EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6270 ISSN 1307-5543 – ejpam.com Published by New York Business Global A Topological Structure on D-Algebras Maha W. Abdulqader1, Alias B. Khalaf1,∗ 1 Department of Mathematics, College of Science, University of Duhok, Kurdistan Region, Iraq Abstract. The primary aim of this paper is applying the concept of b-open sets in topological spaces to explore the idea of b-topological d- algebra (Tbd-algebra), which is a d-algebra equipped with a specific type of topology that ensured the binary operation that is defined on them to be d-topologically continuous. This idea generalizes the notion of topological d-algebras. In addition, we present some relations between open sets and b-open sets in a Tbd-algebra. We also construct some relations between Ti and b-Ti-spaces for (i=0,1,2). Finally, we use left maps on positive implicative d-algebras to establish some Tbd-algebras. 2020 Mathematics Subject Classifications: 03G25, 22A30, 54A05 Key Words and Phrases: b-open set, d-algebra, positive implicative d-algebra, edge d-algebra, b-T2 space, Tbd-algebra 1. Introduction Topology and Algebra are two significant areas of pure mathematics. Topology fo- cuses on concepts such as continuity and convergence, while Algebra explores various operations, forming the foundation for calculations and algorithms. A key principle that links algebraic operations and topology is the requirement that these operations be con- tinuous topologically, either jointly continuous or in the first or second variable, this field is known as Topological Algebra. In recent years, numerous researchers have advanced this area of study. Since the early twentieth century, many mathematicians have made substantial contributions to progress of these interrelated subjects. Generalized open sets are fundamental to general topology and are acknowledged as significant research areas by topologists around the world. These sets are central to major themes in both general topology and real analysis, particularly in connection with various modified forms of con- tinuity, separation axioms, and other related concepts. One of the forms of generalized open sets in topological spaces, known as b-open sets, was introduced by Andrijevic [1] in 1996. These sets were presented by Al-Etik [2] under the name λ-open sets. Further- more, Caldas and Jafari [3] used open sets b to define separation axioms b-Ti (i ∈ 0, 1, 2) ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6270 Email addresses: maha.waleed@uod.ac (M. W. Abdulqader), alias.khalaf@uod.ac (A. B. Khalaf) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 2 of 17 in topological spaces. The concept of d-algebra was introduced by Neggers and Kim in [4]. The topological BCK-algebra and the topological d-algebras were defined in [5] and [6] respectively. Recently, Khalaf in [7–9] introduced some topological notions defined on BCK-algebras. We can notice that the concept of b-open was included in many topics and here we review some of them, in [10] the supra-b limit points and supra-b separation axioms were investigated, also in [11] b-open sets via infra soft topological spaces were studied. Limit points and separation axioms with respect to supra semi-open sets were established in [12]. By (W,Ω) we mean a topological space and if M is any subset of a topological space (W,Ω), then the interior and closure of M are denoted by Int(M) and Cl(M), respec- tively. 2. Preliminaries In this section, we recall some definitions and results that are needed in the next section. Definition 1. In a topological space W, a subset M is called b-open [1] (resp., regular open [13], semi-open [14], pre-open [15]) if M ⊆ Int(Cl(M)) ∪ Cl(Int(M)) ( resp., M = Int(Cl(M)), M ⊆ Cl(Int(M)), M ⊆ Int(Cl(M)) ) . Lemma 1. [1] The intersection of an open and a b-open set is a b-open set. Lemma 2. [1] In a topological space W, a subset M is b-open if and only if M =( M ∩ Int(Cl(M)) ) ∪ ( M ∩ Cl(Int(M))) Definition 2. [16] A topological space (W,Ω) is called: (i) locally indiscrete if every open set is closed. (ii) extremally disconnected if Cl(U) ∈ Ω for every U ∈ Ω. (iii) submaximal if every dense subset of W is open. Lemma 3. [16] (i) In a locally indiscrete space, every subset of W is pre-open. (ii) In a submaximal space, every pre-open set is open. Lemma 4. [2] If (W,Ω) is an extremally disconnected, then (i) every b-open set is pre-open. (ii) a subset M of W is b-open if and only if W \M is dense. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 3 of 17 Lemma 5. [2] (i) If M is both open and b-closed subset in W, then it is semi-closed. (ii) If M is a b-closed subset of a b-compact space W, then it is b-compact. Definition 3. [17] A topological space W is semi-disconnected if it can be partitioned into two non-empty open sets U and V such that Cl(U) ∩ V = ∅ and Cl(V ) ∩ U = ∅. Definition 4. [3, 13] A topological space (W,Ω) is said to be: (i) b-T0 (resp., T0) ∀ ζ, η ∈ W, ∃ a b-open (open) set containing one of them but not the other. (ii) b-T1 (resp., T1) if ∀ ζ, η ∈ W, ∃ b-open (open) sets G,H such that ζ ∈ G, η /∈ G and η ∈ H, ζ /∈ H. (iii) b-T2 (resp., T2) if ∀ ζ, η ∈ W, ∃ b-open (open) sets G,H such that ζ ∈ G, η ∈ H, G ∩H = ϕ. Definition 5. [4] d-algebra is an algebra (W,⊙, 0) of type (2, 0) such that ⊙ is a binary operation and 0 is a fixed element, satisfy the axioms listed below: for each ζ, η, θ ∈ W, (i) ζ ⊙ ζ = 0, (ii) ζ ⊙ η = 0 and η ⊙ ζ = 0 ⇒ ζ = η, (iii) 0⊙ ζ = 0. Moreover, if a d-algebra satisfies the following axioms (i) (B1) ((ζ ⊙ η)⊙ (ζ ⊙ θ))⊙ (θ ⊙ η) = 0, (ii) (B2) (ζ ⊙ (ζ ⊙ η))⊙ η = 0. Then it is a BCK-algebra. We indicate a partial order relation (≤) by ζ ≤ η ⇐⇒ ζ ⊙ η = 0. Definition 6. [18] A d-algebra (W,⊙, 0) is said to be edge d-algebra, if ζ ⊙W = {ζ, 0} for all ζ ∈ W Lemma 6. [4] If (W,⊙, 0) is an edge d-algebra, then it satisfies condition (B2). Definition 7. [6] A nonempty subset I of a d-algebra (W,⊙, 0) is said to be an ideal of W if both of the following conditions are met: (i) 0 ∈ I , (ii) ∀ ζ ∈ W, ∀η ∈ I, if ζ ⊙ η ∈ I, then ζ ∈ I. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 4 of 17 Definition 8. [19] Let α be an element of a d-algebra W. α is said to be an atom in W if, for any ζ ∈ W, α⊙ ζ = 0 implies α = ζ. Definition 9. [4] Let Z be a subset of a d-algebra W, Z is called a d-subalgebra if it is also a d-algebra. Definition 10. [6] Let W be a d-algebra , α ∈ W. A left map Lα : W → W defined by, Lα(ζ) = α⊙ ζ,∀ζ ∈ W and a right map Rα : W → W by Rα(ζ) = ζ ⊙ α ∀ζ ∈ W. L(W) represents the family of all left maps on W, while R(W) represents the family of all right maps. W is denoted by and the family of all right maps If S ⊆ W then Lα(S) = α⊙S and Rα(S) = S ⊙ α. Definition 11. [20] A d-algebra W is known as a positive implicative d-algebra , if (η ⊙ ζ)⊙ (θ ⊙ ζ) = (η ⊙ θ)⊙ ζ for all ζ, η, θ ∈ W. Definition 12. [5] A BCK-algebraW with a topology Ω is a TBCK-algebra if the function f : W × W → W known as f(ζ, η) = ζ ⊙ η possesses the attribute that for every open set O having ζ ⊙ η, there exist open sets U, V having ζ, η correspondingly such that f(U, V ) = U ⊙ V ⊆ O ∀ ζ, η ∈ W. Definition 13. [6] A d-algebraW with a topology Ω is called an Td-algebra if the function f : W × W → W known as f(ζ, η) = ζ ⊙ η possesses the attribute that for every open set O having ζ ⊙ η, there exist open sets U, V having ζ, η correspondingly such that f(U, V ) = U ⊙ V ⊆ O,∀ ζ, η ∈ W. 3. b-topological d-algebras This section presents the idea of b-topological d-algebras and covers some of its prop- erties. Definition 14. A d-algebra W equipped with a topology Ω is known as b-topological d-algebra (Tbd-algebra) if f : W × W → W defined by f(ζ, η) = ζ ⊙ η has the property that for each open set O having ζ ⊙ η,, there are b-open sets U, V having ζ, η respectively, such that f(U, V ) = U ⊙ V ⊆ O ∀ ζ, η ∈ W. By the definitions of TBCK-algebra and Td-algebra, and since every BCK-algebra is a d-algebra, we deduce that every TBCK-algebra is a Td-algebra and every Td-algebra is a Tbd-algebra. The following example shows that the implication is not reversible in general. Example 1. Let W = {0, α, β, γ} and ⊙ be given as in the Cayley table as follows: It is easy to verify that From Table 2. Next think about the topology Ω on W given as: Ω = {ϕ, {α}, {γ}, {α, γ},W}. Then W is not a Td-algebra because γ ⊙ β = γ, and the only open set having β is W and {γ} × W ̸⊆ {γ}. It is easy to confirm that the b-open sets in (W,Ω) are P (W) \ ({0}, {β}, {0, β}) and we can show by basic computation that (W,Ω) is a Tbd-algebra. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 5 of 17 ⊙ 0 α β γ 0 0 0 0 0 α α 0 0 α β β β 0 0 γ γ γ γ 0 Table 1: A Tbd-algebra which is not Td-algebra Proposition 1. For each subset M of a Tbd-algebra W and any element ζ ∈ W, the following statements are true: (i) Clb(M)⊙ ζ ⊆ Cl(M ⊙ ζ). (ii) If Clb(M)⊙ ζ is closed, then Clb(M)⊙ ζ = Cl(M ⊙ ζ). Proof. (i) Suppose that η = α ⊙ ζ ∈ Clb(M) ⊙ ζ where α ∈ Clb(M) and U is any open set having η. Since W is a Tbd-algebra, so there exist b-open sets V that have α and G having ζ such that V ⊙G ⊆ U . Also, we have α ∈ Clb(M) implies that M ∩ V ̸= ϕ. Now assume that θ ∈ M ∩ V , so θ ⊙ ζ ∈ M ⊙ ζ and θ ⊙ ζ ∈ V ⊙ ζ ⊆ V ⊙ G ⊆ U . Hence θ⊙ζ ∈ U∩(M⊙ζ) implies that η ∈ Cl(M⊙ζ). Thus, Clb(M)⊙ζ ⊆ Cl(M⊙ζ). (ii) Suppose that Clb(M)⊙ ζ is closed, we have M ⊙ ζ ⊆ Clb(M)⊙ ζ and hence Cl(M ⊙ ζ) ⊆ Clb(M)⊙ ζ. Therefore, by (i) we get the equality. In general, the inclusion of (i) can not replaced by equality as it can be seen in Example 1, if we take M = {o, α}, then Clb(M)⊙β = {0} and Cl(M⊙β) = {0, β}, so Clb(M)⊙β ̸= Cl(M ⊙ β). Proposition 2. If M is any subset of a Tbd-algebra W and any element ζ ∈ W, , the statements listed below are accurate: (i) ζ ⊙ Clb(M) ⊆ Cl(ζ ⊙M). (ii) If ζ ⊙ Clb(M) is closed, then ζ ⊙ Clb(M) = Cl(ζ ⊙M). Proof. (i) Suppose that η ∈ W ⊙Clb(M) and U be any open set having η. So η = ζ ⊙α where α ∈ Clb(M). Since W is a Tbd-algebra, there exist open sets V having ζ and G having α such that V ⊙G ⊆ U . Also, we have α ∈ Clb(M) implies that M ∩G ̸= ϕ. Now assume that θ ∈ M ∩G, so ζ ⊙ θ ∈ W ⊙M and ζ ⊙ θ ∈ W ⊙G ⊆ V ⊙G ⊆ U . Hence we obtain that η ∈ Cl(M ⊙ ζ). M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 6 of 17 (ii) Assume that ζ ⊙ Clb(M) is closed and let ζ ⊙M ⊆ ζ ⊙ Clb(M). Therefore, Cl(ζ ⊙ M) ⊆ ζ ⊙ Clb(M). From (1), we get ζ ⊙ Clb(M) = Cl(ζ ⊙M). In general the inclusion in (i) can not be replaced by equality, as shown in Example 1, if we take M = {o, α}, then β⊙Clb(M) = {β} and Cl(β⊙M) = {0, β}, so β⊙Clb(M) ̸= Cl(β ⊙M). Proposition 3. For each subsets M and N of a Tbd-algebra W, the statements listed below are accurate.: (i) Clb(M)⊙ Clb(N) ⊆ Cl(M ⊙N). (ii) If Clb(M)⊙ Clb(N) is closed, then Clb(M)⊙ Clb(N) = Cl(M ⊙N). Proof. (i) Let ζ = α ⊙ β ∈ Clb(M) ⊙ Clb(N) and U be any open set having ζ. Since W is a Tbd-algebra, so there exist open sets V having α and G having β such that V ⊙ G ⊆ U . Also, we have α ∈ Clb(M) and β ∈ Clb(N), implies that M ∩ V ̸= ϕ and N ∩G ̸= ϕ. assume that α1 ∈ M ∩V and β1 ∈ N ∩G, so α1 ⊙ β1 ∈ M ⊙N and α1 ⊙ β1 ∈ V ⊙G ⊆ U . Hence we get ζ ∈ Cl(M ⊙N). (ii) Assume that Clb(M) ⊙ Clb(N) is closed. We have M ⊆ Clb(M) and N ⊆ Clb(N) and hence M ⊙ N ⊆ Clb(M) ⊙ Clb(N). Therefore, by hypothesis, Cl(M ⊙ N) ⊆ Clb(M)⊙ Clb(N). Hence, from (i), the equality holds. From Proposition 1 and Proposition 2, we obtain the following result: Corollary 1. For a subset M of a Tbd-algebra W and an element ζ ∈ W, the statements listed below are accurate: (i) If M ⊙ ζ is closed, then Clb(M)⊙ ζ = M ⊙ ζ. (ii) If ζ ⊙M is closed, then ζ ⊙ Clb(M) = ζ ⊙M . Proof. (i) by Proposition 1, we have Clb(M) ⊙ ζ ⊆ Cl(M ⊙ ζ) = M ⊙ ζ and M ⊙ ζ ⊆ Clb(M)⊙ ζ. Hence, the proof. (ii) is similar. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 7 of 17 Definition 15. Let W be a d-algebra, U be a non-empty subset of W and α ∈ W. We define the following subsets. Uα and αU : Uα = {ζ ∈ W : ζ ⊙ α ∈ U} and αU = {ζ ∈ W : α⊙ ζ ∈ U}. Also if K ⊆ W we define KU = ⋃ α∈K αU & UK = ⋃ α∈K Uα . Proposition 4. Let W be an d-algebra and M,N,W,K are subsets of W then: (i) MW ⊆ NW , if M ⊆ N . (ii) MW ⊆ MK, if W ⊆ K. (iii) (Fα) c = (F c)α and (αF )c =α (F c) for each α ∈ W, If F ⊆ W. Proof. These results follow from Definition 15. Proposition 5. Let W be a Tbd-algebra and α ∈ W and U be any nonempty subset of W, , then the statements listed below are accurate: (i) Uα and αU are b-open sets, if U is an open set. (ii) Fα and αF are b-closed sets, if F is closed set. (iii) If U is open, then KU and UK are b-open sets for every subset K of W. Proof. (i) Let ζ ∈ Uα, so ζ ⊙ α ∈ U . Since W is Tbd-algebra and U is open, so there exists a b-open set G having ζ such that G ⊙ α ⊆ U , ζ ⊙ α ∈ G ⊙ α ⊆ U . Therefore, for every θ ∈ G, we have θ ⊙ α ⊆ U implies that θ ∈ Uα. Hence, G ⊆ Uα implies that Uα is b-open. To show that αU is b-open, let ζ ∈ αU implies that α ⊙ ζ ∈ U . Since W is Tbd- algebra, then there exists a b-open set H that has ζ such that α ⊙ H ⊆ U , so for each θ ∈ H, we have α⊙ θ ∈ U . Hence, θ ∈ H ⊆ αU . Therefore, αU is a b-open set. (ii) Assuming F be a closed set, then F c is open. Hence, according (1), (F c)α and α(F c) are b-open. using Proposition 4, (Fα) c = (F c)α and (αF )c =α (F c). Hence, (Fα) c and (αF )c are b-open. Consequently, Fα and αF are b-closed. (iii) Follows from (i) and the fact that arbitrary union of b-open sets is b-open. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 8 of 17 Corollary 2. Let W be Tbd–algebra, U and M be two non–empty subset of W, then the statements listed below are accurate: i) The sets MU and UM are b-open sets if U is open. ii)The sets MU and UM are closed sets if U is closed set and M is finite. Example 2. In Example 1, we have U = {α, γ} ∈ Ω and by simple calculation we can see that MU and UM are b-open sets for any subset M of W. Proposition 6. Let W be Tbd–algebra and W be T2 b-compact space. If S is a compact subset of W, then Sα and αS are b-compact sets for all α ∈ W. Proof. Let S be a compact subset of W. Since W is T2, then S is closed set in W. Thus by Proposition 5 Sα and αS are b-closed sets in W for all ζ ∈ W. Then, by Lemma 5, Sα and αS are b-compact sets in W for all ζ ∈ W. In the following example, we see that Sα and αS are b-compact sets while S is no compact. Example 3. Consider the set of real numbers R and an operation ⊙ defined as: x⊙ y= { 0 if x ≤ y 1 otherwise. Then, (R,⊙, τ) is a Tbd–algebra T2-space where τ is the discrete topology. Suppose that S = (1, 5), then obviously, S is not compact, but we have Sα and αS are empty sets for every α ∈ R and hence they are b-compact. Proposition 7. If {0} is open in a Tbd-algebra W, then it is b-T1. Proof. Assume that {0} is open and ζ, η ∈ W be any two distinct points. Since ζ ⊙ ζ = 0 for all ζ ∈ W and W is Tbd-algebra, then there exist b-open sets H and G containing ζ such that H ⊙ G ⊆ {0}. Hence, either η /∈ H or η /∈ G. Also, we have η ⊙ η = 0, so the exist two b-open sets U, V containing η such that U ⊙ V ⊆ {0}. Hence, either ζ /∈ U or ζ /∈ V .Therefore, we obtain that a b-open set containing ζ but not η and a b-open set containing η but not ζ. Therefore, W is a b-T1-space. Remark 1. The space (W,Ω) in Example 1 is b-T1 but {0} is not b-open. Corollary 3. If {0} is an open set in a Tbd-algebra W, then every open singular set is regular open. Proof. From Proposition 7, we have W is b-T1. Hence, every singleton set is b-closed. If {ζ} is an open set, so we have Int(Cl({ζ} ⊆ {ζ} ⊆ Int(Cl({ζ} which implies that {ζ} is regular open. Corollary 4. Suppose that {0} is an open set in a Tbd-algebra W. If W is a door space, then every singular set is either closed or regular open. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 9 of 17 Proof. Follows from Corollary 3 and the definition of door space. Proposition 8. In a Tbd-algebra W which is both submaximal and extremally discon- nected space. If {0} is open, then the space is discrete. Proof. Assume that {0} is open and let ζ be any point in W. Since ζ ⊙ ζ = 0 for all ζ ∈ W and W is Tbd-algebra, so there exist b-open sets U and V having ζ such that U⊙V ⊆ {0}. Since W is extremally disconnected, so by using Lemma 4, U, V are pre-open sets. Also, W is submaximal, so by Lemma 3, U, V are open. Hence W = U ∩V is an open set having ζ. Now if W having another point η, then we obtain ζ ⊙ η = 0 and η ⊙ ζ = 0 which is a contradiction. Hence W is an open set having ζ only. therefore, {ζ} is open for each ζ ∈ W. Thus, the space W is discrete. Proposition 9. In a Tbd-algebra (W,⊙,Ω). If W is an open set not containing 0, then for each distinct elements ζ, η ∈ W with ζ ⊙ η ∈ W there exist two disjoint b-open sets containing ζ and η. Proof. Let ζ ⊙ η ∈ W , Since (W,⊙,Ω) is a Tbd-algebra, then there exist b-open sets U and V of containing ζ and η such that U ⊙ V ⊆ W . If U ∩ V ̸= ϕ, then there is a point θ ∈ U ∩ V which implies 0 = θ ⊙ θ ∈ U ⊙ V ⊆ W which contradicts itself.therefore, U ∩ V = ϕ. Proposition 10. In a Tbd-algebra W, if {0} is closed, then W is b-T2. Proof. Let {0} is closed and let ζ and η be any two distinct points in W, then either ζ⊙ η ̸= 0 or η⊙ ζ ̸= 0 without loss of generality assume that ζ⊙ η ̸= 0. Hence, there exist b-open sets U and V having ζ and η respectively such that U ⊙ V ⊆ ζ \ {0} and hence U ∩ V = ϕ. Therefore, W is b-T2. Example 4. Consider any infinite d-algebra (W,⊙,Ω) where Ω is the indescrete topology on W. Then, bO(W,Ω) = P (W) and hence (W,Ω) is a b-T2-space but {0} is not closed. Corollary 5. If a Tbd-algebra (W,⊙,Ω) is T1, then it is b-T2. Proof. Since (W,⊙,Ω) is T1, so {0} is closed. Hence, by Proposition 10, (W,⊙,Ω) is b− T2. Proposition 11. If a Tbd-algebra (W,⊙,Ω) is T0, then it is b-T1. Proof. Suppose that ζ, η ∈ W and ζ ̸= η. Then either ζ ⊙ η ̸= 0 or η ⊙ ζ ̸= 0. Now let ζ ⊙ η ̸= 0. Since W is T0 space, then there is an open set W that contains one of them but not the other. Case 1. Assume that W contains ζ ⊙ η and 0 /∈ W . Since (W,⊙,Ω) is a Tbd-algebra, then there exist b-open sets U of ζ and V of η such that M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 10 of 17 U ⊙ V ⊆ W . Then U is a b-open set and V is a b-open set having η . If U ∩ V ̸= ϕ, that is mean there is a point θ ∈ U ∩V . Thus 0 = η⊙ θ ∈ U ⊙V ⊆ W that is a contradiction. Case 2. now if 0 ∈ W and ζ⊙η /∈ W . Then we have, ζ⊙ ζ = 0 ∈ W , so there exist b-open sets U1, U2 having ζ such that U1 ⊙ U2 ∈ W . Obviously, U2 is a b-open set containing ζ and does not contains η. Again η ⊙ η = 0 ∈ W , so there exist b-open sets Uη and Vη containing η such that Uη ⊙ Vη ⊆ W . Hence, Uη is a b-open set containing η but not ζ. Therefore, (W,⊙,Ω) is a b-T1 space. Remark 2. The space (W,⊙,Ω) in Example 4 is a b-T1-space which is not T0. Definition 16. We say that a topological space (W,Ω) is Bζ-space if U1, U2 are any b-open subsets of W containing ζ, then there exists a b-open set U such that ζ ∈ U ⊆ U1 ∩ U2. Proposition 12. In a Bζ Tbd-algebra (W,⊙,Ω). If W is an open set having 0, then for all ζ ∈ W there exists a b-open set U having ζ such that U ⊙ U ⊆ W . Proof. Let W be an open set having 0. We have ζ ⊙ ζ = 0 for all ζ ∈ W and W is Tbd-algebra, so there exist b-open sets U1, U2 having ζ such that U1 ⊙ U2 ⊆ W . Since W is a Bζ space, so there exists a b-open set U ⊆ U1 ∩ U2. Therefore, we obtain that U ⊙ U ⊆ W . Proposition 13. In a Bζ Tbd-algebra, if (W,⊙,Ω) is an extremally disconnected sub- maximal T0 space, then it is b-T2. Proof. Let ζ, η ∈ W, so either ζ⊙ η ̸= 0 or η⊙ ζ ̸= 0. Assume that ζ⊙ y ̸= 0. Since W is T0, so there exists an open set W having either ζ ⊙ η and not having 0 or conversely. If ζ⊙ η ∈ W and 0 /∈ W , so by Proposition 9, there exist two disjoint b-open sets containing ζ and η. If ζ ⊙ η /∈ W and 0 ∈ W , then we have, ζ ⊙ ζ = 0 ∈ W , so there exist b-open sets U1, U2 containing ζ such that U1 ⊙ U2 ∈ W . Since W is a Bζ space, so there exists a b-open set U such that ζ ∈ U ⊆ U1 ∩ U2 implies that U is a b-open set containing ζ such that U ⊙ U ⊆ W . Again η ⊙ η = 0 ∈ W , so there exists a b-open setV containing η such that V ⊙ V ⊆ W . Hence, we have η /∈ U and ζ /∈ V because we get a contradiction. Since W is extremally disconnected, so by Lemma 4, W \ U and W \ V are dense in W. Since W is submaximal, so by Definition 2, W \U and W \ V are open. Thus, ζ ∈ U ∩ (W \ V ) and η ∈ V ∩ (W \ U). Obviously, U ∩ (W \ V ) and V ∩ (W \ U) are disjoint b-open sets in W. Hence, W is b-T2. Proposition 14. If Z is an open d-subalgebra of a Tbd-algebra W, then Z is also a Tbd-algebra. Proof. Suppose that ζ, η ∈ Z and let U be an open set in the subspace Z containing ζ ⊙ η. Since Z is open in W, so U is open in W. Since W is a Tbd-algebra, so there exist b-open sets H,G in W having ζ and η respectively such that H ⊙G ⊆ U . Then, by Lemma 1, we have O1 = H ∩ Z and O2 = G ∩ Z are b-open sets in Z having ζ and η respectively and obviously, O1 ⊙O2 ⊆ H ⊙G ⊆ U .given the proof. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 11 of 17 Proposition 15. If I is an ideal in a Tbd-algebra W with 0 ∈ int(I), then I is b-open. Proof. Suppose that ζ ∈ I. Since 0 ∈ int(I), then there is an open set U such that 0 ∈ U ⊆ I. Since W is a Tbd-algebra, then there exists a b-open set V having ζ such that V ⊙ ζ ⊆ U . If there is a point η ∈ V ∩ (W \ I), so we get η ⊙ ζ ∈ I. Since ζ ∈ I and I is an ideal, so η ∈ I that is a contradiction. therefore ζ ∈ V ⊆ I means that I is b-open. Example 5. Consider the d-algebra (W,⊙, 0) in Example 1. we define a topology Ω on W as follows: Ω = {ϕ, {α, β},W}. Then, I = {0, α, β} is an ideal which is b-open but 0 /∈ int(I). Proposition 16. If I is an open proper ideal in a Tbd-algebra W, then I is b-closed and regular open. Proof. Suppose that ζ ∈ W \ I. Since I is an ideal so ζ ⊙ ζ = 0 ∈ I. Since W is a Tbd-algebra, so there exist b-open sets V and U having ζ such that V ⊙ U ⊆ I. If there exists η ∈ (U ∩ I), then we have ζ ⊙ η ∈ I and η ∈ I. Since I is an ideal, then we get ζ ∈ I which is contradiction. Hence, ζ ∈ U ⊆ W \ I and so ζ \ I is b-open. Therefore, I is b-closed. Since I is open, so by Lemma 5, we find I is semi-closed which implies that I is regular open. Corollary 6. In a Tbd-algebra (W,⊙, 0), if {0} is open, then W is semi-disconnected. Proof. Since {0} is an ideal inW, by using Proposition 16, {0} is regular open. thus, W is semi-disconnected so it is having a proper non-empty set which is open and semi-closed. Proposition 17. In a Tbd-algebra (W,⊙, 0). If I is an ideal such that every sequence in I converging to 0 contains 0, then I is b-closed. Proof. Let ζ ∈ Clb(I), so there exists a sequence ≪ ζn ≫ in I which is b-convergent to ζ. Now, we claim that the sequence ≪ ζ ⊙ ζn ≫ converges to 0. For this, Assume that U is any open set that contains 0 and we have ζ ⊙ ζ = 0. Since, W is a Tbd- algebra, so there exists a b-open set V containing ζ such that ζ ⊙ V ⊆ U . Again, we have ≪ ζn ≫ b-converges to ζ, so there is K ∈ N such that ζn ∈ V for all n ≫ K. Therefore, ζ ⊙ ζn ∈ W ⊙ V ⊆ U for all n ≫ K. Hence, ≪ ζ ⊙ ζn ≫ converges to 0 and by hypothesis the sequence ≪ ζ ⊙ ζn ≫ contains 0. Thus, there exists n ∈ N such that ζ ⊙ ζn = 0. Since ζn ∈ I and I is an ideal, so ζ ∈ I. Hence, I is b-closed. Proposition 18. In an edge Tbd-algebra (W,⊙, 0). If all elements of W are atoms and ϕ ̸= S ⊆ W not containing 0, then 0 ∈ Cl(S). Proof. Suppose that 0 ̸= ζ ∈ Clb(S), then there exists a sequence ≪ sn ≫ in S which b-converges to ζ. Hence, as in Proposition 17, we obtain that ≪ sn ⊙ ζ ≫ converges to 0. Since all elements of W are atoms and 0 ̸= ζ ̸= sn, so sn ⊙ ζ ̸= 0 implies that sn ⊙ ζ = sn. Therefore, the sequence, ≪ sn ≫ in S converges to 0 implies that 0 ∈ Cl(S). M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 12 of 17 Proposition 19. In an edge Tbd-algebra (W,⊙, 0). If S ⊆ W and there is a sequence of atom elements in S b-converges to an element ζ /∈ S, then the sequence converges to 0 and 0 ∈ Cl(S). Proof. Let ≪ ζn ≫ be a sequence of atom elements in S which is b-convergent to ζ /∈ S, so as in Proposition 16, we obtain the sequence ≪ ζn ⊙ ζ ≫ which converges to 0. Since W is an edge d-algebra, so ζn ⊙ ζ = {0, ζn} but ζn are atoms and ζ /∈ S, so ζn ⊙ ζ = ζn. Therefore, ≪ ζn ≫ converges to 0 and hence 0 ∈ Cl(S). Definition 17. Let W be a d-algebra. We can define the binary operation ◦ on L(W) by (Lα ◦ Lβ)(ζ) = Lα(ζ)⊙ Lβ(ζ) for each ζ ∈ W. Theorem 1. Let W be a positive implicative d-algebra, then (L(W), ◦,L0) is a d-algebra. Proof. Assume thatLα, Lβ ∈ L(W). Then, by using the definition of the binary operation ◦ on L(W) we get (Lα ◦ Lβ)(ζ) = Lα(ζ) ⊙ Lβ(ζ) = (α ⊙ ζ) ⊙ (β ⊙ ζ). Since W is a positive implication d-algebra, then (α ⊙ ζ) ⊙ (β ⊙ ζ) = (α ⊙ β) ⊙ ζ. therefore, (Lα ◦ Lβ)(ζ) = Lα⊙β(ζ) which implies that Lα ◦ Lβ = Lα⊙β ∀ α, β ∈ W. Furthermore, the following statements are correct. (i) Lζ ◦ Lζ = Lζ⊙ζ = L0, (ii) Lζ ◦ Lη = L0 and Lη ◦ Lζ = L0, then Lζ⊙η = L0 and Lη⊙ζ = L0 which implies that ζ ⊙ η = 0 and η ⊙ ζ = 0 ⇒ ζ = η and hence, Lζ = Lη , (iii) L0 ◦ Lζ = L0⊙ζ = L0. Therefore, L(W) is a d-algebra. Proposition 20. Let W be a Tbd-algebra, then every left map on W is b-continuous. Proof. Let ζ ∈ W and W be any open set having Lα(ζ) = α ⊙ ζ. Since W is a Tbd- algebra, so there exists a b-open set V having ζ such that a⊙V ⊆ W . Hence, Lα(V ) ⊆ W . Thus, we get Lα is b-continuous. Proposition 21. Let W be an edge Tbd-algebra. If α ∈ W such that α ⊙ (α ⊙ ζ) is an atom for every ζ ∈ W, then the left map Lα on W is b-open. Proof. Let W be any open set in W and α ∈ W. We have to prove that Lα(W ) is b-open. Let ζ ∈ Lα(W ), so there exists η ∈ W such that ζ = Lα(η) = α ⊙ η. Since W, is an edge d-algebra, so it satisfies condition (B2), that is (α ⊙ (α ⊙ η)) ⊙ η = 0 for each η ∈ W. Also, by hypothesis, we have α ⊙ (α ⊙ η) is an atom. Hence, we get α⊙ ζ = α⊙ (α⊙ η) = η. Therefore, α⊙ ζ ∈ W . Since W is a Tbd-algebra, so there exists a b-open set V containing ζ such that a ⊙ V ⊆ W . Since W, satisfies (B2), so we get V = α⊙ (α⊙ V ) ⊆ α⊙W = Lα(W ). This implies that Lα(W ) is b-open. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 13 of 17 Proposition 22. Let W be a Tbd-algebra, then every right map on W is b-continuous. Proof. The proof is similar to the proof of Proposition 20. Proposition 23. If W is an edge d-algebra in which condition (B1) does not hold for all ζ, η ∈ W, then (ζ ⊙ η)⊙ y = ζ for all ζ, η ∈ W. Proof. Since W is an edge d-algebra, so ζ ⊙ y = {ζ, 0}. Now if ζ ⊙ y = 0, and by hypothesis, we have (ζ ⊙ η) ⊙ (ζ ⊙ θ)) ⊙ (θ ⊙ η) ̸= 0 for all ζ, η, θ ∈ W. Hence, we get 0 ̸= (ζ ⊙ η) ⊙ (ζ ⊙ θ)) ⊙ (θ ⊙ η) = 0 ⊙ (θ ⊙ η) = 0 which is contradiction. Therefore, ζ ⊙ η = ζ and thus (ζ ⊙ η)⊙ η = ζ. Corollary 7. If W is an edge d-algebra in which all elements of W are atoms, then (ζ ⊙ η)⊙ η = ζ for all ζ, η ∈ W. Proof. Since each element of W is an atom, so if ζ, y ∈ W, then ζ ⊙ y ̸= 0. Since W is an edge d-algebra, so ζ ⊙ η = ζ. Hence, (ζ ⊙ η)⊙ η = ζ for all ζ, η ∈ W. Proposition 24. Let W be an edge Tbd-algebra not satisfying condition (B1) for all ζ, η, θ ∈ W, then every right map on W is b-open. Proof. Let W be any open set in W and α ∈ W. We have to prove that Rα(W ) is b-open. Let ζ ∈ Rα(W ), then ∃ η ∈ W such that ζ = Rα(η) = η ⊙ α. Hence, we have ζ ⊙ α = (η ⊙ α) ⊙ α and by Proposition 23, we get ζ ⊙ α = (η ⊙ α) ⊙ α = η. Hence, ζ ⊙α ∈ W . Since W is a Tbd-algebra, so ∃ a b-open set V having ζ such that V ⊙α ⊆ W . Again by Proposition 23, we get V = (V ⊙ α)⊙ α ⊆ W ⊙ α = Rα(W ). Thus, Rα(W ) is b-open. Corollary 8. Let W be an edge Tbd-algebra such that every element of W is an atom, then every right map on W is b-open. Proof. Follows from Proposition 24 and Corollary 7. Definition 18. Let W be a d-algebra, we define a map F : ζ → L(W) by F(W) = Lζ for all ζ ∈ W. If M ⊆ W, then F(M) = {Lζ : ζ ∈ M}. Proposition 25. If W is a d-algebra, then the following statements are true: (i) If M ⊆ N , then F(M) ⊆ F(N). (ii) F(M c) = (F(M))c. (iii) If {Mλ : λ ∈ Λ} is any family of subsets of W, then F( ⋃ λ∈ΛMλ) = ⋃ λ∈ΛF(Mλ) and F( ⋂ λ∈ΛMλ) = ⋂ λ∈ΛF(Mλ). Proof. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 14 of 17 (i) Let M ⊆ N and let Lζ ∈ F(M). Hence, ζ ∈ M ⊆ N implies Lζ ∈ F(N). Thus, F(M) ⊆ F(N). (ii) Lζ ∈ F(M c) if, and only if ζ ∈ M c implies ζ /∈ M if, and only if Lζ /∈ F(M) if, and only if Lζ ∈ (F(M))c. (iii) We shall prove for the union, the other proof is similar. Let Lζ ∈ F( ⋃ λ∈ΛMλ), then ζ ∈ ⋃ λ∈ΛMλ that is there is λ ∈ Λ such that ζ ∈ Mλ, so Lζ ∈ F(Mλ) for some λ ∈ Λ. Hence, Lζ ∈ ⋃ λ∈ΛF(Mλ). Proposition 26. Let W be a positive implicative d-algebra, then the map F : ζ → L(W) is a d-isomorphism. Proof. It is obvious that F is a bijection. We have F(ζ ⊙ η) = Lζ⊙η and Lζ⊙η(θ) = (ζ ⊙ η) ⊙ θ. Since W is positive implicative, we have (ζ ⊙ η) ⊙ θ = (ζ ⊙ θ) ⊙ (η ⊙ θ). Therefore, Lζ⊙η(θ) = Lζ(θ) ◦ Lη(θ) = (Lζ ◦ Lη)(θ). Hence, F(ζ ⊙ η) = F(ζ) ◦ F(η) for all ζ, η ∈ W, so F is a d-isomorphism. Proposition 27. Let W be a positive implicative d-algebra and Ω be a topology on W, Consequently, the following statements are correct: (i) The family σ = {F(G) ⊆ L(W) : G ∈ Ω} is a topology on L(W). (ii) For every subset M of W, LCl(M) = σCl(LM ). (iii) For any subset M of W, F(Int(M)) = σInt(F(M)). (iv) If M is any pre-open set in (W,Ω), then F(M) is a pre-open set in (L(W), σ). (v) If M is any semi-open set in (W,Ω), then F(M) is a semi-open set in (L(W), σ). Proof. (i) The proof of σ being a topology is clear. (ii) For every subset M of W, we have M ⊆ Cl(M). Hence, LM ⊆ LCl(M) and Cl(M) is closed in W, so by definition of σ, we have LCl(M) is σ-closed in L(W). There- fore, we obtain σCl(LM ) ⊆ σCl(LCl(M)) = LCl(M). To prove LCl(M) ⊆ σCl(LM ), let Lζ ∈ LCl(M), then ζ ∈ Cl(M) and let F(G) be any σ-open set containing Lζ . Hence G is an open set that contains ζ, so M ∩ G ̸= ϕ and by Proposition 25, F(M ∩ G) = F(M) ∩ F(G). Therefore, F(M) ∩ F(G) ̸= ϕ. Implies that Lζ ∈ σCl(LM ), so LCl(M) ⊆ σCl(LM ) and hence LCl(M) = σCl(LM ). M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 15 of 17 (iii) We have Int(M) ⊆ M , so F(Int(M)) ∈ σ and F(Int(M)) ⊆ F(M). Hence F(Int(M)) ⊆ σInt(F(M)). Now if Lζ ∈ σInt(F(M)), so there exists F(G) ∈ σ such that Lζ ∈ F(G) ⊂ F(M). Hence, ζ ∈ G ⊆ M implies that ζ ∈ Int(M). Therefore, Lζ ∈ F(Int(M)). Thus, F(Int(M)) = σInt(F(M)). (iv) Let M be any pre-open set in W, so there exists an open set V in W such that M ⊆ V ⊆ Cl(M). Hence F(M) ⊆ F(V ) ⊆ F(Cl(M)) and by (2, 3), we have F(M) ⊆ F(V ) ⊆ σCl(F(M)) and F(V ) is σ-open. Hence, F(M) is pre-open in (L(W), σ). (v) The proof is similar to the proof of (iv). Corollary 9. If M is any subset of (W,Ω), then F(Int(Cl(M))) = σInt(σCl(F(M))) and F(Cl(Int(M))) = σCl(σInt(F(M))) where (L(W), σ) is defined as in Proposition 27. Proof. The proof follows from (ii, iii) of Proposition 27. Proposition 28. A subset M is b-open in (W,Ω) if and only if F(M) is a b-open set in (L(W), σ). Proof. Suppose that M is a b-open set, then by Lemma 2, M = M ∩ (Int(Cl(M)) ∩ Cl(Int(M)). From Proposition 25, we get F(M) = F(M)∩ ( F(Int(Cl(M)) ) ∩ ( FCl(Int(M)) ) . By Corollary 9, we get F(M) = F(M)∩ ( σInt(σCl(F(M)) ) ∩ ( σCl(σInt(F(M)) ) . Hence, F(M) is a b-open set in (L(W), σ). Reversing the statement, we get the result. Proof. Suppose that M is a b-open set, then by Lemma 2, M = M ∩ (Int(Cl(M)) ∩ Cl(Int(M)). From Proposition 25, we get F(M) = F(M)∩ ( F(Int(Cl(M)) ) ∩ ( FCl(Int(M)) ) . By Corollary 9, we get F(M) = F(M)∩ ( σInt(σCl(F(M)) ) ∩ ( σCl(σInt(F(M)) ) . Hence, F(M) is a b-open set in (L(W), σ). Proposition 29. Let W be a positive implicative Tbd-algebra. Then (L(W), ◦, σ) is a Tbd-algebra. Proof. Suppose that Lζ , Lη are any elements in L(W) and F(W ) is a σ-open set having Lζ ◦ Lη = Lζ⊙η. Then, W is an open set having ζ ⊙ η in W, since W is a Tbd-algebra, so there exist b-open sets U and V containing ζ and η respectively such that U ⊙ V ⊆ W . Therefore, F(U ⊙ V ) ⊆ F(W ). Since W is positive implicative, and by Proposition 26, F(U ⊙ V ) = F(U) ◦ F(V ) ⊆ F(W ). By Proposition 27, F(U) and F(V ) are b-open sets in (L(W), ◦, σ) containing Lζ and Lη respectively, hence the proof. M. W. Abdulqader, A. B. Khalaf / Eur. J. Pure Appl. Math, 18 (3) (2025), 6270 16 of 17 Example 6. Consider the d-algebra (W,⊙,Ω) in Example 1, the operation (◦) defined on L(W) is given as follows: ◦ L0 Lα Lβ Lγ L0 L0 L0 L0 L0 Lα Lα L0 L0 Lα Lβ Lβ Lβ L0 L0 Lγ Lγ Lγ Lγ L0 Table 2: The d-algebra (L(W), ◦, σ) From Proposition 27 and the definition of Ω, we obtain that σ = {Φ,Lα,Lγ ,L{α,γ},L(W)}. By routine calculation, we can show that (L(W), ◦, σ) is a Tbd-algebra but (W,⊙,Ω) is not positive implicative because (α⊙ γ)⊙ (β ⊙ γ) = α ̸= (α⊙ β)⊙ γ = 0. 4. Conclusions In this paper, we applied the family of b-open sets in topological spaces to define the concept of b-topological d-algebra. Through this work, we reviewed the basics necessary for studying this new topological algebra. 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