EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 3, 2024, 1691-1704 ISSN 1307-5543 – ejpam.com Published by New York Business Global Hierarchy Elements in an Almost Distributive Lattice S. Ramesh1, G. Chinnayya1, G. Jogarao1, Ravikumar Bandaru2, Aiyared Iampan3,∗ 1 Department of Mathematics, GITAM School of Science, GITAM (Deemed to be University), Visakhapatnam-530045, India 2 Department of Mathematics, School of Advanced Sciences, VIT-AP University, Andhra Pradesh-522237, India 3 Department of Mathematics, School of Science, University of Phayao, Mae Ka, Mueang, Phayao 56000, Thailand Abstract. In this paper, we introduce hierarchy elements in an almost distributive lattice with respect to a non-empty set and obtain some of their algebraic properties. We characterize initial segments, ideals, and maximal sets in almost distributive lattices in terms of hierarchy sets and prove that the class of hierarchy sets forms a distributive lattice, which is not an induced sublattice. Also, we characterize hierarchy sets using compatible sets in an almost distributive lattice. 2020 Mathematics Subject Classifications: 06D99, 06D75 Key Words and Phrases: Hierarchy elements, ideals, principal ideals, almost distributive lattices 1. Introduction One of both the lattice theoretic and ring theoretic generalizations of a distributive lat- tice (Boolean algebra) [7], led by Swamy and Rao in 1981 and called an almost distributive lattice [8]. Almost distributive lattices is an algebraic structure (L,∨,∧, 0) that satisfies almost every axiom of a distributive lattice with zero, not including the three identities (the commutativity of ∧,∨ and the right distributivity of ∨ over ∧). Each of these three identities is equivalent to each other in an almost distributive lattice, and an almost dis- tributive lattice with any of the above identities gives a distributive lattice. The structure of an almost distributive lattice is not even distributive; the lattice and the associativity of ∨ are not yet to be known. Hence, it is difficult to deal with it. For example, fix an element x0 in a non-empty set L. Given x, y ∈ L, define x ∧ y = y, x0 ∧ y = x0, x ∨ y = x and x0 ∨ y = y. Then (L,∧,∨, x0) is an almost distributive lattice. This L is neither lattice nor distributive. It is called a discrete [8] almost distributive lattice. Given a, b in ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i3.5226 Email addresses: ramesh.sirisetti@gmail.com (S. Ramesh), cgondu@gitam.in (G. Chinnayya), jogarao.gunda@gmail.com (G. Jogarao), ravimaths83@gmail.com (R. Bandaru), aiyared.ia@up.ac.th (A. Iampan) https://www.ejpam.com 1691 © 2024 EJPAM All rights reserved. A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1692 an almost distributive lattice L, we say that a ≤ b if a∧ b = a (or equivalently, a∨ b = b). Then, ≤ is a partial ordering on L. An element m ∈ L is called a maximal element [8] if m ∧ x = x for all x ∈ L. The authors [3–5, 8, 9] developed the theory of initial segments, ideals, and filters in almost distributive lattices analogous to the concepts in distributive lattices. It is well- known that “every distributive lattice is a one-to-one correspondence with the class of principal ideals” in it [1]. The above-mentioned property does not hold in an almost distributive lattice L because “given a, b ∈ L, (a] = (b] does not imply a = b” [8], where (a] = {a ∧ x | x ∈ L} is the smallest ideal containing a. Given an initial segment [0, a] = {x ∈ L | 0 ≤ x ≤ a} in an almost distributive lattice L, the authors [8] observed that given b ∈ L, b ∈ [0, a] if and only if a ∧ b = b ∧ a. Further, the authors extended the above properties to the concept of principal ideals and stated that “given b ∈ L, b ∈ (a] if and only if a ∧ b = b” [8]. In 2024, Noorbhasha et al. [2] proved some properties of prime σ-ideals of a normal almost distributive lattice topologically. Srikanth et al. [6] looked into why the binary operation ρ does not work in semi-Brouwerian almost distributive lattices. They found that it does not work associatively or commutatively. With this motivation, we find an arbitrary non-empty set S in an almost distributive lattice L with the property that for any h ∈ L, s ∧ h = h, for some s ∈ S. We derive algebraic properties from the class of hierarchy elements with respect to a set S. Also, we characterize the class of hierarchy sets in an almost distributive lattice and provide suffi- cient counterexamples. Mainly, we observe that the class of hierarchy sets is a distributive lattice with respect to the operations ∪ and ∧ where S1 ∧ S2 = {s1 ∧ s2 | s1 ∈ S1 and s2 ∈ S2}, for all non-empty subsets S1, S2 of L. 2. Hierarchy sets in almost distributive lattices In this section, we define hierarchy elements with respect to a non-empty set in an almost distributive lattice. We prove several algebraic properties in the class of hierarchy elements and hierarchy sets. Finally, we obtain that the class of hierarchy sets forms a distributive lattice, which is not an induced sub-distributive lattice of the class of ideals of almost distributive lattices. Finally, we derive some equivalent conditions for a hierarchy set in an almost distributive lattice to become an ideal. Definition 1. An element h ∈ L is said to be a hierarchy with respect to a non-empty subset S of L if s ∧ h = h, for some s ∈ S. The set of hierarchy elements with respect to S is denoted by HS. It is easy to observe that HS ̸= ∅ (since 0 ∈ HS). Proposition 1. For any ∅ ≠ S ⊆ L, we have (i) S ⊆ HS. (ii) HS is closed under ∧. (iii) HL = L and H{0} = {0}, where 1 is the greatest element in L. A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1693 (iv) If 1 ∈ HS, then HS = L. (v) If m ∈ HS, then HS = L, where m is a maximal element in L. Proof. (i) Let s ∈ S. Now, s ∧ s = s. Therefore, s ∈ HS . Hence, S ⊆ HS . (ii) Let h1, h2 ∈ HS . Then we can find s1, s2 ∈ S such that s1∧h1 = h1 and s2∧h2 = h2. Now, s1 ∧ (h1 ∧ h2) = (s1 ∧ h1) ∧ h2 = h1 ∧ h2. Therefore, h1 ∧ h2 ∈ HS . Hence, HS is closed under ∧. (iii) From (i), it is easy to observe that HL = L. Let x ∈ H{0}. Then 0 ∧ x = x. Therefore, x = 0. Hence, H{0} = {0}. (iv) If 1 ∈ HS , then there is an element s ∈ S such that 1 = s ∧ 1 = s ∈ S. For this 1 ∈ S, 1 ∧ x = x for all x ∈ L. Therefore, L ⊆ HS . Hence, HS = L. (v) For m ∈ HS , there is an element s ∈ S such that s ∧ m = m. Let x ∈ L. Now, s ∧ x = m ∧ (s ∧ x) = (m ∧ s) ∧ x = (s ∧m) ∧ x = m ∧ x = x. Therefore, x ∈ HS . Hence, L ⊆ HS . Thus, HS = L. Remark 1. For any ∅ ̸= S ⊆ L, HS need not be closed under ∨ by the following coun- terexample: Example 1. Let L = {0, a, b, c, 1}, whose Hasse diagram is given below: c a 0 b 1 For S1 = {a, b}, it is easy to verify that a∨ b = c /∈ HS1 = {0, a, b}. Therefore, HS1 is not closed under ∨. Proposition 2. For any ∅ ≠ S ⊆ L, and a, b, h ∈ L, we have (i) a ≤ b implies Ha ⊆ Hb, (ii) a ≤ b and b ∈ HS implies a ∈ HS, (iii) h ∈ HS implies (h] ⊆ HS, (iv) Ha is an ideal of L. A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1694 Proof. (i) Assume that a ≤ b. Let x ∈ Ha. Then a∧ x = x. Now, b∧ x = b∧ (a∧ x) = (b ∧ a) ∧ x = a ∧ x = x. Therefore, x ∈ Hb. Hence, Ha ⊆ Hb. (ii) Assume that a ≤ b and b ∈ HS . For this b ∈ HS , there is an element s ∈ S such that s ∧ b = b. Now, s ∧ a = s ∧ (a ∧ b) = (s ∧ a) ∧ b = a ∧ (s ∧ b) = a ∧ b = a. Therefore, a ∈ HS . (iii) Let h ∈ HS . Then s ∧ h = h for some s ∈ S. For any x ∈ (h], x = h ∧ x. Now, s ∧ x = s ∧ (h ∧ x) = (s ∧ h) ∧ x = h ∧ x = x. Therefore, x ∈ HS . Hence, (h] ⊆ HS . (iv) Let x, y ∈ Ha. Then a∧x = x and a∧y = y. Now, a∧(x∨y) = (a∧x)∨(a∧y) = x∨y. Therefore, x ∨ y ∈ Ha. Let l ∈ L. Then a ∧ (l ∧ x) = (a ∧ l) ∧ x = l ∧ (a ∧ x) = l ∧ x. Therefore, l ∧ x ∈ Ha and x ∧ l ∈ Ha. Thus, Ha is an ideal of L. Remark 2. For any non-empty subsets S of a discrete almost distributive lattice X, either HS = {0} or HS = L. Suppose HS ̸= {0}. Then there exists a non-zero element h in HS such that s ∧ h = h for some non-zero element s ∈ S (if s = 0, then h = 0). Let y ∈ L. Now, s ∧ y = (s ∨ h) ∧ y = (s ∧ y) ∨ (h ∧ y) = (s ∧ y) ∨ y = y (since h ̸= 0). Therefore, y ∈ HS. Hence, L ⊆ HS. Thus, HS = L. Proposition 3. For any ∅ ≠ S1, S2 ⊆ L, we have (i) S1 ⊆ S2 implies HS1 ⊆ HS2, (ii) HS1∪S2 = HS1 ∪HS2, (iii) HS1∩S2 ⊆ HS1 ∩HS2. Proof. (i) Let x ∈ HS1 . Then s1 ∧ x = x for some s1 ∈ S1. Since S1 ⊆ S2, x ∈ HS2 . Hence, HS1 ⊆ HS2 . (ii) Since S1, S2 ⊆ S1∪S2, we haveHS1 , HS2 ⊆ HS1∪S2 . Therefore,HS1∪HS2 ⊆ HS1∪S2 . Let h ∈ HS1∪S2 . Then there is an element s ∈ S1 ∪S2 such that s∧ h = h. If s ∈ S1, then h ∈ HS1 ; if s ∈ S2, then h ∈ HS2 ; and if s ∈ S1 ∩ S2, then h ∈ HS1 ∩ HS2 . Therefore, h ∈ HS1 ∪HS2 . So that HS1∪S2 ⊆ HS1 ∪HS2 . Hence, HS1∪S2 = HS1 ∪HS2 . (iii) Let h ∈ HS1∩S2 . Then there is an element s ∈ S1 ∩ S2 such that s ∧ h = h. Therefore, h ∈ HS1 ∩HS2 (since s ∈ S1 ∩ S2). Hence, HS1∩S2 ⊆ HS1 ∩HS2 . Remark 3. For any non-empty subsets S1, S2 of L, HS1∩S2 need not be equal to HS1∩HS2 by the following counterexample: Example 2. Let L = {0, a, b, 1}, whose Hasse diagram is given below: 1 a 0 b A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1695 For S1 = {a, b} and S2 = {a, 1}, it is clearly to observe that S1∩S2 = {a}, HS1 = {0, a, b}, HS2 = {0, a, b, 1}, HS1∩S2 = {0, a} and HS1∩HS2 = {0, a, b}. Hence, HS1∩S2 ̸= HS1∩HS2. Theorem 1. If ∅ ≠ S ⊆ L, is closed under ∨, then (i) HS is closed under ∨, (ii) HS is a sub-almost distributive lattice of L, (iii) HS is an ideal of L, (iv) HS is the smallest ideal generated by S(HS = (S]). Proof. Suppose that S is closed under ∨. (i) Let h1, h2 ∈ HS . Then there exist s1, s2 ∈ S such that s1∧h1 = h1 and s2∧h2 = h2. Now, (s1 ∨ s2) ∧ (h1 ∨ h2) = (s2 ∨ s1) ∧ (h1 ∨ h2) = [(s2 ∨ s1) ∧ h1] ∨ [(s2 ∨ s1) ∧ h2)] = [(s2 ∨ s1) ∧ h1] ∨ [(s1 ∨ s2) ∧ h2)] = [(s2 ∧ h1) ∨ (s1 ∧ h1] ∨ [(s1 ∧ h2) ∨ (s2 ∧ h2)] = [(s2∧h1)∨h1]∨ [(s1∧h2)∨h2] = h1∨h2. Since S is closed under ∨, h1∨h2 ∈ HS . Hence, HS is closed under ∨. (ii) By Proposition 1(ii) and Definition 1, we can clearly observe that the set HS is a sub-almost distributive lattice of L. (iii) Let l ∈ L and h ∈ HS . Then s∧h = h for some s ∈ S. Now, s∧(h∧l) = (s∧h)∧l = h∧ l and s∧ (l ∧ h) = (s∧ l)∧ h = l ∧ (s∧ h) = l ∧ h. Therefore, h∧ l, l ∧ h ∈ HS . Hence, HS is an ideal of L (since HS is a sub-almost distributive lattice of L). (iv) Since S ⊆ HS and HS is an ideal of L, (S] ⊆ HS . Let h ∈ HS . Then s∧ h = h for some s ∈ S. Therefore, h = s ∧ h ∈ (S]. Hence, HS ⊆ (S]. Thus, HS = (S]. Let us denote S1 ∧ S2 = {s1 ∧ s2 ∈ L | s1 ∈ S1 and s2 ∈ S2}, where S1 and S2 are two non-empty subsets of L. Proposition 4. For any ∅ ≠ S1, S2, S3 ⊆ L, we have (i) S1 ∧ S2 ̸= ∅, (ii) S1 ⊆ S1 ∧ S1, (iii) (S1 ∧ S2) ∧ S3 = S1 ∧ (S2 ∧ S3). Proof. (i) For any s1 ∈ S1 and s2 ∈ S2, we have s1∧s2 ∈ L. Therefore, s1∧s2 ∈ S1 ∧ S2. Hence, S1 ∧ S2 ̸= ∅. (ii) For any s1 ∈ S1, s1 = s1 ∧ s1 ∈ S1 ∧ S1. Therefore, S1 ⊆ S1 ∧ S1. (iii) Let s ∈ (S1 ∧ S2) ∧ S3. Then s = (s1 ∧ s2) ∧ s3 for some s1 ∈ S1, s2 ∈ S2, s3 ∈ S3. Since L is ∧-associative, s = s1 ∧ (s2 ∧ s3). Therefore, s ∈ S1 ∧ (S2 ∧ S3). Hence, (S1 ∧ S2) ∧ S3 ⊆ S1 ∧ (S2 ∧ S3). Similarly, we can prove the converse. Thus, (S1 ∧ S2)∧ S3 = S1 ∧ (S2 ∧ S3). Remark 4. For any ∅ ≠ S ⊆ L, S need not be idempotent under the operation ∧ . In Example 1, S1 ̸= S1 ∧ S1 (assume S1 = {a, b}). A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1696 Remark 5. Any ∅ ≠ S1, S2 ⊆ L need not be commutative under ∧ . In a discrete almost distributive lattice X, let S1 = {x} and S2 = {y}, then it can be easy to verify that S1 ∧ S2 ̸= S2 ∧ S1. Proposition 5. For any ∅ ≠ S1, S2, S3 ⊆ L, we have (i) HS1 ∧ HS2 = HS1 ∧ S2 , (ii) HS1 ∧ HS1 = HS1 (Idempotent law), (iii) HS1 ∧ HS2 = HS2 ∧ HS1 (Commutative law), (iv) HS1 ∧ (HS2 ∧ HS3) = (HS1 ∧ HS2) ∧ HS3 (Distributive law). Proof. (i) Let h ∈ HS1 ∧ HS2 . Then h = h1 ∧ h2 for some h1 ∈ HS1 and h2 ∈ HS2 . For this h1 ∈ HS1 and h2 ∈ HS2 , there exist s1 ∈ S1, s2 ∈ S2 such that s1 ∧ h1 = h1 and s2 ∧ h2 = h2. Now, s1 ∧ s2 ∧ h = s1 ∧ s2 ∧ h1 ∧ h2 = s1 ∧ h1 ∧ s2 ∧ h2 = h1 ∧ h2 = h where s1 ∧ s2 ∈ S1 ∧ S2. Therefore, h ∈ HS1 ∧ S2 . Hence, HS1 ∧ HS2 ⊆ HS1 ∧ S2 . Let h ∈ HS1 ∧ S2 . Then there exist s1 ∈ S1, s2 ∈ S2 such that s1 ∧ s2 ∧ h = h. Now, s1 ∧ h = s1∧s1∧s2∧h = s1∧s2∧h = h and s2∧h = s2∧ (s1∧s2∧h) = s1∧s2∧h = h. Therefore, h ∈ HS1 and h ∈ HS2 . Hence, h = h ∧ h = HS1 ∧ HS2 . Thus, HS1 ∧ HS2 = HS1 ∧ S2 . (ii) Let h ∈ HS1 ∧ HS1 = HS1 ∧ S1 (from (i)). Then there exist s1, s ′ 1 ∈ S1 such that s1 ∧ s′1 ∧ h = h. Now, s1 ∧ h = s1 ∧ (s1 ∧ s′1 ∧ h) = s1 ∧ s′1 ∧ h = h. Therefore, h ∈ HS1 . Hence, HS1 ∧ HS1 ⊆ HS1 . By Proposition 4(ii), S1 ⊆ S1 ∧ S1. Therefore, HS1 ⊆ HS1 ∧ HS1 . Thus, HS1 = HS1 ∧ HS1 . (iii) Let h ∈ HS1 ∧ HS2 = HS1 ∧ S2 (from (i)). Then s1 ∧ s2 ∧ h = h = s2 ∧ s1 ∧ h for some s1 ∈ S1 and s2 ∈ S2. Therefore, h ∈ HS2 ∧ HS1 . Hence, HS1 ∧ HS2 ⊆ HS2 ∧ HS1 . Similarly, we can prove the converse. Hence, HS1 ∧ HS2 = HS2 ∧ HS1 . (iv) By Proposition 4(iii), we haveHS1 ∧ (HS2 ∧ HS3) = HS1 ∧ HS2 ∧ S3 = HS1 ∧ (S2 ∧ S3) = H(S1 ∧ S2) ∧ S3 = H(S1 ∧ S2) ∧ HS3 = (HS1 ∧ HS2) ∧ HS3 . Proposition 6. For any ∅ ≠ S1, S2 ⊆ L, we have (i) HS1 ∪HS2 = HS1∪S2, (ii) HS1 ∪HS1 = HS1 (Idempotent), (iii) HS1 ∪HS2 = HS2 ∪HS1 (Commutative), (iv) HS1 ∪ (HS2 ∪HS3) = (HS1 ∪HS2) ∪HS3 (Associative). Proposition 7. For any ∅ ≠ S ⊆ L and a is an element in L, we have (i) Ha = (a], (ii) HS = ⋃ s∈S (s]. A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1697 Proof. (i) By Proposition 2(iv), Ha is an ideal of L and a ∈ Ha. Then (a] ⊆ Ha. Let h ∈ Ha. Then a ∧ h = h. Therefore, h = a ∧ h ∈ (a]. So that Ha ⊆ (a]. Hence, Ha = (a]. (ii) Let h ∈ HS . Then, we have an element s ∈ S such that h = s∧h ∈ (s]. Therefore, HS ⊆ ⋃ s∈S (s]. Let y ∈ ⋃ s∈S (s]. Then y ∈ (s] for some s ∈ S. Therefore, y = s ∧ y. So that y ∈ HS . Hence, ⋃ s∈S (s] ⊆ HS . Thus, HS = ⋃ s∈S (s]. Proposition 8. For any ∅ ≠ S ⊆ L, S ⊆ HS ⊆ (S]. Remark 6. For any ∅ ≠ S ⊆ L, HS need not be equal to (S]. In Example 2, let S = {a, b}, then HS = {0, a, b} and (S] = {0, a, b, 1} = L. Therefore, HS ⊂ (S] but HS ̸= (S]. Proposition 9. For any ∅ ≠ S1, S2, S3 ⊆ L, we have (i) S1 ⊆ S1 ∪ (S2 ∧ S1), (ii) S1 ⊆ S1 ∧ (S2 ∪ S1), (iii) S1 ∪ (S2 ∧ S3) ⊆ (S1 ∪ S2) ∧ (S1 ∪ S3), (iv) (S1 ∧ S2) ∪ S3 ⊆ (S1 ∪ S3) ∧ (S2 ∪ S3), (v) S1 ∧ (S2 ∪ S3) = (S1 ∧ S2) ∪ (S1 ∧ S3) (Distributive law), (vi) (S1 ∪ S2) ∧ S3 = (S1 ∧ S3) ∪ (S2 ∧ S3) (Distributive law). Remark 7. For any non-empty subsets S1, S2 of L, the following absorption laws need not be held: (i) S1 ∪ (S2 ∧ S1) = S1, (ii) S1 ∧ (S2 ∪ S1) = S1. In Example 1, (i) take S1 = {a} and S2 = {b}. Then S2 ∧ S1 = {0}. Therefore, S1 ∪ (S2 ∧ S1) = {0, a} ≠ S1. (ii) Take S2∪S1 = {a, b}. Then S1 ∧ (S2∪S1) = {0, a}. Therefore, S1 ∧ (S2∪S1) ̸= S1. Remark 8. For any ∅ ≠ S1, S2, S3 ⊆ L, the following distributive laws need not be held: (i) S1 ∪ (S2 ∧ S3) = (S1 ∪ S2) ∧ (S1 ∪ S3) (Distributive law), (ii) (S1 ∧ S2) ∪ S3 = (S1 ∪ S3) ∧ (S2 ∪ S3) (Distributive law). In Example 1, (i) take S1 = {a, b}, S2 = {1} and S3 = {c}. Then S2 ∧ S3 = {c}, S1 ∪ S2 = {a, b, 1} and S1 ∪S3 = {a, b, c}. Now, S1 ∪ (S2 ∧ S3) = {a, b, c} and (S1∪S2) ∧ (S1∪ S3) = {0, a, b, c}. Therefore, S1 ∪ (S2 ∧ S3) ̸= (S1 ∪ S2) ∧ (S1 ∪ S3). (ii) Take S1 = {a}, S2 = {1} and S3 = {b, c}. Then S1 ∧ S2 = {a}, S1 ∪ S3 = {a, b, c} and S2∪S3 = {b, c, 1}. Now, (S1 ∧ S2)∪S3 = {a, b, c} and (S1∪S3) ∧ (S2∪S3) = {0, a, b, c}. Therefore, (S1 ∧ S2) ∪ S3 ̸= (S1 ∪ S3) ∧ (S2 ∪ S3). A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1698 Proposition 10. For any ∅ ≠ S1, S2, S3 ⊆ L, we have (i) HS1 ∪ (HS2 ∧ HS1) = HS1 (Absorption law), (ii) HS1 ∧ (HS2 ∪HS1) = HS1 (Absorption law), (iii) HS1 ∪ (HS2 ∧ HS3) = (HS1 ∪HS2) ∧ (HS1 ∪HS3) (Distributive law), (iv) HS1 ∧ (HS2 ∪HS3) = (HS1 ∧ HS2) ∪ (HS1 ∧ HS3) (Distributive law). Proof. (i) Let x ∈ HS1 ∪ (HS2 ∧ HS1) = HS1 ∪ HS2 ∧ S1 . If x ∈ HS1 , then HS1 ∪ (HS2 ∧ HS1) ⊆ HS1 . If x ∈ HS2 ∧ S1 , then there exists s = s2 ∧ s1 ∈ S2 ∧ S1 such that s ∧ x = s2 ∧ s1 ∧ x = x for some s2 ∈ S2 and s1 ∈ S1, then s1 ∧ x = s1 ∧ s2 ∧ s1 ∧ x = s2∧s1∧x = x. Therefore, x ∈ HS1 . Hence, HS1∪(HS2 ∧ HS1) ⊆ HS1 . By Proposition 9(i), S1 ⊆ S1 ∪ (S2 ∧ S1). Therefore, HS1 ⊆ HS1∪(S2 ∧ S1). Hence, HS1 ⊆ HS1 ∪ (HS2 ∧ HS1). Thus, HS1 ∪ (HS2 ∧ HS1) = HS1 . (ii) Let x ∈ HS1 ∧ (HS2 ∪ HS1) = HS1 ∧ (HS2∪S1) = HS1 ∧ (S2∪S1). Then there exists s ∈ S1 ∧ (S2∪S1) such that s∧x = x. Therefore, for this s ∈ S1 ∧ (S2∪S1), s = s1∧ s′ for some s1 ∈ S1 and s′ ∈ S2∪S1. Now, s1∧x = s1∧ (s∧x) = s1∧ (s1∧ s′)∧x = s1∧ s′∧x = s ∧ x = x. Therefore, x ∈ HS1 . Hence, HS1 ∧ (HS2 ∪HS1) ⊆ HS1 . By Proposition 9(ii), S1 ⊆ S1 ∧ (S2∪S1). Therefore, HS1 ⊆ HS1 ∧ (S2∪S1) = HS1 ∧ H(S2∪S1) = HS1 ∧ (HS2∪HS1). Hence, HS1 ∧ (HS2 ∪HS1) = HS1 . (iii) Let x ∈ (HS1 ∪ HS2) ∧ (HS1 ∪ HS3) = H(S1∪S2) ∧ H(S1∪S3) = H(S1∪S2) ∧ (S1∪S3). Then there exists s ∈ (S1 ∪ S2) ∧ (S1 ∪ S3) such that s ∧ x = x. For this s ∈ (S1 ∪ S2) ∧ (S1 ∪ S3), s = a ∧ b for some a ∈ (S1 ∪ S2), b ∈ (S1 ∪ S3). If a ∈ S1 and b ∈ S1, then a ∧ x = a ∧ s ∧ x = a ∧ a ∧ b ∧ x = a ∧ b ∧ x = s ∧ x = x. Therefore, x ∈ HS1 . If a ∈ S2 and b ∈ S1, then b∧ x = b∧ s∧ x = b∧ a∧ b∧ x = a∧ b∧ x = s∧ x = x. Hence, x ∈ HS1 . If a ∈ S2 and b ∈ S3, then s = a ∧ b ∈ S2 ∧ S3 and s ∧ x = x. Therefore, x ∈ HS2 ∧ S3 . Hence, (HS1 ∪ HS2) ∧ (HS1 ∪ HS3) ⊆ HS1 ∪ (HS2 ∧ HS3). By Proposition 9(iii), S1 ∪ (S2 ∧ S3) ⊆ (S1 ∪ S2) ∧ (S1 ∪ S3). Therefore, HS1∪(S2 ∧ S3) ⊆ H(S1∪S2) ∧ (S1∪S3). Hence, HS1 ∪ (HS2 ∧ HS3) ⊆ (HS1 ∪ HS2) ∧ (HS1 ∪ HS3). Thus, HS1 ∪ (HS2 ∧ HS3) = (HS1 ∪ HS2) ∧ (HS1 ∪HS3). (iv) We can prove similarly by Proposition 9(v). Let us denote the hierarchy sets of L by HS(= {HS | S ⊆ L and S ̸= ∅}). We have the final conclusion from Proposition 10, below. Theorem 2. (HS , ∧ ,∪, {0}, L) becomes a bounded distributive lattice. Proof. By the above Propositions 5, 6, and 10, we observe that the set of hierarchy sets HS is a distributive lattice with the operations ∧ and ∪ and hence, H{0} = {0} is the least element and HL = L is the greatest element in HS . A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1699 Theorem 3. For any non-empty subsets S0, S1 of L, HS0 ∧ HS1 = {0} and HS0∪HS1 = L if and only if S0 ∧ S1 = {0} and S0 ∪ S1 contains a maximal element if and only if S0 = {0} ⇒ S1 contains a maximal element or S1 = {0} ⇒ S0 contains a maximal element. Theorem 4. For any non-empty subset S of L, the following are equivalent: (i) HS is closed under ∨, (ii) HS is a subADL of L, (iii) HS is an ideal of L, (iv) HS is a smallest ideal containing S. Lemma 1. Let S be a non-empty subset and I be an ideal of L. If HS = HI , then (S] = I. Remark 9. The converse of Lemma 1 need not be true. We can see the following coun- terexample: Example 3. Let L = {0, a, b, c, d, e, f, 1} be an ADL whose Hasse diagram is given below: 1 d b f e a 0 c Let S1 = {a, b}. Then (S1] = {0, a, b, d} := I. But HS1 = {0, a, b} ≠ {0, a, b, d} = HI . Lemma 2. For any non-empty subset S of L, we have (i) ⋂ S⊆L HS = {0}, (ii) ⋃ S⊆L HS = L, (iii) HHS = HS. Let us denote the set SM = set of maximal elements of S, and we can prove the following theorem on an ADL with maximal elements. A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1700 Theorem 5. For any non-empty subset S of L, we have (i) (S] = (SM ], (ii) ⋃ sm∈SM Hsm = HS. Proof. (i) Let h ∈ (S]. Then h = ( m∨ i=1 si ) ∧ x for some x ∈ L and si ∈ S for all 1 ≤ i ≤ m. Let sm ∈ SM . Then sm ∧ si = si for all 1 ≤ i ≤ m. Therefore, sm ∧ si ∈ (SM ] for all 1 ≤ i ≤ m. So that m∨ i=1 si = m∨ i=1 (sm ∧ si) ∈ (SM ]. Hence, h = ( m∨ i=1 si ) ∧ x ∈ (SM ]. Since SM ⊆ S, (SM ] ⊆ (S]. Therefore, (S] = (SM ]. (ii) Let h ∈ ⋃ sm∈SM Hsm . Then h ∈ Hsm for some sm ∈ SM and sm ∧ h = h. Since SM ⊆ S, h ∈ HS . Therefore, ⋃ sm∈SM Hsm ⊆ HS . Let h ∈ HS . Then there exists s ∈ S such that s ∧ h = h. Let sm ∈ sM . Then sm ∧ s = s. Now, sm ∧ h = sm ∧ s ∧ h = s ∧ h = h. Therefore, h ∈ Hsm for some sm ∈ SM . So that h ∈ ⋃ sm∈SM Hsm . Hence, HS ⊆ ⋃ sm∈SM Hsm . Thus, HS = ⋃ sm∈SM Hsm . 3. Characterization of hierarchy sets with respect to a compatible set In this section, we characterize the class of hierarchy sets in terms of compatible sets, introduce a new class of sets in an almost distributive lattice, and study rigorously. A non-empty subset S of L is said to be compatible if for each s1, s2 ∈ S, s1∧s2 = s2∧s1 or equivalently, s1 ∨ s2 = s2 ∨ s1. Lemma 3. Let h ∈ L and a non-empty subset S of L. If there exists s ∈ S such that h ≤ s, then h ∈ HS. Proof. If h ≤ s for some s ∈ S, then s ∧ h = h = h ∧ s. Therefore, h ∈ HS . Remark 10. The converse of Lemma 3 need not be true. In a discrete ADL X, let S = {s} for some s ∈ X \ {0}. Then HS = X. Let h ∈ X. Then h ∈ HS and h ∧ s = s and s ∧ h = h for all s ∈ S. Therefore, h ⩽̸ s. Remark 11. If HS is compatible, then the converse of Lemma 3 is true. Proof. Let h ∈ HS for some h ∈ L. Then, an element s ∈ S exists such that s∧h = h. Since S ⊆ HS , s ∧ h = h ∧ s = h. Hence, h ≤ s. Definition 2. For any non-empty subset S of L, define a set Ŝ = {x ∈ L | x ≤ s, for some s ∈ S}. Remark 12. For any non-empty subset S of L, we can observe that Ŝ ̸= ∅ and S ⊆ Ŝ. A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1701 Lemma 4. For any non-empty subset S of L, we have the following: (i) Ŝ is closed under ∧, (ii) h ∧ ŝ1 ∈ Ŝ, for any h ∈ L, (iii) Ŝ ⊆ HS, (iv) HS = H Ŝ . Proof. (i) Let ŝ1, ŝ2 ∈ Ŝ. Then there exist s1, s2 ∈ S such that ŝ1 ≤ s1 and ŝ2 ≤ s2. Now, (ŝ1 ∧ ŝ2)∧ s2 = ŝ1 ∧ (ŝ2 ∧ s2) = ŝ1 ∧ ŝ2. Therefore, ŝ1 ∧ ŝ2 ≤ s2. So that ŝ1 ∧ ŝ2 ∈ Ŝ. Hence, Ŝ is closed under ∧. (ii) Let h ∈ L and ŝ1 ∈ Ŝ. Then there exists s1 ∈ S such that ŝ1 ≤ s1. Now, (h ∧ ŝ1) ∧ s1 = h ∧ (ŝ1 ∧ s1) = h ∧ ŝ1. Then h ∧ ŝ1 ≤ s1. Therefore, h ∧ ŝ1 ∈ Ŝ. (iii) Let ŝ1 ∈ Ŝ. Then ŝ1 ∧ s1 = s1 ∧ ŝ1 = ŝ1. Therefore, ŝ1 ∈ HS (since s1 ∈ S). Hence, Ŝ ⊆ HS . (iv) By Definition 2, S ⊆ Ŝ. Then HS ⊆ H Ŝ . Let h ∈ H Ŝ . Then ŝ1 ∧ h = h for some ŝ1 ∈ Ŝ. For this ŝ1 ∈ Ŝ, there exists s1 ∈ S such that ŝ1 ≤ s1. Now, s1 ∧ h = s1 ∧ (ŝ1 ∧ h) = (s1 ∧ ŝ1) ∧ h = ŝ1 ∧ h = h. Then h ∈ HS (since s1 ∈ S). Hence, H Ŝ ⊆ HS . Thus, HS = H Ŝ . Remark 13. Ŝ need not be closed under ∨. In Example 2, let S = {a, b}. Then Ŝ = {0, a, b}. Now, a ∨ b = c /∈ Ŝ. Hence, Ŝ is not closed under ∨. Remark 14. For any h ∈ L and ŝ1 ∈ Ŝ, ŝ1 ∧ h need not be in Ŝ. In a discrete ADL X, let S = {a} for some non-zero element a ∈ X. Then Ŝ = {0, a}. Let x ∈ X. Then a ∧ x = x /∈ Ŝ. Remark 15. For any non-empty subset S of L, Ŝ need not be equal to H Ŝ . In a discrete ADL X, let S = {a}, for some non-zero element a ∈ X. Then HS = Ha = X and Ŝ = {0, a}. Therefore, HS ̸= Ŝ. Lemma 5. If HS is compatible, then HS = Ŝ, where S ⊆ L and S ̸= ∅. Proof. Let h ∈ HS . Then, an element s ∈ S exists such that s ∧ h = h. Since S ⊆ HS , s ∧ h = h ∧ s = h. Therefore, h ≤ s and s ∈ S. So that h ∈ Ŝ. Hence, HS ⊆ Ŝ. By Lemma 4(iii), HS = Ŝ. Theorem 6. For any compatible subset S of L, we have the following: (i) Ŝ is compatible, (ii) For each h ∈ HS, there exists unique ŝ ∈ Ŝ such that ŝ ∧ h = h and h ∧ ŝ = ŝ. A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1702 Proof. (i) Let ŝ1, ŝ2 ∈ Ŝ. Then there exist s1, s2 ∈ S such that ŝ1 ≤ s1 and ŝ2 ≤ s2. Now, ŝ1∧ŝ2 = (ŝ1∧s1)∧(ŝ2∧s2) = ŝ2∧ŝ1∧(s1∧s2) = ŝ2∧ŝ1∧(s2∧s1) = (ŝ2∧s2)∧(ŝ1∧s1) = ŝ2 ∧ ŝ1 (since S is compatible). Therefore, Ŝ is compatible. (ii) Let h ∈ HS . By Lemma 4(iv), h ∈ H Ŝ = HS . For this h ∈ H Ŝ , there exists ŝ ∈ Ŝ such that ŝ ∧ h = h and for this ŝ ∈ Ŝ, there exists s ∈ S such that ŝ ≤ s. Let s0 = h∧s. Than s0∧s = (h∧s)∧s = h∧s = s0. Therefore, s0 ≤ s. So that s0 ∈ Ŝ. Now, h∧s0 = h∧h∧s = h∧s = s0 and s0∧h = h∧s∧h = s∧h∧h = s∧h = h. Let â ∈ Ŝ such that h∧ â = â and â∧ h = h. Now, s0 ∧ â = h∧ s∧ â = h∧ â∧ s = â∧ h∧ s = h∧ s = s0 (since Ŝ is compatible). Therefore, s0 ≤ â and â∧ s0 = â∧ h∧ s = h∧ â∧ s = h∧ s∧ â = s ∧ h ∧ â = h ∧ â = â. Therefore, â ≤ s0. Hence, s0 is unique. Theorem 7. Let S be a non-empty compatible subset of L. For any h1, h2 ∈ HS, we have (i) ̂(h1 ∨ h2) = ĥ1 ∨ ĥ2, (ii) ̂(h1 ∧ h2) = ĥ1 ∧ ĥ2, (iii) h ∈ HS ⇔ h = ĥ, for any ∅ ≠ S ⊆ L, (iv) 0̂ = 0. Proof. Let h1, h2 ∈ HS . By Theorem 6(ii), there exists ĥ1, ĥ2 ∈ Ŝ such that h1 ∧ ĥ1 = ĥ1, h2 ∧ ĥ2 = ĥ2 and ĥ1 ∧ h1 = h1, ĥ2 ∧ h2 = h2. (i) Now, (ĥ1 ∨ ĥ2) ∧ (h1 ∨ h2) = [ĥ1 ∨ ĥ2) ∧ h1] ∨ [ĥ1 ∨ ĥ2) ∧ h2] = [(ĥ1 ∧ h1) ∨ (ĥ2 ∧ h1)] ∨ [(ĥ1 ∧ h2) ∨ (ĥ2 ∧ h2)] = [h1 ∨ (ĥ2 ∧ h1)] ∨ [(ĥ1 ∧ h2) ∨ h2] = [(h1 ∨ (ĥ2 ∧ h1)] ∨ h2 = [(h1 ∨ ĥ2) ∧ h1)] ∨ h2 = [(ĥ2 ∨ h1) ∧ h1] ∨ h2 = h1 ∨ h2, (h1 ∨ h2) ∧ (ĥ1 ∨ ĥ2) = [(h1 ∨ h2) ∧ ĥ1] ∨ [(h1 ∨ h2) ∧ ĥ2] = [(h1 ∧ ĥ1) ∨ (h2 ∧ ĥ1)] ∨ [(h1 ∧ ĥ2) ∨ (h2 ∧ ĥ2)] = [ĥ1 ∨ (h2 ∧ ĥ1)] ∨ [(h1 ∧ ĥ2) ∨ ĥ2] = [(ĥ1 ∨ h2) ∧ ĥ1] ∨ ĥ2 = [(h2 ∨ ĥ1) ∧ ĥ1] ∨ ĥ2 = ĥ1 ∨ ĥ2. Therefore, ̂(h1 ∨ h2) = ĥ1 ∨ ĥ2. (ii) Now, (ĥ1 ∧ ĥ2) ∧ h1 ∧ h2 = ĥ1 ∧ h1 ∧ ĥ2 ∧ h2 = h1 ∧ h2, A. Iampan et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1691-1704 1703 h1 ∧ h2 ∧ (ĥ1 ∧ ĥ2) = h1 ∧ ĥ1 ∧ h2 ∧ ĥ2 = ĥ1 ∧ ĥ2. Therefore, ̂(h1 ∧ h2) = ĥ1 ∧ ĥ2. (iii) Let h ∈ S. Then h ∈ HS . By Theorem 6(iii), there exists a unique ĥ ∈ Ŝ such that h∧ ĥ = ĥ and ĥ∧ h = h. Since h, ĥ ∈ Ŝ and Ŝ is compatible, h = ĥ. The converse is trivial. (iv) It’s clear from (iii). Theorem 8. Let S be a compatible subset of L and HS = L. Then the following are equivalent: (i) HS = L ⇒ Ŝ has large element, (ii) L has a maximal element. Proof. (i) ⇒ (ii) Suppose that HS = L implies Ŝ has the largest element. Let l be the largest element in Ŝ. Let x ∈ L = HS . By Theorem 6(ii), there exists x̂ ∈ Ŝ such that x̂ ∧ x = x and x ∧ x̂ = x. Now, l ∧ x = l ∧ (x̂ ∧ x) = (l ∧ x̂) ∧ x = x̂ ∧ x = x. Therefore, l has a maximal element of L. (ii) ⇒ (i) Suppose that L has a maximal element, say m. If HS = L, for some S ⊆ L and S ̸= ∅, then m ∈ HS . By Theorem 6(ii), there exists m̂ ∈ Ŝ such that m̂ ∧ m = m and m ∧ m̂ = m̂. 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