Abstract Algebra Theory and Applications - Computer Science ...

304 CHAPTER 17 LATTICES AND BOOLEAN ALGEBRASLemma 17.10 Let B be a Boolean algebra and b and c be elements in Bsuch that b ⋠ c. Then there exists an atom a ∈ B such that a ≼ b and a ⋠ c.Proof. By Lemma 17.9, b ∧ c ′ ≠ O. Hence, there exists an atom a suchthat a ≼ b ∧ c ′ . Consequently, a ≼ b and a ⋠ c.□Lemma 17.11 Let b ∈ B and a 1 , . . . , a n be the atoms of B such that a i ≼ b.Then b = a 1 ∨· · ·∨a n . Furthermore, if a, a 1 , . . . , a n are atoms of B such thata ≼ b, a i ≼ b, and b = a ∨ a 1 ∨ · · · ∨ a n , then a = a i for some i = 1, . . . , n.Proof. Let b 1 = a 1 ∨ · · · ∨ a n . Since a i ≼ b for each i, we know that b 1 ≼ b.If we can show that b ≼ b 1 , then the lemma is true by antisymmetry. Assumeb ⋠ b 1 . Then there exists an atom a such that a ≼ b and a ⋠ b 1 . Since a isan atom and a ≼ b, we can deduce that a = a i for some a i . However, this isimpossible since a ≼ b 1 . Therefore, b ≼ b 1 .Now suppose that b = a 1 ∨ · · · ∨ a n . If a is an atom less than b,a = a ∧ b = a ∧ (a 1 ∨ · · · ∨ a n ) = (a ∧ a 1 ) ∨ · · · ∨ (a ∧ a n ).But each term is O or a with a ∧ a i occurring for only one a i . Hence, byLemma 17.8, a = a i for some i.□Theorem 17.12 Let B be a finite Boolean algebra. Then there exists a setX such that B is isomorphic to P(X).Proof. We will show that B is isomorphic to P(X), where X is the setof atoms of B. Let a ∈ B. By Lemma 17.11, we can write a uniquely asa = a 1 ∨ · · · ∨ a n for a 1 , . . . , a n ∈ X. Consequently, we can define a mapφ : B → P(X) byφ(a) = φ(a 1 ∨ · · · ∨ a n ) = {a 1 , . . . , a n }.Clearly, φ is onto.Now let a = a 1 ∨· · ·∨a n and b = b 1 ∨· · ·∨b m be elements in B, where eacha i and each b i is an atom. If φ(a) = φ(b), then {a 1 , . . . , a n } = {b 1 , . . . , b m }and a = b. Consequently, φ is injective.The join of a and b is preserved by φ sinceφ(a ∨ b) = φ(a 1 ∨ · · · ∨ a n ∨ b 1 ∨ · · · ∨ b m )= {a 1 , . . . , a n , b 1 , . . . , b m }= {a 1 , . . . , a n } ∪ {b 1 , . . . , b m }= φ(a 1 ∨ · · · ∨ a n ) ∪ φ(b 1 ∧ · · · ∨ b m )= φ(a) ∪ φ(b).