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 welldefined. (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 nonempty. 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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