S(A) X α for all α Λ. Consequently, S(A) X, by the definition of intersection. Therefore, X is inductive.

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1 MA 274: Exam 2 Study Guide (1) Know the precise definitions of the terms requested for your journal. (2) Review proofs by induction. (3) Be able to prove that something is or isn t an equivalence relation. (4) Be able to prove that something is or isn t a partial order. (5) Understand what it means to prove that a function on equivalence classes is well-defined. (6) Be able to prove all or portions of the following facts. You should also study other homework problems (a) The intersection of inductive sets is inductive. Proof. Recall that a set X is inductive if the following hold: X For all A X, the set S(A) = A {{A}} is an element of X. Suppose that X α is an inductive set for all α in some index set Λ. Let X = α Λ X α. We desire to show that X α is inductive. Since for all α Λ, X α, by the definition of intersection, we also have X α. Let A X. By the definition of intersection, for all α Λ, A X α. Since each X α is inductive, S(A) X α for all α Λ. Consequently, S(A) X, by the definition of intersection. Therefore, X is inductive.. (b) There exists a unique smallest inductive set. Proof. We begin by showing that if U is a set, then if U has an inductive subset, there is a smallest one X U. By the power set axiom and the axiom of specification, there exists a set S(U) = {A U : A is inductive}. By the assumption that U has an inductive subset, S(U) is non-empty. By the previous problem, 1

2 the set X U = S(U) is inductive. Furthermore, since every inductive subset of A is contained in S(U), X U is contained in each inductive subset of U. Hence, U has a smallest inductive subset. If X U were another such, then we would have X U X U and X U X U and so X U is the unique smallest inductive subset of U. Now we show that if U and V are each sets having an inductive subset, then X U = X V. Assume that U and V each have an inductive subset. By the previous paragraph they each have a unique smallest such subset, called X U and X V respectively. Similarly, since U V has inductive subset, it has a unique smallest one X U V. Since U (U V ) and since X U is inductive and since X U V is the smallest inductive subset of U V, we have X U V X U. Similarly, X U V X V. Since X U U, we ahve X U V X U U. The set X U is defined to be the unique smallest inductive subset of U and so X U X U V. Hence, X U = X U V. Similarly, X V = X U V. We conclude that X U = X V as desired. Finally, if W is an inductive set, then we have X U = X W W and so there is a smallest inductive set. Since it is contained in every inductive set and is itself inductive, it is unique. 2 (c) There does not exist a set of all sets. Proof. Suppose, to the contrary, that X is a set such that every set is an element of X. Then by the axiom of specification, the set is a set. Y = {x X : x x} Either Y Y or Y Y and not both, since those statements are negations of each other. But if Y Y, then Y satisfies the criterion for entry in Y and so Y Y, a contradiction. Hence, Y Y. But then Y satisfies the criterior for entry in Y and so Y Y, another contradiction. Consequently, the set X cannot exist. (d) = 4

3 Proof. The number 3 is defined to be the successor of 2. By definition of addition, this means that = S(2 + 0) = S(2) = 3. Hence, by the definition of addition, = 2 + S(1) = S(2 + 1) = S(3) = 4. (e) If a, b, c N 0, then b + a = c + a implies b = c. Proof. We prove this by induction on a. Base Case: a = 0. In this case, by definition of addition, b + 0 = b and c + 0 = c. Hence, b = c. Inductive Step: Assume that b + a = c + a implies b = c. We prove that implies b = c. b + (a + 1) = c + (a + 1) To that end, assume that b + (a + 1) = c + (a + 1). By the definition of + 1, this means that b + S(a) = c + S(a). By the definition of addition, this implies that S(b + a) = S(c + a). Since N 0 satisfies the Peano axioms, by axiom 4, we have b + a = c + a. By the inductive hypothesis, b = c. Hence, by induction for all a N 0, b + a = c + a implies that b = c. (f) If X is a set, then the relation (A B) (A B) is a partial order on P(X). Proof. Let X be a set. We must prove that is a reflexive, antisymmetric, and transitive relation on P(X). Reflexive: Assume that a P(X). Then a a and so a a. Hence, is reflexive. 3

4 Antisymmetric: Assume that a, b P(X) and that a b and that b a. Then a b and b a and so a = b. Hence, is antisymmetric. Transitive: Assumet that a, b, c P(X) and that a b and that b c. Then a b and b c. Since the subset of a subset is a subset, a c. Hence, a c. Consequently, is transitive. 4 (g) Prove that equivalence classes form a partition. Proof. Suppose that A is a set with an equivalence relation. We prove that A/ is a partition of A. Since is reflexive, x x for all x A. Hence, for all x A, x [x]. Thus, A/ covers A. Suppose that [x], [y] A/. We must prove that either [x] = [y] or [x] [y] =. Suppose that [x] [y]. Let z [x] [y]. Then, by definition of equivalence relation, x z and y z. Since is symmetric, z y. Since is transitive, x y. We now show that [y] [x]. Let w [y]. Then y w. Since x y and since is transitive, x w. Hence w [x]. Therefore, [y] [x]. Now we show that [x] [y]. Let w [x]. Then x w. Since x y and since is symmetric, y x. Since is transitive, y w. Hence, w [y]. Consequently, [y] [x]. Since [x] [y] and [y] [x], [x] = [y], as desired. Thus, the elements of A/ are pairwise disjoint. Therefore, A/ is a partition of A. (h) If G is a group and if H is a subgroup, then is an equivalence relation on G where (x y) h H such that x = y h Proof. We must prove that is reflexive, symmetric, and transitive. Reflexive: Let x G. Since H is a group, the identity 1 H. Since, x = x 1, we have x x.

5 Symmetric: Suppose that x y and that x = y h with h H. Since H is a subgroup, h 1 H. Thus, since is associative: Consequently, y x. x = y h x h 1 = (y h) h 1 x h 1 = y (h h 1 ) x h 1 = y 1 x h 1 = y. Transitive: Suppose that x y and that y z. Let h 1, h 2 H such that x = y h 1 and y = z h 2. Then Since is associative, x = (z h 2 ) h 1. x = z (h 2 h 1 ). Since H is a group and since h 2, h 1 H we also have h 2 h 1 H. Thus, x z. (i) Using the previous equivalence relation, for all x G, prove that there exists a bijection from [x] to H. Proof. Let x G. Consider f : H [x] defined by f(h) = x h. Clearly, f is a function. We show that f is injective. Suppose that f(h 1 ) = f(h 2 ). Then x h 1 = x h 2. Consequently, x 1 (x h 1 ) = x 1 (x h 2 ). By the associativity of and the definition of x 1, it follows that h 1 = h 2. Thus, f is injective. We show that f is surjective. Suppose that y [x]. Then, there exists h H such that x = y h. Thus, there exists h H such that y = x h 1. Since H is a group, h 1 H. Thus, f(h 1 ) = y and so f is surjective. Since is f is injective and surjective, it is bijective. (j) If G is a finite group and if H is a subgroup, then the number of elements in G is a multiple of the the number of elements in H. 5

6 Proof. Let be the equivalence relation defined above. By a previous problem, the equivalence classes of partition G. Since H = [1], the subgroup H is one of the equivalence classes. By the previous problem, each equivalence class is in bijection with H. Hence, all equivalence classes have the same number of elements as H. Consequently, #G = (#A/ )(#H). Since all factors are natural numbers, the number of elements in G is a multiple of the number of elements in H. (k) A function f : X Y is a bijection if and only if there exists an inverse function f 1 : Y X such that for all y Y and x X: f f 1 (y) = y f 1 f(x) = x Proof. Suppose that f : X Y is a bijection. Define g : Y X as follows: Let y Y. Since f is a surjection, there exists x X such that f(x) = y. Define g(y) = x. Since f is an injection, g is a function. Let f 1 = g. Then, f f 1 (y) = f(f 1 (y)) = f(x) = y and f 1 f(x) = f 1 (f(x)) = f 1 (y) = x. Now suppose that there exists an inverse function f 1 : Y X. We start by showing that f is an injection. Suppose that f(x 1 ) = f(x 2 ). Then f 1 (f(x 1 )) = f 1 (f(x 2 )). By our assumption on f 1 this implies that x 1 = x 2. Hence f is injective. We now show that f is surjective. Suppose that y Y. Let x = f 1 (y). Then f(x) = f(f 1 (y)) = y, so f is surjective. (l) Suppose that f : X Y is a function and that A, B are subsets of X. Prove that f(a B) = f(a) f(b). Give an example to show that f(a B) need not be equal to f(a) f(b). Proof. We show that f(a B) f(a) f(b). Let x f(a B). Then there exists w A B so that f(w) = x. Since w A B either w A or w B. If w A, then x = f(w) f(a). If w B, then x = f(w) f(b). Hence, in either case, x f(a) f(b). Thus, f(a B) f(a) f(b). Now we show that f(a) f(b) f(a B). Let x f(a) f(b). Then either x f(a) or x f(b). In the first case, 6

7 there exists a A such that f(a) = x. In the second case, there exists b B such that f(b) = x. Hence, there exists c A B such that f(c) = x. Consequently, f(a) f(b) f(a B). Therefore, f(a) f(b) = f(a B). Let A = {1, 2, 3} and let B = { 1, 2, 3}. Let f : A B R be defined by f(x) = x 2. Since A B =, f(a B) =. But f(a) = {1, 4, 9} = f(b). Thus, f(a) f(b) = {1, 4, 9} is not equal to f(a B) =. (7) Here are some new facts for you to try to prove: (a) If X is a set with n elements, then P(X) has 2 n elements. Proof. We prove this by induction on n. Base Case: n = 0. In this case X = and P(X) = { }. Thus P(X) has 2 0 = 1 elements. Inductive Case: Assume that all sets with n elements have power sets with 2 n elements. Let X be a set with n+1 elements. We prove that P(X) has 2 n+1 elements. Since X has n elements, X. Thus, there exists x X. Let A = {A P(X) : x A}. Let B = {B P(X) : x B. Then A B = and A B = P(X). Let Y = X {x}. Since Y X, P(Y ) B. If U B, then x U and so U Y. Hence, P(Y ) = B. Since Y has n elements, by the inductive hypothesis, B has 2 n elements. Consider the function f : A B defined by f(u) = U {x}. Since two sets are equal if and only if they have the same elements, f is injective. Suppose that V B. Then x V by definition of B and so V {x} A. Thus, f(v {x}) = V and so f is surjective. Since there is a bijection between A and B, they have the same number of elements. Thus, A also has 2 n elements. Thus, P(X) has 2 n + 2 n = 2 n+1 elements. By induction, the result holds for all n N 0. (b) Prove using induction that the number of permutations of a set of n elements is n!. Proof. Recall that the set of permutations, B(X) is defined to be the set of bijections of X to itself. We prove the result by induction on n, the number of elements of X. Base Case: Suppose that n = 0. Then X =. The only function on the empty set is the empty function, which is a 7

8 bijection, and so the number of bijections of X to itself is 1 = 0!. Inductive Step: Suppose that all sets with n 0 elements have n! bijections. Let X be a set with n + 1 elements. We will show that X has (n + 1)! bijections. Let y X. Let A y = {f B(X) : f(x) = y}. Restricting the domain of each f A y to be X {x, y} produces a function that is a bijection of X {x, y} to itself. Similarly, each bijection g of X {x, y} can be promoted to a bijection of X by declaring that g(x) = y and g(y) = x. Hence A y is in bijection with the set of bijections of X {x, y} to itself. By the inductive hypothesis, there are n! such bijections. Hence A y has n! such bijections. Since the X has n+1 elements, there are n+1 sets A y. Since the A y partition B(X), the number of bijections of X to itself is equal to (n + 1)n! = (n + 1)!. By induction, the statement holds for all n. (c) Suppose that X is a set with a partial order. Define (x y) (y x). Prove that is a partial order on X. (d) Suppose that f : X X is a bijection on a set with n elements. Prove that there exist transpositions f 1,..., f k of X such that f = f k f k 1... f 2 f 1. (A transposition is a bijection that simply swaps two elements and leaves all other elements unchanged.) Hint: Induct on n. Proof. We prove this by induction on n. Base Case: n = 1. In this case, X has a single element and so there is only one bijection of X to itself. It is the composition of zero transpositions and so the result holds. Inductive Step: Assume that the result holds for all sets with n 1 elements. We prove that it holds for a set X with n + 1 elements. Suppose that f : X X is a bijection. If for each x X, f(x) = x, then f is the composition of zero transpositions and the result holds. Assume therefore that there exists x X such that f(x) = y x. Let T be the transposition that swaps x and y. Then T f is a bijection of X such that T f(x) = x and T f(y) = y. Thus, T f is a bijection of the set X {x} to itself. By the inductive hypothesis, there is a finite sequence of transpositiosn T 1,..., T k such that T f = T 1 T 2... T k. 8

9 Each T i can be extended to be a bijection of X by declaring T i (x) = x. Thus, as bijections of X: T f = T 1... T k. We have that T T = 1 (i.e. the bijection of X such that 1(x) = x for all x x.) Thus, T T f = T T 1... T k. implies f = T T 1... T k. Consequently, f is the composition of finitely many transpositions as desired. (e) Give an example of a permutation of N which is not the composition of a finite number of transpositions. Solution: Define f(n) = n + 1 if n is odd and define f(n) = n 1 if n is even. Then the set X = {n N : f(n) n} has infinitely many elements. If f were the composition of a finite number of transpositions, then X would have only finitely many elements since each transposition moves only two elements. 9

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