Sets+ combinatorics

Lecture notes Sets and Combinatorics These lecture notes are based on earlier texts by Marcel van de Vel, Andr´e Ran, Wouter Kager, and other staff members of the Department of Mathematics at the Vrije Universiteit Amsterdam Contents 1.1 1.2 1.3 1.4 Basic concepts of set theory Principles of set theory Fundamental set operations The algebra of sets Collections of sets 1 11 2.1 2.2 2.3 2.4 2.5 Standard sample spaces Sets in probability theory Draws without replacement / with order Draws without replacement / without order Draws with replacement / with order Draws with replacement / without order 15 15 17 20 22 24 3.1 3.2 3.3 Combinatorial analysis The principle of mathematical induction The binomial and multinomial theorems Combinatorial principles of counting 29 29 32 37 4.1 4.2 4.3 4.4 Functions and cardinality Product sets Functions and graphs Equipotence and cardinality Countable and uncountable sets 40 40 43 48 50 Answers to selected assignments 55 Index 57 Basic concepts of set theory 1.1 Principles of set theory In mathematics, a set is an (unordered) collection of elements, such as the collection of all integers greater than and less than 10, or that of all even numbers Such a set is often indicated by a listing of the elements inside curly brackets, with the elements being separated by commas Thus we can write the above two sets as {1, 2, 3, , 9} and {2, 4, 6, } In accordance with good practice, we are using “ ” here to indicate that the listing of the foregoing elements is to be continued in the same pattern We agree that in such lists, the ellipsis (the three dots) represents all intermediate integers, unless the (partial) listing which precedes the ellipsis makes it clear that some other continuation is intended By {1, , 6} we thus mean all integers from through 6, but by {0, 2, 4, , 12} only the even numbers from through 12 The following sets are assumed to be known from Pre-University Education: • • • • N = {1, 2, 3, } Z = { , −2, −1, 0, 1, 2, } Q R the set of natural numbers; the set of integers; the set of rational numbers (the fractions); the set of real numbers Note that here we not consider as belonging to the natural numbers There is no general consensus about this: some textbooks and others not include in the set of natural numbers To designate the set of non-negative integers {0, 1, 2, }, we will use the notation N0 or, alternatively, Z+ : • • N0 := {0, 1, 2, } Z+ := {0, 1, 2, } the set of non-negative integers; the set of non-negative integers The assignment symbol := indicates that the expression to the left of the symbol (on the side of the colon) is defined by the expression on the right side (the side of the equal sign) Basic concepts of set theory In our examples so far, the elements of our sets have always been numbers, but that need not necessarily be so The elements of a set can be anything, for example, vectors or (ordered) sequences of numbers, matrices, operators, functions, What is more, they may even be sets themselves, resulting in a set of sets, et cetera A set can generally be conceived of as an (imaginary) totality of elements That a is an element of V is expressed by the notation a ∈ V If a is not an element of V then we write a V For example, ∈ {1, 2, 3} N 10 ∈ R − 10 ∈ Z If we have a series of data such as a ∈ Z, b ∈ Z, c ∈ Z, then to save space, this can be represented as a, b, c ∈ Z A second, common way to designate a set makes use of a description, like this: {prototype : description} or {prototype| description} (1.1) You can read such an expression as “the set of all elements of the form prototype, for which it holds that description” This can best be clarified with a few examples • • • N0 = {z ∈ Z : z ≥ 0} The prototype here is z ∈ Z and the description is “z ≥ 0”: it is the set of all numbers z in Z, for which it holds that z ≥ The set of even natural numbers is {n ∈ N : n is even} The prototype is a number n ∈ N and the description is “n is even”: it is the set of all numbers n in N, for which it holds that n is even An alternative representation of the previous set is {2k : k ∈ N} The prototype is now a number of the form 2k with description “k ∈ N”: it is the set of all numbers of the form 2k for which it holds that k is in N The last two examples show that in general, there is more than one way to specify a given set using a description Two sets V and W are termed equal, notation V = W, if they contain exactly the same elements, more precisely: if each element of V is also in W, and each element of W is also in V Take for example V = {1, 1, 2, 3} and W = {3, 1, 2} Then each element of V is also in W, and each element of W is also in V (verify this), so V = W It follows that it does not matter in what order the elements of a set are listed, or how often each element appears in the listing A set W is a subset of V if each element of W is also in V We write this as W ⊂ V, which you can read as “W is contained in V” An alternative way of writing this is V ⊃ W, read as “V contains W” For example, {1, 2, 3} ⊂ N Q ⊃ Z Note: the statement “V = W” is equivalent to the statement “V ⊂ W and W ⊂ V” If W is not a subset of V, then we can write that as W V The set W is called a proper subset of V if W is a subset of V (W ⊂ V), but V and W are not equal (W V) Thus, for example, the set {3, 1, 2, 3} is a proper subset of {1, 2, 3, 4} A set which has no elements is called an empty set For two empty sets, the statement “if x is in one set, then x is also in the other set” is (vacuously) true, 1.1 Principles of set theory because the premise is always false Two empty sets are thus equal Hence there exists just one empty set, which we call the empty set and denote by ∅ Note that ∅ = { } (a set formed by an empty listing) We conclude this section with the notations commonly used in mathematics for the intervals of the real line: (a, b) := {x ∈ R : a < x < b} (a, ∞) := {x ∈ R : x > a} [a, b) := {x ∈ R : a ≤ x < b} (−∞, b) := {x ∈ R : x < b} (a, b] := {x ∈ R : a < x ≤ b} [a, b] := {x ∈ R : a ≤ x ≤ b} [a, ∞) := {x ∈ R : x ≥ a} (−∞, b] := {x ∈ R : x ≤ b} The set consisting of the non-negative real numbers is also written as R+ := [0, ∞) Assignments We have two real intervals V := (1, 3) and W := [1, π) Check which of the following statements are true and which are not: (a) ∈ V; (b) ∈ W; (c) V; (d) π ∈ V; (e) π ∈ W; ( f ) V ⊂ W; (g) W ⊂ V The sets A, B and V are defined as follows: A := {x ∈ Z : x is even}; B := {x ∈ Z : x is a multiple of three}; V := {x ∈ Z : x is odd or x is a multiple of three} Check which of the following statements are true and which are not: (a) ∈ V; (b) −111 ∈ V; (c) 110 ∈ V; (d) 31 ∈ V; (e) A ⊂ V; ( f ) B ⊂ V; (g) Z ⊂ V Represent the following sets in the format using a prototype and a description as in the formula (1.1): (a) The set of all negative integers; (b) The set of all irrational (real) numbers; (c) The set of all integers which are divisible by 7; (d) The set of all positive irrational numbers; (e) The set of all real roots of the polynomial x3 − 6x2 + 11x − Basic concepts of set theory Let A := (0, 10), B := {x ∈ R : x2 + x + = 0} and C := {x ∈ R : x3 − = 0} Formulate each of the following statements in English, and check whether the statement is true or not: (a) A ⊂ B; (b) B ⊂ C; (c) C ⊂ A For each of the following statements, introduce appropriate sets A and B, and then formulate the statement in the form “A ⊂ B”: (a) Every integer that is divisible by 21, is also divisible by (b) All real solutions of the equation x2 − x + = lie between and 10 (c) The polynomial x3 − 6x2 + 11x − has no positive roots Write down all the subsets of the sets V := {1} and W := {1, 2} Determine the total number of subsets of V := {1, 2, 3, 4} 1.2 Fundamental set operations In what follows, we consider all the sets we are talking about as subsets of a special set U, which we call the universe This universe determines the context in which we work, for example U = R if we only want to discuss sets of real numbers Given two sets A and B in a universe U, we can construct new sets in the following manner: A ∩ B := {x ∈ U : x ∈ A and x ∈ B} the intersection of A and B; A \ B := {x ∈ U : x ∈ A and x the set-theoretic difference of A and B, or A ∪ B := {x ∈ U : x ∈ A or x ∈ B} B} the union of A and B; the relative complement of B in A In words: • • • A ∩ B is the set of all elements of U which are both in A and in B; A ∪ B is the set of all elements of U which are in at least one of the sets A and B; A \ B is the set of all elements of U which are in A but not in B Note that we can also write A ∩ B = {x ∈ A : x ∈ B} or A ∩ B = {x ∈ B : x ∈ A} and A \ B = {x ∈ A : x B} Thus, strictly speaking, we not need the universe in order to define the intersection and the relative complement of two sets To visualize these sets, it is often convenient to use so-called Venn diagrams1 In these diagrams, by convention we represent the universe as a rectangular frame, and subsets thereof as portions of the rectangle demarcated by an ellipse Various basic properties of sets can be visualised using such a diagram See for example the illustration at the top of the next page Example (Throwing a die) To mathematically describe throwing a die, we want to work within a universe that describes all possible outcomes of the throw For this purpose, we can take U := {1, , 6}, where each number represents the number of John Venn, English logician and mathematician, 1834–1923 1.2 Fundamental set operations A B A Intersection Union A B B A Symmetric difference A Relative complement A Complement B B (A ∪ B)c = Ac ∩ Bc dots thrown The subsets A := {2, 4, 6} and B := {5, 6} contain, respectively, the even outcomes of U, and the outcomes greater than Then A ∩ B = {6}, A ∪ B = {2, 4, 5, 6}, A \ B = {2, 4} and B \ A = {5} These sets consist successively of the outcomes which are even and greater than 4, the outcomes that are even or greater than 4, the outcomes that are even but not greater than 4, and the outcomes that are greater than but not even If A ∩ B = ∅, then we call A and B disjoint In words: A and B are disjoint if they have no common element at all We further define Ac := U \ A = {x ∈ U : x A} the complement of A Note that although Ac depends on the universe U, this is not expressed in the notation Thus, it is implicitly assumed that complements are always taken with respect to U Note that we can now also write the relative complement A \ B as A \ B = A ∩ Bc It is not difficult to see that in general A \ B B \ A (see example 1) In other words, the operation \ is not commutative Therefore one sometimes uses the symmetric difference A B of two sets A and B: A B := (A \ B) ∪ (B \ A) The equality A B=B the symmetric difference of A and B A does always hold true The fundamental set operations can be combined For example, if we have three sets A, B and C, we can first remove the elements of B that lie in C, and then take the intersection with A to obtain the set A ∩ (B \ C) Alternatively, we could Basic concepts of set theory A B A B A B C C C B\C A ∩ (B \ C) = (A ∩ B) \ C A∩B have first taken the intersection of A and B, and then removed the elements that lie in C: it turns out that the two results are always the same That is, the equality A ∩ (B \ C) = (A ∩ B) \ C holds true for all choices of the sets A, B and C Equalities like this one can be verified using Venn diagrams; see the illustration at the top of this page In drawing such Venn diagrams for three sets, we have to take care that all possible overlaps between the sets A, B and C are represented Thus there should be different regions: one region of points that lie in all three sets, three regions of points that lie in exactly two of the sets, but not in the third, three regions of points that lie in exactly one of the three sets, and one region of points that lie in none of the three sets Depending on A, B and C, any of these regions could of course be empty, but it would be wrong to draw conclusions from a Venn diagram about arbitrary A, B and C if any of these regions was omitted at the outset Assignments Express in words (as on page 4) what elements of A and B generally constitute the set A B Determine the symmetric difference A B of the sets in example Check, using Venn diagrams, that the following equations are correct: (a) A B = B A (b) A B = (A ∪ B) \ (A ∩ B) Check, using Venn diagrams, whether the following expressions are correct The possible answers are “yes”, “no” or “maybe, but it depends on A and B” If the last is the case, give an example of sets A and B for which the expression is true, and an example for which the expression is not true: (a) U = A ∪ Ac (b) A = (B ∩ A) ∪ (Bc ∩ A) (c) U = (B \ A) ∪ (A \ B) For problems involving real numbers, Venn diagrams are often less appropriate as an aid In this assignment, we will instead use a representation with the aid of a coordinate system The following sets are given: A = {(x, y) : x, y ∈ R, x + y ≤ 1}; B = {(x, y) : x, y ∈ R, x2 + y2 ≤ 1} 1.3 The algebra of sets (a) Draw the sets A and B in an (x, y) system of coordinates (b) Indicate in the drawing the sets A ∪ B, A ∩ B, A \ B and B \ A Write each of the sets described below as an intersection, union or relative complement of the two given sets A and B: (a) The set of all x ∈ R with ≤ x ≤ or ≤ x ≤ Use A = [0, 1], B = [2, 3] (b) The set of all integers which are divisible by 7, but not by 11 Use A = {7n : n ∈ Z}, B = {11n : n ∈ Z} √ (c) The set of all real √ numbers x such that x2 > Use A = {x ∈ R : x > 2}, B = {x ∈ R : x < − 2} (d) The set of all non-negative, irrational numbers Use A = R+ , B = Q (e) The set of all negative rational numbers Use A = R+ , B = Q Using Venn diagrams, check whether the expression is true or false: (a) A ∪ (B \ C) = (A ∪ B) \ C (b) (A \ B) ∪ (A \ C) ⊃ A \ (B ∪ C) (c) (A \ B) ∪ (A \ C) ⊂ A \ (B ∪ C) Check using Venn diagrams that in general: (a) A \ B B \ A (b) (A \ B) \ C A \ (B \ C) (c) A \ (B \ C) (A ∪ C) \ B 1.3 The algebra of sets In the previous section we used Venn diagrams to check equalities and inclusions between sets In this section we provide more targeted methods for dealing with such comparisons The first method is an algebraic method, whereby a complex formula is analysed in terms of a limited number of fundamental sub-formulas We can organize the fundamental sub-formulas into three groups: Basic laws of the algebra of sets A∩B=B∩A A∪B=B∪A commutativity of ∩ commutativity of ∪ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) distributivity of ∩ over ∪ distributivity of ∪ over ∩ A ∩ Ac = ∅ A ∪ Ac = U complement law for ∩ complement law for ∪ A ∩ (B ∩ C) = (A ∩ B) ∩ C A ∪ (B ∪ C) = (A ∪ B) ∪ C A∩U =A A∪∅=A associativity of ∩ associativity of ∪ identity law for ∩ identity law for ∪ Laws that follow from the basic laws A∩A=A A∪A=A idempotency of ∩ idempotency of ∪ Basic concepts of set theory A∩∅=∅ A∪U =U domination law for ∩ domination law for ∪ (Ac )c = A involution of complement (A ∩ B)c = Ac ∪ Bc (A ∪ B)c = Ac ∩ Bc De Morgan’s law for ∩ De Morgan’s law for ∪ c complement law for ∅ complement law for U ∅ =U Uc = ∅ Definitions A \ B = A ∩ Bc A B = (A ∪ B) ∩ (A ∩ B)c definition of \ definition of Remark: Since ∩ and ∪ are associative, we can henceforth simply write A ∩ B ∩ C and A ∪ B ∪ C (i.e without using parentheses), without giving rise to confusion about what we mean Moreover, since ∩ and ∪ are also commutative, it is also clear that the order in which we write the sets A, B and C in these expressions does not matter: A ∪ B ∪ C = A ∪ C ∪ B = B ∪ A ∪ C, et cetera However, in a formula in which both ∩ and ∪ appear, it is necessary to use parentheses and heed the order! We will treat the above equalities as axioms from which we can discover new formulas by means of calculation This gives rise to a real algebra of sets By the remark above, in using this algebra, we can freely omit parentheses and change the order of the sets A, B and C in expressions such as A ∪ B ∪ C and A ∩ B ∩ C Example This example is similar to the familiar “banana formula” for the product (a + b)(c + d): (A ∪ B) ∩ (C ∪ D) = (A ∪ B) ∩ C ∪ (A ∪ B) ∩ D = (A ∩ C) ∪ (B ∩ C) ∪ (A ∩ D) ∪ (B ∩ D) = (A ∩ C) ∪ (B ∩ C) ∪ (A ∩ D) ∪ (B ∩ D) On the second line, the parts (A ∪ B) ∩ C and (A ∪ B) ∩ D were replaced by equivalent parts according to the distributive laws On the third line, we have used the associativity of ∪ to get rid of the outermost parentheses Example Show that A \ (B ∩ C) = (A \ B) ∪ (A \ C) Solution According to the definition of \ and De Morgan’s first law, it holds that A \ (B ∩ C) = A ∩ (B ∩ C)c = A ∩ (Bc ∪ Cc ) Next, we use the distributivity of ∩ over ∪ to see that A ∩ (Bc ∪ Cc ) = (A ∩ Bc ) ∪ (A ∩ Cc ) = (A \ B) ∪ (A \ C) Augustus De Morgan, British mathematician and logician, 1806–1871 Functions and cardinality 44 2 1 (0, 0) (0, 1) (1, 0) (1, 1) Graph of the heads function −1 Graph of the square function Whenever we wish to represent a concrete function f : V → W explicitly, we often this by means of a formula that tells us how to compute f (x) In doing so, it should be clearly indicated from which domain V the argument x is chosen and in which codomain W the value f (x) is expected to lie.2 Two functions f and g are said to be equal, notation f = g, if and only if i) f and g have the same domain V, ii) f and g have the same codomain W, and iii) for all x ∈ V it holds that f (x) = g(x) Example 32 (Square function) We define the function f : R → R by f (x) := x2 So here V = W := R To every every function f there corresponds a graph G f , which is the set of all pairs (x, y) ∈ V × W such that y is the image of x under f : G f := (x, y) ∈ V × W : y = f (x) = (x, f (x)) : x ∈ V Conversely, we can define a graph in V × W as a subset G of V × W with the property that for every x ∈ V there is exactly one y ∈ W such that (x, y) ∈ G It then follows from the definitions that every graph is actually the graph of some function g : V → W That is, every graph specifies a function, and we can formally identify every function of V to W with its graph (a set) in V × W Example 33 (Heads function, continued) The graph of the heads function X is GX = ((0, 0), 0), ((0, 1), 1), ((1, 0), 1), ((1, 1), 2) Example 34 (Square function, continued) The graph of the square function f is the well-known U-shaped parabola G f = (x, x2 ) : x ∈ R • • • A function f : V → W is called an injection (injective) if f (x) = f (y) implies x = y for all x, y ∈ V; a surjection (surjective) if for every y ∈ W there is an x ∈ V such that f (x) = y; a bijection (bijective) if f is both injective and surjective In calculus books, the domains and codomains of functions are often not specified explicitly 4.2 Functions and graphs 45 In other words, a function is injective if different elements of the domain are mapped to different elements of the codomain For this reason, injective functions are also called one-to-one Surjectivity signifies that every point in W is the image under f of at least one point in V A surjective function f : V → W is therefore also called a function of V onto W: every value in W is actually attained by a surjective function f Example 35 (Injections, surjections and bijections) • • • • • The function f : R+ → R given by f (x) = x2 is injective, because f (x) is strictly increasing in x on the domain R+ (i.e f (y) > f (x) if x, y ∈ R+ and y > x) The function f : R+ → R given by f (x) = (x − 1)2 is not injective, because f maps (for example) x = and x = to the same image: f (0) = f (2) = The function f : R → R+ given by f (x) = x2 is surjective, because for y ∈ R+ , √ the equation f (x) = x2 = y has the solution x = y ∈ R The function f : R → R+ given by f (x) = x2 + is not surjective, because the equation f (x) = x2 + = has no solution for x ∈ R The function f : R+ → R+ given by f (x) = x2 is bijective, because for every √ y ∈ R+ , the equation f (x) = x2 = y has a solution for x in R+ , namely x = y (surjectivity), and this is in fact the only solution in R+ (injectivity) Again, consider a function f : V → W Let A be a subset of the domain V We define the image f (A) of A under f to be the set of images of all elements of A: f (A) := { f (x) : x ∈ A} The image f (V) of the domain V is called the range R f of f : R f := f (V) = { f (x) : x ∈ V} Observe that a function f : V → W is a bijection precisely when its range R f coincides with the codomain W We can also map sets in the opposite direction Let B be a subset of the codomain W Then the inverse image (also called the (complete) preimage) of the set B is the set f −1 (B) consisting of all x ∈ V such that f (x) ∈ B, that is, f −1 (B) := {x ∈ V : f (x) ∈ B} If B = {y} (a singleton set, i.e., a set with one element), we usually speak of the inverse image of y (rather than of the set {y}), and write f −1 (y) instead of f −1 ({y}) Example 36 (Heads function, continued) Let A := {(0, 0), (1, 0)} Then X(A) consists of the elements X(0, 0) and X(1, 0), and consequently X(A) = {0, 1} The range of X is RX = X(U) = {0, 1, 2} The toss (1, 0) is one preimage of under X The complete preimage (or inverse image) of is X−1 (1) = {x ∈ U : X(x) = 1} = {(0, 1), (1, 0)}, and thus has two elements On the other hand, X−1 (3) = ∅ For B := {0, 1} we find X−1 (B) = {(0, 0), (0, 1), (1, 0)} Functions and cardinality 46 Example 37 (Square function, continued) If A := (−1, 2], then f (A) = [0, 4] The range of f is R f = f (V) = [0, ∞) The inverse image of B := [−2, 1] is f −1 (B) = [−1, 1] √ √ Furthermore, f −1 (0) = {0} and f −1 (5) = {− 5, 5} Example 38 (Sum of the dice) The standard sample space for tossing two dice is U := {1, , 6}2 Let X denote the function that assigns to every outcome in U the sum of the two numbers of dots thrown, that is, X(n, k) = n + k for every pair (n, k) ∈ U Then X−1 (7), for instance, is the set of outcomes for which the sum of the throws is 7, and X−1 ({3, 4}) consists of the outcomes for which the sum of the throws is either or Hence X−1 (7) = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, X−1 ({3, 4}) = {(1, 2), (2, 1), (1, 3), (2, 2), (3, 1)} In the general case of a function f : V → W it can be checked that f −1 (W) = V and f −1 (R f ) = V If B ⊂ W is disjoint from R f , then f −1 (B) = ∅ Finally, we mention below a series of properties that apply to various constructions involving images and inverse images This list reveals that inverse images behave a little bit better than (direct) images (compare properties iii vs v, and iv vs vi): i) ii) f −1 ( f (A)) ⊃ A f(f −1 (A ⊂ V); (B)) ⊂ B (B ⊂ W); iii) f (A1 ∩ A2 ) ⊂ f (A1 ) ∩ f (A2 ) (A1 , A2 ⊂ V); iv) f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ) f −1 vi) f −1 v) (B1 ∩ B2 ) = f (B1 ∪ B2 ) = f −1 −1 (B1 ) ∩ f (B1 ) ∪ f −1 (B2 ) −1 (B2 ) (A1 , A2 ⊂ V); (B1 , B2 ⊂ W); (B1 , B2 ⊂ W) Proof of property vi) Let x ∈ V be such that x ∈ f −1 (B1 ∪ B2 ) Then f (x) ∈ B1 ∪ B2 (definition of the inverse image), hence f (x) ∈ B1 or f (x) ∈ B2 The definition of the inverse image then gives x ∈ f −1 (B1 ) or x ∈ f −1 (B2 ) Thus x ∈ f −1 (B1 ) ∪ f −1 (B2 ) Next assume that x ∈ f −1 (B1 ) ∪ f −1 (B2 ) The definition of the inverse image then gives f (x) ∈ B1 or f (x) ∈ B2 In other words: f (x) ∈ B1 ∪ B2 , and so (again by the definition of the inverse image) x ∈ f −1 (B1 ∪ B2 ) If f : V → W is an injection, then the inverse image of each singleton set {y} with y ∈ f (V) = R f consists of exactly one element: f −1 (y) = {x} for a certain x ∈ V In this case, we usually leave out the curly braces and simply write f −1 (y) = x This gives us, in a natural way, a new function f −1 : R f → V, which we call the inverse function or simply the inverse of f The inverse of f is thus by definition the unique mapping f −1 : R f → V with the property f −1 (y) = x ⇐⇒ y = f (x) for all x ∈ V and y ∈ f (V) It is only defined (as a function) if f is injective; after all, the inverse function must assign a unique value x ∈ V to every y in the range of f The value f −1 (y) assigned to y is that unique element x of V for which it holds that f (x) = y 4.2 Functions and graphs 47 If B is a subset of R f , then with the above definition of the inverse function f −1 , the image of B under f −1 is precisely the set that we defined as the inverse image f −1 (B) of B before This shows that the notation f −1 for the inverse function is consistent with our previous notation for inverse images, and that no confusion can arise about what we mean by f −1 (B) Assignments Let f : (−1, 3) → R be given by f (x) := x2 (a) What is the domain of f and what is the codomain? (b) What is the range R f ? Let f : (−1, 1) → R be given by f (x) := |x| (a) Draw the graph G f of f (b) What is the image f ((− 23 , 12 )) of the open interval (− 32 , 12 ) ? (c) What is the inverse image f −1 ([0, 12 ]) of the closed interval [0, 12 ] ? (d) Give two sets A1 , A2 ⊂ (−1, 1) such that f (A1 ∩ A2 ) f (A1 ) ∩ f (A2 ) (e) True or false: f (A ∪ B) = f (A) ∪ f (B) for every A, B ⊂ (−1, 1) ? ( f ) True or false: f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B) for every A, B ⊂ R ? Check whether the given set G can be seen as the graph G f of an R-valued function f on a subset of R If so, give the domain and range of f : (a) G := {(x, y) ∈ R2 : x2 + y2 = 1} (b) G := {(x, y) ∈ R × R+ : x2 + y2 = 1} (c) G := {(x, y) ∈ R2 : x · y = 1} (d) G := {(x, y) ∈ R2 : x · y = 0} Determine the image of A under f : (a) A := [−2, 0) and f : R → R with f (x) := x2 (b) A := [0, 1]2 and f : R2 → R with f (x, y) := x2 + y2 (c) A := {1, , 10} and f : N → R with f (n) := 1/n2 (d) A := {1, 2, 3} × {4, 5} and f : N2 → N with f (m, n) := m + n Determine the inverse image of B under f : (a) B := {0} and f : R → R with f (x) := sin(x) (b) B := {−1} and f : R → R with f (x) := x2 + x (c) B := Z and f : R → R with f (x) := x2 (d) B := {1, 2, 3, 4} and f : N2 → N with f (m, n) := m + n (e) B := {2n : n ∈ Z} and f : Z → N with f (n) := (3n − 1)2 Let f be a function of V into W (a) Prove: f (Ac ) ⊃ ( f (A))c ∩ R f for all A ⊂ V (b) Prove: f −1 (Bc ) = ( f −1 (B))c for all B ⊂ W Are the following functions injective? surjective? bijective?: (a) f : Z → N with f (n) := |n| + (b) f : Z → N with f (n) := (3n − 1)2 (c) f : R → R with f (x) := x3 (d) f : R → R with f (x) := x3 − x (e) f : R+ → [0, 1) with f (x) := x/(1 + x) ( f ) f : R → R2 with f (x) := (2x + 1, x − 1) Show that for any function f : V → W it holds that: (a) f is injective ⇐⇒ f −1 ( f (A)) = A for all A ⊂ V (b) f is surjective ⇐⇒ f ( f −1 (B)) = B for all B ⊂ W (c) f is injective ⇐⇒ f (A1 ∩ A2 ) = f (A1 ) ∩ f (A2 ) for all A1 , A2 ⊂ V Functions and cardinality 10 11 12 13 48 Two functions f1 : V → W1 and f2 : V → W2 are given Now construct a new function f : V → W1 × W2 , given by f (x) := ( f1 (x), f2 (x)) (a) Show: f is injective if f1 and f2 are injective (b) Show, using an example, that the result in (a) is false if “injective” is replaced by “surjective” or “bijective” Is the given function injective? surjective? bijective?: (a) f : R → R defined by f (x) := ex √ (b) g : (0, ∞) → R2 defined by g(x) := ( x, ln(x)) (c) h : R2 → R defined by h(x, y) = 2y Consider the function h : R2 → R2 with h(x, y) := (2xy, x2 + y2 ) Show that h is not injective and not surjective Consider the function h : R × (0, ∞) → R with h(x, y) := 2x + ln(y) Is h injective? surjective? bijective? Is the given function injective? surjective? bijective?: (a) f : R2 → R2 defined by f (x, y) := (x2 , y2 ) (b) f : R+2 → R+2 defined by f (x, y) := (x2 , y2 ) (c) g : R → R2 defined by g(x) := (sin(x), cos(x)) (d) g : [0, 2π) → R2 defined by g(x) := (sin(x), cos(x)) (e) h : R2 → R defined by h(x, y) := x + y 4.3 Equipotence and cardinality Recall Example 18 from Section 2.2 In that example, we counted the elements of the sample space U by dividing the interval (0, 1] into subintervals, and associating to each subinterval a unique element of U In other words, we in fact constructed a bijection between the collection of sub-intervals of (0, 1] and the set U, and claimed that the number of elements of U must be equal to the number of sub-intervals This approach is precisely how, in mathematics, following Cantor3 , we define the size of a set: we call two sets V and W equipotent, or of the same size, if a bijection of V onto W exists Furthermore, for each m ∈ N, we define the size of the set {1, 2, , m} as m It then follows that by definition, #V = m ⇐⇒ there exists a bijection f : {1, , m} → V In this way, the size of each finite set is precisely defined and determined That this is a natural definition is also illustrated by the following example: Example 39 (Musical chairs) Suppose you want to seat a group of people in a room with a limited number of chairs, and you want to know if there are enough chairs In that case you can admit the people one by one, and have each person take the next available chair You thus construct an injective map from the group of people to the set of chairs If, at the end, there are chairs left over (the map is injective but not surjective), then you can conclude that the number of people was less than the number of chairs If, at the end, all chairs are occupied, and everyone has found a place (the image is both injective and surjective), there were exactly the same Georg Ferdinand Ludwig Philipp Cantor, German mathematician, 1845–1918 4.3 Equipotence and cardinality 49 number of chairs as people And if, at a certain moment, all the chairs are occupied, while there are still people standing outside, then the remaining people can only be seated by having them sit on other people’s laps (the map is surjective but not injective) In the last case, there were more people than chairs As an application of the definition of equipotence, in the following example, we will determine the size of the power set of a set V with a given finite number of elements The outcome should not come as a surprise if you have done all the assignments (see assignment from section 2.4) Example 40 (Size of a power set) The total number of subsets of a set V which has m elements is 2m Proof We construct a bijection f : P(V) → {0, 1}m , as follows Let a1 , a2 , , am be the elements of V If X ⊂ V then we define f (X) := (b1 , b2 , , bm ) with the “bit” bi ∈ {0, 1} at position i given by    1 if ∈ X; bi :=   0 if X The bit at position i thus indicates whether the point is or is not in the subset X Hence for every m-tuple (b1 , , bm ), there is a preimage X ⊂ V that is mapped by f onto this m-tuple, namely the set X consisting of exactly those for which the associated bit bi is This preimage is unique, because for every Y ⊂ V with Y X, f (Y) will differ at least at one position from f (X) Therefore f : P(V) → {0, 1}m is a bijection, and thus it holds that #P(V) = #({0, 1}m ) = 2m , where the last equality follows from the second combinatorial principle (Section 3.3) The beauty of the definition of equipotence is that this definition can also be used perfectly well for sets that contain infinitely many elements The definition enables us to compare the sizes of such infinite sets with each other This was the main reason why Cantor introduced his definition of equipotence Example 41 We claim that the sets Q and (−1, 1) ∩ Q are equipotent Indeed, verify for yourself that a bijection f of Q to (−1, 1) ∩ Q is given by the function x f (x) := + |x| Instead of the somewhat vague term “size”, in mathematics we usually prefer to speak of the cardinality of a set The cardinality of a set A is often denoted by |A| instead of #A It thus holds by Cantor’s definition that • |A| ≤ |B| if there exists an injection f : A → B; • |A| = |B| if there exists a bijection f : A → B The ordering relation “≤” between cardinalities introduced here satisfies the usual property that it follows from |A| ≤ |B| and |B| ≤ |A| that |A| = |B| That this is indeed so, is a direct consequence of the Schroder Bernstein5 theorem: ă SchroderBernstein ă theorem Let A and B be two sets If injections f : A → B and g : B → A exist, then there exists a bijection h : A → B Ernst Schroder, German mathematician, 18411902 ă Felix Bernstein, German mathematician, 1878–1956 Functions and cardinality 50 Proof We call an element a of A of type if a ∈ g(B), g−1 (a) ∈ f (A), f −1 (g−1 (a)) ∈ g(B), g−1 ( f −1 (g−1 (a))) ∈ f (A), et cetera For an element a ∈ A to be of type 1, we must thus be able to apply f −1 and g−1 infinitely repeatedly and alternately to a If a is not of type 1, then this procedure falters either at a certain element a ∈ A \ g(B), or else at a certain element b ∈ B \ f (A) In these cases we call a of type or of type 3, respectively Now define the map h : A → B by    if a is of type or 2;  f (a) h(a) :=    g−1 (a) if a is of type We show successively that this map is both surjective and injective Surjectivity: Let b ∈ B, and let a := g(b) If a is of type 3, then it holds that b = g−1 (a) = h(a) If, on the contrary, a is of type or 2, then it holds (by definition of types and 2) that b ∈ f (A), and then f −1 (b) must be of the same type as a (verify this!) In that case, it therefore holds that b = f (a ) = h(a ) with a = f −1 (b) Injectivity: Let a1 , a2 ∈ A and suppose that h(a1 ) = h(a2 ) =: b If b ∈ f (A), then f −1 (b) and g(b) are necessarily of the same type Therefore, it must be true that a1 and a2 are either both of type 3, or else both are of type or (verify this!) In both cases, it follows from h(a1 ) = h(a2 ) that a1 = a2 , because both f and g are one-to-one Assignments How many subsets does {1, , 6} have? How many subsets does {1, 2}3 have? (a) If V ⊂ W and V W, can V be equipotent with W? (b) If V ⊂ W, W is finite and V W, can V be equipotent with W? Given are two sets V and W Construct a concrete bijection f : V → W to prove that the two sets are equipotent: (a) V := {3n + : n ∈ N} and W := {2n : n ∈ N} (b) V := [0, 1] and W := [3, 8] (c) V := R and W := (− 12 π, 21 π) (d) V := Q and W := (0, 1) ∩ Q (e) V := R and W := (0, 1) Given are two sets V and W Construct a concrete bijection f : V → W to prove that the two sets are equipotent: (a) V := [0, 1] × (0, 1] and W := (1, 5] × [2, 3] (b) V := [0, 1] × [0, 1] and W := {(x, y) ∈ [0, 2] × [0, 1] : x − ≤ y ≤ x} (c) V := [0, 1) and W := {(x, y) ∈ R2 : x2 + y2 = 1} 4.4 Countable and uncountable sets For every number n ∈ N, however large it may be, {1, , n} is a finite set Therefore, it appears that the limit of this (increasing) sequence of sets, which is the set N := {1, 2, }, should be the “smallest possible” infinite set But this way of looking at things is not very adequate, as the next example shows 4.4 Countable and uncountable sets 51 Example 42 The set V of natural numbers greater than 100 is a proper subset of N, yet V is equipotent to N, because the function f : N → V given by f (n) := n + 100 is a bijection Example 43 (Even numbers) The set E of all even natural numbers is an infinite proper subset of N Intuitively speaking, E is only half as big as N, because the complement of E in N is just as big as E (there is a bijection f : Ec → E given by f (n) := n + 1) Still, E and N are equipotent To demonstrate this, we construct the bijection g : N → E given by g(n) := 2n A set V which is equipotent to N is called countably infinite You can enumerate the elements of such a set by means of a bijection f : N → V with the natural numbers Expressed differently: you can put them in a row in a complete enumeration (or infinite listing) of V, V = {x1 , x2 , x3 , }, where the nth element of the enumeration is given by xn := f (n) The bijection g from example 43, for instance, results in the following enumeration of the even natural numbers E: E = {2, 4, 6, } Example 44 (Integers) The set Z is countably infinite After all, an explicit enumeration of Z is given by Z = {0, 1, −1, 2, −2, 3, −3, } We claim that the associated bijection h : N → Z is given by    for n even; 2n h(n) :=   − 12 (n − 1) for n odd We can prove that h is indeed a bijection as follows: Surjectivity: Let z ∈ Z be arbitrary If z > 0, then take n = 2z If z ≤ 0, then take n = −2z + In both cases, n ∈ N and h(n) = z Therefore h is surjective Injectivity: Suppose h(n) = h(m) for certain m, n ∈ N If h(n) > 0, then it follows from h(n) = h(m) that 12 n = 12 m and therefore n = m If, on the contrary, h(n) ≤ 0, then it follows from h(n) = h(m) that − 12 (n − 1) = − 12 (m − 1), so again n = m We conclude that h(n) = h(m) implies n = m in all cases Therefore h is injective A set V is called countable if V is finite or countably infinite Non-countable sets are called uncountably infinite or simply uncountable In many ways, countable sets are easier than uncountable sets This is evident from the fact that it is not possible to put the elements of an uncountable set “in a row” The set Q of rational numbers seems to be a candidate: after all, Q would seem to be quite a lot bigger than N and Z Moreover, we know that between two different rational numbers there is always a third rational number That makes it difficult to put these numbers in a row Nonetheless, we can enumerate Q, as the following example shows Functions and cardinality 52 Example 45 (Rational numbers) An enumeration of the rational interval (0, 1) ∩ Q is given by (0, 1) ∩ Q = 1 3 , , , , , , , , , 3 4 5 5 Here, the irreducible fractions are enumerated according to ascending denominators, and where the denominators are the same, by ascending numerators The countability of Q can be derived from this result by constructing a bijection from Q onto (0, 1) ∩ Q, which is an assignment Because every positive rational number x can be written as x = m/n for certain (m, n) ∈ N2 , we suspect that N2 is also countable That suspicion is correct (see the assignments) It follows more generally that the product of two countable sets is countable The set R of all real numbers is not countable In the following example, we show that the real interval (0, 1) is not countable, from which this result follows It is an assignment to construct a bijection from the interval (0, 1) to all of R Example 46 (The interval (0, 1) is not countable) Every point x ∈ (0, 1) has a unique decimal representation 0.d1 (x) d2 (x) d3 (x) which does not end in zeros only (for example, we write 1/4 = 0.24999 · · · instead of 0.25000 · · · ), and such that x= ∞ i=1 di (x) 10i Suppose there does exist an enumeration {x1 , x2 , x3 } of (0, 1) Consider the decimal representations of the xi s in this supposed enumeration: x1 = 0.d1 (x1 ) d2 (x1 ) d3 (x1 ) · · · x2 = 0.d1 (x2 ) d2 (x2 ) d3 (x2 ) · · · x3 = 0.d1 (x3 ) d2 (x3 ) d3 (x3 ) · · · et cetera We will show that there is a point x ∈ (0, 1) which differs from each of the xi s We construct this number x by defining every decimal in its decimal representation 0.d1 (x) d2 (x) d3 (x) · · · explicitly as follows: if (for i = 1, 2, 3, ) di (xi ) > 5, then we set di (x) := 1, and otherwise we set di (x) := The point x constructed in this way lies in (0, 1) because none of the decimals of x is or Moreover, x differs by construction from every number xi , because the ith decimals of x and xi differ by at least Therefore x is a number from the interval (0, 1) that is not in the infinite list x1 , x2 , x3 , , and so {x1 , x2 , x3 , } cannot have been an enumeration of (0, 1) Theorem The power set P(V) of a set V is not equipotent with V Proof The empty set ∅ and its power set {∅} are clearly not equipotent, so we can restrict our attention to nonempty sets V Assume that the sets V and P(V) are equipotent, i.e that there exists a bijection f : V → P(V) We will show that this leads to a contradiction Indeed, define A := {x ∈ V : x f (x)} Then A is a subset of V, so if f is a bijection, then there must be an x ∈ V such that f (x) = A Suppose 4.4 Countable and uncountable sets 53 that x ∈ A Based on the definition of A, it must then hold that x f (x), but since f (x) = A, this is impossible Hence x A But since x ∈ V, it then follows from the definition of A that x ∈ f (x) = A, which contradicts x A As a corollary of the theorem, it follows that the set P(N) is uncountably infinite In the assignments, it will be proven that P(N) is equipotent with (0, 1), and hence with R The set P(R) again has greater potency than R The set P(P(R)) has an even greater potency, et cetera Hence by repeatedly taking power sets, starting from N, one can construct ever “more potent” sets: there is no “most potent” set More loosely phrased, there exist infinitely many different kinds of infinities Assignments Consider the set A := {1, 2} × N (a) Show that A is countable by providing an explicit enumeration, as in Examples 44 and 45 (b) Now prove formally that A is countable by providing a concrete bijection f : A → N (tricky) The aim of this assignment is to show in a few steps that every infinite subset A of N is countable Note first that for every a ∈ A the set [1, a] ∩ N is finite So the subset [1, a] ∩ A is also finite Now let g : A → N be the function with the rule g(a) := #{x ∈ A : x ≤ a} (a) Show that if a1 < a2 , then g(a1 ) < g(a2 ) for all a1 , a2 ∈ A (b) Show: for every a ∈ A and n ∈ N it holds that: if n ≤ g(a), then n ∈ g(A) Hint: there are exactly g(a) elements x in A with x ≤ a, and for every such x, g(x) ≤ g(a); now use part (a) (c) Use parts (a) and (b) to show that g : A → N is a bijection (laborious assignment) Let f : N2 → N be defined by f (m, n) := (2m − 1)2n−1 (a) (b) (c) (d) Calculate the values of f (m, n) for m, n ≤ For which (m, n) ∈ N2 does it hold that f (m, n) = 112? Successively determine f −1 ({i}) for i = 1, 2, , 10 Show that f is bijective Hint: show in one go that every natural number has one preimage under f (e) What implication does this result have for the “size” of N2 ? Consider the set P(N) of all subsets of N and the set {0, 1}∞ of all infinite sequences of zeros and ones Define f : P(N) → {0, 1}∞ by    1 if i ∈ X; f (X) := (a1 , a2 , ) with :=   0 if i X Show that f is bijective, and thus that P(N) and {0, 1}∞ are equipotent (continuation of the previous assignment; tricky) Just as in example 46, every number x from the interval (0, 1) also has a unique binary expansion with “bits” i bi (x) ∈ {0, 1}, which does not end in zeros only, and such that x = ∞ i=1 bi (x)/2 By mapping every number from (0, 1) onto its infinite sequence of bits, we obtain in a natural manner an injection f : (0, 1) → {0, 1}∞ Functions and cardinality 54 (a) Why isn’t f a bijection? (b) Define g : {0, 1}∞ → (0, 1) by g((b1 , b2 , b3 , )) = + ∞ i=1 bi 3i Show that g is an injection into (0, 1), and conclude using Schroder ă Bernstein that P(N) and (0, 1) (and thus also R) are equipotent Answers to selected assignments 1.1 1a: false, 1b: true, 1c: true, 1d: false, 1e: false, 1f: true, 1g: false, 2a: true, 2b: true, 2c: false, 2d: false, 2e: false, 2f: true, 2g: false, 3a: {z ∈ Z : z < 0}, 3b: {x ∈ R : x Q}, 3c: {z ∈ Z : 7z ∈ Z} or {7z : z ∈ Z}, 3d: {x ∈ R : x > and x Q}, 3e: {x ∈ R : x3 − 6x2 + 11x − = 0}, 4a: “Every real number (strictly) between and 10 is a root of x2 + x + = 0”; false, 4b: “If x2 + x + = and x is real, then x3 − = 0”; true, 4c: “All real roots of x3 − = lie (strictly) between and 10”; true, 5a: A = {21z : z ∈ Z} and B = {7z : z ∈ Z}, 5b: A = {x ∈ R : x2 − x + = 0} and B = (0, 10), 5c: A = {x ∈ R : x3 − 6x2 + 11x − = 0} and B = {x ∈ R : x ≤ 0}, 6: of V: ∅ and {1} = V, of W: ∅, {1}, {2} and {1, 2} = W, 7: 16 1.2 1: A B consists of those elements of U that are elements of exactly one of the two sets A and B, 2: A B = {2, 4, 5}, 4a: yes, 4b: yes, 4c: depends on A and B; take A = U and, respectively, B = U or B = ∅, 6a: A ∪ B, 6b: A \ B, 6c: A ∪ B, 6d: A \ B, 6e: B \ A, 7a: false, 7b: true, 7c: false 1.3 4a: true, 4b: true, 4c: true, 6a: take e.g A = B = D ∅ and C = ∅, 6b: take e.g B = C = ∅ and A = D ∅, 9a: false, true, true, 9b: true, false, false 1.4 3a: {1}, [1, 10], 3b: [1, 6], R, 4a: {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}, 4b: {{1}, {2, 3}}, {{2}, {1, 3}}, {{3}, {1, 2}}, 5: e.g An = (0, n], n ∈ N, 6: ∅, {0}, R, 7a: ( 21 , 54 ), ( 12 , 98 ), ( 13 , 32 ), ( 41 , 32 ), 7b: ( 12 , 1], (0, 23 ), {0} ∪ [ 32 , 2], [0, 21 ] ∪ (1, 2] 2.1 1a: A ∪ B occurs if and only if at least one of the events A or B occurs, 1b: A \ B occurs if and only if A occurs but B does not occur, 1c: A B occurs if and only if exactly one of the events A or B occurs, 1d: Ac occurs if and only if A does not occur, 2: w/ replacement w/ order (38 balls), 3a: w/ replacement w/ order (365 balls), 3b: w/o replacement w/ order (365 balls) 2.2 1: A ∩ B = {(x1 , , x5 ) ∈ U : x1 , , x5 ∈ {1, , 10}, x1 < x2 < · · · < x5 }, 2a: 544 320, 2b: 36, 2c: 890, 3: 64 584 000, 4a: (n − 1)!, 4b: (n − k + 1)!, 5a: 52!, 5b: 26! · 39!/13! 2.3 2: A = {(y1 , , y20 ) ∈ U : y1 + · · · + y10 = 5} with #A = 10 , 3: let yi be the number of the draw in which you grab the ball numbered i (let yi = in case you Answers to selected assignments 56 never grab ball i); then U = {(y1 , , yb ) : y1 , , yb ∈ {0, 1, , n} and for every j ∈ 13 13 {1, , n} there is exactly one i ∈ {1, , b} such that yi = j}, 4: 52 = 13 and 656 369, 5a: 356, 5b: 450 2.4 4: 81 and 64, 5: 59 049 and 720 087, 6a: 251 528, 6b: 78 125, 6c: 700, 6d: 420, 7: take balls and draws, 8: there are nk subsets with k elements; on the other hand, you could throw a coin n times to decide for each of the n elements whether it should be included in the subset or not 2.5 3a: 73 , 5: 640, 6a: 840, 6b: 360, 6c: 30 240, 6d: 300, 7a: 15, 7b: 90, 7c: 450, 7d: 800, 8a: 720, 8b: 600, 8c: 800, 8d: 200, 8e: 300, 8f: 150, 8g: 3.1 d n (n) 2b: no, 6c: dx (x) = cos(x + 12 nπ), 7b: f (n) (x) = n sin(2x) = sin(2x + nπ), 7a: f n −(n+1) (n) n−1 −n (−1) n!(1 + x) , 7c: f (x) = (−1) (n − 1)!x 3.2 7a: 27, 7b: 261 − 230 , 8a: 253 440, 8b: 920 3.3 4: #W = 15 4.1 1: A × B = {(1, 2), (1, 3), (2, 2), (2, 3)} and B × A = {(2, 1), (2, 2), (3, 1), (3, 2)}; not equal, 2a: A2 = {(0, 0), (0, 1), (1, 0), (1, 1)} and A3 = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)}, 2b: e.g (0, 0, 0, 0), (0, 0, 0, 1) and (0, 0, 1, 0), 2c: A4 consists of 4-tuples, A2 × A2 of pairs of pairs, 3: no; yes, 6a: R+ × R+ , 6b: [2, ∞) × R, 6c: {1, 2, 4} × {3, 5}, 9: no, take for instance A = C := {0} and B = D := {1}, 11a: false, 11b: true, 11c: true 4.2 1a: (−1, 3) and R, 1b: [0, 9), 2b: [0, 32 ), 2c: [− 21 , 12 ], 2d: A1 := (−1, 0) and A2 := (0, 1), √ 2e: true, 2f: true, 3a: no, 3b: yes, f : [−1, 1] → [0, 1] with f (x) := − x2 , 3c: yes, 1 f : R \ {0} → R \ {0} with f (x) := x1 , 3d: no, 4a: (0, 4], 4b: [0, 2], 4c: {1, 14 , 19 , 16 , , 100 }, √ √ 4d: {5, 6, 7, 8}, 5a: {kπ : k ∈ Z}, 5b: ∅, 5c: { n : n ∈ N0 } ∪ {− n : n ∈ N}, 5d: {(1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1)}, 5e: {2n + : n ∈ Z}, 7a: surjective, not injective, 7b: injective, not surjective , 7c: bijective, 7d: surjective, not injective, 7e: bijective, 7f: injective, not surjective, 9b: take V = W1 = W2 = {0, 1} and the identity maps f1 (x) = f2 (x) = x on V, 10a: injective, not surjective, 10b: injective, not surjective, 10c: surjective, not injective, 12: surjective, not injective, 13a: not injective, not surjective, 13b: bijective, 13c: not injective, not surjective, 13d: injective, not surjective, 13e: surjective, not injective 4.3 1: 64 and 256, 2a: yes, 2b: no, 3a: f (k) = 32 (k − 1), 3b: f (x) = 5x + 3, 3c: f (x) = arctan(x), x 3d: f (x) = 21 1+|x| + , 3e: f (x) = π1 arctan(x) + 12 , 4a: f (x, y) = (4y + 1, x + 2), 4b: f (x, y) = (x + y, y), 4c: f (x) = (cos(2πx), sin(2πx)) 4.4 1a: {(1, 1), (2, 1), (1, 2), (2, 2), (1, 3), (2, 3), }, 1b: f (a, n) = 2(n − 1) + a, 3a: f (1, 1) = 1, f (2, 1) = 3, f (3, 1) = 5, f (1, 2) = 2, f (2, 2) = 6, f (3, 2) = 10, f (1, 3) = 4, f (2, 3) = 12, f (3, 3) = 20, 3b: (4, 5), 3c: (1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2), (4, 1), (1, 4), (5, 1), (3, 2), 3e: |N2 | = |N| n Index algebra of sets, argument, 43 assignment symbol, associativity, banana formula, Bernstein, Felix, 49 bijection, 44, 45, 48 binomial coefficient, 21, 32, 34 binomial expansion, 33, 34 binomial theorem, 32 bridge (card game), 16, 20 Cantor, Georg, 48 cardinality, 49 Cartesian product, 40, 42 codomain, 44 collection, 13 combinations, 21 combinatorial principle, 37, 38 combinatorics, 37 commutativity, complement, 5, 7, relative —, complete preimage, 45 component, 42 coordinate, 38, 40–42 countable, 51 countably infinite, 51 De Morgan’s law, 8, 9, 12 De Morgan, Augustus, decreasing sequence, 12 Descartes, Ren´e, 41 description, 2, 16 dice, throwing —, 4, 15, 16, 37 difference set-theoretic —, disjoint, distributivity, 7, 13, 41 divisibility, 30 domain, 44 domination, draws, 15–27 types of —, 17 w/ replacement w/ order, 22, 23 w/ replacement w/o order, 24, 27 w/o replacement w/ order, 17–19 w/o replacement w/o order, 20, 21 element, 1, empty set, enumeration, 51 equality, 2, 40, 42, 44 equipotence, 48–50 Euclidean space, 42 even numbers, 1, 2, 51 event, 15 factorial, 19 family, 13 formula, 44 fractions, 1, 52 function, 43 one-to-one —, 45 Gauss’ trick, 29 Index Gauss, Carl Friedrich, 29 graph, 44 homogeneous probability model, 17 idempotency, identity, image, 43, 45 increasing sequence, 12 index set, 12 induction, 29–31 base case, 29 inductive step, 29 inductive assumption, 30 inductive hypothesis, 30 injection, 44, 45 integers, 1, 51 non-negative —, intersection, 4, 11, 12 intervals, inverse function, 46 inverse image, 45 involution, Leibniz formula, 36 Leibniz, Gottfried Wilhelm (von), 36 limit, 12 listing, 1, 3, 51 map or mapping, 43 one-to-one —, 45 mathematical induction, 29–31 base case, 29 inductive step, 29 monotony, 41 multinomial coefficient, 26 multinomial expansion, 35 multinomial theorem, 26, 35 musical chairs, 48 natural numbers, Newton’s binomial theorem, 32 Newton, Isaac, 32 ordered n-tuple, 42 ordered pair, 40 ordered sequence, 16, 40, 42 ordered triple, 42 part, 13 partition, 13, 14 58 Pascal’s triangle, 35 Pascal, Blaise, 35 permutations, 19 power set, 13, 49, 52 preimage, 43, 45 product set, 17, 40–43 proper subset, prototype, 2, 16 random variable, 43 range, 45 rational numbers, 1, 52 real numbers, 1, 52 non-negative —, relative complement, rule of product, 38 rule of sum, 37 sample space, 15 standard —, 15–27 w/ replacement w/ order, 22 w/ replacement w/o order, 24 w/o replacement w/ order, 18 w/o replacement w/o order, 20, 21 Schroder, Ernst, 49 ă SchroderBernstein theorem, 49 ă set, set-theoretic difference, singleton, 45 size, 48, 49 subset, proper —, surjection, 44, 45 symmetric difference, uncountable, 51, 52 uncountably infinite, 51, 52 union, 4, 11, 12 universe, 4, 15 value, 43 vector, 42 Venn diagram, Venn, John, ... which deals with the determination of the number of elements of all kinds of finite sets is called combinatorics In this section we will look more closely at the size of finite sets In principle,