EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 3730-3742 ISSN 1307-5543 – ejpam.com Published by New York Business Global δ(τ1, τ2)-Continuous Functions Chatchadaporn Prachanpol1, Chawalit Boonpok1, Chokchai Viriyapong1,∗ 1 Mathematics and Applied Mathematics Research Unit, Department of Mathematics, Faculty of Science, Mahasarakham University, Maha Sarakham, 44150, Thailand Abstract. This paper introduces a new class of functions called δ(τ1, τ2)-continuous functions. Several characterizations of δ(τ1, τ2)-continuous functions are investigated. The relationships be- tween δ(τ1, τ2)-continuity and the other types of δ(τ1, τ2)-continuity are also discussed. 2020 Mathematics Subject Classifications: 54C05, 54C08, 54E55 Key Words and Phrases: δ(τ1, τ2)-open set, δ(τ1, τ2)-continuous function 1. Introduction The field of the mathematical science which goes under the name of topology is con- cerned with all questions directly or indirectly related to continuity. By using various forms of open sets many authors introduced and studied various types of continuity. In 1968, Veličko [22] introduced a new class of open sets in topological spaces called δ-open sets and investigated some properties of δ-closed sets and δ-open sets. The class of open sets including the class of δ-open sets. In 1980, Noiri [17] introduced and studied the notion of δ-continuous functions. Munshi and Bassan [16] defined and developed the concept of super-continuity. The concept has been investigated further by Reilly and Vamanamurthy [21] where super-continuity is characterized in terms of the semi-regularization topology. Super-continuity is related to the concepts of δ-continuity and strong θ-continuity devel- oped by Noiri [17]. In particular, super-continuity is strictly between strong θ-continuity and δ-continuity and strictly between complete continuity [1] and δ-continuity. Raychaud- huri and Mukherjee [20] introduced the concept of δ-preopen sets which is weaker than that of preopen sets and used this concept to define the notion of δ-almost continuous func- tions as a generalization of precontinuous functions due to Mashhour et al. [15]. Baker [2] introduced and investigated the notion of weakly δ-continuous functions. The class of weakly δ-continuous functions is a generalization of δ-continuous functions. In 1997, Park et al. [19] introduced the notion of δ-semiopen sets by using δ-open sets due to Valičko [22]. In 2005, Ekici and Navalagi [13] introduced and investigated ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5335 Email addresses: prachanpol.ch@gmail.com (C. Prachanpol), chawalit.b@msu.ac.th (C. Boonpok), chokchai.v@msu.ac.th (C. Viriyapong) https://www.ejpam.com 3730 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3731 the concept of δ-semicontinuous functions. The class of δ-semicontinuous functions is a weaker form of the classes of perfectly continuous functions [18], strongly θ-continuous functions [14] and super-continuous functions. Ekici [12] introduced the notion of almost δ- semicontinuous functions which generalize R-maps [11] and δ-continuous functions. Yüksel et al. [25] extended the concept of δ-open sets to ideal topological spaces and defined δ- I-continuous functions. Moreover, some characterizations of θ-I -continuous functions, ⋆-continuous functions and θ(⋆)-precontinuous functions were presented in [3], [5] and [6], respectively. In [9], the present authors introduced and studied the concept of (τ1, τ2)- continuous functions. Futhermore, several characterizations of almost (τ1, τ2)-continuous functions and weakly (τ1, τ2)-continuous functions were established in [8] and [7], respec- tively. In this paper, we introduce the notions of δ(τ1, τ2)-continuous functions, almost δ(τ1, τ2)-continuous functions, and weakly δ(τ1, τ2)-continuous functions. We also investi- gate several characterizations of δ(τ1, τ2)-continuous functions, almost δ(τ1, τ2)-continuous functions, and weakly δ(τ1, τ2)-continuous functions. Finally, the relationships among δ(τ1, τ2)-continuous functions, almost δ(τ1, τ2)-continuous functions, and weakly δ(τ1, τ2)- continuous functions are discussed. 2. Preliminaries Throughout the present paper, spaces (X, τ1, τ2) and (Y, σ1, σ2) (or simply X and Y ) always mean bitopological spaces on which no separation axioms are assumed unless ex- plicitly stated. Let A be a subset of a bitopological space (X, τ1, τ2). The closure of A and the interior of A with respect to τi are denoted by τi-Cl(A) and τi-Int(A), respec- tively, for i = 1, 2. A subset A of a bitopological space (X, τ1, τ2) is called τ1τ2-closed [10] if A = τ1-Cl(τ2-Cl(A)). The complement of a τ1τ2-closed set is called τ1τ2-open. The intersection of all τ1τ2-closed sets of X containing A is called the τ1τ2-closure [10] of A and is denoted by τ1τ2-Cl(A). The union of all τ1τ2-open sets of X contained in A is called the τ1τ2-interior [10] of A and is denoted by τ1τ2-Int(A). A subset A of a bitopological space (X, τ1, τ2) is called (τ1, τ2)r-open [23] (resp. (τ1, τ2)s-open [4], (τ1, τ2)p-open [4], (τ1, τ2)β-open [4], α(τ1, τ2)-open) [24]) if A = τ1τ2-Int(τ1τ2-Cl(A)) (resp. A ⊆ τ1τ2-Cl(τ1τ2-Int(A)), A ⊆ τ1τ2-Int(τ1τ2-Cl(A)), A ⊆ τ1τ2-Cl(τ1τ2-Int(τ1τ2-Cl(A))), A ⊆ τ1τ2-Int(τ1τ2-Cl(τ1τ2-Int(A)))). The complement of a (τ1, τ2)r-open (resp. (τ1, τ2)s- open, (τ1, τ2)p-open, (τ1, τ2)β-open, α(τ1, τ2)-open) set is called (τ1, τ2)r-closed (resp. (τ1, τ2)s-closed, (τ1, τ2)p-closed, (τ1, τ2)β-closed, α(τ1, τ2)-closed). Let A be a subset of a bitopological space (X, τ1, τ2). A point x of X is called a δ(τ1, τ2)-cluster point of A if V ∩ A ̸= ∅ for every (τ1, τ2)r-open set V containing x. The set of all δ(τ1, τ2)-cluster points of A is called the δ(τ1, τ2)-closure of A and is denoted by δ(τ1, τ2)-Cl(A). A subset A of a bitopological space (X, τ1, τ2) is called δ(τ1, τ2)-closed if A = δ(τ1, τ2)-Cl(A). The complement of a δ(τ1, τ2)-closed set is called δ(τ1, τ2)-open (τ1τ2-δ-open [8]). The fam- ily of all δ(τ1, τ2)-open (resp. δ(τ1, τ2)-closed) sets of a bitopological space (X, τ1, τ2) is denoted by δ(τ1, τ2)O(X) (resp. δ(τ1, τ2)C(X)). The δ(τ1, τ2)-interior of A denoted by C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3732 δ(τ1, τ2)-Int(A) is defined as follows: δ(τ1, τ2)-Int(A) = ∪{G ⊆ X | G ∈ δ(τ1, τ2)O(X) and G ⊆ A}. Lemma 1. For a subset A of a bitopological space (X, τ1, τ2), x ∈ δ(τ1, τ2)-Cl(A) if and only if V ∩A ̸= ∅ for every V ∈ δ(τ1, τ2)O(X) containing x. Lemma 2. For a subset A of a bitopological space (X, τ1, τ2), the following properties hold: (1) δ(τ1, τ2)-Int(X −A) = X − δ(τ1, τ2)-Cl(A). (2) δ(τ1, τ2)-Cl(X −A) = X − δ(τ1, τ2)-Int(A). 3. On δ(τ1, τ2)-continuous functions In this section, we introduce the notion of δ(τ1, τ2)-continuous functions. We also discuss several characterizations of δ(τ1, τ2)-continuous functions. Definition 1. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called δ(τ1, τ2)-continuous at x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ V . A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be δ(τ1, τ2)-continuous if f is δ(τ1, τ2)-continuous at each point of X. Example 1. Let X = {a, b, c} with topologies τ1 = {∅, {a}, {b}, {a, b}, {a, c}, X} and τ2 = {∅, {a}, {b}, {a, b}, X}. Let Y = {1, 2, 3} with topologies σ1 = {∅, {1}, {3}, {1, 2}, {1, 3}, Y } and σ2 = {∅, {1}, {3}, {1, 3}, Y }. Define a function f : (X, τ1, τ2) → (Y, σ1, σ2) as follows: f(a) = f(c) = 2 and f(b) = 3. Then, f is δ(τ1, τ2)-continuous. Theorem 1. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is δ(τ1, τ2)-continuous at x; (2) x ∈ δ(τ1, τ2)-Int(f −1(V )) for every σ1σ2-open set V of Y containing f(x); (3) x ∈ f−1(σ1σ2-Cl(f(A))) for every A ⊆ X such that x ∈ δ(τ1, τ2)-Cl(A); (4) x ∈ f−1(σ1σ2-Cl(B)) for every B ⊆ Y such that x ∈ δ(τ1, τ2)-Cl(f −1(B)); (5) x ∈ δ(τ1, τ2)-Int(f −1(B)) for every B ⊆ Y such that x ∈ f−1(σ1σ2-Int(B)); (6) x ∈ f−1(F ) for every σ1σ2-closed set F of Y such that x ∈ δ(τ1, τ2)-Cl(f −1(F )). Proof. (1) ⇒ (2): Let V be any σ1σ2-open set of Y containing f(x). By (1), There exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ V . Hence, x ∈ U ⊆ f−1(V ). Therefore, x ∈ δ(τ1, τ2)-Int(f −1(V )). (2) ⇒ (3): Let A ⊆ X such that x ∈ δ(τ1, τ2)-Cl(A) and V be any σ1σ2-open set of Y containing f(x). By (2), x ∈ δ(τ1, τ2)-Int(f −1(V )). Then, there exists a δ(τ1, τ2)-open C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3733 set U of X containing x such that U ⊆ f−1(V ). By Lemma 1, we have U ∩ A ̸= ∅ and hence ∅ ≠ f(U ∩ A) ⊆ f(U) ∩ f(A) ⊆ V ∩ f(A). Thus, f(x) ∈ σ1σ2-Cl(f(A)) and so x ∈ f−1(σ1σ2-Cl(f(A))). (3) ⇒ (4): Let B be any subset of Y and x ∈ δ(τ1, τ2)-Cl(f −1(B)). Then by (3), we have x ∈ f−1(σ1σ2-Cl(f(f −1(B)))) ⊆ f−1(σ1σ2-Cl(B)). (4) ⇒ (5): Let B be any subset of Y and x ∈ f−1(σ1σ2-Int(B)). Suppose that x /∈ δ(τ1, τ2)-Int(f −1(B)). Then, x ∈ X − δ(τ1, τ2)-Int(f −1(B)). Since X − δ(τ1, τ2)-Int(f −1(B)) = δ(τ1, τ2)-Cl(f −1(Y −B)) and by (4), we obtain that x ∈ f−1(σ1σ2-Cl(Y −B)) = X − f−1(σ1σ2-Int(B)) and hence x /∈ f−1(σ1σ2-Int(B)), which is a contradiction that x ∈ f−1(σ1σ2-Int(B)). Therefore, x ∈ δ(τ1, τ2)-Int(f −1(B)). (5) ⇒ (6): Let F be any σ1σ2-closed set of Y and x ∈ δ(τ1, τ2)-Cl(f −1(F )). Suppose that x /∈ f−1(F ). Since Y − F is σ1σ2-open in Y , x ∈ X − f−1(F ) = f−1(Y − F ) = f−1(σ1σ2-Int(Y − F )). By (5) and Lemma 2 (1), we have x ∈ δ(τ1, τ2)-Int(f −1(Y − F )) = X − δ(τ1, τ2)-Cl(f −1(F )) and hence x ̸∈ δ(τ1, τ2)-Cl(f −1(F )). This is a contradiction. Therefore, x ∈ f−1(F ). (6) ⇒ (2): Let V be any σ1σ2-open set of Y containing f(x). Then, x ∈ f−1(V ). Suppose that x ̸∈ δ(τ1, τ2)-Int(f −1(V )). By Lemma 2 (2), x ∈ X−δ(τ1, τ2)-Int(f −1(V )) = δ(τ1, τ2)-Cl(f −1(Y − V )). Since Y − V is σ1σ2-closed in Y and by (6), we have x ∈ f−1(Y − V ) = X − f−1(V ). This implies that x /∈ f−1(V ), which is a contradiction. Thus, x ∈ δ(τ1, τ2)-Int(f −1(V )). (2) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). By (2), we have x ∈ δ(τ1, τ2)-Int(f −1(V )). Then, there exists a δ(τ1, τ2)-open set U such that x ∈ U ⊆ f−1(V ). Thus, f(U) ⊆ V . This shows that f is δ(τ1, τ2)-continuous at x. Theorem 2. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is δ(τ1, τ2)-continuous; (2) f−1(V ) is δ(τ1, τ2)-open in X for every σ1σ2-open set V of Y ; (3) f(δ(τ1, τ2)-Cl(A)) ⊆ σ1σ2-Cl(f(A)) for every subset A of X; (4) δ(τ1, τ2)-Cl(f −1(B)) ⊆ f−1(σ1σ2-Cl(B)) for every subset B of Y ; (5) f−1(σ1σ2-Int(B)) ⊆ δ(τ1, τ2)-Int(f −1(B)) for every subset B of Y ; (6) f−1(F ) is δ(τ1, τ2)-closed in X for every σ1σ2-closed set F of Y . C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3734 Proof. (1) ⇒ (2): Let V be any σ1σ2-open of Y and x ∈ f−1(V ). By (1), there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ V . Thus, x ∈ U ⊆ f−1(V ) and hence x ∈ δ(τ1, τ2)-Int(f −1(V )). This implies that f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(V )). Therefore, f−1(V ) is δ(τ1, τ2)-open in X. (2) ⇒ (3): Let A be any subset of X and let x ∈ δ(τ1, τ2)-Cl(A). Then, we have f(x) ∈ f(δ(τ1, τ2)-Cl(A)). Let V be any σ1σ2-open set of Y containing f(x). By (2), f−1(V ) is δ(τ1, τ2)-open in X. Therefore, x ∈ f−1(V ) = δ(τ1, τ2)-Int(f −1(V )). Then, there exists a δ(τ1, τ2)-open set U of X containing x such that U ⊆ f−1(V ). Since x ∈ δ(τ1, τ2)-Cl(A), then U ∩ A ̸= ∅. Hence, ∅ ̸= f(U ∩ A) ⊆ f(U) ∩ f(A) ⊆ V ∩ f(A). Thus, f(x) ∈ σ1σ2-Cl(f(A)) and so f(δ(τ1, τ2)-Cl(A)) ⊆ σ1σ2-Cl(f(A)). (3) ⇒ (4): Let B be any subset of Y . Then by (3), f(δ(τ1, τ2)-Cl(f −1(B))) ⊆ σ1σ2-Cl(f(f −1(B))) ⊆ σ1σ2-Cl(B). Therefore, δ(τ1, τ2)-Cl(f −1(B)) ⊆ f−1(σ1σ2-Cl(B)). (4) ⇒ (5): Let B be any subset of Y . By (4) and Lemma 2 (2), we have X − δ(τ1, τ2)-Int(f −1(B)) = δ(τ1, τ2)-Cl(f −1(Y −B)) ⊆ f−1(σ1σ2-Cl(Y −B)) = X − f−1(σ1σ2-Int(B)) and hence f−1(σ1σ2-Int(B)) ⊆ δ(τ1, τ2)-Int(f −1(B)). (5) ⇒ (6): Let F be any σ1σ2-closed set of Y . Then, Y −F is σ1σ2-open in Y . By (5) and Lemma 2 (1), we obtain that X − f−1(F ) = f−1(Y − F ) = f−1(σ1σ2-Int(Y − F )) ⊆ δ(τ1, τ2)-Int(f −1(Y − F )) = X − δ(τ1, τ2)-Cl(f −1(F )). Therefore, δ(τ1, τ2)-Cl(f −1(F )) ⊆ f−1(F ). This shows that f−1(F ) is δ(τ1, τ2)-closed in X. (6) ⇒ (2): Let V be any σ1σ2-open set of Y . Then, Y −V is σ1σ2-closed in Y . By (6), we have X−f−1(V ) = f−1(Y −V ) is δ(τ1, τ2)-closed in X. Thus, f−1(V ) is δ(τ1, τ2)-open in X. (2) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). By (2), f−1(V ) is δ(τ1, τ2)-open in X. Hence, x ∈ f−1(V ) = δ(τ1, τ2)-Int(f −1(V )). Then, there exists a δ(τ1, τ2)-open set U such that x ∈ U ⊆ f−1(V ). Therefore, f(U) ⊆ V . Thus, f is δ(τ1, τ2)-continuous at x. This shows that f is δ(τ1, τ2)-continuous. 4. On almost δ(τ1, τ2)-continuous functions In this section, we introduce the notion of almost δ(τ1, τ2)-continuous functions and investigate some characterizations of almost δ(τ1, τ2)-continuous functions. Moreover, the relationships between δ(τ1, τ2)-continuous functions and almost δ(τ1, τ2)-continuous func- tions are considered. Definition 2. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be almost δ(τ1, τ2)- continuous at x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ σ1σ2-Int(σ1σ2-Cl(V )). A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called almost δ(τ1, τ2)-continuous if f is almost δ(τ1, τ2)-continuous at each point of X. C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3735 Remark 1. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following implication holds: δ(τ1, τ2)-continuity ⇒ almost δ(τ1, τ2)-continuity. The converse of the implication is not true in general. We give an example for the implication as follows. Example 2. Let X = {1, 2, 3, 4} with topologies τ1 = {∅, {1}, {2}, {1, 2}, X} and τ2 = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, X}. Let Y = {a, b, c} with topologies σ1 = {∅, {a}, {b}, {a, b}, Y } and σ2 = {∅, {a}, {b}, {a, b}, {a, c}, Y }. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is defined as follows: f(1) = a, f(2) = b and f(3) = f(4) = c. Then f is almost δ(τ1, τ2)-continuous, but f is not δ(τ1, τ2)-continuous. Theorem 3. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is almost δ(τ1, τ2)-continuous at x; (2) x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) for every σ1σ2-open set V of Y con- taining f(x); (3) x ∈ δ(τ1, τ2)-Int(f −1(V )) for every (σ1, σ2)r-open set V of Y containing f(x); (4) for each x ∈ X and each (σ1, σ2)r-open set V of Y containing f(x), there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ V . Proof. (1) ⇒ (2): Let V be any σ1σ2-open set of Y containing f(x). Since f is almost δ(τ1, τ2)-continuous at x ∈ X. There exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ σ1σ2-Int(σ1σ2-Cl(V )). Thus, x ∈ U ⊆ f−1(σ1σ2-Int(σ1σ2-Cl(V ))). Therefore, x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))). (2) ⇒ (3): Let V be any (σ1, σ2)r-open set of Y containing f(x). Then, V is σ1σ2-open in Y . By (2), x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) = δ(τ1, τ2)-Int(f −1(V )). (3) ⇒ (4): Let V be any (σ1, σ2)r-open set of Y containing f(x). Thus by (3), we have x ∈ δ(τ1, τ2)-Int(f −1(V )). Then, there exists a δ(τ1, τ2)-open set U of X such that x ∈ U ⊆ f−1(V ). Therefore, f(U) ⊆ V . (4) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since σ1σ2-Int(σ1σ2-Cl(V )) is a (σ1, σ2)r-open set and by (4), there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ σ1σ2-Int(σ1σ2-Cl(V )). Consequently, f is almost δ(τ1, τ2)-continuous at x. Theorem 4. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3736 (1) f is almost δ(τ1, τ2)-continuous; (2) f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) for every σ1σ2-open set V of Y ; (3) δ(τ1, τ2)-Cl(f −1(σ1σ2-Cl(σ1σ2-Int(F )))) ⊆ f−1(F ) for every σ1σ2-closed set F of Y ; (4) δ(τ1, τ2)-Cl(f −1(σ1σ2-Cl(σ1σ2-Int(σ1σ2-Cl(B))))) ⊆ f−1(σ1σ2-Cl(B)) for every sub- set B of Y ; (5) f−1(σ1σ2-Int(B)) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(σ1σ2-Int(B))))) for every sub- set B of Y ; (6) f−1(V ) is δ(τ1, τ2)-open in X for every (σ1, σ2)r-open set V of Y ; (7) f−1(F ) is δ(τ1, τ2)-closed in X for every (σ1, σ2)r-closed set F of Y . Proof. (1) ⇒ (2): Let V be any σ1σ2-open set of Y and x ∈ f−1(V ). Since f is almost δ(τ1, τ2)-continuous, there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ σ1σ2-Int(σ1σ2-Cl(V )). Then, we have x ∈ U ⊆ f−1(σ1σ2-Int(σ1σ2-Cl(V ))) and hence x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))). This implies that f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))). (2) ⇒ (3): Let F be any σ1σ2-closed set of Y . Then, Y −F is σ1σ2-open in Y . Thus by (2), we have X− f−1(F ) = f−1(Y −F ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(Y −F )))) = δ(τ1, τ2)-Int(f −1(Y−σ1σ2-Cl(σ1σ2-Int(F )))) = X−δ(τ1, τ2)-Cl(f −1(σ1σ2-Cl(σ1σ2-Int(F )))) and hence δ(τ1, τ2)-Cl(f −1(σ1σ2-Cl(σ1σ2-Int(F )))) ⊆ f−1(F ). (3) ⇒ (4): Let B be any subset of Y . Then, σ1σ2-Cl(B) is σ1σ2-closed in Y and by (3), δ(τ1, τ2)-Cl(f −1(σ1σ2-Cl(σ1σ2-Int(σ1σ2-Cl(B))))) ⊆ f−1(σ1σ2-Cl(B)). (4) ⇒ (5): Let B be any subset of Y . Then by (4), f−1(σ1σ2-Int(B)) = X − f−1(σ1σ2-Cl(Y −B)) ⊆ X − δ(τ1, τ2)-Cl(f −1(σ1σ2-Cl(σ1σ2-Int(σ1σ2-Cl(Y −B))))) = δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(σ1σ2-Int(B))))). Therefore, f−1(σ1σ2-Int(B)) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(σ1σ2-Int(B))))). (5) ⇒ (6): Let V be any (σ1, σ2)r-open set of Y . Then, V is a σ1σ2-open set of Y and V = σ1σ2-Int(σ1σ2-Cl(σ1σ2-Int(V ))). By (5), f−1(V ) = f−1(σ1σ2-Int(V )) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(σ1σ2-Int(V ))))) = δ(τ1, τ2)-Int(f −1(V )) and so f−1(V ) is δ(τ1, τ2)-open in X. (6) ⇒ (7): The proof is obvious. (7) ⇒ (1): Let x ∈ X and V be any (σ1, σ2)r-open set of Y containing f(x). Then, Y − V is (σ1, σ2)r-closed in Y . Thus by (7), we have X − f−1(V ) = f−1(Y − V ) = δ(τ1, τ2)-Cl(f −1(Y −V )) = X−δ(τ1, τ2)-Int(f −1(V )) and hence x ∈ δ(τ1, τ2)-Int(f −1(V )). Then, there exists a δ(τ1, τ2)-open set U of X containing x such that U ⊆ f−1(V ). Thus, f(U) ⊆ V . By Theorem 3 (4), f is almost δ(τ1, τ2)-continuous at x. This shows that f is almost δ(τ1, τ2)-continuous. C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3737 Theorem 5. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is almost δ(τ1, τ2)-continuous; (2) δ(τ1, τ2)-Cl(f −1(V )) ⊆ f−1(σ1σ2-Cl(V )) for every (σ1, σ2)β-open set V of Y ; (3) δ(τ1, τ2)-Cl(f −1(V )) ⊆ f−1(σ1σ2-Cl(V )) for every (σ1, σ2)s-open set V of Y ; (4) f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) for every (σ1, σ2)p-open set V of Y . Proof. (1) ⇒ (2): Let V be any (σ1, σ2)β-open set of Y . Then, σ1σ2-Cl(V ) is (σ1, σ2)r-closed in Y . By Theorem 4 (7), f−1(σ1σ2-Cl(V )) is δ(τ1, τ2)-closed in X. Hence, δ(τ1, τ2)-Cl(f −1(V )) ⊆ δ(τ1, τ2)-Cl(f −1(σ1σ2-Cl(V ))) = f−1(σ1σ2-Cl(V )). (2) ⇒ (3): The proof is obvious since every (σ1, σ2)s-open set is (σ1, σ2)β-open. (3) ⇒ (1): Let F be any (σ1, σ2)r-closed set of Y . Then, F is (σ1, σ2)s-open in Y . By (3), we have δ(τ1, τ2)-Cl(f −1(F )) ⊆ f−1(σ1σ2-Cl(F )) = f−1(F ) and hence f−1(F ) is δ(τ1, τ2)-closed in X. By Theorem 4 (7), f is almost δ(τ1, τ2)-continuous. (1) ⇒ (4): Let V be any (σ1, σ2)p-open set of Y . Then, σ1σ2-Int(σ1σ2-Cl(V )) is (σ1, σ2)r-open in Y . By Theorem 4 (6), f−1(σ1σ2-Int(σ1σ2-Cl(V ))) is δ(τ1, τ2)-open in X. Thus, f−1(σ1σ2-Int(σ1σ2-Cl(V ))) = δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) and hence f−1(V ) ⊆ f−1(σ1σ2-Int(σ1σ2-Cl(V ))) = δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))). (4) ⇒ (1) : Let V be any (σ1, σ2)r-open set of Y . Then, V = σ1σ2-Int(σ1σ2-Cl(V )) and V is (σ1, σ2)p-open in Y . By (4), f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) = δ(τ1, τ2)-Int(f −1(V )). Therefore, f−1(V ) is δ(τ1, τ2)-open in X. By Theorem 4 (6), f is almost δ(τ1, τ2)-continuous. 5. On weakly δ(τ1, τ2)-continuous functions In this section, we introduce and investigate the concept of weakly δ(τ1, τ2)-continuous functions. Furthermore, we discuss the relationships between almost δ(τ1, τ2)-continuous functions and weakly δ(τ1, τ2)-continuous functions. Definition 3. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be weakly δ(τ1, τ2)- continuous at x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ σ1σ2-Cl(V ). A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called weakly δ(τ1, τ2)-continuous if f is weakly δ(τ1, τ2)- continuous at each point of X. Remark 2. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following implication holds: almost δ(τ1, τ2)-continuity ⇒ weak δ(τ1, τ2)-continuity. The converse of the implication is not true in general. We give an example for the implication as follows. C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3738 Example 3. Let X = {a, b, c} with topologies τ1 = {∅, {a}, {b}, {a, b}, {a, c}, X} and τ2 = {∅, {a}, {b}, {a, b}, X}. Let Y = {1, 2, 3} with topologies σ1 = {∅, {1}, {3}, {1, 3}, Y } and σ2 = {∅, {1}, {3}, {1, 2}, {1, 3}, Y }. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is defined as follows: f(a) = f(b) = 2 and f(c) = 1. Then f is weakly δ(τ1, τ2)-continuous, but f is not almost δ(τ1, τ2)-continuous. Theorem 6. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is weakly δ(τ1, τ2)-continuous at x ∈ X if and only if x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))) for every σ1σ2-open set V of Y containing f(x). Proof. Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since f is weakly δ(τ1, τ2)-continuous at x, so there exists a δ(τ1, τ2)-open set U of X containing x such that f(U) ⊆ σ1σ2-Cl(V ). Therefore, x ∈ U ⊆ f−1(σ1σ2-Cl(V )). This implies that x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). Conversely let V be any σ1σ2-open set of Y containing f(x). By the hypothesis, we obtain that x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). Then, there exists a δ(τ1, τ2)-open set U of X containing x such that U ⊆ f−1(σ1σ2-Cl(V )). Thus, f(U) ⊆ σ1σ2-Cl(V ). This shows that f is weakly δ(τ1, τ2)-continuous at x. Theorem 7. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is weakly δ(τ1, τ2)-continuous if and only if f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))) for every σ1σ2-open set V of Y . Proof. Let V be any σ1σ2-open set of Y and x ∈ f−1(V ). Since f is weakly δ(τ1, τ2)- continuous, by Theorem 6, x ∈ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). Therefore, f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). Conversely, let x ∈ X and V be any σ1σ2-open set of Y containing f(x). By the hypothesis, x ∈ f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). By Theorem 6, f is weakly δ(τ1, τ2)-continuous at x. This shows that f is weakly δ(τ1, τ2)-continuous. Theorem 8. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is weakly δ(τ1, τ2)-continuous; (2) f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))) for every σ1σ2-open set V of Y ; (3) δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(F ))) ⊆ f−1(F ) for every σ1σ2-closed F of Y ; (4) δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(B)))) ⊆ f−1(σ1σ2-Cl(B)) for every subset B of Y ; (5) f−1(σ1σ2-Int(B)) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(σ1σ2-Int(B)))) for every subset B of Y ; (6) δ(τ1, τ2)-Cl(f −1(V )) ⊆ f−1(σ1σ2-Cl(V )) for every σ1σ2-open set V of Y ; (7) δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(F ))) ⊆ f−1(F ) for every (σ1, σ2)r-closed set F of Y ; C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3739 (8) δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) ⊆ f−1(σ1σ2-Cl(V )) for every (σ1, σ2)β-open set V of Y ; (9) δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) ⊆ f−1(σ1σ2-Cl(V )) for every (σ1, σ2)s-open set V of Y ; (10) δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) ⊆ f−1(σ1σ2-Cl(V )) for every (σ1, σ2)p-open set V of Y ; (11) δ(τ1, τ2)-Cl(f −1(V )) ⊆ f−1(σ1σ2-Cl(V )) for every (σ1, σ2)p-open set V of Y ; (12) f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))) for every (σ1, σ2)p-open set V of Y . Proof. (1) ⇒ (2): Let V be any σ1σ2-open set of Y . It follows from Theorem 7, we obtain that f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). (2) ⇒ (3): Let F be any σ1σ2-closed set of Y . Then, Y − F is σ1σ2-open in Y . Thus by (2), we have X − f−1(F ) = f−1(Y − F ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(Y − F ))) = δ(τ1, τ2)-Int(f −1(Y − σ1σ2-Int(F ))) = δ(τ1, τ2)-Int(X − f−1(σ1σ2-Int(F ))) = X − δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(F ))) and hence δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(F ))) ⊆ f−1(F ). (3) ⇒ (4): Let B be any subset of Y . Since σ1σ2-Cl(B) is σ1σ2-closed in Y and by (3), δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(B)))) ⊆ f−1(σ1σ2-Cl(B)). (4) ⇒ (5): Let B be any subset of Y . Then by (4), f−1(σ1σ2-Int(B)) = X − f−1(σ1σ2-Cl(Y −B)) ⊆ X − δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(Y −B)))) = δ(τ1, τ2)-Int(X − f−1(σ1σ2-Int(σ1σ2-Cl(Y −B)))) = δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(σ1σ2-Int(B)))). Thus, f−1(σ1σ2-Int(B)) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(σ1σ2-Int(B)))). (5) ⇒ (6): Let V be any σ1σ2-open set of Y . Suppose that x /∈ f−1(σ1σ2-Cl(V )). Then, f(x) ̸∈ σ1σ2-Cl(V ). There exists a σ1σ2-open set U of Y containing f(x) such that U ∩ V = ∅. Hence, σ1σ2-Cl(U) ∩ V = ∅. By (5), x ∈ f−1(U) = f−1(σ1σ2-Int(U)) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(σ1σ2-Int(U)))) = δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(U))). Then, there exists a δ(τ1, τ2)-open set G of X such that x ∈ G ⊆ f−1(σ1σ2-Cl(U)). Therefore, f−1(V ) ∩G ⊆ f−1(V ) ∩ f−1(σ1σ2-Cl(U)) = f−1(V ∩ σ1σ2-Cl(U)) = ∅. Thus, x ̸∈ δ(τ1, τ2)-Cl(f −1(V )) and hence δ(τ1, τ2)-Cl(f −1(V )) ⊆ f−1(σ1σ2-Cl(V )). (6) ⇒ (7): Let F be any (σ1, σ2)r-closed set of Y . Then, σ1σ2-Int(F ) is σ1σ2-open in Y . By (6), δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(F ))) ⊆ f−1(σ1σ2-Cl(σ1σ2-Int(F ))) = f−1(F ). C. Prachanpol, C. Boonpok, C. Viriyapong / Eur. J. Pure Appl. Math, 17 (4) (2024), 3730-3742 3740 (7) ⇒ (8): Let V be any (σ1, σ2)β-open set of Y . Then, we have V ⊆ σ1σ2-Cl(σ1σ2-Int(σ1σ2-Cl(V ))) and so σ1σ2-Cl(V ) is (σ1, σ2)r-closed. By (7), δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) ⊆ f−1(σ1σ2-Cl(V )). (8) ⇒ (9): The proof is obvious since every (σ1, σ2)s-open set is (σ1, σ2)β-open. (9)⇒ (10): Let V be any (σ1, σ2)p-open set of Y . Then, V ⊆ σ1σ2-Int(σ1σ2-Cl(V )) and σ1σ2-Cl(V ) ⊆ σ1σ2-Cl(σ1σ2-Int(σ1σ2-Cl(V ))). Therefore, σ1σ2-Cl(V ) is (σ1, σ2)s-open in Y . Thus by (9), δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) = δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(σ1σ2-Cl(V ))))) ⊆ f−1(σ1σ2-Cl(σ1σ2-Cl(V ))) = f−1(σ1σ2-Cl(V )). (10)⇒ (11): Let V be any (σ1, σ2)p-open set of Y . Then by (10), δ(τ1, τ2)-Cl(f −1(V )) ⊆ δ(τ1, τ2)-Cl(f −1(σ1σ2-Int(σ1σ2-Cl(V )))) ⊆ f−1(σ1σ2-Cl(V )). (11) ⇒ (12): Let V be any (σ1, σ2)p-open set of Y . Thus by (11), f−1(V ) ⊆ f−1(σ1σ2-Int(σ1σ2-Cl(V ))) = X − f−1(σ1σ2-Cl(Y − σ1σ2-Cl(V ))) ⊆ X − δ(τ1, τ2)-Cl(f −1(Y − σ1σ2-Cl(V ))) = δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). (12) ⇒ (1): Let V be any σ1σ2-open set of Y . Then, V is (σ1, σ2)p-open in Y and by (12), f−1(V ) ⊆ δ(τ1, τ2)-Int(f −1(σ1σ2-Cl(V ))). It follows from Theorem 7 that f is weakly δ(τ1, τ2)-continuous. 6. Conclusion This paper deals with the concepts of δ(τ1, τ2)-continuous functions, almost δ(τ1, τ2)- continuous functions, and weakly δ(τ1, τ2)-continuous functions. Moreover, some charac- terizations and several properties concerning δ(τ1, τ2)-continuous functions, almost δ(τ1, τ2)- continuous functions, and weakly δ(τ1, τ2)-continuous functions are obtained. 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