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Mixed θ -continuity on generalized topological spaces
Mathematical and Computer Modelling 54 (2011) 2597–2601
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Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Mixed θ -continuity on generalized topological spaces Won Keun Min Department of Mathematics, Kangwon National University, Chuncheon 200-701, Republic of Korea
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Article history: Received 25 March 2011 Received in revised form 13 June 2011 Accepted 14 June 2011 Keywords: Mixed weak continuity Mixed θ -continuity Mixed faint continuity (ν1 , ν2 )-regular G-regular
abstract We introduce and investigate the notions of mixed θ (µ, ν1 ν2 )-continuous functions and mixed faintly (µ, ν1 ν2 )-continuous functions between a generalized topology µ and two generalized topologies ν1 , ν2 . We investigate relationships between such two continuities and another mixed continuity on generalized topological spaces. © 2011 Elsevier Ltd. All rights reserved.
1. Introduction Császár [1] introduced the notions of generalized topology and generalized open sets. He also introduced the notions of continuous functions and associated interior and closure operators on generalized topological spaces. In [2], he introduced and investigated the notions of mixed generalized open sets ((ν1 , ν2 )-semiopen, (ν1 , ν2 )-preopen, (ν1 , ν2 )-β ′ -open) on two generalized topologies. Császár and Makai jr. [3] modified the notions of δ and θ by mixing two generalized topologies. In the same way, the author introduced and investigated the notion of mixed weak (µ, ν1 ν2 )-continuity [4] between a general topology µ and two generalized topologies ν1 , ν2 . The purpose of this paper is to introduce and investigate the notions of mixed θ (µ, ν1 ν2 )-continuity and mixed faint (µ, ν1 ν2 )-continuity between a generalized topology µ and two generalized topologies ν1 , ν2 . In particular, we investigate characteristics of the continuities and relationships among mixed θ -continuity, mixed faint continuity and mixed weak continuity on general topological spaces. 2. Preliminaries Let X be a nonempty set, and let µ be a collection of subsets of X . Then µ is called a generalized topology (briefly GT) [1] on X if ∅ ∈ µ and Gi ∈ µ for i ∈ I ̸= ∅ implies G = ∪i∈I Gi ∈ µ. We say that the GT µ is strong [3] if X ∈ µ. We call the pair (X , µ) a generalized topological space (briefly GTS) on X . The elements of µ are called µ-open sets and the complements are called µ-closed sets. The generalized-closure of a subset A of X , denoted by cµ (A), is the intersection of generalized closed sets including A. The interior of A, denoted by iµ (A), is the union of generalized open sets included in A. Let µ be a GT on a nonempty set X and P (X ) the power set of X . Let us define the collection θ (µ) ⊆ P (X ) by A ∈ θ (µ) iff for each x ∈ A, there exists M ∈ µ such that cµ M ⊆ A [5]. Then θ (µ) is also a GT included in µ [5]. The elements of θ (µ) are called θ -open sets and the complements are called θ -closed sets. Simply, θ (µ) is denoted by θ . Let (X , µ) be a GTS and A ⊆ X . We mention here the following notations: cθ (A) = ∩{F ⊆ X : A ⊆ F , F is θ -closed F in X } [6]; iθ (A) = ∪{V ⊆ X : V ⊆ A, V is θ -open V in X } [6];
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ιθ (A) = {x ∈ X : cµ (U ) ⊆ A for some µ-open set U containing x} [6]; γθ (A) = {x ∈ X : cµ (U ) ∩ A ̸= ∅ for all µ-open set U containing x} [1]. Theorem 2.1 ([5]). Let (X , µ) be a GTS and A ⊆ X . Then A is θ -closed if and only if A = γθ (A). Lemma 2.2 ([5,6]). Let (X , µ) be a GTS and A ⊆ X . Then (1) iθ (A) ⊆ ιθ (A) ⊆ iµ (A) ⊆ A ⊆ cµ (A) ⊆ γθ (A) ⊆ cθ (A) [5,6]. (2) If A ∈ µ, then cµ (A) = γθ (A) [5]. Theorem 2.3 ([1,6]). Let (X , µ) be a GTS and A ⊆ X . Then (1) (2) (3)
iθ (A) = X − cθ (X − A) and cθ (A) = X − iθ (X − A) [6]. ιθ (A) = X − γθ (X − A) and γθ (A) = X − ιθ (X − A) [6]. cµ (A) = X − iµ (X − A) and iµ (A) = X − cµ (X − A) [1]. Let µ and ν be GT’s on nonempty sets X and Y , respectively. Then a function f : (X , µ) → (Y , ν) is said to be
(1) (µ, ν)-continuous [1] if G ∈ ν implies that f −1 (G) ∈ µ; (2) θ(µ, ν)-continuous [1] if for x ∈ X and each V ∈ ν such that f (x) ∈ V , there exists U ∈ µ such that x ∈ U and f (cµ (U )) ⊆ cν (V ). 3. Mixed θ(µ, ν1 ν2 )-continuous functions Definition 3.1. Let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then a function f : X → Y is said to be mixed θ (µ, ν1 ν2 )-continuous at x ∈ X if for each ν1 -open set V containing f (x), there exists a µ-open set U containing x such that f (cµ (U )) ⊆ cν2 (V ). Then f is said to be mixed θ (µ, ν1 ν2 )-continuous (briefly, θ (µ, ν1 ν2 )-continuous) if it is mixed θ (µ, ν1 ν2 )-continuous at every point of X . Remark 3.2. Let µ be a GT on a nonempty set X . Consider ν and ν two GT’s on a nonempty set Y . Then the θ (µ, νν)continuous function f : X → Y is a θ (µ, ν)-continuous function. Theorem 3.3. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) (2) (3) (4) (5) (6)
f is θ (µ, ν1 ν2 )-continuous. f −1 (V ) ⊆ ιθ (f −1 (cν2 (V ))) for every ν1 -open subset V of Y . γθ (f −1 (iν2 (F ))) ⊆ f −1 (F ) for every ν1 -closed subset F of Y . γθ (f −1 (iν2 (cν1 (B)))) ⊆ f −1 (cν1 (B)) for every subset B of Y . f −1 (iν1 (B)) ⊆ ιθ (f −1 (cν2 (iν1 (B)))) for every subset B of Y . γθ (f −1 (U )) ⊆ f −1 (cν1 (U )) for every ν2 -open subset U of Y .
Proof. (1) ⇒ (2) If V is a ν1 -open subset of Y and x ∈ f −1 (V ), then there exists a µ-open set U containing x such that f (x) ∈ f (cµ (U )) ⊆ cν2 (V ). Since x ∈ U ⊆ cµ (U ) ⊆ f −1 (cν2 (V )), from the definition of the operator ιθ , we have x ∈ ιθ (f −1 (cν2 (V ))), and so f −1 (V ) ⊆ ιθ (f −1 (cν2 (V ))).
(2) ⇒ (1) Let V be a ν1 -open subset of Y and x ∈ f −1 (V ). Then by (2), since x ∈ ιθ (f −1 (cν2 (V ))), there exists a µ-open set U containing x such that x ∈ U ⊆ cµ (U ) ⊆ f −1 (cν2 (V )). This implies f (cµ (U )) ⊆ cν2 (V ) for some µ-open set U containing x. Hence f is θ (µ, ν1 ν2 )-continuous. (2) ⇔ (3) It follows from Theorem 2.3. (3) ⇔ (4) For B ⊆ Y , since cν1 (B) is ν1 -closed, it is obvious. (4) ⇒ (5) It follows from Theorem 2.3 (5) ⇒ (6) For a ν2 -open subset U of Y , suppose x ̸∈ f −1 (cν1 (U )). Then since f (x) ̸∈ cν1 (U ), there exists a ν1 -open set W containing f (x) such that U ∩ W = ∅. This implies cν2 (W ) ∩ U = ∅. Moreover, by (5), x ∈ f −1 (W ) ⊆ ιθ (f −1 (cν2 (W ))), and so there exists a µ-open set G satisfying x ∈ G ⊆ cµ (G) ⊆ f −1 (cν2 (W )). From cν2 (W ) ∩ U = ∅ and f (cµ (G)) ⊆ cν2 (W ), it follows cµ (G) ∩ f −1 (U ) = ∅, and so x ̸∈ γθ (f −1 (U )). Consequently, γθ (f −1 (U )) ⊆ f −1 (cν1 (U )). (6) ⇒ (4) For B ⊆ Y , since iν2 cν1 (B) is a ν2 -open set in Y , from (6), it follows γθ (f −1 (iν2 cν1 (B))) ⊆ f −1 (cν1 iν2 cν1 (B)) ⊆ f −1 (cν1 (B)). Hence we have the statement (4). Corollary 3.4. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If ν1 = ν2 , then the following are equivalent:
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(1) (2) (3) (4) (5) (6)
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f is θ (µ, ν1 ν1 )-continuous. f −1 (V ) ⊆ ιθ (f −1 (cν1 (V ))) for every ν1 -open subset V of Y . γθ (f −1 (iν1 (A))) ⊆ f −1 (A) for every ν1 -closed subset A of Y . γθ (f −1 (iν1 (cν1 (B)))) ⊆ f −1 (cν1 (B)) for every subset B of Y . f −1 (iν1 (B)) ⊆ ιθ (f −1 (cν1 (iν1 (B)))) for every subset B of Y . γθ (f −1 (U )) ⊆ f −1 (cν1 (U )) for every ν1 -open subset U of Y .
Remark 3.5. Let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then a function f : X → Y is said to be mixed weakly (µ, ν1 ν2 )-continuous (briefly, weakly (µ, ν1 ν2 )-continuous) [4] if for x ∈ X and each ν1 -open set V containing f (x), there exists a µ-open set U containing x such that f (U ) ⊆ cν2 (V ). Every θ (µ, ν1 ν2 )-continuous function is weakly (µ, ν1 ν2 )-continuous. But the converse is not true in general as shown in the next example. Example 3.6. Let X = {1, 2, 3} and a generalized topology µ = {∅, {1, 2}}. Let Y = {a, b, c , d} and two generalized topologies ν1 , ν2 as the following:
ν1 = {∅, {a, b}, {b, c }, {a, b, c }};
ν2 = {∅, {d}}.
Let us consider a function f : X → Y defined as f (1) = a, f (2) = b, f (3) = d. Then obviously f is weakly (µ, ν1 ν2 )continuous. For x = 1, consider a ν1 -open set V = {a, b} containing a = f (1). Note that U = {1, 2} is the only µ-open set containing 1 and cµ (U ) = X . Since f (cµ (U )) = {a, b, d}, f is not θ (µ, ν1 ν2 )-continuous at 1. Consequently, f is not θ(µ, ν1 ν2 )-continuous Let ν1 and ν2 be two GT’s on a nonempty set X . Let θ (ν1 , ν2 ) [3] be composed of the sets A ⊆ X such that x ∈ A implies the existence of a set M ∈ ν1 satisfying x ∈ M ⊆ cν2 (M ) ⊆ A. Then the family θ (ν1 , ν2 ) is a GT contained in ν1 on X [3]. The element of θ (ν1 , ν2 ) is said to be θ (ν1 , ν2 )-open and the complement of θ (ν1 , ν2 )-open set is called a θ (ν1 , ν2 )-closed set. cθ (ν1 ,ν2 ) (A) = ∩{F ⊆ X : A ⊆ F for θ (ν1 , ν2 )-closed set F in X } [4]; iθ (ν1 ,ν2 ) (A) = ∪{V ⊆ X : V ⊆ A for θ (ν1 , ν2 )-open set V in X } [4]; γθ (ν1 ,ν2 ) (A) = {x ∈ X : cν2 (M ) ∩ A ̸= ∅ for every M ∈ ν1 containing x} [3]. Lemma 3.7 ([3,4]). Let ν1 and ν2 be two GT’s on a nonempty set X and A ⊆ X . Then the following hold: (1) (2) (3) (4)
A ⊆ γθ(ν1 ,ν2 ) (A) ⊆ cθ(ν1 ,ν2 ) (A) [3]. A is θ (ν1 , ν2 )-closed iff A = γθ(ν1 ,ν2 ) (A) [3]. x ∈ iθ(ν1 ,ν2 ) (A) if and only if there exists a ν1 -open set M containing x such that x ∈ M ⊆ cν2 (M ) ⊆ A [4]. If A is ν2 -open in X , then γθ(ν1 ,ν2 ) (A) = cν1 (A) [4].
Theorem 3.8 ([4]). Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) (2) (3)
f is weakly (µ, ν1 ν2 )-continuous. f (cµ (A)) ⊆ γθ(ν1 ,ν2 ) (f (A)) for every subset A of X . cµ (f −1 (B)) ⊆ f −1 (γθ(ν1 ,ν2 ) (B)) for every subset B of Y .
Theorem 3.9. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If f is (µ, ν2 )-continuous and a weakly (µ, ν1 ν2 )-continuous function, then it is θ (µ, ν1 ν2 )-continuous. Proof. For U ∈ ν2 , since f is (µ, ν2 )-continuous, f −1 (U ) is µ-open and so by Lemma 2.2(2) and Lemma 3.7(4), cµ (f −1 (U )) = γθ (f −1 (U )) and γθ(ν1 ,ν2 ) (U ) = cν1 (U ). By Theorem 3.8, γθ (f −1 (U )) = cµ (f −1 (U )) ⊆ f −1 (γθ (ν1 ,ν2 ) (U )) = f −1 (cν1 (U )). Hence by Theorem 3.3(6), f is θ (µ, ν1 ν2 )-continuous. Theorem 3.10. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) f is θ (µ, ν1 ν2 )-continuous. (2) f (γθ (A)) ⊆ γθ(ν1 ,ν2 ) (f (A)) for every subset A of X . (3) γθ (f −1 (B)) ⊆ f −1 (γθ(ν1 ,ν2 ) (B)) for every subset B of Y . Proof. (1) ⇒ (2) For A ⊆ X , let x ∈ γθ (A) and V any ν1 -open set containing f (x). Then since f is θ (µ, ν1 ν2 )-continuous, there exists a µ-open set Ux containing x such that f (cµ (Ux )) ⊆ cν2 (V ). Moreover, since x ∈ γθ (A), for the µ-open set Ux , cµ (Ux ) ∩ A ̸= ∅. So cν2 (V ) ∩ f (A) ⊇ f (cµ (Ux )) ∩ f (A) ⊇ f (cµ (Ux ) ∩ A) ̸= ∅. This implies f (x) ∈ γθ (ν1 ,ν2 ) (f (A)) and hence, f (γθ (A)) ⊆ γθ(ν1 ,ν2 ) (f (A)). (2) ⇒ (3) Obvious. (3) ⇒ (1) Let U be any ν2 -open set. Then from Lemma 3.7(4), we know that γθ (ν1 ,ν2 ) (U ) = cν1 (U ), and so γθ (f −1 (U )) ⊆ f −1 (γθ (ν1 ,ν2 ) (U )) = f −1 (cν1 (U )). Hence by Theorem 3.3(6), f is θ (µ, ν1 ν2 )-continuous.
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Let ν1 and ν2 be two GT’s on a nonempty set X . Then X is said to be (ν1 , ν2 )-regular [4] on X if for x ∈ X and ν1 -closed set F with x ̸∈ F , there exist U ∈ ν1 , V ∈ ν2 such that x ∈ U, F ⊆ V and U ∩ V = ∅. Theorem 3.11 ([4]). Let ν1 and ν2 be two GT’s on a nonempty set X . Then (1) X is (ν1 , ν2 )-regular if and only if for x ∈ X and a ν1 -open set U containing x, there exists a ν1 -open set V containing x such that x ∈ V ⊆ cν2 (V ) ⊆ U. (2) If X is (ν1 , ν2 )-regular, every ν1 -open set is θ (ν1 , ν2 )-open. Theorem 3.12. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) (2) (3)
f is θ (µ, ν1 ν2 )-continuous. f −1 (B) is θ -closed for every θ (ν1 , ν2 )-closed subset B of Y . f −1 (V ) is θ -open for every θ (ν1 , ν2 )-open subset V of Y .
Proof. (1) ⇒ (2) Let B be a θ (ν1 , ν2 )-closed subset of Y . Then from Lemma 3.7 and Theorem 3.10, it follows γθ (f −1 (B)) ⊆ f −1 (γθ(ν1 ,ν2 ) (B)) = f −1 (B). Hence f −1 (B) is θ -closed. (2) ⇒ (3) Obvious. (3) ⇒ (1) For any ν1 -open subset V of Y , since Y is (ν1 , ν2 )-regular, V is also θ (ν1 , ν2 )-open. By hypothesis, f −1 (V ) = ιθ (f −1 (V )) ⊆ ιθ (f −1 (cν2 (V ))). Thus from Theorem 3.3(2), f is θ (µ, ν1 ν2 )-continuous. 4. Mixed faintly (µ, ν1 ν2 )-continuous functions Definition 4.1. Let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then a function f : X → Y is said to be mixed faintly (µ, ν1 ν2 )-continuous at x ∈ X if for each θ (ν1 , ν2 )-open set V containing f (x), there exists a µ-open set U containing x such that f (U ) ⊆ V . Then f is said to be mixed faintly (µ, ν1 ν2 )-continuous (briefly, faintly (µ, ν1 ν2 )-continuous) if it is mixed faintly (µ, ν1 ν2 )-continuous at every point of X . Theorem 4.2. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) (2) (3) (4) (5)
f is faintly (µ, ν1 ν2 )-continuous. For every θ (ν1 , ν2 )-open subset V of Y , f −1 (V ) ∈ µ. f −1 (iθ(ν1 ,ν2 ) (B)) ⊆ iµ (f −1 (B)) for every subset B of Y . cµ (f −1 (B)) ⊆ f −1 (cθ(ν1 ,ν2 ) (B)) for every subset B of Y . f (cµ (A)) ⊆ cθ(ν1 ,ν2 ) (f (A)) for every subset A of X .
Proof. (1) ⇔ (2) It is obvious. (2) ⇒ (3) For B ⊆ Y , since iθ(ν1 ,ν2 ) (B) is θ (ν1 , ν2 )-open, obviously it is obtained from (2). (3) ⇒ (4) Since θ (ν1 , ν2 ) is a GT in Y , from Theorem 2.3, (4) is easily obtained. (4) ⇒ (5) Obvious. (5) ⇒ (1) Let V be a θ (ν1 , ν2 )-open set in Y . Then since Y − V is θ (ν1 , ν2 )-closed, from (5), it follows f (cµ (f −1 (Y − V ))) ⊆ cθ (ν1 ,ν2 ) (f (f −1 (Y − V ))) ⊆ cθ(ν1 ,ν2 ) (Y − V ) = Y − V . This implies cµ (f −1 (Y − V )) = f −1 (Y − V ), and so f −1 (V ) is µ-open in X . Hence by (2), f is faintly (µ, ν1 ν2 )-continuous. Theorem 4.3. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If f is weakly (µ, ν1 ν2 )-continuous, then it is faintly (µ, ν1 ν2 )-continuous. Proof. Let V be any θ (ν1 , ν2 )-open subset in Y and f (x) ∈ V . Then there exists a ν1 -open set W containing f (x) such that W ⊆ cν2 (W ) ⊆ V . For the ν1 -open set W of f (x), since f is weakly (µ, ν1 ν2 )-continuous, there is a µ-open set U of x satisfying f (U ) ⊆ cν2 (W ), and so we have f (U ) ⊆ V . Thus f is faintly (µ, ν1 ν2 )-continuous. In the above Theorem 4.3, the converse may not be true as shown in the next example. Example 4.4. Let X = Y = {a, b, c , d}. Consider a generalized topology µ on X and two generalized topologies ν1 and ν2 on Y as follows
µ = {∅, {b, c }, {b, c , d}}; ν1 = {∅, {a}, {b, c }, {a, b, c }};
ν2 = {∅, {a, d}, {b, d}, {a, b, d}}.
Let us consider a function f : X → Y defined as f (a) = d, f (b) = b, f (c ) = c, f (d) = a. Then since U = {b, c } is the only nonempty θ (ν1 , ν2 )-open set in Y and f −1 (U ) is µ-open in X , by Theorem 4.2(2), f is faintly (µ, ν1 ν2 )-continuous. But for the point d ∈ X , we can easily show that f is not weakly (µ, ν1 ν2 )-continuous at d, and so f is not weakly (µ, ν1 ν2 )-continuous.
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Lemma 4.5. Let ν1 , ν2 be two GT’s on a nonempty set X . If X is (ν1 , ν2 )-regular, then the following things hold. (1) θ(ν1 , ν2 ) = ν1 [4]. (2) cν1 (B) = γθ(ν1 ,ν2 ) (B) = cθ(ν1 ,ν2 ) (B) for B ⊆ X . Proof. (1) See Lemma 3.12(2) of [4]. (2) By (1), cν1 (B) = cθ(ν1 ,ν2 ) (B), and so from Lemma 3.7(1), it is obvious.
Theorem 4.6. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) f is faintly (µ, ν1 ν2 )-continuous. (2) f is weakly (µ, ν1 ν2 )-continuous. (3) f is (µ, ν1 )-continuous. Proof. (1) ⇔ (2) It follows from Theorem 3.8, Theorem 4.2 and Lemma 4.5(2). (2) ⇔ (3) See Theorem 3.18 of [4]. Let µ1 and µ2 be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . A function f : X → Y is said to be (θ (µ1 , µ2 ), θ (ν1 , ν2 ))-continuous [3] if G ∈ θ (ν1 , ν2 ) implies that f −1 (G) ∈ θ (µ1 , µ2 ) Theorem 4.7. Let f : X → Y be a function, let µ1 and µ2 be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If X is (µ1 , µ2 )-regular, then the following are equivalent: (1) f is (θ (µ1 , µ2 ), θ (ν1 , ν2 ))-continuous. (2) f is faintly (µ1 , ν1 ν2 )-continuous. Proof. Since X is (µ1 , µ2 )-regular, by Lemma 4.5, θ (µ1 , µ2 ) = µ1 . So from Theorem 4.2(2), it is proved.
Corollary 4.8. Let f : X → Y be a function, let µ1 and µ2 be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If X is (µ1 , µ2 )-regular and Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) f is (θ (µ1 , µ2 ), θ (ν1 , ν2 ))-continuous. (2) f is weakly (µ1 , ν1 ν2 )-continuous. (3) f is faintly (µ1 , ν1 ν2 )-continuous. (4) f is (µ1 , ν1 )-continuous. Proof. It follows from Theorems 4.6 and 4.7.
Let (X , µ) be a generalized topological space and Mµ = ∪{G ⊆ X : G ∈ µ}. Then X is said to be relative G-regular (simply, G-regular) [7] on Mµ if for x ∈ Mµ and µ-closed set F with x ̸∈ F , there exist U , V ∈ µ such that x ∈ U, F ∩ Mµ ⊆ V and U ∩ V = ∅. Theorem 4.9 ([7]). Let (X , µ) be a GTS. Then X is G-regular if and only if for x ∈ Mµ and a µ-open set U containing x, there exists a µ-open set V containing x such that x ∈ V ⊆ cµ V ∩ Mµ ⊆ U. Lemma 4.10. Let (X , µ) be a generalized topological space. If X is G-regular and strong, then every µ-open set is θ -open. Proof. Let U be a µ-open set in X . For each x ∈ U, from the above Theorem 4.9, there exists a µ-open set V such that x ∈ V ⊆ cµ V ∩ Mµ ⊆ U. Then since X is strong, it implies x ∈ V ⊆ cµ V ⊆ U. Therefore, U is θ -open. Theorem 4.11. Let f : X → Y be a function, let µ be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If X is G-regular and strong, and if Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) f is θ (µ, ν1 , ν2 )-continuous. (2) f is faintly (µ, ν1 ν2 )-continuous. Proof. It follows from Theorem 3.12 and Lemma 4.10.
Remark 4.12. Let (X , µ) and (Y , ν) be GTS’s. Then a function f : X → Y is said to be weakly θ (µ, ν)-continuous [6] if for x ∈ X and each V ∈ θ (ν) containing f (x), there exists U ∈ µ such that x ∈ U and f (U ) ⊆ V . Obviously the mixed faintly θ (µ, νν)-continuous function f is weakly θ (µ, ν)-continuous. Moreover, if the generalized topologies µ and ν are topologies, then the mixed faintly θ (µ, νν)-continuous function f is faintly continuous [8]. References [1] [2] [3] [4] [5] [6] [7] [8]
Á. Császár, Generalized topology, generalized continuity, Acta Math. Hungar. 96 (2002) 351–357. Á. Császár, Mixed constructions for generalized topologies, Acta Math. Hungar. 122 (1–2) (2009) 153–159. Á. Császár, E. Makai jr., Further remarks on δ - and θ -modificatons, Acta Math. Hungar. 123 (3) (2009) 223–228. W.K. Min, Mixed weak continuity on generalized topological spaces, Acta Math. Hungar. (2011), doi:10.1007/s10474-011-0078-2, online first. Á. Császár, δ - and θ -modificatons of generalized topologies, Acta Math. Hungar. 120 (3) (2008) 275–279. W.K. Min, A note on θ(g , g ′ )-continuity on generalized topological spaces, Acta Math. Hungar. 125 (4) (2009) 387–393. W.K. Min, (δ, δ ′ )-continuity on generalized topological spaces, Acta Math. Hungar. 129 (4) (2010) 350–356. P.E. Long, L.L. Herrington, The Tθ -topology and faintly continuous functions, Kyungpook Math. J. 22 (1982) 7–14.
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Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Mixed θ -continuity on generalized topological spaces Won Keun Min Department of Mathematics, Kangwon National University, Chuncheon 200-701, Republic of Korea
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Article history: Received 25 March 2011 Received in revised form 13 June 2011 Accepted 14 June 2011 Keywords: Mixed weak continuity Mixed θ -continuity Mixed faint continuity (ν1 , ν2 )-regular G-regular
abstract We introduce and investigate the notions of mixed θ (µ, ν1 ν2 )-continuous functions and mixed faintly (µ, ν1 ν2 )-continuous functions between a generalized topology µ and two generalized topologies ν1 , ν2 . We investigate relationships between such two continuities and another mixed continuity on generalized topological spaces. © 2011 Elsevier Ltd. All rights reserved.
1. Introduction Császár [1] introduced the notions of generalized topology and generalized open sets. He also introduced the notions of continuous functions and associated interior and closure operators on generalized topological spaces. In [2], he introduced and investigated the notions of mixed generalized open sets ((ν1 , ν2 )-semiopen, (ν1 , ν2 )-preopen, (ν1 , ν2 )-β ′ -open) on two generalized topologies. Császár and Makai jr. [3] modified the notions of δ and θ by mixing two generalized topologies. In the same way, the author introduced and investigated the notion of mixed weak (µ, ν1 ν2 )-continuity [4] between a general topology µ and two generalized topologies ν1 , ν2 . The purpose of this paper is to introduce and investigate the notions of mixed θ (µ, ν1 ν2 )-continuity and mixed faint (µ, ν1 ν2 )-continuity between a generalized topology µ and two generalized topologies ν1 , ν2 . In particular, we investigate characteristics of the continuities and relationships among mixed θ -continuity, mixed faint continuity and mixed weak continuity on general topological spaces. 2. Preliminaries Let X be a nonempty set, and let µ be a collection of subsets of X . Then µ is called a generalized topology (briefly GT) [1] on X if ∅ ∈ µ and Gi ∈ µ for i ∈ I ̸= ∅ implies G = ∪i∈I Gi ∈ µ. We say that the GT µ is strong [3] if X ∈ µ. We call the pair (X , µ) a generalized topological space (briefly GTS) on X . The elements of µ are called µ-open sets and the complements are called µ-closed sets. The generalized-closure of a subset A of X , denoted by cµ (A), is the intersection of generalized closed sets including A. The interior of A, denoted by iµ (A), is the union of generalized open sets included in A. Let µ be a GT on a nonempty set X and P (X ) the power set of X . Let us define the collection θ (µ) ⊆ P (X ) by A ∈ θ (µ) iff for each x ∈ A, there exists M ∈ µ such that cµ M ⊆ A [5]. Then θ (µ) is also a GT included in µ [5]. The elements of θ (µ) are called θ -open sets and the complements are called θ -closed sets. Simply, θ (µ) is denoted by θ . Let (X , µ) be a GTS and A ⊆ X . We mention here the following notations: cθ (A) = ∩{F ⊆ X : A ⊆ F , F is θ -closed F in X } [6]; iθ (A) = ∪{V ⊆ X : V ⊆ A, V is θ -open V in X } [6];
E-mail address: [email protected]. 0895-7177/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2011.06.032
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ιθ (A) = {x ∈ X : cµ (U ) ⊆ A for some µ-open set U containing x} [6]; γθ (A) = {x ∈ X : cµ (U ) ∩ A ̸= ∅ for all µ-open set U containing x} [1]. Theorem 2.1 ([5]). Let (X , µ) be a GTS and A ⊆ X . Then A is θ -closed if and only if A = γθ (A). Lemma 2.2 ([5,6]). Let (X , µ) be a GTS and A ⊆ X . Then (1) iθ (A) ⊆ ιθ (A) ⊆ iµ (A) ⊆ A ⊆ cµ (A) ⊆ γθ (A) ⊆ cθ (A) [5,6]. (2) If A ∈ µ, then cµ (A) = γθ (A) [5]. Theorem 2.3 ([1,6]). Let (X , µ) be a GTS and A ⊆ X . Then (1) (2) (3)
iθ (A) = X − cθ (X − A) and cθ (A) = X − iθ (X − A) [6]. ιθ (A) = X − γθ (X − A) and γθ (A) = X − ιθ (X − A) [6]. cµ (A) = X − iµ (X − A) and iµ (A) = X − cµ (X − A) [1]. Let µ and ν be GT’s on nonempty sets X and Y , respectively. Then a function f : (X , µ) → (Y , ν) is said to be
(1) (µ, ν)-continuous [1] if G ∈ ν implies that f −1 (G) ∈ µ; (2) θ(µ, ν)-continuous [1] if for x ∈ X and each V ∈ ν such that f (x) ∈ V , there exists U ∈ µ such that x ∈ U and f (cµ (U )) ⊆ cν (V ). 3. Mixed θ(µ, ν1 ν2 )-continuous functions Definition 3.1. Let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then a function f : X → Y is said to be mixed θ (µ, ν1 ν2 )-continuous at x ∈ X if for each ν1 -open set V containing f (x), there exists a µ-open set U containing x such that f (cµ (U )) ⊆ cν2 (V ). Then f is said to be mixed θ (µ, ν1 ν2 )-continuous (briefly, θ (µ, ν1 ν2 )-continuous) if it is mixed θ (µ, ν1 ν2 )-continuous at every point of X . Remark 3.2. Let µ be a GT on a nonempty set X . Consider ν and ν two GT’s on a nonempty set Y . Then the θ (µ, νν)continuous function f : X → Y is a θ (µ, ν)-continuous function. Theorem 3.3. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) (2) (3) (4) (5) (6)
f is θ (µ, ν1 ν2 )-continuous. f −1 (V ) ⊆ ιθ (f −1 (cν2 (V ))) for every ν1 -open subset V of Y . γθ (f −1 (iν2 (F ))) ⊆ f −1 (F ) for every ν1 -closed subset F of Y . γθ (f −1 (iν2 (cν1 (B)))) ⊆ f −1 (cν1 (B)) for every subset B of Y . f −1 (iν1 (B)) ⊆ ιθ (f −1 (cν2 (iν1 (B)))) for every subset B of Y . γθ (f −1 (U )) ⊆ f −1 (cν1 (U )) for every ν2 -open subset U of Y .
Proof. (1) ⇒ (2) If V is a ν1 -open subset of Y and x ∈ f −1 (V ), then there exists a µ-open set U containing x such that f (x) ∈ f (cµ (U )) ⊆ cν2 (V ). Since x ∈ U ⊆ cµ (U ) ⊆ f −1 (cν2 (V )), from the definition of the operator ιθ , we have x ∈ ιθ (f −1 (cν2 (V ))), and so f −1 (V ) ⊆ ιθ (f −1 (cν2 (V ))).
(2) ⇒ (1) Let V be a ν1 -open subset of Y and x ∈ f −1 (V ). Then by (2), since x ∈ ιθ (f −1 (cν2 (V ))), there exists a µ-open set U containing x such that x ∈ U ⊆ cµ (U ) ⊆ f −1 (cν2 (V )). This implies f (cµ (U )) ⊆ cν2 (V ) for some µ-open set U containing x. Hence f is θ (µ, ν1 ν2 )-continuous. (2) ⇔ (3) It follows from Theorem 2.3. (3) ⇔ (4) For B ⊆ Y , since cν1 (B) is ν1 -closed, it is obvious. (4) ⇒ (5) It follows from Theorem 2.3 (5) ⇒ (6) For a ν2 -open subset U of Y , suppose x ̸∈ f −1 (cν1 (U )). Then since f (x) ̸∈ cν1 (U ), there exists a ν1 -open set W containing f (x) such that U ∩ W = ∅. This implies cν2 (W ) ∩ U = ∅. Moreover, by (5), x ∈ f −1 (W ) ⊆ ιθ (f −1 (cν2 (W ))), and so there exists a µ-open set G satisfying x ∈ G ⊆ cµ (G) ⊆ f −1 (cν2 (W )). From cν2 (W ) ∩ U = ∅ and f (cµ (G)) ⊆ cν2 (W ), it follows cµ (G) ∩ f −1 (U ) = ∅, and so x ̸∈ γθ (f −1 (U )). Consequently, γθ (f −1 (U )) ⊆ f −1 (cν1 (U )). (6) ⇒ (4) For B ⊆ Y , since iν2 cν1 (B) is a ν2 -open set in Y , from (6), it follows γθ (f −1 (iν2 cν1 (B))) ⊆ f −1 (cν1 iν2 cν1 (B)) ⊆ f −1 (cν1 (B)). Hence we have the statement (4). Corollary 3.4. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If ν1 = ν2 , then the following are equivalent:
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(1) (2) (3) (4) (5) (6)
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f is θ (µ, ν1 ν1 )-continuous. f −1 (V ) ⊆ ιθ (f −1 (cν1 (V ))) for every ν1 -open subset V of Y . γθ (f −1 (iν1 (A))) ⊆ f −1 (A) for every ν1 -closed subset A of Y . γθ (f −1 (iν1 (cν1 (B)))) ⊆ f −1 (cν1 (B)) for every subset B of Y . f −1 (iν1 (B)) ⊆ ιθ (f −1 (cν1 (iν1 (B)))) for every subset B of Y . γθ (f −1 (U )) ⊆ f −1 (cν1 (U )) for every ν1 -open subset U of Y .
Remark 3.5. Let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then a function f : X → Y is said to be mixed weakly (µ, ν1 ν2 )-continuous (briefly, weakly (µ, ν1 ν2 )-continuous) [4] if for x ∈ X and each ν1 -open set V containing f (x), there exists a µ-open set U containing x such that f (U ) ⊆ cν2 (V ). Every θ (µ, ν1 ν2 )-continuous function is weakly (µ, ν1 ν2 )-continuous. But the converse is not true in general as shown in the next example. Example 3.6. Let X = {1, 2, 3} and a generalized topology µ = {∅, {1, 2}}. Let Y = {a, b, c , d} and two generalized topologies ν1 , ν2 as the following:
ν1 = {∅, {a, b}, {b, c }, {a, b, c }};
ν2 = {∅, {d}}.
Let us consider a function f : X → Y defined as f (1) = a, f (2) = b, f (3) = d. Then obviously f is weakly (µ, ν1 ν2 )continuous. For x = 1, consider a ν1 -open set V = {a, b} containing a = f (1). Note that U = {1, 2} is the only µ-open set containing 1 and cµ (U ) = X . Since f (cµ (U )) = {a, b, d}, f is not θ (µ, ν1 ν2 )-continuous at 1. Consequently, f is not θ(µ, ν1 ν2 )-continuous Let ν1 and ν2 be two GT’s on a nonempty set X . Let θ (ν1 , ν2 ) [3] be composed of the sets A ⊆ X such that x ∈ A implies the existence of a set M ∈ ν1 satisfying x ∈ M ⊆ cν2 (M ) ⊆ A. Then the family θ (ν1 , ν2 ) is a GT contained in ν1 on X [3]. The element of θ (ν1 , ν2 ) is said to be θ (ν1 , ν2 )-open and the complement of θ (ν1 , ν2 )-open set is called a θ (ν1 , ν2 )-closed set. cθ (ν1 ,ν2 ) (A) = ∩{F ⊆ X : A ⊆ F for θ (ν1 , ν2 )-closed set F in X } [4]; iθ (ν1 ,ν2 ) (A) = ∪{V ⊆ X : V ⊆ A for θ (ν1 , ν2 )-open set V in X } [4]; γθ (ν1 ,ν2 ) (A) = {x ∈ X : cν2 (M ) ∩ A ̸= ∅ for every M ∈ ν1 containing x} [3]. Lemma 3.7 ([3,4]). Let ν1 and ν2 be two GT’s on a nonempty set X and A ⊆ X . Then the following hold: (1) (2) (3) (4)
A ⊆ γθ(ν1 ,ν2 ) (A) ⊆ cθ(ν1 ,ν2 ) (A) [3]. A is θ (ν1 , ν2 )-closed iff A = γθ(ν1 ,ν2 ) (A) [3]. x ∈ iθ(ν1 ,ν2 ) (A) if and only if there exists a ν1 -open set M containing x such that x ∈ M ⊆ cν2 (M ) ⊆ A [4]. If A is ν2 -open in X , then γθ(ν1 ,ν2 ) (A) = cν1 (A) [4].
Theorem 3.8 ([4]). Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) (2) (3)
f is weakly (µ, ν1 ν2 )-continuous. f (cµ (A)) ⊆ γθ(ν1 ,ν2 ) (f (A)) for every subset A of X . cµ (f −1 (B)) ⊆ f −1 (γθ(ν1 ,ν2 ) (B)) for every subset B of Y .
Theorem 3.9. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If f is (µ, ν2 )-continuous and a weakly (µ, ν1 ν2 )-continuous function, then it is θ (µ, ν1 ν2 )-continuous. Proof. For U ∈ ν2 , since f is (µ, ν2 )-continuous, f −1 (U ) is µ-open and so by Lemma 2.2(2) and Lemma 3.7(4), cµ (f −1 (U )) = γθ (f −1 (U )) and γθ(ν1 ,ν2 ) (U ) = cν1 (U ). By Theorem 3.8, γθ (f −1 (U )) = cµ (f −1 (U )) ⊆ f −1 (γθ (ν1 ,ν2 ) (U )) = f −1 (cν1 (U )). Hence by Theorem 3.3(6), f is θ (µ, ν1 ν2 )-continuous. Theorem 3.10. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) f is θ (µ, ν1 ν2 )-continuous. (2) f (γθ (A)) ⊆ γθ(ν1 ,ν2 ) (f (A)) for every subset A of X . (3) γθ (f −1 (B)) ⊆ f −1 (γθ(ν1 ,ν2 ) (B)) for every subset B of Y . Proof. (1) ⇒ (2) For A ⊆ X , let x ∈ γθ (A) and V any ν1 -open set containing f (x). Then since f is θ (µ, ν1 ν2 )-continuous, there exists a µ-open set Ux containing x such that f (cµ (Ux )) ⊆ cν2 (V ). Moreover, since x ∈ γθ (A), for the µ-open set Ux , cµ (Ux ) ∩ A ̸= ∅. So cν2 (V ) ∩ f (A) ⊇ f (cµ (Ux )) ∩ f (A) ⊇ f (cµ (Ux ) ∩ A) ̸= ∅. This implies f (x) ∈ γθ (ν1 ,ν2 ) (f (A)) and hence, f (γθ (A)) ⊆ γθ(ν1 ,ν2 ) (f (A)). (2) ⇒ (3) Obvious. (3) ⇒ (1) Let U be any ν2 -open set. Then from Lemma 3.7(4), we know that γθ (ν1 ,ν2 ) (U ) = cν1 (U ), and so γθ (f −1 (U )) ⊆ f −1 (γθ (ν1 ,ν2 ) (U )) = f −1 (cν1 (U )). Hence by Theorem 3.3(6), f is θ (µ, ν1 ν2 )-continuous.
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Let ν1 and ν2 be two GT’s on a nonempty set X . Then X is said to be (ν1 , ν2 )-regular [4] on X if for x ∈ X and ν1 -closed set F with x ̸∈ F , there exist U ∈ ν1 , V ∈ ν2 such that x ∈ U, F ⊆ V and U ∩ V = ∅. Theorem 3.11 ([4]). Let ν1 and ν2 be two GT’s on a nonempty set X . Then (1) X is (ν1 , ν2 )-regular if and only if for x ∈ X and a ν1 -open set U containing x, there exists a ν1 -open set V containing x such that x ∈ V ⊆ cν2 (V ) ⊆ U. (2) If X is (ν1 , ν2 )-regular, every ν1 -open set is θ (ν1 , ν2 )-open. Theorem 3.12. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) (2) (3)
f is θ (µ, ν1 ν2 )-continuous. f −1 (B) is θ -closed for every θ (ν1 , ν2 )-closed subset B of Y . f −1 (V ) is θ -open for every θ (ν1 , ν2 )-open subset V of Y .
Proof. (1) ⇒ (2) Let B be a θ (ν1 , ν2 )-closed subset of Y . Then from Lemma 3.7 and Theorem 3.10, it follows γθ (f −1 (B)) ⊆ f −1 (γθ(ν1 ,ν2 ) (B)) = f −1 (B). Hence f −1 (B) is θ -closed. (2) ⇒ (3) Obvious. (3) ⇒ (1) For any ν1 -open subset V of Y , since Y is (ν1 , ν2 )-regular, V is also θ (ν1 , ν2 )-open. By hypothesis, f −1 (V ) = ιθ (f −1 (V )) ⊆ ιθ (f −1 (cν2 (V ))). Thus from Theorem 3.3(2), f is θ (µ, ν1 ν2 )-continuous. 4. Mixed faintly (µ, ν1 ν2 )-continuous functions Definition 4.1. Let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then a function f : X → Y is said to be mixed faintly (µ, ν1 ν2 )-continuous at x ∈ X if for each θ (ν1 , ν2 )-open set V containing f (x), there exists a µ-open set U containing x such that f (U ) ⊆ V . Then f is said to be mixed faintly (µ, ν1 ν2 )-continuous (briefly, faintly (µ, ν1 ν2 )-continuous) if it is mixed faintly (µ, ν1 ν2 )-continuous at every point of X . Theorem 4.2. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . Then the following are equivalent: (1) (2) (3) (4) (5)
f is faintly (µ, ν1 ν2 )-continuous. For every θ (ν1 , ν2 )-open subset V of Y , f −1 (V ) ∈ µ. f −1 (iθ(ν1 ,ν2 ) (B)) ⊆ iµ (f −1 (B)) for every subset B of Y . cµ (f −1 (B)) ⊆ f −1 (cθ(ν1 ,ν2 ) (B)) for every subset B of Y . f (cµ (A)) ⊆ cθ(ν1 ,ν2 ) (f (A)) for every subset A of X .
Proof. (1) ⇔ (2) It is obvious. (2) ⇒ (3) For B ⊆ Y , since iθ(ν1 ,ν2 ) (B) is θ (ν1 , ν2 )-open, obviously it is obtained from (2). (3) ⇒ (4) Since θ (ν1 , ν2 ) is a GT in Y , from Theorem 2.3, (4) is easily obtained. (4) ⇒ (5) Obvious. (5) ⇒ (1) Let V be a θ (ν1 , ν2 )-open set in Y . Then since Y − V is θ (ν1 , ν2 )-closed, from (5), it follows f (cµ (f −1 (Y − V ))) ⊆ cθ (ν1 ,ν2 ) (f (f −1 (Y − V ))) ⊆ cθ(ν1 ,ν2 ) (Y − V ) = Y − V . This implies cµ (f −1 (Y − V )) = f −1 (Y − V ), and so f −1 (V ) is µ-open in X . Hence by (2), f is faintly (µ, ν1 ν2 )-continuous. Theorem 4.3. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If f is weakly (µ, ν1 ν2 )-continuous, then it is faintly (µ, ν1 ν2 )-continuous. Proof. Let V be any θ (ν1 , ν2 )-open subset in Y and f (x) ∈ V . Then there exists a ν1 -open set W containing f (x) such that W ⊆ cν2 (W ) ⊆ V . For the ν1 -open set W of f (x), since f is weakly (µ, ν1 ν2 )-continuous, there is a µ-open set U of x satisfying f (U ) ⊆ cν2 (W ), and so we have f (U ) ⊆ V . Thus f is faintly (µ, ν1 ν2 )-continuous. In the above Theorem 4.3, the converse may not be true as shown in the next example. Example 4.4. Let X = Y = {a, b, c , d}. Consider a generalized topology µ on X and two generalized topologies ν1 and ν2 on Y as follows
µ = {∅, {b, c }, {b, c , d}}; ν1 = {∅, {a}, {b, c }, {a, b, c }};
ν2 = {∅, {a, d}, {b, d}, {a, b, d}}.
Let us consider a function f : X → Y defined as f (a) = d, f (b) = b, f (c ) = c, f (d) = a. Then since U = {b, c } is the only nonempty θ (ν1 , ν2 )-open set in Y and f −1 (U ) is µ-open in X , by Theorem 4.2(2), f is faintly (µ, ν1 ν2 )-continuous. But for the point d ∈ X , we can easily show that f is not weakly (µ, ν1 ν2 )-continuous at d, and so f is not weakly (µ, ν1 ν2 )-continuous.
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Lemma 4.5. Let ν1 , ν2 be two GT’s on a nonempty set X . If X is (ν1 , ν2 )-regular, then the following things hold. (1) θ(ν1 , ν2 ) = ν1 [4]. (2) cν1 (B) = γθ(ν1 ,ν2 ) (B) = cθ(ν1 ,ν2 ) (B) for B ⊆ X . Proof. (1) See Lemma 3.12(2) of [4]. (2) By (1), cν1 (B) = cθ(ν1 ,ν2 ) (B), and so from Lemma 3.7(1), it is obvious.
Theorem 4.6. Let f : X → Y be a function, let µ be a GT on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) f is faintly (µ, ν1 ν2 )-continuous. (2) f is weakly (µ, ν1 ν2 )-continuous. (3) f is (µ, ν1 )-continuous. Proof. (1) ⇔ (2) It follows from Theorem 3.8, Theorem 4.2 and Lemma 4.5(2). (2) ⇔ (3) See Theorem 3.18 of [4]. Let µ1 and µ2 be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . A function f : X → Y is said to be (θ (µ1 , µ2 ), θ (ν1 , ν2 ))-continuous [3] if G ∈ θ (ν1 , ν2 ) implies that f −1 (G) ∈ θ (µ1 , µ2 ) Theorem 4.7. Let f : X → Y be a function, let µ1 and µ2 be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If X is (µ1 , µ2 )-regular, then the following are equivalent: (1) f is (θ (µ1 , µ2 ), θ (ν1 , ν2 ))-continuous. (2) f is faintly (µ1 , ν1 ν2 )-continuous. Proof. Since X is (µ1 , µ2 )-regular, by Lemma 4.5, θ (µ1 , µ2 ) = µ1 . So from Theorem 4.2(2), it is proved.
Corollary 4.8. Let f : X → Y be a function, let µ1 and µ2 be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If X is (µ1 , µ2 )-regular and Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) f is (θ (µ1 , µ2 ), θ (ν1 , ν2 ))-continuous. (2) f is weakly (µ1 , ν1 ν2 )-continuous. (3) f is faintly (µ1 , ν1 ν2 )-continuous. (4) f is (µ1 , ν1 )-continuous. Proof. It follows from Theorems 4.6 and 4.7.
Let (X , µ) be a generalized topological space and Mµ = ∪{G ⊆ X : G ∈ µ}. Then X is said to be relative G-regular (simply, G-regular) [7] on Mµ if for x ∈ Mµ and µ-closed set F with x ̸∈ F , there exist U , V ∈ µ such that x ∈ U, F ∩ Mµ ⊆ V and U ∩ V = ∅. Theorem 4.9 ([7]). Let (X , µ) be a GTS. Then X is G-regular if and only if for x ∈ Mµ and a µ-open set U containing x, there exists a µ-open set V containing x such that x ∈ V ⊆ cµ V ∩ Mµ ⊆ U. Lemma 4.10. Let (X , µ) be a generalized topological space. If X is G-regular and strong, then every µ-open set is θ -open. Proof. Let U be a µ-open set in X . For each x ∈ U, from the above Theorem 4.9, there exists a µ-open set V such that x ∈ V ⊆ cµ V ∩ Mµ ⊆ U. Then since X is strong, it implies x ∈ V ⊆ cµ V ⊆ U. Therefore, U is θ -open. Theorem 4.11. Let f : X → Y be a function, let µ be two GT’s on a nonempty set X , and let ν1 , ν2 be two GT’s on a nonempty set Y . If X is G-regular and strong, and if Y is (ν1 , ν2 )-regular, then the following are equivalent: (1) f is θ (µ, ν1 , ν2 )-continuous. (2) f is faintly (µ, ν1 ν2 )-continuous. Proof. It follows from Theorem 3.12 and Lemma 4.10.
Remark 4.12. Let (X , µ) and (Y , ν) be GTS’s. Then a function f : X → Y is said to be weakly θ (µ, ν)-continuous [6] if for x ∈ X and each V ∈ θ (ν) containing f (x), there exists U ∈ µ such that x ∈ U and f (U ) ⊆ V . Obviously the mixed faintly θ (µ, νν)-continuous function f is weakly θ (µ, ν)-continuous. Moreover, if the generalized topologies µ and ν are topologies, then the mixed faintly θ (µ, νν)-continuous function f is faintly continuous [8]. References [1] [2] [3] [4] [5] [6] [7] [8]
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