2 Ordinal arithmetic, Cantor normal form, Hessenberg sum, rank of On On,

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1 2 Ordinal arithmetic, Cantor normal form, Hessenberg sum, rank of On On, In this lecture-note we deal with four matters. (1) Present more facts in ordinal arithmetic. (2) Present the Cantor normal form of an ordinal, and the more general form: the α-polynomial representation of an ordinal. (3) Define the Hessenberg sum of two ordinals, and prove a basic equivalence. (4) Solve an exercise on the rank function of On On,. Definition 2.1. α,β,γ,δ denote ordinals. 1 denotes S(0). (a) Ordinal Addition: (1) α+0 = α, (2) α+s(β) = S(α+β), (3) If δ is a limit ordinal, then α+δ = sup({α+β β < δ}). Equivalently, (1) α+0 = α, (2) If β > 0, then α+β = sup + ({α+γ γ < β}). (b) Ordinal Multiplication: (1) α 0 = 0, (2) α S(β) = α β +α, (3) If δ is a limit ordinal, then α δ = sup({α β β < δ}). Equivalently, (1) α 0 = 0, (2) If β > 0, then α β = sup({α γ +α γ < β}). 1

2 (c) Ordinal Exponentiation: For the time being let us denote α to the β by α β. (1) If α > 0, then α 0 = 1, and 0 0 = 0. (2) α S(β) = (α β) α, (3) If δ is a limit ordinal, then α δ = sup({α β β < δ}). Equivalently, (1) If α > 0, then α 0 = 1, and 0 0 = 0. (2) If β > 0, then α β = sup({(α γ) α γ < β}). The proof of the following facts are left to the students. Proposition 2.2. (a) For every ordinal α: (i) The function β α+β, is strictly increasing. (ii) If α > 0, then the function β α β is strictly increasing. and (iii) If α > 1, then the function β α β is strictly increasing. (b) Let α be an ordinal and β be a set of ordinals, then sup({α+β β β} = α+sup(β), sup({α β β β} = α sup(β) and sup({α β β β} = α sup(β). The following proposition deals with identities of On,+,,. Theorem 2.3. Let α,β,γ be ordinals. Then the following hold. (a) Associativity of + and. (α+β)+γ = α+(β +γ) and (α β) γ = α (β γ). 2

3 (b) Distributivity. α (β +γ) = α β +α γ. (c) Exponontiation rules. α (β +γ) = (α β) (α γ) and (α β) γ = α (β γ). From Proposition 2.2(a) we get the following left-cancellation laws. Part(b) says that whenver α β, the the equation α+x = β is solvable. Proposition 2.4. (a) Let α,β,γ On. Then the following hold. (1) If α+β = α+γ, then β = γ. (2) Suppose that α 0. Then if α β = α γ, then β = γ. (3) Suppose that α 0,1. If α β = α γ, then β = γ. (b) Let α,β On. If α β, then there is γ On such that α+γ = β. The polynomial representation of ordinals. We denote α β by α β. Let α,ε,ν On be such that ν < α. Then the expression α ε ν is called a monomial in α or an α-monomial. Let α On and n ω. For every i = 0,...,n 1 let ε i,ν i On be such that ε n 1 > ε n 2 >... > ε 0 and ν 0,...,ν n 1 < α. The expression α ε n 1 ν n 1 +α ε n 2 ν n α ε0 ν 0 is called a polynomial in α or an α-polynomial. We write the above expression also as 0 n 1 αεi ν i. If for every i < n, ν i 0, then the expression 0 n 1 αεi ν i is called a proper α-polynomial. The expression p:= 0 n 1 αεi ν i is called an α-polynomial representation of β; and if p is proper, the p is called a proper α-polynomial representation of β. We make the convention that the expression 0 1expression denotes 0. So 0 is considerd to be a proper α-polynomial. 3

4 Proposition 2.5. (a) Let n > 0 and p:= 0 n 1 αεi ν i be an α-polynomial. Then the following hold. α ε n 1+1 > p. (b) Let n > 0 and p:= 0 n 1 αεi ν i be an α-polynomial. Then the following hold. α ε n 1 (ν n 1 +1) > p. (c) Let M be a set of α-monomials. Then sup(m) is an α-monomial. Proof (a) By induction on n. For n = 1 we have α ε 0+1 = α ε0 α > α ε0 ν 0 = p. Suppose that the induction hypothesis holds for n, and let p:= 0 n αεi ν i be an α-polynomial. So Then by the induction hypthesis, α εn α ε n 1+1 > n 1 α εi ν i. α εn+1 =α εn α α εn (ν n +1)=α εn ν n +α εn >α εn ν n + (b) Let p:= 0 n 1 αεi ν i be an α-polynomial. Then α εi ν i =p. α ε n 1 (ν n 1 +1) = α ε n 1 ν n 1 +α ε n 1 α ε n 1 ν n 1 +α εn 2+1 > α ε n 1 ν n 1 + α εi ν i = p. n 2 (c) If M has a maximum, then max(m) = sup(m), and hence sup(m) is an α-monomial. Suppose that M does not have a maximum. Let E = {ε there is ν such that α ε ν M}. Suppose first that E has no maximum. Let M 0 = {α ε+1 ε E}. Then (i) for every m 0 M 0 there is m M such that m 0 m, and (ii) for every m M there is m 0 M 0 such that m < m 0. Hence sup(m) = sup(m 0 ) = α sup(e). 4 n 1

5 So sup(m) is an α-monomial. Even though (i) and (ii) are trivial, let us verify them. (i) Let m 0 M 0. Then m 0 has the form α ε+1, where ε E. Since E does not have a maximum, there is η E such that η > ε. Let 0 < ρ < α be such that α ε ρ M. Then α ε+1 α η α η ρ M. (ii) Let m M. So m has the form α ε ρ, where 0 < ρ < α. Then m = α ε ρ < α ε α = α ε+1 M 0. Next assume the E has a maximum ε. Let N = {ν α ε ν M}. Let µ = sup(n). Then sup(m) = sup({α ε ν ν N}) = α ε µ. Clearly, µ α. So if µ < α, then α ε µ is an α-monomial, and if µ = α, then α ε µ = α ε+1. In this case, too, we obtain an α-monomial. Theorem 2.6. (a) Let α,β On and α 2. Then β has an α-polynomial representation. (b) Let α,β On be such that α 2. Then β has a unique proper α-polynomial representation. Proof (a) By induction on β. The claim is clearly true for every β α. So let β be an ordinal bigger than α, and assume that for every γ < β, γ has an α-polynomial representation. Let M = {α ε ν α ε ν is an α-monomial and α ε ν β} Then sup(m) is an α-monomial, and sup(m) β. So M has a maximum α ε ν. If α ε ν = β, we are done. So suppose that α ε ν < β. Let γ On be such that α ε ν +γ = β. Since α ε ν +α ε = α ε (ν +1) > β, 5

6 and α ε ν +γ = β, it follows that α ε > γ. So γ < α ε < α ε ν < β. By the induction hypothesis, γ has a proper α-polynomial representation 0 n 1 αεi ν i. Since γ < α ε, it follows that ε n 1 < ε. So α ε ν + 0 n 1 αεi ν i is an α-polynomial. Also, β = α ε ν + α εi ν i. So β has an α-polynomial representation. (b) Let α 2 and p = 0 n 1 αεi ν i and q = 0 m 1 αρi µ i be two proper α-polynomials. Suppose that either m n or m = n and for some i < n, ε i,ν i ρ i,µ i. We may assume that m n. Then either (i) m < n and for every i = 0,...,m 1, n 1 ρ i = ε i+(n m) and µ i = ν i+(n m) or (ii) there is 0 j m 1 such that for every j < i m 1, and ρ j,µ j ε j+(n m),ν j+(n m). Suppose that (i) happens. Then p ρ i = ε i+(n m) and µ i = ν i+(n m) α ρi µ i +α ε n m 1 ν n m 1 = q +α ε n m 1 ν n m 1 > q. m 1 Suppose that (ii) happens. The lexicographic order between two pairs of ordianls η l,ζ l, l = 0,1, is defined as follows: η 0,ζ 0 η 1,ζ 1, if η 0 < η 1 or ( η 0 = η 1 and ζ 0 < ζ 1 ). If ρ j,µ j ε j+(n m),ν j+(n m) then q > p, and if ρ j,µ j ε j+(n m),ν j+(n m) then q < p. result. It follows that two different proper α-polynomials do not yield the same The (unique) proper ω-polynomial representation of an ordinal is called the Cantor normal form of that ordinal. 6

7 Definition 2.7. (a) An ordinal α is closed under addition, if for every β,γ α, β +γ α. (b) An ordinal α is closed under multiplication, if for every β,γ α, β γ α. Let A be a class. Suc(A):=A {α On α is a successor ordinal}. If α,β On, then [α,β]:={γ α γ β}. Proposition 2.8. (a) Let α On. The following are equivalent. (1) α is closed under addition. (2) For every β < α, β +α = α. (3) Either α = 0, or there is ε On such that α = ω ε. (b) Let α On. The following are equivalent. (1) α is closed under multiplication. (2) For every 0 β < α, β α = α. (3) Either α 2, or there is an ordinal ε such that α = ω ωε. Proof (a) (1) (2). Suppose that α is closed under addition. If α = 0 or α = 1, then indeed, for every β < α, β+α = α. Assume that α > 1, then α is a limit ordinal. For if α = α +1, then α +α α +1 = α. Hence α +α α, a contradiction. So α Lim(On), and hence α = sup({γ γ < α}). Let β < α. For every γ < α, β +γ < α. So β +α = β +sup({γ γ < α}) = sup({β +γ γ < α}) = sup([β,α]) = sup(α) = α. (2) (1). Suppose that for every β < α, β+α = α. Let β,γ α. Then β +γ < β +α = α. So α is closed under addition. (3) (1). Clearly, 0 is closed under addition. Let α = ω ε. If ε = 0, then α = 1, and hence α is closed under addition. Suppose that ε 1. 7

8 Case 1 ε Lim(On). Let β,γ < α. There is η < ε such that β,γ < ω η. Hence β +γ < ω η +ω η = ω η 2 < ω η ω = ω η+1 < ω ε = α. Case 2 ε is a successor ordinal. Let ε = η + 1. Let β,γ < α. There is m {0,1} and β 0 < ω η such that β = ω η m + β 0. Similarly, γ can be written as γ = ω η n+γ 0. So β +γ = ω η m+β 0 +ω η n+γ 0 < ω η m+ω η +ω η n+ω η = ω η (m+1+n+1) < ω η ω = ω η+1 = ω ε. (1) (3). Suppose that α is closed under addition and α > 0. If α = 1, then α = ω 0. So suppose that α > 1. Let S = {η On ω η α}. Clearly, S has a maximum. Denote max(s) by ε. Assume by contradiction that ω ε < α. Since α is closed under addition, for every n ω, ω ε n < α. So ω ε+1 = ω ε ω = ω ε sup({n n ω}) = sup({ω ε n n ω}). For every n ω, ω ε n < α. So sup({ω ε n n ω}) α. That is, ω ε+1 α, contradicting the maximality of ε. (b) (1) (2). Suppose that α is closed under multiplication. If α 2, then indeed, for every 0 β < α, β α = α. Assume that α > 2, then α is a limit ordinal. For if α = α + 1, then α 2 = α + α > α + 1 = α. Hence α + α α, a contradiction. So α Lim(On), and hence α = sup({γ γ < α}). Let β < α. For every γ < α, β γ < α. So β α = β sup({γ γ < α}) = sup({β γ γ < α}) α. Since β 0, β α 1 α = α. So, β α = α. (2) (1). Suppose that for every β < α, β α = α. Let β,γ α. Then β γ < β α = α. So α is closed under addition. (3) (1). Clearly, every ordinal 2 is closed under multiplication. Let α = ω ωε. Ifε = 0, thenα = ω, andhenceαisclosedunderaddition. Suppose that ε 1. Then ω ε Lim(On). So α = sup({ω ρ ρ < ω ε }). 8

9 Let β,γ < α. Then there is ρ < ω ε such that β,γ < ω ρ. So β γ < ω ρ ω ρ = ω ρ+ρ. By (3) (1) of Part (a), ρ + ρ < ω ε. So ω ρ+ρ < ω ωε. Hence β γ < ω ωε. (1) (3). Let α be closed under multiplication and α > 2. Then α is closed under addition. So by (3) (1) of Part (a), α has the form α = ω ρ. If ρ = 1, then α = ω ω0. So assume ρ > 1. Suppose by contradiction that ρ is not closed under addition. Let β,γ < ρ be such that β +γ ρ. Then ω β ω γ = ω β+γ ω ρ = α. However, ω β < ω ρ = α, and the same holds for γ. So α is not closed under multiplication. A contradiction. Proposition 2.9. Let α = 0 i=k 1 ωεi n i and α = 0 i=k 1 ωε i n i be two ω-polynomial representations of α. Similarly, let β = 0 i=k 1 ωεi m i and β = 0 i=k 1 ωε i m i be two ω- polynomial representations of β. Then i=k 1 ω εi (n i +m i ) = i=k 1 ω ε i (n i +m i) Proof This proof is left to the students. Definition (a) Let α,β be ordinals. Set Sum(α,β):={γ On there is a partition {A,B} of γ such that A, A = α, α and B, B = β, β }. Define α β:=sup(sum(α,β)). The operation is called the Hessenberg sum. Note that the definition of applies also to the case that α = 0 or that β = 0. In order for the definition to really apply, we have to regard, as a linearly ordered set. Regarding, as a linearly ordered set makes the statement A, A = 0, 0 meaningful. (b) We define α β. Let α = 0 i=k 1 ωεi n i and β = 0 i=k 1 ωεi m i be ω-polynomial representations of respectively α and β. (Note that some of 9

10 the n i s and m i s may be 0.) Then α β:= i=k 1 ω εi (n i +m i ). (c) We define the relation on On On. Let α,β,γ,δ On Then α,β γ,δ if α γ, β δ and α,β γ,δ. Set α,β γ,δ if α,β γ,δ or α,β = γ,δ Proposition (a) Let α,β On. Then α β = β α. (b) Let α,β On. Then (α+1) β = (α β)+1 and α (β +1) = (α β)+1. (c) For every α,β, γ,δ On On: if α,β γ,δ, then α β < γ δ. Proof (a) This part follows trivially from the definition of. (b) Let 0 i=k 1 ωεi nα i, 0 i=k 1 ωεi n β i, be ω-polynomial representions of respectively α and β such that ε 0 = 0. Then 1 i=k 1 ωεi nα i +(n α 0 +1) is an ω-polynomial representatipn of α + 1. So (α+1) β = 1 i=k 1 1 i=k 1 ω εi (n α i +n β i )+((nα 0 +1)+n β 0) = ω εi (n α i +n β i )+(nα 0 +n β 0)+1 = (α β)+1. The second part of (b) follows from the first part and the commutativity of. (c) It suffices to prove the following two facts. (1) Let α,β,γ On and suppose that α < γ. Then α β γ β. 10

11 (2) Let α,β,δ On and suppose that β < δ. Then α β α β. Let us see why (1) and (2) above suffice. Suppose that α,β γ,δ. If α = γ, then it follows from (2) that α β γ β. Similarly, if β = δ, then by (1), α β γ β. Suppose that α < γ and β < δ. Then by (1) and (2), α β γ β γ δ. We still need to show that if α,β γ,δ, then α β < γ δ. If α < γ, then α β < (α+1) β γ δ. And if If β < δ, then α β < α (β +1) γ δ. Theorem For every α,β On, α β = α β. Proof Clearly, α β Sum(α,β). So α β α β. We show that if γ Sum(α,β), then γ α β. Clearly, On On, is well founded. We prove by induction on On On, that for every α,β On On: if γ Sum(α,β), then γ α β. Let α,β On On, and suppose that the claim is true for every α,β α,β. Let γ Sum(α,β). Let {A,B} be a partition of γ such that A, A = α, α and B, B = β, β. Case 1 γ = 0. Since γ Sum(α,β), it follows that α = β = 0. Hence γ = 0 0 = α β. Case 2 γ Lim(On). Let δ < γ. Define α δ to be the ordinal such that A δ, A δ = α δ, α δ. Similarly, define β δ to be the ordinal such that B δ, B δ = β δ, β δ. 11

12 Then α δ,β δ α,β. Clearly, δ Sum(α δ,β δ ). So by the induction hypothesis, δ α δ β δ. Since α δ,β δ α,β, α δ β δ < α β. Hence δ α β. Since γ is a limit ordinal, γ = sup({δ δ < γ}). For every δ < γ, δ α β. So γ α β. Case 3 γ Suc(On). Let γ = δ +1. Either δ A or δ B. Without loss of generality we may assume that δ A. So there is α such that α = α +1. Then clearly, δ Sum(α,β). Also, α,β α,β. So by the induction hypothesis, δ α β. Hence γ = δ +1 (α β)+1 = (α +1) β = α β. Theorem For every α,β On, ρ On On, ( α,β ) = α β. Proof By induction on On On,. Denote ρ On On, by σ. Let α,β On. Case 1 α = β = 0. Then σ( 0,0 ) = 0 = 0 0. Case 2 α = 0 or β = 0 and α,β = 0,0. We may assume that α 0 and β = 0. Then { γ,δ γ,δ α,β }, = α, α. So σ( α,0 ) = ρ(α) = α = α 0. Case 3 α,β Suc(On). Let α = α + 1 and β = β + 1. Then σ( α,β ) = max(σ( α,β ),σ( α,β ) + 1. By the induction hypothesis, σ( α,β ) = α β = α β +1. Similarly, σ( α,β ) = α β +1. So σ( α,β ) = α β +2 = α β. Case 4 α Suc(On) and β Lim(On) or α Lim(On) and β Suc(On). We may assume that α Suc(On) and β Lim(On). Let α = α +1. Then (1) σ( α,β ) = max ( ) (α β)+1, sup + ({σ( α,γ ) γ < β}). 12

13 We compute sup + ({σ( α,γ ) γ < β}). Now, σ( α,γ ) = σ( α +1,γ ) = (α +1) γ = (α γ)+1 Since γ α γ is strictly increasing and β is a limit ordinal, the second equality in the formula below holds. sup + ({σ( α,γ ) γ < β}) = sup + ({(α γ)+1 γ < β}) = sup({(α γ) γ < β}). For every γ < β, α γ < α β < (α β)+1. So the maximum in Equality (1) is indeed (α β)+1. But (α β)+1 = (α +1) β = α β. So σ( α,β ) = α β. Case 5 α,β Lim(On). For every γ,δ α,β there are γ,δ On such that γ,δ γ,δ α,β. So σ( α,β ) = sup({σ( γ,δ ) γ,δ α,β }). By the increasingness of, for every γ,δ α,β, γ δ α β. So ( ) σ( α,β ) α β. It remains to show that α β σ( α,β ). Let I = {α} β and J = α {β}. For every γ,δ α,β there is p I J such that γ,δ p α,β. So ( ) σ( α,β ) = max sup({σ(p) p I}), sup({σ(p) p J}). Let 0 i=k 1 ωεi n i, 0 i=l 1 ωηi m i, be proper ω-polynomial representions of respectively α and β. Without loss of generality, η 0 ε 0. We prove that ( ) sup({σ(p) p I}) = α β. 13

14 By the induction hypothesis, sup({σ(p) p I}) = sup({α γ γ < β}). Case 5.1 η 0 < ε 0. Let β = 1 i=l 1 ωηi m i. So α β = α β +ω η0 m 0. Recall that m 0,η 0 0. It follows that sup({α γ γ < β}) = sup({α β +ω η0 (m 0 1)+δ δ < ω η 0 }) = α β +sup({ω η0 (m 0 1)+δ δ < ω η 0 }) = α β +ω η0 m 0 = α β. Hence sup({σ(p) p I}) = sup({α γ γ < β}) = α β. So ( ) holds. Case 5.2 η 0 = ε 0. Let α = 1 i=k 1 ωεi n i and β = 1 i=l 1 ωηi m i. So α β = α β +ω η0 (n 0 +m 0 ). It follows that sup({α γ γ < β}) = sup({α β +ω η0 (n 0 +m 0 1)+δ δ < ω η 0 }) = α β +sup({ω η0 (n 0 +m 0 1)+δ δ < ω η 0 }) = α β +ω η0 n 0 +m 0 ) = α β. Hence sup({σ(p) p I}) = sup({α γ γ < β}) = α β. So ( ) holds. Clearly, ( ) implies ( ) α β σ( α,β ) and ( ) and ( ) imply that α β = σ( α,β ). 14

15 Symbol index by order of appearance α+β. Ordinal addition 1 α β. Ordinal multiplication 1 α β. Ordinal Exponentiation 2 α β :=α β. Ordinal exponentiation 3 0 n 1 αεi ν i. α-polynomial 3 [α,β]:={γ α γ β} 7 α β:=sup(sum(α,β)) 9 α β 10 Notation index by alphabetic order Suc(A):=A {α On α is a successor ordinal} 7 Sum(α,β) 9 Definition Index by alphabetic order Cantor normal form of an ordinal 6 Hessenberg sum 9 α-monomial 3 α-polynomial representation 4 proper α-polynomial 3 proper α-polynomial representation 4 15