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STRONGLY -GENERALISED IRRESOLUTE MAPS, STRONGLY SEMI -GENERALISED IRRESOLUTE MAPS AND ALMOST -GENERALISED IRRESOLUTE MAPS R.Devi and K.Bhuvaneswari Department of Mathematics, Kongunadu Arts and Science College, Coimbatore-641 029 Abstract: The objective of the present paper is to introduce three new classes of functions called strongly -generalised irresolute functions, strongly semi -generalised irresolute functions and almost -generalised irresolute functions in a topological space. We obtain some characterizations of this classes and several properties and investigate the relationships with other classes of functions between topological spaces. Keywords: Strongly -generalised irresolute functions; strongly semi -generalised irresolute functions; almost -generalised irresolute functions. Introduction: O. Njåstad[14] and N. Levine [6] introduce the notion of -open sets and generalised closed sets in topological spaces respectively .H.Maki et al [10] defined the concept of generalised - closed sets and -generalised closed sets and investigated their topological properties . G. LoFaro [8] defined and investigated the notion of a strong form of irresolute functions. In section 1 we define and investigate the notions of new classes of function namely strongly -generalised irresolute function . In section 2 we introduce strongly semi -generalised irresolute function and almost -generalised irresolute function. In section 3 we study final remarks. Preliminaries: Throughout this paper a space means a topological space which lacks any separation axioms unless explicitly stated. Let ( X, ) and ( Y,) be topological spaces. Definition(i).[10] A function f : (X, ) (Y, ) is said to be - generalised -1 irresolute if f (V) is - generalised open in X for every -generalised open set V of Y. (ii).[8] A function f : (X, ) (Y, ) is said to be strongly - irresolute if f -1 (V) is open in X for every - open set V of Y. (iii).[17] A function f : (X, ) (Y, ) is said to be g* -continuous if f -1 (V) is g* closed set of ( X, ) for every closed set V of (Y, ). (iv)[1]A function f : (X, ) (Y, ) is said to be - continuous(resp. -irresolute ) if f -1 (V) is -open in ( X, ) for every open (resp.-open) set V of (Y, ). (v)[16] A function f: ( X, ) (Y , ) is said to be strongly generalised -irresolute if f -1 (V) is open in X for every generalised -open set V of Y. (vi) [2] . A topological space ( X, ) is said to be an Tb –space if every -generalised closed set in it is closed. 1. STRONGLY -GENERALISED IRRESOLUTE MAP Definition 1.1 A function f : (X, ) (Y, ) is said to be strongly -generalised irresolute if f -1 (V) is open in X for every - generalised open( written as g-open) set V of Y. Theorem 2.1 If f: (X, ) (Y, ) is strongly -generalised irresolute map then (i) f is -generalised irresolute map. (ii) f is strongly -irresolute map. (iii) f is g* -continuous map. (iv) f is strongly generalised - irresolute map. Proof : (i) Let V be - generalised open set in Y. Since f is strongly -generalised irresolute map, f -1 (V) is open in X , and by [10] it is -generalised -open in X. (ii) Let V be -open set in Y, by [10] it is -generalised open in Y since f is strongly generalised irresolute map, f -1 (V) is open in X. and hence it is -open in X. (iii) Let V be open set in Y and hence - generalised - open set in Y. Since f is strongly -generalised irresolute map, f -1 (V) is open in X and by [17] it is g* open in X. (iv) Let V be generalised -open set in Y , by [2] it is -generalised-open in Y. Since f is strongly -generalised irresolute map, f -1 (V) is open in X. and hence it is -generalised -open in X. Example 3.1 : The converse of the above theorem need not be true. (i) Let X = Y = {a ,b, c } and let = { X,,{a},{a, b}} and = { Y, ,{a}}. Let f: (X, ) (Y, ) be defined as identity map. The mapping f is -generalised irresolute map but not strongly -generalised irresolute map. Since the inverse image of every -generalised open set in Y is generalised open in X, but the inverse image of -generalised open set {a,c } is {a,c}, which is not open in X (ii) Let X = Y = {a, b, c} with = { X,,{a},{a,b}} and = { Y, ,{a, c}}.Define a function f: (X, ) (Y, ) by setting f(a) =a, f(b) =c and f(c) =b. Then f is strongly -irresolute map but not strongly -generalised irresolute map. (iii) Let X = Y = {a, b, c } With = { X,,{a},{a, c}} and = { Y, ,{a},{b},{a, b}}. Let f: (X, ) (Y, ) be defined by f(a) = a, f(b) = c and f(c) = b. In this example. the mapping f is g* continuous but not strongly -generalised irresolute map. Since the inverse image of -generalised open set is {b} in Y is {c}, which is not open in X..(iv) Let X = Y = {a, b, c } and let = { X,,{b}, {b,c},{a, b}} and = { Y, ,{a},{a,b}} .Define a function f: (X, ) (Y, ) by setting f(a) =b, f(b)=a and f(c)=c. Then f is strongly generalised -irresolute but not strongly -generalised irresolute. Since the inverse image of all generalised - open set is open in X. but the inverse image of -generalised open set {b} is {a} which is not open in X. Theorem 2.2 If a map f: (X, ) (Y, ), is continuous and Y is Tb , then f is strongly generalised irresolute map. Proof : Let V be -generalised open set in Y. since Y is Tb, V is open in Y. Hence f -1 (V) is open in X. Therefore f is strongly -generalised irresolute map. Theorem 2.3 Let f : (X, ) (Y, ), g : (Y, ) (Z, ) be any two functions, then the composition g o f : (X, ) (Z, ) is (i) Strongly -irresolute if f is strongly -generalised irresolute and g is -irresolute map. (ii) Continuous if f is strongly -generalised irresolute and g is -generalised irresolute map. (iii) g* continuous if f is g*-irresolute and g is strongly -generalised irresolute map. From the Theorems 2.1 and Examples 3.1 we obtain the following diagram -generalised irresolute map Strongly-irresolute map Strongly -generalised irresolute map g*- continuous Strongly generalised irresolute map Theorem 2.4 The following are equivalent for a function f: (X, ) (Y, ) : (a) f is strongly -generalised irresolute; (b) For each x X and each -generalised open set V of Y containing f(x) , there exists a open set U of X containing x such that f(U) V; (c) f -1(V) Int (f -1(V )) for each -generalised open set V of Y; (d) f -1(F) is closed in X for every -generalised open set F of Y; proof: (a) (b) Let x X and let V be any -generalised open set of Y containing f(x) . By definition 1.1 f -1(V) is open in X and contains x. Set U = f -1(V) , then by (a) U is open in X containing x and f(U) V. (b) (c) Let V be any -generalised open set of Y and x f -1(V) -----------(1) By (b) there exist an open set U of X containing x such that f(U) V. Thus we have x U Int (U) Int (f -1(V)) ----(2) and hence from (1) and (2) f -1(V) Int (f -1(V)). (c) (d) Let F be any -generalised closed subset of Y. Set V = Y – F, then V is generalised open in Y. By (c) we obtain f -1(V) Int (f -1(V)) and hence f -1(F) = X – f -1(Y F) = X - f -1(V) is closed in X. (d) (a) Let V be any -generalised open in Y set F = Y – V, by (d) f -1(F) is closed in X that is f -1(F) = f -1(Y - V) = X - f -1(V) is closed in X and hence f -1(V) is open in X. Therefore f is strongly -generalised irresolute. Theorem 2.5 A function f: (X,) (Y ,) is strongly -generalised irresolute if the graph function g: X X Y denoted by g(x) = ( x, f(x) ) for each x X is strongly generalised irresolute. Proof : Let x X and V be any -generalised open set in Y containing f(x) . Then X V be g-open set of X Y containing g(x) by lemma [2]. Since g is strongly -generalised irresolute , there exist an open set U of X containing x such that g(U) X V and hence f(U) V. Thus f is strongly -generalised irresolute by Theorem 2.4. Theorem 2.6 If f : (X,) (Y,) is strongly -generalised irresolute , then the restriction f/A :(A, /A) (Y,) is strongly -generalised irresolute. Proof : Let V be any -generalised open set of Y. Since f is strongly -generalised irresolute, f -1 (V) is open in X. Since A is open in X , (f / A) –1 (V) = A f –1 (V) is open in A and hence f/A is strongly -generalised irresolute. Theorem 2.7 Let f: (X, ) (Y, ) be a function and {Ai / i } be a cover of X by open sets of ( X, ). Then f is strongly -generalised irresolute, if f / A i : (A i , / A i) (Y,) is strongly -generalised irresolute for each i . Proof : Let V be -generalised open set of Y. Since f /A i is strongly -generalised irresolute map, ( f / A i ) -1 (V ) is open in A i . Since A i is open in X, ( f / A i ) -1 (V ) is open in X for every i . Therefore f -1 (V) = X f -1(V) = {A i f -1(V): i } = { (f / A i) -1 (V) : i } is open in X, because the union of open sets is a open set. Hence f is strongly - generalised irresolute. Definition 1.2 (i)A collection { A i / i } of -generalised open sets in a topological space X is called -generalised open cover of a subset S if S { Ai / i } holds. (ii)A topological space ( X, ) is -generalised compact if every -generalised open cover of X has a finite sub cover. (iii) A subset S of a topological space X is said to be -generalised compact relative to X, if for every collection { A i / i } of -generalised open subset of X such that S { Ai / i } there exists a finite subset 0 of such that S { A i / i 0 } (iv) A subset S of a topological space X is said to be -generalised compact if S is generalised compact as a subspace of X. Theorem 2.8 Let f: (X, ) (Y, ) be a surjective, strongly -generalised irresolute map. If X is compact, then Y is -generalised compact. Proof : Let { Ai / i A } be a -generalised open cover of Y. Then {f –1 (Ai ) / i } is a open cover of X. Since X is compact, it has a finite sub cover say { f –1 (A1), f –1 (A2), …… f –1 (An)} surjectiveness of f is implies {A1, A2, …… An} is -generalised open cover of Y and hence Y is -generalised compact. Theorem 2.9 If a map f: (X, ) (Y, ) is strongly -generalised irresolute and a subset S of X is compact relative X, then the image f(S) is -generalised compact relative to Y. Proof : Let { A i / i A } be a collection of -generalised open sets in Y such that f(S) {(Ai )/ i }. Then S {f –1 (A i ) / i } where f –1 (A i ) is open in X for each i. Since S is compact relative to X, there exists a finite sub collection {A1, A2, …… An} such that S {f –1 (A i ) / i = 1 to n } that is f(S) {A i / i = 1 to n }. Hence f(S) is -generalised compact relative to Y. 2. STRONGLY SEMI -GENERALISED IRRESOLUTE MAP AND ALMOST -GENERALISED IRRESOLUTE MAP Definition 1.3 Let ( X, ) and ( Y, ) be two topological spaces and f: (X, ) (Y, ) (i) A function f is said to be strongly semi -generalised irresolute if f -1 (V) is semiopen in X for every -generalised open set V in ( Y, ). (ii) A function f is said to be almost -generalised irresolute , if f -1 (V) is -open in ( X,) for every -generalised open set V in ( Y, ). Theorem 2.10(i) Every strongly -generalised function is strongly semi -generalised irresolute. (ii) Every strongly semi- generalised irresolute function is almost - generalised irresolute. (iii) Every almost -generalised irresolute function is -continuous. Example 3.2 The converse of Theorem 2.10(i) is not true, Let X = {a, b, c} and = {, X, {a}, {b} , {a, b} } The identity function f: ( X,) ( X, ) is strongly semi -generalised irresolute. The function f is not strongly -generalised irresolute , because the inverse image of { a,c} is { a, c} which is a -generalised open set in X but not open in( X, ) The converse of Theorem 2.10 (ii) is not true .Let X= { a, b, c} =Y, = { , {a}, { b, c }, X} and = { , Y, {a}, { a, b} } . Define a function f: ( X, ) ( Y, ) by setting f(a) = b, f(b)= c, f(c)= a. Then f is almost -generalised irresolute. The function f is not strongly semi -generalised irresolute, because {a} is -generalised open in ( Y, ) and f - 1 ( {a} ) = {c} is not semi open in ( X, ) . The converse of the Theorem 2.10 (iii) is not true. Let X= { a, b, c} and = { , { a} , X} . Define a function f: (X,) ( Y,) by setting f(a)= f(b)= b , f(c) =c. Then f -1( V) is -open for every open set V. i.e., f is -continuous .How ever f is not almost generalised irresolute. Theorem 2.11 Let f: (X, ) (Y, ), g : (Y, ) (Z, ) be any two functions, then the composition g o f : (X, ) (Z, ) is (i) Semi continuous if f is strongly semi -generalised irresolute and g is -generalised irresolute map. (ii) continuous if f is strongly semi -generalised irresolute and g is -generalised irresolute map. (iii) Semi continuous if f is semi continuous and pre-open and g is strongly semi generalised irresolute map. Theorem 2.12 Let f: (X, ) Y, ), g : (Y, ) (Z, ) be any two functions, then the composition g o f : (X, ) (Z, ) is Strongly semi -generalised irresolute (i) if f is strongly -generalised irresolute and g is -generalised irresolute. (ii) if f is strongly semi generalised -continuous and g is -generalised irresolute. (iii) if f is strongly -pre irresolute and g is strongly -generalised irresolute. (iv) if f is semi continuous and g is strongly -generalised irresolute. (v) if f is -continuous and g is strongly -generalised irresolute. (vi) if f is strongly semi -irresolute and g is strongly -generalised irresolute. Theorem 2.13 Let f: (X, ) (Y, ) be a function (i) The following properties are equivalent a) f is strongly semi - generalised irresolute. b) For each x X, and each - generalised open set V of Y containing f(x) there exits a semi-open set U of X containing x, such that f(U) V c) f -1 (V) Cl (Int ( f -1(V))) for every -generalised open set V of Y d) f -1(F) is semi-closed in X for every -generalised closed set F of Y (ii) The following properties are equivalent: a) f is almost -generalised irresolute. b) For each x X, and each - generalised open set V of Y containing f(x) there exits a -open set U of X containing x, such that f(U) V c) f -1 (V) Cl (Int(Cl( f -1(V)))) for every - generalised open set V of Y d) f -1(F) is -closed in X for every generalised closed set F of Y Proof: (i) (a) (b) Let x X and let V be an -generalised open set of Y containing f(x).By definition 1.3 (i) f -1 (V) is semi-open in X and contains x. Set U = f -1(V) then by (a) U is semi-open subset of X containing x and f(U) V. (b) (c) Let V be any -generalised open set of Y and x f -1(V). ----------------- (1) By (b) there exits a semi-open set U of X containing x such that f(U) V. U f -1(V). Thus we have x U Cl (Int (U)) Cl (Int (f -1(V))) ---------------------- (2) -1 and U Cl (Int (f (V))) and hence f -1(V) Cl (Int (f -1(V))) [ by (1) and (2) ] (c) (d) Let F be any -generalised closed subset of Y. Set V = Y – F, then V is generalised open in Y. By (c) we obtain, f -1(V) Cl (Int (f -1(V))) and hence f -1(F) = X – f -1(Y – F) = X – f -1(V) is semi-closed in X. (d) (a) Let V be any -generalised open in Y set F = Y – V, by (d) f -1(F) is semi - closed in X that is f -1(F) = f -1(Y - V) = X - f -1(V) is semi - closed in X and hence f -1(V) is semi open in X. Therefore f is strongly semi - generalised irresolute. (ii) (a) (b) Let x X and let V be an - generalised open set of Y containing f(x).By definition 1.3(ii) , f -1 (V) is -open in X and contains x. Set U = f -1(V) then by (a) U is -open subset of X containing x and f(U) V. (b) (c) Let V be any -generalised open set of Y and x f -1(V). ----------------- (1) By (b) there exits a -open set U of X containing x such that f (U) V. U f -1(V). Thus we have x U Cl (Int(Cl(U))) Cl (Int (Cl (f -1(V)))) ---------------------- (2) -1 Using (1) and (2) f (V) Cl (Int (f -1(V))) (c) (d) Let F be any -generalised closed subset of Y. Set V = Y – F, then V is generalised open in Y. By (c) we obtain, f -1(V) Cl (Int (Cl (f -1(V)))) and hence f -1(F) = X – f -1(Y – F) = X – f -1(V) is -closed in X. (d) (a) Let V be any -generalised open in Y set F = Y – V, by (d) f -1(F) is - closed in X that is f -1(F) = f -1(Y - V) = X - f -1(V) is -closed in X and hence f -1(V) is -open in X. Therefore f is almost -generalised -irresolute. Theorem 2.14 (i) A function f: X Y is strongly semi -generalised irresolute if the graph function g: X X Y denoted by g (x) = ( x, f(x) ) for each x X is strongly semi generalised irresolute. (ii) A function f: X Y is almost -generalised irresolute if the graph function g: X X Y denoted by g(x) = ( x, f(x) ) for each x X is almost -generalised irresolute. Proof: Using Theorem 2.13 (i) and (ii) the proof is similar to 2.5. Theorem 2.15 (i) If a function f: X Y is strongly semi - -generalised irresolute. Then P f: X Y is strongly semi - -generalised irresolute for each where, P is the projection of Y onto Y. (ii) If a function f: X Y is almost -generalised irresolute. Then P f: X Y is almost -generalised irresolute for each where P is the projection of Y onto Y. Proof: (i) Let V be any -generalised open set of Y. Since P is continuous and open it is generalised irresolute and hence P -1 (V) is -generalised open in Y. Also f is strongly generalised irresolute f –1 (P -1 (V)) = (P f) –1 (V) is semi open in Y. Hence (P f) is strongly semi -generalised irresolute for each . (ii) Similar to (i). Theorem 2.16 If f: (X,) (Y, ) is strongly semi -generalised irresolute and A is a pre open subset of X. Then the restriction f/A: AY is strongly semi -generalised irresolute. Proof: Let V be any -generalised open set of Y. Since f is strongly semi -generalised irresolute. f -1 (V) is semi - open in X. By Lemma 1.1[12] and since A is open in X , (f / A) –1 (V) = A f generalised irresolute. –1 (V) is semi - open in A and hence f/A is strongly semi - - Theorem 2.17 Let f: (X,) (Y,) be a function and {A : } be a cover of X by semi-open sets of ( X, ). Then f is strongly semi--generalised irresolute, if f / A : A Y is strongly semi--generalised irresolute for each . Proof: Let V be generalised -open set of Y. Since f / A is strongly semi -generalised irresolute map, (f / A ) -1 (V ) is semi-open in A. Since A is semi-open in X, by Theorem 5 of [15] f / A ) -1 (V) is semi-open in X for every . f -1 (V) = X f -1(V) = {A f -1(V): } = {(f / A) -1 (V) : } is semi-open in X, because the union of semi-open sets is a semi-open set. Hence f is strongly semi- - generalised irresolute. Theorem 2.18. If f: (X,) (Y, ) is almost -generalised irresolute and A is an -open subset of X. Then the restriction f/A: A Y is almost -generalised irresolute. Proof: Let V be any - generalised open set of Y. Since f is almost -generalised irresolute, f -1 (V) is - open in X. By Lemma 3.2(b) of [13] and since A is -open in X, (f / A) –1 (V) = A f –1 (V) is - open in A and hence f/A is almost -generalised irresolute. Theorem 2.19 Let f: (X,) (Y,) be a function and {A: } be a cover of X by -open sets of (X, ). Then f is almost -generalised irresolute, if f / A : A. Y is almost generalised irresolute for each Proof: Let V be - generalised open set of Y. Since f / A is almost -generalised irresolute map, (f / A) -1 (V) is -open in A. By Lemma3.3 (b) of [13] and since A O(X), then ( f / A ) -1 (V ) is -open in X for every . Therefore f -1 (V) = X f -1(V) = {A f -1(V): } = {(f / A) -1 (V): } is -open in X, because the union of -open sets is a -open set. Hence f is almost -generalised irresolute. Theorem 2.20 If a function f: (X, ) (Y, ) is almost -generalised irresolute, then f -1(B) is -closed in X for any nowhere dense set B of Y Proof: Let B be any nowhere dense subset of Y. Then Y-B is -open in Y and hence by [10] it is -generalised open in Y. Since f is almost -generalised irresolute, then f -1(Y –B) = -1 -1 X- f (B) is -open in X and hence f (B) is -closed in X. Theorem 2.21 Let f: (X, ) (Y, ), g : (Y, ) (Z, ) be any two functions, then the composition g o f : (X, ) (Z, ) is almost -generalised irresolute (i) if f is almost -generalised irresolute and g is -generalised irresolute. (ii) if f is almost -irresolute and g is strongly -generalised irresolute. (iii) if f is -irresolute and g is almost -generalised irresolute. (iv) if f is -irresolute and g is strongly -generalised irresolute. (v) if f is -continuous and g is strongly -generalised irresolute. (vi) if f is -irresolute and g is strongly semi -generalised irresolute. (vii) if f is almost-irresolute and g is strongly semi -generalised irresolute. Theorem 2.22 Let (X,) be a sub maximal and extremely disconnected space. Then the following are equivalent for a function f: (X,) (Y,) (i) f is strongly -generalised -irresolute (ii) f is strongly semi -generalised irresolute (iii) f is almost -generalised irresolute Proof: This follows from the fact that if (X,) is sub maximal and extremely disconnected, then = SO(X) = O (X) [13] 3.FINAL REMARKS Definition 1.4 A function f: (X,) (Y,) is said to be (i) Strongly -generalised c-homeomorphism if f is bijective, strongly -generalised irresolute function and f –1 is strongly -generalised irresolute, (ii) Strongly semi -generalised c- homeomorphism if f is bijective, strongly semi generalised irresolute and f –1 is strongly semi -generalised irresolute. (iii) Almost -generalised c-homeomorphism if f is bijective, almost -generalised irresolute function and f –1 is almost -generalised irresolute. Theorem 2.23. If a bijection map f: X Y is strongly -generalised c-homeomorphism then it is -generalised c- homeomorphism. Proof: Since a bijection map f is strongly -generalised c-homeomorphism, f and f –1 are strongly -generalised –irresolute. By Theorem 2.1 f and f –1 are -generalised irresolute. Thus f is -generalised c-homeomorphism. Remark : Using Example 3.1 it is clear that -generalised c-homeomorphism need not be strongly -generalised c-homeomorphism. Theorem 2.24 If f: X Y is strongly -generalised c-homeomorphism then it is generalised homeomorphism. Proof: Since every strongly -generalised irresolute map is -generalised continuous the proof follows. Remark :(i) Since every open set is semi open and -open, every semi open set is -open, it is clear that strongly -generalised c-homeomorphism is (a) Strongly semi -generalised chomeomorphism (b) Almost -generalised c-homeomorphism, (ii)Every Strongly semi -generalised c-homeomorphism is almost -generalised chomeomorphism (iii) The family of all strongly -generalised c-homeomorphism from (X, ) on to itself is denoted by strongly -generalised ch (X, ). Theorem 2.25 Let X, Y and Z be topological spaces and f: X Y, g: Y Z be strongly generalised c-homeomorphisms then their composition g o f: X Z is a strongly generalised c-homeomorphism. Proof: Let V be a -generalised open set in Z. Consider (g o f) –1(V) = f –1 (g –1(V)) = f –1(U) where U = g –1(V). As g is strongly -generalised c-homeomorphism, g is strongly generalised irresolute, g –1(V) is open in Y. by [10] g –1(V) is -generalised open in Y. 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