup(

αβ | 0 < β < γ)

.

Example 3ω = sup (3n | n < ω) = ω < ω3.

Example If β ≥ ω2, then ω + β = β. To see this, first note that

ω2 = ω · ω = ω · (1 + ω) = (ω · 1) + (ω · ω) = ω + ω2.


Now suppose that β ≥ ω2. By Lemma 3.37, there is an ordinal δsuch that β = ω2 + δ. Thus,

β = ω2 + δ = (ω + ω2) + δ = ω + (ω2 + δ) = ω + β.

Exercises

Exercise 3.1 Let (A,≺A) be a wellordering such that A = ∅.For each y ∈ A, define

pred(A,≺A )(y) = x ∈ A | x ≺A y.

Suppose that S A and, for all x, y ∈ A, if y ∈ S and x ≺A y,then x ∈ S. Prove that there exists y ∈ A such that

S = pred(A,≺A )(y).

Exercise 3.2 Let (A,≺A) and (B,≺B) be wellorderings. UseTheorem 3.28 to prove that exactly one of the following conditionsholds:

• (A,≺A) (B,≺B).• There exists y ∈ B,

(A,≺A) pred(B,≺B )(y) ordered by ≺B .

• There exists y ∈ A,

(B,≺B) pred(A,≺A )(y) ordered by ≺A .

Exercise 3.3 Prove that, for every ordinal β,

α ∈ Vβ | α is an ordinal = β.

As a hint, see the last part of Exercise 2.3.

Exercise 3.4 Prove that the following facts about ordinal addi-tion hold for all ordinals α, β and γ.

1. 0 + α = α = α + 0.2. β ≤ α + β.3. If β < γ, then α + β < α + γ.4. If α ≤ β, then α + γ ≤ β + γ.

Exercise 3.5 Complete the proof of Lemma 3.36.

50 Order

Exercise 3.6 Prove that the following facts about ordinal mul-tiplication hold for all ordinals α, β and γ.

1. 0 · α = 0 = α · 0.2. α · 1 = 1 · α.3. If 0 < α and β < γ, then α · β < α · γ.4. If α ≤ β, then α · γ ≤ β · γ.5. (α · β) · γ = α · (β · γ). (That is, prove Lemma 3.41.)

Exercise 3.7 Prove Lemma 3.39.

Exercise 3.8 This exercise is on distributive laws for ordinaladdition and multiplication.

1. Prove Lemma 3.42.2. Give an example of ordinals α, β and γ such that

(α + β) · γ = (α · γ) + (β · γ).

Exercise 3.9 Prove Lemma 3.43.

Exercise 3.10 Prove that the following facts about ordinal ex-ponentiation hold for all ordinals α, β and γ.

1. If β = 0, then 0β = 0.2. 1β = 1.3. If 1 < α and β < γ, then αβ < αγ .4. If α ≤ β, then αγ ≤ βγ .5. If 1 < α, then β ≤ αβ .

Exercise 3.11 Prove that the following facts about ordinal arith-metic hold for all ordinals α, β and γ.

1. αβ+γ = αβ · αγ .2. (αβ)γ = αβ·γ .

Exercise 3.12 Complete the proof of Lemma 3.37.

Exercise 3.13 Let α and β be ordinals. Prove that if β > 0,then there are unique ordinals δ and ρ such that ρ < β and

α = (β · δ) + ρ.


Exercise 3.14 Let α be an ordinal such that α = 0. Prove thatthere are unique n, β1, . . . , βn, 1, . . . , n such that

• 1 ≤ n < ω,• α ≥ β1 > · · · > βn,• 1 ≤ i < ω for every i = 1, . . . n, and• α = ωβ1 · 1 + · · ·+ ωβn · n.

This is called Cantor normal form.

Exercise 3.15 For each function x : ω → ω, define the supportof x to be the set

n < ω | x(n) = 0.

Recall thatωω = x | x is a function from ω to ω.

Let

A = x ∈ ωω | x has finite support.

Given x, y ∈ A such that x = y, there exists a largest n < ω suchthat x(n) = y(n) and we define

x ≺A y ⇐⇒ x(n) < y(n).

Prove that (A,≺A) is a wellordering and

type(A,≺A) = ωω (ordinal exponentiation).

Exercise 3.16 Find two functions

f : ω → ω + ω

and

g : ω + ω → ω + ω + ω

such that

sup(f [ω]) = ω + ω

and

sup(g[ω + ω]) = ω + ω + ω

but if h = g f is the composition, then

sup(h[ω]) < ω + ω + ω.

52 Order

Exercise 3.17 Let κ < λ < µ be three limit ordinals and

f : κ → λ

andg : λ → µ

be two functions such that

sup(f [κ]) = λ

andsup(g[λ]) = µ.

Assume that g is non-decreasing in the sense that if α ≤ β < λ,then g(α) ≤ g(β). Let h = g f be the composition. Prove that

sup(h[κ]) = µ.

4

Cardinality

We now turn from the study of order to that of cardinality, whichis a fancy word for size. Cardinal numbers will be defined to becertain kinds of ordinal numbers. Not every ordinal is a cardinalthough. The theory here builds on that of the previous chapter.

4.1 Cardinal numbers

Definition 4.1 We say that A and B have the same cardinalityand write A ≈ B iff there is a bijection from A to B.

Granted, it is strange to say that two sets have the same cardi-nality without having said what cardinality means. But we needa lemma before giving that definition.

Lemma 4.2 For every set A, there exists an ordinal γ such thatγ ≈ A.

Proof The rough idea is to let f(0) be an element of A, then letf(1) be an element of A other than f(0), etc. We keep going untilwe list all the elements of A as f(α) for some α < γ. Now we haveto make rigorous mathematical sense of this idea. Let

F = X ⊆ A | X = ∅ .

By the Axiom of Choice, there exists a choice function c : F → Asuch that, for every X ∈ F ,

c(X) ∈ X.

54 Cardinality

Define f(α) by recursion on ordinals α as follows. If

A− f [α] ∈ F ,

then let

f(α) = c (A− f [α]) .

On the other hand, if

A− f [α] ∈ F ,

then leave f(β) undefined for every β ≥ α.Observe that if α < β and both f(α) and f(β) are defined,

then f(α) ∈ f [β] and f(β) ∈ A − f [β], hence f(α) = f(β). Thiscalculation almost shows that f is an injection; what is missing isa proof that f has a set domain.

First suppose there is an ordinal γ such that f(γ) is undefined.Let γ be the least such ordinal. Then f is an injection with γ =dom(f) and ran(f) ⊆ A. Also,

A− f [γ] ∈ F ,

which means exactly that A − f [γ] = ∅. Equivalently, it meansthat f [γ] = A. Therefore, f is a bijection from γ to A as desired.

It remains to see that there is an ordinal γ such that f(γ) isundefined. Suppose otherwise. Then, for every ordinal α, f(α) isdefined. Let

S = a ∈ A | there exists α such that f(α) = a.

Then, for every a ∈ S, there exists a unique α such that f(α) = a.Apply the Replacement Scheme to this property to conclude thatthere is a set Ω and a function g : S → Ω such that, for everya ∈ S, f(g(a)) = a. Because f(α) is defined for every ordinal α,and because f(α) = f(β) whenever α = β, it must be that Ω isthe set of all ordinals. But we proved earlier, in Lemma 3.25, thatthere is no set of all ordinals.

We remark that the use of the Axiom of Choice cannot be elim-inated from the proof of Lemma 4.2. Elaborating on the meaningof this remark: If you remove the Axiom of Choice from ZFC theresult is known as ZF. It turns out that ZF + Lemma 4.2 impliesthe Axiom of Choice. Prove this as a practice problem!

At last, we define cardinal numbers and the cardinality of sets.

4.1 Cardinal numbers 55

Definition 4.3 κ is a cardinal iff κ is an ordinal and, for everyη < κ,

η ≈ κ

Definition 4.4 |A| is the least ordinal κ such that A ≈ κ.

You should convince yourself of the following facts, whose proofsamount to composing bijections.

Lemma 4.5 |A| is a cardinal.

Lemma 4.6 A ≈ B ⇐⇒ |A| = |B|.Every natural number is both an ordinal and a cardinal. Also,

ω is a cardinal. However, the ordinals

ω + 1, ω + 2, ω + 3, . . .

are all countably infinite, which is to say that their cardinality is ω.(For us, countable means finite or countably infinite. Uncountablemeans not countable.) In particular, the ordinals displayed aboveare not cardinals. Moreover, the ordinals

ω · 2, ω · 3, ω · 4, . . .

are not cardinals. Nor are the ordinals

ω2, ω3, ω4, . . . .

The ordinalsωω , ωωω

, ωωω ω

. . .

are not cardinals either. All of these are ordinals are countable, asyou should try to verify. We need another idea to reach uncount-able sets.

Theorem 4.7 (Cantor) There is no surjection from A to P(A).

Proof Consider an arbitrary function f : A → P(A). Let

C = x ∈ A | x ∈ f(x).For every x ∈ A,

x ∈ C ⇐⇒ x ∈ f(x)

henceC = f(x).

56 Cardinality

In particular,

C ∈ f [A].

Therefore f : A → P(A) is not a surjection.

The proof we just gave is an example of a diagonal argument.This is an imprecise term that you will see used in more and moregeneral ways throughout the book. Here, the idea is that if youvisualize the graph of the relation

(x, y) ∈ A×A | x ∈ f(y),

then what we call the diagonal is

D = x | x ∈ f(x).

To come up with a set C missing from the range of f , we take thecomplement of the diagonal,

C = A−D.

The reason C = f(x) is that one of C and f(x) has x as an element,and the other does not. We can express the previous sentence interms of the symmetric difference:

x ∈ C f(x)

hence

C f(x) = ∅

thus

C = f(x).

Corollary 4.8 P(ω) is uncountable.

Proof Clearly P(ω) is infinite. By Theorem 4.7, there is no sur-jection from ω to P(ω). Hence there is no bijection.

Bijections are used in the definition of cardinality but sometimesonly surjections or injections are easily available. Here are somebasic facts that relate these notions.

Lemma 4.9 If κ < θ and there is a surjection f : κ → θ, thenθ is not a cardinal.


Proof Suppose that κ < θ and f : κ → θ is a surjection. Let

S = β < κ | f(α) = f(β) for every α < β.Let g = f S. Then g is a bijection from S to θ. Let σ = type(S)and π : (S, <) σ be the Mostowski collapse of (S, <). ByLemma 3.31, since S ⊆ κ, σ ≤ κ. So σ < θ. Let

h = g π−1.

Then h : σ → θ is a bijection. In other words σ ≈ θ. Thus θ is nota cardinal.

Lemma 4.10 Let κ and λ be cardinals. Then

κ ≤ λ ⇐⇒ there is an injection f : κ → λ.

Proof If κ ≤ λ, then the identity function is an injection fromκ to λ. For the reverse direction, suppose that f : κ → λ is aninjection. Let

S = f [κ] = f(α) | α < κand π : (S, <) σ be the Mostowski collapse. Then the composi-tion

π f : κ → σ

is a bijection. Thus |κ| = |σ|. Since κ is a cardinal, |κ| = κ. BecauseS ⊆ λ, by Lemma 3.31, σ ≤ λ. Putting these facts together wehave that

κ = |κ| = |σ| ≤ σ ≤ λ.

Corollary 4.11 For every cardinal κ, there exists a cardinal λsuch that λ > κ.

Proof Let λ = |P(κ)|. First note that the identity function is aninjection from κ to P(κ), which implies that κ ≤ λ by Lemma4.10. But Theorem 4.7 implies that κ = λ, so κ < λ.

Definition 4.12 κ+ is the least cardinal strictly greater than κ.

The proof of Corollary 4.11 shows that κ+ ≤ |P(κ)|. You shouldnot assume that equality holds. The discussion after Corollary 4.27explains why.

Sometimes, the following theorem is covered in courses other

58 Cardinality

than set theory but with a very different proof. (See Exercise 4.14.)The proof here is extremely short because it builds on the theoryof ordinals and cardinals, which we have at hand.

Theorem 4.13 (Cantor–Bernstein–Schroeder) Suppose that

• there is an injection from A to B, and• there is an injection from B to A.

Then A ≈ B.

Proof First recall that A ≈ |A| and B ≈ |B|. By Lemma 4.10and the hypothesis of the theorem, |A| ≤ |B| and |B| ≤ |A|. So|A| = |B|. Hence A ≈ B.

By Corollary 4.11, there is no largest cardinal. The next resultis a kind of continuity for cardinal numbers.

Lemma 4.14 If A is a set of cardinals, then sup(A) is a cardinal.

Proof We may assume that A does not have a maximum element,as otherwise

sup(A) = max(A) ∈ A.

For contradiction, suppose that sup(A) is not a cardinal. Let κ <sup(A) and f : κ → sup(A) be a surjection. Since A does not havea maximum element, there exists λ ∈ A such that κ < λ. Let

S = α < κ | f(α) ∈ λand g = f S. Then g : S → λ is a surjection. Let σ = type(S)and

π : (S, <) σ

be the Mostowski collapse. Because S ⊆ κ, by Lemma 3.31, σ ≤ κ.So σ < λ. Let h = gπ−1. Then h : σ → λ is a surjection. Togetherwith Lemma 4.9, this shows that λ is a not a cardinal. But weassumed that every element of A is a cardinal.

Next we list the infinite cardinals in increasing order using or-dinals as indices:

ℵ0, ℵ1, ℵ2, . . . , ℵω , ℵω+1, ℵω+2, . . .

The letter ℵ is read aleph and is the first letter of the Hebrew al-phabet. Here is the formal recursive definition of our list of infinitecardinals.


Definition 4.15 Letℵ0 = ω.

By recursion on β > 0, define ℵβ to be the least cardinal greaterthan ℵα for all α < β.

It is tempting and correct to write

ℵβ = min(

κ | κ is a cardinal and κ > ℵα for all α < β)

but keep in mind that what we have inside min(·) is not a set.

Corollary 4.16 We have that

• ℵ0 = ω,• ℵα+1 = (ℵα)+ for every ordinal α, and• ℵβ = sup (ℵα | α < β) for every limit ordinal β.

Proof The first two clauses are obvious. The last clause followsfrom Lemma 4.14.

Definition 4.17 We say that λ is a successor cardinal iff thereis a cardinal κ such that λ = κ+. If λ = 0 and λ is not a successorcardinal, then we say that λ is a limit cardinal.

It is important to note that the only successor ordinals that arecardinals are the natural numbers. Every infinite cardinal (includ-ing every infinite successor cardinal) is a limit ordinal. Corollary4.16 implies the following result, which spells out how these con-cepts are related.

Corollary 4.18 For every ordinal α,

• ℵα is a limit cardinal iff either α = 0 or α is a limit ordinal,and

• ℵα is a successor cardinal iff α is a successor ordinal.

Corollary 4.18 covers all infinite cardinals by the following fact.

Lemma 4.19 Let λ be an infinite cardinal. Then there is anordinal β ≤ λ such that λ = ℵβ .

Proof By induction on infinite cardinals λ.

60 Cardinality

Base case λ = ω.

Then λ = ℵ0 and 0 < ω = λ.

Successor case λ = κ+.

By the induction hypothesis, there is an α ≤ κ such that κ = ℵα.Then

λ = κ+ = (ℵα)+ = ℵα+1

by Corollary 4.16, and

α + 1 ≤ κ + 1 < κ+ = λ.

Limit case λ is a limit cardinal.

Let

β = sup(α | ℵα < λ).

By the case hypothesis and the induction hypothesis,

β ≤ sup(ℵα | ℵα < λ)= sup(κ < λ | κ is a cardinal)= λ

and β is a limit ordinal. Similarly,

λ = sup(κ < λ | κ is a cardinal)= sup(ℵα | ℵα < λ)= sup(ℵα | α < β)= ℵβ .

The last line is by Corollary 4.16.

4.2 Cardinal arithmetic

Every cardinal is an ordinal but cardinal arithmetic is completelydifferent from ordinal arithmetic when it comes to infinite cardi-nals. It is important to keep track of which kind of arithmetic youare doing. Usually, it is clear from the context.

4.2 Cardinal arithmetic 61

Definition 4.20 For all cardinals κ and λ,

κ⊕ λ = |(0 × κ) ∪ (1 × λ)|

and

κ⊗ λ = |κ× λ| .

Unlike ordinal addition and multiplication, cardinal additionand multiplication are commutative. The main point in seeingthat

κ⊗ λ = λ⊗ κ

is that if A and B are sets, then (x, y) → (y, x) is a bijection fromA×B to B ×A, hence |A×B| = |B ×A|.

The next result says that, for natural numbers, cardinal additionand multiplication coincide with ordinal addition and multiplica-tion. The reader should work out the proofs as an exercise.

Lemma 4.21 If m, n < ℵ0, then m⊕n = m+n and m⊗n = m·n.

Remember that ordinal addition and multiplication for infiniteordinals were interesting operations with subtle properties. Bycontrast, cardinal addition and multiplication for infinite cardinalsturn out to be trivial to calculate by the following two results.

Lemma 4.22 Let λ be an infinite cardinal. Then λ⊗ λ = λ.

Proof By induction on λ. The induction hypothesis is that

µ⊗ µ = µ

whenever µ is a cardinal such that ℵ0 ≤ µ < λ. We will use twodifferent orderings of the Cartesian product λ× λ. First define(

α, β)

<lex (α, β)

iff either

α < α

or

α = α and β < β.

Then define

(α, β) (α, β)

62 Cardinality

iff eithermax

(α, β

)< max (α, β)

ormax

(α, β

)= max (α, β) and

(α, β

)<lex (α, β) .

Figure 4.1 is a picture of λ×λ ordered by . For a given α < λ,the order increases across the horizontal arrow (1) leaving out(α, α), then increases up the vertical arrow (2) until it reaches(α, α). Next, increases across (3) leaving out (α+1, α+1), thenup (4) including (α + 1, α + 1). And so on.

We claim that is a wellordering of λ×λ. It is obviously a strictlinear ordering. Towards seeing that is wellfounded, consider anarbitrary S ⊆ λ× λ such that S = ∅. Let

γ = min (max(α, β) | (α, β) ∈ S) .

Let

α = min (α | there is β with (α, β) ∈ S and max (α, β) = γ)and

β = min (β | (α, β) ∈ S) .

...

(0, α + 1)(3) (α + 1, α + 1)

(0, α)(1) (α, α)

...

(0, 0) · · · (α, 0)

(2)

(α + 1, 0)

(4)

· · ·

Figure 4.1 λ × λ ordered by


It is easy to check that (α, β) is the -least element of S. Now thatwe know is a wellordering, we can talk about its order type.

Claim 4.22.1 type (λ× λ, ) = λ.

Proof First note that type (λ× λ, ) ≥ λ. This is because

(α, 0) (α, 0)

whenever α < α < λ. So it is enough to see that

type (λ× λ, ) ≤ λ.

For this, it is enough to see that, for every (α, β) ∈ λ× λ,

type (pred (α, β) , ) < λ

where, by definition,

pred (α, β) =(

α, β)|(α, β

) (α, β)

.

If λ = ℵ0, then α and β are natural numbers and pred (α, β) isfinite, hence

type (pred (α, β) , ) < ℵ0 = λ

as desired. So we may assume that λ ≥ ℵ1. Let

γ = max(α, β,ℵ0) + 1

andµ = |γ|.

Then µ is a cardinal and ℵ0 ≤ µ < λ, so by the induction hypoth-esis,

µ⊗ µ = µ.

Therefore

type (pred (α, β) , ) ≤ type (γ × γ, ) < µ+ ≤ λ

as desired.

From Claim 4.22.1, it follows that

(λ× λ, ) (λ, <).

Since isomorphisms are bijections,

|λ× λ| = |λ| = λ.

This completes the proof of Lemma 4.22.

64 Cardinality

With regard to the proof of Lemma 4.22, we remark that if λ isan ordinal, then the lexigraphic ordering on λ×λ is a wellorderingof type λ · λ (ordinal product). In other words,

type(λ× λ, <lex) = λ · λ.

This is by the definition of ordinal multiplication, Definition 3.40.In particular, if λ is an infinite ordinal, then

|λ| ≤ λ < λ · λ < |λ|+.

In particular, λ · λ is not a cardinal.Lemma 4.22 is a special case of the following general result.

Theorem 4.23 If 0 < κ ≤ λ are cardinals and ℵ0 ≤ λ, then

κ⊕ λ = κ⊗ λ = λ.

Proof The theorem can be verified easily if κ = 1. If κ ≥ 2, thenby Lemma 4.22,

λ ≤ κ⊕ λ ≤ λ⊕ λ = 2⊗ λ ≤ κ⊗ λ ≤ λ⊗ λ = λ

so equality holds throughout.

Theorem 4.23 tells us that if at least one of κ and λ is infinite,then

κ⊕ λ = κ⊗ λ = max(κ, λ).

So cardinal addition and multiplication really are easy to calculate!Now we define cardinal exponentiation. The notation is the

same as for ordinal exponentiation but the meaning is different.For this definition, recall from Exercise 2.8 that if A and B aresets, then

AB = f | f is a function from A to B.

Definition 4.24 For all cardinals κ and λ,

λκ = |κλ|

We repeat our warning that writing λκ is ambiguous. Do youmean ordinal exponentiation or cardinal exponentiation? Alwaysmake sure it is clear which, either from the context, or by sayingso explicitly.

Our first fact about cardinal exponentiation is that it is the same


as ordinal exponentiation when restricted to the natural numbers.The reader should work out the proof as an exercise.

Lemma 4.25 If m, n < ℵ0, then nm is the same whether com-puted as ordinal exponentiation or cardinal exponentiation.

Cardinal exponentiation becomes quite interesting when we lookat infinite powers.

Lemma 4.26 |P(A)| = 2|A|

Proof Define a function

char : P(A) → A2

by setting

char(X)(a) =

0 if a ∈ X

1 if a ∈ X

for every X ⊆ A and a ∈ A. Note that char is a function whoseoutputs are themselves functions. The function

char(X) : A → 2

is called the characteristic function of X in A.1 To see that charis an injection, note that if X, Y ⊆ A and X = Y , then, for anya ∈ XY ,

char(X) (a) = char(Y ) (a) ,

so

char(X) = char(Y ).

To see that char is a surjection, note that if f ∈ A2, then

char (a ∈ A | f(a) = 1) = f.

We have shown that P(A) ≈ A2. From this, Lemma 4.26 is clear.

Earlier, we established that κ+ ≤ |P(κ)|. Thus, the following isa consequence of Lemma 4.26.

Corollary 4.27 If κ is a cardinal, then 2κ ≥ κ+.1 Elsewhere, the characteristic function of X is written χX , where χ is the lower

case Greek letter chi, and X is a subscript. But this is hard to read.

66 Cardinality

This brings up an important question, namely, what is the valueof 2κ? Focusing on the most basic case, what is the value of 2ℵ0 ?Is 2ℵ0 = ℵ1? Perhaps 2ℵ0 = ℵ2? Could it be that 2ℵ0 = ℵω·7+4?We know that, for some ordinal α ≥ 1,

2ℵ0 = ℵα

but it turns out that the value of α cannot be determined usingonly the axioms of ZFC because of deep theorems of Kurt Godeland Paul Cohen.2 This is interesting because

|R| = 2ℵ0 ,

so really we are asking how many real numbers there are. Thisproblem was posed by Georg Cantor in the late 1800s, who asked:

Is there is an uncountable A ⊆ R such that A ≈ R?

It was also first on the most famous list of open problems, whichDavid Hilbert compiled at the start of the twentieth century. Theanswer no to Cantor’s question is known as the Continuum Hy-pothesis, or CH, which says 2ℵ0 = ℵ1. The answer yes says that2ℵ0 ≥ ℵ2. Cantor conjectured CH is true, but an informal poll sug-gests that most set theorists today who have an opinion believeCH is counterintuitive. Few have strong feelings about what theactual value of 2ℵ0 should be although some feel it should be ℵ2.As we mentioned already, there are theorems due to Godel and Co-hen which together say roughly that ZFC is not powerful enoughto answer Cantor’s question, so it is unknown what methodologycould lead to an answer. Additional explanation would be beyondthe scope of this book; it should be the topic of your next settheory course!

Remember that, for finite numbers, (m )k = m k . Do not con-fuse this with the standard convention m k

= m( k ). The firstequation generalizes to all cardinal numbers as the following lemmashows. Other basic facts about cardinals can be found in the ex-ercises.2 The hypothesis of these Godel and Cohen theorems is that ZFC is a consistent

theory, meaning there is no proof that 0 = 1 using only the axioms of ZFC.Godel proved that ZFC is consistent with CH in 1940. Cohen proved that ZFC isconsistent with the negation of CH in 1963. The combination of these resultssays CH is independent of ZFC. These results can be found in many graduatelevel textbooks on set theory.

4.3 Cofinality 67

Lemma 4.28 Let κ, λ and µ be cardinals. Then

(µλ)κ = µλ⊗κ.

Proof It is easy to see that(µλ

)κ=

∣∣∣κ (λµ

)∣∣∣and

µλ⊗κ =∣∣∣κ×λµ

∣∣∣ .

There is a bijection

F : κ(

λµ)→ κ×λµ

defined by

F (g)(α, β) = g(α)(β).

Putting together these observations, we are done.

Definition 4.29 (Strange notation) ωα = ℵα for ordinals α ≥ 1.

We write ωα when we want to emphasize that it is an ordinal.We write ℵα when we want to emphasize that it is a cardinal.However, it is difficult to keep the notation consistent with suchintentions when we simultaneously consider cardinal and ordinalproperties of ℵα = ωα. Ultimately, whether we write ωα or ℵα

reduces to a matter of style.Note that ω0 is not defined; we always write either ω or ℵ0.

4.3 Cofinality

Recall that ℵω is the least cardinal greater than every ℵn for n < ω.In this sense, ℵω feels rather large. On the other hand, the function

n → ℵn

maps ω to ℵω and has range unbounded in ℵω , i.e.,

ℵω = supn<ω

ℵn.

Let us point out that ω = ℵ0 < ℵω . The fact that ℵω can bereached from below in this way makes it feel somewhat smaller

68 Cardinality

than before. Here, we examine this phenomenon and tie it upwith cardinal arithmetic.

Definition 4.30 If λ is a limit ordinal, then the cofinality of λ,

cf(λ),

is the least ordinal κ such that there exists a function f : κ → λwith

sup(f [κ]) = λ.

We say λ is singular if cf(λ) < λ. Otherwise, we say λ is regular.

The notions above are defined only for limit ordinals, not for 0 orsuccessor ordinals α+1. Keep in mind that every infinite cardinalis a limit ordinal. (Recall the reason is that |A ∪ A| = |A| forevery infinite set A.) Here is a list of examples of cofinalities tothink about now and as you read.

cf (ℵ0) = ℵ0

cf (ℵ1) = ℵ1

cf (ℵ2) = ℵ2

cf (ℵω) = ℵ0

cf (ℵω+1) = ℵω+1

cf (ℵω+2) = ℵω+2

cf (ℵω+ω) = ℵ0

cf (ℵω1 ) = ℵ1

Here is a second list with more examples to think about.

cf (ω + ω) = ℵ0

cf (ω · ω) = ℵ0

cf (ωω) = ℵ0 (ordinal exponentiation)cf (ω1 + ω) = ℵ0

cf (ω1 + ω1) = ℵ1

cf (ω1 + ω1 + ω) = ℵ0

By the end of this chapter, you should be able to explain the

4.3 Cofinality 69

equations listed above. Currently, you should see that, for everylimit ordinal λ,

cf(λ) ≤ |λ| ≤ λ.

This is because every surjection onto λ has range unbounded in λ.It follows from these inequalities that if cf(λ) = λ, then |λ| = λ.In other words, if λ is a regular limit ordinal, then λ is a cardinal.

It is worth doing Exercises 3.16 and 3.17 before reading theproof of the next result.

Lemma 4.31 cf(λ) is a regular cardinal.

Proof Let f : cf(λ) → λ be a function whose range is unboundedin λ. We know there is such a function f by the definition ofcofinality.

The following claim records slightly more information than weneed for the rest of the proof but the extra information is usefulelsewhere.

Claim 4.31.1 There is a function g : cf(λ) → λ such that

• ran(g) is unbounded in λ,• g is non-decreasing in the sense that if α ≤ β < cf(λ), then

g(α) ≤ g(β),• g is continuous in the sense that for every limit ordinal β <

cf(λ),

g(β) = supα<β

g(α).

Proof of claim Define g with dom(g) = cf(λ) by

g(β) = supα<β

f(α).

By the definition of cofinality, if β < cf(λ), then g(β) < λ. Thatis,

g : cf(λ) → λ.

To see that g is continuous, observe that, for every limit ordinalγ < cf(λ),

g(γ) = supα<γ

f(α) = supβ<γ

(supα<β

f(α)

)= sup

β<γg(β).

70 Cardinality

Clearly, cf(λ) is a limit ordinal. Using this, we see that the rangeof g is unbounded in λ because

supβ<cf(λ)

g(β) = supβ<cf(λ)

(supα<β

f(α)

)= sup

α<cf(λ)f(α) = λ.

Continuing with the proof of Lemma 4.31, let g : cf(λ) → λ beas in Claim 4.31.1. Since every regular limit ordinal is a cardinal,in order to finish proving the lemma, it suffices to show that cf(λ)is a regular ordinal. That is, given an ordinal κ < cf(λ) and afunction h : κ → cf(λ), we must conclude that the range of h isbounded in cf(λ). Let S = h[κ]. For contradiction, assume that

sup(S) = cf(λ).

Using this assumption and the fact that g is non-decreasing andunbounded, we see that

supη<κ

g(h(η)) = supα∈S

g(α) = supα<cf(λ)

g(α) = λ.

We have shown that the composition

g h : κ → λ

has range unbounded in λ. Because κ < cf(λ), this contradicts thedefinition of cofinality.

The following result implies that cf(ℵα+1) = ℵα+1 for everyordinal α.

Lemma 4.32 Let λ be an infinite cardinal and µ = λ+. Then µis a regular cardinal.

Proof We must show that cf(µ) = µ. Let κ ≤ λ and f : κ → µbe a function. For each α < κ, f(α) < µ = λ+, so |f(α)| ≤ λ.Thus

|sup(f [κ])| ≤∣∣∣∣∣⋃α<κ

f(α)

∣∣∣∣∣ ≤ λ⊗ κ = λ < µ.

In particular, sup(f [κ]) < µ. This shows that cf(µ) ≥ µ. Butobviously cf(µ) ≤ µ for every limit ordinal µ. Hence cf(µ) = µ,which means µ is regular.

4.3 Cofinality 71

Now that we understand successor cardinals, let us look at afew examples of singular cardinals.

Example Using the fact that

ℵω = supn<ω

ℵn

we see thatcf(ℵω) = ℵ0.


ℵω+ω = supn<ω

ℵω+n,

we see thatcf(ℵω+ω) = ℵ0.


ℵω1 = supα<ω1

ℵα,

we see thatcf(ℵω1 ) ≤ ℵ1.

We claim thatcf(ℵω1 ) = ℵ0.

Consider an arbitrary function

f : ω → ℵω1 .

Define g : ω → ω1 by letting

g(n) = the least α < ω1 such that f(n) < ℵα.

Then, for every n < ω,

|f(n)| ≤ f(n) < ℵg(n) < ℵω1 .

Since ℵ1 is a regular cardinal, there exists α < ω1 such that, forevery n < ω, g(n) < α. Hence f(n) < ℵα for every n < ω. So

supn<ω

f(n) ≤ ℵα < ℵω1 .

This proves our claim. We conclude that

cf(ℵω1 ) = ℵ1.

72 Cardinality

The solution to Exercise 4.13 involves calculations similar tothose in our examples above.

The following theorem is an extension of Theorem 4.7. Its proofis a more elaborate diagonal argument.

Theorem 4.33 (G. Konig) If λ is an infinite cardinal, then

λcf(λ) > λ.

Proof Let κ = cf (λ). Fix f : κ → λ such that the range of f isunbounded in λ. Consider an arbitrary function G : λ → κλ. It isenough to see that G is not a surjection. For each α < κ, let

Aα = G(η)(α) | η < f(α).Then, for every α < κ,

Aα ⊆ λ

and|Aα| ≤ f(α) < λ.

In particular, for every α < κ,

λ−Aα = ∅.For each α < κ, let h(α) be the least element of λ−Aα. Then, forevery α < κ and every η < f(α),

h(α) = G(η)(α).

Recall that, for every η < λ, there exists α < κ such that η < f(α).Therefore, for every η < λ,

h = G(η).

This shows that G is not a surjection.

Corollary 4.34 If κ is an infinite cardinal, then

cf (2κ) > κ.

Proof Apply the previous theorem with λ = 2κ to see that

(2κ)cf(2κ ) > 2κ.

But, if µ ≤ κ, then

(2κ)µ = 2κ⊗µ = 2κ

by Lemmas 4.28 and Theorem 4.23. Corollary 4.34 follows.

4.3 Cofinality 73

An interesting special case of Corollary 4.34 is the fact that

cf(2ℵ0 ) > ℵ0.

Notice, also, that we recover Theorem 4.7 from Corollary 4.34because

2κ ≥ cf (2κ) > κ.

Exercises

Exercise 4.1 Let<ω2 =

⋃n<ω

n2.

1. Prove that <ω2 is countable.2. Let

F = x n | n < ω | x ∈ ω2 .

Prove that |F| = 2ℵ0 .3. Prove that there exists a family G ⊆ P(ω) such that

|G| = 2ℵ0

and for all A, B ∈ G, if A = B, then A ∩B is finite.Hint: Observe that F ⊆ P(<ω2).

Exercise 4.2 Let<ωω =

⋃n<ω

nω

and

ℵ0<ℵ0 =

∣∣<ωω∣∣ .

Show that ℵ<ℵ00 = ℵ0.

Exercise 4.3 Prove the following equations.

1. ℵℵ00 = 2ℵ0

2. ℵℵ01 = 2ℵ0

3. ℵℵ10 = 2ℵ1

4. ℵℵ11 = 2ℵ1

74 Cardinality

Exercise 4.4 Let<κλ =

⋃α<κ

αλ

andλ<κ =

∣∣<κλ∣∣

whenever κ and λ are infinite cardinals. Show that κ(<κ+ ) = 2κ.

Exercise 4.5 Prove that if κ ≤ λ are infinite cardinals, then

|X ⊆ λ | |X| = κ| = λκ.

Then explain why

|X ⊆ ω2 | |X| = ℵ0| = max(ℵ2, 2ℵ0 ),

|X ⊆ ω2 | |X| = ℵ1| = 2ℵ1

and|X ⊆ ω2 | |X| = ℵ2| = 2ℵ2 .

Exercise 4.6 Prove that, for every ordinal ξ, there is a cardinalλ > ξ such that

cf(λ) = ω

andλ = ℵλ.

Hint: First recall that α ≤ ℵα for every ordinal α by Lemma 4.19.Now consider the sequence of cardinals 〈κn | n < ω〉 defined byinduction according to κ0 = ℵξ+1 and κn+1 = ℵκn .

Exercise 4.7 Express the cardinality of the sets below in theform

ℵα, 2ℵα , 22ℵα , . . .

and explain your calculations. For your solutions, you may usefacts about Q, R, C and continuous functions from calculus courses.

1. Q = x | x is a rational number2. R = x | x is a real number3. R−Q = x ∈ R | x is irrational4. x ∈ R | 0 < x < 15. x ∈ C | x is a root of a polynomial with rational coefficients

4.3 Cofinality 75

6. RR = f | f is a function from R to R7. QR = f | f is a function from Q to R8. RQ = f | f is a function from R to Q9. f | f is a continuous function from R to R

Exercise 4.8 Let <R be the usual ordering of R. For everyA ⊆ R, we may abuse notation by writing

(A, <R)

when we really mean

(A, (x, y) ∈ A×A | x <R y).

You may use facts from calculus courses in your solutions to:

1. Prove that, for every A ⊆ R, if (A, <R) is a wellordering, then

type(A, <R) < ω1.

2. Prove that, for every α < ω1, there exists Aα ⊆ R such that

type(Aα, <R) = α.

Exercise 4.9 Let f : B → B be a function and X ⊆ B. Provethat there exists A ⊆ B such that X ⊆ A, f [A] ⊆ A and

|A| ≤ |X| ⊗ ℵ0.

Exercise 4.10 If 〈Sn | n < ω〉 is a sequence of sets, then define

∏n<ω

Sn =

ff is a function with dom(f) = ω

and f(n) ∈ Sn for all n < ω

.

1. Show that

|P(ℵω)| ≥∣∣∣∣∣∏n<ω

P(ℵn)

∣∣∣∣∣ .

2. Show that

|P(ℵω)| ≤∣∣∣∣∣∏n<ω

P(ℵn)

∣∣∣∣∣ .

Hint: Consider the function

A → 〈A ∩ ℵn | n < ω〉

76 Cardinality

We remark that, from Exercise 4.10, it is immediate that

2ℵω =

∣∣∣∣∣∏n<ω

2ℵn

∣∣∣∣∣ ,

which is usually abbreviated

2ℵω =∏n<ω

2ℵn .

Exercise 4.11 Prove that, for every n < ω,

(ℵn)ℵ0 = max(ℵn, (ℵ0)ℵ0 ).

Hint: It is obvious that the left side is at least as large as the rightside. To prove the other direction, use induction on n < ω and thefact that ℵn is a regular cardinal.

Exercise 4.12 This exercise is about ordinal exponentiation.

1. Show that if α < ω1, then ωα < ω1.2. Show that ωω1 = ω1.3. Show that

β < ω1 | ωβ = β

is uncountable.

Exercise 4.13 Prove that, for every limit ordinal α,

cf(ℵα) = cf(α).

Exercise 4.14 (Cantor–Bernstein–Schroeder theorem) The proofwe gave of Theorem 4.13 used the Axiom of Choice because it usedthe fact that every set has a cardinality. Complete the followingoutline of a proof that avoids the Axiom of Choice. Let

f : A → B

and

g : B → A

be injections. By recursion, define A0 = A, B0 = B,

An+1 = g[Bn]

and

Bn+1 = f [An].

4.3 Cofinality 77

LetAeven =

⋃n<ω

(A2n −A2n+1) ,

Aodd =⋃n<ω

(A2n+1 −A2n+2)

andA∞ =

⋂n<ω

An.

Define

h(x) =

⎧⎪⎨⎪⎩

f(x) if x ∈ Aeven

g−1(x) if x ∈ Aodd

f(x) if x ∈ A∞.

Prove that h is well-defined and h is a bijection from A to B.

Exercise 4.15 As in Exercises 2.10 and 2.13, let E be the equiv-alence relation on P(ω) defined by

xEy ⇐⇒ x y is finite.

Prove that P(ω)/E has cardinality 2ℵ0 .

Exercise 4.16 (Zorn’s lemma) Let (P, ) be a partial ordering.By definition this means that is a relation on P that is reflexiveand transitive. A subset C ⊆ P is called a chain iff (C, ) is alinear ordering. Since we already know that is reflexive andtransitive, if C ⊆ P , then C is a chain iff for all x, y ∈ C, eitherx y or y x. Assume that every chain has an upper bound in(P, ). In other words, assume that, for every chain C, there existsy ∈ P such that, for every x ∈ C, x y. Prove that (P, ) hasa maximal element. In other words, prove that there exists y ∈ Psuch that, for every x ∈ P , y x.

Hint: Suppose otherwise. Let κ = |P |. By recursion on α < κ,build a chain C = xα | α < κ such that C does not have anupper bound to get a contradiction.

Remark: This proof of Zorn’s lemma uses the Axiom of Choice toknow that the partial ordering has a cardinality. It also turns outthat ZF together with Zorn’s lemma implies AC. Therefore, Zorn’slemma and AC are equivalent. This is another good exercise!

78 Cardinality

Exercise 4.17 (Boolean algebras of truth tables) For each pos-itive n < ω, define

Tn = (Tn,∨n,∧n,¬n, 0, 1)

as follows.

• Tn is the set of all functions f from ω2 to 2 with the propertythat, for all a, b ∈ ω2, if a n = b n, then f(a) = f(b).

• If f, g ∈ Tn and a ∈ ω2, then

(f ∨n g)(a) = 1 ⇐⇒ f(a) = 1 or g(a) = 1,

(f ∧n g)(a) = 1 ⇐⇒ f(a) = 1 and g(a) = 1

and

(¬nf)(a) = 1 ⇐⇒ f(a) = 0.

• For every a ∈ ω2, 0(a) = 0 and 1(a) = 1.

Notice that 0, 1 ∈ Tn for every n < ω.You may take it for granted that each Tn is a Boolean algebra.

It is helpful to think of Tn as the Boolean algebra of truth tables inn variables. For example, a typical element f of T2 can be thoughtof as the truth table

0 0 f(〈0, 0, . . . 〉)

0 1 f(〈0, 1, . . . 〉)

1 0 f(〈1, 0, . . . 〉)

1 1 f(〈1, 1, . . . 〉)

and, if f happens to be an element of T1, then f can be thought

4.3 Cofinality 79

of as the simpler truth table

0 f(〈0, . . . 〉)

1 f(〈1, . . . 〉)

1. How many elements does Tn have? Explain.2. Find a finite Boolean algebra B such that, for every n < ω,

B Tn.

3. The Boolean algebra relation for Tn is given by

f n g ⇐⇒ f ∧n g = f.

Give a more practical description of n in terms of entries intruth tables.

4. Figure 4.2 shows the elements of T2 organized into levels withsome arrows drawn between some truth tables. What is thesignificance of the arrows? Which arrows are missing? Copythe figure and add all the missing arrows between truth tableson neighboring levels.

5. List all the atoms of T1 using truth table notation. Where arethey on Figure 4.2? List all the atoms of T2. Where are theyon Figure 4.2? How many atoms does T3 have?

6. DefineT∞ = (T∞,∨∞,∧∞,¬∞, 0, 1)

by setting

T∞ =⋃n<ω

Tn,

∨∞ =⋃n<ω

∨n,

∧∞ =⋃n<ω

∧n

and¬∞ =

⋃n<ω

¬n.

80 Cardinality

To make sure you understand the definition of T∞, convinceyourself that if f ∈ Tm and g ∈ Tn where m < n < ω, thenf ∈ Tn and f ∨∞ g = f ∨n g.(a) It is a fact that T∞ is a Boolean algebra. Pick any three of

the ten defining equations for Boolean algebras and showthat they hold for T∞. You may use the fact that Tn is aBoolean algebra for n < ω.

(b) Prove that T∞ has no atoms.(c) Explain why T∞ is countable.(d) Give a specific example of a function f : ω → 2 that does

not belong to T∞.

00

10

11

10

11

11

00

00

11

10

11

11

00

10

10

10

11

11

00

10

11

10

01

11

00

10

11

10

11

10

00

00

10

10

11

11

0

00

01

11

00

11

1

00

10

10

10

01

11

00

00

11

10

11

10

00

10

10

10

11

10

00

10

11

10

01

10

00

00

10

10

01

11

00

00

10

10

11

10

0

00

01

11

00

11

0

00

10

10

10

01

10

00

00

10

10

01

10

Fig

ure

4.2

Wha

tis

the

sign

ifica

nce

ofth

ispi

ctur

e?Se

eExe

rcis

e4.

17.

5

Trees

As you might expect, trees play important roles in many partsof mathematics. Most of this chapter is concerned with trees ofheight at most ω but the last section goes into trees of heightω1. We will look at trees in various contexts: topology, analysis,combinatorics and games.

5.1 Topology fundamentals

To get started, we go over some elementary definitions and factsabout topological spaces and metric spaces.

Definition 5.1 A topological space is a pair (S, T ) such that

• T ⊆ P(S),• S ∈ T ,• for every non-empty finite F ⊆ T ,⋂

F ∈ T ,

• for every F ⊆ T , ⋃F ∈ T .

We also say that T is a topology on S.

The most important example of a topological space (S, T ) has

S = R

andT = U | U is an open subset of R

5.1 Topology fundamentals 83

where U is an open subset of R iff U is a union of open intervals.For the record, open intervals of R are sets of the form

x ∈ R | a < x < bwhere a < b are real numbers ordered in the usual way. In thissection, we will use the notation (a, b) for the open interval froma to b even though it conflicts with our notation for ordered pairs.Other basic notation from calculus may also be used here. Someexamples of open subsets of R are

R =⋃(−n, n) | n = 1, 2, 3, . . . ,

∅ =⋃∅,

(0, 1) =⋃(0, 1),

(0,∞) =⋃(0, n) | n = 1, 2, 3, . . . ,

(−∞, 0) =⋃(−n, 0) | n = 1, 2, 3, . . . ,

R− 0 = (−∞, 0) ∪ (0,∞)

andR− Z =

⋃(n, n + 1) | n ∈ Z.

The following fact is left as an exercise; we will give a similar proofin the next section.

Lemma 5.2 The family of open subsets of R is a topology on R.

Topological spaces are related to metric spaces, which we definenext.

Definition 5.3 A metric space is a pair (S, d) where

d : S × S → [0,∞) = x ∈ R | 0 ≤ xis a function from S × S to the set of non-negative real numberssuch that, for all x, y, z ∈ S,

d(x, y) = d(y, x),

d(x, y) = 0 ⇐⇒ x = y

84 Trees

and

d(x, z) ≤ d(x, y) + d(y, z).

We also say that d is a metric on S.

The last clause in the definition is called the triangle inequality.The most important example of a metric space (S, d) has

S = R

and

d = |x− y|.

In this context, |x− y| means the absolute value of the differencebetween x and y. This is the usual distance function for R. Hereare some well-known facts:

|x− y| ≥ 0,

|x− y| = |y − x|,

|x− y| = 0 ⇐⇒ x = y

and

|x− z| ≤ |x− y|+ |y − z|.

The following lemma is immediate from these facts.

Lemma 5.4 The usual distance function for R is a metric onR.

Notice that each open interval (a, b) of R has the form

x ∈ R | |x− c| < r

for some c ∈ R (the center) and positive r ∈ R (the radius). Justtake c = (b + a)/2 and r = (b − a)/2 to see this. In this sense,the topology of R comes from the metric on R. One says that thetopology and the metric are compatible when they are related inthis manner. Not every topological space has a compatible metricbut many interesting ones do.

5.2 The Baire space 85

x

s

Figure 5.1 A basic open neighborhood Ns and x ∈ Ns

5.2 The Baire space

In this section, we endow ωω with a topology and a metric, whichturn out to be compatible. Throughout this and subsequent sec-tions, it is very important to keep in mind the distinction between<ωω (the set of finite sequences of natural numbers) and ωω (theset of infinite sequences of natural numbers).

Definition 5.5 If n < ω and s ∈ nω, then

Ns = x ∈ ωω | x n = s.

These are the basic open subsets of ωω.

Figure 5.1 is an attempt to illustrate the basic open set Ns,which consists of all the infinite branches x that pass through s.The following easy observation is often useful.

Lemma 5.6 If s, t ∈ <ωω, then

• if s ⊆ t, then Nt ⊆ Ns,• if t ⊆ s, then Ns ⊆ Nt, and• otherwise, Ns ∩Nt = ∅.

Definition 5.7 U is an open subset of ωω iff there is a family Fof basic open subsets of ωω such that

U =⋃F

86 Trees

Note that, because <ωω is countable, U is an open subset of ωωiff there is a sequence 〈sn | n < ω〉 from <ωω such that

U =⋃n<ω

Nsn .

Lemma 5.8 U | U is an open subset of ωω is a topology onωω.

This is the Baire topological space.

Proof Everything is obvious except the fact that the intersectionof finitely many open sets is open. It is enough to show that theintersection of two open sets is open. The general statement followsby induction because we can add parentheses as follows:

U0 ∩ U1 ∩ · · · ∩ Un = U0 ∩ (U1 ∩ · · · ∩ Un) .

Say

A =⋃F

and

B =⋃G,

where F and G are families of basic open sets. We must show thatA ∩ B is an open set. Let H be the collection of basic open setsNt for which there are r, s ∈ <ωω such that

• Nr ∈ F and Ns ∈ G,• r ⊆ s or s ⊆ r, and• t = r ∪ s.

The last two clauses say that either

r = s dom(r)

or

s = r dom(s)

(we say r and s are comparable in this case), and t is the longerof the two (which is the union because they are comparable). Wewill be done when we show that

A ∩B =⋃H.


First suppose that x ∈ A∩B. Then there are r, s ∈ <ωω such that

x ∈ Nr ∈ F

and

x ∈ Ns ∈ G.

Then the finite sequences r and s are comparable because theyare both restrictions of the same infinite sequence x. That is,

r = x dom(r)

and

s = x dom(s).

Let t = r ∪ s be the longer of the two. Then

x ∈ Nt ∈ H,

so

x ∈⋃H.

This shows that

A ∩B ⊆⋃H.

We leave the easier reverse inclusion to the reader.

Definition 5.9 C is a closed subset of ωω iff ωω −C is an opensubset of ωω.

It is time for some examples. Consider an arbitrary x ∈ ωω. Thesingleton x is closed since its complement is open:

ωω − x =⋃Ns | s ∈ <ωω but s ⊂ x.

However, x is not open since, for every s ∈ <ωω,

Ns ⊆ x

because Ns has infinitely many elements while x has just one.Thus x is closed but not open. It follows easily that ωω − xis open but not closed.

Most sets are neither open nor closed. One way to see this is toobserve that ∣∣Ns | s ∈ <ωω

∣∣ =∣∣<ωω

∣∣ = ℵ0,

88 Trees

and

|C | C is a closed subset of ωω| = |U | U is an open subset of ωω|=

∣∣ω (<ωω

)∣∣= ℵℵ0

0

= 2ℵ0 ,

which is strictly smaller than

|P(ωω)| = 22ℵ0.

Since there are strictly more subsets of the Baire space than thereare open or closed subsets, there must be subsets which are neitheropen nor closed.

It is also easy to come up with specific examples of sets whichare neither open nor closed. Given n < ω, s ∈ nω and x ∈ ωω, let

sx = s ∪ (n + k, x(k)) | k < ω.

A less precise but somehow clearer way to write this is

sx = 〈s(0), . . . , s(n− 1), x(0), x(1), x(2), . . . 〉

where if n = 0, then sx = x. Suppose that U is open but notclosed and C is closed but not open. (We already gave examplesof such sets.) Let

A = 〈0〉x | x ∈ U ∪ 〈1〉y | y ∈ C.

We will prove that A is not open and leave the verification that Ais not closed to the reader. As C is not open, there exists x ∈ Csuch that, for every n < ω,

Nxn ⊆ C.

Let

y = 〈1〉x.

Then, for every n < ω,

Nyn ⊆ A.

This implies that A is not open.

Definition 5.10 Clopen means both closed and open.


If (S, T ) is a topological space, then S ∈ T by definition and ∅ ∈T because ∅ =

⋃∅ is the union of the empty family of open sets.

Thus both ∅ and S are always clopen. In the standard topologyon R, the only clopen sets are ∅ and R. In the Baire space, thereare clopen sets other than ∅ and ωω. For example, if n < ω ands ∈ nω, then Ns is clopen since Ns is obviously open and

ωω −Ns =⋃Nt | t ∈ nω but t = s

is also open. We get more examples by noting that a union offinitely many clopen sets is also clopen. For example,

N〈0〉 ∪N〈1〉 = x ∈ ωω | x(0) = 0 or x(0) = 1is clopen.

Definition 5.11 For x, y ∈ ωω, the distance between x and y is

d(x, y) =

1/2n if n is least such that x(n) = y(n)0 if x = y.

For example,

d (〈0, 7, 4, 3, . . . 〉, 〈0, 7, 9, 9, . . . 〉) = 1/22 = 1/4.

The proofs of the following lemmas are left to the reader.

Lemma 5.12 d is a metric on ωω.

Lemma 5.13 For every A ⊆ ωω, the following are equivalent.

1. There exists s ∈ <ωω such that A = Ns.2. There exist c ∈ ωω and a real number r > 0 such that

A = x ∈ ωω | d(x, c) < r.The previous lemma says that the topology on ωω is compatible

with the metric d. It is worth observing that d is not the onlymetric with this property. For example, if we define e so that

e(x, y) =

1/(n + 1) if n is least such that x(n) = y(n)0 if x = y,

then e is also a metric compatible with the Baire topology on ωω.It is also worth observing that, although we wrote c for center

and r for radius in Lemma 5.13, the Baire metric space is different

90 Trees

from the real line in that the center and radius of a basic open setare not unique. For example,

x ∈ ωω | d(x, c) < r = N〈〉 = ωω

for every c ∈ ωω and real number r > 1. We are using the notation〈〉 to denote the empty sequence. Technically, 〈〉 = ∅ = 0, so wehave three names for the same thing. Another example is

x ∈ ωω | d(x, c) < r = N〈0〉

for every c ∈ ωω with c(0) = 0 and 1/2 < r ≤ 1.Next we explain what this has to do with trees, the title of this

chapter. The following definition of tree is not the most generalbut it suffices for all but the last section of this chapter.

Definition 5.14 Let Ω be a set. Then T is a tree on Ω iff

T ⊆ <ωΩ

and, for all m, n ∈ ω and s ∈ nω, if s ∈ T and m < n, thens m ∈ T .

We will focus on the cases Ω = ω and Ω = 2. As an example,

T = 〈〉, 〈2〉, 〈7〉, 〈2, 8〉, 〈2, 9〉, 〈7, 1〉, 〈7, 5〉, 〈7, 7〉,〈2, 8, 1〉, 〈2, 8, 1, 1〉, 〈2, 8, 1, 5〉

is a tree on ω. As part of checking that T is a tree, note that〈2, 8, 1, 5〉 ∈ T and so are all of its restrictions: 〈2, 8, 1〉, 〈2, 8〉, 〈2〉and 〈〉.Definition 5.15 If T is a tree on ω, then the set of infinitebranches of T is

[T ] = x ∈ ωω | x n ∈ T for every n < ω.Example [<ωω] = ωω.

Example [<ω2] = ω2.

Example If s ∈ <ωω, then [r ∈ <ωω | r ⊆ s or s ⊆ r] = Ns.

Example If x ∈ ωω, then [x n | n < ω] = x.It is easy to see that these examples of sets of the form [T ]

are closed subsets of the Baire space. This is no accident as thefollowing result explains.


Lemma 5.16 Let C ⊆ ωω. Then C is a closed subset of ωω iffthere is a tree T on ω such that C = [T ].

Proof First we prove the reverse direction. Assume T is a treeon ω and C = [T ]. Let U = ωω − C. To see that C is closed weshow that U is open. For this, simply observe that

U = ωω − [T ]= x ∈ ωω | there exists n < ω such that x n ∈ T=

⋃Ns | s ∈ <ωω − T

is a union of basic open sets.For the forward direction of Lemma 5.16, consider an arbitrary

closed subset C of ωω. Put

T = x n | n < ω and x ∈ C.Clearly C ⊆ [T ]. We finish by showing that [T ] ⊆ C. For contra-diction, suppose

y ∈ [T ]− C.

Let U = ωω−C. Then U is open and y ∈ U . So there exists n < ωsuch that

Nyn ⊆ U.

In other words,Nyn ∩ C = ∅.

But, since y ∈ [T ], there exists x ∈ C such that x n = y n.Thus

x ∈ Nyn ∩ C.

This contradiction completes the proof.

Exercises

Exercise 5.1 Let

I = x ∈ ωω | x is an injection from ω to ωand

S = x ∈ ωω | x is a surjection from ω to ω.Answer the following questions and prove your answer is correct.

92 Trees

1. Is I open?2. Is I closed?3. Is S open?4. Is S closed?

Exercise 5.2 Let A ⊆ ωω. Put

T = x n | n < ω and x ∈ A.

Prove that [T ] is the closure of A in the Baire space. By this wemean that [T ] is closed and, for every closed set C, if A ⊆ C, then[T ] ⊆ C.

Exercise 5.3 A topological space (S, T ) is said to be a Lindelofspace iff for every F ⊆ T , if

S =⋃F ,

then there is a countable G ⊆ F such that

S =⋃G.

Prove that the Baire space is a Lindelof space.

Exercise 5.4 A topological space (S, T ) is said to be compactiff for every F ⊆ T , if

S =⋃F ,

then there is a finite G ⊆ F such that

S =⋃G.

1. Prove that the Baire space is not compact.2. The Cantor space is the topological space on ω2 whose open

sets are exactly those of the form ω2 ∩ U where U is an opensubset of the Baire space. You could say that the Cantor spacetopology is inherited from the Baire space.

Prove that the Cantor space is compact.Hint: Let F be a family of open subsets of ω2. Assume there

is no finite G ⊆ F such that⋃G = ω2.


Prove that there is an x ∈ ω2 such that

x ∈⋃F .

Use recursion to define x(n) in terms of x n. Along the recur-sion, maintain that there is no finite G ⊆ F such that⋃

G ⊇ Nxn ∩ ω2.

Exercise 5.5 If (S, T ) is a topological space and D ⊆ S, thenD is said to be dense iff for every non-empty U ∈ T ,

D ∩ U = ∅.

A topological (S, T ) space is said to be separable iff it has a count-able dense subset. Show that the Baire space is separable.

Exercise 5.6 Let D be the set of x ∈ ωω such that, for everym < ω, there exists n < ω such that m < n and x(n) = 0.

1. Prove that D is dense.2. Prove that D is not open.3. Prove that D is not closed.4. Find a sequence 〈Un | n < ω〉 of subsets of ωω such that

D =⋂n<ω

Un

and, for every n < ω, Un is open and dense.

Exercise 5.7 We need three definitions before stating the exer-cise. Consider an arbitrary metric space (S, d).

• Let 〈xi | i < ω〉 be a sequence of elements of S and y ∈ S.

– We say that 〈xi | i < ω〉 converges to y and write

limi→ω

xi = y

iff for every r ∈ R, if r > 0, then there exists i < ω such that,for every j < ω, if j > i, then d(xj, y) < r.

– We call 〈xi | i < ω〉 a Cauchy sequence iff for every r ∈ R, ifr > 0, then there exists i < ω such that, for all j, k < ω, ifj, k > i, then d(xj, xk) < r.

94 Trees

• We say that (S, d) is complete iff for every Cauchy sequence〈xi | i < ω〉 from S, there is y ∈ S such that

limi→ω

xi = y.

The following exercises are about the Baire space with the metricdefined by

d(x, y) = 1/2n ⇐⇒ (x n = y n but x(n) = y(n))

and

d(x, y) = 0 ⇐⇒ x = y

but we remark that parts 1 and 2 hold in every metric space.

1. Let C ⊆ ωω. Prove that C is closed iff for every sequence

〈xi | i < ω〉

from C and every y ∈ S, if

limi→ω

xi = y,

then y ∈ C. This says that a set is closed iff it has all its limitpoints.

2. Prove that if limi→ω xi = y, then 〈xi | i < ω〉 is a Cauchysequence.

3. Prove that the Baire space is complete.

Exercise 5.8 (Baire category theorem) Let 〈Dn | n < ω〉 be asequence of subsets of ωω. Assume that, for every n < ω, Dn isboth open and dense in the Baire space. Let

E =⋂n<ω

Dn.

Prove that E is dense in the Baire space.

Exercise 5.9 A tree T on ω is called perfect iff for every r ∈ T ,there are s, t ∈ T such that r ⊆ s, r ⊆ t, s ⊆ t and t ⊆ s. Provethat if T is a non-empty perfect tree on ω, then T has 2ℵ0 manybranches, that is,

|[T ]| = 2ℵ0 .


Exercise 5.10 Let 〈Dn | n < ω〉 be a sequence of subsets of ωω.Assume that, for every n < ω, Dn is both open and dense in theBaire space. Let

E =⋂n<ω

Dn.

Prove that

|E| = 2ℵ0 .

Hint: By Exercise 5.9, it is enough to show that there is a perfecttree T such that [T ] ⊆ E. Construct T using ideas similar to thesolution to Exercise 5.8.

Exercise 5.11 (Cantor perfect set theorem) Let C be a closedsubset of the Baire space and T be a tree on ω such that C = [T ].The Cantor–Bendixon derivative of T is defined to be

T ′ = s ∈ T | Ns ∩ [T ] has at least two elements.

By recursion, define

T 0 = T,

Tα+1 = (Tα)′

whenever α is an ordinal, and

Tβ =⋂α<β

Tα.

whenever β is a limit ordinal.

1. By induction on all ordinals β, prove that Tβ is a tree on ωand, for every α < β,

Tβ ⊆ Tα.

2. Prove that there exists δ < ω1 such that T δ+1 = T δ .3. Let δ be least such that

T δ+1 = T δ.

(This is the Cantor–Bendixon rank of T .)

(a) Prove that T δ is a perfect tree and [T δ ] ⊆ C.(b) Prove that if T δ = ∅, then |C| ≤ ℵ0.

96 Trees

Notice that the combination of Exercises 5.9 and 5.11 shows thatclosed subsets of the Baire space are either countable or havecardinality 2ℵ0 .

Exercise 5.12 Prove by induction that, for every δ < ω1, thereis a tree Tδ on ω whose Cantor–Bendixon rank is δ and (Tδ)δ = ∅.Hint: Obviously, T0 = ∅ and T1 = <ω0 work. Next define T2 andT3. Once you see a pattern for natural numbers, try Tω . Then youwill be on your way to constructing Tδ by recursion.

Exercise 5.13 Prove that there exists a set A ⊆ ωω such that,for every non-empty perfect tree T , neither [T ] ⊆ A nor [T ] ⊆ωω −A. Hint: Let

〈Tα | α < 2ℵ0 〉

enumerate the non-empty perfect trees. Recursively define

〈xα | α < 2ℵ0 〉

and

〈yα | α < 2ℵ0 〉

such that, for every β < 2ℵ0 ,

yβ ∈ [Tβ ]− xα | α < β

and

xβ ∈ [Tβ ]− yα | α ≤ β.

Then let A = xα | α < 2ℵ0 .Remark: Exercise 5.13 is another example of a diagonal argu-

ment. Intuitively, we diagonalize over all non-empty perfect treesto make sure none of them work.

5.3 Illfounded and wellfounded trees

The first result of this section is that if T is a tree on ω withinfinite height and finite levels, then T has an infinite branch.

5.3 Illfounded and wellfounded trees 97

Theorem 5.17 (D. Konig) Let T be a tree on ω. Assume that,for every n < ω,

T ∩ nω = ∅

and

|T ∩ nω| < ℵ0.

Then

[T ] = ∅.

Corollary 5.18 Let T be a tree on 2. Assume that, for everyn < ω,

T ∩ n2 = ∅.

Then

[T ] = ∅.

Corollary 5.18 is an immediate consequence of Theorem 5.17,which we will prove after some discussion and an example. Wealready used the words level and height informally. Now let usofficially define them.

Definition 5.19 Let T be a tree on ω. Then, for every n < ω,

leveln(T ) = T ∩ nω

and

height(T ) = n < ω | leveln(T ) = ∅.

Notice that if T is a tree on ω, then height(T ) is an ordinal and

height(T ) ≤ ω.

This is because trees are closed downward: if s ∈ leveln(T ) andm < n, then s m ∈ levelm(T ). As height(T ) is a transitive set ofnatural numbers, it is itself an ordinal ≤ ω.

In Theorem 5.17, we cannot drop the hypothesis that all levelsof T are finite. Consider the tree depicted in Figure 5.2. It consistsof all restrictions sn m of sequences

sn = 〈n, 0, . . . , 0〉

where, in the displayed sequence, n < ω and there are n zeros.Then T has infinite height but no infinite branch.

98 Trees

. ..

〈3, 0, 0, 0〉

〈2, 0, 0〉〈3, 0, 0〉

〈1, 0〉〈2, 0〉〈3, 0〉

〈0〉〈1〉〈2〉〈3〉 · · ·

〈〉

Figure 5.2 An infinite tree with no infinite branches

Proof of Theorem 5.17 We will need the following notation. Givenr ∈ <ωω, let

Tr = s ∈ T | r ⊆ s or s ⊆ r.Notice that Tr ⊆ T , Tr is a tree on ω and Tr has finite levels.

Define a function x : ω → ω by recursion as follows. Assumethat x n has been defined so that x n ∈ T and Txn has infiniteheight. Since Txn has finite levels, we can write

Txn =⋃i<j

Tsi

where j < ω and, for every i < j,

si ∈ n+1ω ∩ Txn.

From the equation above and the fact that Txn has infinite height,it follows that there is at least one i < j such that Tsi has infiniteheight. Define x (n+1) = si for the least such i. In other words,put x(n) = si(n) for this i. This completes the definition of x.Clearly, x ∈ [T ], which proves the theorem.


Recall that the Cantor space is compact by Exercise 5.4. It ispossible to derive Corollary 5.18 directly from the fact that theCantor space is compact. This alternative proof uses Lemma 5.16,which, in the case of the Cantor space, tells us that C is closedsubset of ω2 iff there is a tree T on 2 such that C = [T ]. Considerthis a hint for Exercise 5.15.

Theorem 5.17 gives conditions that imply a tree has infinitebranches. Now we want to understand when a tree does not haveinfinite branches. It may help to picture trees growing downwardinstead of upward for this discussion. Not having infinite branchesmakes a tree wellfounded according to the following definition.

Definition 5.20 If T is a tree on ω, then T is wellfounded iff[T ] = ∅. Otherwise, T is illfounded. We call s a terminal node ofT iff s ∈ T and there is no t ∈ T with t s.

This terminology makes sense if you think of the tree as growingdownward because if x ∈ [T ], then

· · · x 2 x 1 x 0.

Theorem 5.17 says that an infinite tree with finite levels is ill-founded. We will characterize wellfounded trees in terms of rankfunctions.

Definition 5.21 Let T be a tree on ω. A rank function for T isfunction f with domain T such that, for all s, t ∈ T ,

• f(s) is an ordinal and• if s t, then f(s) > f(t).

Lemma 5.22 If T is a tree on ω and T has a rank function,then T is wellfounded.

Proof For contradiction, suppose that [T ] = ∅. Let x ∈ [T ]. Then

· · · < f(x 2) < f(x 1) < f(x 0)

is an infinite descending sequence of ordinals.

The converse of Lemma 5.22 is also true. In fact, if T is a well-founded tree on ω, then there is a natural way to define a rankfunction for T , which is what the following theorem explains.

100 Trees

Theorem 5.23 Let T be a wellfounded tree on ω. Then there isa unique function

ρT : T −→ ω1

such that, for every s ∈ T ,

• if s is a terminal node of T , then ρT (s) = 0, and• if s is not a terminal node of T , then

ρT (s) = sup (ρT (t) + 1 | t s) .

In particular, ρT is a rank function for T .

We call ρT the rank function associated to T . The two clausesdetermining ρT in Theorem 5.23 look like a recursive definitionbut it is not immediately clear which wellordering underlies therecursion. This is sorted out in Exercises 5.18 and 5.19.

Corollary 5.24 Let T be a tree on ω. Then T is wellfounded iffT has a rank function.

Definition 5.25 If T is a non-empty wellfounded tree on ω, thenwe let the rank of T be

rank(T ) = ρT (〈〉).

Do not confuse height with rank. Every tree on ω has height atmost ω. But only wellfounded trees have ranks, and these ranksare sometimes strictly greater than ω.

Example For every n < ω, the trees

s | dom(s) ≤ n and s(m) = 1 for every m < dom(s)

and

s | dom(s) ≤ n and s(m) < ω for every m < dom(s)

both have rank n.

Example Figure 5.2 shows an example of a wellfounded tree ofrank ω. Figure 5.3 shows the same tree with the nodes labeledaccording to their rank values.


. ..

0

0 1

0 1 2

0 1 2 3 · · ·

ω

Figure 5.3 A wellfounded tree of rank ω (labels are ranks)

Example If we let T be the tree in Figure 5.2 and define

U = 〈0〉s | s ∈ T,

then

rank(U) = ω + 1.

See Figure 5.4 for a picture of U with its nodes labeled accordingto their rank values. We should explain the notation we are usingin the definition of U . For r ∈ mω and s ∈ nω, let

rs = 〈r(0), . . . , r(m− 1), s(0), . . . , s(n− 1)〉 ∈ m+nω.

Put another way,

(rs) (i) =

r(i) if i < m

s(i−m) if m ≤ i < m + n.

The examples above beg the question: which ordinals are theranks of trees on ω? The answer is exactly the countable ordinalsby Theorem 5.23 and Exercise 5.20.

102 Trees

. ..

0

0 1

0 1 2

0 1 2 3 · · ·

ω

ω + 1

Figure 5.4 A wellfounded tree of rank ω + 1 (labels are ranks)

Exercises

Exercise 5.14 Let T be a tree on ω. Assume that [T ] = ∅. Provethat there exists a unique x ∈ [T ] such that, for all y ∈ [T ] andn < ω, if y n = x n, then x(n) ≤ y(n). We call x the left-mostbranch of T .

Exercise 5.15 You should notice that the proof of Theorem 5.17is similar to the solution to Exercise 5.4(2). This exercise explainswhy.

1. Use Lemma 5.16 and Corollary 5.18 to derive the fact that theCantor space is compact.

2. Use Lemma 5.16 and the fact that the Cantor space is compactto derive Corollary 5.18.


Exercise 5.16 Recall that clopen means closed and open.

1. Prove that, for every C ⊆ ω2, if C is a clopen subset of theCantor space, then C is a union of finitely many basic opensubsets of the Cantor space. In other words, there is a finite set

s0, . . . , sn−1 ⊂ <ωω

such thatC =

⋃i<n

Nsi ∩ ω2.

2. Find an example of a set C such that C is a clopen subset ofthe Baire space but neither C nor ωω − C is a finite union ofbasic open sets. Explain why your example has this property.

3. Prove that the only two clopen subsets of R are ∅ and R.

Exercise 5.17 Let T be the wellfounded tree consisting of de-scending sequences of natural numbers. In other words,

T = s ∈ <ωω | for all m, n ∈ dom(s), if m < n, then s(m) > s(n).Calculate rank(T ).

Exercise 5.18 (Kleene–Brouwer ordering) Define a relation <KBon <ωω by declaring that, for all s, t ∈ <ωω,

t <KB s

iff either t s or there exists n < ω such that t n = s n butt(n) < s(n).

1. Prove that <KB is a strict linear ordering of <ωω.2. Let T be a tree on ω.

(a) Prove that if the restriction of <KB to T is a wellordering,then T is wellfounded.

(b) Prove that if T is wellfounded, then the restriction of <KBto T is a wellordering.

Exercise 5.19 Prove Theorem 5.23 in the following two steps.

1. Explain why the properties of ρT listed in the statement ofTheorem 5.23 form a legitimate definition by recursion on therestriction of <KB to T .

2. Explain why the range of ρT is a countable ordinal.

104 Trees

Exercise 5.20 Prove by induction on α < ω1 that there existsa wellfounded tree T on ω with rank(T ) = α.

Exercise 5.21 Let T be a non-empty wellfounded tree on 2.Prove that rank(T ) < ω.

Exercise 5.22 Let B be the Boolean algebra of clopen subsetsof the Cantor space. That is,

B = X | X is a clopen subset of ω2,

X ∨B Y = X ∪ Y,

X ∧B Y = X ∩ Y,

¬BX = ω2−X,

⊥B = ∅

and

B = ω2.

Prove B is a countable atomless Boolean algebra.

Exercise 5.23 Let B be the Boolean algebra of clopen subsetsof the Baire space. That is,

B = X | X is a clopen subset of ωω,

X ∨B Y = X ∪ Y,

X ∧B Y = X ∩ Y,

¬BX = ωω −X,

⊥B = ∅

and

B = ωω.

Prove B is an atomless Boolean algebra of cardinality 2ℵ0 .

5.4 Infinite games 105

5.4 Infinite games

Let A ⊆ ωω. We describe a game, which is called GA. The gamehas two players, I and II, who take turns playing natural numbersx0, x1, etc. A run of the game GA looks as follows.

I x0 x2 x4· · ·

II x1 x3 x5

If x = 〈xn | n < ω〉 is a run of GA, then player I wins the run iffx ∈ A. Otherwise, player II wins the run.

This is a very general sort of game. Notice that finite games alsofit this scheme because we may ignore moves after a winner hasbeen declared. (There are really two kinds of finite length games.Either the length is fixed in advance or else the length dependson exactly how the players move. Both kinds of finite games canbe modeled with our infinite games.) One difference between ourgames and some familiar games like chess is that we do not allow arun of the game to end in a draw. This is because either x ∈ A orx ∈ A. An arbitrary way to get around this objection is to declarethat draws go to player II. See Exercise 5.24 for more about chess.

Many properties in mathematics can be expressed in terms ofgames, so general theorems about games can be quite useful. Alongthese lines, a series of exercises at the end of this section illustrateone of many ways in which games and mathematical analysis arerelated.

Naturally, we are more interested in winning strategies than weare in the player who wins a particular run of a particular game.So let us continue making definitions associated to the game GA.

A strategy is a function σ : <ωω → ω. If σ is a strategy andb ∈ ωω, then σ ∗ b is the run that results if I uses σ and II plays b.Formally, σ ∗ b is defined by recursion according to the equations

(σ ∗ b)2n+1 = bn

and(σ ∗ b)2n = σ((σ ∗ b) 2n).

See Figure 5.5 for another way of depicting the run σ ∗ b. We callσ a winning strategy for player I iff for every b ∈ ωω,

σ ∗ b ∈ A.

Iσ(〈〉)

σ(〈

σ(〈〉)

,b0〉

)σ(〈

σ(〈〉)

,b0,

σ(〈

σ(〈〉)

,b0〉

),b 1〉)

···

IIb 0

b 1b 2

Fig

ure

5.5

The

run

σ∗b

Ia

0a

1a

2a

3···

IIσ(〈

a0〉

)σ(〈

a0,

σ(〈

a0〉

),a

1〉)

σ(〈

a0,

σ(〈

a0〉

),a

1,σ(〈

a0,

σ(〈

a0〉

),a

1〉),

a2〉

)

Fig

ure

5.6

The

run

a∗σ

108 Trees

Similarly, if σ is a strategy and a ∈ ωω, then a ∗ σ is the runthat results if II uses σ and I plays a. Formally, a ∗σ is defined byrecursion according to the equations

(a ∗ σ)2n = an

and(a ∗ σ)2n+1 = σ((a ∗ σ) (2n + 1)).

Figure 5.6 depicts the run a ∗ σ another way. We call σ a winningstrategy for player II iff for every a ∈ ωω,

a ∗ σ ∈ A.

We say that A is determined iff either player I has a winningstrategy or player II has a winning strategy. Otherwise, A is un-determined. Obviously, it is not possible for both players to havewinning strategies because otherwise we could play the two strate-gies against each other to get a contradiction.

Example Let A be the set of surjections from ω to ω. Let σ bethe strategy such that

σ(〈xi | i < 2n〉) = n

andσ(〈xi | i < 2n + 1〉) = 0.

Then, for every b ∈ ωω,

σ ∗ b ∈ A.

Thus σ is a winning strategy for player I in GA. So A is determined.Observe that σ is not the only winning strategy for player I in GA;there are infinitely many others.

This section includes two theorems about the determinacy ofgames. The first, Theorem 5.26, says that some games are unde-termined. It will be apparent from our proof that there are 2ℵ0

many strategies and 22ℵ0 many games. In particular, there arestrictly more games than strategies. By itself, this is not an argu-ment that there are undetermined games because some strategieswin in more than one game. For example, if σ is a winning strategyfor player I in GA and A ⊆ B, then σ is also a winning strategyfor player I in GB . So more work than mere counting is needed tosee that there are undetermined games.


Theorem 5.26 (Gale–Stewart) There is an undetermined subsetof ωω.

Proof First we claim that if σ is a strategy, then the two functions

b → σ ∗ b

and

a → a ∗ σ

are injections from ωω to itself. This claim is clear because b isthe sequence of odd values of σ ∗ b and a is the sequence of evenvalues of a ∗ σ.

Next observe that

|ωω| = ℵℵ00 = 2ℵ0

and

|σ | σ is a strategy| =∣∣∣(< ω ω)ω

∣∣∣ = ℵ(ℵ<ℵ00 )

0 = ℵℵ00 = 2ℵ0 .

Say

σ | σ is a strategy =

σα | α < 2ℵ0

.

Now choose aθ and bθ by recursion on θ < 2ℵ0 as follows. Assumethat aη and bη have been selected for η < θ. Let

Aθ = aη ∗ ση | η < θ

and

Bθ = ση ∗ bη | η < θ.

Notice that

|Aθ | ≤ θ < 2ℵ0 .

Because

b → σθ ∗ b

is an injection, we can pick bθ ∈ ωω such that

σθ ∗ bθ ∈ Aθ.

This determines

Bθ+1 = ση ∗ bη | η ≤ θ.

110 Trees

Notice that

|Bθ+1| ≤ θ + 1 < 2ℵ0 .

Because

a → a ∗ σθ

is an injection, we can pick aθ ∈ ωω such that

aθ ∗ σθ ∈ Bθ+1.

That completes the definition of 〈aθ | θ < 2ℵ0 〉 and 〈bθ | θ < 2ℵ0 〉.Let

A = aη ∗ ση | η < 2ℵ0

and

B = ση ∗ bη | η < 2ℵ0 .

From the recursive definition, it is clear that

A ∩B = ∅.

Let σ be an arbitrary strategy. Say σ = σθ . Then σ is not a winningstrategy for player II in GA because

aθ ∗ σθ ∈ A.

And σ is not a winning strategy for player I in GA because

σθ ∗ bθ ∈ B.

Therefore, A is undetermined.

It is worth noting how the Axiom of Choice was used in theprevious proof. Without it, we would not necessarily be able toindex all the strategies with ordinals at the start. The proof wasyet another example of a diagonal argument. Intuitively, we diag-onalized over all strategies to make sure that none of them work,handling the θth strategy at stage θ.

We have seen that some games are undetermined. But manyof the games that come up in practice, open games for example,turn out to be determined. This is more important because it hasmathematical applications.

Theorem 5.27 (Gale–Stewart) Every open subset of ωω is de-termined.


Proof Let U be an open subset of ωω. Suppose that player I doesnot have a winning strategy in GU . We must show that player IIhas a winning strategy in GU . Let

C = ωω − U.

Then C is closed, so by Lemma 5.16, there is a tree T on ω suchthat C = [T ]. For s ∈ <ωω, let

Us = x | sx ∈ U.Notice that

U〈〉 = U.

Let W be the set

s ∈ <ωω | dom(s) is even and I has a winning strategy in GUs .We refer to the elements of W as winning positions for player I inGU . The following conditions are obviously true.

1. 〈〉 ∈W .2. Let s ∈ <ωω such that dom(s) is even but s ∈ W . Then:

(a) For every k < ω, there is < ω such that s〈k, 〉 ∈ W .(b) There exists x ∈ C such that x dom(s) = s.

From conditions (1) and (2a) we can read off a certain strategy

τ : <ωω → ω

such that, for all a ∈ ωω and n < ω,

(a ∗ τ) 2n ∈W.

Just to be specific, given s ∈ <ωω such that dom(s) is even buts ∈ W , for every k < ω, define

τ(s〈k〉) = the least < ω such that s〈k, 〉 ∈ W.

The other values of τ are irrelevant, so make them zero. Some-times this τ is called a non-losing strategy for player II because itavoids winning positions for player I. We claim that τ is a winningstrategy for player II in GU . To see this, let a ∈ ωω and y = a ∗ τ .We must show that y ∈ U . By condition (2b), for every n < ω,there exists xn ∈ C such that

xn 2n = y 2n.

112 Trees

This can be expressed by the inequality

d(xn, y) ≤ 1/22n.

Hence

limn→∞

xn = y.

Since C is closed,

y ∈ C = ωω − U.

A more direct way to argue this last part is to note that, for everyn < ω,

y 2n = xn 2n ∈ T

hence

y ∈ [T ] = C = ωω − U.

Exercises

Exercise 5.24 Use Theorem 5.27 to explain why, in chess, eitherWhite has a winning strategy, or Black has a strategy to avoidlosing.

Exercise 5.25 Prove there is an undetermined set B ⊆ ωω suchthat ωω − B is determined. Hint: By Theorem 5.26, there is anundetermined set A. Do not worry about how A was constructed.Rather, take A as given and define B from A in a way that takesadvantage of the asymmetry that player I goes first.

Exercise 5.26 Let C be a closed subset of ωω. Prove that Cis determined. Hint: One approach is to model the proof on thatof Theorem 5.27. An easier approach is to use the statement ofTheorem 5.27 and derive closed determinacy as a corollary. Butthe proof is not completely trivial because, by Exercise 5.25, thereare determined sets whose complements are not determined.


For the following series of exercises, we define another kind ofgame, called a perfect set game for reasons that will become ap-parent. For A ⊆ ω2, let G∗

A be the game whose runs have thefollowing pattern.

I s0 s1 s2 · · ·II n0 n1 n2 · · ·

At stage 2i, player I must play si ∈ <ω2 or else he loses. At stage2i + 1, player II must play ni < 2 or else he loses. At the end ofthe run, player I wins if

s0〈n0〉s1

〈n1〉s2〈n2〉 · · · ∈ A.

Otherwise, player II wins the run.

Exercise 5.27 Provide formal definitions of the following ter-minology.

1. σ is a strategy for player I in G∗A.

2. σ is a winning strategy for player I in G∗A.

3. τ is a strategy for player II in G∗A.

4. τ is a winning strategy for player II in G∗A.

Exercise 5.28 Let A ⊆ ω2. Suppose that there exists a perfecttree T on 2 such that [T ] ⊆ A. Prove that player I has a winningstrategy in G∗

A.

Exercise 5.29 Let A ⊆ ω2. Suppose that player I has a winningstrategy in G∗

A. Prove that there exists a perfect tree T on 2 suchthat [T ] ⊆ A.

Exercise 5.30 Let A ⊆ ω2. Suppose that A is countable. Provethat player II has a winning strategy in G∗

A.

Exercise 5.31 This exercise is harder than the previous threebut it is the most important. Let A ⊆ ω2. Suppose that playerII has a winning strategy in G∗

A. Prove that A is countable bycompleting the following outline.

Let τ be a winning strategy for player II in G∗A. Suppose that

p is a position of even length 2j. If 2j = 0, then

p = 〈〉,whereas if 2j > 0, then we may specify that

p = 〈s0, n0, . . . , sj−1, nj−1〉.

114 Trees

Saying that p has even length is the same as saying that, startingfrom p, it is player I’s turn to move. Assume, in addition, that pis consistent with τ , by which we mean that

n0 = τ(〈s0〉),

n1 = τ(〈s0, n0, s1〉),

n1 = τ(〈s0, n0, s1, n1, s2〉),and so on for every ni with i < j. Now define

p∗ = s0〈n0〉 . . . sj−1

〈nj−1〉.Notice that p∗ ∈ <ω2 and the domain of p∗ is its finite cardinality

|p∗| = j +∑i<j

|si|.

For x ∈ ω2, we say that p rejects x iff

• p∗ ⊂ x and• if q is a position such that

– q extends p,– q has even length (so it is player I’s turn to move after q),

and– q is consistent with τ ,then q∗ ⊂ x.

1. Prove that each p as above rejects exactly one x ∈ ω2. Hint:Define x(m) by recursion on m < ω. Start by setting x(m) =p∗(m) for all m < |p∗|. Now suppose that m ≥ |p∗|. There is aunique s ∈ <ω2 such that

x m = p∗s.

Letn = τ(p〈s〉).

If n = 0, then put x(m) = 1. Otherwise n = 1, in which caseput x(m) = 0. Show that p rejects x and, if p rejects y, theny = x.

2. Prove that if x ∈ A, then there exists p as above that rejects x.Hint: Suppose otherwise and contradict the assumption that τis a winning strategy for player II.

5.5 Ramsey theory 115

3. Use the previous results 1 and 2 to conclude that A is countable.Hint: Count positions.

Exercise 5.32 Let C ⊆ ω2 and assume that C is closed. Sketcha proof that G∗

C is determined. In other words, prove that eitherplayer I has a winning strategy in G∗

C , or else player II does.Hint: All of the ideas are contained in Theorem 5.27 and Exercise5.26 but writing up a complete proof is challenging because thenotation is complicated.

Exercise 5.33 Use Exercises 5.29, 5.31 and 5.32 to prove thatif C is a closed subset of ω2, then either C is countable or C hasa perfect subset. Notice this also follows from the Cantor perfectset theorem, which was the subject of Exercise 5.11, but the twoproofs are very different.

5.5 Ramsey theory

Ramsey theory is often introduced with the following scenario.Imagine that you are hosting a party and you would like to inviteenough people so that either there is a trio of guests who havemet the other two in the trio, or there is a trio of guests whohave met neither of the other two in the trio. This can be modeledmathematically as follows. Let I represent the set of invited guestsand

[I]2 = a, b | a, b ∈ I and a = bbe the set of pairs of guests. Let the function

F : [I]2 → 2

be given by

F (a, b) =

0 a and b have not met each other1 a and b have met each other.

Such a function is referred to as a coloring of pairs from I bytwo colors. The two colors are 0 and 1. So 2 = 0, 1 is the set ofcolors. A subset H ⊆ I is called homogeneous for F iff there existsk ∈ 2 such that, for all a, b ∈ H, if a = b, then

F (a, b) = k.

116 Trees

Your goal as host is to choose I large enough so that, for every Fas above, there is an H as above with |H| = 3. By the followingtwo exercises, you should invite at least six guests.

Exercise 5.34 Prove that for every function

F : [6]2 → 2,

there exists H ⊆ 6 and k ∈ 2 such that |H| = 3 and, for alla, b ∈ H, if a = b, then

F (a, b) = k.

Exercise 5.35 Find a function

F : [5]2 → 2

such that, for every H ⊆ 5 and k ∈ 2, if |H| = 3, then there existsa, b ∈ H such that a = b and

F (a, b) = k.

The phenomenon described above has many important exten-sions. Given m < ω and a set I, define

[I]m = p ⊆ I | |p| = m.The finite Ramsey theorem says that, for all positive , m, n < ω,there exists r < ω such that, given

• a set I such that |I| ≥ r, and• a function F : [I]m → ,

there exist

• a subset H ⊆ I such that |H| ≥ n, and• a number k <

with the property that, for every p ∈ [H]m ,

F (p) = k.

Exercises 5.34 and 5.35 show that, with m = 2 (colorings of pairs), = 2 (two colors) and n = 3 (homogeneous set with three ele-ments), the least witness to the finite Ramsey theorem is r = 6.The finite Ramsey theorem is typically proved in a basic courseon discrete mathematics; we will not prove it here. The follow-ing infinite Ramsey theorem and results like it are of significant

5.5 Ramsey theory 117

importance in set theory and its applications. The way we haveorganized the proof explains why it sits in our chapter on trees.

Theorem 5.28 (Ramsey) Let 0 < < ω and F be a function ofthe form

F : [ω]2 → .

Then there exists k < and an infinite H ⊆ ω such that, for everyp ∈ [H]2,

F (p) = k.

Proof Let T be the set of strictly increasing s ∈ <ωω such that,for every m < dom(s), there exists k < such that, for everyn < dom(s), if m < n, then

F (s(m), s(n)) = k.

Then T is a tree on ω. In this context, strictly increasing meansthat, for all m < n < dom(s),

s(m) < s(n).

We will prove that T has an infinite branch. But first let us showwhy the existence of such a branch suffices to prove the theorem.Suppose that x is an infinite branch of T . For each m < ω, let km

be the unique k < such that, for every n < ω, if m < n, then

F (x(m), x(n)) = k.

Since km | m < ω ⊆ is finite, there exists an infinite S ⊆ ωand k < such that, for every m ∈ S,

km = k.

Let

H = x(m) | m ∈ S.

Since S is infinite and x is increasing, H is infinite. Moreover, forevery a, b ∈ H,

F (a, b) = k.

Thus H witnesses the conclusion of the theorem.Now we construct an infinite branch x through T . By recursion

118 Trees

on n < ω, we define x(n) and, simultaneously, kn < and aninfinite In ⊆ ω. Start by defining

I0 = ω

andx(0) = 0.

Now assume we are given In and it is an infinite subset of ω. Let

x(n) = min(In).

This is consistent with our having set x(0) = 0. For each k < ,let

Jk = i ∈ In | i > x(n) and F (x(n), i) = k.Then

In − x(n) = J0 ∪ · · · ∪ J −1.

Since In − x(n) is infinite, there exists k < such that Jk isinfinite. Let kn be the least such k and

In+1 = Jkn .

That completes the recursive construction. By induction on n < ω,it is obvious that

x(n) = min(In)

and, for every m < n,

In ⊆ Im+1 ⊆ Im − x(m),

x(m) < x(n)

andF (x(m), x(n)) = km.

Therefore, x is an infinite branch through T .

Exercise 5.36 Let 0 < m < ω. Show that Theorem 5.28 remainstrue if “2” is replaced by “m” in its statement. Hint: Use inductionon m. The case m = 1 follows from the pigeonhole principle. (Ifyou partition an infinite set into finitely many pieces, then one ofthe pieces must be infinite.) Think of the proof of Theorem 5.28 asshowing that the case m = 1 implies the case m = 2. Generalizethis to see that case m implies case m + 1.

5.6 Trees of uncountable height 119

Exercise 5.37 Prove that Theorem 5.28 becomes false if “ω”is replaced by “ω1” and “infinite” is replaced by “uncountable”using the following example. Fix an injection

g : ω1 → R.

In order to avoid possible confusion, we write <R for the usualorder of R here. Define

F : [ω1]2 → 2

by

F (α, β) =

0 if α < β and g(α) >R g(β)1 if α < β and g(α) <R g(β).

Prove that there is no uncountable set that is homogeneous forF . In other words, prove that if H is an uncountable subset of ω1and k < 2, then there exists p ∈ [H]2 such that F (p) = k. Hint:Argue by contradiction and use the fact that between any two realnumbers there is a rational number.

5.6 Trees of uncountable height

So far, all the trees we have looked at have at most ω many levels.This section is on trees of height ω1. We might expect that ourresults about trees of height ω would lift to theorems about treesof height ω1. For example, recall that Theorem 5.17 says that if Tis a subtree of <ωω and, for every n < ω,

0 < |T ∩ nω| < ℵ0,

then T has an ω-branch, i.e., there exists b : ω → ω such that, forevery n < ω,

b n ∈ T.

Does this statement remain true if we replace ω by ω1 and ℵ0 byℵ1? It turns out that the answer is no; counterexamples are calledAronszajn trees and the point of this section is to construct one.This is a big topic; in order to keep this section manageable, westill do not give the most general definition of tree.

120 Trees

Definition 5.29 A subtree of <ω1 ω is a subset T ⊆ <ω1 ω suchthat, for all s ∈ T and α < dom(s),

s α ∈ T.

An ω1-branch of T is a function b : ω1 → ω such that, for everyα < ω1,

b α ∈ T.

Theorem 5.30 (Aronszajn) There exists a subtree T of <ω1 ωsuch that

0 < |T ∩ αω| < ℵ1

for every α < ω1 but T has no ω1-branch.

Proof LetI = s ∈ <ω1 ω | s is an injection.

The good news is that I is obviously a subtree of <ω1 ω and I hasno ω1-branch because there is no injection from ω1 to ω. The badnews is that, whenever ω ≤ α < ω1,

|I ∩ αω| = |s ∈ αω | s is an injection| = 2ℵ0 ≥ ℵ1,

which is too large. We will find an appropriate subtree T ⊆ I. Forβ < ω1 and s, t ∈ βω, define

s ∼β t ⇐⇒ |α < β | s(α) = t(α)| < ℵ0.

It is easy to see that ∼β is an equivalence relation on βω.

Claim 5.30.1 There is a sequence 〈sα | α < ω1〉 so that, forevery β < ω1,

sβ ∈ I ∩ βω

and, for every α < β,

sα ∼α sβ α.

Assuming Claim 5.30.1, if we define

T =⋃

α<ω1

t ∈ I ∩ αω | t ∼α sα

then T witnesses the statement of the theorem. To see this, firstobserve that T is a subtree of <ω1 ω. The point is that T is closed


downward under restriction. This is true because if α < β < ω1and

t ∈ T ∩ βω,

then

t ∼β sβ,

hence

t α ∼α sβ α ∼α sα,

so

t α ∈ T ∩ αω.

Second, observe that the levels of T are countable. This is because,for every γ < ω1 and

t ∈ T ∩ γω,

there are n < ω and α0 < · · · < αn−1 < γ such that t(β) = sγ(β)for every β < γ except if β = αm for some m < n. Since there areat most countably many ways to make these sorts of finite changesto sγ , we have that, for every γ < ω1,

1 ≤ |T ∩ γω| ≤ ℵ0.

It remains to prove Claim 5.30.1. We define sβ by recursion onβ < ω1. In addition to maintaining the two requirements of theclaim, we also maintain the extra property that, for every β < ω1,

|ω − ran(sβ)| = ℵ0.

This is so that the recursion does not run out of steam, as you willsee. We have no choice but to set s0 = ∅. If sα has been defined,then pick k ∈ ω − ran(sβ) and define

sα+1 = sα ∪ (α, k).

Now suppose that γ < ω1 is a limit ordinal and we have

〈sα | α < γ〉

satisfying the three requirements. The definition of sγ is somewhattricky in this case. Since cf(γ) = ω, there are ordinals

β0 < β1 < · · ·βi < · · · < γ = supi<ω

βi.

122 Trees

By recursion on i < ω, define a sequence 〈ti | i < ω〉 from I suchthat, for every j < ω,

tj ∼βj sβj

and, for every i < j,

ti = tj βi.

Start with

t0 = sβ0 .

Suppose we are given ti with

ti ∼βi sβi .

Let us say that a pair (α, α′) is bad iff

α < βi ≤ α′ < βi+1

and

ti(α) = sβi+1 (α′).

The reason we call them bad is because their existence preventsus from setting

ti+1(η) =

ti(η) if η < βi

sβi+1 (η) if η ∈ βi+1 − βi.

Remember that ti+1 is supposed to be an injection! However, weclaim that there are only finitely many bad pairs. First note thatif (α, α′) is a bad pair, then

ti(α) = sβi+1 (α).

This is because sβi+1 is an injection, so it cannot take on the samevalue, ti(α), twice. From this and the fact that

ti ∼βi sβi+1 βi,

it follows that there are only finitely many first coordinates of badpairs. Finally, observe that if (α, α′) and (α, α′′) are bad pairs,then

sβi+1 (α′) = ti(α) = sβi+1 (α

′′),

so α′ = α′′, again because sβi+1 is an injection. This shows thatthere are only finitely many bad pairs. Say the number of bad


pairs is n where n < ω. List the bad pairs as (αm, α′m) for m < n.

Then pick distinct

km ∈ ω − (ran(ti) ∪ ran(sβi+1 ))

for m < n. This is possible by the extra property on sβi and sβi+1 ,and the fact that ti ∼βi sβi ∼βi sβi+1 βi. Our solution to theproblem of bad pairs is to define

ti+1(η) =

⎧⎪⎨⎪⎩

ti(η) if η < βi

sβi+1 (η) if η ∈ βi+1 − βi and η = α′m for every m < n

km if η = α′m .

Obviously, this satisifies our requirements for ti+1. Now, havingcompleted the definition of 〈ti | i < ω〉, we put

t =⋃i<ω

ti.

The good news is that t meets the first and second requirementsfor sγ of Claim 5.30.1, namely that

t ∈ I ∩ γω

and, for every α < γ,

sα ∼α t α.

The problem is that it might not satisfy the extra property that

|ω − ran(t)| = ℵ0.

To fix up this potential problem, put

sγ(α) =

t(β2i) if α = βi for some i < ω

t(α) if α ∈ γ − βi | i < ω.

Then all three requirements on sγ are met.

Exercises

Exercise 5.38 Let T be the tree constructed in the proof ofTheorem 5.30. Suppose that α < β < ω1 and u ∈ T ∩ αω. Explainwhy there exists v ∈ T ∩ βω such that u = v α. We say that uhas extensions to every level of T .

124 Trees

Exercise 5.39 Let T be the tree constructed in the proof ofTheorem 5.30. Suppose that α < ω1 and u ∈ T ∩ αω. Explain whythere exist β > α and v, w ∈ T ∩ βω such that

u = v α = w α

butv = w.

We say that u splits in T .

Exercise 5.40 Let T be the tree constructed in the proof ofTheorem 5.30. Find a sequence 〈uα | α < ω1〉 of members of Tsuch that, for all α < β < ω1,

uα ⊆ uβ

anduβ ⊆ uα.

In this case, we say that 〈uα | α < ω1〉 is an antichain of Tbecause uα and uβ are incomparable (cannot be compared using⊆) whenever α = β.

Exercise 5.41 Let T be a subtree of <ω1 ω with an ω1-branch

b : ω1 → ω.

Assume that every u ∈ T splits in T . Prove that T has an un-countable antichain.

6

Dense linear orderings

This chapter is mainly about two theorems, one due to Cantor, theother to Dedekind, which are characterizations of the rationals, Q,and the reals, R, in terms of their respective orderings.

6.1 Definitions and examples

Recall that (A,≺) is a strict linear ordering iff ≺ is a transitive,irreflexive and total relation on A. As usual, we write x y forx ≺ y or x = y.

Definition 6.1 (A,≺) is a dense linear ordering iff (A,≺) is astrict linear order with at least two elements and, for all x, y ∈ A,if x ≺ y, then there exists z ∈ A such that x ≺ z ≺ y.

We required A to have at least two elements because, otherwise,(0, <) and (1, <) would be dense linear orderings, which would becounterintuitive. It follows easily from Definition 6.1 that everydense linear ordering is infinite.

Definition 6.2 If (A,≺) is a strict linear ordering and L, R ∈ A,then we say:

• L is a left endpoint of (A,≺) iff L x for every x ∈ A.• R is a right endpoint of (A,≺) iff x R for every x ∈ A.

In this chapter we are mainly interested in dense linear orderingswithout endpoints. Here are several examples, each referring to theusual ordering of real numbers.

• Q = m/n | m, n ∈ Z and n = 0

126 Dense linear orderings

• R = the set of real numbers• the open interval (0, 1)• (0, 1) ∩Q

The reader has not yet seen definitions of Q and R in this bookbut has intuition about these based on doing mathematics sincechildhood. Temporarily, we rely only on that intuition.

Continuing with our introduction, we make the following keydefinition, which is really just a repetition of Definition 3.27.

Definition 6.3 We say that f is an isomorphism from (A,≺A)to (B,≺B) and write

f : (A,≺A) (B,≺B)

iff f is a bijection from A to B and, for all x, y ∈ A,

x ≺A y ⇐⇒ f(x) ≺B f(y).

We say that (A,≺A) is isomorphic to (B,≺B) and write

(A,≺A) (B,≺B)

iff there is an isomorphism f : (A,≺A) (B,≺B).

For example, with the usual ordering on the real line, the openinterval (−π/2, π/2) is isomorphic to R as witnessed by the func-tion

x → tan(x).

So,(−π/2, π/2) R.

It also turns out that

(−π/2, π/2) ∩Q Q

but since arctan(1) = π/4 ∈ Q, a function other than tangent isneeded. We will see why there is such an isomorphism later.

As an example of non-isomorphism, observe that Q R because

|Q| = ℵ0 = 2ℵ0 = |R|,so there is not even a bijection between Q and R. A more subtleexample is the fact that

RQ R.

6.1 Definitions and examples 127

f (z)(f (z ) ,∞)

z

(0 ,∞) ∩ Q

Figure 6.1 R Q R

Here, RQ is the concatenation of R followed by Q. It is easyto see that RQ is a dense linear ordering without endpoints.Suppose for contradiction that

f : RQ R.

Let z be the zero of Q in the concatenation RQ.1 Then, f(z) ∈ Rand

(0,∞) ∩Q (f(z),∞).

The set on the left has cardinality ℵ0 and the set on the right hascardinality 2ℵ0 , so there is no bijection between them. Figure 6.1illustrates the point.

This brings up an interesting question: what about Q versusQQ? In the next section, we will state and prove Theorem 6.5,which implies that

Q QQ.

Theorem 6.5 also implies that

(−π/2, π/2) ∩Q Q,

which we already mentioned.

1 Formally, the underlying set of the strict linear ordering R Q is(0 × R) ∪ (1 × Q) and z is the ordered pair (1, 0).


6.2 Rational numbers

This section has two theorems, the first of which should come asno surprise.

Theorem 6.4 There is a countable dense linear ordering withoutendpoints.

Exercise 6.1 outlines a proof of Theorem 6.4 based only on whatwe already know about natural numbers, their ordering and theirarithmetic. Obviously, (ω, <) is not dense and has a left endpoint,so some work is needed. It would be tempting to blurt out thatQ with its ordering already witnesses Theorem 6.4 but this wouldbe cheating since the point here is to say what Q is up to isomor-phism.

The next theorem tells us that in a certain sense it does notmatter which countable dense linear ordering without endpointswe work with since they are all the same up to isomorphism. Itsays we can take our pick and make it Q. The technique usedin the proof, a back-and-forth construction, is also important inother parts of mathematics. An application of this technique tothe theory of Boolean algebras is the topic of Exercise 6.6.

Theorem 6.5 (Cantor) Let (A,≺A) and (B,≺B) be countabledense linear orderings without endpoints. Then

(A,≺A) (B,≺B).

Proof Say A = a0, a1, . . . and B = b0, b1, . . . . Warning! It isdefinitely not true that i < j ⇐⇒ ai ≺A aj because (ω, <) is awellordering whereas (A,≺A) is an illfounded relation.

By recursion on n < ω, we will define finite bijections

fn : dom(fn) → ran(fn)

such that

ai | i < n ⊆ dom(fn) ⊆ A

and

bi | i < n ⊆ ran(fn) ⊆ B

and

fn : (dom(fn),≺A) (ran(fn),≺B).

6.2 Rational numbers 129

We also maintain that, for all m < n < ω,

fm ⊆ fn.

In other words,

fn dom(fm) = fm.

Mere success in this construction is enough because if

f =⋃n<ω

fn,

then it is easy to verify that

f : (A,≺A) (B,≺B).

For example, to verify that f defined this way is order preserving,note that if

i, j < n < ω,

then

ai ≺A aj ⇐⇒ fn(ai) ≺B fn(aj) ⇐⇒ f(ai) ≺B f(aj)

because ai, aj ∈ dom(fn), fn is order preserving and

fn = f dom(fn).

Here is the recursive definition of fn. For the base step, setf0 = ∅. Now assume that fn has been defined. The definition offn+1 is a two-step process: back and forth.

Forth step We define a finite isomorphism

g : (dom(g),≺A) (ran(g),≺B).

with

dom(g) = fn ∪ an.

Case 1 an ∈ dom(fn).

Set g = fn.

Case 2 an ∈ dom(fn).

We claim that there exists i < ω such that if

g = fn ∪ (an, bi),


then

g : (dom(g),≺A) (ran(g),≺B).

The proof of the claim breaks up into three subcases.

Subcase 1 an ≺A x for every x ∈ dom(fn).

Since (B,≺B) does not have a left endpoint, we may pick i < ωsuch that bi ≺B y for every y ∈ ran(fn).

Subcase 2 x ≺A an for every x ∈ dom(fn).

Since (B,≺B) does not have a right endpoint, we may pick i < ωsuch that y ≺B bi for every y ∈ ran(fn).

Subcase 3 Otherwise.

Since dom(fn) is finite, we can pick , r ∈ dom(fn) such that

≺A an ≺A r

and, for every a ∈ dom(fn), either

a A

or

r A a.

Then,

fn( ) ≺B fn(r)

and, for every b ∈ ran(fn), either

b B fn( )

or

fn(r) B b.

Because (B,≺B) is a dense linear ordering, we can pick i < ω suchthat

fn( ) ≺B bi ≺B fn(r).

In each of the three subcases, it is clear that the prescribedchoice of i works, so the claim is proved, case 2 has been handledand the forth step is complete.

6.3 Real numbers 131

Back step We define a finite isomorphism

h : (dom(h),≺A) (ran(h),≺B).

withran(h) = ran(g) ∪ bn.

The process for defining h from g is like the process of definingg from fn except that we reverse the roles of A and B.

Finally, having completed both the back and forth steps, letfn+1 = h.

6.3 Real numbers

Having characterized Q with its ordering up to isomorphism, weturn our attention to R. What is R and which criteria characterizeR with its ordering up to isomorphism? As motivation, considerthe set

S = x ∈ Q | x2 < 2.Notice that S is bounded above by the rational number 3/2. Thisis because, if x is a rational number and x2 < 2, then x < 3/2.(If x ≥ 3/2, then x2 ≥ 9/4 > 2.) Based on the mathematics youknew before starting to read this book, you would have said that√

2 is also an upper bound for S, in fact,√

2 is the least upperbound for S, which we write

lub(S) =√

2.

This is still correct except that√

2 ∈ Q and we have not definedR so it is not fair to mention

√2 yet. The moral of this paragraph

is that, in passing from Q to R, we want to include least upperbounds for all bounded sets.

Definition 6.6 Let (A,≺A) be a strict linear ordering. If S ⊆ Aand y ∈ A, then y is an upper bound for S iff x A y for everyx ∈ S.

To be clear, x A y means that either x ≺A y or x = y.

Definition 6.7 Let (A,≺) be a strict linear ordering. If S ⊆ Aand y ∈ A, then y is a least upper bound for S iff y is an upperbound for S and, for every upper bound z for S, y A z.


It is easy to see that S has at most one least upper bound in(A,≺A). For if it had two, y and z, then y A z and z A y soy = z.

Now we arrive at the property that distinguishes R from Q.

Definition 6.8 A strict linear ordering (A,≺A) has the leastupper bound property iff for every non-empty S ⊆ A, if S hasan upper bound in (A,≺A), then S has a least upper bound in(A,≺A).

We indicated above that Q with its usual ordering does not havethe least upper bound property since

lub(x ∈ Q | x2 < 2) ∈ Q.

Of course, this is not the only example of a subset of Q withouta least upper bound in Q. By contrast, the least upper boundproperty is one of the key properties that characterize R with itsusual ordering.

Another essential property about the usual ordering of R isthat Q ⊆ R and between any two real numbers there is a rationalnumber. In other words, Q is dense in R.

Definition 6.9 If (B,≺B) is a strict linear ordering and A ⊆ B,then A is dense in (B,≺B) iff for all x, y ∈ B, there exists z ∈ Asuch that x ≺B z ≺B y.

Theorem 6.5 tells us that in a certain sense it does not matterwhich countable dense linear ordering without endpoints we taketo be Q since they are all isomorphic. The next theorem tells usthat, once we settle on a choice of Q, there is a way to extendQ to obtain an appropriate choice for R, and this choice for R isunique up to isomorphisms that fix Q.

Theorem 6.10 (Dedekind) Let (A,≺A) be a countable denselinear ordering without endpoints. Then there exists (B,≺B) suchthat:

• (B,≺B) is a dense linear ordering without endpoints;• (A,≺A) is a subordering of (B,≺B), that is, A ⊆ B and, for all

x, y ∈ A,x ≺A y ⇐⇒ x ≺B y;

• A is dense in (B,≺B);


• (B,≺B) has the least upper bound property.

Moreover, if (B′,≺B ′) has the same four properties as (B,≺B),then there is an isomorphism

f : (B,≺B) (B′,≺B ′)

such that, for every x ∈ A,

f(x) = x.

Proof The moreover part of Theorem 6.10 is proved by observingthat, for every y ∈ B,

y = the lub in (B,≺B) of x ∈ A | x ≺B y

and setting

f(y) = the lub in (B′,≺B ′) of x ∈ A | x ≺B y.

The details are left to the reader. See Exercise 6.4.Now we prove the first part of Theorem 6.10. The commutative

diagram in Figure 6.2 summarizes a big part of our plan. First, wewill define (D,≺D) and prove it is a dense linear ordering withoutendpoints that has the least upper bound property. Then, we willfind an isomorphic copy (C,≺C ) of (A,≺A) sitting densely insideof (D,≺D). Finally, we will find (B,≺B) sitting above (A,≺A)the way that (D,≺D) sits above (C,≺C ). What exactly the linearorderings and isomorphisms in the diagram are will be explainedsoon.

The proof uses the following key definition. We say that a setL is a left-cut iff

• L ⊆ A,• L = ∅,• L = A,• for every x ∈ A and y ∈ L, if x ≺A y, then x ∈ L, and• for every x ∈ L, there exists y ∈ L such that x ≺A y.

The next to last condition says that left-cuts are closed to the left.The last condition says that left-cuts do not have a right endpoint.

(B,≺

B)

is

omor

phis

mg

(D,≺

D)

(A,≺

A)

coun

tabl

ede

nse

subo

rder

ing

isom

orph

ism

x→

Lx

(C,≺

C)coun

tabl

ede

nse

subo

rder

ing

Fig

ure

6.2

Pla

nfo

rth

epr

oofof

The

orem

6.10


Before continuing, we give a couple of examples of left-cuts inthe most interesting case where (A,≺A) = (Q, <Q). Both

x ∈ Q | x < 2

and

x ∈ Q | x < 0 or x2 < 2

are left-cuts of (Q, <Q). The first example has a least upper boundin Q, namely

2 = lub(x ∈ Q | x < 2).

But the second example does not have a least upper bound in Qbecause

√2 ∈ Q. It is worth keeping these examples in mind in

what follows.Returning to the proof of the theorem, define

D = L ⊆ A | L is a left-cut

and

L ≺D M ⇐⇒ L M.

We claim that (D,≺D) is a strict linear ordering. It is obviouslytransitive and irreflexive, so all that remains is to see that it istotal. Suppose that L and M are left-cuts. We must show thatM ⊆ L or L M . For this, we assume M ⊆ L and prove L M .Pick y ∈ M such that y ∈ L. Consider an arbitrary x ∈ L. Since(A,≺A) is a strict linear ordering, either x A y or y ≺A x. Ifx A y, then x ∈ M since y ∈ M and M is closed to the left. Ify ≺A x, then y ∈ L since x ∈ L and L is closed to the left, but thisis a contradiction since y ∈ L. Therefore x ∈ M . Since x was anarbitrary element of L, we have seen that L ⊆M . Clearly, L Msince y ∈ M − L. This shows that (D,≺D) is total.

Now we verify that (D,≺D) has no endpoints. Let M ∈ D. SinceM is a left-cut, M = ∅ and M = A. Pick y ∈ M and z ∈ A−M .Let L = x ∈ A | x ≺A y and N = x ∈ A | x ≺A z. Routinechecking shows that L and N are left-cuts and L M ⊆ N .Since (A,≺A) has no right endpoint, there exists z′ ∈ A such thatz ≺A z′. Let N ′ = x ∈ A | x ≺A z′. It is easy to see thatL M N ′. Therefore, M is not an endpoint of (D,≺D).

Before proving that (D,≺D) has the least upper bound prop-erty, we pause again to give a motivating example in the case


(A,≺A) = (Q, <Q). Let

S =x ∈ Q | x < y | y ∈ Q and y2 < 2

.

Then S is a non-empty family of left-cuts of (Q, <Q). Also, S isbounded in the sense that every element of S is contained in theleft-cut z ∈ Q | z < 3/2. Finally, notice that⋃

S = x ∈ Q | x < 0 and x2 < 2.

This example gives a big hint for what is going on.Back to the proof of the theorem, we are ready to show that

(D,≺D) has the least upper bound property. Suppose that S isa non-empty subset of D and S has an upper bound in (D,≺D).We must prove that S has a least upper bound in (D,≺D). LetM =

⋃S. It suffices to show that M ∈ D and M = lub(S).

Leaving some of the details to the reader, here are the main ideasfor showing that M is a left-cut and an upper bound for S.

• If L ∈ S, then L ⊆ A. Thus M ⊆ A.• S = ∅ and, if L ∈ S, then L = ∅. Thus M = ∅.• If L ∈ S, then L has no right endpoint. Therefore M has no

right endpoint.• If L ∈ S, then L is closed to the left. Therefore M is closed to

the left.• If N is an upper bound for S, then L ⊆ N for every L ∈ S, so

M ⊆ N . Because N = A, also M = A.

Now we explain why M is the least upper bound of S. Considerany upper bound N for S. We must show that M ⊆ N . Supposeotherwise. Then N M since both are left-cuts. Pick y ∈ M−N .Since M =

⋃S, there exists L ∈ S such that y ∈ L. Then L ⊆ N

because N is an upper bound for S. Thus y ∈ N , which is acontradiction.

Technically, (A,≺A) is not a subordering of (D,≺D) as we can-not expect A ⊆ D. However, we can find a subordering (C,≺C )of (D,≺D) such that

(A,≺A) (C,≺C ).

To see this, for each y ∈ A, let

Ly = x ∈ A | x ≺A y.


Then letC = Ly | y ∈ A.

It is easy to see that the map

y → Ly

is an order-preserving injection from (A,≺A) to (D,≺D) whoserange is C.

We must prove that C is dense in (D,≺D). Suppose that K ≺D

M . Recall this means that K M . Pick y ∈ M −K. Since M is aleft-cut, it does not have a largest element, so we may pick z ∈Msuch that y ≺A z. Clearly,

K D Ly ≺D Lz ≺D M.

Observe that Lz is strictly between K and M . This shows that Cis dense in (D,≺D).

Towards defining (B,≺B) as in the statement of Theorem 6.10,first define an injection g with domain D as follows.

• If x ∈ A, then g(Lx) = x.• If M ∈ D − C, then g(M) = (A, M).

In the first case, notice that if Lx = Ly , then x = y, so g reallyis a function. In the second case, one point is that (A, M) ∈ Abecause otherwise

A ∈ A ∈ A, A, M = (A, M) ∈ A,

which contradicts the Foundation Axiom. The other point is that

(A, M) = (A, N) ⇐⇒ M = N.

The two points combined are used to see that g really is an injec-tion. Finally, define

B = g[D]

and define ≺B by

g(L) ≺B g(M) ⇐⇒ L ≺D M.

Clearly,g : (D,≺D) (B,≺B)

andg C : (C,≺C ) (A,≺A),


which completes the proof of Theorem 6.10.

To describe the relationship between (A,≺A) and (B,≺B) inTheorem 6.10, we say that (B,≺B) is a Dedekind completion of(A,≺A). We may define R to be a Dedekind completion of Q. Bythe moreover part of Theorem 6.10, it is not particularly importantwhich Dedekind completion we choose.

Throughout this chapter, we have been thinking about Q andR as certain kinds of linear orderings. But there is much moreto numbers than how they are ordered! The rational numbersalso come equipped with an arithmetic structure, by which wemean addition, subtraction, multiplication, division, exponentia-tion, etc. Moreover, the arithmetic structure of the rational num-bers lifts nicely to the familiar arithmetic structure for the realnumbers. It is possible to explain how this lifting is achieved interms of left-cuts and Dedekind completions. We already saw ahint of this in our discussion of

√2 earlier in the chapter. Curious

readers might enjoy working out the formal development as anindependent project.

Exercises

Exercise 6.1 The point of this exercise is to define Q. Therefore,you may not assume anything about Q, not even that Q exists,in your solution. However, you may use standard properties of ωwith its order, addition and multiplication. For example, writing32 = 6

4 is unacceptable at this point because we have not yet definedfractions but writing 3 ·4 = 12 = 2 ·6 is acceptable because it onlyrefers to natural numbers and multiplication.

Let

S = ω × (ω − 1).

Here ω− 1 = ω−0 = 1, 2, 3, . . . . Define a relation E on S by

(a, b)E (c, d) ⇐⇒ ad = bc.


1. Prove that E is an equivalence relation on S.2. Write [(a, b)]E for the E-equivalence class of (a, b). That is,

[(a, b)]E =(a′, b′) | (a′, b′)E(a, b)

.

Let B = S/E. That is

B = [(a, b)]E | (a, b) ∈ S .

Explain why the formula

[(a, b)]E ≺B [(c, d)]E ⇐⇒ ad < bc

defines a relation ≺B on B. Hint: If you think there is nothingto check, think again.

3. Prove that (B,≺B) is a countable dense linear ordering withno right endpoint and whose left endpoint is [(0, 1)]E .

4. LetA = B − [(0, 1)]E

and ≺A be the restriction of ≺B to A. Prove that (A,≺A) is acountable dense linear ordering without endpoints.

5. Prove that (A,#A) is a countable dense linear ordering withoutendpoints. (Notice the order is reversed.)

6. Define(Q, <Q) = (A,#A)(B,≺B).

Prove that (Q, <Q) is a countable dense linear ordering withoutendpoints.

Exercise 6.2 Let

Z = . . . ,−2,−1, 0, 1, 2, . . . have the usual order on the integers. Prove that Z ω.

Exercise 6.3 Find a family F such that

• every element of F is a countable dense linear ordering, and• for every countable dense linear ordering (A,≺A), there exists

a unique (B,≺B) ∈ F such that (A,≺A) (B,≺B).

Exercise 6.4 Complete the proof of the moreover part of The-orem 6.10. See the hint given there.

Exercise 6.5 Prove that RR R.


Exercise 6.6 If

B = (B,∨B,∧B,¬B,⊥B,B)

is a Boolean algebra, then we call B countable iff B is countablyinfinite. Prove that if A and B are countable Boolean algebraswith no atoms, then A and B are isomorphic. Hint: The solutionis quite involved and breaks up into two main components:

• Use a back-and-forth style argument to build an order isomor-phism

f : (A, A) (B,B).

• Prove that every order isomorphism between Boolean algebrasis also a Boolean algebra isomorphism. Hence,

f : A B.

This second part is a general fact about Boolean algebras, soyour proof should not use the countable and atomless assump-tions.

Here are a couple of useful lemmas you should prove for the back-and-forth part of the argument:

• If f is a finite partial Boolean algebra isomorphism from A to B,then there exists a finite partial Boolean algebra isomorphismg from A to B such that the domain and range of g are finiteBoolean algebras. To see this, take dom(g) to be the Booleansubalgebra of A generated by dom(f), and ran(g) to be theBoolean subalgebra of B generated by ran(f). Then extend fin the obvious way to define g.

• The ordering of an atomless Boolean algebra is dense in thesense that if x ≺ z, then there exists y such that x ≺ y ≺ z.Keep in mind that the ordering is not linear!

We remark that two examples of a countable atomless Booleanalgebras were given in Exercises 4.17 and 5.22. By this exercise,they are isomorphic.

7

Filters and ideals

Filters and ideals come up in just about every area of modernmathematics. After some preliminaries, in Section 7.1, we proveTarski’s ultrafilter existence theorem and touch on the theory ofultraproducts. Filters and ideals are particularly important in ad-vanced set theory, where the filter generated by the closed un-bounded subsets of an uncountable regular cardinal plays a majorrole. We give the reader a taste of this sort of infinitary combina-torics in Section 7.2. The main results there are Fodor’s theoremand an interesting special case of Solovay’s splitting theorem.

7.1 Motivation and definitions

There are many mathematical contexts in which we are given aset X and we talk about large subsets of X and small subsets ofX. This is so common that it is worth writing down what thesesituations share.

Definition 7.1 Let X be a non-empty set and F ⊆ P(X). ThenF is a filter over X iff the following conditions hold:

• ∅ ∈ F .• X ∈ F .• For all A, B ⊆ X, if A ∈ F and A ⊆ B, then B ∈ F .• For all A, B ⊆ X, if A, B ∈ F , then A ∩B ∈ F .

To understand the motivation for filters, one can paraphrase thedefining conditions as follows.

142 Filters and ideals

• The empty set is not large.• X is large.• If A is large and A ⊆ B, then B is also large.• If A and B are large, then so is A ∩B.

If you are not convinced by the last clause, replace large byalmost everything to make it even more believable.

Ideals are to small sets what filters are to large sets.

Definition 7.2 Let X be a non-empty set and I ⊆ P(X). ThenI is an ideal over X iff the following conditions hold:

• ∅ ∈ I.• X ∈ I.• For all A, B ⊆ X, if B ∈ I and A ⊆ B, then A ∈ I.• For all A, B ⊆ X, if A, B ∈ I, then A ∪B ∈ I.

In reading the list above, where you see a set belongs to I, youcan say out loud that it is small (or, maybe better, almost nothing)to understand the motivation for the condition.

The next two results explain how filters and ideals are related.

Lemma 7.3 If F is a filter over X, then

X −A | A ∈ F

is an ideal over X.

Lemma 7.4 If I is an ideal over X, then

X −A | A ∈ I

is a filter over X.

So the operation A → X −A takes filters over X to ideals overX and vice-versa as described by the lemmas. This is an exampleof what we call duality in mathematics.

Is every subset A of X either large or small? It depends on thesituation. This thought leads to ultrafilters and prime ideals.

Definition 7.5 Let F be a filter over X. Then F is an ultrafilterover X iff for every A ⊆ X, either A ∈ F or X −A ∈ F .

Definition 7.6 Let I be an ideal over X. Then I is a primeideal over X iff for every A ⊆ X, either A ∈ I or X −A ∈ I.

7.1 Motivation and definitions 143

It is about time we introduced some examples!

Example If p ∈ X, then A ⊆ X | p ∈ A is a principal ultrafilterover X. This is the least interesting kind of ultrafilter.

Example A ⊆ ω | ω −A is finite is the Frechet filter over ω. Itis not an ultrafilter over ω. For example, neither

Even = 2n | n < ωnor

Odd = 2n + 1 | n < ωare members of the Frechet filter over ω.

Example For A ⊆ ω, define

density(A) = limn→∞

|A ∩ n |n

if the limit exists. Then

A ⊆ ω | density(A) = 0is the density ideal over ω.

Example Let I = x ∈ R | 0 ≤ x ≤ 1. For the reader who knowsabout Lebesgue measure, we mention that an important topic inanalysis and probability is the ideal of null sets,

A ⊆ I | A has Lebesgue measure 0,and its dual filter,

A ⊆ I | A has Lebesgue measure 1.This is not an ultrafilter because there are subsets of I whoseLebesgue measure is strictly between 0 and 1. For example, theinterval

x ∈ R | 0 ≤ x ≤ 1/2has Lebesgue measure 1/2.

Exercise 7.1 Let P and X be non-empty sets with P ⊆ X, and

F = A ⊆ X | P ⊆ A.1. Prove that F is a filter over X.2. Prove that the following are equivalent:


(a) F is an ultrafilter over X.(b) P is a singleton.(c) F is a principal ultrafilter over X.

Exercise 7.2 Let F be the Frechet filter over ω. Suppose thatG is an ultrafilter over ω such that G ⊇ F . Prove that G is notprincipal.

Exercise 7.3 Let X be a non-empty set and F be an ultrafilterover X. Prove that, for every n < ω and sequence 〈Ai | i < n〉 ofsubsets of X, if ⋃

i<n

Ai ∈ F ,

then there exists i < n such that Ai ∈ F .

The first important result on this topic is that every filter ex-tends to an ultrafilter. You should notice that the first sentenceof the proof uses the Axiom of Choice to know that P(X) has acardinality.

Theorem 7.7 (Tarski) Let F be a filter over X. Then thereexists an ultrafilter G over X such that F ⊆ G.

Proof Let κ = |P(X)| and 〈Aα | α < κ〉 be a sequence such that

P(X) = Aα | α < κ.

We will define a sequence 〈Gα | α < κ〉 by recursion. After eachcase in our definition of Gβ , we will verify that if α < β, then:

• Gβ is a filter,• Aα ∈ Gβ or X −Aα ∈ Gβ , and• Gα ⊆ Gβ .

Base case β = 0.

Define G0 = F .

Successor case β = α + 1.

We break up this case into three subcases. The first subcase is,for every B ∈ Gα,

B ∩Aα = ∅.


In the first subcase, let

Gα+1 = C ⊆ X | there exists B ∈ Gα such that B ∩Aα ⊆ C.Obvserve that Gα ∪ Aα ⊆ Gα+1 because:

• Gα ⊆ Gα+1 since, if B ∈ Gα, then B ∩ Aα ⊆ B, so B ∈ Gα+1,and

• Aα ∈ Gα+1 since X ∈ Gα, so X ∩Aα ∈ Gα+1.

Observe that Gα+1 is a filter because:

• ∅ ∈ Gα+1 by the first subcase hypothesis,• X ∈ Gα since X ∩Aα ⊆ X,• Gα+1 is clearly closed upward under ⊆, and• Gα+1 is closed under pairwise intersections since if B, C ∈ Gα,

then

(B ∩Aα) ∩ (C ∩Aα) = (B ∩ C) ∩Aα ∈ Gα+1.

The second subcase is that the first subcase fails and, for everyB ∈ Gα,

B ∩ (X −Aα) = ∅.In the second subcase, let

Gα+1 = C ⊆ X | there exists B ∈ Gα such that B∩(X−Aα) ⊆ C.Much like in the first subcase, one shows that

Gα ∪ X −Aα ⊆ Gα+1

and Gα+1 is a filter. The reader should complete the verification.The third subcase is that the first and second subcases fail. We

will show this does not happen. For contradiction, suppose

B, C ∈ Gα,

B ∩Aα = ∅and

C ∩ (X −Aα) = ∅.Let D = B ∩ C. Then

D ∈ Gα,

D ∩Aα = ∅


andD ∩ (X −Aα) = ∅.

Therefore D = ∅. But ∅ ∈ Gα since Gα is a filter.

Limit case β is a limit ordinal.

DefineGβ =

⋃α<β

Gα.

It is easy to check that Gβ is a filter because it is the union of a⊆-increasing sequence of filters. Also, if α < β, then either

Aα ∈ Gα+1 ⊆ Gβ

or(X −Aα) ∈ Gα+1 ⊆ Gβ .

That completes the recursive definition of the sequence

〈Gα | α < κ〉and the verification that it has the desired properties. Now let

G =⋃α<κ

Gα.

As in the limit case, we see that G is a filter. Clearly, F = G0 ⊆ G.Finally, G is an ultrafilter because, for every α < κ, either

Aα ∈ Gα+1 ⊆ Gor

(X −Aα) ∈ Gα+1 ⊆ G.

Exercise 7.4 Consider the special case of Theorem 7.7 in whichX = ω and F is the Frechet filter over ω. Notice that in the proof

κ = |P(ω)| = 2ℵ0 .

Let 〈Gα | α < 2ℵ0 〉 be the sequence of filters extending F that wasrecursively constructed in the proof of Theorem 7.7.

1. Prove by induction on α < 2ℵ0 that |Gα| < 2ℵ0 and Gα is notan ultrafilter over ω.


2. Use ideas similar to the proof of Theorem 7.7 to show that thereare 22ℵ0 many non-principal ultrafilters over ω. Hint: This is aslightly challenging exercise. View 〈Gα | α < 2ℵ0 〉 as a branchthrough a certain kind of tree with 2ℵ0 many levels. Arguethat the tree has 22ℵ0 many distinct branches each of whichcorresponds to a different ultrafilter over ω.

Exercise 7.5 below introduces the reader to a certain construc-tion, which is known as taking an ultrapower by an ultrafilter.Intuitively, the idea is to start with a sequence of structures (inthe exercise, the structures are linear orderings) and an ultrafilter,F , and to form a new structure by averaging out according to F .Our meaning will become clear when the reader does the exerciseand reads the discussion that follows. Before starting, recall howproducts of sets were defined in Exercise 4.10. In particular, givena sequence 〈An | n < ω〉, we define

∏n<ω An to be the set of

functions f such that dom(f) = ω and, for every n < ω,

f(n) ∈ An.

In the special case where all the Ans are the same, say An = B,we end up with∏

n<ωAn =∏

n<ωB = f | f is a function from ω to B = ωB.

Immediately after Exercise 7.5, there is a long discussion of thesignificance of the ultrapower construction.

Exercise 7.5 Let 〈An | n < ω〉 be a sequence and

P =∏

n<ωAn.

Let F be an ultrafilter over ω.

1. Define a relation ∼ on P by

f ∼ g ⇐⇒ n < ω | f(n) = g(n) ∈ F .

Prove that ∼ is an equivalence relation on P .2. For f ∈ P , let

[f ] = g ∈ P | f ∼ g .

Also, let

A = [f ] | f ∈ P .


Assume that, for each n < ω, we are given a relation

Rn ⊆ An ×An.

Prove that we may define a relation

R ⊆ A×A

by setting

[f ] R [g] ⇐⇒ n < ω | f(n) Rn g(n) ∈ F .

In other words, prove that the definition of R does not dependon the choice of representatives for the equivalence classes [f ]and [g].

3. Assume that (An, Rn) is a strict linear ordering for every n < ω.Prove that (A, R) is also a strict linear ordering.

4. Now assume that, for every n < ω,

(An, Rn) = (ω, <)

where < is the usual order on the natural numbers. Supposethat F is a non-principal ultrafilter over ω. Prove that (A, R)is not a wellordering. Hint: Notice that, in this case, P = ωω.For c ∈ Z, consider the function

fc : ω → ω

defined by

fc(n) =

n + c if n + c ≥ 00 otherwise.

Prove that

· · · R [f−3] R [f−2] R [f−1] R [f0].

5. Again assume that, for every n < ω,

(An, Rn) = (ω, <).

But suppose instead that F is a principal ultrafilter over ω.Prove that

(A, R) (ω, <).

Hint: Most but not all of the details are contained in the dis-cussion after this exercise, so read it first if you get stuck.


That completes the instructions for Exercise 7.5 but there ismuch more we should tell the reader about the construction donethere. The pair (A, R) is called the ultraproduct of the sequence〈(An, Rn) | n < ω〉 by F . A popular way to express the definitionof R is

[f ] R [g] ⇐⇒ f(n) Rn g(n) for F-almost every n < ω.

Often, it helps to think about ultraproducts using this alternativelanguage. This language makes it clearer what we meant by aver-aging out in the paragraph preceding Exericse 7.5. In this exercise,we took ultraproducts of linear orderings but it is also possible totake ultraproducts of other kinds of structures. For example, thereader should see how the ultraproduct of a sequence of Booleanalgebras would be defined.

The example of an ultraproduct (A, R) given in the last twoparts of Exercise 7.5 is called the ultrapower of (ω, <) by F . Insteadof saying ultrapower, we use the term ultraproduct in this casebecause all of the pairs (An, Rn) are the same. The main pointof part 4 of Exercise 7.5 is that the ultraproduct of wellorderingsneed not be a wellordering. Figure 7.1 is a rough picture of what(A, R) looks like in this case. The initial segment of (A, R) that isisomorphic to ω is really:

[n → 0] R [n → 1] R [n → 2] R · · ·

Keep in mind that, for a fixed c < ω, the function n → c is thefunction with the constant value c. The chain of relations that wejust displayed says that, for every constant c < ω,

[n → c] R [n → c + 1].

This is because

n < ω | c < c + 1 = ω ∈ F .

But we are also claiming that there is no equivalence class strictlybetween each [n → c] and [n → c + 1]. To see this, let b = c + 1and suppose f : ω → ω is a function such that

[f ] R [n → b].

Then

n < ω | f(n) < b ∈ F .


(isomorphic to ω)

...

(isomorphic to Z)

...

(isomorphic to Z)

...

Figure 7.1 An ultrapower of (ω, <) by a non-principal ultrafilter

Clearly,

n < ω | f(n) < b =⋃a<b

n < ω | f(n) = a.

By Exercise 7.3, there exists a < b such that

n < ω | f(n) = a ∈ F .

By definition, this just says that

[f ] = [n → a].

Finally, observe that a ≤ c.Looking again at Figure 7.1, after the initial segment of (A, R)

that is isomorphic to ω, there are infinitely many pieces each ofwhich is isomorphic to Z. We call these Z-chains. In terms of thefunctions fc defined in part 4 of Exercise 7.5, here is one example


of a Z-chain:

· · · R [f−2] R [f−1] R [f0] R [f1] R [f2] R · · · .

Using the ideas from the previous paragraph, the reader shouldprove that [fc] R [fc+1] and there are no equivalence classes offunctions strictly between [fc] and [fc+1]. In other words, thatthis really is an example of a Z-chain.

Now define

hc(n) =

3n + c if 3n + c ≥ 00 otherwise.

Here is a second example of a Z-chain:

· · · R [h−2] R [h−1] R [h0] R [h1] R [h2] R · · ·

The reader should verify that this is indeed a Z-chain. We claimthat this Z-chain lies entirely after our first example of a Z-chain.In other words, for all a, c ∈ Z,

[fa] R [hc].

By the definition of R, this just says that

n < ω | fa(n) < hc(n) ∈ F .

The main observation needed to see why this is true is that

n < ω | n + a < 3n + c = n < ω | (a− c)/2 < n= ω − n < ω | n ≤ (a− c)/2∈ F

because it is the complement of a finite set and F is non-principal.We give a third example of a Z-chain. For b ∈ Z, define

gb(n) =

2n + b if 2n + b ≥ 00 otherwise.

Our third Z-chain

· · · R [g−2] R [g−1] R [g0] R [g1] R [g2] R · · ·

lies strictly between the other two as the reader should verify.Building on these observations, one sees that the Z-chains them-

selves form a dense linear ordering without endpoints. We leave it


to the reader to fill in the details and continue this investigationas an extremely worthwhile project.

Ultraproducts are used in many branches of mathematics, notjust set theory. A famous and intriguing example is AbrahamRobinson’s theory of non-standard analysis, which rehabilitatedGottfried Wilhelm Leibniz’s seventeenth-century infinitesimal cal-culus. Infinitesimals are supposed to be numbers ε > 0 such thatε < x for every positive number x ∈ R. Of course, there are no suchε ∈ R. Nevertheless, without really knowing what he meant by in-finitesimals, Leibniz developed recipes for working with them thatyielded correct answers to questions about geometry and physics.While this represented tremendous intuition, Leibniz’s theory wasconsidered controversial and was eventually abandoned in favor ofthe rigorous development of calculus provided by Augustin-LouisCauchy in the eighteenth-century. Much later, in the 1960s, Robin-son vindicated Leibniz by saying what infinitesimals really are andexplaining why it was legitimate to derive calculus formulas usingthem. To get the idea, let F be a non-principal ultrafilter overω and take the ultrapower of (R, <) by F . Call this ultrapower(R∗, <∗). Pretty much like in our elaboration on Exercise 7.5, wesee that, for all constants a < b in R,

[n → a] <∗ [n → b].

So there is a copy of (R, <) sitting inside of (R∗, <∗). But there arenew points to the left, to the right and in between. For example,for every c ∈ R,

[n → c] <∗ [n → n],

which shows there are new positive infinite members of R∗ entirelyto the right of our copy of R. Similarly, for every c ∈ R,

[n → −n] <∗ [n → c],

so there are new negative infinite members of R∗ entirely to theleft of our copy of R. Even more interesting is the fact that, forevery positive c ∈ R,

[n → 0] <∗ [n → 1/n] <∗ [n → c].

In other words, [n → 1/n] is greater than our copy of 0 and lessthan our copy of c for every positive real number c. For this reason,it is reasonable to say that the equivalence class [n → 1/n] is an

7.2 Club and stationary sets 153

example of an infinitesimal member of R∗. (Technically, when wewrite n → 1/n here, we really mean

n →

1/n if n = 00 if n = 0

because the domain must be ω but we cannot divide by zero.)There is more to understanding why one can reason about (R∗, <∗)and come to certain correct conclusions about (R, <) but thatrequires a basic background in mathematical logic, which we donot presume. Our intent was merely to introduce the reader tothis fascinating and historically significant subject.

7.2 Club and stationary sets

This section builds on the previous section and Chapters 3 and 4.Given a limit ordinal κ, there is a interesting and very useful filterover κ called the club filter. Its dual is called the non-stationaryideal. Before we can say what these are, we need some definitions.

Definition 7.8 Let κ be a limit ordinal and C be a set. Then:

• C is unbounded in κ iff sup(C ∩ κ) = κ.• C is closed in κ iff for every α < κ, if C ∩ α = ∅, then

sup(C ∩ α) ∈ C.

• C is club in κ iff C is closed and unbounded in κ.

Here are some examples with κ = ω1.

Example If α < ω1, then β < κ | α < β is club in ω1.

Example α < ω1 | α is a limit ordinal is club in ω1. It is closedbecause a limit of limit ordinals is also a limit ordinal. It is un-bounded because if α is a countable ordinal, then α+ω is a count-able limit ordinal.

Example α < ω1 | α is a successor ordinal is not club in ω1.While it is unbounded, it is not closed. For example,

ω = supn<ω

(n + 1)

but ω is not a successor ordinal.


Example α < ω1 | ωα = α is club in ω1. Here ωα is ordinalexponentiation. The proof is a little more than what you wereasked to show in Exercise 4.12.

Definition 7.8 applies to all limit ordinals but most often weapply it specifically to uncountable regular cardinals such as ω1.Remember that κ is regular iff cf(κ) = κ. And remember that allinfinite successor cardinals are regular.

Exercise 7.6 Let κ be a regular cardinal and C ⊆ κ. Prove Cis unbounded in κ iff |C| = κ.

Now we say what this has to do with filters.

Lemma 7.9 Assume that cf(κ) > ω. Let

F = A ⊆ κ | there exists a club C in κ such that C ⊆ A.Then F is a filter over κ.

Proof The only condition in the definition of filter that is notobvious is closure under intersections. It is enough to show that ifC and D are club in κ, then so is C ∩D.

First we show that C ∩D is closed in κ. Let β < κ and assumethat C ∩D ∩ β = ∅. Let

α = sup(C ∩D ∩ β).

We must show that α ∈ C∩D. Easily, we see that α = sup(C∩α)and α = sup(D∩α). Since C and D are closed, α ∈ C and α ∈ D.

To finish we show that C ∩ D is unbounded in κ. Let α < κ.We must find δ ∈ C ∩D such that α < δ. By recursion on n < ω,define ordinals βn, γn < κ as follows. Pick β0 ∈ C with β0 > α.Given βn, pick γn ∈ D with γn > βn. Given γn, pick βn+1 ∈ Cwith βn+1 > γn. We can do all this picking because C and D areunbounded in κ. Now let δ be the supremum of either sequence;it is the same because of the interleaving. That is,

δ = supn<ω

βn = supn<ω

γn.

Then δ < κ because cf(κ) > ω. Since C and D are closed,

δ = sup(C ∩ δ) ∈ C

andδ = sup(D ∩ δ) ∈ D,


hence δ ∈ C ∩D as desired.

The filter in Lemma 7.9 is called the club filter over κ. Themain point of the proof is that the intersection of two clubs isclub if κ has uncountable cofinality. This is not necessarily true ifκ has countable cofinality. For instance, Even = 2n | n < ω andOdd = 2n + 1 | n < ω are disjoint unbounded subsets of ω, andthese sets are closed for trivial reasons.

Exercise 7.7 Let κ be an uncountable regular cardinal and

f : κ → κ

be a function. Prove that

α < κ | f [α] ⊆ α

is club in κ. Remark: This implies that α < ω1 | ωα = α is clubin ω1 in the special case κ = ω1 and f : α → ωα. The specialcase was the subject of Exercise 4.12 and the general argument issimilar.

Exercise 7.8 Let κ be an uncountable regular cardinal. Provethat if θ < κ and 〈Cα | α < θ〉 is a sequence of club subsets of κ,then the set ⋂

Cα | α < θ

is club in κ. Hint: Use induction on θ < κ. The successor case,θ = η+1, is immediate from the induction hypothesis and the caseθ = 2, which was handled in the proof of Lemma 7.9. Suppose θ isa limit ordinal. In this case, the proof of closure is straightforward(similar to the case θ = 2). For the proof of unboundedness, givenβ0 < κ, define an increasing sequence 〈βη | η < θ〉 such that, forevery η < θ,

βη ∈⋂Cζ | ζ < η.

Exercise 7.9 Let κ be a regular cardinal. Give an example of asequence 〈Cα | α < κ〉 such that, for every α < κ, Cα is club in κbut ⋂

Cα | α < κ = ∅.


Let F be the club filter over κ. Recall that

F = S ⊆ κ | there exists a club C in κ such that C ⊆ S.As we indicated before, we intuitively think of members of F aslarge subsets of κ. Let I be the ideal dual to F . Then

I = S ⊆ κ | κ− S ∈ F.We think of members of I as small subsets of κ. Observe that ifS ⊆ κ, then

S ∈ I ⇐⇒ there is a club C in κ such that S ∩ C = ∅.We are also interested in subsets of κ that are not small. Noticethat if S ⊆ κ, then

S ∈ I ⇐⇒ for every club C in κ, S ∩ C = ∅.Intuitively, this says that a subset of κ is not small iff it meetsevery large subset of κ. We give such sets the following name.

Definition 7.10 Let κ be a limit ordinal and S ⊆ κ. Then S isstationary in κ iff for every club C in κ,

S ∩ C = ∅.Notice that if C is club, then C is stationary. This is because if

D is club, then C∩D is club, in particular, C∩D = ∅. Intuitively,this says that if a set is large, then it is not small.

You can get additional intuition for Definition 7.10 if you hap-pen to know about Lebesgue measure on the unit interval

x ∈ R | 0 ≤ x ≤ 1.The relevant analogies are:

contains a clubmeasure 1

=stationary

positive measure=

not stationarymeasure 0

The following exercise gives an important example with κ = ω2.

Exercise 7.10 Let

C = α < ω2 | α is a limit ordinal,

E = α ∈ C | cf(α) = ωand

C − E = α ∈ C | cf(α) = ω1.


1. Prove that C is club in ω2. (This is pretty obvious.)2. Prove that E is stationary in ω2.3. Prove that C − E is stationary in ω2.4. Use parts 1, 2 and 3 to prove that the club filter over ω2 is not

an ultrafilter.

Now we come to one of the most fundamental tools for studyingclub and stationary sets.

Theorem 7.11 (Fodor) Let κ be uncountable regular cardinaland f : κ → κ be a function such that

α < κ | f(α) < αis stationary in κ. Then there exists θ < κ such that

α < κ | f(α) = θis stationary in κ.

We will derive Theorem 7.11 from Lemma 7.12, which is inter-esting in its own right. Exercise 7.9 tells us that the intersectionof κ many club sets might not be club. Lemma 7.12 says that thediagonal intersection of κ many club sets is club.

Lemma 7.12 Let κ be an uncountable regular cardinal. Supposethat

〈Cα | α < κ〉is a sequence of club subsets of κ. Let

D = α < κ | α ∈ Cθ for every θ < α.Then D is club in κ.

We call D the diagonal intersection of 〈Cα | α < κ〉. Mostcommonly, you will see it written ∆α<κCα.

Proof of Lemma 7.12 First we show that D is closed in κ. Letγ < κ. Assume that D ∩ γ = ∅ and let β = sup(D ∩ γ). Wemust show that β ∈ D. For contradiction, suppose that β ∈ D. Itfollows easily that β is a limit ordinal and D ∩ β = D ∩ γ. Hence

β = sup(D ∩ β).

By the definition of D, there exists θ < β such that β ∈ Cθ . SinceCθ is closed, there are two cases:


1. Cθ ∩ β = ∅.2. Cθ ∩ β = ∅ and sup(Cθ ∩ β) ∈ Cθ .

In the second case, sup(Cθ ∩ β) < β because β ∈ Cθ . In eithercase, we may pick α ∈ D such that θ < α < β. In the secondcase, we can also make sure that sup(Cθ ∩ β) < α. Then, in bothcases, α ∈ D and θ < α but α ∈ Cθ . This directly contradicts thedefinition of D.

Now we show that D is unbounded in κ. By recursion on θ < κdefine βθ as follows. Let β0 = 1. Given βθ < κ, pick βθ+1 > βθ

such that

βθ+1 ∈⋂

η<βθ

Cη.

This is possible by Exercise 7.8. If θ is a limit ordinal, then let

βθ = supη<θ

βη.

That completes the recursive definition of 〈βθ | θ < κ〉. By induc-tion on θ < κ, one sees that, for every η < θ,

βη < βθ

and if ζ < βη , then

βθ ∈ Cζ.

Suppose that θ < κ is a limit ordinal. Then, for every ζ < βθ ,

βη | ζ + 1 < η < θ ⊆ Cζ

and

βθ = sup(βη | ζ + 1 < η < θ) = sup(Cζ ∩ βθ)

so

βθ ∈ Cζ

since Cζ is closed. We have seen that

βθ | θ < κ and θ is a limit ordinal ⊆ D.

The set on the left is unbounded in κ and hence so is D.

Proof of Theorem 7.11 Let

S = α < κ | f(α) < α.


Our assumption is that S is stationary in κ. For each θ < κ, let

Tθ = α ∈ S | f(α) = θ.

For contradiction, suppose that no Tθ is stationary in κ. For eachθ < κ, pick Cθ club in κ such that Tθ ∩ Cθ = ∅. Let

D = α < κ | α ∈ Cθ for every θ < α.

By Lemma 7.12, D is club in κ. Pick α ∈ D ∩ S. Then, for everyθ < α, f(α) = θ. In other words, f(α) ≥ α. But f(α) < α sinceα ∈ S.

In Exercise 7.10, we saw that the club filter over ω2 is not an ul-trafilter. The proof outlined there generalizes to regular cardinalsλ ≥ ω2. To see this, note that, for every regular cardinal κ < λ,

α < λ | cf(α) = κ

is stationary in λ. Moreover, these sets are disjoint for differentκ. But Exercise 7.10 does not generalize to λ = ω1 because ωis the only regular cardinal less than ω1. However, the followingtheorem implies that the club filter is not an ultrafilter over ω1. Itis a special case of a more powerful result known as the Solovaysplitting theorem.

Theorem 7.13 There is S ⊆ ω1 such that S and ω1 − S arestationary in ω1.

Proof Suppose Theorem 7.13 is false. For each positive α < ω1,pick a surjection

fα : ω → α.

This is obviously possible because α is countable. Intuitively, wewill reach a contradiction by using the club filter over ω1 to averageout the sequence 〈fα | α < ω1〉 and obtain a new surjection fromω onto ω1. This will be a contradiction because ω1 is uncountableby definition.

Claim 7.13.1 For every n < ω, there exists θ < ω1 such that

α < ω1 | fα(n) = θ

is stationary in ω1.


Proof Fix n < ω. Consider the function

g : α → fα(n).

Then g(α) < α whenever 0 < α < ω1. By Fodor’s theorem, thereexists θ < ω1 such that

α < ω1 | g(α) = θ


Claim 7.13.2 For every n < ω, there is at most one θ < ω1such that

α < ω1 | fα(n) = θ


Proof Fix n < ω. For θ < ω1, let

Sθ = α < ω1 | fα(n) = θ.

Clearly, if η < θ < ω1, then

Sη ∩ Sθ = ∅.

Suppose η < θ < ω1 and both Sη and Sθ are stationary. Thenω1 − Sη is also stationary because

Sθ ⊆ ω1 − Sη.

This means the statement of Theorem 7.13 holds with S = Sη . Butwe assumed that Theorem 7.13 is false, so we have a contradiction.

Claims 7.13.1 and 7.13.2 allow us to define a function g : ω → ω1by setting g(n) equal to the unique θ < ω1 such that

α < ω1 | fα(n) = θ


Claim 7.13.3 g is a surjection from ω to ω1.

Proof Let θ < ω1. For contradiction, suppose that θ ∈ ran(g).Then, for every n < ω,

α < ω1 | fα(n) = θ


is non-stationary in ω1. For each n < ω, pick a club Cn in ω1 suchthat

Cn ∩ α < ω1 | fα(n) = θ = ∅.In other words, for every n ∈ ω and every α ∈ Cn,

fα(n) = θ.

LetD =

⋂Cn | n < ω.

Then D is club in ω1 and, for every n < ω and every α ∈ D,

fα(n) = θ.

Since D is unbounded in ω1, there exists α ∈ D such that θ < α.Since fα is a surjection from ω to α, there exists n < ω such that

fα(n) = θ.

This is a contradiction.

Claim 7.13.3 contradicts the fact that ω1 is uncountable.

Exercise 7.11 Use Theorem 7.13 to prove that the club filterover ω1 is not an ultrafilter.

Solovay’s splitting theorem says that if κ is a regular uncount-able cardinal and S is a stationary in κ, then there is a sequence〈Sα | α < κ〉 of stationary subsets of S such that, for all α < β < κ,

Sα ∩ Sβ = ∅.In other words, S can be split into κ many disjoint stationarypieces. This is more powerful than Theorem 7.13, which says thatω1 can be split into two disjoint stationary pieces.

Appendix

Summary of exercises on Boolean algebra

Boolean algebras were defined just before Exercises 2.12. Thatexercise gave a characterization of finite Boolean algebras up toisomorphism, namely, they all look like P(S) for some finite setS. The proof involved looking at atoms.

An example of an infinite atomless Boolean algebra, P(ω)/Finite,was given in Exercise 2.13. By Exercise 4.15, P(ω)/Finite is un-countable. Another example of an uncountable atomless Booleanalgebra, the family of clopen subsets of the Baire space, was thesubject of Exercise 5.23.

The finite Boolean algebras of truth tables, Tn for n < ω, andthe infinite Boolean algebra T∞ where discussed in Exercise 4.17.We saw that T∞ is a countable atomless Boolean algebra in part 6of that exercise. Another example of a countable atomless Booleanalgebra was given in Exercise 5.22. This was the family of clopensubsets of the Cantor space. The fact that all countable atomlessBoolean algebras are isomorphic was the topic of Exercise 6.6.This is an important theorem whose proof uses a back-and-forthconstruction.

Selected further reading

There are several other undergraduate textbooks on set theory.For example, some of the material that we covered in our coursecan also be found in Enderton (1977) and Hrbacek and Jech(1999).1 These books are different from each other and from ours,which certainly benefits the reader.

Those who would like to go on to more advanced set theoryshould first learn basic mathematical logic. Again, there are manyoptions. To name two, Enderton (2001) is an excellent startingpoint, while Goldstern and Judah (1998) is a bit more advanced.

This course and an elementary background in logic prepare thereader for graduate level set theory. Two indispensable texts areKunen (1983) and Jech (2003). Our reader who enjoyed ordinaland cardinal arithmetic and infinitary combinatorics, especiallySections 4.3, 5.5, 5.6 and 7.2, and would like to learn the rela-tive consistency results of Godel and Cohen on the ContinuumHypothesis, which were mentioned in Section 4.2, will be particu-larly drawn to these wonderful classics.

The material covered in Sections 5.1 through 5.4 is part of abroad subject called descriptive set theory, which is a certain com-bination of set theory, analysis and logic. To continue in this direc-tion, the reader would want to know the fundamentals of analysis.At the advanced undergraduate level, two analysis textbooks toconsider are Rudin (1976) and Royden (1988). For descriptive settheory, two beginning graduate level texts are Kechris (1995) andMoschovakis (2009). These have very different emphases. Roughly,

1 Here and below, we cite only the most recent edition available.

Selected further reading 165

the former ties set theory to analysis more than logic, while forthe latter it is the other way around.

Yet another subject that is intertwined with set theory and logicis model theory. In Chapter 6, we saw examples of classification upto isomorphism. This idea is important throughout mathematicsbut especially in model theory. We also touched on ultraproductsin Section 7.1. This is a model-theoretic construction that hasapplications in many fields, particularly in set theory. A classicbeginning graduate model theory text is Chang and Keisler (1990).

Set theory is a vast topic of current mathematical research.The enormous Handbook of set theory, edited by Foreman andKanamori (2010), comes in three volumes with a total of twenty-four chapters by various authors. It suffices to give the reader anaccurate impression of the many directions the subject has takenin recent decades.

Bibliography

Chang, C. C., and Keisler, H. J. 1990. Model theory. Third edn. Studies inLogic and the Foundations of Mathematics, vol. 73. Amsterdam: North-Holland.

Enderton, Herbert B. 1977. Elements of set theory. New York: AcademicPress [Harcourt Brace Jovanovich Publishers].

Enderton, Herbert B. 2001. A mathematical introduction to logic. Second edn.Harcourt/Academic Press, Burlington, MA.

Foreman, Matthew, and Kanamori, Akihiro (eds). 2010. Handbook of set the-ory. New York: Springer-Verlag. In three volumes.

Goldstern, Martin, and Judah, Haim. 1998. The incompleteness phenomenon.Natick, MA: A. K. Peters Ltd. Reprint of the 1995 original.

Hrbacek, Karel, and Jech, Thomas. 1999. Introduction to set theory. Thirdedn. Monographs and Textbooks in Pure and Applied Mathematics, vol.220. New York: Marcel Dekker Inc.

Jech, Thomas. 2003. Set theory. Springer Monographs in Mathematics. Berlin:Springer-Verlag. The third millennium edition, revised and expanded.

Kechris, Alexander S. 1995. Classical descriptive set theory. Graduate Textsin Mathematics, vol. 156. New York: Springer-Verlag.

Kunen, Kenneth. 1983. Set theory. Studies in Logic and the Foundations ofMathematics, vol. 102. Amsterdam: North-Holland. Reprint of the 1980original.

Moschovakis, Yiannis N. 2009. Descriptive set theory. Second edn. Mathe-matical Surveys and Monographs, vol. 155. Providence, RI: AmericanMathematical Society.

Royden, H. L. 1988. Real analysis. Third edn. New York: Macmillan.Rudin, Walter. 1976. Principles of mathematical analysis. Third edn. New

York: McGraw-Hill. International Series in Pure and Applied Mathe-matics.

Index

≺-least element, 24

aleph (ℵ), 58antichain, 124Aronszajn tree, 120atom, 18–20, 78, 104, 139

back-and-forth construction, 128, 129,140, 163

Baire category theorem, 94Baire space, 84bijection, 4Boolean algebra, 18–20, 77, 78, 104,

139, 163

Cantor normal form, 51Cantor perfect set theorem, 95, 113, 115Cantor space, 92Cantor theorem, 55Cantor–Bendixon derivative, 95Cantor–Bernstein–Schroeder theorem,

58, 76cardinal arithmetic

addition, 61cofinality, 68exponentiation, 64multiplication, 61

Cartesian product, 2, 11, 15Cauchy, 152Cauchy sequence, 93chain in a partial ordering, 77characteristic function, 65choice function, 14clopen, 88, 103closed set of ordinals, 153closure, 92club filter, 154club set, 153cofinality, 68Cohen, 66, 164collection, 7coloring, 115

compact topological space, 92, 102complete metric space, 94concatenation of wellorderings, 41Continuum Hypothesis (CH), 66, 164convergent sequence, 93countable, 55countably infinite, 55

Dedekind completion, 138DeMorgan laws, 16density ideal, 143determined game, 105diagonal argument, 56, 72, 96, 110diagonal intersection, 157domain, 3duality, 142

equivalence relation, 5, 17, 20, 77, 120,139

family, 7filter, 141Fodor’s theorem, 157Frechet filter, 143, 146function, 3

Gale–Stewart theorems, 108, 110Godel, 66, 164graph of a function, 4

Hilbert, 66homogeneous set, 115

ideal, 142image, 3indexed family, 3induction, viii, 25, 34inductive set, 11, 15inherited topology, 92injection, 4inverse function, 4irreflexive relation, 23isomorphism

of Boolean algebras, 19, 139

168 Index

of relations, 37, 126

Kleene–Brouwer ordering, 103Konig infinity lemma, 97Konig lemma, 72

language of set theory, 7Lebesgue measure, 143, 156left-most branch of a tree, 102Leibniz, 152limit cardinal, 59limit ordinal, 34Lindelof space, 92linear ordering, 24

map, 3maximum, 34metric compatible with a topology, 84,

89metric space, 83Mostowski collapse, 37, 40

non-standard analysis, 152

one-to-one correspondence, 4one-to-one function, 4onto function, 4operation, 3ordered pair, 2ordinal, 30ordinal arithmetic

addition, 42exponentiation, 48multiplication, 46

partial function, 26partial ordering, 77partition, 5perfect set game, 113perfect tree, 94pigeonhole principle, 118prime ideal, 142principal ultrafilter, 143

Ramsey theorem, 117range, 3recursion, viii, 26, 35regular cardinal, 68relation, 2restriction of a function, 4Robinson, 152Russell paradox, 16

separable topological space, 93singular cardinal, 68Solovay splitting theorem, 159, 161stationary set, 156strategy, 105strict linear ordering, 24subset, 7successor cardinal, 59successor ordinal, 34

supremum, 34surjection, 4symmetric difference, 16Tarski ultrafilter existence theorem, 144topological space, 82total relation, 23transitive closure, 17transitive relation, 23transitive set, 15, 28tree on Ω, 90truth table, 78type of a wellordering, 40ultrafilter, 142ultrapower, 149ultraproduct, 149unbounded set of ordinals, 153uncountable, 55V hierarchy, 15, 30, 36, 37, 49wellfoundedness, 24wellordering, 24ZF, 54, 77ZFC

Axiom of Choice, 14, 16, 53, 54, 57,76, 77, 110, 144

Comprehension Scheme, 10Empty Set Axiom, 7Extensionality Axiom, 8Foundation Axiom, 13, 16, 30, 36, 37,

39, 137Infinity Axiom, 11Pairing Axiom, 8Power Set Axiom, 10Replacement Scheme, 12, 54Union Axiom, 9

Zorn lemma, 77

A Course on Set Theory

Documents

Transcript of A Course on Set Theory