Well Ordered Sets (Continued) Theorem 8 Given Any Two Well-Ordered Sets, Either They Are Isomorphic, Or One Is Isomorphic to an Initial Segment of the Other

Home , Order type

Well Ordered Sets (continued) Theorem 8 Given any two well-ordered sets, either they are isomorphic, or one is isomorphic to an initial segment of the other.

Proof Let ha, < i and hb, ⊳i be well-ordered sets. Let d be such that d∈ / b (take e.g. d = b). Deﬁne f : a → b ∪ { d} to satisfy

the ⊳−least element in b\Rng (f ↾a ) if b\Rng (f ↾a ) 6= ∅ f(x) = x x ½ d otherwise (12) The existence of such f is guaranteed by the previous theorem 1 since for each x from the well ordered set a, f(x) is uniquely determined in terms of x and f ↾ax.

Observe that if x1, x 2 ∈ a and x1 < x 2 then ax1 ⊂ ax2 so

b\Rng (f ↾ax2 ) ⊆ b\Rng (f ↾ax1 ) and hence we have

either both f(x1) and f(x2) are in b and f(x1) ⊳ f(x2),

(f(x2) ∈ b\Rng (f ↾ax2 ) and f(x1) ∈ Rng (f ↾ax2 ) so f(x1) 6= f(x2))

or only f(x1) is in b and f(x2) = d,

or f(x1) = f(x2) = d.

Deﬁne g = f ∩ (a × b), i.e. g(x) = f(x) except when f(x) = d in which case x∈ / Dom (g). Let s = Dom (g), t = Rng (g). If x1, x 2 ∈ a, x2 ∈ s and x1 < x 2 then x1 ∈ s by the above observation. Also, if y1, y 2 ∈ b, y2 ∈ t and y1 ⊳ y2 then y1 ∈ t since if y2 = f(x) for some x in a then y1 must be in Rng (f ↾ax). It follows by Example 6.2 that s = a or s = ax for some x ∈ a, and t = b or t = by for some y ∈ b. Note that s is well ordered by the restriction of < to s and t is well ordered by the restriction of ⊳ to t. g : s → t is a bijection since it is surjective by deﬁnition of t and if x1, x 2 ∈ s, x1 6= x2 then assuming without loss of generality that x1 < x 2 we have f(x1) ⊳ f(x2) by the above observation, so g an order preserving bijection and as such an isomorphism between s and t by Corollary 3.

1For θ(x, u, y ) we can take e.g. the formula

[x ∈ a & F n (u) & Dom (u) = ax & b\Rng (u) 6= ∅ & y = Lst (b\Rng (u))]

∨ [¬(x ∈ a & F n (u) & Dom (u) = ax & b\Rng (u) 6= ∅) & y = d] where y = Lst (w) stands for y ∈ w & ( ∀z ∈ w) y E z. It is easy to check that ∀x∀u∃!yθ (x, u, y ), and that for x ∈ a, (12) corresponds to θ(x, f ↾ax, f (x)).

48 Since s = a or s = ax for some x ∈ a, and t = b or t = by for some y ∈ b, and since the case of both s = ax and y = by cannot occur (it would mean f(x) = d but b\Rng (f ↾ax) 6= ∅), the theorem follows. ¥

Corollary 4 If sets a and b can be well ordered then either a ¹ b or b ¹ a.

Consequently, if all sets can be well ordered then for any two sets a, b we have a ¹ b or b ¹ a. Later, we will show the converse: if a ¹ b or b ¹ a for any two sets a, b, then every set can be well ordered! The statement that all sets can be well ordered is usually referred to as the Well Ordering Theorem (WOT) and the statement that for any two sets a, b we have a ¹ b or b ¹ a is referred to as the Law of Trichotomy (LOT). Hence we have shown that it follows from out axioms that WOT implies LOT and we will soon show that is also follows that LOT implies WOT. (However, neither can be proved in ZF. We will return to this later.)

Proposition 45 (i) Let a be a set well-ordered by ⊳ and let f : a ¹ a be order-preserving. Then x E f(x) for each x ∈ a. (ii) There can be no order-preserving injective function from a well ordered set to an initial segment of it.

Proof (i) Assume that the statement does not hold and let c = {x ∈ a | f(x) ⊳ x}. Note that c is non-empty since for some x ∈ a, x E f(x) does not hold, and ⊳ is total. Let x0 be the least element of c. We have f(x0) ⊳ x0 so since f is injective and order preserving also f(f(x0)) ⊳ f(x0) but that means f(x0) ∈ c and f(x0) ⊳ x0, contradicting the choice of x0 as the least element in c. (ii) Let a be well ordered by ⊳, x ∈ a. If f was an order preserving injective function from a to ax then f(x) ⊳ x which contradicts (i). ¥

Corollary 5 The isomorphism in Theorem 8 is unique.

Proof See Example 6.4.

Ordinals

Well-ordered sets generalize the notion of counting as represented by natural numbers, when counting is understood as arranging elements in order, starting from a ﬁrst one and proceeding in discrete steps. Assuming again we have N for the moment, note that for any ﬁnite set there is (up to isomorphism) only one way to count it: if n ∈ N and a set a has n elements and ha, ⊳i is well ordered then ha, ⊳i is isomorphic to hNn

49 of elements. However, we have seen that on N there can be essentially different well-orderings (cf. Example 5.4(ii)). Non-isomorphic well-orderings correspond to essentially different outcomes we may get when counting sets. Isomorphic well orderings share an important feature, similarly to equinumerous sets having the same cardinality. We say that isomorphic well ordered sets have the same order type. We could proceed similarly as before and talk about an order type being that which is shared by isomorphic well-ordered sets. However, we are now in a position to say exactly what we mean, and define a unique set representing each such class of isomorphic well orderings. When we have done it, we will proceed to use it to define what we mean by cardinal numbers, too.

Deﬁnition 20 Let a be a set. Then ∈a is the relation on a deﬁned by

∈a= {h x, y i | x, y ∈ a & x ∈ y} .

∈a is irreﬂexive on a by virtue of the Axiom of Foundation, but in general it is not transitive nor total. We shall be interested in special sets which are totally ordered by ∈a. First we point out some properties of sets a totally ordered by ∈a.

Lemma 9 Let a be a set such that ∈a totally orders a, and let b ⊆ a and w ∈ b. Then w is the least element of b (with respect to ∈a) if and only if w ∩ b = ∅.

Proof w is the least element of b just when ( ∀y ∈ b) ( y = w ∨ w ∈a y) which, by virtue of ∈a being total, is equivalent to any of the following

(∀y ∈ b) ¬y ∈a w ⇐⇒ (∀y ∈ b) ¬y ∈ w ⇐⇒ w ∩ b = ∅.

Corollary 6 A set a is totally ordered by ∈a if and only if it is well ordered by it.

Proof This follows by the above Lemma and the Axiom of Foundation. ¥

Corollary 7 A set a is totally ordered by ∈a if and only if

For all x, y ∈ a, exactly one of x = y, x ∈ y, y ∈ x holds . (OT )

Proof By the previous Corollary and Proposition 42 (since ( OT ) implies the existence of a least elements for any non-empty set subset of a, as in the proof of Lemma 9).

Deﬁnition 21 A set a is transitive if it contains all elements of its elements, that is, (∀x ∈ a) x ⊆ a.

50 Warning The terminology is established but potentially confusing: to say that a relation is transitive is not the same thing as saying that a set is transitive (even though any relation is also a set!). It is always clear from the context what is meant.

Deﬁnition 22 A set a is an ordinal if it is transitive and totally ordered by

∈a= {h x, y i | x, y ∈ a & x ∈ y}

Proposition 46 If a is an ordinal then a is well ordered by ∈a.

Proof From Corollary 6. ¥

We remark that when Axiom of Foundation is not assumed, an ordinal is usually deﬁned as a transitive set a which is well ordered by ∈a. As Proposition 46 shows, in ZF this is equivalent to our deﬁnition.

Proposition 47 A set a is an ordinal if and only if it is transitive and

For all x, y ∈ a, exactly one of x = y, x ∈ y, y ∈ x holds . (OT )

Proof From Corollary 7. ¥

For example, ∅, {∅} , {∅ , {∅}} , {∅ , {∅} , {∅ , {∅}}} are ordinals.

We will use lower case Greek letters α,β,γ,δ,... for ordinals. Let Ord stand for the class of all ordinals. We will write Ord (x) to mean x is an ordinal. We will need the following properties of ordinals. Recall that x ⊂ y stands for x ⊆ y and x 6= y.

Proposition 48 Let α, β be ordinals, d a set. (i) If d ∈ α then d ⊂ α. (ii) If d ∈ α then d ∈ Ord . (iii) β ∈ α ⇐⇒ β ⊂ α. (iv) Exactly one of α ∈ β, α = β, β ∈ α holds.

Proof (i) By transitivity of α, if d ∈ α then d ⊆ α. Equality would mean d ∈ d, which cannot be by the Axiom of Foundation.

(ii) Suppose that d ∈ α, and y ∈ d and z ∈ y. By transitivity of α we know that z, y, d ∈ α, so by transitivity of ∈α, z ∈ d and hence d is a transitive set. Suppose that z, y ∈ d. By transitivity of α we also have z, y ∈ α and consequently, ( OT ) holds for d because it holds for α. The result follows by Proposition 47.

51 (iii) The implication = ⇒ follows from (i). Suppose β ⊂ α. Then α\β 6= ∅ so let δ ∈ α\β be such that δ ∩ (α\β) = ∅. (Note that δ is an ordinal by (ii).) We shall show that β = δ (and hence since δ ∈ α, also β ∈ α.) If λ ∈ δ then λ ∈ α (by transitivity of α). Also, λ∈ / α\β because δ ∩ (α\β) = ∅, so we must have λ ∈ β. If λ ∈ β then since β ⊂ α, we have λ ∈ α. Hence by ( OT ) for α, exactly one of λ ∈ δ, λ = δ, δ ∈ λ holds. It cannot be λ = δ since λ ∈ β ⊂ α so λ∈ / (α\β) but δ ∈ (α\β) and it cannot be δ ∈ λ since that would mean δ ∈ β by transitivity of β but δ ∈ (α\β). Hence λ ∈ δ, as required.

(iv) It is impossible to have two of them holding by the Axiom of Foundation. Hence it suﬃces to prove that it cannot be the case that none of them holds. Assume α∈ / β and α 6= β. Then by (iii) α * β so α\β is a non-empty subset of α. Let δ ∈ α\β be such that δ ∩ (α\β) = ∅. If γ ∈ δ then γ ∈ α by transitivity of α, so γ ∈ β since δ ∩ (α\β) = ∅. Consequently, δ ⊆ β and hence by part (iii), δ = β or δ ∈ β. The latter is impossible since δ ∈ α\β, so δ = β and hence β ∈ α, as required. ¥ Theorem 9 The class of all ordinals Ord is a proper class. Proof Assume that Ord is a set. Then Ord is an ordinal by Propositions 47 and 48(ii),(iv) so Ord ∈ Ord which cannot be by the Axiom of Foundation. ¥

Like for a set, deﬁne ∈Ord to be {h β, α i | α, β ∈ Ord & β ∈ α}. ∈Ord is a proper class (exercise). For α ∈ Ord let Ord α = {β | β ∈ Ord & β ∈Ord α}. By Proposition 48 (ii), Ord α = {x | x ∈ α} = α, that is, the predecessors of α in Ord are just the elements of α. If γ ∈ Ord then for α, β ∈ γ we have α, β ∈ Ord and

β ∈γ α ⇐⇒ β ∈ α ⇐⇒ β ∈Ord α. It follows that α = Ord α = γα.

∈Ord is usually replaced by <. So for α, β ∈ Ord ,

α ∈Ord β, α ∈ β, α<β all stand for the same thing. As usual, α ≤ β stands for α < β ∨ α = β.

Proposition 49 (i) Let α be an ordinal. Then α ∪ { α} is an ordinal, denoted α + 1 . (ii) Let b be a set of ordinals. Then b is an ordinal. (iii) Let b be a non-empty set of ordinals.S Then b is an ordinal and it is equal to the least element of b. T

52 Proof Example 6.5. ¥

Theorem 10 Let b be a set well ordered by ⊳. Then there exists a unique ordinal α such that hb, ⊳i is isomorphic to hα, ∈αi.

Proof Let f be the unique function with domain b guaranteed by Theorem 7 satisfying

f(x) = Rng (f ↾bx) for all x ∈ b.

First note that f is a bijection between b and Rng (f): it is obviously surjective. If x, y ∈ b, x ⊳ y then x ∈ by so f(x) ∈ Rng (f ↾ by) = f(y) and hence f is injective (since x 6= y means x ⊳ y or y ⊳ x so f(x) ∈ f(y) or f(y) ∈ f(x), either of which incompatible with f(x) = f(y) by virtue of the Axiom of Foundation). Also, as above, for x, y ∈ b, x ⊳ y =⇒ f(x) ∈ f(y). Conversely, if f(x) ∈ f(y) then since f(y) = Rng (f ↾ by) = {f(z)| z ∈ b & z ⊳ y}, and f is injective, we have x ⊳ y. Hence f is an isomorphism between hb, ⊳i and hRng (f), ∈Rng (f)i. Consequently, Rng (f) satisﬁes ( OT ) because ⊳ is total. It is clearly a transitive set so by Proposition 47 it is an ordinal. The uniqueness of α follows from the fact that there can be no isomorphism between two distinct ordinals (given two distinct ordinals, one must be an initial segment of the other and by Proposition 45(ii) no well-ordered set is isomorphic to an initial segment of itself). If hb, ⊳i was isomorphic to ordinals α and β then α and β would be isomorphic. ¥

Deﬁnition 23 If hb, ⊳i is a well ordered set then the unique ordinal α isomorphic to hb, ⊳i is called the order type of hb, ⊳i.

Proposition 50 If b is a set, γ an ordinal and b ¹ γ then b can be well ordered (so that the order type of this ordering is less or equal to γ).

Proof We can obtain a well ordering ⊳ for b from f : b ¹ γ as in the Proposition 44. Note that f is then order preserving. Let α be the order type of hb, ⊳i and let g be an isomorphism between hα, < i and hb, ⊳i . Then we have α ≤ γ, since if not then α > γ , so γ = αγ and f ◦ g is an order preserving injection from α to an initial segment of itself which cannot be by Lemma 45(ii).

Theorem 11 (Hartogs’ Theorem) Given any set b there is an ordinal α such that not α ¹ b.

Proof First notice that

d = { h x, ⊳i | x ⊆ b & ⊳ is a well ordering of x }

53 is a set (please check it as an exercise, using the Separation and other Axioms), and by Theorem 10 for every hx, ⊳i ∈ d there is a unique ordinal α isomorphic to it. By the Replacement Axiom, the collection of such ordinals is a set, call it c. Now assume that for every ordinal α we have α ¹ b, so for every ordinal α there is a subset x of b such that x ∼ α. As in Proposition 44, we can deﬁne a well ordering ⊳ on x so that hx, ⊳i is isomorphic to α. But that means that every ordinal is in c, which contradicts Theorem 9. ¥

As a consequence, we get the promised result about the Law of Trichotomy implying the Well Ordering Theorem: Assume LOT holds. Let b be a set and α is an ordinal such that α ¹ b does not hold, then b ¹ α and hence b can be well ordered by Proposition 50.

Deﬁnition 24 Let α be an ordinal, α 6= ∅. If there is an ordinal β such that α = β ∪ { β} then we say that α is a successor ordinal and we write α = β + 1 . Otherwise α is a limit ordinal .

Note that if α = β ∪ { β} then β ∈ α, i.e β < α and there is no γ such that β < γ < α (otherwise β ∈ γ and γ ∈ β ∪{ β} which by β 6= γ gives β ∈ γ, γ ∈ β, contradicting the Axiom of Foundation). It follows that β + 1 is the immediate successor of β.

Proposition 51 If γ, β are ordinals such that γ + 1 = β + 1 then γ = β.

Proof Assume γ + 1 = β + 1 and γ 6= β. Then γ < β or β < γ so assume without loss of generality that γ < β . It follows that γ < β < β + 1 = γ + 1 which contradicts γ + 1 being the immediate successor of γ. ¥ If α = β ∪ { β} then β is denoted α − 1 and referred to as the immediate predecessor of α. Each successor ordinal has an immediate predecessor but limit ordinals do not.

Proposition 52 If α is a limit ordinal then α = α = {β | β < α }. S S Proof By Proposition 49(ii), γ = {β | β < α } is an ordinal, and since α is transitive, γ ⊆ α If δ ∈ α then Sδ ∪ { δ} ∈ α since α is not a successor so δ ∈ α = γ, that is α ⊆ γ. It follows that γ = α as required. ¥ S We have seen examples of successor ordinals ( ∅, {∅} , {∅ , {∅}} ) {∅ , {∅} , {∅ , {∅}}} . . . ). Now we will use the Axiom of Inﬁnity to prove that there is a limit ordinal.

Deﬁnition 25 A set x is called an inductive set if ∅ ∈ x and

∀y (y ∈ x =⇒ y ∪ { y} ∈ x).

54 Let ω = {x | x is an inductive set } = {z | ∀ x (x is an inductive set = ⇒ z ∈ x)}. ThenT ω is a set by the Separation Axiom since if a is an inductive set (which is guaranteed to exist by the Axiom of inﬁnity) then

ω = {z ∈ a | ∀ x (x is an inductive set = ⇒ z ∈ x)}.

Clearly, ω is an inductive set.

Proposition 53 ω is an ordinal.

Proof Since ∅ is an ordinal, we can see from Proposition 49(i) that ω ∩ Ord is an inductive set, so since ω is the intersection of all inductive sets, we must have ω ⊆ Ord . Hence by Proposition 48(iv) ω satisﬁes ( OT ). The set c = {α ∈ ω | α ⊆ ω} is an inductive set since clearly ∅ ∈ c and if α ∈ c then α ⊆ ω, α ∈ ω so α ∪ { α} ⊆ ω and α ∪ { α} ∈ ω (as ω is inductive) and hence α ∪ { α} ∈ c. It follows that c = ω, so ω is transitive. ¥

Note that ω is not a successor ordinal since if it was equal to γ ∪ { γ} for some γ then γ would be an element of ω so since ω is inductive, we would have ω = γ ∪ { γ} ∈ ω. Every element of ω other than ∅ is a successor ordinal, since otherwise the predecessors of a limit ordinal α ∈ ω would form an inductive set strictly included in ω, which cannot be. We will now postpone further study of larger ordinals and concentrate on ω showing that it provides the promised representation of N in ZF and using it to obtain Z, Q and R, too.