ORDINAL ARITHMETIC JULIAN J. SCHLODER¨ Abstract. These are verbose tutorial notes on the fundamentals of ordinal arithmetic. We define ordinal arithmetic and give proofs for laws of Left-Monotonicity, Associativity, Distributivity, some minor related properties and the Cantor Normal Form. 1. Ordinals Definition 1.1. A set x is called transitive iff 8y 2 x8z 2 y : z 2 x. Definition 1.2. A set α is called an ordinal iff α transitive and all β 2 α are transitive. Write α 2 Ord. Lemma 1.3. If α is an ordinal and β 2 α, then β is an ordinal. Proof. β is transitive, since it is in α. Let γ 2 β. By transitivity of α, γ 2 α. Hence γ is transitive. Thus β is an ordinal. Definition 1.4. If a is a set, define a + 1 = a [ fag. Remark 1.5. ; is an ordinal. Write 0 = ;. If α is an ordinal, so is α + 1. Definition 1.6. If α and β are ordinals, say α < β iff α 2 β. Lemma 1.7. For all ordinals α, α < α + 1. Proof. α 2 fαg, so α 2 α [ fαg = α + 1. Notation 1.8. From now on, α; β; γ; δ; "; ζ; η denote ordinals. Theorem 1.9. The ordinals are linearly ordered i.e. i. 8α : α 6< α (strictness). ii. 8α8β8γ : α < β ^ β < γ ! α < γ (transitivity). iii. 8α8β : α < β _ β < α _ α = β (linearity). Proof. \i." follows from (Found). \ii." follows from transitivity of the ordinals. \iii.": Assume this fails. By (Found), choose a minimal α such that some β is neither smaller, larger or equal to α. Choose the minimal such β. Show towards a contradiction that α = β: 1 2 JULIAN J. SCHLODER¨ Let γ 2 α. By minimality of α, γ < β _ β < γ _ β = γ. If β = γ, β < α . If β < γ then by \ii." β < α . Thus γ < β, i.e. γ 2 β. Hence α ⊆ β . Let γ 2 β. By minimality of β, γ < α _ α < γ _ α = γ. If α = γ, α < β . If α < γ then by \ii" α < β . Thus γ < α, i.e. γ 2 α. Hence β ⊆ α . Lemma 1.10. If α 6= 0 is an ordinal, 0 < α, i.e. 0 is the smallest ordinal. Proof. Since α 6= 0, by linearity α < 0 or 0 < α, but α < 0 would mean α 2 ; . Definition 1.11. An ordinal α is called a successor iff there is a β with α = β + 1. Write α 2 Suc. An ordinal α 6= ; is called a limit if it is no successor. Write α 2 Lim. Remark 1.12. By definition, every ordinal is either ; or a successor or a limit. Lemma 1.13. For all ordinals α, β: If β < α+1, then β < α_β = α, i.e. β ≤ α. Proof. Let β < α + 1, i.e. β 2 α [ fαg. By definition of [, β 2 α or β 2 fαg, i.e. β 2 α _ β = α. Lemma 1.14. For all ordinals α, β: If β < α, then β + 1 ≤ α. Proof. Suppose this fails for some α; β. Then by linearity, β + 1 > α, hence by the previous lemma α ≤ β. Hence by transitivity β < α ≤ β, contradicting strictness. Lemma 1.15. For all α, there is no β with α < β < α + 1. Proof. Assume there are such α and β. Then, since β < α + 1, β ≤ α, but since α < β, by linearity α < α, contradicting strictness. Lemma 1.16. For all α; β, if there is no γ with α < γ < β, then β = α + 1. Proof. Suppose β 6= α + 1. Since α < β, α + 1 ≤ β, so α + 1 < β. Then α + 1 is some such γ. Lemma 1.17. The operation +1 : Ord ! Ord is injective. Proof. Let α 6= β be ordinals. Wlog α < β. Then by the previous lemmas, α + 1 ≤ β < β + 1, i.e α + 1 6= β + 1. Lemma 1.18. α 2 Lim iff 8β < α : β + 1 < α and α 6= 0. ORDINAL ARITHMETIC 3 Proof. Let α 2 Lim, β < α. By linearity, β + 1 < α _ α < β + 1 _ β + 1 = α. The last case is excluded by definition of limits. So suppose α < β + 1. Then α = β _ α < β. Since β < α, α = β implies α < α, contradicting strictness. By linearity, α < β implies β < β, contradicting strictness. Thus β + 1 < α. Now suppose α 6= 0 and 8β < α : β + 1 < α. Assume α 2 Suc. Then there is β such that α = β + 1. Then β < β + 1 = α, thus β < α, i.e. β + 1 < α. Then α = β + 1 < α, contradicting strictness. Hence α is a limit. Theorem 1.19 (Ordinal Induction). Let ' be a property of ordinals. Suppose the following holds: i. '(;) (base step). ii. 8α : '(α) ! '(α + 1) (successor step). iii. 8α 2 Lim : (8β < α : '(β)) ! '(α) (limit step). Then '(α) holds for all ordinals α. Proof. Suppose i, ii and iii hold. Assume there is some α such that :'(α). By (Found), take the smallest such α. Suppose α = ;. This contradicts i. Suppose α 2 Suc. Then there is β such that α = β + 1, since β < β + 1, β < α and hence by minimality of α, '(β). By ii, '(α) . Suppose α 2 Lim. By minimality of α, all β < α satisfy '(β). Thus by iii, '(α) . Hence there can't be any such α. Definition 1.20. Let ! be the (inclusion-)smallest set that contains 0 and is closed under +1, i.e. 8x 2 ! : x + 1 2 !. More formally, ! = Tfw j 0 2 w ^ 8v 2 w : v + 1 2 wg. Remark 1.21. ! is a set by the Axiom of Infinity. Theorem 1.22. ! is an ordinal. Proof. Consider ! \ Ord. This set contains 0 and is closed under +1, as ordinals are closed under +1. So ! must by definition be a subset of ! \ Ord, i.e. ! contains only ordinals. Hence it suffices to show that ! is transitive. Consider !0 = fx j x 2 ! ^ 8y 2 x : y 2 !g. Clearly, 0 2 !0. Let x 2 !0 and show that x + 1 2 !0. By definition, x + 1 2 !. Let y 2 x + 1, i.e. y = x _ y 2 x. If y = x, y 2 !. If y 2 x then y 2 ! by definition of !0. Hence x + 1 2 !0. Thus !0 contains 0 and is closed under +1, i.e. ! ⊆ !0. But !0 ⊆ ! 0 by defintion, hence ! = ! , i.e. ! is transitive. 4 JULIAN J. SCHLODER¨ Theorem 1.23. ! is a limit, in particular, it is the smallest limit ordinal. Proof. ! 6= 0, since 0 2 !. Let α < !. Then α + 1 < ! by definition. Assume γ < ! is a limit ordinal. Since γ 6= ;, 0 2 γ. Also, as a limit, γ is closed under +1. Hence γ contradicts the minimality of !. 2. Ordinal Arithmetic Definition 2.1. Define an ordinal 1 := 0 + 1 = f;g. Lemma 2.2. 1 2 !. Proof. 0 2 ! and ! is closed under +1. Definition 2.3. Let α; β be ordinals. Define ordinal addition recur- sively: i. α + 0 = α. ii. If β 2 Suc, β = γ + 1, define α + β = (α + γ) + 1. S iii. If β 2 Lim, define α + β = γ<β(α + γ). Remark 2.4. By this definition, the sum α + 1 of an ordinal α and the ordinal 1 = f0g is the same as α + 1 = α [ fαg. Definition 2.5. Let α; β be ordinals. Define ordinal multiplication recursively: i. α · 0 = 0. ii. If β 2 Suc, β = γ + 1, define α · β = (α · γ) + α. S iii. If β 2 Lim, define α · β = γ<β(α · γ). Definition 2.6. Let α; β be ordinals. Define ordinal exponentiation recursively: i. α0 = 1. ii. If β 2 Suc, β = γ + 1, define αβ = (αγ) · α. β S γ iii. If β 2 Lim and α > 0, define α = γ<β(α ). If α = 0, define αβ = 0. Lemma 2.7. If A is a set of ordinals, S A is an ordinal. Proof. Let A be a set of ordinals, define a = S A. Let x 2 y 2 a, then there is an α 2 A such that x 2 y 2 α, so x 2 α hence x 2 a. Thus, a is transitive. Let z 2 a. There is α 2 A such that z 2 α, hence z is transitive. Thus a is transitive and every element of a is transitive, i.e. a is an ordinal. ORDINAL ARITHMETIC 5 Remark 2.8. By induction and this lemma, the definitions of +, · and exponentiation above are well-defined, i.e. if α, β are ordinals, α + β, α · β and αβ are again ordinals. Definition 2.9. Let A be a set of ordinals. The supremum of A is definied as: sup A = minfα j 8β 2 A : β ≤ αg. Lemma 2.10. Let A be a set of ordinals, then sup A = S A. Proof. Since sup A is again an ordinal, it is just the set of all ordinals smaller than it. Hence by linearity, sup A = fα j 9β 2 A : α < βg. S Which equals A by definition. Lemma 2.11. Let A be a set of ordinals. If sup A is a successor, then sup A 2 A. Proof. Assume sup A = α + 1 2= A, then for all β 2 A, β < α + 1, i.e.