4_aljarrah.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 5, No. 2, 2012, 129-140 ISSN 1307-5543 – www.ejpam.com On ωβ−Continuous Functions Heyam Hussein Aljarrah∗, Mohd Salmi Md Noorani School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 UKM Bangi, Selangor, MALAYSIA Abstract. A subset A of topological space (X ,τ) is said to be ωβ−open [3] if for every x ∈ A there exists an β−open set U containing x such that U − A is a countable. In this paper, we introduce and study new class of function which is ωβ−continuous functions by using the notion of ωβ−open sets. This new class of function defines as a function f : (X ,τ)→ (Y,σ) from a topological space (X ,τ) into a topological space (Y,σ) is ωβ−Continuous function if and only if for each x ∈ X and each open set V in (Y,σ) containing f (x) there exists an ωβ−open set U containing x such that f (U)⊆ V . We give some characterizations of ωβ−Continuous functions, define ωβ−irresolute and ωβ−open function. Finally, we find relationship between these type of function. 2010 Mathematics Subject Classifications: 54C05, 54C08, 54C10 Key Words and Phrases: ωβ−open set, ωβ−Continuous, ωβ−irresolute, ωβ−open functions, ωβ−cloesd functions. 1. Introduction Throughout the present paper, a space mean topological space on which no separation axiom is assumed unless explicitly stated. Let A be a subset of a space (X ,τ). The closure of A and interior of A in (X ,τ) are denoted by Int(A) and cl(A), respectively. A subset A of a space (X ,τ) is said to be b−open [4], (reps. β−open [7]) if A ⊆ Int(cl(A)) ∪ cl(Int(A)), (resp. A⊆ cl(Int(cl(A)))). Recall that a subset A of a space (X ,τ) is said to be ωβ−open [3] (resp. ωb−open [9], ω−open [2]) set if for every x ∈ A there exists an β−open (resp. b−open, open) set U containing x such that U − A is a countable. We write ωβO(X ,τ) (resp. ωbO(X ,τ), βO(X ,τ), ωO(X ,τ), bO(X ,τ)) to denote the family of all ωβ−open (resp. ωb−open, β−open, ω−open, b−open) subsets of (X ,τ). ∗Corresponding author. Email addresses: hiamaljarah�yahoo. om (H. Aljarrah), msn�ukm.my (M. Noorani) http://www.ejpam.com 129 c© 2012 EJPAM All rights reserved. H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 130 Definition 1. A function f : (X ,τ)→ (Y,σ) is called ω−continuous [6] (resp. ωb−continuous [9]) if for every x ∈ X and each open set V in (Y,σ) containing f (x) there exists an ωO(X ,τ) (resp. ωbO(X ,τ)) set U containing x such that f (U)⊆ V . Lemma 1. [3] Let (X ,τ) be a topological space: i. The union of any family of ωβO(X ,τ) sets is ωβO(X ,τ). ii. The intersection of an ωβO(X ,τ) set and open set is ωβO(X ,τ). Theorem 1. [3] Let (Y,τY )be a subspace of (X ,τ), A⊆ Y and Y is βO(X ,τ) sets. Then A∈ωβO(X ,τ) if and only if A∈ωβO(Y,τY ). Theorem 2. [3] Let A be a subset of a topological space (X ,τ). Then x ∈ωβ cl(A) if and only if for every ωβO(X ,τ) set U containing x, A∩ U 6= φ. Theorem 3. [5] If f : (X ,τ)→ (Y,σ) is an open continuous function, then f −1(cl(A)) =cl( f −1(A)). 2. ωβ−Continuous Functions Definition 2. A function f : (X ,τ)→ (Y,σ) is called ωβ−continuous at a point x ∈ X , if for every open set V in (Y,σ) containing f (x) there exists an ωβO(X ,τ) set U containing x such that f (U) ⊆ V . If f is ωβ−continuous at each point of X then f is said to be ωβ−continuous on X . Definition 3. Let (X ,τ) be any space, a set A⊆ X is said to be ωβ−neighborhood of a point x in X if and only if there exists a ωβO(X ,τ) set U containing x such that U ⊆ A. Theorem 4. Let f : (X ,τ)→ (Y,σ) be a function, where X and Y are topological space. Then the following are equivalent: i. The function f is ωβ−continuous. ii. For each open set V ⊂ Y , f −1(V ) is ωβO(X ,τ). iii. For each x ∈ X , the inverse of every neighborhood of f (x) is an ωβ−neighborhood of x. iv. For each x ∈ X and each neighborhood Nx of f (x), there is an ωβ−neighborhood V of x such that f (U)⊆ Nx . v. For each closed set M ⊂ Y , f −1(M) is ωβ−closed in X . vi. For each subset A⊂ X , f (ωβ cl(A)) ⊂ cl( f (A)). vii. For each subset B ⊂ Y , ωβ cl( f −1(B))⊆ ( f −1(cl(B))). H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 131 Proof. (i → ii) Let V be open in Y and x ∈ f −1(V ) then f (x) ∈ V , by (i), there ex- ists an ωβO(X ,τ) set Ux in X containing x and f (Ux) ⊆ V . Then x ∈ Ux ⊆ f −1(V ) and hence f −1(V ) = ∪ x∈ f −1(V ) Ux . By Lemma 1(i), f −1(V ) ∈ ωβO(X ,τ), which implies that f is ωβ−continuous. (ii→ iii) For x ∈ X , let V be the neighborhood of f (x) then f (x) ∈W ⊆ V , where W is open in Y . By (ii), f −1(W ) ∈ωβO(X ,τ), and x ∈ f −1(W )⊆ f −1(V ). Then by Definition 3, f −1(V ) is ωβ−neighborhood of x . (iii→ iv) For x ∈ X and Nx be a neighborhood of f (x). Then V = f −1(Nx) is anωβ−neighborhood of x and f (V ) = f ( f −1(Nx))⊂ Nx . (iv → v) For any x ∈ X − f −1(M), f (x) ∈ Y − M . Since M is closed, the set Y − M is neighborhood of f (x), hence there is a ωβ−neighborhood V of x such that f (V ) ⊂ Y −M , there exists an ωβO(X ,τ) set Ux in X containing x and Ux ⊆ V ⊆ X − f −1(M), take (X − f −1(M)) = ∪ x∈ f −1(Y−M) Ux . By Lemma 1(i), the set (X − f −1(M)) ∈ ωβO(X ,τ), which implies f −1(M) is ωβC(X ,τ). (v→ vi) Let A⊆ X , since cl( f (A)) is a closed set in Y by (vi), f −1(cl( f (A))) is an ωβC(X ,τ) set containing A, then f (ωβ cl(A)) ⊂ cl( f (A)). (vi→ vii) Let B ⊂ Y . By (vi), f (ωβ cl( f −1(B)))⊆ cl(B), so ωβ cl( f −1(B)) ⊆ f −1(cl(B)). (vii → i) Suppose on the contrary that f is not ωβ−continuous. So there exist x ∈ X and V ∈ σ with f (x) ∈ V such that for all ωβO(X ,τ) sets U with x ∈ U and f (U) 6⊂ (V ) i.e. f (U) ∩ (Y − V ) 6= φ. Therefore, by Theorem 2, x ∈ ωβ cl( f −1(Y − V )) and so by (vii), f (x) ∈ cl(Y − V ), thus for all open sets V in (Y,σ) containing f (x), the set V ∩ (Y − V ) 6= φ, a contradiction. Therefore, f is ωβ−continuous. Definition 4. For any subset A of a topological space (X ,τ) the frontier of A, denoted byωβFr(A), is define as ωβ cl(A)∩ωβ cl(X − A). Theorem 5. Let (X ,τ), (Y,σ) be a topological space and f : (X ,τ) → (Y,σ) be a function. Then X −ωβ c( f ) = ∪{ωβFr ( f −1(V )) : V ∈ σ, f (x) ∈ V, x ∈ X } where ωβ c( f ) denotes the set of points at which f is ωβ−continuous. Proof. Let x ∈ X −ωβ c( f ). Then for every ωβO(X ,τ) set U containing x there exists open sets V in (Y,σ) containing f (x) such f (U) 6⊂ V , Hence U ∩ (X − f −1(V )) 6= φ for every ωβO(X ,τ) set U containing x . Therefore, by Theorem 2 x ∈ωβ cl(X − f −1(V )). Then x ∈ f −1(V )∩ωβ cl(X − f −1(V ))⊆ωβFr( f −1(V )). Hence, X −ωβ c( f ) ⊆ ∪{ωβFr ( f −1(V )), V ∈ σ, f (x) ∈ V, x ∈ X }. Conversely, let x /∈ X −ωβ c( f ). Then for each open sets V in (Y,σ) containing f (x), f −1(V ) is ωβO(X ,τ) containing x , thus for every V ∈ σ containing f (x), x ∈ ωβ Int( f −1(V )) and hence x /∈ ωβFr( f −1(V )). So ∪{ωβFr ( f −1(V )) : V ∈ σ, f (x) ∈ V, x ∈ X } ⊆ X −ωβ c( f ). Corollary 1. A function f : (X ,τ)→ (Y,σ) is ωβ−continuous if and only if f −1(int(G))⊆ωβ int( f −1(G)), for any subset G ⊆ Y . Proof. NECESSITY. Let G be any subset of Y . Since f is ωβ−continuous, f −1(int(G)) is ωβO(X ,τ) set. As f −1(int(G))⊆ f −1(G), then f −1(int(G))⊆ωβ int( f −1(G)). H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 132 SUFFICIENCY. Let x ∈ X and V ∈ σ with f (x) ∈ V . Then x ∈ f −1(V ) and so by assump- tion x ∈ ωβ Int( f −1(V )). There exists an ωβO(X ,τ) such that x ∈ U ⊆ f −1(V ). Hence f (x) ∈ f (U)⊆ V and the result follows. Note that if X is a countable set then every function f : (X ,τ)→ (Y,σ) isωβ−continuous. The following diagram follows immediately from the definitions in which none of the impli- cations is reversible. continuous → b−continuous → β−continuous ↓ ↓ ↓ ω−continuous → ωb−continuous → ωβ−continuous Example 1. Let X = R with the topology τ = τu and Y = {0,1} with the topology σ = {φ, Y, {0}}. Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( 1 x ∈ R−Q 0 x ∈Q Then f is ωβ−continuous but it is neither continuous nor ω−continuous. Example 2. Let X = {1,2,3} with the topology τ = {X ,φ, {1}, {2}, {1,2}} and Y = {a, b} with the topology σ = {φ, Y, {a}}. Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( b x = {1,2} a x = 3 Then f is not β−continuous, but it can be easily seen that f is ωβ−continuous. Example 3. Let X = R with the topology τ = τu and Y = {a, b} with the topology σ = {φ, Y, {a}}. Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( a x ∈ [0,2)∩R−Q b x ∈ [0,2)∩Q Then f is ωβ−continuous, but it is not ωb−continuous. Proposition 1. If f : (X ,τ)→ (Y,σ) is an ωβ−continuous function and A is an open set in X , then the restriction f |A : (A,τA)→ (Y,σ) is ωβ−continuous. Proof. Since f is an ωβ−continuous, for any open set V in Y , f −1(V ) is a ωβO(X ,τ) set. Hence by Lemma 1(ii), f −1(V )∩A is a ωβO(X ,τ) since A is an open set. Therefore, by Theo- rem 1, ( f |A)−1(V ) = f −1(V )∩A is ωβO(A,τA) sets, which implies that f |A is ωβ−continuous function. Observe that the above theorem is not true if A were taken to be βO(X ,τ) sets or ωO(X ,τ), as it shown in the next examples. H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 133 Example 4. Let X = R with the topology τ = τcoc and Y = {0,1} with the topology σ = {φ, Y, {1}}. Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( 1 x ∈ (0,1] 0 x /∈ (0,1] It can be easily seen that f is ωβ−continuous. We take A= (0,1]. Then A ∈ βO(X ,τ) and f |A is not ωβ−continuous since ( f |A)−1(1) = {1} /∈ωβO(A,τA) . Example 5. Let X = R with the topology τ = τu and Y = {0,1} with the topology σ = {φ, Y, {1}}. Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( 1 x = p 2 0 x ∈Q It can be easily seen that f is ωβ−continuous. We take A= R−Q. Then A∈ ωO(X,τ) and f |A is not ωβ−continuous since ( f |A)−1(Y ) = {p2} /∈ωβO(A,τA). Definition 5. [7] A cover υ = {Uα : α ∈ ∆} of subset of X is called a βO(X ,τ) cover if Uα is βO(X ,τ) for each α ∈∆. Now we prove the following proposition. Proposition 2. Let f : (X ,τ)→ (Y,σ) be any function and A = {Aα : α ∈ ∆} be a βO(X ,τ) cover of X . If the restriction, f |Aα : (Aα,τAα )→ (Y,σ) is ωβ−continuous for each α ∈ ∆, then f is ωβ−continuous. Proof. Let V be any open set in Y . Since f |Aα is ωβ−continuous, then for each α ∈∆, we have( f |A)−1(V ) = f −1(V )∩Aα ∈ωβO(Aα,τAα ). So by Theorem 1, f −1(V )∩Aα ∈ωβO(X ,τ) for each α ∈∆. Take f −1(V ) = ∪ α∈∆ ( f −1(V )∩ Aα). By Lemma 1(i) f −1(V ) ∈ωβO(X ,τ). Corollary 2. Let f : (X ,τ) → (Y,σ) be any function and A = {Aα : α ∈ ∆} a open cover of X . If the restriction, f |Aα : (Aα,τAα ) → (Y,σ) is ωβ−continuous for each α ∈ ∆, then f is ωβ−continuous. The composition g ◦ f : (X ,τ)→ (Z ,ρ) of a continuous function f : (X ,τ)→ (Y,σ) and an ωβ−continuous function g : (Y,σ) → (Z ,ρ) is not necessarily ωβ−continuous function as the following example shows. Thus, the composition of ωβ−continuous functions need not be ωβ−continuous. Example 6. Let X = R with the topology τ = τcoc, Y={1,2} with the topology σ = {φ, Y, {1}} and Z = {a, b} with the topology ρ = � φ, Z , {a} . Let f : (X ,τ) → (Y,σ) be the function defined by f (x) = ( 1 x ∈ R−Q 2 x ∈Q H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 134 and g : (X ,σ)→ (Y,ρ) be the function defined by g(x) = ( a x = 2 b x = 1 Then f is continuous ( hence ωβ−continuous) and g is ωβ−continuous. However g ◦ f is not ωβ−continuous, because (g ◦ f )−1({a}) = Q /∈ωβO(X ,τ). Proposition 3. If f : (X ,τ)→ (Y,σ) is ωβ−continuous and g : (Y,σ)→ (Z ,ρ) is continuous, then g ◦ f : (X ,τ)→ (Z ,ρ) is ωβ−continuous. Proof. Let x ∈ X and V ∈ ρ with (g ◦ f )(x) ∈ V and f (x) ∈ Y , since g is continuous, there exists open sets W in (Z ,ρ) with f (x) ∈ W and g(W ) ⊆ V . Moreover f is ωβ−continuous there existsωβO(X ,τ) say U containing x such that f (U)⊆W . Now (g ◦ f )(U)⊆ g(W )⊆ V . We note that Proposition 3 is not true if g is assumed to be only ω−continuous or β−continuous as it is shown in the next example. Example 7. Consider X = R with the topology τ = τcoc, Y = {a, b, c} with the topology σ = {φ, Y, {a}, {b}, {a, b}} and Z = {1,2,3,4}with the topologyρ = {φ, Z , {1}, {1,2}, {1,2,3}}. Let f : (X ,τ)→ (Y,σ) be the function define by f (x) = ( a x ∈ R−Q c x ∈Q and g : (Y,σ)→ (Z ,ρ) be the function define by g(x) =    1 x = a 3 x = b 2 x = c Then f is ωβ−continuous, g is ω−continuous and β−continuous function but g ◦ f is not ωβ−continuous since (g ◦ f )−1(2) = Q /∈ωβO(X ,τ). Corollary 3. If f : (X ,τ)→ ∏ α∈∆ Xα is an ωβ−continuous function from a space (X ,τ) into a product space ∏ α∈∆ Xα, then Pα ◦ f is ωβ−continuous for each α ∈ ∆, where Pα is the projection function from the product space ∏ α∈∆ Xα onto the space Xα for each α ∈∆. Theorem 6. Let X and Y be a topological spaces, let f : (X ,τ) → (Y,σ) be a function and g : (X ,τ)→ (X × Y,τ×σ) be the graph function of f given by g(x) = (x , f (x)) for every point x ∈ X . Then g is ωβ−continuous if and only if f is ωβ−continuous. Proof. Assume that g is ωβ−continuous. Now f = PY ◦ g where PY : X ×Y → Y , then f is ωβ−continuous by Corollary 3. Conversely, assume that f is ωβ−continuous. Let x ∈ X and H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 135 W be any open set in X × Y containing g(x). Then there exist open sets U ⊆ X and V ⊆ Y such that g(x) = (x , f (x)) ∈ U × V ⊆W . Since f is ωβ−continuous, there exists ωβO(X ,τ) sets U1 in containing x such that f (U1) ⊆ V . Put H = U ∩ U1. Then H ∈ ωβO(X ,τ), by Lemma 1(ii), such that x ∈ H and f (H)⊆ V . Therefore we have g(H)⊆ U × V ⊆W . Thus g is ωβ−continuous. Definition 6. [8] A function f : (X ,τ)→ (Y,σ) is called pre-semi-preopen if the image of each semi-preopen set in X is a semi-preopen set in Y . Theorem 7. Let f : (X ,τ)→ (Y,σ) be an pre-semi-preopen surjection and let g : (Y,σ)→ (Z ,ρ) such that g◦ f : (X ,τ)→ (Z ,ρ) isωβ−containuous, then g isωβ−containuous. Proof. At first we show if f : (X ,τ)→ (Y,σ) be an pre-semi-preopen function and U ∈ ωβO(X ,τ), then f (U) ∈ ωβO(Y,σ). So let U ∈ ωβO(X ,τ) then for all x ∈ U there exists βO(X ,τ) sets U1 in (X ,τ) containing x and U1 − U ⊆ C where C is a countable set. Thus f (U1)− f (U)⊆ f (C) where f (C) is a countable set. This implies f (U) ∈ωβO(Y,σ). Now, Let y ∈ Y and let V ∈ ρ with g(y) ∈ V . Choose x ∈ X such that f (x) = y. Since g ◦ f is ωβ−continuous there exists U ∈ ωβO(X ,τ) with x ∈ U and g( f (U)) ⊆ V . But f is pre-semi-preopen function therefore, by assumption, f (U) ∈ ωβO(Y,σ) with f (x) ∈ f (U). So we get the result. Corollary 4. Let fα : (Xα,τα)→ (Yα,τα) be a function for each α ∈ ∆. If the product function f = ∏ α∈∆ fα : ∏ α∈∆ Xα→ ∏ α∈∆ Yα is ωβ−continuous, then fα is ωβ−continuous. Proof. At first we prove that any projection function is pre-semi-preopen function. Let U ∈ βO(X ,τ) hence f (U) ⊆ f (cl(int(cl(U)))), by using the assumption that f is open and continuous surjective, f (U)⊆ cl(int(cl( f (U)))). Thus f (U) ∈ βO(Y,σ). Now For each β ∈∆, let pβ : ∏ α∈∆ Xα→ Xβ and qβ : ∏ α∈∆ Yα→ Yβ be the projections, then we have qβ ◦ f = fβ ◦ pβ for each β ∈ ∆. Since f is ωβ−continuous and qβ is continuous, by Propo- sition 3 qβ ◦ f is ωβ−continuous and hence fβ ◦ pβ is ωβ−continuous function. Since pβ is pre-semi-preopen function it follows from Theorem 7 that fβ is ωβ−continuous function. Theorem 8. [3] For any space X , the following properties are equivalent: i. X is β−Lindelőf. ii. Every ωβO(X ,τ) cover of X has a countable subcover. Proposition 4. Let f : (X ,τ) → (Y,σ) be an ωβ−continuous surjective function. If X is β−Lindelőf, then Y is Lindelőf. Proof. Let {Vα : α ∈∆} be an open cover of Y . Then { f −1(Vα) : α ∈∆} isωβO(X ,τ) cover of X (since f is ωβ− continuous). Since X is β−Lindelőf, by Theorem 8, X has a countable subcover, say f −1(Vα1 ), f −1(Vα2 ), . . . , f −1(Vαn ), . . ., Thus Vα1 , Vα2 , . . . , Vαn , . . . is a subcover of {Vα : α ∈∆} of Y . This shows that Y is Lindelőf. Corollary 5. Let f : (X ,τ)→ (Y,σ) be a β−continuous (or ω−continuous) surjective function. If X is β−Lindelőf, then Y is Lindelőf. H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 136 3. ωβ−Irresolute Functions Definition 7. A function f : (X ,τ)→ (Y,σ) is called ωβ−irresolute if the inverse image of each ωβO(Y,σ) set is an ωβO(X ,τ) set. Note that every ωβ−irresolute function is ωβ−continuous but the converse is not true, which is shown by the following example. Example 8. Let X = R with the topologies τ = τcoc and let Y = {1,2} with the topology σ = � φ, Y, {2} . Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( 1 x ∈Q 2 x ∈ R−Q Then f is ωβ−continuous but not ωβ−irresolute since f −1({1}) = Q /∈ωβO(X ,τ). Theorem 9. Let f : (X ,τ)→ (Y,σ) be a function. Then the following conditions are equivalent: i. The function f is ωβ−irresolute. ii. For each x ∈ X and V ∈ ωβO(Y,σ) containing f (x), there exists U ∈ ωβO(X ,τ) con- taining x and f (U)⊆ V . iii. For each x ∈ X , the inverse of every ωβ−neighbourhood of f (x) is ωβ− neighbourhood of x. iv. For each x ∈ X and ωβ−neighbourhood V of f (x), there exists ωβ−neighbourhood U of x such that f (U)⊆ V . Proof. (i→ ii) Assume x ∈ X and V isωβO(Y,σ) containing f (x), since f isωβ−irresolute then f −1(V ) ∈ωβO(X ,τ) containing x and hence f ( f −1(V ))⊆ V . (ii → iii) Assume x ∈ X and V is ωβ−neighbourhood of f (x), by Definition 3 there ex- ists V1 ∈ ωβO(Y,σ) such that f (x) ∈ V1 ⊆ V , there exists U ∈ ωβO(X ,τ) containing x and f (U) ⊆V1, x ∈ U ⊆ f −1(V1) ⊆ f −1(V ). Hence by use Definition 3, f −1(V ) is ωβ−neighbourhood of x . (iii→ iv) Let V is ωβ−neighbourhood of f (x), by (iii), f −1(V ) is ωβ−neighbourhood of x and f ( f −1(V ))⊆ V . (iv→ i) For each x ∈ X , let V ∈ωβO(Y,σ) containing f (x). Put A= f −1(V ), let x ∈ A. Then f (x) ∈ V . Since V ∈ ωβO(Y,σ) then V is a ωβ−neighbourhood of f (x). So by hypothesis, A= f −1(V ) is ωβ−neighbourhood of x . Hence by Definition 3 there exists Ax ∈ ωβO(X ,τ) such that x ∈ Ax ⊆ A. Thus, by Lemma 1(i) A = ∪ x∈A Ax is ωβO(X ,τ) set. Therefore, f is ωβ−irresolute. Theorem 10. The following conditions are equivalent for a function f : (X ,τ)→ (Y,σ): i. f is ωβ−irresolute. H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 137 ii. For each ωβC(Y,σ) subset C of Y , f −1(C) is ωβC(X ,τ). iii. For each subset A of X , f (ωβ cl(A))⊆ωβ cl( f (A)). Proof. (i → ii) Let C be ωβC(Y,σ) subset of Y . Then X − f −1(C) ∈ ωβO(X ,τ), which implies that f −1(C) is ωβC(X ,τ). (ii → iii) Let A be a subset of X , Since A ⊂ f −1( f (A)), we have A ⊂ f −1(ωβ cl( f (A))). Now by (ii), f −1(ωβ cl( f (A))) is ωβC(X ,τ) set containing A then ωβ cl(A) ⊆ f −1(ωβ cl( f (A))), which implies f (ωβ cl(A))⊆ωβ cl( f (A)). (iii → iv) Let B ⊂ Y , by (iii) f (ωβ cl( f −1(B))) ⊆ ωβ cl( f ( f −1(B))) ⊆ ωβ cl(B), hence ωβ cl( f −1(B)) ⊆ f −1(ωβ cl(B)). (iv → i) Suppose f is not ωβ−irresolute. So there exist x ∈ X and V ∈ ωβO(Y,σ) with f (x) ∈ V such that for allωβO(X ,τ) set U with x ∈ U and f (U) 6⊂ (V ) i.e. f (U)∩(Y−V ) 6= φ. Therefore, by (vii), x ∈ f −1(ωβ cl(Y − V )). So by Theorem 2, f (x) ∈ωβ cl(Y − V ). Thus for all ωβO(Y,σ) sets V containing f (x), so V ∩ (Y − V ) 6= φ, a contradiction. Therefore, f is ωβ−irresolute. Theorem 11. Let f : (X ,τ) → (Y,σ) be a function. Then f is ωβ−irresolute if and only if f −1(ωβ Int(B)) ⊆ωβ Int( f −1(B)). Proof. NECESSITY. Let B be any subset of Y . Since f isωβ−irrrsolute, we have f −1(ωβ Int(B)) is ωβO(X ,τ) set. As f −1(ωβ Int(B)) ⊆ f −1(B), then f −1(ωβ Int(B)) ⊆ωβ Int( f −1(B)). SUFFICIENCY. Let x ∈ X and V ∈ ωβO(Y,σ) with f (x) ∈ V . Then x ∈ f −1(V ) and so by assumption x ∈ ωβ Int( f −1(V )). There exists an ωβO(X ,τ) sets such that x ∈ U ⊆ f −1(V ). Hence f (x) ∈ f (U)⊆ V and the result follows. Proposition 5. If f : (X ,τ)→ (Y,σ) isωβ−irresolute and g : (Y,σ)→ (Z ,ρ) isωβ−continuous, then g ◦ f is ωβ−continuous. Proof. Let x ∈ X and let V be any open set in (Z ,ρ) containing g( f (x)). Since g is ωβ−continuous, there exists an ωβO(Y,σ) set W containing f (x) such that g(W ) ⊆ V . Put U = f −1(W ) since f is ωβ−irresolute, then U ∈ωβO(X ,τ) such that x ∈ U and g( f (U))⊆ g(W )⊆ V . Hence g ◦ f is ωβ−continuous. Corollary 6. If f : (X ,τ)→ (Y,σ) isωβ−irresolute and g : (Y,σ)→ (Z ,ρ) isωb−continuous, then g ◦ f is ωβ−continuous. Recall that a function f : (X ,τ) → (Y,σ) is said to be ω−irresolute [1] if the inverse image of each ωO(Y,σ) set is an ωO(X ,τ). Theorem 12. If f : (X ,τ)→ (Y,σ) isω−irresolute and every βO(Y,σ) set is closed in the space (Y,σ) then f is ωβ−irresolute. Proof. Let U be any ωβO(Y,σ) set, then for all y ∈ Y , there exists βO(Y,σ) sets U1 containing x such that U1 − U is a countable, thus by assumption U1 ⊆ cl(Int(cl(U1)))⊆ Int(U1), so U1 is open sets in (Y,σ), hence U ∈ ωO(Y,σ). Since f is ω−irresolute, then f −1(U) ∈ωO(X ,τ)⊆ωβO(X ,τ). H. Aljarrah , M. Noorani / Eur. J. Pure Appl. Math, 5 (2012), 129-140 138 Proposition 6. Let f : (X ,τ)→ (Y,σ) be an open continuous function and every ωβO(Y,σ) is closed in the space (Y,σ) then f is ωβ−irresolute. Proof. Let U ∈ωβO(Y,σ), by Theorem 3, ωβ cl( f −1(U)) ⊆ cl( f −1(U)) = f −1(cl(U))⊆ f −1(ωβ cl(U)), hence f is ωβ−irresolute, by Theorem 10. In [3], Aljarrah And Noorani define the ωβ − T2 as if for each two distinct point x , y ∈ X , there exists U , V ∈ωβO(X ,τ) such that x ∈ U , y ∈ V and U ∩ V = φ. Theorem 13. If f : (X ,τ) → (Y,σ) is an ωβ−irresolute injective function and the space Y is ωβ − T2, then X is ωβ − T2. Proof. Let x1 and x2 be two distinct points of X . Since f is injective and Y isωβ−T2, there exist V1, V2 ∈ωβO(Y,σ) such that f (x1) ∈ V1, f (x2) ∈ V2 and V1∩V2 = φ. Now x1 ∈ f −1(V1), x2 ∈ f −1(V2) and f −1(V1 ∩ V2) = f −1(V1) ∩ f −1(V2) = φ. Since f is ωβ−irresolute then f −1(V1), f −1(V2) is ωβO(X ,τ). Hence X is ωβ − T2. Definition 8. A space X is said to be ωβ−connected if there exist disjoint ωβO(X ,τ) sets A and B such that A∪ B = X . Proposition 7. If f : (X ,τ)→ (Y,σ) is anωβ−irresolute surjective function and X isωβ−connected, then Y is ωβ−connected. Proof. Suppose Y is not ωβ−connected. Then there exist disjoint ωβO(Y,σ) sets A and B such that A ∪ B = Y . Since f is ωβ−irresolute surjective, f −1(A) and f −1(B) are nonempty ωβO(X ,τ) sets. Moreover f −1(A) ∪ f −1(B) = X . This is show that (X ,τ) is not ωβ−connected, which is a contradiction. Hence (Y,σ) is ωβ−connected. 4. ωβ−Open and ωβ−Closed Functions Definition 9. A function f : (X ,τ)→ (Y,σ) is called ωβ−open (resp. ωβ−closed) if the image of each open (resp. closed) set in (X ,τ) is an ωβO(Y,σ) (resp. ωβC(Y,σ)). Note that every open (closed) function is ωβ−open (resp. ωβ−closed) function, but the converse is not true, which is shown by the following example. Example 9. Let X = {a, b} with the topology τ = {φ, X , {a}} and Y = {1,2,3} with the topology σ = {φ, X , {1}, {2}, {1,2}}. Let f : (X ,τ)→ (Y,σ) be the function define by f (x) = 3 for all x ∈ X . Then f is ωβ−open and ωβ−closed function, but it is neither open nor closed function. Proposition 8. A function f : (X ,τ) → (Y,σ) is ωβ−open if and only if for each x ∈ X and each open set U of X containing x, there exists an ωβO(Y,σ) set W containing f (x) such that W ⊂ f (U). Theorem 14. Let f : (X ,τ)→ (Y,σ) be a function from space (X ,τ) into a space (Y,σ). Then f is ωβ−closed if and only if ωβ cl( f (A))⊆ f (ωβ cl(A)) for each set A subset of (X ,τ). REFERENCES 139 Proof. Let f is ωβ−closed function and A any subset of X . Then f (A) ⊂ f (ωβ cl(A)) ∈ ωβC(Y,σ), it follows that ωβ cl( f (A)) ⊂ f (ωβ cl(A)). Conversely, assume that B ∈ ωβC(X ,τ). Then ωβ cl( f (B)) ⊂ f (ωβ cl(B)) = f (B). Thus we obtain that ωβ cl( f (B)) = f (B), so f is ωβ−closed function. Proposition 9. Let f : (X ,τ)→ (Y,σ) be a continuous surjection function and let g : (Y,σ) → (Z ,ρ) be such that g ◦ f : (X ,τ) → (Z ,ρ) is ωβ−open function, then g is ωβ−open. Proof. Let y ∈ Y and let V ∈ ρ with g(y) ∈ V . Choose x ∈ X such that f (x) = y. Since g ◦ f is ωβ−open function, then g(V ) = g ◦ f ( f −1(V )) ∈ ωβO(Z ,ρ). This is show that g is ωβ−open function. The following examples show that theωβ−open function is independent withωβ−irresolute and ωβ−continuous function. Example 10. Let X = R with the topologies τ = τcoc and let Y = {1,2} with the topology ρ = � φ, Y, {2} . Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( 1 x ∈ R−Q 2 x ∈Q Then f is not ωβ−continuous, but it can easily seen that f (x) is ωβ−open function. Example 11. Let X = {1,2} with the topology τ = � φ, X , {1} and let Y = R with the topologies σ = τcoc. Let f : (X ,τ)→ (Y,σ) be the function defined by f (x) = ( R−Q x = 2 Q x = 1 Then f is not ωβ−open, but it can easily seen that f is ωβ−continuous and ωβ−irresolute function. Example 12. Consider the function f in the Example 8 which isωβ−open, but notωβ−irresolute. ACKNOWLEDGEMENTS: This work is financially supported by the Malaysian Ministry of Science, Technology and Environment, Science Fund Grant no. UKM-ST-06-FRGS0146-2010. References [1] K Al-Zoubi. Semi ω− containuous functions. Abhath Al-yarmouk, 12(1):119–131, 2003. [2] K Al-Zoubi and B Al-Nashef. The topology of ω− open subsets. Al-Manarah Journal, 9(2):169–179, 2003. [3] H Aljarrah and M Noorani. On ωβ− open sets. submitted. REFERENCES 140 [4] D Andrijevic. On b-open sets. Mat. Vesnik, 48:59–64, 1996. [5] S Crossley and S Hildebrand. Semi-topological properties. Fund. Math., 74(3):233–254, 1972. [6] H Hdeib. ω− continuous functions. Dirasat, XVI:136–142, 1996. 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