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 <i byTheorem 8, and because finite sets can only be equinumerous if they have the same number 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.