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MAGIC SET THEORY LECTURE NOTES (SPRING2014)

DAVID ASPERO

Contents

1. Introduction 22. The axiomatic method: A crash course in first order logic 53. Axiomatic set theory: ZFC 93.1. The axioms 103.2. ZFC vs PA 153.3. The consistency question 174. Ordinals 205. Cardinals 266. Foundation, recursion and induction. The cumulative

hierarchy 347. Inner models and relativization 387.1. Our first relative consistency proof: Con(ZF \{Foundation})

implies Con(ZF) 418. Elementary substructures and Skolem closures 428.1. Reflection 468.2. �–systems 479. Forcing 499.1. The method of forcing: Partial orders and genericity (an

example) 519.2. Formal development of forcing 529.3. The forcing relation �P 5510. Two forcing constructions 5910.1. Adding many Cohen reals 5910.2. Chain condition and cardinal preservation 6010.3. �–closure and not adding new reals 6211. Improving ZFC? 64References 66

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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 = b

0).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 = a

0 and b = b

0.

[Easy exercise: Check]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 notion

of 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.In general, we can define n+1 = n[{n}. With this definition n is a

set with exactly n many members, namely all natural numbers m suchthat m < n.

With this definition, each n is an ordinal � which is either ; or of theform ↵ [ {↵} for some ordinal ↵ and all of whose members are either

MAGIC Set Theory lecture notes (Spring 2014) 3

the empty set or of the form ↵[ {↵} for some ordinal ↵, and every or-dinal which is either ; or of the form ↵[{↵} and all of whose membersare either the empty set or of the form ↵ [ {↵} for some ordinal ↵ isa natural number (the notion of ordinal, which we will see later on, isdefined only in terms of sets).

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

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 �.

+ 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 + b

0 = a

0 + 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.

Also: We can derive everything we know about the higher level ob-jects (like, say, the fact that ⇡ is transcendental) from elementary factsabout 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, ...).

Some elementary facts about sets: Given sets A, B, we say thatA is of cardinality at most that of B, and write

|A| |B|,

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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)| < . . .

. . . < |S

n2N Pn(N)| < |P(S

n2N 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?),

(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.

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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 f

M : M �! M it has arity 1, if it ismeant to express a function f

M : 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 R

M ✓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, namely

a 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 v

j

) or (vi

= v

j

), with v

i

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 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:A pair M = (M,R), where M is a set and R ✓M ⇥M , is called an

L–structure.Given an assignment ~a : Var �!M :

• M |= (vi

2 v

j

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

),~a(vj

)) 2 R.• M |= (v

i

= v

j

)[~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 |='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= v

i

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].If ' is a sentence, then M |= '[~a] for some assignment ~a if and only

if M |= '[~a] for every assignment ~a. In that case we say that M is amodel of �.

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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. Thisframework is mostly due to the logician A. Tarski (1930’s).

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,1 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

.

If ' is a theorem from T , we write

T ` '` is often called the relation of logical derivability.

Theorem 2.1. (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. A theory is inconsistent if and only ifit is trivial, in the sense that it proves everything.

By the completeness theorem the following are equivalent:

• T is consistent.

1A logical axiom is a member of a certain fixed infinite list – which certainlydoes not depend on the theory – expressing universal (logical) truths (e.g., ↵ _ ¬↵for every L-formula ↵, which of course expresses the fact that if something is nottrue then its negation is true).


• 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 an axiom guaranteeing the existence of sets witha given property, even if these sets are not definable: The Axiom ofChoice. 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.

(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.

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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 + b

n = c

n

, 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.

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 given so far gives us already the existence of infinitely

many sets! For example ;, {;}, {{;}}, {{{;}}}, {{{{;}}}}, {;, {;}},


{;, {;, {;}}}, {{;}, {;, {;}}}, {;, {;, {;, {;}}}}, ... With the definitionof 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))S

x is the set consisting of all the members of members of x,SS

x

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• ...

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T is consistent: Let (Xn

)n2N be defined recursively by

• X0 = {;}• X

n+1 = X

n

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

n

} [ {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 $ '(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 of n⇥m ={(a, b) : a 2 n, b 2 m}, and much more.

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)”.

2

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 all

2Which can be expressed in our language since it has =.


a 2 X.

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

Exercise: Write down a sentence expressing the Axiom of Choice.AC is needed in a lot of mathematics. For example, to prove that

every 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,3 but ifX is infinite, where did the choice function for X come from? Also,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!4 This result is knownas the Banach–Tarski paradox.

AC has interesting equivalent formulations (module 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 X

such 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

, x

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

n+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, X is no longer X.

3Try to see why.4The pieces are not Lebesgue measurable, though.

14 D. ASPERO

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 +Foundation, is consistent. 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).

Axiom of Infinity: There is an infinite set.

Definition 3.1. 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.Note: Every inductive set is infinite (in our present sense). To see

this, suppose X is inductive, let a 2 X, and let f : X \ {a} �! X bethe function sending every set of the form S

n(a) (for n 2 N, n > 0)to S

n�1(a), and every set x 2 X which is not of the above form to x

itself. 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 someordinal y.

The Axiom of Infinity is then equivalent to:

Axiom of Infinity’: The class of all natural numbers is a set.In other words: There is some x such that for all y, y 2 x if and onlyy is a natural number.

Note that Axiom of Infinity’ is a constructive set–existence axiom,whereas Axiom of Infinity was not, strictly speaking. Axiom of Infinity’and Axiom of Infinity are equivalent modulo the other axioms.

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.5 Now, withthe addition of Infinity, it’s not so obvious that ZFC is consistent... .

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)• 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!

5Since (S

n2N Vn,2) |= T .

16 D. ASPERO

Theorem 3.2. (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.

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 .”

HP can be easily expressed by a sentence, which I will call HP, inthe language of arithmetic.

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

Theorem 3.3. (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 3.4. 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 3.5. (Godel’s Incompleteness Theorems) Suppose T is a firstorder theory such that

• T recursively enumerable (i.e., there is an algorithm deciding,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 recursively enumerable. Hence, IF they

are consistent, THEN they are incomplete and they cannot prove their

18 D. ASPERO

own consistency. It follows that if we adopt, say, ZFC as our meta-theory, we won’t be able to prove any statement of the form “ZFC+�is consistent”. What we can do is prove relative consistency state-ments of the form “If ZFC is consistent, then ZFC+� is consistent”(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, T1 has consistency strength at most that of T0 if and onlyif we can prove that if T1 is consistent then T0 is consistent. T1 hasconsistency strength strictly larger than T0 if and only if we can prove“if T1 is consistent, then T0 is consistent”, but we cannot prove ”if T0 isconsistent, then T1 is consistent” unless we can prove “T0 inconsistent.”And, similarly, we define “T0 and T1 are equiconsistent.”

For example, although ZFC does not prove Con(ZFC), if ZFC isconsistent, it proves that ZFC+� and ZFC+¬� are equiconsistent formany interesting choice of � (we will hopefully see examples !).

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.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.

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 L

M ✓M such that (LM

, R \ L

M ⇥ L

M) |= 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 cumulative

20 D. ASPERO

hierarchy 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.

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.

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= v

0. Wlog v <1 v

0.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.

22 D. ASPERO

Similarly one can show that for all u, u0, u 0 u

0 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 letu

0<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 v

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

, v

0) 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 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, ↵ = �.

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.

24 D. ASPERO

[Exercise.]

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 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 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= ;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

f

z

: (pred((L,), z),) �! (↵z

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

0< x, then ↵

z

2 ↵z

0 and f

z

= f

z

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) = f

y

0(x) for any y

0 such that y < y

0< x. This function

is 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 L

there 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. ⇤

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 to

26 D. ASPERO

2. This well–order has order type ! + 1.

[Exercise: Characterize the sets that can be well–ordered in di↵erentways.]

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

can be coded by a subset of ! ([Exercise]). Hence � = ! [ {↵ : ↵ =ot((!,)) for some coded by some X ✓ !} is transitive and is aset since it is ! [ range(F ), where F : P(!) �! Ord is the function


sending X to ot((!,)) if X codes and to 0 otherwise. Hence � isan 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. ⇤

It is easy to see that the ordinal � in the above proof is precisely !1;that is,

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

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.

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: Say that ✓ ⇥ is awell–order if (,) is a well–order on .Every initial segment of a well–order on of order type at least is

order–isomorphic to a well–order on ([Exercise]). Hence,

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

{(↵,↵0) 2 ⇥ : f�1(↵) 2 f

�1(↵0)}would be a well–order of . But then � 2 �, which is impossible forordinals. ⇤

It is easy to see that the ordinal � in the above proof is in fact +;that is,

+ = [ {↵ : ↵ = ot((,)) for some well–order on }[Exercise]

Theorem 5.8. (Cantor–Bernstein–Schroder Theorem) (ZF) For allsets X and Y , the following are equivalent.

28 D. ASPERO

(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 X

and 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 and

that 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 X

or 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 �! X

are 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 bijection


h : ! �! {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 5.9. (ZFC) Dual C–B–S is true.

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

and 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 5.10. It is not known whether or not, modulo ZF, DualC–B–S is equivalent to the Axiom of Choice.

Definition 5.11. 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 5.12. (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 given a definable sequence (fn

)1n<!

30 D. ASPERO

where f

n

: ! �! 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 Z

and 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 < a1

or else a0 = b1 and b0 < b1 (where < refers to the usual order on thereal line). ⇤Proposition 5.13. (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! X0| 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, f

n

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

k

. 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 5.14. |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: |C| = |P(!)|.] Hence, since the set of algebraic num-bers is countable, most complex numbers are transcendental (i.e., non–algebraic). In fact

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

Definition 5.15. (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 5.16. 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 5.17. Cantor’s Continuum Hypothesis (CH): 2@0 = @1.

Theorem 5.18. (ZF) The following are equivalent.

32 D. ASPERO

(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

↵

: ↵ < �}. 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 5.19. (ZF) The following are equivalent:

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

Corollary 5.20. (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 are


models 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 fbe a choice function for the quotient set R/ ⌘ (i.e., f picks an elementout of each equivalence class of ⌘). Then the range(f) is not Lebesguemeasurable. It’s called a Vitali set.

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.

We have seen that |!| < |P(!)|. This implies in particular that acollection of pairwise disjoint subsets of ! has to be finite or countable.

Definition 5.21. Two sets X, Y are almost disjoint if X \Y is finite.

Theorem 5.22. (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).

34 D. ASPERO

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. ⇤

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

↵

())|)• 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?

36 D. ASPERO

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 ↵.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 . ⇤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

X

n

So TC(x) = x [S

x [SS

x [SSS

x [ . . .

[Exercise: TC(x) is the ✓–least transitive set y such that x ✓ y. Inother 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 ↵.

38 D. ASPERO

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, a 2-minimal in y, such that

a /2 WF. For every z 2 a, it follows that z 2 y (by transitivityof y) and therefore z 2 V

↵

for some ↵ 2 Ord. 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 2 V

�

✓ V

↵

. Hence a ✓ V

↵

and therefore a 2 P(V↵

) = V

↵+1.Contradiction with a /2WF. ⇤

The fact that V = WF realises the idea that a set is any collectionbuilt 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.

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 b

if 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 2M)'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 M

N 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} 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:

40 D. ASPERO

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.

[Proofs: Exercises.]

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). 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 2M 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 2M consisting of nonempty sets there is a choicefunction f for a, f 2M . 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 2M). Then M |= Separation.

Lemma 7.8. 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 Mis 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 7.9. 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 2WF, 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 2WF,

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 |= �

WFMfor every � 2 ZFC. Hence,

Con(ZFC \{Foundation}) implies Con(ZFC).

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

42 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 also

consistent, 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.6 From now on we will mostly aim at “extending” models ofgiven models of (fragments) of ZFC (I believe). But first we will needa string of preliminaries. These preliminaries are in fact very centralnotions in set theory and other areas of mathematical logic.

8. Elementary substructures and Skolem closures

Let M and N be two structures (in a first order language L) and let

j : M �! N

We say that j is an elementary embedding if and only if for everyL–formula '(x0, . . . , xn�1) and all a0, . . . , an�1 2M ,

M |= '(a0, . . . , an�1)

if and only ifN |= '(j(a0), . . . , j(an�1))

Note: If f : M �! N is an isomorphism, then f is an elementaryembedding between M and N . 7 The converse is not true: There areelementary embeddings j : M �! N such that j is not an isomorphism.

6What 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.

7This is of course the reason why people are interested in structures modulo theirisomorphism types (i.e., two isomorphic structures are regarded as being essentiallythe same structure).


If M ✓ N and the identity on M is an elementary embedding fromM to N , we say that M is an elementary substructure of N andwrite

M 4 N

For example: If M ✓ N , a 2M ,

M |= a is uncountable,

butN |= a is countable,

then M is certainly a substructure of N but it is not an elementarysubstructure of N .

Given a first order language L, we assign a rank to L–formulas inthe following way: We say that

• an atomic formula has rank 0,• if ' has rank n, then ¬' has rank n+ 1,• if '0 has rank n0 and '1 has rank n1, then '0 _ '1 has rankmax{n0, n1}+ 1, and

• if '(x, ~x) has rank n, then (9x)('(x, ~x)) has rank n+ 1.

Thus, the rank of a formula measures its complexity.

Lemma 8.1. (Tarski–Vaught Lemma) Let L be a first order language,let M ✓ N be L–structures, and suppose for all a0, . . . , an�1 2 M andevery L–formula '(x, x0, . . . , xn�1), if

N |= (9x)'(x, a0, . . . , an�1),

then there is some a 2M such that

N |= '(a, a0, . . . , an�1)

Then

M 4 N

Proof. We prove, by induction on the complexity of L–formulas, thatthe conclusion of the lemma holds for every L–formula. For atomicformulas there is nothing to prove.Suppose ' = ¬'0. Given a0, . . . , an�1 2M , we want to see thatM |=

¬'0(a0, . . . , an�1) i↵ N |= ¬'0(a0, . . . , an�1). But we have that M |=¬'0(a0, . . . , an�1) i↵ M 6|= '0(a0, . . . , an�1) i↵ N 6|= '0(a0, . . . , an�1)(by induction hypothesis) i↵ N |= ¬'0(a0, . . . , an�1).Now suppose ' = '0 _ '1. Given a0, . . . , an�1 2 M , we want to

see that “M |= '0(a0, . . . , an�1) or M |= '1(a0, . . . , an�1)” holds i↵“N |= '0(a0, . . . , an�1) or N |= '1(a0, . . . , an�1)” holds. But “M |='0(a0, . . . , an�1) or M |= '1(a0, . . . , an�1)” holds if and only if “N |=

44 D. ASPERO

'0(a0, . . . , an�1) or N |= '1(a0, . . . , an�1)” holds (by induction hypoth-esis with '0 and '1, resp.), so we are done.

Finally suppose ' = (9x)('0(x, ~x)). Given a0, . . . , an�1 2 M , wewant to see that there is some a 2M such that M |= '0(a, a0, . . . an�1)if and only if there is some a 2 N such that N |= '0(a, a0, . . . , an�1).If there is some a 2 M such that M |= '0(a, a0, . . . , an�1), then N |='0(a, a0, . . . , an�1) by induction hypothesis with '0. Finally, if thereis some a 2 N such that N |= '0(a, a0, . . . , an�1), then there is somea 2M such that N |= '0(a, a0, . . . , an�1) by our assumption. But thenM |= '0(a, a0, . . . an�1) again by induction hypothesis with '0. ⇤Definition 8.2. Suppose '(x, x0, . . . , xn�1) is a formula, M is a struc-ture, and F : n

M �! M . F is a Skolem function for ' if for alla0, . . . , an�1 2M , if there is some a 2M such that

M |= '(a, a0, . . . , an�1),

thenM |= '(F (a0, . . . , an�1), a0, . . . , an�1)

That is, F chooses a witness to the existential statement

(9x)'(x, a0, . . . , an�1)

whenever this is possible.

Sometimes we denote a function as in the above definition by F

M

'

.Working in ZFC (so, with Choice): Let L be a first order lan-

guage and let FmlL denote the set of L–formulas. Suppose N is anL–structure and X ✓ N . Our next goal will be to build a “small” M

such that

• X ✓M ✓ N and• M 4 N

Small will mean |M | = |X [ FmlL |.

The construction: Let {FN

'

: '(x, x0, . . . xn�1) 2 FmlL} be acollection of Skolem functions for N ; in other words, for every for-mula '(x, x0, . . . xn�1) and for all a0, . . . , an�1 2 N , if we have N |=(9x)'(x, a0, . . . , an�1), then N |= '(FN

'

(a0, . . . , an�1), a0, . . . , an�1).For every formula '(x, x0, . . . xn�1) there is some Skolem function as

above by the Axiom of Choice: Simply pick a choice function of F',where

F' = {F'

a0,...,an�1: a0, . . . , an�1 2 N, F'

a0,...,an�16= ;},

and where

F'

a0,...,an�1= {b 2 N : N |= '(b, a0, . . . , an�1)}


for all a0, . . . , an�1 2 N .We can then of course pick one Skolem function for each ' by using

again AC and that gives the existence of

{FN

'

: '(x, x0, . . . xn�1) 2 FmlL}The idea is to take M to be the “closure of X under Skolem func-

tions.” This closure will be of size |X[FmlL | and will be an elementarysubstructure of N by Tarski–Vaught:

Let M =S

i2! Xi

, where

• X0 = X,• for all i < !,

X

i+1 = X

i

[ {FN

'

(~a) : ~a 2 <!

X

i

, '(x, ~x) 2 FmlL}Clearly X

j

✓ X

j

0 ✓ M for all j < j

0< ! and X ✓

Si<!

X

i

= M .Also, for every '(x, ~x) 2 FmlL and all a0, . . . , an�1 2

Si<!

X

i

, if N |=(9x)'(x, a0, . . . , an�1), then N |= '(FN

'

(a0, . . . , an�1), a0, . . . , an�1).Let i

⇤< ! be such that a0, . . . , an�1 2 X

i

⇤ . This i

⇤ exists sincea

j

2 X

ij for all j < n and since X

j

✓ X

j

0 for all j < j

0< !. But then

F

N

'

(a0, . . . , an�1) 2 X

i

⇤+1 ✓M . Hence, by the Tarski–Vaught Lemma,M 4 N .

Smallness of M : Let us prove the following special case (which willbe enough for our purposes):

Claim 8.3. Suppose |L| @0 and |X| @0. Then |FmlL | = |M | [email protected]. Note: By |L| @0 we may identify L with a subset of !. Thenwe may code L–formulas in some natural way by members of <!

!. But|<!

!| = @0.We prove by induction on i < ! that |X

i

| @0: True for i = 0 byassumption.

Let i < ! and suppose |Xi

| @0. Then|X

i+1| |<!

X

i

⇥ FmlL | |<!

! ⇥ !|since |FmlL | = @0. But |<!

! ⇥ !| = |<!

!| = @0. Let fi : Xi

�! ! bean injective function for each i < !. Finally, f :

Si<!

X

i

�! ! ⇥ ! isan injective function, where f(x) = (i⇤, f

i

⇤(x)) if i⇤ = min{i : x 2 X

i

},and therefore |

Si

X

i

| @0 since |! ⇥ !| = @0. ⇤We have proved:

Theorem 8.4. (ZFC) (Downward Lowenheim–Skolem Theorem, spe-cial case) Let L be a language which is either finite or countable. LetN be an L–structure and let X ✓ N be such that |X| @0. Then thereis some M such that

46 D. ASPERO

• X ✓M ✓ N

• M 4 N

• |M | @0.

8.1. 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 8.5. (ZF) (Reflection) Let '(x0, . . . , xn�1) be a formula inthe language of set theory. There is a proper class C

'

of ordinals suchthat:

(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.(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 < i

0.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 anyi

0< i and any variable x, then C

'i = C

'i�1 .Finally, suppose i > 0 and '

i

= (9x)'(x, x0, . . . , xm�1) for somei

0< 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 8.6. 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.

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 everyj n, sup

i<!

↵

i

is a limit of ordinals in C

'j and therefore in C

'j . ⇤

8.2. �–systems.

Definition 8.7. A set X is a �–system if and only if there is a set Rsuch that for all a, a0 2 X, if a 6= a

0, then a \ a

0 = R.

R is called the root of X.

Theorem 8.8. (ZFC) Let X be an uncountable set such that everya 2 X is finite. Then there is an uncountable set X 0 ✓ X such that X 0

is a �–system.

Proof. We may fix a one–to–one enumeration (a⌫

: ⌫ < !1) of (anuncountable subset of) X.

Let F be a certain finite set of formulas. Exactly which formulas arein F can be decided by looking at the rest of the proof and identifyingexactly which correctness facts we will be using. By Reflection we mayfind an ordinal ↵ such that V

↵

is correct about ' for every ' 2 F , i.e.,for all ' 2 F and Z 2 V

↵

, V↵

|= '(Z) if and only V |= '(Z).

48 D. ASPERO

Let M be an elementary substructure of (V↵

,2) such that (a⌫

:⌫ < !1) 2 M and |M | @0 (M exists by the Lowenheim–SkolemTheorem we have seen). In fact it will typically be the case that actually|M | = @0, for example if F contains a sentence saying that ! exists.

We may assume thatM is correct about countable sets, and thereforex ✓ M for every countable x 2 M : For every x 2 M , x is countableif and only if M |= x is countable (if F contains “x is countable”).Since we may assume that ! 2 M and ! ✓ M , if x is countable, thenthere is, in M , a bijection f : ! �! x. But then x = f [!] ✓ M . Inparticular, �

M

:= M \ !1 2 !1: If ↵ 2 � and � 2M \ !1, then � ✓M

since � is countable, and therefore ↵ 2M . Hence �\M is a countabletransitive set of ordinals and therefore it is an ordinal.

Since X is countable, we may take ⌫⇤ 2 !1 \ �M .We may also assume that M |= Axiom of unordered pairs. Hence,

R := a

⌫

⇤ \M 2M (since a

⌫

⇤ \M is a finite subset of M).

Claim 8.9. For every ⌫ 2 �M

there is some ⌫ 0 2 !1, ⌫0> ⌫, such that

R ✓ a

⌫

0 and

a

⌫

0 \[

⌫<⌫

a

⌫

✓ R

Proof. Let ⌫ 2 �M

. The statement “there is some ⌫ 0 2 !1, ⌫ 0 > ⌫, suchthat R ✓ a

⌫

0 and a

⌫

0 \S

⌫<⌫

a

⌫

✓ R” – with ⌫, R and (a⌫

: ⌫ < !1) asparameters – is witnessed by a

⌫

⇤ : Certainly R ✓ a

⌫

⇤ . Also, for every⌫ < ⌫, a

⌫

✓M , and hence a

⌫

⇤ \ a

⌫

✓ a

⌫

⇤ \M = R.But since we may assume M to be correct about this statement, it

holds in M , which gives that M thinks that there is some ⌫ 0 2 !1,⌫

0> ⌫, such that R ✓ a

⌫

0 and a

⌫

0 \S

⌫<⌫

a

⌫

✓ R. But again we mayassume that M is correct about this fact, so in V it is true that thereis some ⌫ 0 2 !1, ⌫ 0 > ⌫, such that R ✓ a

⌫

0 and a

⌫

0 \S

⌫<⌫

a

⌫

✓ R. ⇤Claim 8.10. For every ⌫ < !1 there is some ⌫ 0 < !1, ⌫

0> ⌫, such

that R ✓ a

⌫

0 and such that

a

⌫

0 \[

⌫<⌫

a

⌫

✓ R

Proof. Suppose otherwise. We may assume that M correctly believesthe negation of the claim. But then there is some ⌫ 2 �

M

such thatM thinks that there is no ⌫

0< !1, ⌫ 0 > ⌫, such that R ✓ a

⌫

0 anda

⌫

0 \S

⌫<⌫

a

⌫

✓ R. Since we may assume as well that M is correctabout this fact about a

⌫

, this fact is true in V. But this contradictsClaim 8.9. ⇤

Using Claim 8.10 we can build a �–system (a⌫i)i<!1 with root R by

recursion on !1:


a

⌫0 is any ⌫ such that R ✓ a

⌫

(which exists by Claim 2). Giveni, assuming a

⌫i0 has been defined for all i0 < i, let ⌫i

be the least⌫ > sup

i

0<i

⌫

i

0 such that

• R ✓ a

⌫

, and• a

⌫

\S

i

0<i

a

⌫i0 ✓ R.

This ⌫ exists by Claim 2. This completes the proof of the theorem. ⇤Simplification: It would have been enough to take any ordinal ↵

such that (a⌫

: ⌫ < !1) 2 V

↵

and then take any countable M 4 V

↵

such that (a⌫

: ⌫ < !1) 2 M . The reason this su�ces is that V

↵

iscorrect about all the relevant facts about (a

⌫

: ⌫ < !1).

9. Forcing

Forcing is a method, devised by Paul Cohen (1934–2007) in 1963 inorder to prove relative consistency results in the context of ZF and ZFC.In other words: For some statement � in the language of set theory,prove Con(ZFC) ! Con(ZFC+�) (and similarly with ZF instead ofZFC, or with certain extensions of ZFC, etc.). In particular, Cohenproved

• Con(ZF)! Con(ZF+¬AC)• Con(ZF)! Con(ZFC+2@0 = @2), Con(ZF)! Con(ZFC+2@0 =@27), Con(ZF)! Con(ZFC+2@0 = @

!2+3456), etc.

These results earned him the Fields medal in 1966. The consistencyof ZFC and of ZFC+2@0 = @1 (in other words, ZFC+CH) relative tothe consistency of ZF had been proved already in the late 1930’s by K.Godel; in fact he had proved, in ZF, that L |= ZFC+CH.

Forcing has proved to be an extremely powerful method; a cornucopiaof di↵erent models of set theory have appeared using this method inthese 50 years. In fact, forcing is essentially the only method we havefor producing consistency (with ZF or ZFC or ZFC + large cardinals) ofstatements with interesting mathematical content (i.e., not ‘artificiallybuilt’ statements as in the proof of Godel’s incompleteness theorems).It might even be that there is an actual theorem behind the aboveobservation, but this is still one of the main open questions in settheory (known as the ⌦ Conjecture).

The main informal idea of forcing can be described as the wish tobuild an extension V[G] of the universe V, where

• V[G] is the “minimal” extension W of V satisfying ZF andsuch that G 2W, and

50 D. ASPERO

• G is a set that has been chosen in such a way that V[G] neces-sarily satisfies some statement � of interest (V[G] is forced tosatisfy �).

Strictly speaking this is of course nonsense: V is the universe, itcontains all sets!One way to make sense of the above idea (not the only way, but

conceptually the cleanest for us now) is the following: Working in V:We will start with a countable model N of a su�ciently large finitefragment F of ZFC. Exactly which theory F is will depend on exactlywhich facts in the metatheory the argument at hand is going to use.Our argument will be finite (all arguments are finite), so it will usea finite set of axioms. We can find such a model N in V, withoutassuming Con(T ) in V, thanks to Reflection + Downward Lowenheim–Skolem (!).

We will then be able to find a transitive model (M,2) isomorphic to(N,2) (M is the transitive collapse of N). Why?

Say that a relation R is set–like i↵ for all x, {y : yRx} is a set.

Theorem 9.1. (Mostowski) Suppose (X,R) is a well–founded set–likerelation such that (X,R) |= Axiom of Extensionality. Then there is aunique transitive class (a set or a proper class) M for which there isan isomorphism ⇡ : (X,R) �! (M,2).

[Proof: Exercise and/or easy to find.]Using the theory of forcing we will be able to find a set G (in V, of

course) such that

• M [G] is the ✓–minimal transitive modelM 0 such thatM ✓M

0,G 2M and such that M 0 |= F .

(when building M [G] we say that we have forced over M). If we areskilled enough, we will be able to ensure that M [G] |= � for somestatement � of interest.

What is this good for?Now suppose we want to prove Con(ZFC)! Con(ZFC+�). Assume

Con(ZFC). Suppose, towards a contradiction that there is a proof of¬� from ZFC.

Let F0 be the finite set of axioms of ZFC involved in this proof. LetF ◆ F0, F ✓ ZFC, be still finite and such that the all relevant points inthe theory of forcing that we are about to see can be developed withinF . Since ZFC is consistent, let (M, E) |= ZFC. Working inside (M, E),argue as before: Find, in (M, E), a countable transitive model M of F(F can be coded by a natural number in the meta-theory, but (M, E)contains (interpretations of) all natural numbers from the meta-theory,


so F 2 M and F is, in M, a finite set of ZFC axioms). Note that Mis only transitive from the point of view of (M, E), of course.

Suppose in M we can force over M and build M [G] 2M such thatM [G] |= F +� (this last step will obviously depend of the statement �at hand, if � is 0 = 1 this of course cannot be done, if it is 2@0 = @27,then yes).

Now we are done: Since F0 ` ¬�, M [G] |= ¬�. But we have shownM [G] |= �. Contradiction.

Note that the above construction scheme does not work to provestatements of the form Con(ZF)! Con(ZF+�) in general (i.e., withoutAC): One reason is that inside of (M, E) we might not be able ingeneral to find a countable transitive models M of F . And even if wecould find it, we might in general not be able to find a suitable genericfilter G overM within (M, E) (see Subsection 9.2). There are, however,other constructions to make things work also in the ZF–context.

9.1. The method of forcing: Partial orders and genericity (anexample). We take a closer look at the idea behind the constructionof forcing extensions of our countable transitive model M . Let us seea specific example (and the simplest one):

Adding a Cohen real: For us here, a real will be a function f :! �! 2 (equivalent, a subset of !). We want to add a new real x to M

and build the ✓–minimal extension M

0 such that M ✓M

0 and x 2M .We want to make sure that x is not in M . In particular, for every

real r 2 M there must be some n 2 ! such that x � n 6= r � n. Thenotion of genericity over <!2 will guarantee that:

Let us consider the partial order C = (<!2, ) 2 M , where q p

if and only if p ✓ q (intuitively, q p means that q “carries moreinformation about the generic real” (see below)). We say that q isstronger than p. We call members of C conditions in C and say that Cis a forcing notion.

So C is the partial order consisting of finite sequences of 0’s and 1’s,and given two conditions p, q in C, q is stronger than p if and only ifq extends p. Reals correspond to infinite branches through C (growingdownwards, that is).

We say that D ✓ <!2 is dense in C if and only if for every p 2 <!2there is some q 2 D such that q p. For example, for every r : ! �! 2,r 2M ,

D

r

= {p 2 <!2 : p 6= r � n for some n < !} 2M

is s dense set.

52 D. ASPERO

We say that x : ! �! 2 is C–generic over M if and only if the branchb = {x � n : n < !} is such that for every dense D ✓ <!2, D 2 M ,b \ D 6= ;. The idea is that x is generic over M if x is “completelyrandom” from the point of view of M , in the sense that it avoids every“regularity pattern” specifiable in M . Here is a specific example: theset

X = {y 2 !2 : y(n) = y(n+ 1) for every even n < !}

is such that

D = {p 2 <!2 : p 2 r for some r 2 !2 \X} 2M

is dense. Hence, the generic real x will be such that x(n) 6= x(n+1) forsome even n. And similarly, it will also be such that x(n) = x(n + 1)for some even n. And so on.

Remarkable fact: If x is C–generic over M , M transitive, andM |= F for a su�ciently large finite fragment F of ZFC, then there isM [x] such that M [x] is the ✓–minimal M 0 such that

• M

0 is a transitive model of F ,• M ✓M

0, and• x 2M

0.

In other words, M 0 is the ✓–minimal obtained from adding the newreal x to M . We call x a Cohen real over M (because it is generic forC over M and C is known as Cohen forcing).

9.2. Formal development of forcing. A forcing notion is a partialorder (P,P) with a P–maximum, which we will sometimes denote1P. (1P is the trivial condition or weakest condition). Members of Pare called P–conditions.

We say that two P–conditions p0, p1 are compatible if there is a P–condition p such that p P p0 and p P p1. Otherwise they are calledincompatible.

(P,P) is non-atomic if for every p 2 P there are q0 P p and q1 P p

such that q0 and q1 are incompatible.D ✓ P is dense if and only if for every p 2 P there is some q 2 D,

q P p.A nonempty set G ✓ P is a filter if the following holds:

• If q 2 G and p 2 P, q P p, then p 2 G.• For all p1 2 G and p0 2 G there is some p 2 G such that p P p0

and p P p1.


Let M be a model of (enough of) ZFC. Let (P,) 2M be a partialorder in M . Let G ✓ P, G 2 V, be a filter. We say that G is P–genericover M if G \D 6= ; for every dense set D ✓ P such that D 2M .

Note: If P is non-atomic and G is P–generic over M , then G /2 M :D = P \G is dense ([Exercise]). But now, if G 2M , then D 2M , andso G \D = ;. Contradiction.

Given a transitive M , a forcing notion P 2M and a P–generic filterG over M , we will build the forcing extension (or generic extension)M [G]. M ✓ M [G] and G 2 M [G], so M 6= M [G] if P is non-atomic.However, M will have a certain amount of access to M [G] (this will bethe reason why M [G] |= F if M |= F ). In fact, members x of M [G]will have names x in M . x will be the interpretation of x by G:Given a partial order P, we define the class of P–names by recursion

on the rank, as follows: A set x is a P–name if and only if it is a set ofordered pairs (p, y) such that

• p 2 P and• y is a P–name.

For example, ; is the simplest name. And if p, p0 2 P, then both of⌧ = {(p, ;)} and � = {(p, {(p, ;)}), (p0, ;)} are P–names.

The following names play a distinguished role: Given a forcing notionP, we define x (for all x), again by recursion on the rank, as follows:For every x,

x = {(1P, y) : y 2 x}x is called the canonical P–name for x.Interpreting names: Given a P–name x and a filter G ✓ P, the

interpretation of x by G, denoted val

G

(x) or xG

, is

x

G

= {yG

: there is some p 2 G such that (p, y) 2 x}For example, the interpretation of the name ; is of course always ;

itself. Now suppose ⌧ and � are the above names and G ✓ P is a filter.If p 2 G, then ⌧

G

= {;} = 1, but if p /2 G, then ⌧G

= ; (so di↵erentnames, like ; and ⌧ , can sometimes have the same interpretation).Also, if p 2 G, then �

G

= {{;}} = {1} if p0 /2 G and �G

= {{;}, ;} = 2if p0 2 G. Finally, if p /2 G, then �

G

= ; if p0 /2 G, and �G

= {;} = 1 ifp

0 2 G. We thus see in particular that � can be interpreted as one of{1}, 2, 0 and 1.The canonical name of any x is always interpreted as x:

Lemma 9.2. For every forcing notion P, every set x, and every filterG ✓ P, x

G

= x.

Proof. By induction on the rank: Suppose y

G

= y for all y 2 x. Thenx

G

= {⌧G

: (p, ⌧) 2 x for some p 2 G such that (p, ⌧) 2 x} = {yG

:

54 D. ASPERO

y 2 x}, where this equality is true since necessarily 1P 2 G (as G 6= ;).But {y

G

: y 2 x} = {y : y 2 x} by induction hypothesis, and{y : y 2 x} is of course x. ⇤

Now we can define M [G]: Given a transitive model M , a forcingnotion P 2M , and a P–generic filter G over M ,

M [G] = {xG

: x 2M, x a P–name}

LetG = {(p, p) : p 2 P}

Lemma 9.3. For every transitive M , every forcing notion P 2M andevery P–generic filter G over M , if G is the above name, then G

G

= G.

Proof. GG

= {pG

: (p, p) 2 G for some p 2 G} = {p : p 2 G} =G. ⇤Corollary 9.4. Let M be a transitive model of F , let P 2 M be aforcing notion, and let G ✓ P be a filter. Then the following hold.

(1) M ✓M [G] and G 2M .(2) M [G] is transitive and Ord\M = M [G] \Ord.(3) If N is a transitive model of F and M[{G} ✓ N , then M [G] ✓

N .

Proof. (1): By the first of the above two lemma, for every x 2 M ,x

G

= x 2M [G]. By the second one, since G 2M , GG

= G 2M [G].(2): It x 2 M and y 2 x

G

, then in particular y = y

G

for some P–name y such that (p, y) 2 x for some p 2 G. But y 2 M since M istransitive ({{p}, {p, y}} 2M , which implies {p, y} 2M , and thereforey 2M). Hence M [G] is transitive.

Ord\M ✓ Ord\M [G] since a 2 M for every ordinal ↵ 2 M and↵

G

= ↵.

Also: rank(xG

) rank(x) for every P–name x [Exercise.] But then, if↵ = x

G

2 OrdM [G] for some P–name x 2M , then ↵ rank(x) 2 OrdM

(because every transitive model of ZF is correct about rank).(3): Since G 2 N , N can correctly build x

G

from x and G by recur-sion. Hence x

G

2 N for every P–name x 2M . ⇤We will see that if P 2M is a forcing notion in M and G is P–generic

over M , then M |= F (i.e., if M satisfies enough of ZFC, then M [G]satisfies enough of ZFC). But why can we find G?

Since M is countable, we can fix an enumeration (Dn

: n < !) of alldense subsets D of P such that D 2M (this enumeration is obviously


not in M). Now we can build G recursively, using AC, as follows. Let(q

n

)n<!

be a P–decreasing sequence such that

• q0 2 D0, and• for all n, q

n+1 2 D

n+1, qn+1 P q

n

.

Given q

n

, qn+1 may be found since M knows that D

n+1 is dense. Thesequence (q

n

)n<!

is not in M , but each q

n

is. Finally, let G = {p 2 P :q

n

P p for some n}.

9.3. The forcing relation �P.

Definition 9.5. Given a transitive model M of (enough of) ZF, aforcing notion P 2 M , a formula '(x0 . . . , xn�1) in the language of settheory, P–names ⌧ 0, . . . , ⌧n�1 in M , and p 2 P,

p �P '(⌧0, . . . , ⌧

n�1)

if and only if for every G, if G is a P–generic filter over M and p 2 G,then M [G] |= '(⌧ 0

G

, . . . , ⌧

n�1G

).We will say that p forces '(⌧ 0, . . . , ⌧n�1) and will call �P the forcing

relation for P.Strictly speaking we should have included M in the definition (M

will be understood or irrelevant).

We will see that �P is actually definable overM . In fact we will definea certain relation �⇤

P over M by recursion on complexity of formulas,and will prove

�P =�⇤P

Given D ✓ P and p 2 P, D is dense below p if and only if for everyq 2 P, if q P p, then there is some r 2 D, r P q.

Lemma 9.6. If G ✓ P is P–generic over M , p 2 P, D ✓ P is densebelow p, and D 2M , then G \D 6= ;.

[Proof: Exercise.]

Definition 9.7. Let p 2 P.• Given P–names � and ⌧ , p �⇤

P � 2 ⌧ i↵

{r 2 P : r �⇤P � = ⇡ for some (q, ⇡) 2 ⌧ such that r P q}

is dense below p.• p �⇤

P � = ⌧ if and only if for all (q, ⇡) 2 � [ ⌧ and all r suchthat r P p and r P q, r �⇤

P ⇡ 2 ⌧ i↵ r �⇤P ⇡ 2 �.

• Given a formula '(x0, . . . , xn�1) and P–names ⌧ 0, . . . , ⌧n�1,

p �⇤P ¬'(⌧ 0, . . . , ⌧n�1)

i↵ for all q P p, q 1⇤P '(⌧

0, . . . , ⌧

n�1).

56 D. ASPERO

• Given formulas '0(x0, . . . , xn�1) and '1(x0, . . . , xn�1) and P–names ⌧ 0, . . . , ⌧n�1,

p �⇤P '(⌧

0, . . . , ⌧

n�1) ^ '(⌧ 0, . . . , ⌧n�1)

i↵ p �⇤P '0(⌧ 0, . . . , ⌧n�1) and p �⇤

P '1(⌧ 0, . . . , ⌧n�1).• Given a formula '(x, x0, . . . , xn�1) and P–names ⌧ 0, . . . , ⌧n�1,

p �⇤P (8x)('(⌧ 0, . . . , ⌧n�1))

i↵ for every P–name ⌧ , p �⇤P '(⌧, ⌧

0, . . . , ⌧

n�1).

Note that there is no circularity in the definition of �⇤P for atomic

formula as it is by recursion on the rank of the corresponding names.And neither it is for non-atomic formulas since the definition is byrecursion on their complexity.

Lemma 9.8. Let '(x0, . . . , xn�1) be a formula, let ⌧ 0, . . . , ⌧n�1 be P–names, and let p 2 P. Then:

(1) If p �⇤P '(⌧

0, . . . , ⌧

n�1) and q P p, then q �⇤P '(⌧

0, . . . , ⌧

n�1).(2) p �⇤

P '(⌧0, . . . , ⌧

n�1) if and only if {q 2 P : q �⇤P '(⌧

0, . . . , ⌧

n�1)}is dense below p.

(3) If p 1⇤P '(⌧ 0, . . . , ⌧n�1), then there is some q P p such that

q �⇤P ¬'(⌧ 0, . . . , ⌧n�1).

[Proof: Exercise.]

Theorem 9.9. (Forcing Theorem) Suppose M is a transitive model of(a fragment of) ZF, P 2 M is a forcing notion, and for every p 2 Pthere is a P–generic filter G over M with p 2 G. Let G be a P–genericfilter over M , '(x0, . . . , xn�1) a formula, and let ⌧ 0, . . . , ⌧n�1 be P–names in M . Then the following are equivalent.

(1) M [G] |= '(⌧ 0G

, . . . , ⌧

n�1G

)(2) There is some p 2 G such that M |= p �⇤

P '(⌧0, . . . , ⌧

n�1).

Proof. By induction on the rank of names for atomic formulas and onthe complexity of '(x0, . . . , xn�1) for non-atomic '(x0, . . . , xn�1).� 2 ⌧ : (=)). If �

G

2 ⌧G

, choose (p, ⇡) 2 ⌧ such that �G

= ⇡

G

andp 2 G. By induction we can choose q 2 G such that q �⇤

P � = ⇡. Bythe lemma, {r P q : r �⇤

P � = ⇡} is dense below q. Then q �⇤P � 2 ⌧

by definition of “q �⇤P � 2 ⌧”.

((=) Suppose p 2 G and

D = {r 2 P : r �⇤P � = ⇡ for some (q, ⇡) 2 ⌧, r P q}

is dense below p. Then by genericity we may choose r 2 G \ D and(q, ⇡) witnessing r 2 D. By induction, �

G

= ⇡

G

. But then q 2 G, and


therefore �G

= ⇡

G

2 ⌧G

(by definition of ⌧G

).

� = ⌧ : (=)) Suppose �G

= ⌧

G

. Let D be the set of p 2 P suchthat p �⇤

P � = ⌧ or for some (r, ⇡) 2 � [ ⌧ , p P r and p �⇤P ⇡ /2 �

or p �⇤P ⇡ /2 ⌧ . Then D is dense, by the definiton of �⇤

P � = ⌧ andthe previous lemma. Hence there is some q 2 G \ D, and necessarilyq �⇤

P � = ⌧ (using the induction hypothesis).((=) Suppose p 2 G and p �⇤

P � = ⌧ . By induction hypothesistogether with the definition of p �⇤

P � = ⌧ , for all (r, ⇡) 2 � [ ⌧ suchthat r 2 G we have that ⇡

G

2 �G

if and only if ⇡G

2 ⌧G

. Hence �G

= ⌧

G

.

¬'(⌧ 0, . . . , ⌧n�1): (=)) Consider the set

D = {p 2 P : p �⇤P '(⌧

0, . . . , ⌧

n�1) or p �⇤P ¬'(⌧ 0, . . . , ⌧n�1)}

D 2 M is dense by definition of p �⇤P ¬'(⌧ 0, . . . , ⌧n�1). Hence there

is some p 2 G \ D by density. If p �⇤P '(⌧

0, . . . , ⌧

n�1), then M [G] |='(⌧ 0

G

, . . . , ⌧

n�1G

) by induction hypothesis. It follows that necessarilyp �⇤

P ¬'(⌧ 0, . . . , ⌧n�1).((=) Suppose p 2 G, M |= p �⇤

P ¬'(⌧ 0, . . . , ⌧n�1), but M [G] |='(⌧ 0

G

, . . . , ⌧

n�1G

). By induction hypothesis there is p

0 2 G such thatM |= p

0 �⇤P '(⌧

0, . . . , ⌧

n�1). Let q P p, p0, and note that M |= q �⇤P

'(⌧ 0, . . . , ⌧n�1) and M |= q �⇤P ¬'(⌧ 0, . . . , ⌧n�1) by the lemma. But

this contradicts the definition of q �⇤P ¬'(⌧ 0, . . . , ⌧n�1).

'0(⌧ 0, . . . , ⌧n�1) ^ '1(⌧ 0, . . . , ⌧n�1): For this case we note that

M [G] |= '0(⌧0G

, . . . , ⌧

n�1G

) ^ '1(⌧0G

, . . . , ⌧

n�1G

)

i↵ M [G] |= '0(⌧ 0G

, . . . , ⌧

n�1G

) and M [G] |= '1(⌧ 0G

, . . . , ⌧

n�1G

) i↵ thereare p0 2 G and p1 2 G such that M |= p0 �⇤

P '0(⌧0, . . . , ⌧n�1) andM |= p1 �⇤

P '1(⌧0, . . . , ⌧n�1) (by induction hypothesis) i↵ there is somep 2 G such that M |= p �⇤

P '0(⌧ 0, . . . , ⌧n�1) ^ '1(⌧ 0, . . . , ⌧n�1) (�! istrue since p0, p1 2 G and G is a filter, and � is true since p �⇤

P '0^'1

implies p �⇤P '0 and p �⇤

P '1).

(9x)('(x, x0, . . . , xn�1)): (=)) For this case suppose that

M [G] |= (9x)('(x, ⌧ 0G

, . . . , ⌧

n�1G

))

and let ⌧ 2 M be a P–name such that M [G] |= '(⌧G

, ⌧

0G

, . . . , ⌧

n�1G

).By induction hypothesis we may fix some p 2 G such that M |= p �⇤

P'(⌧, ⌧ 0, . . . , ⌧n�1). But then

{q 2 P : M |= q �⇤P '(�, ⌧

0, . . . , ⌧

n�1) for some P–name � 2M} 2M

is dense below p, and therefore M |= p �⇤P (9x)('(x, ⌧ 0, . . . , ⌧n�1)).

58 D. ASPERO

((=) The set D of p such that M |= p �⇤P '(⌧,~⌧) for some P–name

⌧ 2 M or M |= p �⇤P ¬'(⌧,~⌧) for all P–names ⌧ 2 M is dense and in

M . By genericity there is some p

0 2 G \D, and necessarily there is aP–name ⌧ 2 M such that M |= p

0 �⇤P '(⌧,~⌧) as otherwise we reach a

contradiction as in previous case. But thenM [G] |= '(⌧G

, ⌧

0G

, . . . , ⌧

n�1G

)and hence M [G] |= (9x)('(⌧ 0

G

, . . . , ⌧

n�1G

)). ⇤Theorem 9.10. Let M be a transitive model of ZF, let P 2 M be aforcing notion, and let G be P–generic over M . Then M [G] |= ZF. If,in addition, M |= AC, then M [G] |= AC.

Proof. As M [G] is transitive and ! 2M [G], M [G] satisfies Extension-ality, Infinity, and of course Foundation.

Axiom of unordered pairs: Given P–names ⌧ 0, ⌧ 1, consider ⌧ ={(1P, ⌧ 0), (1P, ⌧ 1)}. Then ⌧G = {⌧ 0

G

, ⌧

1G

}.Union: Given a P–name ⌧ , let

� = {(p, ⇡) : p �⇤P ⇡ 2

[⌧, rank(⇡) < rank(⌧)}

By the forcing theorem,

�

G

=[

⌧

G

\ {⇡G

: ⇡ 2M a P–name, rank(⇡) < rank(⌧)}

But any member ofS⌧

G

is ⇡G

for some name ⇡ 2 M such thatrank(⇡) < rank(⌧). Hence �

G

=S⌧

G

.Power set: Given a P–name ⌧ 2 M , let ↵ = max{rank(P), rank(⌧)}

and � = ↵ + !. Let

� = {(p, ⇡) : p �⇤P ⇡ ✓ ⌧, ⇡ 2M, rank(⇡) < �} 2M

Then

�

G

= P(⌧G

) \ {⇡G

: ⇡ 2M a P–name, rank(⇡) < �}again by the forcing theorem. Now let ⇡ 2 M be a name such that⇡

G

✓ ⌧

G

. Let

⇡

⇤ = {(p, x) : (p, x) 2 ⇡, rank(x) < ↵, p �⇤P x = x}

Then rank(⇡⇤) < � and it is easy to see that ⇡⇤G

= ⇡

G

(since ⇡G

✓ ⌧

G

and hence every member of ⇡G

is xG

for some x with (q, x) 2 ⌧ for someq, and therefore such that rank(x) < ↵). Hence, �

G

= P(⌧G

) \M [G],so M [G] |= �

G

= P(⌧G

).Replacement: Let ⌧ 2 M be a name, and let f : ⌧

G

�! M [G] bea function definable over M [G] possibly with parameters. Given anordinal ↵ 2M , let

�

↵ = {(p, ⇡) : rank(⇡) < ↵, p �⇤P ⇡ 2 range(f)}


(we are abusing notation here; f really refers to the definition of f).Then

�

↵

G

= range(f) \ {⇡G

: ⇡ 2M a P–name, rank(⇡) < ↵}once again by the forcing theorem. By the right instance of Replace-ment in M there is an ordinal ↵ 2M such that if p 2 P, rank(⇡0) < ↵,p �⇤

P ⇡0 2 ⌧ , and p �⇤P f(⇡0) = ⇡, then there is such a ⇡ such that

rank(⇡) < ↵ (given such ⇡0 and p consider the minimal ↵ such thatthere is a name ⇡ such that rank(⇡) = ↵ and p �⇤

P f(⇡0) = ⇡). Now itis easy to check that �↵

G

= range(f).Finally suppose the Axiom of Choice holds in M . Let ⌧ 2 M be a

name and let < be a well–order on the set of names � 2 M such thatrank(�) < rank(⌧).

Now we can well–order ⌧G

as follows: Given x, y 2 ⌧

G

, x <

G

y i↵x < y, where x is the <–first name z such that z

G

= x and y is the<–first name z such that z

G

= y. ⇤The following is a corollary of the above proof.

Corollary 9.11. Suppose F ✓ ZF is finite. Then there is a finite F 0 ✓ZF such that for every transitive M |= F

0, every forcing notion P 2M ,and every P–generic filter G over M , M [G] |= F . And similarly withZFC instead of ZF.

10. Two forcing constructions

The first of our constructions will be adding many Cohen reals to amodel of (a fragment) of ZFC. With this construction we will show theconsistency of the continuum being arbitrarily large.

10.1. Adding many Cohen reals.

Definition 10.1. Let ↵ be an ordinal and let Add(!,↵) be the fol-lowing forcing notion. A condition in Add(!,↵) is a finite functionp ✓ (↵⇥ !)⇥ 2. Given Add(!,↵)–conditions p0, p1, p1 Add(!,↵) p0 i↵p0 ✓ p1.

Lemma 10.2. Let M be a transitive model of (enough of) ZF, let↵ 2 M be an ordinal and let � < ↵. If G is generic for Add(!, ↵)M

(= Add(!, ↵)) over M , then

c

G

�

= {(n, ✏) 2 ! ⇥ 2 : ((↵, n), ✏) 2 p for some p 2 G}

is a Cohen real over M (in other words, cG�

is 2<!–generic over M).

Furthermore, if � < �

0< ↵, then c

G

�

6= c

G

�

0.

60 D. ASPERO

Proof. Let D 2 M be a dense subset of 2<!. Then D = {p 2Add(!, ↵) : {(n, ✏) : ((↵, n), ✏) 2 p} 2 D} is obviously dense. Hence,c

G

↵

\D 6= ;. Now suppose �0< ↵, �0 6= �.

Claim 10.3. The set

D

�,�

0 = {p 2 Add(!,↵) : p(�, n) 6= p(�0, n) for some n 2 !} 2M

is dense.

Proof. Obvious, since conditions in Add(!, ↵) are finite functions andthus involve only finitely many n 2 !. ⇤

By the claim, there is some n < ! such that cG�

(n) 6= c

G

�

0(n), and soc

G

�

6= c

G

�

0 . ⇤Corollary 10.4. If M is a transitive model of enough of ZFC, ↵ 2M

is an ordinal, and G is Add(!, ↵)–generic over M , then M [G] |= 2@0 �|↵|.

In this corollary, if in particular ↵ = @1000, then M [G] |= 2@0 �|@V1000|. Is this enough to conclude the consistency of ZFC+2@0 � @1000relative to that of ZFC? Not quite yet. For all we know so far it couldbe that M [G] |= |@M1000| < @1000. It could even be that @M1000 has becomecountable in M . We will see next that this does not happen, and infact @M1000 = @

M [G]1000 .

10.2. Chain condition and cardinal preservation.

Definition 10.5. Given a partial order P, A ✓ P is an antichain if pand p

0 are incompatible (i.e., there is no q 2 P, q P p, q P p

0) for allp, p0 2 A, p 6= p

0. A ✓ P is a maximal antichain if A is an antichainand it is ✓–maximal with respect to ✓ (equivalently, for every p 2 Pthere is some p

0 2 A and some q 2 P such that q P p, p0).

Definition 10.6. Given a cardinal , a partial order P has the –chaincondition (P has the –c.c.) i↵ there are no antichains A of P such that|A| = . If P has the @1–c.c., then we also say that P has the countablechain condition (P is c.c.c.).

Fact 10.7. (ZFC) Suppose P is a forcing notion and D ✓ P is dense.Then there is a maximal antichain A of P such that A ✓ D.

Proof. Let (p⇠

)⇠<µ

be an enumeration of D, where µ = |D|. Build(⇠

i

)i<µ

0 , for some cardinal µ0 µ, as follows:

• ⇠0 = 0• Let i > 0 be given and suppose (⇠

i

0)i

0<i

defined.


– If there is no p 2 D such that p is incompatible with allp

0 2 {p⇠i0 : i

0< i}, then we let A = {p

⇠i0 : i

0< i} and

stop.– If there is some p 2 D such that p is incompatible with allp

0 2 {p⇠i0 : i

0< i}, then we let i be least such that p

⇠i isincompatible with all p

⇠i0 .

A is an antichain of P by construction. Suppose it is not maximal.Let p 2 P be such that p is incompatible with all p0 2 A. Since D isdense, let q 2 D, q P p. By construction of A there is some p

0 2 A

and some q0 2 P such that q0 P p

0, q. But then q

0 P p

0, p, and so p iscompatible with some p

0 2 A. Contradiction.⇤

Lemma 10.8. Suppose P is a forcing notion, � < are infinite cardi-nals, f is a P–name, and p 2 P is a condition such that

p �P f is a function, f : � �!

Suppose P is c.c.c. Then p �P f is not onto .

Proof. For every ⇠ 2 � let D⇠

= {p 2 P : p �P f(⇠) = ↵ for some ↵ 2}.

Claim 10.9. D⇠

is dense.

Proof. Given p 2 P, let G be P–generic, p 2 G. Since p �P f : � �! is a function, f

G

: � �! is a function. Let ↵ < be such thatf

G

(⇠) = ↵. By the forcing theorem there is some p

0 2 G such thatp

0 �P f(⇠) = ↵. Since p, p0 2 G, we may find q 2 P, q p, p0. Butthen q �P f(⇠) = ↵. ⇤

Given ⇠ < �, let A

⇠

✓ D

⇠

be a maximal antichain in P. By ourhypothesis, |A

⇠

| @0. Given ⇠ < �, let

A⇠

= {↵ < : p �P p(⇠) = ↵ for some p 2 D

⇠

}

Claim 10.10. |A⇠

| @0.

Proof. For every ↵ 2 A⇠

let q

↵

2 A

⇠

compatible with some conditionp

↵

2 P forcing f(⇠) = ↵. Since p

↵

and q

↵

are compatible and q

↵

�Pf(⇠) = ↵

0 for some ↵0, it must be that ↵0 = ↵. Hence A⇠

✓ {↵ : p �Pf(⇠) = ↵ for some p 2 A

⇠

}. But |{↵ : p �P f(⇠) = ↵ for some p 2A

⇠

}| @0. ⇤

Finally, by the forcing theorem, range(fG

) ✓S

⇠<�

A⇠

for every P–generic G over M , p 2 G. But M |= |

S⇠<�

A⇠

| |� ⇥ @0| = � <

62 D. ASPERO

impliesS

⇠<�

A⇠

6= (it is easy to see that |!⇥�| = � for every infinite

cardinal � [Exercise.]). Hence fG

cannot be onto for any such G. ⇤Corollary 10.11. Suppose M is a transitive model of enough of ZFC,P 2M and M |= P is c.c.c. Suppose G is P–generic over M . Then, forevery 2 M such that M |= is a cardinal, M [G] |= is a cardinal.

In particular, for every ordinal ↵ 2M , @M↵

= @M [G]↵

.

Lemma 10.12. (ZFC) For every ordinal ↵, Add(!, ↵) has the count-able chain condition.

Proof. Suppose, towards a contradiction, that (p⌫

)⌫<!1 enumerates an

uncountable antichain of Add(!, ↵). Since {dom(p⌫

) : ⌫ < !1} is acollection of finite sets, by the �–system lemma we know that there isX ✓ !1, |X| = @1, such that {dom(p

⌫

) : ⌫ 2 X} forms a �–systemwith root R; in other words, for all distinct ⌫, ⌫ 0 in X, dom(p

⌫

) \dom(p

⌫

0) = R. Note that there are only finitely many functions fromthe finite set R to 2 (in fact there are exactly 2|R| such functions). Since|X| = @1 and 2|R| is finite, it follows that there are distinct ⌫, ⌫ 0 2 X

such that p⌫

� R = p

⌫

0 � R. But then,

p

⌫

[ p

⌫

0 : dom(p⌫

) [ dom(p⌫

0) �! 2

is a function. Since q := p

⌫

[ p⌫

0 extends both p

⌫

and p

⌫

0 , we have thatq Add(!,↵) p⌫ , p⌫0 . A contradiction. ⇤Corollary 10.13. Suppose M is a transitive model of enough of ZFC, 2 M is such that M |= “ is a cardinal”, and G is an Add(!, )–generic filter over M . Then

(1) for every ordinal � 2 M , M |= “� is a cardinal” if and only ifM [G] |= “� is a cardinal”, and

(2) M [G] |= 2@0 �

In particular, if ↵ 2 Ord\M and = @M↵

, then M [G] |= 2@0 � @↵

.

10.3. �–closure and not adding new reals. We see a companionresult next: The consistency of ZFC+CH, relative to the consistencyof ZFC, using forcing (the original proof of this result is due to Godel,long before forcing was invented: he built the constructible universe,L, and proved that L |= ZFC+CH. We don’t have time here to saymuch about L except for its definition, which we have already seen (cf.after Corollary 7.10).

Definition 10.14. A partial order P is �–closed i↵ for every P–decreasing sequence (p

n

)n<!

of P–conditions there is some q 2 P suchthat q P p

n

for all n.


Lemma 10.15. Suppose M is a transitive model of enough of ZFC,P 2 M is a forcing notion, M |= P is �–closed, G is P–generic overM , and f : ! �! Ord, f 2M [G]. Then f 2M .

Proof. For every n < !, Dn

= {p 2 P : p �P f(n) = ↵ for some ↵} 2M is dense. Also, of course (D

n

)n<!

2M . It su�ces to see that thereis an ordinal ✓ 2M such that

D = {p 2 P : p �P f = g for some function g 2 V

M

✓

} 2M

is dense.For this let p 2 P. Build, in M , a P–decreasing sequence (p

n

)n<!

of conditions in P extending p and such that pn+1 2 D

n

for all n.Let p0 = p. Given p

n

, we can find p

n+1 since D

n

is dense. Finally,let q P p

n

for all n. q exists since P is �–closed in M . But thenq 2 D. ⇤Definition 10.16. Let Coll(P(!),!1) be the following partial order:p 2 Coll(P(!),!1) i↵ p ✓ !1 ⇥ P(!) is a function such that |p| @0.Given Coll(P(!),!1)–conditions p0, p1, p1 Coll(P(!),!1) p0 i↵ p0 ✓ p1.

Lemma 10.17. Coll(P(!),!1) is �–closed.

Proof. Obvious, since the union of a decreasing !–sequence in the par-tial order Coll(P(!),!1) is a countable function. ⇤Lemma 10.18. For every transitive model M of enough of ZFC andevery r ✓ !, r 2M , D

r

= {p 2 Coll(P(!),!1)M : r 2 range(p)} 2M

is a dense subset of Coll(P(!),!1)M .

Proof. If p 2 Coll(P(!),!1)M and r /2 range(p), then p[ {(↵, r)} 2 D,where ↵ is any ordinal in !M

1 \ dom(p) 6= ;. ⇤Similarly we can prove:

Lemma 10.19. For every transitive model M of enough of ZFC andevery ↵ 2 !M

1 , E↵

= {p 2 Coll(P(!),!1)M : ↵ 2 dom(p)} 2 M is adense subset of Coll(P(!),!1)M .

Corollary 10.20. Let M be a transitive model of enough of ZFC andlet G be a generic filter for Coll(P(!),!1)M . Then M [G] |= 2@0 = @1.

Proof. By the last two lemmas above,SG : !M

1 �! P(!)M is a sur-jection. Also, by the previous two lemmas, P(!)M [G] = P(!)M sinceColl(P(!),!M

1 ) is �–closed in M and therefore does not add new func-tions f : ! �! 2. But then M [G] thinks that

SG : !1 �! P(!) is

onto all of P(!). ⇤

64 D. ASPERO

By the above corollaries, together with the metamathematical con-siderations at the beginning of this section on forcing, we have provedthe following:

• Con(ZFC) implies Con(ZFC+CH).• Con(ZFC) implies Con(ZFC+2@0 � @2), Con(ZFC+2@0 � @3),Con(ZFC+2@0 � @2454536254465), Con(ZFC+2@0 � @@@785665

),etc.

It is not di�cult to prove, by the same forcing construction we haveseen (adding many Cohen reals) using a slightly more refined analysis,that 2@0 can consistently be exactly @2, @3, @2454536254465, @@@785665

, etc.It is also easy to change the value of 2@1 , 2@2 , 2@!+1 , and so on,

by forcing. And, in fact, any constellation of values is possible forcardinals as the above (and many others), subject to some very simplerules. For example, it is consistent – modulo Con(ZFC) of course –that 2@0 = @1 + 2@1 = @27 + 2@2 = 2@3 = @28 + 2@!+1 = @

!1 , etc.On the other hand, there are very deep and surprising cardinal arith-

metical facts one can prove in ZFC for other cardinals! For example:

Theorem 10.21. (Shelah) (ZFC) Suppose 2@n< @

!

for all n < !.Then 2@!

< @!4.

A big open question in “pcf theory” is whether the above bound @!4

can be improved to something like @!1 .

Forcing has applications not only in cardinal arithmetical. In fact itcan be used to build models of set theory with many di↵erent proper-ties.

One example (J. Moore): If M is a transitive countable model of (afragment of) ZFC+LC, where LC is some (relatively weak) “largecardinal axiom”, then there is a forcing extension M [G] such thatM [G] thinks that there is a basis of exactly 5 elements for the un-countable linear orders; in other orders, in M [G] it holds that there isB = {L0, L1, L2, L3, L4} such that each L

i

is an uncountable linear or-der, and such that if L is any uncountable linear order, then L containsan isomorphic copy of at least one L

i

. In this model, 2@0> @1. In fact,

2@0 = @1 implies that if B is a basis for the uncountable linear orders,then |B| = 2@1 .

11. Improving ZFC?

We have seen that ZFC does not decide the value of 2@0 (i.e., does notdecide the ordinal ↵ such that 2@0 = @

↵

). Also, ZFC does not decidewhether or not there is a 5–element basis for the uncountable linear


orders, as well as infinitely many other questions in mathematics (hun-dreds or thousands of these questions have been explicitly consideredby mathematicians). Given the independence phenomenon, should wegather these questions as meaningless?

Not necessarily. We can hope to find natural axioms which, whenadded to ZFC, decide these questions. The most successful and so farbest understood family of such axioms come under the name of “largecardinal axioms”. The weakest such axiom asserts that there is anuncountable cardinal such that

• is a limit cardinal, and• is regular (i.e., is not the union of less than –many sets ofsize less than ).

Such a is called a (strongly) inaccessible cardinal.It is not possible to prove

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

because ZFC+“There is an inaccessible cardinal” proves Con(ZFC) (un-less we can prove ¬Con(ZFC), that is): The reason is that if is in-accessible, then V

|= ZFC. Now suppose towards a contradiction thatwe can prove the arithmetical statement

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

in, say, Peano Arithmetic or ZFC. But

ZFC+“There is an inaccessible cardinal”

proves Con(ZFC). Hence


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


is inconsistent by Godel’s 2nd Incompleteness Theorem. Hence ZFCwould be inconsistent.

We say that ZFC + “There is an inaccessible cardinal” is strictlystronger than ZFC.

Similarly:

• 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”.

66 D. ASPERO

• 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”.

• ...

The theories of the form ZF + LC (or ZFC+LC), where LC is a largecardinal axiom, form a hierarchy of theories natural linearly orderedby consistency strength.

It is a remarkable empirical fact that all ‘natural’ theories arisingin mathematics can be interpreted relative to some such theory, andthat in many cases they can be proved to be equiconsistent with sucha theory. In this sense, large cardinal axioms provide a very naturaltemplate for extending ZF or ZFC.

Su�ciently strong large cardinal axioms have also remarkable “astro-logical” properties: They prove things about ‘simply definable’ sets ofreals that ZFC does not decide, if consistent. It is surprising that largecardinal axioms, that assert the existence of certain objects very highin the cumulative hierarchy with properties that can only be recognisedby looking very high, have direct influence on objects lying very lowin the hierarchy. For example: If there are infinitely many Woodincardinals, then every projective set of reals has the usual regularityproperties (it is Lebesgue measurable, has the Baire property, and hasthe perfect set property). On the other hand, in the absence of suchlarge cardinals it is possible that there be a closed set X ✓ R3 suchthat Y is not Lebesgue measurable, where Y is the projection to Rof the complement, in R2, of the projection of X to R2. This Y is,by construction, a projective set, in fact quite low in the projectivehierarchy.

(Un)fortunately one can prove that large cardinal axioms do notdecide 2@0 . For this one has to look at other families of axioms, com-patible with large cardinal axioms (for example “forcing axioms”).

Acknowledgements: Thanks to Daoud Siniora for a list of correc-tions.

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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