EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 2, 2017, 312-322 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Contra δgb-Continuous Functions in Topological Spaces S.S.Benchalli1, P.G.Patil 2,∗, J.B.Toranagatti3, S.R.Vighneshi4 1,2 Department of Mathematics, Karnatak University, Dharwad, India 3 Department of Mathematics, Karnatak University’s Karnatak College, Dharwad, India 4 Department of Mathematics, R.L.S College, Dharwad, India Abstract. In this paper,the notion of δgb-open sets in topological spaces is applied to study a new class of functions called contra δgb - continuous functions as a new generalization of contra continuity and obtain their characterizations and properties. 2010 Mathematics Subject Classifications: 54C08, 54C10 Key Words and Phrases: δgb-open, δgb-closed, δgb-connected, contra δgb-continuous, δgb- continuous. 1. Introduction and Preliminaries In 1996, Dontchev[6] introduced contra continuous functions. Nasef [10] introduced and studied contra b-continuous functions.Al-Omari and Noorani[1] introduced the concept of contra gb-continuous functions. Recently Benchalli et.al.[5] introduced and studied δgb- continuous functions.These concepts motivated us to define a new class of functions called contra δgb-continuous functions. Throughout this paper,(X, τ),(Y,σ) and (Z,η)(or simply X,Y and Z ) represent topological spaces on which no separation axioms are assumed unless explicitly stated. For a subset A of a space X, the closure of A, interior of A and complement of A are denoted by cl(A), int(A) and Ac respectively. Definition 1. A subset A of a topological space X is called a (i) pre-closed [9] if cl(int(A))⊆A (ii) b-closed [2] if cl(int(A))∩int(cl(A))⊆A (iii) regular-closed [14] if A=cl(int(A)) (iv) δ-closed [17] if A=clδ(A)where clδ(A)={x∈X: int(cl(U))∩A 6=φ, U∈τ and x∈U} ∗Corresponding author. Email addresses: benchalliss@gmail.com(S.S.Benchalli),pgpatil01@gmail.com(P.G.Patil), jagadeeshbt2000@gmail.com(J.B.Toranagatti),vighneshisr@gmail.com (S.R.Vighneshi) http://www.ejpam.com 312 c© 2017 EJPAM All rights reserved. S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 313 (v) delta generalized b-closed (briefly,δgb-closed) [4] if bcl(A)⊆G whenever A⊆G and G is δ-open in X. The complements of the above mentioned closed sets are their respective open sets. The b-closure of a subset A of X is the intersection of all b-closed sets containing A and is denoted by bcl(A). Definition 2. A function f:X→Y from a topological space X into a topological space Y is called a, (i) contra continuous [6] if f−1(G) is closed in X for every open set G of Y. (ii) contra b-continuous [10] if f−1(G) is b-closed in X for every open set G of Y. (iii) contra rgb-continuous [13] if f−1(G) is rgb-closed in X for every open set G of Y. (iv) δgb-continuous [5] if f−1(G) is δgb-open in X for every open set G of Y. (v) completely-continuous [3] if f−1(G) is regular-open in X for every open set G of Y. (vi) perfectly-continuous [12] if f−1(G) is clopen in X for every open set G of Y. (vii) δ∗-continuous if f−1(G) is δ-open in X for every open set G of Y. (viii) contra gb-continuous [1] if f−1(G) is gb-closed in X for every open set G of Y. (ix) pre-closed [7]if for every closed subset A of X f(A) is pre-closed in Y. Definition 3. [5] A topological space X is said to be a, (i) Tδgb-space if every δgb-closed subset of X is closed. (ii) δgbT 1 2 -space if every δgb-closed subset of X is b-closed. 2. Contra δgb-Continuous Functions. Definition 4. A function f:X→Y is called contra δgb-continuous if f−1(V) is δgb-closed in X for each open set V of Y. Clearly, f:X→Y is contra δgb-continuous if and only if f−1(G) is δgb-open in X for every closed set G in Y. Theorem 1. If f:X→Y is contra gb-continuous then it is contra δgb-continuous. Proof: Follows from the fact that every gb-closed set is δgb-closed. Theorem 2. If f:X→Y is contra b-continuous then it is contra δgb-continuous. Proof:Follows from the fact that every contra b-continuous function is contra gb-continuous and Theorem 1. Remark 1. The converse of Theorem 1 and Theorem 2 need not be true as seen from the following example. S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 314 Example 1. Let X=Y={a,b,c}. Let τ={X,φ,{a}} and σ={X,φ,{a},{b},{a,b}} be topolo- gies on X and Y respectively. Then the identity function f:X→Y is contra δgb-continuous but neither contra b-continuous and nor contra gb-continuous, since {a} is open in Y but f−1({a})={a} is not gb-closed in X and hence not b-closed in X. Theorem 3. If f:X→Y is contra δgb-continuous then it is contra rgb-continuous. Proof: Follows from the fact that every δgb-closed set is rgb-closed. Remark 2. The converse of Theorem 3 need not be true as seen from the following example. Example 2. Let X=Y={a,b,c}. Let τ={X,φ,{a},{b},{a,b}} and σ={X,φ,{a}} be topolo- gies on X and Y respectively.Let f:X→Y be a function defined by f(a)=a=f(b) and f(c)=c. Then f is contra rgb-continuous but not contra δgb-continuous,since {a} is open in Y but f−1({a})={a,b} is not δgb-closed in X . Theorem 4. Let f:X→Y be a function. (i) If X is Tδgb-space then f is contra δgb-continuous if and only if it is contra continuous. (ii) If X is δgbT 1 2 -space then f is contra δgb-continuous if and only if it is contra b- continuous. Proof:(i) Suppose X is Tδgb-space and f is contra δgb-continuous. Let G be an open set in Y. Then by hypothesis f−1(G) is δgb-closed in X and hence f−1(G) is closed in X.Therefore f is contra continuous. Converse is obvious . (ii)Suppose X is δgbT 1 2 -space and f is contra δgb-continuous. Let G be an open set in Y then f−1(G) is δgb-closed in X and hence f−1(G) is b-closed in X.Therefore f is contra b-continuous. Converse is follows from the Theorem 2. Theorem 5. [5] Let A⊆X.Then x ∈ δgbcl(A) if and only if U∩A 6= Φ, for every δgb-open set U containing x. Lemma 1. [8] The following properties are hold for subsets A and B of a space X : (i) x∈ker(A) if and only if A∩F=φ for any closed set F of X containing x. (ii) A⊆ker(A) and A = ker(A) if A is open in X. (iii) If A⊆B then ker(A)⊆ker(B). Theorem 6. Suppose that δGBC(X) is closed under arbitrary intersections.Then the fol- lowing are equivalent for a function f:X→Y: (i) f is contra δgb-continuous S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 315 (ii) For each x ∈ X and each closed set B of Y containing f(x) there exists an δgb-open set A of X containing x such that f(A)⊆B (iii) For each x∈X and each open set G of Y not containing f(x) there exists an δgb-closed set H in X not containing x such that f−1(G)⊆H (iv) f(δgbcl(A))⊆ker(f(A)) for every subset A of X (v) δgbcl(f−1(B))⊆f−1(ker(B)) for every subset B of Y. Proof:(i)→(ii) Let B be a closed set in Y containing f(x) then x ∈f−1(B). By (i),f−1(B) is δgb-open set in X containing x. Let A =f−1(F) then f(A)=f(f−1(B))⊆B. (ii)→(i) Let F be a closed set in Y containing f(x) then x ∈f−1(F). From (ii), there exists δgb-open set Gx in X containing x such that f(Gx)⊂F which implies Gx⊆f−1(F ).Thus f−1(F)=∪{Ux: x ∈f−1(F )} which is δgb-open. Hence f−1(F) is δgb-open set in X. (ii)→(iii) Let G be an open set in Y not containing f(x). Then Y-G is closed set in Y containing f(x). From (ii), there exists a δgb-open set F in X containing x such that f(F)⊆Y-G . This implies F⊆f−1(Y-G)=X-f−1(G). Hence f−1(G)⊆X-F. Set H=X-F,then H is δgb-closed set not containing x in X such that f−1(G )⊆H. (iii)→(ii) Let F be a closed set in Y containing f(x). Then Y-F is an open set in Y not containing f(x). From (iii), there exists δgb-closed set K in X not containing x such that f−1(Y-F)⊆K.This implies X-K ⊆f−1(F) that is f(X-K)⊆F.Set U=X-K then U is δgb-open set containing x in X such that f(U)⊆F. (i)→(iv) Let A be any subset of X. Suppose y /∈ ker(f(A)). Then by Lemma 1, there exists a closed set F in Y containing y such that f(A)∩F=φ. Hence we have A∩f−1(F)=φ and δgb-cl(A)∩f−1(F)=φ which implies f(δgbcl(A))∩F=φ and hence y /∈ δgbcl(A). Therefore f(δgbcl(A))⊂ker(f(A)) (iv)→(v) Let B⊆Y then f−1(B)⊆X. By (iv), f (δgbcl(f−1(B)))⊆ ker(f (f−1(B)))⊆Ker(B). Thus δgbcl(f−1(B))⊆f−1(ker(B)) . (v)→(i) Let V be any open subset of Y. Then by (v) and Lemma 1, δgbcl(f−1(V)⊆f−1(ker(V))=f−1(V) and δgbcl(f−1(V ))=f−1(V). Therefore f−1(V) is δgb-closed set in X Lemma 2. [16] For a subset A of a space X,the following are equivalent: (i) A is open and gb-closed (ii) A is regular open. Theorem 7. [4]If A⊆X is both δ-open and δgb-closed then it is b-closed. Theorem 8. If A⊆X is regular open then it is b-closed. Lemma 3. For a subset A of a space X the following are equivalent: (i) A is δ-open and δgb-closed (ii) A is regular open S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 316 (iii) A is open and b-closed. Proof:(i)→(ii):Let A be an δ-open and δgb-closed set.Then by Theorem 7, A is b-closed that is bcl(A)⊆A and so int(cl(A))⊆A. Since A is δ-open then A is pre-open and thus A⊆int(cl(A)).Hence A is regular open. (ii)→(i): Follows from the fact that every regular open set is δ-open and by Theorem 8. (ii)→(iii): Follows from the fact that every regular open set is open and Theorem 8. (iii)→(ii): Let A be an open and b-closed set then bcl(A)⊆A and so int(cl(A))⊆A. since A is open,then A is pre-open and thus A⊆int(cl(A)),which implies A=int(cl(A)). As a consequence of the above lemma, we have the following result: Theorem 9. The following statements are equivalent for a function f:X→Y: (i) f is completely continuous (ii) f is contra δgb-continuous and δ∗-continuous (iii) f is contra b-continuous and continuous. Definition 5. [16] A subset A of X is said to be Q-set if int(cl(A))=cl(int(A)). Definition 6. [16] A function f:X→Y is Q-continuous if f−1(V) is Q-set in X for every open set V of Y. Theorem 10. For a subset A of a space X the following are equivalent: (i) A is clopen (ii) A is δ-open and δ-closed (iii) A is regular-open and regular-closed. Theorem 11. For a subset A of a space X the following are equivalent: (i) A is clopen (ii) A is δ-open,Q-set and δgb-closed (iii) A is open,Q-set and b-closed. Proof:(i)→(ii):Let A be clopen then by Theorem 10 we have A=int(cl(A))=cl(int(A)). Hence A is Q-set.Again by Theorem 10, A is δ-open and δ-closed. Since every δ-closed set is δgb-closed. Therefore (ii) holds. (ii)→(iii): Follows from the Theorem 7. (iii)→(i):Let A be an open,Q-set and b-closed set then by Lemma 3, A is regular open. Since A is Q-set,then A=int(cl(A))=cl(int(A)) which implies A is regular closed. Hence by Theorem 10, A is clopen. S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 317 Theorem 12. The following statements are equivalent for a function f:X→Y: (i) f is perfectly continuous (ii) f is δ∗-continuous,Q-continuous and contra δgb-continuous (iii) f is continuous,Q-continuous and contra b-continuous. Definition 7. A space X is called locally δgb-indiscrete if every δgb-open set is closed in X. Theorem 13. If f:X→Y is a contra δgb-continuous and X is locally δgb-indiscrete space then f is continuous. Proof: Let G be a closed set in Y.Since f is contra δgb-continuous and X is locally δgb- indiscrete space then f−1(G)is a closed set in X. Hence f is continuous Definition 8. [11] A space X is called locally indiscrete if every open set is closed in X. Theorem 14. If f:X→Y is a contra δgb-continuous preclosed surjection and X is Tδgb- space then Y is locally indiscrete . Proof: Let V be an open set in Y. Since f is contra δgb-continuous and X is Tδgb-space then f−1(G)is closed in X. Also f is preclosed then V is preclosed in Y. Now we have cl(V)=cl(int(V))⊆V. This means V is closed in Y and hence Y is indiscrete. Theorem 15. Suppose that δGBC(X) is closed under arbitrary intersections. If f:X→Y is contra δgb-continuous and Y is regular then f is δgb-continuous. Proof: Let x∈X and V be an open set of Y containing f(x). Since Y is regular,there exists an open set G in Y containing f(x) such that cl(G)⊆V. Since f is contra δgb-continuous,there exists an δgb-open set U in X containing x such that f(U)⊆cl(G).Then f(U)⊆cl(G)⊆V. Hence f is δgb-continuous. Recall that for a function f:X→Y the subset {(x,f(x)): x∈X}⊆X×Y is called the graph of f and is denoted by G(f). Definition 9. The graph G(f) of a function f:X→Y is said to be contra δgb-closed if for each (x,y)∈(X×Y)-G(f) there exists U∈δgbO(X,x) and V∈C(Y,y) such that (U×V)∩G(f)=φ. Theorem 16. The graph G(f) of a function f:X→Y is contra δgb-closed in X×Y if and only for each (x,y)∈(X×Y)-G(f) there exists U∈δgbO(X,x) and V∈C(Y,y) such that f(U)∩V=φ. Theorem 17. If f:X→Y is contra δgb-continuous and Y is Urysohn then G(f) is contra δgb-closed in the product space X×Y. Proof: Let (x,y)∈(X×Y)-G(f), then y 6=f(x) and there exist open sets A and B such that f(x)∈A, y∈B and cl(A)∩cl(B)=φ. Since f is contra δgb-continuous then there exists U∈δgbO(X,x) such that f(U)⊆cl(A). Therefore we obtain f(U)∩cl(B)=φ. This shows that G(f) is contra δgb-closed. S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 318 Theorem 18. If f:X→Y is δgb-continuous and Y is T1 then G(f) is contra δgb-closed in X×Y. Proof: Let (x,y)∈(X×Y)-G(f) then y 6=f(x) and there exists open set U such that f(x)∈U and y/∈U. Since f is δgb-continuous, then there exists V∈δgbO(X,x) such that f(V)⊆U.Therefore we obtain f(V)∩(Y-U)=φ and Y-U∈C(Y,y).This shows that G(f) is contra δgb-closed. Theorem 19. Let f:X→Y be a function and g:X→X×Y be the graph function of f defined by g(x)=(x,f(x)) for each x∈X. If g is contra δgb-continuous then f is contra δgb-continuous. Proof: Let U be an open set in Y then X×U is an open set in X×Y. Since g is contra δgb-continuous. It follows that f−1(U)=g−1(X×U) is δgb-closed in X. Thus f is contra δgb-continuous. Theorem 20. If f:X→Y is contra δgb-continuous then for each x∈X and for each closed set V in Y with f(x)∈V there exists a δgb-open set U in X containing x such that f(U)⊆V. Proof: Let x∈X and V is a closed set in Y with f(x)∈V then x∈f−1(V). Since f is contra δgb- continuous,f−1(V) is δgb-open in X. Put U=f−1(V) then x∈U and f(U)=f(f−1(V))⊆V. Definition 10. [16] A space X is submaximal and extremally disconnected if every b-open set is open. Theorem 21. If A and B are δgb-closed sets in submaximal and extremally disconnected space X then A∪B is δgb-closed in X. Proof: Let A∪B⊆G where G is δ-open in X. Since A⊆G,B⊆G, A and B are δgb-closed sets then bcl(A)⊆G and bcl(B)⊆G. As X is submaximal and extremally disconnected,bcl(M)=cl(M) for any M⊆X.Therefore bcl(A∪B)=bcl(A)∪bcl(B)⊆G and hence A∪B is δgb-closed. Corollary 1. If A and B are δgb-open sets in submaximal and extremally disconnected space X then A∩B is δgb-open in X. Theorem 22. :Suppose that δGBC(X) is closed under arbitrary intersections then A⊆X is δgb-closed if and only if A=δgbcl(A) Theorem 23. Suppose that δGBC(X) is closed under arbitrary intersections. If f:X→Y and g:X→Y are contra δgb-continuous ,Y is Urysohn and X is submaximal and extremally disconnected, then K={x∈X:f(x)=g(x)} δ is gb-closed in X. Proof: Let x∈X-K.Then f(x) 6=g(x).Since Y is Urysohn there exist open sets U and V such that f(x)∈U,g(x)∈V and cl(U)∩cl(V)=φ. Since f and g are contra δgb-continuous,f−1(cl(U)) and g−1(cl(V)) are δgb-open sets in X. Let A=f−1(cl(U)) and B=g−1(cl(V)). Then A and B are δgb-open sets containing x.Set C=A∩B, then C is δgb-open set in X. Hence f(C)∩g(C)=f(A∩B)∩g(A∩B)⊆f(A)∩g(B)=cl(U)∩cl(V)=φ. Therefore C∩K=φ. By Theorem 5, x/∈δgbcl(K). Hence K is δgb-closed in X. Definition 11. A space X is called δgb-connected provided that X is not the union of two disjoint nonempty δgb-open sets. Theorem 24. If f:X→Y is a contra δgb-continuous function from a δgb- connected space X onto any space Y then Y is not a discrete space. S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 319 Proof: Since f is contra δgb-continuous and X is δgb-connected space. Suppose Y is a discrete space. Let V be a proper non empty open and closed subset of Y. Then f−1(V) is proper nonempty δgb-open and δgb-closed subset of X, which contradicts the fact that X is δgb-connected space. Hence Y is not a discrete space. Theorem 25. If f:X→Y is a contra δgb-continuous surjection and X is δgb-connected space then Y is connected. Proof: Let f:X→Y is a contra δgb-continuous and X is δgb-connected space. Suppose Y is not connected. Then there exist disjoint open sets U and V in Y such that Y=U∪V. Therefore U and V are clopen in Y. Since f is contra δgb-continuous f−1(U) and f−1(V) are δgb-open sets in X. Further f is surjective implies, f−1(U) and f−1(V) are non empty disjoint and X =f−1(U)∪f−1(V). This contradicts the fact that X is δgb-connected space. Therefore Y is connected. Theorem 26. Let X be a δgb-connected and Y be T1-space. If f:X→Y is contra δgb- continuous then f is constant. Proof: Since Y is T1-space, U={f−1(y): y∈Y} is a disjoint δgb-open partition of X. If |U|≥2 then X is the union of two nonempty δgb-open sets.This contradicts the fact that X is δgb-connected. Therefore |U|=1 and hence f is constant. Definition 12. [4] A topological space X is said to be δgb-T2 space if for any pair of distinct points x and y there exist disjoint δgb-open sets G and H such that x∈G and y∈H. Theorem 27. Let X and Y be topological spaces. If (i) for each pair of distinct points x and y in X there exists a function f:X→Y such that f(x)6=f(y), (ii) Y is Urysohn space and (iii) f is contra δgb-continuous at x and y. Then X is δgb-T2 . Proof: Let x and y be any distinct points in X and f is a function such that f(x) 6=f(y). Let a=f(x) and b=f(y) then a 6=b. Since Y is an Urysohn space there exist open sets V and W in Y containing a and b respectively such that cl(V)∩cl(W)=φ. Since f is contra δgb-continuous at x and y then there exist δgb-open sets A and B in X containing x and y respectively such that f(A)⊆cl(V) and f(B)⊆cl(W). We have A∩B⊆f−1(cl(V))∩f−1(cl(W))=f−1(φ)=φ. Hence X is δgb-T2. Corollary 2. Let f:X→Y be a contra δgb-continuous injective function from a space X into Urysohn space Y then X is δgb-T2. Definition 13. [14] A topological space X is called Ultra Hausdorff space if for every pair of distinct points x and y in X there exist disjoint clopen sets U and V in X containing x and y respectively. S.S.Benchalli, P.G.Patil, J.B.Toranagatti, S.R.Vighneshi / Eur. J. Pure Appl. Math, 10 (2) (2017), 312-322 320 Theorem 28. If f:X→Y be contra δgb-continuous injective function from space X into a Ultra Hausdorff space Y then X is δgb-T2 . Proof: Let x and y be any two distinct points in X . Since f is injective f(x) 6=f(y) and Y is Ultra Hausdorff space implies there exist disjoint clopen sets U and V of Y containing f(x) and f(y) respectively. Then x∈f−1(U) and y∈f−1(V) where f−1(U) and f−1(V) are disjoint δgb-open sets in X . Therefore X is δgb-T2 . Definition 14. [14] A space X is called Ultra normal space if each pair of disjoint closed sets can be separated by disjoint clopen sets. Definition 15. [4] A topological space X is said to be δgb-normal if each pair of disjoint closed sets can be separated by disjoint δgb-open sets. Theorem 29. If f:X→Y be contra δgb-continuous closed injection and Y is ultra normal then X is δgb-normal. Proof: Let E and F be disjoint closed subsets of X. Since f is closed and injective f(E) and f(F) are disjoint closed sets in Y. Since Y is ultra normal there exists disjoint clopen sets U and V in Y such that f(E)⊆U and f(F)⊆V. This implies E⊆f−1(U) and F⊆f−1(V). Since f is contra δgb-continuous injection,f−1(U) and f−1(V) are disjoint δgb-open sets in X . This shows X is δgb-normal. Remark 3. The composition of two contra-δgb-continuous functions need not be contra- δgb-continuous as seen from the following example. Example 3. Let X=Y=Z={a,b,c}, τ={X,φ,{a},{b},{a,b}},σ={Y,φ,{a}} and η={Z,φ,{b,c}} be topologies on X,Y and Z respectively.Then the identity function f:X→Y and a function g:Y→Z defined by g(a)=b,g(b)=c and g(c)=a are contra δgb-continuous but g◦f:X→Z is not contra δgb-continuous, since there exists a open set {b,c} in Z such that (g◦f)−1{b,c}={a,b} is not δgb-closed in X. Theorem 30. Let f:X→Y and g:Y→Z be any two functions. (i) If f is contra δgb-continuous and g is continuous then g◦f is contra δgb-continuous. (ii) If f is contra δgb-continuous and g is contra continuous then g◦f is δgb-continuous. (iii) If f is δgb-continuous and g is contra continuous then g◦f is contra δgb-continuous. (iv) If f is δgb-irresolute and g is contra δgb-continuous then g◦f is contra δgb-continuous. Proof:(i) Let h=g◦f and V be an open set in Z. Since g is continuous, g−1(V) is open in Y. Therefore f−1[g−1(V)]=h−1(V) is δgb-closed in X because f is contra δgb-continuous. Hence g◦f is contra δgb-continuous. The proofs of (ii),(iii) and (iv) are similar to (i). Theorem 31. Let f:X→Y be contra δgb-continuous and g:Y→Z be δgb-continuous. If Y is Tδgb-space, then g◦f:X→Z is contra δgb- continuous. Proof: Let V be any open set in Z . Since g is δgb-continuous g−1(V) is δgb-open in Y and since Y is Tδgb-space, g−1(V) open in Y. Since f is contra δgb-continuous, then f−1(g−1(V )) = (g◦f)−1(V ) is δgb-closed set in X. Therefore g◦f is contra δgb-continuous. 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