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					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 ab )

 Let A = The set of all positive integers.
       Let x, y, z are any three positive integers.
  (i) We know that, xx           x  A
  The relation  is reflexive on A.
  (ii) Let xy and yx .
          x = y
 The relation  is anti symmetric on A.
  (iii) Let x  y and y  z .
          xz
 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 .
         AC
 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
    yx )
   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       aa=a                  (Idempotent laws)
    Theorem: Let [L,  , ] be a lattice, then
          prove that    ab = a  ab =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         ab = 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
              aI = I,            aI = a,
              aO= 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
           bc
    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)
                aI           = 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.

				
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