Theorem 3P ≡ Elements of Set Theory Let be an equivalence relation on a nonempty set A. The quotient set A=≡ = f[x]: x 2 Ag is a partition of A. Chapter 3: Relations and Functions Proof. 1 For any x; y 2 A, by Proposition 3.13 (iii), January 28 & 30, 2014 [x] 6= [y] =) [x] \ [y] = ;: [ 2 A = [x]. Indeed, [x] = fa 2 A : x ≡ ag⊆ A for any x 2 A, which x2A [ Teacher: Fan Yang leads to [x] ⊆ A. x2A [ On the other hand, A ⊆ [x] since x2A [ [ y 2 A =) y 2 [y] ⊆ [x] =) y 2 [x]: x2A x2A 1/36 2/36 The converse of the preceding theorem also holds: Theorem 3.16 Let Π be a partition of a nonempty set A. The relation ≡ defined as follows is an equivalence relation on A: for any x; y 2 A, x ≡ y () there exists X 2 Π such that x; y 2 X: X Y Functions Z Proof. By definition, ≡ is clearly reflexive and symmetric. It remains to show that ≡ is transitive. For any x; y; z 2 A such that x ≡ y and y ≡ z, by the definition of ≡, there exist X; Y 2 Π such that x; y 2 X and y; z 2 Y : Hence y 2 X \ Y 6= ;, which implies X = Y as Π is a partition. It then follows that x; z 2 X = Y , thereby x ≡ z. 3/36 4/36 (x1; x2) (y1; y2) (z1; z2) Definition 3.17 A function is a binary relation F such that for each x 2dom F, there is The squaring function f : R ! R is defined as f (x) = x2. The action of a unique y such that (x; y) 2 F. f on particular elements of R can be described by writing p The unique y is called the value of F at x, denoted byF (x). −2 7! 4; 2 7! 2; 2 7! 4; 3 7! 9; etc. If dom F = X and ran F ⊆ Y , then we write F : X ! Y and say that F is Each individual action can be represented by an ordered pair: a function from X into Y , or that F maps X into Y. p (−2; 4); ( 2; 2); (2; 4); (3; 9); etc. X Y X Y The set y1 y1 2 F = f(x; y) 2 R × R j y = x g x1 y2 x1 y2 x2 y3 x2 y3 of all such ordered pairs adequately represents the squaring function x3 y4 x3 y4 f , and is called the graph of f . In set theory, we simply take this set F y5 y5 of ordered pairs to be the function f . F R Clearly, the set F ⊆ R × R is a relation such that for any x 2dom F, there is a unique y such that (x; y) 2 F. For example: The relation F = f(x; y) 2 R × R j y = x2g is a function, + dom F = R and ran F = R [ f0g. We write F : R ! R. The relation R = f(x; y) 2 R × R j x2 + y 2 = 4g is not a function, 5/36 since e.g., (0; 2); (0; −2) 2 R. 6/36 The domain of a function can be a set of ordered pairs or n-tuples. For Definition 3.19 example, addition is a function + : R × R ! R. In this case, in place of Let F : X ! Y be a function from X into Y . +((a; b)), we write either +(a; b) or a + b. If ran F = Y , then we say that F is surjective or F is a function from X ontoY. Fact 3.18 If for any x ; x 2 X, If F and G are functions for which dom F = dom G and 1 2 x1 6= x2 =) F(x1) 6= F(x2); F(x) = G(x) then we say that F is one-to-one or injective. for every x 2 dom F, then F = G. If F is both surjective and injective, then F is called a bijection. Proof. Indeed, by the definition of functions, we have that X Y X Y X Y y (x; y) 2 F () x 2 dom F and y = F(x) 1 x1 y21 x1 y1 x1 y1 () x 2 dom G and y = G(x) (by assumption) x2 y3 x2 y2 x2 y2 x y x y x y () (x; y) 2 G: 3 42 3 3 3 3 y5 y4 F F F Hence F = G. surjective injective bijection 7/36 8/36 (x1; x2) (y1; y2) (z1; z2) Note: A function F : X ! Y is surjective iff for each y 2 Y , there exists Definition 3.20 x 2 X such that F(x) = y. A relation R is said to be single-rooted iff for each y 2 ranR, there is only one x such that xRy. For example: Consider the function F : R ! R, defined as F(x) = x2. X Y + y1 F is not surjective, since ran F = R [ f0g 6= R. x1 y2 F is not injective, since e.g. F(2) = 4 = F(−2). x2 y3 x3 y4 3 Consider the function G : R ! R, defined as G(x) = x . y p 5 G is surjective, since for each y 2 R, we have that G( 3 y) = y. R 3 3 not single-rooted G is injective, since G(x1) = G(x2) =) x1 = x2 =) x1 = x2. Thus G is a bijection. Clearly, if F is a function, then F is single-rooted iff F is one-to-one. 9/36 10/36 Theorem 3K Definition 3.21 Let F be a relation. Let F be an arbitrary relation, and A a set. (a)F [A [ B] = F[A] [ F[B]. The restriction of F to A is the set (b)F [A \ B] ⊆ F[A] \ F[B]. Equality holds if F is single-rooted. F A = f(x; y) j xFy ^ x 2 Ag: (b)F [A] − F[B] ⊆ F[A − B]. Equality holds if F is single-rooted. The image of A under F is the set F[A] = ran (F A) = fy j 9x 2 A(xFy)g: X Y x y In particular, if F is a function, then F[A] = fF(x) j x 2 Ag. 1 1 y2 x2 y X Y X Y 3 a0 a0 x3 y4 0 a1 0 a1 F 1 a2 1 a2 2 a3 2 a3 Proof. (a). y 2 F[A [ B] () 9x 2 A [ B s.t. xFy a a 3 4 3 4 () (9x 2 A s.t. xFy ) or (9x 2 B s.t. xFy) F F () y 2 F[A] or y 2 F[B] () y 2 F[A] [ F[B]: 11/36 12/36 Definition 3.22 Let F be a relation. The inverse of F is the set (b). y 2 F[A \ B] () 9x 2 A \ B s.t. xFy F −1 = f(y; x) j xFyg: =) (9x 2 A s.t. xFy) and (9x0 2 B s.t. x0Fy) () y 2 F[A] and y 2 F[B] Note: (F −1)−1 = F () y 2 F[A] \ F[B]: In general, even if F is a function, F −1 is not necessarily a If F is single-rooted, then in the second line of the above expression, function, but F −1 is a single-rooted relation, i.e., for any x = x0, which means that the second arrow is reversible. x 2 ran F −1, there is only one y such that yF −1x. (c). y 2 F[A] − F[B] () y 2 F[A] and y 2= F[B] Since the inverse of a function is always single-rooted, the following () (9x 2 A s.t. xFy) and (:9x 2 B(xFy)) corollary is an immediate consequence of the Theorem 3K. =) 9x 2 A − B s.t. xFy Corollary 3L () y 2 F[A − B]: For any function G and sets A, B: If F is single-rooted, then the thrid arrow is reversible. (a)G −1[A [ B] = G−1[A] [ G−1[B]. (b)G −1[A \ B] = G−1[A] \ G−1[B]. (b)G −1[A] − G−1[B] = G−1[A − B] 13/36 14/36 0 Definition 3.23 Since F is a function, we derive that y = y . Furthermore, since G is a function, yGz and yGz0 implies that z = z0, as desired. Let F; G be arbitrary relations. The composition of F and G is the set G ◦ F = f(x; z) j 9y(xFy ^ yGz)g: For any x 2 X, we have that (x; F(x)) 2 F and (F(x); G(F(x))) 2 G, thus (x; G(F(x))) 2 G ◦ F, which implies that G ◦ F(x) = G(F(x)), as X Y Z G ◦ F is a function. y1 z1 x1 z2 Theorem 3I x2 y2 z3 x3 For any relations F and G, y3 z4 (G ◦ F)−1 = F −1 ◦ G−1: F G Proof. For any ordered pair (z; x), Theorem 3H If F : X ! Y and G : Y ! Z are functions, then G ◦ F : X ! Z is a (z; x) 2 (G ◦ F)−1 () (x; z) 2 G ◦ F function, and G ◦ F(x) = G(F(x)) for all x 2 X. () xFy ^ yGz; for some y − − Proof. We first check that G ◦ F is a function. Suppose that x(G ◦ F)z () yF 1x ^ zG 1y; for some y 0 0 and x(G ◦ F)z .