VIEWS: 0 PAGES: 49 CATEGORY: Education POSTED ON: 10/3/2012
Unit -3 Relations Posets Lattices Boolean algebras Functions Groups Relations Cartesian Product: Cartesian Product of two sets A and B, written as A x B, is defined as A x B = { (a, b) a A and b B } If A ={a1 , a2} and B = { b1, b2, b3} then A x B = {(a1 , b1), (a1 , b2), (a1 , b3), (a2 , b1), (a2 , b2), (a2 , b3)} a1 b1 a2 b2 b3 Note: 1) In general, (A x B) (B x A) 2) If A x B = B x A then either A = B or A= or B = ) 3) If A = m, and B = n then A x B = m n = Cardinality of A x B Relations (Contd.,) Binary Relation: If A and B are two sets, then every sub set of A x B is called a relation from A to B . i.e., Every relation from A to B is a subset of A x B. If A = B , then we say R is a relation on A. We some times write, (x,y) R as x R y which reads ‘x relates y’ Inverse Relation: Let R be a relation from a set A to a set B. The inverse of R , denoted by R-1 is the relation from B to A which consists of those ordered pairs, which when reversed belongs to R. i.e , R-1 = { (b, a) (a, b) R} Relations (Contd.,) Equivalence Relation: Let R be relation on A. R is said to be an equivalence relation if the following conditions are satisfied. 1) x R x x A ( R is reflexive) i.e., (x , x) R x A 2) If x R y then y R x x, y A ( R is symmetric) i.e., if (x , y) R then (y , x) R for all x, y A 3) If x R y and y R z then x R z for all x, y, z A ( R is transitive) i.e., if (x , y) R and (y , z) R then (x , z) R for all x, y, z A Ex: If A = {1,2,3}, find A x A. Show that A x A is an equivalence relation on A. Solution: Given that A = {1,2,3} Let R = A x A = {1,2,3} x {1,2,3} = {(1,1),(1,2), (1,3),(2,1), (2,2),(2,3), (3,1),(3,2), (3,3)} Here, (x , x) R for all x A R is reflexive on A. Further, R is symmetric, because if (x , y) R then (y , x) R for all x, y A Finally, the relation is transitive,because if (x , y) R and (y , z) R then (x , z) R for all x, y, z A Hence, A x A is an equivalence relation on A Note: For any set A, AxA is the largest equivalence relation on A. Ex: Show that the relation R = {(x,y) (x – y) is an integer } is an equivalence relation on set of all real numbers. Let A = Set of all real numbers. Let x, y, z A Case 1 : Now, x – x = 0 and 0 is an integer i.e., (x , x) R for all x A R is reflexive. Case 2 : Let (x , y) R where x and y are any two real numbers. x – y is an integer y – x is an integer (y , x) R R is Symmetric. Contd., Case 3 : Let (x , y) R and (y , z) R . (x – y) is an integer and (y – z) is an integer. Now, (x – z ) = (x - y) + (y - z) is an integer. (x , z) R for all x,y,z A R is transitive Hence, R is an equivalence relation on A Ex: Let m be a positive integer greater than 1. Show that the relation R = { (x , y) x y (mod m)} is an equivalence relation on the set of integers. Proof : x y (mod m) iff (x - y) is divisible by m. Case 1: x – x = 0 is divisible by m. x x (mod m) (x , x) R for all x A R is reflexive. Case 2 : Let (x , y) R where x and y are any two integers. x y (mod m) (x - y) is divisible by m. (y - x) is divisible by m. y x (mod m) (y , x) R R is Symmetric. Contd., Case 3: Let (x , y) R and (y , z) R . x y (mod m) and y z (mod m). (x - y) is divisible by m and (y - z) is divisible by m. (x - y) = k1 m and (y - z) = k2 m where k1 and k2 are some integers. Now, x – z = (x – y) +(y - z). = k1 m + k2 m = (k1 + k2 ) m where (k1 + k2 ) is an integer. (x - z) is divisible by m. (x , z) R R is transitive. Hence, R is an equivalence relation on A. Ex: Show that the relation R = { (x , y) x -y is divisible by 5} is an equivalence relation on the set of all real numbers. Ex: If R is an equivalence relation on a set A, then show that R-1 is also an equivalence relation . Proof: Let R be an equivalence relation. R-1 = { (b,a) (a,b) R} Let x, y, z be any three elements of A. Case 1 : Now, (x , x) R ( Since, R is reflexive) (x , x) R-1 ( By the def. Of inverse relation) R-1 is reflexive. Case 2 : Let (x , y) R-1 (y , x) R ( By the def. Of inverse relation) (x , y) R ( Since, R is symmetric) (y , x) R-1 ( By the def. Of inverse relation) R-1 is Symmetric. Contd., Case 3: Let (x , y) R-1 and (y , z) R-1 (z , y) R and (y , x) R ( By the def. Of inverse relation) ( z , x ) R ( since R is transitive) ( x , z ) R-1 ( By the def. Of inverse relation) R-1 is transitive Hence, R-1 is an equivalence relation on A. More on relations Anti symmetric relation : A relation R on a set A is anti symmetric relation, if {a R b and b R a} then a = b i.e, whenever a R b and b R a then a = b Note: 1) The properties of being symmetric and being anti symmetric are not negatives of each other 2) For the sets A and B, A x B is called universal relation from A to B and is called the Empty relation from A to B. Irreflexive relation: A relation R on a set A is irreflexive, if (x , x) R for all x A Note: Any relation which is not reflexive, need not be irreflexive. Relations (Contd.,) Asymmetric relation: A relation R on a set A is said to be asymmetric, if (x , y) R then (y , x) R x, y A Diagonal Relation : Let A be any set. The diagonal relation on A consists of all ordered pairs (a , b) such that a = b i.e, A = {(a , a) a A} Complementary relation : If R is a relation from A to B then RC = {(a , b) (a , b) R} = (A x B) – R Partial ordering relation: A relation R on a set A is said to be a Partial ordering relation (or partial order), if R is reflexive, antisymmetric and transitive. Poset and Power set Partially ordered set: A set ‘A’ with a partial order R defined on it, is called a partially ordered set (or Poset). It is denoted by [A ; R]. Power set : If A is any set, then the set of all subsets of A is called Power set of A . It is denoted by P(A). Ex: If A = { a, b, c } then P(A) = { , { a }, { b }, {c}, {a , b}, {b, c}, {a, c}, A } Note: If a set A contains n elements, then its power set P(A) contains 2n elements. Ex: Show that the relation ‘less than or equal to’ on a set of real numbers is a partial ordering relation. Let R = The set of all real numbers. Let x, y, z are any three real numbers. (i) We know that, x x x R The relation is reflexive on R. (ii) Let x y and y x . x=y The relation is anti symmetric on R. (iii) Let x y and y z . x z The relation is transitive on R. Hence, is a Partial ordering relation on R. Ex: Show that the relation ‘greater than or equal to’ on a set of real numbers is a partial ordering relation. Ex: Show that the relation ‘divides’ on a set of positive integers is a partial ordering relation. ( ‘a divides b’ is denoted by ab ) Let A = The set of all positive integers. Let x, y, z are any three positive integers. (i) We know that, xx x A The relation is reflexive on A. (ii) Let xy and yx . x = y The relation is anti symmetric on A. (iii) Let x y and y z . xz The relation is transitive on A. Hence, is a Partial ordering relation on A. Ex: Show that the relation ‘set inclusion’ on a collection of sets is a partial ordering relation. Let S = a collection of sets. Let A, B, C are any three sets. (i) We know that, A A for all A S The relation is reflexive on S. (ii) Let A B and B A . A=B The relation is anti symmetric on S. (iii) Let A B and B C . AC The relation is transitive on S. Hence, is a Partial ordering relation on S. Ex: Let X be any non empty set and P(X) is power set of A. Show that [ P(X) ; ] is a poset. Try yourself. Ex:A relation R is defined over set of all integers as a R b b = ar for some positive integer r Show that the relation R is a partial ordering relation. Solution: Let A be set of all integers. (i) We know that, a = a1 for all a A aRa for all a A The relation R is reflexive on A. (ii) Let a R b and b R a . b = ar …(1) and a = bs …(2) where r and s are positive integers. a = (ar )s from (1) and (2) rs =1 r = 1 and s = 1 a=b from (1) and (2) Contd., The relation R is anti symmetric on A. iii) Let a R b and b R c. b = ar …(1) and c = bs …(2) where r and s are positive integers. c = (ar )s = ar s from (1) and (2) a Rc ( Since rs is a positive integer ) The relation R is transitive on A. Hence, R is a Partial ordering relation on A. Hasse diagram (Poset diagram) Poset Diagram (Hasse diagram) Let [A; R] be a finite poset . On a poset diagram i) There is a vertex for each element of A ii) All loops are omitted, thus eliminating the explicit representation of reflexive property. iii) An edge is not present in a poset diagram, if it is implied by transitivity of the relation. iv) An edge connects a vertex x to a vertex y y covers x i.e., x R y and there is no element z such that x R z and z R y Totally ordered sets Comparability: Two elements a and b in a set A are said to be comparable under the relation R, if either a R b or b R a . Otherwise, they are not comparable. If every pair of elements of a poset A are comparable, then we say [A ; R] is a Totally ordered set (or) Linearly ordered set (or) Chain . Note: If n is a positive integer, then Dn = set of all positive divisors of n. Ex: D12 = {1, 2, 3, 4, 6, 12} Ex: D6 = {1, 2, 3, 6} Ex : D8 = {1, 2, 4, 8} Note: If n is a positive integer, then [ Dn ; ] is a poset Examples: Ex: If A is any set of all real numbers, then the poset [A , ] is a totally ordered set. ( If x and y are any two real numbers, then either x y or yx ) Ex: If A is the set of all positive integers, then Show that the poset [A , ] is not a totally ordered set. Proof: For example, consider the positive integers 2 and 3 Here, ‘2 cannot divide 3’ and ‘3 cannot divide 2’ 2 and 3 are not comparable Hence, A is not a totally ordered set with respect to the relation . Ex : [ D8 ; ] = [ {1, 2, 4, 8} ; ] is a totally ordered set. Ex : [ D12 ; ] = [ {1, 2, 3,4, 6,12} ; ] is not a totally ordered set. Ex: Let A = {a , b} and P(A) is power set of A. Show that [ P(A) ; ] is not a totally ordered set. Solution: A = { a, b } then P(A) = { , { a }, { b }, {a , b} } Here, { a } is not a subset of { b } and { b } is not a subset of { a} { a } and { b } are not comparable. Hence, A is not a totally ordered set with respect to the relation . Least upper bound (Join) and Greatest lower bound (Meet) Least upper bound (lub) (Join or Supremum) The lub of two elements a and b is denoted by a b If c = a b then c satisfies i) c a and c b ( c is upper bound of a and b) ii) If d a and d b then d c ( i.e., c is lub of {a , b}) Greatest lower bound (glb) (Meet or infimum) The g.l.b of two elements a and b is denoted by a b If c = a b then c satisfies i) c a and c b ii) If d a and d b then d c lub and glb (contd.,) Both join and meet operations are commutative and associative In a poset, the lub( glb) of any two elements, if exists, is unique . Ex: In the poset [R ; ], where R is the set of all real numbers a b = max{a, b} and a b = min{a, b} Ex: For the poset [N ; ], where N is set of all positive integers a b = L.C.M of a and b a b = G.C.D of a and b Contd., Ex: If S is any collection of sets, then [S; ] is a poset. For any two sets A,B S The lub of A and B = A B and The glb of A and B = A B . Join semi lattice: A poset [A; R] in which each pair of elements a and b of A has a least upper bound is called a join semi lattice. Meet Semi Lattice: A poset [A; R] in which each pair of elements a and b of A has a glb (meet) is called ‘meet semi lattice’. Lattice Lattice: A lattice is a poset [A; R] in which each pair of elements has a lub and a glb. In other words, a lattice is both a join semi lattice and a meet semi lattice. A lattice is often denoted by [L, , ] The following laws hold in L i) a b = b a and a b = b a ii) (a b) c = a ( b c) and (a b) c = a (b c) iii) a (a b) = a and a (a b) = a (Absorption laws) iv) a a = a and aa=a (Idempotent laws) Theorem: Let [L, , ] be a lattice, then prove that ab = a ab =b Proof: Case(i) Let a b = a …(1) By Absorption law, we have a (a b) = a …(2) Since (2) is an identity, Interchanging a and b in(2), we have b (b a) = b …(3) Substituting (1) in (3), we have b a= b …(4) a b= b (By commutative law) Case(ii) Let a b = b …(5) By Absorption law, we have a (a b) = a …(6) From (5) and (6), we have ab = a (proved). Sub lattices and Bounded lattices Sub Lattice : Suppose M is a non empty subset of a lattice L. We say M is a sub lattice of L, if M is a lattice by itself. Bounded lattice: A lattice L is said to have lower bound O if O x x L. Similarly L is said to have an upper bound I , if x I for all x L We say L is bounded if L has both a lower bound O and upper bound I. Note : In a bounded lattice, for each element a L we have aI = I, aI = a, aO= a, a O = O. Note: Every finite lattice L is bounded. Distributive lattice and Complemented lattice Distributive Lattice: A lattice (L, , ) is said to be distributive if the following distributive Laws hold a, b, c L . a (b c) = (a b) (a c) a (b c) = (a b) (a c) Complement : Let L be a bounded lattice with lower bound O and upper bound I . Let a be an element of L . An element x in L is called a complement of a , if a x = I and a I = O Note: In a lattice, complement of an element need not exist and need not be unique. Def: A lattice L is said to be a Complemented lattice, if L is bounded and every element in L has a complement. Boolean Algebra Note: Let L be a bounded distributive lattice. Then complements are unique if they exist. Boolean Algebra (Boolean Lattice) A Lattice which is both distributive and complemented is called a Boolean Algebra. Maximal Element: An element of a poset which is not related to any other element of the poset. Minimal Element: An element of a poset to which no other element of the poset is related. More on lattices 1) Every finite non empty poset has a minimal (maximal)element. 2) If a lattice has a universal lower bound ( universal upper bound ) it is unique. 3) Every totally ordered set is a distributive lattice. 4) Every finite totally ordered set has a least element and a greatest element Examples Ex: Let A = {2, 5, 9, 14} . Draw the Hasse diagram for the poset [ A ; ] Ex. Let X = {2, 3, 6, 12, 24, 36}. Draw the Hasse diagram for the poset [ X ; ] . Ex. Let A = {1, 2, 3, 4, 6, 9}. Draw the Hasse diagram for the poset [ A ; ] . Ex. Let A = { 2, 3, 4, 9, 12, 18}. Draw the Hasse diagram for the poset [ A ; ] . Examples Ex: Let A = {2, 3, 5, 30, 60, 120, 180, 360} . Draw the Hasse diagram for the poset [ A ; ] Ex. Draw the Hasse diagram for the poset [ D6 ; ] Ex. Draw the Hasse diagram for the poset [ D8 ; ] Ex. Draw the Hasse diagram for the poset [ D12 ; ] Ex. Draw the Hasse diagram for the poset [ D30 ; ] Ex. Draw the Hasse diagram for the poset [ D45 ; ] Ex. Draw the Hasse diagram for the poset [ D210 ; ] Ex. Let A be a given finite set and P(A) its power set. Draw the Hasse diagrams of [P(A) ; ] for (a) A = {a}; b) A = {a,b}; (c) A = {a, b, c} (d) A = {a,b,c,d} Solution: (a) A = {a} P(A) = { , {a} } • {a} • (b) P(A) = { , {a}, {b}, {a,b} } * {a, b} {a} * * {b} * Contd., (c) A = {a, b, c} P(A) = { , {a}, {b},{c},{a,b},{b,c},{c,a} {a,b,c} } A * * {a,b} * {a,c} *{b,c} *{a} * {b} *{c} * Ex. Let C be a collection of sets which are closed under intersection and union. Verify whether (C,,) is a lattice. Solution: We know that [C ; ] is a poset , Because the relation is reflexive, anti symmetric and transitive on C. Let A and B are any two sets in C Least upper bound of A and B = A B and A B C (Since C is closed under union) Greatest lower bound of A and B = A B and A B C (Since C is closed under intersection) For any two elements in C , the lub and glb exists. Hence, (C, , ) is a lattice. Ex: Show that, the poset [P(S) ; ] is a lattice where S is a finite set. Try yourself Ex. Show that, The poset [Z+, ] a lattice where Z+ is set of all positive integers. Solution: We know that [Z+ ; ] is a poset, Because the relation is reflexive,anti symmetric and transitive on Z+ . Let a and b are any two positive integers Now a b = (LCM of a and b) Z+ and a b = (GCD of a and b) Z+ For any two elements in Z+, the lub and glb exists. Hence, [Z+ ; ] is a lattice. _ Ex: In a distributive lattice if b c = O then b c Let b,c L where L is a distributive lattice. _ Given, b c = O _ Now, (b c ) c = O c _ (b c ) (c c ) = c (b c ) I = c (b c ) = c bc Ex: Show that the lattice L1 given below is not distributive I * a * *b * c *O Consider, a (b c) = (a b) (a c) a O = I I a = I which is not true. Distributive law does not hold good. Hence, the given lattice is not distributive. Ex: Show that the lattice L2 given below is not distributive I * a * c * *b * O Consider, a (b c) = (a b) (a c) aI = O c a = c Which is not true. Distributive law does not hold good. Hence, the given lattice is not distributive. Theorem: A lattice L is not distributive iff L has a sub lattice isomorphic to L1 or L2 ( Where L1 and L2 are lattices given in the previous two examples.) Ex: Draw Hasse diagram for the poset [D12 ; ]. Find complements of each element of the poset. Is the lattice Distributive? Justify your answer. Solution : [D12 ; ] = [{1,2,3,4,6,12}; ] * 12 4* *6 2* *3 *1 Contd., Complement of 1 = 12 Complement of 3 = 4 Complement of 4 = 3 Complement of 12 = 1 Complements of 2 and 6 do not exist. Here, the complement of each element is unique whenever it exists. Further, the given lattice does not have any sub lattice isomorphic to L1 or L2 Hence, the lattice [D12 ; ] is distributive.