Initial Ordinals We now return to ordinals in general and use them to give a more precise meaning to the notion of a cardinal. First we make some observations. Note that if there is an ordinal with a certain property (expressible by a formula φ(x) of ZF) then there is a least such ordinal since the if φ(α) holds for some ordinal α then the set {β α + 1 φ(β)} is non-empty and its least element is also the least ordinal satisfying φ. Definition by Transfinite Recursion on Ord. If θ(x,u, y) is as in Theorem 7 (see (4), page 45 ) then there is a unique class function F such that for all ordinals α we have θ(α, F α, F(α)). This is because by Theorem 7 for each δ Ord there is a unique f δ with domain δ such that for each α δ we have θ(α, f δ α,f δ (α)), and since it is easy to see that for γ, δ Ord, δ < γ we must have f γ δ = f δ, we can take F to be the union of all f δ over δ Ord (so F δ = f δ for each δ Ord). In order to apply Theorem 7 when dealing with ordinals, it is often convenient to specify the values of the function that is being defined separately: for 0, for successor ordinals in terms of the values for the immediate predecessors, and for limit ordinals. Definition 26 We say that an ordinal κ is an initial ordinal, or a cardinal, if for each ordinal β such that β < κ we have β κ (that is, β κ and β κ). The class of all initial ordinals is denoted Card. Notes 1 If β < κ then β κ by Proposition 48(iii), so β κ holds obviously. The distinguishing feature is that β κ for all β < κ. Note that for example, ω + 1 = ω {ω} is not an initial ordinal since f : ω ω {ω} defined by f(0) = ω, f(m + 1) = m for m ω is a bijection. More generally, no infinite successor ordinal is an initial ordinal by Example 8.1. 2 We have an apparent clash in terminology here since we have already used the term cardinal (number) in the naive part for something that is shared by equinumerous sets. We will explain the connection shortly. Proposition 57 For every ordinal α there is an initial ordinal κ such that κ α and α κ. Proof Let α be an ordinal and let c = {γ γ is an ordinal &γ α &α γ}. Clearly α c so c is a nonempty set of ordinals, and as such it has a least element δ. δ is an initial ordinal since if not then there would be some λ < δ such that λ δ but then also λ α so λ c but that cannot be because δ is the least element of c. Hence we can take δ for κ. 63
Proposition 58 (i) Each n ω is an initial ordinal. (ii) ω is an initial ordinal. Proof (i) follows by the Pigeonhole Principle for natural numbers (Proposition 5), since if n ω = N and k < n then n k would mean an injection from n to k < n which cannot be. (ii) follows by Proposition 7. Proposition 59 If b is a set of initial ordinals then b is an initial ordinal. Proof By Proposition 49(ii) α = b is an ordinal. Let β < α, that is, β α. By the definition of α there is some γ b such that β γ. Hence β < γ and γ α. If α β was true then we would have γ β (since the restriction of a bijection between α and β to γ is an injection from γ to β) but clearly also β γ, so we would have β γ which contradicts γ being an initial ordinal. Proposition 60 If κ is an initial ordinal then there is an initial ordinal κ + such that κ < κ + and if α < κ + then α κ. Proof Let C = {γ γ is an ordinal & γ κ}. C is non-empty by Theorem 11 and the least element of it has the required properties. Hence we have initial ordinals 0, 1, 2,3,, }{{} ω, ω + + + 0, ω1, ω2,, ω n, ω ω + }{{}}{{}}{{} 0,,ω ω1, =ω 0 n N }{{} =ω 1 =ω 2 =ω 3 }{{} =ω ω0 +1 =ω ω0 The subscripts 1 are ordinals, for a successor ordinal ω α+1 = ω + α and for a limit ordinal ω α = β<α ω β. The ω α are well defined by Theorem 7 and the above observation about using it for Ord. In more detail, for each ordinal δ there is a unique function f δ with Dom(f δ ) = δ (and Rng(f δ ) Card) such that f δ (0) = ω and { (f f δ δ (α 1)) + if α is a successor (α) = β<α fδ (β) if α is a limit ordinal By Example 8.2, if 0 < δ < γ then f γ δ = f δ so w α is f γ (α) where γ is any ordinal greater than α. By the same Example, any infinite initial ordinal is equal to ω α for some α. Hence we have { Pow(N) R N = ω = ω 0 1 Note that ω n for n ω could also stand for the initial segment N n as we used it before. It is always clear from the context what is meant. ω 1 64
The Continuum Hypothesis, which we have mentioned before, asserts that ω 1 R. It cannot be proved, nor shown false, from the ZF axioms. We remark that ω α in their role as initial ordinals are sometimes also written as ℵ α. Recall that we have defined sets a, b to have the same cardinality, a = b if a b, but we have not said what cardinality is. For sets that are equinumerous to an initial ordinal we take their cardinality to be that initial ordinal, and we identify ℵ α and ℵ α. Hence ω α = ℵ α and since ω = ω 0 = N, we have ω = N = ℵ 0. Expressions like a = ℵ 0, b = 2 ℵ0 mean a N, b 2 N as before. We will not use this notation except for ℵ 0 as we have already done in the early part of the course. Theorem 13 For κ an initial ordinal such that ω κ we have κ κ κ. Proof Assume that the theorem fails and let λ be the least infinite initial ordinal such that λ λ λ. Let be the well-ordering of λ λ defined by γ,δ α,β [γ δ < α β (γ δ = α β & (γ < α (γ = α & δ < β)))] (please see Coursework 2 & solutions). By Theorem 8 there is an isomorphism between (λ,<) and λ λ, or between one of them and an initial segment of the other. Assume that λ is isomorphic to some initial segment (λ λ) α,β where α, β λ. Note that α β is equal to the greater of α and β so since λ is an infinite initial ordinal and as such a limit ordinal, γ = (α β) + 1 λ. γ is infinite (exercise) and by definition of we have (λ λ) α,β γ γ so the assumption of λ being isomorphic to (λ λ) α,β means that λ γ γ. Let µ γ be the initial ordinal such that γ µ (it exists by Proposition 57). Then µ < λ so we must have µ µ µ because λ is the first infinite initial ordinal for which it fails, so λ γ γ µ µ µ and hence (using the Schröder Bernstein theorem) λ µ but that contradicts λ being an initial ordinal. It follows that there must be an isomorphism between (λ λ, ) and either λ or an initial segment of it but that means that λ λ λ so since also obviously λ λ λ, λ λ λ by the Schröder Bernstein theorem again, but that contradicts the choice of λ. The result follows. Corollary 8 For every infinite set b which can be well ordered we have b b b. Proof This follows from by the previous theorem, Theorem 10 and Proposition 57 (exercise). 65
The Axiom of Choice The Axiom of Choice, AC For b a set, / b there is a function f : b b such that for all x b, f(x) x. In this case we say that f is a choice function on b since for each x b f picks an element of x. Notes 1 From work of Gödel (1937) it is known that AC cannot be disproved from ZF whilst from work of Cohen (1963) (for which he won the Field s Medal), AC cannot be proved from ZF. So AC is independent of ZF (as is the Continuum Hypothesis). 2 The axioms of ZF with AC added to them are referred to as ZFC. If a theorem, lemma or proposition uses AC in its proof we write Theorem (AC) etc. 3 We have already seen an example of AC when we showed that for a countable set u such that all elements of u are countable the set u is countable, which is the contents of Proposition 16: assume that u is a non-empty and countable set of non-empty countable sets. 1 So u = {a 0,a 1,a 2,...} and there is a function t with domain N such that 2 t(n) = { g g is a surjective function from N to a n } for n N Let b = Rng(t) = {t(n) n N}. Each element of b is non-empty since each a n is non-empty and countable, so by AC there is a function f with domain b such that for y = t(n) b, f(y) is a surjective function from N to a n, that is, Defining by f(t(n)) : N a n for n N. h : N N n=0 a n h(n,m) = [f(t(n))](m) for n,m N, noting that h is a surjection and recalling that N N N gives N n=0 a n so the result follows. 4 It is not necessarily always the case that we need to appeal to AC in order to guarantee the existence of a choice function. If the set b is finite then we 1 Recall that u being non-empty and countable means that u = {a 0, a 1, a 2,...}, where a n = r(n) for some surjective function with domain N 2 t exists by Proposition 31. 66
can prove that a choice function exists, see Example 8.3. Even when b is infinite there may be some way of picking out a distinguished element of each x b, and thus we may be able to show that a choice function exists just from ZF alone 1. The following is an important example of this: Lemma 10 If / b and b can be well ordered then b has a choice function. Proof Let be a well ordering of b. Notice that if x b then = x b so x has a (unique) least element with respect to. Defining f : b b by equivalently f(x) = the least element of x wrt, f = { x,y x b & y x & ( z x)y z }, gives the required choice function without using AC. The main purpose of this section is to prove the equivalence of AC with several other well known possible axioms. The first of these is: The Well Ordering Theorem, WOT Every set can be well ordered. Theorem 14 WOT AC. Proof In the direction this is an immediate consequence of Lemma 10. In the other direction assume AC and let c be a set. We need to define a well ordering of c. Let b be the set of non-empty subsets of c and let g be a choice function for b, that is, g : b b and g(x) x for all x b. Note that b = c. Given an ordinal α, by Theorem 7 there is a function f α : α c {c} such that for β α { f α g(c\rng(f (β) = α β)) if c\rng(f α β), c otherwise (13) Notice that if α < λ Ord then f α = f λ α. For if not there would be a β < α such that f α (β) f λ (β). But then there would be a least such β and for this we must have f α β = f λ β and hence from (13), f α (β) = f λ (β), contradiction. Clearly for γ < β < α and f α (β) c, f α (β) f α (γ) since f α (γ) Rng(f α β) and f α (β) = g(c\rng(f α β)) c\rng(f α β) so f α (β) / Rng(f α β). Hence if c / Rng(f α ) then f α is an injection from α into c. 1 Bertrand Russel gave an example of an infinite set of pairs of shoes on the one hand and an infinite set of pairs of socks on the other hand. We can argue that within the former there clearly exists the subset of left shoes but why should there be a subset of the set of socks containing just one from each pair? 67
By Theorem 11 there is some ordinal λ such that λ c, so for this λ it must be the case that c Rng(f λ ). Hence for some α < λ, c\rng(f λ α) =. For the least such α then c Rng(f λ α) and for γ < α so f λ (γ) c and Rng(f λ α) = c. c\rng(f λ γ) Recalling that f λ α = f α and putting this together with the fact established above that if c / Rng(f α ) then f α is an injection from α into c we see that f α is a bijection from α to c. By Proposition 44 (applied to (f α ) 1 ) we can well order c, as required. Cardinals By Theorem 10 and Proposition 57 every set which can be well ordered is equinumerous to some, necessarily unique, initial ordinal. Under AC any set can be well ordered so cardinal numbers as we intuitively described them early on can all be identified with initial ordinals (cardinals). A second equivalent of AC is: Law of Trichotomy, LOT For any two sets a, b either a b or b a. Theorem 15 AC LOT. Proof Since WOT is equivalent to AC it is enough to show that WOT is equivalent to LOT. We have done this already, see Corollary 4 and the remark following Hartog s Theorem (Theorem 11). Our third equivalent is commonly applied in mainline Pure Mathematics. It concerns the existence of maximal elements in posets. If (x, ) is a poset and y x we say that y is a chain if for all u, v y, we have u v or v v, that is, if y is a subset of x that is totally ordered by the restriction of. We say that w x is an upper bound for y if for all u y we have u w. Recall that s is a maximal element of x if ( u x) s u. Zorn s Lemma, ZL Let x, be a poset such that every chain has an upper bound. Then x has a maximal element. Theorem 16 ZL AC. Proof Assume ZL. We shall show LOT. Let a,b be sets and let x be the set of bijections from a subset of a to a subset of b, that is x = {f ( c a)( d b)f : c d}. 68
Please check that it is a set as an exercise. For f, g x let f g f g. Then x, clearly is a poset, and if y x is a chain then y is also bijection from a subset of a to a subset of b, so y x, and y is an upper bound for y. See Example 8.5 for details. Hence by ZL there is a maximal element f x. It must be the case that either Dom(f) = a, in which case f : a b, or Rng(f) = b, in which case f 1 : b a. Otherwise we would have Dom(f) a and Rng(F) b so there would exist some u a\dom(f) and v b\rng(f) which would mean f { u, v } x and f f { u,v }, contradicting the maximality of f. It follows that ZL implies LOT, and hence AC. For the other direction assume WOT. Let x, be a poset such that every chain has an upper bound. By WOT x can be well ordered and hence by Theorem 10 there is an ordinal α and a function g such that g : α x. By Theorem 7 there is a function h : α x such that { g(β) if ( γ < β) [ h(γ) g(β) g(β) h(γ)], h(β) = g(0) otherwise. Notice that the first case holds when β = 0, and that h(0) = g(0). We shall show that Rng(h) x is a chain. Let u, v Rng(h) and suppose u = h(γ), v = h(δ) with, say, γ < δ. Then by the choice of h(δ) either h(δ) = g(δ) and h(γ) h(δ) h(δ) h(γ), or h(δ) = g(0) = h(0) so again, if h(γ) = g(0) then h(δ) = g(0) = h(γ) and otherwise h(γ) = g(γ) and h(δ) = h(0) h(γ) h(γ) h(0) = h(δ). Since Rng(h) is a chain, by the assumption on x, it has an upper bound. Let s x be an upper bound for Rng(h). This s must be a maximal element of x (with respect to ). Otherwise there would be some u x such that s u, say u = g(β), β α. Then for all v Rng(h), v s u so v u. But since Rng(h β) Rng(h) this means that ( γ < β)h(γ) g(β) so h(β) = g(β) = u and hence u Rng(h) and u s, contradiction. Hence s is indeed a maximal element of x. It follows that WOT, and hence AC, implies ZL. Zorn s Lemma has a number of applications outside of Set Theory, for (nonexaminable) example: Proposition 61 (AC) Every vector space has a basis. 69
Proof Let V be a vector space over a field F 1 and let x be the set of linearly independent subsets of V. Consider the poset x, (somewhat abusing the notation, we mean x with the restriction of to x). If y x is a chain then the vectors in y are also linearly independent (exercise) so y x, and y is an upper bound for y. Hence x satisfies the assumption of ZL. By ZL there is a maximal element b of x. Since the vectors in b are linearly independent to show that b is a basis it is enough to show that b spans V. Suppose not, say u V but u cannot be written in the form r 1 v 1 + r 2 v 2 +... + r n v n for any r 1,r 2,...,r n F and v 1,v 2,...,v n b. Then the vectors in b {u} are still linearly independent, so this set is in x but b b {u} which contradicts b being a maximal element of x. So b is the required base. This is obviously a result which algebraists would welcome. However it already hints at a darker side to AC: Corollary 9 There is a non-linear function f : R R satisfying the equation f(x + y) = f(x) + f(y), By non-linear we mean that f is not of the form f(x) = ax for any constant a R. Proof Using Proposition 61 let b R be a basis for the vector space of the reals over the field of rationals (a Hamel Basis). Pick v 0 b. Then any x R can be written as x = r 0 v 0 + r 1 v 1 + r 2 v 2 +... + r m v m for some r 0,r 1,r 2,...,r m Q and v 1,v 2,...v m b, where none of v 1,v 2,...,v m is equal to v 0. Furthermore, r 0 in this representation is unique (it may be 0). Define f(x) = r 0. Noticing that if x = r 0 v 0 + r 1 v 1 + r 2 v 2 +... + r m v m, y = p 0 v 0 + p 1 w 1 + p 2 w 2 +... + p n w n, with r 0,r 1,...,r m,p 0,p 1,...,p n Q and v 1,v 2,...,v m,w 1,w 2,...,w n b and v 0 distinct from any of v 1, v 2,...,v m, w 1,w 2,...,w n then x + y = (r 0 + p 0 )v 0 + r 1 v 1 +... + r m v m + p 1 w 1 +... + p n w n 1 Recall that a vector space over a field F (with operations + F, F) is a set V that is an Abelian group under vector addition (which we denote +), with a scalar multiplication assigning rv V to each r F and v V, such that that the appropriate properties of distributivity, and of compatibility of the multiplications hold. A set a V is linearly independent if for any finite subset {v 1,...,v n} of a (with v 1,...,v n distinct) and r 1,... r n F, if r 1 v 1 +... + r n v n = 0 then r 1 = = r n = 0. A set a V is a basis for V if it is linearly independent and any u V can be expressed as a linear combination of elements of a, i.e. as r 1 v 1 +... + r nv n for some v 1,...,v n a and r 1,, r n F. 70
so f(x + y) = r 0 + p 0 = f(x) + f(y) though f is clearly not linear (it only takes rational values and is not identically zero). Just for interest we now give another, rather surprising equivalent of AC. Theorem 17 AC a a a for every infinite set a. Proof Left to right follows from Corollary 8 In the other direction given a set b it is enough to show that assuming the right hand side b can be well ordered. We can obviously assume that b is non-empty. Let α be such that not α b (in Theorem 11 such an ordinal α has been shown to exist, without using AC). Clearly we may assume that α ω 0. To simplify the notation without loss of generality assume that b α = (otherwise we can take e.g. b {α} in place of b; there is an obvious bijection between this set and b, so showing that it can be well ordered implies that b can be well ordered and (b {α}) α = - exercise ). We first show that b α b α using the Schröder-Bernstein Theorem. We have b α (b b) (b α) (α b) (α α) = (b α) (b α) b α. Now pick x 0 b. The function f : b α b (α α) defined by f(x) = x, 0,0 for x b, f(β) = x 0, β,1 is an injection, so b α b (α α). This gives for β α b α b (α α) b α. Hence we have some function g : b α b α. There are two possibilities, either for some x 0 b we have g(x 0,β) b for all β α, or for every x b there is some β α such that g(x,β) α. In the first case the function f : α b defined by f(β) = g(x 0,β) for β α gives an injection from α into b, contradicting the choice of α. So assume that for every x b there is some β α such that g(x,β) α, and define z : b α by z(x) = the least β such that g(x,β) α for x b. Then the function f : b α defined by f(x) = g(x,z(x)) for x b gives an injection from b into α, so b can be well ordered (Proposition 44). 71