Thus X is nonempty by supposition. By (i), let x be a minimal element of X. Then let

4. Ordinals July 26, 2011 In this chapter we introduce the ordinals, prove a general recursion theorem, and develop some elementary ordinal arithmetic. A set A is transitive iff x A y x(y A); in other words, iff every element of A is a subset of A. This is a very important notion in the foundations of set theory, and it is essential in our definition of ordinals. An ordinal number, or simply an ordinal, is a transitive set of transitive sets. We use the first few Greek letters to denote ordinals. Note that for any set x, x / x by the foundation axiom. If α, β, γ are ordinals and α β γ, then α γ since γ is transitive. These two facts justify writing α < β instead of α β when α and β are ordinals. This helps the intuition in thinking of the ordinals as kinds of numbers. We also define α β iff α < β or α = β. It is useful in what follows to introduce some of the terminology of ordered sets and prove a lemma before proceeding with our development of ordinals. A poset or partially ordered set is an ordered pair (A, <) such that < is a binary relation contained in A A, it is irreflexive (there is no x such that x < x), and it is transitive (x < y and y < z imply that x < y). Frequently we will simply write A for the poset, with the relation < understood. Then we define x y iff x < y or x = y. Given a subset X of A, a minimal element of X is an element x X such that there is no y X such that y < x. Elements x, y of A are comparable iff x < y, y < x, or x = y. A poset is a simple ordered set or linearly ordered set iff any two of its elements are comparable. Lemma 4.1. Suppose that (A, <) is a poset satisfying the following conditions: (i) Any nonempty subset of A has a minimal element. (ii) If x, y A, x y, and for all z A, z < y implies that z < x, then there is a z < x such that z y. Then (A, <) is linearly ordered. Proof. Suppose not; so there are incomparable elements in A. Let X = {x A : there is a b A which is not comparable with x}. Thus X is nonempty by supposition. By (i), let x be a minimal element of X. Then let Y = {y A : x and y are not comparable}. Hence Y is nonempty by the fact that x X. By (i), let y be a minimal element of Y. Suppose that z < y. Then z / Y, and so x and z are comparable. If x z, then x < y, contradicting the fact that y Y. It follows that z < x. Thus we have shown that z < y implies that z < x. Hence by (ii) there is a z < x such that z y. By the choice of x we thus have that z / X, and so in particular z is comparable with y. So y z, hence y < x, contradiction. This lemma will be used shortly. Now we return to the discussion of ordinals; The following properties of ordinals are easy to prove: 14

is an ordinal. Because of this fact, the empty set is a number; it will turn out to be the first nonnegative integer, and the first cardinal number. For this reason, we will use 0 and interchangably, trying to use 0 when numbers are involved, and when they are not. If α is an ordinal, then so is α {α}. We denote α {α} by α + 1. After introducing addition of ordinals, it will turn out that α + 1 = α + 1 for every ordinal α, so that the prime can be dropped. This ordinal α + 1 is the successor of α. We define 1 = 0 + 1, 2 = 1 + 1, etc. (up through 9; no further since we do not want to try to justify decimal notation). If A is a set of ordinals, then A is an ordinal. We sometimes write sup(a) for A. In fact, A is the least ordinal each member of A. We prove this shortly. Every member of an ordinal is an ordinal. There does not exist a set which has every ordinal as a member. In fact, suppose to the contrary that A is such a set. Let B = {x A : x is an ordinal}. Then B is a set of transitive sets and B itself is transitive. Hence B is an ordinal. So B A, and even B B, contradicting the foundation axiom. This fact is what happens in our axiomatic framework to the Burali-Forti paradox. Now we can prove that the ordinals are simply ordered. More precisely, any set of ordinals is linearly ordered. Theorem 4.2. If α and β are ordinals, then α = β, α β, or β α. Proof. Let α and β be ordinals. Let A = (α+ 1) (β + 1). So (A, <) is a poset. We are going to verify the conditions of Lemma 4.1. Suppose that X is a nonempty subset of A. By the regularity axiom, choose γ X such that γ X =. Thus for any δ < γ we have δ / X, since < is the same as among ordinals. So γ is a minimal element of X, as desired in 4.1(i). For 4.1(ii), suppose that δ, ε A, δ ε, and θ < δ implies that θ < ε. Then δ ε since < is. Since δ ε, choose τ ε\δ. Thus τ < ε but τ δ, as desired in 4.1(ii). So we apply 4.1 and infer that (A, <) is a linear order. Since α, β A, they are comparable. We give some more simple properties of ordinals. For some of them, it is convenient to use Theorem 4.2 to check them. α β iff α β. α < β iff α β. α < β iff α + 1 β. These facts can be used to check the implicit statement above that A is the least upper bound of A, for any set A of ordinals. In fact, if α A, then α A, so α A. If α β for all α A, then α β for each α A, hence A β, hence A β. 15

The following theorem is also fundamental. Theorem 4.3. If Γ is a nonempty set of ordinals, then Γ is an ordinal, Γ Γ, and Γ α for every α Γ. Proof. The members of Γ are clearly ordinals, so for the first statement it suffices to show that Γ is transitive. Suppose that α β Γ; and suppose that γ Γ. Then β γ, and hence α γ since γ is transitive. This argument shows that α Γ. Since α is arbitrary, it follows that Γ is transitive, and hence is an ordinal. For every α Γ we have Γ α, and hence Γ α by a previous fact. Suppose that Γ / Γ. For any α Γ we have Γ α, hence Γ α, hence Γ < α since α Γ but we are assuming that Γ / Γ. But this means that α Γ[ Γ α]. So Γ Γ, contradiction. A linearly ordered set (A, <) is a well-ordered set iff every nonempty subset X of A has a least element, i.e., an element x X such that x y for all y X. So 4.3 says that for any ordinal α, (α, <) is a well-ordered set. Later in this chapter we will show that every well-ordered set is isomorphic to an ordinal; this is the fundamental theorem about ordinals. Ordinals are divided into three classes as follows. First there is 0, the empty set. An ordinal α is a successor ordinal if α = β + 1 for some β. Finally, α is a limit ordinal if it is nonzero and is not a successor ordinal. Thus 1, 2, etc. are successor ordinals. To prove the existence of limit ordinals, we need the infinity axiom. Let x be as in the statement of the infinity axiom. Thus 0 x, and y {y} x for all y x. We define ω = {z x : 0 z and y {y} z for all y z}. This definition does not depend on the choice of x. The members of ω are natural numbers, and the usual induction principle is built into the definition. It is easy to check that ω is an ordinal, and in fact is the least limit ordinal. Proposition 4.4. The following conditions are equivalent: (i) α is a limit ordinal; (ii) α 0, and for every β < α there is a γ such that β < γ < α. (iii) α = α 0. Proof. (i) (ii): suppose that α is a limit ordinal. So α 0, by definition. Suppose that β < α. Then β + 1 < α, since α is not a successor ordinal. Thus γ = β + 1 works as indicated. (ii) (iii): if β α, choose γ α such that β γ. Then β α since α is an ordinal. If β α, choose γ with β < γ < α. Thus β α. This proves that α = α, and α 0 is given. (iii) (i): suppose that α = β + 1. Then β α = α, so choose γ α such that β γ. Thus β < γ β, so β < β, contradiction. Also note that if α = β + 1, then α = β. 16

Classes and sets Although expressions like {x : x = x} and {α : α is an ordinal} are natural, they cannot be put into the framework of our logic for set theory. These collections are too big. It is intuitively indispensible to continue using such expressions. One should understand that when this is done, there is a rigorous way of reformulating what is said. These big collections are called classes; their rigorous counterparts are simply formulas of our set theoretic language. We can also talk about class functions, class relations, the domain of class functions, etc. Most of the notions that we have introduced so far in our sketch of a development from the axioms have class counterparts. In particular, we have the important classes V, the class of all sets, and On, the class of all ordinals. They correspond to the formulas x = x and α is an ordinal. We attempt to use bold face letters for classes; in some cases the classes in question are actually sets. A class which is not a set is called a proper class. Sequences of ordinals We say that F is an ordinal class function iff F is a class function whose domain is an ordinal, or the whole class On, and whose range is contained in On. If its domain is an ordinal, then of course it is a set. We consider three properties of an ordinal class function F with domain A: F is strictly increasing iff for any ordinals α, β A, if α < β then F(α) < F(β). F is continuous iff for every limit ordinal α A, F(α) = β<α F(β). F is normal iff it is continuous and strictly increasing. Proposition 4.5. If F is a strictly increasing ordinal class function with domain A, then α F(α) for every ordinal α A. Proof. Suppose not, and let α be the least member of A such that F(α) < α. Then F(F(α)) < F(α), so that F(α) is an ordinal β smaller than α such that F(β) < β, contradiction. Proposition 4.6. If F is a continuous ordinal class function with domain A, and F(α) < F(α + 1) for every ordinal α such that α + 1 A, then F is strictly increasing. Proof. Fix an ordinal γ A, and suppose that there is an ordinal δ A with γ < δ and F(δ) F(γ); we want to get a contradiction. Take the least such δ. Case 1. δ = θ + 1 for some θ. Thus γ θ. If γ = θ, then F(γ) < F(δ) by the hypothesis of the proposition, contradicting our supposition. Hence γ < θ. Hence F(γ) < F(θ) by the minimality of δ, and F(θ) < F(δ) by the assumption of the proposition, so F(γ) < F(δ), contradiction. Case 2. δ is a limit ordinal. Then there is a θ < δ with γ < θ, and so by the minimality of δ we have F(γ) < F(θ) ε<δf(ε) = F(δ), contradiction. 17

Proposition 4.7. Suppose that F is a normal ordinal class function with domain A, and ξ A is a limit ordinal. Then F(ξ) is a limit ordinal too. Proof. Suppose that γ < F(ξ). Thus γ η<ξ F(η), so there is a η < ξ such that γ < F(η). Now F(η) < F(ξ). Hence F(ξ) is a limit ordinal. Proposition 4.8. Suppose that F and G are normal ordinal class functions, with domains A, B respectively, and the range of F is contained in B. Then also G F is also normal. Proof. Clearly G F is strictly increasing. Now suppose that ξ A is a limit ordinal. Then F(ξ) is a limit ordinal by 4.7. Suppose that ρ < ξ. Then F(ρ) < F(ξ), so G(F(ρ)) η<f(ξ) G(η) = G(F(ξ)). Thus ( ) G(F(ρ)) G(F(ξ)). ρ<ξ Now if η < F(ξ), then by the continuity of F, η < ρ<ξ F(ρ), and hence there is a ρ < ξ such that η < F(ρ); so G(η) < G(F(ρ)). So for any η < F(ξ) we have G(η) G(F(ρ)). Hence ρ<ξ G(F(ξ)) = η<f(ξ)g(η) G(F(ρ)); ρ<ξ together with ( ) this gives the continuity of G F. Transfinite induction and recursion The following theorem generalizes the familiar principle of complete induction for natural numbers. Theorem 4.9. (Complete transfinite induction) Suppose that A is an ordinal, or is On, B A, and ( ) For all β A, if γ B for all γ < β, then β B. Then B = A. Proof. Suppose not. Then A\B is nonempty, and we let β be the least element of it. Thus γ B for all γ < β, so by the assumption ( ), also β B, contradiction. There is also an ordinary principle of transfinite induction, in which the argument goes step-by-step, except for limit ordinals, where we have to do complete induction again. This generalizes the usual induction principle for ω, which as we stated is really built in to the definition of ω. Theorem 4.10. (Ordinary transfinite induction) Suppose that A is an ordinal or is On, B A, and the following three conditions hold: (i) If 0 A, then 0 B. 18

(ii) If β + 1 A and β B, then β + 1 B. (iii) If β A is a limit ordinal, and if γ B for all γ < β, then β B. Under these assumptions, B = A. Proof. Again, suppose not, and let β be the least element of A\B. Then β 0 by (i). Suppose that β is a successor ordinal γ + 1. Then γ B, and (ii) is contradicted. Finally, if β is a limit ordinal, then (iii) is contradicted. We need the following lemma before formulating and proving the transfinite recursion principle. Lemma 4.11. Suppose that F is a ordinal class function with domain On, and α is a particular ordinal. Then there is a unique function f with domain α such that f(ξ) = F(ξ) for every ξ < α. Note that f here is a set. More formally, Lemma 4.11 looks like this: For any formula ϕ(x, y), we can prove the following in ZFC: [ x[ yϕ(x, y) x is an ordinal] and x[x is an ordinal!yϕ(x, y)]] x[x is an ordinal!f[f is a function and dmn(f) = x and y z[ y, z f ϕ(y, z)]]. Proof. By the axiom of replacement, the set A def = {x : F(ξ) = x for some ξ < α} exists. Let f = {(ξ, η) α A : F(ξ) = η}. Clearly f is as desired, and it is unique. The function f asserted to exist in this lemma will be denoted by F α. Theorem 4.12. (Class version of the transfinite recursion principle) Suppose that G is a class function with domain the class of all (ordinary) functions. Then there is a unique class function F with domain On such that for every ordinal α we have F(α) = G(F α). Before beginning the proof it may clarify things to formulate the theorem more precisely. The existence part of the theorem says that for any formula ϕ(x, y) there is another formula ψ(x, y) such that it is provable that if ϕ(x, y) represents a function G defined for all (ordinary) functions, then ψ(x, y) represents a function F as indicated. The uniqueness part says that if ψ (x, y) is any other formula that works as indicated, then one can prove x y[ψ(x, y) ψ (x, y)]. Proof of 4.12. We consider the following condition: 19

(*) f is a function with domain α, and for every ξ < α we have f(ξ) = G(f ξ). First we show (1) If f, α satisfy (*) and g, β satisfy (*) and α β, then f = g α. To prove this, we prove by transfinite induction on ξ that if ξ < α then f(ξ) = g(ξ). Suppose that this is true for all η < ξ, where ξ < α. Then f ξ = g ξ, so f(ξ) = G(f ξ) = G(g ξ) = g(ξ), finishing the inductive proof. (2) For every ordinal α there is a function f such that (*) holds. We prove this by transfinite induction. It trivially holds for α = 0. Assume that it holds for α, and f is a function such that (*) holds for α. Let h = f {(α,g(f))}. Clearly (*) holds for h and α + 1. Finally, suppose that α is a limit ordinal, and (*) holds for every β < α. By (1), for each β < α there is a unique f satisfying (1); we denote it by f β (the replacement axiom is being used). Let g = β<α f β. Then g is a function by (1), and its domain is clearly α. (3) For any β < α we have f β = g β, and g(β) = G(g β). In fact, the first condition is clear. For the second, g(β) = f β+ 1(β) = G(f β+ 1 β) = G(g β). So, (3) holds; hence (*) holds for α. This finishes the inductive proof of (2). Now for any ordinal α we let F(α) = f(α), where f is chosen so that (*) holds for α + 1 and f. This definition is unambiguous by (1). Also by (1), we have F α = f α. Hence F(α) = f(α) = G(f α) = G(F α). This finishes the proof of existence. For uniqueness, suppose that H also satisfies the conditions of the theorem. We prove that F(α) = H(α) for every ordinal α by induction. Suppose that this is true for all β < α. Then F α = H α, and hence F(α) = G(F α) = G(H α) = H(α). This finishes the inductive proof. More on well-ordered sets Now we prove some more facts about well-ordered sets, leading up to the fundamental fact that each such set is similar to an ordinal, in a certain well-defined sense. Let (A, <) be a well-ordered set. An initial segment of A under < is a subset B of A such that a A b B(a < b a B). An initial segment is proper iff it is different from A itself. If a A, then the initial segment determined by a under < is the set {b A : b < a}. Clearly this really is an initial segment of A under <. Then ({b A : b < a}, <) is a well-ordered set which we denote by (A a, <). Note here the slight sloppiness: we really should talk about {(b, c) : b, c < a and b < c} instead of < here. Sometimes we omit the phrase under < when talking about initial segments, if the < involved is clear. 20

Proposition 4.13. A proper initial segment of a well-ordered set is determined by one of its elements. Proof. Suppose that B is a proper initial segment of the well-ordered set (A, <). Let a be the least element of A\B. We claim that B is determined by a. For, if b B, then it cannot happen that a b, since this would imply that a B; so b < a. And if b < a, then b B by the minimality of a. Suppose that (A, <) and (B, ) are linearly ordered sets. A function f : A B is strictly increasing iff a 0, a 1 A[a 0 < a 1 f(a 0 ) f(a 1 )]. Note that such a function is necessarily one-one. The following easy fact will be useful. Proposition 4.14. If (A, <) and (B, ) are linearly ordered sets and f : A B is strictly increasing, then a 0, a 1 A[a 0 < a 1 f(a 0 ) f(a 1 )]. Proof. The direction is given by the definition. Now suppose that it is not true that a 0 < a 1. Then a 1 a 0, so f(a 1 ) f(a 0 ). So f(a 0 ) < f(a 1 ) is not true. We already know the following fact for ordinals, but now we need it for arbitrary wellordered sets. Proposition 4.15. If (A, <) is a well-ordered set and f : A A is strictly increasing, then x f(x) for all x A. Proof. Suppose not. Then then set B def = {x A : f(x) < x} is nonempty. Let b be the least element of B. Thus f(b) < b. Hence by the choice of b, we have f(b) f(f(b)). Hence by 4.14, b f(b), contradiction. Proposition 4.16. If (A, <) is a well-ordered set, then there does not exist a strictly increasing function from A onto a proper initial segment of A. Proof. Suppose that f is such a function. By 4.13, the proper initial segment is determined by some element x. By 4.15, x f(x), contradicting the assumption that f maps into the given proper initial segment. Let (A, <) and (B, ) be linearly ordered sets. We say that they are similar if there is a a strictly increasing function f mapping A onto B. We call f a similarity mapping, or an order-isomorphism. Proposition 4.17. If (A, <) and (B, ) are similar well-ordered sets, then there is a unique strictly increasing function f mapping A onto B. Proof. The existence of f follows from the definition. Suppose that both f and g are strictly increasing functions mapping A onto B. Then f 1 g is a strictly increasing function from A into A, so by 4.15 we get x (f 1 g)(x) for every x A; so f(x) g(x) for every x A. Similarly, g(x) f(x) for every x A, so f = g. Corollary 4.18. If α β, then α and β are not similar. Proof. Suppose to the contrary that f is a similarity mapping from α onto β, with β < α. This contradicts 4.16. 21

Proposition 4.19. Suppose that f is a one-one function mapping an ordinal α onto a set X. Define R = {(f(β), f(γ)) : β < γ < α}. Then (X, R) is a well-ordered set. Proof. Completely straightforward. The following theorem is fundamental. The proof is also of general interest; it can be followed in outline form in many other situations. Theorem 4.20. Every well-ordered set is similar to an ordinal. Proof. Let (A, ) be a well-ordered set. We may assume that A. We define a class function G whose domain is the class of all functions. For any function h, { G(h) = -least element of A\rng(h) if this set is nonempty, A otherwise. Now by the recursion theorem let F be a ordinal class function with domain On such that F(β) = G(F β) for each ordinal β. (1) If β < γ and F(β) = A, then F(γ) = A. For, A\rng(F γ) A\rng(F β), so (1) is clear. (2) if β < γ α and F(γ) A, then F(β) A and F(β) F(γ). The first assertion follows from (1). For the second assertion, note that A\rng(F γ) A\rng(F β), hence F(γ) A\rng(F β), so F(β) F(γ) by definition. Also F(β) rng(f γ), and F(γ) / rng(f γ), so F(β) F(γ), as desired in (2). By (2), F(γ) cannot be A for all γ, because then F would be a one-one function mapping On into A, contradicting the replacement axiom. Choose γ minimum such that F(γ) = A. (Note that F(0) A, since A is nonempty and so has a least element.) By (2), F γ is one-one and maps onto A. So γ is similar to A. By 4.17 and 4.20, every well-ordered set (A, ) is similar to a unique ordinal. This ordinal is called the order type of (A, ), and is denoted by o.t.(a, ). Ordinal addition We define ordinal addition by transfinite recursion: α + 0 = α; α + (β + 1) = (α + β) + 1; α + β = (α + γ) for β limit. γ<β Proposition 4.21. α + 1 = α + 1 for any ordinal α. Proof. α + 1 = α + (0 + 1) = (α + 0) + 1 = α + 1. Now we can stop using the notation α + 1. 22

We state the simplest properties of ordinal addition in the following theorem, but only prove a couple of representative parts of it. Weakly increasing means that a < b implies that F(a) F(b). An ordinal α is infinite iff ω α. Theorem 4.22. (i) If m, n ω, then m + n ω. (ii) For any ordinal α, the class function F which takes each ordinal β to α + β is a normal function. (iii) For any ordinal β, the class function F which takes each ordinal α to α + β is continuous and weakly increasing. (iv) α + (β + γ) = (α + β) + γ. (v) β α + β. (vi) 0 + α = α. (vii) α β iff there is a δ such that α + δ = β. (viii) α < β iff there is a δ > 0 such that α + δ = β. (ix) α is infinite iff 1 + α = α. Proof. (i) is easily proved by ordinary induction on n; this is the induction given by the very definition of ω. We prove (iv) by fixing α and β and proceeding by induction on γ. The case γ = 0 is obvious. Assume that α + (β + γ) = (α + β) + γ. Then α + (β + (γ + 1)) = α + ((β + γ) + 1) = (α + (β + γ)) + 1 = ((α + β) + γ) + 1 = (α + β) + (γ + 1). Finally, suppose that γ is a limit ordinal and we know our result for all δ < γ. Let F,G,H be the ordinal class functions such that, for any ordinal δ, F(δ) = α + δ; G(δ) = (α + β) + δ; H(δ) = β + δ. Thus according to (i), all three of these functions are normal. Hence, using 4.8, α + (β + γ) = F(H(γ)) = δ<γ F(H(δ)) = δ<γ(α + (β + δ)) = δ<γ((α + β) + δ = G(δ) δ<γ = G(γ) = (α + β) + γ. 23

Note, with reference to 4.22(i), that addition, restricted to natural numbers, is ordinary addition familiar to the reader. Also note that ordinal addition is not commutative in general. For example, ω = 1 + ω ω + 1. There is an equivalent definition of ordinal addition which is more intuitive and direct: Theorem 4.23. For any ordinals α, β let α β = (α {0}) (β {1}). We define a relation on α β as follows. For any x, y α β, x y iff one of the following three conditions holds: (i) There are ξ, η < α such that x = (ξ, 0), y = (η, 0), and ξ < η. (ii) There are ξ, η < β such that x = (ξ, 1), y = (η, 1), and ξ < η. (ii) There are ξ < α and η < β such that x = (ξ, 0) and y = (η, 1). Then (α β, ) is a well-order which is order-isomorphic to α + β. A simple picture helps to explain the construction in this theorem: α α β β Thus a copy of β is put to the right of a copy of α. The purpose of the definition of α β is to make the copies of α and β disjoint. Proof. Clearly is a well-order. We show by transfinite induction on β, with α fixed, that (α β, ) is order isomorphic to α + β. For β = 0 we have α + β = α + 0 = α, while α β = α 0 = α {0}. Clearly ξ (ξ, 0) defines an order-isomorphism from α onto (α {0}, ). So our result holds for β = 0. Assume it for β, and suppose that f is an order-isomorphism from α + β onto (α β, ). Now the last element of α (β + 1) is (β, 1), and the last element of α + (β + 1) is α + β, so the function f {(α + β, (β, 1))} is an order-isomorphism from α + (β + 1) onto α (β + 1). Now assume that β is a limit ordinal, and for each γ < β, the ordinal α + γ is isomorphic to α γ. For each such γ let f γ be the unique isomorphism from α + γ onto α γ. Note that if γ < δ < β, then f δ γ is an isomorphism from α +γ onto α γ; hence f δ γ = f γ. It follows that is an isomorphism from α + β onto α β, finishing the inductive proof. γ<β f γ 24

Ordinal multiplication We define ordinal multiplication by recursion: α 0 = 0; α (β + 1) = α β + α; α β = (α γ) for β limit. γ<β Here are some basic properties of ordinal multiplication: Theorem 4.24. (i) If m, n ω, then m n ω. (ii) If α 0, then α β < α (β + 1); (iii) If α 0, then the class function assigning to each ordinal β the product α β is normal. (iv) 0 α = 0; (v) If α, β 0, then α β 0; (vi) α (β + γ) = (α β) + (α γ); (vii) α (β γ) = (α β) γ; (viii) If α 0, then β α β; (ix) If α < β then α γ β γ; (x) α 1 = 1 α = α. (xi) α 2 = α + α. Proof. We only give the proof of (vi). Fix α and β. By (iv) we may assume that α 0; we then proceed by induction on γ. We define some ordinal class functions F,F,G: for any γ, F(γ) = β + γ; F (γ) = α β + γ; G(γ) = α γ. These are normal functions. First of all, α (β + 0) = α β = (α β) + 0 = (α β) + (α 0), so (v) holds for γ = 0. Now assume that (v) holds for γ. Then α (β + (γ + 1)) = α ((β + γ) + 1) = α (β + γ) + α = (α β) + (α γ) + α = (α β) + (α (γ + 1)), as desired. 25

Finally, suppose that δ is a limit ordinal and we know (v) for all γ < δ. Then α (β + δ) = G(F(δ)) = (G F)(δ) = γ<δ(g F)(γ) = γ<δ(α (β + γ)) = γ<δ((α β) + (α γ)) = γ<δ F (G(γ)) as desired. This completes the proof of (v). = γ<δ(f G)(γ) = (F G)(δ) = (α β) + (α δ), The following is a generalization of the division algorithm for natural numbers. That algorithm is very useful for the arithmetic of natural numbers, and the ordinal version is a basic result for more advanced arithmetic of ordinals. Theorem 4.25. (Division algorithm) Suppose that α and β are ordinals, with β 0. Then there are unique ordinals ξ, η such that α = β ξ + η with η < β. Proof. First we prove the existence. Note that α < α + 1 β (α + 1). Thus there is an ordinal number ρ such that α < β ρ; take the least such ρ. Obviously ρ 0. If ρ is a limit ordinal, then because β ρ = σ<ρ (β σ), it follows that there is a σ < ρ such that α < β σ, contradicting the minimality of ρ. Thus ρ is a successor ordinal ξ + 1. By the definition of ρ we have β ξ α. Hence there is an ordinal η such that β ξ + η = α. We claim that η < β. Otherwise, α = β ξ + η β ξ + β = β (ξ + 1) = β ρ, contradicting the definition of ρ. This finishes the proof of existence. For uniqueness, suppose that also α = β ξ + η with η < β. Suppose that ξ ξ. By symmetry, say ξ < ξ. Then α = β ξ + η < β ξ + β = β (ξ + 1) β ξ β ξ + η = α, contradiction. Hence ξ = ξ. Hence also η = η. We now give, similarly to the case of ordinal addition, an equivalent definition of ordinal multiplication which is somewhat more intuitive than the definition above. Given ordinals α, β, we define the following relation on α β: (ξ, η) (ξ, η ) iff ((ξ, η) and (ξ, η ) are in α β and: η < η, or (η = η and ξ < ξ ). 26

We may say that this is the anti-dictionary or anti-lexicographic order. Theorem 4.26. For any two ordinals α, β, the set α β under the anti-lexicographic order is a well-ordering which is isomorphic to α β. Proof. We may assume that α 0. It is straightforward to check that is a well-order. Now we define, for any (ξ, η) α β, f(ξ, η) = α η + ξ. We claim that f is the desired order-isomorphism from α β onto α β. If (ξ, η) α β, then f(ξ, η) = α η + ξ < α η + α = α (η + 1) α β. Thus f maps into α β. To show that f is one-one, suppose that (ξ, η), (ξ, η ) α β and f(ξ, η) = f(ξ, η ). Then by Theorem 4.25, (ξ, η) = (ξ, η ). So f is one-one. To show that f maps onto α β, let γ < α β. Choose ξ and η so that γ = α η + ξ with ξ < α. Now η < β, as otherwise γ = α η + ξ α η α β. It follows that f(ξ, η) = (α η) + ξ = γ. so f is onto. Finally, we show that the order is preserved. Suppose that (ξ, η) (ξ, η ). Then one of these cases holds: Case 1. η < η. Then f(ξ, η) = α η + ξ < α η + α = α (η + 1) α η α η + ξ = f(ξ, η ), as desired. Case 4. η = η and ξ < ξ. Then f(ξ, η) < f(ξ, η ). Now it follows that f is the desired isomorphism. Ordinal exponentiaton We define exponentiation of ordinals by recursion: α 0 = 1; α β+1 = α β α; α β = α γ γ<β for β limit. Now we give the simplest properties of exponentiation. Theorem 4.27. (i) If m, n ω, then m n ω. (ii) 0 0 = 1; 27

(iii) 0 β+1 = 0; (iv) 0 β = 1 for β a limit ordinal; (v) 1 β = 1; (vi) If α 0, then α β 0; (vii) If α > 1 then α β < α β+1 ; (viii) If α > 1, then the ordinal class function assigning to each ordinal β the value α β is normal; (ix) If α > 1, then β α β ; (x) If 0 < α < β, then α γ β γ ; (xi) For α 0, α β+γ = α β α γ ; (xii) For α 0, (α β ) γ = α β γ. The following is another kind of division algorithm for ordinals. Theorem 4.28. (Extended division algorithm) Let α and β be ordinals, with α 0 and 1 < β. Then there exist unique ordinals γ, δ, ε such that the following conditions hold: (i) α = β γ δ + ε. (ii) γ α. (iii) 0 < δ < β, (iv) ε < β γ. Proof. We have α < α + 1 β α+1 ; so there is an ordinal ϕ such that α < β ϕ. We take the least such ϕ. Clearly ϕ is a successor ordinal γ + 1. So we have β γ α < β γ+1. Now β γ 0, since β > 1. Hence by the division algorithm there are ordinals δ, ε such that α = β γ δ + ε, with ε < β γ. Now δ < β; for if β δ, then α = β γ δ + ε β γ β = β γ+1 > α, contradiction. We have δ 0, as otherwise α = ε < β γ, contradiction.. Also, γ α, since α = β γ δ + ε β γ γ. This proves the existence of γ, δ, ε as called for in the theorem. Suppose that γ, δ, ε also satisfy the indicated conditions; thus (1) α = β γ δ + ε, (2) γ α, (3) 0 < δ < β, (4) ε < β γ. Suppose that γ γ ; by symmetry, say that γ < γ. Then α = β γ δ + ε < β γ δ + β γ = β γ (δ + 1) β γ β = β γ+1 β γ α, contradiction. Hence γ = γ. Hence by the ordinary division algorithm we also have δ = δ and ε = ε. 28

We can obtain an interesting normal form for ordinals by re-applying 4.28 to the remainder ε over and over again. That is the purpose of the following definitions and results. This generalizes the ordinary decimal and binary systems of notation, by taking β = 10 or β = 2 and restricting to natural numbers. For infinite ordinals it is useful to take β = ω; this gives the Cantor normal form. To abbreviate some long expressions, we let N(β, m, γ, δ) stand for the following statement: β is an ordinal > 1, m is a positive integer, γ and δ are sequences of ordinals each of length m, and: (1) γ(0) > γ(1) > > γ(m 1); (2) 0 < δ(i) < β for each i < m. If N(β, m, γ, δ), then we define k(β, m, γ, δ) = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1). Lemma 4.29. Assume that N(β, m, γ, δ) and N(β, n, γ, δ ). Then (i) k(β, m, γ, δ) γ(0). (ii) k(β, m, γ, δ) < β γ(0) (δ(0) + 1) β γ(0)+1. (iii) If γ(0) γ (0), then k(β, m, γ, δ) < k(β, n, γ, δ ) iff γ(0) < γ (0). (iv) If γ(0) = γ (0) and δ(0) δ (0), then k(β, m, γ, δ) < k(β, n, γ, δ ) iff δ(0) < δ (0). (v) If γ(j) = γ (j) and δ(j) = δ (j) for all j < i, while γ(i) γ (i), then k(β, m, γ, δ) < k(β, n, γ, δ ) iff γ(i) < γ (i). (vi) If γ(j) = γ (j) and δ(j) = δ (j) for all j < i, while γ(i) = γ (i) and δ(i) δ (i), then k(β, m, γ, δ) < k(β, n, γ, δ ) iff δ(i) < δ (i). (vii) If γ γ, δ δ, and m < n, then k(β, m, γ, δ) < k(β, n, γ, δ ) Proof. (i): k(β, m, γ, δ) β γ(0) γ(0). (ii): We prove this by induction on m. It is clear for m = 1. Now assume that it holds for m 1, where m > 1. Then β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1) < β γ(0) δ(0) + β γ(1)+1 β γ(0) δ(0) + β γ(0) = β γ(0) (δ(0) + 1) β γ(0) β = β γ(0)+1, finishing the inductive proof. For (iii), assume the hypothesis, and suppose that γ(0) < γ (0). Then k(β, m, γ, δ) < β γ(0) (δ(0) + 1) β γ(0)+1 29 β γ (0) by (ii) k(β, n, γ, δ ) by (i).

By symmetry (iii) now follows. For (iv), assume the hypothesis, and suppose that δ(0) < δ (0). Then k(β, m, γ, δ) < β γ(0) (δ(0) + 1) β γ (0) (δ(0) + 1) by (ii) β γ (0) δ (0) k(β, n, γ, δ ) By symmetry (iv) now follows. (v) is clear from (iii), by deleting the first i summands of the sums. (vi) is clear from (iv), by deleting the first i summands of the sums. (vii) is clear. Theorem 4.30. Let α and β be ordinals, with α 0 and 1 < β. Then there exist a unique m ω and finite sequences γ(i) : i < m and δ(i) : i < m of ordinals such that the following conditions hold: (i) α = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1). (ii) α γ(0) > γ(1) > > γ(m 1). (iii) 0 < δ(i) < β for each i < m. Proof. For the existence, with β > 1 fixed we proceed by induction on α. Assume that the theorem holds for every α < α such that α 0, and suppose that α 0. By Theorem 4.28, let ϕ, ψ, θ be such that (1) α = β ϕ ψ + θ, (2) ϕ α, (3) 0 < ψ < β, (4) θ < β ϕ. If θ = 0, then we can take our sequences to be γ(0) and δ(0), with γ(0) = ϕ and δ(0) = ψ. Now assume that θ > 0. Then θ < β ϕ β ϕ ψ + θ = α; so θ < α. Hence by the inductive assumption we can write with θ = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1) (5) θ γ(0) > γ(1) > > γ(m 1). (6) 0 < δ(i) < β for each i < m. Then our desired sequences for α are ϕ, γ(0), γ(1),..., γ(m 1) and ψ, δ(0), δ(1),..., δ(m 1). 30

To prove this, we just need to show that ϕ > γ(0). If ϕ γ(0), then β ϕ β γ(0) θ, contradiction. This finishes the existence part of the proof. For the uniqueness, we use the notation introduced above, and proceed by induction on α. Suppose the uniqueness statement holds for all nonzero α < α, and now we have N(β, m, γ, δ), N(β, n, γ, δ ), and α = k(β, m, γ, δ) = k(β, n, γ, δ ). We suppose that the uniqueness fails. Say m n. Then there is an i < m such that γ(i) γ (i) or δ(i) δ (i); we take the least such i. Then we have a contradiction of Lemma 4.29. Now we give an equivalent definition of exponentiation similar to those given above for addition and multiplication. A set X is finite iff there is a bijection from some natural number m onto X. Theorem 4.31. Suppose that α and β are ordinals, with β 0. We define α β w = {f α β : {ξ < α : f(ξ) 0} is finite}. For f, g α β w we write f g iff f g and f(ξ) < g(ξ) for the greatest ξ < α for which f(ξ) g(ξ). Then ( α β w, ) is a well-order which is order-isomorphic to the ordinal exponent β α. Proof. If α = 0, then β α = 1, and α β w also has only one element, the empty function (= the emptyset). So, assume that α 0. If β = 1, then α β w has only one member, namely the function with domain α whose value is always 0. This is clearly order-isomorphic to 1, as desired. So, suppose that β > 1. Now we define a function f mapping β α into α β w. Let f(0) be the member of α β w which takes only the value 0. Now suppose that 0 < ε < β α. By 4.30 write ε = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1), where ε γ(0) > γ(1) > > γ(m 1) and 0 < δ(i) < β for each i < m. Note that β γ(0) ε < β α, so γ(0) < α. Then we define, for any ζ < α, (f(ε))(ζ) = { 0 if ζ / {γ(0),..., γ(m 1)}, δ(i) if ζ = γ(i) with i < m. Clearly f(ε) α β w. To see that f maps onto α β w, suppose that x α β w. If x takes only the value 0, then f(0) = x. Suppose that x takes on some nonzero value. Let {ξ < α : x(ξ) 0} = {γ(0), γ(1),..., γ(m 1)}, 31

where γ(0) > γ(1) > > γ(m 1). Let δ(i) = x(γ(i)) for each i < m, and let ε = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1). Clearly then f(ε) = x. Now we complete the proof by showing that for any ε, θ < β α, ε < θ iff f(ε) < f(θ). This equivalence is clear if one of ε, θ is 0, so suppose that both are nonzero. Write ε = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1), where α γ(0) > γ(1) > > γ(m 1) and 0 < δ(i) < β for each i < m, and θ = β γ (0) δ (0) + β γ (1) δ (1) + + β γ (n 1) δ (n 1), where α γ (0) > γ (1) > > γ (n 1) and 0 < δ (i) < β for each i < n. By symmetry we may suppose that m n. Note that N(β, m, γ, δ), k(β, m, γ, δ) = ε, N(β, n, γ, δ ), and k(β, n, γ, δ ) = θ. We now consider several possibilities. Case 1. ε = θ. Then clearly f(ε) = f(θ). Case 4. γ γ, δ δ, and m < n. Thus ε < θ. Also, γ (m) is the largest ξ < α such that (f(ε))(ξ) (f(θ))(ξ), and (f(ε))(ξ) = 0 < δ (m) = (f(θ))(γ (m)), so f(ε) < f(θ). Case 3. There is an i < m such that γ(j) = γ (j) and δ(j) = δ (j) for all j < i, while γ(i) γ (i). By symmetry, say that γ(i) < γ (i). Then we have ε < θ. Since γ (i) is the largest ξ < α such that (f(ε))(ξ) (f(θ))(ξ), and (f(ε))(γ (i)) = 0 < δ (i) = (f(θ))(γ (i)), we also have f(ε) < f(θ). Case 4. There is an i < m such that γ(j) = γ (j) and δ(j) = δ (j) for all j < i, while γ(i) = γ (i) and δ(i) δ (i). By symmetry, say that δ(i) < δ (i). Then we have ε < θ. Since γ(i) is the largest ξ < α such that (f(ε))(ξ) (f(θ))(ξ), and (f(ε))(γ (i)) = δ(i) < δ (i) = (f(θ))(γ (i)), we also have f(ε) < f(θ). We finish this chapter with two important characterizations of absorption properties of ordinals. Lemma 4.32. If α < ω β, then α + ω β = ω β. Proof. First we prove (1) If γ < β, then ω γ + ω β = ω β. In fact, suppose that γ < β. Then there is a nonzero δ such that γ + δ = β. Then ω γ + ω β = ω γ + ω γ+δ = ω γ + ω γ ω δ = ω γ (1 + ω δ ) = ω γ ω δ = ω β. By an easy ordinary induction, we obtain from (1) (2) If γ < β and m ω, then ω γ m + ω β = ω β. Now we turn to the general case. If β = 0 or α < ω, the desired conclusion is clear. So suppose that ω α and β > 0. Then we can write α = ω γ m+δ with m ω and δ < ω γ. Then ω β α + ω β = ω γ m + δ + ω β ω γ (m + 1) + ω β = ω β 32

Now we have the following characterization of absorption under addition: Theorem 4.33. The following conditions are equivalent, for any ordinal α: (i) β + α = α for all β < α. (Absorption under addition) (ii) For all β, γ < α, also β + γ < α. (iii) α = 0, or α = ω β for some β. Proof. (i) (ii): Assuming (i), if β, γ < α, then β + γ < β + α = α. (ii) (iii): Assume (ii). If α = 0 or α = 1, condition (iii) holds, so suppose that 2 α. Then clearly (ii) implies that α ω. Choose β, m, γ such that m ω, α = ω β m + γ, and γ < ω β. If γ 0, then ω β m < ω β m + γ = α, and also γ < ω β < α, so that (ii) is contradicted. So γ = 0. If m > 1, write m = n + 1 with n 0. Then α = ω β m = ω β (n + 1) = ω β n + ω β, and ω β n, ω β < α, again contradicting (ii). Hence m = 1, as desired in (iii). Finally, (iii) (i) by Lemma 4.32. Theorem 4.34. For any ordinal α the following conditions are equivalent: (i) For all β, if 0 < β < α then β α = α. (ii) For all β, γ < α, also β γ < α. (iii) α {0, 1, 2} or there is a β such that α = ω (ωβ). Proof. (i) (ii): Assume (i), and suppose that β, γ < α. If β = 0, then β γ = 0 < α. If β 0, then β γ < β α = α. (ii) (iii): Assume (ii), and suppose that α / {0, 1, 2}. Clearly then ω α. Now if β, γ < α, then β + γ < α. In fact, if β γ, then β + γ γ + γ = γ 2 < α by (ii); and if γ < β then β + γ < β + β = β 2 < α. Hence by 4.33 there is a γ such that α = ω γ. Now if δ, ε < γ, then ω δ, ω ε < ω γ = α, and hence ω δ+ε = ω δ ω ε < α = ω γ, so that δ + ε < γ. Hence by 4.33, γ = ω β for some β. (iii) (i): Assume (iii). Clearly 0, 1, 2 satisfy (i), so assume that α = ω (ωβ). Take any γ < α with γ 0. If γ < ω, clearly γ α = α. So assume that ω γ. Write γ = ω δ m + ε with m ω and ε < ω δ. Then δ < β, and so α = ω (ωβ) γ ω (ωβ) = (ω δ m + ε) ω (ωβ ) (ω δ m + ω δ ) ω (ωβ ) = ω δ (m + 1) ω (ωβ ) ω δ+1 ω (ωβ ) = ω δ+1+ωβ = ω (ωβ ) = α 33

EXERCISES E4.1. Prove that the following conditions are equivalent: (i) x is an ordinal; (ii) x is transitive, and for all y, z x, either y z, y = z, or z y; (iii) x is transitive, and for all y, if y x and y is transitive, then y x; (iv) for all y x, either y {y} = x or y {y} x; and for all y x, either y = x or y x; (v) x is transitive and {(y, z) x x : y z} well-orders x. Hint: prove this in the following order: (i) (v) [easy]; (v) (ii) [obvious]; (ii) (iii) [Assume that y x, y transitive; apply the foundation axiom to x\y]; (iii) (i) [Let y = {z x : z is an ordinal}, and get a contradiction from assuming that y x]; (i) (iv) [easy]; (iv) (i) [Let α be an ordinal not in x, and take the least ordinal β α {α} such that β / x. Work with β to show that x is an ordinal. E4.2. Show that (2 2) ω 2 ω 2 ω. E4.3. Use the transfinite recursion principle to justify the definition of ordinal addition. E4.4. Use the transfinite recursion principle to justify the definition of ordinal multiplication. E4.5. Use the transfinite recursion principle to justify the definition of ordinal exponentiation. E4.6. Formulate and prove a set version of the transfinite recursion principle. E4.7. Suppose that α, β, γ are ordinal numbers with α < ω γ. Prove that α + β + ω γ = β + ω γ. E4.8. Show that for every nonzero ordinal α there are only finitely many ordinals β such that α = γ β for some γ. E4.9. Prove that n (ωω) = ω (ωω) for every natural number n > 1. E4.10. Assume that γ 0, 0 < m < ω, and β < ω γ. Prove that β + ω γ m = ω γ m. E4.11. Suppose that 0 < k ω and α 0. Then k ω α = ω α. E4.12. Suppose that 0 < k ω, α 0, and 0 < m ω. Then k ω α m = ω α m. E4.13. Suppose that α = ω δ k + ε with δ 0, k ω, and ε < ω δ, β < ω δ, and m ω\1. Show that (α + β) m = α m + β. E4.14. Suppose that α = ω δ k + ε with δ 0, k ω, and ε < ω δ, and β < ω δ. Show that (α + β) ω = α ω. E4.15. Show that (α + β) γ α γ + β γ for any ordinals α, β, γ.hint: write α and β using Theorem 4.28 and γ using Theorem 4.25. E4.16. Show that the following conditions are equivalent for any ordinals α, β: (i) α + β = β + α. 34

(ii) There exist an ordinal γ and natural numbers k, l such that α = γ k and β = γ l. References Sierpiński, W. Cardinal and ordinal numbers. Pań. Wyd. Naukowe, 1958, 487 pp. 35