EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 2, 2022, 443-453 ISSN 1307-5543 – ejpam.com Published by New York Business Global A note on strongly δθ-I-continuous functions José Sanabria1,∗, Rafael Lozada-Yavina2,3, José Tormet4 1 Departamento de Matemáticas, Facultad de Educación y Ciencias, Universidad de Sucre, Sincelejo, Colombia 2 Facultad de Ciencias Básicas, Universidad Católica del Maule, Talca, Chile 3Facultad de Ciencias e Ingeniería, Universidad Tecnológica del Perú, Lima, Perú 4 Departamento de Ciencias, Unidad de Estudios Básicos, Universidad de Oriente, Puerto La Cruz, Venezuela Abstract. In this article, we investigate some properties of strongly δθ-I-continuous functions and other types of related functions. Specially, we characterize strongly δθ-I-continuous, we investigate their relationship with other types of functions, and we study the behavior of certain topological notions under the action of these functions. 2020 Mathematics Subject Classifications: 54A05, 54C10. Key Words and Phrases: Ideals, δ-local function, δθ-I-open set. 1. Introduction The concept of an ideal on a topological space (nowadays called a topological ideal) has played a fundamental role in several of the advances in general topology. In the last century, a large number of works have arisen that have enriched the literature related to the concept of topological ideal. Very recently, topological ideals have again received special attention for their versatility in tackling topology problems and in studying rough set models, as we can see in the references [19], [7], [3], [12], [16], [5], [9], [10]. In 2014, Hatir and Al-Omari [8] introduced the concept of δ-local function and studied some of its most relevant properties. The study carried out in [8] served as motivation to define the class of the δθ-I -open sets in [11], which was later used in [14] to introduce new variants of continuous functions, called δθ-I-continuous, weakly δ-J -continuous and strongly δθ-I-continuous functions. In this article, we study and characterize the strongly δθ-I-continuous functions, we investigate their relationship with other types of functions, and also, we explore the behavior of some topological notions under these classes of func- tions. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i2.4317 Email addresses: jesanabri@gmail.com, jose.sanabria@unisucre.edu.co (J. Sanabria), trobuyo@gmail.com (R. Lozada-Yavina), ajosetormet@gmail.com (J. Tormet) https://www.ejpam.com 443 © 2022 EJPAM All rights reserved. J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 444 2. Preliminaries Throughout this paper, (X, τ) always means a topological space on which no separation axioms are assumed unless explicitly stated. If A is a subset of X, we denote the closure of A and the interior of A by Cl(A) and Int(A), respectively. The definitions and results that we present below can be consulted in [8], [14] and [11]. A point x ∈ X is called a δ-cluster (resp. θ-cluster) point of A if Int(Cl(U))∩A 6= ∅ (resp. Cl(U)∩A 6= ∅) for each open set U containing x. The set of all δ-cluster (resp. θ-cluster) points of A is called the δ-closure (resp. θ-closure) of A and is denoted by δCl(A) (resp. Clθ(A)). A subset A of X is said to be δ-closed (resp. θ-closed) if A = δCl(A) (resp. A = Clθ(A)). The complement of a δ-closed (resp. θ-closed) set is said to be a δ-open (resp. θ-open) set. Similarly, the θ-interior of a subset A of X, denoted by Intθ(A), consists of all points x in X such that for some open set U containing x, Cl(U) ⊂ A. It is well known that, a subset A of X is θ-open if and only if A = Intθ(A). It follows that the collection of all δ-open (resp. θ-open) sets in a topological space (X, τ) forms a topology on X which is denoted by τδ (resp. τθ). From the definitions it follows that τθ ⊂ τδ ⊂ τ . The topology τδ is called the semi-regularization of τ . Observe that δCl is the closure operator with respect to τδ, but Clθ is not the closure operator with respect to τθ. An ideal I on a topological space (X, τ) is a nonempty collection of subsets of X which satisfies the following two properties: (i) A ∈ I and B ⊂ A implies B ∈ I; (ii) A ∈ I and B ∈ I implies A ∪ B ∈ I. An ideal topological space (or simply a space) is a topological space (X, τ) together with an ideal I on X and is denoted by (X, τ, I). If (X, τ, I) is a space, then for any A ⊂ X, the local function (rep. δ-local function) of A with respect to I and τ , denoted by A?(I, τ) (resp. Aδ?(I, τ)), is defined as A?(I, τ) = {x ∈ X : A ∩ U /∈ I for every open set U containing x} (resp. Aδ?(I, τ) = {x ∈ X : A ∩ U /∈ I for every δ-open set U containing x}). We simply write A? (resp. Aδ?) in case there is no chance for confusion. In general, X? is a proper subset of X. The hypothesis X = X? is equivalent to the hypothesis τ ∩ I = {∅}. We call the ideals which satisfy this condition τ -boundary ideals. A Kuratowski closure operator Cl?(.) (resp. δCl?(.)) for a topology τ? (res. τ δ?) finer than τ (resp. τδ), is defined by Cl?(A) = A∪A? (resp. δCl?(A) = A ∪ Aδ?). It is well known that τδ ⊂ τ ⊂ τ? and τδ ⊂ τ?δ ⊂ τ?. A point x ∈ X is called a δθ-I-cluster point of A if δCl?(U) ∩ A 6= ∅ for every open subset U of X containing x. The set of all δθ-I-cluster points of A is called the δθ-I-closure of A and is denoted by δCl?θ (A). A subset A of X is said to be δθ-I-closed if δCl?θ (A) = A. The complement of a δθ-I-closed set is said to be δθ-I-open set. A point x ∈ X is called a δθ-I-interior point of a subset A of X if there exists an open set U such that x ∈ U ⊂ δCl?(U) ⊂ A. The set of all δθ-I-interior points of A is called the δθ-I-interior of A and is denoted by δInt?θ (A). In [11, Proposition 4.1] it was shown that a subset A of X is δθ-I-open if and only if A = δInt?θ (A). Since the main objective of this article is to study strongly δθ-I-continuous functions, now we will recall some variants of continuity related to the type of functions that we will study. In what follows, we consider that (X, τ, I) and (Y, σ,J ) are spaces. Definition 1. A function f : (X, τ) → (Y, σ) is said to be strongly θ-continuous [18], J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 445 (resp. weakly continuous [13], θ-continuous [15]) if for each x ∈ X and each open set V in Y containing f(x), there exists an open set U in X containing x such that f(Cl(U)) ⊂ V (resp. f(U) ⊂ Cl(V ), f(Cl(U)) ⊂ Cl(V )). Definition 2. A function f : (X, τ) → (Y, σ,J ) is said to be weakly J -continuous [1] (resp. weakly δ-J -continuous [14]), if for each x ∈ X and each open set V in Y containing f(x), there exists an open set U in X containing x such that f(U) ⊂ Cl?(V ) (resp. f(U) ⊂ δCl?(V )). Definition 3. A function f : (X, τ, I) → (Y, σ,J ) is said to be δθ-I-continuous [14] (resp. θ-I-continuous [17]), if for each x ∈ X and each open set V in Y containing f(x), there exists an open set U containing x such that f(δCl?(U)) ⊂ δCl?(V ) (resp. f(Cl?(U)) ⊂ Cl?(V )). Definition 4. A function f : (X, τ, I) → (Y, σ) is said to be strongly θ-I-continuous [17] (resp. strongly δθ-I-continuous [14]), if for each x ∈ X and each open set V in Y containing f(x), there exists an open set U in X containing x such that f(Cl?(U)) ⊂ V (resp. f(δCl?(U)) ⊂ V ). Remark 1. The following diagram shows the relationship between the types of functions given in Definitions 1, 2, 3 and 4. In general none of the implications is reversible. weaklyδ-J -continuous weakly continuous δθ-I-continuous weakly J -continuous θ-continuous strongly δθ-I-continuous θ-I-continuous continuous strongly θ-I-continuous strongly θ-continuous 3. Properties related to strongly δθ-I-continuous functions In this section, we characterize strongly δθ-I-continuous. Also, we looking for some topological conditions in order to find the relationship between the strongly δθ-I-continuous functions, δθ-I-continuous functions and weakly δ-J -continuous functions. First, we characterize strongly δθ-I-continuous functions using δθ-I-open sets and δθ-I-closed sets. Theorem 1. Let f : (X, τ, I) → (Y, σ). Then, the following properties are equivalent: (i) f is strongly δθ-I-continuous. (ii) f−1(V ) is a δθ-I-open set in X, for each open set V ⊂ Y . J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 446 (iii) f−1(F ) is a δθ-I-closed set in X, for each closed set F ⊂ Y . (iv) f(δCl?θ (A)) ⊂ Cl(f(A)), for each A ⊂ X. (v) δCl?θ (f −1(B)) ⊂ f−1(Cl(B)), for each B ⊂ Y . Proof. (i) ⇒ (ii) Suppose that V is an open set in Y and x ∈ f−1(V ). Since f is strongly δθ-I-continuous, there exists an open set U in X containing x such that f(δCl?(U)) ⊂ V . Thus, x ∈ U ⊂ δCl?(U) ⊂ f−1(V ). Consequently, f−1(V ) is a δθ-I-open set in X. (ii) ⇒ (iii) Let F be a closed set in Y . Then, V = Y − F is an open set in Y and by (ii), f−1(V ) = f−1(Y − F ) = f−1(Y ) − f−1(F ) = X − f−1(F ) is a δθ-I-open set in X. Therefore, f−1(F ) is a δθ-I-closed set in X. (iii) ⇒ (iv) Let A ⊂ X. Since Cl(f(A)) is a closed set in Y , by hypothesis, it follows that f−1(Cl(f(A))) is a δθ-I-closed set. Then, we have δCl?θ (A) ⊂ δCl?θ (f −1(f(A))) ⊂ δCl?θ (f −1(Cl(f(A)))) = f−1(Cl(f(A))), which implies that f(δCl?θ (A)) ⊂ Cl(f(A)) for each A ⊂ X. (iv) ⇒ (v) Let B ⊂ Y . By hypothesis, we have f(δCl?θ (f −1(B))) ⊂ Cl(f(f−1(B))) ⊂ Cl(B). Consequently, δCl?θ (f−1(B)) ⊂ f−1(Cl(B)). (v) ⇒ (i) Let x ∈ X and V ∈ σ be such that f(x) ∈ V . Then, Y − V is a closed set in Y and by hypothesis, δCl?θ (f−1(Y − V )) ⊂ f−1(Cl(Y − V )) = f−1(Y − V ), which tells us that f−1(Y − V ) is a δθ-I-closed set in X. Since f−1(Y − V ) = X − f−1(V ), we conclude that f−1(V ) is a δθ-I-open set containing x. Thus, there exists U ∈ τ such that x ∈ U ⊂ δCl?(U) ⊂ f−1(V ). Therefore, f(δCl?(U)) ⊂ V and so, f is a strongly δθ-I-continuous function. According to [14], we say that a function f : (X, τ, I) → (Y, σ,J ) is δθ-I-irresolute, if f−1(V ) is a δθ-I-open set in X, for each δθ-J -open set V ⊂ Y . This type of functions is used in the following theorem, where we analyze when the composition of functions is strongly δθ-I-continuous. Theorem 2. Let f : (X, τ, I) → (Y, σ,J ) and g : (Y, σ,J ) → (Z,ϕ) be two functions. Then, the following properties hold: (i) If f is strongly δθ-I-continuous and g is continuous, then g ◦ f : (X, τ, I) → (Z,ϕ) is strongly δθ-I-continuous. (ii) If f is δθ-I-irresolute and g is strongly δθ-J -continuous, then g ◦ f : (X, τ, I) → (Z,ϕ) is strongly δθ-I-continuous. Proof. (i) Let V ∈ ϕ. Since g is continuous, g−1(V ) is an open set in Y , and as f is strongly δθ-I-continuous, by Theorem 1, it follows that (g ◦ f)−1(V ) = f−1(g−1(V )) is a δθ-I-open set in X. Again, by Theorem 1, we get that g ◦ f is a strongly δθ-I-continuous function. The proof of (ii) is similar to that of (i). J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 447 Theorem 3. Let {(Yλ, σλ) : λ ∈ Λ} be a collection of topological spaces and let Y =∏ {Yλ : λ ∈ Λ} with product topology σ = ∏ {σλ : λ ∈ Λ} induced by the projections pλ : Y → Yλ, λ ∈ Λ. A function f : (X, τ, I) → (Y, σ) is strongly δθ-I-continuous if and only if every composition pλ ◦ f is strongly δθ-I-continuous. Proof. Let f : (X, τ, I) → (Y, σ) be a strongly δθ-I-continuous function and let λ ∈ Λ. Since the projection pλ : Y → Yλ is continuous, by Theorem 2, we get that pλ ◦ f : (X, τ, I) → (Yλ, σλ) is strongly δθ-I-continuous for every λ ∈ Λ. Conversely, suppose that every composition pλ ◦ f : (X, τ, I) → (Yλ, σλ) is strongly δθ-I-continuous. Since the collection τδθ−I of all δθ-I-open sets in X is a topology, it suffices to show that the inverse image under f of each subbasic open set of Y = ∏ {Yλ : λ ∈ Λ} is a δθ-I-open set in X. Let p−1 λ (Vλ) be a subbasic open set in Y . Then, Vλ ∈ σλ and so, f−1 ( p−1 λ (Vλ) ) = (pλ ◦ f)−1 (Vλ) is a δθ-I-open set in X. Therefore, f is strongly δθ-I- continuous. Corollary 1. Let (X, τ, I) be a space, {(Yλ, σλ) : λ ∈ Λ} be a collection of topological spaces and fλ : (X, τ, I) → (Yλ, σλ) be a function for every λ ∈ Λ. Let σ = ∏ {σλ : λ ∈ Λ} be the product topology on Y = ∏ {Yλ : λ ∈ Λ} and f : (X, τ, I) → (Y, σ) be the function defined by f(x) = (fλ(x))λ∈Λ for each x ∈ X. Then, f is strongly δθ-I-continuous if and only if fλ is strongly δθ-I-continuous for every λ ∈ Λ. Next, we present some topological notions to establish some properties related to strongly δθ-I-continuous functions. Definition 5. A space (X, τ, I) is said to be δ?-regular [14] (resp. ?-regular [2]), if for each pair consisting of a closed set F and a point x /∈ F , there exist V ∈ τ and U ∈ τ δ? (resp. U ∈ τ?) such that x ∈ V , F ⊂ U and U ∩ V = ∅. The concepts of δ?-regular space and ?-regular space are related as follows. Lemma 1. Every δ?-regular space is a ?-regular. Proof. Suppose that (X, τ, I) is a δ?-regular space. Let F be a closed set and x /∈ F . Then, there exist V ∈ τ and U ∈ τ δ? such that x ∈ V , F ⊂ U and U ∩ V = ∅. Since τ δ? ⊂ τ?, we have U ∈ τ? and hence, (X, τ, I) is a ?-regular space. The following example shows that, in general, the converse of Lemma 1 is not true. Example 1. A ?-regular space need not be δ?-regular space. Consider the space (X, τ, I), where X = {a, b, c, d}, τ = {∅, X, {a, b, c}, {a, b, d}, {a, b}} and I = {∅, {a}, {b}, {a, b}}. Observe that the collection of all closed sets is {∅, X, {c}, {d}, {c, d}}. Also, τ? = P(X), τδ = {∅, X} and τ δ? = {∅, X, {c, d}, {b, c, d}, {a, c, d}}. Then, we have: (i) For F1 = {d} and a /∈ F1, there exist V1 = {a, b} ∈ τ and U1 = {c, d} ∈ τ? such that a ∈ V1, F1 ⊂ U1 and V1 ∩ U1 = ∅. (ii) For F1 = {d} and b /∈ F1, there exist V1 = {a, b} ∈ τ and U1 = {c, d} ∈ τ? such that b ∈ V1, F1 ⊂ U1 and V1 ∩ U1 = ∅. J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 448 (iii) For F1 = {d} and c /∈ F1, there exist V2 = {a, b, c} ∈ τ and U2 = {d} ∈ τ? such that c ∈ V2, F1 ⊂ U2 and V2 ∩ U2 = ∅. (iv) For F2 = {c} and a /∈ F2, there exist V1 = {a, b} ∈ τ and U1 = {c, d} ∈ τ? such that a ∈ V1, F2 ⊂ U1 and V1 ∩ U1 = ∅. (v) For F2 = {c} and b /∈ F2, there exist V1 = {a, b} ∈ τ and U1 = {c, d} ∈ τ? such that b ∈ V1, F2 ⊂ U1 and V1 ∩ U1 = ∅. (vi) For F2 = {c} and d /∈ F2, there exist V3 = {a, b, d} ∈ τ and U3 = {c} ∈ τ? such that d ∈ V3, F2 ⊂ U3 and V3 ∩ U3 = ∅. (vii) For F3 = {c, d} and a /∈ F3, there exist V1 = {a, b} ∈ τ and U1 = {c, d} ∈ τ? such that a ∈ V1, F3 ⊂ U1 and V1 ∩ U1 = ∅. (viii) For F3 = {c, d} and b /∈ F3, there exist V1 = {a, b} ∈ τ and U1 = {c, d} ∈ τ? such that b ∈ V1, F3 ⊂ U1 and V1 ∩ U1 = ∅. By (i)-(viii), we deduce that (X, τ, I) is a ?-regular space. Now, we will show that (X, τ, I) is not a δ?-regular space. Indeed, let F = {d}. Then, F is a closed set and c /∈ F . Observe that the only open sets containing c are V2 = {a, b, c} and V4 = X, and the only τ δ?- open sets containing F are W1 = {c, d}, W2 = {b, c, d} and W3 = {a, c, d}. In addition, V2 ∩ W1 = {c} 6= ∅, V2 ∩ W2 = {b, c} 6= ∅, V2 ∩ W3 = {a, c} 6= ∅, V4 ∩ W1 = W1 6= ∅, V4 ∩W2 = W2 6= ∅, V4 ∩W3 = W3 6= ∅. Therefore, (X, τ, I) is not a δ?-regular space. Lemma 2. [14] A space (X, τ, I) is δ?-regular if and only if for each x ∈ X and each open set V containing x, there exists an open set U such that x ∈ U ⊂ δCl?(U) ⊂ V . The following result shows that the strongly δθ-I-continuity and the continuity are equivalent under the assumption that the domain is a δ?-regular space. Theorem 4. Let (X, τ, I) be a δ?-regular space. A function f : (X, τ, I) → (Y, σ) is strongly δθ-I-continuous if and only if it is continuous. Proof. By Remark 1, if f is strongly δθ-I-continuous then it is continuous. Conversely, suppose that f is continuous. Let x ∈ X and V be any open set in Y containing f(x). Then, there exists an open set U containing x such that f(U) ⊂ V . Since (X, τ, I) is a δ?- regular space, by Lemma 2, there exists an open set W such that x ∈ W ⊂ δCl?(W ) ⊂ U . Thus, f(x) ∈ f(δCl?(W )) ⊂ f(U) ⊂ V , which implies that f(δCl?(W )) ⊂ V . Therefore, f is strongly δθ-I-continuous. Theorem 5. Let f : (X, τ, I) → (Y, σ) be a function and (X, τ, I) be a δ?-regular space. If f is strongly θ-continuous, then it is strongly δθ-I-continuous. Proof. The proof is similar to the part of the proof of Theorem 4, where it was proved that if (X, τ, I) is δ?-regular and f is continuous, then f is strongly δθ-I-continuous. The following example shows that a strongly δθ-I-continuous function is not necessarily strongly θ-continuous, even though the domain is a δ?-regular space. J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 449 Example 2. Consider X = {a, b, c, d} with the topology τ = {X, ∅, {a, b}, {a, b, c}, {a, b, d}} and the ideal I = {∅, {a}, {b}, {a, b}}. Let us define a function f : (X, τ, I) → (X, τ) as follows: f(a) = a, f(b) = b, f(c) = c, f(d) = b. Note that: (i) The only open sets in X containing f(a) = a are V1 = {a, b}, V2 = {a, b, c}, V3 = {a, b, d} and V4 = X. In addition, V1 satisfies that f(δCl?(V1)) = f(δCl?({a, b})) = f({a, b}) = {a, b} = V1, f(δCl?(V1)) = V1 ⊂ V2, f(δCl?(V1)) = V1 ⊂ V3 and f(δCl?(V1)) = V1 ⊂ V4. (ii) Using the same argument from part (1), we get the result for f(b) = b. (iii) The only open sets in X containing f(c) = c are V2 = {a, b, c} and V4 = X. In addi- tion, V2 satisfies that f(δCl?(V2)) = f(δCl?({a, b, c})) = f({a, b, c}) = {a, b, c} = V2 and f(δCl?(V2)) = V2 ⊂ V4. (iv) The only open sets in X containing f(d) = b are V1 = {a, b}, V2 = {a, b, c}, V3 = {a, b, d} and V4 = X. In addition, V3 = {a, b, d} is an open set in X containing d such that f(δCl?(V3)) = f(δCl?({a, b, d})) = f({a, b, d}) = {a, b} = V1 which is contained in V1, V2, V3 and V4. By (i)-(iv), we conclude that f is strongly δθ-I-continuous. On the other hand, since Cl(V1) = Cl(V2) = Cl(V3) = Cl(V4) = X and f(X) = V2 6⊂ V1, we conclude that f is not strongly θ-continuous. Observe that (X, τ, I) is a δ?-regular space that is not regular. Recall that a function f : (X, τ) → (Y, σ) is super-continuous [6] if for each x ∈ X and each open set V in Y containing f(x), there exists an open set U in X containing x such that f(Int(Cl(U))) ⊂ V . This type of function is characterized by the property that the inverse image of each open set in Y is a δ-open set in X. Clearly, every super-continuous function is continuous, but the converse, in general, is not true. Theorem 6. Let (Y, σ) be a regular space and f : (X, τ, I) → (Y, σ) be a function. If f is super-continuous, then it is strongly δθ-I-continuous. Proof. Let x ∈ X and V be an open set in Y containing f(x). Since Y is regular, there exists an open set U such that f(x) ∈ U ⊂ Cl(U) ⊂ V . On the other hand, as f is super-continuous, there exists a δ-open set W containing x such that f(W ) ⊂ U . We will show that f(δCl?(W )) ⊂ Cl(U). Indeed, suppose that y /∈ Cl(U). Then, we choose an open set O such that y ∈ O and O ∩ U = ∅. By the super-continuity of f , we have f−1(O) is a δ-open set such that f−1(O) ∩ f−1(U) = ∅, which implies that f−1(O) ∩W = ∅. We affirm that f−1(O) ∩ δCl?(W ) = ∅. Otherwise, there exists a point z ∈ f−1(O) ∩ δCl?(W ) and so, z ∈ f−1(O) and z ∈ δCl?(W ). It follows that z ∈ W or z ∈ W δ?. If z ∈ W then z ∈ f−1(O) ∩ W , which is a contradiction. If z ∈ W δ? then G ∩ W /∈ I for each δ-open set G containing z; in particular, f−1(O) ∩ W /∈ I, which implies that f−1(O)∩W 6= ∅, so again we get a contradiction. Therefore, we deduce that f−1(O) ∩ δCl?(W ) = ∅. Thus, O ∩ f(δCl?(W )) = ∅ and hence, y /∈ f(δCl?(W )). This shows that f(δCl?(W )) ⊂ Cl(U) ⊂ V and so, f is strongly δθ-I-continuous. J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 450 In the following example, we show that there exists a strongly δθ-I-continuous function that is not super-continuous. Example 3. The function f given in Example 2 is strongly continuous, but it is not super-continuous, because in this case τδ = {∅, X} and {a, b} is an open set such that f−1({a, b}) = {a, b} /∈ τδ. Theorem 7. Let f : (X, τ, I) → (Y, σ) be any function and g : (X, τ, I) → (X × Y, τ × σ) be the graph function of f defined by g(x) = (x, f(x)) for every x ∈ X, where τ × σ is the product topology on X × Y . Then, g is strongly δθ-I-continuous if and only if f is strongly δθ-I-continuous and (X, τ, I) is δ?-regular. Proof. Clearly, g(x) = (1X(x), f(x)) for every x ∈ X, where 1X : (X, τ, I) → (X, τ) is the identity function on X. Then, by Corollary 1, g is strongly δθ-I-continuous if and only if 1X and f are strongly δθ-I-continuous. In addition, 1X is strongly δθ-I-continuous if and only if for each x ∈ X and each open set V in X containing x, there exists an open set U in X containing x such that δCl?(U) ⊂ V , but by Lemma 2, the latter is equivalent to that (X, τ, I) is a δ?-regular space. Definition 6. [14] A space (X, τ, I) is said to be δ?-Urysohn, if for each pair of distinct points x and y in X, there exist two open subsets U and V of X containing x and y respectively, such that δCl?(U) ∩ δCl?(V ) = ∅. Theorem 8. If f : (X, τ, I) → (Y, σ) is a strongly δθ-I-continuous injective function and (Y, σ) is T2, then (X, τ, I) is δ?-Urysohn. Proof. Let x and y be distinct points of X. Then, f(x) 6= f(y) and as (Y, σ) is T2, there exist disjoint open sets V and W in Y containing f(x) and f(y), respectively. Since f is strongly δθ-I-continuous, there exist two open sets G and H in X containing x and y, respectively, such that f(δCl?(G)) ⊂ V and f(δCl?(H)) ⊂ W . It follows that δCl?(G) ∩ δCl?(H) ⊂ f−1(f(δCl?(G))) ∩ f−1(f(δCl?(H))) ⊂ f−1(V ) ∩ f−1(W ) = ∅, and hence, δCl?(G) ∩ δCl?(H) = ∅. This shows that (X, τ, I) is δ?-Urysohn. Theorem 9. If f : (X, τ, I) → (Y, σ) is a strongly δθ-I-continuous injective function and (Y, σ) is T0, then (X, τ?) is T2. Proof. Let x and y be distinct points of X. Then, f(x) 6= f(y) and as (Y, σ) is T0, there exists an open set V in Y containing one the points f(x) and f(y) but not both. Without loss of generality, we assume that f(x) ∈ V and f(y) /∈ V . Since f is a strongly δθ-I-continuous function, there exists an open set U in X containing x such that f(δCl?(U)) ⊂ V . Thus, we obtain that x ∈ U ⊂ δCl?(U) ⊂ f−1(f(δCl?(U))) ⊂ f−1(V ) and y /∈ δCl?(U), which implies that U and X \ δCl?(U) are two disjoint τ?-open sets in X containing x and y, respectively. Therefore, (X, τ?) is a T2-space. Definition 7. A strongly δθ-I-continuous retraction is a strongly δθ-I-continuous func- tion f : (X, τ, I) → (Y, σ), where Y ⊂ X and f |Y = 1Y the identity function on Y . J. Sanabria, R. Lozada-Yavina, J. Tormet / Eur. J. Pure Appl. Math, 15 (2) (2022), 443-453 451 Theorem 10. If f : (X, τ, I) → (Y, σ) is a strongly δθ-I-continuous retraction and (Y, σ) is T2, then Y is a δθ-I-closed set in X. Proof. We will show that X \A is a δθ-I-open set in X. Let x ∈ X \A. Then, f(x) ∈ A and x /∈ A because f is a strongly δθ-I-continuous retraction. Since (Y, σ) is T2, there exist two disjoint open sets V and W in Y containing x and f(x), respectively. By the strongly δθ-I-continuity of f , we have f−1(W ) is a δθ-I-open set in X containing x. Thus, there exists an open set G in X such that x ∈ G ⊂ δCl?(G) ⊂ f−1(W ). Let us observe that U = G ∩ V is an open set in X containing x. We affirm that δCl?(U) ⊂ X \ A and so, X \ A is a δθ-I-open set in X. Indeed, if y ∈ δCl?(U) then y ∈ δCl?(G) ⊂ f−1(W ), which implies that f(y) ∈ W , and as V and W are disjoint sets, it follows that f(y) /∈ V , but since y ∈ V , we get that y 6= f(y) and hence, y /∈ A. Definition 8. The graph G(f) of a function f : (X, τ, I) → (Y, σ) is said to be strongly δθ-I-closed with respect to X, if for each (x, y) ∈ (X × Y ) \ G(f), there exist two open sets U and V containing x and y, respectively, such that (δCl?(U)× V ) ∩G(f) = ∅. Lemma 3. The graph G(f) of a function f : (X, τ, I) → (Y, σ) is strongly δθ-I-closed with respect to X if and only if for each (x, y) ∈ (X × Y ) \G(f), there exist two open sets U and V containing x and y, respectively, such that f(δCl?(U)) ∩ V = ∅. Theorem 11. If f : (X, τ, I) → (Y, σ) is strongly δθ-I-continuous and (Y, σ) is T2, then G(f) is strongly δθ-I-closed with respect to X. Proof. Let (x, y) ∈ (X × Y ) \ G(f). Then, f(x) 6= y and as (Y, σ) is T2, there exist open sets V and W in Y containing f(x) and y, respectively, such that V ∩W = ∅. Since f is strongly δθ-I-continuous, there exists an open sets U in X containing x such that f(δCl?(U)) ⊂ V , which implies that f(δCl?(U))∩W = ∅. By Lemma 3, we conclude that G(f) is strongly δθ-I-closed with respect to X. 4. Conclusion The notion of a continuous function and its generalizations have important applications in various areas of mathematics and related sciences; for example, this notion is widely used in physics and information systems, as described in [4]. In this article we have studied a generalization of continuous functions using concepts recently derived from the theory of topological ideals, such as δθ-I-open set and δθ-I-closure operator. This class of functions can have applications in computation and image design, especially in digital topology, as well as in information systems and quantum physics. 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