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Well-Ordering and Induction Instructor: Goutam Paul Scribe: Soham Das

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Well-Ordering and Induction Instructor: Goutam Paul Scribe: Soham Das Discrete Mathematics 16/01/2019 Lecture 3: Well-Ordering and Induction Instructor: Goutam Paul Scribe: Soham Das Definition 3.1. A relation R on X is called a partial order if R is reflexive, transitive and anti-symmetric. Generic notation for any partial order R is `≤'. Given (a; b) 2 ≤, we write a ≤ b. Thus, ≤ is a partial order on X if it satifies (i) x ≤ x for all x 2 X , (ii) x ≤ y; y ≤ z implies x ≤ z and (iii) if x ≤ y and y ≤ x then x = y: If a ≤ b and a 6= b then we write a < b. If a partial order ≤ is defined on a set X then (X; ≤) is called partial ordered set or poset. Examples: 1. X = fa; b; cg; P(X) = fφ, fag; fbg; ··· ; fa; b; cgg. Then ⊆ operation induces a partial order ≤ on P(X) : we write x ≤ y if x ⊆ y. For example, (fbg; fb; cg) 2 ≤, but (fag; fb; cg) 2= ≤. 2. Consider the set of natural numbers N. We write (a; b) 2 ≤ if a j b. For example, (1; 3); (2; 8) 2 ≤, but neither (3,5) nor (5,3) belongs to ≤. One can easily check that this ≤ is a partial order on N: Definition 3.2. A partial order relation ≤ on X is called a total order if for any two elements x; y 2 X we have either x ≤ y or y ≤ x. Examples: 1. The set of real numbers with the usual ordering ≤. It is a property of real numbers that for any two x; y 2 R, one and exactly one of the following is true: x < y, x = y or y < x: 2. Consider the set complex numbers C. We can define a dictionary ordering on it as follows: (a + ib; c + id) 2 ≤ if (i) a < c or (ii) a = c and b ≤ d. Then (1 + i; 2 − i); (0; 2 − 3i) 2 ≤. One can easily show that this ≤ is a total order on C: 3. Consider the set of natural numbers N. We write (a; b) 2 ≤ if a j b. Note that neither (3,5) nor (5,3) belongs to ≤. Hence this ≤ is not a total order on N: Definition 3.3. Consider a partial order ≤ on X, (i) x is the least(smallest) element, if 8 a 2 X, x ≤ a. (ii) x is the greatest(largest) element, if 8 a 2 X, a ≤ x. (iii) x is a minimal element, if @ a 2 X, such that a < x. (iv) x is a maximal element, if @ a 2 X, such that x < a. 3-1 Example: Consider the partial order ≤ on N defined by x ≤ y iff x j y: Then, note that 1 ≤ n for all n 2 N as 1 divides every other N; whereas there does not exist any n < 0 which could divide 0: Thus, 1 is the least element and 0 is a minimal element of the partially ordered set (N; ≤): (Note that (N; ≤) is not a total order. Thus, in order to have a least/greatest element, the set need not be totally ordered.) Remark. Note that, least (and greatest) element is unique if it exists. However, minimal and maximal element may not be unique. For example, if we define a partial order ≤ on fa; bg by ≤ = f(a; a); (b; b)g then a; b are both minimal elements. Definition 3.4. A set is called well-ordered if every non-empty subset of X has a least element. Remark. (Well-order implies total order but total order does not imply well- order.) Suppose a set X is well-ordered, with respect to the partial order ≤. Then, for any x; y 2 X, the set fx; yg must have a least element, which means either x ≤ y or y ≤ x is true. Thus, if (X; ≤) is well-ordered then ≤ must be a total order on X: However, the converse is not true. Suppose we take Q with the usual ordering ≤. Then ≤ is a total order on Q but it does not make Q an well-ordered set. Theorem 3.5 (Zermelo's well ordering theorem). Every non-empty set can be well- ordered. That is, given any non-empty set X, there exists a partial order on it which makes X an well-ordered set. A1. Well-ordering principle (WOP) Every non-empty subset of N has a least element. A2. Principle of weak induction (WIP) If B ⊆ N such that 0 2 B and n 2 B ) n+ 2 B then B = N. A3. Principle of strong induction (SIP) If B ⊆ N such that 0 2 B and 8a ≤ n; a 2 B ) n+ 2 B then B = N. Theorem 3.6. A1, A2, A3 are equivalent. WOP ) WIP: Assume WOP holds. Suppose B ⊆ N such that 0 2 B and if n 2 B then n+ 2 B. Let if possible, B 6= N then NnB 6= φ. Define J = fx 2 N : x2 = Bg 6= φ. By WOP, J has a least element say, l. Now, l 6= 0 since 0 2 B. Then, l has a predecessor, i.e. 9m 2 N such that l = m+. Since we assumed l is the smallest member of J, so m2 = J. Hence m 2 B: But m 2 B implies l = m+ 2 B, which contradicts that l2 = B. Thus, our initial assumption that B 6= N is wrong. Hence proved. WIP ) SIP: Consider B ⊆ N such that 0 2 B and if for all a ≤ n, a 2 B then n+ 2 B: Assume that WIP holds. We shall show that B = N: Construct C as follows: n 2 C iff a 2 B for all a ≤ n: 3-2 First note that 0 2 B, so 0 2 C: Next, n 2 C ) a 2 B 8 a ≤ n ) n+ 2 B as per the definition of B: This in turn implies that n+ 2 C: Thus, 0 2 C and n 2 C ) n+ 2 C: Therefore WIP allows us to conclude that C = N: This yeilds B = N since C ⊆ B ⊆ N: SIP ) WOP: Assume that SIP holds. Consider B to be a non-empty subset of N. Let if possible, B does not have any least element. Define J = fx : x2 = Bg. Note, 0 2= B otherwise 0 will be the least element of B, so 0 2 J. Suppose, 0; 1; : : : ; n 2 J. Now if n + 1 2 B, then again it will be the least element in B, and hence n + 1 must belong to J. Thus, 0 2 J and 8 a ≤ n; a 2 J implies n+ 2 J: Hence, by SIP, J = N ) B = φ which is a contradiction. Therefore, B must have a least element. Definition 3.7 (Transfinite induction). Consider a well-ordered set X and a non- empty subset B ⊆ X. If for any x 2 X, α < x and α 2 B ) x 2 B, then X = B. In a well-ordered set, every non-empty subset contains a distinct smallest element. Given the axiom of dependent choice, this is equivalent to just saying that the set is totally ordered and there is no infinite decreasing sequence, something perhaps easier to visualize. In practice, the importance of well-ordering is justified by the possibility of applying transfinite induction, which says, essentially, that any property that passes on from the predecessors of an element to that element itself must be true of all elements (of the given well-ordered set). If the states of a computation (computer program or game) can be well-ordered in such a way that each step is followed by a \lower" step, then the computation will terminate. It is inappropriate to distinguish between two well-ordered sets if they only differ in the \labeling of their elements", or more formally: if the elements of the first set can be paired off with the elements of the second set such that if one element is smaller than another in the first set, then the partner of the first element is smaller than the partner of the second element in the second set, and vice versa. Such a one-to-one correspondence is called an order isomorphism and the two well-ordered sets are said to be order-isomorphic, or similar (obviously this is an equivalence relation). Definition 3.8 (Order isomorphism). Two posets (X; ≤) and (X0; ≤0) are called order isomorphic if and only if 9 a bijection between X and X0 that preserves the 0 0 order, i.e., 9f : X ! X such that if (x1; x2) 2 ≤ then (f(x1); f(x2)) 2 ≤ . That is, 0 x1 ≤ x2 =) f(x1) ≤ f(x2): Example: Define ≤ on X = N as: x ≤ y iff x j y: Take X0 = set of all natural numbers which are perfect squares, with the ordering ≤0 defined as: x ≤0 y iff x j y: Then f(x) = x2 sets up an order isomorphism between (N; ≤) and (X0; ≤0): Note that, if there exists an order isomorphism between two well-ordered sets, the order isomorphism is unique: this makes it quite justifiable to consider the two sets as essentially identical, and to seek a \canonical" representative of the isomorphism type 3-3 (class). This is exactly what the ordinals provide, and it also provides a canonical labeling of the elements of any well-ordered set. Essentially, an ordinal is intended to be defined as an isomorphism class of well- ordered sets: that is, as an equivalence class for the equivalence relation of \being order-isomorphic".
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