Chapter 3: Ordinal Numbers


 Ἀριστοφάνης Ēᾍιδης Δουμπιώτης
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1 Chapter 3: Ordinal Numbers There are two kinds of number.. Ordinal numbers (0th), st, 2nd, 3rd, 4th, 5th,..., ω, ω +,... ω2, ω2+,... ω 2... answers to the question What position is... in a sequence? What position is... in an ordered set? 2. Cardinal numbers 0,, 2, 3, 4, 5,..., ℵ 0, ℵ, ℵ 2,... answers to the question How may... s are there? How many elements in a set? Finite ordinals correspond to finite cardinals, but the two notions differ in the transfinite. Ordinal numbers this chapter Cardinal numbers chapter 4 2 Ordinal numbers Cantor introduced them as labels for stages in a transfinite iteration. In 882 he became interested in them as numbers in their own right. How to represent them as sets? Finite ordinals = natural numbers 0 = = 0 {0}, 2 = {}, 3 = 2 {2},... n = {0,, 2,... n } Extend this into the transfinite: each ordinal number is the set of all its predecessors. ω = {0,, 2, 3,...} = N ω + = ω {ω} = {0,, 2, 3,..., ω} ω + 2 = (ω + ) {ω + } = {0,, 2, 3,..., ω, ω + } ω + ω (= ω2) = {0,, 2, 3,..., ω, ω +, ω + 2, ω + 3,...} Note: every ordinal is either 0, a successor ordinal, + = {}, a limit ordinal, e.g., ω = lim(0,, 2, 3, 4,...); ω ω = lim(ω, ω 2, ω 3, ω 4,...)
2 Ordering of ordinals: < iff iff. 3 CAUTION: this is not a proper definition of ordinal number. First step: examine the distinctive ordering that occurs in ordinals any non empty subset has a least element ( well ordered ). DEF n. A set A is well ordered by a binary relation iff () x A x x (irreflexive) (2) x, y, z A (x y y z x z) (transitive) (3) S A (S y S ( x S x y y x)) (any non empty subset S has a least element y) We also say (A, ) is a well ordered set (woset), or is a well ordering on A. If (A, ) merely satisfies () and (2), it is called a partially ordered set (poset). Note that () (2) x, y A (x y y x) (anti symmetry). 4 Examples of wosets. (N,<) is a woset. Proof of (3) by induction (Z,<) is a poset but not a woset Likewise, Q and R are partially, but not well, ordered by <. 4. Ordinals are well ordered by < (can t prove it yet). E.g., take ω + 5: ω ω+ ω+2 ω+3 ω+4 Can prove it for ω + 5.
3 5. Order N by m n { (m, n 2 m < n) (m, n > 2 m < n) (m > 2 n 2) 6. Order N by m n iff n < m 7. Order N N by (m, n ) (m 2, n 2 ) iff n < n 2 (n = n 2 m < m 2 )
4 8. Order N N by (m, n ) (m 2, n 2 ) iff m + n < m 2 + n 2 (m + n = m 2 + n 2 m < m 2 ) 9. Order N N by (m, n ) (m 2, n 2 ) iff max{m, n } < max{m 2, n 2 } (m < n = m 2 > n 2 ) (m < m 2 n 2 = n ) (n 2 < n m = m 2 )
5 0. Order N N by 9. Order N by (m, n ) (m 2, n 2 ) iff m < m 2 n < n 2 m n iff { (m, n are even m < n) (m, n are odd m < n) (m is even n is odd) THEOREM 2. A woset (A, ) is linearly ordered, that is, x, y A (x y x = y y x). THEOREM 3. Any subset of a woset is also well ordered. 0 Similar orderings DEF n. Let (A, ) and (B, 2 ) be posets. (A, ) is similar to (B, 2 ), i.e., (A, ) (B, 2 ), iff there exists a function f: A B such that (a) f is a bijection (b) f preserves order: x, y A (x y f(x) 2 f(y)) (f is called a similarity or order isomorphism; we write f: (A, ) (B, 2 ) or just f: A B.) Note that is an equivalence relation on posets.
6 Examples 2. ( π 2, π 2 ) R (usual orderings on both). tan: ( π 2, π 2 ) R 3. ( π 2, π 2 ) (0, ). f: ( π 2, π 2 ) (0, ) defined by x ( π 2, π 2 ) f(x) = x π Consequently, (0, ) R. tan f : (0, ) R. 5. [0, ] R. Assume f: [0, ] R and consider f (f(0) ). 6. Revisiting old examples: Ex 5 ω + 3 Ex 7 ω 2 Ex 8 ω Ex 9 ω Ex ω + ω THEOREM 4. Let (A, ) and (B, 2 ) be similar posets. Then (A, ) is a woset iff (B, 2 ) is. Conjecture. Every woset is similar to a unique ordinal. We need a rigorous definition of ordinal. Burali Forti s Paradox (897) 2 Let W be the set of all ordinals. CLAIM. W is an ordinal. Proof. (Recall that an ordinal is a set that is transitive and well ordered by.) W is transitive since if x W then x W (theorem 5). W is well ordered by since W / (theorem ),, W (since is transitive) any non empty set of ordinals has a least element (theorem 9). Therefore W W, i.e., W < W, contrary to theorem. Contradiction! Burali Forti s explanation: W is only partially ordered, not linearly ordered. Russell s explanation: W is linearly ordered but not well ordered. Jourdain, Russell and Zermelo subsequently decided there was no set W of all ordinals. This resolution is followed in modern set theory.
7 3 Morals () We need a clear definition of ordinal. (2) We need a clear notion of what a set is. We are about to give (). [See separate document.] Do we have (2)? 4 Ordinal Arithmetic What do + and mean? Let and be ordinals. Let S = { (0, x) x } { (, y) y } with order relation defined by (0, x ) (0, x 2 ) iff x < x 2 (, y ) (, y 2 ) iff y < y 2 (0, x) (, y) [Convenient notation: for any set C, C 0 = { (0, x) x C}, C = { (, x) x C}.] Thus S = 0. THEOREM 25. (S, ) is a woset. DEF n. + = the unique ordinal similar to (S, ) (by theorem 23). EXAMPLES. 7. ω + = ω {ω} > ω. Define f: (ω 0, ) (ω {ω},<) by n < ω f(0, n) = n f(, 0) = ω
8 In general, + = {} > ω = ω. Define f: ( 0 ω, ) (ω,<) by f(0, 0) = 0 n < ω f(, n) = n+ Likewise 2+ω = ω: define f: (2 0 ω, ) (ω,<) by f(0, 0) = 0 f(0, ) = n < ω f(, n) = n+2 In general, n < ω n+ω = ω. 9. ω + ω = {0,, 2,..., ω, ω +, ω + 2,...} > ω. Define f: (ω 0 ω, ) ({0,, 2,..., ω, ω +, ω + 2,...},<) by n < ω f(0, n) = n n < ω f(, n) = ω + n Note that + +, in general. 6 THEOREM 26. For any ordinals,,, + ( + ) = ( + )+. Proof. The construction of + ( + ) gives us similarities f: 0 +, g: 0 ( + ) + ( + ) The construction of ( + )+ gives us similarities h: 0 +, i: ( + ) 0 ( + )+ 0 + f f 0 + ( + ) g ( + ) g ( + )+ ( + ) 0 i i + 0 h h
9 Define j: + ( + ) ( + )+ by 7 x j ( g(0, x) y j ( z j ( g(, f(0, y)) g(, f(, z)) ) = i(0, h(0, x)) ) = i(0, h(, y)) ) = i(, z) Then by theorem 20 + ( + ) = ( + )+. What about? Let P = (product of sets), ordered by (x, y ) (x 2, y 2 ) iff (y < y 2 ) (y = y 2 x < x 2 ) THEOREM 27. (P, ) is a woset. DEF n. (Ordinal multiplication) = the unique ordinal similar to (P, ). 2 =, n+ = n for n < ω. 8 EXAMPLES. 20. ω2 = ω + ω. Define f: (ω 2, ) (ω 0 ω, ) by n < ω f(n, 0) = (0, n) n < ω f(n, ) = (, n) Then (ω2,<) (ω 2, ) (ω 0 ω, ) (ω + ω,<) so by theorem 20 ω2 = ω + ω. 2. 2ω = ω. Define f: (2 ω, ) (ω,<) by n < ω f(0, n) = 2n n < ω f(, n) = 2n+ Thus, by definition, 2ω = ω. In general, n < ω nω = ω. Note: in general.
10 9 THEOREM 28. For any ordinals,,, () = (). Proof. The construction of () gives us similarities f: The construction of () gives us similarities h: g: () () i: () () g ( ) X X h X ( ) i ( ) ( ) f X Define j: () () by x y z j ( i(x, h(y, z)) ) = g(f(x, y), z) So by theorem 20 () = (). 20 THEOREM 29. For any ordinals,,, + = ( + ). Proof. The construction of + gives us similarities f:, g:, h: () 0 () +. The construction of ( + ) gives us similarities i: 0 +, j: ( + ) ( + ). X X g f ( ) ( ) 0 h h j X( + ) i + i 0 + ( + )
11 Define a similarity k: + ( + ) by x y k ( ) h(0, f(x, y)) = j(x, i(0, y)) x z k ( ) h(, g(x, z)) = j(x, i(, z)) 2 Then by theorem 20 + = ( + ). BEWARE: ( + ) + in general. THEOREM 30. (ω + )ω = ω 2. Proof. The construction of (ω + )ω gives us similarities f: ω 0 ω + g: (ω + ) ω (ω + )ω The construction of ω 2 gives us a similarity h: ω ω ω 2 22 ω ω ω ( +)X g h ω ω Xω ω ω 0 f ω+ f ( ω +) ω ω 2 ω Define a similarity i: (ω + )ω ω 2 by m i(g(f(0, m), 0)) = h(m, 0) m n i(g(f(0, m), n+)) = h(m+, n+) n i(g(f(, 0), n)) = h(0, n+) So by theorem 20 (ω + )ω = ω 2.
12 23 THEOREM 3. (ω + )ω < ω 2 + ω. Proof. Theorem 30 shows (ω + )ω = ω 2. We must just show ω 2 < ω 2 + ω. The construction of ω 2 + ω gives us a similarity f: (ω 2 ) 0 ω ω 2 + ω We can then define a function g: ω 2 ω 2 + ω by x ω 2 g(x) = f(0, x) This is a similarity from ω 2 to an initial segment of ω 2 + ω (i.e., everything below f(, 0)). So (ω + )ω < ω 2 + ω. THEOREM 32. n < ω (ω + n)ω = ω 2. Proof. As in theorem 30. THEOREM 33. m, n < ω (0 < m (ω + n)m = ωm+n). Proof. Similar to theorem 30. Or, (ω + n)m = (ω + n)(++ + ) 24 = (ω + n)+(ω + n)+ + (ω + n) = ω + (n+ω)+(n+ω)+ +(n+ω)+n = ω + ω + ω + +ω + n by theorem 29 by theorem 26 by example 8 = ω+ω+ω+ +ω+n = ω(+++ + )+n by theorem 29 = ωm+n. EXAMPLE. 22. Reduction to normal form. (Use associativity tacitly.) (ω + )(ω + 2)(ω + 3) = (ω + )(ω + 2)ω + (ω + )(ω + 2)3 = (ω + )ω 2 + (ω + )(ω3+2) = ω 3 + (ω + )ω3+(ω + )2 = ω 3 + ω 2 3+ω2+ by theorem 29 by theorems 32, 33 by theorems 30, 29 by theorems 30, 33
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