A generalization of a contra pre semi-open maps

Raf. J. of Comp. & Math’s. , Vol. 3, No. 2, 2006 A Generalization of A Contra Pre Semi-Open Maps Abdullah M. Abdul-Jabbar College of Science University of Salahaddin Received on:14/8/2005 Accepted on: 26/12/2005 ‫ﺍﻟﻤﻠﺨﺹ‬ ‫ ﻓﻲ ﺍﻟﻔﻀﺎﺀ ﺍﻟﺘﺒﻭﻟﻭﺠﻲ ﻓﻲ‬θ ‫ ﻤﻔﻬﻭﻡ ﻤﺠﻤﻭﻋﺔ ﺸﺒﻪ ﻤﻔﺘﻭﺤﺔ ﻤﻥ ﺍﻟﻨﻤﻁ‬T. Noiri ‫ﻋﺭﻑ‬ ‫ ﻓﻲ ﻫﺫﺍ ﺍﻟﺒﺤﺙ ﻨﻌﺭﻑ ﻭ ﻨﺩﺭﺱ ﺘﻌﻤﻴﻡ ﻟﻠﺩﻭﺍل ﺸﺒﻪ ﻤﻔﺘﻭﺤﺔ ﻤـﻥ‬. 1986 ‫ ﻭ‬1984 ‫[ ﺴﻨﺔ‬9, 10] ‫ ﺍﻟﺘﻲ ﺘـﺴﻤﻰ ﺍﻟﺩﺍﻟـﺔ ﺸـﺒﻪ‬، [3] ‫ ﻓﻲ‬Baker ‫ ﻭ‬Caldas ‫ ﺍﻟﻤﻘﺩﻤﺔ ﻤﻥ ﻗﺒل‬contra Pre ‫ﺍﻟﻨﻤﻁ‬ ‫ ﻭﻫﻲ ﺍﻟﺩﻭﺍل ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﺼﻭﺭ ﺍﻟﻤﺠﻤﻭﻋﺎﺕ ﺸﺒﻪ ﺍﻟﻤﻔﺘﻭﺤﺔ‬. contra Pre θ ‫ﺍﻟﻤﻔﺘﻭﺤﺔ ﻤﻥ ﺍﻟﻨﻤﻁ‬ ‫ ﻜﻤﺎ ﻨﻌﺭﻑ ﻭ ﻨﺩﺭﺱ ﻨﻤﻁ ﹰﺎ ﺠﺩﻴﺩﹰﺍ ﻤﻥ ﺍﻟـﺩﻭﺍل ﺍﻟﻤﻐﻠﻘـﺔ‬، θ ‫ ﺸﺒﻪ ﺍﻟﻤﻐﻠﻘﺔ ﻤﻥ ﺍﻟﻨﻤﻁ‬، θ ‫ﻤﻥ ﺍﻟﻨﻤﻁ‬ ‫ ﻭﻫﺫﻩ ﺍﻟﺩﻭﺍل ﺃﻗﻭﻯ ﻤﻥ ﺍﻟﺩﻭﺍل ﺸﺒﻪ ﺍﻟﻤﻐﻠﻘـﺔ‬، contra Pre θ ‫ﺘﺴﻤﻰ ﺍﻟﺩﻭﺍل ﺸﺒﻪ ﻤﻐﻠﻘﺔ ﻤﻥ ﺍﻟﻨﻤﻁ‬ ‫ ﻭﻫـﻲ ﺍﻟـﺩﻭﺍل ﺍﻟﺘـﻲ ﺘﻜـﻭﻥ ﺼـﻭﺭ‬، [2] ‫( ﻓﻲ‬caldas) ‫ ﺍﻟﻤﻘﺩﻤﺔ ﻤﻥ ﻗﺒل‬contra ‫ﻤﻥ ﺍﻟﻨﻤﻁ‬ . θ ‫ ﺸﺒﻪ ﻤﻔﺘﻭﺤﺔ ﻤﻥ ﺍﻟﻨﻤﻁ‬، θ ‫ﺍﻟﻤﺠﻤﻭﻋﺎﺕ ﺸﺒﻪ ﺍﻟﻤﻐﻠﻘﺔ ﻤﻥ ﺍﻟﻨﻤﻁ‬ ABSTRACT The concept of θ-semi-open sets in topological spaces was introduced in 1984 and 1986 by T. Noiri [9, 10]. In this paper we introduce and study a generalization of a contra pre semi-open maps due to (Caldas and Baker) [3], it is called contra pre θs-open maps, the maps whose images of a θ-semi-open sets is θ-semi-closed. Also, we introduce and study a new type of closed maps called contra pre θs-closed maps, which is stronger than contra pre semi-closed due to Caldas [2], the maps whose image of a θsemi-closed sets is θ-semi-open.1991 Math. Subject Classification: 54 C10, 54 D 10. Keywords and Phrases: θ-semi-open sets, Contra pre θs-open and Contra pre θs-closed maps. 59 Abdullah M. Abdul-Jabbar 1. Introduction The concept of θ-semi-open set in topological spaces was introduced in 1984 and 1986 by T. Noiri [9, 10], which depends on semi-open sets due to N. Levine [8]. When semi-open sets are replaced by θ-semi-open sets, new results are obtained. M. Caldas and C. Baker defined and studied the concept of contra pre semi-open maps [3], where the maps whose images of semi-open sets are semi-closed. In this direction we shall define the concept of Pre θs-open maps. In this paper we introduce two new types of open and closed maps called contra pre θs-open and contra pre θs-closed maps via the concept of θsemi-open sets and study some of their basic properties. We also establish relationships a mong these maps with other types of continuity, openness and closedness. 2. Preliminaries Throughout the present paper, spaces always mean topological spaces on which no separation axiom is assumed unless explicitly stated. Let S be a subset of a space X. The closure and the interior of S are denoted by Cl(S) and Int(S), respectively. A subset S is said to be regular open(resp. semi-open[8]) if S = Int(Cl(S)) (resp. S ⊂ Cl(Int((S))). A subset S is said to be θ-semi-open [9] if for each x ∈ S, there exists a semi-open set U in X such that x ∈ U ⊂ Cl(U) ⊂ S. The complement of each regular open (resp. semi-open and θ-semi-open) set is called regular closed (resp. semi-closed and θ-semi-closed). The family of all semi-open (resp. semi-closed, θ-semiopen and θ-semi-closed) sets of X is denoted by SO(X) (resp. SC(X), θSO(X) and θSC(X)). A point x is said to be in the θ-semi-closure [10] of S, denoted by sClθ(S), if S ∩ Cl(U) ≠ φ for each U∈SO (X) containing x. If S = sClθ(S), then S is called θ-semi-closed. A point x is said to be in the θsemi-interior [10] of S denoted by sIntθ(S), if Cl(U) ⊂ S for some U ∈ SO (X) containing x. If S = sIntθ(S), then S is called θ-semi-open. For each U ∈SO (X), Cl(U) is θ-semi-open and hence every regular closed set is θsemi-open. Therefore, x∈sClθ(S) if and only if S ∩ A ≠ φ for each θ-semiopen set A containing x. A function f : X→Y is said to be contra pre semiopen [3] (resp. contra pre semi-closed [2]) if for each semi-open (resp. semiclosed) set U of X, f (U) ∈ SC(Y) (resp. f (U) ∈ SO(Y)). 3. Contra pre θs-open and contra pre θs-closed maps Let f : (X, τ) →(Y, σ) be a map from a topological space (X, τ) into a topological space (Y, σ). 60 A Generalization of A Contra… Definition 3.1: A map f : (X, τ) →(Y, σ) is said to be contra pre θs-open (resp. contra pre θs-closed) if f (A) is θ-semi-closed (resp. θ-semi-open ) in (Y, σ), for each set A ∈ θSO(X, τ) ( resp. A ∈ θSC(X, τ)). The proof of the following two lemmas follows directly from their definitions and, therefore, they are omitted. Lemma 3.1: Every contra pre semi-open map is contra pre θs-open. Lemma 3.2: Every contra pre θs-closed map is contra pre semi-closed. The converse of the above lemmas is not true in general as it is shown by the following examples. Example 3.1: Let X = {a, b, c} and τ = {φ, X, {a}, {b}, {a, b}, {a, c}}. Then the family of all semi-open subsets of X with respect to τ is: SO(X) = {φ, X, {a}, {b}, {a, b}, {a, c}} and the family of all θ-semi-open subsets of X with respect to τ is θSO(X) = {φ, X, {b}, {a, c}}. The identity map f : (X, τ) → (X, τ) is contra pre θs-open map, but it is not contra pre semiopen maps. Example 3.2: Let X = {a, b, c} and τ = {φ, X, {a}, {b}, {a, b}}. Then, the family of all semi-open subsets of X with respect to τ is: SO(X) = {φ, X, {a}, {b}, {a, b}, {a, c}, {b, c}} and the family of all θ-semiopen subsets of X with respect to τ is : θSO(X) = {φ, X, {a, c}, {b, c}}. Define a function f : (X, τ) → (X, τ) as follows: f (a) =b, f (b) = f (c) = a. Then f is contra pre semi-closed, but it is not contra pre θs-closed. Remark 3.1: Contra pre θs-openness and contra pre θs-closedness are equivalent if the map is bijective. Theorem 3.1: For a map f : X → Y the following are equivalent: i) f is contra pre θs-open; ii) for every subset D of Y and for every θ-semi-closed subset G of X with f f -1 -1 (D) ⊂ G, there exists a θ-semi-open subset B of Y with D ⊂ B and (B) ⊂ G; iii) for every y ∈Y and for every θ-semi-closed subset G of X with 61 Abdullah M. Abdul-Jabbar f -1(y) ⊂ G, there exists a θ-semi-open subset B of Y with y∈B and -1 f (B) ⊂ G. Proof: (i)⇒(ii). Let D be a subset of Y and let G be a θ-semi-closed subset of X with f -1(D) ⊂ G. Set, B = Y \ f (X \ G). Since f is contra pre θsopen, then B is a θ-semi-open set of Y and since f (D) ⊂ G we have -1 f (X \ G) ⊂ Y \ D and hence D ⊂ B. Also, f -1 (B) = X \ [ f -1 ( f (X \ G))] ⊂ X \ (X \ G) = G. (ii)⇒(iii). It is sufficient, set D = {y}, we get the result. (iii)⇒(i). Let A be a θ-semi-open subset of X with y ∈ Y \ f (A) and let G = X \ A. By(iii) , there exists a θ-semi-open subset By of Y with y ∈By and f -1 (By) ⊂ G. Then, y ∈ By ⊂ Y \ f (A). Hence Y \ f (A) = ∪ { By : y ∈ Y \ f (A)}. Therefore, by [6, Lemma 2.2] that Y \ f (A) is θ-semi-open. Thus, f (A) is a θ-semi-closed subset in Y. Theorem 3.2: For a map f : X → Y the following are equivalent: i) f is contra pre θs-closed; ii) for every subset D of Y and for every θ-semi-open subset A of X with f - 1 (D) ⊂ A, there exists a θ-semi-closed subset H of Y with D ⊂ H and f - 1 (H) ⊂ A. Proof: (i)⇒(ii). Let D be a subset of Y and let A be a θ-semi-open subset of X with f -1 (D) ⊂ A. Set, H = Y \ f (X \ A). Since f is contra pre θs-closed, therefore, H is a θ-semi-closed set of Y and since f -1 we have f (X \ A) ⊂ X \ D and hence D ⊂ H. Also, f (ii)⇒(i). Let G be a θ-semi-closed subset of X. Set, D = Y \ f (G) and let A = X \ G. 62 (D) ⊂ A, -1 (H) ⊂ A. A Generalization of A Contra… Hence f -1 (D) = f -1 (Y \ f (G)) = X \ f -1 ( f (G)) ⊂ X \ G = A. By assumption, there exists a θ-semi-closed set H ⊂ Y for which D ⊂ H and f -1 (H) ⊂ A. It follows that D = H. If y ∈ H and y ∉ D, then y ∈ f (G). therefore, y = f (x) for some x ∈ G and we have x ∈ f -1 (H) ⊂ A = X \ G which is a contradiction. Since D = H, that is, Y \ f (G) = H, which implies that f (G) is θ-semi-open and hence f is contra pre θs-closed. Taking the set D in Theorem 3.2 to be {y} for y ∈ Y we obtain the following result. Corollary 3.1: If f : X → Y is contra pre θs-closed map, then for every y ∈ Y and every θ-semi-open subset A of X with f -1(y) ⊂ A, there exists a θsemi-closed subset H of Y with y ∈ H and f -1(H) ⊂ A. Theorem 3.3: A map f : X → Y is contra pre θs-open if and only if for each x ∈ X and each semi-open set S in X containing x, there exists a θsemi-closed set H in Y containing f (x) such that H ⊂ f (Cl(S)). Corollary 3.2: A map f : X → Y is contra pre θs-open if and only if for each x ∈ X and each θ-semi-open subset A of X containing x, there exists a θ-semi-closed subset H of Y containing f (x) such that H ⊂ f (A). Corollary 3.3: A map f : X → Y is contra pre θs-open, then for each x ∈ X and each regular closed subset R of X containing x, there exists a θ-semiclosed subset H of Y containing f (x) such that H ⊂ f (R). Theorem 3.4: A map f : X → Y is contra pre θs-closed if and only if for each x ∈ X and each θ-semi-closed subset G of X containing x, there exists a semi-open subset W of Y containing f (x) such that Cl(W) ⊂ f (G). Corollary 3.4: A map f : X → Y is contra pre θs-closed if and only if for each x ∈ X and each θ-semi-closed subset G of X containing x, there exists a θ-semi-open subset B of Y containing f (x) such that B ⊂ f (G). Theorem 3.5: For a map f : X → Y, the following are equivalent: a) f is contra pre θs-open; b) f (sIntθ (A)) ⊂ sClθ ( f (A)) for each subset A of X; 63 Abdullah M. Abdul-Jabbar c) sIntθ ( f -1 (B)) ⊂ f –1 (sClθ (B)) for each subset B of Y; d) f –1 (sIntθ (B)) ⊂ sClθ ( f –1 (B)) for each subset B of Y. Proof: (a)⇒(b). Suppose f is contra pre θs-open maps and A ⊂ X. Since sIntθ (A) ⊂ A, f (sIntθ (A)) ⊂ f (A) and hence f (sIntθ (A)) ⊂ sClθ ( f (A)). (b)⇒(c). Let B be any subset of Y. Then f –1 (B) ⊂ X. Therefore, we apply (b), we obtain f (sIntθ ( f –1 (B))) ⊂ sClθ ( f ( f –1 (B))) ⊂ sClθ (B). Thus, sIntθ ( f -1 (B)) ⊂ f –1 (sClθ (B)). (c)⇒(d). In (c), we take Y \ B instead of B, we get sIntθ ( f –1 (Y \ B)) ⊂ f –1 (sClθ (Y \ B)). Then, sIntθ (X \ f –1 (B)) ⊂ f –1 ( Y \ sClθ (B)), which implies that X \ sClθ ( f –1 (B)) ⊂ X \ f –1 (sIntθ (B)). Hence f –1 (sIntθ (B)) ⊂ –1 sClθ ( f (B)). (d)⇒(a). Let A be any θ-semi-open subset of X and set B = Y \ f (A) = f (X \ A). By (d), f –1 (sIntθ ( ( f (X \ A))) ⊂ sClθ ( f –1 ( f (X \ A))) = sClθ (X \ A) = X \ A. Therefore, f (X \ A) = Y \ f (A) is θ-semi-open and hence f (A) is θ-semi-closed subset of Y. Thus, f is contra pre θs-open map. The proof of the following theorem is similar to the above theorem for the contra pre θs-closed maps. Theorem 3.6: For a map f : X → Y, the following are equivalent: a) f is contra pre θs-closed; b) f (sClθ (A)) ⊂ (sIntθ f (A)) for each subset A of X; c) sClθ ( f -1 (B)) ⊂ f –1 (sIntθ (B)) for each subset B of Y; d) f –1 (sClθ (B)) ⊂ sIntθ ( f –1 (B)) for each subset B of Y. Theorem 3.7: Let f : (X, τ) → (Y, σ) be a map. Then, i) If f is contra pre θs-open, then sClθ ( f (A)) ⊂ f (sClθ (A)) for every θsemi-open subset A of X. ii) If f is contra pre θs-closed, then f (A) ⊂ sIntθ ( f (sClθ (A))) for every subset A of X. Proof: i) Since f is contra pre θs-open, then sClθ ( f (A)) = f (A) ⊂ f (sClθ (A)) for every A ∈ θSO(X, τ). ii) Since f is contra pre θs-closed and since sClθ (A) is θ-semi-closed, then f (A) ⊂ f (sClθ (A)) = sIntθ ( f (sClθ (A))) for every subset A of X. A map f : (X, τ) → (Y, σ) is said to be pre θs-open, if f (A) is θsemi-open in (Y, σ), for every A ∈ θSO(X, τ). 64 A Generalization of A Contra… Recall, that a map f : (X, τ) → (Y, σ) is called S-closed [4] if sClθ ( f (A)) ⊂ f (sClθ (A)) for every subset A of X. Theorem 3.8: For a map f : (X, τ) → (Y, σ) ,the following properties hold, i) f is S-closed, whenever f is contra pre θs-closed and sClθ (sIntθ ( f (A))) ⊂ f (A) for every θ-semi-closed set A of X. ii) f is pre θs-open, whenever f is contra pre θs-open and f (A) ⊂ sIntθ (sClθ ( f (A))) for every θ-semi-open set A of X. Proof: i) Let G be a θ-semi-closed subset of X. Since sClθ (sIntθ ( f (G))) ⊂ f (G) and f (G) is θ-semi-open, then sClθ (sIntθ ( f (G))) = sClθ ( f (G)) ⊂ f (G). So, by [1, Remark 1.2.6], f (G) is θ-semi-closed. Therefore, by [10, Theorem 3.1], f is S-closed map. ii) Let A be a θ-semi-open subset of X. But f (A) ⊂ sIntθ (sClθ (f (A))) and f (A) is θ-semi-closed, then sIntθ (sClθ ( f (A))) = sIntθ ( f (A)) and hence f (A) ⊂ sIntθ ( f (A)). Therefore, f (A) = sIntθ ( f (A)). So, by [1, Proposition 1.2.2(4)], f (A) is θ-semi-open. Lemma 3.3[7]: If Y is a regular closed subset of a space X and A ⊂ Y, then A is θ-semi-open in X if and only if A is θ-semi-open in Y. Regarding the restriction f | R of a map f : (X, τ) → (Y, σ) to a subset R of X we have the following: Theorem 3.9: If f : (X, τ) → (Y, σ) is contra pre θs-open and R is a regular closed set of (X, τ), then the map f | R : (R, τ R) → (Y, σ) is also contra pre θs-open. Proof: Let A be a θ-semi-open set in the subspace R. Since R is regular closed in X, then by Lemma 3.3, A is θ-semi-open set in X. Since f is contra pre θs-open. Therefore, f (A) is θ-semi-closed in Y. Thus, f | R is contra pre θs-open map. The proof of the following result is not hard, therefore, it is omitted. Theorem 3.10: Let f : (X, τ) → (Y, σ) and g: (Y, σ) → (Z, γ) be two maps such that g o f : (X, τ) →(Z, γ). Then, a) g o f is contra pre θs-open, if f is pre θs-open and g is contra pre θsopen. b) g o f is contra pre θs-open, if f is contra pre θs-open and g is S-closed. 65 Abdullah M. Abdul-Jabbar c) g o f is contra pre θs-closed, if f is S-closed and g is contra pre θsclosed. d) g o f is contra pre θs-closed, if f is contra pre θs-closed and g is pre θsopen. Recall, that a map f : (X, τ) → (Y, σ) is S-continuous [10], if and only if for each θ-semi-open subset A of Y, f -1 (A) is θ-semi-open in X. Theorem 3.11: Let f : (X, τ) → (Y, σ) and g: (Y, σ) → (Z, γ) be two maps such that g o f : (X, τ) →(Z, γ) is contra pre θs-closed. a) If f is S-continuous surjection, then g is contra pre θs-closed. b) If g is S-continuous injection, then f is contra pre θs-closed. Proof: a) Suppose G is any arbitrary θ-semi-closed set in Y. Since f is Scontinuous. Therefore, by [10, Theorem 1.1], f -1 (G) is θ-semi-closed in X. Since g o f is contra pre θs-closed and f is surjective (g ο f ) ( f –1 (G)) = g (G) is θ-semi-open in Z. This implies that g is a contra pre θs-closed map. b) Suppose G is any arbitrary θ-semi-closed set in X. Since g o f is contra pre θs-closed, (g ο f )(G) is θ-semi-open in Z. Since g is S-continuous injection, g –1 ( (g ο f ))(G) = f (G) is θ-semi-open in Y. This implies that f is a contra pre θs-closed map. Arguing as in the proof of Theorem 3.11, we obtain the following result. Theorem 3.12: Let f : (X, τ) → (Y, σ) and g: (Y, σ) → (Z, γ) be two maps such that g o f : (X, τ) →(Z, γ) is contra pre θs-open. a) If f is S-continuous surjection, then g is contra pre θs-open. b) If g is S-continuous injection, then f is contra pre θs-open. Lemma 3.4[10]: Let (X, τ) be a topological space and D be a subset of X. Then x ∈ sClθ (D) if and only if for every θ-semi-open A of x such that A ∩ D ≠ φ. Definition 3.2[5]: A subset D of a topological space (X, τ) is called θ-semidense if sClθ (D) = X. 66 A Generalization of A Contra… Theorem 3.13: For a map f : (X, τ) → (Y, σ), the following properties hold: a) If f is contra pre θs-open and B ⊂ Y has the property that B is not contained in proper θ-semi-open sets, then f –1 (B) is θ-semi-dense in X. b) If f is contra pre θs-closed and A is θ-semi-dense subset of Y, then f –1 (A) is not contained in a proper θ-semi-dense set. Proof: a) Let x ∈ X and let A be a θ-semi-open subset of X containing x. Then f (A) is θ-semi-closed and Y \ f (A) is a proper θ-semi-open subset of Y. Thus, B ⊄Y \ f (A) and hence there exists y ∈ B such that y ∈ f (A). Let z ∈ A for which y = f (z). Then z ∈ A ∩ f –1 (B). Hence A ∩ f –1 (B) ≠ φ and thus by Lemma 3.4, x ∈ sClθ ( f –1 (B)). Hence f –1 (B) is θ-semi-dense in X. b) Assume that f –1 (A) ⊂ O where O is a proper θ-semi-open subset of X. Then, we have that f (X \ O) is a non-empty θ-semi-open set such that f (X \ O) ∩ A = φ, which a contradicts the fact that A is θ-semi-dense. Lemma 3.5[6]: Let X1 and X2 be two topological spaces and X = X1 × X2. Let Ai ∈ θSO(Xi) for i = 1,2, then A1 × A2 ∈ θSO(X1 × X2). Definition 3.3[7]: A space X is said to be strongly semi-T2 if and only if for each two distinct points x and y in X, there exists two disjoint θ-semi-open sets A and B in X containing x and y, respectively. Theorem 3.14: If X is a strongly semi-T2 space and f :X→Y is contra pre θs-open map, then the set A = {(x1, x2) : f (x1) = f (x2)} is θ-semi-closed in the product space X × X. 67 Abdullah M. Abdul-Jabbar REFERENCES [1] Abdul-Jabbar A. M. (2000) “θs-Continuity, Openness and Closed graphs in topological spaces”, M. Sc. Thesis, College of Science, Salahaddin- Erbil Univ. [2] Caldas M., “Weak and strong forms of irresolute maps”, Internat J. Math. & Math. Sci. (to appear). [3] Caldas M. and C. W. 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