Sets and Motivation for Boolean algebra

Transcription

1 SET THEORY Basic concepts Notations Subset Algebra of sets The power set Ordered pairs and Cartesian product Relations on sets Types of relations and their properties Relational matrix and the graph of a relation Partitions Equivalence relations Partial ordering Poset Hasse diagram Lattices and their properties Sub lattices Boolean algebra Homomorphism. 1

2 Sets and Motivation for Boolean algebra 1 Sets 2 Algebra of Sets 3 Cartesian Products 4 Motivation of Boolean Algebra Sets We devote this section to brief some set theoretic notions, which will play an essential role in the lattice theory. Set is a collection of well-defined objects. An object belonging to a set is called member or element of the set. In most cases, set will be defined by means of a characteristic property of the objects belonging to the set. That is, for a given property P(x), let {x : P(x)} denote the set of all objects x such that P(x) is true. A set with no member is said to be an empty set. We use upper case letters such as A, B, C, etc., to denote sets and lower case letters such as a, b, c, x, y, z, etc, to denote members of the sets. We denote the fact that x is an element of the set S by x S, while x is not an element of S by x S. Two sets A and B are equal if and only if A and B have the same members. Equality of A and B is denoted by A=B. We say that A is a subset of B if and only if every member of A is also a member of B. We write A B as an abbreviation for A is a subset of B. It is interesting to observe that, for all sets A, B and C, we have (i) A A [Reflexive] (ii) A B and B A if and only if A=B (iii) If A B, B C then A C [transitive] P(A) denote the set of all subset of A, we shall call P(A), the power set of A. That is, P(A) = {B : B A} 2

3 Example 1: Let A = {1,2,3} Then P(A) = {φ, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}. Let A be a finite set of n elements, say A= {a 1, a 2,, a n }, then P(A) consists of φ, the n sets {a i } containing single element, nc 2 sets {a i, a j / i j}containing two elements, etc. In general P(A) contains nc i subsets containing i distinct elements of A, for 1 i n. Therefore the number of elements in P(A) is 1+ nc 1 + nc nc n = (1+1) n = 2 n. Algebra of Sets In this section we define, union of two sets, intersection of two sets, complement of a set and we brief some of their basic properties and their operations. Given sets A and B, their union A B consists of all elements in A or B or both. That is, A B = {x / x A or x B}. It is interesting to observe that the union operation on sets has properties: i. A A = A (Idempotent) ii. A B = B A (Commutative) iii. A φ = A iv. (A B) C = A (B C) (Associative) v. A B = B if and only if A B vi. A (A B) and B (A B) Given sets A and B, their intersection A B consists of all objects which are in both A and B. Thus, A B = {x / x A and x B}.Two sets A and B are said to be disjoint if and only if A B = φ. The intersection operation on sets has the following evident properties: i. A A = A (Idempotent) ii. A B = B A (Commutative) iii. A φ = φ iv. (A B) C = A (B C) (Associative) v. A B = A if and only if A B vi. (A B) A and (A B) B 3

4 An important property connecting and is the distributive law. (i) A (B C) = (A B) (A B) (ii) A (B C) = (A B) (A C) Proof: A (B C) = {x / x A and x B C} = {x / x A and (x B or x C)} = {x / (x A and x B) or (x A and x C)} = {x / x A B or x A C} = (A B) (A C) Similarly, A (B C) = (A B) (A C) can be proved. It is very interesting to observe that A (A B) = A, for A (A B) and A A, therefore, A (A B) = A. Similarly, A (A B) = A, for (A B) A and A A, hence, A (A B) = A. Thus, for sets A and B, we have, A (A B) = A; A (A B) = A [Absorption law]. By the difference, B\A, of the sets B and A, we mean the set of all those objects in B which are not in A. Thus, B\A = {x : x B and x A}. The symmetric difference, A B, of sets A and B is the set (A\B) (B\A). Let X be given set. If we deal with subset of X then we call X is a universal set. If A X, then the complement Thus, = {x / x X and x A} of A is defined to be X\A. It is clear that whenever we use complements, it is assumed that we are dealing only with subset of some fixed universal set X. 4

5 The following properties can be easily verified: Cartesian Products In this section we define Cartesian product of sets. Cartesian product set A B of two arbitrary sets A and B is the set of all pairs (a, b) such that a A and b B. That is, A B = {(a,b) / a A and b B}. Note that the sets A and B need not be distinct in the Cartesian product. In the product A B, the element (a 1, b 1 ) and (a 2, b 2 ) are regarded as equal if and only if a 1 = a 2 and b 1 = b 2. Thus, if A consists of m elements a 1, a 2,, a m and B consists of n elements b 1, b 2,,b n, then A B consists of mn elements (a i, b j ), 1 i m, 1 j n. 5

6 Motivation of Boolean Algebra In this section we shall discuss the motivation of Boolean algebra. From the Section Algebra of Sets, it is interesting to observe that, if A is any set, then the binary operations, and on the set P(A) satisfy the following properties i. Commutative law, ii. Associative law, iii. Absorption law, iv. Idempotent property and v. Distributive law. and the uninary " - " operation on P(A) satisfies the De Morgan's law. So it is tempting to ask the question: "Are there other sets like P(A) and binary operations like and satisfying commutative law, associative law, absorption law, idempotent property and distributive law and uninary operation " - " satisfying De Morgan's law?". Indeed there are many such sets and such binary and uninary operations on such sets satisfying the above mentioned laws and property. In fact, one of the main motivations of Boolean Algebra is a search of such sets and understanding their algebraic structure. Try Yourself: 1. Give proofs for all the properties about the set inclusion " ", union of sets intersection of sets and complement of a set mentioned in the Sections 1.1and1.2. 6

7 Partially Ordered Set and Hasse Diagram 1 Relation and Poset 2 Hasse Diagram 3 Totally Ordered Set and Dual Poset 4 Extremal Elements of Partially Ordered Sets Relation and Poset In this section we define relation and special type of relation called reflexive, symmetric, anti-symmetric and transitive relation and we discuss examples on these relations. Further, we define partial ordering relation and posets and discuss some examples of posets. Let A and B the non-empty sets. A relation R from A to B is a subset of A B. Relation from A to A is called relation on A. If (a, b) R then we write arb and say that "a is in relation R to b". Also, if a is not in relation R to b, we write a b. Example: Let A = B = Z +, the set of all positive integers. The relation R is defined on Z + in the following way arb if and only if a divides b. So, 6 R 18, but 3 8 A relation R on a set A is reflexive if (a, a) for every a A, that is, if ara for all a A. Example: Let A= {1,2,3} and let R = {(1,1), (2,2), (3,3), (1,2) (1,3)} Then R is reflexive, while R' = {(1,1), (2,2), (2,1), (1,2)} is not a reflexive, for (3,3) R', i.e., 3 3. A relation R, on a set A, is anti-symmetric if whenever arb and bra, then a = b. That is, R is anti-symmetric whenever a b we must have either a b or b a. 7

8 Example 1: Let X be a set and let A X and B X. Then from the Section 1.1, it is clear that A = B. Therefore, " " is anti-symmetric on the set of subsets of X. Example 2: Let A=Z +, the set of all positive integer. Define R on A by arb if and only if a b that is "a divides b". Then, if a R b and b R a, i.e., a b and b a, then there exist integers c, d Z + such that b = ca and a = db. Therefore, b = cbd. So, 1 = cd. Therefore c = d = 1. Hence a = b. Therefore, R is anti-symmetric on Z + Example 3: Let A = Z, the set of integers and let R = {(a, b) A A / a < b}, i.e., R is usual "less than". If a b, then either a < b or b < a, that is, equivalently, if a b then either b a or a b. Hence,if a b then either a b or b a is true. Therefore, "<", usual "less than", is anti-symmetric. [Note that in the anti-symmetric relation symmetryness will happen only with equal elements and symmetryness will never happen between the unequal elements]. We say that a relation R on a set A is transitive if arb and brc then arc. Example: Let A={1,2,3,4} and let R={(1,2), (2,1), (1,1), (2,2), (2,3), (2,3)}. Then R is transitive. Let R={(1,2) (1,3), (4,2)}. Then R is also transitive, because we cannot find elements a, b, and c in A such that arb and brc, but a c. So, we conclude that R is transitive. A relation R on a set A is called a partial order relation or partial ordering if R is reflexive, anti-symmetric and transitive. A non-empty set A together with a partial order relation R, (A, R), is called partially ordered set or poset. Partial order relations are "hierarchical relations", usually we write instead of R for partial ordering. Thus, (A, ) denote a partially ordered set or poset. 8

9 Example 1: Let A be a non empty set. Consider the power set P(A), the set of subsets of A, together with the relation set inclusion, " ". Then (P(A), ) is a poset. For, S S, for all S P(A). Therefore, is reflexive. If S T and T S, for S, T P(A), then S = T (since S = T if and only if S T and T S). Therefore, is anti-symmetric. Also, if S T and T U, then by definition of, it is clear that S U. Therefore, is transitive. Hence, (P(A), ) is a partially ordered set. Example 2: Let A = Z + and let a b if and only if a b then (A, ) is a poset. For, since a = 1.a, a Z +, i.e., a a, a Z +. Therefore " " is reflexive. If a b and b c, then there exists d 1 and d 2 in Z +, such that b = d 1 a and c = d 2 b. So, we have c = d 2 d 1 a. As d 1, d 2 Z +, d 2 d 1 Z + Then a c. Hence, " " is transitive. Already we have seen that " " is anti-symmetric. Thus, " " is partial ordering. Hence, (A, ) is a poset. Example 3: (R, ) is a poset, where R is the set of all real number and " " is "usual less than or equal to " (Prove!). Example 4: (P(S), ) is not a poset, for, the relation is anti-symmetric and transitive but not reflexive so it is not a poset. Example 5: Let S be the set of all subgroups of a group G. Then (S, ) is a poset (Prove!) Example 6: Let S be the set of all normal subgroups of G. then (S, ) is a poset (Prove!). Example 7: Let S = {( i 1, i 2,,i n ) / i r = 0 or 1, 1 r n} Define (i 1,i 2,,i n ) R ( j 1,j 2,,j n ) if and only if i 1 j 1, i 2 j 2,,i n j n. Then (S,R) is a poset (Prove!). 9

10 Hasse Diagram In this section we discuss the diagrammatic representation of a poset. In a poset (A, ), if a b and a b then we write a < b. Further, in a poset (A, ), we say that a is a cover of b if a < b and there exists no u such that a < u < b. A finite poset can be represented graphically, such a diagram of a poset is called Hasse diagram. We represent the elements of A by points in the plane. If b is a cover of a then we place b above a and connect the two points a and b by a straight line (actually directing from a to b). Thus, if a < b if and only if there is a ascending broken line (a path) connecting a to b. If no line connecting a and b and a b, then a and b are not related or not comparable, that is, we have neither a b nor b a. Example: 1. Hasse diagram of the poset ({1,2,3,4,5}, ), where is "usual less than or equal to" is given below. 10

11 2. Consider the poset (P(A 3 ), ), where A 3 = {a, b, c}. The Hasse diagram of (P(A 3 ), ) is given below : Note that actually all the lines should have upward directions. Since all the lines are having only one direction, it is a convention to draw without direction in the lines. Thus, in the Hasse diagram every path has only upward direction, i.e., there cannot be any path connecting top to bottom in the downward direction. In the above example we see that {a} and {a, b, c} are related, since there is an ascending path from {a} to {a, b, c}, while {a} and {b} are not related or not comparable, since there is no ascending path connecting {a} and {b}. 3. Let D n denote the set of all positive divisiors of a positive integer n. Hasse diagram of the poset (D 12, ) is given below. 11

12 Conversely, if we have a Hasse diagram then we can describe the poset, that is, we can describe the partial order relation. For example, consider the Hasse diagram given below : Then the set is {a, b, c, d, e, f, g} and the relation is {(a, a) (a, c) (a, d), (a,e), (a, f), (a, g), (b, b), (b, c), (b, d), (g, e), (g, f), (b, g), (c, c), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g), (e, e), (e, g), (f, f), (f, g), (g, g)}. One can check that is indeed partial ordering. So, we conclude that for the finite poset it is enough to just give the Hasse diagram to describe the poset. Remark: From the above examples of the posets, we observe that in a poset it is not necessary that all the pair of elements are related or comparable, that is, in a poset not necessarily all the elements are ordered. That is the reason we call such sets (posets) as partially ordered set. Totally Ordered Set and Dual Poset In this section we define totally ordered set and dual poset and discuss about them. A partial order relation on a set A is called a total order (or linear order) if, for each a, b A either a b or b a. If is a totally order on a set A, then (A, ) is called totally ordered set or linearly ordered set or chain. 12

13 Example 1: The poset (N, ), where N is the set of all natural numbers and is the usual "less than or equal to", is a totally ordered set, since for any two a, b N, we have either a b or b a. In fact, ( Z, ), ( Q, ) ( R, ), are all totally ordered set, where Z, the set of all integers, Q, the set of all rational numbers and R, the set of all real number and is usual less than or equal to. Also, any finite subset of either Z or Q or R with usual "less than or equal to" are also totally ordered sets. Example 2: ( D 30, ) is not a totally ordered set (Prove!) Try yourself: Let L be the set of complex numbers z = x + iy, where x and y are rationals. Define a partial order " " on L by: x 1 + iy 1 x 2 + iy 2 if and only if y 1 y 2. Is ( L, ) a totally ordered set. If not, what is the additional condition needed in order to make ( L, ) into a chain? 13

14 Remark 1: From the definition of totally ordered set, it is clear that any two elements in a totally ordered set are related or comparable. That is, in a totally ordered set all the elements are ordered or related. That is the reason we call such sets as totally ordered set. If R is a relation from A to B then R -1 defined by (a, b) R -1 if and only if (b, a) R, is a relation from B to A, called the converse relation of R. Remark 2: If (A, R) is a partially ordered set then (A, R -1 ) is a partial ordered set. Proof: Let (A, R) be a poset. Then R is a partial ordering on A. Therefore, R is reflexive, antisymmetric and transitive. Now we shall prove that R -1 is reflexive, anti-symmetric and transitive. Since R is reflexive, we have, (a, a) R,for all a A. Therefore, (a, a) R -1, for all a A also. Thus, R -1 is reflexive. If ar -1 b and br -1 a, then by definition of R -1 we have bra and arb. Since R is anti-symmetric, we have a = b. Therefore, R -1 is anti-symmetric. Let ar -1 b and br -1 c. Then by definition of R -1, we have bra and crb. Since R is transitive, we have cra. Therefore, ar -1 c. Thus, R -1 is transitive. Hence (A, R -1 ) is a partially ordered set. Note: Since we use for denoting partial ordering we denote for its converse. Thus, if (A, ) is a partial ordered set then (A, ) is also partial ordered set. The poset (A, ) is called dual poset to the poset (A, ). Remark 3: (Duality Principle) Every statement or formula or expression on an ordered set (A, ) remains correct if everywhere in the statement the relation is replaced by its converse relation. 14

15 Extremal Elements of Partially Ordered Sets In this section we discuss about the extremal elements of a poset. Let (A, ) be a poset. An element a A is called a maximal element of A if there is no element c in A such that a < c. An element b A is called a minimal element of A if there is no element c in A such that c < b. An element a A is called a greatest element of A if x a, for all x A. Similarly, an element a A is called a least element of A if a x, for all x A. Example: Consider the following poset represented by the following Hasse diagram In this poset, a, f and i are minimal elements c, h, k are maximal elements, there is a no greatest element and there is no least element. 15

16 Consider the following posets represented by Hasse diagrams. In the poset (i), a is the least and minimal element and d is the greatest and maximal element. In the poset (ii), a is the least and minimal element and d and e are maximal elements but there is no greatest element. Remark: In a poset, least element or greatest element need not always exist. It is clear that least element is a minimal element and greatest element is a maximal element but not conversely. Try yourself: 1. Let (A, ) be a finite non empty poset. Then prove that A has at least one maximal element and at least one minimal element. 2. Let (A, ) be a poset. Then prove that if least element or greatest element exist then they are unique. 16

17 Let (A, ) be a poset and B A. i. a A is called an upper bound of B if and only if b a, for all b B. ii. a A is called a lower bound of B if and only if a b, for all b B. iii. An lower bound g of B is called a greatest lower bound or infimum if and only if h g for every lower bound h of B in A and it is denoted by inf B or GLB of B. iv. An upper bound v of B is called least upper bound or superimum if and only if v u for all upper bound u of B in A and it is denoted by sup B or LUB of B. Example: Consider the poset (D 30, ), i..e ({30, 15, 10, 6, 5, 3, 2, 1}, ). Let B = {2, 3, 6}. Then inf B = 1, upper bounds of B are 6 and 30 but supb = 6. Try yourself: 1.Let (A, ) be a poset and let B A. Prove that if inf B or sup B exist then they are unique. 2. On the set A = {a, b, c}, find all partial orders in which a b? 3. If (A, ) is a poset and A' is a subset of A. check whether (A', ' ) is also a poset, where ' is the restriction of to A. 4. Construct the Hasse diagram of (D 30, ) and (P(A 3 ), ), where A 3 = {1,2,3} Do they have structural similarity? 17

18 5. Find ( i ) all the lower bound of B. ( ii ) all the upper bound of B. ( iii ) the least upper bound of B. ( iv ) the greatest lower bound of B, where B = {c, d, e} in the poset represented by the following poset. 6. Whether every finite poset has maximal element? If yes, justify your answer 18

19 Lattices 1 Lattice Ordered Sets 2 Algebraic Lattice 3 Sub lattices, Direct Product and Homomorphism 3.1 Sub lattices 3.2 Direct Product of Lattices 3.3 Homomorphism 3.4 Special Lattices Isotone, Distributive and Modular Inequalities Modular Lattices Distributive Lattices Complemented Lattices Lattices In this section we introduce lattices as special type of partial ordered set and we discuss basic properties of lattices and some important type of special lattices. Lattice Ordered sets In this section we define lattice ordered sets and see some examples. A poset (L, ) is called lattice ordered set if for every pair of elements x, y L, the sup (x, y) and inf (x, y) exist in L. Example 1: Let S be a nonempty set. Then (P(S), ) is a lattice ordered set. For (P (S), ) is a poset. Further for any subsets A, B of S, inf (A, B) = A B P(S) and sup (A, B) = A B P(S). 19

20 Example 2: Every totally ordered set is a lattice ordered set (Prove!). Example 3: Consider the set of all positive integer Z + with divisor as a relation, i.e., a b if and only if a b.then (Z +, ) is a poset. For, if a, b Z +, then inf (a, b) = GCD(a, b) Z + and sup (a, b) = LCM(a, b) Z +. Thus, inf (a, b) and sup (a,b) exist in Z + for any two element a, b Z +. Hence (Z +, ) is a lattice ordered set. In fact (D n, ) ( D n denotes the set of all positive divisors of positive number n ) is also a lattice ordered set. Example 4: Consider the set B, where B n = {(l 1, l 2,, l n ) / l i = 0 or 1, for 1 r n}. Define the relation ' by (i 1, i 2,, i n ) ' (j 1, j 2,, j n ) if and only if i r j r, 1 r n. Note that here in the expression i r j r, is usual less than or equal to. We have already seen in Example 7 of Section that (B n, ') is a poset. Observe that inf [ (i 1, i 2,..,i n ), (j 1, j 2,, j n )] = (min (i 1, j 1 ), min (i 2,j 2 ),., min (i n, j n ) ) and sup [ (i 1, i 2,, i n ), (j 1, j 2,, j n )] = (max (i 1, j 1 ), max (i 2,j 2 ),., max (i n, j n ) ) Since min (i r, j r ) and max (i r, j r ) is either 0 or 1, so, inf { (i 1, i 2,, i n ), (j 1,j 2,..,j n ) } and sup { (i 1, i 2,, i n ), (j 1, j 2,, j n ) } exist in B. Thus, (B n, ) is a lattice ordered set. Example 5: Poset represented by the Hasse diagram is not a lattice ordered set since inf (a, b) does not exist. 20

21 Example 6: Poset represented by the Hasse diagram is not a lattice ordered set as sup (f, g) does not exist. Try yourself: 1. Prove that if (L, ) and (M, ' ) are lattice ordered sets. Then (L M, R) is a lattice ordered set, where (a, b) R (x, y) if and only if a x in L and b ' y in M. 2. Check whether the poset represented by the following Hasse diagram that is lattice ordered set or not? 21

22 Remark 1: Let (L, ) be a lattice ordered set and let x, y L. Then the following are equivalent. (i) x y (ii) sup (x, y) = y (iii) inf (x, y) = x Proof: ( i ) ( ii ) Let x y (1) We have, y y, for all y L. (2) From (1) and (2), we have y is an upper bound of (x, y). Therefore, sup (x, y ) y (by definition of superimum). But, y sup (x, y). Therefore, y = sup (x, y) (since is anti - symmetric). ( ii ) ( iii ) Given that sup (x, y) = y. Therefore we have x y. Also, we have x x. Therefore, x is a lower bound for x and y. Thus, x inf (x, y). But, we have inf (x, y) x. Hence, x = inf (x, y) ( since is anti-symmetric). ( iii ) ( i ) Given that x = inf (x, y). Therefore, by the definition of infimum, x y. Algebraic Lattice In this section we define algebraic lattice by using two binary operations * (meet) and (join). Then we shall prove that lattice ordered sets and algebraic lattices are equivalent. An algebraic lattice (L, *, ) is a non empty set L with two binary operations * (meet) and (join), which satisfy the following conditions for all x, y, z L. 22

23 L1. x * y = y * x, x y = y x (Commutative) L2. x * (y*z) = (x * y) * z, x (y z) = (x y) z (Associative) L3. x * (x y) = x, x (x * y) = x (Absorption) L4. x * x = x, x x = x (Idempotent) Theorem: Let (L, ) be a lattice ordered set. If we define x * y = inf (x, y), x y = sup (x, y) then (L, *, ) is an algebraic lattice. Proof: Given that (L, ) is a lattice ordered set and x * y = inf (x, y) and x y = sup (x, y). Now we shall prove that * and satisfy the commutative, associative, absorption and idempotent laws. Commutative x * y = inf (x, y) = inf (y, x) = y * x. x y = sup (x, y) = sup (y, x) = y x. Associative x * (y * z) = inf (x, (y * z)) = inf (x, inf (y,z)) = inf (x,y,z) = inf ( inf (x, y), z)) = inf ((x*y), z) = (x*y) *z. 23

24 Now, x (y z) = sup (x, (y z)) = sup (x, sup (y,z)) = sup (x, y,z) = sup (sup (x,y), z) = sup ((x y), z) = (x y) z. Absorption Now, x * (x y) = inf (x, x y) = inf (x, sup (x, y)) = x [since x sup (x, y)] and x (x * y) = sup (x, x * y) = sup (x, inf (x, y)) = x [since inf (x, y) x ]. Idempotent We have, x * x = inf (x, x) = x and x x = sup (x, x) = x. Hence (L, *, ) is an algebraic lattice. 24

25 Theorem: Let (L, *, ) be an algebraic lattice. If we define x y x * y = x or x y x y = y, then (L, ) is a lattice ordered set. Proof: Given that (L, *, ) is an algebraic lattice and x y x * y = x or x y = y. We shall now prove that (L, ) is a poset and inf (x, y) and sup (x, y) exist in L, for all x, y in L. is reflexive Since x * x = x, for all x L (by indempotent of *). We have by definition of, x x, for all x L. Therefore, is reflexive. is anti-symmetric If x y and y x in L, then by definition of, we have x * y = x and y * x = y. But * satisfies commutative law, so, we have x * y = y * x. Therefore, x = y. Hence, is anti-symmetric. is transitive If x y and y z, then by the definition of, we have x * y = x and y * z = y. Therefore, x * z = (x * y) * z = x * (y * z) [by associativity of *] = x * y [by definition )] = x [by definition of ] Hence, by definition of, we have x z. Thus, is transitive. 25

26 sup (x, y) and inf (x, y) exist in L We shall now show that inf (x, y) = x * y and sup (x, y) = x y. Now by absorption law, we have x = x * (x y) and y = y * (x y) Then by the definition of, we have x x y and y x y Therefore, x y is an upper bound for x and y. Let z be any upper bound for x and y in L. Then x z and y z. So, by definition of, we have x z = z and y z = z ( 1 ) Therefore, (x y) z = x (y z) [by associative law] = x z [by ( 1 )] = z [by ( 1 )]. Thus, by definition of, we have x y z. that is, x y is related to every upper bound of x and y. Hence sup (x, y) = x y. Similarly, we can show that inf (x, y) = x * y. Thus, x * y and x y exists for every x, y L. Hence, we have (L, ) is lattice ordered set. Remark 1: From the Theorem and the Theorem 3.2.2, we see that if we have lattice ordered set (L, ) then we can get an algebraic lattice (L, *, ) and conversely. Hence, we conclude that the algebraic lattice and a lattice ordered sets are equivalent system. Thus, here after we shall say simply lattice to mean both. In an algebraic system it is better convenience for imposing further conditions on the binary operations. Hence developing structural concepts will be much easier than the ordered system. In fact, it is one of the motivation to view a lattice ordered set as an algebraic lattice. 26

27 Sublattice, Direct Product and Homomorphism In this section we discuss sublattices of a lattice, direct product of two lattices and homomorphism between two lattices. Sublattices Substructure helps to know more about the whole structure. So, here we discuss about sublattice of a lattice. Let (L, *, ) be a lattice and let S be subset of L. The substructure (S, *, ) is a sublattice of (L, *, ) if and only if S is closed under both operations * and. Remark 1: A subset S in a lattice (L, *, ) is said to be sublattice, for a, b a b = d in L, then c, d must necessary exist in S also. S, if a * b = c in L and Example 1: Consider the lattice L represented by the following Hasse diagram. 27

28 Here the substructure S 1 represented by the Hasse diagram given below is not a sublattice, for inf (a, b) = 0 in L, which does not belong to S 1. It is clear that the substructure S 2 represented by the Hasse diagram given below is sublattice of L. It is interesting to note that the substructure S 3 of L represented by the in the following Hasse diagram is not a sublattice but it is a lattice on its own. So, it is a lattice without being a sublattice. 28

29 Try Yourself: 1. Find all the sub lattices of (D 30, ). 2. Let L be a lattice and let a < b. Then [a, b] is called an interval of the lattice, where [a, b] = {x L / a x b }. Prove that [a,b] is a sublattice. 29

30 Direct Product of Lattices From given two lattices, we can always construct a new lattice by taking Cartesian product of the given two lattices. So, in this section we discuss about product of two lattices. Let (L, *, ) and (M,, ) be two lattices. Consider the Cartesian product of L and M, that is, L M = {(x, y) / x L, y M}. Define operations and in L M, by (x, y) (a, b) = (x * a, y b) and (x, y) (a, b) = (x a, y b), then we shall prove that ( L M,, ) is a lattice. and are commutative. By definition (x, y) (a, b) = (x * a, y b) = (a * x, b y) (since * and are commutative) = (a, b) (x, y) (by definition ). Similarly, (x, y) (a, b) = (x a, y b) = (a x, b y) = (a, b) (x, y). Hence, commutative law holds good for both operations and. and are associative [ ( x, y) (a, b) ) (u, v)] = [ (x * a, y b) ] (u, v) (by definition of ) = [ (x * a ) * u, (y b) v] (again by definition of ) = [ x * (a * u), y (b v) ] (since * and are associative) = [ (x, y) (a * u, b v) ] (by definition of ) = [ (x, y) [ (a, b) (u, v) ] (by definition of ) Similarly we can show that [ (x, y) (a, b) ] (u, v) = (x, y) [ (a, b) (u, v) ] Thus, associative law hold good for both operations and in L M. 30

31 Absorption (x, y) [ (x, y) (a, b) ] = (x, y) [ x a, y b] = [ x * (x a), y (y b) ] (by definition of and ) = (x, y ) (by absorption law in L and M). Therefore, absorption law hold good in L M. Idempotent (x, y) (x, y) = [ x * x, y y] = (x, y) (since * and satisfy idempotent law) Similarly, (x, y) (x, y) = (x, y). Hence, (L M,, ) is an algebraic lattice. Thus, (L M,, ) is a lattice. Remark 1: If (L, *, ) is a lattice, then L² = L x L is a lattice. In general one can show that L n = L L L. L (n times) is a lattice. In the finite lattices, B n is a very important lattices, where B = {0,1}, which has rich structural property and will play very important role in the applications. Let (L, ) and (M, ' ) be two lattices, then we have already seen that (L M, R) where (x 1, y 1 ) R (x 2, y 2 ) if and only if x 1 x 2 and y 1 ' y 2, is a poset. It can be proved that (L M, R) is a lattice ordered set. [ Prove!] 31

32 Homomorphism In this section to understand the structural similarity between two lattices we define homomorphism between two lattices and we discuss about it. Let (L, *, ) and (M,, ) be two lattices. A function f : L M is called a Lattice homomorphism if f (a * b) = f (a) f (b) and f (a b) = f (a) f (b), for all a, b L, and it is called order preserving if x y in L implies f (x) ' f (y), where is an order relation in L and ' is an order relation in M. A bijecitve homomorphism is called isomorphism. Example 1: Consider the lattices D 6 = {1,2,3,6} and D 30 = {1,2,3,5,6,10,15,30}. We can show that there exist a homomorphism f between D 6 and D 30. Define a mapping f from D 6 into D 30 by f (1) = 1, f(2) = 6, f (3) = 15 and f (6) = 30. Then f (1 * 2) = f(1) = 1. f (1) * f(2) = 1 * 6 = 1. [Note that in both D 6 and D 30 a * b and a b are GCD and LCM of two element a, b] 32

33 Similarly, f (1 * 3) = f(1) = f(1) * f(3). f(1 * 6) = f(1) * f (6) = f(1). f(2 * 3) = f(2) * f(3) = f(1). f(2 * 6) = f(2) * f(6) = f(2). f(3 * 6) = f(3) * f(6) = f(3). Similarly we can prove that f (x y) = f (x) f (y) for all x, y D 6. Thus, f is homomorphism. Note that f is not onto but f is one-one. Hence f is not an isomorphism. Try yourself: It is very interesting to prove that the mapping f defined from B n into P(A n ), where A n = {1,2,., n} by f (i 1, i 2,., i n ) = { k / i k 0} is a homomorphism. 33

34 Special Lattices In this section we shall discuss some of the special types of lattices which in turn help to define the Boolean algebra. Isotone, Distributive and Modular Inequalities In this subsection we shall prove that in every lattice the operation * and are isotone and distributive and modular inequalities holds good. Lemma: In every lattice L the operation * and are isotone, i.e., if y z x * y x * z and x y x z. Proof: x * y = (x * x) * (y * z) (since x * x = x and since y z, y * z = y) = (x * y) * (x * z) (by associative and commutative of *). x * y x * z (since a b a * b = a). Also, x z = (x x) (y z) = (x y) (x z) x y x z. Hence the lemma. 34

35 Theorem: The elements of an arbitrary lattice satisfy the following inequalities i. x * (y z) (x * y) (x * z) x (y * z) (x y) * (x z) (distributive inequalities.) ii. x z x * (y z) (x * y) (x * z) = (x * y) z x z x (y * z) (x y) * (x z) = (x y) * x (modular inequalities.) Proof: Claim: Distributive inequalities are true in a lattice. By definition of, y y z and z y z. By isotone property we have x * y x * (y z)... (1) and x * z x * (y z)... (2) From (1) and (2) we observe that x * (y z) is an upper bound for x * y and x * z. Hence, (x * y) (x * z) x * (y z). By duality principle, we have (x y) * (x z) x (y * z). ii) It is given that z x, then by Theorem z * x = z. From the inequality ( i ) we have x* (y z) (x*y) (x z). Therefore, x * (y z) (x * y) z. Since x z, we have x z = z. Thus, from the inequality x (y * z) (x y) * (x z), we have x (y * z) (x y) * z. Hence the theorem. Modular Lattices In this section we shall define and discuss about the modular lattices. A lattice L is said to be modular if for all x, y, z L, x z x * (y z) = (x * y) z. Example 1: (P(A),, ) is a modular lattice [ Proof left as an exercise]. 35

36 Example 2: The set of all normal subgroups of a group form a modular lattice. [Recall that a subgroup H of a group G is said to be normal if ghg -1 = H, for all g G]. It can be easily shown that M, the set of normal subgroups of a group G, with `set inclusion' relation is a poset. Now for any two normal subgroups H 1 and H 2 of G, we have H 1 * H 2 = inf (H 1, H 2 ) = H 1 H 2 and H 1 H 2 = sup (H 1, H 2 ) = H 1 H 2. Since H 1 H 2 and H 1 H 2 are normal subgroups if both H 1 and H 2 are normal subgroups. Therefore, (M, ) is a lattice ordered set. Hence (M, *, ) is an algebraic lattice. Let H 1, H 2, H 3, M such that H 1 H 3. Since in every lattice modular inequality holds good, we have H 1 (H 2 * H 3 ) (H 1 H 2 ) * H 3. (1) Now we shall prove that (H 1 H 2 ) * H 3 H 1 (H 2 * H 3 ). Let a (H 1 H 2 ) * H 3 i.e., a (H 1 H 2 ) H 3 a H 1 H 2 and a H 3 a = h 1 h 2 and a = h 3, for some h 1 H 1,h 2 H 2 and h 3 H 3. Therefore, h 1 h 2 = h 3 h 2 = h 1-1 h 3 H 3 [ h 3 H 3 and h 1 H 1 H 3, therefore, h 1-1 h 3 H 3 ] h 2 H 2 and h 2 H 3 Thus, h 2 H 2 H 3 (2) Since a = h 1 h 2, we have a H 1 (H 2 H 3 ) (by (2)) That is, a H 1 (H 2 * H 3 ) That is, (H 1 H 2 ) * H 3 H 1 (H 2 * H 3 ). (3) From (1) and (3) we have H 1 (H 2 * H 3 ) = (H 1 H 2 ) * H 3. Hence (M, *, ) is a modular lattice. 36

37 Theorem: A lattice L is modular if and only if x, y, z L, x (y * (x z)) = (x y) * (x z). Proof : Let (L, *, ) be a modular lattice Then, if x z implies x (y * z) = (x y) * z (1) But, for all x, z L, x x z, So, by (1) we have x (y * (x z)) = (x y) * (x z), for all x, y, z L Conversely, suppose x (y * (x z)) = (x y) * (x z). ( 2 ) Then we shall prove that L is modular. If x z then x z = z. (3) Substitute (3) in (2), we have if x z x (y * z) = (x y) * z Thus, (L, *, ) is modular. If we have a characteristic result in terms of Hasse diagram for the modular lattice then it would effectively help in deciding a given lattice is modular or not. So, we shall prove the following lemma. Lemma : The "Pentagon Lattice" represented by the Hasse diagram given below is not modular. 37

38 Proof: From the structure of the pentagon lattices we see that c a. Now c (b * a) = c 0 = c. On the other hand, (c b) * a = 1 * a = a. Definitely c a, thus, pentagon lattice is not a modular lattice. On the other hand, it can be proved that if any lattice whose substructure is isomorphic to a pentagon lattice cannot be a modular lattice. So, we have the following theorem. Theorem: A lattice L is modular if and only if none of its sublattice is isomorphic to the "pentagon lattice" [ for the detailed proof refer [2]] Try yourself: Prove that the intervals [x, x y] and [ x * y, y] are isomorphic in a modular lattice. [ By interval [u,v] we mean the set { t L / u t v }] Try yourself: Prove that if a b and if there exists c L with a c = b c and a * c = b * c then a = b. Distributive Lattices In this section we will define and discuss distributive lattices. A lattice (L, *, ) is called a distributive lattice if for any a, b, c L, a * (b c) = (a * b) (a * c) a (b * c) = (a b)*(a c) Example 1: (P(A),, ) is a distributive lattice. [ For proof refer Section1.2] 38

39 Example 2: Every totally ordered set is a distributive lattice. Proof: In a totally ordered set (T, ) for any two elements a, b in T, we have either a b or b a. Therefore, for any three elements a, b, c in T, the following are the possible situations in the structure of (T, ). All these possible choices can be covered in the following two cases. Case 1: a b or a c [ the first four choices ] Case 2: a b and a c [ the last two choices ] For the case 1, we have, a*(b c) = a and (a * b) (a * c) = a a = a For the case 2, we have, a*(b c ) = b c and (a * b) (a * c) = b c Hence any totally ordered set (T, ) is a distributive lattice. 39

40 Example 3: The set of all positive integers Z +, ordered by divisibility is a distributive lattice. Proof: We know that (Z +, ) is lattice, where x * y = GCD(x, y) and x y = LCM(x, y), for x, y Z +. Since every positive integer can be expressed as product of powers of primes, Let x =, y = and z =. Note that x i, y i, z i may be zero also. Now, y * z = x (y * z) = = = * = (x y) * (x z) 40

41 Remark 1: It is clear from the definition of distributive lattice that if a c then a * c = c and a c = a. Therefore, a * (b c) = (a * b) (a * c) = (a * b) c. If a c then a * c = a and a c = c. Therefore we have a (b * c) = (a b) * (a c) = (a b) * c. Thus, every distributive lattice is modular. On the other hand every modular lattice need not be distributive. For example, the following modular lattice called diamond lattice is not distributive lattice. Diamond Lattice: For, a (b * c) = a 0 = a (a b) * (a c) = 1 * 1 = 1 Therefore, a (b * c) (a b) * (a c). Hence, a diamond lattice is not a distributive lattice. 41

42 It is interesting to observe that if a lattice is not modular it cannot be a distributive lattice. It follows from the Theorem and the above Remark1, we have the following theorem which will effectively decide the given lattice is distributive or not. Theorem: A lattice is distributive iff none of its sublattice is isomorphic to either the pentagon lattice or diamond lattice. [ For the proof refer [ 2 ] ] Try yourself: Prove that the direct product of two distributive lattices is a distributive lattice. Theorem : Let (L, *, ) be a distributive lattice. Then for any a, b, c L, a * b = a * c and a b = a c b = c [cancellation law]. Proof: (a * b) c = (a * c) c = c (since a * b = a * c and by absorption law, (a * c) c = c) Now, (a * b) c = (a c) * (b c) (by distributive law) = (a b) * (b c) (since a c = a b) = b (a * c) (by distributive law) Hence, b = c. = b (a * b) (since a * c = a * b) = b (by absorption law) Complemented Lattices In this section we shall define complemented lattices and discuss briefly. In a lattice (L, *, ), the greatest element of the lattice is denoted by 1 and the least element is denoted by 0. 42

43 If a lattice (L, *, ) has 0 and 1, then we have, x * 0 = 0, x 0 = x, x * 1 = x, x 1 = 1, for all x L. A lattice L with 0 and 1 is complemented if for each x in L there exists atleast one y L such that x * y = 0 and x y = 1 and such element y is called complement of x. Note: It is customary to denote complement of x by x'. Remark 1: It is clear that complement of 1 is 0 and vice versa. Example 1: Consider the lattice (P(A), ) Here 0 = φ and 1 = A. Then, for every S P(A), that is, S A, complement of S is A\ S, i.e. S'. Example 2: Consider the lattice L described in the Hasse diagram given below. Here, c does not have a complement. For, c * 0 = 0, but, c 0 = c. Therefore, 0 can not be a complement of c. Since a c = c, therefore, a can not be a complement of c. 43

44 Further, since c * b = c, b is not a complement of c. Also, c * 1 = c, 1 is not a complement of c. Hence, c does not have any complement in L. Therefore, L is not a complemented lattice. Example 3: In a totally ordered set, except 0 and 1, all the other elements do not have complements. Example 4: Consider the lattice L described by the Hasse diagram given below. Here, for c, we have, c * a = 0 and c a =1 and also, c * b = 0 and c b = 1. Thus, c has a and b as its complements. From this example, it is clear that complement of an element in a complemented lattice need not be unique. Theorem: In a distributive lattice L with 0 and 1, if a complement of an element exists then it is unique. 44

45 Proof: Let L be a distributive lattice with 0 and 1. Let a L. Suppose a' and a" be two complement of a. Then by the definition of complement of an element, we have a * a' = 0 and a a' = 1 a a" = 0 and a a" = 1 Therefore, a * a' = a * a" and a a' = a a". Therefore by cancellation law in a distributive lattice, we have a' = a". Thus, complement of an element in a distributive lattice is unique. Try yourself: 1. Is the poset A = {2, 3, 6, 12, 24, 36, 72} under the relation of divisibility a lattice. 2. If L 1 and L 2 are the lattices shown in the following figure, draw the Hasse diagram.of L 1 L 2 with product partial order. 3. Show that a subset of a totally ordered set is a sublattice. 4. Find all the sublattices of D 24 that contains at least five elements. 5. Shows that sublattice of a distributive lattice is distributive. 6. Show that the lattice given below is not a distributive lattice but modular. 45

46 7. Find the complement of each element of D Determine whether the lattice given below is distributive, complemented or both. 46

47 Boolean Algebra Boolean Algebra Boolean algebra is one of the most interesting and important algebraic structure which has significant applications in switching circuits, logic and many branches of computer science and engineering. Boolean algebra can be viewed as one of the special type of lattice. A complemented distributive lattice with 0 and 1 is called Boolean algebra. Generally Boolean algebra is denoted by (B, *,, ', 0, 1). Example 1 : ( P(A),,, ', φ, Α) is a Boolean algebra. This is an important example of Boolean algebra [In fact the basic properties of the (P (A),,, ' ) led to define the abstract concept of Boolean algebra]. Further, it can be proved that every finite Boolean algebra must be isomorphic to (P (A),,, ', φ, A) for a suitably chosen finite set A. [refer [2]]. Example 2: The structure ( B n = {0,1} n, *,, 1, 0 ) is a Boolean algebra, where B n is an n-fold Cartesian product of {0,1} and the operations *,, are defined below. We have B n = {( l 1, l 2,, l n) / l r = 0 or 1, 1 r n} (i 1, i 2,, i n ) * (j 1, j 2,, j n ) = (min (i 1, j 1 ), min (i 2, j 2 ),.., min (i n, j n ) ) (i 1, i 2,, i n ) (j 1, j 2,, j n ) = (max ( i 1, j 1 ), max ( i 2, j 2 ),, max ( i n, j n ) ) (l 1, l 2, l 3,, l n)' = (l 1', l 2',., l n'), 1 = (1,1,, 1) is the greatest element of B n. 0 = (0, 0,0,,0) is the least element of B n. Since B is distributive, B n is distributive. From the definition of unary operation " ' ", it is clear that B n is complemented. Further, it has 0 and 1. Thus, (B n, *,, ', 0,1) is a Boolean algebra. 47

48 For the case n = 3 we have, B 3 = {000, 100, 010, 001, 110, 101, 011, 111}. The structure of the B 3 is given in the following Hasse diagram. The Boolean algebra (B n, *,, ', 0,1) plays an important role in the construction of switching circuits, electronic circuits and other applications. Also it can be proved that every finite Boolean algebra is isomorphic to the above Boolean algebra (B n, *,, ', 0,1), for some n. Thus, it is interesting to observe that number of elements in any finite Boolean algebra must be always 2 n, for some n. Let (B, *,, ', 0,1) be a Boolean algebra and S B. If S contains the elements 0 and 1 and is closed under the operation *, and ' then (S, *,, ', 0,1) is called sub Boolean algebra. Example 1: Consider the Boolean algebra (P ({1,2,3}),,, ',φ, {1,2,3}) Then (S = {φ, {1}, {2,3}, {1,2,3}},,, ', φ, {1,2,3}) is also sub Boolean algebra. Similarly, S = ({φ,{3},{1,2},{1,2,3}},,, ', φ, {1,2,3}) is also sub Boolean algebra. 48

49 But (S = ({φ, {1}, {2,3}, {1,2,3}},,,', φ, {1,2,3})) is not a sub Boolean algebra. [Find why it is not a sub Boolean algebra]. If we have two Boolean algebras (B 1, * 1, 1, ', 0 1, 1 1 ) and (B 2, * 2, 2, '', 0 2, 1 2 ) then we can get new Boolean algebra by taking direct product of these two Boolean algebras. The direct product of these Boolean algebra is the Boolean algebra (B 1 B 2, * 3, 3, ''', 0 3, 1 3 ); where for any two elements (a 1, b 1 ) and (a 2, b 2 ) B 1 B 2, (a 1,b 1 ) * (a 2,b 2 ) = (a 1 * 1 a 2, b 1 * 2 b 2 ) (a 1,b 1 ) (a 2,b 2 ) = (a 1 1 a 2, b 1 2 b 2 ) (a 1,b 1 ) ''' = (a 1 ', b 1 '') 0 3 =(0 1,0 2 ) and 1 3 =(1 1,1 2 ). Let (B, *,, ', 0 B, 1 B ) and (A,,, -, 0 A, 1 A ) be two Boolean algebra. A mapping f : B A is called a Boolean homomorphism if f preserves all the Boolean operations, that is, f (a * b) = f (a) f (b). f (a b) = f (a) f (b). f (0 B ) = 0 A. f (1 B ) = 1 A. A bijective Boolean homomorphism is called Boolean isomorphism. 49

50 Try yourself: 1. Prove that mapping f defined in the exercise 1 of the Section is a Boolean isomorphism. 2. In every Boolean algebra, prove that (x * y)' = x' y' and (x y)' = x' * y', for all x, y. 3. In a Boolean algebra B, for all x, y B, prove that x y x * y = 0 x' y = 1 x * y = x x y = y. Try yourself: 1. Are there any Boolean algebra having three elements? Justify your answer? 2. Show that in a Boolean algebra, for any a and b, b * (a (a' * (b b' ))) = b? 3. Determine whether the posets given below are Boolean algebra? fig ( i ) fig ( ii ) fig ( iii ) 4. Show that in a Boolean algebra the following are equivalent for any a and b. ( i ) a' b = 1 ( ii ) a b' = The Boolean algebras (P (A 3 ),,, ', A 3, φ) and D 30, where A 3 = {1,2,3} are (Boolean) isomorphic? Justify your answer. 50