Int. J. Anal. Appl. (2023), 21:88 Some Properties of Generalized (Λ,α)-Closed Sets Chawalit Boonpok, Montri Thongmoon∗ Mathematics and Applied Mathematics Research Unit, Department of Mathematics, Faculty of Science, Mahasarakham University, Maha Sarakham, 44150, Thailand ∗Corresponding author: montri.t@msu.ac.th Abstract. The aim of this paper is to introduce the concept of generalized (Λ,α)-closed sets. More- over, we investigate some characterizations of Λα-T 1 2 -spaces, (Λ,α)-normal spaces and (Λ,α)-regular spaces by utilizing generalized (Λ,α)-closed sets. 1. Introduction The concept of generalized closed sets was first introduced by Levine [7]. Moreover, Levine defined a separation axiom called T1 2 between T0 and T1. Dontchev and Ganster [3] introduced the notion of T3 4 -spaces which are situated between T1 and T1 2 and showed that the digital line or the Khalimsky line [5] (Z,κ) lies between T1 and T3 4 . As a modification of generalized closed sets, Palaniappan and Rao [10] introduced and studied the notion of regular generalized closed sets. As the further modification of regular generalized closed sets, Noiri and Popa [9] introduced and investigated the concept of regular generalized α-closed sets. Park et al. [11] obtained some characterizations of T3 4 spaces. Dungthaisong et al. [4] characterized µ(m,n)-T1 2 spaces by utilizing the concept of µ(m,n)- closed sets. Torton et al. [12] introduced and studied the notions of µ(m,n)-regular spaces and µ(m,n)- normal spaces. Buadong et al. [1] introduced and investigated the notions of T1-GTMS spaces and T2-GTMS spaces. Caldas et al. [2] by considering the concepts of α-open sets and α-closed sets, introduced and investigated Λα-sets, (Λ,α)-closed sets, (Λ,α)-open sets and the (Λ,α)-closure operator. Khampakdee and Boonpok [6] studied some properties of (Λ,α)-open sets. In the present paper, we introduce the concept of generalized (Λ,α)-closed sets. Furthermore, some properties of Received: Jun. 13, 2023. 2020 Mathematics Subject Classification. 54A05, 54D10. Key words and phrases. generalized (Λ,α)-closed set; Λα-T 1 2 -space; (Λ,α)-normal space; (Λ,α)-regular space. https://doi.org/10.28924/2291-8639-21-2023-88 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-88 2 Int. J. Anal. Appl. (2023), 21:88 generalized (Λ,α)-closed sets are discussed. In particular, several characterizations of Λα-T1 2 -spaces, (Λ,α)-normal spaces and (Λ,α)-regular spaces are established. 2. Preliminaries Let A be a subset of a topological space (X,τ). The closure of A and the interior of A are denoted by Cl(A) and Int(A), respectively. A subset A of a topological space (X,τ) is said to be α-open [8] if A ⊆ Int(Cl(Int(A))). The complement of an α-open set is called α-closed. The family of all α-open sets in a topological space (X,τ) is denoted by α(X,τ). A subset Λα(A) [2] is defined as follows: Λα(A) = ∩{O ∈ α(X,τ)|A ⊆ O}. Lemma 2.1. [2] For subsets A, B and Ai (i ∈ I) of a topological space (X,τ), the following properties hold: (1) A ⊆ Λα(A). (2) If A ⊆ B, then Λα(A) ⊆ Λα(B). (3) Λα(Λα(A)) = Λα(A). (4) Λα(∩{Ai|i ∈ I}) ⊆∩{Λα(Ai )|i ∈ I}. (5) Λα(∪{Ai|i ∈ I}) = ∪{Λα(Ai )|i ∈ I}. Recall that a subset A of a topological space (X,τ) is said to be a Λα-set [2] if A = Λα(A). Lemma 2.2. [2] For subsets A and Ai (i ∈ I) of a topological space (X,τ), the following properties hold: (1) Λα(A) is a Λα-set. (2) If A is α-open, then A is a Λα-set. (3) If Ai is a Λα-set for each i ∈ I, then ∩i∈IAi is a Λα-set. (4) If Ai is a Λα-set for each i ∈ I, then ∪i∈IAi is a Λα-set. A subset A of a topological space (X,τ) is called (Λ,α)-closed [2] if A = T∩C, where T is a Λα-set and C is an α-closed set. The complement of a (Λ,α)-closed set is called (Λ,α)-open. The collection of all (Λ,α)-open (resp. (Λ,α)-closed) sets in a topological space (X,τ) is denoted by ΛαO(X,τ) (resp. ΛαC(X,τ)). Let A be a subset of a topological space (X,τ). A point x ∈ X is called a (Λ,α)-cluster point of A [2] if for every (Λ,α)-open set U of X containing x we have A∩U 6= ∅. The set of all (Λ,α)-cluster points of A is called the (Λ,α)-closure of A and is denoted by A(Λ,α). Lemma 2.3. [2] Let A and B be subsets of a topological space (X,τ). For the (Λ,α)-closure, the following properties hold: (1) A ⊆ A(Λ,α) and [A(Λ,α)](Λ,α) = A(Λ,α). (2) A(Λ,α) = ∩{F |A ⊆ F and F is (Λ,α)-closed}. (3) If A ⊆ B, then A(Λ,α) ⊆ B(Λ,α). Int. J. Anal. Appl. (2023), 21:88 3 (4) A is (Λ,α)-closed if and only if A = A(Λ,α). (5) A(Λ,α) is (Λ,α)-closed. Definition 2.1. [6] Let A be a subset of a topological space (X,τ). The union of all (Λ,α)-open sets of X contained in A is called the (Λ,α)-interior of A and is denoted by A(Λ,α). Lemma 2.4. [6] Let A and B be subsets of a topological space (X,τ). For the (Λ,α)-interior, the following properties hold: (1) A(Λ,α) ⊆ A and [A(Λ,α)](Λ,α) = A(Λ,α). (2) If A ⊆ B, then A(Λ,α) ⊆ B(Λ,α). (3) A is (Λ,α)-open if and only if A(Λ,α) = A. (4) A(Λ,α) is (Λ,α)-open. (5) [X −A](Λ,α) = X −A(Λ,α). (6) [X −A](Λ,α) = X −A(Λ,α). 3. Generalized (Λ,α)-closed sets In this section, we introduce the notion of generalized (Λ,α)-closed sets. Moreover, some properties of generalized (Λ,α)-closed sets are discussed. Definition 3.1. A subset A of a topological space (X,τ) is said to be generalized (Λ,α)-closed (briefly g-(Λ,α)-closed) if A(Λ,α) ⊆ U and U is (Λ,α)-open in (X,τ). The complement of a generalized (Λ,α)- closed set is said to be generalized (Λ,α)-open (briefly g-(Λ,α)-open). Definition 3.2. A topological space (X,τ) is said to be Λα-symmetric if for x and y in X, x ∈{y}(Λ,α) implies y ∈{x}(Λ,α). Theorem 3.1. A topological space (X,τ) is Λα-symmetric if and only if {x} is g-(Λ,α)-closed for each x ∈ X. Proof. Assume that x ∈ {y}(Λ,α) but y 6∈ {x}(Λ,α). This implies that the complement of {x}(Λ,α) contains y. Therefore, the set {y} is a subset of the complement of {x}(Λ,α). This implies that {y}(Λ,α) is a subset of the complement of {x}(Λ,α). Now the complement of {x}(Λ,α) contains x which is a contradiction. Conversely, suppose that {x} ⊆ V ∈ ΛαO(X,τ), but {x}(Λ,α) is not a subset of V . This means that {x}(Λ,α) and the complement of V are not disjoint. Let y belongs to their intersection. Now, we have x ∈{y}(Λ,α) which is a subset of the complement of V and x 6∈ V . This is a contradiction. � Theorem 3.2. A subset A of a topological space (X,τ) is g-(Λ,α)-closed if and only if A(Λ,α) − A contains no nonempty (Λ,α)-closed set. 4 Int. J. Anal. Appl. (2023), 21:88 Proof. Let F be a (Λ,α)-closed subset of A(Λ,α)−A. Now, A ⊆ X−F and since A is g-(Λ,α)-closed, we have A(Λ,α) ⊆ X −F or F ⊆ X −A(Λ,α). Thus, F ⊆ A(Λ,α) ∩ [X −A(Λ,α)] = ∅ and hence F is empty. Conversely, suppose that A ⊆ U and U is (Λ,α)-open. If A(Λ,α) * U, then A(Λ,α) ∩ (X −U) is a nonempty (Λ,α)-closed subset of A(Λ,α) −A. � Definition 3.3. Let A be a subset of a topological space (X,τ). The (Λ,α)-frontier of A, ΛαFr(A), is defined as follows: ΛαFr(A) = A(Λ,α) ∩ [X −A](Λ,α). Theorem 3.3. Let A be a subset of a topological space (X,τ). If A is g-(Λ,α)-closed and A ⊆ V ∈ ΛαO(X,τ), then ΛαFr(V ) ⊆ [X −A](Λ,α). Proof. Let A be g-(Λ,α)-closed and A ⊆ V ∈ ΛαO(X,τ). Then, A(Λ,α) ⊆ V . Suppose that x ∈ ΛαFr(V ). Since V ∈ ΛαO(X,τ), ΛαFr(V ) = V (Λ,α) −V . Therefore, x 6∈ V and x 6∈ A(Λ,α). Thus, x ∈ [X −A](Λ,α) and hence ΛαFr(V ) ⊆ [X −A](Λ,α). � Theorem 3.4. Let (X,τ) be a topological space. For each x ∈ X, either {x} is (Λ,α)-closed or g-(Λ,α)-open. Proof. Suppose that {x} is not (Λ,α)-closed. Then, X−{x} is not (Λ,α)-open and the only (Λ,α)- open set containing X−{x} is X itself. Thus, [X−{x}](Λ,α) ⊆ X and hence X−{x} is g-(Λ,α)-closed. Therefore, {x} is g-(Λ,α)-open. � Theorem 3.5. Let A be a subset of a topological space (X,τ). Then, A is g-(Λ,α)-open if and only if F ⊆ A(Λ,α) whenever F ⊆ A and F is (Λ,α)-closed. Proof. Suppose that A is g-(Λ,α)-open. Let F ⊆ A and F be (Λ,α)-closed. Then, we have X −A ⊆ X −F ∈ ΛαO(X,τ) and X −A is g-(Λ,α)-closed. Thus, X −A(Λ,α) = [X −A](Λ,α) ⊆ X −F and hence F ⊆ A(Λ,α). Conversely, let X −A ⊆ U and U ∈ ΛαO(X,τ). Then, X −U ⊆ A and X −U is (Λ,α)-closed. By the hypothesis, X −U ⊆ A(Λ,α) and hence [X −A](Λ,α) = X −A(Λ,α) ⊆ U. This shows that X −A is g-(Λ,α)-closed. Thus, A is g-(Λ,α)-open. � Theorem 3.6. A subset A of a topological space (X,τ) is g-(Λ,α)-closed if and only if A∩{x}(Λ,α) 6= ∅ for every x ∈ A(Λ,α). Proof. Let A be a g-(Λ,α)-closed set and suppose that there exists x ∈ A(Λ,α) such that A∩{x}(Λ,α) = ∅. Therefore, A ⊆ X − {x}(Λ,α) and so A(Λ,α) ⊆ X − {x}(Λ,α). Hence x 6∈ A(Λ,α), which is a contradiction. Int. J. Anal. Appl. (2023), 21:88 5 Conversely, suppose that the condition of the theorem holds and let U be any (Λ,α)-open set containing A. Let x ∈ A(Λ,α). Then, by the hypothesis A∩A(Λ,α) 6= ∅, so there exists y ∈ A∩{x}(Λ,α) and so y ∈ A ⊆ U. Thus, {x}∩U 6= ∅. Hence x ∈ U, which implies that A(Λ,α) ⊆ U. This shows that A is g-(Λ,α)-closed. � Definition 3.4. A subset A of a topological space (X,τ) is said to be locally (Λ,α)-closed if A = U∩F, where U ∈ ΛαO(X,τ) and F is a (Λ,α)-closed set. Theorem 3.7. For a subset A of a topological space (X,τ), the following properties are equivalent: (1) A is locally (Λ,α)-closed; (2) A = U ∩A(Λ,α) for some U ∈ ΛαO(X,τ); (3) A(Λ,α) −A is (Λ,α)-closed; (4) A∪ [X −A(Λ,α)] ∈ ΛαO(X,τ); (5) A ⊆ [A∪ [X −A(Λ,α)]](Λ,α). Proof. (1) ⇒ (2): Suppose that A = U ∩ F, where U ∈ ΛαO(X,τ) and F is a (Λ,α)-closed set. Since A ⊆ F, we have A(Λ,α) ⊆ F (Λ,α) = F. Since A ⊆ U, A ⊆ U ∩ A(Λ,α) ⊆ U ∩ F = A. Thus, A = U ∩A(Λ,α) for some U ∈ ΛαO(X,τ). (2) ⇒ (3): Suppose that A = U ∩A(Λ,α) for some U ∈ ΛαO(X,τ). Then, we have A(Λ,α) −A = [X −U ∩A(Λ,α)] ∩A(Λ,α) = (X −U) ∩A(Λ,α). Since (X −U) ∩A(Λ,α) is (Λ,α)-closed, A(Λ,α) −A is (Λ,α)-closed. (3) ⇒ (4): Since X − [A(Λ,α) −A] = [X −A(Λ,α)] ∪A and by (3), A∪ [X −A(Λ,α)] ∈ ΛαO(X,τ). (4) ⇒ (5): By (4), we obtain A ⊆ A∪ [X −A(Λ,α)] = [A∪ [X −A(Λ,α)]](Λ,α). (5) ⇒ (1): We put U = [A∪ [X −A(Λ,α)]](Λ,α). Then, U ∈ ΛαO(X,τ) and A = A∩U ⊆ U ∩A(Λ,α) ⊆ [A∪ [X −A(Λ,α)]](Λ,α) ∩A (Λ,α) = A∩A(Λ,α) = A. Thus, A = U ∩A(Λ,α), where U ∈ ΛαO(X,τ) and A(Λ,α) is a (Λ,α)-closed set. This shows that A is locally (Λ,α)-closed. � Theorem 3.8. A subset A of a topological space (X,τ) is (Λ,α)-closed if and only if A is locally (Λ,α)-closed and g-(Λ,α)-closed. Proof. Let A be (Λ,α)-closed. Then, A is g-(Λ,α)-closed. Since X ∈ ΛαO(X,τ) and A = X ∩A, A is locally (Λ,α)-closed. Conversely, suppose that A is locally (Λ,α)-closed and g-(Λ,α)-closed. Since A is locally (Λ,α)- closed, by Theorem 3.7, A ⊆ [A∪ [X −A(Λ,α)]](Λ,α). Since [A∪ [X −A(Λ,α)]](Λ,α) ∈ ΛαO(X,τ) and A is g-(Λ,α)-closed, A(Λ,α) ⊆ [A∪ [X−A(Λ,α)]](Λ,α) ⊆ A∪ [X−A(Λ,α)] and hence A(Λ,α) = A. Thus, by Lemma 2.3, A is (Λ,α)-closed. � 6 Int. J. Anal. Appl. (2023), 21:88 4. Applications of generalized (Λ,α)-closed sets We begin this section by introducing the concept of Λα-T1 2 -spaces. Definition 4.1. A topological space (X,τ) is called a Λα-T1 2 -space if every g-(Λ,α)-closed set of X is (Λ,α)-closed. Lemma 4.1. Let (X,τ) be a topological space. For each x ∈ X, the singleton {x} is (Λ,α)-closed or X −{x} is g-(Λ,α)-closed. Proof. Let x ∈ X and the singleton {x} be not (Λ,α)-closed. Then, X −{x} is not (Λ,α)-open and X is the only (Λ,α)-open set which contains X −{x} and X −{x} is g-(Λ,α)-closed. � Let A be a subset of a topological space (X,τ). A subset Λ(Λ,α)(A) [6] is defined as follows: Λ(Λ,α)(A) = ∩{U | A ⊆ U,U ∈ ΛαO(X,τ)}. Lemma 4.2. [6] For subsets A,B of a topological space (X,τ), the following properties hold: (1) A ⊆ Λ(Λ,α)(A). (2) If A ⊆ B, then Λ(Λ,α)(A) ⊆ Λ(Λ,α)(B). (3) Λ(Λ,α)[Λ(Λ,α)(A)] = Λ(Λ,α)(A). (4) If A is (Λ,α)-open, Λ(Λ,α)(A) = A. A subset A of a topological space (X,τ) is called a Λ(Λ,α)-set if A = Λ(Λ,α)(A). The family of all Λ(Λ,α)-sets of (X,τ) is denoted by Λ(Λ,α)(X,τ) (or simply Λ(Λ,α)). Definition 4.2. A subset A of a topological space (X,τ) is called a generalized Λ(Λ,α)-set (briefly g-Λ(Λ,α)-set) if Λ(Λ,α)(A) ⊆ F whenever A ⊆ F and F is (Λ,α)-closed. Lemma 4.3. Let (X,τ) be a topological space. For each x ∈ X, the singleton {x} is (Λ,α)-open or X −{x} is g-Λ(Λ,α)-set. Proof. Let x ∈ X and the singleton {x} be not (Λ,α)-open. Then, X −{x} is not (Λ,α)-closed and X is the only (Λ,α)-closed set which contains X −{x} and X −{x} is g-Λ(Λ,α)-set. � Theorem 4.1. For a topological space (X,τ), the following properties are equivalent: (1) (X,τ) is a Λα-T1 2 -space. (2) For each x ∈ X, the singleton {x} is (Λ,α)-open or (Λ,α)-closed. (3) Every g-Λ(Λ,α)-set is a Λ(Λ,α)-set. Proof. (1) ⇒ (2): By Lemma 4.1, for each x ∈ X, the singleton {x} is (Λ,α)-closed or X −{x} is g-(Λ,α)-closed. Since (X,τ) is a Λα-T1 2 -space, we have X −{x} is (Λ,α)-closed and hence {x} is (Λ,α)-open in the latter case. Thus, the singleton {x} is (Λ,α)-open or (Λ,α)-closed. Int. J. Anal. Appl. (2023), 21:88 7 (2) ⇒ (3): Suppose that there exists a g-Λ(Λ,α)-set A which is not a Λ(Λ,α)-set. Then, there exists x ∈ Λ(Λ,α)(A) such that x 6∈ A. In case the singleton {x} is (Λ,α)-open, A ⊆ X −{x} and X −{x} is (Λ,α)-closed. Since A is a g-Λ(Λ,α)-set, Λ(Λ,α)(A) ⊆ X −{x}. This is a contradiction. In case the singleton {x} is (Λ,α)-closed, A ⊆ X −{x} and X −{x} is (Λ,α)-open. By Lemma 4.2, Λ(Λ,α)(A) ⊆ Λ(Λ,α)(X −{x}) = X −{x}. This is a contradiction. Therefore, every g-Λ(Λ,α)-set is a Λ(Λ,α)-set. (3) ⇒ (1): Suppose that (X,τ) is not a Λα-T1 2 -space. There exists a g-(Λ,α)-closed set A which is not (Λ,α)-closed. Since A is not (Λ,α)-closed, there exists a point x ∈ A(Λ,α) such that x 6∈ A. By Lemma 4.3, the singleton {x} is (Λ,α)-open or X −{x} is a g-Λ(Λ,α)-set. (a) In case {x} is (Λ,α)-open, since x ∈ A(Λ,α), {x}∩A 6= ∅ and x ∈ A. This is a contradiction. (b) In case X−{x} is a Λ(Λ,α)-set, if {x} is not (Λ,α)-closed, X−{x} is not (Λ,α)-open and Λ(Λ,α)(X−{x}) = X. Thus, X −{x} is not a Λ(Λ,α)-set. This contradicts (3). If {x} is (Λ,α)-closed, A ⊆ X −{x}∈ ΛαO(X,τ) and A is g-(Λ,α)-closed. Thus, A(Λ,α) ⊆ X −{x}. This contradicts that x ∈ A(Λ,α). Therefore, (X,τ) is a Λα-T1 2 -space. � Definition 4.3. A topological space (X,τ) is said to be (Λ,α)-normal if for any pair of disjoint (Λ,α)- closed sets F and H, there exist disjoint (Λ,α)-open sets U and V such that F ⊆ U and H ⊆ V . Lemma 4.4. Let (X,τ) be a topological space. If U is a (Λ,α)-open set, then U(Λ,α)∩A ⊆ [U∩A](Λ,α) for every subset A of X. Theorem 4.2. For a topological space (X,τ), the following properties are equivalent: (1) (X,τ) is (Λ,α)-normal. (2) For every pair of (Λ,α)-open sets U and V whose union is X, there exist (Λ,α)-closed sets F and H such that F ⊆ U, H ⊆ V and F ∪H = X. (3) For every (Λ,α)-closed set F and every (Λ,α)-open set G containing F , there exists a (Λ,α)- open set U such that F ⊆ U ⊆ U(Λ,α) ⊆ G. (4) For every pair of disjoint (Λ,α)-closed sets F and H, there exist disjoint (Λ,α)-open sets U and V such that F ⊆ U and H ⊆ V and U(Λ,α) ∩V (Λ,α) = ∅. Proof. (1) ⇒ (2): Let U and V be a pair of (Λ,α)-open sets such that X = U∪V . Then, X−U and X −V are disjoint (Λ,α)-closed sets. Since (X,τ) is (Λ,α)-normal, there exist disjoint (Λ,α)-open sets G and W such that X −U ⊆ G and X −V ⊆ W. Put F = X −G and H = X −W. Then, F and H are (Λ,α)-closed sets such that F ⊆ U, H ⊆ V and F ∪H = X. (2) ⇒ (3): Let F be a (Λ,α)-closed set and G be a (Λ,α)-open set containing F. Then, X −F and G are (Λ,α)-open sets whose union is X. Then by (2), there exist (Λ,α)-closed sets M and N such that M ⊆ X − F, N ⊆ G and M ∪ N = X. Then, F ⊆ X − M, X − G ⊆ X − N and (X −M) ∩ (X −N) = ∅. Put U = X −M and V = X −N. Then U and V are disjoint (Λ,α)-open 8 Int. J. Anal. Appl. (2023), 21:88 sets such that F ⊆ U ⊆ X −V ⊆ G. As X −V is a (Λ,α)-closed set, we have U(Λ,α) ⊆ X −V and hence F ⊆ U ⊆ U(Λ,α) ⊆ G. (3) ⇒ (4): Let F and H be two disjoint (Λ,α)-closed sets of X. Then, F ⊆ X −H and X −H is (Λ,α)-open and hence there exists a (Λ,α)-open set U of X such that F ⊆ U ⊆ U(Λ,α) ⊆ X −H. Put V = X −U(Λ,α). Then, U and V are disjoint (Λ,α)-open sets of X such that F ⊆ U, H ⊆ V and U(Λ,α) ∩V (Λ,α) = ∅. (4) ⇒ (1): The proof is obvious. � Theorem 4.3. For a topological space (X,τ), the following properties are equivalent: (1) (X,τ) is (Λ,α)-normal. (2) For every pair of disjoint (Λ,α)-closed sets F and H of X, there exist disjoint g-(Λ,α)-open sets U and V of X such that F ⊆ U and H ⊆ V . (3) For each (Λ,α)-closed set F and each (Λ,α)-open set G containing F, there exists a g-(Λ,α)- open set U such that F ⊆ U ⊆ U(Λ,α) ⊆ G. (4) For each (Λ,α)-closed set F and each g-(Λ,α)-open set G containing F, there exists a (Λ,α)- open set U such that F ⊆ U ⊆ U(Λ,α) ⊆ G(Λ,α). (5) For each (Λ,α)-closed set F and each g-(Λ,α)-open set G containing F, there exists a g- (Λ,α)-open set U such that F ⊆ U ⊆ U(Λ,α) ⊆ G(Λ,α). (6) For each g-(Λ,α)-closed set F and each (Λ,α)-open set G containing F, there exists a (Λ,α)- open set U such that F (Λ,α) ⊆ U ⊆ U(Λ,α) ⊆ G. (7) For each g-(Λ,α)-closed set F and each (Λ,α)-open set G containing F, there exists a g- (Λ,α)-open set U such that F (Λ,α) ⊆ U ⊆ U(Λ,α) ⊆ G. Proof. (1) ⇒ (2): The proof is obvious. (2) ⇒ (3): Let F be a (Λ,α)-closed set and G be a (Λ,α)-open set containing F. Then, we have F and X−G are two disjoint (Λ,α)-closed sets. Hence by (2), there exist disjoint g-(Λ,α)-open sets U and V of X such that F ⊆ U and X−G ⊆ V . Since V is g-(Λ,α)-open and X−G is (Λ,α)-closed, by Theorem 3.5, X−G ⊆ V(Λ,α). Thus, [X−V ](Λ,α) = X−V(Λ,α) ⊆ G and hence F ⊆ U ⊆ U(Λ,α) ⊆ G. (3) ⇒ (1): Let F and H be two disjoint (Λ,α)-closed sets of X. Then, F is a (Λ,α)-closed set and X −H is a (Λ,α)-open set containing F. Thus by (3), there exists a g-(Λ,α)-open set U such that F ⊆ U ⊆ U(Λ,α) ⊆ X − H. By Theorem 3.5, F ⊆ U(Λ,α), H ⊆ X − U(Λ,α), where U(Λ,α) and X −U(Λ,α) are two disjoint (Λ,α)-open sets. (4) ⇒ (5) and (5) ⇒ (2): The proofs are obvious. (6) ⇒ (7) and (7) ⇒ (3): The proofs are obvious. (3) ⇒ (5): Let F be a (Λ,α)-closed set and G be a g-(Λ,α)-open set containing F. Since G is g-(Λ,α)-open and F is (Λ,α)-closed, by Theorem 3.5, F ⊆ G(Λ,α) and by (3), there exists a g-(Λ,α)-open set U such that F ⊆ U ⊆ U(Λ,α) ⊆ G(Λ,α). Int. J. Anal. Appl. (2023), 21:88 9 (5) ⇒ (6): Let F be a g-(Λ,α)-closed set and G be a (Λ,α)-open set containing F . Then, F (Λ,α) ⊆ G. Since G is g-(Λ,α)-open, by (6), there exists a g-(Λ,α)-open set U such that F (Λ,α) ⊆ U ⊆ U(Λ,α) ⊆ G. Since U is g-(Λ,α)-open and F (Λ,α) ⊆ U, by Theorem 3.5, F (Λ,α) ⊆ U(Λ,α). Put V = U(Λ,α). Then, V is (Λ,α)-open and F (Λ,α) ⊆ V ⊆ V (Λ,α) = [U(Λ,α)](Λ,α) ⊆ U(Λ,α) ⊆ G. (6) ⇒ (4): Let F be a (Λ,α)-closed set and G be a g-(Λ,α)-open set containing F. Then by Theorem 3.5, F (Λ,α) = F ⊆ G(Λ,α). Since F is g-(Λ,α)-closed and G(Λ,α) is (Λ,α)-open, by (6), there exists a (Λ,α)-open set U such that F (Λ,α) = F ⊆ U ⊆ U(Λ,α) ⊆ G(Λ,α). � Definition 4.4. A topological space (X,τ) is said to be (Λ,α)-regular if for each (Λ,α)-closed set F of X not containing x, there exist disjoint (Λ,α)-open sets U and V such that x ∈ U and F ⊆ V . Theorem 4.4. For a topological space (X,τ), the following properties are equivalent: (1) (X,τ) is (Λ,α)-regular. (2) For each x ∈ X and each U ∈ ΛαO(X,τ) with x ∈ U, there exists V ∈ ΛαO(X,τ) such that x ∈ V ⊆ V (Λ,α) ⊆ U. (3) For each (Λ,α)-closed set F of X, ∩{V (Λ,α) | F ⊆ V ∈ ΛαO(X,τ)} = F. (4) For each subset A of X and each U ∈ ΛαO(X,τ) with A∩U 6= ∅, there exists V ∈ ΛαO(X,τ) such that A∩V 6= ∅ and V (Λ,α) ⊆ U. (5) For each nonempty subset A of X and each (Λ,α)-closed set F of X with A∩F = ∅, there exist V,W ∈ ΛαO(X,τ) such that A∩V 6= ∅, F ⊆ W and V ∩W = ∅. (6) For each (Λ,α)-closed set F of X and x 6∈ F, there exist U ∈ ΛαO(X,τ) and a g-(Λ,α)-open set V such that x ∈ U, F ⊆ V and U ∩V = ∅. (7) For each subset A of X and each (Λ,α)-closed set F with A∩F = ∅, there exist U ∈ ΛαO(X,τ) and a g-(Λ,α)-open set V such that A∩U 6= ∅, F ⊆ V and U ∩V = ∅. Proof. (1) ⇒ (2): Let G ∈ ΛαO(X,τ) and x 6∈ X −G. Then, there exist disjoint U,V ∈ ΛαO(X,τ) such that X −G ⊆ U and x ∈ V . Thus, V ⊆ X −U and so x ∈ V ⊆ V (Λ,α) ⊆ X −U ⊆ G. (2) ⇒ (3): Let X −F ∈ ΛαO(X,τ) with x ∈ X −F. Then by (2), there exists U ∈ ΛαO(X,τ) such that x ∈ U ⊆ U(Λ,α) ⊆ X −F. Thus, F ⊆ X −U(Λ,α) = V ∈ ΛαO(X,τ) and hence U ∩V = ∅. Then, we have x 6∈ V (Λ,α). This shows that F ⊇∩{V (Λ,α) | F ⊆ V ∈ ΛαO(X,τ)}. (3) ⇒ (4): Let A be a subset of X and U ∈ ΛαO(X,τ) such that A∩U 6= ∅. Let x ∈ A∩U. Then, x 6∈ X −U. Hence by (3), there exists W ∈ ΛαO(X,τ) such that X −U ⊆ W and x 6∈ W (Λ,α). Put V = X −W (Λ,b) which is a (Λ,α)-open set containing x and A∩V 6= ∅. Now, V ⊆ X −W and so V (Λ,α) ⊆ X −W ⊆ U. (4) ⇒ (5): Let A be a nonempty subset of X and F be a (Λ,α)-closed set such that A∩F = ∅. Then, X −F ∈ ΛαO(X,τ) with A∩ (X −F ) 6= ∅ and hence by (4), there exists V ∈ ΛαO(X,τ) such that A∩V 6= ∅ and V (Λ,α) ⊆ X −F. If we put W = X −V (Λ,α), then F ⊆ W and W ∩V = ∅. 10 Int. J. Anal. Appl. (2023), 21:88 (5) ⇒ (1): Let F be a (Λ,α)-closed set not containing x. Then, F ∩{x} = ∅. Thus by (5), there exist V,W ∈ ΛαO(X,τ) such that x ∈ V,F ⊆ W and V ∩W = ∅. (1) ⇒ (6): The proof is obvious. (6) ⇒ (7): Let A be a subset of X and F be a (Λ,α)-closed set such that A∩F = ∅. Then, for x ∈ A, x 6∈ F and by (6), there exist U ∈ ΛαO(X,τ) and a g-(Λ,α)-open set V such that x ∈ U, F ⊆ V and U ∩V = ∅. Thus, A∩U 6= ∅, F ⊆ V and U ∩V = ∅. (7) ⇒ (1): Let F be a (Λ,α)-closed set such that x 6∈ F. Since {x}∩F = ∅, by (7), there exist U ∈ ΛαO(X,τ) and a g-(Λ,α)-open set W such that x ∈ U, F ⊆ W and U ∩ W = ∅. Since W is g-(Λ,α)-open, by Theorem 3.5, we have F ⊆ W(Λ,α) = V ∈ ΛαO(X,τ) and hence U ∩V = ∅. This shows that (X,τ) is (Λ,α)-regular. � Acknowledgements: This research project was financially supported by Mahasarakham University. Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi- cation of this paper. References [1] S. Buadong, C. Viriyapong, C. Boonpok, On Generalized Topology and Minimal Structure Spaces, Int. J. Math. Anal. 5 (2011), 1507–1516. [2] M. Caldas, D. N. Georgiou, S. 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