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ISSN XXXX XXXX © 2016 IJESC Research Article Volume 6 Issue No. 7 On Generalized *+b-Continuous and Irresolute Maps B. Kanchana1, F. Nirmala Irudayam2 Research Scholar1, Assistant Professor 2 Department of Mathematics, Nirmala College for Women, Coimbatore, India Abstract: In this paper, we introduce a new class of maps named g *+b-continuous and g*+b-irresolute maps in simple extended topological spaces. We have intended to study some of its basic properties and relations among them. Keywords: g*+b-closed set, g*+b-continuous, g*+b-irresolute maps. I. INTRODUCTION Levine [9] and Andrijevic [1] initiated the concept of generalized closed (open) sets and b-open sets respectively in topological spaces. The class of b-open sets is contained in the class of semipre-open sets and contains the class of semi-open and the class of pre-open sets. A.A.Omari and M.S.M.Noorani [14] introduced generalized b-closed sets and gb-continuous maps in topological spaces. Ekici [6] presented the concept of b-continuous functions in topological spaces. El-Etik [5] discussed gb-continuous functions with the aid of b-open sets. Crossley and Hildebrand [4] have introduced and investigated the idea of irresolute functions which are stronger than semicontinuous maps but are independent of continuous maps. Munshi and Bassan [11] proposed the notion of generalized continuous functions which are called in as g-irresolute functions. M.K.R.S.Veerakumar [16] defined the concept of gc-irresolute function in topological spaces. Vidhya and Parimelazhagan [17, 18] studied the properties of g*b-closed sets and g*b-continuous maps in topological spaces. In 1963 Levine [8] formulated the concept of simple extension of a topology as / . B.Kanchana and F.Nirmala Irudayam [7] coined the concept of g*+b-closed sets in extended topological spaces. The aim of this paper is to introduce and study the concepts of new class of maps namely g*+b-continuous maps and g*+b-irresolute maps. Throughout this paper X, Y and Z (or ( ), and ) are simple extension topological space in which no separation axioms are assumed unless and otherwise stated. For any subset A of X, the interior of A is same as the interior in usual topology and the closure of A is newly defined in simple extension topological spaces. The complement of A is denoted by Ac respectively. II. PRELIMINARIES We recall the following definitions. Definition 2.1: A subset A of a topological space ( called a, ) is (i) b-open set [1], if A ⊆ cl(int(A)) ∪ int(cl(A)). (ii) generalized closed set (briefly g-closed) [10], if cl(A) ⊆ U, whenever A ⊆ U and U is open in X. (iii) g*-closed set[16], if cl(A) ⊆ U whenever A ⊆ U and U is g-open in X. International Journal of Engineering Science and Computing, July 2016 (iv) generalized b-closed set (briefly gb-closed) [14], if bcl(A) ⊆ U whenever A ⊆ U and U is open in X. (v) g*b-closed set [17], if bcl(A) ⊆ U whenever A ⊆ U and U is g-open in X. Definition 2.2: A subset A of a topological space ( called a, ) is (i) b+-open set[12], if A ⊆ cl+(int(A)) ∪ int(cl+(A)). (ii) generalized closed+ set (briefly g-closed+) [7], if cl+(A) ⊆ U whenever A ⊆ U and U is open in X. (iii) g*+-closed set [7], if cl+(A) ⊆ U whenever A ⊆ U and U is g+-open in X. (iv) generalized b+-closed set (briefly gb+-closed) [12], if bcl+(A) ⊆ U whenever A ⊆ U and U is open in X. (v) g*+b-closed set [7], if bcl+(A) ⊆ U whenever A ⊆ U and U is g+-open in X. Definition 2.3: A map from a topological space X into a topological space Y is called, (i) b-continuous [5], if is b-closed in X for every closed set V of Y. (ii) gb-continuous [14], if is gb-closed in X for every closed set V of Y. (iii) g*b-continuous [18], if is g*b-closed in X for every closed set V of Y. (iv) irresolute [4], if is semi-closed in X for every semi-closed set V of Y. (v) gc-irresolute [2], if is g-closed in X for every gclosed set V of Y. (vi) g*b-irresolute [18], if is g*b-closed in X for * every g b-closed set V of Y. Definition 2.4: A map from a topological space X into a topological space Y is called, (i) b+-continuous [15], if is b+-closed in X for every closed set V of Y. (ii) gb+-continuous [15], if is gb+-closed in X for every closed set V of Y III. g*+b-CONTINUOUS MAPS In this section we introduce the concept of g*+b-continuous maps in topological spaces. 1667 http://ijesc.org/ Definition 3.1: A map is said to be g*+b-continuous *+ if is g b-closed in X for every closed set V of Y. Theorem 3.2: If a map continuous but not conversely. is continuous, then it is g*+b- Proof: Let be continuous. Let F be any closed set in Y. Then the inverse image is closed in X. Since every closed set is g*+b-closed, is g*+b-closed in X. Therefore f is g*+b-continuous. Remark 3.3: The converse of the theorem need not be true as seen from the following example. Example 3.4: Let X = Y = {a, b, c} with topologies and . Let be a map defined by f (a) = b, f (b) = a, f (c) = c. Here f is g*+bcontinuous. But f is not continuous since for the closed set F = {b} in Y, = {a} is not closed in X. Theorem 3.5: If a map is b+-continuous, then it is *+ g b-continuous but not conversely. Proof: Let be b+-continuous. Let F be any closed set in Y. Then the inverse image is b+-closed in X. Since + *+ every b -closed set is g b-closed, is g*+b-closed in X. *+ Therefore f is g b-continuous. Remark 3.6: The converse of the theorem need not be true as seen from the following example. Example 3.7: Let X = Y = {a, b, c, d} with topologies and . Let be the identity function. Here f is g*+b-continuous. But f is not b+continuous since for the closed set F = {a,b,d} in Y, = {a,b,d} is not b+-closed in X. Theorem 3.8: If a map is g*+b-continuous, then it is + gb -continuous but not conversely. Proof: Let be g*+b-continuous. Let F be any closed set in Y. Then the inverse image is g*+b-closed in X. *+ + Since every g b-closed set is gb -closed, is gb+-closed + in X. Therefore f is gb -continuous. Remark 3.9: The converse of the theorem need not be true as seen from the following example. (i) The following statements are equivalent: (a) f is g*+b-continuous. (b) The inverse image of each open set in Y is g*+b-open in X. (ii) If is g*+b-continuous, then f (b+*(A)) ⊂ (A)) for every subset A of X. (iii) The following statements are equivalent: cl+(f (a) For each point x ∈ X and each open set V in Y with f (x) ∈ V, there is a g*+b-open set U in X such that U f (U) ⊂ V. (b) For every subset A of X, f (b+*(A)) ⊂ cl+(f (A)) holds. (c) For each subset B of Y, b+*(f -1(B)) ⊂ f -1(cl+(B)). Proof: (i) Assume that f : X → Y be g*+b-continuous. Let G be open in Y. Then Gc is closed in Y. Since f is g*+b-continuous, f −1 (Gc) is g*+b-closed in X. But f −1(Gc) = X − f −1(G). Thus X − −1 f (G) is g*+b-closed in X and so f −1(G) is g*+b-open in X. Therefore (a) implies (b). Conversely assume that the inverse image of each open set in Y is g*+b-open in X. Let F be any closed set in Y. The Fc is open in Y. By assumption, f −1(Fc) is g*+b-open in X. But f −1 c (F ) = X − f −1(F). Thus X − f −1(F) is g*+b-open in X and so f −1 (F) is g*+b-closed in X. Therefore f is g*+b-continuous. Hence (b) implies (a). Thus (a) and (b) are equivalent. (ii) Assume that f is g*+b-continuous. Let A be any subset of X. Then cl+(f (A)) is closed in Y. Since f is g*+b-continuous, f −1 (cl+(f (A))) is g*+b-closed in X and it contains A. But b+*(A) is the intersection of all g*+b-closed sets containing A. Therefore b+*(A) ⊂ f −1(cl+ ( f (A))) and so f (b+*(A)) ⊂ + cl (f (A)). (iii) (a) ⇒ (b) Let y b+*(A) and x X , f (x) V. Let V be any neighbourhood of y. Then there exists a point x X and a g*+b-open set U such that f (x) = y, x U, x b+*(A) and f (U) V. Since x b+*(A), U ∩ A holds and hence f (A) ∩ V . Therefore we have y = f (x) cl+(f (A)). (b) ⇒ (a) Let x X and V be any open set containing f (x). Let A = f −1 (V c), then x ∉ A. Since f (b+*(A)) ⊂ cl+(f (A)) ⊂ V c. Then, since x ∉ b+*(A), there exist a g*+b-open set U containing x such that U ∩A and hence f (U ) ⊂ f (Ac) ⊂ V. (b) ⇒ (c) Let B be any subset of Y. Replacing A by f −1(B) we get from (b), f (b+*(f −1(B))) ⊆ cl+(f f −1(B)) ⊂ B. Hence b+*(f −1(B)) ⊂ f −1(cl+(B)). Example 3.10: Let X = Y = {a, b, c} with topologies and . Let be the identity map. Here f is gb+-continuous. But f is not g*+bcontinuous since for the closed set F = {a,c} in Y, = {a,c} is not closed in X. (c) ⇒ (b) Let B = f (A) where A is a subset of X. Then b+*(A) ⊂ b+*(f −1 (B)) ⊂ f −1(cl+(f (A))). Therefore f (b+*(A)) ⊂ cl+(f (A)). Theorem 3.11: If a map into a topological space Y, Example 3.13: Let X = Y = {a, b, c} with topologies and from a topological space X International Journal of Engineering Science and Computing, July 2016 Remark 3.12: The converse of the statement (ii) need not be true as seen from the following example. 1668 http://ijesc.org/ . Let be a map defined by f (a) = b, f (b) = a, f (c) = c. Then for every subset A of X, f (b+*(A)) ⊂ cl+(f (A)) holds but it is not g*+b-continuous, since for the closed set {a} in Y, = {b} is not g*+bclosed in X. Theorem 3.14: If and be any two functions, then is g*+b-continuous if g is continuous and f is g*+b-continuous. Proof: Let V be any closed set in Z. Since g is continuous, is closed in Y and since f is g*+b-continuous, is g*+b-closed in X. Hence (g f)-1(V) is g*+bclosed in X. Thus g f is g*+b-continuous. Remark 3.15: The composition of two g *+b-continuous map need not be g*+b-continuous. Let us prove the remark by the following example. Example 3.16: Let X = Y = Z = {a, b, c} with topologies ; and . Let be a map defined by g(a) = a, g(b) = c, g(c) = b. Let be a map defined by f(a) = b, f(b) = a, f(c) = c. Both f and g are g*+b-continuous. Define . Here {b} is a closed set of . Therefore = {a} is not g*+b-continuous. Theorem 3.17: Let be a g*+b-continuous map from a topological space X into a topological space Y and let H be a closed subset of X. Then the restriction is g*+bcontinuous where H is endowed with the relative topology. Proof: Let F be any closed subset in Y. Since f is g*+bcontinuous, is g*+b-closed in X. If ∩ H = H1 *+ then H1 is a g b-closed set in X, since intersection of two g*+bclosed set is g *+b-closed set. Since (f / H)-1(F) = H1, it is sufficient to show that H1 is g*+b-closed set in H. Let G1 be any open set of H such that G1 contains H1. Let G1 = G∩H where G is open in X. Now H1 ⊂ G∩H ⊂ G. Since H1 is g*+b-closed set in X. ⊂ G. Now (H1) = ∩ H ⊂ G∩H = G1 where (A) is the closure of a subset A of the subspace H of X. Therefore f / H is g*+b-continuous. Remark 3.18: In the above theorem, the assumption of closedness of H cannot be removed as seen from the following example. Example 3.19: Let Let X = Y = {a, b, c} with topologies and . Let be a map defined by f (a) = b, f (b) = a, f (c) = c. Now H = {a, b} is not closed in X. Then f is g*+b-continuous but the restriction f / H is not g*+b-continuous. Since for the closed set F = {b, c} in Y, and is not g*+b-closed in H. International Journal of Engineering Science and Computing, July 2016 PASTING LEMMA FOR g*+b – CONTINUOUS MAPS Theorem 3.20: Let X = A B be a topological space with topology + and Y be a topological space with topology σ +. Let f : (A, A) → (Y, σ+) and g : (B, B) → (Y, σ+) be g*+bcontinuous maps such that f (x) = g(x) for every x A ∩ B. Suppose that A and B are g*+b-closed sets in X. Then the combination α : (X, + ) → (Y, σ+) is g*+b-continuous. Proof: Let F be any closed set in Y. Clearly α−1(F ) = f −1 (F ) g− 1(F ) = C D where C = f −1(F ) and D = g−1(F ). But C is g*+b-closed in A and A is g*+b-closed in X and so C is g*+b-closed in X. Since we have proved that if B ⊆ A ⊆ X, B is g*+b-closed in A and A is g*+b-closed in X then B is g*+bclosed in X. Also C D is g*+b-closed in X. Therefore α−1(F ) is g*+b-closed in X. Hence α is g*+b-continuous. IV. g*+b-IRRESOLUTE MAPS In this section we introduce the concept of g*+b-irresolute maps in topological spaces. Definition 4.1: A map is said to be irresolute, if the inverse image of every semi+-closed in Y is semi+-closed in X. Definition 4.2: A map is said to be gc+-irresolute, if + the inverse image of every g -closed in Y is g+-closed in X. Definition 4.3: A map is said to be g*+b-irresolute if *+ the inverse image of every g b-closed in Y is g*+b-closed in X. Definition 4.4: A topological space ( ) is said to be Tg*+bspace if every g*+b-open set of X is open in X. Theorem 4.5: A map is g*+b-irresolute if and only if *+ the inverse image of every g b-open set in Y is g*+b-open in X. Proof: Assume that f is g*+b-irresolute. Let A be any g*+b-open set in Y. Then Ac is g*+b-closed set in Y. Since f is g*+birresolute, f −1(Ac) is g*+b-closed in X. But f −1(Ac) = X − −1 −1 *+ f (A) and so f (A) is g b-open in X. Hence the inverse image of every g*+b-open set in Y is g*+b-open in X. Conversely assume that the inverse image of every g*+b-open set in Y is g*+b-open in X. Let A be any g*+b-closed set in Y. Then Ac is g*+b-open in Y. By assumption, f −1(Ac) is *+ −1 c −1 g b-open in X. But f (A ) = X − f (A) and so f −1(A) is g*+b-closed in X. Therefore f is g*+b-irresolute. Theorem 4.6: If a map f : X → Y is g*+b-irresolute, then it is g*+b-continuous but not conversely. Proof: Assume that f is g*+b-irresolute. Let F be any closed set in Y. Since every closed set is g*+b-closed, F is g*+b-closed in Y. Since f is g*+b-irresolute, f −1(F) is g*+b-closed in X. Therefore f is g*+b-continuous. Remark 4.7: The converse of the theorem need not be true as seen from the following example. 1669 http://ijesc.org/ Example 4.8: Let X = Y = {a, b, c} with topologies and . Let be the identity map. Here f is g*+b-continuous. But {a,c} is g*+bclosed in Y but = {a,c} is not g*+b-closed in X. Therefore f is not g*+b-irresolute. Example 4.15: Let X = Y = {a, b, c} with topologies and . Let be the identity map. Here f is gc+-irresolute. But {a} is g*+b-closed in Y but = {a} is not g*+b-closed in X. Therefore f is not g*+b-irresolute. Theorem 4.9: Let X be any topological space, Y be a T g*+bspace and f : X → Y be a function. Then the following are equivalent: (i) f is g*+b-irresolute (ii) f is g*+b-continuous Remark 4.16: The irresolute maps and g*+b-irresolute maps are independent and can be seen from the following examples. Proof: (i) ⇒ (ii): obvious by the definition. (ii) ⇒ (i): Let F be an g*+b-closed set in Y. Since Y is a Tg*+bspace, F is a closed set in Y and by hypothesis, f -1(F) is *+ *+ g b-closed in X. Therefore f is g b-irresolute. Theorem 4.10: If f : X → Y is bijective, b+-open and g*+bcontinuous then f is g*+b-irresolute. Proof: Let V be any g*+b-closed set in Y and let ⊂U where U is b+-open in X. Then V ⊂ f (U). Since f (U) is b+open in Y and V is g*+b-closed in Y, then cl+(V) ⊂ f (U) implies that ⊂ U. Since f is g*+b-continuous, *+ is g b-closed in X. Hence ⊂ U. Therefore ⊂ ⊂ U. That is ⊂ U. This shows that f is g*+b-irresolute. Example 4.17: Let X = Y = {a, b, c} with topologies and . Let be a map defined by f (a) = b, f (b) = a, f (c) = c. Here f is irresolute. But {a,c} is g*+b-closed in Y but = {b,c} is not g*+b-closed in X. Therefore f is not g*+b-irresolute. Example 4.18: Let X = Y = {a, b, c} with topologies and . Let be a map defined by f (a) = b, f (b) = a, f (c) = c. Here f is g*+b-irresolute. But {b,c} is semi+-closed in Y but = {a,c} is not semi+-closed in X. Therefore f is not irresolute. [1] [2] K.Balachandran, P.Sundaram and H.Maki, “On generalized continuous maps in topological spaces”, Mem. Fac. Sci. Kochi Univ. Ser. A. Math, 12 (1991), 5-13. [3] S.Bharathi, “Perfectly g *b-continuous functions”, International Journal of scientific Research in Science and Technology, Vol.1, Issue.5, 2015, 17-20. [4] S.G.Crossley and S.K.Hildebrand, “Semi-topological properties”, Fund Math, 74 (1972), 233-254. [5] A.A.El-Etik, “A study of some types of mappings on topological spaces”, M.Sc thesis, Tanta University, Egypt, (1997). [6] E.Ekici and M.Caldas, “Slightly γ-continuous functions”, Bol. Soc. Parana. Mat, 22(3) (2004), 6374. [7] B.Kanchana and F.Nirmala Irudayam, “On the generalization of *+b sets in simple extended topological spaces”,International Journal of Innovative Research in Science, Engineering and Technology, Vol.5, Issue.5, may 2016, 7107-7114. [8] N.Levine, “Simple extension of topology”, Amer. Math. Monthly, 71 (1964), 22-105. [9] N.Levine, “Generalized closed sets in topology”, Rend. Circ. Mat. Palermo, (2), 19 (1970), 89-96. Theorem 4.11: If f : X → Y is bijective, closed and irresolute then the inverse function is g*+b-irresolute. Proof: Let A be g*+b-closed in X. Let ⊆ U, where U is g+-open in Y. Then ⊆ U holds. Since U is g+-open in X and A is g*+b-closed in X, b+cl(A) ⊆ U and hence f (b+cl(A)) ⊆ U. Since f is b+-closed in Y. Therefore b+cl(f (b+cl(A))) ⊆ U and hence b+cl(f (A)) ⊆ U. Thus f (A) is g*+b-closed in Y and so is g*+b-irresolute. Theorem 4.12: Let X, Y and Z be any topological spaces. For any g*+b-irresolute map f : X → Y and any g*+b-continuous map g : Y → Z, the composition g ◦ f : (X, τ+ ) → (Z, η+) is g*+b-continuous. Proof: Let F be any closed set in Z. Since g is g*+b-continuous, g− 1(F) is g*+b-closed in Y. Since f is g*+b-irresolute, f −1 −1 (g (F)) is g*+b-closed in X. But f −1(g−1(F)) = (g ◦ f )−1(F). Therefore g ◦ f : X → Z is g*+b-continuous. Remark 4.13: The following examples shows that the notion of gc+-irresolute maps and g*+b-irresolute maps are independent. Example 4.14: Let X = Y = {a, b, c} with topologies and . Let be the identity map. Here f is g*+b-irresolute. But {a,b} is g+-closed in Y but = {a,b} is not g+-closed in X. Therefore f + is not gc -irresolute. International Journal of Engineering Science and Computing, July 2016 V. REFERENCES D.Andrijevic, “On b-open sets”, Mat. Vesink, 48 (1996), 59-64. 1670 http://ijesc.org/ [10] N.Levine, “Strong continuity in topology”, Amer. Math. Monthly, 67 (1960), 269. [11] B.M.Munshi and D.S.Bassan, “g-continuous mappings”, Vidya J. Gujarat Univ. B. Sci, 24 (1981), 63-68. [12] F.Nirmala Irudayam and Sr.I.Arockiarani, “A note on the weaker form of bI set and its generalization in SEITS”, International Journal of Computer Application, Issue 2, Vol.4 (Aug 2012), 42-54. [13] T.Noiri, “A Generalization of some forms of girresolute functions”, European journal of pure and Applied Mathematics, Vol.2, no.4, 2009, 473-493. [14] A.A.Omari and M.S.M.Noorani, “On generalized bclosed sets”, Bull. Malays. Math. Sci. Soc(2), 32 (1) (2009), 19-30. [15] S.Reena and F.Nirmala Irudayam, “A new weaker form of πgb-continuity”,International Journal of Innovative Research in Science, Engineering and Technology, Vol.5, Issue.5, may 2016, 8676-8682. [16] M.K.R.S.Veerakumar, “Between closed sets and gclosed sets”, Mem. Fac. Sci. Kochi. Univ. Ser. A. Math, 21 (2000), 1-19. [17] D.Vidhya and R.Parimelazhagan, “g*b-closed sets in topological spaces”, Int. J. Contemp. Math. Sciences, 7 (2012), 1305-1312. [18] D.Vidhya and R.Parimelazhagan, “g*b-continuous maps and pasting lemma in topological spaces”, Int. Journal of Math. Analysis, 6 (47), 2012, 2307-2315. International Journal of Engineering Science and Computing, July 2016 1671 http://ijesc.org/