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


 Alaina Margaret Scott
 2 years ago
 Views:
Transcription
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
f(x) is a singleton set for all x A. If f is a function and f(x) = {y}, we normally write
Math 525 Chapter 1 Stuff If A and B are sets, then A B = {(x,y) x A, y B} denotes the product set. If S A B, then S is called a relation from A to B or a relation between A and B. If B = A, S A A is called
More informationClassical Analysis I
Classical Analysis I 1 Sets, relations, functions A set is considered to be a collection of objects. The objects of a set A are called elements of A. If x is an element of a set A, we write x A, and if
More informationSets, Relations and Functions
Sets, Relations and Functions Eric Pacuit Department of Philosophy University of Maryland, College Park pacuit.org epacuit@umd.edu ugust 26, 2014 These notes provide a very brief background in discrete
More informationSets and functions. {x R : x > 0}.
Sets and functions 1 Sets The language of sets and functions pervades mathematics, and most of the important operations in mathematics turn out to be functions or to be expressible in terms of functions.
More informationINTRODUCTORY SET THEORY
M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H1088 Budapest, Múzeum krt. 68. CONTENTS 1. SETS Set, equal sets, subset,
More informationReview for Final Exam
Review for Final Exam Note: Warning, this is probably not exhaustive and probably does contain typos (which I d like to hear about), but represents a review of most of the material covered in Chapters
More informationChapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics.
Chapter Three Functions 3.1 INTRODUCTION In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics. Definition 3.1: Given sets X and Y, a function from X to
More informationCartesian Products and Relations
Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special
More informationIntroducing Functions
Functions 1 Introducing Functions A function f from a set A to a set B, written f : A B, is a relation f A B such that every element of A is related to one element of B; in logical notation 1. (a, b 1
More informationCourse 421: Algebraic Topology Section 1: Topological Spaces
Course 421: Algebraic Topology Section 1: Topological Spaces David R. Wilkins Copyright c David R. Wilkins 1988 2008 Contents 1 Topological Spaces 1 1.1 Continuity and Topological Spaces...............
More informationTOPOLOGICAL PROOFS OF THE EXTREME AND INTERMEDIATE VALUE THEOREMS. Contents
TOPOLOGICAL PROOFS OF THE EXTREME AND INTERMEDIATE VALUE THEOREMS JAMES MURPHY Abstract. In this paper, I will present some elementary definitions in Topology. In particular, I will explain topological
More informationProblem Set. Problem Set #2. Math 5322, Fall December 3, 2001 ANSWERS
Problem Set Problem Set #2 Math 5322, Fall 2001 December 3, 2001 ANSWERS i Problem 1. [Problem 18, page 32] Let A P(X) be an algebra, A σ the collection of countable unions of sets in A, and A σδ the collection
More informationBasic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011
Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely
More informationStructure of Measurable Sets
Structure of Measurable Sets In these notes we discuss the structure of Lebesgue measurable subsets of R from several different points of view. Along the way, we will see several alternative characterizations
More informationMA651 Topology. Lecture 6. Separation Axioms.
MA651 Topology. Lecture 6. Separation Axioms. This text is based on the following books: Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology by Nicolas Bourbaki Counterexamples
More informationFoundations of Mathematics I Set Theory (only a draft)
Foundations of Mathematics I Set Theory (only a draft) Ali Nesin Mathematics Department Istanbul Bilgi University Kuştepe Şişli Istanbul Turkey anesin@bilgi.edu.tr February 12, 2004 2 Contents I Naive
More informationThis chapter describes set theory, a mathematical theory that underlies all of modern mathematics.
Appendix A Set Theory This chapter describes set theory, a mathematical theory that underlies all of modern mathematics. A.1 Basic Definitions Definition A.1.1. A set is an unordered collection of elements.
More information0 ( x) 2 = ( x)( x) = (( 1)x)(( 1)x) = ((( 1)x))( 1))x = ((( 1)(x( 1)))x = ((( 1)( 1))x)x = (1x)x = xx = x 2.
SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 8. Prove that if x and y are real numbers, then xy x + y. Proof. First we prove that if x is a real number, then x 0. The product of two positive
More informationPOWER SETS AND RELATIONS
POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty
More informationProblem Set 1 Solutions Math 109
Problem Set 1 Solutions Math 109 Exercise 1.6 Show that a regular tetrahedron has a total of twentyfour symmetries if reflections and products of reflections are allowed. Identify a symmetry which is
More informationIn mathematics you don t understand things. You just get used to them. (Attributed to John von Neumann)
Chapter 1 Sets and Functions We understand a set to be any collection M of certain distinct objects of our thought or intuition (called the elements of M) into a whole. (Georg Cantor, 1895) In mathematics
More informationCARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE
CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE With the notion of bijection at hand, it is easy to formalize the idea that two finite sets have the same number of elements: we just need to verify
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationCourse 221: Analysis Academic year , First Semester
Course 221: Analysis Academic year 200708, First Semester David R. Wilkins Copyright c David R. Wilkins 1989 2007 Contents 1 Basic Theorems of Real Analysis 1 1.1 The Least Upper Bound Principle................
More informationMathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson
Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement
More information1 if 1 x 0 1 if 0 x 1
Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or
More informationSolutions to Homework Problems from Chapter 3
Solutions to Homework Problems from Chapter 3 31 311 The following subsets of Z (with ordinary addition and multiplication satisfy all but one of the axioms for a ring In each case, which axiom fails (a
More informationMath 421, Homework #5 Solutions
Math 421, Homework #5 Solutions (1) (8.3.6) Suppose that E R n and C is a subset of E. (a) Prove that if E is closed, then C is relatively closed in E if and only if C is a closed set (as defined in Definition
More informationRELATIONS AND FUNCTIONS
Chapter 1 RELATIONS AND FUNCTIONS There is no permanent place in the world for ugly mathematics.... It may be very hard to define mathematical beauty but that is just as true of beauty of any kind, we
More informationSets and Cardinality Notes for C. F. Miller
Sets and Cardinality Notes for 620111 C. F. Miller Semester 1, 2000 Abstract These lecture notes were compiled in the Department of Mathematics and Statistics in the University of Melbourne for the use
More information6.2 Permutations continued
6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of
More informationMAT2400 Analysis I. A brief introduction to proofs, sets, and functions
MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take
More informationLECTURE NOTES ON RELATIONS AND FUNCTIONS
LECTURE NOTES ON RELATIONS AND FUNCTIONS PETE L. CLARK Contents 1. Relations 1 1.1. The idea of a relation 1 1.2. The formal definition of a relation 2 1.3. Basic terminology and further examples 2 1.4.
More information1. Prove that the empty set is a subset of every set.
1. Prove that the empty set is a subset of every set. Basic Topology Written by MenGen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since
More information3. Equivalence Relations. Discussion
3. EQUIVALENCE RELATIONS 33 3. Equivalence Relations 3.1. Definition of an Equivalence Relations. Definition 3.1.1. A relation R on a set A is an equivalence relation if and only if R is reflexive, symmetric,
More informationRELATIONS AND FUNCTIONS
Chapter 1 RELATIONS AND FUNCTIONS 1.1 Overview 1.1.1 Relation A relation R from a nonempty set A to a non empty set B is a subset of the Cartesian product A B. The set of all first elements of the ordered
More informationFIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.
FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is
More informationIntroduction to Relations
CHAPTER 7 Introduction to Relations 1. Relations and Their Properties 1.1. Definition of a Relation. Definition: A binary relation from a set A to a set B is a subset R A B. If (a, b) R we say a is related
More informationCHAPTER 5: MODULAR ARITHMETIC
CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called
More informationFinite Sets. Theorem 5.1. Two nonempty finite sets have the same cardinality if and only if they are equivalent.
MATH 337 Cardinality Dr. Neal, WKU We now shall prove that the rational numbers are a countable set while R is uncountable. This result shows that there are two different magnitudes of infinity. But we
More information3. Recurrence Recursive Definitions. To construct a recursively defined function:
3. RECURRENCE 10 3. Recurrence 3.1. Recursive Definitions. To construct a recursively defined function: 1. Initial Condition(s) (or basis): Prescribe initial value(s) of the function.. Recursion: Use a
More informationPractice with Proofs
Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using
More informationFinite and Infinite Sets
Chapter 9 Finite and Infinite Sets 9. Finite Sets Preview Activity (Equivalent Sets, Part ). Let A and B be sets and let f be a function from A to B..f W A! B/. Carefully complete each of the following
More informationTHE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. Saturday, March 5, 2016
THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION In Memory of Robert Barrington Leigh Saturday, March 5, 2016 Time: 3 1 2 hours No aids or calculators permitted. The grading is designed
More information5. ABSOLUTE EXTREMA. Definition, Existence & Calculation
5. ABSOLUTE EXTREMA Definition, Existence & Calculation We assume that the definition of function is known and proceed to define absolute minimum. We also assume that the student is familiar with the terms
More informationSelected problems and solutions
Selected problems and solutions 1. Prove that (0, 1) = R. Solution. Define f : (0, 1) R by 1 x 0.5 + 2, x < 0.5 f(x) = 1 x 0.5 2, x > 0.5 0, x = 0.5 Note that the image of (0, 0.5) under f is (, 0), the
More information1.3 Induction and Other Proof Techniques
4CHAPTER 1. INTRODUCTORY MATERIAL: SETS, FUNCTIONS AND MATHEMATICAL INDU 1.3 Induction and Other Proof Techniques The purpose of this section is to study the proof technique known as mathematical induction.
More informationMath 563 Measure Theory Project 1 (Funky Functions Group) Luis Zerón, Sergey Dyachenko
Math 563 Measure Theory Project (Funky Functions Group) Luis Zerón, Sergey Dyachenko 34 Let C and C be any two Cantor sets (constructed in Exercise 3) Show that there exists a function F: [,] [,] with
More informationEquivalence relations
Equivalence relations A motivating example for equivalence relations is the problem of constructing the rational numbers. A rational number is the same thing as a fraction a/b, a, b Z and b 0, and hence
More informationLemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S.
Definition 51 Let S be a set bijection f : S S 5 Permutation groups A permutation of S is simply a Lemma 52 Let S be a set (1) Let f and g be two permutations of S Then the composition of f and g is a
More informationDiscrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set.
Discrete Mathematics: Solutions to Homework 2 1. (12%) For each of the following sets, determine whether {2} is an element of that set. (a) {x R x is an integer greater than 1} (b) {x R x is the square
More informationThe BanachTarski Paradox
University of Oslo MAT2 Project The BanachTarski Paradox Author: Fredrik Meyer Supervisor: Nadia S. Larsen Abstract In its weak form, the BanachTarski paradox states that for any ball in R, it is possible
More informationMATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:
MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,
More informationCardinality. The set of all finite strings over the alphabet of lowercase letters is countable. The set of real numbers R is an uncountable set.
Section 2.5 Cardinality (another) Definition: The cardinality of a set A is equal to the cardinality of a set B, denoted A = B, if and only if there is a bijection from A to B. If there is an injection
More informationLogic & Discrete Math in Software Engineering (CAS 701) Dr. Borzoo Bonakdarpour
Logic & Discrete Math in Software Engineering (CAS 701) Background Dr. Borzoo Bonakdarpour Department of Computing and Software McMaster University Dr. Borzoo Bonakdarpour Logic & Discrete Math in SE (CAS
More informationp 2 1 (mod 6) Adding 2 to both sides gives p (mod 6)
.9. Problems P10 Try small prime numbers first. p p + 6 3 11 5 7 7 51 11 13 Among the primes in this table, only the prime 3 has the property that (p + ) is also a prime. We try to prove that no other
More informationClimbing an Infinite Ladder
Section 5.1 Climbing an Infinite Ladder Suppose we have an infinite ladder and the following capabilities: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder,
More informationDiscrete Mathematics, Chapter 5: Induction and Recursion
Discrete Mathematics, Chapter 5: Induction and Recursion Richard Mayr University of Edinburgh, UK Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. Chapter 5 1 / 20 Outline 1 Wellfounded
More informationChapter 1. SigmaAlgebras. 1.1 Definition
Chapter 1 SigmaAlgebras 1.1 Definition Consider a set X. A σ algebra F of subsets of X is a collection F of subsets of X satisfying the following conditions: (a) F (b) if B F then its complement B c is
More informationREAL ANALYSIS I HOMEWORK 2
REAL ANALYSIS I HOMEWORK 2 CİHAN BAHRAN The questions are from Stein and Shakarchi s text, Chapter 1. 1. Prove that the Cantor set C constructed in the text is totally disconnected and perfect. In other
More informationBasic Set Theory. Chapter Set Theory. can be written: A set is a Many that allows itself to be thought of as a One.
Chapter Basic Set Theory A set is a Many that allows itself to be thought of as a One.  Georg Cantor This chapter introduces set theory, mathematical induction, and formalizes the notion of mathematical
More informationChapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1
Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide Recall that denotes
More informationMath 230a HW3 Extra Credit
Math 230a HW3 Extra Credit Erik Lewis November 15, 2006 1 Extra Credit 16. Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = p q. Let E be the set of all p Q such that 2 < p
More informationIf f is a 11 correspondence between A and B then it has an inverse, and f 1 isa 11 correspondence between B and A.
Chapter 5 Cardinality of sets 51 11 Correspondences A 11 correspondence between sets A and B is another name for a function f : A B that is 11 and onto If f is a 11 correspondence between A and B,
More information2.3. Relations. Arrow diagrams. Venn diagrams and arrows can be used for representing
2.3. RELATIONS 32 2.3. Relations 2.3.1. Relations. Assume that we have a set of men M and a set of women W, some of whom are married. We want to express which men in M are married to which women in W.
More informationMath 507/420 Homework Assignment #1: Due in class on Friday, September 20. SOLUTIONS
Math 507/420 Homework Assignment #1: Due in class on Friday, September 20. SOLUTIONS 1. Show that a nonempty collection A of subsets is an algebra iff 1) for all A, B A, A B A and 2) for all A A, A c A.
More informationPolish spaces and standard Borel spaces
APPENDIX A Polish spaces and standard Borel spaces We present here the basic theory of Polish spaces and standard Borel spaces. Standard references for this material are the books [143, 231]. A.1. Polish
More information13 Infinite Sets. 13.1 Injections, Surjections, and Bijections. mcsftl 2010/9/8 0:40 page 379 #385
mcsftl 2010/9/8 0:40 page 379 #385 13 Infinite Sets So you might be wondering how much is there to say about an infinite set other than, well, it has an infinite number of elements. Of course, an infinite
More informationSETS, RELATIONS, AND FUNCTIONS
September 27, 2009 and notations Common Universal Subset and Power Set Cardinality Operations A set is a collection or group of objects or elements or members (Cantor 1895). the collection of the four
More informationDay 11. Wednesday June 6, Today, we begin by investigating properties of relations. We use a few toy relations as examples.
Day 11 Wednesday June 6, 2012 1 Relations Today, we begin by investigating properties of relations. We use a few toy relations as examples. 1.1 Examples of Relations There are a few relations that particularly
More informationCourse 214 Section 1: Basic Theorems of Complex Analysis Second Semester 2008
Course 214 Section 1: Basic Theorems of Complex Analysis Second Semester 2008 David R. Wilkins Copyright c David R. Wilkins 1989 2008 Contents 1 Basic Theorems of Complex Analysis 1 1.1 The Complex Plane........................
More informationMath/CSE 1019: Discrete Mathematics for Computer Science Fall Suprakash Datta
Math/CSE 1019: Discrete Mathematics for Computer Science Fall 2011 Suprakash Datta datta@cse.yorku.ca Office: CSEB 3043 Phone: 4167362100 ext 77875 Course page: http://www.cse.yorku.ca/course/1019 1
More informationBOREL SETS, WELLORDERINGS OF R AND THE CONTINUUM HYPOTHESIS. Proof. We shall define inductively a decreasing sequence of infinite subsets of N
BOREL SETS, WELLORDERINGS OF R AND THE CONTINUUM HYPOTHESIS SIMON THOMAS 1. The Finite Basis Problem Definition 1.1. Let C be a class of structures. Then a basis for C is a collection B C such that for
More informationMetric Spaces Joseph Muscat 2003 (Last revised May 2009)
1 Distance J Muscat 1 Metric Spaces Joseph Muscat 2003 (Last revised May 2009) (A revised and expanded version of these notes are now published by Springer.) 1 Distance A metric space can be thought of
More information18.312: Algebraic Combinatorics Lionel Levine. Lecture 8
18.312: Algebraic Combinatorics Lionel Levine Lecture date: March 1, 2011 Lecture 8 Notes by: Christopher Policastro Remark: In the last lecture, someone asked whether all posets could be constructed from
More informationMath 320 Course Notes. Chapter 7: Countable and Uncountable Sets
Math 320 Course Notes Magnhild Lien Chapter 7: Countable and Uncountable Sets Hilbert s Motel: Imagine a motel with in nitely many rooms numbered 1; 2; 3; 4 ; n; : One evening an "in nite" bus full with
More informationChap2: The Real Number System (See Royden pp40)
Chap2: The Real Number System (See Royden pp40) 1 Open and Closed Sets of Real Numbers The simplest sets of real numbers are the intervals. We define the open interval (a, b) to be the set (a, b) = {x
More informationMath 317 HW #7 Solutions
Math 17 HW #7 Solutions 1. Exercise..5. Decide which of the following sets are compact. For those that are not compact, show how Definition..1 breaks down. In other words, give an example of a sequence
More informationLimits and convergence.
Chapter 2 Limits and convergence. 2.1 Limit points of a set of real numbers 2.1.1 Limit points of a set. DEFINITION: A point x R is a limit point of a set E R if for all ε > 0 the set (x ε,x + ε) E is
More informationGeometric Transformations
Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted
More informationChapter 1. Informal introdution to the axioms of ZF.
Chapter 1. Informal introdution to the axioms of ZF. 1.1. Extension. Our conception of sets comes from set of objects that we know well such as N, Q and R, and subsets we can form from these determined
More informationThis asserts two sets are equal iff they have the same elements, that is, a set is determined by its elements.
3. Axioms of Set theory Before presenting the axioms of set theory, we first make a few basic comments about the relevant first order logic. We will give a somewhat more detailed discussion later, but
More informationGROUPS ACTING ON A SET
GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for
More informationDiscrete Mathematics. Hans Cuypers. October 11, 2007
Hans Cuypers October 11, 2007 1 Contents 1. Relations 4 1.1. Binary relations................................ 4 1.2. Equivalence relations............................. 6 1.3. Relations and Directed Graphs.......................
More informationWeek 5: Binary Relations
1 Binary Relations Week 5: Binary Relations The concept of relation is common in daily life and seems intuitively clear. For instance, let X be the set of all living human females and Y the set of all
More informationLecture 16 : Relations and Functions DRAFT
CS/Math 240: Introduction to Discrete Mathematics 3/29/2011 Lecture 16 : Relations and Functions Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT In Lecture 3, we described a correspondence
More information5.1 Commutative rings; Integral Domains
5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following
More informationIt is not immediately obvious that this should even give an integer. Since 1 < 1 5
Math 163  Introductory Seminar Lehigh University Spring 8 Notes on Fibonacci numbers, binomial coefficients and mathematical induction These are mostly notes from a previous class and thus include some
More informationCourse Notes for Math 320: Fundamentals of Mathematics Chapter 3: Induction.
Course Notes for Math 320: Fundamentals of Mathematics Chapter 3: Induction. February 21, 2006 1 Proof by Induction Definition 1.1. A subset S of the natural numbers is said to be inductive if n S we have
More informationProblems on Discrete Mathematics 1
Problems on Discrete Mathematics 1 ChungChih Li 2 Kishan Mehrotra 3 Syracuse University, New York L A TEX at January 11, 2007 (Part I) 1 No part of this book can be reproduced without permission from
More informationvertex, 369 disjoint pairwise, 395 disjoint sets, 236 disjunction, 33, 36 distributive laws
Index absolute value, 135 141 additive identity, 254 additive inverse, 254 aleph, 466 algebra of sets, 245, 278 antisymmetric relation, 387 arcsine function, 349 arithmetic sequence, 208 arrow diagram,
More informationIntegral Domains. As always in this course, a ring R is understood to be a commutative ring with unity.
Integral Domains As always in this course, a ring R is understood to be a commutative ring with unity. 1 First definitions and properties Definition 1.1. Let R be a ring. A divisor of zero or zero divisor
More informationTHE BANACH CONTRACTION PRINCIPLE. Contents
THE BANACH CONTRACTION PRINCIPLE ALEX PONIECKI Abstract. This paper will study contractions of metric spaces. To do this, we will mainly use tools from topology. We will give some examples of contractions,
More information3. Mathematical Induction
3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 19967 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More informationA LEBESGUE MEASURABLE SET THAT IS NOT BOREL
A LEBESGUE MEASURABLE SET THAT IS NOT BOREL SAM SCHIAVONE Tuesday, 16 October 2012 1. Outline (1 Ternary Expansions (2 The Cantor Set (3 The Cantor Ternary Function (a.k.a. The Devil s Staircase Function
More informationNotes on counting finite sets
Notes on counting finite sets Murray Eisenberg February 26, 2009 Contents 0 Introduction 2 1 What is a finite set? 2 2 Counting unions and cartesian products 4 2.1 Sum rules......................................
More informationA set is a Many that allows itself to be thought of as a One. (Georg Cantor)
Chapter 4 Set Theory A set is a Many that allows itself to be thought of as a One. (Georg Cantor) In the previous chapters, we have often encountered sets, for example, prime numbers form a set, domains
More informationCS 103X: Discrete Structures Homework Assignment 3 Solutions
CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On wellordering and induction: (a) Prove the induction principle from the wellordering principle. (b) Prove the wellordering
More informationThe Real Numbers. Here we show one way to explicitly construct the real numbers R. First we need a definition.
The Real Numbers Here we show one way to explicitly construct the real numbers R. First we need a definition. Definitions/Notation: A sequence of rational numbers is a funtion f : N Q. Rather than write
More information