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NOTES ON BASIC SET THEORY

RODRIGO HERNANDEZ-GUTIERREZ

This notes are basically the contents of the course on Set Theory that I have given atFacultad de Ciencias, UNAM. The official programs can be found in the following:

www.fciencias.unam.mx/estudiosProfesionales/asignaturas/0760.pdf

www.fciencias.unam.mx/estudiosProfesionales/asignaturas/0779.pdf

Proofs are only drafted and details are left for the reader to prove.

Contents

The Axioms 2

Set Formation 2

Replacement schema 4

Foundation 4

The Axiom of Choice 5

How to use the axioms 5

1. General concepts 5

1.1. Basic Constructions 5

1.2. Binary Relations 7

1.3. Partial orders 8

1.4. Lattices 11

1.5. Equivalence Relations 11

1.6. Functions 11

1.7. Classes 12

2. Natural numbers 12

2.1. Definition of Natural numbers 12

2.2. Recursion 14

3. Construction of the Reals 14

4. Equinumerosity 14

5. Ordinal numbers 15

Date: February 17, 2011.

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2 RODRIGO HERNANDEZ-GUTIERREZ

5.1. Well-ordered sets 15

5.2. Ordinals 18

5.3. Transfinite recursion 21

5.4. Ordinal arithmetic 22

5.5. Indecomposable ordinals 26

6. Cardinality 29

6.1. Cardinal numbers 29

6.2. Axiom of choice 32

6.3. Cofinality 32

6.4. Infinite operations 32

Appendix A. The class WF 33

References 34

The Axioms

In this notes, we use the Set Theory of Zermelo and Fraenkel, a system oftenrepresented by ZF. We list the axioms here and give a brief explanation.

First we make a note what type of language we use in set theory. We only talkabout sets and membership ∈. That is, in one hand, if we write x then it must bethat x is a set. So there are no ‘sets of apples’ or ‘sets of cats’ because apples andcats are not sets. We only consider sets whose elements are sets, this is possible asyou will notice as you continue reading. On the other hand, when we talk aboutx, y (which have to be sets), we can only say whether x ∈ y or not. So we can justtalk about set membership. Our axioms will tell us what the rules for playing withsets are.

But we must also be able to distinguish what type of logical statements we canread in set theory. First, given two sets x, y, we can say whether x = y or whetherx ∈ y, these are called the ‘atomic’ formulas. Given formulas φ, θ, we can consider¬φ, φ ∨ θ and the quantification ∃x(φ). By inductively iterating this process, weobtain all the formulas we can consider in set theory. Following our intuition, wehave some abbreviations: we write x /∈ y for ¬(x ∈ y), φ ∧ θ for ¬(¬φ ∨ ¬θ), ∀x(φ)for ¬∃x(¬φ). The meaning (semantics) of formulas is the usual, we will not give theprecise mathematical definition of them. The interested reader should read aboutModel Theory (for example see ??).

Set Formation. First, let’s state the axioms that tell us how to construct sets.This axioms were proposed by Zermelo (see ??).

Axiom 1: Extensionality. A set can be caracterized by its elements

∀x, y[(∀z(z ∈ x↔ z ∈ y))→ x = y]

NOTES ON BASIC SET THEORY 3

We define x ⊂ y if for every z ∈ x, we have z ∈ y.

Exercise 1. Prove the well-known and always used equivalence: x = y if and onlyif x ⊂ y and y ⊂ x.

Axiom 2: Existence of sets. There exists a set which has no elements.

∃Θ(∀x(x /∈ Θ))

We call this Θ the empty set and always write it as ∅.

Exercise 2. Prove that the empty set is unique.

Axiom 3: Pair formation. For each pair of sets, a, b, there is a set whose elementsare precisely a and b.

∀a, b[∃x(∀z(z ∈ x↔ (z = a ∨ z = b)))]

We call this set {a, b}.

Exercise 3. Prove that {x} is a set whenever x is a set.

Axiom 4: Union. For each set x, we can consider the set whose elements arez ∈ y such that y ∈ x.

∀x[∃X(z ∈ X ↔ ∃y ∈ x(z ∈ y))]

We call such a set⋃

x.

We define, as usual, the abbreviation A∪B =⋃

{A,B}. By induction (see ??), wealso write

⋃

{Ak : k < n+ 1} = A0 ∪ · · · ∪An when n ∈ ω.

Axiom 5: Infinity. There exists an inductive set.

∃W [∅ ∈W ∧ ∀x ∈W (x ∪ {x}) ∈W ]

This axiom implies the existence of an infinite set, in particular the set of naturalnumbers ω = {0, 1, . . .}. It requires more work to formalize these notions. Theywill be developed in Section ??.

Axiom 6: Power Set. If x is a set, the class of all subsets of x is a set.

∀x[∃X(y ∈ X ↔ y ⊂ x)]

We call this the power set of x and denote it by P(x).

One of the fundamental problems in Set Theory is to determine how big P(X) isin relation to x. It turns out that this question is independent of the axioms of thissection (see ??).

In the early stages of set theory, it was purposed that if φ is a formula with one freevariable, the class {x : φ(x)} should be a set. However, there is an contradiction ifwe accept this axiom. This is known as Russel’s paradox : let R = {x : x /∈ x}.

Exercise 4. Assume R is a set and prove R ∈ R if and only if R /∈ R.


However, restricting to subsets of a fixed set, we remove this contradiction.

Axiom 7: Specification schema. Consider φ(x) a formula in the language ofset theory and A a set such that φ does not talk about A. Then {y ∈ A : φ(y)} isa set.

We must now comment on the technical parts of this axiom. It is called an schemabecause it is really not only one axiom. For each φ(x) as described, we have theSpecification Axiom for φ(x). So this axiom is really a countable collection ofaxioms, one for each formula in the language of set theory. We will not say whythere are countably many formulas, if the reader is interested, she should read???. Also, we ask that φ does not “talk” about A (precisely speaking, φ is freeon A) because we don’t want to consider properties of A whose definition dependson A itself: this would lead to circular definitions which do not make sense. Forexample, consider B = {x ∈ A : x ∈ B} which is not well-defined. Other names forSpecification are Separation and Comprehension.

Replacement schema. The replacement schema is, as in the case of the Specifi-cation schema, a countable family of axioms indexed by the formulas of set theory.It basically says that if we have a function f : X →M from a set X to a class M(see subsection 1.7), the image is a set.

Axiom 8: Replacement schema. Consider φ(x, y) a formula in the language ofset theory and S a set such that for each s ∈ S there is a unique set ys such thatφ(s, ys) holds. Then, {ys : s ∈ S} is a set.

The first time this axiom appears is in Theorem 5.2.13. It is not essential for thedevelopment of mainstream mathematics but it is in the theorem of ordinals andhigher set theory. The Replacement schema was first proposed by Fraenkel andSkolem (??).

As an example of the use of this axiom, we can define inductively (see ??) W0 = ωand Wn+1 = P(Wn) for n < ω. Then, by the replacement schema {Wn : n < ω} isa set.

Foundation. The foundation axiom is another technical part of Set Theory whichdoes not really play an important role in the rest of mathematics.

Axiom 9: Axiom of Foundation. If x is non-empty set, then there is a y ∈ xwhich is ∈-minimal (see ??):

∀x 6= ∅[∃y ∈ x(x ∩ y = ∅))]

Exercise 5. Prove that the axiom of fundation implies that there is no set {xn :n ∈ ω} such that xn+1 ∈ xn for all n ∈ ω.

We will say that a set x is well-founded if either x = ∅ or there is a y ∈ x such thatx ∩ y = ∅. The foundation axiom can be shown to be consistent with axioms 1 to8. An outline of this proof can be found in Appendix A.


The Axiom of Choice.

Axiom 10: Axiom of Choice (AC). Every set can be well-ordered.

How to use the axioms. We will always assume axioms 1 through 8 and callthis system ZF−, Zermelo-Fraenkel without Foundation. Axiom 9 is refered as theaxiom of foundation and can be shown to be consistent with ZF− (see AppendixA). We refer to the theory ZF− plus foundation as ZF. The axiom of choice ACcan be proved to be consistent with ZF but will will not assume it until the chapteron cardinals (see ??). We call ZFC the theory of ZF+AC.

1. General concepts

In this section, we will give construction of specific types of sets used in Mathe-matics. We will also give examples of how these sets are used. In some of theseexamples we will assume that we know about specific sets that will be constructedin later sections but are well-known (for example the set of real numbers R definedin ??).

1.1. Basic Constructions. We have defined the union of sets in Axiom 4. It iscommon to define the intersection too.

1.1.1. Definition Let A be a nonempty set. Define the intersection of A as⋂

A = {a : ∀A ∈ A, a ∈ A}.

Exercise 6. Prove that⋂

A is a set. How is the fact that A 6= ∅ used?

As in the case of the union, one may define A ∩ B =⋂

{A,B} and⋂

{Ak : k <n+ 1} = A0 ∪A1 ∪ . . . An.

One may also define the complement of a set inside another set. However, thisconcept is relative.

1.1.2. Definition Let A and B be sets. Then the complement of B in A is

A−B = {x ∈ A : x /∈ B}.

Clearly, A−B is a set by the Schema of Specification.

A common misconception in elementary education is to define the “universal set”,that is, a set that contains all sets as elements. This “universal set” would be whatwe call the class V of all sets (see subsection 1.7 on classes). If V were a set, givenanother set A we should be able to define an absolute complement {x ∈ V : x /∈ A}(commonly written as AC). However, then we could also form R = {x : x /∈ x}which gives Russel’s paradox (see Exercise 4). Another strange thing one can do isin the following Exercise.

Exercise 7. Assume that V is a set and prove that using the definition of inter-section,

⋂

∅ is well-defined and equals V.


Thus, in Mathematics we are not allowed to consider the universal set or the abso-lute complement.

We also call A − B the difference. Notice that properties like the so-called De-Morgan laws hold.

Exercise 8. Let A, B be sets such that A 6= ∅. Show

B − (⋃

A) =⋂

{B −A : A ∈ A}

B − (⋂

A) =⋃

{B −A : A ∈ A}.

Another construction used in specialized mathematics is that of the symmetriccomplement or symmetric difference.

1.1.3. Definition Given sets A,B, we define their symmetric complement

A∆B = (A ∪B)− (A ∩B).

Exercise 9. Show A∆B = (A−B) ∪ (B −A).

The next construction we define is the Cartesian product of two sets. To define aproduct, we must first define what is meant by ordered pair. The definition of or-dered pair we use is one proposed by Kazimierz Kuratowski, a Polish mathematician(1896-1980).

1.1.4. Definition If x, y are sets, we define

(x, y) = {{x}, {x, y}}.

The importance of ordered pairs is precisely, that they are ordered. More formally,we have the following. t

1.1.5. Theorem For every x, y, z, w such that (x, y) = (z, w), we have x = z andy = w.

Exercise 10. Prove Theorem 1.1.5. (Hint: for a more organized proof, considerthe cases x = y and x 6= y.)

Notice that we may write a formula for “the set z is an ordered pair” as ∃x, y(z =(x, y)) so that we can define the set of ordered pairs in the following way.

Exercise 11. Prove that if x ∈ X and y ∈ Y , then the ordered pair (x, y) is anelement of P(P(X ∪ Y )).

By Exercise 11, we may define sets of ordered pairs.

1.1.6. Definition For each pair X,Y we define their Cartesian product in thefollowing way:

X × Y = {z ∈ P(P(X ∪ Y )) : ∃x, y(z = (x, y))}.

A trivial example of this is when one of the “factors” is the empty set.


Exercise 12. Show that A×B = ∅ if and only if A = ∅ or B = ∅.

Notice that the definition of Cartesian product is not associative.

Exercise 13. Prove that A× (B × C) 6= (A×B)× C in general.

Although we may not state distributivity of Cartesian product, we can identify twoforms of writing a Cartesian product of three factors as the same. This is done bymeans of a bijection (see ??).

Exercise 14. Although we haven’t studied functions yet, show a “natural” one-to-one function from A× (B × C) onto (A×B)× C (that is, a bijection).

The definition of products with more than two factors could be done inductively(see ??) using the idea of Exercise 14. However, it is not clear how to define infiniteproducts (although we yet haven’t defined what “infinite” means). In ?? we willgive a more general definition of Cartesian product of an arbitrary collection ofsets. We need to develop the machinery of Binary Relations first.

1.2. Binary Relations. The informal, non-mathematical idea of defining a re-lation is to compare two different things in some way. There are basically threediferent ways in which one may do that:

(1) Consider when something is bigger/better/greater than another thing, inthis case we are considering order relations,

(2) Divide things in groups such that the things in a group are similar in someway, in this case we are considering equivalence relations,

(3) Define a rule that assigns to each thing another thing in a unique way, inthis case we are considering functions.

The mathematical definition is far more general than this

1.2.1. Definition A binary relation (or relation for short) is a subset R ⊂ A×Bfor some sets A, B.

Trivially, the empty set is a relation, we call it the empty relation when we wantto emphasize its role as a relation. Given a set A, another “natural” relationwhich we can always define on A × A, we call it the identity and it is defined asidA = {(a, a) : a ∈ A}.

We now give definitions of many aspects of relations that will be useful for ourtreatment of the subject.

1.2.2. Definition If R is a relation, we define the domain and image as

dom(R) = {x : ∃y (x, y) ∈ R}im(R) = {y : ∃x (x, y) ∈ R}

Exercise 15. Show that dom(R) and im(R) are sets.

Notice that when we say R is a relation, it implies there exist sets A,B such thatR ⊂ A × B (trivially, by definition). However, from a set-theoretic point of view,


we define R first so that A and B must exist, contrary to the common view ofmainstream mathematics that defines A and B first. This is because this R is asubset of A×B but it is also a subset of dom(R)× im(R) and when we define R wewould like to forget about which A and B to choose. However, this will not applyto Orders, Equivalence Relations and Functions, as we will see in the followingsections.

1.2.3. Definition If R and S are relations, define their composition

R ◦ S = {(x, z) : ∃y (x, y) ∈ S ∧ (y, z) ∈ R}.

Exercise 16. Show that R◦S is a set. Also show that in general R◦S is not equalto S ◦R.

1.2.4. Definition Given a relation R, we define the inverse of R as

R−1 = {(x, y) : (y, x) ∈ R}.

Exercise 17. Show that R−1 is a set.

Finally, we define the operation of taking direct and inverse image of a set. Noticethat the image of a set under a relation R is different from the set im(R) in general.

1.2.5. Definition Given a relation R and an arbitrary set A, define the directand inverse image of A as

R[A] = {y : ∃x ∈ A (x, y) ∈ R},R←[A] = {x : ∃y ∈ A (x, y) ∈ R},

respectively.

Note. The sets R[A] and R←[A] are commonly written as R(A) and R−1(A),respectively. However, the author of this notes prefers to write them as in thedefinition given because of the possible confusion presented when one defines thenotation y = f(x) for functions. See ??.

The definition of image of a set does not mention the domain of R so one may take,for example, A 6= ∅ such that A ∩ dom(R) = ∅, we obtain R[A] = ∅.

In the following three sections, we will study the three special kind of relations(which are most used in mainstream mathematics): order relations, equivalencerelations and functions.

1.3. Partial orders. There are two types of orders one has to deal with in Math-ematics. We now define both and state their relationship.

1.3.1. Definition A partial order on a nonempty set A is a relation R ⊂ A × Asuch that

(1) R is reflexive, that is, (a, a) ∈ R for all a ∈ A,(3) R is antisymmetric, that is, if (a, b) ∈ R and (b, a) ∈ R, then a = b.(2) R is transitive, that is, if (a, b), (b, c) ∈ R, then (a, c) ∈ R.


1.3.2. Definition A strict order on a nonempty set A is a relation R ⊂ A × Asuch that

(1) R is antireflexive, that is, (a, a) /∈ R for all a ∈ A,(2) R is transitive, that is, if (a, b), (b, c) ∈ R, then (a, c) ∈ R.

We will say that a set X is partially ordered (strictly ordered) if there is R ⊂ X×Xthat is a partial order (strict order) on X. In both cases, one usually abbreviates“X is partially/strictly ordered by R” by “(X,R) is a partial/strict” and writesxRy to mean (x, y) ∈ R.

Exercise 18. Let A be a nonempty set and R ⊂ A × A. Prove that R is a strictorder if and only if R ∩ idA = ∅ and R ∪ idA is a partial order.

For example, for any set A, idA is a partial order for A and ∅ is its correspondingstrict order. According to exercise 18 we sometimes denote a partial order by ≤and its induced strict order by <.

1.3.3. Example (The inclusion order). For a set X, we may define a partial orderR in P(X) by R = {(A,B) : A ⊂ B}. We call this the inclusion order and simplydenote it by (P(X),⊂). The inclusion order is in many ways a cannonical ordering,as will be seen in the examples that will follow. �

Exercise 19. Prove that the inclusion order (P(X),⊂) is indeed a partial order.

We next define concepts for special elements associated to subsets.

1.3.4. Definition Let (X,≤) be a partial order and A ⊂ X.

• We say a ∈ X is an upper bound (lower bound, respectively) if for eachx ∈ A, x ≤ a (a ≤ x, respectively).• An element a ∈ X is the greatest (least, respectively) element of A if a ∈ Aand a is an upper bound (lower bound, respectively) of A.• An element a ∈ X is the supremum (infimum, respectively) of A if a is theleast element (greatest element, respectively) of the set of upper bounds(lower bounds, respectively) of A.• An element a ∈ X is a maximal element (minimal element, respectively) ofA if a ∈ A and every time x ∈ A is such that a ≤ x (x ≤ a, respectively),then a = x.

Exercise 20. Let (X,≤) be a partially ordered set and A ⊂ X. Show that “thegreatest” is well-defined. That is, show that if a, b ∈ X are such that the definitionof greatest element of A holds for both a and b, then a = b. Do the same for thedefinitions of the least, supremum and infimum.

Let (X,≤) be a partial order and A ⊂ X. We usually denote the greatest (max-imum), least (minimum), supremmum and infimum of A by maxA, minA, supAand inf A, respectively (when they exist).

Now we show some examples of the previously defined concepts in concrete partiallyordered sets.


Exercise 21. Consider the inclusion order in P(X) (Example 1.3.3) and let A ⊂P(X). Show that the supremum of A exist and is equal to

⋃

A. What about theinfimum of A?

1.3.5. Example (maximals may not be unique). A concrete example which showsmaximal elements may not be unique is the following. Let X = {∅, {0}, {1}}and consider (X,⊂) (the inclusion order from Example 1.3.3). Notice X has twomaximal elements: {0} and {1}

1.3.6. Example (Vector space basis). Let k be a field (for example k = R ork = C) and V a vector space over k. Consider ℓ the set of all nonempty familiesof linearly independent elements of V . We know that maximal elements of ℓ areprecisely those subsets of V that are basis of V . We may thus consider (ℓ,⊂) as apartial order with the inclusion relation (Example 1.3.3). It can be shown in ZFC(using Zorn’s Lemma ??) that every element F ⊂ ℓ is contained in a maximal G ∈ ℓ,this is comonly said “every linearly independent subset can be enlarged to a basis”.More concretely, in R2, one may start with {(0, 1)} and {(0, 2)} and enlarge themto {(0, 1), (1, 0)} and {(0, 2), (1, 0)} which are both basis but one is not containedin the other. Thus, a partially ordered set may contain several maximal elements,which must be pairwise incomparable. �

For the next example, we will use the definition of ω. For the time being, we willconsider 0 = ∅ and for each n ∈ ω, n + 1 = n ∪ {n}. Thus, one may think thatn+1 = {0, 1, . . . , n}. The formalization of this concept will be given in Chapter 2.

1.3.7. Example (Divisibility). If m,n ∈ ω, we define the partial order m ⊳ n ifand only if m divides n (the exists k ∈ ω such that n = m · k).

Notice the following properties of this partial order.

(1) The greatest element is 0 because every natural number divides 0.(2) The least element is 1 because every natural number is divisible by 1.(3) The minimal elements of ω − {1} are precisely the prime numbers.(4) There are no maximal elements in the subset ω − {0}.

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Exercise 22. Show that (ω,⊳) is indeed a partial order and the properties of Ex-ample 1.3.7 hold.

We end this section with one special kind of partial orders.

1.3.8. Definition Let (X,≤) be a partial order. We say that it is a linear orderif for every two x, y ∈ X, we have x ≤ y or y ≤ x.

Examples of linear orders are (ω,≤) and its extensions such as R. For the following,it is easier to think about R or the rationals Q.


1.3.9. Definition Let X be a partially ordered set. An interval is a subset J ⊂ Xsuch that if x, y ∈ J and z ∈ X is such that x < z < y, then z ∈ J . We also definethe following intervals, for x, y ∈ X:

• (x, y) = {z ∈ X : x < z < y},• (x, y] = {z ∈ X : x < z ≤ y},• [x, y) = {z ∈ X : x ≤ z < y},• [x, y] = {z ∈ X : x ≤ z ≤ y},• (←, y) = {z ∈ X : z < y},• (←, y] = {z ∈ X : z ≤ y},• (x,→) = {z ∈ X : x < z},• [x,→) = {z ∈ X : x ≤ z},

Exercise 23. Show that not all intervals are like the ones defined in definition1.3.9. (Hint: consider Q)

1.4. Lattices.

1.4.1. Definition A partially ordered set (L,≤) is called a lattice if for everya, b ∈ L, both sup {a, b} and inf {a, b} exist.

If (L,≤) is a lattice, we denote a ∨ b = sup {a, b} and a ∧ b = inf {a, b}. These twosymbols may be thought as binary “operations” on L.

1.4.2. Example . By exercise 21, the inclusion order in P(X) forms a lattice.

1.4.3. Lemma Let L be a lattice. Then the following properties hold for everya, b, c ∈ L.

(1a) a ∨ a = a, (1b) a ∧ a = a,(2a) a ∨ b = b ∨ a, (2b) a ∧ b = b ∧ a,(3a) a ∨ (b ∨ c) = (a ∨ b) ∨ c, (3b) a ∧ (b ∧ c) = (a ∧ b) ∧ c.

Exercise 24. Prove Lemma 1.4.3.

pendiente hablar de distributividad, retıculas completas y algebras booleanas

1.5. Equivalence Relations.

1.5.1. Definition Let A be a nonempty set. An equivalence relation on A is arelation R ⊂ A×A such that

(1) R is reflexive, that is, (a, a) ∈ R for all a ∈ A,(3) R is symmetric, that is, if (a, b) ∈ R then (b, a) ∈ R.(2) R is transitive, that is, if (a, b), (b, c) ∈ R, then (a, c) ∈ R.

falta mucho de aqui

1.6. Functions.


1.6.1. Definition Let (X,≤1) and (Y,≤2) be two partially ordered sets andf : X → Y a function.

(a) f is a order-preserving function if x ≤1 y implies f(x) ≤2 f(y),(b) f is a isomorphism if it is an order-preserving function bijective function.(c) if f is a isomorphism and X = Y , then we call f an automorphism.

Exercise 25. Prove that if f is a isomorphism between linearlly ordered sets, thenthe inverse function f−1 is also an isomorphism.

1.6.2. Definition Let f, g be functions. We say that g extends f if f ⊂ g.

1.6.3. Lemma Let F a set of functions such that if f, g ∈ F, then either f ⊂ g org ⊂ f . Then

⋃

F is a function with dom(⋃

F) =⋃

{dom(F ) : F ∈ F}.

Exercise 26. Prove Lemma ??.

poner el ejemplo del conjunto de Cantor como el conjunto de anticadenasde <ω2.

1.7. Classes. In ZF−, classes have no formal existence. However, for each formulaof set theory φ, we would like to consider the “class” of all sets x such that φ(x),so we write

M = {x : φ(x)}

as an abbreviation. For example, x ∈M means “for x, φ(x) holds”. If N is anotherclass, then N ⊂M is an abbreviation of “every set x such that x ∈ N, we also havex ∈M”. Notice, however, that statements such as N ∈M are meaningless.

The biggest class we deal with is the universe of all sets V = {x : x = x} (see ??).We may think that for all classes M, M ⊂ V accorging to last paragraph.

Somtimes we would like to compare elements of some class by some relational. Forexample, in V we may want to say that x is “smaller” than y if x ∈ y (in particular,when we define ordinal numbers). We do this in the following way. Start with aclass M and define M×M = {(x, y) : x, y ∈M} (that is, an element is in M×Mif it is an ordered pair (x, y) so that both x and y are in M). A relational is a classR ⊂M×M such that the axioms of partial order, definition 1.3.1, hold exchangingA by M and R by R (or definition 1.3.2 if we want to make it strict).

A relational F ⊂ M ×M is called a functional if every time (a, b), (a, c) ∈ F,then b = c (compare to the definition of function, ??). We call dom(F) the class{x : ∃y such that(x, y) ∈ F}.

2. Natural numbers

2.1. Definition of Natural numbers. Our objective in this section is to definethe natural numbers from the axioms of set theory. First we will give a definitionof the kind “a set is a natural number if...” and then with the axiom of infinity wewill prove the natural numbers form a set.

We want that the natural numbers are exactly 0 = ∅ and n + 1 = {0, . . . , n}. Wewill refer to these sets as our “intuitive natural numbers” (because we have not


yet formalized their existence and behavior). We will model these natural numbersusing set-theoretic properties.

2.1.1. Definition A set x is said to be transitive if every time y ∈ x, we also havey ⊂ x.

2.1.2. Example (Transitive sets). Obviously the empty set is transitive. If therewere a set a such that a = {a} (which is consistent with ZF−), then this set wouldbe transitive. Also, a set such as {∅, {∅}, {{∅}}} is transitive.

Notice that our intuitive natural numbers are transitive. However, by Example2.1.2, we need stronger conditions. Since n + 1 = {0, 1, . . . , n + 1, n} and 0 ∈ 1 ∈· · · ∈ n − 1 ∈ n ∈ n + 1, we would like that natural numbers are linearly orderedby the ∈ relation. This is not enough as the set a in Example 2.1.2 shows, becausea ∈ a. Thus, we want ∈ to be an strict linear order. It turns our that we need toask that ∈ is a well-order.

2.1.3. Definition Let (X,<) be a strict order. We say that (X,<) is a well-orderif every time A ⊂ X is non-empty, X has a <-least element.

2.1.4. Example (Well-ordered sets). Obviously the empty set is well-ordered by∈. Every finite linealy ordered set is also well-ordered, of course we need to waituntil the definition of finite before we prove this. The set of natural numbers(assuming it exists as we imagine it) is also well-ordered.

Note. Notice that for an arbitrary set A, the relation ∈ may not even be an order.For example, consider {∅, {∅}, {{∅}}}, in this case the relation ∈ is not transitive.

Because the set of natural numbers is well-ordered, we have to somehow restrict tofinite sets. This leads to the definition of natural number in the following way.

2.1.5. Definition We call a set x a natural number if: (a) x is a transitive set;(b) (x,∈) is a well-order; (c) every time ∅ 6= A ⊂ x, there exists a ∈ A a ∈-greatestelement.

aqui me quede redactando

2.1.6. Lemma If x is a natural number and y ∈ x, then y is also a natural number.

2.1.7. Definition For any set x, we define its succesor S(x) = x ∪ {x}.

2.1.8. Lemma If x is a natural number, then S(x) is also a natural number.

2.1.9. Proposition Let x be a natural number, x 6= ∅. If y = max (x), thenx = S(y).


We now use the axiom of infinity to prove that the natural numbers form a set.We first define in a set-theoretical way what it means that the famous principle ofmathematical induction holds in a set.

2.1.10. Definition A set W is inductive if ∅ ∈ W and every time x ∈ W , thenS(x) ∈W .

Acording to Axiom 5, there exists an inductive set so we may take the intersectionof all inductive sets.

2.1.11. Definition

ω =⋂

{W : W is inductive}

2.1.12. Lemma The set ω is inductive.

Our next step is proving that ω is in fact the set of natural numbers. Let N ={x : x is a natural number}, ?? To acomplish this, we first prove that every naturalnumber must be an element of ω so the class of natural numbers... ??

2.1.13. Proposition If x is a natural number, then x ∈ ω.

2.1.14. Theorem The natural numbers form a set a moreover N = ω.

poner notacion de 0 y de n+1

2.1.15. Corollary (The Principle of Mathematical Induction) Let A ⊂ ω benonempty. Assume 0 ∈ A and every time n ∈ A we also have n + 1 ∈ A. ThenA = ω.

2.1.16. Corollary (Peano Axioms) Redactar axiomas de Peano

aqui falta poner que los naturales estan bien ordenados

2.2. Recursion.

2.2.1. Theorem Let A be a set, a ∈ A and f : A×ω → A a function. Then thereis an unique function g : ω → A such that g(0) = a and g(S(n)) = f(g(n), n).

3. Construction of the Reals

4. Equinumerosity

If X,Y are sets, we say that they are equinumerous, X =∗ Y , if there is a bijectionh : X → Y . This concept corresponds to the idea of two sets having the samenumber of elements: the bijection h gives a correspondence between the elementsof X and the elements of Y . We can also say whether some set is smaller thananother in this sense. For sets X,Y we say X ≤∗ Y if there is an injective function


f : X → Y . We can think that this is a way to say that X can be ‘put inside’ Ywithout distorting it.

4.0.2. Theorem (Cantor-Schroeder-Bernstein) If X,Y are sets such that X ≤∗ Yand Y ≤∗ X, then X =∗ Y .

4.0.3. Proposition For every X, X2 =∗ P(X).

4.0.4. Proposition ω × ω =∗ ω.

5. Ordinal numbers

5.1. Well-ordered sets. We would like to extend the notion of mathematicalinduction to arbitrarily “big” sets. Let’s recall the definition of well-ordered set.

5.1.1. Definition Let (X,<) a strictly ordered set. We say that X is well-orderedby < if for every A ⊂ X with A 6= ∅, there exists a ∈ A a <-least element of A.

Exercise 27. Prove that a well-ordered set is totally ordered.

5.1.2. Example Some well-ordered sets. As examples of well-ordered sets we haveall natural numbers n ∈ ω and the set ω itself. We can also consider the followingsets, defined by recursion (we don’t have the tools to formalize this, we have towait for Theorem 5.3.1):

ω + 1 = ω ∪ {ω},ω + (n+ 1) = (ω + n) ∪ {ω + n}, (n ∈ ω)

ω · 2 =⋃

n∈ω (ω + n),

and ordered by ∈. Notice n ∈ ω ∈ ω + n ∈ ω · 2 for all n ∈ ω. It can easily beshown that these sets are well-ordered. �

Exercise 28. Prove that the sets defined in example 5.1 are well-ordered by ∈ andthat they are pairwise not isomorphic as ordered sets.

We can easily see that a well-order is preserved under some set-theoretic operations.For example, it is easy to see that the restriction of a well-order to a subset is againa well-order. From this, we get the following

5.1.3. Proposition If X can be well-ordered and Y ≤∗ X, then Y can be well-ordered.

5.1.4. Proposition Let A a well-ordered set and f : A → B is a surjective func-tion, then B is well-ordered.

Proof. Let g : B → A be defined as g(x) = min f←(x). g shows that B ≤∗ A so weare done by Proposition 5.1.3. �

Exercise 29. Do the details of Propositions 5.1.3 and 5.1.4.


Start with an infinite well-ordered set X and let x0 = minX. Defining inductivelyxn+1 = min(X − {x0, . . . , xn}) we can find an isomorphic copy of the naturalnumbers inside X. If X is not isomorphic to ω, then we can also find xω =min(X−{xn : n ∈ ω}). If the set X is “big” enough (in terms of order) we can findcopies of ω + n for n ∈ ω or ω · 2.

With this construction we can infer that a well-ordered set behaves just like ourexamples above. We will conlude that this is the case in the next section (see??). The first thing we have to notice is that there are two types of points in awell-ordered set.

5.1.5. Definition Let X be well-ordered and x ∈ X.

• If there is a y ∈ X such that x = min{z ∈ X : z > y}, then we say that xis the successor of y.• If x 6= minX and x is not a successor, then we say that x is a limit.

For example, in ω · 2, all elements different from 0, ω are successors and ω is theonly limit point.

Now we would like to classify well-ordered sets by order-preserving maps. The nextresult is fundamental for our study of this classification.

5.1.6. Lemma Let W be a well-ordered set and f : W → W a one-to-one order-preserving function. Then, for all x ∈W we have f(x) ≥ x.

Proof. Consider the set A = {x ∈ W : f(x) < x}, it is sufficient to prove thatA = ∅. Assume not and let x1 = minA. Then for each x < x1, we have x ≤ f(x).Since f is order-preserving and injective, we have that x ≤ f(x) < f(x1) for allx < x1. Thus, (←, x1) ⊂W −A is a set bounded above by f(x1).

Exercise 30. Prove that x1 = sup(←, x1).

By exercise 30, x1 ≤ f(x1) so we are done. �

From this apparently naive lemma we get the following facts

5.1.7. Corollary Let W a well-ordered set. Then the only automorphism of Wis the identity function.

5.1.8. Corollary Let W1 and W2 be two well-ordered sets. Then there is at mostone isomorphism from W1 onto W2.

Exercise 31. Prove corollaries 5.1.7 and 5.1.8 using lemma 5.1.6.

So well-ordered sets have an unique way to be “embeded” one in another. Now wewill prove that there is a sort of triconomy in the class of well-ordered sets. To thisend, we first define initial segments of well-orders:

5.1.9. Definition Let W be a well-ordered set. A subset J ⊂ W is an initialsegment of W if every time x ∈ J and y ∈ X is such that y < x, then y ∈ J .


Exercise 32. Show that if J is an initial segment of a well-ordered set W , theneither J = (←, x) for some x ∈ J or J = W .

We now see that any well-ordered set cannot be isomorphic to an initial segmentof itself:

5.1.10. Corollary Let W be a well ordered set and J ⊂ W an initial segment ofW . Then W is isomorphic to J if and only if W = J .

Exercise 33. Prove Corollary 5.1.10

Before the main theorem of this section, we need a technical lemma:

5.1.11. Lemma Let W1,W2 be a well-ordered sets, x, y ∈ W1, z ∈ W2 such thaty < x and there is an isomorphism h : (←, x)→ (←, z). Prove that h[(←, y)] = (←, h(y)).


We now prove the characterization theorem:

5.1.12. Theorem Let W1 and W2 be two well-ordered sets. Then one and onlyone of the following holds:

(1) W1 is isomorphic to W2,(2) W1 is isomorphic to a proper initial segment of W2,(3) W2 is isomorphic to a proper initial segment of W1.

Proof. Let

f = {(x, y) ∈W1 ×W2 : (←, x) is isomorphic to (←, y)}.

Exercise 35. Prove that if (x, y), (x, z) ∈ f , then y = z.

Exercise 36. Prove that if (y, x), (z, x) ∈ f , then y = z.

Thus, f : dom(f)→ im(f) is a bijection.

Take x, y ∈ dom(f) with y < x and assume f(x) < f(y). Then there are isomor-phisms h1 : (←, x) → (←, f(x)) and h2 : (←, y) → (←, f(y)). By lemma 5.1.11,h3 = h1|(←,y) : (←, y) → (←, h1(x)) is an isomorphism. Notice that since h1 pre-

serves order, (←, h1(x)) is a proper initial segment of (←, f(x)). Then h3 ◦ h−12 is

an isomorphism of (←, f(y)) with one of its proper initial segments. This is notpossible by Lemma 5.1.10. Thus, f(y) < f(x) so f is order-preserving.

Therefore, f : dom(f)→ im(f) is an isomorphism.

Next, take x ∈ dom(f) and y ∈ W1 such that y < x. So there is an isomorphismh : (←, x) → (←, f(x)). By Lemma 5.1.11, h|(←,y) : (←, y) → (←, h(y)) is an iso-morphism. This proves that y ∈ dom(f). Thus, dom(f) is an initial segment ofW1. By a similar argument, im(f) is an initial segment of W2.


Now assume that dom(f) 6= W1 and im(f) 6= W2 and let

x0 = min (W1 − dom(f)),y0 = min (W2 − im(f)).

Exercise 37. Prove that (←, x0) = dom(f) and (←, y0) = im(f).

Thus, (x0, y0) ∈ f which is a contradiction. We obtain that either W1 = dom(f)or W2 = im(f).

If W1 = dom(f), then f is an isomorphism of W1 with an initial segment of W2. IfW2 = im(f), then f−1 is an isomorphism of W2 with an initial segment of W2.

That no two of (a), (b), (c) can hold simultaneously follows from Lemma 5.1.10.

Exercise 38. Prove this

�

5.2. Ordinals. In this section we shall work in ZF−, that is, without the axiomof foundation. Ordinal numbers will be well-ordered sets that represent all “equiv-alence classes” of well-ordered sets. The examples of well-orders ω+n (n ∈ ω) andω · 2 are examples of ordinal numbers..

5.2.1. Definition An ordinal number α is a transitive set such that (α,∈) is awell-order.

The axiom of foundation is embedded in the definition of ordinal α because ∈ isdefined to be a strict well-order. So when one is asked to prove a set is an ordinalnumber, one must prove that ∈ is a strict well-order.

Notice that all natural numbers are ordinals, but also ω is an ordinal as well as thesets ω + n (n ∈ ω) and ω · 2.

5.2.2. Lemma If α is an ordinal and β ∈ α, then β is an ordinal.


In (??) we defined successors inside well orders but we now define the successor ofan ordinal number

5.2.3. Definition If α is an ordinal, then its successor is S(α) = α ∪ {α}.

5.2.4. Lemma If α is an ordinal, then S(α) is an ordinal.


Our objective is to prove that the class of ordinals is in fact, well-ordered. A firstexample of this is the next result.

5.2.5. Lemma If A is a set of ordinals, then⋃

A is an ordinal.



Now we prove a technical result which will help us in our main theorem but is alsointeresting in its own right.

5.2.6. Lemma Let α, β ordinals. Then α ⊂ β if and only if α = β or α ∈ β

Proof. We only need to prove necessity. Assume α 6= β and take γ = min (β − α).

Exercise 42. Prove that α = γ.

Therefore, α ∈ β. �

Consider the class of ordinal numbers ON with the relational ∈. By Lemma 5.2.2,∈ is “transitive” in the class of ordinal numbers and clearly α /∈ α for each α ∈ ON(by definition of ordinal number). So in the class ON, ∈ behaves like an strictorder. We now see that it is in fact linearlly ordered.

5.2.7. Theorem If α and β are ordinals, then one and only one of the followinghold: (a) α ∈ β, (b) β ∈ α or (c) β = α.

Proof. Let γ = α ∩ β.

Exercise 43. Prove that γ is an ordinal.

If either γ = α or γ = β, then α ⊂ β or β ⊂ α so by Lemma 5.2.5, we are done.Assume then that α 6= γ 6= β. Since γ ⊂ α and γ ⊂ β, by Lemma 5.2.6, we havethat γ ∈ α ∩ β. But this means that γ ∈ γ which contradicts the definition ofordinal. Thus, we have that one of (a), (b) or (c) hold.

Exercise 44. Prove that no two of (a), (b) or (c) can hold simoultaneously.

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We now prove that in fact ∈ “well-orders” ON.

5.2.8. Corollary Let A ⊂ ON. Then A has a ∈-least element.

Proof. Let α ∈ A. If α ∩ A = ∅, then by Theorem 5.2.7, for every β ∈ A we musthave α = β or α ∈ β, so α is the ∈-least element. Assume then that α ∩ A isnonempty and take γ = min (α ∩ A).

Exercise 45. Prove that γ is the ∈-least element of A, using Theorem 5.2.7.

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According to this facts, we will say that ON is well-ordered by ∈. Remmember thisis not a precise formulation in the language of set theory but only an abbreviationof the preceeding results (see subsection 1.7 on classes). We will interchangeblywrite α < β or α ∈ β when α, β ∈ ON.

A natural question one must ask is if ON is really a proper class and not a set.After all, the definition of natural number (see ??) was a little more than thedefinition of ordinal and the class of natural numbers turned out to be a set. Thenext results shows this is not the case.


5.2.9. Theorem The class ON is not a set

Proof. Assume it is.

Exercise 46. Prove that with this assumption, ON is an ordinal.

Thus, ON ∈ ON, which is a contradiction to the definition of ordinal. �

So the class ON is well-odered. We now show that one can find suprema in theclass of ordinals.

5.2.10. Proposition If A is a set of ordinals, then⋃

A is the supremmum of A.

Exercise 47. Prove Proposition 5.2.10.

From now on, we will understand that if A is a set of ordinals, then we will denoteits suppremum

⋃

A by supA.

5.2.11. Corollary If α is a limit ordinal, then⋃

α = α.

Exercise 48. Prove Proposition 5.2.10 and Lemma 5.2.11.

We can also notice that the only finite ordinals are the natural numbers:

5.2.12. Proposition Let α be an ordinal. Then α ∈ ω if and only if α is finite.

Proof. By definition, every natural number is finite. Conversely, if α =∗ n for somen ∈ ω, since n ∈ ω ⊂ α (Theorem 5.2.7 and Lemma 5.2.6), we have that n =∗ ωby the Cantor-Schoeder-Bernstein Theorem 4.0.2. This contradicts (??) so we aredone. �

Now we arrive to the result that tells us that ordinal numbers are representativesof “equivalence classes” of well-orderness.

5.2.13. Theorem Every well-ordered set is isomorphic to a unique ordinal.

Proof. We first notice two facts:

Fact 5.1. (Corollary 5.1.8) If W is well-ordered, α ∈ ON and h : W → α is aisomorphism, then h is the only isomorphism between W and α.

Fact 5.2. If W is well-ordered and α, β ∈ ON are such that W is isomorphic toboth α and β, then α = β.

Exercise 49. Prove fact 5.2.

Let X be a well-ordered set and consider

Y = {x ∈ X : there is an ordinal α such that (←, x) and α are isomorphic}.

By facts 5.1 and 5.2, for each x ∈ Y , we have a unique isomorphism hx : (←, x)→α(x) such that α(x) ∈ ON. Notice that by the schema of replacement (see 8),{α(x) : x ∈ Y } is a set.


Now we prove X = Y . Assume not and let x0 = min (X − Y ).

Exercise 50. x0 6= minX

So x0 is a successor or a limit of elements of X.

First, assume that x0 is the succesor of y ∈ X. By the definition of x0, there is anordinal α(y) and an isomorphism hy : (←, y)→ α(y).

Exercise 51. Prove that hy ∪{(y, α(y))} is an isomorphism from (←, x0) to S(α).

Thus, y ∈ Y which is a contradiction.

Now assume that x0 is a limit. For each x < x0, we have isomorphisms hx : (←, x)→ α(x) by the definition of x0.

Claim 5.1. If x, y ∈ Y are such that x ≤ y, then hx ⊂ hy.

Exercise 52. Prove Claim 5.1.(Hint: use facts 5.1 and 5.2)

Let h0 =⋃

x<x0hx.

Exercise 53. Prove that h0 is an isomorphism h0 : (←, x0) →⋃

{α(x) : x < x0}.(Hint: use Claim 5.1)

By Lemma 5.2.5,⋃

{α(x) : x < x0} is an ordinal, which contradicts the definitionof x0. Thus, X = Y .

Finally, let h =⋃

x∈X hx.

Exercise 54. Prove that h is an isomorphism from X to an ordinal. (Hint: useLemma ?? lema de la union seccion funciones Claim 5.1 and Lemma 5.2.5)

This finishes the proof of the theorem. �

So ordinal numbers represent all ‘equivalence classes’ of well-ordered sets. We canthus give an equivalent version of the axiom of choice.

5.2.14. Corollary AC is equivalent (in ZF−) to the statement that for every setX, there is α ∈ ON such that X =∗ α.

5.3. Transfinite recursion. In this section, we give a generalization of Theorem2.2.1 for arbitrary ordinals. Remmember that ON is not a set (Theorem 5.2.9)so we will have to talk about classes to be able to give this generalization. Seesubsection 1.7 for the basic stuff on classes we need.

5.3.1. Theorem (Transfinite Recursion) Let F be a functional with dom(F) ={x : x = x} (the class of all sets). Then there exists a functional G with dom(G) =ON such that for all α ∈ ON, G(α) = F(G|α).

Note. Notice that if a functional G with dom(G) = ON exists, since for eachα ∈ ON the class {β ∈ ON : β ∈ α} is a set, then for each α ∈ ON, G|α is a setby the axiom of replacement. Thus, we can talk about F(G|α), the ‘value’ that Fassigns to the set G|α.


Proof. For α ∈ ON, we shall say that g is a α-approximation if g is a functionwith dom(g) = α and g(β) = F (g|β) for all β ∈ α. Notice that this definition of α-approximation will only be used inside the proof of this result and has no relevancefor the rest of the notes.

Claim 5.2. If f is an α-approximation, g is a β-approximation for some α, β ∈ON, then f ⊂ g if and only if α ⊂ β.

Exercise 55. Prove this.

Claim 5.3. For each α ∈ ON, there exists an α-approximation.

Exercise 56. Use induction to prove this.

Thus, for each α ∈ ON, there is a unique α-approximation gα. Let G =⋃

α∈ONgα.

Exercise 57. Prove that G is the class we are looking for and use induction toprove it is unique.

�

5.3.2. Example The class of well-ordered sets. We now present the definition ofthe class of all well-founded sets WF as an application to Theorem 5.3.1. We wouldlike to define a class of sets {Vα : α ∈ ON} such that:

(i) V0 = ∅,(ii) Vα+1 = P(Vα) for each α ∈ ON,(iii) Vβ =

⋃

α<β Vα if β ∈ ON is a limit.

Let us define a functional F defined in the class of all sets. For X a set, we define:

• F(X) = ∅ if X = ∅,• F(X) = P(X(α)) if X is a function, dom(X) = α+ 1 for some α ∈ ON,• F(X) =

⋃

α<β X(α) if X is a function, dom(X) = β for some β ∈ ONlimit,• F(X) = ∅ in any other case.

By using the functional F just defined in the statement of Theorem 5.3.1, we obtaina functional G with dom(G) = ON. Define G(α) = Vα.

58. Exercise . Prove that properties (i), (ii) and (iii) hold for Vα.

We now define WF =⋃

α∈ONVα. If you are interested in what WF is for, take a

look at appendix A. �

5.4. Ordinal arithmetic. esta seccion esta muy incompleta

Now we would like to generalize the definition of addition and multiplication ofnatural numbers to the class of ordinals. This will give us an interesting source forexercises.

5.4.1. Definition For an ordinal number α, we define recursively:

• α+ 0 = α,


• α+ S(β) = S(α+ β) for each β ∈ ON,• α+ γ = supβ<γ(α+ β) for each limit ordinal γ.

Now with the sum defined, we can now define the multiplication so that it is dis-tributive:

5.4.2. Definition For an ordinal number α, we define recursively:

• α · 0 = 0,• α · S(β) = α · β + α for each β ∈ ON,• α · γ = supβ<γ(α · β) for each limit ordinal γ.

A first question is whether this operations behave in some way like the ones innatural numbers.

5.4.3. Lemma If α, β, γ ∈ ON are such that β < γ, then

(a) α+ β < α+ γ,(b) α · β < α · γ

5.4.4. Lemma If α, β ∈ ON and β is a limit, then α+β y α ·β are limit ordinals.

Exercise 59. Use induction to prove Lemma 5.4.3.

Using this we can now prove the associativity of both operations:

5.4.5. Proposition If α, β, γ ∈ ON, then α+ (β + γ) = (α+ β) + γ.

Proof. The proof will be by induction on γ.

Exercise 60. Prove (a) in the cases γ = 0 and γ = S(δ) for some δ ∈ ON.

Now assume that γ is a limit ordinal. Notice that by inductive hypothesis:

(α+ β) + δ = supδ<γ

[(α+ β) + δ] = supδ<γ

[α+ (β + δ)](1)

now notice that by lemma 5.4.4, we have that β+γ is a limit ordinal so, by definition,

α+ (β + γ) = supη<β+γ

(α+ η) = supδ<γ

[α+ (β + δ)].(2)

Using equations 1 and 2, we get (α+ β) + δ = α+ (β + γ). �

To prove the associativity of the product, we must first prove the distributive law.

5.4.6. Proposition Let α, β, γ ∈ ON. Then, α · (β + γ) = α · β + α · γ.

With this, we can finally prove the associativity of the product.

5.4.7. Proposition If α, β, γ ∈ ON, then α · (β · γ) = (α · β) · γ.


Now, the next results show us another way to ‘see’ the sum and product of ordinalsin some geometric way. Our first definition will be for the sum of ordinals:

5.4.8. Definition Let (X,≤0) and (Y,≤1) be two linearly ordered sets. In X⊕Y ,we define an order ≤2 in the following way: for x, y ∈ X ⊕ Y we have x <2 yprovided that one of the following conditions holds:

(1) both x, y ∈ X and x <0 y,(2) both x, y ∈ Y and x <1 y,(3) x ∈ X and y ∈ Y .

We call this new order ≤2 the sum order for (X,Y ).

To imagine what the sum order for (X,Y ) looks like, one might imagine that theposet X is placed before (to the left of) the poset Y . poner dibujo de esto

Exercise 61. Prove that the sum order for a pair of lineal orders (X,Y ) is a linealorder.

5.4.9. Lemma If α, β ∈ ON, then α ⊕ β with the sum order for (α, β) is a wellorder.


Now we prove that the sum of two ordinal coincides with their sum order. Sincetwo non-zero ordinals are never disjoint (see Theorem 5.2.7), we will use the precisedefinition of disjoint union of ??, α⊕ β = (α× {0}) ∪ (β × {1}) so that we do notget confused in the proof of the following theorem.

5.4.10. Theorem If α, β ∈ ON, then α + β is isomorphic to the sum order of(α, β).

Proof. We do our proof by induction on β.

For β = 0, h0 : α→ α×{0} is a isomorphsim. If β = S(ξ) for some ξ ∈ ON, thereis an isomorphism hξ : α+ ξ → (α×{0})∪ (ξ×{1}). Define hβ : α+ ξ → α⊕ β by

hβ(x) =

{

hξ(x) if x ∈ α+ ξ,(ξ, 1) if x ∈ α+ ξ.

Exercise 63. Prove that such an hβ is an isomorphism.

Now assume that β is a limit and for each ξ < β, hξ : α + ξ → α ⊕ ξ is anisomorphism.

Exercise 64. Prove that if ξ < γ < β, then hξ ⊂ hγ . (Use the techniques ofLemma 5.1.6 and Subsection 5.1.)

Define hβ =⋃

{hξ : ξ ∈ β}.

Exercise 65. Show that hβ : α→ β → α⊕ β is an isomorphism.

This completes the induction.

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With the sum order in mind, it is now easy to see that the sum is non-conmutative:

Exercise 66. Prove that 1 + ω is isomorphic to ω. Thus, ω + 1 6= 1 + ω.

Now we would like to make an interpretation of multiplication like the sum order,this is the famous lexicographic order.

5.4.11. Definition Let X and Y be linearly ordered sets. In X ×Y , we define anorder <lex in the following way: for (a, b), (c, d) ∈ X × Y , we have (a, b) <lex (c, d)provided that one of the following conditions hold:

(1) a < c,(2) a = c and b < d.

We call this new order the lexicographic order ≤lex and denote it by X ×lex Y .

Exercise 67. Prove that if X, Y are lineal orders, then X×lex Y is a lineal order.

Explicacion del cuadradito... de alguna manera

5.4.12. Lemma If α, β ∈ ON, then α×lex β is a well order.


We now prove that the lexicographic order for the cartesian product of ordinalscoincides with the product of ordinals.

5.4.13. Theorem Let α, β ∈ ON. Then α · β is isomorphic to β ×lex α.

Exercise 69. Prove Theorem 5.4.13. (Hint: use induction and in the inductivestep for successor ordinals, use Theorem 5.4.6.)

As in the case of the sum, we now notice that the product is not commutativeeither:

Exercise 70. Prove that 2 ·ω is isomorphic to ω and thus not isomorphic to ω · 2.

One can also define ordinal exponentiation in the obvious way:

5.4.14. Definition Let α ∈ ON, we define recursively:

• α·0 = 1,• α·S(β) = α·β · α for each β ∈ ON,• α·γ = supβ<γ α

·β for each limit ordinal γ.

The next results will be used later.

5.4.15. Lemma If α, β ∈ ON, then

(∗) if β ∈ {0, 1}, then β·α = 1,(∗) if β > 1, then β·α ≥ α.


5.4.16. Lemma Let α, β, γ ∈ ON with β < γ. Then α·β < α·γ .

Exercise 71. Use induction to prove Lemmas 5.4.15 and 5.4.16.

poner referencias a ejercicios(?)

5.5. Indecomposable ordinals. We begin this section with a motivational exam-ple.

5.5.1. Example . The first indecomposable ordinal.

Consider the ordinal ω·2. Recall ω·2 = ω ·ω = supk∈ω ω · k and ω ·(k+1) = ω ·k+ω.Thus, by Theorem 5.4.10 we can write

ω·2 =⋃

k<ω

Wk,

where the Wk are pairwise disjoint copies of ω, such that if a ∈ Wr, b ∈ Ws andr < s, then a < b. Specifically, Wk = {ω · k +m : m ∈ ω} for each k ∈ ω. hacerdibujo

It is easy to prove that β + ω·2 = ω·2, but here we present an easy geometricargument of why this is true. Every ordinal α smaller than ω·2 can be written inthe form ω · n + m for some m,n ∈ ω. Notice that α + ω·2 may be thought, byTheorem 5.4.10, as the sum order of ω · n+m and the Wk’s. The sum order of mand W0 is isomorphic to ω (because we are adding m ‘points’ just before 0 ∈ W0)so we may think W0 ‘absorbs’ the rightmost part m of α (see figure hacer dibujo).

So now we only have to find the sum order of ω ·n and⋃

n∈ω Wn. Graphically, thisis just adding n copies of ω + 1 to the left of the Wk’s. So just translate each Wk

to Wk+n and place ω · n isomorphically in⋃

k<n Wk (see figure hacer dibujo).

Exercise 72. Formalize the argument that β + ω·2 = ω·2 for each β < ω·2.

This motivates the following definition.

5.5.2. Definition We call α ∈ ON indecomposable if for every β < α, we haveβ + α = α.

We can show a nice characterization of indecomposability that shows example 5.5is typical.

5.5.3. Lemma Let α ∈ ON be nonzero. Then the following conditions areequivalent.

(a) for all β, γ < α, β + γ < α,(b) for every β < α, β + α = β,(c) for each X ⊂ α, either X is isomorphic to α or α−X is isomorphic to α,(d) there exists γ ∈ ON such that α = ω·γ .

Proof. Notice that (c) implies (b) and (b) implies (a) easily.

Exercise 73. Do the details of this


Now we prove (a) implies (d). Let Γ = {β ∈ ON : ω·β ≤ α}. By Lemma 5.4.15,0 ∈ Γ ⊂ α so it is a set. Let γ = supΓ. By our definition, we have that ω·γ ≤ α.

Exercise 74. Prove this.

Assume ω·γ < α. Then, we can inductively prove that ω·γ ·n < α for all n ∈ ω−{0}.

Exercise 75. Prove it.

Thus, ω·(γ+1) = ω·γ · ω = supn∈ω ω·γ · n ≤ α. This contradicts the choice of γ soω·γ = α.

Finally, we prove (d) implies (c). We do this by induction on γ. If γ = 0, α = 1whose only subsets are ∅ and 1 itself, so the conclusion holds trivially.

First assume that γ = δ + 1 for some δ ∈ ON and consider a set X ⊂ ω·γ . Sinceω·γ = ω·δ · ω, by lemma 5.4.12, we can instead consider X ⊂ ω ×lex ω·γ . For eachn ∈ ω, let Xn = X ∩ ({n} × ω·δ) and Yn = ({n} × ω·δ) − Xn. Notice that each{n} × ω·δ is order-isomorphic to ω·δ so by inductive hypothesis, either Xn or Yn isisomorphic to ω·δ and only one. That is, if we define

N1 = {n ∈ ω : Xn is isomorphic to ω·δ},

N2 = {n ∈ ω : Yn is isomorphic to ω·δ},

then ω = N1 ∪ N2 and N1 ∩ N2 = ∅. We may assume that N1 is infinite, in thiscase define

X∗ =⋃

n∈N1

({n} ×Xn).

Exercise 76. Prove that X∗ is isomorphic to ω ×lex ω·δ.

Since X∗ ⊂ X ⊂ ω·γ , by (falta un lema que diga esto!!) we get X is isomorphicto ω·γ .

Exercise 77. What happens if N1 is finite? Why is it equivalent?

Now take the case γ is a limit ordinal and let X ⊂ ω·γ . For each β < γ, letXβ = X ∩ β and Yβ = β −X. Now define the sets

J1 = {β < γ : Xβ is isomorphic to ω·β},

J2 = {β < γ : Yβ is isomorphic to ω·β}.

Exercise 78. Prove that either sup J1 = γ or sup J2 = γ.

We may assume that sup J1 = γ. Thus X =⋃

β∈J1Xβ and each Xβ is isomorphic

to ω·β , so X is isomorphic to ω·γ by (falta la misma referencia de antes).

Exercise 79. Prove this. Also, what happens if sup J1 < γ?

This proves (c). �

With this lemma we are just one step from proving Cantor’s representation.


5.5.4. Theorem Let α ∈ ON. There exist n ∈ ω, a1, . . . , an ∈ ω, γ1 > · · · > γnordinals such that

(3) α = ω·γ1 · a1 + . . .+ ω·γn · an

Note: Equation 3 is called the normal form of α.

Proof. The proof is by induction being 0 = ω · 0 the base case. If α is not indecom-posable, we may find γ and k ∈ ω such that ωγ · k < α < ωγ · k + 1. Then, there isβ < α such that ωγ · k + β = α. By induction, β has such a normal form.

Exercise 80. Fill in the details.

�

For the next concept we present another example.

Exercise 81. The first epsilon.

poner algun rollo filosofico de como se van construyendo los ordinaleshasta ǫ0

Formally, define ω(0) = ω and ω(n + 1) = ω·ω(n). Let ǫ0 = supn∈ω ω(n). Noticethe following properties of ǫ0:

(a) for all n ∈ ω, ω(n) < ω(n+ 1),(b) if n < ω, ω·n 6= n,(c) if ω(n) < α < ω(n+ 1), then ω·α 6= α,(d) ω·ǫ0 = ǫ0.

Exercise 82. Prove that these properties hold.

We can generalize this notion in the following way.

5.5.5. Definition We call α ∈ ON an epsilon ordinal if α = ω·α.

Notice that properties (a)− (d) of ǫ0 imply the following result.

5.5.6. Proposition ǫ0 is the first epsilon ordinal.

Exercise 83. Show why this is true.

Intuitively, an epsilon ordinal is an ordinal that cannot be represented as the resultof applying sum, product and exponentiation (of base ω) to smaller ordinals. Also,by lemma 5.5.3, every epsilon ordinal is indecomposable. However, ω is indecom-posable but not an epsilon, so being an epsilon ordinal is strictly stronger thanbeing indecomposable.

So we can study the ordinals by their complexity. However, as complex as theymay seem, all examples of ordinals we have seen are countable.

5.5.7. Theorem ǫ0 is countable.


Exercise 84. Prove Theorem 5.5.7. (Hint: use induction to prove that iteratingordinal sums and products of countable ordinals, we get countable ordinals)

In the next chapter we will break the wall of countability and start studying specialordinals called cardinals.

6. Cardinality

6.1. Cardinal numbers. Recall from ?? that sets can be given “sizes” with theequivalence relation =∗. It has been proved that all natural numbers have differentsizes (see ??) and the size of a natural number is different from the size of the setof all natural numbers ω ( ω is infinite, by ??). As seen in ??, all ordinals from ω toǫ0 are countable, so in the sense of =∗, they have the same size. So ω a a naturalrepresentative for the “class” of countable ordinals, which we will see is in fact aset in ??. We extend this notion in the following way.

6.1.1. Definition An ordinal number κ is a cardinal number if every time α < κis an ordinal, we have α 6=∗ κ.

In this sense, we have that cardinal numbers are representatives of sizes.

6.1.2. Lemma Let κ, τ be cardinals. Then κ <∗ τ if and only if κ < τ .

Exercise 85. Prove lemma 6.1.2.

By lemma 6.1.2, we have that the relation<∗ is a total order in the class of cardinals.This will be important when we use the axiom of choice, see Theorem 6.2.1. Letus first show that for every cardinal we have a strictly bigger cardinal. This isnot obvious: we had a hard time constucting ǫ0 is still a countable ordinal (seeProposition 5.5.7) so we don’t know if there are uncountable cardinals in ZF. Thenext result says we in fact can

6.1.3. Theorem For every X, there exists a cardinal ℵ(X) such that

(∗) ℵ(X) = min{α ∈ ON : α �∗ X}.

Proof. We first consider the set of well-orders for subsets of X.

W (X) = {(Y,R) : Y ⊂ X,R well-orders Y }.

Exercise 86. Prove that W (X) is a set by showing it is subset of P(X)×P(P(X)).

As seen in ??, each well order is isomorphic to some unique ordinal so we may definefor each (Y,R) an ordinal α(Y,R) such that (Y,R) ≈ (α(Y,R),∈). Let

B = {α(Y,R) : (Y,R) ∈W (X)},

which is a set of ordinals by the axiom of replacement. Notice this set is non-emptybecause (∅, ∅) ∈W (X).

We define ℵ(X) = sup{α + 1 : α ∈ B}. Notice that α(Y,R) < ℵ(X) for each(Y,R) ∈W (X).


Exercise 87. Prove that ℵ(X) �∗ X.

Now let’s prove that ℵ(X) satisfies (∗). Let β < ℵ(X).

Case 1. ℵ(X) = α+ 1 for some α ∈ B.

Since α is isomorphic to some well-order for some subset of X, we get α ≤∗ X.Since β ≤ α, we get β ≤∗ X.

Case 2. not case 1

Let γ ∈ B such that β < γ + 1. As in the first case, γ ≤∗ X and β ≤ γ so β ≤∗ X.This proves (∗).

These two cases prove (∗).

Exercise 88. Prove that (∗) implies ℵ(X) is a cardinal.

This completes the proof. �

Exercise 89. Let κ be a cardinal. In the proof of Theorem 6.1.3 for X = κ, provethat Case 1 holds if and only if κ if finite.

The cardinal ℵ(X) is called Hartog’s aleph for X.

6.1.4. Definition Let κ be a cardinal. Denote κ+ = ℵ(κ).

Since cardinals are “linearly ordered” by <, by Lemma 6.1.2, we get that κ+ is thefirst cardinal that is strictly bigger than κ. This is why we call κ+ the succesor ofκ. For n ∈ ω, n+ = n+ 1 by ??, but this is not true in general.

6.1.5. Lemma If κ is an infinite ordinal, κ+ 1 =∗ κ.

Exercise 90. Prove Lemma 6.1.5. (Hint: move the last point to the begining)

We can continue the operation of taking cardinals in a transfinite manner.

6.1.6. Definition We define by transfinite recursion the Alephs:

• ℵ0 = ω0 = ω,• ℵα+1 = ωα+1 = (ωα)

+ if α ∈ ON,• ℵγ = ωγ = supβ<γ ωβ for every limit γ ∈ ON.

6.1.7. Lemma Each Aleph is a cardinal.


Exercise 92. Let β, γ ∈ ON such that β < γ. Prove that ℵβ < ℵγ .

A natural question one asks is why we use two different notations (alephs andomegas) for the same object. It is common that when one writes ℵα, one emphasizesthe size of the cardinal and when one writes ωα, one emphasizes the order structureas an ordinal.

6.1.8. Lemma Every infinite cardinal is an Aleph



For the next example, let’s use the axiom of choice. By ??, the axiom of choiceimplies that every set has the same size than an ordinal number. This will allowus to give a definition of cardinality.

6.1.9. Definition (AC) For every X, let

|X| = min{α ∈ ON : X =∗ α},

which is a cardinal. We call |X| the cardinality of X.

Clearly, if κ is a cardinal, |κ| = κ and we don’t need the axiom of choice to defineit.

6.1.10. Definition (AC) If κ is a cardinal, we define 2κ to be the cardinality ofthe set κ2 of functions κ→ 2 or equivalently (see ??), the set P(κ).

6.1.11. Example . Let c = 2ω. Since ω <∗ R, we have ω < c. The questionto determine the exact α ∈ ON for which c = ℵα was one of Hilbert’s problemsexplicar bien esto. The continuum hypothersis CH is the statement that c = ℵ1.In ZFC we cannot decide whether CH is true or false. In fact, it has been shownthat it is consistent that c can be any cardinal which is not one of the type of (citarcofinalidad).

Notice that for every cardinal κ, κ+ ≤ 2κ by ??. In particular, for every n < ω,n + 1 = n+ < 2n. The Generalized Continuum Hypothesis (GCH) is the equality2κ = κ+ for every infinite cardinal κ.

Our last result for this section is that for every infinite cardinal κ, κ × κ =∗ κ inZF. To acomplish this, we will define a “well-order” for the class ON.

6.1.12. Definition For α, β, γ, δ ∈ ON, we define (α, β)⊳(γ, δ) if either max{α, β} <max{γ, δ} or max{α, β} = max{γ, δ} and (α, β) <lex (γ, δ).

poner rollito de como se ve esto en el cuadrado con la diagonal

Since we already know the lexicographic order is a well-order, it is easy to provethe following result.

6.1.13. Proposition Let α ∈ ON. Then the relation ⊳ well-orders α× α.

Exercise 94. Prove Proposition 6.1.13.

Let us give examples of this result.

dar ejemplos cuando α es finito, ω, ω + 1

With this we can state the theorem that computes the cardinality of the squareκ× κ in ZF.

6.1.14. Theorem For each infinite cardinal κ, (κ× κ,⊳) is isomorphic to (κ,∈).


Proof. We prove this by induction. Assume κ be the minimal infinite cardinal suchthat (κ× κ,⊳) is not isomorphic to (κ,∈). Let α ∈ ON be such that (κ× κ,⊳) isisomorphic to (α,∈).

Consider the diagonal ∆ = {(β, γ) ∈ κ× κ : β = γ}.

Exercise 95. Prove that (∆,⊳) is isomorphic to (κ,∈).

Thus, κ < α by ?? lema que no esta escrito en la seccion de ordinales.

Let h : α → κ × κ be the unique isomorphism and let h(κ) = (β1, β2). Considerthe ordinal β = max{β1, β2, ω}+ 1 so that β < κ because κ is a limit ordinal (seeLemma 6.1.5). We know that κ is a cardinal so τ =∗ β < κ.

By minimality, (τ × τ,⊳) is isomorphic to (τ,∈) so β×β =∗ τ . Since h(κ) ∈ β×β,we get that

κ ≤∗ β × β =∗ τ <∗ κ

which is a contradiction to the choice of κ. This completes the proof.

�

6.2. Axiom of choice.

6.2.1. Theorem In ZF, the following are equivalent:

(a) the axiom of choice(b) for every x, y, we have x <∗ y, x =∗ y or y <∗ x and only one,• poner otros

6.3. Cofinality. In this section, working in ZFC, we give a small step towardsfinding the value of c, namely that cf (c) > ω (see the definition below). This ispractically the only result one can get in ZFC without using any other set-theoreticassumptions (see ??).

6.3.1. Definition Let f : α→ β a function between ordinals. We say f is cofinalin β if sup{f(γ) : γ < α} = β.

Notice that in this definition, β must be a limit ordinal. Thus, a cofinal function islike a well-ordered sequence in a limit ordinal β that “converges” in some way toβ. One can ask what is the minimal lenght of such sequence.

6.3.2. Definition For a limit α ∈ ON, we define the cofinality of α, cf (α) to bethe least ordinal γ such that there is a cofinal function f : γ → α.

Notice that the cofinality is well-defined because the identity function α → α iscofinal and thus cf (α) ≤ α.

6.4. Infinite operations. Now we use the Axiom of Choice to define arbitrarysums and products.


Appendix A. The class WF

In this section, we study the class WF defined in Example 5.3. We notice twofundamental properties of the class WF:

A.0.1. Lemma For all α ∈ ON, Vα is a transitive set.

A.0.2. Lemma If α ∈ β ∈ ON, then Vα ⊂ Vβ .

Exercise 96. Prove lemmas A.0.1 and A.0.2 by transfinite induction. (Hint: Provelemma A.0.1 first and use it for lemma A.0.2)

We give two examples of the importance of WF to set theory:

(a) The class WF is a “model” for all the axioms ZF, including well-foundness.(b) The set Vω is a “model” for the axioms of ZF− {infinity}+ {¬infinity}.

Statement (a) means that WF is closed under all set-theoretic operations and iscomposed only of well-founded sets (see below). Statement (b) means that theset Vω is closed under all set-theoretic operations that do not include the infinityaxiom and does not include infinite sets. Notice that all the construction of ordinalnumbers was done assuming only the axioms ZF−. With this in mind, we maythink this has the following consequences:

• Statement (b) implies that if we assume the consistency of ZF−, then wecan prove the consistency of ZF.• Statement (b) implies that if we assume the consistency of ZF−, then wecan prove the consistency of ZF− {infinity}+ {¬infinity}.

Our treatment will be informal and we will be far from proving these claims. Wewill just give an informal proof of why (a) and (b) are true. Our argument is basedon the following claims:

Claim A.1. The axiom of extensionality is valid on WF.

This fact is not trivial but its proof would require technical material we don’t wantto cover. A little thought will convince you that it is true.

Claim A.2. ∅ ∈ V1

This is clear for the definition of V1.

Claim A.3. If x ∈ Vα, y ∈ Vβ and β ≤ α, then {x, y} ∈ Vα+1.

Since Vβ ⊂ Vα (lemma A.0.2), x, y ∈ Vα. Now, {x, y} ⊂ Vα so {x, y} ∈ Vα+1.

Claim A.4. If A ∈ Vα, then⋃

A ∈ Vα+1.

Since Vα is transitive (lemma A.0.1), for each x ∈ A, we have x ∈ Vα and thusx ⊂ Vα. Thus,

⋃

A ⊂ Vα so⋃

A ∈ Vα+1.

Claim A.5. If n ∈ ω, then n ∈ Vm for some m ∈ ω. Thus, ω ∈ Vω+1.


The proof of this claim is by induction. Notice 0 ∈ V1. If n ∈ Vm, then {n} ∈ Vm+1

by claim A.3. Using claim A.3 again, {n, {n}} ∈ Vm+2. Then n+1 =⋃

{n, {n}} ∈Vm+3 by claim A.4. For the second part, notice ω ⊂ Vω and thus w ∈ Vω + 1.

Claim A.6. If x ∈ Vα, then P(x) ∈ Vα+2.

If x ∈ Vα, by lemma A.0.1, x ⊂ Vα. Thus, for each y ∈ P(x), y ⊂ Vα so y ∈ Vα+1.So P(x) ⊂ Vα+1 implies P(x) ∈ Vα+2.

Claim A.7. The specification schema holds in the class WF.

This follows at once by claim A.6.

Notice that axioms 1 to 8 follow inWF by the claims. For the axioms of replacementand foundation we need some more tools.

A.0.3. Definition If x ∈WF, we define the height of x (in the class WF) to bethe ordinal number h(x) = min{α ∈ ON : x ∈ Vα}.

Exercise 97. Prove that h(x) is a succesor ordinal for each x ∈WF.

With this definition we are ready to prove that the last two axioms hold in WF:

Claim A.8. The replacement schema holds in the class WF.

Let f : α → WF a function defined on an ordinal α ∈ ON. Consider B ={h(f(β)) : β ∈ α}. Then B is a set of ordinals that show the levels for the imageof f . It follows that the set f [α] ⊂ Vβ , where β =

⋃

B ∈ ON . Thus, f [α] ∈ Vβ+1.

Claim A.9. If x ∈WF, then x is well-founded.

To prove this, notice that if x 6= ∅, then we can define α = min{h(y) : y ∈ x}. Ifz ∈ x is such that h(z) = α, we get x ∩ z = ∅.

Thus, the axioms of ZF hold in the class WF. This shows (a). For (b), we wouldonly have to prove that every set in WF is finite because the other axioms holdby our claims (this is not direct but we will ommit the details). Claim A.5 showsthat ω appears in step ω + 1. However, this is not enough because it could appearat some finite step. We show this is not the case and in fact we show a little more.

A.0.4. Lemma If n ∈ ω and x ∈ Vn, then x is a finite set.

Note that if x ∈ Vn, then x ⊂ Vn by lemma A.0.1. So it is enough to prove that foreach n ∈ ω, Vn is finite. We do this by induction: clearly V0 is finite and if Vn isfinite, then Vn+1 = P(Vn) is also finite by ??. This proves Lemma A.0.4 and thus(b) is true. Notice that all sets that are elements of Vω are finite but ω ⊂ Vω so Vω

is infinite.

References

[1] Hernandez-Hernandez, Fernando, “Teorıa de conjuntos. Una introduccion”. Aporta-ciones Matematicas Textos, 13. Sociedad Matematica Mexicana, Mexico, 1998. x+342 pp.

ISBN 968-36-6854-2


[2] Herrlich, Horst, “Axiom of choice”. Lecture Notes in Mathematics, 1876. Springer-Verlag,Berlin, 2006. xiv+194 pp. ISBN: 978-3-540-30989-5; 3-540-30989-6

[3] Kunen, Kenneth, “Set theory. An introduction to independence proofs”. Reprint of the1980 original. Studies in Logic and the Foundations of Mathematics, 102. North-HollandPublishing Co., Amsterdam, 1983. xvi+313 pp. ISBN: 0-444-86839-9

[4] Moschovakis, Yiannis, “Notes on set theory. Second edition”. Undergraduate Texts in

Mathematics. Springer, New York, 2006. xii+276 pp. ISBN: 978-0387-28722-5; 0-387-28722-1[5] Behounek, Libor , “Ordinal calculator”, http://www.volny.cz/behounek/logic/papers/ordcalc/index.html

Prague, Faculty of Arts, Department of Logic.

E-mail address: [email protected]

Instituto de Matematicas, UNAM

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