EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6017 ISSN 1307-5543 – ejpam.com Published by New York Business Global On Strongly θ(τ1, τ2)-continuous Functions Prapart Pue-on1, Supunnee Sompong2, Chawalit Boonpok1,∗ 1 Mathematics and Applied Mathematics Research Unit, Department of Mathematics, Faculty of Science, Mahasarakham University, Maha Sarakham, 44150, Thailand 2 Department of Mathematics and Statistics, Faculty of Science and Technology, Sakon Nakhon Rajbhat University, Sakon Nakhon, 47000, Thailand Abstract. This paper deals with the concept of strongly θ(τ1, τ2)-continuous functions. Fur- thermore, some characterizations and several properties concerning strongly θ(τ1, τ2)-continuous functions are considered. 2020 Mathematics Subject Classifications: 54C08; 54E55 Key Words and Phrases: (τ1, τ2)θ-open set, strongly θ(τ1, τ2)-continuous function 1. Introduction In 1941, Fomin [1] introduced the concept of θ-continuous functions. Noiri [2] stud- ied some properties of θ-continuous functions. Popa [3] investigated several character- izations of θ-continuous functions. Arya and Bhamini [4] introduced the notion of θ- semi-continuous functions. Jafari and Noiri [5] investigated several characterizations of θ-semi-continuous functions. Noiri [6] introduced and investigated the concept of θ- precontinuous functions. In 1981, Long and Herrington [7] investigated some charac- terizations of strongly θ-continuous functions. Jafari and Noiri [8] introduced and in- vestigated the concept of strongly θ-semi-continuous functions. Noiri [9] introduced and studied the notion of strongly θ-precontinuous functions. Noiri and Popa [10] introduced and investigated the concept of strongly θ-β-continuous functions. On the other hand, Di Maio and Noiri [11] introduced the concept of strongly irresolute functions. Pal and Bhattacharyya [12] introduced and investigated the notion of strongly preirresolute func- tions. Noiri [13] investigated several characterizations of strongly β-irresolute functions. Jafari and Noiri [14] studied some characterizations of strongly sober θ-continuous func- tions. These classes of functions have characterizations similar to the class of strongly θ-continuous functions. In 2005, Noiri and Popa [15] introduced a new class of func- tions called strongly θ-M -continuous functions as functions defined between sets satisfying ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6017 Email addresses: prapart.p@msu.ac.th (P. Pue-on), s−sompong@snru.ac.th (S. Sompong), chawalit.b@msu.ac.th (C. Boonpok) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 2 of 11 some minimal conditions and obtained several characterizations and some properties of such functions. Furthermore, the present authors [15] defined and studied the notions of strongly θ-M -closed graphs and m-closed spaces. Thongmoon and Boonpok [16] intro- duced and studied the notion of strongly θ(Λ, p)-continuous functions. Quite recently, the present authors [17] introduced and investigated the notion of almost strongly θ(Λ, p)- continuous functions. Moreover, several characterizations of (τ1, τ2)-continuous functions, almost (τ1, τ2)-continuous functions, weakly (τ1, τ2)-continuous functions, almost weakly (τ1, τ2)-continuous functions, faintly (τ1, τ2)-continuous functions, weakly quasi (τ1, τ2)- continuous functions, almost quasi (τ1, τ2)-continuous functions, δ(τ1, τ2)-continuous func- tions and quasi θ(τ1, τ2)-continuous functions were established in [18], [19], [20], [21], [22], [23], [24], [25] and [26], respectively. In this paper, we introduce the concept of strongly θ(τ1, τ2)-continuous functions. We also investigate several characterizations of strongly θ(τ1, τ2)-continuous functions. 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 explicitly 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), respectively, for i = 1, 2. A subset A of a bitopological space (X, τ1, τ2) is called τ1τ2-closed [27] 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 [27] 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 [27] of A and is denoted by τ1τ2-Int(A). Lemma 1. [27] Let A and B be subsets of a bitopological space (X, τ1, τ2). For the τ1τ2- closure, the following properties hold: (1) A ⊆ τ1τ2-Cl(A) and τ1τ2-Cl(τ1τ2-Cl(A)) = τ1τ2-Cl(A). (2) If A ⊆ B, then τ1τ2-Cl(A) ⊆ τ1τ2-Cl(B). (3) τ1τ2-Cl(A) is τ1τ2-closed. (4) A is τ1τ2-closed if and only if A = τ1τ2-Cl(A). (5) τ1τ2-Cl(X −A) = X − τ1τ2-Int(A). A subset A of a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)r-open [28] (resp. (τ1, τ2)s-open [29], (τ1, τ2)p-open [29], (τ1, τ2)β-open [29]) 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)))). The complement of a (τ1, τ2)r-open (resp. (τ1, τ2)s-open, (τ1, τ2)p-open, (τ1, τ2)β-open) set is called (τ1, τ2)r-closed (resp. (τ1, τ2)s-closed, (τ1, τ2)p-closed, (τ1, τ2)β-closed). A subset A of a bitopological space (X, τ1, τ2) is said to be α(τ1, τ2)-open [30] if A ⊆ τ1τ2-Int(τ1τ2-Cl(τ1τ2-Int(A))). The complement of an α(τ1, τ2)-open set is said to be P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 3 of 11 α(τ1, τ2)-closed. For a subset A of a bitopological space (X, τ1, τ2), a point x ∈ X is called a (τ1, τ2)θ-cluster point [28] of A if τ1τ2-Cl(U)∩A ̸= ∅ for every τ1τ2-open set U containing x. The set of all (τ1, τ2)θ-cluster points of A is called the (τ1, τ2)θ-closure [28] of A and is denoted by (τ1, τ2)θ-Cl(A). A subset A of a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)θ-closed [28] if A = (τ1, τ2)θ-Cl(A). The complement of a (τ1, τ2)θ-closed set is said to be (τ1, τ2)θ-open. The union of all (τ1, τ2)θ-open sets contained in A is called the (τ1, τ2)θ-interior [28] of A and is denoted by (τ1, τ2)θ-Int(A). Lemma 2. [28] For a subset A of a bitopological space (X, τ1, τ2), the following properties hold: (1) If A is τ2τ2-open in X, then τ1τ2-Cl(A) = (τ1, τ2)θ-Cl(A). (2) (τ1, τ2)θ-Cl(A) is τ1τ2-closed in X. 3. On strongly θ(τ1, τ2)-continuous functions In this section, we introduce the concept of strongly θ(τ1, τ2)-continuous functions. Moreover, some characterizations of strongly θ(τ1, τ2)-continuous functions are discussed. Definition 1. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be strongly θ(τ1, τ2)- continuous at a point x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a τ1τ2-open set U of X containing x such that f(τ1τ2-Cl(U)) ⊆ V . A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be strongly θ(τ1, τ2)-continuous if f is strongly θ(τ1, τ2)-continuous at each point x of X. Theorem 1. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is strongly θ(τ1, τ2)-continuous at x ∈ X if and only if for each σ1σ2-open set V of Y containing f(x), x ∈ (τ1, τ2)θ-Int(f −1(V )). Proof. Suppose that f is strongly θ(τ1, τ2)-continuous at x ∈ X. Let V be any σ1σ2- open set of Y containing f(x). Then, there exists a τ1τ2-open set U of X containing x such that f(τ1τ2-Cl(U)) ⊆ V . Thus, τ1τ2-Cl(U) ⊆ f−1(V ) and hence x ∈ (τ1, τ2)θ-Int(f −1(V )). Conversely, let V be any σ1σ2-open set of Y containing f(x). Then, by the hypothesis we have x ∈ (τ1, τ2)θ-Int(f −1(V )). There exists a τ1τ2-open set U of X such that x ∈ U ⊆ τ1τ2-Cl(U) ⊆ f−1(V ); hence f(τ1τ2-Cl(U)) ⊆ V . This shows that f is strongly θ(τ1, τ2)-continuous at x ∈ X. Theorem 2. For a function (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equiva- lent: (1) f is strongly θ(τ1, τ2)-continuous; (2) f−1(V ) is (τ1, τ2)θ-open in X for every σ1σ2-open set V of Y ; P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 4 of 11 (3) f−1(F ) is (τ1, τ2)θ-closed in X for every σ1σ2-closed set F of Y ; (4) f((τ1, τ2)θ-Cl(A)) ⊆ σ1σ2-Cl(f(A)) for every subset A of X; (5) (τ1, τ2)θ-Cl(f −1(B)) ⊆ f−1(σ1σ2-Cl(B)) for every subset B of Y . Proof. (1) ⇒ (2): Let V be any σ1σ2-open set of Y and x ∈ f−1(V ). Then, f(x) ∈ V . Since f is strongly θ(τ1, τ2)-continuous, by Theorem 1 we have x ∈ (τ1, τ2)θ-Int(f −1(V )). Thus, f−1(V ) ⊆ (τ1, τ2)θ-Int(f −1(V )) and hence f−1(V ) = (τ1, τ2)θ-Int(f −1(V )). This shows that f−1(V ) is (τ1, τ2)θ-open in X. (2) ⇒ (3): The proof is obvious. (3) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). By (3), f−1(Y − V ) is (τ1, τ2)θ-closed and so f−1(V ) is (τ1, τ2)θ-open. Then, there exists a τ1τ2- open set U of X such that x ∈ U ⊆ τ1τ2-Cl(U) ⊆ f−1(V ). Thus, f(τ1τ2-Cl(U)) ⊆ V . This shows that f is strongly θ(τ1, τ2)-continuous. (1) ⇒ (4): Let A be any subset of X. Let x ∈ (τ1, τ2)θ-Cl(A) and V be any σ1σ2- open set of Y containing f(x). Since f is strongly θ(τ1, τ2)-continuous, there exists a τ1τ2-open set U of X such that f(τ1τ2-Cl(U)) ⊆ V . Since x ∈ (τ1, τ2)θ-Cl(A), we have τ1τ2-Cl(U) ∩ A ̸= ∅. It follows that ∅ ̸= f(τ1τ2-Cl(U)) ∩ f(A) ⊆ V ∩ f(A). Thus, f(x) ∈ σ1σ2-Cl(f(A)). (4) ⇒ (5): Let B be any subset of Y . By (4), we have f((τ1, τ2)θ-Cl(f −1(B))) ⊆ σ1σ2-Cl(f(f −1(B))) ⊆ σ1σ2-Cl(B) and hence (τ1, τ2)θ-Cl(f −1(B)) ⊆ f−1(σ1σ2-Cl(B)). (5) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since V ∩(Y −V ) = ∅, f(x) ̸∈ σ1σ2-Cl(Y −V ) and so x ̸∈ f−1(σ1σ2-Cl(Y −V )). By (5), we have x ̸∈ (τ1, τ2)θ-Cl(f −1(Y −V )) = X−(τ1, τ2)θ-Int(f −1(V )). Thus, x ∈ (τ1, τ2)θ-Int(f −1(V )). Then, there exists a τ1τ2-open set U of X such that τ1τ2-Cl(U) ⊆ f−1(V ); hence f(τ1τ2-Cl(U)) ⊆ V. This shows that f is strongly θ(τ1, τ2)-continuous. Definition 2. [18] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called (τ1, τ2)-continuous at a point x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a τ1τ2-open set U of X containing x such that f(U) ⊆ V . A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called (τ1, τ2)-continuous if f has this property at each point of X. Lemma 3. [18] For a function (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equiv- alent: (1) f is (τ1, τ2)-continuous; (2) f−1(V ) is τ1τ2-open in X for every σ1σ2-open set V of Y ; (3) f(τ1τ2-Cl(A)) ⊆ σ1σ2-Cl(f(A)) for every subset A of X; P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 5 of 11 (4) τ1τ2-Cl(f −1(B)) ⊆ f−1(σ1σ2-Cl(B)) for every subset B of Y ; (5) f−1(σ1σ2-Int(B)) ⊆ τ1τ2-Int(f −1(B)) for every subset B of Y ; (6) f−1(K) is τ1τ2-closed in X for every σ1σ2-closed set K of Y . Definition 3. [20] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be weakly (τ1, τ2)- continuous at a point x ∈ X if for each τ1τ2-open set V of Y containing f(x), there exists a τ1τ2-open set U of X containing x such that f(U) ⊆ σ1σ2-Cl(V ). A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be weakly (τ1, τ2)-continuous if f has this property at each point of X. Lemma 4. [20] For a function (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equiv- alent: (1) f is weakly (τ1, τ2)-continuous; (2) f(τ1τ2-Cl(A)) ⊆ (σ1, σ2)θ-Cl(f(A)) for every subset A of X; (3) τ1τ2-Cl(f −1(B)) ⊆ f−1((σ1, σ2)θ-Cl(B)) for every subset B of Y . Definition 4. [31] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called faintly (τ1, τ2)-continuous at a point x ∈ X if for each (σ1, σ2)θ-open set V of Y containing f(x), there exists a τ1τ2- open set U of X containing x such that f(U) ⊆ V . A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called faintly (τ1, τ2)-continuous if f has this property at every point of X. Lemma 5. [31] For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is faintly (τ1, τ2)-continuous; (2) f−1(V ) is τ1τ2-open in X for each (σ1, σ2)θ-open set V of Y ; (3) f−1(K) is τ1τ2-closed in X for each (σ1, σ2)θ-closed set K of Y . Recall that a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)-regular [32] if for each τ1τ2-closed set F and each x ̸∈ F , there exist disjoint τ1τ2-open sets U and V such that x ∈ U and F ⊆ V . Lemma 6. [33] A bitopological space (X, τ1, τ2) is (τ1, τ2)-regular if and only if for each x ∈ X and each τ1τ2-open set U containing x, there exists a τ1τ2-open set V such that x ∈ V ⊆ τ1τ2-Cl(V ) ⊆ U . Lemma 7. [33] Let (X, τ1, τ2) be a (τ1, τ2)-regular space. Then, the following properties hold: (1) τ1τ2-Cl(A) = (τ1, τ2)θ-Cl(A) for every subset A of X. (2) Every τ1τ2-open set is (τ1, τ2)θ-open. P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 6 of 11 Theorem 3. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), where (Y, σ1, σ2) is (σ1, σ2)- regular, the following properties are equivalent: (1) f is (τ1, τ2)-continuous; (2) f is weakly (τ1, τ2)-continuous; (3) f is faintly (τ1, τ2)-continuous; (4) f is strongly θ(τ1, τ2)-continuous. Proof. (1) ⇒ (2): The proof is obvious. (2) ⇒ (3): Let K be any θ(σ1, σ2)-closed set of Y . By Lemma 4, we have τ1τ2-Cl(f −1(K)) ⊆ f−1((σ1, σ2)θ-Cl(K)) = f−1(K) and hence f−1(K) is τ1τ2-closed in X. Thus by Lemma 5, f is faintly (τ1, τ2)-continuous. (3) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since (Y, σ1, σ2) is (σ1, σ2)-regular, by Lemma 7 we have V is a θ(τ1, τ2)-open set of Y . Since f is faintly (τ1, τ2)-continuous, by Lemma 5 we have f−1(V ) is τ1τ2-open in X. Then by Lemma 3, f is (τ1, τ2)-continuous. (1) ⇒ (4): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since (Y, σ1, σ2) is (σ1, σ2)-regular, by Lemma 6 there exists a σ1σ2-open set W of Y such that f(x) ∈ W ⊆ σ1σ2-Cl(W ) ⊆ V . Since f is (τ1, τ2)-continuous, there exists a τ1τ2-open set U of X containing x such that f(U) ⊆ V . Now, we shall show that f(τ1τ2-Cl(U)) ⊆ σ1σ2-Cl(W ). Suppose that y ̸∈ σ1σ2-Cl(W ). Then, there exists a σ1σ2-open set G of Y containing y such that G ∩ W = ∅. Since f is (τ1, τ2)-continuous, by Lemma 3 we have f−1(G) is τ1τ2-open in X and f−1(G) ∩ U = ∅, which implies that f−1(G) ∩ τ1τ2-Cl(U) = ∅. If f−1(G) ∩ τ1τ2-Cl(U) ̸= ∅, then τ1τ2-Int(f −1(G)) ∩ τ1τ2-Cl(U) ̸= ∅. Let z ∈ τ1τ2-Int(f −1(G)) ∩ τ1τ2-Cl(U). Then, z ∈ τ1τ2-Int(f −1(G)) and z ∈ τ1τ2-Cl(U). There exists a τ1τ2-open set U0 of X containing x such that U0 ⊆ f−1(G). Since z ∈ τ1τ2-Cl(U), we have U0∩U ̸= ∅ and so f−1(G)∩U ̸= ∅. This is a contradiction. Therefore, f−1(G) ∩ τ1τ2-Cl(U) = ∅ which implies that G ∩ f(τ1τ2-Cl(U)) = ∅. Thus, y ̸∈ f(τ1τ2-Cl(U)) and hence f(τ1τ2-Cl(U)) ⊆ σ1σ2-Cl(W ) ⊆ V . This shows that f is strongly θ(τ1, τ2)-continuous. (4) ⇒ (1): The proof is obvious. Theorem 4. Let (X, τ1, τ2) be (τ1, τ2)-regular. Then, a function f : (X, τ1, τ2) → (Y, σ1, σ2) is strongly θ(τ1, τ2)-continuous if and only if f is (τ1, τ2)-continuous. P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 7 of 11 Proof. We prove only the sufficiency. Suppose that f is (τ1, τ2)-continuous. Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Then, there exists a τ1τ2-open set G of X containing x such that f(G) ⊆ V . Since (X, τ1, τ2) is (τ1, τ2)-regular, by Lemma 6 there exists a τ1τ2-open set U of X such that x ∈ U ⊆ τ1τ2-Cl(U) ⊆ G. Thus, f(τ1τ2-Cl(U)) ⊆ V . This shows that f is strongly θ(τ1, τ2)-continuous. 4. Some results on strong θ(τ1, τ2)-continuity Recall that a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)-T0 [34] if for any pair of distinct points in X, there exists a τ1τ2-open set of X containing one of the points but not the other. Definition 5. [35] A bitopological space (X, τ1, τ2) is said to be (τ1, τ2)-T2 if for any pair of distinct points x, y in X, there exist disjoint τ1τ2-open sets U and V of X containing x and y, respectively. Theorem 5. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a strongly θ(τ1, τ2)-continuous injection and (Y, σ1, σ2) is (σ1, σ2)-T0, then (X, τ1, τ2) is (τ1, τ2)-T2. Proof. Suppose that (Y, σ1, σ2) is (σ1, σ2)-T0. Let x and y be any distinct points of X. Since f is injective, f(x) ̸= f(y). Since (Y, σ1, σ2) is (σ1, σ2)-T0, there exists a σ1σ2- open set V of Y which either contains f(x) and not f(y) or contains f(y) and not f(x). If the first case holds, then there exists a τ1τ2-open set U of X containing x such that f(τ1τ2-Cl(U)) ⊆ V . Thus, f(y) ̸∈ f(τ1τ2-Cl(U)) and hence y ∈ X − τ1τ2-Cl(U) = τ1τ2-Int(X − U). Then, there exists a τ1τ2-open set W of X such that y ∈ W ⊆ X − U . Therefore, U ∩W = ∅. This shows that (X, τ1, τ2) is (τ1, τ2)-T2. Definition 6. [36] A bitopological space (X, τ1, τ2) is said to be τ1τ2-Urysohn if for each pair of distinct points x and y in X, there exist τ1τ2-open sets U and V such that x ∈ U , y ∈ V and τ1τ2-Cl(U) ∩ τ1τ2-Cl(V ) = ∅. Theorem 6. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a strongly θ(τ1, τ2)-continuous injection and (Y, σ1, σ2) is (σ1, σ2)-T2, then (X, τ1, τ2) is τ1τ2-Urysohn. Proof. Suppose that (Y, σ1, σ2) is (σ1, σ2)-T2. Let x and y be any distinct points of X. Since f is injective, f(x) ̸= f(y). Since (Y, σ1, σ2) is (σ1, σ2)-T2, there exist σ1σ2-open sets V and W of Y containing f(x) and f(y), respectively, such that V ∩W = ∅. Since f is strongly θ(τ1, τ2)-continuous, there exist τ1τ2-open sets U and G of X containing x and y, respectively, such that f(τ1τ2-Cl(U)) ⊆ V and f(τ1τ2-Cl(G)) ⊆ W . It follows that τ1τ2-Cl(U) ∩ τ1τ2-Cl(G) = ∅. Thus, (X, τ1, τ2) is τ1τ2-Urysohn. P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 8 of 11 Definition 7. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to have a strongly θ(τ1, τ2)- closed graph with respect to X if for each (x, y) ∈ (X × Y )−G(f), there exist a τ1τ2-open set U of X containing x and a σ1σ2-open set V of Y containing y such that [τ1τ2-Cl(U)× V ] ∩G(f) = ∅. Lemma 8. A function f : (X, τ1, τ2) → (Y, σ1, σ2) has a strongly θ(τ1, τ2)-closed graph with respect to X if and only if for each (x, y) ∈ (X × Y )−G(f), there exist a τ1τ2-open set U of X containing x and a σ1σ2-open set V of Y containing y such that f(τ1τ2-Cl(U)) ∩ V = ∅. Theorem 7. If f : (X, τ1, τ2) → (Y, σ1, σ2) is strongly θ(τ1, τ2)-continuous and (Y, σ1, σ2) is (σ1, σ2)-T2, then G(f) is strongly θ(τ1, τ2)-closed graph with respect to X. Proof. Let (x, y) ∈ (X × Y ) − G(f). Then, y ̸= f(x). Since (Y, σ1, σ2) is (σ1, σ2)-T2, there exist σ1σ2-open sets V and W of Y containing f(x) and f(y), respectively, such that V ∩ W = ∅. Since f is strongly θ(τ1, τ2)-continuous, there exist a τ1τ2-open set U of X containing x such that f(τ1τ2-Cl(U)) ⊆ W . This implies that f(τ1τ2-Cl(U)) ∩ V = ∅ and by Lemma 8, G(f) is strongly θ(τ1, τ2)-closed graph with respect to X. Definition 8. [37] Let A be a subset of a bitopological space (X, τ1, τ2). The (τ1, τ2)θ- frontier of A, (τ1, τ2)θ-fr(A), is defined by (τ1, τ2)θ-fr(A) = (τ1, τ2)θ-Cl(A) ∩ (τ1, τ2)θ-Cl(X −A). Theorem 8. The set of all points x ∈ X at which a function f : (X, τ1, τ2) → (Y, σ1, σ2) is not strongly θ(τ1, τ2)-continuous is identical with the union of the (τ1, τ2)θ-frontier of the inverse images of σ1σ2-open sets containing f(x). Proof. Suppose that f is not strongly θ(τ1, τ2)-continuous. Then, there exists a σ1σ2- open set V of Y containing f(x) such that f(τ1τ2-Cl(U)) is not contained in V for every τ1τ2-open set U of X containing x. Then, τ1τ2-Cl(U) ∩ (X − f−1(V )) ̸= ∅ for every τ1τ2- open set U of X containing x. Thus, x ∈ (τ1, τ2)θ-Cl(X − f−1(V )). On the other hand, we have x ∈ f−1(V ) ⊆ (τ1, τ2)θ-Cl(f −1(V )) and hence x ∈ (τ1, τ2)θ-fr(f −1(V )). Conversely, suppose that f is strongly θ(τ1, τ2)-continuous at x ∈ X. Let V be any σ1σ2-open set of Y containing f(x). By Theorem 1, x ∈ (τ1, τ2)θ-Int(f −1(V )). Thus, x ̸∈ (τ1, τ2)θ-fr(f −1(V )) for every σ1σ2-open set V of Y containing f(x). This completes the proof. Recall that a bitopological space (X, τ1, τ2) is said to be quasi (τ1, τ2)-H -closed [38] if every τ1τ2-open cover {Uγ | γ ∈ Γ}, there exists a finite subset Γ0 of Γ such that X = ∪{τ1τ2-Cl(Uγ) | γ ∈ Γ0}. A subset K of a bitopological space (X, τ1, τ2) is said to be quasi (τ1, τ2)-H -closed relative to (X, τ1, τ2) if for any cover {Vγ | γ ∈ Γ} by τ1τ2-open sets of X, there exists a finite subset Γ0 of Γ such that K ⊆ ∪{τ1τ2-Cl(Vγ) | γ ∈ Γ0}. A subset K of a bitopological space (X, τ1, τ2) is said to be τ1τ2-compact relative to (X, τ1, τ2) if for P. Pue-on, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6017 9 of 11 any cover {Vγ | γ ∈ Γ} by τ1τ2-open sets of X, there exists a finite subset Γ0 of Γ such that K ⊆ ∪{Vγ | γ ∈ Γ0}. If X is τ1τ2-compact relative to (X, τ1, τ2), then (X, τ1, τ2) is said to be τ1τ2-compact [27]. Theorem 9. If f : (X, τ1, τ2) → (Y, σ1, σ2) is strongly θ(τ1, τ2)-continuous and K is quasi (τ1, τ2)-H -closed relative to (X, τ1, τ2), then f(K) is σ1σ2-compact relative to (Y, σ1, σ2). Proof. Let K be quasi (τ1, τ2)-H -closed relative to (X, τ1, τ2). Let {Vγ | γ ∈ Γ} be any cover of f(K) by σ1σ2-open sets of Y . For each x ∈ K, there exists γ(x) ∈ Γ such that f(x) ∈ Vγ(x). Since f is strongly θ(τ1, τ2)-continuous, there exists a τ1τ2-open set U(x) of X containing x such that f(τ1τ2-Cl(U(x))) ⊆ σ1σ2-Cl(Vγ(x)). The family {U(x) | x ∈ K} is a cover of K by τ1τ2-open sets of X. Since K is quasi (τ1, τ2)-H -closed relative to (X, τ1, τ2), there exists a finite number of points, say, x1, x2, x3, ..., xn in K such that K ⊆ ∪{τ1τ2-Cl(U(xk)) | xk ∈ K; 1 ≤ k ≤ n}. Thus, f(K) ⊆ ∪{f(τ1τ2-Cl(U(xk))) | xk ∈ K; 1 ≤ k ≤ n} ⊆ ∪{Vγ(xk) | xk ∈ K; 1 ≤ k ≤ n}. This shows that f(K) is σ1σ2-compact relative to (Y, σ1, σ2). Acknowledgements This research project was financially supported by Mahasarakham University. 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