International Journal of Mathematical Archive-5(6), 2014, Available online through ISSN

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1 International Journal of Mathematical Archive-5(6), 2014, Available online through ISSN CLOSURE OPERATORS ON COMPLETE ALMOST DISTRIBUTIVE LATTICES-I G. C. Rao Department of Mathematics, Andhra University, Visakhapatnam, Andhra Pradesh, India Venugopalam Undurthi* Department of Mathematics, Andhra University, Visakhapatnam, Andhra Pradesh, India (Received on: ; Revised & Accepted on: ) ABSTRACT In this paper, we introduce the concept of a closure operator, dual atom, meet-distributive closure operator, joindistributive closure operator, distributive closure operator and derive their important properties. AMS 2000 subject Classification: 06D99, 06E15, 06B23. Keywords: Almost Distributive Lattice, Complete Almost Distributive Lattice, Closure Operator, Dual atom, Meet- Distributive Closure Operator, Join-Distributive Closure Operator, Distributive Closure Operator. 1. INTRODUCTION In [7], Swamy and Rao introduced the concept of an Almost Distributive Lattice (ADL) as a common abstraction of most of the existing lattice theoretic and ring theoretic generalizations of a Boolean algebra and they observed that the set PI (L) of all principal ideals of an ADL L = (L,,, 0, m) with a maximal element m, forms a bounded distributive lattice. Through this lattice PI(L) many concepts existing in the class of distributive lattices were extended to the class of ADL by different authors. Prominent among Them are Pseudo-Complementation on ADLs [4], Stone ADLs [8], Normal ADLs [5], Heyting ADLs [3]. In our paper [6], we have introduced the concept of a complete Almost Distributive Lattice as an ADL in which the lattice PI(L) of all principal ideals of L is a complete sublattice of the lattice I(L) of all ideals of L. In [2], Jose Morgado introduced the concept of a closure operator on a complete lattice and studied some important properties of closure operators. In this paper, we introduce the concept of a closure operator on a Complete Almost Distributive Lattice on the lines of Jose Morgado and study its properties. We define the concepts of meet-distributive closure operator, join-distributive closure operator, and distributive closure operator. We prove that the set Φ (L) of all closure operators on a complete Almost Distributive Lattice L is a complete lattice. Finally we derive a necessary and sufficient condition for a closure operator ϕ on L to be a (meet/join) distributive closure operator in terms of the closed elements of L with respect to ϕ. 2. PRELIMINARIES In this section, we give some fundamental concepts and results on ADLs for the sake of ready reference. Definition 2.1: [7] An algebra L = (L,,, 0) of type (2, 2, 0) is called an Almost Distributive Lattice (ADL), if it satisfies the following axioms: (1) x 0 = x (2) 0 x = 0 (3) ( x y) z = (x z) (y z) (4) x (y z) = (x y) (x z) (5) x (y z) = (x y) (x z) (6) (x y) y = y For all x, y, z L. Corresponding author: Venugopalam Undurthi* International Journal of Mathematical Archive- 5(6), June

2 Let L = (L,,, 0) be an ADL. For x, y L, define x y if and only if x y = x or equivalently x y = y. Then is a partial order on L. A non-empty subset I of L is called an ideal if (i) a, b I, then a b I (ii) a I, x L, then a x I. The set of all ideals of L is denoted by I (L) and (I (L), ) is a complete distributive lattice. For any x L, (x] = {y x y L} is an ideal of L and is called the principal ideal generated by x. The set PI (L) of all principal ideals of L is a sublattice of the lattice I (L). Lemma 2.2: [7] If L = (L,,, 0) is an ADL and x, y, z L. Then we have the following: (1) x x = x (2) x x = x (3) x y y (4) x x y (5) x y z = y x z. Theorem 2.3: [7] Let L be an ADL, then for any m L the following are equivalent: (1) m is maximal (2) m x = m, for all x L (3) m x = x, for all x L. Definition 2.4: [6] An ADL L = (L,,, 0) with a maximal element m is said to be complete, if PI(L) is a complete sublattice of I(L). Theorem 2.5: [6] Let L = (L,,, 0) be an ADL with a maximal element m, then L is a complete ADL if and only if [0,m] is a complete distributive lattice. Example 2.6: [7] Let X be a non-empty set and x 0 X. Define binary operations, on X by: x y = y if x = x 0 and x y = x 0, if xx = x 0 x, if x x 0 y, if xx x 0. Then (X,,, x 0 ) is an ADL, with x 0 as its zero element. This ADL is called a Discrete ADL. For all standard definitions and results in distributive lattices we refer to Gratzer, G. [1]. 3. CLOSURE OPERATORS In this section, we introduce the concepts of a closure operator on a complete almost distributive lattice, and derive many important properties of closure operators. We begin with the following definition of a closure operator. Definition 3.1: Let L be a complete ADL with a maximal element m. Then a mapping ϕ: L L is said to be a closure operator of L if, for any x, y L, the following conditions hold: (1) ϕ(x) m (2) ϕ(x) x = x (3) If x y, then ϕ(x) ϕ(y) (4) ϕ(x y) = ϕ(y x) (5) ϕ (ϕ(x)) = ϕ(x). Let L be a complete ADL with a maximal element m, then we denote set of all closure operators on L by Φ (L). In the following we give two important closure operators on a complete almost distributive lattice. Example 3.2: Let L be a complete ADL with a maximal element m, define t: L L by t(x) = x m for all x L. Then t is a closure operator of L. Example 3.3: Let L be a complete ADL with a maximal element m, define ωω: L L by ω(x) = m, for all x L. Then ω is a closure operator of L. Theorem 3.4: If L = (L,, 0, m) is a discrete ADL with a maximal element m. Then there are only two closure operators on L. Proof: Suppose L = (L,,, 0, m) is a discrete ADL with a maximal element m. Let ϕ be a closure operator on L. Let x, y L, if x 0 and y 0, then ϕ(x y) = ϕ(y x) and hence ϕ(y) = ϕ(x). Suppose ϕ (0) = 0, since ϕ(x) x = x, for x 0. Therefore ϕ(x) 0. Also, since ϕ(x) m, we get that m = ϕ(x). Hence ϕ(x) = x m for all x L. Thus ϕ = t. Suppose ϕ (0) 0, then ϕ (0) m implies ϕ (0) = m. Also, ϕ (0) ϕ(x) implies ϕ(x) = m for all x L. Thus ϕ = ωω. 2014, IJMA. All Rights Reserved 120

3 Definition 3.5: Let L be a complete ADL with a maximal element m, and ϕ a closure operator of L. Then an element x L is said to be closed under ϕ, if ϕ(x) = x. The following result can be verified routinely. Lemma 3.6: Let ϕ be a closure operator of a complete ADL L. Then an element y L is closed under ϕ if and only if y = ϕ(x) for some x L. Lemma 3.7: L be a complete ADL with a maximal element m. Then m is closed under every closure operator of L. Proof: Let ϕ be any closure operator of L. Then, by (2) of the definition 3.1, we get that ϕ (m) m = m. Therefore ϕ (m) = m, since ϕ (m) m. Hence m is closed under any closure operator of L. Lemma 3.8: Let L be a complete ADL with a maximal element m, let ϕ be a closure operator of L and {x αα m αα J} aa family of closed elements under ϕ in L. Then αα JJ (x α m) is also a closed element under ϕ in L. Proof: Suppose {x α m α J} is a family of closed elements under ϕ in L. Let x = αα JJ (x α m). Clearly, x L and x x αα m for all α J. Then ϕ(x) ϕ (x α m) = x α m for all α J. Therefore ϕ(x) αα JJ (x α m) = x and hence ϕ(x) = ϕ(x) x = x. Therefore x is closed under ϕ. Lemma 3.9: Let L be a complete ADL with a maximal element m, ϕ a closure operator of L and x L. Then ϕ(x) = inf {y m L y m is closed under ϕ and y x = x}. Proof: Let A = {y m L y m is closed under ϕ and y x = x}. Let y 0 = y m A (y m). Now, we prove that ϕ(x) = y 0. Let y m A. Then y x = x. Thus x m y m for all y m A. Therefore ϕ(x) ϕ(y) = y m for all y m A and hence ϕ(x) αα JJ (x α m). = y 0. Since, ϕ (ϕ(x)) = ϕ(x) and ϕ(x) x = x, we get that ϕ(x) A, and hence y 0 ϕ(x). Thus ϕ(x) = y 0. Definition 3.10: Let Φ(L) be the set of all closure operators of L and for any ϕ, ψ Φ(L), define ϕ ψ if and only if ϕ(x) ψ(x) for all x L. Lemma 3.11: Let L be a complete ADL with a maximal element m. Then is a partial order on Φ(L). Definition 3.12: Let {ϕ αα α J} Φ(L), define ( α J ϕ αα )(x) = αα JJ (ϕ αα (x)) for all x L. Theorem 3.13: Let { ϕ αα α J} Φ(L). Then αα JJ ϕ αα set (Φ(L), ). Φ (L) is the g.l.b of {ϕ αα α J} in the partially ordered Proof: Write ϕ = ϕ αα (1) For x L, ϕ(x) ϕ αα (x) m for all α J. (2) Let x L. Then ϕ αα (x) x = x and hence x m ϕ αα (x) m for all α J. Thus x m αα JJ (ϕ αα (x)) = ϕ (x) and hence x m ϕ(x) = x m. Therefore x ϕ(x) x = x m x. Hence ϕ(x) x = x. (3) Let x, y L such that x y. Then ϕ αα (x) ϕ αα (y) for all α J. Since, ϕ ϕ αα for all α J, we get that ϕ(x) ϕ αα (x) ϕ αα (y) for all α J and hence ϕ(x) αα JJ (ϕ αα (y)). Thus ϕ(x) ϕ(y). (4) Let x, y L. Now, ϕ(x y) = α J ϕ αα ((x y)) = α J ϕ αα ((y x)) = ϕ(y x). (5) Let x L. Then ϕ(ϕ(x)) ϕ(x) = ϕ(x) and hence ϕ(x) ϕ(ϕ(x)), we have ϕ(x) ϕ αα (x) for all α J, then ϕ(ϕ(x)) ϕ(ϕ αα (x) ) ϕ αα (ϕ αα (x) ) = ϕ αα (x) for all α J and hence ϕ(ϕ(x)) inf { ϕ αα (x) α J } = ϕ(x). Thus ϕ(ϕ(x)) = ϕ(x). Therefore ϕ is a closure operator of L. Suppose ψ Φ (L) and ψ ϕ αα for all α J. Then ψ(x) ϕ αα (x) for all α J. Hence ψ(x) (x)). Thus ψ(x) ϕ(x). α J. Therefore ψ ϕ and hence ϕ is the g.l.b of {ϕ αα α J}. (ϕ αα αα JJ Since ω is the greatest element of the poset (Φ(L), ), by above theorem, we get the following. Theorem 3.14: The poset (Φ(L), ) is a complete lattice. Theorem 3.15: Let {ϕ αα α J} Φ(L) and ψ = α J. α J ϕ αα. Then, for any x L, ψ(x) = x if and only if ϕ αα (x) = x for all 2014, IJMA. All Rights Reserved 121

4 Proof: Let {ϕ αα α J} Φ(L) and ψ = α J ϕ αα. Suppose x L and ψ(x) = x. Since ϕ αα ψ for α J, we get that ϕ αα (x) ψ(x) = x for all α J. Hence ϕ αα (x) = ϕ αα (x) x = x for all α J. Conversely, suppose that ϕ αα (x) = x for all α J. Define η: L L by x, if x y = yy η(y) = m, otherwise First we prove that η is a closure operator of L. (1) Let y L and x y = y. Then η(y) = x = = ϕ αα (x) m. Otherwise η(y) = m. Therefore η(y) m for all y L. (2) Let y L. If x y = y, then η(y) y = x y = y. Otherwise η(y) y = m y = y. Therefore η(y) y = y for all y L. (3) Let y, z L such that y z. Suppose x z = z, then x z y = z y. Hence x y = y. Therefore η(z) = x = η(y). If x z = z, then η(z) = m. Therefore η(y) η(z). (4) Let y, z L. Since x y z = y z if and only if x z y = z y, we get that η(y z) = η (z y). (5) Let y L. If x y = y, then η(y) = x. Now, η(η(y)) = η(x) = x = η(y). Suppose x y y, then η(y) = m. Therefore η (η(y)) = η(m) = m = η(y). Hence η is a closure operator of L. Now, we prove that ϕ αα η for all α J. Let y L, suppose x y = y, then ϕ αα (x) = ϕ αα (x y) = ϕ αα (y x) = ϕ αα (x) = x = η(y). Now, suppose x y y, then ϕ αα (x) m = η(y). Hence ϕ αα η. Since α is arbitrary, we get that ψ = α J ϕ αα. Therefore ψ(x) η(x) = x. Hence ψ(x) = x. Lemma 3.16: Let L be a complete ADL with a maximal element m, a L such that a m m and define ϕ a: L L by ϕ a(x) = a m, if a x = x and ϕ a (x) = m, if a x x for all x L. Then ϕ aa is a closure operator of L. Proof: (1) For any x L, ϕ a (x) = a m or m and hence ϕ a (x) m for all x L. (2) Let x L. If a x = x. Then ϕ a (x) x = a m x = a x = x. If a x x. Then ϕ a (x) x = m x = x. Therefore ϕ a(x) x = x for all x L. (3) Let x, y L such that x y, if a y y, then by (1), ϕ a (x) m = ϕ a (y). Suppose a y = y, then a x = x and hence ϕ a (x) = a m = ϕ a (y). Thus ϕ a (x) ϕ a (y), whenever x y. (4) Let x, y L. Now, a x y = x y iff a x y x = x y x iff a y x = y x. Therefore ϕ a (x y) = ϕ a (y x). (5) Let x L. If a x = x. Then ϕ a (x) = a m. Since a a m = a m, we get that ϕ a (a m) = a m m = a m and hence ϕ a (ϕ a (x)) = ϕ a (x). If a x x, then a m m. Now, ϕ a (x) = m and ϕ a (ϕ a (x)) = ϕ a (m) = m = ϕ a (x). Thus ϕ a is a closure operator of L. Definition 3.17: Let L be a complete ADL with a maximal element m. Then a closure operator ϕ of L is said to be a dual atom of Φ(L) if there exists no closure operator ψ of L such that ϕ < ψ < ω. In other words, ψ is a closure operator of L such that ϕ ψ ω. Then either ϕ = ψ or ψ = ω. Lemma 3.18: Let L be a complete ADL with a maximal element m and a L such that a m m. Then ϕ a is a dual atom of L. Proof: Let ψ be a closure operator of L such that ϕ a ψ ω. Case (i) if ψ (a) = m. If a x = x, then a m = ϕ a (x) ψ(x). Now, m = ψ (a) = ψ (a x) ψ(x). If a x x, then m = ϕ a (x) ψ(x). Hence ψ(x) = m. Thus ψ = ω. Case (ii) if ψ (a) < m. If a x = x, then x m a m and hence ψ(x) ψ (a). Also, a m = ϕ a (x) ψ(x) implies ψ (a) ψ(x). Thus ψ(x) = ψ (a). Now, ϕ a (ψ (a)) ψ (ψ (a)) = ψ (a) < m. Therefore a ψ (a) = ψ (a). So that, ψ (a) = a m. Hence ψ(x) = a m if a x = x. If a x x, then ϕ a (x) = m and hence ψ(x) = m. Therefore ψ = ϕ a. Thus ψ = ω or ψ = ϕ a. Therefore ϕ a is a dual atom of Φ(L). Lemma 3.19: If ϕ is a dual atom of Φ(L). Then there exists b L such that b m m and ϕ = ϕ b. Proof: Suppose ϕ is a dual atom of Φ(L). Then, ϕ ω. Therefore there exists x ( m) L such that ϕ(x) m. If b = ϕ(x), then b m = ϕ(x) m = ϕ(x). Therefore b m m. Now, we prove that ϕ b = ϕ. Let y L. If b y = y, then ϕ b (y) = b m and ϕ(y) ϕ (b) = b. Hence ϕ(y) b m = ϕ b (y). Thus ϕ(y) = ϕ b(y). If b y y, then ϕ b (y) = m. Therefore ϕ(y) ϕ b (y). Hence ϕ(y) = ϕ b (y), if b y y. Therefore ϕ(y) = ϕ b (y), for all y L. Thus ϕ b = ϕ. Lemma 3.20: Letϕ 1, ϕ 2 be two closure operators of a complete ADL L and x L. Then ϕ 1 ϕ 2 if and only if for any x L, ϕ 2 (x) = x implies ϕ 1 (x) = x. Proof: Suppose ϕ 1 ϕ 2, x L and ϕ 2 (x) = x. Then ϕ 1 (x) ϕ 2 (x) = x. Therefore ϕ 1 (x) = ϕ 1 (x) x = x. Conversely, suppose for any x L, ϕ 2 (x) = x implies ϕ 1 (x) = x. Since x m ϕ 2 (x), we get that ϕ 1 (x) = ϕ 1 (x m) ϕ 1 (ϕ 2 (x)) = ϕ 2 (x) by our assumption since ϕ 2 (ϕ 2 (x)) = ϕ 2 (x). 2014, IJMA. All Rights Reserved 122

5 4. DISTRIBUTIVE CLOSURE OPERATORS In this section, we introduce the concept of meet-distributive closure operator, join-distributive closure operator and distributive closure operator on complete almost distributive lattice and prove some important properties of these closure operators. Definition 4.1: An element x of a lattice X is said to be (1) meet-distributive if for any y, z X, x (y z) = (x y) (x z) (2) join-distributive if for any y, z X, x (y z) = (x y) (x z) (3) distributive if it is both meet-distributive and join-distributive. Theorem 4.2: Let L be a complete ADL with a maximal element m, then the element ϕ Φ(L) is meet-distributive if and only if the following condition is satisfied: If a, b L such that a m is closed under ϕ and a m, b m are incomparable then the element c = a b m is closed under ϕ. Proof: Suppose ϕ Φ (L) is meet-distributive. Let a, b L such that a m is closed under ϕ and b m is incomparable with a m. We prove that c = a b m is closed under ϕ. Since c = a b m, b m m and c m m, ϕ b, ϕ c are dual atoms of Φ(L) and ϕ (ϕ b ϕ c ) = (ϕ ϕ b ) (ϕ ϕ c ), then [ϕ (ϕ b ϕ cc )](c) = [(ϕ ϕ b ) (ϕ ϕ c )](c). Now, (ϕ ϕ b )(c) = ϕ(c) ϕ b (c) ϕ(a) ϕ b (b) = a b m = c. Clearly, c = c m (ϕ ϕ b )(c). Hence (ϕ ϕ b )(c) = c. Now, (ϕ ϕ cc )(c) = ϕ(c) ϕ c (c) = ϕ(c) c m = ϕ(c) c = c. Thus [(ϕ ϕ b ) (ϕ ϕ cc )](c) = c and hence c = [ϕ (ϕ b ϕ c )](c) = ϕ(c) ω(c) = ϕ(c). Therefore c is closed under ϕ. Conversely, assume the condition. Let ϕ 1, ϕ 2 Φ(L). Then, Clearly, (ϕ ϕ 1 ) (ϕ ϕ 2 ) ϕ (ϕ 1 ϕ 2 ). Let x L and suppose [(ϕ ϕ 1 ) (ϕ ϕ 2 )](x) = x. Then ϕ(x) ϕ 1 (x) = x and ϕ(x) ϕ 2 (x) = x. If ϕ(x), ϕ 1 (x) are not comparable, then by our assumption, we get that ϕ(x) = x and hence ϕ(x) ϕ 1 (x) = ϕ(x). Therefore, ϕ(x) ϕ 1 (x), which is a contradiction. Hence ϕ(x), ϕ 1 (x) are comparable. Similarly, ϕ(x), ϕ 2 (x) are comparable. If ϕ(x) ϕ 1 (x) (or ϕ(x) ϕ 2 (x)), then ϕ(x) = ϕ(x) ϕ 1 (x) = x. Now, [ϕ (ϕ 1 ϕ 2 )](x) = ϕ(x) (ϕ 1 ϕ 2 )(x) = (ϕ 1 ϕ 2 )(x) x = x. If ϕ 1 (x) ϕ(x) and ϕ 2 (x) ϕ(x), then ϕ 1 (x) = ϕ 2 (x) ϕ(x) = x. Similarly, ϕ 2 (x) = x. Hence (ϕ 1 ϕ 2 )(x) = x. Now, [ϕ (ϕ 1 ϕ 2 )](x) = ϕ(x) (ϕ 1 ϕ 2 )(x) = ϕ(x) x = x. Therefore in all the cases, we proved that [ϕ (ϕ 1 ϕ 2 )](x) = x. Hence ϕ (ϕ 1 ϕ 2 ) (ϕ ϕ 1 ) (ϕ ϕ 2 ). Therefore ϕ is meet-distributive. Lemma 4.3: If ϕ Φ(L) is join-distributive. Let a, b L such that a m and b m are incomparable and ϕ (a b) = a b m. Then ϕ ϕ a ω and ϕ ϕ b ω. Proof: Suppose ϕ ϕ a = ω and let c = a b m. Then ϕ(c) = c. Now, (ϕ a ϕ b )(c) = ϕ a (c) ϕ b (c) ϕ a (a m) ϕ b (b m) = a b m = c. Hence (ϕ a ϕ b )(c) = c. Therefore c = [ϕ (ϕ a ϕ b )](c) = [(ϕ ϕ a ) (ϕ ϕ b )](c) = (ϕ ϕ b )(c). Thus ϕ b (c) = c. Therefore c = b m or m. Hence a m and b m are comparable, which is a contradiction. Therefore ϕ ϕ a ω. Similarly, ϕ ϕ b ω. Theorem 4.4: Let L be a complete ADL with a maximal element m. Then the element ϕ Φ(L) is join-distributive if and only if the following condition is satisfied: If a, b L such that a m and b m are incomparable and if c = a b m is closed under ϕ, then a m and b m are closed under ϕ. Proof: Suppose ϕ Φ(L) is join-distributive. Let a, b L such that a m, b m are incomparable and c = a b m is closed under ϕ. Clearly, a m m and b m m. Thereforeϕ a, ϕ b are dual atoms of Φ(L). By above Lemma, (ϕ ϕ a ) = ϕ a. Thus (ϕ ϕ a )(a) = ϕ a (a) = a m and hence ϕ(a) = a m. Similarly, ϕ (b) = b m. Conversely, assume the condition. Let ϕ 1, ϕ 2 Φ(L). Then clearly, ϕ (ϕ 1 ϕ 2 ) (ϕ ϕ 1 ) (ϕ ϕ 2 ). Suppose x L and [ϕ (ϕ 1 ϕ 2 )](x) = x. Then ϕ(x) = x and ϕ 1 (x) ϕ 2 (x) = x. Suppose ϕ 1 (x), ϕ 2 (x) are comparable. If ϕ 1 (x) ϕ 2 (x), then ϕ 1 (x) = ϕ 1 (x) ϕ 2 (x). Hence (ϕ ϕ 1 ) (x) = x. Therefore [(ϕ ϕ 1 ) (ϕ ϕ 2 )](x) = (ϕ ϕ 1 )(x) (ϕ ϕ 2 )(x) = x. Now, suppose ϕ 1 (x), ϕ 2 (x) are incomparable. Since ϕ 1 (x) ϕ 2 (x) = x is closed under ϕ, we get ϕ(ϕ 1 (x)) = ϕ 1 (x) and ϕ(ϕ 2 (x)) = ϕ 2 (x). Since x m ϕ 1 (x) m, we get ϕ 1 (x) (ϕ ϕ 1 )(x) (ϕ ϕ 1 )( ϕ 1 (x)) = ϕ 1 (x). Since ϕ (ϕ 1 (x)) = ϕ 1 (x) = ϕ 1 (ϕ 1 (x)). Hence ϕ 1 (x) = (ϕ ϕ 1 )(x). Similarly, we can prove (ϕ ϕ 2 )(x) = ϕ 2 (x). Hence, we have [(ϕ ϕ 1 ) (ϕ ϕ 2 )](x) = (ϕ ϕ 1 )(x) (ϕ ϕ 2 )(x) = ϕ 1 (x) ϕ 2 (x) = x. Therefore (ϕ ϕ 1 ) (ϕ ϕ 2 ) ϕ (ϕ 1 ϕ 2 ). Thus ϕ is a join-distributive. 2014, IJMA. All Rights Reserved 123

6 Theorem 4.5: Let L be a complete ADL with a maximal element m. Then the element ϕ Φ(L) is distributive if and only if the following condition is satisfied: If a, b L such that a m and b m are incomparable, and if some element of the set {a m, b m, a b m} is closed under ϕ, then every element of the set is closed under ϕ. Proof: Suppose ϕ Φ(L) is distributive and x L. Let a, b L such that a m and b m are incomparable. Hence, by Theorems 4.2 and 4.4, we get that all elements of the set {a m, b m, a b m} are closed under ϕ, if any one of them is closed under ϕ. Conversely, assume the condition. Suppose a, b L such that a m and b m are incomparable and ϕ (a m) = a m, then by our assumption a b m is closed under ϕ. Therefore ϕ is meetdistributive. Suppose a b m is closed under ϕ, then by our assumption, a m and b m are closed under ϕ. Hence by above Theorem, we get that ϕ is join-distributive. Therefore ϕ is distributive. ACKNOWLEDGEMENT This Research is supported by U. G. C. Major Research Project Ref. No /2012(SR), REFERENCES [1] Grätzer, G.: Lattice Theory: Foundation. Birkhäuser Verlag, Basel, XXIX + 613P. ISBN: [2] Jose Morgado.: Note on the Distributive Closure Operators of a Complete Lattice, Portugaliae Mathematica, 23(1964), [3] Rao, G. C., Berhanu Assaye and M. V. Ratnamani.: Heyting Almost Distributive Lattices, Southeast Asian Bulletin of Mathematics, 27(2004), [4] Rao, G. C. and Nanaji Rao, G.: Pseudo-Complementation on Almost Distributive Lattices, Southeast Asian Bulletin of Mathematics, 24,(2000), [5] Rao, G. C. and Ravi Kumar. S.: Normal Almost Distributive Lattices, Southeast Asian Bulletin of Mathematics, 32(2008), [6] Rao, G. C. and Venugopalam Undurthi.: Complete Almost Distributive Lattices (Accepted for Publication in Asian European Journal of Mathematics). [7] Swamy, U. M., Rao, G. C.: Almost Distributive Lattices, J. Aust. Math. Soc. (Series-A), 31(1981), [8] Swamy, U. M., Rao, G. C. and Nanaji Rao, G.: Stone Almost Distributive Lattices, Southeast Asian Bulletin of Mathematics, 27(2003), [9] Szasz, G.: Introduction to Lattice Theory, Academic Press New York, (1963). Source of support: U. G. C. Major Research Project, India, Conflict of interest: None Declared [Copy right 2014 This is an Open Access article distributed under the terms of the International Journal of Mathematical Archive (IJMA), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.] 2014, IJMA. All Rights Reserved 124