MAGIC SET THEORY LECTURE NOTES(AUTUMN 2018)

DAVID ASPERÓ

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–Schröder Theorem 315.2. Countable and uncountable sets 325.3. Almost disjoint families 355.4. Cantor’s Continuum Hypothesis 355.5. AC vs. the Well–Ordering Principle 366. Foundation, recursion and induction. The cumulative

hierarchy 387. Inner models and relativization 437.1. Our first relative consistency proof: Con(ZF \{Foundation})

implies Con(ZF) 457.2. Reflection 478. The constructible universe 529. �–systems 5510. Forcing 5610.1. The method of forcing: Partial orders and genericity 5810.2. Formal development of forcing 5910.3. Adding many Cohen reals 6610.4. �–closedness and not adding new reals 6911. Improving ZFC? 71

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References 73

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 any 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.

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

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 Qfrom the natural arithmetical 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 confident

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with the existence of R and know how to build C from R in a verysimple set–theoretic way.

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–Schröder 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}.

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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 objectsin our meta–theory. Presumably they will obey laws expressible insome meta–meta–theory (perhaps the same laws the theory T is, inour understanding, 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)• (, ), =

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= 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

andv

j

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 forevery variable v and every atomic subformula '0 of ', if v occurs in'0, then '0 is a subformula of some subformula of ' 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 |= (v

i

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(vi

) = ~a(vi

) if v 6= vi

and ~b(v) = b.• M |= (8v')[~a] if and only if for every b 2M , 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 derivation� = (�0, �1, . . . �n)

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

MAGIC Set Theory lecture notes (Autumn 2018) 9

• �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 Gödel in his PhD thesis.

Theorem 2.2. (Completeness theorem for first order logic) (K. Gödel,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 theory

is 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.

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(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 its

members:

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 Axiom

of 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. In fact this theory is consistent together with thesentence that says “Every set has at most two elements” (starting from; and closing under unordered pair gives rise to a model of it where“Every set has at most two elements” also holds). Also, this theoryproves 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, all

MAGIC Set Theory lecture notes (Autumn 2018) 13

particular 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 Xn+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.

MAGIC Set Theory lecture notes (Autumn 2018) 15

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 xn+1 2x

n

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. ASPERÓ

Foundation, is consistent.9 In fact

([

n

V

n

,2) |= T

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

([

n

V

n

,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

V

n

,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?

MAGIC Set Theory lecture notes (Autumn 2018) 17

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. ASPERÓ

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

for every first order formula '(x, ȳ) 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. (Gödel, 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 ¬�

Gödel’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.

MAGIC Set Theory lecture notes (Autumn 2018) 19

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 HPConsider the theory T = (ZFC \{Infinity}) [ {¬Infinity}. It turns

out 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 a well–defined class–function, and thatS

range(F ) exists, i.e., as-suming something 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 encode formulas, proofs,and other syntactical notions as natural numbers and reduce a state-ment like “PA is consistent” to an arithmetical statement (some spe-cific, but extremely complex, diophantine equation p(x̄) = 0 does nothave solutions). It then makes sense to ask whether T proves thatp(x̄) = 0 does not have solutions.

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

20 D. ASPERÓ

• 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

V

n

,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 T0if 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 “T0inconsistent.” And, similarly, we define “T0 and T1 are equiconsistent.”

It is important to bear in mind that we need to be careful withwhat 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

MAGIC Set Theory lecture notes (Autumn 2018) 21

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'(x̄) in the language of set theory,

9x8y(y 2 x$ '(y, x̄))

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. ASPERÓ

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 (Gödel).

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.

MAGIC Set Theory lecture notes (Autumn 2018) 23

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,

24 D. ASPERÓ

Fact 4.4. If (L,

MAGIC Set Theory lecture notes (Autumn 2018) 25

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 andx = 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 we

have 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. ASPERÓ

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.

MAGIC Set Theory lecture notes (Autumn 2018) 27

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 bijective 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 relation2 |↵ 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 uniquecardinal.

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. ASPERÓ

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 letf

z

: (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),) �! Ordbe given by f(y) = f

y

0(y) 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

f

x

: 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

MAGIC Set Theory lecture notes (Autumn 2018) 29

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

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

Definition 5.2. A cardinal is an ordinal such that is not bijectivewith 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 bijective with a cardinal , we say that is thecardinality of X and write |X| = .

We will see that ZFC proves that every set is bijective with an or-dinal. 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 bijective with !).

Proposition 5.4. !1 exists.

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

{(n,m) 2 ! ⇥ ! : 2n+13m+1 2 X}

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

! can be encoded by a subset of !. Hence

� = {↵ : ↵ = ot() for some encoded by some X ✓ !}

is transitive and is a set since it is range(F ), where F : P(!) �! Ordis the function sending X to ot() if X encodes 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 order type �. But then� 2 �, which is impossible for ordinals. ⇤

30 D. ASPERÓ

Exercise 5.1. Prove that the ordinal � in the above proof is precisely!1; that is,

!1 = {↵ : ↵ = ot() for some encoded 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:

Definition 5.5. Given an ordinal ↵, @↵

, also denoted !↵

, is the ↵–thinfinite cardinal.

Notation 5.6. The !↵

notation is usually used in contexts where thefocus is the ordinal structure. The @

↵

notation is instead used when thefocus is the cardinality of this set (i.e., the property of being bijectiveto @

↵

).

Definition 5.7. Given a cardinal , +, the successor of , is the leastcardinal strictly bigger than .

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

Proposition 5.8. (ZF) For every infinite cardinal 2 Ord, + exists.

This follows immediately from:

Theorem 5.9. (Hartogs) Given any set X there is an ordinal ↵ forwhich there is no injective function f : ↵ �! X.

Proof. (of Hartogs’ Theorem): Let ↵ be the collection of ordinals �such that there is an injective function f : � �! X. ↵ is the collectionof of ordinals of the form ot(Y,), where Y ✓ X and is an ordertype of Y . Hence ↵ is a set by Replacement (being the image of thefunction with domain P(X ⇥X) sending W to ot(W ) if W is a well–order of some Y ✓ X, and to ; otherwise), and it is trivial to see thatit is transitive. Hence ↵ is an ordinal. Finally, there is no injectivef : ↵ �! X; otherwise ↵ 2 ↵, which is impossible for ordinals. ⇤Exercise 5.2. Prove that t if X, in the proof of Hartogs’ Theorem, isa cardinal , then the ordinal ↵ in that proof is in fact +; that is,

+ = {� : � = 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.

MAGIC Set Theory lecture notes (Autumn 2018) 31

Definition 5.10. A successor cardinal is a cardinal of the form + forsome cardinal . A limit cardinal is a cardinal which is not a successorcardinal.

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 5.11. An ordinal is regular if and only if there is no↵ < for which there is a function f : ↵ �! with range cofinal in. An ordinal 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–Schröder Theorem.

Theorem 5.12. (Cantor–Bernstein–Schröder 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) = c̄ if c̄ 2 X issuch that f(c̄) = �

c

(z) (if �c

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

c

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

(z) (if �c

(z) 2 Xand if there is such a c̄). 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 c̄ 2 Y with g(c̄) = �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.

32 D. ASPERÓ

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 !}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 5.13. (ZFC) Dual C–B–S is true.

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

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

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

Question 5.14. 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 5.15. A set X is countable if and only if |X| = @0. A setis uncountable if it is not finite or countable.

MAGIC Set Theory lecture notes (Autumn 2018) 33

Let us see some examples of countable sets.

Proposition 5.16. (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.)

•

n2!n

!. (the set

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 induc-tion on n since |n+1!| = |(n!) ⇥ !|. This gives a definable sequence(f

n

)1n

n+1(m). Since there is a bijectiong : ! �! ! ⇥ !, the composition f � g : ! �!

n

: [!]n �! ! for every n � 1, sendx 2 [!]n to f�1

n

((x0, . . . , xn�1)), where (x0, . . . , xn�1) is the strictly in-creasing enumeration of x. We can also encode all (the inverses of)these bijections together into a bijection h : ! �! [!]

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

34 D. ASPERÓ

Proposition 5.17. (ZFC) The union of every countable collection ofcountable sets is countable: If (X

n

)n2! is such that each Xn is count-

able, thenS

n2! Xn is countable.

Proof. @0 |S

n2! Xn| is clear: There is a bijection f : ! �! X0, andf : ! �!

Sn

X

n

is an injection.|S

n

X

n

| @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, fn

(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

X

n

�! ! is an injection.Since |

Sn

X

n

| @0 and @0 |S

n

X

n

|, by C–B–S we get |S

n

X

n

| =@0. ⇤

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

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

MAGIC Set Theory lecture notes (Autumn 2018) 35

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

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

Theorem 5.20. (ZF) There is a collection A ✓ P(!) of size 2@0 con-sisting of pairwise almost disjoint sets.

Proof. Let

Now consider any two distinct infinite branches b, b0 through

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

Definition 5.21. (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.

36 D. ASPERÓ

Question 5.22. 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. Later Iwill probably say something on how to “solve” this problem.

Definition 5.23. Cantor’s Continuum Hypothesis (CH): 2@0 = @1.

5.5. AC vs. the Well–Ordering Principle.

Theorem 5.24. (ZF) The following are equivalent.

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

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

↵

: ↵ < �} (and to, say, ; otherwise). Since P(X)is a set, by Replacement range(F ) is a set of ordinals so it cannot beall of Ord. Hence this construction has to stop at some point (theremust be � such that X \{x

↵

: ↵ < �} = ;). But then {(x↵

, x

↵

0) : ↵ 2↵

0 2 �} is a well–order of X.Now assume the Well–ordering Principle. Let X be a set consisting

of nonempty sets and let be a well–order ofS

X. Now, given a 2 Xlet 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 5.25. (ZF) The following are equivalent:

MAGIC Set Theory lecture notes (Autumn 2018) 37

(1) AC(2) Every set is bijective with a unique cardinal.

Corollary 5.26. (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

X

n

) =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 all

38 D. ASPERÓ

sets 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).

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↵())|)

MAGIC Set Theory lecture notes (Autumn 2018) 39

• 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 a

first-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:P

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↵

.

40 D. ASPERÓ

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

=S

�

MAGIC Set Theory lecture notes (Autumn 2018) 41

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 � ↵) for all ↵, where G(x, y) is:

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

X

n

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}.

42 D. ASPERÓ

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

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

y /2WF: 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 /2WF}. 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 /2WF. ⇤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 V orthe Russell class {x : x /2 x}. Take for example V. Certainly, if V isa set, then V 2 V. But this goes against the iterative conception ofset, 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). This

MAGIC Set Theory lecture notes (Autumn 2018) 43

picture 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 'M0 _ 'M1 .• (¬')M is ¬('M).• ((8x)('(x̄))M is 8x(⇥(x)! 'M(x̄)). We may also write some-thing like (8x 2M)'M(x).

Note: Given a formula '(x0, . . . , xn) and a0, . . . , an 2M , 'M(a0, . . . , an)holds if and only if M |= '[a0, . . . , an].

[This is easy to prove by induction on the complexity of '.]Notation: If (N,E) is a structure in the language of set theory, M

is 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).(This note is pertinent under ZF \{Foundation}, in which context V =WF need not be true.)

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).

44 D. ASPERÓ

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} 2M). 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. ⇤

Similarly:

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

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

Lemma 7.4. Suppose M is a transitive set or a transitive proper class.Suppose for every a 2M there is some b 2M 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 2M such thatP(a)M is strictly included in P(a) (i.e., there are subsets b of a suchthat b /2M).Lemma 7.5. Suppose M is a transitive set or a transitive proper class.If ! 2M , 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):

A formula ' is restricted if all its quantifiers are restricted. In otherwords, they appear in subformulas of the form (8x)(x 2 y �! ) and(9x)(x 2 y ^ ). We generally abbreviate the above by (8x 2 y) and(9x 2 y) , respectively.

The following general fact is not di�cult.

MAGIC Set Theory lecture notes (Autumn 2018) 45

Fact 7.6. Suppose '(~x) is a restricted formula and M is transitive.The following are equivalent for every ~a 2M

(1) '(~a)(2) M |= '(~a)

Note: The notion of finiteness is also absolute with respect to tran-sitive models containing !. On the other hand, the notion of count-ability is highly non–absolute with respect to transitive models: Thereare transitive models M and a 2M such that

M |= a is uncountablebut 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 transi-tive models: If M is transitive, a 2 M consists of nonempty sets, andf 2 M , then f is a choice function for a if and only if we have thatM |= “f is a choice function for a”. Hence:

Lemma 7.7. Let M be a transitive set or a transitive proper class.Suppose for every a 2M consisting of nonempty sets there is a choicefunction f for a, f 2M . Then M |= AC.

Lemma 7.8. 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 2M). Then M |= Separation.

Lemma 7.9. Let M be a transitive set or a transitive proper class.Suppose F [a] 2M whenever a 2M , and F is a class–function over M(in other words, if F is definable by a formula '(x, y, ~z) which, over M ,is functional, ~p 2M , and a 2M , 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 7.10. Let M |= ZF \{Foundation}. Then M |= �WFM forevery � 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].

46 D. ASPERÓ

Then

WF |= rank(b) = min{rank(z) : z 2 Z}by absoluteness of the relevant notions.

To see that M |= FoundationWF holds, let us work in M : LetX 2WF, and let b 2 X such that rank(b) = min{rank(x) : x 2 X}.

By definition of rank and definition of WF =S

↵2Ord V↵ there is thenno y 2 X such that y 2 b. Since WF is transitive, this means thatthere is no y 2 WF\X such that WF |= y 2 b. Hence, WF thinksthat the restriction of 2 to X is well–founded. Since this is true for allX 2WF,

WF |= Foundation⇤

Corollary 7.11. 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 7.12. 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:

• 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

↵ 0 is a limit ordinal.

L =S

↵2Ord L↵.This construction is due to Gödel. 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