Ordinal numbers and Ordinal arithmetic Ordinal numbers and Ordinal arithmetic Cardinal and ordinal numbers Cardinal numbers are counting numbers @ one, two, three, ··· , @0; 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 = f1; 3; 5; ··· ; 2; 4; 6; · · · g 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 9f :(N; ≤ ) ! (N; ≤), an order-isomorphism. Then 0 fa 2 N j a < 2g is infinite, but fb 2 N j b < f (2)g is finite, a contradiction. Ordinal numbers and Ordinal arithmetic Axiom of Ordinality Remark The order-isomorphic relation ≈ is an equivalence relation. AXIOM OF 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) $ fcardinal numbersg n Well-ordered Sets o ≈(order-isom) $ fordinal numbersg NOTATION ! := ord(N; ≤) = ordf1; 2; 3; · · · g Remark that ! 6= ordf1; 3; 5; ··· ; 2; 4; 6; · · · g 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 2 S, x ≤ y with x 2 A ) x 2 S. Remark Let (A; ≤) be a well-ordered set. For x 2 A, let Ax := fa 2 A j a < xg. 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; ≤ ) = f1; 3; ··· ; 2; 4; · · · g, 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 ! 4; 6= ordf1; 3; 5; ··· ; 2; 4; 6; · · · g 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: 8 a 2 A; b 2 B or <> a ≤∗ b , a; b 2 A; a ≤ b or > :a; b 2 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 + ! = ! Why? 1 + ! = ord(f0g Sf1; 2; · · · g) = ordf1; 2; 3; · · · g = ! (2) ! 4; 6= ! + 1 Why? ∗ (i) f1; 2; 3 · · · g (A; ≤ ) = f1; 2; 3; ··· ; 0g, order-isomorphic to the segment A0, Hence 4. (ii) Let f :(A; ≤∗) = f1; 2; 3 ··· ; 0g ! f1; 2; 3; · · · g 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) 2! = ! Why? f1; 2; 3; · · · g × f0; 1g = f(1; 0); (1; 1); (2; 0); (2; 1); · · · g ≈ f1; 2; 3; · · · g (2) !2 = ! + ! Why? f0; 1g × f1; 2; 3; · · · g = f(0; 1); (0; 2); (0; 3); ··· ;(1; 1); (1; 2); (1; 3); · · · g ≈ f1; 3;; 5 ··· ; 2; 4; 6; · · · g 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 fβ j β 4; 6= αg is a well-ordered set, and 2 α = ordfβ j β 4; 6= αg Can identify α ≡ fβ j β 4; 6= αg. Ordinal numbers and Ordinal arithmetic Example 0 ≡ ; 1 ≡ f0g ··· k ≡ f0; 1; ··· ; k − 1g ··· ! ≡ f0; 1; 2; · · · g ∼ N ! + 1 ≡ f0; 1; 2; ··· ; !g ∼ N ! + 2 ≡ f0; 1; 2; ··· ; !; ! + 1g ∼ N ··· !2 ≡ f0; 1; 2; ··· ; !; ! + 1;! + 2; · · · g ∼ N !2 + 1 ≡ f0; 1; 2; ··· ; !; ! + 1;! + 2; ··· ; !2g ∼ N ··· Ordinal numbers and Ordinal arithmetic Initial ordinal Recalling that α ≡ fβ j β 4; 6= αg, we consider the set N = fα j α ∼ Ng: The fact is (N ; 4) is a well-ordered set. By the definition of well-order relation, N has the least element, say α0 2 N , and hence α0 ∼ N. The element α0 is called the initial ordinal for N. In fact, α0 = !. Ordinal numbers and Ordinal arithmetic.