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MA532 Lecture MA532 Lecture Timothy Kohl Boston University April 9, 2020 Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 1 / 24 Order Types and Order Type Arithmetic Back in calculus, one of the points that was frequently made was that ’∞’ is not a number! This point is made, in particular, to prevent one from trying to (carelessly) try to make sense of statements like ’∞ +1= ∞’ which of course lead to the non-sensical conclusion that 0 = 1. In the setting of the natural numbers this would be translated into the statement that ’ω’ isn’t a number, but can we make sense of ω +1 in any way? However, we will see that one can make sense of expressions like ’1 + ω’ and ’ω +1’ which is a formal kind of arithmetic, but does not imply non-sensical statements like ’1 = 0’. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 2 / 24 Before we get to this ordinal (actually order type) arithmetic, we need to define what are known as order types, which effectively allow us to distinguish between different order relations. Recall the following. Definition Let (A, A) and (B, B ) be ordered sets then (A, A) and (B, B ) are said to be order isomorphic if there is a bijection f : A → B such that x A y implies f (x) B f (y), and we write (A, A) =∼ (B, B ) to denoted this isomorphism. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 3 / 24 For a finite set A with n elements, there are n! rearrangements (permutations) of A and each such permutation is representable by a bijection σ : A → A. If we regard each of these as a distinct ordering of the set, then these n! orderings are actually all of the same order type on the set A with |A| =c |n|. Note, each of these orderings is a total order, so indeed every total ordering of a finite set is a permutation, and each of these permutations are order isomorphic to each other. Of course, not all orderings on a set are total orderings, since for some orderings, a given pair of elements may not be comparable. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 4 / 24 If N = {0, 1, 2,... } and 2N = {0, 2, 4, 6}˙ then if < is the usual (numerical) order relation on N then 2N inherits the same order. But one can show that the function f : N → 2N given by f (n) = 2n yields (N,<) =∼ (2N,<). Similarly Z with the order {0, 1, −1, 2, −2,... } is order isomorphic to N. On the mid-term we had the following, which we recall here. Theorem If (A, A) and (B, B ) are ordered sets and if (A, A) =∼ (B, B ) then if A has a first (resp. last) element, then so does B. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 5 / 24 It is clear that if (A, A) =∼ (B, B ) by means of some bijection f , then f induces |A| =c |B|. Conversely, if |A| =c |B| via a bijection f and if A is ordered by A then − one can induce an ordering B on B by means of the inverse f 1 : B → A − − so that for x, y ∈ B, x B y iff f 1(x) A f 1(y). With such an induced relation, we have the following. Theorem If f : A → B is a bijection and A is an ordering then for the induced ordering B one has (A, A) =∼ (B, B ). Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 6 / 24 Definition The order type of an ordered set (A, A) is defined to be the equivalence class of (A, A). We denote the order type of (A, A) by ot(A, A). So if we look at the natural numbers as (N,<) with the usual ’numeric’ ordering that we’re familiar with, then some authors define ω in terms of this by ω = ot(N,<) which, operationally, is the same as our usual definition of ω with its ordering induced by the successor function n → n+. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 7 / 24 We saw that for disjoint ordered sets (A, A) and (B, B ) that one can define orderings on A ∪ B and B ∪ A as follows: If x, y ∈ A where x A y then x A∪B y and x B∪A y. If x, y ∈ B where x B y then x A∪B y and x B∪A y. If x ∈ A and y ∈ B then x A∪B y and y B∪A x. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 8 / 24 And the example we considered were orderings of A = {0, 1, 2} (inherited from ω) and B = {3, 4, 5, 6 . } (also inherited from ω) where now A ∪ B = {0, 1, 2, 3, 4, 5, 6 . .} B ∪ A = {3, 4, 5, 6 . ., 0, 1, 2} which are demonstrably different since B ∪ A has a maximum element whereas A ∪ B does not. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 9 / 24 This construction with the union is the (beginning) of the ’arithmetic’ of order types. Definition If α = ot(A, A) and β = ot(B, B ) then α + β = ot(A ∪ B, A∪B ). As we’ve just seen α + β is not necessarily β + α. As such the arithmetic of order types is quite a bit different than the arithmetic in ω. One immediate bit of ’familiarity’ we have is how for finite order types, the arithmetic is the same as that in ω. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 10 / 24 For example if m = {0, 1, 2,..., m − 1} and n = {0, 1,..., n − 1} then m + n = {0, 1, 2,..., m − 1, 0, 1,..., n − 1} n + m = {0, 1,..., n − 1, 0, 1, 2,..., m − 1} and we observe that m + n = n + m indeed holds since there is an obvious order preserving bijection between them, i.e. 0 7→ 0 1 7→ 1 . n − 2 7→ m − 2 n − 1 7→ m − 1 Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 11 / 24 For example, if we let n = {0, 1,..., n − 1} and ω = {0, 1, 2,... } then n + ω = {0, 1,..., n − 1, 0, 1, 2,... } ω + n = {0, 1, 2,..., 0, 1,..., n − 1} which highlights that n + ω 6= ω + n since one has a maximum while the other does not. Moreover, we see that n + ω = ω since the first n elements on the ’left’ in n + ω still lead to a set with an initial element 0 and with successors yielding a linearly ordered set. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 12 / 24 We saw that n + ω = ω and that ω + n is quite different. And similarly n + α 6= n + α generally. We can give a particular interpretation of α + 1, especially in the setting where α is well ordered such as α = ω. If we view 1 as {0} then ω +1 is {0, 1,..., {0}} but it’s not hard to see that 1 = {0} is the same order type as {ω} so that we can interpret ω +1= ω+ = ω ∪ {ω} = {0, 1,..., {ω}} namely the successor of ω. So in particular, we have total ordering on sets that properly contain ω. We’ll explore this a bit more later. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 13 / 24 How different is the addition of order types from the arithmetic in ω? One familiar fact we can show is the following, based on the (relatively) obvious fact that ∅ has a unique ordering by default as it has no elements. Lemma For any order type α we have α + ∅ = α and ∅ + α = α. Proof. If α = (A, A) then since A ∪ ∅ = A then A∪∅=A, and similarly ∅∪ A = A. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 14 / 24 We can show other properties we expect for an arithmetic system. Lemma For α, β, γ order types, one has (α + β)+ γ = α + (β + γ). Proof. This is based on the associativity of set union, namely that if α = (A, A), β = (B, B ), and γ = (C, C ) then (A ∪ B) ∪ C = A ∪ (B ∪ C) implies that (A∪B)∪C =A∪(B∪C). Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 15 / 24 We can even make sense of an expression like ω + ω, which can be written as {0, 1, 2,..., 0′, 1′, 2′,... } which is not the same as ω since in ω every element besides 0 is the successor of another element, whereas in ω + ω both 0 and 0′ fail to be successors of another element. So ω + ω 6= ω. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 16 / 24 There is also a multiplication of order types that we can develop. Definition Let (A, A) and (B, B ) be ordered sets where α = ot(A, A) and β = (B, B ), we can define an order B×A by hb1, a1iB×A hb2, a2i if b1 B b2 or, if b1 = b2 then a1 A a2. We define α · β = ot(B × A, B×A). That is, we think of α · β as β copies of α. Timothy Kohl (Boston University) MA532 Lecture April 9, 2020 17 / 24 As an example, consider 2 = {0, 1} and ω = {0, 1, 2,... } then the order relations on 2 × ω and ω × 2 ω × 2= {h0, 0i, h0, 1i, h1, 0i, h1, 1i, h2, 0i, h2, 1i,..
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