(D) Introduction to order types and ordinals

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1 (D) Introduction to order types and ordinals Linear orders are one of the mathematical tools that are used all over the place. Well-ordered sets are a special kind of linear order. At first sight well-orders seem to be a mere curiosity. However, they are an important measuring device in many parts of mathematics. Contents 1 Basic material Exercises: basic material Problems: basic material Equivalence and order types Exercises: Equivalence and order types Problems: Equivalence and order types Addition of order types and ordinals Exercises: Addition of order types and ordinals Problems: Addition of order types ordinals Multiplication of order types and ordinals Exercises: Multiplication of order types and ordinals Problems: Multiplication of order types and ordinals Comparison of ordinals Basic material You will have met the following notions before, but it will do no harm to introduce them in a precise way. 1.1 DEFINITION. A linear set (linearly ordered set) is a set A with a given linear comparison. In other words A carries a binary relation such that (Reflexive) a a (Transitive) a b c = a c (Antisymmetric) a b & b a = a = b (Linear) a b or b a for all a, b, c A. We sometimes use the strict version < of, that is a < b a b and a b for a, b A. There are some standard examples of linear sets. For instance N Z Q R 1

2 are such linear sets with their standard comparisons. However, the way we usually look at C it is not a linear set. Of course, this does not mean that C can not be linearly ordered in some way or other. Strictly speaking a linear set is a pair (A, ), a set A with the carried comparison. The standard convention when dealing with algebraic structures such linear sets, groups, rings, fields,..., is to refer to the structure by its carrier. Thus most of the time we say the linear set A when we mean the linear set (A, ). There are times when we should be more precise, and use a notation to distinguish between A and (A, ). Let s look at an example of that. 1.2 DEFINITION. Let A = (A, ) be a linear sets. We let A = (A, ) where is the reverse of, that is the relation on A given by a b b a for all a, b A. We call A the reverse of A. It doesn t take long to show that the reverse of a linear set is also a linear set. Let s look at some examples of that. 1.3 EXAMPLES. Consider the standard linear sets N Z Q R of various kinds of numbers. Each can be reversed N Z Q R and then we find that N is a copy of the negative integers, but Z = Z Q = Q R = R hold. Most of the time we will not use the notation A to stand for (A, ). Normally we will simply write A and assume that the carried comparison is understood. However, there are times when it is useful to name the comparison explicitly. In the Example above the symbol = indicates that that the two linear sets are isomorphic. We make this notion precise in the next section. Before that we look at a special kind of linear set. When first seen the following notion might look a bit odd, but once we start to formalize the notion of counting item by item we see it is just what we want. 1.4 DEFINITION. A linear set A is well-ordered if each non-empty subset X A has a least, that is first, member of X. As always, a few examples will make this notion clearer. 2

3 1.5 EXAMPLES. (a) The empty set, viewed as a linear set, is well-ordered. This is because satisfies the required condition vacuously. There are no non-empty subsets X. (b) Each finite linear set A is well-ordered. Start counting A from the beginning and at some stage you will meet a member of the non-empty subset X A. (c) The naturals N, viewed as a linear set, is well-ordered. The same counting trick that works for a finite linear set also works for N. (d) Neither the rationals not the reals are well-ordered. For instance the rational interval (0, 1) Q has no first element in either linear set. (e) Suppose A is any well-ordered set. We may form a new well-ordered set A + = A { } by adjoining a new element at the top end, right hand end. A few moments thought shows how to verify this. We leave it as an exercise. 1.1 Exercises: basic material 1.1 Show that the reverse of a linear set is also a linear set. 1.2 Show that if A is well-ordered then so is A Consider a linear set A where both A and its reverse A are well ordered. Show that A is finite. 1.2 Problems: basic material 1.4 Consider the following algorithm we may apply to any well-ordered set A. It helps us to numerate the elements of A and give A a name. In short, we start to generate the ordinals. If A = then it has no elements, so we name A as 0. Otherwise A and so has a least element which we call a(0). If A = {a(0)} then we name A as 1. Otherwise A {a(0)} is non-empty, and so has a least element, which we call a(1). Thus we have A = {a(0) < a(1) } and we don t yet know what goes on above a(1). Let A(2) = {a(0) < a(1)} the set of elements we have already enumerated. If A(2) is the whole of A then we name A as 2. Otherwise A A(2) is non-empty, and so has a least element, which we call a(2). We let A(3) = {a(0) < a(1) < a(2)} 3

4 the initial part we have already enumerated. If A(3) is the whole of A then we name A as 3. We can continue this idea through the natural numbers. At some stage we may find that A = {a(0) < a(1) < < a(n)} = A(n + 1) in which case we have enumerated the whole of A and then we name A as n + 1. Otherwise we have A(ω) = {a(0) < a(1) < < a(n) < n N} as an initial section of A. This may be the whole of A, or not. If A = A(ω) then we name A as ω. If A A(ω) is non-empty then this set has a least element, which we call a(ω). We let A(ω + 1) = {a(0) < a(1) < < a(n) < < a(ω)} be this initial part of A. If A(ω + 1) is the whole of A then we name A as ω + 1. Otherwise,..., and so on. (a) You problem is to sort out as many new names as possible. 2 Equivalence and order types For much of what we do here we are not concerned with a particular linear order, only its order type. To make that more precise we need an auxiliary notion. 2.1 DEFINITION. Let be a pair of linear sets. An isomorphism (A, A ) (B, B ) (A, A ) f (B, B ) from (A, A ) to (B, B ) is a bijection f : A B such that for all a 1, a 2 A. a 1 A a 2 f(a 1 ) B f(a 2 ) You can see here why it is often convenient to refer to a linear set (A, ) by its carrier. If we always named the carried comparison then the notation would get a bit cluttered. Sometimes we are concerned with the nature of a particular isomorphism. At other times all we want is that two linear sets A, B are isomorphic and we don t need to know a particular isomorphism. In those cases we use the fairly standard notation and write A = B to indicate that there is an isomorphism, that is an isomorphism does exists. When we are not concerned with what happens inside a linear set we can replace it by any isomorphic copy. In fact, we invent a name, notation, for all of the linear sets in a particular isomorphism class. 4

5 2.2 DEFINITION. The order type ι(a) of a linear set A is a representation of the isomorphism class to which it belongs. Thus the order type of A does not distinguish between A and all of its isomorphic mates. Thus ι(a) = ι(b) A = B for linear sets A, B. By convention we use small Greek letters α, β, γ,... as order types. This is not quite a definition for it doesn t say how such a representative ι(a) should be set up. This can be done, but it takes quite a bit of time, and once it is done we forget the construction and work informally as we do here. So we don t need to worry about those details just yet. You can save that problem for later, to help you not do the washing up. Let s look at some examples, with some hints as how the formal construction works. 2.3 EXAMPLES. (a) Each finite linear set looks like a finite initial part of the natural numbers. That is we can start at its beginning and count it. Thus we use 0, 1, 2, 3,... as the finite order types. Notice that we do view the empty set as a linear set. More formally, we may set 0 = 1 = {0} 2 = {0, 1} (n + 1) = {0, 1,..., n} to obtain the von Neumann natural numbers. As an aside Zermelo, one of the founders of Axiomatic Set Theory, suggested that we could formally defined the natural numbers as { } {{ }} {{{ }}}... and so on. These are like a present from Ebeneezer. You strip off the wrapping paper until you are left with [insert your own phrase]. (b) We use ω η ζ N Q R for each of the indicated infinite linear sets. The notation ω for the order type of N and η for the order type of Q was introduced by Cantor. The notation ω is now standard, and η is fairly standard. There does not seem to be a standard notation for the order type of R. Here I will use ζ, as indicated. As I said, the notation ω was chosen by Cantor, and it has stuck ever since. I don t know why he chose this letter. It may be something to do with the phrase From alpha to omega meaning From beginning to end. Then again, it might just have been the name of his cat. I also don t know why he choose η. Perhaps the explanation is food for thought. He was pleased with the notation, and a happy η user. However, that explanation doesn t work in German. 5

6 It isn t hard to see that if two linear sets are isomorphic then their reverses are isomorphic. Thus we have an operation on order types. 2.4 DEFINITION. For each order type α we let α be the order type of the reverse of α. That is if α is the order type of A then α is the order type of A. Of course, some example will make this a bit clearer. 2.5 EXAMPLES. Viewing each natural number n as an order type we have n = n, for if we count a finite set backwards then we get the same answer. Since ω is the order type of N, we see that ω is the order type of the negative integers, and hence ω ω. We have η = η and ζ = ζ, for if we reverse Q or R then we get the same linear set (up to an isomorphism). There is another 1-placed operation on order types, and this is more important. Recall that in Example 1.5(e) we introduced the construction A A + of linear orders which simple adds an new end point at the top, right hand end. We can now make this a bit more formal. 2.6 DEFINITION. For each linear set A we let A + = A {A} to produce a new linear order with an new end point. For each order type α we let α + = ι(a + ) where A is any linear order with ι(a) = α. More formally, once we have produced α as a particular set we let α + = α {α} to produce the successor of α. To help to explain this idea let s have another look at some of our earlier examples. 2.7 EXAMPLES. (a) Consider the way we generated the von Neumann natural numbers in Example 2.3(a). This can be condensed to 0 = n + 1 = n + for each natural number n. Yes, I know there is a bit of confusion here about the external natural numbers which are assumed to exists already, and the formal versions of these. This construction is the way natural numbers are produced in Axiomatic Set Theory. (b) Observe that the formal construction gives n = {0 < 1 < < n 1} with n + 1 = n + 6

7 for each natural number. This produces an ascending chain n n + 1 and then we may define ω = {n n a vn natural} to obtain an order type just beyond each n. (c) We then let ω + 1 = ω {ω} = ω + ω + 2 = ω + {ω + } and so on. This produces another ascending chain ω ω + 1 ω + 2 ω + n and then we may let ω 2 = ω + ω = {ω + n n < ω} to produce a bigger order type. (d) We can keep on repeating this trick to produce ω ω 2 ω 3 ω n and then we may let ω 2 = ω ω = {ω n n < ω} to produce an even bigger order type. (e) Where does this end? There is something going on here and it is not entirely clear what that is. Don t worry about that. Even Cantor was a bit mystified when he first invented all this stuff. In this document we try to sort out some of the hidden secrets. Have another look at the linear orders generated in Examples 2.7. Each one is a wellordering. In due course we will concentrate on these, and the associated order types have a special name. 2.8 DEFINITION. An ordinal is the order type of a well-ordered set. We know that each finite linear order is a well-order, and so the natural numbers are the finite ordinals. After that ω is the next ordinal, and then Examples 2.7 begins to generate the ordinals that follow. 2.1 Exercises: Equivalence and order types 2.1 Show that the underlying function of each isomorphism between two linear sets A, B is also an isomorphisms between the reverse linear sets A, B. 2.2 Show that if both A and A are well-ordered then A is finite. 7

8 2.3 Let A f g B be isomorphism between a pair of well-ordered sets. Show that f = g. 2.4 Write out 0 = 1 = {0} 2 = {0, 1} (n + 1) = {0, 1,..., n} entirely in terms of and nested uses of { and }. Once you have reach 42 you can get yourself a drink. 2.5 Let A be an arbitrary linear set. A lower section is a subset X A such that holds. For each a A we let y x X = y X a = {x A x a} to obtain the principal lower section generated by a. Let LA be the family of all lower sections of A. (a) Show that LA is linearly ordered by inclusion. (b) Show that if A is well ordered then so is LA. (c) For the well-ordered case, produce another well-ordered set that is isomorphic to LA in a canonical fashion. 2.2 Problems: Equivalence and order types 2.6 This problem shows how the set of natural numbers are produced in Axiomatic Set Theory. For an arbitrary set a we let a + = a {a} to produce another set. We say a set I is inductive if I and a I = a + I for each set I. (This terminology is not very good, but it seems to be the standard one.) (a) Starting with a = write down what happen when we start to apply the operation ( ) + on sets. (b) Show that if I is a non-empty family of inductive sets then I is also inductive. (c) Show there is an inductive set that is included in every inductive set. (d) Can you describe the inductive set of part (c). (e) Can you suggest a non obvious axiom that is needed in Axiomatic Set Theory. 8

9 2.7 Let L, M be two arbitrary non-empty well-ordered sets. For a function f : L M we define the support spt(f) of f to be that subset of L given by x spt(f) f(x) 0 M where 0 M is the least (first) element of M. Let [L, fs, M] be the set of all function f : L M of finite support. One member of [L, fs, M] is the function zero, that function with zero(x) = 0 M for every x L. The support of this function is empty. Consider the comparison on [L, fs, M] given by f g f = g or ( a L) [ f(a) < g(a) & ( x L)[a < x = f(x) = g(x)] ] for f, g [L, fs, M]. It turns out that this comparison well-orders [L, fs, M]. The aim of this problem is to investigate some of the properties of is well-ordered set. (a) Show that the comparison linearly orders [L, fs, M]. (b) Show that zero is the first member of this linear set. (c) Show that each member f of [L, fs, M] can be described as 0 if a 0 < x p 0 if x = a 0 0 if a 1 < x < a 0 p 1 if x = a 1 f(x) = 0 if a 2 < x < a 1 for some descending chain p m if x = a m 0 if x < a m a m < < a 2 < a 1 < a 0 from L and some list p 0, p 1, p 2,..., p m from M. (d) Consider two members f, g of [L, fs, M] with f given as above and g given by the chain b n < < b 1 < b 0 and the list q 0,..., q n. Show how the comparison between f, g can be tested. (e) Describe an informal algorithm that shows that [L, fs, M] is well-ordered. 3 Addition of order types and ordinals Given two linear orders we can form a new one by sticking one after the other. To do that we have to be careful. If the carrying sets overlap then we might confuse the new linear order, or ourselves. To avoid this we take isomorphic copies of the given linear sets where the new copies don t overlap, and then stick these together. There is a neat way of doing this. We tag the original elements so we know where the came from. 9.

10 3.1 DEFINITION. Let A, B be a pair of sets. The disjoint union of the pair is the set B A = (B {0}) (A {1}) where 0, 1 are a pair of distinct tags. In other words the elements of B A are the pairs (x, i) where { x A if i = 1 x B if i = 0 and each x A B does appear twice but with two different tags. Of course, it is not important which tags we use, but here it is useful that 0 < 1. You may wonder why we write B A and not A B which perhaps seem more natural. The reason is an important aspect that occurs later, and was introduced by Cantor. We do it here to get used to the idea. 3.2 DEFINITION. Given two linear sets (B, 0 ) (A, 1 ) the sum is given by where (B, 0 ) + (A, 1 ) = (B A, ) (x, i) (y, j) i < j or i = j and x i y for x, y A B and tags i, j {0, 1}. Usually we write B + A for (B, 0 ) + (A, 1 ) by hiding the carried comparisons. Notice how the tags are used. Each pair with tag 1 goes to the right hand end of the sum, and each pair with tag 0 goes to the left hand end of the sum. Thus we could use a different pair of tags, such as right and left, but the pair 0, 1 are fairly standard. However, there are situations where different tags make life a bit easier. Of course, there is something to check here. The proof of the following is an easy exercise. 3.3 LEMMA. (a) The sum of two linear sets is a linear set. (b) The sum of two well-ordered sets is well-ordered. (c) Given four linear sets A, A, B, B the implication holds. A = A and B = B = B + A = B + A 10

11 You should note part (b) of this result. This will become important later. This result enables us to transfer the addition of linear sets to order types. 3.4 DEFINITION. For each pair α, β of order types the sum β + α is the order type of the sum B + A where A and B are any linear sets of types α and β, respectively. In other words this definition says ι(b) + ι(a) = ι(b + A) and this is justified by Lemma 2.3(b). Note that, in particular, the sum of two ordinals is an ordinal. Let s look at some examples of this sum. There are one or two things you might not have spotted 3.5 EXAMPLES. (a) The sum m+n of any two finite order types m, n is what you expect it to be, the sum as natural numbers. (b) The order type 0 is the type of the empty linear set. Thus 0 + α = α = α + 0 for each order type α. (c) The order type 1 is the type of a singleton viewed as linear set. Thus and the three order types are different. For each order type α the two types 1 + ω = ω ω + 1 ω 1 + η η η α α + 1 are formed by sticking a new point at the beginning or at the end. This can make a difference, and sometimes it makes no difference. (d) The two order types ω + ω ω + ω are very different. The left hand one is the order type of Z. The right hand one has a huge chasm in the middle. (e) We have η + η = η ζ + ζ ζ which is perhaps a bit of a surprise. The left hand one can be seen by splitting the rationals in two at some irrational. The right hand one can be seen by removing the number 0 from R. 11

12 The examples (c, d) shows that, in general, addition of order types is not commutative. Other aspects are not too bad. The proof of the following is left as an exercise 3.6 THEOREM. (a) Addition of order types is associative, that is for order types α, β, γ. (b) We have for each pair of order types α, β. (γ + β) + α = γ + (β + α) (β + α) = α + β And we haven t even started multiplying yet. 3.1 Exercises: Addition of order types and ordinals 3.1 Sort out the proofs missing from this section, namely the proofs of Lemma 3.3 and Theorem 3.6. For part (a) of the Theorem you will need two layers of tags, or something better. 3.2 (a) What are the order types of the following subsets of R. (i) {1/(n + 1) n N} (ii) {2 n n N} (iii) {m/2 n m, n N} (iv) {e x x Q} (b) Can you find any strange subsets of R with order types you can t yet name but which ought to have a simple name. 3.3 Let η, ζ, ξ be the order types of Q, R, and the irrationals I. Using these describe the possible order types of all intervals of Q, R, I. 3.4 Using η, ζ, ξ as in Exercise 2.3, which of the following are true and which are false. In each case give an informal justification of your answer. (i) η + η = η (ii) ζ + ζ = ζ (iii) ξ + ξ = ξ (iv) η η = η (v) ζ ζ = ζ (vi) ξ ξ = ξ 3.2 Problems: Addition of order types ordinals 3.5 Show that for each order type α, we have α = α if and only if for some order type β. α = β + β or α = β β 3.6 Find six (not necessarily different) order types α α, β β, γ, δ such that the following hold. (i) α = α + γ (ii) β = γ + β (iii) α + β = α + β (iv) α = α + γ (v) β = γ + β (vi) β + γ = γ + α 3.7 Let A be the set of all functions N 2 which are eventually zero. Show that each of the following linearly orders A, and determine the order type of each. (i) f g f = g or ( m)( n < m) [ f(m) < g(m) & f(n) = g(n) ] (ii) f g f = g or ( m)( n > m) [ f(m) < g(m) & f(n) = g(n) ] 12

13 4 Multiplication of order types and ordinals In some ways the addition of order types is obvious and easy to understand. We now come to the multiplication of order types, and the multiplication of ordinals in particular. At first this can look at bit odd, but once we get into it the idea is not too bad. As with addition, we first set up the product of two linear sets, and this is carried by the cartesian produce of the two carrying sets. However, this product is not the categorical product. 4.1 DEFINITION. Given two linear sets the product, as linear sets, is given by (A, A ) (B, B ) (B, B ) (A, A ) = (B A, ) where the comparison on B A is given by a 1 < A a 2 (b 1, a 1 ) (b 2, a 2 ) or a 1 = a 2 and b 1 B b 2 for a 1, a 2 A and b 1, b 2 B. You might think I am being a bit awkward here because I am using the alphabet backwards. I am using the alphabet backwards, but I am not being awkward. There are good reasons for setting up the product in this way. One is historical, and I will tell you that story later (and this does have something to do with a right to left alphabet), and the other is that once we get into it we find that this way is a bit more natural. As with addition we often hide the carried comparisons and write B A for (B, B ) (A, A ) = (B A, ) the formal product. This won t be a problem once you get used to it. To view B A (the product linear set) first take a copy of A and think of this as a washing line. Here I have drawn the linear set as though it may have gaps. That is merely to help me draw the next picture. Now take many disjoint copies of B. Let me use as a typical example of such a linear order. Now dangle from each point of A a copy of B. Think of this as lots of socks hanging from the line

14 These dangling disjoint copies of form a linear sets B A. Within each copy of B the elements are ordered as in the copy. Two elements in different copies are ordered by the two elements in A from which these dangle. (Of course, for the pictorial representation you must assume that there is no more than a light breeze, and the washing line is not one of those new fangled things that goes round in a triangle.) Of course, there is something to check here. The following is the analogue of Lemma 3.3, and the proof of the following is an easy exercise. 4.2 LEMMA. (a) The product of two linear sets is a linear set. (b) The product of two well-ordered sets is well-ordered. (c) Given four linear sets A, A, B, B the implication holds. A = A and B = B = B A = B A With this we can define the product of two linear types. 4.3 DEFINITION. For each pair α, β of order types the product β α is the order type of the product B A where A and B are any linear sets of types α and β, respectively. In other words this definition says ι(b) ι(a) = ι(b A) and this is justified by Lemma 4.2(b). Let s take a look at some examples. Some of these assertions need proofs, and these are left as an exercise. Notice that the product of two ordinals is an ordinal, and some instances of this occur in the following. 4.4 EXAMPLES. (a) The product n m of any two finite ordinals m, n is what you expect it is, the product as natural numbers. (b) The ordinal 0 is the type of the empty linear set. Thus 0 α = 0 = α 0 for each order type α. (c) The ordinal 1 is the type of a singleton viewed as linear set. Thus 1 α = α = α 1 for each order type α. These two equalities hold for slightly different reasons which you should sort out. (d) The ordinal 2 can be thought of as 0 < 1. (In fact, formally it is {0 < 1}). Then α 2 = α + α 14

15 but 2 α can be very different. We have 2 ω = ω but 2 η is nothing like η, since it is full of gaps. (e) For each countable non-zero order type α we have by the characterization of Q. (f) We have η α = η (ζ + 1) ω = ζ for consider the order type of the real interval (n, n + 1] for each n N. However, we have ζ ω ζ by thinking of a similar picture. From these and similar examples we see that, in general, multiplication of order types is not commutative. Other aspects are not too bad. The proof of the following is left as an exercise 4.5 THEOREM. (a) Multiplication of order types is associative, that is for order types α, β, γ. (b) We have for each pair of order types α, β. (γ β) α = γ (β α) (β α) = β α We now have two binary operations on order types, addition and multiplication. With these we can start to form polynomial compounds. We find that the two compounds (γ + β) α γ α + β α need not be equal. For instance α = ω and β = γ = 1 gives an example of this difference. However, we do have the following. 4.6 THEOREM. We have for all order types α, β, γ. γ (β + α) = γ β + γ α Informal proof. Think of the construction of both sides in terms of the washing line idea. On the left we have a washing line which is divided into a left part β and a right part α. We dangle from each point a copy of γ. These dangling bits split into a left part, below β, and a right part, below α. That split is just the right hand side. We now have a new form of arithmetic, that of order types and ordinals in particular. This extends the arithmetic of natural numbers. This new arithmetic looks more complicated, for the two operations are not commutative and we have only one distributive law. However, in some ways it is easier, for in a calculation many components can disappear. 15

16 4.7 DEFINITION. For each ordinal β we set β 0 = 1 β a+1 = β a β for each finite ordinal a. For instance β 0 = 1 β 1 = β 0 β = 1 β = β β 2 = β 1 β = β β β 3 = β 2 β = β β β and so on. Notice how we have used the associative law in the last calculation. Here is one of those disappearing acts. 4.8 LEMMA. We have for all natural numbers a < b. ω a + ω b = ω b The proof of this is a bit finicky, and you should worry about it for a while, As a clue, notice that if b = a + n where n 0, then ω a + ω b = ω a + ω (a+n) = ω a 1 + ω a ω n = ω a (1 + ω n ) so it suffices to attack the compound in brackets. With this result and some informal arguments we can carry out other calculations. 4.9 EXAMPLE. We have (ω b + ω a ) ω = ω b+1 for all natural numbers a < b. To see this we think of the washing line picture. We have (ω b + ω a ) ω = (ω b + ω a ) + (ω b + ω a ) + (ω b + ω a ) + = ω b + (ω a + ω b ) + (ω a + ω b ) + (ω a + = ω b + ω b + ω b + ω b + ω b + = ω b ω = ω b+1 where at each step we re-organize the washing line, and at the third step we use Lemma 4.8. There are several other calculations of this kind in the exercises. 4.1 Exercises: Multiplication of order types and ordinals 4.1 Sort out the proofs missing from this section, namely the proofs of 4.2, 4.5, 4.6, and Which of the following are true, and which are false. In each case give an informal justification of your answer. (i) ω η = η (ii) ω ζ = ζ (iii) ω ξ = ξ (iv) η ω = η (v) ζ ω = ζ (vi) ξ ω = ξ Remember that ζ is the order type of R and ξ is the order type of I. 16

17 4.3 Compute the possible values of where α is any countable order type. 4.4 Show that are distinct. η α (1 + η) α (η + 1) α ζ n (n < ω) ζ ω ζ ω ζ η 4.5 Find an order type α such that α ω α but α (n + 1) = α for each n < ω. 4.6 Let θ = ω + ω and α = ω + θ η + ω. Show that α = α α ω = ω + θ η α ω = θ η + ω α θ = θ η and hence compute α 2 where this is α α. 4.7 Consider the following (i) γ + 0 = γ (ii) 0 + γ = γ (iii) γ + 1 = γ + (iv) 1 + γ = γ + (v) γ 0 = 0 (vi) 0 γ = 0 (vii) γ 1 = γ (viii) 1 γ = γ (ix) γ 0 = 1 (x) 0 γ = 0 (xi) γ 1 = γ (xii) 1 γ = 1 where γ is an arbitrary ordinal. Which of these valid, and which not. For those that are false provide a counter-example. 4.8 Consider any two pairs 0 < a < b 0 < c < d of natural numbers, and let n be an arbitrary natural number. Verify the following. (i) 1 + ω n+1 = ω n+1 (ii) ω a + ω b = ω b (iii) (ω b + ω a ) ω = ω b+1 (iv) (ω b + ω a ) ω n+1 = ω b+n+1 (v) (ω b + ω a ) (ω d + ω c ) = ω b+d + ω a+d (vi) (ω b + ω a ) n+1 = ω (n+1)b + ω bn+a 4.9 Calculate each of the following (i) (ω 3 + ω) 2 (v) (ω 5 + ω 3 ) 2 (ii) (ω 3 + ω) 3 (vi) (ω 5 + ω 3 ) 3 (iii) (ω 3 + ω) 4 (vii) (ω 5 + ω 3 ) 4 (iv) (ω 3 + ω) 5 (viii) (ω 5 + ω 3 ) 5 and put the answer in some kind of normal form. 17

18 4.2 Problems: Multiplication of order types and ordinals 4.10 For each n {1, 2, 3, 4, 5} find order types α, β, γ such that the six possible sums of these give exactly n values Find order types α, β such that α β but α 2 = β 2 = β (a) Show that for all r, k N with k > 0. (b) Show that for all m, n N with m < n. (c) For r, s, t N compute r + ω k = ω k ω m + ω n = ω n (ω + r) (ω + s) (ω + t) and put into some kind of normal form. (d) Let α = ω n + β where n > 0 and β is an order type with β + ω n = ω n. Compute α s α ω for arbitrary s N. (e) For r 0, r 1,..., r n N compute and put into some kind of normal form. α = (ω + r 0 ) (ω + r 1 ) (ω + r n ) 4.13 Is it true that ω ω = ω ω? What about ζ 2 = ζ? 4.14 Let θ = ω + ω, and for each n N let α n = ω + ω η + n, β n = θ α n. For each m N compute β m n. 5 Comparison of ordinals We have set up the ordinals as the order types of well-ordered sets. From what we have seen so far it seem that these ordinals come in a natural linear sequence. In this section we make that idea precise. 5.1 DEFINITION. Let Ord be the whole class of ordinals. The word class here has a technical meaning. The class Ord is certainly a collection of things, but it is too big to be a set. We can not assign to it a cardinality. This means that there are times when we have to be careful. We can not treat Ord in all the ways we can treat sets. However, for what we do here that won t matter too much. The aim of this section is to indicate that the following can be proved. 18

19 5.2 THEOREM. The whole class Ord of ordinals is well-ordered. By this we mean that Ord is linearly ordered in a natural way, and that ordering is well-founded. Notice also that I say indicate that the following can be proved. I don t say that we will prove the results, because some of the details will be a bit sketchy. Our first job is to set up the comparison between ordinals. To do that we look first at the corresponding comparison between well-ordered sets. 5.3 DEFINITION. Given two well-ordered sets A, B, a well-ordered embedding A f B is a monotone injection such that the range f[a] is an initial section of B. We sometimes refer to such a gadget as a woe (pronounced as you like). We write A B to indicate there is an embedding from A to B. Let s look at the properties of a woe f in more detail. We have a function f : A B with x y = f(x) f(y) f(x) = f(y) = x = y for x, y A. The left hand implication says that f is monotone, and the right hand implication says that f is injection. These combine to give x y f(x) f(y) for x, y A. In some ways this is a more convenient property of f. Thus a woe is an isomorphism to its range, and this is an initial section of its target. You should check that this is an equivalent way of defining the notion. With this we see that A B means that A is isomorphic to an initial section of B. Each instance A B is witnessed by a woe from A to B. It turns out that there is always a unique such witness, and this has a big impact on the nature of ordinals. 5.4 LEMMA. Let A, B be a pair of well-ordered sets, and let A f g B be a pair of woes with ranges L = f[a] M = g[a] respectively. Then f = g, and in particular L = M. 19

20 Proof. By way of contradiction suppose f g. Thus f(a) g(a) for some a A. Since A is well ordered there is a least such a A. Thus we have an element a 0 A with and by symmetry we may suppose that f(x) = g(x) for all x < a 0 f(a 0 ) < g(a 0 ) holds. We have g(a 0 ) M and M, the range of g, is an initial section of B, so that f(a 0 ) M and hence f(a 0 ) = g(a 1 ) for some a 1 A. Since g(a 1 ) = f(a 0 ) < g(a 0 ) and g is an isomorphism we have a 1 < a 0. But now, by choice of a 0, we have f(a 0 ) = g(a 1 ) = f(a 1 ) and hence a 0 = a 1 since f is an injection. This is the contradiction. By taking B = A we obtain the following simple consequence. 5.5 COROLLARY. Let A be any well-ordered set. The only woe on A is the identity automorphism id A. A f A We are now in a position to obtain the first crucial result. 5.6 THEOREM. Consider a pair of well-ordered sets A B B A comparable in both ways. Then and this isomorphism is unique. A = B Proof. We are given a pair of woes A f B A g B going in opposite directions. Consider the composite h = g f on A. This is certainly a monotone injection. We show that its range is initial in A, and hence it is a woe. With this, by Corollary 5.5, we have g f = id A. In a similar fashion we have f g = id B, and hence f and g are an inverse pair of isomorphism, which leads to the required result. 20

21 It remains to show that h[a] is initial in A. To this end consider element x, a A such that x h(a) so we required x h[a]. We have and g[b] is initial in A, so that for some y B. We have and hence x h(a) = g(f(a)) g[b] x = g(y) g(y) = x g(f(a)) y f(a) f[a] since g is a woe. Since f[a] is initial in B this gives y = f(z) for some z A. But now for the required result. x = g(y) = h(z) By Lemma 5.4 and Theorem 5.6 any two well-ordered set are comparable in at most one way. What we do not yet know is that any two well-ordered sets are always comparable. That is our next result. And it takes just a little bit more work. 5.7 THEOREM. For each pair A, B of well-ordered sets we have A B or B A or both, in which case they are uniquely isomorphic. Proof. Since the empty well-ordered set is embedded (trivially) in every well-ordered set, we may assume the both A and B are non-empty. We consider the partial embeddings from A to B or from B to A. That is we consider the situations A B X f B A g Y where X is an initial section of A viewed as a well-ordered set, Y is an initial section of B viewed as a well-ordered set, and each of f, g is an embedding to the other side. There is at least one example of f and one example of g. Since A, B have least elements 0 A, 0 B we may let X = {0 A } Y = {0 B } and send 0 A to 0 B or 0 B to 0 A. 21

22 By the uniqueness of Lemma 5.4 each such partial embedding f is uniquely determined by its source X. Similarly, each such partial embedding g is uniquely determined by its source Y. Thus we may consider those collections X of initial sections X of A Y of initial sections Y of B from which there is an embedding to the other side. Also let L = X R = Y to produce two special initial sections, L of A and R of B. Consider any member a L. This is a member of at least one X X. Consider any two X 1, X 2 X which contain a. Since X 1, X 2 are initial sections of A we see that one is included in the other. Thus we may suppose X 1 X 2. Thus we have a situation A X 2 X 1 f 2 f 1 B B where f 1, f 2 are the associated embeddings. But now f 2 X1 is an embedding of X 1 into A, and hence must be f 1. This show that f 1 (a) = f 2 (a) and more importantly, it show there is a unique function f L : L B with f X (a) = f L (a) for each a X X. A few moment s thought shows that f L is an embedding. Thus f L is the unique largest partial embedding from A to B. In the same way we obtain a unique partial embedding from B to A. In other words, we have a pair of partial embeddings A B L f L B A g R R where each is the largest possible partial embedding from one side to the other. We show that either L = A or R = B, so that either f L is an embedding of A into B or g R is an embedding of B into A. Consider the nature of f L. We have A B L f L f L [L] 22

23 where we have displayed the range of f L. This bottom function is an isomorphism from L to f + L[L], and so has an inverse isomorphism. A L B f L f L [L] g as shown. By the maximality of R and g R we have f L [L] R and g is a restriction of g R. Thus. for each a L we have (g R f L )(a) = (g f L )(a) = a since g is a restriction of g R and g is the inverse function of f L. There is a similar argument with left and right swapped. Thus we have f L [L] R g R [R] L and ( a L) [ (g R f L )(a) = a ] ( b R) [ (f L g R )(b) = b ] where the left hand information is that obtained above, and the right hand information is the mirror of that. To show that g R [R] = L consider any a L. We require a g R [R]. But and hence f L (a) f L [L] R a = (g R f L )(a) g R [R] L as required. We obtain g R [R] = L by a similar argument. With this we can show that one of L = A R = B must hold. By way of contradiction suppose neither holds. Then each of the sets A L B R is non-empty, so each has a minimal element a A L b B R since both A and B are well ordered. Consider the function f : L {a} R {b} which is the extension of f L given by a b. By a bit of fiddking about we see that this is an isomorphism between the two sets. This contradiction the maximality of f L. This result show that if two well-ordered sets A, B have the same ordinal, then they are uniquely isomorphic. This suggest that it might be better to concentrate on ordinals rather than well-ordered sets. That is what we may do later [if there is time]. 23

24 We now transfer this comparison to the ordinals. Recall that an ordinal ι(a) is the order type of a well-ordered set A, the name of the block of well-ordered sets all of which are isomorphic. Thus ι(a) = ι(b) A = B for well-ordered sets A, B. As before we let α, β, γ,... range over typical ordinals. We use the same idea to impose a comparison on Ord. For that we first make an observation. 5.8 LEMMA. Let A = A B = B be two pairs of isomorphic well-orders. Then holds. B A B A With this we see that the following does make sense. 5.9 DEFINITION. Let α, β be the ordinals of the well-ordered set A, B, that is α = ι(a) β = ι(b) hold. Then we let to obtain a comparison on ordinals. β α B A At first sight this comparison on Ord looks as thought it depends on which well-ordered sets we pick to represent the ordinals. However, Lemma 5.8 show that the comparison is independent of the representatives. What are the properties of this ordinal comparison? 5.10 LEMMA. The class Ord of ordinals is linearly ordered by the comparison induced by embeddings between well-ordered sets. Proof. Let us first check that the comparison is a partial order, that is it is reflexive, transitive, and antisymmetric. In other words, we require (r) α α (t) α β γ = α γ (a) α β α = α = β for ordinals α, β, γ. These translate into properties of well-ordered sets as follows. (R) A A (T ) A B C = A C (A) A B A = A = B Property (R) is immediate. Verifying (T ) takes a few moment s thought. We need to verify that the composite of two woes is a woe. Property (A) is Theorem

25 Finally, we must show that the comparison is linear. But that is precisely what Theorem 5.7 gives. This is the start of the proof of Theorem 5.2. It remains to check that the linear ordering on Ord is a well-ordering. Proof of Theorem 5.2. As remarked above it remains to show that Ord is well-ordered. In other words, we must check that for an arbitrary ordinal α the linear set of ordinals α = {β Ord β α} is well-ordered. To do that we use Exercise 2.5. Consider any well-ordered set A with α = ι(a). By parts (a, b) of Exercise 2.5 we know that the family LA of lower sections of A is well-ordered by inclusion. By the solution to part (c) we know that the order type of LA is α +, the ordinal successor of α. More generally, there is a isomorphism LA {A} X α ι(x) to show that α is well-ordered. 5.1 Prove Lemma 5.8. We may return to ordinals later if there is time 25

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