-θ-semiopen Sets and

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1 NTMSCI 6, No 3, 6775 (2018) 67 New Trends in Mathematical Sciences semiregular, θsemiopen Sets and ( )θsemi Continuous Functions Hariwan Z Ibrahim Department of Mathematics, Faculty of Education, University of Zakho, KurdistanRegion, Iraq Received: 24 January 2018, Accepted: 18 June 2018 Published online: 29 July 2018 Abstract: The purpose of the present paper is to introduce and study two strong forms of semiopen sets called semiregular sets and θsemiopen sets And also introduce a new class of functions called ( )θsemi continuous functions and obtain several properties of such functions Keywords: open sets, semiregular sets, θsemiopen sets, ( )θsemi continuous functions 1 Introduction Njastad [4] defined αopen sets in a space X and discussed many of its properties Ibrahim [3] introduced and discussed an operation on a topology αo(x) into the power set P(X) and introduced open sets in topological spaces and studied some of its basic properties [1] And also in [2] the author introduced the notion of semiopen sets in a topological space and studied some of its properties In this paper, the author introduce and study the notion of θsemiclosed sets, and introduce( )θsemi continuous functions and investigate some important properties 2 Preliminaries Throughout the present paper, (X, τ) and (Y, σ) represent nonempty topological spaces on which no separation axioms are assumed, unless otherwise mentioned The closure and the interior of a subset A of X are denoted by Cl(A) and Int(A), respectively Definition 1[4] A subset A of a topological space(x,τ) is called αopen, if A Int(Cl(Int(A))) The family of all αopen sets in a topological space(x,τ) is denoted by αo(x,τ) (or αo(x)) Definition 2[3] Let(X,τ) be a topological space An operation γ on the topology αo(x) is a mapping from αo(x) into the power set P(X) of X such that V V γ for each V αo(x), where V γ denotes the value of γ at V It is denoted by γ : αo(x) P(X) Definition 3[3] An operation γ on αo(x, τ) is said to be αregular if for every αopen sets U and V containing x X, there exists an αopen set W of X containing x such that W γ U γ V γ Corresponding author hariwan math@yahoocom

2 68 Hariwan Z Ibrahim: semiregular, θsemiopen Definition 4[1] Let (X,τ) be a topological space and γ,γ be operations on αo(x,τ) A subset A of X is said to be open if for each x Athere exist αopen sets U and V of X containing x such that U γ V γ A A subset of(x,τ) is said to be closed if its complement is open The family of all open sets of (X,τ) is denoted by αo(x,τ) (γ,γ Proposition 1 [1] Let γ and γ be αregular operations If A and B are open, then A B is open Definition 5[2] A subset A of X is said to be α (γ,γ ) semiopen, if there exists an α (γ,γ ) open set U of X such that U A α (γ,γ ) Cl(U) A subset A of X is α (γ,γ ) semiclosed if and only if X\ A is α (γ,γ ) semiopen The family of all semiopen sets of a topological space(x,τ) is denoted by αso(x,τ) (γ,γ, the family of all semiopen sets of (X,τ) containing x is denoted by αso(x,x) (γ,γ Also the family of all semiclosed sets of a topological space(x,τ) is denoted by αsc(x,τ) (γ,γ Definition 6 Let A be a subset of a topological space(x,τ) Then: (1) Cl(A)= F : F is closed and A F} [1] (2) Int(A)= U : U is open and U A} [1] (3) scl(a)= F : F is semiclosed and A F} [2] (4) sint(a)= U : U is semiopen and U A} [2] 3 semiregular Sets and θsemiopen Sets Definition 7 A subset A of a topological space (X,τ) is said to be semiregular, if it is both semiopen and semiclosed The family of all semiregular sets in X is denoted by αsr(x) (γ,γ Lemma 1 The following properties hold for a subset A of a topological space(x,τ): (1) If A αso(x) (γ,γ, then scl(a) αsr(x) (γ,γ (2) If A αsc(x) (γ,γ, then sint(a) αsr(x) (γ,γ Proof (1) Since α (γ,γ ) scl(a) is α (γ,γ ) semiclosed, we have to show that α (γ,γ ) scl(a) αso(x) (γ,γ ) Since A αso(x) (γ,γ ), then for α (γ,γ ) open set U of X, U A α (γ,γ ) Cl(U) Therefore we have, U α (γ,γ ) scl(u) α (γ,γ ) scl(a) α (γ,γ ) scl(α (γ,γ ) Cl(U)) = α (γ,γ ) Cl(U) or U α (γ,γ ) scl(a) α (γ,γ ) Cl(U) and hence α (γ,γ ) scl(a) αso(x) (γ,γ ) (2) This follows from (1) Definition 8 A point x X is said to be θsemiadherent point of a subset A of X if scl(u) A φ for every semiopen set U containing x The set of all θsemiadherent points of A is called the θsemiclosure of A and is denoted by scl θ (A) A subset A is called θsemiclosed if scl θ (A)=A A subset A is called θsemiopen if and only if X\ A is θsemiclosed Definition 9 A point x X is said to be θadherent point of a subset A of X if Cl(U) A φ for every open set U containing x The set of all θadherent points of A is called the θclosure of A and is denoted by Cl θ (A) A subset A is called θclosed if Cl θ (A) = A The complement of an θclosed set is called an θopen set Corollary 1 Let x X and A X If x scl θ (A), then x Cl θ (A)

3 NTMSCI 6, No 3, 6775 (2018) / wwwntmscicom 69 Proof Let x scl θ (A) Then, scl(u) A φ for every semiopen set U containing x Since scl(u) Cl(U), so we have φ scl(u) A Cl(U) A Hence, Cl(U) A φ for every open set U containing x Therefore, x Cl θ (A) Lemma 2 The following properties hold for a subset A of a topological space(x,τ): (1) If A αso(x) (γ,γ, then scl(a)= scl θ (A) (2) If A αsr(x) (γ,γ if and only if A is both θsemiclosed and θsemiopen (3) If A αo(x) (γ,γ, then Cl(A)= Cl θ (A) Proof (1) Clearly scl(a) scl θ (A) Suppose that x / scl(a) Then, for some semiopen set U containing x, A U = φ and hence A scl(u) = φ, since A αso(x) (γ,γ This shows that x / scl θ (A) Therefore scl(a)= scl θ (A) (2) Let A αsr(x) (γ,γ Then, A αso(x) (γ,γ, by (1), we have A= scl(a)= scl θ (A) Therefore, A is θsemiclosed Since X\ A αsr(x) (γ,γ, by the argument above, X\ A is θsemiclosed and hence A is θsemiopen The converse is obvious (3) This similar to (1) Theorem 1 Let(X,τ) be a topological space and A X Then, A is θsemiopen in X if and only if for each x A there exists U αso(x,x) (γ,γ such that sc(u) A Proof Let A be θsemiopen and x A Then, X \ A is θsemiclosed and X \ A = scl θ (X \ A) Hence, x / scl θ (X\ A) Therefore, there exists U αso(x,x) (γ,γ such that scl(u) (X \ A)=φ and so scl(u) A Conversely, let A X and x A From hypothesis, there exists U αso(x,x) (γ,γ such that scl(u) A Therefore, scl(u) (X\ A)= φ Hence, X\ A= scl θ (X\ A) and A is θsemiopen Theorem 2 For a subset A of a topological space(x,τ), we have scl θ (A)= V : A V and V αsr(x) (γ,γ } Proof Let x / scl θ (A) Then, there exists an semiopen set U containing x such that scl(u) A = φ Then A X \ scl(u) = V (say) Thus V αsr(x) (γ,γ such that x / V Hence x / V : A V and V αsr(x) (γ,γ } Again, if x / V : A V and V αsr(x) (γ,γ }, then there exists V αsr(x) (γ,γ containing A such that x / V Then (X\V) (= U, say) is an semiopen set containing x such that scl(u) V = φ This shows that scl(u) A=φ, so that x / scl θ (A) Corollary 2 A subset A of X is θsemiclosed if and only if A= V : A V αsr(x) (γ,γ } Proof Obvious Theorem 3 Let A and B be any two subsets of a space X Then, the following properties hold: (1) x scl θ (A) if and only if U A φ for each U αsr(x) (γ,γ containing x (2) If A B, then scl θ (A) scl θ (B) Proof Clear Theorem 4 For any subset A of X, scl θ ( scl θ (A))= scl θ (A) Proof Obviously, scl θ (A) scl θ ( scl θ (A)) Now, let x scl θ ( scl θ (A)) and U αso(x,x) (γ,γ Then, scl(u) scl θ (A) φ Let y scl(u) scl θ (A) Since scl(u) αso(x,y) (γ,γ, then scl( scl(u)) A φ, that is scl(u) A φ Thus, x scl θ (A)

4 70 Hariwan Z Ibrahim: semiregular, θsemiopen Corollary 3 For any A X, scl θ (A) is θsemiclosed Proof Obvious Theorem 5 Intersection of arbitrary collection of θsemiclosed sets in X is θsemiclosed Proof Let A i : i I} be any collection of θsemiclosed sets in a topological space (X,τ) and A= i I A i Now, using Definition 8, x scl θ (A), in consequence, x scl θ (A i ) for all i I Follows that x A i for all i I Therefore, x A Thus, A= scl θ (A) Corollary 4 For any A X, scl θ (A) is the intersection of all θsemiclosed sets each containing A Proof Obvious Corollary 5 Let A and A i (i I) be any subsets of a space X Then, the following properties hold: (1) A is θsemiopen in X if and only if for each x Athere exists U αsr(x) (γ,γ such that x U A (2) If A i is θsemiopen in X for each i I, then i I A i is θsemiopen in X Proof Obvious Remark The following example shows that the union of θsemiclosed sets may fail to be θsemiclosed Example 1 Let X =1,2,3} and τ =φ,1},3},1,3},x} be a topology on X For each A αo(x,τ), we define two operations γ and γ, respectively, by A γ = Int(Cl(A)) and A γ = X, if A=1,3} A, if A 1,3} Then, the subsets A=1} and B=3} are θsemiclosed, but their union1,3}=a B is not θsemiclosed Example 2 Let X =1,2,3} and τ =φ,1},3},1,3},x} be a topology on X For each A αo(x,τ), we define two operations γ and γ, respectively, by A γ A, if A 1,3} = X, if A=1,3}, and A γ = A The subsets2} is θsemiclosed, but not semiregular Remark From Lemma 2 (2), we have semiregular set is θsemiclosed set In the above example, 2} is θsemiclosed, but not semiregular Remark For a subset A, we always have A scl(a) scl θ (A) Therefore, every θsemiopen set is semiopen The following example shows that the converse is not true in general Example 3 Let X = 1,2,3} and τ = φ,1},2},1,2},2,3},x} be a topology on X For each A αo(x,τ), we define two operations γ and γ, respectively, by A γ = A, if A 1} X, if A=1}, and A γ = A, if A 2} X, if A=2} Then,1,2} is semiopen set but not an θsemiopen set

5 NTMSCI 6, No 3, 6775 (2018) / wwwntmscicom 71 Remark The notions openness and θsemiopenness are independent In Example 2, 1,2} is an θsemiopen set but not an open set, whereas in Example 3,1,2} is an open set but not an θsemiopen set Remark Every θopen set is open 4 ( )θsemi continuous Throughout this section, let f :(X,τ) (Y,σ) be a function and γ,γ : αo(x,τ) P(X) be operations on αo(x,τ) and β,β : αo(y,σ) P(Y) be operations on αo(y,σ) Definition 10 A function f : (X,τ) (Y,σ) is said to be ( )θsemi continuous if for each point x X and each α (β,β semiopen set V of Y containing f(x), there exists an open set U of X containing x such that f(u) α (β,β scl(v) Example 4 Let X =r,m,n}, Y =1,2,3}, τ =φ,r},m},r,m},r,n},x} and σ =φ,3},1,2},y} For each A αo(x,τ) and B αo(y,σ), we define the operations γ : αo(x,τ) P(X), γ : αo(x,τ) P(X), β : αo(y,σ) P(Y) and β : αo(y,σ) P(Y), respectively, by A γ A, if n / A = X, if n A, A, if m Aor A=φ A γ = A m}, if m / A, B β Y, if 2 / B = B, if 2 Bor B=φ, and B β = Y, if 1 / B B, if 1 Bor B=φ Define a function f :(X,τ) (Y,σ) as follows: 1, if x=r f(x)= 1, if x=m 3, if x=n Clearly, αo(x,τ) (γ,γ = φ,m},r,m},x} and αso(y,σ) (β,β = φ,1,2},y} Then, f is ( )θsemi continuous Theorem 6 The following statements are equivalent for a function f :(X, τ) (Y, σ): (1) f is ( )θsemi continuous (2) For each x X and V αsr(y) (β,β containing f(x), there exists an open set U containing x such that f(u) V (3) f 1 (V) is clopen (That is, open as well as closed) in X for every V αsr(y) (β,β (4) f 1 (V) Int( f 1 (α (β,β scl(v))) for every V αso(y) (β,β (5) Cl( f 1 (α (β,β sint(v))) f 1 (V) for every α (β,β semiclosed set V of Y (6) Cl( f 1 (V)) f 1 (α (β,β scl(v)) for every V αso(y) (β,β

6 72 Hariwan Z Ibrahim: semiregular, θsemiopen Proof(1) (2): Let x X and V αsr(y) (β,β containing f(x) By (1), there exists an open set U containing x such that f(u) α (β,β scl(v)= V (2) (3): Let V αsr(y) (β,β and x f 1 (V) Then, f(u) V for some open set U of X containing x, hence x U f 1 (V) This shows that f 1 (V) is open in X Since Y \ V αsr(y) (β,β, f 1 (Y \ V) is also open and hence f 1 (V) is clopen in X (3) (4): Let V αso(y) (β,β Since V α (β,β scl(v) and by Lemma 1, we have α (β,β scl(v) αsr(y) (β,β By (3), we have f 1 (V) f 1 (α (β,β scl(v)) and f 1 (α (β,β scl(v)) is open in X Therefore, we obtain f 1 (V) Int( f 1 (α (β,β scl(v))) (4) (5): Let V be an α (β,β semiclosed subset of Y By (4), we have f 1 (Y \ V) Int( f 1 (α (β,β scl(y \ V))) = Int( f 1 (Y \ α (β,β sint(v))) = X \ Cl( f 1 (α (β,β sint(v))) Therefore, we obtain Cl( f 1 (α (β,β sint(v))) f 1 (V) (5) (6): Let V αso(y) (β,β By Lemma 1, α (β,β scl(v) αsr(y) (β,β By (5), we obtain Cl( f 1 (V)) Cl( f 1 (α (β,β scl(v)))= Cl( f 1 (α (β,β sint(α (β,β scl(v)))) f 1 (α (β,β scl(v)) (6) (1): Let x X and V αso(y, f(x)) (β,β By Lemma 1, we have α (β,β scl(v) αsr(y) (β,β and f(x) / Y \ α (β,β scl(v) = α (β,β scl(y \ α (β,β scl(v)) Thus, by (6) we obtain x / Cl( f 1 (Y \ α (β,β scl(v))) There exists an open set U of X containing x such that U f 1 (Y \ α (β,β scl(v)) = φ Therefore, we have f(u) (Y \ α (β,β scl(v)) = φ and hence f(u) α (β,β scl(v) This shows that f is( )θsemi continuous Proposition 2 The following statements are equivalent for a function f :(X, τ) (Y, σ): (1) f is ( )θsemi continuous (2) For each x X and V αsr(y) (β,β containing f(x), there exists an clopen set U containing x such that f(u) V (3) For each x X and V αso(y) (β,β containing f(x), there exists an open set U containing x such that f( Cl(U)) α (β,β scl(v) Proof (1) (2): Let x X and V αsr(y) (β,β containing f(x) By Theorem 6, f 1 (V) is clopen in X Put U = f 1 (V), then x U and f(u) V (2) (3): Let V αso(y, f(x)) (β,β By Lemma 1, we have α (β,β scl(v) αsr(y) (β,β and by (2), there exists an clopen set U containing x such that f( Cl(U))= f(u) α (β,β scl(v) (3) (1): Let x X and V αso(y, f(x)) (β,β By (3), there exists an open set U containing x such that f( Cl(U)) α (β,β scl(v) implies that f(u) f( Cl(U)) α (β,β scl(v) This shows that f is( )θsemi continuous Theorem 7 The following statements are equivalent for a function f :(X, τ) (Y, σ): (1) f is ( )θsemi continuous (2) Cl( f 1 (B)) f 1 (α (β,β scl θ (B)) for every subset B of Y (3) f( Cl(A)) α (β,β scl θ ( f(a)) for every subset A of X (4) f 1 (F) is closed in X for every α (β,β θsemiclosed set F of Y (5) f 1 (V) is open in X for every α (β,β θsemiopen set V of Y Proof (1) (2): Let B be any subset of Y and x / f 1 (α (β,β scl θ (B)) Then, f(x) / α (β,β scl θ (B) and there exists V αso(y, f(x)) (β,β such that α (β,β scl(v) B=φ By (1), there exists an open set U containing x such that f(u) α (β,β scl(v) Hence, f(u) B=φ and U f 1 (B)=φ Consequently, we obtain x / Cl( f 1 (B)) (2) (3): Let A be any subset of X By (2), we have Cl(A) Cl( f 1 ( f(a))) f 1 (α (β,β scl θ ( f(a))) and hence f( Cl(A)) α (β,β scl θ ( f(a))

7 NTMSCI 6, No 3, 6775 (2018) / wwwntmscicom 73 (3) (4): Let F be any α (β,β θsemiclosed set of Y Then, by (3), we have f( Cl( f 1 (F))) α (β,β scl θ ( f(f 1 (F))) α (β,β scl θ (F) = F Therefore, we have Cl( f 1 (F)) f 1 (F) and hence Cl( f 1 (F))= f 1 (F) This shows that f 1 (F) is closed in X (4) (5): Obvious (5) (1): Let x X and V αso(y, f(x)) (β,β By Lemmas 1 and 2 (2), α (β,β scl(v) is α (β,β θsemiopen in Y Put U = f 1 (α (β,β scl(v)) Then by (5), U is open containing x and f(u) α (β,β scl(v) Thus, f is ( )θsemi continuous Proposition 3 The following statements are equivalent for a function f :(X, τ) (Y, σ): (1) f is ( )θsemi continuous (2) Cl θ ( f 1 (B)) f 1 (α (β,β scl θ (B)) for every subset B of Y (3) f( Cl θ (A)) α (β,β scl θ ( f(a)) for every subset A of X (4) f 1 (F) is θclosed in X for every α (β,β θsemiclosed set F of Y (5) f 1 (V) is θopen in X for every α (β,β θsemiopen set V of Y Proof (1) (2): Let B be any subset of Y and x / f 1 (α (β,β scl θ (B)) Then, f(x) / α (β,β scl θ (B) and there exists V αso(y, f(x)) (β,β such that α (β,β scl(v) B = φ By Proposition 2 (3), there exists an open set U containing x such that f( Cl(U)) α (β,β scl(v) Hence, f( Cl(U)) B = φ and Cl(U) f 1 (B)=φ Consequently, we obtain x / Cl θ ( f 1 (B)) (2) (3): Let A be any subset of X By (2), we have Cl θ (A) Cl θ ( f 1 ( f(a))) f 1 (α (β,β scl θ ( f(a))) and hence f( Cl θ (A)) α (β,β scl θ ( f(a)) (3) (4): Let F be any α (β,β θsemiclosed set of Y Then, by (3), we have f( Cl θ ( f 1 (F))) α (β,β scl θ ( f(f 1 (F))) α (β,β scl θ (F) = F Therefore, we have Cl θ ( f 1 (F)) f 1 (F) and hence Cl θ ( f 1 (F)) = f 1 (F) This shows that f 1 (F) is θclosed in X (4) (5): Obvious (5) (1): Let x X and V αso(y, f(x)) (β,β By Lemmas 1 and 2 (2), α (β,β scl(v) is α (β,β θsemiopen in Y Put U = f 1 (α (β,β scl(v)) Then, by (5), U is θopen containing x and by Remark 3, U is open such that f(u) α (β,β scl(v) Thus, f is( )θsemi continuous Proposition 4 The following statements are equivalent for a function f :(X, τ) (Y, σ): (1) f is ( )θsemi continuous (2) Cl( f 1 (α (β,β sint(α (β,β scl(b)))) f 1 (α (β,β scl(b)), for every subset B of Y (3) f 1 (α (β,β sint(b)) Int( f 1 (α (β,β scl(α (β,β sint(b)))), for every subset B of Y Proof (1) (2): Let B be any subset of Y Then, α (β,β scl(b) is α (β,β semiclosed in Y and by Theorem 6 (5), we have Cl( f 1 (α (β,β sint(α (β,β scl(b)))) f 1 (α (β,β scl(b)) (2) (3): Let B be any subset of Y and x f 1 (α (β,β sint(b)) Then, we have x X \ f 1 (α (β,β scl(y \ B)) So, x / f 1 (α (β,β scl(y \ B)) and by (2), we have x X \ Cl( f 1 (α (β,β sint(α (β,β scl(y \ B)))) = Int( f 1 (α (β,β scl(α (β,β sint(b)))) (3) (1): Let V be any α (β,β semiopen set of Y Suppose that z / f 1 (α (β,β scl(v)) Then, f(z) / α (β,β scl(v) and there exists an α (β,β semiopen set W containing f(z) such that W V = φ and hence scl(w) V = φ By(3), we have z Int( f 1 (α (β,β scl(w))) and hence there exists U αo(x) (γ,γ such that z U f 1 (α (β,β scl(w)) Since α (β,β scl(w) V = φ, then U f 1 (V) = φ and so, z / Cl( f 1 (V)) Therefore, Cl( f 1 (V)) f 1 (α (β,β scl((v))) for every V αso(y) (β,β Hence, by Theorem 6, f is( )θsemi continuous Proposition 5 The following statements are equivalent for a function f :(X, τ) (Y, σ):

8 74 Hariwan Z Ibrahim: semiregular, θsemiopen (1) f is ( )θsemi continuous (2) Cl( f 1 (α (β,β sint(α (β,β scl θ (B)))) f 1 (α (β,β scl θ (B)), for every subset B of Y (3) Cl( f 1 (α (β,β sint(α (β,β scl(b)))) f 1 (α (β,β scl θ (B)), for every subset B of Y (4) Cl( f 1 (α (β,β sint(α (β,β scl(o)))) f 1 (α (β,β scl(o)), for every α (β,β semiopen set O of Y Proof (1) (2): Let B be any subset of Y Then, α (β,β scl θ (B) is α (β,β semiclosed in Y Then by Theorem 6 (5), if x Cl( f 1 (α (β,β sint(α (β,β scl θ (B)))), then x f 1 (α (β,β scl θ (B)) (2) (3): This is obvious since α (β,β scl(b) α (β,β scl θ (B) for every subset B (3) (4): By Lemma 2 (1), we have α (β,β scl(o)=α (β,β scl θ (O) for every α (β,β semiopen set O (4) (1): Let V be any α (β,β semiopen set of Y and x Cl( f 1 (V)) Then, V is α (β,β semiopen and x Cl( f 1 (α (β,β sint(α (β,β scl(v)))) By (4), x f 1 (α (β,β scl(v)) It follows from Theorem 6, that f is ( )θsemi continuous Corollary 6 A function f : (X,τ) (Y,σ) is ( )θsemi continuous if and only if f 1 (α (β,β scl(v)) is open set in X, for each α (β,β semiopen set V in Y Proof Since, if V αso(y) (β,β, then, α (β,β scl(v) αsr(y) (β,β, so by Theorem 6 (3), f 1 (α (β,β scl(v)) is clopen, which is open Conversely, if V αso(y) (β,β, then by hypothesis, f 1 (V) f 1 (α (β,β scl(v)) = sint( f 1 (α (β,β scl(v))), so by Theorem 6 (4), f is ( ) θsemi continuous Corollary 7 A function f :(X,τ) (Y,σ) is ( ) θsemi continuous if and only if f 1 (α (β,β sint(f)) is an closed set in X, for each α (β,β semclosed set F of Y Proof It follows from Corollary 6 Corollary 8 Let f :(X,τ) (Y,σ) be a function If f 1 (α (β,β scl θ (B)) is closed in X for every subset B of Y, then f is( )θsemi continuous Proof Let B Y Since f 1 (α (β,β scl θ (B)) is closed in X, then Cl( f 1 (B)) Cl( f 1 (α (β,β scl θ (B)))= f 1 (α (β,β scl θ (B)) By Theorem 7, f is( )θsemi continuous Proposition 6 The following statements are equivalent for a function f :(X, τ) (Y, σ): (1) f is ( )θsemi continuous (2) Cl( f 1 (V)) f 1 (α (β,β scl(v)), for every V α (β,β sint(α (β,β scl(v)) (3) f 1 (V) Int( f 1 (α (β,β scl(v))), for every V α (β,β sint(α (β,β scl(v)) Proof (1) (2): Let V α (β,β sint(α (β,β scl(v)) such that x Cl( f 1 (V)) Suppose that x / f 1 (α (β,β scl(v)) Then there exists an α (β,β semopen set W containing f(x) such that W V = φ Hence, we have W α (β,β scl(v) = φ and hence α (β,β scl(w) α (β,β sint(α (β,β scl(v)) = φ Since V α (β,β sint(α (β,β scl(v)) and we have V α (β,β scl(w)=φ Since f is ( )θsemi continuous at x X and W is an α (β,β semopen set containing f(x), there exists U αo(x) (γ,γ containing x such that f(u) α (β,β scl(w) Then, f(u) V = φ and hence U f 1 (V)=φ This shows that x / Cl( f 1 (V)) This is a contradiction Therefore, we have x f 1 (α (β,β scl(v)) (2) (3): Let V α (β,β sint(α (β,β scl(v)) and x f 1 (V) Then, we have f 1 (V) f 1 (α (β,β sint(α (β,β scl(v))) = X \ f 1 (α (β,β scl(y \ α (β,β scl(v))) Therefore, x / f 1 (α (β,β scl(y \ α (β,β scl(v))) Then by (2), x / Cl( f 1 (Y \ α (β,β scl(v))) Hence, x X\ Cl( f 1 (Y \ α (β,β scl(v)))= Int( f 1 (α (β,β scl(v))) (3) (1): Let V be any α (β,β semiopen set of Y Then, V = α (β,β sint(v) α (β,β sint(α (β,β scl(v)) Hence, by (3) and Theorem 6, f is( )θsemi continuous

9 NTMSCI 6, No 3, 6775 (2018) / wwwntmscicom 75 Proposition 7 If f : (X,τ) (Y,σ) is ( )θsemi continuous at x X, then for each α (β,β semiopen set B containing f(x) and each open set A containing x, there exists a nonempty open set U A such that U Cl( f 1 (α (β,β scl(b))) Where γ and γ are αregular operations Proof Let B be any α (β,β semiopen set containing f(x) and A be an open set of X containing x By Lemma 1 and Theorem 6, x Int( f 1 (α (β,β scl(b))), then A Int( f 1 (α (β,β scl(b))) φ Take U = A Int( f 1 (α (β,β scl(b))) Thus, U is a nonempty open set by Proposition 1, and hence U A and U Int( f 1 (α (β,β scl(b))) Cl( f 1 (α (β,β scl(b))) Competing interests The authors declare that they have no competing interests Authors contributions All authors have contributed to all parts of the article All authors read and approved the final manuscript References [1] H Z Ibrahim, On open Sets in Topological Spaces, New Trends in Mathematical Sciences, 6 (2) (2018), [2] H Z Ibrahim, semiopen Sets in Topological Spaces, (submitted) [3] H Z Ibrahim, On a class of α γ open sets in a topological space, Acta Scientiarum Technology, 35 (3) (2013), [4] O Njastad, On some classes of nearly open sets, Pacific J Math 15 (1965),