p. 1 Math 490 Notes 4 We continue our examination of well-ordered sets. If (X, ≤) is a w.o.set and x ∈ X, we ¯ define Sx = {y ∈ X ¯ y < x}. Sx is called the initial section of X determined by x. Recall from a previous example the well-ordered set A ∪ B = {a0,a1,...,an, . , b0, b1, . , bm,...,...} , formed from the union of two disjoint well-ordered sets A and B. For this set, we have Sa0 = φ, Sa2 = {a0,a1}, San = {a0,a1,...,an−1}, Sb0 = A, etc. Proposition N4.1 There exists an uncountable well-ordered set SΩ such that every initial section of SΩ is countable. Proof : Start with any uncountable set X and use the Well Ordering Theorem to well-order X. Let Y be another uncountable well-ordered set obtained by setting Y = X × X with the dictionary order. If x1 is not the least element of X, then the initial segment of Y determined ¯ by (x1, x1) is clearly uncountable, and so A = {y ∈ Y ¯ Sy is uncountable } is a non-empty set in Y , which by assumption contains a least element; call it Ω. Note that by assumption, SΩ ≤ Sy for all y ∈ Y . Claim 1: Every initial segment of SΩ countable. If not, then there would exists a z < Ω such that the initial section (SΩ)z = Yz is uncountable. By definition of A and Ω, we’d then have z ∈ A and z ≥ Ω, which is a contradiction. Claim 2: SΩ is uncountable. This follows immediately, since Ω ∈ A. ¥ This uncountable w.o.set SΩ plays an important role as a source of examples in topology, so we’ll explore it further. p. 2 Prop. N4.2 If z ∈ SΩ, the set U(z) of upper bounds of z is uncountable. Proof : This follows because SΩ = U(z) ∪ (SΩ)z, and by Prop. N4.1, SΩ is uncountable, whereas (SΩ)z is countable. ¥ ¯ Prop. N4.3 If A = {zn ¯ n ∈ N} ⊆ SΩ, then z = sup A ∈ SΩ, and (SΩ)z is countable. ¯ N Proof : First note that B = A ∪ {∪(SΩ)zn ¯ n ∈ } is a countable union of countable sets, and is therefore countable. Thus SΩ − B is uncountable, by Prop. N4.1. But every element of SΩ − B is an upper bound of A, so z = sup A ∈ SΩ follows by Proposition 5, and (SΩ)z is countable by Prop. N4.1. ¥ The next important theorem about well-ordered sets generalizes the well-known induction theorem for the natural numbers. Transfinite Induction Theorem Let A be a well-ordered set, and let B ⊆ A have the property that Aa ⊆ B =⇒ a ∈ B for all a ∈ A. Then B = A. Proof : If B = A, then A − B = φ, and we can let y be the least element of A − B. If z < y, then z ∈ B, and hence Ay ⊆ B. It follows by assumption that y ∈ B, a contradiction. Since A − B = φ yields a contradiction, we have A = B. ¥ Two well-ordered sets A and B are similar (we’ll write A≈B) iff there is an order-preserving bijection ψ from A to B. This is a bijection ψ : A → B such that x ≤ y in A iff ψ(x) ≤ ψ(y) in B. Note that similar w.o.sets have the same cardinal number. If there is a b ∈ B such that A ≈ Bb, we write A ≺ B. p. 3 Consider the following w.o.sets, ordered from left to right: N = {0, 1, 2,...,n,...} E = {0, 2, 4, 6, 8,...} D = {d0,d1,d2,...,dn,...,e0,e1,e2} N N Note that ≈ E ≈ De0 , so ≺ D and E ≺ D. One can show that every uncountable well-ordered set A for which every initial segment is countable is similar to SΩ as constructed in Prop. 6. Suppose one wishes to find an order-preserving bijection (an isomorphism) ψ between two w.o.sets A and B. Clearly, ψ must map the least element of A to the least element of B, the successor of the least element of A to the successor of the least element of B, and so on. If we denote the immediate successor of a ∈ A by a+, then in general, we must have ψ(a+) = ψ(a)+ for all a ∈ A. Furthermore, if A′ ⊆ A and a = sup A′, then ¯ ′ ′ ψ(a) = sup{ψ(x) ¯ x ∈ A } = sup ψ(A ). That is, ψ preserves sup’s. These basic facts, along with the Transfinite Induction Theorem, lead to the following result. Trichotomy Theorem If A and B are well-ordered sets, then exactly one of the fol- lowing is true: (a) A ≺ B, (b) B ≺ A, (c) A ≈ B. Furthermore, in all three cases, the order-preserving bijection is unique. Sketch of proof : If a0 is the least element of A and b0 is the least element of B, define ψ(a0) = b0. Next, for a ∈ A, assume the “induction hypothesis” that ψ : Aa → Bψ(a) is an order-preserving bijection, and for all x ∈ Aa, there exists a b ∈ B such that b > ψ(x). If a has an immediate predecessor y (i.e. y+ = a), then ψ(y) is already defined by hypothesis, and we let ψ(a) = ψ(y)+ (note that we know ψ(y)+ exists, because ψ(y) can not be the greatest element of B by assumption). If a has no immediate predecessor, then let p. 4 ¯ ′ ′ ψ(a) = sup{ψ(x) ¯ x ∈ Aa}. It should be relatively easy to see that a<a =⇒ ψ(a) < ψ(a ), so the function ψ constructed in this way is both one-to-one and order-preserving. This construction ends when either all elements of A or all elements of B (or both) have been exhausted. Suppose the elements of A are exhausted first. In other words, after ψ(a) defined for all a ∈ A, there exists a b ∈ B such that b > ψ(a) for all a ∈ A. If A has a greatest element z, ¯ then A ≈ Bψ(z)+ . If A has no greatest element, then A ≈ Bω, where ω = sup{ψ(a) ¯ a ∈ A}. Either way, A ≺ B. If the elements of B are exhausted first, then we may apply the same reasoning to ψ−1 to show B ≺ A. Finally, if the elements of A and B are exhausted simul- taneously, then A ≈ B. ¤ Trichotomy Corollary 1 If α and β are cardinal numbers, then either α ≤ β or β ≤ α. Proof : Choose A ∈ α, B ∈ β, and well-order both A and B. By the preceding theorem, exactly one of the following holds: A ≺ B, B ≺ A, or A ≈ B. If A ≺ B or A ≈ B, then α ≤ β. Otherwise, β ≤ α. ¥ Trichotomy Corollary 2 A well-ordered set is not similar to any of its initial sections. Proof : If A is well-ordered and a ∈ A, the identity injection from Aa to A is the unique order-preserving map which establishes Aa ≺ A. Thus Aa ≈ A is impossible by the Tri- chotomy theorem. ¥ p. 5 Ordinal Numbers Recall that cardinal numbers are defined by partitioning the class S of all sets into equiva- lence classes, with A ∼ B iff there is a bijection from A to B. Similarly, we define ordinal numbers to be the equivalence classes obtained when the class W of all well-ordered sets is partitioned into equivalence classes relative to the equivalence relation similarity. If L ∈ W, let L denote the ordinal number of the equivalence class containing L. As before, we denote by |L| the cardinal number for L. For each ordinal number λ, there is an associated cardinal number |λ|, which is the cardinality of any set L ∈ λ. Let O denote the class of ordinal numbers. A partial order on O is defined by λ ≤ µ iff there exists L ∈ λ and M ∈ µ such that L ≺ M or L ≈ M, with λ < µ corresponding to L ≺ M. Note that this definition is independent of the representatives L and M chosen from the given equivalence classes. It follows by the Trichotomy Theorem that ≤ is a simple order on O, and also λ ≤ µ =⇒ |λ| ≤ |µ|. The finite ordinal numbers are those ordinals λ for which |λ| is finite. Note that the empty set φ is a well-ordered set (vacuously), and the ordinal containing φ is naturally de- noted 0 (zero). Now consider all well-ordered sets with exactly n elements for some n ∈ N. It should be easy to see that all such well-ordered sets are similar to each other, and thus they all belong to the same equivalence class in W. So there is a unique ordinal number corresponding to an n-element set, which it is natural to denote by n. Since there is an obvious order-preserving bijection between finite ordinals and finte cardinals, it is customary to make no distinction between finite ordinal numbers and finite cardinal numbers. When we consider the infinite ordinals (those ordinals λ for which |λ|≥ℵ0), the situation is drastically different. We define denumerable ordinals (resp., countable ordinals) to be those of cardinality ℵ0 (resp., ≤ℵ0). We can consider the denumerable ordinals to p. 6 correspond to the distinct ways to well-order a denumerable set. For instance, N is already well-ordered in its natural ordering. Some other w.o.sets defining denumerable ordinals are listed here: ω = {a0,a1,...,an,...} = N ω +1= {a0,a1,...,an, . , b0} ω + ω = {a0,a1,...,an, . , b0, b1, . , bm,...} ω × ω = N × N, with dictionary order How many distinct denumerable ordinals are there? Recall the uncountable w.o.set SΩ such that each of its initial sections is countable.