NOTES ON BASIC SET THEORY

Transcription

1 NOTES ON BASIC SET THEORY RODRIGO HERNÁNDEZ-GUTIÉRREZ ThisnotesarebasicallythecontentsofthecourseonSetTheorythatIhavegivenat Facultad de Ciencias, UNAM. The official programs can be found in the following: 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 Basic Constructions Binary Relations Partial orders Lattices Equivalence Relations Functions Classes Natural numbers Definition of Natural numbers Recursion Construction of the Reals Equinumerosity Ordinal numbers 15 Date: February 17,

2 2 RODRIGO HERNÁNDEZ-GUTIÉRREZ 5.1. Well-ordered sets Ordinals Transfinite recursion Ordinal arithmetic Indecomposable ordinals Cardinality Cardinal numbers Axiom of choice Cofinality 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 often represented 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 talk about sets and membership. That is, in one hand, if we write x then it must be that x is a set. So there are no sets of apples or sets of cats because apples and cats are not sets. We only consider sets whose elements are sets, this is possible as you will notice as you continue reading. On the other hand, when we talk about x,y (which have to be sets), we can only say whether x y or not. So we can just talk about set membership. Our axioms will tell us what the rules for playing with sets are. But we must also be able to distinguish what type of logical statements we can read in set theory. First, given two sets x,y, we can say whether x = y or whether x 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, we obtain all the formulas we can consider in set theory. Following our intuition, we have 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 the precise mathematical definition of them. The interested reader should read about Model 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]

3 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 only if 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. Foreachpairofsets,a,b,thereisasetwhoseelements are precisely a and b. We call this set {a,b}. a,b[ x( z(z x (z = a z = 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 are z y such that y x. We call such a set x. x[ X(z X y x(z y))] We define, as usual, the abbreviation A B = {A,B}. By induction (see??), we also write {A k : k < n+1} = A 0 A n 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 natural numbers ω = {0,1,...}. It requires more work to formalize these notions. They will 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) is in relation to x. It turns out that this question is independent of the axioms of this section (see??). In the early stages of set theory, it was purposed that if φ is a formula with one free variable, the class {x : φ(x)} should be a set. However, there is an contradiction if we 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.

4 4 RODRIGO HERNÁNDEZ-GUTIÉRREZ However, restricting to subsets of a fixed set, we remove this contradiction. Axiom 7: Specification schema. Consider φ(x) a formula in the language of set theory and A a set such that φ does not talk about A. Then {y A : φ(y)} is a set. We must now comment on the technical parts of this axiom. It is called an schema because it is really not only one axiom. For each φ(x) as described, we have the Specification Axiom for φ(x). So this axiom is really a countable collection of axioms, one for each formula in the language of set theory. We will not say why there 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 free on A) because we don t want to consider properties of A whose definition depends on A itself: this would lead to circular definitions which do not make sense. For example, consider B = {x A : x B} which is not well-defined. Other names for Specification are Separation and Comprehension. Replacement schema. The replacement schema is, as in the case of the Specification 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 of set theory and S a set such that for each s S there is a unique set y s such that φ(s,y s ) holds. Then, {y s : s S} is a set. The first time this axiom appears is in Theorem It is not essential for the development of mainstream mathematics but it is in the theorem of ordinals and higher set theory. The Replacement schema was first proposed by Fraenkel and Skolem (??). As an example of the use of this axiom, we can define inductively (see??) W 0 = ω and W n+1 = P(W n ) for n < ω. Then, by the replacement schema {W n : n < ω} is a set. Foundation. The foundation axiom is another technical part of Set Theory which does 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 x which is -minimal (see??): x [ y x(x y = ))] Exercise 5. Prove that the axiom of fundation implies that there is no set {x n : n ω} such that x n+1 x n for all n ω. We will say that a set x is well-founded if either x = or there is a y x such that x y =. The foundation axiom can be shown to be consistent with axioms 1 to 8. An outline of this proof can be found in Appendix A.

5 NOTES ON BASIC SET THEORY 5 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 call this system ZF, Zermelo-Fraenkel without Foundation. Axiom 9 is refered as the axiom of foundation and can be shown to be consistent with ZF (see Appendix A). We refer to the theory ZF plus foundation as ZF. The axiom of choice AC can be proved to be consistent with ZF but will will not assume it until the chapter on 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 Mathematics. We will also give examples of how these sets are used. In some of these examples we will assume that we know about specific sets that will be constructed in later sections but are well-known (for example the set of real numbers R defined in??) Basic Constructions. We have defined the union of sets in Axiom 4. It is common to define the intersection too 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 used? As in the case of the union, one may define A B = {A,B} and {A k : k < n+1} = A 0 A 1...A n. One may also define the complement of a set inside another set. However, this concept is relative 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 what we call the class V of all sets (see subsection 1.7 on classes). If V were a set, given another set A we should be able to define an absolute complement {x V : x / A} (commonly written as A C ). However, then we could also form R = {x : x / x} which gives Russel s paradox (see Exercise 4). Another strange thing one can do is in the following Exercise. Exercise 7. Assume that V is a set and prove that using the definition of intersection, is well-defined and equals V.

6 6 RODRIGO HERNÁNDEZ-GUTIÉRREZ Thus, in Mathematics we are not allowed to consider the universal set or the absolute 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. Show B ( A) = {B A : A A} B ( A) = {B A : A A}. Another construction used in specialized mathematics is that of the symmetric complement or symmetric difference 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 a product, we must first define what is meant by ordered pair. The definition of ordered pair we use is one proposed by Kazimierz Kuratowski, a Polish mathematician ( ) 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 Theorem For every x,y,z,w such that (x,y) = (z,w), we have x = z and y = w. Exercise 10. Prove Theorem (Hint: for a more organized proof, consider the cases x = y and x 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 an element of P(P(X Y)). By Exercise 11, we may define sets of ordered pairs Definition For each pair X, Y we define their Cartesian product in the following 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.

7 NOTES ON BASIC SET THEORY 7 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) (A B) C in general. Although we may not state distributivity of Cartesian product, we can identify two forms of writing a Cartesian product of three factors as the same. This is done by means of a bijection (see??). Exercise 14. Although we haven t studied functions yet, show a natural one-toone 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 infinite products (although we yet haven t defined what infinite means). In?? we will give a more general definition of Cartesian product of an arbitrary collection of sets. We need to develop the machinery of Binary Relations first Binary Relations. The informal, non-mathematical idea of defining a relation is to compare two different things in some way. There are basically three diferent ways in which one may do that: (1) Consider when something is bigger/better/greater than another thing, in this case we are considering order relations, (2) Divide things in groups such that the things in a group are similar in some way, in this case we are considering equivalence relations, (3) Define a rule that assigns to each thing another thing in a unique way, in this case we are considering functions. The mathematical definition is far more general than this Definition A binary relation (or relation for short) is a subset R A B for some sets A, B. Trivially, the empty set is a relation, we call it the empty relation when we want to emphasize its role as a relation. Given a set A, another natural relation which we can always define on A A, we call it the identity and it is defined as id A = {(a,a) : a A}. We now give definitions of many aspects of relations that will be useful for our treatment of the subject 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 that R A B (trivially, by definition). However, from a set-theoretic point of view,

8 8 RODRIGO HERNÁNDEZ-GUTIÉRREZ we define R first so that A and B must exist, contrary to the common view of mainstream mathematics that defines A and B first. This is because this R is a subset of A B but it is also a subset of dom(r) im(r) and when we define R we would like to forget about which A and B to choose. However, this will not apply to Orders, Equivalence Relations and Functions, as we will see in the following sections 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 equal to S R Definition Given a relation R, we define the inverse of R as Exercise 17. Show that R 1 is a set. R 1 = {(x,y) : (y,x) R}. Finally, we define the operation of taking direct and inverse image of a set. Notice that the image of a set under a relation R is different from the set im(r) in general Definition Given a relation R and an arbitrary set A, define the direct and inverse image of A as respectively. R[A] = {y : x A (x,y) R}, R [A] = {x : y A (x,y) R}, 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 the definition given because of the possible confusion presented when one defines the notation 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 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, equivalence relations and functions Partial orders. There are two types of orders one has to deal with in Mathematics. We now define both and state their relationship Definition A partial order on a nonempty set A is a relation R A A such 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.

9 NOTES ON BASIC SET THEORY Definition A strict order on a nonempty set A is a relation R A A such 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 X that 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 writes xry to mean (x,y) R. Exercise 18. Let A be a nonempty set and R A A. Prove that R is a strict order if and only if R id A = and R id A is a partial order. For example, for any set A, id A is a partial order for A and is its corresponding strict order. According to exercise 18 we sometimes denote a partial order by and its induced strict order by < Example (The inclusion order). For a set X, we may define a partial order R in P(X) by R = {(A,B) : A B}. We call this the inclusion order and simply denote 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 Definition Let (X, ) be a partial order and A X. We say a X is an upper bound (lower bound, respectively) if for each x A, x a (a x, respectively). An element a X is the greatest (least, respectively) element of A if a A and a is an upper bound (lower bound, respectively) of A. An element a X is the supremum (infimum, respectively) of A if a is the least 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) of A 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 the greatest is well-defined. That is, show that if a,b X are such that the definition of greatest element of A holds for both a and b, then a = b. Do the same for the definitions of the least, supremum and infimum. Let (X, ) be a partial order and A X. We usually denote the greatest (maximum), least (minimum), supremmum and infimum of A by maxa, mina, supa and inf A, respectively (when they exist). Now we show some examples of the previously defined concepts in concrete partially ordered sets.

10 10 RODRIGO HERNÁNDEZ-GUTIÉRREZ 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 the infimum of A? Example (maximals may not be unique). A concrete example which shows maximal 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 two maximal elements: {0} and {1} Example (Vector space basis). Let k be a field (for example k = R or k = C) and V a vector space over k. Consider l the set of all nonempty families of linearly independent elements of V. We know that maximal elements of l are precisely those subsets of V that are basis of V. We may thus consider (l, ) as a partial order with the inclusion relation (Example 1.3.3). It can be shown in ZFC (usingzorn slemma??) thateveryelementf liscontainedinamaximalg l, this is comonly said every linearly independent subset can be enlarged to a basis. More concretely, in R 2, one may start with {(0,1)} and {(0,2)} and enlarge them to {(0,1),(1,0)} and {(0,2),(1,0)} which are both basis but one is not contained in 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 will consider 0 = and for each n ω, n + 1 = n {n}. Thus, one may think that n+1 = {0,1,...,n}. The formalization of this concept will be given in Chapter Example (Divisibility). If m,n ω, we define the partial order m n if and 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}. Exercise 22. Show that (ω, ) is indeed a partial order and the properties of Example hold. We end this section with one special kind of partial orders Definition Let (X, ) be a partial order. We say that it is a linear order if 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.

11 NOTES ON BASIC SET THEORY Definition Let X be a partially ordered set. An interval is a subset J X such that if x,y J and z X is such that x < z < y, then z J. We also define the 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 definition (Hint: consider Q) 1.4. Lattices Definition A partially ordered set (L, ) is called a lattice if for every a,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 two symbols may be thought as binary operations on L Example. By exercise 21, the inclusion order in P(X) forms a lattice Lemma Let L be a lattice. Then the following properties hold for every a,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 pendiente hablar de distributividad, retículas completas y algebras booleanas 1.5. Equivalence Relations Definition Let A be a nonempty set. An equivalence relation on A is a relation 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.

12 12 RODRIGO HERNÁNDEZ-GUTIÉRREZ Definition Let (X, 1 ) and (Y, 2 ) be two partially ordered sets and f: 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, then the inverse function f 1 is also an isomorphism Definition Let f,g be functions. We say that g extends f if f g Lemma Let F a set of functions such that if f,g F, then either f g or g 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 anticadenas de <ω Classes. In ZF, classes have no formal existence. However, for each formula of 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 another class, then N M is an abbreviation of every set x such that x N, we also have x 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. For example, 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 a class M and define M M = {(x,y) : x,y M} (that is, an element is in M M if it is an ordered pair (x,y) so that both x and y are in M). A relational is a class R M M such that the axioms of partial order, definition 1.3.1, hold exchanging A by M and R by R (or definition 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 define the natural numbers from the axioms of set theory. First we will give a definition of the kind a set is a natural number if... and then with the axiom of infinity we will prove the natural numbers form a set. We want that the natural numbers are exactly 0 = and n+1 = {0,...,n}. We will refer to these sets as our intuitive natural numbers (because we have not

13 NOTES ON BASIC SET THEORY 13 yet formalized their existence and behavior). We will model these natural numbers using set-theoretic properties Definition A set x is said to be transitive if every time y x, we also have y x Example (Transitive sets). Obviously the empty set is transitive. If there were a set a such that a = {a} (which is consistent with ZF ), then this set would be transitive. Also, a set such as {,{ },{{ }}} is transitive. Notice that our intuitive natural numbers are transitive. However, by Example 2.1.2, 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 ordered by the relation. This is not enough as the set a in Example shows, because a a. Thus, we want to be an strict linear order. It turns our that we need to ask that is a well-order Definition Let (X,<) be a strict order. We say that (X,<) is a well-order if every time A X is non-empty, X has a <-least element 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 wait until 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 to finite sets. This leads to the definition of natural number in the following way 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 A x, there exists a A a -greatest element. aqui me quede redactando Lemma Ifxisanaturalnumberandy x, theny isalsoanaturalnumber Definition For any set x, we define its succesor S(x) = x {x} Lemma If x is a natural number, then S(x) is also a natural number Proposition Let x be a natural number, x. If y = max(x), then x = S(y).

14 14 RODRIGO HERNÁNDEZ-GUTIÉRREZ 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 of mathematical induction holds in a set Definition A set W is inductive if W and every time x W, then S(x) W. Acording to Axiom 5, there exists an inductive set so we may take the intersection of all inductive sets Definition ω = {W : W is inductive} 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 natural number must be an element of ω so the class of natural numbers...?? Proposition If x is a natural number, then x ω Theorem The natural numbers form a set a moreover N = ω. poner notacion de 0 y de n Corollary (The Principle of Mathematical Induction) Let A ω be nonempty. Assume 0 A and every time n A we also have n + 1 A. Then A = ω Corollary (Peano Axioms) Redactar axiomas de Peano aqui falta poner que los naturales estan bien ordenados 2.2. Recursion Theorem Let A be a set, a A and f : A ω A a function. Then there is 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 bijection h : X Y. This concept corresponds to the idea of two sets having the same number of elements: the bijection h gives a correspondence between the elements of X and the elements of Y. We can also say whether some set is smaller than another in this sense. For sets X,Y we say X Y if there is an injective function

15 NOTES ON BASIC SET THEORY 15 f : X Y. We can think that this is a way to say that X can be put inside Y without distorting it Theorem (Cantor-Schroëder-Bernstein) If X,Y are sets such that X Y and Y X, then X = Y Proposition For every X, X 2 = P(X) Proposition ω ω = ω. 5. Ordinal numbers 5.1. Well-ordered sets. We would like to extend the notion of mathematical induction to arbitrarily big sets. Let s recall the definition of well-ordered set Definition Let (X, <) a strictly ordered set. We say that X is well-ordered by < if for every A X with A, there exists a A a <-least element of A. Exercise 27. Prove that a well-ordered set is totally ordered Example Some well-ordered sets. As examples of well-ordered sets we have all natural numbers n ω and the set ω itself. We can also consider the following sets, defined by recursion (we don t have the tools to formalize this, we have to wait 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 be shown that these sets are well-ordered. Exercise 28. Prove that the sets defined in example 5.1 are well-ordered by and that 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 again a well-order. From this, we get the following Proposition If X can be well-ordered and Y X, then Y can be wellordered Proposition Let A a well-ordered set and f: A B is a surjective function, then B is well-ordered. Proof. Let g: B A be defined as g(x) = minf (x). g shows that B A so we are done by Proposition Exercise 29. Do the details of Propositions and

16 16 RODRIGO HERNÁNDEZ-GUTIÉRREZ Start with an infinite well-ordered set X and let x 0 = minx. Defining inductively x n+1 = min(x {x 0,...,x n }) we can find an isomorphic copy of the natural numbers inside X. If X is not isomorphic to ω, then we can also find x ω = min(x {x n : n ω}). If the set X is big enough (in terms of order) we can find copies of ω +n for n ω or ω 2. With this construction we can infer that a well-ordered set behaves just like our examples 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 a well-ordered set 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 x is the successor of y. If x 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 the only limit point. Now we would like to classify well-ordered sets by order-preserving maps. The next result is fundamental for our study of this classification Lemma Let W be a well-ordered set and f: W W a one-to-one orderpreserving 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 that A =. Assume not and let x 1 = mina. Then for each x < x 1, we have x f(x). Since f is order-preserving and injective, we have that x f(x) < f(x 1 ) for all x < x 1. Thus, (,x 1 ) W A is a set bounded above by f(x 1 ). Exercise 30. Prove that x 1 = sup(,x 1 ). By exercise 30, x 1 f(x 1 ) so we are done. From this apparently naive lemma we get the following facts Corollary Let W a well-ordered set. Then the only automorphism of W is the identity function Corollary Let W 1 and W 2 be two well-ordered sets. Then there is at most one isomorphism from W 1 onto W 2. Exercise 31. Prove corollaries and using lemma So well-ordered sets have an unique way to be embeded one in another. Now we will prove that there is a sort of triconomy in the class of well-ordered sets. To this end, we first define initial segments of well-orders: Definition Let W be a well-ordered set. A subset J W is an initial segment of W if every time x J and y X is such that y < x, then y J.

17 NOTES ON BASIC SET THEORY 17 Exercise 32. Show that if J is an initial segment of a well-ordered set W, then either J = (,x) for some x J or J = W. We now see that any well-ordered set cannot be isomorphic to an initial segment of itself: Corollary Let W be a well ordered set and J W an initial segment of W. Then W is isomorphic to J if and only if W = J. Exercise 33. Prove Corollary Before the main theorem of this section, we need a technical lemma: Lemma Let W 1,W 2 be a well-ordered sets, x,y W 1, z W 2 such that y < x and there is an isomorphism h: (,x) (,z). Prove that h[(,y)] = (,h(y)). Exercise 34. Prove Lemma We now prove the characterization theorem: Theorem Let W 1 and W 2 be two well-ordered sets. Then one and only one of the following holds: (1) W 1 is isomorphic to W 2, (2) W 1 is isomorphic to a proper initial segment of W 2, (3) W 2 is isomorphic to a proper initial segment of W 1. Proof. Let f = {(x,y) W 1 W 2 : (,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 isomorphisms h 1 : (,x) (,f(x)) and h 2 : (,y) (,f(y)). By lemma , h 3 = h 1 (,y) : (,y) (,h 1 (x)) is an isomorphism. Notice that since h 1 preserves order, (,h 1 (x)) is a proper initial segment of (,f(x)). Then h 3 h 1 2 is an isomorphism of (,f(y)) with one of its proper initial segments. This is not possible by Lemma 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 W 1 such that y < x. So there is an isomorphism h: (,x) (,f(x)). By Lemma , h (,y) : (,y) (,h(y)) is an isomorphism. This proves that y dom(f). Thus, dom(f) is an initial segment of W 1. By a similar argument, im(f) is an initial segment of W 2.

18 18 RODRIGO HERNÁNDEZ-GUTIÉRREZ Now assume that dom(f) W 1 and im(f) W 2 and let x 0 = min(w 1 dom(f)), y 0 = min(w 2 im(f)). Exercise 37. Prove that (,x 0 ) = dom(f) and (,y 0 ) = im(f). Thus, (x 0,y 0 ) f which is a contradiction. We obtain that either W 1 = dom(f) or W 2 = im(f). If W 1 = dom(f), then f is an isomorphism of W 1 with an initial segment of W 2. If W 2 = im(f), then f 1 is an isomorphism of W 2 with an initial segment of W 2. That no two of (a),(b),(c) can hold simultaneously follows from Lemma Exercise 38. Prove this 5.2. Ordinals. In this section we shall work in ZF, that is, without the axiom of foundation. Ordinal numbers will be well-ordered sets that represent all equivalence classes of well-ordered sets. The examples of well-orders ω+n (n ω) and ω 2 are examples of ordinal numbers Definition An ordinal number α is a transitive set such that (α, ) is a well-order. The axiom of foundation is embedded in the definition of ordinal α because is defined to be a strict well-order. So when one is asked to prove a set is an ordinal number, 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 the sets ω +n (n ω) and ω Lemma If α is an ordinal and β α, then β is an ordinal. Exercise 39. Prove Lemma In (??) we defined successors inside well orders but we now define the successor of an ordinal number Definition If α is an ordinal, then its successor is S(α) = α {α} Lemma If α is an ordinal, then S(α) is an ordinal. Exercise 40. Prove Lemma Our objective is to prove that the class of ordinals is in fact, well-ordered. A first example of this is the next result Lemma If A is a set of ordinals, then A is an ordinal. Exercise 41. Prove Lemma

19 NOTES ON BASIC SET THEORY 19 Now we prove a technical result which will help us in our main theorem but is also interesting in its own right 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 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 strict order. We now see that it is in fact linearlly ordered Theorem If α and β are ordinals, then one and only one of the following hold: (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 γ β. Since γ α and γ β, by Lemma 5.2.6, we have that γ α β. But this means that γ γ which contradicts the definition of ordinal. 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. We now prove that in fact well-orders ON 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 must have α = β or α β, so α is the -least element. Assume then that α A is nonempty and take γ = min(α A). Exercise 45. Prove that γ is the -least element of A, using Theorem According to this facts, we will say that ON is well-ordered by. Remmember this is not a precise formulation in the language of set theory but only an abbreviation of the preceeding results (see subsection 1.7 on classes). We will interchangebly write 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 the definition of ordinal and the class of natural numbers turned out to be a set. The next results shows this is not the case.

20 20 RODRIGO HERNÁNDEZ-GUTIÉRREZ 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 the class of ordinals Proposition If A is a set of ordinals, then A is the supremmum of A. Exercise 47. Prove Proposition From now on, we will understand that if A is a set of ordinals, then we will denote its suppremum A by supa Corollary If α is a limit ordinal, then α = α. Exercise 48. Prove Proposition and Lemma We can also notice that the only finite ordinals are the natural numbers: 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 some n ω, since n ω α (Theorem and Lemma 5.2.6), we have that n = ω by the Cantor-Schoëder-Bernstein Theorem This contradicts (??) so we are done. Now we arrive to the result that tells us that ordinal numbers are representatives of equivalence classes of well-orderness 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 a isomorphism, 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 to both 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 h x : (,x) α(x) such that α(x) ON. Notice that by the schema of replacement (see 8), {α(x) : x Y} is a set.

21 NOTES ON BASIC SET THEORY 21 Now we prove X = Y. Assume not and let x 0 = min(x Y). Exercise 50. x 0 minx So x 0 is a successor or a limit of elements of X. First, assume that x 0 is the succesor of y X. By the definition of x 0, there is an ordinal α(y) and an isomorphism h y : (,y) α(y). Exercise 51. Prove that h y {(y,α(y))} is an isomorphism from (,x 0 ) to S(α). Thus, y Y which is a contradiction. Now assume that x 0 is a limit. For each x < x 0, we have isomorphisms h x : (,x) α(x) by the definition of x 0. Claim 5.1. If x,y Y are such that x y, then h x h y. Exercise 52. Prove Claim 5.1.(Hint: use facts 5.1 and 5.2) Let h 0 = x<x 0 h x. Exercise 53. Prove that h 0 is an isomorphism h 0 : (,x 0 ) {α(x) : x < x 0 }. (Hint: use Claim 5.1) By Lemma 5.2.5, {α(x) : x < x 0 } is an ordinal, which contradicts the definition of x 0. Thus, X = Y. Finally, let h = x X h x. Exercise 54. Prove that h is an isomorphism from X to an ordinal. (Hint: use Lemma?? 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 can thus give an equivalent version of the axiom of choice Corollary AC is equivalent (in ZF ) to the statement that for every set X, there is α ON such that X = α Transfinite recursion. In this section, we give a generalization of Theorem 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. See subsection 1.7 for the basic stuff on classes we need 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 each α ON the class {β ON : β α} is a set, then for each α ON, G α is a set by the axiom of replacement. Thus, we can talk about F(G α ), the value that F assigns to the set G α.

22 22 RODRIGO HERNÁNDEZ-GUTIÉRREZ Proof. For α ON, we shall say that g is a α-approximation if g is a function with 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 relevance for 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 g α. Let G = α ON g α. Exercise 57. Prove that G is the class we are looking for and use induction to prove it is unique Example The class of well-ordered sets. We now present the definition of the class of all well-founded sets WF as an application to Theorem We would like to define a class of sets {V α : α ON} such that: (i) V 0 =, (ii) V α+1 = P(V α ) for each α ON, (iii) V β = α<β V α 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 ON limit, F(X) = in any other case. By using the functional F just defined in the statement of Theorem 5.3.1, we obtain a functional G with dom(g) = ON. Define G(α) = V α. 58. Exercise. Prove that properties (i),(ii) and (iii) hold for V α. We now define WF = α ON V α. If you are interested in what WF is for, take a look at appendix A Ordinal arithmetic. esta sección esta muy incompleta Now we would like to generalize the definition of addition and multiplication of natural numbers to the class of ordinals. This will give us an interesting source for exercises Definition For an ordinal number α, we define recursively: α+0 = α,

23 NOTES ON BASIC SET THEORY 23 α+s(β) = S(α+β) for each β ON, α+γ = sup β<γ (α+β) for each limit ordinal γ. Now with the sum defined, we can now define the multiplication so that it is distributive: Definition For an ordinal number α, we define recursively: α 0 = 0, α S(β) = α β +α for each β ON, α γ = sup β<γ (α β) for each limit ordinal γ. A first question is whether this operations behave in some way like the ones in natural numbers Lemma If α,β,γ ON are such that β < γ, then (a) α+β < α+γ, (b) α β < α γ 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 Using this we can now prove the associativity of both operations: 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: (1) (α+β)+δ = sup δ<γ [(α+β)+δ] = sup[α+(β +δ)] δ<γ nownoticethatbylemma5.4.4, wehavethatβ+γ isalimitordinalso,bydefinition, (2) α+(β +γ) = sup η<β+γ (α+η) = sup[α+(β +δ)]. δ<γ 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 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 Proposition If α,β,γ ON, then α (β γ) = (α β) γ.

24 24 RODRIGO HERNÁNDEZ-GUTIÉRREZ Now, the next results show us another way to see the sum and product of ordinals in some geometric way. Our first definition will be for the sum of ordinals: 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 y provided 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 the poset 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 lineal order Lemma If α,β ON, then α β with the sum order for (α,β) is a well order. Exercise 62. Prove Lemma Now we prove that the sum of two ordinal coincides with their sum order. Since two non-zero ordinals are never disjoint (see Theorem 5.2.7), we will use the precise definition of disjoint union of??, α β = (α {0}) (β {1}) so that we do not get confused in the proof of the following theorem 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, h 0 : α α {0} is a isomorphsim. If β = S(ξ) for some ξ ON, there is an isomorphism h ξ : α+ξ (α {0}) (ξ {1}). Define h β : α+ξ α β by { hξ (x) if x α+ξ, h β (x) = (ξ,1) if x α+ξ. Exercise 63. Prove that such an h β is an isomorphism. Now assume that β is a limit and for each ξ < β, h ξ : α + ξ α ξ is an isomorphism. Exercise 64. Prove that if ξ < γ < β, then h ξ h γ. (Use the techniques of Lemma and Subsection 5.1.) Define h β = {h ξ : ξ β}. Exercise 65. Show that h β : α β α β is an isomorphism. This completes the induction.

25 NOTES ON BASIC SET THEORY 25 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 1+ω. Now we would like to make an interpretation of multiplication like the sum order, this is the famous lexicographic order Definition Let X and Y be linearly ordered sets. In X Y, we define an order < 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 Lemma If α,β ON, then α lex β is a well order. Exercise 68. Prove Lemma We now prove that the lexicographic order for the cartesian product of ordinals coincides with the product of ordinals Theorem Let α,β ON. Then α β is isomorphic to β lex α. Exercise 69. Prove Theorem (Hint: use induction and in the inductive step for successor ordinals, use Theorem ) As in the case of the sum, we now notice that the product is not commutative either: 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: Definition Let α ON, we define recursively: α 0 = 1, α S(β) = α β α for each β ON, α γ = sup β<γ α β for each limit ordinal γ. The next results will be used later Lemma If α,β ON, then ( ) if β {0,1}, then β α = 1, ( ) if β > 1, then β α α.

26 26 RODRIGO HERNÁNDEZ-GUTIÉRREZ Lemma Let α,β,γ ON with β < γ. Then α β < α γ. Exercise 71. Use induction to prove Lemmas and poner referencias a ejercicios(?) 5.5. Indecomposable ordinals. We begin this section with a motivational example Example. The first indecomposable ordinal. Consider the ordinal ω 2. Recall ω 2 = ω ω = sup k ω ω k and ω (k+1) = ω k+ω. Thus, by Theorem we can write ω 2 = W k, k<ω where the W k are pairwise disjoint copies of ω, such that if a W r, b W s and r < s, then a < b. Specifically, W k = {ω k +m : m ω} for each k ω. hacer dibujo It is easy to prove that β + ω 2 = ω 2, but here we present an easy geometric argument of why this is true. Every ordinal α smaller than ω 2 can be written in the form ω n + m for some m,n ω. Notice that α + ω 2 may be thought, by Theorem , as the sum order of ω n+m and the W k s. The sum order of m and W 0 is isomorphic to ω (because we are adding m points just before 0 W 0 ) so we may think W 0 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 W n. Graphically, this is just adding n copies of ω +1 to the left of the W k s. So just translate each W k to W k+n and place ω n isomorphically in k<n W k (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 Definition We call α ON indecomposable if for every β < α, we have β +α = α. We can show a nice characterization of indecomposability that shows example 5.5 is typical Lemma Let α ON be nonzero. Then the following conditions are equivalent. (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