EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6686 ISSN 1307-5543 – ejpam.com Published by New York Business Global Hierarchy Sets in Almost Distributive Lattices and Their Structural Properties G. Chinnayya1, Ramesh Sirisetti2, G. Jogarao3, Ravikumar Bandaru4, Aiyared Iampan5,∗ 1 Department of Mathematics, GITAM School of Science, GITAM (Deemed to be a University), Visakhapatnam 530045, India 2 Department of Mathematics, Aditya University, Surampalem, Kakinada, Andhra Pradesh- 533437, India 3 Department of BS & H, Aditya Institute of Technology and Management, Tekkali 532001, Srikakulam, India 4 Department of Mathematics, School of Advanced Sciences, VIT-AP University, Amaravati 522237, Andhra Pradesh, India 5 Department of Mathematics, School of Science, University of Phayao, Mae Ka, Mueang, Phayao 56000, Thailand Abstract. This paper studies hierarchy sets in almost distributive lattices, focusing on two key types: prime and maximal hierarchy sets. We show that every maximal hierarchy set is prime, but not vice versa, and use Zorn’s Lemma to prove the existence of prime hierarchy sets extending a given one. We also introduce inverted-hierarchy sets, defined via join operations, and analyze their relation to filters. The results provide structural insights and extend ideal and filter theory within almost distributive lattices. 2020 Mathematics Subject Classifications: 06D99, 06D75 Key Words and Phrases: almost distributive lattices, ideals, filters, hierarchy sets, prime hier- archy sets, maximal hierarchy sets, inverted-hierarchy sets 1. Introduction In lattice theory, the study of special subsets such as ideals, filters, and related con- structions [1–6] plays a crucial role in understanding structural properties of the lattice. In particular, hierarchy sets [7], defined by their closure under the meet operation with a fixed generating set, offer a useful framework for analyzing elements in almost distributive ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6686 Email addresses: cgondu@gitam.in (G. Chinnayya), ramesh.sirisetti@gmail.com (S. Ramesh), jogarao.gunda@gmail.com (G. Jogarao), ravimaths83@gmail.com (R. Bandaru), aiyared.ia@up.ac.th (A. Iampan) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 2 of 12 lattices (ADLs) [8]. This framework builds upon foundational work by Birkhoff [9], and re- lates to established studies in the structure of complemented and quasi-complemented lat- tices [10, 11], relatively complemented distributive lattices [12, 13], and the general theory of ideal-like constructs. Important extensions to the ADL setting have also been explored through weak relative complements, normal filters, and weakly complemented structures [14–17], offering alternative perspectives on distributivity and absorption in non-classical environments. Recently, Ramesh et al. introduced and developed the notion of hierarchy elements in ADLs [18], establishing foundational properties such as closure under meet, behavior with maximal elements, and conditions under which hierarchy elements generate ideals or sublattices. A natural extension of that work led to the formalization of hierarchy sets [7], shown to form a distributive lattice under union and a meet-based operation. In parallel, Khabyah’s work on star filters and starlets enriched the lattice-theoretic landscape by introducing new substructures under relaxed closure properties. Additionally, Rafi et al. contributed to this growing body of work by introducing prime E-ideals [19], which refine the concept of absorption and primeness within lattice ideals, and by developing w- filters [20], a generalization of filters based on weakened closure under joins. Collectively, these developments underscore the need to investigate more nuanced classes of subsets in ADLs that extend classical notions like prime ideals and filters. Within this context, two important classes of hierarchy sets are introduced in this paper: prime hierarchy sets and maximal hierarchy sets. Prime hierarchy sets generalize the notion of prime ideals [21] by requiring that the meet of two elements lies in the set only if at least one of the elements does. Maximal hierarchy sets, on the other hand, represent the largest proper hierarchy sets under inclusion. We establish several structural properties, show that every maximal hierarchy set is necessarily prime, and use Zorn’s Lemma to demonstrate the existence of prime hierarchy sets extending a given one while avoiding a particular subset. We provide examples to clarify the distinctions between these types of sets and highlight the fact that the converse of some implications does not hold in general. Also, we introduce inverted- hierarchy sets HS , defined for a non-empty subset S of an almost distributive lattice L with maximal elements. Each HS consists of elements in L that are idempotent under join with some u ∈ S. These sets exhibit notable algebraic properties, including closure under join and, in some cases, meet, forming filters or related substructures. We explore their characterizations, relationships to filters, and criteria under which HS aligns with or differs from the filter generated by S. 2. Preliminaries The study of lattice-theoretic structures has played a pivotal role in abstract algebra and its applications to computer science, logic, and information systems. In this aspect, the concept of an almost distributive lattice [8] was introduced by Swamy and Rao in 1981 as a common abstraction of both lattice-theoretic and ring-theoretic generalizations of a Boolean algebra (ring). It is an algebraic structure which satisfies all axioms of a distributive lattice (L,∨,∧, 0) with the zero element 0 except the commutativity of the binary operations ∨,∧, the right distributivity of ∨ over ∧, and the associativity of ∨. G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 3 of 12 Given an almost distributive lattice L and a non-empty subset S ⊆ L, an element h ∈ L is said to be a hierarchy with respect to S [7] if it satisfies a specific absorption condition with at least one element of S. The set of all such elements, denoted by HS , exhibits several interesting structural properties. In particular, this set is non-empty, contains S, and is closed under the meet operation. In this context, the notion of hierarchy elements and their associated hierarchy sets provides a useful framework for examining the internal organization of elements within an almost distributive lattice. The study of hierarchy sets not only enhances our understanding of ideal-theoretic constructions in almost distributive lattices but also bridges concepts related to sub-almost distributive lattices, closure operators, and distributivity. Several results characterize the algebraic and order-theoretic behavior of hierarchy sets, including their stability under inclusion, intersection, and union. Additionally, under certain conditions, HS forms an ideal and even a sub-almost distributive lattice. Definition 1. [7] Given a non-empty subset S of L, an element h ∈ L is said to be hierarchy with respect to S if s∧ h = h, for some s ∈ S. It is observed that the set HS of hierarchy elements with respect to S is non-empty, containing S, and it is closed under ∧. Lemma 1. [7] For any non-empty subsets S1, S2 of L, (i) S1 ⊆ S2 implies HS1 ⊆ HS2 (ii) HS1∪S2 = HS1 ∪HS2 (iii) HS1∩S2 ⊆ HS1 ∩HS2. Lemma 2. [7] For any non-empty subsets S of L, (i) If m ∈ HS, then HS = L, where m is a maximal element in L (ii) a ≤ b implies Ha ⊆ Hb, where Ha = {h ∈ L | a ∧ h = h} (iii) a ≤ b and b ∈ HS imply a ∈ HS (iv) h ∈ HS implies (h] ⊆ HS, where (h] = {h ∧ a | a ∈ L} (v) Ha is an ideal of L. Theorem 1. [7] For any non-empty subset S of L, the following are equivalent; (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]). Theorem 2. [7] The set HS of hierarchy sets in L forms a distributive lattice with respect to the operations; HS1 ∪HS2 = HS1∪S2 and HS1 ∧ HS2 = {s1 ∧ s2 | s1 ∈ S1, s2 ∈ S2}. G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 4 of 12 3. Prime Hierarchy Sets in Almost Distributive Lattices In this section, we investigate structural properties of hierarchy sets in an abstract distributive lattice. A hierarchy set HS generated by a non-empty subset S of an almost distributive lattice L plays a central role in the algebraic and order-theoretic analysis of L. We begin by establishing foundational closure properties of hierarchy sets under meet operations. Building on this, we introduce and characterize two important classes of hierarchy sets: prime and maximal hierarchy sets. A prime hierarchy set satisfies the absorption condition that whenever a meet a ∧ b belongs to the set, then at least one of the elements a or b must also belong to it. A maximal hierarchy set, on the other hand, is one that is not properly contained in any other proper hierarchy set. Through illustrative examples, we demonstrate that while every maximal hierarchy set is prime, the converse need not hold. We further employ Zorn’s Lemma to show the existence of prime hierarchy sets ex- tending a given one, while avoiding a specific closed set under the meet operation. The section concludes by demonstrating that every hierarchy set can be represented as the intersection of all prime hierarchy sets that contain it. Lemma 3. If S is a non-empty subset and a is an element in L, then (i) h ∧ a ∈ HS, for all h ∈ HS (ii) a ∧ h ∈ HS, for all h ∈ HS. Proof. Let h ∈ HS . Then there exists s ∈ S such that s ∧ h = h. Given a ∈ L. (i) s ∧ (h ∧ a) = (s ∧ h) ∧ a = h ∧ a. Therefore, h ∧ a ∈ HS . (ii) s∧ (a∧ h) = (s∧ a)∧ h = (a∧ s)∧ h = a∧ (s∧ h) = a∧ h. Therefore, a∧ h ∈ HS . A hierarchy set HS in L is said to be proper if HS ̸= L. Definition 2. A proper hierarchy set HS of L is said to be prime, if given a, b ∈ L, a ∧ b ∈ HS implies a ∈ HS or b ∈ HS. Remark 1. Every hierarchy set does not need to be prime. For example, see the following example: Example 1. Let L = {0, a, b, 1} be an almost distributive lattice whose Hasse diagram is given below: 1 a 0 b G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 5 of 12 For S = {0}, HS = {0} is not prime. Definition 3. A proper hierarchy set HS of L is said to be maximal, if, given a proper hierarchy set HT in L, HS ⊆ HT implies HS = HT . Theorem 3. Every maximal hierarchy set is prime. Proof. Let HS be a maximal hierarchy set in L and a, b ∈ L such that a ∧ b ∈ HS . If a /∈ HS , then HS ⊆ HS ∪ Ha = HS∪{a} = L (since HS is maximal). Now, for this b ∈ L = HS ∪ Ha. we have b ∈ HS , or b ∈ H{a}, or b ∈ HS ∩ H{a}. If b ∈ HS , or b ∈ HS ∩H{a}, then nothing else to do. If b ∈ H{a}, then b = a ∧ b ∈ HS . Therefore, HS is prime. Remark 2. The converse of Theorem (3) need not be true. That is, every prime hierarchy set need not be maximal. For example, see the following example: Example 2. Let L = {0, a, b, c, 1} be an almost distributive lattice whose Hasse diagram is given below: c a 0 b 1 Let S1 = {a, b} and S2 = {b, c}. Then HS1 = {0, a, b} and HS2 = {0, a, b, c}. Therefore, HS1 and HS2 are two prime hierarchy sets and HS1 ⫋ HS2. Hence, HS1 is prime but not maximal. Theorem 4. Let HS be a hierarchy set and K is a non-empty subset of L which is closed under ∧ such that HS ∩K = ∅. Then there exists a prime hierarchy set HP of L such that HS ⊆ HP and HP ∩K = ∅. Proof. Let HS be a hierarchy set and K be a non-empty subset of L and closed under ∧ such that HS ∩ K = ∅. Consider Q = {HT | HS ⊆ HT and HT ∩ K = ∅}. Then Q ̸= ∅ (since HS ∩ K = ∅) and (Q,⊆) is a partially ordered set with respect to the inclusion order. Let {0} = HS1 ⊆ HS2 ⊆ HS3 ⊆ . . . be an increasing chain in Q. Then ⋃ i∈I HSi = HS1 ∪ HS2 . . . = H( ⋃ i∈I Si), H( ⋃ i∈I Si) ∩ K = ⋃ i∈I (HSi ∩ K) = ⋃ i∈I ∅ = ∅, and G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 6 of 12 HS ⊆ HSi , for all i ∈ I. Therefore, H( ⋃ i∈I Si) is an upper bound of the chain in Q. By Zorn’s lemma, Q has maximal element, say HP . Let a, b ∈ L such that a /∈ HP and b /∈ HP . Then a /∈ P and b /∈ P . Now, HP ∪ Ha = HP∪{a} and HP ∪ Hb = HP∪{b}. Since HP is maximal, HP∪{a} ∩ K ̸= ∅ and HP∩{b} ∩ K ̸= ∅. Let x ∈ HP∪{a} ∩ K and y ∈ HP∪{b} ∩ K. By Lemma (3), x ∧ y ∈ HP∪{a} ∩ HP∪{b} ∩ K. Then x ∧ y ∈ (HP ∪Ha)∩ (HP ∪Hb)∩K = [HP ∪ (Ha ∧Hb)]∩K = (HP ∪H a∧b)∩K = HP∪{a∧b} ∩K. If a ∧ b ∈ HP , then HP∪{a∧b} = HP . So that x ∧ y ∈ HP ∩ K and hence HP ∩ K ̸= ∅. Which is a contradiction. Therefore, a∧ b /∈ HP and hence HP is a prime hierarchy set of L. Corollary 1. Let HS be a hierarchy set in L and a ∈ L such that a /∈ HS. Then there exists a prime hierarchy set HP of L such that HS ⊆ HP and a /∈ HP . Proof. Let HS be a hierarchy set in L and a ∈ L such that a /∈ HS . If K = {a}, then HS ∩K = ∅ and K is closed under ∧. By Theorem 4, there exists a prime hierarchy set HP in L such that HS ⊆ HP , and HP ∩ K = ∅. Hence, a /∈ HP and HS ⊆ HP . Since P ⊆ HP , a /∈ P . Theorem 5. If HS is a hierarchy set in L, then HS is the intersection of all prime hierarchy sets containing HS in L. Proof. Let S be a non-empty subset of L and a ∈ L such that a /∈ HS . Consider a set Q = {HT | a /∈ HT and HS ⊆ HT }. Then Q ≠ ∅ (since a /∈ HS) and it is a poset with the inclusion order. Let HS1 ⊆ HS2 ⊆ . . . be an increasing chain in Q. Then⋃ i∈I HSi = H( ⋃ i∈I Si). If a ∈ H( ⋃ i∈I Si), then a ∈ ⋃ i∈I HSi . Therefore, a ∈ HSi , for some i ∈ I. Which is a contradiction to a /∈ HSi . So that a /∈ H( ⋃ i∈I Si). Since HS ⊆ HSi , for all i ∈ I, HS ⊆ H( ⋃ i∈I Si). Therefore, H( ⋃ i∈I Si) ∈ Q and it is an upper bound for the chain. By Zorn’s lemma, Q has a maximal element, say HP . That is a /∈ HP and HS ⊆ HP . Let a, b ∈ L such that a /∈ HP and b /∈ HP . Then HP ∪ Ha = HP∪{a} and HP ∪ Hb = HP∪{b}. Since HP is maximal in Q, a ∈ HP ∪ Ha and a ∈ HP ∪ Hb, and then a ∈ (HP ∪ Ha) ∩ (HP ∪Hb) = HP ∪ (Ha ∩ Hb) = HP ∪ Ha∧b. If a ∧ b ∈ HP , then {a ∧ b} ⊆ HP , and H{a∧b} ⊆ HHP = HP . Therefore, a ∈ HP . Which is a contradiction. Hence, HP is a prime hierarchy set in L. Thus, HS = ⋂ {HP | HP is a hierarchy set containing HS in L. A non-empty subset F of L is said to be a filter [8] if it is closed under ∧ and given a ∈ L, b ∈ F , a∨b ∈ F . A proper filter F of L is called prime [8] if given a, b ∈ L, a∨b ∈ F implies a ∈ F or b ∈ F . Theorem 6. If HP is a prime hierarchy set of L, where P is a non-empty subset of L, then L \HP is a filter of L. Proof. Let HP be a prime hierarchy set, where P is a non-empty subset of L. Since HP is proper, we can choose a, b ∈ L \HP . If a ∧ b ∈ HP , then a ∈ HP or b ∈ HP (since G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 7 of 12 HP is prime). Which is not possible. Therefore, a∧ b ∈ L\HP and hence L\HP is closed under ∧. Let c ∈ L and a ∈ L \ HP . If c ∨ a ∈ HP , then p ∧ (c ∨ a) = c ∨ a, for some p ∈ HP . So that p ∧ (c ∨ a) ∧ a = (c ∨ a) ∧ a and then p ∧ a = a. It means that a ∈ HP . Which is not true. Therefore, c ∨ a ∈ L \HP and hence L \HP is a filter. Remark 3. If HP is a prime hierarchy set in L, then L \ HP need not be prime. For example, see the following example: Example 3. Let L = {0, a, b, c,m1,m2} be an almost distributive lattice with maximal elements m1,m2, where the operations ∧ and ∨ are defined below: ∧ 0 a b c m1 m2 0 0 0 0 0 0 0 a 0 a 0 a a a b 0 0 b b b b c 0 a b c c c m1 0 a b c m1 m2 m2 0 a b c m1 m2 ∨ 0 a b c m1 m2 0 0 a b c m1 m2 a a a c c m1 m2 b b c b c m1 m2 c c c c c m1 m2 m1 m1 m1 m1 m1 m1 m1 m2 m2 m2 m2 m2 m2 m2 Take P = {a, b}. Then HP = {0, a, b} is a prime hierarchy set in L and L \ HP = {c,m1,m2} is a filter of L, but not prime. 4. Inverted-Hierarchy Sets in an Almost Distributive Lattice with Maximal Elements This section introduces and develops the theory of inverted-hierarchy sets, denoted by HS , where S is a non-empty subset of an almost distributive lattice L with maximal elements. These sets consist of elements in L that are idempotent over at least one element of S under the join operation. Such sets exhibit rich algebraic properties, including closure under join and, under suitable conditions, closure under meet, forming filters and substructures of the lattice. We explore various characterizations and properties of HS , its relationship to filters, and conditions under which it coincides with or differs from the filter generated by S. Definition 4. An element h in L with maximal elements is said to be inverted-hierarchy with respect to a non-empty set S in L, if h ∨ s = h, for some s ∈ S. Let us denote HS as the set of inverted-hierarchy elements with respect to a non-empty set S in L. Then it is easy to observe that HS ̸= ∅ (because m ∨ s = m, for all s ∈ S, where m is a maximal element in L) and S ⊆ HS. Theorem 7. For any non-empty subset S of L, we have (i) S is closed under ∨ (ii) For any h ∈ HS , [h) ⊆ HS G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 8 of 12 (iii) For any h ∈ HS and a ∈ L, a ∨ h, h ∨ a ∈ HS (iv) If a ∈ L and s ∈ S such that s ≤ a, then a ∈ HS. Proof. (i) Let h1, h2 ∈ HS . Then h1 ∨ s1 = h1 and h2 ∨ s2 = h2, for some s1, s2 ∈ S. Now, (h1 ∨ h2) ∨ s2 = h1 ∨ (h2 ∨ s2) = h1 ∨ h2. Then h1 ∨ h2 ∈ HS . Therefore, HS is closed under ∨. (ii) Let h ∈ HS . Then h ∨ s = h, for some s ∈ S. Let a ∈ [h). Then a = b ∨ h, for some b ∈ L. Now, a ∨ s = (b ∨ h) ∨ s = b ∨ (h ∨ s). Then a ∈ HS . Therefore, [h) ⊆ HS . (iii) Let h ∈ HS . Then h∨s = s, for some s ∈ S. Given a ∈ L, (a∨h)∨s = a∨(h∨s) = a ∨ h. Therefore, a ∨ h ∈ HS . Similarly, (h ∨ a) ∧ s = (a ∨ h) ∧ s = (a ∧ s) ∨ (h ∧ s) = (a ∧ s) ∨ s = s. Therefore, (h ∨ a) ∨ s = h ∨ a and hence h ∨ a ∈ HS . (iv) Let a ∈ L and s ∈ S such that s ≤ a. Then a ∨ s = a. Therefore, a ∈ HS . Lemma 4. If S1, S2 are any two non-empty subsets of L, then (i) S1 ⊆ S2 implies HS1 ⊆ HS2 (ii) HS1 ∪HS2 = HS1∪S2 (iii) HS1∩S2 ⊆ HS1 ∩HS2. Proof. Let S1, S2 be two non-empty subsets in L. (i) Let h ∈ HS1 . Then h ∨ s1 = h, for some s1 ∈ S1 ⊆ S2. Therefore, h ∈ HS2 and hence HS1 ⊆ HS2 . (ii) By (i), we have HS1 , HS2 ⊆ HS1∪S2 . Therefore, HS1 ∪ HS2 ⊆ HS1∪S2 . Let h ∈ HS1∪S2 . Then h ∨ s = h for some s ∈ S1 ∪ S2 ⊆ HS1 ∪ HS2 . By Theorem 7 (iii), h = h ∨ sHS1 ∪HS2 and hence HS1∪S2 ⊆ HS1 ∪HS2 . Thus, HS1 ∪HS2 = HS1∪S2 . (iii) We have S1 ∩ S2 ⊆ S1, S2. Then HS1∩S2 ,⊆ HS1 , HS2 (by (ii)). Therefore, HS1∩S2 ⊆ HS1 ∩HS2 . Remark 4. HS1∩S2 need not be equal to HS1 ∩HS2. For, in Example (1); Let S1 = {a, b} and S2 = {0, a}. Then S1 ∩S2 = {a}, HS1 = {a, b, 1}, HS2 = {0, a, b, 1}, HS1∩S2 = {a, 1} and HS1 ∩HS2 = {a, b, 1}. Therefore, HS1 ∩HS2 ̸= HS1∩S2. Lemma 5. If s1 ∈ S1, s2 ∈ S2 and s1 ∧ s2 ∈ S1 ∩ S2, then HS1∩S2 = HS1 ∩HS2. Proof. Let h ∈ HS1 ∩ HS2 . Then h ∨ s1 = h and h ∨ s2 = h, for some s1 ∈ S1 and s2 ∈ S2. Since s1 ∧ s2 ∈ S1 ∩S2, h∨ (s1 ∧ s2) = (h∨ s1)∧ (h∨ s2) = h∧h = h. Therefore, h ∈ HS1∩S2 . Hence, HS1 ∩HS2 ⊆ HS1∩S2 . Thus, HS1 ∩HS2 = HS1∩S2 (by Lemma 7 (iv)). Theorem 8. Let S be a non-empty subset of L and S be closed under ∧. Then (i) HS is closed under ∧ (ii) HS is a sub-almost distributive lattice of L G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 9 of 12 (iii) HS is a filter of L (iv) HS is the smallest filter containing S. Proof. Let S be a non-empty subset of L and S is closed under ∧. (i) Let h1, h2 ∈ HS . Then h1 ∨ s1 = h1 and h2 ∨ s2 = h2, for some s1, s2 ∈ S. Now, h1 ∧h2 ∧ (s1 ∧ s2) = h1 ∧ s1 ∧h2 ∧ s2 = s1 ∧ s2. Then (h1 ∧h2)∨ (s1 ∧ s2) = h1 ∧h2. Since S is closed under ∧, h1 ∧ h2 ∈ HS . Therefore, HS is closed under ∧. (ii) By (i) and Theorem 7 (iii), HS is a sub-almost distributive lattice of L. (iii) By (i) and Theorem 7 (iii), HS is a filter of L. (iv) Let F be a filter of L such that S ⊆ F . By Lemma 4 (i), HS ⊆ HF and F ⊆ HF . Let h ∈ HF . Then h ∨ s = h, for some s ∈ F . For s ∈ F, h ∨ s = h ∈ F (since F is a filter). Therefore, h ∈ F . So that HF ⊆ F . Hence, HS ⊆ F = HF . Thus, HS is the smallest filter containing S. Remark 5. HS need not be closed under ∧. See the following example: Example 4. Let L = {0, a, b, c, 1} be an almost distributive lattice with maximal element 1, whose Hasse diagram is given below: 1 c b a 0 Let S1 = {b, c}. Then HS = {b, c, 1}. Let b, c ∈ HS. Then b ∧ c = a /∈ HS. Therefore, HS is not closed under ∧. Given a non-empty set S of L, it is known that [S) = {a ∨ ( n∧ i=1 si) | a ∈ L and si ∈ S} is the smallest filter containing S. Lemma 6. For any non-empty subset S of L, HS ⊆ [S). Proof. Let h ∈ HS . Then h∨ s = h and h∧ s = s, for some s ∈ S. Since [S) is a filter generated by S, h ∨ s = h ∈ [S). Therefore, HS ⊆ [S). Remark 6. HS need not be equal to [S). See the following example: G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 10 of 12 Example 5. Let L = {0, a, b, c, d, e, f, 1} be an almost distributive lattice whose Hasse diagram is given below: 1 d b f e a 0 c Let S = {e, f}. Then [S) = {c, e, f, 1} is a filter of L and HS = {e, f, 1}. Therefore, HS ̸= [S). Lemma 7. Let S be a non-empty subset of L and HS = HF , for some filter F of L. Then [S) = F . Proof. Suppose HS = HF , for some filter F of L. Since F ⊆ HF = HS and F is closed under ∧, F ⊆ HF = HS ⊆ [S) and HF is a filter of L. Therefore, F = HS ⊆ [S). Now, S ⊆ HS = HF = F . Then [S) ⊆ HF = F . Hence, F = [S). Remark 7. The converse of Lemma 7 need not be true. For, in Example (5); let L = {0, a, b, c, d, e, f, 1} be an almost distributive lattice whose Hasse-diagram is given in Example 5. Let S = {d, e}. Then [S) = F = {a, d, e, 1} is a filter of L and HS = {d, e, 1}, but HS ̸= HF = F . Theorem 9. For any non-empty subsets S1, S2 of L, we have (i) HS1 ∪HS1 = HS1 (Idempotent Law) (ii) HS1 ∪HS2 = HS1∪S2 = HS2 ∪HS1 = HS2∪S1(Commutative Law) (iii) (HS1 ∪HS2) ∪HS3 = H(S1∪S2)∪S3 = HS1 ∪ (HS2 ∪HS3)(Associative Law). Proof. (i) It is easy to observe that HS1 ∪HS1 = HS1 , by Lemma 4 (i). (ii) By Lemma 4 (ii), we can prove HS1 ∪HS2 = HS1∪S2 = HS2∪S1 = HS2 ∪HS1 . (iii) Since the set union satisfies associative law and by (ii), we have (HS1 ∪ HS2) ∪ HS3 = H(S1∪S2)∪S3 = HS1 ∪ (HS2 ∪HS3). Theorem 10. For any non-empty subset S of L and SM is the set of maximal elements in L, we have G. Chinnayya et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6686 11 of 12 (i) ⋃ S⊆L HS = L (ii) SM ⊆ HS (iii) HSM = SM (iv) ⋂ S⊆L HS = SM (v) HHS = HS. Proof. (i) Since L ⊆ HL, HL = L and ⋃ S⊆L HS = L. (ii) Let m ∈ SM . Then m ∨ s = m, for all s ∈ S. Therefore, m ∈ HS and SM ⊆ HS . (iii) Since SM is a filter L, HSM = SM . (iv) From (ii), SM ⊆ HS , for all S ⊆ L, so that SM ⊆ ⋂ S⊆L HS . Let h ∈ ⋂ S⊆L HS . Then h ∨ s = h, for some s ∈ S and for all S ⊆ L. Let m ∈ SM . Since SM is a filter of L, h = h ∨m ∈ SM . Therefore, ⋂ S⊆L HS ⊆ SM and hence ⋂ S⊆L HS = SM . (v) Since S ⊆ HS , HS ⊆ HHS . Let h ∈ HHS . Then h ∨ t = h, for some t ∈ HS . For this t ∈ HS , t ∨ s = t, for some s ∈ S. Now, h ∧ s = h ∧ (t ∧ s) = (h ∧ t) ∧ s = t ∧ s = s. Then h ∨ s = h, for some s ∈ S. Therefore, h ∈ HS and HHS ⊆ HS . Thus, HHS = HS . 5. Conclusion These findings not only clarify the distinctions and relationships between prime, max- imal, and inverted-hierarchy sets but also enrich the broader theory of lattices by high- lighting how such structures interact with classical notions, such as ideals and filters, in almost distributive lattices. Acknowledgements This research was supported by University of Phayao and Thailand Science Research and Innovation Fund (Fundamental Fund 2026, Grant No. 2252/2568). References [1] N. H. McCoy and D. Mantgomery. A representation of generalized boolean rings. Duke. Math. 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