Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

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1 Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1

2 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal Exponentiation Ordinal Limit Continuity Stanford, 5/11/2001 2

3 Definition 0.1 (Limit Ordinal). An ordinal is said to be a limit ordinal if it has no immediate predecessor. Example is a limit ordinal. 2. ω is a limit ordinal. 3. n are not limit ordinals whence n > 0. Remark 0.1. is not deducible from A E. y(y 0 x(y = Sx)) Stanford, 5/11/2001 3

4 Theorem 0.1 (Cantor Normal Form). If an ordinal α 0 then there exists a natural number n and sequences α 1... α n such that α = n ω α i = ω α ω α n. i=1 where α 1... α n. Notation def = ω 0 2. n def = ω ω 0 }{{} n 3. ω def = ω 1 Stanford, 5/11/2001 4

5 Definition 0.2. If α = ω α ω α n. and β = ω β ω β m. are two ordinals, then α β iff for some k n, α 1 = β 1,..., α k 1 = β k 1 and either α k β k or m = k 1 < n. Notation α β def = β α 2. α β def = α β or α = β Stanford, 5/11/2001 5

6 Exercise 0.1. Which of the following is the canonical representation form? 1. ω ω 3. ω ω2 +ω + ω ω ω + 1 Exercise 0.2. Which of the following relations is true? 1. ω ωω ω ω10 +ω 10 +ω ω ω5 + ω ω4 + ω ω3 ω ω6 3. ω 100 ω Stanford, 5/11/2001 6

7 Definition 0.3 (Ordinal Addition). 1. α + 0 def = 0 + α def = α 2. (ω α ω α k + ω α k ω α n ) + (ω β ω β m ) def = (ω α ω α k + ω β ω β m ) where k is the maximal number such that k n and α k β 1. Exercise (ω 5 + ω 4 + ω 2 + ω 2 + ω + 5) + (ω 3 + ω 2 ) = ω 5 + ω 4 + ω 3 + ω 2 2. (ω 5 +ω 4 +ω 2 +ω 2 +ω+5)+(ω 2 +ω 2 ) = ω 5 +ω 4 +ω 2 +ω 2 +ω 2 +ω 2 3. (ω 2 +ω 2 )+(ω 5 +ω 4 +ω 2 +ω 2 +ω +5) = ω 5 +ω 4 +ω 2 +ω 2 +ω +5 Stanford, 5/11/2001 7

8 Definition 0.4 (Ordinal Multiplication). 1. α 0 def = 0 α def = 0 2. α ω x = ω α 1+x where x 1 and α is of canonical form n i=1 ωα i. 3. α n def = α α }{{} n 4. α (β + γ) def = α β + α γ Remark 0.2. We can also write n α = ω αi a i = ω α1 a ω αn a n where i=1 α 1... α n and a 1,..., a n ω Stanford, 5/11/2001 8

9 Exercise (ω 2 + ω + 1) (ω 3 + ω) = (ω 2 + ω + 1) ω 3 + (ω 2 + ω + 1) ω 1 = ω 5 + ω 3 2. (ω ω+1 + ω ω + 1) (ω ω+1 + ω ω + ω) = (ω ω+1 + ω ω + 1) ω ω+1 + (ω ω+1 + ω ω + 1) ω ω +(ω ω+1 + ω ω + 1) ω = ω (ω+1)+(ω+1) + ω (ω+1)+ω + ω (ω+1)+1 = ω ω+ω+1 + ω ω+ω + ω ω+2 = ω ω ω ω 2 + ω ω+2 Stanford, 5/11/2001 9

10 Definition 0.5 (Ordinal Exponentiation). 1. α 0 def = 1 def = ω 0 2. α 1 def = α 3. 0 α def = 0 for α 0 4. α β def = ω α 1 β where β is a limit ordinal and α is of canonical form n i=1 ωα i and α ω. β+γ def 5. α = α β α γ ω x def 6. n = ω x Stanford, 5/11/

11 Exercise 0.5. Prove that (ω 3 + ω) 5 = (ω 5 + ω 3 ) 3. Proof. (ω 5 + ω 3 ) 3 = (ω 5 + ω 3 ) (ω 5 + ω 3 ) (ω 5 + ω 3 ) = ((ω 5 + ω 3 ) ω 5 + (ω 5 + ω 3 ) ω 3 ) (ω 5 + ω 3 ) = (ω 10 + ω 8 ) (ω 5 + ω 3 ) = (ω 10 + ω 8 ) ω 5 + (ω 10 + ω 8 ) ω 3 = ω 15 + ω 13 Stanford, 5/11/

12 (ω 3 + ω) 5 = (ω 3 + ω) (ω 3 + ω) (ω 3 + ω) 3 = ((ω 3 + ω) ω 3 + (ω 3 + ω) ω) (ω 3 + ω) 3 = (ω 6 + ω 4 ) (ω 3 + ω) 3 = (ω 6 + ω 4 ) (ω 3 + ω) (ω 3 + ω) 2 = ((ω 6 + ω 4 ) ω 3 + (ω 6 + ω 4 ) ω) (ω 3 + ω) 2 = (ω 9 + ω 7 ) (ω 3 + ω) 2 = (ω 9 + ω 7 ) (ω 3 + ω) (ω 3 + ω) = ((ω 9 + ω 7 ) ω 3 + (ω 9 + ω 7 ) ω) (ω 3 + ω) = (ω 12 + ω 10 ) (ω 3 + ω) = (ω 12 + ω 10 ) ω 3 + (ω 12 + ω 10 ) ω = ω 15 + ω 13 Stanford, 5/11/

13 Exercise ω = 2 ω 1 = ω 1 = ω 2. 2 ω2 = 2 ω ω = ω ω 3. 2 ωω = 2 ω1+ω = 2 ω1 ω ω = 2 ω ωω = ω ωω 4. (ω + 1) ω = (ω 1 + 1) ω = ω 1 ω = ω ω 5. (ω ω ) ω = ω ω ω = ω ω2 Stanford, 5/11/

14 Exercise 0.7. (ω + 1) n = (ω + 1) (ω + 1) (ω + 1) n 2 = (ω 2 + ω + 1) (ω + 1) n 2 = (ω 2 + ω + 1) (ω + 1) (ω + 1) n 3 = (ω 3 + ω 2 + ω + 1) (ω + 1) (ω + 1) n 3 =... = ω n + ω n ω + 1 Stanford, 5/11/

15 Theorem 0.2. The set W (α) consisting of all ordinals less than α is well ordered by relation. Moreover, the type of W (α) is α. Theorem 0.3. Every set of ordinals is well ordered by the relation. In other words, in any non-empty set Z of ordinals there exists a smallest ordinals. Theorem 0.4. For every set Z of ordinals there exists an ordinal greater than all ordinals belongin g to Z. Corollary 0.1. There exist no set of all ordinals. Corollary 0.2. There exists a smallest ordinal not belonging to a given set Z. Stanford, 5/11/

16 Definition 0.6 (Transfinite Sequence). A transfinite sequence (α sequence) is a function φ whose domain is W (α) and whose reange is also a set of ordinals. If β γ α implies φ(β) φ(γ), then we say that the α sequence is increasing. Definition 0.7 (Ordinal Limit). Given an α sequence φ, if α is a limit ordinal, there exist ordinals greater than all the ordinals φ(β) where β α. We call the smallest such ordinal the limit of the α sequence and denote it by lim β<α φ(β). Example lim n<ω n = ω 2. lim n<ω 2 n = ω 3. lim n<ω n n = ω Stanford, 5/11/

17 Lemma 0.1. If α β then there exists exactly one ordinal γ such that α = β + γ. Proof. Let A = α, let B be a initial segment of A of type β and let γ = A B. Clearly, α = β + γ. The uniqueness follows from trichotomy and Lemma 0.2. Definition 0.8 (Ordinal Subtraction). The difference of the ordinals α and β (α β) is defined to be the unique ordinal γ such that α = β + γ. The ordinal is denoted by α β. Stanford, 5/11/

18 Lemma 0.2 (Monotonic Laws of Addition). 1. (α β) = (γ + α γ + β). 2. (0 β) = (γ γ + β). 3. (α β) = (α + γ β + γ). 4. γ β + γ. Lemma 0.3 (Monotonic Laws of Subtraction). 1. α = β + (α β) if α β. 2. (α + β) α = β. 3. (α β) = (α γ β γ). 4. (γ β) = (α β α γ). Stanford, 5/11/

19 Lemma 0.4 (Monotonic Laws of Multiplication). 1. (0 α β) = (γ α γ β). 2. (α β) = (α γ β γ). 3. (α + β) γ α γ + β γ. Lemma 0.5 (Monotonic Laws of Exponentiation). 1. (0 α β) = γ α γ β if γ 1. Stanford, 5/11/

20 Theorem 0.5 (Continuity of Addition). Assume that λ is a limit ordinal and the φ is an increasing λ-sequence. Then we have lim (α + φ(ξ)) = α + lim φ(ξ). ξ λ ξ λ Proof. Note that α + φ(ξ) is an increasing λ-sequence by Lemma 0.2. Thus the lefthand side is well-defined. Let β = lim ξ λ φ(ξ). If ξ λ, then φ(ξ) β and therefore α + φ(ξ) α + β by Lemma 0.2 again. Let ζ α + β; we need to show that there exists ξ λ such that ζ α + φ(ξ). If ζ α, then ζ α + φ(0). On the other hand, if ζ α, then ζ = α + (ζ α) and ζ α (α + β) α = β by Lemma 0.3. It follows that for some ξ λ we have ζ α φ(ξ) (since β is the limit of φ(ξ) where ξ λ), thus ζ α + φ(ξ) by Lemma 0.2 and 0.3. Hence the ordinal α + β is the smallest ordinal greater than all ordinals α + φ(ξ) for ξ λ. Stanford, 5/11/

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