2MA105 Algebraic Structures I Per-Anders Svensson http://homepage.lnu.se/staff/psvmsi/2ma105.html Lecture 12
Partially Ordered Sets Lattices Bounded Lattices Distributive Lattices Complemented Lattices Boolean Algebras December 19, 2011 2(34)
Partially Ordered Sets Definition (Partial Ordering, Partially Ordered Set; Poset) A relation on a set X is called a partial ordering on X, if it is reflexive, i.e. a a for all a X, antisymmetric, i.e. if a b and b a imply a = b for all a,b X, and transitive, i.e. if a b and b c imply a c for all a,b,c X. We will sometimes write x y to point out that x y but x y. If is a partial ordering on X, then the pair (X, ) is called a partially ordered set (or poset, for short). December 19, 2011 3(34)
Example Properties of the divisibility relation make (Z +, ) into a poset, since for all positive integers a,b,c we have a a (reflexivity) a b and b a = a = b (antisymmetry) a b and b c = a c (transitivity). Example Let M be a set. Then the power set P(M ), i.e. the set of all subsets of M, is a poset with respect to. This is due to A A (reflexivity) A B and B A = A = B (antisymmetry) A B and B C = A C (transitivity) for all subsets A,B,C of M. December 19, 2011 4(34)
A finite poset X can be illustrated by a Hasse diagram; a graph, in which every node represents an element of X. If x,y X are such that x y, and if there are no elements z X such that x z y, then the graph has an edge between the two nodes that represent x and y. Furthermore, the node of y is placed higher up in the diagram, than the node of x (since x y). Example Let M = {a,b,c}. Then a Hasse diagram of the poset (P(M ), ) is {a,b,c} {a,b} {a,c} {b,c} {a} {b} {c} December 19, 2011 5(34)
We may use a Hasse diagram to define a partial ordering on a finite set. Example Let X = {a,b,c,d,e,f,g,h}. The Hasse diagram below defines a partial ordering on this set. a b c d e f g According to the diagram we have for instance e b and h a (we can travel from h to a, only by moving upwards in the Hasse diagram). However, e d, since when going from e to d, you must at some occasion move downwards in the diagram. h December 19, 2011 6(34)
Definition (Upper Bound, Supremum, Lower Bound, Infimum) Let (X, ) be a poset and Y a subset of X. An element s X is called an upper bound of Y, if y s for all y Y. If this s are such that s t for all upper bounds t of Y, then s is called a supremum of Y. An element i X is called a lower bound of Y, if i y for all y Y. If for this i, every lower bound j of Y fulfills j i, then i is called an infimum of Y. Theorem Let (X, ) be a poset and Y a subset of X. Then there are at most one supremum and at most one infimum of Y in X. Proof. Suppose s 1 and s 2 are two suprema of Y. Then s 1 and s 2 are upper bounds of Y. By regarding s 1 as a supremum and s 2 as an upper bound of Y, we obtain s 1 s 2. If we instead regard s 2 as a supremum and s 1 as an upper bound, we get s 2 s 1. Thus s 1 s 2 and s 2 s 1. Since is antisymmetric, s 1 = s 2 follows. The uniqueness of infimum is proved similarly. December 19, 2011 7(34)
Since supremum and infimum of a subset Y of a poset are uniquely determined, in case they exist, we may introduce the notations sup Y for the supremum of Y and infy for the infimum of Y. If Y = {y 1,y 2,...,y n } is finite, we will write sup(y 1,y 2,...,y n ) instead of sup({y 1,y 2,...,y n }), and similarly for inf. Example Let X = {a,b,c,d,e,f,g,h} and consider the poset (X, ), where is defined by the adjoining Hasse diagram. Let Y = {c,e,f }. (The elements of Y are written in red in the diagram.) There are two upper bounds of Y, namely a and c, since these elements are both larger than or equal to every element of Y. The least upper bound is c, so supy = c. There are no lower bounds of Y, and therefore no infimum. a b c d e g f h December 19, 2011 8(34)
Example Let M = {a,b,c} and consider the subset Y = { {a,b}, {b,c} } of the poset (P(M ), ). The upper bounds of Y are those elements in P(M ) that contain {a,b} as well as {b,c}. The only upper bound is M, so supy = M. An element of P(M ) is a lower bound of Y, if and only if it is a subset of both {a,b} and {b,c}. There are two such lower bounds: {b} and, so infy = {b}. {a,b,c} {a,b} {a,c} {b,c} {a} {b} {c} Note that supy = {a,b,c} = {a,b} {b,c} is the union and infy = {b} = {a,b} {b,c} is the intersection of the two members of Y. December 19, 2011 9(34)
Lattices A closer inspection of the previous example reveals that supremum and infimum of any pair (A,B) of subsets of M exist and are equal to A B and A B, respectively. This is true in (P(M ), ), no matter what set M is, i.e. we have sup(a,b) = A B and inf(a,b) = A B for all A,B P(M ): That inf(a,b) = A B is due to A B being the largest subset of M, that is a subset of both A and B; A B is the greatest lower bound of A and B. That sup(a,b) = A B is due to A B being the smallest subset of M, that contains both A and B; A B is the least upper bound of A and B. Definition (Lattice) Let L be a poset. If sup(a,b) and inf(a,b) exist for every a,b L, then L is said to be a lattice. Example Let M be a set. By the discussion above, (P(M ), ) is a lattice. December 19, 2011 10(34)
Example The poset (R, ) is a lattice, since for each pair (a,b) of real numbers, sup(a,b) and inf(a,b) exist and are given by max(a,b) and min(a,b), respectively. Example The poset (Z +, ) is also a lattice. To prove this: let a and b be two positive integers. Suppose c is a lower bound of a and b. Then c a and c b, i.e. c is a common divisor of a and b. If d = inf(a,b) exists, then on one hand d a and d b (i.e. d is a common divisor of a and b), and on the other hand c d (i.e. d is larger then any common divisor c, with respect to ). Hence inf(a,b) = gcd(a,b). By a similar argument, sup(a, b) = lcm(a, b). December 19, 2011 11(34)
Since supremum and infimum of a set are uniquely determined (once they exist), and since we in a lattice L require that supremum and infimum exist, for each pair of elements in L, we can in a natural way define two binary operations on L. Definition (Join, Meet) Let L be a lattice. The binary operations and on L, defined by a b = inf(a,b) and a b = sup(a,b) for all a,b L, are called the meet and join, respectively, of a and b. Example When considering the power set P(M ) as a lattice, the meet corresponds to intersection of sets, and the join corresponds to union of sets. December 19, 2011 12(34)
Example On the lattice (Z +, ) we have a b = gcd(a,b) and a b = lcm(a,b) (see the previous example). Hence we have, for instance, 3 (5 6) = 3 lcm(5,6) = 3 30 = gcd(3,30) = 3. Example The Hasse diagram to the right defines a lattice on the set X = {a 1,a 2,...,a 9 }. Using the diagram, we may compute e.g. a 2 a 3 = a 5, a 2 a 3 = a 1 a 5 a 6 = a 3, a 5 a 6 = a 9 a 7 a 6 = a 9, a 7 a 6 = a 3 a 3 a 8 = a 3, a 3 a 8 = a 8 a 2 (a 3 a 4 ) = a 2 a 1 = a 2 (a 2 a 3 ) (a 2 a 4 ) = a 5 a 9 = a 5 a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 December 19, 2011 13(34)
The following theorem establishes a connection between the binary operations and, and the partial ordering, and will therefore frequently be referred to. Theorem Let a and b be elements of a lattice L. Then a b a b = a a b = b. Corollary In a lattice, each element is idempotent with respect to both and, meaning that a a = a a = a for all a L. December 19, 2011 14(34)
Theorem Let L be a lattice. Then (i) and are commutative (ii) and are associative (iii) a (a b) = a and a (a b) = a for all a,b L, i.e. the so-called absorption laws are fulfilled. Given a purely algebraic structure (L,, ), where and are binary operations on L that satisfies the laws in the theorem above, we can on L define a partial ordering that makes (L, ) into a lattice. Theorem Let (L,, ) be an algebraic structure, such that and are commutative, associative, and fulfills the two absorption laws. Then the relation on L, defined by a b a b = a for all a,b L, is a partial ordering on L. For we have that sup(a,b) and inf(a,b) both exist, for all a,b L, and are equal to a b and a b, respectively. In other words, (L, ) is a lattice. December 19, 2011 15(34)
Bounded Lattices When dealing with sets, we recognize among other things, the following rules: Suppose A,B,C P(M ). Then (i) A (B C ) = (A B) C; A (B C ) = (A B) C (ii) A B = B A; A B = B A (iii) A M = A; A = A (iv) A = ; A M = M (v) A (A B) = A; A (A B) = A (vi) A A = M; A A = (vii) A (B C ) = (A B) (A C ); A (B C ) = (A B) (A C ). Here we recognize the associative laws (i), commutative laws (ii), and absorption laws (v), which together imply that (P(M ),, ) is a lattice. We now intend to generalize the remaining laws above, to any lattice (L,, ), starting with (iii) and (iv). December 19, 2011 16(34)
Definition (Universal Upper/Lower Bound) Let (X, ) be a poset. An element a X is called a universal upper bound of X, if x a for all x X. Similarly, if a x for all x X, we say that a is a universal lower bound of X. Universal upper and lower bounds are uniquely determined, in case they exist. Example In the power set P(M ) we have that M is the universal upper bound, since A M for all A P(M ). The universal lower bound of P(M ) is, since A for all A P(M ). a Example b In the poset defined by the adjoining Hasse diagram (taken from a previous example), a is the universal c d upper bound. There is no universal lower bound. (A reason that e.g. h is not the universal lower bound, is that h e.) e g f h December 19, 2011 17(34)
Definition (Bounded Lattice) A lattice L is said to be bounded, if it contains both universal upper and lower bounds. If this is the case, these elements are denoted 1 L (upper bound) and 0 L (lower bound). Theorem A lattice (L,, ) is bounded, if and only if it contains identity elements with respect to and, in shape of 1 L and 0 L, respectively. Example In the lattice (P(M ),, ) we have that M (the universal upper bound) is an identity element with respect to, while (the universal lower bound) is an identity element with respect to. December 19, 2011 18(34)
Example Consider the lattice (Z +, ). In an earlier example, we found that a b = lcm(a,b) and a b = gcd(a,b). According to the theorem above, an identity element with respect to should be the universal lower bound, which is 1, since 1 a for all a Z +. This is actually the case, since a 1 = lcm(a,1) = a for all a Z +. Since there is no universal upper bound in Z + (there is no b Z + such that a b for all a Z + ), there is no identity element with respect to. December 19, 2011 19(34)
Distributive Lattices Among the rules of arithmetics for union and intersection of sets, we find the distributive laws A (B C ) = (A B) (A C ); A (B C ) = (A B) (A C ). These are now to be generalized to arbitrarily lattices. Definition (Distributive Lattice) A lattice L is said to be distributive, if the distributive laws are fulfilled for any a,b,c L. a (b c) = (a b) (a c) a (b c) = (a b) (a c) Example (P(M ),, ) is a distributive lattice. December 19, 2011 20(34)
Example Suppose A,B,C P(M ). Is it then true that the cancellation laws holds, i.e. A B = A C = B = C (1) for, and for? A B = A C = B = C (2) Let M = {1,2,3,4,5}, A = {1,2,3}, B = {3,4}, and C = {3,5}. Then A B = A C = {3} but B C, which contradicts (1). (Find your own counterexample of (2).) But if we know that both A B = A C and A B = A C, then we must have B = C. December 19, 2011 21(34)
Theorem (Cancellation Law) Let a, b, and c be elements of a distributive lattice L. Suppose a b = a c and a b = a c. Then b = c. Proof. On one hand b = b (a b) while on the other hand, Absorption Law = b (a c) a b = a c = (b a) (b c) Distributive Law, c = c (a c) Absorption Law = c (a b) a b = a c = (c a) (c b) Distributive Law. But since a b = a c (and is commutative), the two right-hand sides above are equal. Hence b = c. December 19, 2011 22(34)
Complemented Lattices Among the different laws that are valid in the lattice (P(M ),, ), we have left to formulate A A = ; A A = M for general lattices. We note that this cannot be done for all possible lattices; in the formulae above we find M and, the universal upper and lower bounds of P(M ), which means that the lattice in question has to be bounded. Definition (Complement, Complementary Lattice) Let L be a bounded lattice and a an element of L. An element b L is called a complement of a, if a b = 0 L and a b = 1 L. If each element of L have a complement, we say that L is a complementary lattice. Example The power set P(M ) is a complementary lattice. Each A P(M ) has a uniquely determined complement, namely A = M \ A, since A A = och A A = M. December 19, 2011 23(34)
Example Let L = {1,2,3,4,6,12} be all positive divisors of 12. Then (L,, ) is a bounded lattice, if a b = gcd(a,b) and a b = lcm(a,b); we have 1 L = 12 and 0 L = 1, see the adjoining Hasse diagram. The only elements of L, having a complement, are 1, 3, 4, and 12; we see that 1 and 12 are each others complements, as well as 3 and 4. The elements 6 and 2 have no complements. For example 4 cannot be a complement of 6; certainly 6 4 = 12 = 1 L but on the other hand, 6 4 = 2 0 L. This lattice is thus not complementary. Example An element of a bounded lattice has not necessarily a uniquely determined complement. Consider e.g. the lattice defined by the Hasse diagram to the right. Here a has two complements: b and c. Is the lattice complementary? 12 6 4 3 2 1 a 1 L 0 L b c December 19, 2011 24(34)
As we saw in the previous example, an element of a bounded lattice has not necessarily a uniquely determined complement. What must be fulfilled for this to be the case? Theorem An element in a bounded lattice has at most one complement, whenever the lattice is distributive. Proof. Let L be a bounded distributive lattice. Suppose a L has two complements b and c. Then a b = a c = 0 L and a b = a c = 1 L. But in a distributive lattice, the Cancellation Law holds (i.e. if a b = a c and a b = a c, then b = c). Hence the two complements of a are equal. December 19, 2011 25(34)
Boolean Algebras Definition (Boolean Algebra) A Boolean algebra is a lattice that is both distributive and complementary. A Boolean algebra is a bounded lattice and therefore contains universal upper and lower bounds In a Boolean algebra, the Cancellation Law is valid Each element of a Boolean algebra has exactly one complement Sometimes we write (B,,,,0 B,1 B ) for a Boolean algebra. Here B is a set and are the two binary operations meet (a b = inf(a,b)) and join (a b = sup(a,b)) on B denotes the mapping on B that maps each element a B onto its uniquely determined complement (which we denote by a ) 0 B and 1 B are the universal lower and upper bounds of B. December 19, 2011 26(34)
Example The classical example of a Boolean algebra is the power set (P(M ),,,,,M). We have defined a Boolean algebra as a certain kind of poset. But we may also define a Boolean algebra as a purely algebraic structure: Theorem Let B be a set, furnished with two binary operations and such that (i) and are commutative (ii) and are associative (iii) and are distributive over one another (iv) B contains an identity element 1 B with respect to and an identity element 0 B with respect to (v) for each a B there is an a B such that a a = 0 B and a a = 1 B. Then the relation on B, defined by a b a b = a for all a,b B, is a partial ordering on B, and the poset (B, ) is a Boolean algebra. December 19, 2011 27(34)
From elementary set theory we recall De Morgan s Laws: (A B) = A B and (A B) = A B. De Morgan s Laws are thus valid in the Boolean algebra (P(M ),,,,,M). We will now show that it can be generalized to any Boolean algebra. Theorem (De Morgan s Laws) Let a,b B, where B is a Boolean algebra. Then (a b) = a b and (a b) = a b. December 19, 2011 28(34)
Proof. We will prove one of the two laws: (a b) = a b. In the left-hand side, we have the complement of a b. If we can show that a b also is a complement of a b, the proof will be finished, since each element of a Boolean algebra has a uniquely determined complement. To show that a b is a complement of a b, we have to show that (a b ) (a b) = 0 B and (a b ) (a b) = 1 B, by the definition of a complement. That (a b ) (a b) = 1 B follows from (a b ) (a b) = (a b a) (a b b) = (1 B b ) (a 1 B ) = 1 B 1 B = 1 B, and similar computations yields (a b ) (a b) = 0 B. December 19, 2011 29(34)
Example 105 Let B = {1,3,5,7,15,21,35,105} be the set of all positive divisors of 105. Defining meet and 35 join as a b = gcd(a,b) and a b = lcm(a,b), 21 15 respectively, B will be a bounded lattice (with 0 B = 1 and 1 B = 105). To the right we see a 7 5 3 Hasse diagram for this lattice. Is B a Boolean algebra? {a,b,c} 1 {a,b} {a,c} To the left is shown a Hasse diagram for the Boolean algebra P(M ), where M = {a,b,c}. {b,c} Except for how the nodes of the diagram are labelled, the diagram looks exactly the same {a} {b} {c} as the Hasse diagram for B. This indicates that B actually is a Boolean algebra; in fact a Boolean algebra that is isomorphic to P(M ). December 19, 2011 30(34)
Definition (Boolean Homomorphism, Boolean Isomorphism) Let B and C be Boolean algebras. A mapping φ: B C is called a Boolean homomorphism, if (i) φ(a b) = φ(a) φ(b) (ii) φ(a b) = φ(a) φ(b) (iii) φ(a ) = φ(a) If φ in addition is bijective, then φ is a Boolean isomorphism. The Boolean algebras B and C are then isomorphic; we write B C. Theorem Let φ : B C be a Boolean homomorphism. Then (i) If a,b B fulfills a b, then φ(a) φ(b) in C (ii) φ(0 B ) = 0 C and φ(1 B ) = 1 C. Unlike e.g. groups, there are not that many Boolean algebras, up to isomorphism. December 19, 2011 31(34)
Theorem Let B be a finite Boolean algebra. Then B P(M ) for some finite set M. Sketch of proof. Let {x 1,x 2,...,x n } be a non-empty subset of M. Then we can write {x 1,x 2,...,x n } = {x 1 } {x 2 } {x n } = n {x i }. (3) For each i, there is no set A P(M ) such that A {x i }, i.e. the sets {x i } are so small as possible, without being equal to the universal lower bound of P(M ). The representation (3) is also unique, if you not consider the succession of the sets {x i }. One can show that each b B, where b 0 B, has a unique representation in the form n b = a 1 a 2 a n = a i, (4) where each a i B is as small as possible, without being equal to 0 B, i.e. for each i there is no c B such that 0 B c a i. We say that a i is a atom of B. December 19, 2011 32(34) i=1 i=1
Let M be the set of all atoms in B. We define a mapping φ: B P(M ) in the following manner: Take any b B. If b = 0 B, we put φ(b) =. If b 0 B, we know that there is a uniquely determined set {a 1,a 2,...,a n } of atoms in B (i.e. a subset of M and thereby an element of P(M )) such that b = n i=1 a i. We put φ(b) = {a 1,a 2,...,a n }. One can now show that φ: B P(M ) is a Boolean isomorphism. Corollary Let B be a finite Boolean algebra. Then B = 2 n for some n N. Proof. We know that B P(M ) for some finite set M. Suppose M = n. When constructing an element of P(M ), i.e. a subset of M, we have 2 choices for each one of the n elements in M: either we include it in the subset we want to construct, or we do not. Due to this, there are 2 n possible subsets of M, whence B = P(M ) = 2 n. December 19, 2011 33(34)
Example In a previous example we concluded that the bounded lattice (L,, ), where L = {1,2,3,4,6,12} consists of all positive divisors of 12, and where 12 a b = gcd(a,b), a b = lcm(a,b), is not complementary (2 and 6 have no complements). Therefore L cannot be a Boolean algebra. Alternatively, we can make the same conclusion 6 4 about L, by counting the number of elements in L; we 3 2 have in this case L = 6. If L is a Boolean algebra, then L must be a power of 2, which is not the case. 1 Exercise Let B be the set of all positive divisors of 252, and construct a bounded lattice of B in the same spirit as in the example above. Is B a Boolean algebra? If this is the case, find a set M for which B P(M ). December 19, 2011 34(34)