MAGIC SET THEORY LECTURE NOTES (FALL 2014)bfe12ncu/MAGIC-set... · MAGIC SET THEORY LECTURE NOTES (FALL 2014) DAVID ASPERO´ Contents 1. Introduction 2 1.1. Some elementary facts

MAGIC SET THEORY LECTURE NOTES(FALL 2014)

DAVID ASPERO

Contents

1. Introduction 21.1. Some elementary facts about sets 42. The axiomatic method: A crash course in first order logic 63. Axiomatic set theory: ZFC 103.1. The axioms 113.2. ZFC vs PA 173.3. The consistency question 194. Ordinals 235. Cardinals 285.1. The Cantor–Bernstein–Schroder Theorem 315.2. Countable and uncountable sets 325.3. Almost disjoint families 345.4. Cantor’s Continuum Hypothesis 355.5. AC vs. the Well–Ordering Principle 356. Foundation, recursion and induction. The cumulative

hierarchy 377. Inner models and relativization 427.1. Our first relative consistency proof: Con(ZF \{Foundation})

implies Con(ZF) 457.2. Reflection 468. Trees 488.1. Suslin’s problem 519. Infinite Ramsey theory 5210. The measure problem 5511. Ultraproducts 5911.1. Measurable cardinals and elementary embeddings 6312. Beyond measurable cardinals 6412.1. Reinhardt cardinals 65

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2 D. ASPERO

References 66

1. Introduction

Set theory plays a dual role. It provides a foundation for mathema-tics and it is itself a branch of mathematics with applications to otherareas of mathematics.

Reducing everything to sets: Set theory was developed / dis-covered / instigated by Georg Cantor in the second half of the 19thcentury, as a result of his investigations of trigonometric series ratherthan out of foundational considerations. However, set theory wouldsoon become the prevalent foundation of mathematics. In fact, it wasborn at a time when mathematicians saw the need to define things care-fully (i.e., define the object of their study in a mathematical languagereferring to reasonably ‘simple’ and well–understood entities) and settheory provided the means to do exactly that.

Example: What is a di↵erentiable function? What is a continuousfunction? What is a function?

A case example: A relation is a set of ordered pairs (a, b). And afunction f is a functional relation (i.e., (a, b), (a, b0) 2 f implies b = b0).

What is an ordered pair (a, b)? Well, given a, b, we can define

(a, b) = {{a}, {a, b}}(this definition is due to Kuratowski).

Fact 1.1. Given any ordered pairs (a, b), (a0, b0), (a, b) = (a0, b0) if andonly if a = a0 and b = b0.

Exercise 1.1. Prove Fact 1.1.

Similarly, for given n, we can define the n–tuple (a0, . . . , an, an+1) =((a0, . . . , an), an+1).

So we can successfully define the notion of function from the notionof set (and the membership relation 2, of course). And the notion ofset is presumably easier to grasp than the notion of function.

What about natural numbers, integers, rational, reals and so on?We can define 0 = ; (the empty set, the unique set with no elements).

The set ; has 0 members.We can define 1 = {0} = {;}. The set {;} has 1 member.We can define 2 = {0, 1} = {;, {;}}. The set {;, {;}} has 2 mem-

bers.

MAGIC Set Theory lecture notes (Fall 2014) 3

In general, we can define n+1 = n[{n}. With this definition n is aset with exactly n many members, namely all natural numbers m suchthat m < n.

With this definition, each natural number n is an ordinal � whichis either ; or of the form ↵ [ {↵} for some ordinal ↵ and all of whosemembers are either the empty set or of the form ↵ [ {↵} for some or-dinal ↵, and every ordinal which is either ; or of the form ↵[ {↵} andall of whose members are either the empty set or of the form ↵ [ {↵}for some ordinal ↵ is a natural number (the notion of ordinal, which wewill see later on, is defined only in terms of sets and the membershiprelation).

What is nice about this is that it gives a definition of the set N ofnatural numbers involving only the notion of set (and the membershiprelation):

N is the set of all those ordinals ↵ such that ↵ is either the emptyset or of the form � [ {�} for some ordinal � and such that each ofits members is either the empty set or of the form � [ {�} for someordinal �.

In particular, we may want to say that a set x is finite i↵ there is abijection between x and some member of N. By the above definitionof N we thus have a definition of finiteness purely in terms of sets andthe membership relation.

+ and · on N can be defined also in a satisfactory way using thenotion of set. Then we can define Z in the usual way as the set ofequivalence classes of the equivalence relation ⇠ on N ⇥ N defined by(a, b) ⇠ (a0, b0) if and only if a + b0 = a0 + b, and we can define also Qand the corresponding operations from Z in the usual way.

We can define R as the set of equivalence classes of the equivalencerelation ⇠ on the set of Cauchy sequences f : N �! Q where f ⇠ g ifand only if lim

n!1 h = 0, where h(n) = f(n)� g(n). And so on.All these constructions involve only notions previously defined to-

gether with the notion of set and the membership relation. So theyultimately involve only the notion of set and the membership relation.

If there is nothing fishy with the notion of set and the operationswe have used to build more complicated sets out of simpler ones, thenthere cannot be anything fishy with these higher level objects.

Similarly: We feel confident with the existence of C (which, by theway, contains “imaginary numbers” like i) once we become confidentwith the existence of R and know how to build C from R in a verysimple set–theoretic way.

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Also: We can derive everything (?) we know about the higher levelobjects (like, say, the fact that ⇡ is transcendental) from elementaryfacts about sets.

And, presumably, we would expect that the combination of elemen-tary facts about sets can ultimately answer every question we are in-terested in (is e+ ⇡ transcendental?, Goldbach’s conjecture, ...). Thiswould reduce mathematics to considerations of sets and their (elemen-tary) properties.

1.1. Some elementary facts about sets. Given sets A, B, we saythat A is of cardinality at most that of B, and write

|A| |B|,

if there is an injective (or one–to–one) function f : A �! B (remember,a function is a special kind of set!).

We say that that A and B have the same cardinality, and write

|A| = |B|,

if and only if there is a bijection f : A �! B.We say that A has cardinality strictly less than B, and write

|A| < |B|,

if and only if there is an injective function f : A �! B but there is nobijection f : A �! B.

Clearly |A| |B| and |B| |C| together imply |A| |C|. Also, itis true, but not a trivial fact, that |A| = |B| holds if and only if both|A| |B| and |B| |A| hold (Cantor–Bernstein–Schroder theorem, wewill see this later on).

The notion of cardinality captures the notion of “size” of a set. (Ex-ample: |5| < |6|).

Notation: Given a set X, P(X) is the set of all sets Y such thatY ✓ X. (P(X) is the power set of X).

The following theorem arguably marks the beginning of set theory.

Theorem 1.2. (Cantor, December 1873) Given any set X, |X| <|P(X)|.

Proof. There is clearly an injection f : X �! P(X): f sends x to thesingleton of x, i.e., to {x}.


Now suppose f : X �! P(X) is a function. Let us see that f cannotbe a surjection: Let

Y = {a 2 X : a /2 f(a)}

Y 2 P(X). But if a 2 X is such that f(a) = Y , then a 2 Y if andonly if a /2 f(a) = Y . This is a logical impossibility, so there is no sucha. ⇤

This theorem immediately yields that not all infinite sets are of thesame size, and in fact there is a whole hierarchy of infinities! (whichwas not known at the time):

|N| < |P(N)| < |P(P(N))| = |P2(N)| < . . .. . . < |Pn(N)| < |Pn+1(N)| < . . .. . . < |

Sn2N Pn(N)| < |P(

Sn2N Pn(N))| < . . .

More elementary facts:Let R be the collection of all those sets X such that

X /2 X

R is a collection of objects, and so (we would naturally say that) itis therefore a set.

R contains many sets. For instance, ; 2 R, 1 2 R, every naturalnumber is in R, N 2 R, R 2 R, etc.

Question 1.3. Does R belong to R?

Well, R 2 R if and only if R /2 R, which is the same kind ofcontradiction that we obtained at the end of the proof of Cantor’stheorem! So R cannot be a set!! (Russell’s paradox)

So, our naive “theory” of sets is inconsistent and maybe it’s not sogood a foundation of mathematics after all...

Is this the end of the story for set theory?Well, we like to think in terms of objects built out of sets and like

the simplicity of the foundations set theory was intending to provide.Also, we find the multiplicities of infinities predicted by set theory anexciting possibility, and there was nothing obviously contradictory inCantor’s theorem.

A retreat: A valid move at this point would be to retreat to a moremodest theory T such that

(1) T should express true facts about sets (or should we say plau-sible, desirable instead of true?),

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(2) T enables us to carry out enough constructions so as to build allusual mathematical objects (real numbers, spaces of functions,etc.),

(3) T gives us an interesting theory of the infinite (|N| < |P(N)|,etc.), and such that

(4) we can prove that T is consistent; or, if we cannot prove that,such that we have good reasons to believe that T is consistent.

First questions:

(1) What is a theory?(2) Which should be our guiding principles for designing T?

We answer (1) first.

2. The axiomatic method: A crash course in first orderlogic

For us a theory will be a first order theory. A theory T will alwaysbe a theory in a given language L. It will be a set (!) of L–sentencesexpressing facts about our intended domain of discourse.

Talk of “sets” of L–sentences before we have even defined T (whichmight end up being an intended theory of sets)? Well, those sets ofsentences, as well as the sentences, the language L, etc., are objects inour meta–theory. Presumably they will obey laws expressible in somemeta–meta–theory (perhaps the same laws the same theory T is, in ourunderstanding, trying to express!).

A language L consists of

• a (possible empty) set of constant symbols c, d, ...• a (possibly empty) set of functional symbols f , g, ..., togetherwith their arities (this arity is a natural number; if f is meantto represent a function fM : M �! M it has arity 1, if it ismeant to express a function fM : M ⇥M �! M it has arity 2,etc.)

• a (possibly empty) set of relational symbols R, S, .... togetherwith their arities (this arity is again a natural number; if R ismeant to represent a subset RM ✓ M , then it has arity 1, if itis meant to express a binary relation RM ✓ M ⇥M , then it hasarity 2, etc.)

These are the non-logical symbols and completely determine L.We also have logical symbols, which are independent from L:

• ^, _, ¬, !, $ (connectives)• 8, 9 (quantifiers)• (, ), =


= is sometimes omitted. Also, many of these symbols are not neces-sary; we could actually do with just ¬, _ and 9.

Finally, we have a su�ciently large supply of variables : V ar ={v0, v1, . . . , vn, . . .}. For most uses it is enough to take the set of vari-ables to have the same size as the natural numbers.

The language of set theory has only one non–logical symbol, namelya relational symbol 2 of arity 2.

Let us focus on the language of set theory from now on:

(1) Every expression of the form (vi

2 vj

) or (vi

= vj

), with vi

andvj

variables, is a formula (an atomic formula).(2) If ' and � are formulas, then (¬'), (' _ ), (' ^ _), (' ^ ),

(' ! ), (' $ �) are formulas. Also, if v is a variable, then(8v') and (9v') are formulas.

(3) Something is a formula if and only if it is an atomic formula oris obtained from formulas as in (2).

When referring to a formula, we often omit parentheses to improvereadability (these expressions are not actual o�cial formulas but referto them in a clear way).

A sentence is a formula ' without free variables, i.e., such that everyvariable v occurring in ' occurs in some subformula of the form 8v or of the form 9v .

Examples of formulas are the formulas abbreviated as:

8x8y(x = y $ 8z(z 2 x $ z 2 y))

(The axiom of Extensionality)

8x8y9z8w(w 2 z $ (w = x _ w = y))

or, even more abbreviated,

“for all x, y, {x, y} exists”

(Axiom of unordered pairs).Another example:9a9b8y(y 2 x $ ((8w(w 2 y $ (w = a _ w = b))) _ (8w(w 2 y $

w = a)))))(x is in ordered pair)

The first two formulas are sentences. The third one is not.

Satisfaction:This takes place of course in the meta–theory:

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A pair M = (M,R), where M is a set and R ✓ M ⇥M , is called anL–structure.

Given an assignment ~a : Var �! M :

• M |= (vi

2 vj

)[~a] if and only if (~a(vi

),~a(vj

)) 2 R.• M |= (v

i

= vj

)[~a] if and only if ~a(vi

) = ~a(vj

).• M |= (¬')[~a] if and only if M |= '[~a] does not hold.• M |= ('0 _ '1)[~a] if and only if M |= '0[~a] or M |= '1[~a]; andsimilarly for the other connectives.

• M |= (9v')[~a] if and only if there is some b 2 M such thatM |= '[~a(v/b)], where ~a(v/b) is the assignment ~b such that~b(v

i

) = ~a(vi

) if v 6= vi

and ~b(v) = b.• M |= (8v')[~a] if and only if for every b 2 M , M |= '[~a(v/b)].

We say that M satisfies ' with the assignment ~a if M |= '[~a].

Easy fact: If ' is a sentence, then M |= '[~a] for some assignment ~aif and only if M |= '[~a] for every assignment ~a. In that case we saythat M is a model of �.

Definition 2.1. Given a set T of formulas and a formula ', we write

T |= '

if and only if for every L–structure M = (M,R) and every assignment~a : Var �! M , IF M |= �[~a] for every � 2 T , THEN M |= '[~a].

The relation |= aims at capturing the notion of ‘logical consequence’:' follows logically from T if and only if ' is true in every world in whichT is true. |= is often called the relation of logical consequence.

‘First order’ in ‘first order logic’ refers to the fact that variables rangein the above definition only over the individuals of the universe of therelevant L–structures M. In second order logic we can have variablesthat range over (arbitrary) subsets of the universe of the relevant L–structures M. Etc.

Syntactical deductionLet T be a set of formulas. We will view T as a set of axioms and

deduce theorems from T : A theorem of T will be the final member �n

of a deduction� = (�0, �1, . . . �n)

from T , where we say that � = (�0, �1, . . . �n) is a deduction from T ifit is a finite sequence of L–formulas and for every i,


• �i

is either in T , or• �

i

is a logical axiom of first order logic, or• �

i

is obtained form �j

and �k

, for some j, k < i, by the rule ofModus Ponens “If ' ! and ', then ” (for all L–formulas', ). In other word, there are j, k < i and an L–formula 'such that �

j

is ' and �k

is '! �i

.

Here, a logical axiom is a member of a certain infinite easily specifi-able list, independent of the theory, consisting of formulas that expresslogical / completely universal truths. Typical members of this list arefor example, ' ! ( ! ') for all formulas ', , or ' _ ¬' for allformulas '. Indeed, we see it as a general truth that if ' is true, thenit is true that if is true then ' is true. Other typical members ofthis sequence are all instances of the schema ' _ ¬', for ' being anyformula. Again, we see it as a general truth that for every ' either 'is true or ¬' is true.1

This list of axioms is not unique: Many di↵erent lists of axioms giverise to the same system of logic.

If ' is a theorem from T , we write

T ` '` is often called the relation of logical derivability.On the face of their definition, |= and ` are quite di↵erent relations,

aimed at capturing two apparently di↵erent notions: The notion oflogical (semantical) consequence and the notion of deducibility in areasonable calculus. However, we do have the following remarkablefact, proved by Kurt Godel in his PhD thesis.

Theorem 2.2. (Completeness theorem for first order logic) (K. Godel,1930’s) |==`

A theory T is consistent if no contradiction (say, 9x¬(x = x)) canbe derived from it:

T 0 9x¬(x = x)

Otherwise, it is inconsistent. In classical first oder logic, a theoryis inconsistent if and only if it is trivial, in the sense that it proveseverything.2

1If we are classical logicians. There are weakening / versions of classical firstorder logic in which '_¬', also known as Law of Excluded Middle, is not true forsome choices of '.

2There are other logics, so called para–consistent logics, which may allow thepresence of contradictions but which nevertheless may not be trivial, i.e., whichmay not prove everything to be a theorem.

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By the completeness theorem the following are equivalent:

• T is consistent.• There is an L–structure M such that M |= T (T is true insome world).

We will be interested in whether or not T ` � for various choices oftheories T and sentences �. The following are equivalent again by (thecontrapositive of) the completeness theorem:

• T 0 �• There is an L–structure M such that M |= T but M |= ¬�.

3. Axiomatic set theory: ZFC

Z is for Ernst Zermelo, F is for Abraham Fraenkel, C is for theAxiom of Choice.

The objects of set theory are sets. As in any axiomatic theory, theyare not defined (they are feature–less objects; in the context of thetheory there is nothing to them apart from what the theory says).

ZFC expresses facts about sets expressible in the first order languageof set theory. The same is true for any other first order theory inthe language of set theory, like ZF, ZFC+“There is a supercompactcardinal” + ZFC+GCH, ZFC+V = L, ZFC+PFA, ...

Most ZFC axioms will be axioms saying that certain “classes” (builtout of given sets) are actual sets (they are objects in the set–theoreticuniverse): Axiom 0, The Axiom of unordered pairs, Union set Axiom,Power set Axiom, Axiom Scheme of Separation, Axiom Scheme of Re-placement and Axiom of Infinity will be of this kind. Here, a classis any collection of objects, where this collection is definable possiblywith parameters. For example the class of all sets. A proper classwill be a class which is not a set.

ZFC will also have one axiom guaranteeing the existence of sets witha given property, even if these sets are not definable: The Axiom ofChoice.3 We will also have two “structural” axioms, namely the Axiomof Extensionality and the Axiom of Foundation.

A classification of the ZFC axioms.

(1) Structural axioms: Axioms of Extensionality, Axiom of Foun-dation.

3There are strengthenings of ZFC incorporating more non-constructive set exis-tence axioms; for example, the extensions of ZFC one gets by adding to it forcingaxioms are of this sort.


(2) Constructive set–existence axioms: Axiom 0, The Axiomof unordered pairs, Union set Axiom, Power set Axiom, Ax-iom Scheme of Separation, Axiom Scheme of Replacement andAxiom of Infinity.

(3) Non–constructive set–existence axiom: Axiom of Choice.

3.1. The axioms. The following is the list of the ZFC axioms.

Axiom of Extensionality: Two sets are equal if and only if theyhave the same elements:

8x8y(x = y $ 8z(z 2 x $ z 2 y))

In other words: the identity of a set is completely determined by itsmembers:

The sets

• ;• {(a, b, c, n) : an + bn = cn, a, b, c, n 2 N, a, b, c � 2, n � 3}

are the same set.

Axiom 0: ; exists.

9x8y(y 2 x $ y 6= y)

(of course y 6= y abbreviates ¬(y = y)).

Strictly speaking this axiom is not needed: It follows from the otheraxioms. It is convenient to postulate it at this point, though.

In the theory given by the Axiom of Extensionality together withAxiom 0 we can only prove the existence of one set:

;

So this theory is not so interesting yet. The theory T = {Axiom 0,Axiom of Extensionality} surely is consistent: For any set a,

({a}, ;) |= T

On the other hand, note that ({a, b}, ;) 6|= T if a 6= b.

Axiom of unordered pairs: For any sets x, y there is a set whosemembers are exactly x and y; in other words, {x, y} exists.

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8x8y9z8w(w 2 z $ (w = x _ w = y))

Of course, if x = y, then {x, y} = {x} [prove this using the Axiomof Extensionality.]

Recall that we defined the ordered pair (x, y) as the set {{x}, {x, y}}.The theory laid down so far gives us already the existence of in-

finitely many sets! For example ;, {;}, {{;}}, {{{;}}}, {{{{;}}}},{;, {;}}, {;, {;, {;}}}, {{;}, {;, {;}}}, {;, {;, {;, {;}}}}, ... With thedefinition of the natural numbers we have adopted these sets are: 0,1, {1} = (0, 0), {{1}} = {(0, 0)}, ((0, 0), (0, 0)), 2 = (0, 1), {0, 2},{1, 2} = (0, 1), {0, {0, 2}}, ...

All sets whose existence is proved by the theory given so far have atmost two elements (!). Also, this theory proves the existence of (a, b)for all a, b.

Union set Axiom: For every set x,[

x = {y : (9w)(w 2 x ^ y 2 w)}

exists:

8x9v8y(y 2 v $ (9w)(w 2 x ^ y 2 w))Sx is the set consisting of all the members of members of x,

SSx

is the set of all the members of members of members of x, etc.

Notation: Given sets x, y, x [ y = {a : a 2 x _ a 2 y} =S{x, y}.

Note: Given sets x, y, x[ y exists (by the Axiom of unordered pairsand the Union set Axiom).

With the theory given so far we can prove the existence of: {0} [{1, 2} = {0, 1, 2} = 3, {0, 1, 2} [ {3} = {0, 1, 2, 3} = 4, {0, 1, 2, 3} [{4} = {0, 1, 2, 3, 4} = 5, ....

So we can prove the existence of every individual natural number!Similarly, we can prove the existence of every finite set of natural num-bers, every ordered pair of natural numbers, every tuple of naturalnumbers, every finite set of tuples of natural numbers, ... However, allparticular sets proved to exist by the theory given so far are finite.


Notation: z ✓ x means: Every member of z is a member of x.

Power set Axiom: For every x there is y whose elements are exactlythose z which are a subset of x:

8x9y8z(z 2 y $ (8w)(w 2 z ! w 2 x))

Notation: For every a, P(a) = {z : z ✓ a}.

The Power set Axiom says that P(a) is a set whenever a is a set.With the theory T laid down so far we can prove the existence of

P(n) for any particular n 2 N.For example:

• P(0) = {;} = 1• P(1) = {;, {;}} = 2• P(2) = {;, {;}, {{;}}, {;, {;}}} 6= 4• ...

T is consistent:4 Let (Xn

)n2N be defined recursively by

• X0 = {;}• X

n+1 = Xn

[ {{a, b} : a, b 2 Xn

} [ {Sa : a 2 X

n

} [ {P(a) :a 2 X

n

}Then (

Sn2N Xn

,2) |= T .Actually it would be enough to start with ; and take X

n+1 = P(Xn

)at each stage n+ 1.

Note: All particular sets proved to exist by T are still finite.

Axiom Scheme of Separation: Given any set X and any firstorder property P ,

{y 2 X : P (y)}exists; in other words: any definable subclass of a set exists as a set.

8x8v0, . . . , vn9y8z(z 2 y $ (z 2 x ^ '(x, z, v0, . . . vn)))

for every L–formula '(x, z, v0, . . . vn) such that y does not occur asbound (i.e., non–free) variable in it, and where x, y, z, v0, . . . , vn aredistinct variables.

In the theory laid down so far we can prove the existence, for all x,y, of

x⇥ y = {(a, b) : a 2 x, b 2 y},and much more.

4Isn’t it?

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Fact 3.1. The theory we have so far proves the existence of x⇥ y forall y.

Proof. Work in the theory. Let x and y be given, Let z = x [ y,which we know exists in our theory. Note that x ⇥ y is a definablesub-collection of P(P(x[y)). Hence x⇥y exists using Power set twiceand Separation once. ⇤

For a formula '(v0, . . . vn, u, v), ‘'(v0, . . . vn, u, v) is functional ’ is anabbreviation of the formula expressing “for all u there is at most onev such that '(v0, . . . vn, u, v)”.

5

Axiom Scheme of Replacement: Given any set X and any de-finable (class)–function F , range(F � X) is a set:

“For all x, v0, . . . , vn, if '(v0, . . . vn, x, u, v) is functional, then thereis y such that for all v, v 2 y if and only if there is some u 2 x suchthat '(v0, . . . vn, x, u, v),”

for every formula '(v0, . . . vn, x, u, v) such that y does not occur asbound variable, and where x, y, u, v, v0, . . . , vn are distinct variables.

Caution: The Axiom schemes of Separation and Replacement arenot axioms but infinite sets of axioms (!). However, it is obviouslypossible to write down a computer program which, given a sentence �,recognises whether or not � belongs to either of these schemes.

Given a set X such that a 6= ; for all a 2 X, a choice functionfor X is a function f with dom(f) = X and such that f(a) 2 a for alla 2 X.

Axiom of Choice (AC): Every set consisting of nonempty setshas a choice function.

Exercise 3.1. Write down a sentence expressing the Axiom of Choice.

AC is needed in a lot of mathematics. For example, to prove thatevery vector space has a basis, that there are sets of reals which arenot Lebesgue measurable, etc. Nevertheless, historically AC has beenseen with suspicion: Finite sets clearly have choice functions,6 but ifX is infinite, where did the choice function for X come from? Also,

5Which can be expressed in our language since it has =.6Try to see why.


AC has “strange consequences”: For example, it is possible to decom-pose a sphere S into finitely many pieces and rearrange them, withoutchanging their volumes – in fact by moving them around and rotatingthem, and without running into one another –, in such a way that weobtain two spheres with the same volume as S!7 This result is knownas the Banach–Tarski paradox.8

AC has interesting equivalent formulations (modulo the rest of ZFC).For example AC is equivalent to “For every two nonempty sets A, B,either |A| |B| or |B| < |A|”. AC is also equivalent to “Every productof compact topological spaces is compact”.

The Axiom of Foundation: If X 6= ; is a set, there is some a 2 Xsuch that b /2 a for every b 2 X.

In other word: Every nonempty sets has some 2–minimal element.Modulo the other axioms (in particular AC), the following are equiv-

alent:

• Foundation• There are no x0, x1, . . . , xn

, xn+1, . . . such that . . . 2 x

n+1 2xn

2 . . . 2 x1 2 x0.

The idea behind Foundation is that sets are generated at di↵erentstages. If a set X is generated at stage ↵, then all members of X havebeen generated at some stage before ↵.

Foundation, together with Extensionality, of course, is perhaps themost fundamental axiom in set theory.

As with AC, one could perhaps also complain: Where did the 2–minimal element a of X come from? But wait. a was already in X. Ifyou remove a from X, what you get is no longer X!

In fact, most people like Foundation: It says that the universe isgenerated in an orderly fashion. And it provides a very convenient toolto use in proofs, which we will be using all the time: Induction.

Let (Vn

)n2N be defined by recursion as follows.

• V0 = ;• V

n+1 = P(Vn

)

The theory laid down so far, T = Ax0+ Extensionality + UnorderedPairs + Union + Power Set + Separation + Replacement + AC +

7The pieces are not Lebesgue measurable, though.8The Banach–Tarski is not an actual paradox, in the sense that Russel’s paradox

is, but a counterintuitive fact.

16 D. ASPERO

Foundation, is consistent.9 In fact

([

n

Vn

,2) |= T

Still, all sets proved by T to exist are finite. In fact,

([

n

Vn

,2) |= “Every set is finite”

What do we mean by finite? For the moment let us say that a set Xis finite if and only if for every a 2 X, |X \ {a}| < |X|. Correspond-ingly, let us say the a set is infinite if and only if it is not finite. Thisis not the o�cial definition of ‘finite’ but is equivalent to the o�cialdefinition. But it makes things easier to deal with the above ‘definition’(which does not involve the notion of ordinal, which we haven’t definedyet). In any case, (

Sn

Vn

,2) thinks that every set is finite in this sense.

Axiom of Infinity: There is an infinite set.

Definition 3.2. Given a set x,

S(x) = x [ {x}

(the successor of x).

So, S(0) = 1, S(1) = 2, ... S(n) = n+ 1.The Axiom of Infinity is equivalent to:

(9x)(; 2 x ^ (8y)(y 2 x ! S(y) 2 x))

This is also phrased as: There is an inductive set.

Proposition 3.3. Every inductive set is infinite (in our present sense).

Proof. Suppose X is inductive, let a 2 X, and let f : X \ {a} �! Xbe the function sending every set of the form Sn(a) (for n 2 N, n > 0)to Sn�1(a), and every set x 2 X which is not of the above form to xitself. It is easily checked that this function f is a bijection. ⇤

One could also define “↵ is an ordinal” (which we will do soon).Then we would define a natural number as an ordinal ↵ such that

(1) ↵ is either ; or of the form S(y) for some ordinal y and(2) for every x 2 ↵, x is either ; or of the form S(y) for some

ordinal y.

9Isn’t it?


The Axiom of Infinity is then equivalent to:

Axiom of Infinity’: The class of all natural numbers is a set.In other words: The Axiom of Infinity’ says that there is some x suchthat for all y, y 2 x if and only y is a natural number.

Remark 3.4. Note that Axiom of Infinity’ is a constructive set–exis-tence axiom, whereas Axiom of Infinity was not, strictly speaking (itjust says that there is an infinite, without defining it). However, Axiomof Infinity’ and Axiom of Infinity are equivalent modulo the other ax-ioms (and we don’t need many of them for this; in particular, we don’tneed the Axiom of Choice). This observation shows that one shouldbe carefully when specifying what a classification of set–existence ax-ioms into constructive set–existence axioms and non–constructive set–existence axioms would be. Indeed, depending on the context, twoaxioms which apparently belong to di↵erent classes can actually beequivalent.

Another observation that shows that one should be careful aboutthe above classification is the following: Given any sentence �, � isequivalent to a (seemingly) constructive set–existence axiom over anytheory that proves, say, that ; exists as a set and that R = {y : y /2 y}does not exist as a set. This axiom is 9x8y(y 2 x $ (y /2 y ^ ¬�)).The point of course is that if � is true then the above set x is the emptyset, and if � is false, then the above set x is R.

The Axiom of Infinity completes the list of ZFC axioms.

Notice the big leap when adding Infinity to the list of axioms. ZFCcertainly proves the existence of infinite sets, by design! Before addingInfinity we had a theory T which ‘surely’ was consistent.10 Now, withthe addition of Infinity, it’s not so obvious that ZFC is consistent... .

Challenge 3.1. Construct a model of ZFC.

3.2. ZFC vs PA. Peano Arithmetic, also known as PA, is the followingfirst order theory for (N, S,+, ·, 0), where S(n) = n+1 (in the languageof arithmetic, i.e., the language with S, +, ·, 0):

• 8x(S(x) 6= 0)• 8x, y, (S(x) = S(y) $ x = y)• 8x(x+ 0 = x)• 8x, y(x+ S(y) = S(x+ y)

10Since (S

n2N Vn,2) |= T .

18 D. ASPERO

• 8x(x · 0 = 0)• 8x, y, x · S(y) = x · y + x• 8y(('(0, y) ^ (8x('(x, y) ! '(S(x), y))) ! 8x'(x, y))

for every first order formula '(x, y) in the language of arith-metic

(First order Induction Axiom Scheme)

First order arithmetical facts can be expressed in this language, likefor example “· is distributive with respect to +”, Fermat’s last theorem,Goldbach’s conjecture, ...

PA does prove many facts about (N, S,+, ·, 0). But it does not proveeverything!

Theorem 3.5. (Godel, 1930’s, Incompleteness Theorem (special case))If PA is consistent then there is a sentence � in the language of arith-metic such that

• PA 0 � and• PA 0 ¬�

Godel’s Incompleteness theorem(s), in their general formulation, arevery profound facts that we will look back into in a moment.

The sentence � in the Incompleteness Theorem does not express anyfact that mathematicians would have looked into prior to proving theincompleteness theorem. � is designed for the purpose of the proofonly.11

Notation: Given a set X and n 2 N, let[X]n = {a ✓ X : |a| = |n|}

Consider the following statement HP:“For all n, k, m there is some N such that for every colouring f : [N]ninto k colours there is some Y ✓ N such that Y has at least m manymembers and at least min(Y ) many members and such that all mem-bers of [Y ]n have the same colour under f .”

Here, n, k, m and N range over natural numbers.HP can be easily expressed by a sentence, which I will call HP, in thelanguage of arithmetic.

11In its intended interpretation, � says about itself that it cannot be provedin PA. This type of self–reference may sound strange at first; in particular, itmay seem doubtful that it even makes sense. However, there is a perfectly soundmathematical way to make sense of this, and in fact such a sentence can be writtendown in the language of arithmetic.


ZFC proves that (N, S,+, ·, 0) |= HP. On the other hand:

Theorem 3.6. (L. Harrington and J. Paris, 1977): If PA is consis-tent, then

PA 0 HP

Consider the theory T = (ZFC \{Infinity}) [ {¬Infinity}. It turnsout that T and PA are essentially the same theory: There are e↵ectivetranslation procedures

' �! �(')

between the sentences in the language of set and the sentences in thelanguage of arithmetic and

�! �( )

between the sentences in the language of arithmetic and the sentencesin the language of set theory such that for all ', ,

• T ` ' if and only if PA ` �(')• PA ` if and only if T ` �( )

The Harrington–Paris theorem gives an example of a simple “natu-ral” (purely combinatorial) statement � talking only about finite setswhich is true if there is an infinite set but need not be true if there areno infinite sets (!). Other examples have been found since then.

3.3. The consistency question. We pointed out that the theoryT = ZFC \{Infinity} was ‘surely’ consistent, based on the fact that(S

n2N Vn

,2) |= T (assuming, in our metatheory, that P(a) exists forevery a, that N exists, that the recursive construction of F = (V

n

)n2N is

well–defined class–function, and thatS

range(F ) exists, i.e., assumingsomething like ZFC in our metatheory!)

Question 3.7. Can we prove, in T (equivalently, in PA), that T isconsistent? Can we prove, in ZFC, that ZFC is consistent?

The above questions do make sense: Both T and PA have enoughexpressive power to make “T is consistent”, “PA is consistent”, etc. ex-pressible in the theory: For example, we can code formulas, proofs, andother syntactical notions as natural numbers and reduce a statementlike “PA is consistent” to an arithmetical statement (some specific, butextremely complex, diophantine equation p(x) = 0 does not have so-lutions). It then makes sense to ask whether T proves that p(x) = 0does not have solutions.

Theorem 3.8. (Godel’s Incompleteness Theorems) Suppose T is a firstorder theory such that

20 D. ASPERO

• T is computable (in the sense that there is an algorithm decid-ing, for any given sentence �, whether or not � 2 T ),

• T interprets PA and• T is consistent.

Then:

(1) There is a sentence � such that• T 0 � and• T 0 ¬�

(First Incompleteness Theorem)

(2) T does not prove that T is consistent (T 0 Con(T ))

(Second Incompleteness Theorem)

A theory T as in (1) is said to be incomplete.Note: Both ZFC and PA are computable (in the above sense). Hence,

IF they are consistent, THEN they are incomplete and they cannotprove their own consistency. It follows that if we adopt, say, ZFC asour meta-theory, we won’t be able to prove any statement of the form“ZFC+� is consistent”. What we can do is prove relative consistencystatements of the form “If ZFC is consistent, then ZFC+� is consis-tent” (Con(ZFC) ! Con(ZFC+�)).

On the other hand, note that ZFC ` Con(ZFC \{Infinity}) (equiv-alently, ZFC ` Con(PA)): Working within ZFC we can build the setS

n2N Vn

and we can prove

([

n2N

Vn

,2) |= ZFC \{Infinity}

We express the above fact by saying that ZFC has consistency strengthstrictly larger than ZFC \{Infinity}.

In general, we say that T1 has consistency strength at least that of T0

if and only if we can prove that if T1 is consistent then T0 is consistent.T1 has consistency strength strictly larger than T0 if and only if we canprove “if T1 is consistent, then T0 is consistent”, but we cannot prove“if T0 is consistent, then T1 is consistent” unless we can prove “T0

inconsistent.” And, similarly, we define “T0 and T1 are equiconsistent.”It is important to bear in mind that we need to be careful with

what we understand by the informal ‘proving’ in the above definitionor otherwise we might render the notion of consistency strength unin-teresting. In fact, if we identify ‘proving’ with ‘being true’, then allconsistent theories would have the same consistency strength, and all


inconsistent theories would have the same consistency strength too.This is not so interesting, so we instead interpret ‘proving’ as ‘prov-ing within some reasonable theory, like PA or ZFC.’ For example, weshould understand a statement of the form “T1 has consistency strengthstrictly larger than T0” as meaning, for example, that we can prove inZFC the arithmetical statement Con(T1) ! Con(T0) and we can provein ZFC that if T0 and ZFC are both consistent, then ZFC does notprove the arithmetical statement Con(T0) ! Con(T1) .

For example, although ZFC does not prove Con(ZFC), if ZFC isconsistent, it proves that ZFC+� and ZFC+¬� are equiconsistent(i.e., it proves Con(ZFC+�) $ Con(ZFC+¬�)) for many interestingchoices of � (we will hopefully see examples of this).

If T1 has consistency strength strictly larger than T0, then T1 is more“daring” than T0. There is a whole natural hierarchy of theories orderedby consistency strength:

• ZFC is equiconsistent with ZF (= ZFC \{AC}) and is strictlystronger than ZFC \{Infinity}.

• ZFC + “There is an inaccessible cardinal” is strictly strongerthan ZFC.

• ZFC + “There is a weakly compact cardinal” is strictly strongerthan ZFC + “There is an inaccessible cardinal”.

• ZFC + “There is a measurable cardinal” is strictly strongerthan ZFC + “There is a weakly compact cardinal”.

• ZFC + “There is a Woodin cardinal” is strictly stronger thanZFC + “There is a measurable cardinal”.

• ZFC + “There is a supercompact cardinal” is strictly strongerthan ZFC + “There is a Woodin cardinal”.

• ZFC + “There is a huge cardinal” is strictly stronger than ZFC+ “There is a supercompact cardinal”.

• ...

Later on I will hopefully say a bit more about the above hierarchyof so–called ‘large cardinal theories (or axioms).’

Let the Axiom Scheme of Comprehension be: For every formula'(x) in the language of set theory,

9x8y(y 2 x $ '(y, x))

Frege’s set theory T consists of the Axiom of Extensionality togetherwith all instances of the Axiom Scheme of Comprehension. (This wasFrege’s bold attempt to reduce all of mathematics to logic).

T is of course inconsistent by Russell’s paradox.

22 D. ASPERO

In which way does ZFC (or ZF) neutralise Russell’s paradox? Well,ZF proves that there is no R such that for every x, x 2 R if and onlyif x /2 x: If there was such an R, then R 2 R if and only if R /2 R.

So R = {x : x /2 x} is, in ZFC, a proper class but not a set.ZFC is not the only theory of sets that people have considered as

a foundation for mathematics and which neutralises Russell’s paradox(and other related paradoxes). There are also: Type theories, Quine’sNew Foundations (NF), etc. However, ZFC is the most well–suitedfor developing mathematics. Incidentally, it is worth pointing out thatNF is not known to be consistent relative to any natural extension ofZFC.12

We cannot prove that ZFC is consistent. So why should we feelconfident about its consistency?

The first observation is that the question on the consistency of ZFCis reducible to the question on the consistency of the smaller theoryZF:

• ZFC is equiconsistent with ZF: Given any (M,R) |= ZF thereis a LM ✓ M such that (LM , R \ LM ⇥ LM) |= ZFC (Godel).

OK, why should we trust ZF then? I will give three reasons next.

• All set–existence axioms of ZF assert the set–hood only of “smallclasses.” (This is perhaps vague at this point, but in a littlewhile you’ll get a clearer picture.) Compare with the AxiomScheme of Comprehension, which says that EVERY class is aset!

• All axioms of ZF are “reasonable” assertions about sets: ZFsays that the set–theoretic universe is exactly the “cumulativehierarchy”, which provides a very appealing picture of the ‘gen-eration of sets from previously generated sets’ (see later). Thisis perhaps the best intrinsic justification for ZF and, as a by–product, speaks in favour of it consistency: The cumulativehierarchy looks so natural that it should be a “real object”. Itsatisfies the axioms of ZF. Therefore, ZF should not be incon-sistent.

• History: No inconsistency has ever been detected within ZF.

We will be working in ZFC until further notice.

12Placing this comment here is, admittedly, a bit ZFC–centric. One could believein NF instead and, working within NF, could try to prove whether or not ZFC, orsome extension of ZFC by large cardinals, say, is consistent.


4. Ordinals

Definition 4.1. A partial order is an ordered pair (X,R) such that

• R ✓ X ⇥X,• for every x 2 X, (x, x) 2 R (R is reflexive),• for all x, y 2 X, (x, y) 2 R and (y, x) 2 R together imply x = y(R is anti–symmetric), and

• for all x, y, z 2 X, (x, y) 2 R and (y, z) 2 R together imply(x, z) 2 R (R is transitive).

We say that R is a partial order on X. Also, we often write xRy for(x, y) 2 R.

(X,R) is a total ordering (or linear order) if for all x, y 2 R, eitherxRy or yRx.

For example, (N, <), (Q, <) and Q⇥Q (with the product order) arepartial orders, and (N, <) and (Q, <) are linear but Q⇥Q is not.

Definition 4.2. A binary relation R ✓ X ⇥X is well–founded if andonly if for every nonempty Y ✓ X there is some a 2 Y which is R–minimal, i.e., such that (b, a) /2 R for every b 2 Y , b 6= a.

For example, the Axiom of Foundation says that 2 \X ⇥X is well–founded for every set X).

Definition 4.3. A well–order is a well–founded linear order.

For example, (5, <) and (N, <) are well–orders, but (R, <) is not.Note: If (L,) is a well–order and A ✓ L is nonempty, then min(A)

exists.Notation: Given a partial order (L,), < is the relation on L given

by y < x if and only if y x and y 6= x.We will sometimes abuse language and say that a pair (L,<) is a

well–order if

• < is transitive,• < is irreflexive (i.e., x < x fails for all x 2 dom(<))

(i.e., < is a strict order) and

• < is total (i.e., for all x, y, x < y, y < x or x = y) and• every nonempty subset of dom(<) has an <–minimal member.

Given a partial order (L,) and x 2 L,

pred(L, x) = {y 2 L : y < x}A ✓ L is an initial segment of (L,) i↵ for all x < y 2 L, if

y 2 A, then x 2 A. A ✓ L is a proper initial segment of (L,) ifA = pred((L,), a) for some a 2 L.

The following fact is immediate.

24 D. ASPERO

Fact 4.4. If (L,<) is a well–order and A ✓ L, then (A,<� A) is awell–order.

Given two partial orders (X0,0), (X1,1) a bijection f : X0 �! X1

is an order–isomorphism if and only if for all x, y 2 X0,

x 0 y if and only if f(x) 1 f(y).

If there is such an order–isomorphism we write

(X0,0) ⇠= (X1,1)

Proposition 4.5. Let (L0,0) and (L1,1) be two well–orders. Thenexactly one of the following holds.

(1) (L0,0) ⇠= (L1,1)(2) (L0,0) is order–isomorphic to some proper initial segment of

(L1,1).(3) (L1,1) is order–isomorphic to some proper initial segment of

(L0,0).

(Trichotomy)

Proof. Let

f = {(u, v) 2 L0 ⇥ L1 : pred((L0,0), u) ⇠= pred((L1,1), v)}f is a function: Suppose (u, v), (u, v0) 2 f , v 6= v0. Wlog v <1 v0.Since pred((L0,0), u) ⇠= pred((L1,1), v) and pred((L0,0), u) ⇠=pred((L1,1), v0), by composing these order–isomorphisms we obtainan order–isomorphism

g : pred((L1,1), v) �! pred((L1,1), v0)

Let v <1 v such that g(v) = v. The existence of v shows that

A = {z 2 L1 : z <1 g(z)} 6= ;Let z⇤ = min(A). Let z be such that g(z) = z⇤. Then g(z) = z⇤ <1

g(z⇤) implies z < z⇤ and therefore z /2 A. Hence, z⇤ = g(z) 1 z <1 z⇤

and therefore z⇤ <1 z⇤. Contradiction.

Similarly one can show that for all u, u0, u 0 u0 i↵ f(u) 1 f(u0).In particular, f is injective.

dom(f) is an initial segment of (L0,0): Let u 2 dom(f) and letu0 <0 u. There is some v 2 L1 such that there is an order–isomophism

h : pred(L0,0), u) �! pred((L1,1), v)

Let v0 = h(u0). Then h � pred((L0,0), u0) is an order–isomorphismbetween pred((L0,0), u0) and pred((L1,1), v0). This shows (u0, v0) 2f .


Similarly one shows range(f) is an initial segment of (L1,1).If either dom(f) = L0 or range(f) = L1, then we are done: In the

first case, either range(f) = L1 and therefore f is an order–isomorphismbetween (L0,0) and (L1,1) or else min(L1\range(f)) = v exists andf is an order–isomorphism between (L0,0) and pred((L1,1), v). Inthe second case one proceeds similarly.

Suppose towards a contradiction that L0 \dom(f) and L1 \ range(f)are both nonempty. Let u = min(L0 \ dom(f)) and v = min(L1 \range(f)). Then f is an order–isomorphism between (pred(L0,0), u)and (pred(L1,1), v) and therefore (u, v) 2 f . But then u 2 dom(f)and v 2 range(f). A contradiction.

Finally: It is easy to check that (1)–(3) are mutually exclusive. ⇤Definition 4.6. A set x is transitive i↵ y ✓ x for every y 2 x. Inother words, x is transitive i↵ y 2 x and z 2 y imply z 2 x.

Examples:

• Every natural number is transitive.• N is transitive.• P(N) is transitive.• {1} is not transitive.

Definition 4.7. A set ↵ is an ordinal if and only if ↵ is transitive andwell–ordered under 2. In other words, letting 2 |↵ be the restriction of2 to ↵⇥ ↵, i.e., the relation on ↵ given by x 2 |↵ y i↵ x 2 y, (↵,2 |↵)is a well–order.

Fact 4.8. If ↵ is an ordinal and x 2 ↵, then x is an ordinal and

x = pred((↵,2 |↵), x)Proof. Let ↵ be an ordinal and x 2 ↵.

x is transitive: Let z 2 y 2 x. Using the transitivity of ↵ twice wehave that z 2 y 2 ↵ and therefore z 2 ↵. Since 2 |↵ is a transitiverelation (as ↵ is an ordinal), x, y and z are in ↵, and both y 2 x andz 2 y hold, we must have that z 2 x.

The proof that x = pred((↵,2 |↵), x) is then trivial.Since both x and ↵ are transitive, (2 |↵) \ (x ⇥ x) =2 |x. To see

this, note that the following are equivalent for all sets y, z:

• z 2 y 2 x• y 2 ↵ and z 2 ↵ and z 2 y 2 x.

But then 2 |x is a well–order on x since it is the restriction of thewell–order 2 |↵ to x. ⇤Fact 4.9. If ↵ and � are ordinals and f : (↵,2) �! (�,2) is anorder–isomorphism, then f is the identity on ↵. In particular, ↵ = �.

26 D. ASPERO

Proof. Suppose towards a contradiction that there is a minimal ⇠ 2 ↵such that f(⇠) 6= ⇠. Since f � ⇠ is the identity on ⇠,

f(⇠) = {f(⇠0) : ⇠0 2 ⇠} = {⇠0 : ⇠0 2 ⇠} = ⇠

which is a contradiction, where the first equality holds since the func-tion f : (↵,2) �! (�,2) is an isomorphism. ⇤Corollary 4.10. (Trichotomy for ordinals) Suppose ↵ and � are ordi-nals. Then exactly one of the following holds.

(1) ↵ = �(2) ↵ 2 �(3) � 2 ↵

Corollary 4.11. For every ordinal ↵, ↵ /2 ↵.

Corollary 4.12. For all ordinals ↵, �, �, if ↵ 2 � and � 2 �, then↵ 2 �.

Corollary 4.13. If A is a nonempty set of ordinals, then A has an2–minimal element.

Proof. Let ↵ 2 A. If ↵ is not 2–minimal, then A \ ↵ 6= ;. But then� = min(A \ ↵) exists, and then � is an 2–minimal member of A. ⇤

Notation: Ord denotes the class of all ordinals.The previous corollary says that the relation 2 well–orders Ord. So,

if Ord were a set, it would be an ordinal. But then Ord 2 Ord, and wehave seen that ↵ /2 ↵ for every ordinal ↵. Hence we have the following.

Theorem 4.14. Ord is not a set. (Burali–Forti Paradox)

On the other hand:

Fact 4.15. Every transitive set of ordinals is an ordinal.

Exercise 4.1. Prove Fact 4.15.

Notation: In the context of ordinals, we will often use < to denote2. For example, if ↵ and � are ordinals, ↵ < � means ↵ 2 �.

The Burali–Forti Paradox indicates that there should be many ordi-nals. Here is one:

Fact 4.16. ; is an ordinal.

The following fact shows how to generate the least ordinal biggerthan a given ordinal.

Fact 4.17. If ↵ is an ordinal, then S(↵) = ↵ [ {↵} is an ordinal.


Proof. If y 2 S(↵), then either y 2 ↵ or y = ↵. In the first case,y ✓ ↵ [ S(↵). In the second case, y = ↵ ✓ ↵ [ {↵}. Hence S(↵) istransitive.

Every member of S(↵) is either a member of ↵ or is ↵, and hence is anordinal and therefore transitive. It follows that 2 |S(↵) is a transitiverelation, and it can be shown similarly that it is linear.

Finally, ifX ✓ ↵[{↵} is nonempty andX\↵ 6= ;, then a 2–minimalmember of X \ ↵ (which exists since ↵ is an ordinal) is 2–minimal inX. The other case is when X = {↵}. Then ↵ is 2–minimal. ⇤Definition 4.18. An ordinal is a successor ordinal if and only if it isof the form S(x). It is a limit ordinal if and only if it is not a successorordinal (so, ; is a limit ordinal).

Definition 4.19. A natural number is an ordinal which is either ;or a successor ordinal and such that all its members are either ; or asuccessor ordinal.

A set is finite if and only if it bijectable with a natural number.

Notation: Given any ordinal ↵, ↵ + 1 = S(↵).The set of all natural numbers is denoted by !. ! exists by the

Axiom of Infinity.! is an ordinal since it is a transitive set of ordinals. It is the least

nonzero limit ordinal.!+1 = S(!) = ![{!}, (!+1)+1 = S(!+1) = (![{!})[{![{!}},

etc. are successor ordinals.We will automatically view ordinals ↵ as embedded with the relation

2 |↵ well–ordering them.

Lemma 4.20. Let (L,) be a well–order and ↵ an ordinal. Then thereis at most one order–isomorphism f : (L,) �! (↵,2).

This lemma is immediate since the composition of order–isomorphismsis an order–isomorphism, the inverse of an order–isomorphism is anorder–isomorphism, and since the identity is the only order–isomorphismbetween (↵,2) and itself.

Theorem 4.21. Every well–order (L,) is order–isomorphic to a uniqueordinal.

Proof. By what we have seen it su�ces to prove that (L,) is order–isomorphic to some ordinal.

Suppose, for a contradiction, that

{y 2 L : pred((L,), y) 6⇠= (↵,2) for any ↵ 2 Ord} 6= ;

28 D. ASPERO

and let

x = min{y 2 L : pred((L,), y) 6⇠= (↵,2) for any ↵ 2 Ord}By the lemma, for all z < x let ↵

z

be the unique ordinal such that(pred((L,), z),) ⇠= (↵

z

,2) and let

fz

: (pred((L,), z),) �! (↵z

,2)be the corresponding unique order–isomorphism. Then, again by thelemma, if z < z0 < x, then ↵

z

2 ↵z

0 and fz

= fz

0 � pred((L,), z).Assume max(pred((L,), x) does not exist (the proof in the other

case is similar [Exercise]). Let now

f : (pred((L,), x),) �! Ord

be given by f(y) = fy

0(x) for any y0 such that y < y0 < x. This functionis well–defined by the above and it is easy to see that it is an order–isomorphism between pred((L,), x) and (range(f),2). But range(f)is a set, by Replacement, and it is transitive. Therefore it is an ordinal.This contradicts the choice of x. We thus have that for every x 2 Lthere is a unique ordinal ↵

x

such that there is an order–isomorphism

fx

: pred(L,), x) �! (↵,2),and this isomorphism is unique.

Now, arguing as above, we can glue together all these isomorphismsinto an isomorphism f : (L,) �! (X,2), where X is a transitive setof ordinals and therefore an ordinal. ⇤Exercise 4.2. Complete the proof of Theorem 4.21

Given a well–order (L,), the unique ordinal ↵ such that (L,) ⇠=(↵,2 |↵) is the order type of (L,), denoted ot((L,)).

Many sets can be well–ordered in di↵erent ways (so that the corre-sponding well–orders have di↵erent order types). For example, ! canbe well–ordered by 2 in order type !. And it can be well—ordered byputting 0 on top of every n > 0 and well-ordering ! \ {0} according to2. This well–order has order type ! + 1.

Exercise 4.3. Characterize the sets that can be well–ordered in di↵er-ent ways.

5. Cardinals

We have seen the ordinals ! + 1, (! + 1) + 1, ((! + 1) + 1) + 1, etc.,aka !, !+1, !+2, etc. The set consisting of all natural numbers and! + n for every n < ! is a transitive set of ordinals and therefore also


an ordinal. It is called ! + !. We can then build (! + !) + 1, and soon. All these ordinals are countable (i.e., they are bijectable with !).

Question 5.1. Is there an infinite ordinal which is not bijectible with!?

Definition 5.2. A cardinal is an ordinal such that is not bijectiblewith any ordinal ↵ < .

So, each natural number is a cardinal, ! is a cardinal, but no !+ n,is a cardinal. And the same goes for ! + !, (! + !) + 1, etc.

When regarded as a cardinal, ! is also denoted @0 (so N = ! = @0).Notation: If X is bijectable with a cardinal , we say that is the

cardinality of X and write |X| = .We will see that ZFC proves that every set is bijectable with an

ordinal. Caveat : In a context without AC one can extend the notion ofcardinals to things that are not ordinals in a perfectly meaningful way.We don’t need to do that for the moment. So, for us, at least for themoment, cardinals are ordinals. Cardinals, in our sense, are sometimesalso called ‘alephs ’.

Definition 5.3. !1, also denoted @1, is the first uncountable cardinal(in other words, the first infinite cardinal not bijectable with !).

Proposition 5.4. !1 exists.

Proof. Say that X ✓ ! codes a well–order if

{(n,m) 2 ! ⇥ ! : 2n+13m+1 2 X}

is a well–order.Every infinite initial segment of a well–order coded by a subset of !

can be coded by a subset of !. Hence � = {↵ : ↵ = ot() for some coded by some X ✓ !} is transitive and is a set since it is range(F ),where F : P(!) �! Ord is the function sending X to ot() if X codes and to 0 otherwise. Hence � is an ordinal.� is not countable: If f : � �! ! were a bijection, {2n+13m+1 :

f�1(n) 2 f�1(m)} would be a subset of ! coding a well–order of ordertype �. But then � 2 �, which is impossible for ordinals. ⇤Exercise 5.1. Prove that the ordinal � in the above proof is precisely!1; that is,

!1 = {↵ : ↵ = ot() for some coded by some X ✓ !}

Similarly, one can prove in ZF that there is a least cardinal strictlybigger than !1. It is called !2, or @2. In general, we define:

30 D. ASPERO

Definition 5.5. Given an ordinal ↵, @↵

, also denoted !↵

, is the ↵–thinfinite cardinal.

Definition 5.6. Given a cardinal , +, the successor of , is the leastcardinal strictly bigger than .

Hence, (@0)+ = @1, (@1)+ = @2, and in general, (@↵

)+ = @↵+1.

Proposition 5.7. (ZF) For every infinite cardinal , + exists.

Proof. Similar to the proof that @1 exists:Every initial segment R0 of a well–order R ✓ ⇥ is a well–order

and of course R0 ✓ ⇥ . Hence,

� = {↵ : ↵ = ot(R) for some well–order R ✓ ⇥ }is transitive and is a set since it is range(F ), where F : P(⇥) �! Ordis the function sending X to ot(X) if X is a well-order on , and to 0otherwise. Hence � is an ordinal.� is not bijectible with : If f : �! � were a bijection,

{(↵0,↵1) 2 ⇥ : f(↵0) 2 f(↵1)}would be a well–order of order type �. But then � 2 �, which isimpossible for ordinals. ⇤Exercise 5.2. Prove that

+ = {↵ : ↵ = ot(R) for some well–order R ✓ ⇥ }Exercise 5.3. We have seen that @0 exists, and so does @1 and, ingeneral, @

n

for every n. Prove thatS{@

n

: n < !} is the least cardinalbigger than @

n

for all n < !. Hence @!

=S{@

n

: n < !}. In general,prove that if X is a set of cardinals, then

SX is a cardinal and is the

least cardinal � such that � � for every 2 X.

We will need the following notion later on.

Regular cardinals: Let (P,) be a partial order and let X ✓ P .X is cofinal i↵ for every a 2 P there is some b 2 X such that a b.

Definition 5.8. An ordinal is regular if and only if there is no ↵ < for which there is a function f : ↵ �! with range cofinal in . Anordinal is singular i↵ it is not regular.

Exercise 5.4. Prove that:

(1) Every regular ordinal is a cardinal.(2) 0 and 1 are the only regular natural numbers.(3) ! is regular.(4) !1 is regular in ZFC.(5) @

!

is a singular cardinal.


5.1. The Cantor–Bernstein–Schroder Theorem.

Theorem 5.9. (Cantor–Bernstein–Schroder Theorem) (ZF) For allsets X and Y , the following are equivalent.

(1) |X| |Y | and |Y | |X|.(2) |X| = |Y |

Proof. The implication from (2) to (1) is of course trivial, so we onlyneed to prove that (1) implies (2). For this, let f : X �! Y andg : Y �! X be injective functions. By replacing if necessary X andY by, for example, X ⇥ {0} and Y ⇥ {1}, respectively, we may assumethat X and Y are disjoint in the first place. Given c 2 X [ Y , let�c

be the ✓–maximal sequence with domain included in Z such that�c

(0) = c, �c

(z + 1) = f(�c

(z)) or �c

(z + 1) = g(�c

(z)) depending onwhether �

c

(z) 2 X or �c

(z) 2 Y , such that �c

(z � 1) = c if c 2 X issuch that f(c) = �

c

(z) (if �c

(z) 2 Y and if there is such a c), and suchthat �

c

(z � 1) = c if c 2 Y is such that g(c) = �c

(z) (if �c

(z) 2 Xand if there is such a c). We will call �

c

the orbit of c. We say that�c

starts in X if there is some z 2 Z in the domain of �c

such that�c

(z) 2 X and such that there is no c 2 Y with g(c) = �c

(z) (so �c

(z)is the first member of �

c

). Similarly we define ‘�c

starts in Y ’. Andwe say that �

c

does not start in the remaining case (i.e., if and only ifdom(�

c

) = Z). We also say that a set � is an orbit if � is (the rangeof) the orbit of some c 2 X [ Y in the above sense.

The first observation is that every two distinct orbits are disjoint andthat the orbits partition X [ Y . The second observation is that if � isan orbit, then

• f � � is a bijection between � \X and � \ Y if � starts in X,• g � � is a bijection between � \ Y and � \X if � starts in Y ,and

• f � � is a bijection between � \ X and � \ Y and g � � is abijection between � \ Y and � \X if � does not start.

Using these two observations we can now define a bijection h : X �!Y by ‘gluing together’ suitable restrictions of f and/or of the inverseof g: Given a 2 X, if the unique orbit to which a belongs starts in Xor does not start, then h(a) = f(a). And if this orbit starts in Y , leth(a) be the unique b 2 Y such that g(b) = a. ⇤

As we have seen in this proof, if f : X �! Y and g : Y �! Xare injective functions, then there is a bijection h : X �! Y thatcan be e↵ectively constructed from f and g. For example, let f bethe identity on {2n : n 2 !} and let g : ! �! {2n : n 2 !}

32 D. ASPERO

be given by g(n) = 4n. These are injective non–surjective functionsbetween ! and {2n : n 2 !}, and the above proof produces a bijectionh : ! �! {2n : n 2 !} that can be e↵ectively constructed from f andg.

Let us consider the following statement:Dual C–B–S : For all sets X, Y , the following are equivalent:

(1) |X| = |Y |(2) There is a surjection f : X �! Y and there is a surjection

g : Y �! X.

Proposition 5.10. (ZFC) Dual C–B–S is true.

Proof. Suppose (2) holds. Using AC we find functions f : Y �! Xand g : X �! Y as follows:

• For every b 2 Y , f(b) is some a 2 X such that f(a) = b.• For every a 2 X, g(a) is some b 2 Y such that g(b) = a.

Then f and g are injective functions, so by C–B–S, |X| = |Y |. ⇤The following question is apparently open.

Question 5.11. It is not known whether or not, modulo ZF, DualC–B–S is equivalent to the Axiom of Choice.

5.2. Countable and uncountable sets.

Definition 5.12. A set X is countable if and only if |X| = @0. A setis uncountable if it is not finite or countable.

Let us see some examples of countable sets.

Proposition 5.13. (ZF) The following sets are countable.

• ! ⇥ !; in general, n! := {s : s : n �! !} for any n 2 !,n � 1. (the set n! is often denoted !n.)

• <!! :=S

n2!n!. (the set <!! is also denoted !<!.)

• [!]n := {X ✓ ! : |X| = n} for any n 2 !, n � 1.• [!]<! = {X ✓ ! : X finite}• Z• Q• The set of algebraic numbers (x 2 C is algebraic if and only ifit is a root of a polynomial with rational coe�cients).

Proof. We can produce the corresponding bijections h : X �! ! (orh : ! �! X) by showing that there are one–to–one functions f : X �!! and g : ! �! X and then appealing to C–B–S. In many cases theexistence of at least one of these one–to–one functions is immediate.


An injective f : ! ⇥ ! �! ! is given for example by f(n,m) =2n+13m+1 (we used this coding in the proof of Proposition 5.4). Theexistence of a bijection between n! and ! can be proved by induction onn since |n+1!| = |(n!)⇥ !|. This gives a definable sequence (f

n

)1n<!

where fn

: ! �! n! is a bijection for all n. We can then find abijection f : ! ⇥ ! �! <!! by, for example, sending (0, 0) to ; andsending (n,m) 6= (0, 0) to f

n+1(m). Since there is a bijection g : ! �!! ⇥ !, the composition f � g : ! �! <!! is a bijection. To see thatthere is a bijection h

n

: [!]n �! ! for every n � 1, send x 2 [!]n

to f�1n

((x0, . . . , xn�1)), where (x0, . . . , xn�1) is the strictly increasingenumeration of x. We can also code all (the inverses of) these bijectionstogether into a bijection h : ! �! [!]<! exactly as in the proof of|<!!| = |!|.

Using the above bijections and any of the usual representations of Zand Q (as, say, pairs of natural numbers and pairs of integers, respec-tively), we can easily build bijections between ! and Z and between! and Q. Using also the above bijections, we can well–order all poly-nomials with coe�cients in Q in length !. Once this is done, we caneasily find a bijection between a subset of !⇥! and the set of algebraicnumbers (which gives what we want by C–B–S): Given (n, k), if p(x) isthe n–th polynomial with rational coe�cients and p(x) has at least kdistinct roots, then we send (n, k) to the k–th root of p(x) in (say) thelinear order <

lex

of C given by a0 + ib0 <lex

a1 + ib1 i↵ either a0 < a1or else a0 = b1 and b0 < b1 (where < refers to the usual order on thereal line). ⇤

Proposition 5.14. (ZFC) The union of every countable collection ofcountable sets is countable: If (X

n

)n2! is such that each X

n

is count-able, then

Sn2! Xn

is countable.

Proof. @0 |S

n2! Xn

| is clear: There is a bijection f : ! �! X0, andf : ! �!

Sn

Xn

is an injection.|S

n

Xn

| @0: For every n < ! pick, using the Axiom of Choice, abijection f

n

: Xn

�! ! (i.e., let X = {Fn

: n 2 !} where for each n,F

n

is the set of all pairs (n, f), where f : Xn

�! ! is a bijection, letG be a choice function for X, and let f

n

= f if G(Fn

) = (n, f)).Now let F :

Sn2! Xn

�! ! ⇥ ! be the function sending x to(n, f

n

(x)) if n is first k < ! such that x 2 Xk

. F is an injection,and if g : ! ⇥ ! �! ! is a bijection (which exists since |! ⇥ !| = @0),then g � F :

Sn

Xn

�! ! is an injection.Since |

Sn

Xn

| @0 and @0 |S

n

Xn

|, by C–B–S we get |S

n

Xn

| =@0. ⇤

34 D. ASPERO

Some form of Choice is necessary in the above proposition. In fact,this proposition is not necessarily true without the Axiom of Choice: IfZF is consistent, then there are models of ZF in which !1 is a countableunion of countable sets (!)

Let us see some examples of uncountable sets now:

• !1, and in fact all ordinals ↵ � !1.• P(!) (by Cantor’s Theorem 1.2).

Proposition 5.15. |R| = |P(!)|. In particular, R is uncountable.

Proof. Let I be the closed–open interval [0, 1) ✓ R. Since of course|[0, 1)| |R|, by C–B–S it su�ces to show |P(!)| |[0, 1)| and |R| |P(!)|.

Let f : P(!) �! [0, 1) send X ✓ ! to ⌃n2!

✏n2n+1 , where ✏n = 0 if

n /2 X and ✏n

= 1 if n 2 X (i.e., (✏n

)n2! is the characteristic function

of X).Let h : Q �! ! be a bijection and let g : R �! P(!) send x 2 R to

{h(q) : q < x} (this < is of course the natural order on R).f and g are injective functions, so by C–B–S, |[0, 1)| = |R| = |P(!)|.

⇤Remark: Even if |Q| = @0 < |P(!)| = |R|, the rationals are dense in

the reals, i.e., between every two reals there is some (in fact, infinitelymany) rationals (!)

Exercise 5.5. |C| = |P(!)|.

Hence, since the set of algebraic numbers is countable, most complexnumbers are transcendental (i.e., non–algebraic). In fact

|{x 2 C : x transcendental}| = |C| = |R| = |P(!)|

5.3. Almost disjoint families. Note that a collection of pairwise dis-joint subsets of ! has to be finite or countable. We even have thefollowing.

Exercise 5.6. Let A ✓ P(!) and suppose n < ! is such that |a\b| nfor all distinct a, b 2 A. Then A is finite or countable.

Definition 5.16. Two sets X, Y are almost disjoint if X \Y is finite.

We have seen that |!| < |P(!)|. Therefore, the following might looksurprising.

Theorem 5.17. (ZF) There is a collection A ✓ P(!) of size 2@0 con-sisting of pairwise almost disjoint sets.


Proof. Let <!2 be the complete binary tree of height !, that is, thetree of n–sequences of 0’s and 1’s, for n < !. By what have seen, <!2is countable. In fact, there is a simple enumeration of the nodes of T ,where the first member is ;, the next two members are h0i and h1i, thenext fours members are h0, 0i, h0, 1i, h1, 0i and h1, 1i, the next eightmembers are the sequences with exact three members, and so on. Letf : ! �! <!2 be such a bijection.

Now consider any two distinct infinite branches b, b0 through <!2and note that b\ b0 is finite; in fact, if n is the first position where theydisagree, then they have all nodes b � k, for k n, in common, but haveno other nodes in common. Also, there are as many infinite branchesthrough <!2 as there are subsets of !. In fact, there is obviously abijection g from P(!) into the set of such branches sending X ✓ !to the characteristic function �

X

(i.e., the function sending n to 1 ifn 2 X and to 0 if n /2 X).

It follows that {g�1[b] : b an infinite branch through <!2} is a sub-set of P(!) consisting of 2@0–many pairwise almost disjoint sets. ⇤

5.4. Cantor’s Continuum Hypothesis.

Definition 5.18. (ZFC) If is a cardinal, |P()| is denoted by 2.

In particular, |R| = 2@0 . We have seen that @0 < 2@0 (Cantor’sTheorem), and therefore @1 2@0 by definition of @1 as the leastuncountable cardinal (we need the Axiom of Choice to conclude thatthere is an injection from !1 into R; without AC this is not true ingeneral!). The following is therefore a very natural question.

Question 5.19. Is @1 = 2@0? In other words, if X ✓ R is uncountable,does it follow that |X| = |R|?

This is perhaps the most famous question in set theory. Georg Can-tor was certainly obsessed with it, and it is the first question on thefamous list of problems that David Hilbert presented at his address atthe International Congress of Mathematics in 1900 in Paris. We willsee a “solution” later on.

Definition 5.20. Cantor’s Continuum Hypothesis (CH): 2@0 = @1.

5.5. AC vs. the Well–Ordering Principle.

Theorem 5.21. (ZF) The following are equivalent.

(1) AC(2) The Well–ordering Principle: Every set can be well–ordered.

36 D. ASPERO

Proof. Suppose AC holds. Let X be a set. Let f be a choice functionfor P(X) \ {;}. We define enumerations (x

↵

: ↵ < �) of subsets of Xby recursion on the ordinals in such a way that (x

↵

: ↵ < �) = (x↵

:↵ < �0) � � for all � < �0, as follows: Let � be an ordinal and suppose(x

↵

: ↵ < �) has been defined. If {x↵

: ↵ < �} = X, then we aredone. Otherwise X \ {x

↵

: ↵ < �} 6= ;. Setx�

= f(X \ {x↵

: ↵ < �})This defines (x

↵

: ↵ < � + 1). If � is a limit ordinal we let

(x↵

: ↵ < �) =[

{(x↵

: ↵ < �0) : �0 < �}

This gives a class–function F from P(X) to the ordinals, sending Y ✓X to � if Y = X \ {x

↵

: ↵ < �}. Since P(X) is a set, by Replacementrange(F ) is a set of ordinals so it cannot be all of Ord. Hence thisconstruction has to stop at some point (there must be � such thatX \ {x

↵

: ↵ < �} = ;). But then {(x↵

, x↵

0) : ↵ 2 ↵0 2 �} is awell–order of X.

Now assume the Well–ordering Principle. Let X be a set consistingof nonempty sets and let be a well–order of

SX. Now, given a 2 X

let f(a) be the –minimal element of a. Then f is a choice functionfor X. ⇤

We have used recursion on the ordinals to define (x↵

: ↵ < �) inthe first part. Later we will see that this can be done. Read again thisproof then.

Corollary 5.22. (ZF) The following are equivalent:

(1) AC(2) Every set is bijectable with a unique ordinal.

Corollary 5.23. (ZF) The following are equivalent:

(1) AC(2) (Trichotomy for sets) Given any two sets X, Y , exactly one of

the following holds.• |X| = |Y |• |X| < |Y |• |Y | < |X|

Thus, the Axiom of Choice has the following counterintuitive conse-quence: R, and even C, can be well–ordered (of course in length 2@0).This well–order has to be highly non–constructive. In fact there aremodels of ZF in which such a well–order does not exist. The fact thatR can be well–ordered enables one to construct rather pathologicalobjects (Banach–Tarski decompositions, etc.).


Example: A non–Lebesgue measurable set: Let ⌘ be the equivalencerelation on (0, 1) ✓ R given by x ⌘ y if and only if x�y 2 Q. Let f bea choice function for the quotient set (0, 1)/ ⌘ (i.e., f picks an elementout of each equivalence class of ⌘). Then range(f) is not Lebesguemeasurable. It’s called a Vitali set. The reason why X := range(f) isnot Lebesgue measurable is the following. Suppose X were measurableand let r = µ(X). Suppose r > 0. For every q 2 Q let q+X = {q+x :x 2 X}. Then (q+X)\ (q0+X) = ; whenever q 6= q0 are in Q. Henceµ(S

q2Q\(0,1)(q +X)) =P

q2Q\(0,1) µ(X) since

(1) µ is translation invariant (i.e., µ(x+ Y ) = µ(Y ) for every mea-surable Y ✓ R, and

(2) µ is �–additive, meaning that µ(S

n<!

Xn

) =P

n<!

µ(Xn

) when-ever (X

n

)n<!

is a countable sequence of pairwise disjoint mea-surable sets.

It follows that µ(S

q2Q\(0,1)(q + X)) =P

q2Q\(0,1) r = 1. On theother hand, µ(

Sq2Q\(0,1)(q + X)) < 1 since

Sq2Q\(0,1)(q + X)) ✓

(0, 2) and µ((0, 2)) = 2. That is a contradiction, so µ(X) = 0. ButSq2Q\(0,1)(q + X) turns out to be exactly (0, 2) [check]. If µ(X) = 0,

then µ((0, 2)) =P

q2Q\(0,1) µ(q +X) =P

q2Q\(0,1) 0 = 0, which is alsoa contradiction since µ((0, 2)) = 2. It follows that X cannot be mea-surable.

There are extensions of ZF incompatible with AC and which rule outsuch pathological consequences of AC as the Banach–Tarski “paradox”,non–Lebesgue measurable sets, etc. These extensions of ZF say that allsets of reals have nice regularity properties and therefore seem to reflectbetter our intuitions about such sets than ZFC. One such extension isZF + “The Axiom of Determinacy”. Su�ciently strong large cardinalaxioms (these are natural axioms extending ZFC, and in fact the Axiomof Infinity can be regarded as one such axiom) actually imply

L(R) |= ZF+ “The Axiom of Determinacy”,

where L(R) is the minimal inner model of ZF containing all the ordinalsand all the reals.

6. Foundation, recursion and induction. The cumulativehierarchy

We have seen recursive definitions, for example when we proved|n!| = |!| for all n 2 !, n � 1 (this was a recursion on !), or whenwe proved The Well–ordering Principle from the Axiom of Choice (thiswas a recursion on the ordinals).

38 D. ASPERO

Also, many familiar definitions are by recursion: For example n! isdefined by

• 0! = 1• (n+ 1)! = n!(n+ 1)

Another example: For a given n 2 !, we can define the functionf : ! �! ! given by f(x) = n + x (in other words, we can definen+m) as follows:

• n+ 0 = n• n+ (m+ 1) = (n+m) + 1

In fact, for a given ordinal ↵, we define ↵ + � by recursion on theordinals by:

• ↵ + 0 = ↵• ↵+S(�) = S(↵+�) = (↵+�)+1 (recall: �+1 = S(�) = �[{�}by definition).

• ↵ + � =S{↵ + � : � < �} if � is a nonzero limit ordinal.

Note: + is not commutative: ! + 1 6= ! = 1 + ! !Two more examples: @

↵

is defined, by recursion on the ordinals, by

• @0 = !• @

S(↵)(= @↵+1) = (@

↵

)+

• @�

=S{@

�

: � < �} if � is a nonzero limit ordinal.

In ZFC, for every cardinal , i↵

() is defined, by recursion on theordinals, by

• i0() = • i

↵+1() = 2i↵() (= |P(i↵

())|)• i

�

() =S{i

�

() : � < �} for every nonzero limit ordinal �.

Notation: If = !, we write i↵

for i↵

().One last example:

Definition 6.1. We define (V↵

: ↵ 2 Ord) as follows:

• V0 = ;• V

↵+1 = P(V↵

)• V

�

=S{V

�

: � < �} if � is a limit ordinal.

(V↵

: ↵ 2 Ord) is called the cumulative hierarchy.

• V0 = ;• V1 = {;} = 1• V2 = {;, {;}} = 2• V3 = {;, {;}, {{;}}, {{;, {;}}}• |V4| = 24 = 16• |V5| = 216 = 65536


• |V6| = 265536 (which, according to Wikipedia, is much biggerthan the number of atoms of the observable universe!)

• |V7| = 2(265536)

• ...• |V

!

| = @0

• |V!+1| = 2@0 = i1

• |V!+2| = 2i1 = i2

• For every ordinal ↵, |V!+↵

| = i↵

.

Also: We prove things by induction on the ordinals: Let P (x) be afirst-order property. Suppose the following.

(1) P (0) holds.(2) For every ordinal ↵ > 0, if P (�) holds for every ordinal � < ↵,

then P (↵) holds.

Then P (↵) holds for every ordinal ↵.

Example: ⌃kn

k = n(n+1)2

for every n < !.Another example:

Proposition 6.2. (ZF) For every ordinal ↵, V↵

is transitive.

Proof. V0 = ; is transitive.Let ↵ > 0 be an ordinal and suppose V

�

is transitive for every � < ↵.Suppose ↵ is a successor ordinal, ↵ = �+1. Then V

↵

= V�+1 = P(V

�

).Let x 2 V

↵

and let y 2 x. Then x ✓ V�

. It follows that y 2 x ✓ V�

and therefore y 2 V�

. Since V�

is transitive by induction hypothesis,y ✓ V

�

. But then y 2 P(V�

) = V↵

.Finally suppose ↵ > 0 is a limit ordinal. Then V

↵

=S

�<↵

V�

. Let

y 2 x 2 V↵

. Then there is some � < ↵ such that x 2 V�

. Since V�

istransitive by induction hypothesis, y 2 V

�

. But then y 2S

�<↵

V�

=V↵

. ⇤

Why is it ok to make definitions by recursion on the ordinals and toprove things by induction on the ordinals?

Induction is easy: Note that if X is a nonempty class of ordinals,then min(X) exists. Now, suppose P (x) is a first–order property suchthat for every ordinal, if P (�) holds for all � < ↵, then P (↵). We wantto see that P (↵) for all ordinals ↵. Suppose towards a contradictionthat

X = {↵ 2 Ord : ¬P (↵)} 6= ;Let ↵ = min(X). For every ordinal � < ↵, � /2 X by definition ofmin(X). But then P (�). Hence we have that P (�) for all � 2 ↵.

40 D. ASPERO

Therefore P (↵) by our assumption. So ↵ /2 X. This is a contradictionsince ↵ = min(X) 2 X.

What about definitions by recursion?

Theorem 6.3. (ZF) (Recursion (meta)–theorem)

Let G(x, y) be a class–function. Then there is a unique class functionF defined on Ord such that for every ordinal ↵,

F (↵) = G(↵, F � ↵)Proof. We prove, by induction on the ordinals, that for every ordinal ↵there is a unique function f with domain ↵ such that for every � < ↵,

f(�) = G(�, f � �)We show uniqueness first and then existence.

Uniqueness: Suppose f0 and f1 are distinct functions with dom(f0) =dom(f1) = ↵ such that f0(�) = G(�, f0 � �) and f1(�) = G(�, f1 � �)for every � < ↵. Since f0 6= f1, let

� = min{� 2 ↵ : f0(�) 6= f1(�)}Then f0 � � = f1 � �. But then

f0(�) = G(�, f0 � �) = G(�, f1 � �) = f1(�)

Contradiction.Existence: Let ↵ be an ordinal. For every � < ↵ let f� be the unique

function h with domain � such that h(�) = G(�, h � �) for all � < �(which exists by induction hypothesis).

One can easily prove by induction on � < ↵ that if �0 < �, thenf�

0= f� � �0 ([Exercise]).

If ↵ is a limit ordinal, then f↵ =S{f� : � < ↵} is as desired by

the previous line.

If ↵ = ↵ + 1, letf = f ↵ [ {(↵, G(↵, f ↵))}

Let now � < ↵. If � < ↵, then

f(�) = f ↵(�) = G(�, f ↵ � �) = G(�, f � �)If � = ↵, then

f(↵) = G(↵, f ↵) = G(↵, f � ↵)since f ↵ = f � ↵ by definition of f . ⇤Exercise 6.1. Complete the proof of Theorem 6.3


Example: The class–function F sending ↵ 2 Ord to V↵

is such thatF (↵) = G(↵, F � ↵), where G(x, y) is:

• ; if x = 0 or if x is not an ordinal.• P(y(x)) if x is the successor ordinal x+ 1.•S

range(y) if x is a nonzero limit ordinal.

We have seen, by induction on the ordinals, that V↵

is transitive forevery ordinal ↵.

Proposition 6.4. For all ↵ < �, V↵

✓ V�

.

Proof. Again by induction on �. This is vacuously true for � = 0. For� a nonzero limit ordinal, V

�

◆ V↵

by definition of V�

. For � = � + 1,V�

= P(V�

). If ↵ = �, then we are done since every member of V�

is a subset of V�

(as V�

is transitive) and therefore V�

✓ P(V�

). If↵ < �, then V

↵

✓ V�

by induction hypothesis. But V�

✓ P(V�

) by theprevious case, and hence V

↵

✓ P(V�

) = V�

. ⇤Definition 6.5. For every x 2

S↵2Ord V↵

,

rank(x) = min{↵ 2 Ord : x 2 V↵+1}

Definition 6.6. For every set x, the transitive closure of x, denotedby TC(x), is

S{X

n

: n < !} where

• X0 = x• X

n+1 =S

Xn

So TC(x) = x [S

x [SS

x [SSS

x [ . . .

Exercise 6.2. Prove that TC(x) is the ✓–least transitive set y suchthat x ✓ y. In other words, TC(x) =

T{y : y transitive, x ✓ y}.

Let us fix some notation now for the set–theoretic universe and forS↵2Ord V↵

.

Definition 6.7. V denotes the class of all sets; that is,

V = {x : x = x}

Definition 6.8. WF =S{V

↵

: ↵ 2 Ord}: The class of all x suchthat x 2 V

↵

for some ordinal ↵.

In the above definition, WF stands for “well–founded”.Note: WF is a transitive class: y 2 x 2 V

↵

implies y 2 V↵

since V↵

is transitive.

Theorem 6.9. (ZF) V = WF

42 D. ASPERO

Proof. Suppose, towards a contradiction, that there is some set x suchthat x /2 WF. Let y = TC(x).

y /2 WF: Suppose y 2 V↵

. Since x ✓ y ✓ V↵

(where y ✓ V↵

is trueby transitivity of V

↵

), x 2 P(V↵

) = V↵+1. Contradiction.

By Foundation we may find a 2 y [ {y}, a 2–minimal in y [ {y},such that a /2 WF. For every z 2 a, it follows that z 2 y [ {y} (bytransitivity of y [ {y}) and therefore z 2 V

↵

for some ↵ 2 Ord by 2–minimality of a among {w 2 y [ {y} : w /2 WF}. Hence, the functionrank � a sending z 2 a to rank(z) is defined for all z 2 a. But thenrange(rank � a) has to be a set by Replacement and therefore there issome ordinal ↵ such that ↵ > rank(z) for every z 2 a [No set X ofordinals can be cofinal in Ord (i.e., such that for every ↵ 2 Ord thereis some � 2 X with ↵ < �). Why? Otherwise

SX = Ord, which is

not a set (Burali–Forti), butSX is a set if X is a set by Union Axiom.

Contradiction.]

It follows that for every z 2 a there is some � < ↵ such that z 2V�

✓ V↵

. Hence a ✓ V↵

and therefore a 2 P(V↵

) = V↵+1. Contradic-

tion with a /2 WF. ⇤The fact that V = WF realises the idea that a set is any collection

built out of sets already built. This is known as the iterative conceptionof sets. Note that this conception of sets rules out such “sets” as Vor the Russell class {x : x /2 x}: They couldn’t possibly be sets sinceone needs to refer to the totality of sets for their definition, a totalityto which they would belong if they were sets. Take for example, V.Certainly, if V is a set, then V 2 V. But this goes against the iterativeconception of set, whereby a set is built up out of previously built sets.

The picture of the universe provided by V = WF is a very appealingand very natural one (once one has come across it, at least). Thispicture of the universe of all sets, and the fact that ZF implies V =WF, is the main source of intrinsic justifications of the ZF axioms.

7. Inner models and relativization

Let (M,2M) be a submodel, or inner model, defined by a formula⇥(x); in other words, M = {a : ⇥(a)} and, for all a, b 2 M , a 2M bif and only if a 2 b (we usually leave out 2M and write M instead of(M,2M)). (Examples: V, WF, L, HOD, ...).

We define the relativization to M of a formula '(~x), to be denoted'M(~x), in the following manner.

• (x 2 y)M is x 2 y.• (x = y)M is x = y.


• ('0 _ '1)M is 'M

0 _ 'M

1 .• (¬')M is ¬'M .• ((8x)('(x))M is 8x(⇥(x) ! 'M(x)). We may also write some-thing like (8x 2 M)'M(x).

Note: Suppose T is a theory in the language of set theory. Suppose(N,E) is a structure in the language of set theory, and suppose M isan inner model in N . Then (N,E) |= �M for every � 2 T if and onlyif (M,E) |= T .

Notation: If (N,E) is a structure in the language of set theory, Mis an inner model defined by a formula ⇥(x) possibly with parameters(i.e., M = (N,E � (M ⇥M)), where M = {a 2 N : (N,E) |= ⇥(a)}),and we want / need to emphasise that M is the inner model definedby ⇥(x) as defined within (N,E), then we often write MN instead ofM .

Example: WFM

Note: For every ordinal ↵, V WF↵

= V↵

(here V↵

refers to the set,definable from the parameter ↵, with the definition that we have seen).

Many facts about the universe V are inherited by reasonable sub-models. For example:

Lemma 7.1. Suppose M is a transitive set or a transitive proper class.Then M |= Axiom of Extensionality.

Proof. Let a, b 2 M and suppose M |= (8x)(x 2 a $ x 2 b) (this ofcourse is shorthand for

M |= (8x)(x 2 y $ x 2 z)[~a]

where ~a is any assignment sending the variable y to a and the variablez to b).

This means that a\M = b\M . Since M is transitive (in V), everymember of a or of b is a member of M . It follows that a \M = a andb \M = b and therefore a = b. Hence M |= a = b. In sum, M thinksthat for all y, z, if y and z have the same elements, then they are equal.In other words, M |= Axiom of Extensionality. ⇤

Also:

Lemma 7.2. Suppose M is a transitive set or a transitive proper classwhich is closed under unordered pairs (meaning that for all a, b 2 M ,{a, b} 2 M). Then M |= Axiom of Unordered pairs.

Proof. Let c = {a, b} 2 M . Check, as in the previous proof, thatM |= (8x)x 2 c $ x = a _ x = b. ⇤

44 D. ASPERO

Similarly:

Lemma 7.3. Suppose M is a transitive set or a transitive proper class.Suppose

Sa 2 M for every a 2 M . Then M |= Union set Axiom.

Lemma 7.4. Suppose M is a transitive set or a transitive proper class.Suppose for every a 2 M there is some b 2 M such that b = P(a)\M .Then M |= Power set axiom.

Exercise 7.1. Prove Lemmas 7.3 and 7.4.

Note: There are situations in which there are transitive models Mof fragments of ZFC, or even of all of ZFC, and some a 2 M such thatP(a)M is strictly included in P(a) (i.e., there are subsets b of a suchthat b /2 M).

Lemma 7.5. Suppose M is a transitive set or a transitive proper class.If ! 2 M , then M |= Infinity.

Proof idea: As in the previous proofs. The point is that M recog-nises ; correctly, recognises correctly that something is an ordinal, andrecognises correctly that something is the successor of an ordinal.

We say that the notion of ordinal is absolute with respect to tran-sitive models. It is possible to identify large families of properties thatare absolute with respect to transitive models by virtue of their beingdefinable by syntactically ‘simple’ formulas (from the point of view oftheir quantifiers). We don’t need this kind of general analysis at themoment so we won’t go into that now.

Note: The notion of finiteness is also absolute with respect to transi-tive models but, on the other hand, the notion of countability is highlynon–absolute with respect to transitive models: There are transitivemodels M and a 2 M such that

M |= a is uncountable

but there is a bijection f : ! �! a, so a is countable in V. Theproblem of course is that f is not in M . We will soon see that thereare transitive models of (fragments of) ZFC such that all their sets arecountable in V. And even the whole model can be countable in V.

The notion of choice function is also absolute with respect to tran-sitive models: If M is transitive, a 2 M consists of nonempty sets,f 2 M , and f is a choice function for M , then we have that M |=“f is a choice function for a”. Hence:

Lemma 7.6. Let M be a transitive set or a transitive proper class.Suppose for every a 2 M consisting of nonempty sets there is a choicefunction f for a, f 2 M . Then M |= AC.


Lemma 7.7. Let M be a transitive set or a transitive proper class.Suppose b 2 M whenever a 2 M and b ✓ a is definable over M ,possibly from parameters (in other words, b = {c : c 2 a, M |= '(c, ~p)}for some parameters ~p 2 M). Then M |= Separation.

Lemma 7.8. Let M be a transitive set or a transitive proper class.Suppose F [a] 2 M whenever a 2 M , and F is a class–function over M(in other words, if F is definable by a formula '(x, y, ~z) which, over Mis functional, ~p 2 M , and a 2 M , then {c : (9b 2 a)M |= '(b, c, ~p)} 2M). Then M |= Replacement.

7.1. Our first relative consistency proof: Con(ZF \{Foundation})implies Con(ZF).

Theorem 7.9. Let M |= ZF \{Foundation}. Then M |= �WFMfor

every � 2 ZF. Hence, WFM |= ZF.

Proof. By the previous lemmas and the construction of (V↵

: ↵ 2 Ord),WFM |= � for every axiom � of ZF \{Foundation} [go through theseaxioms one by one them and check thatWF is closed under the relevantoperation, then apply the relevant lemma].

To see thatM |= FoundationWF holds, let us work inM : Let a 2 V↵

,let Z ✓ a, Z 2 WF, and let b 2 Z such that rank(b) = min{rank(z) :z 2 Z}. Then

WF |= rank(b) = min{rank(z) : z 2 Z}by absoluteness of the relevant notions. Hence, WF thinks that therestriction of 2 to a is well–founded. Since this is true for all a 2 WF,

WF |= Foundation

⇤Corollary 7.10. If ZF \{Foundation} is consistent, then ZF is con-sistent.

Proof. Suppose ZF \{Foundation} is consistent. By the completenesstheorem we may find a model M |= ZF \{Foundation}. Let M 0 =WFM . By the theoremM 0 |= ZF. Hence, ZF has a model and thereforeit is consistent. ⇤Remark 7.11. By exactly the same argument, if M is a model ofZFC \{Foundation}, then M |= �WFM

for every � 2 ZFC. Hence,Con(ZFC \{Foundation}) implies Con(ZFC).

Similar relative consistency results: One can define “the constructibleuniverse” L:

46 D. ASPERO

• L0 = ;• L

↵+1 = Def(L↵

), where Def(L↵

) is the set of all subsets of L↵

definable over L↵

possibly with parameters, i.e., the collectionof all sets of the form

{b 2 L↵

: L↵

|= '(b, a0, . . . , an�1)}for some formula '(x, ~x) and a0, . . . , an�1 2 L

↵

.• L

�

=S

↵<�

L↵

if � > 0 is a limit ordinal.

L =S

↵2Ord L↵

.This construction is due to Godel. He proved that if we do this con-struction in ZF, then L |= ZF but also L |= AC and L |= CH.

The above results imply that if ZF is consistent, then ZFC is alsoconsistent, and in fact also ZFC+CH. Linking this to the implicationwe have seen we thus have that if ZF \{Foundation} is consistent, thenso is ZFC+CH.

These relative consistency proofs proceed by building suitable innermodels.13

Most relative consistency proofs proceed, on the other hand, bybuilding suitable outer models of some given ground model. The con-struction of these outer models is done with the forcing method. This isan extremely powerful method in set theory. I will try to say somethingmore specific about this method later on.

7.2. Reflection. Given formulas 0, 1, we say that 0 is a subformulaof 1 if and only if

• 0 = 1, or• there is a subformula of 1 such that 0 is a subformula of , or

• either 1 = ¬ 0 or 1 = 0 _ for some or 1 = _ 0 forsome or 1 = (9x)('0) for some variable x.

Theorem 7.12. (ZF) (Levy–Montague Reflection) Let '(x0, . . . , xn�1)be a formula in the language of set theory. There is a proper class C

'

of ordinals such that:

(1) for all ↵ 2 C'

and all a0, . . . , an�1 2 V↵

,

V↵

|= '(a0, . . . , an�1)

if and only if '(a0, . . . , an�1) is true (we may also write V |='(a0, . . . , an�1)). We write V

↵

4'

V.

13What set–theorists understand by “inner models” are usually much more com-plicated than WF or L. However, the construction of L is in fact the paradigm formost of these more complicated constructions.


(2) For every � 2 Ord there is some ↵ 2 C'

such that � < ↵ (C'

is unbounded).(3) For every limit ordinal �, if sup(C

'

\ �) = �, then � 2 C'

(C'

is closed).

Proof. Let ('i

: i < n) list all subformulas of ' in such a way that if'i

is a subformula of 'i

0 , then i < i0.We build a ✓–decreasing sequence C

'i of closed and unboundedproper classes of ordinals (for i < n) as follows.

Let i < n and suppose C'i�1 defined if i > 0. If i = 0 (so '

i

is anatomic formula) then C

'i = Ord. If i > 0 but 'i

6= (9x)('i

0) for anyi0 < i and any variable x, then C

'i = C'i�1 .

Finally, suppose i > 0 and 'i

= (9x)'(x, x0, . . . , xm�1) for somei0 < i and some variable x. We define C

'i = {↵⇠

: ⇠ 2 Ord}, where(↵

⇠

)⇠2Ord is a strictly increasing and continuous – i.e., at limit stages ⇠

we let ↵⇠

= sup⇠

0<⇠

↵⇠

0 – sequence of ordinals defined as follows:

Given an ordinal ⇠, if ↵⇠

has been defined let �⇠

0 = ↵⇠

and let �⇠

1 bethe least � > ↵

⇠

, � 2 C'i�1 , such that for all a0, . . . , am�1 2 V

↵⇠, if

there is some b such that '(b, a0, . . . , am�1), then there is some b 2 V�

such that '(b, a0, . . . , am�1). In general, if �⇠

n

has been defined, let �⇠

n+1

be the least � > �⇠

n

, � 2 C'i�1 , such that for all a0 . . . , a

m�1 2 V�

⇠n, if

there is some b such that '(b, a0, . . . , am�1), then there is some b 2 V�

such that '(b, a0, . . . , am�1).

Let � = supn<!

�⇠

n

, and note that � 2 C'i�1 since C

'i�1 is closed.Let ↵

⇠+1 = �.The construction of ↵0 of course is as above, starting from 0 instead

of ↵⇠

.Since V

↵

4'i0 V for all ↵ 2 C

'i�1 , it follows that V↵

4'i V for all

↵ 2 C'i :

Suppose ↵ 2 C'i and a0, . . . , am�1 2 V

↵

. If there is some b 2 V↵

suchthat V

↵

|= '(b, a0, . . . , am�1), then of course V |= 'i

0(b, a0, . . . , am�1)since V

↵

4'i0 V. And if V |= (9x)'

i

0(x, a0, . . . , am�1), then by con-struction of C

'i there is some b 2 V↵

such thatV |= 'i

0(b, a0, . . . , am�1).But then V

↵

|= 'i

0(b, a0, . . . , am�1) since V↵

4'i0 V. ⇤

Corollary 7.13. Suppose '0, . . . ,'n

are finitely many formulas in thelanguage of set theory. Then there is a proper class C of ordinals suchthat:

(1) For every ↵ 2 C and every j n, V↵

4'j V.

(2) For every � 2 Ord there is some ↵ 2 C such that � < ↵.(3) For every limit ordinal �, if sup(C \ �) = �, then � 2 C.

48 D. ASPERO

Proof. Simply note that C'0 \ . . . \ C

'n is a closed and unboundedclass of ordinals. It is clearly closed. To see that it is unbounded, given� 2 Ord let (↵

i

)i<!

be such that ↵0 > � and such that for all i < !and all j n there is some � 2 C

'j , ↵i

< � < ↵i+1. Then, for every

j n, supi<!

↵i

is a limit of ordinals in C'j and therefore in C

'j . ⇤Moral: The universe cannot be characterised in a first order way:

Every first order fact about it is true about some initial segment of it.

Does reflection hold for infinite sets of sentences in the language ofset theory? For all of them?

Warning: ‘Reflection for all sentences’ cannot be expressed in thefirst order language of set theory. (Tarski’s definability of truth: Thereis no formula '(x) such that for every sentence �, (N,+, ·) |= � if andonly if (N,+, ·) |= '(]�), where ]� is a code for �.)

8. Trees

From now on we will focus on combinatorial aspects of set theory,with an eye on their connections with the theory of large cardinals. Westart o↵ with the basic theory of trees in this section.

Definition 8.1. (T,) is a tree if and only if:

(1) (T,) is a partial order (i.e., is a reflexive, antisymmetricand transitive relation on T ).

(2) For every y 2 T , pred(T, y) = {x 2 T : x < y} is a well–order.

The notion of tree is a weakening of the notion of well–order: Anwell–order is a linearly ordered tree.

Given a tree T and an ordinal ↵,

T↵

= {t 2 T : ot(pred(T, t)) = ↵}(the ↵–level of T ). The first ↵ such that T

↵

= ; is the height of T .

Definition 8.2. Let be a cardinal. A tree T is a –Aronszajn treei↵:

(1) T has height , and for every ↵ < , |T↵

| < (T is a –tree).(2) T does not have any –branch: There is no b : �! T such

that for all ↵ < � < , b(↵) < b(�).

Fact 8.3. (Konig’s Lemma) There is no !–Aronszajn tree.

Proof. We have to show that if T is a tree of height ! with all levelsfinite, then T has an infinite branch.

We build b = (tn

)n<!

recursively as follows:


• t0 is a node in T0 such that

{t 2 T : t0 < t}is infinite (has to exist since a finite union of finite sets isfinite and since T is infinite).

• In general, suppose tn

2 Tn

defined and {t 2 T : tn

< t}infinite. Then let t

n+1 2 Tn+1, tn < t

n+1, such that

{t 2 T : tn+1 < t}

is infinite (has to exist for the same reason).

The construction of (tn

)n

finishes the proof. ⇤What about !1–Aronszajn trees?

Theorem 8.4. (Aronszajn) (ZFC) There is an !1–Aronszajn tree.

Proof. It will be convenient to use the following notation: Let f and gbe functions and ↵ � ordinals such that dom(f) = ↵ and dom(g) =�. Write f ✓fin g if

{⇠ 2 ↵ : f(⇠) 6= g(⇠)}is finite.

Let ~e = (e↵

: ↵ 2 [!,!1)) be such that for all ↵,

• e↵

: ↵ �! ! is a one–to–one function, and• for all ↵ < � in [!, !1), e↵ ✓fin e

�

.

Such a ~e is called coherent.

Fact 8.5. (ZFC) There is a coherent family of one–to–one functions~e = (e

↵

: ↵ 2 [!, !1)).

Proof. By recursion on ↵ 2 [!, !1) we build (e�

: ! � ↵) suchthat for all ↵:

(i) e↵

: ↵ �! ! is one–to–one.(ii) If � < ↵, then {⇠ < � : e

�

(⇠) 6= e↵

(⇠)} is finite.(iii) ! \ range(e

↵

) is infinite.

The successor step is easy: By (iii) we may pick n 2 ! \ range(e↵

).Set

e↵+1 = e

↵

[ {(↵, n)}Then (e

�

: ! � ↵ + 1) satifies (i)–(iii).Finally suppose ↵ nonzero limit ordinal and (e

�

: ! � ↵0)defined for all ↵0 < ↵.Let (↵

n

)n<!

, ↵n

< ↵, such that supn

↵n

= ↵. By recursion we define(t

n

)n<!

such that for all n:

(a) tn

: ↵n

�! ! is one–to–one.

50 D. ASPERO

(b) tn

✓fin e↵n .

(c) tm

= tn

� ↵m

for all m < n.

Set t =S

n

tn

: ↵ �! !.t is one–to–one thanks to (a) and (c) in the construction. Therefore

(i) holds for t. (ii) holds also thanks to (b) in the construction. (iii)may not hold, but we may remedy this by tinkering with t a bit ifnecessary: Let e

↵

: ↵ �! ! such that

• e↵

(⇠) = t(⇠) if ⇠ /2 {↵n

: n < !}, and• e

⇠

(↵n

) = t(↵2n).

Now it is easy to check, with this choice of e↵

, that (e�

: ! � ↵)satisfies (i)–(iii). This finishes the proof of the fact. ⇤

Let now (e↵

: ! ↵ < !1) be a coherent family of one–to–onefunctions, e

↵

: ↵ �! !. Let

T = {e↵

� � : � ↵ 2 [!, !1)}

ordered by ✓.T is clearly a tree. Also, for every ordinal �,

T�

= {e↵

� � : � ↵, ! ↵ < !1}

Claim 8.6. Every level T�

is finite or countable.

Proof. If t 2 T�

, then there is some infinite ↵ < !1, � ↵, such thatt = e

↵

� �. But e�

✓fin e↵

and therefore {⇠ 2 � : t(⇠) 6= e↵

(⇠)} isfinite.

For every finite s ✓ �, let Xs

be the set of all functions f with

• dom(f) = �• f � (� \ s) = e

�

� (� \ s)• range(f) ✓ !

There are at most countably many functions from a finite set into !,so this set is at most countable.

Then T�

✓S{X

s

: s 2 [↵]<!}. Hence T�

is at most countable.Thisconcludes the proof of the claim. ⇤

Finally, it su�ces to show that there is no !1–branch through T .But if b : !1 �! T is a branch,

Srange(b) : !1 �! ! is a one–to–one

function from !1 into !. Impossible. ⇤

It should be pointed out that ZF does not su�ce in the above the-orem: There are models of ZF without Aronszajn trees. Also, !1–Aronszajn trees are also called simply Aronszajn trees.


What about –Aronszajn trees for higher ? For example, one mayask if there are are or not !2–Aronszajn tree? If ZFC is consistent,then is easy to find models of ZFC in which there are !2–Aronszajntrees.14 On the other hand, it turns out that if ZFC + “There is aweakly compact cardinal” is consistent, then so is ZFC + “There is no!2–Aronszajn tree”. In fact these two theories are equiconsistent.

The following question remains open:

Question 8.7. Is it consistent, modulo the consistency of any reason-able extension of ZFC by large cardinal axioms, that ZFC + “!1 isthe only regular cardinal such that there is a –Aronszajn tree” isconsistent?

8.1. Suslin’s problem.

Definition 8.8. A linear order L has the countable chain condition(c.c.c.) if there is no uncountable collection of pairwise disjoint nonemptyintervals of L.

A separable linear order clearly has the c.c.c.Here is a famous old question of Suslin from the 1920’s:

Question 8.9. We know that the real line is the unique separable com-plete linear order without end-points. Can separability be replaced byhaving the c.c.c. in this characterisation?

Definition 8.10. A Suslin line is a complete linear order without end–points which has the c.c.c. but is not separable.

So the answer to Suslin’s question is No i↵ there is a Suslin line.Let T be a tree. A ✓ T is an antichain i↵ for all distinct a, b 2 A,

none of a b, b a holds.Consider the following strengthening of the notion of Aronszajn tree.

Definition 8.11. A Suslin tree is an Aronszajn tree which does nothave uncountable antichains.

Fact 8.12. The following are equivalent.

(1) There is a Suslin line.(2) There is a Suslin tree.

Is there a Suslin tree? The existence of a Suslin tree is independentfrom ZFC: If ZFC is consistent, then both ZFC + “There is a Suslintree” and ZFC + “There are no Suslin trees” are consistent.

14For instance in L.

52 D. ASPERO

9. Infinite Ramsey theory

Let us introduce some notation: Let , �, µ, ⌫ be cardinals. Then

�! (�)µ⌫

denotes the following: For every colouring c : []µ �! ⌫ there is H 2[]� such that c � [H]µ is constant.

Theorem 9.1. (Ramsey) ! �! (!)nm

for all nonzero n, m < !.

Proof. By induction on n. For n = 1: By the pigeon–hole principle.Suppose true for n. Let c : [!]n+1 �! m. Define �0 : [! \ 1]n �! m

by �0(s) = c({k} [ s). By induction hypothesis there is H0 2 [! \ 1]@0

and m0 < m such that �0 � [H0]n has constant value m0.Let n0 = min(H0) and define �1 : [H0 \ (n0 + 1)]n �! m by �1(s) =

c({n0} [ s). By induction hypothesis there is H1 2 [H0 \ (n0 + 1)]@0

and m1 < m such that �1 � [H1]n has constant value m1.And so on: Suppose i < ! and H

i

2 [!]@0 defined. Let ni

= min(Hi

)and define �

i+1 : [Hi

\ (ni

+ 1)]n �! m by �i+1(s) = c({n

i

} [ s). Byinduction hypothesis there is H

i+1 2 [Hi

\ (ni

+ 1)]@0 and mi+1 < m

such that �i+1 � [Hi+1]n has constant value m

i+1.Finally, by the pigeon–hole principle let m < m such that {i :

mi+1 = m} is infinite. Then

H = {ni

: mi+1 = m}

is such that c � [H]n+1 is constant with value m. ⇤Corollary 9.2. (Ramsey’s theorem, finite version) For all nonzero n <!, m < ! and K < ! there is some N < ! such that

N �! [K]nm

Proof. Exercise: By an application of the Compactness theorem of 1storder logic, which follows immediately from the Completeness thoerem:If T is a first order logic, then T has a model if and only if every finitesubset of T has a model. ⇤

For example, the first N such that N �! (3)22 is N = 6. The first Nsuch that N �! (4)22 is N = 18. It is known that the first N such thatN �! (5)22 is such that 43 N 49. The exact value is not known,though (!).

Theorem 9.3. (Sierpinski) (ZFC) !1 �! (!1)22 is false.

Proof. Let (r⇠

: ⇠ < !1) be a one–to–one enumeration of distinct realnumbers. For all ⇠0 < ⇠1 let c({⇠0, ⇠1}<) = 0 i↵ r

⇠0 < r⇠1 (in the natural

real line order). Otherwise set c({⇠0, ⇠1}<) = 1.


c does not have any uncountable homogeneous set: If H ✓ !1 is suchthat c � [H]2 is constant with value ✏ 2 2, then

• {r⇠

: ⇠ 2 H} is a strictly increasing sequence of reals if ✏ = 0,and

• {r⇠

: ⇠ 2 H} is a strictly decreasing sequence of reals if ✏ = 1.

But between any two distinct reals there is a rational. Therefore Hcannot be uncountable. ⇤

Remark: ZF does not su�ce to prove Sierpinski’s theorem.The following more general version of Theorem 9.3 is true.

Theorem 9.4. (ZFC) Let be an infinite cardinal. Then 2 9 (+)22.

Proof. Let (f↵

: ↵ < 2) list 2 = {f | f : �! 2}. Given distinct f ,g in 2, let f <lex g i↵ f(↵) < g(↵), where

↵ = min{� < : f(�) 6= g(�)}(the lexicographical order on 2).

Given ↵ < �, we let c({f↵

, f�

}) = 0 if f↵

<lex f�

; otherwise we letc({f

↵

, f�

}) = 1.

Claim 9.5. There is no strictly <lex–increasing or strictly <lex–decrea-sing sequence (g

⇠

)⇠<

+ of members of 2.

Proof. Let us see there is no strictly <lex–increasing (g⇠

)⇠<

+ (the otherproof is symmetrical to this).

Let � + least such that {g⇠

� � : ⇠ < +} has size +. LetX = {⇠

i

: i < +} ✓ + of size + such that (g⇠i � �)

i<

+ is one–to–one. For each i 2 X let ⌘

i

< � such that f⇠i � ⌘

i

= f⇠i+1 � ⌘

i

,f⇠i(⌘i) = 0 < f

⇠i+1(⌘i) = 1. Let ⌘ < � such that {i < + : ⌘ = ⌘i

} hassize +. If ⇠ = ⇠

i

= ⇠j

and f⇠i � ⌘ = f

⇠j � ⌘, then f⇠j <lex f

⇠i+1 andf⇠i <lex f

⇠j+1 . If follows that ⇠i = ⇠j

.Hence {f

⇠i � ⌘ : i < +} has size +. But this contradicts theminimality of � since ⌘ < �. ⇤

The proof of the theorem now follows immediately. ⇤On the other hand, something like the finite version of Ramsey’s

theorem is true for all cardinals:

Theorem 9.6. (Erdos–Rado) (ZFC) For all cardinals �, µ and n < !there is a cardinal such that

�! (�)nµ

In fact,in

()+ �! (+)n+1

for all and all n < !.

54 D. ASPERO

What about infinite exponents?

Fact 9.7. (ZFC) 9 (!)!2 for every .

Proof. Given X, Y 2 []@0 , let X ⌘ Y i↵ X�Y = (X \ Y ) [ (Y \X)is finite.

For every equivalence class K of []@0/ ⌘ let �(K) 2 K. Let c(X) 2{0, 1} given by c(X) = 0 i↵ |X��([X]⌘)| is even.

Then c has no infinite homogeneous set.⇤

Note: The Axiom of Choice is needed in the above: There aremodels of ZF in which ! �! (!)!2 , for example.

The following is a natural question in view of Ramsey’s and Sierpinski’stheorems:

Question 9.8. Is there a cardinal > ! such that �! ()22?

Definition 9.9. Call a cardinal weakly compact if > ! and �!()22.

So the above question can be rephrased as: Are there weakly compactcardinals?

Definition 9.10. A cardinal > ! is weakly inaccessible i↵

• is regular and• is a limiy cardinal (i.e., 6= �+ for any � < ).

A cardinal is (strongly) inaccessible i↵

• is regular and• 2� < for all � < ( is a strong limit).

Exercise 9.1. Prove that every weakly inaccessible cardinal is stronglyinaccessible.

Proposition 9.11. (ZFC) If is inaccessible, then V

|= ZFC.

[Proof sketch (exercise)]: Since > ! and V

is transitive, V

|=ZFC \{Replacement, Power set Axiom}. Also:Claim 9.12. For every X 2 V

, |X| < .

Exercise 9.2. Justify this.

Since 2� < for all � it follows, by the claim, that Y = P(X) 2 V

for every X 2 V

and V

|= Y is the power set of X. Since is regular,for every � 2 V

and every function F : � �! V

, f 2 V

. It follows,again by the claim, that V

|= Replacement. ⇤


Corollary 9.13. (ZFC) If there is an inaccessible cardinal, then ZFCis consistent. Hence, if ZFC is consistent, then ZFC does not prove theexistence of inaccessible cardinals. In fact we cannot prove

Con(ZFC+“ There is an inaccessible cardinal”)

starting from justCon(ZFC)

(in, say, ZFC+“ There is an inaccessible cardinal”), unless this theoryis inconsistent).

Proof. Suppose T := ZFC+“ There is an inaccessible cardinal” is con-sistent and suppose T proves that Con(ZFC) implies Con(T ). Workingin T we know that Con(ZFC) is true. Still working in T , we knowthat Con(ZFC) implies Con(T ) . Hence T proves Con(T ). By Godel’ssecond incompleteness theorem this implies that T is inconsistent. Butthis implies, working in T , that ZFC is inconsistent. Since T provesthat ZFC is consistent, it follows that T is inconsistent after all. ⇤Proposition 9.14. If is a weakly compact cardinal, then is inac-cessible.

Proof. Since 2� 9 (�+)22 for all � < , we have that is a strong limit:Suppose � < and 2�. Let c : [2�]2 �! 2 witness 2� 9 (�+)22and let c � []2. Then there cannot be any homogeneous H 2 [] forc since �+ . is regular: Suppose � < and f : � �! cofinal in . We may

assume f strictly increasing. Let c : []2 �! 2 given by c({↵, �}) = 0i↵ there is a (unique) ⇠ < � such that both of ↵ and � are in theinterval [f(⇠), , f(⇠ + 1)). Then there is no H 2 [] homogeneous forc. ⇤

More is true: If is weakly compact, then is the -th inaccessiblecardinal.

Weak compactness has a number of natural characterisations. Thefollowing is one.

Theorem 9.15. The following are equivalent for every cardinal .

(1) is weakly compact.(2) is inaccessible and there is no –Aronszajn tree.

10. The measure problem

Let X be a set. µ : P(X) �! R is a (probabilistic) �–additivemeasure on X i↵:

(1) µ(;) = 0 and µ(X) = 1.

56 D. ASPERO

(2) If µ({a}) = 0 for all a 2 X.(3) If (Y

n

)n<!

is a sequence of pairwise disjoint subsets of X, thenµ(S

n

Yn

) =P

n

µ(Yn

).

Exercise 10.1. If µ is a probabilistic �–additive measure on X andY ✓ Y 0 ✓ X, then µ(Y ) µ(Y 0).

Lebesgue measure (restricted to the interval [0, 1]) satisfies (1)–(3).However, not every subset of [0, 1] is Lebesgue measurable (e.g. Vitalisets)!

Question 10.1. (Lebesgue) Is there any �–additive measure on [0, 1]extending Lebesgue measure? Is there any �–additive measure on [0, 1]?

Let X be a set. F ✓ P(X) is a filter on F i↵:

(1) X 2 F and ; /2 X.(2) For all Y ✓ Z ✓ X, if Y 2 F , then Z 2 F (F is closed under

supersets).(3) For all Y0, Y1 2 F , Y0 \ Y1 2 F (F is closed under finite inter-

sections).

A filter F is an ultrafilter i↵:

(4) For all Y ✓ X, either Y 2 F or else X \ Y 2 F .

Examples:

• If a 2 X, Ua

= {Y ✓ X : a 2 Y } is an ultrafilter. Ua

is theprincipal ultrafilter generated by a.

• F = {Y ✓ ! : |! \ Y | < !} is a filter on ! (F is called theFrechet filter).

An ultrafilter on X wich is not of the form {Y ✓ X : a 2 Y } forany a 2 X is a non–principal ultrafilter on X.

Given a filter F on X and a cardinal , F is –complete i↵T

i<�

Yi

2F whenever � < and {Y

i

: i < �} ✓ F . It is very easy to check thatevery filter if !–complete.

Fact 10.2. Let X be a set. The following are equivalent for every filterF on X.

(1) F is an ultrafilter.(2) F is a ✓–maximal filter on F (i.e., there is no filter F 0 ✓ P(X)

such that F ✓ F 0 and F 6= F 0).


Proof. (1) implies (2): Suppose F 0 ✓ P(X) is a filter such that F ✓ F 0

and F 6= F 0. Let Y 2 F 0 \F . Since F is an ultrafilter, X \Y 2 F . Butthen ; = Y \ (X \ Y ) 2 F 0 and so F 0 is not a filter. Contradiction.

(2) implies (1): Let Y ✓ X and suppose none of Y , X \ Y is in F .Suppose

• there is Z 2 F such that Z \ Y = ;, and• there is Z 0 2 F such that Z 0 \ (X \ Y ) = ;.

Then Z \ Z 0 = ; 2 F . Contradiction.Hence, either

(a) for all Z 2 F , Z \ Y 6= ;, or else(b) for all Z 2 F , Z \ (X \ Y ) 6= ;.In case (a),

F 0 = {W ✓ X : Z \ Y ✓ W for some Z 2 F}is a filter on X properly extending F and such that Y 2 F 0.In case (b),

F 0 = {W ✓ X : Z \ Y ✓ W for some Z 2 F}is a filter on X properly extending F and such that X \ Y 2 F 0.

⇤Fact 10.3. (ZFC) For every set X and every filter F on X there is aultrafilter U on X such that F ✓ U .

Proof. Let (Yi

)i<�

, for � 2|X|, enumerate P(X) \ F . Let (Fi

)i<�

bea ✓–increasing sequence of filters on X extending F such that for alli < �,

(a) if Z \ Yi

6= ; for all Z 2 Fi

, then

Fi+1 = {W ✓ X : Z \ Y

i

✓ W for some Z 2 Fi

},and

(b) if Z \ Yi

= ; for some Z 2 Fi

, then

Fi+1 = {W ✓ X : Z \ Y

i

✓ W for some Z 2 Fi

}Now

Si<�

Fi

is an ultrafilter on X extending F . ⇤Exercise 10.2. Every ultrafilter on ! extending the Frechet filter isnon–principal.

Note: ZF does not su�ce to prove the existence of non–principalultrafilters on ! (if ZF is consistent).

Exercise 10.3. Prove, in ZFC, that there are at least 2@0–many non–principal ultrafilters on !.

58 D. ASPERO

Let us go back to the measure problem.A �–additive probabilistic measure µ : P(X) �! [0, 1] is 2–valued

i↵ µ(X) 2 {0, 1}.Note: If µ : P(X) �! {0, 1} is 2–valued, then µ�1(1) is an @1–

complete (a.k.a. �–complete) ultrafilter on X. Conversely, if U is a�–complete non–principal ultrafilter on X, then µ : P(X) �! {0, 1}given by µ(Y ) = 1 i↵ Y 2 U is a 2–valued �–additive probabilisticmeasure.

Let µ : P(X) �! [0, 1] be a measure. Y ✓ X is an atom of µ i↵µ(Y ) > 0 and for all Z ✓ Y , either µ(Z) = 0 or µ(Z) = µ(Y ).

We say that µ is atomless if it has no atoms.Note: If Y ✓ ⇠ is an atom of µ, then {Z ✓ Y : µ(Z) = µ(Y )} is a

�–complete non–principal ultrafilter on Y and µY

: P �! [0, 1] givenby

µY

(Z) =µ(Z)

µ(Y )is a 2–valued �–additive probabilistic measure on Y .

Theorem 10.4. (Ulam) (ZFC) If there is a �–additive measure on X,then at least one of the following holds.

(1) There exists a two–valued �–additive measure on X and thereis a strongly inaccessible cardinal |X|.

(2) There exists an atomless measure on R and there is a weaklyinaccessible cardinal 2@0.

Let us focus now on possibility (1), i.e., let us look at “There existsa two–valued �–additive measure on some set.”

Fact 10.5. Let be the least cardinal such that there is a non-principal�–complete ultrafilter U on . Then U is actually –complete.

Proof. Suppose {Yi

: i < �} ✓ U , for � < , is such thatT

i<�

Yi

/2 U .Let

Xi

= ( \ Yi

) [ (\

i<�

Yi

)

for all i. Then Xi

/2 U for each i and

=[

i<�

Xi

Let f : �! � defined by f(↵) = i i↵ ↵ 2 Xi

. Let D ✓ P(�)defined by Y 2 D i↵ f�1(Y ) 2 U . Then D is a non-principal �–complete ultrafilter on �. This contradicts the minimality of . ⇤Definition 10.6. A cardinal is measurable i↵ there is a –completenon-principal ultrafilter on .


Fact 10.7. Every measurable cardinal is regular.

Proof. Let U be a non–principal –complete ultrafilter on U . Suppose� < and f : � �! is cofinal. We may assume f is strictly increas-ing.

Let Xi

= \ [f(i), f(i+ 1)) for all i < �. Then each Xi

is in U andyet

Ti<�

Xi

= ; /2 U . Contradiction. ⇤Fact 10.8. Every measurable cardinal is a strong limit.

Proof. Suppose there is � < such that 2� � . Let S be a set offunctions f : � �! 2 such that |S| = . Let U be a non-principal–compete ultrafilter on S (which of course exists since |S| = andthere is one on ).

For each ↵ < � let ✏↵

2 2 be such that

X✏↵ := {f 2 S : f(↵) = ✏

↵

} 2 UThen \

↵<�

X✏↵ 2 U

and yetT

↵<�

X✏↵ contains exactly one function. Contradiction. ⇤

We have thus seen that measurable cardinal are inaccessible. Infact, measurable cardinals are weakly compact and, in fact, if ismeasurable, then is the –th weakly compact cardinal.

11. Ultraproducts

LetN be a set or a proper class satisfying enough of ZFC. Let I 2 N.Let U 2 N be an ultrafilter on I.

Given functions f , g 2 N such that dom(f) = dom(g) = I, let

(1) f =U g i↵ {i 2 I : f(i) = g(i)} 2 Uand

(2) f 2U g i↵ {i 2 I : f(i) 2 g(i)} 2 U

Remark 11.1. • =U is an equivalence relation on (IN) \N .• If f 0 =U f 1, g0 =U g1, and f 0 2U g0, then f 1 2U g1.

We can thus define: For f , g 2 (IN) \N ,

(3) [f ]=U 2⇤U [g]=U i↵ f 2U g

2⇤U is well–define since, by the above remark, the truth value of “f 2U

g” does not depend on the representative we pick from the correspond-ing classes.

60 D. ASPERO

Caveat : The fact that the equivalence classes [f ]=U in the aboveconstruction may actually be proper classes (i.e., too large to be sets)may cause problems. Therefore, in what follows we will instead workwith [f ]0=U

rather than [f ]=U , where [f ]0=U

is the set of all g 2 (IN)\Nsuch that

• g =U f , and such that• rank(g) rank(h) for all h 2 (IN) \N such that h =U f .

(This is known as Scott’s trick)We will write [f ]U instead of [f ]0=U

, though (lest the notation willbecome too loaded).

LetUlt(N,U) = ({[f ]U : f 2 (IN) \N},2⇤

U)

Ult(N,U) is the ultraproduct of N by U .

Theorem 11.2. (Los) Let f0, . . . , fn 2 (IN)\N and let '(x0, . . . , xn

)be a formula. The following are equivalent.

(1) Ult(N,U) |= '([f0]U , . . . , [fn]U)(2) {i 2 I : N |= '(f0(i), . . . , fn(i))} 2 U

Proof. By induction on the complexity of ':Case 1. Suppose ' is atomic. If ' is x0 2 x1, then

Ult(N,U) |= [f0]U 2⇤U [f1]U

i↵{i 2 I : f0(i) 2 f1(i)} 2 U

by definiton of 2⇤U .

Similarly, if ' is x0 = x1, then

Ult(N,U) |= [f0]U = [f1]U

i↵{i 2 I : f0(i) = f1(i)} 2 U

by definiton of =U .Case 2. Suppose ' = ¬ . Then

Ult(N,U) |= ¬ ([f0]U , . . . , [fn]U)i↵ it is not true that

Ult(N,U) |= ([f0]U , . . . , [fn]U)

i↵{i 2 I : N |= (f0(i), . . . , fn(i))} /2 U

(by induction hypothesis) i↵

{i 2 I : N |= ¬ (f0(i), . . . , fn(i))} 2 U


(since U is an ultrafilter).Case 3. Suppose ' = 0 ^ 1. Then

Ult(N,U) |= ( 0 ^ 1)([f0]U , . . . , [fn]U)

i↵Ult(N,U) |= ✏([f0]U , . . . , [fn]U)

for ✏ = 0, 1 i↵

X✏

:= {i 2 I : N |= ✏(f0(i), . . . , fn(i))} 2 Ufor ✏ = 0, 1 (by induction hypothesis) i↵

X0 \X1 = {i 2 I : N |= ( 0 ^ 1)(f0(i), . . . , fn(i))} 2 U(since U is a filter).

Case 4. Suppose ' = (9x) (x, x0, . . . , xn

). Suppose

Ult(N,U) |= (9x) (x, [f0]U , . . . , [fn]U)Then there is some [f ]U 2 N such that

Ult(N,U) |= ([f ]U , [f0]U , . . . , [fn]U)

By induction hypothesis,

X := {i 2 I : N |= (f(i), f0(i), . . . , fn(i))} 2 UBut then

X ✓ {i 2 I : N |= (9x) (x, f0(i), . . . , fn(i))} 2 UConversely, suppose

X = {i 2 I : N |= (9x) (x, f0(i), . . . , fn(i))} 2 ULet f 2 N be a function with domain I such that in N it holds thatfor all i 2 I,

• IF there is a set a such that (a, f0(i), . . . , fn(i)),• THEN '(f(i), f0(i), . . . , fn(i)).

f exists by Replacement and Choice in N.

Then

{i 2 I : N |= (f(i), f0(i), . . . , fn(i))} = X 2 UBy induction hypothesis,

Ult(N,U) |= ([f ]U , [f0]U , . . . , [fn]U)

In particular,

Ult(N,U) |= (9x) (x, [f0]U , . . . , [fn]U)⇤

62 D. ASPERO

Given two structures (N,2), (M,2) in the language of set theoryand a function

j : (N,2) �! (M,2)we say that j is an elementary embedding i↵ for every formula '(x0, . . . , xn

)and for all a0, . . . , an 2 N,

(N,2) |= '(a0, . . . , an)

if and only if(M,2) |= '(j(a0), . . . , j(an))

(j preserves satisfaction)

Corollary 11.3. Suppose N is a set or a proper class satisfying enoughof ZFC. Let I 2 N and let U 2 N be an ultrafilter on I. Let

d : N �! Ult(N,U)by given by d(a) = [c

a

]U where ca

is the constant function on I withvalue a. Then d is an elementary embedding.

Proof. For all a0, . . . , an 2 N and for every formula ',

N |= '(a0, . . . , an)

i↵I = {i 2 I : N |= '(c

a0(i), . . . , can(i))} 2 Ui↵

Ult(N,U) |= '([ca0 ]U , . . . , [can ]U)

by Los’s Theorem. ⇤The Mostowski collapse. Recall: A binary relation R ✓ X ⇥X

is well–founded if and only if for every nonempty Y ✓ X there is somea 2 Y which is R–minimal, i.e., such that (b, a) /2 R for every b 2 Y ,b 6= a.

The proof of the following theorem is an exercise and / or easy tofind.

Theorem 11.4. (Mostowski) Suppose (X,R) is a well–founded rela-tion such that {a 2 X : aRb} is a set for all b 2 X and such that(X,R) |= Axiom of Extensionality. Then there is a unique transitiveclass (a set or a proper class) M for which there is an isomorphism⇡ : (X,R) �! (M,2).

Remark 11.5. (In ZFC) (X,R) is well–founded i↵ there is no R–decreasing sequence (x

n

)n<!

.

Fact 11.6. Suppose N satisfies enough of ZFC and U 2 N is a �–complete ultrafilter on I 2 N. Then Ult(N,U) is well–founded.


Proof. Otherwise there is a sequence (fn

)n<!

of functions in (IN) \Nsuch that

. . . 2U fn+1 2U f

n

2U . . . 2U f1 2U f0For each n,

Xn

= {i 2 I : fn+1(i) 2 f

n

(i)} 2 UBy �–completeness of U , we may pick i 2

Tn<!

Xn

. Then

. . . 2 fn+1(i) 2 f

n

(i) 2 . . . 2 f1(i) 2 f0(i)

but that contradicts Foundation. ⇤11.1. Measurable cardinals and elementary embeddings. Wewill be considering elementary embeddings j : (N,2) �! (M,2) whereN = V (!). M will then be a proper class and so will be j, seeing j asa (proper–class sized) collection of ordered pairs.

Fact 11.7. (ZF) Suppose j : (V,2) �! (M,2) is a non–trivial el-ementary embedding (i.e., j 6= id). Then there is a least ordinal such that < j(). This is called the critical point of j, denoted bycrit(j).

Proof. Exercise. Hint: Suppose j does not have critical point (i.e.,j(�) = � for every ordinal �). Then prove by induction on ↵ that if xis a set of rank ↵, then j(x) = x. ⇤Theorem 11.8. (Keisler–Tarski, Scott, early 1960’s) (ZFC) The fol-lowing are equivalent.

(1) is a measurable cardinal.(2) There is a proper class M and an elementary embedding

j : (V,2) �! (M,2),j 6= id, such that

• crit(j) = , and• M is closed under –sequences (i.e., (x

⇠

)⇠<

2 M for every–sequence (x

⇠

)⇠<

such that x⇠

2 M for all ⇠).

Proof. Let us prove first that (1) implies (2). Let U be a –completenon–principal ultrafilter on . We know (Ult(V,U),2⇤

U) is well–founded.By Mostowski’s theorem, let

⇡ : (Ult(V,U) 2⇤U) �! (M,2)

be an isomorphism, where M is transitive. Let

d : (V,2) �! (Ult(V,U),2⇤U)

be the elementary embedding given by d(a) = [ca

]U for all a. Then

j = ⇡ � d : (V,2) �! (M,2)

64 D. ASPERO

is an elementary embedding. = crit(j): Prove by induction on ↵ that, if ↵ < , then j(↵) =

⇡([c↵

]U) = ↵. For this, let f be such that f 2U c↵

, i.e.,

{⇠ < : f(⇠) < ↵} 2 U ,and prove, using the –completness of U , that there is some � < ↵such that

{⇠ < : f(⇠) = �} 2 UIt then follows that ⇡([f ]U) = j(�) = �.

On the other hand, j(↵) = ↵ < ⇡([id

]U) < j() for all ↵ < . Hence

sup{↵ : ↵ < } = ⇡([id

]U) < j()

M is closed under –sequences: Let f↵

: �! V for all ↵ < . Letf : �! V be given by

f(⇠) = (f↵

(⇠) : ↵ < ⇠)

Check that ⇡([f ]U) is a sequence (a↵

)↵<�

, with � � , such that a↵

=⇡([f

↵

]U) for all ↵ < .We are left with proving that (2) implies (1). For this, suppose

j : (V,2) �! (M,2)is an elementary embedding with crit(j) = , and let

U = {X ✓ : 2 j(X)}Then U is a –complete non-principal ultrafilter on . ⇤

12. Beyond measurable cardinals

The characterisation of measurable cardinal in Theorem 11.8 pro-vides a very useful blueprint for generating large cardinal notions. Lookat a statement of the following sort:

“ is the critical point of an elementary embedding

j : (V,2) �! (M,2)such that M is ‘close to V’.”

The closer to V that M is required to be in this definition, thestronger the large cardinal notion is. It is then usually a routine veri-fication to show that if is such a large cardinal, then is a limit ofcardinals � such that � is the critical point of an elementary embedding

j : (V,2) �! (M,2)such that M is close0 to V, where ‘close0 to V’ is now any reasonableweaker notion of being close to V. It is then typically the case that V

,

• satisfies ZFC since is necessarily inaccessible, and


• thinks that there are many cardinals � such that � is the criticalpoint of an elementary embedding j : (V,2) �! (M,2) suchthat M is close0 to V.

This shows that ZFC + “There exists an elementary embedding j :(V,2) �! (M,2) such thatM is close toV” implies the consistency ofZFC + “There exists an elementary embedding j : (V,2) �! (M,2)such that M is close0 to V”.

Strictly speaking what I am saying doesn’t make sense as “Thereexists an elementary embedding j : (V,2) �! (M,2) such that ...”is a second–order statement about the universe (!). However thereare typically first–order characterisations of precisely these situations,exactly like in the above characterisation of measurable cardinal.

A prominent example is the notion of supercompactness:Given an ordinal �, a cardinal is �–supercompact i↵ there is an

elementary embedding j : (V,2) �! (M,2) such that

• crit(j) = , and• M is closed under �–sequences (i.e., (x

⇠

)⇠<�

2 M for every�–sequence (x

⇠

)⇠<�

such that x⇠

2 M for all ⇠).

A cardinal is supercompact if it is �–supercompact for all �.

If is +–supercompact, then is measurable (trivially) and is alsoa limit of measurable cardinals. In fact, there is a –complete non–principal ultrafilter U on and some X 2 U such that every ↵ 2 X isa measurable cardinal.

12.1. Reinhardt cardinals. The following is clearly the natural limitfor this type of large cardinal notions, obtained from requiring that thetarget model M of the postulated elementary embedding j : V �! Mis in fact V itself:

Let us call a Reinhardt cardinal if is the critical point of anelementary

j : (V,2) �! (V,2)This very natural large cardinal axiom was proposed by William

Reinhardt in his 1967 PhD thesis.However:

Theorem 12.1. (Kunen, 1971) (ZFC) There are no Reinhardt cardi-nals.

Kunen’s proof actually shows that there is no elementary embeddingj : V

�+2 �! V�+2 di↵erent from the identity for any ordinal �. This

subsumes the nonexistence of Reinhardt cardinals. This remarkable

66 D. ASPERO

result of Kunen puts an absolute unattainable upper bound to all largecardinal notions, at least in the presence of the Axiom of Choice.

However, Kunen’s proof doesn’t rule out the existence of an elemen-tary embedding

j : V�+1 �! V

�+1

for some ordinal �. In fact it doesn’t even rule out the existence of anelementary embedding

j : L(V�+1) �! L(V

�+1)

for some ordinal �. These axioms have been extensively studied forseveral decades and no contradiction has arisen yet! They are amongthe strongest hypotheses not known to be inconsistent (with ZFC asbackground theory).

The Axiom of Choice figures prominently in Kunen’s inconsistencyproof and in all alternative proofs found since then. The followingquestion remains open.

Question 12.2. Is the existence of a Reinhardt cardinal consistent withZF?

Once again, literally speaking the above question doesn’t make senseas it again involves second–order quantification over the universe. How-ever there are many ways to make good sense of it. For example, onemay ask:

Question 12.3. : Let Lj be the first order language {2, j} and let Tbe the first order theory in L comprising

• the ZF axioms,• the axiom 8x0, . . . xn

,'(x0, . . . , xn

) $ '(j(x0), . . . , j(xn

)) forevery formula ' in the language of set theory,

• (9x)(x 6= j(x)), and• the axiom Scheme of Replacement for Lj–formulas.

Is T a consistent theory?

This question remains open.

References

[1] T. Jech, Set Theory: The Third Millenium Edition, Revised and Expanded,Springer, Berlin (2002).[2] K. Kunen, Set Theory, An introduction to independence proofs, North-HollandPublishing Company, Amsterdam (1980).[3] E. Mendelson, Introduction to mathematical logic, 5th. ed., 2009.


David Aspero, School of Mathematics, University of East Anglia,Norwich NR4 7TJ, UK

E-mail address: [email protected]

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MAGIC SET THEORY LECTURE NOTES (FALL 2014)bfe12ncu/MAGIC-set... · MAGIC SET THEORY LECTURE NOTES (FALL 2014) DAVID ASPERO´ Contents 1. Introduction 2 1.1. Some elementary facts