Ordinal numbers and Ordinal arithmetic

Ordinal numbers and Ordinal arithmetic Cardinal and ordinal numbers

Cardinal numbers are counting numbers ℵ one, two, three, ··· , ℵ0, 2 0 = c, ···

Ordinal numbers are ranking numbers first, second, third, ···

Ordinal numbers and Ordinal arithmetic Well-order relation

Recall that every set can be well-ordered by Zermelo’s Well-ordering Principle.

For example, finite sets can be easily well-ordered. R with usual order ≤ is not well-ordered. N with usual order ≤ is well-ordered. 0 Consider a total order ≤ on N so that

N = {1, 3, 5, ··· ; 2, 4, 6, ···}

0 Then ≤ is a well-order relation on N.

Ordinal numbers and Ordinal arithmetic Order isomorphisms

Definition Two well-ordered sets (A, ≤) and (B, ≤0) are called order-isomorphic, written (A, ≤) ≈ (B, ≤0) or simply A ≈ B, if there is an increasing bijection f : A → B (a ≤ b ⇒ f (a) ≤0 f (b)) Such a function is called an order-isomorphism.

Example 0 (N, ≤) 6≈ (N, ≤ )

0 Suppose ∃f :(N, ≤ ) → (N, ≤), an order-isomorphism. Then 0 {a ∈ N | a < 2} is infinite, but {b ∈ N | b < f (2)} is finite, a contradiction.

Ordinal numbers and Ordinal arithmetic Axiom of Ordinality

Remark The order-isomorphic relation ≈ is an equivalence relation.

AXIOMOF ORDINALITY O1 Each well-ordered set (A, ≤) is assigned an ordinal number ord(A, ≤), and if α is an ordinal number then α = ord(A, ≤) for some well-ordered set (A, ≤). O2 ord(A, ≤) = 0 iff A = ∅ O3 If (A, ≤) is a well-ordered set with cardA = k then ord(A, ≤) = k. O4 For two well-ordered sets (A, ≤) and (B, ≤0), ord(A, ≤) = ord(B, ≤0) iff (A, ≤) ≈ (B, ≤0)

Ordinal numbers and Ordinal arithmetic Briefly speaking

n Sets o ∼(equipotency) ↔ {cardinal numbers}

n Well-ordered Sets o ≈(order-isom) ↔ {ordinal numbers}

NOTATION ω := ord(N, ≤) = ord{1, 2, 3, ···}

Remark that ω 6= ord{1, 3, 5, ··· ; 2, 4, 6, ···}

Ordinal numbers and Ordinal arithmetic Segments

Definition Let (A, ≤) be a toset. A subset S of A is called a segment of A if y ∈ S, x ≤ y with x ∈ A ⇒ x ∈ S.

Remark Let (A, ≤) be a well-ordered set. For x ∈ A, let Ax := {a ∈ A | a < x}. Then Ax is a segment of A.

FACT: Every segment of A is either A itself or a segment of the form Ax .

Example 0 For (N, ≤ ) = {1, 3, ··· ; 2, 4, ···}, 0 we see that (N, ≤) ≈ (N2, ≤ ).

Ordinal numbers and Ordinal arithmetic Ordinal ordering

Definition Let α = ord(A, ≤) and β = ord(B, ≤0) be ordinal numbers. We write α 4 β if (A, ≤) is order-isomorphic to a segment of (B, ≤0).

Example ω 4, 6= ord{1, 3, 5, ··· ; 2, 4, 6, ···}

FACT:

(1) The cardinal ordering 4 is a total order on cardinal numbers. (2) The ordinal ordering 4 is a total order on ordinal numbers.

Ordinal numbers and Ordinal arithmetic Ordinal Sum α + β

Definition Let (A, ≤) and (B, ≤0) be two disjoint well-ordered sets. Define the order relation ≤∗ on A S B as follows:  a ∈ A, b ∈ B or  a ≤∗ b ⇔ a, b ∈ A, a ≤ b or  a, b ∈ B, a ≤0 b

Then it is easy to see that (A S B, ≤∗) is a well-ordered set.

Definition Let α = ord(A, ≤) and β = ord(B, ≤0) where A and B are disjoint. Define the ordinal sum α + β to be α + β := ord(A S B, ≤∗).

Ordinal numbers and Ordinal arithmetic Examples

(1) 1 + ω = ω Why? 1 + ω = ord({0} S{1, 2, ···}) = ord{1, 2, 3, ···} = ω

(2) ω 4, 6= ω + 1 Why? ∗ (i) {1, 2, 3 ···}  (A, ≤ ) = {1, 2, 3, ··· ; 0}, order-isomorphic to the segment A0, Hence 4. (ii) Let f :(A, ≤∗) = {1, 2, 3 ··· ; 0} → {1, 2, 3, ···} be an order-isomorphism. The segment A0 is infinite, by the segment Nf (0) is finite, a contradiction. Hence 6=.

(3) Similarly, we can shows that α 4, 6= α + 1 for any ordinal number α.

Ordinal numbers and Ordinal arithmetic Ordinal product β α

Definition Let α = ord(A, ≤) and β = ord(B, ≤0). Define the ordinal product β α to be

β α := ord(A × B, ≤∗) where ≤∗ is the lexicographic ordering on A × B which is defined as follows: ( a < x or (a, b) ≤∗ (x, y) ⇔ a = x, b ≤0 y

The fact is that the lexicographic ordering ≤∗ is a well-order relation on A × B.

Note that ord(A × B, ≤∗) is β α, not αβ!

Ordinal numbers and Ordinal arithmetic Examples

(1) 2ω = ω Why?

{1, 2, 3, · · · } × {0, 1} = {(1, 0), (1, 1), (2, 0), (2, 1), ···} ≈ {1, 2, 3, ···}

(2) ω2 = ω + ω Why?

{0, 1} × {1, 2, 3, ···} = {(0, 1), (0, 2), (0, 3), ··· ;(1, 1), (1, 2), (1, 3), ···} ≈ {1, 3,, 5 ··· ; 2, 4, 6, ···}

Ordinal numbers and Ordinal arithmetic Conclusion

Theorem For any ordinal number α, 1 the set of ordinal numbers β such that β 4, 6= α

{β | β 4, 6= α} is a well-ordered set, and 2 α = ord{β | β 4, 6= α}

Can identify α ≡ {β | β 4, 6= α}.

Ordinal numbers and Ordinal arithmetic Example

0 ≡ ∅ 1 ≡ {0} ··· k ≡ {0, 1, ··· , k − 1} ··· ω ≡ {0, 1, 2, · · · } ∼ N ω + 1 ≡ {0, 1, 2, ··· ; ω} ∼ N ω + 2 ≡ {0, 1, 2, ··· ; ω, ω + 1} ∼ N ··· ω2 ≡ {0, 1, 2, ··· ; ω, ω + 1, ω + 2, · · · } ∼ N ω2 + 1 ≡ {0, 1, 2, ··· ; ω, ω + 1, ω + 2, ··· ; ω2} ∼ N ···

Ordinal numbers and Ordinal arithmetic Initial ordinal

Recalling that α ≡ {β | β 4, 6= α}, we consider the set

N = {α | α ∼ N}.

The fact is (N , 4) is a well-ordered set. By the definition of well-order relation, N has the least element, say α0 ∈ N , and hence α0 ∼ N. The element α0 is called the initial ordinal for N. In fact, α0 = ω.

Ordinal numbers and Ordinal arithmetic