Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in On gωα -Separation Axioms In Topological Spaces S. S. Benchalli1, P. G. Patil2 & Pushpa M. Nalwad3 123 Department of Mathematics, Karnatak University, Dharwad-580 003, Karnataka, India. Abstract: The aim of this paper is to study and investigate the separation axioms: g - Ti spaces Definition 2.1 A subset A of a space X is called (i) a semiopen set [11] if A cl (int ( A)) . (for i=1, 2, 3) and weaker forms of regular and normal spaces by using the notions of g -open and g -closed sets. Moreover, some of their properties, which belong to them are studied. (ii) an -open set [19] if A int (cl (int ( A))) . (iii) a regular-open set [9] if A = int(cl(A)). The complements of the above mentioned sets are called their respective closed sets. 1. Introduction Definition 2.2 [3] A subset A of X is g -closed if It is observed from literature that there has been a considerable work on different relatively weak forms of separation axioms, like Ti (i = 0,1,2) spaces, X is denoted by regularity and normality axioms in particular, several other neighbouring forms of them have also been studied in many papers. For instance semiTi (i = 0,1,2) spaces which has been defined by Maheshwari and Prasad [14, 15] which are weaker then Ti (i = 0,1, 2) spaces. Maheshwari and Prasad introduced the new class of spaces called s-regular [14] and s-normal [15] spaces using semi-open sets. It was further studied by Noiri and Popa [13], Dorsett [8] and Arya [1]. Munshi [16] introduced g-regular and g-normal spaces using g-closed sets of Levine [12]. Later, El-Deeb et al. [5] and Paul et al. [18] introduced p-regular and p-normal spaces. Later, Benchalli et al. [2], Shik John [10] and P. G. Patil et * * al. [17] studied the concept of g -pre regular, g pre normal, -normal, -regular and regular, -normal spaces in topological spaces. Recently, Benchalli et al. [3] [4] introduced and studied the properties of g - closed sets and g -continuous functions. In this paper, the notion of g -open sets in topological space is used to define new separation axioms and obtained some of their properties. Also, the relationship with some other functions are discussed. cl(A) U whenever A U and U is open in X. The family of all g -closed subsets of the space G C(X). Definition 2.3 [3] The intersection of all g closed sets containing a set A is called g closure of A and is denoted by g -cl(A). A set A is g -closed set if and only if g cl(A) = A. Definition 2.4 [3] The union of all g -open sets contained in A is called g -interior of A and is denoted by g -int(A). A set A is g -open if and only if g -int(A) = A. Definition 2.5 A topological space X is said to be a (i) g-regular [13] (resp. -regular [6], -regular [10], -regular [17]) if for each g-closed (resp. closed, -closed, -closed) set F of X and each point x F , there exist disjoint open (resp. open, open, open) sets U and V such that F U and x V . (ii) s-regular [14] (semi regular [7]) if for each closed (resp. semiclosed) set F of X and each point x F , there exist disjoint semiopen sets U and V such that F V and X U . 2. Preliminaries Throughout this paper (X, ) and (Y, ) (or simply X and Y ) always mean topological spaces on which no separation axioms are assumed unless explicitly stated. For a subset A of a space X the closure and interior of A with respect to are denoted by cl(A) and int(A) respectively. Imperial Journal of Interdisciplinary Research (IJIR) Definition 2.6 A topological space X is said to be a (i) g-normal [13] (resp. -normal [6], -normal [10], -normal [17] if for any pair of disjoint gclosed (resp. closed, -closed, -closed) sets A and B, there exist disjoint open (resp. -open, open, open) sets U and V such that A U and B V . (ii) s-normal [15] (semi normal [8]) if for each pair of disjoint closed (semiclosed) sets A and B in X Page 1873 Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in there exist disjoint semiopen (open) sets U and V such that A U and B V . 3. Separation Axioms In this section, we introduce and study weak separation axioms such as g T0 , g T1 y1 = f ( x1 ) f (G) and y2 = f ( x2 ) f (G) . Since f being strongly g -open function, f (G) is g -open in Y. Thus, there exists a g -open set f (G) in Y such that y1 f (G) and y2 f (G) . Therefore Y is g T0 space. and g T2 spaces and obtain some of their properties. Definition 3.2 A topological space X is said to be g T1 space if for each pair of disjoint points x and y there exist g -open sets G and H such Definition 3.1 A topological space X is said to be a x space if for each pair of disjoint points x and y of X there exists a g -open set containing one point but not the other. that x G , y G and x H , y H . Theorem 3.3 A topological space X is g T1 if and only if singletons are g - closed sets. Theorem 3.1 A topological space X is g T0 - Proof. Let X be a g T1 space and is g T0 -space, there exists a g -open set consequently, space if and only if g -closure of distinct points are distinct. Proof. Let x and y be distinct points of X. Since X G such that x G and y G consequently, X G is a g -closed set containing y but not x , but g -cl( y ) is the intersection of all g closed set containing y . Hence y g -cl( y ) but x g -cl(y) as x X - G. Therefore, g -cl( x ) g -cl( y ). Conversely: Let g -cl( x ) g -cl( y ) for x y . Then there exists at least one point z z g -cl {x} but z g cl {y} . We claim x g -cl {y} , because if x g -cl {y} then {x} g -cl {y} implies g -cl {x} g -cl {y} . So z g -cl {y} , which is contradiction. Hence x g -cl {y} which implies x X - g cl {y} , which is g -open set containing x but not y . Hence X is g T0 space. X such that f : X Y is a bijective strongly g -open and X is g T0 space then Y is also g T0 space. Proof. Let y1 and y 2 be two distinct points of Y. Since f is bijective there exist x1 and x2 of X such that f ( x1 ) = y1 and f ( x2 ) = y2 . Since X is g T0 there exists a g -open set G such that and Therefore x1 G x2 G . Theorem 3.2 If Imperial Journal of Interdisciplinary Research (IJIR) x X . Let y X {x} . Then for x y there exists g open set U y such that y U y and x U y y U y X {x} . X {x} = {U y : y X {x}} , That is which is g -open set. Hence {x} is g -closed set. Conversely. Suppose {x} is g -closed set for every x X . Let x , y X with x y . Now x y implies y X {x} . Hence X {x} is g -open set containing y but not x . Similarly, X {y} is g -open set containing x but not y . Therefore X is g T1 space. Theorem 3.4 The property being g T1 space is preserved under bejection and strongly g open function. Proof. Let f : X Y be bijective and strongly g -open function. Let X be g T1 space and y1 and y 2 be any two distinct points of Y. Since f is bijective there exist distinct points x1 , x2 of X such that y1 = f ( x1 ) and y2 = f ( x2 ) . Now X being a g T1 space, there exist g open sets G and H such that x1 G , x2 G and x1 H , x2 H . Therefore y1 = f ( x1 ) f (G) but and y2 = f ( x2 ) f (G) y2 = f ( x2 ) f ( H ) and y1 = f ( x1 ) f ( H ) . Now f being strongly g -open, f (G) and f (H ) are g -open subset of Y such that Page 1874 Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in y1 f (G) but y2 f (G) and y2 f ( H ) and y1 f ( H ) . Hence Y is g T1 space. f : X Y be bijective and g -open function. If X is g T1 and Tg - Theorem 3.5 Let space then Y is g T1 space. Proof. Let y1 and y 2 be two distinct points of Y. f is bijective there exist disjoint points x1 , x2 of X such that y1 = f ( x1 ) and y2 = f ( x2 ) . Now X being a g T1 space, there exist g open sets G and H such that x1 G , x2 G and x1 H , x2 H . Therefore y1 = f ( x1 ) f (G) but and y2 = f ( x2 ) f (G) y2 = f ( x2 ) f ( H ) and y1 = f ( x1 ) f ( H ) . Not X is Tg -space which implies G and H are open sets in X and f is g -open function, f (G) and f (H ) are g -open subsets of Y. Thus, there exist g -open sets such that y1 f (G) but y2 f (G) and y2 f ( H ) but y1 f ( H ) . Therefore Y is g T1 space. Since f : X Y is g -continuous injection and Y is T1 , then X is g T1 space. Proof. Let f : X Y is g -continuous injection and Y be T1 . For any two distinct points x1 and x2 of X there exist distinct points y1 , y 2 of Y such that y1 = f ( x1 ) and y2 = f ( x2 ) . Since Y is T1 space there exist distinct open sets U and V in Y such that y1 U , y2 U and y1 V , y2 V . That is and x1 f 1 (U ), x1 f 1 (V ) 1 1 x2 f (V ), x2 f (U ) . Since f is g 1 1 continuous f (U ) and f (V ) are g -open sets in X. Thus, for two distinct points x1 , x2 of X 1 1 there exist g -open sets f (U ) and f (V ) such that and x1 f 1 (U ), x1 f 1 (V ) 1 1 x2 f (V ), x2 f (U ) . Therefore X is g T1 space. Theorem 3.6 If f : X Y is g -irresolute injective function and Y is g T1 space, then X is g T1 space. Theorem 3.7 Let Imperial Journal of Interdisciplinary Research (IJIR) Proof. Let x1 , x2 be pair of distinct points of X. f is injective there exists distinct points y1 and y 2 of Y such that y1 = f ( x1 ) and y2 = f ( x2 ) . Since Y is g T1 space, there exist g -open sets U and V in Y such that y1 U , y2 U and y1 V , y2 V . That is and x1 f 1 (U ), x1 f 1 (V ) 1 1 x2 f (V ), x2 f (V ) . Since f is g 1 1 irresolute f (U ) and f (V ) are g -open sets in X. Thus for two distinct points x1 , x2 of X 1 1 there exist g -open sets f (U ) and f (V ) and such that x1 f 1 (U ), x1 f 1 (V ) 1 1 x2 f (V ), x2 f (U ) . Therefore X is g T1 space. Since Definition 3.3 A topological space X is said to be a g T2 space if for any pair of distinct points x and y , there exist disjoint g -open sets G and H such that x G and yH . f : X Y is g -continuous injection and Y is T2 , then X is g T2 space. Proof. Let f : X Y is g -continuous injection and Y is T2 . For any two distinct points x1 , x2 of X there exist distinct points y1 , y2 of Y such that y1 = f ( x1 ) and y2 = f ( x2 ) . Since Y is T2 Theorem 3.8 If space there exist disjoint open sets U and V in Y such that y1 U and y2 V . That is x1 f and x2 f 1 1 (U ) (V ) . Since f is g -continuous f (U ) and f 1 (V ) are g -open sets in X. Further is injective, f 1 f 1 (U ) f 1 (V ) = f 1 (U V ) = f 1 ( ) = . Thus, for two disjoint points x1 , x2 of X there exist disjoint f 1 (V ) g -open sets such that f 1 (U ) and and x1 f 1 (U ) x2 f (V ) . Therefore X is g T2 space. 1 f : X Y is g -irresolute injective function and Y is g T2 space, then X is g T2 space. Theorem 3.9 Let Page 1875 Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in Proof. Let x1 , x2 be pair of distinct points in X. f is injective there exist distinct points y1 and y 2 of Y such that y1 = f ( x1 ) and y2 = f ( x2 ) . Since Y is g T2 space, there exist disjoint g -open sets U and V in Y such that y1 U and y2 V . That is x1 f 1 (U ) and x2 f 1 (V ) . Since f is g -irresolute 1 1 injective f (U ) and f (V ) are disjoint g open sets in X. Thus for two disjoint points x1 , x2 of 1 X there exist disjoint g -open sets f (U ) and such that and f 1 (V ) x1 f 1 (U ) 1 x2 f (V ) . Therefore X is g T2 space. Since Theorem 3.10 In any topological space the followings are equivalent. (i) X is g T2 space. (ii) For each x y , there exists a g -open set U such that x in U and y g cl (U ) . x X , {x} = {g cl (U ) : U is g -open set in X and x U } .. (iii)For each Proof. (i) (ii ) : Assume (i) holds. Let x X and x y , then there exist disjoint g -open y V . Clearly, X - V is g -closed set. Since U V = , There fore U X V . g cl (U ) g cl ( X V ) = X V . Now y X V implies y g cl (U ) . (ii ) (iii ) : For each x y , there exists a g -open set U such that x U and So y g cl (U ) . y {g cl (U ) : U is g -open set in X and x U } = {x} . (iii ) (i) : Let x , y X and x y . By hypothesis there exists a g -open set U such that x U and y g cl (U ) . This implies there exists a g -closed set V such that y V . Therefore y X V and X V is g -open set. Thus, there exist two disjoint g -open sets U and X - V such that x U and y X V . Therefore X is g T2 space. sets U and V such that x U and Imperial Journal of Interdisciplinary Research (IJIR) 4. g -Regular Space In this section, we introduce and study g regular spaces and some of their properties. Definition 4.1 A topological space X is said to be a g -regular if for each closed set F and each x F , there exist disjoint g -open sets U and V such that x U and F V . point Theorem 4.1 Every g -regular T0 space is g T2 . x , y X such that x y Let X be a T0 - space and V be an open set which contains x but not y . Then X - V is a closed set containing y but not x . Now by g -regularity of X there exist disjoint g -open sets U and W such that and Since x U X V W . y X V , y W . Thus for x, y X with x y , there exist disjoint open sets U and W such that x U and y W . Hence X is g T2 . Proof. Let Theorem 4.2 In a topological space X, the following conditions are equivalent: (i) X is g -regular. (ii) For every point x X and open set V containing x there exists a g -open set U such that x U g cl (U ) V . (iii) For every closed set F, F = {g cl (V ) : F V and V is g open set of X } . (iv) For every set A and an open set B such that A B , there exists g -open set O such that A O and g cl (O) B . (v) For every nonempty set A and closed set B such that A B , there exist disjoint g -open sets L and M such that A L and B M . Proof. (i) (ii ) Let V be an open set containing x . Then X - V is closed set not containing x . Since X is g -regular, there exist g -open sets L x U , X L L and U L . This implies U X L . Therefore, g cl (U ) g cl ( X L) = X L , because X - L is g -closed. Hence x U g cl (U ) X L V . That is x U g cl (U ) V . and U such that Page 1876 Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in (ii ) (iii ) Let F be a closed set and x F . Then X - F is an open set containing x. By (ii) there is an set U such that g -open x U g cl (U ) X F . And so, F X g cl (U ) X U . Consequently X - U is g -closed set not containing x . Put V = X g cl (U ) . This implies F V and V is g -open set of X and x g cl (V ) , implies {g cl (V ) : F V and V is g -open set of X } F (i) . But F is closed and every closed set is g -closed. Therefore F {g cl (V ) : F V and V is g open set of X } (ii ) is always true. From (i) and (ii), F = {g cl (V ) : F V and V is g open set of X } . (iii ) (iv ) Let A B and B is open. Let x A B . Then X - B is a closed set not containing x. By (iii), there exists a g -open set V of X such that and X B V x g cl (V ) . Put O = X g cl (V ) , then O is g -open set of X, x A O and g cl (O) g cl ( X V ) = X V B . Hence g cl (O) B . (iv ) (v) If A B = , where A is non empty and B is closed, then A ( X B) and X B is open. Therefore by (iv), there exists g -open set L such that A L and L g cl ( L) X B . (v) (i) Let F be a closed set such that x F , then {x} F = . By (v), there exist disjoint open sets L and M such that {x} L and F M , which implies x L and F M . Hence, X is g -regular. Theorem 4.3 If f : X Y is continuous bijective, g -open function and X is a regular then Y is g -regular. Proof. Let F be a closed set in Y and y F . Take y = f ( x) for some x X . Since f is 1 continuous, f ( F ) is closed set in X such that x f 1 ( F ) . Now X is regular, there exist disjoint Imperial Journal of Interdisciplinary Research (IJIR) x U and f ( F ) V . That is, y = f ( x) f (U ) and F f (V ) . Since f is g -open function f (U ) and f (V ) are g -open sets in Y and is bijective f f (U ) f (V ) = f (U V ) = f ( ) = . Therefore, Y is g -regular. open sets U and V such that 1 Theorem 4.4 If f : X Y is -continuous bijective, g -open function and X is space, then Y is g -regular. * -regular y F . Take y = f ( x) for some x X . Since f is 1 continuous, f ( F ) is -closed set in X and x f 1 ( F ) . Now X is -regular, there exist disjoint -open sets U and V such that x U and f 1 ( F ) V . That is, y = f ( x) f (U ) and F f (V ) . Since f is g * -open function f (U ) and f (V ) are g -open sets in Y and is bijective f f (U ) f (V ) = f (U V ) = f ( ) = . Therefore, Y is g -regular. Proof. Let F be a closed set in Y and Theorem 4.5 If f : X Y is continuous surjective, strongly g -open (resp. quasi g open) function and X is g -regular space then Y is g -regular (resp. regular). y F . Take y = f ( x) for some x X . Since f is 1 continuous surjective, f ( F ) is closed set in X 1 and x f ( F ) . Now since X is g -regular, there exist disjoint g -open sets U and V such that x U and f 1 ( F ) V . That is, y = f ( x) f (U ) and F f (V ) . Since f is strongly g -open (resp. quasi g -open) and bijective, f (U ) and f (V ) are disjoint g open (resp. open) sets in Y. Therefore, Y is g Proof. Let F be a closed set in Y and regular (resp. regular). Theorem 4.6 If f : X Y is g -continuous, closed, injection and Y is regular, then X is g regular. Proof. Let F be a closed set in X and x F . Since f is closed injection f (F ) is closed set in Y such Page 1877 Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in f ( x) f ( F ) . Now Y regular, there exist disjoint open sets G and H such that f ( x) G and f ( F ) H . This implies x f 1 (G) and F f 1 ( H ) . Since f is g -continuous, f 1 (G) and f 1 ( H ) are g -open sets in X. 1 1 Further f (G) f ( H ) = . Hence X is g -regular. that Theorem 4.7 If f : X Y is g -continuous, -closed, injection and Y is -regular, then X is g -regular. * Proof. Let F be a closed set in X and x F . Since f is -closed injection f (F ) is -closed set in Y such that f ( x) f ( F ) . Now Y is -regular, there exist disjoint -open sets G and H such that f ( x) G and f ( F ) H . This implies 1 1 x f (G) and F f ( H ) . Since f is g -continuous, f (G) and f ( H ) are sets in X. Further g -open 1 1 f (G) f ( H ) = . Hence X is g * 1 1 regular. Theorem 4.8 If f : X Y is g -irresolute, closed, injection and Y is g -regular, then X is g -regular. Proof. Let F be a closed set in X and x F . Since f is closed injection f (F ) is closed set in Y such that f ( x) f ( F ) . Now Y is g -regular, there exist disjoint g -open sets G and H such that f ( x) G and f ( F ) H . This implies x f 1 (G) and F f 1 ( H ) . Since f is g -irresolute, f 1 (G) and f 1 ( H ) are sets in X. Further g -open 1 1 f (G) f ( H ) = . Hence X is g regular. 5 g -Normal Spaces In this section, we introduce and study g normal spaces and some of their properties. Definition 5.1 A topological space X is said to be g -normal if for every pair of disjoint closed sets Imperial Journal of Interdisciplinary Research (IJIR) E and F of X there exists disjoint g -open sets U and V such that E U and F V . Theorem 5.1 The following statements are equivalent for a topological space X. (i). X is g -normal. (ii). For each closed set A and for each set U containing A, there exist a g -open set V containing A such that g cl (U ) U . (iii). For each pair of disjoint closed sets A and B there exists a g -open set U containing A such g cl (U ) B = . Proof. (i) (ii ) Let A be closed set and U be an open set containing A. Then A ( X U ) = and that therefore they are disjoint closed sets in X. Since X is g -normal, there exists disjoint g -open sets A U , X U W that is Now V W = , implies V and W such that X W U . Therefore V X W . g cl (U ) g cl ( X W ) = X W , because X - W is g -closed set. Thus A V g cl (V ) X W U . That is A V g cl (U ) U (ii ) (iii ) Let A and B be disjoint closed sets in X, then A X B and X B is an open set containing A. By (ii) there exists a g -open set U such that A U and g cl (U ) X B , which implies g cl (U ) B = . (iii ) (i) Let A and B be disjoint closed sets in X. By (iii) there exists a g -open set U such that g cl (U ) B = and or A U B X g cl (U ) . Now U and X g cl (U ) are disjoint g -open sets of X such that and A U B X g cl (U ) . Hence X is g normal. Theorem 5.2 If X is -normal and F A = , where F is -closed set and A is g -closed set, then there exist disjoint -open sets U and V in X, such that F U and A V . Proof. Let X be an -normal space and F A = , since F is -closed set and A is g -closed set such that A X F and -open. is Therefore X F cl ( A) X F , implies cl ( A) F = . Page 1878 Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in Now F is -closed, hence -closed, which implies F and cl (A) are disjoint -closed sets and X is -normal, therefore there exist disjoint open sets U and V of X such that F U and cl ( A) V . That is, U and V of X are such that F U and A V . Theorem 5.3 If X is -normal, the following statements are true. (i). For each -closed set A and every g open set B such that A B there exists an -open set U such that A U cl (U ) B . (ii). For every g -closed set A and every open set containing A there exists an -open set containing U such that A U cl (U ) B . Proof. (i). Let A be a -closed set and B be a g -open set such that A B . Then A ( X B) = . Since A is -closed set and X B be a g -closed, by theorem 5.2, there exist -open sets U and V such that A U , and Thus U V = , X B V A U X V B . Since X - V is -closed, Therefore, cl (U ) X V . A U cl (U ) B . (ii) Let A be a g -closed set and B be a open set such that A B . Then X B X A . Since X is -normal and X - A is a g -open set containing -closed set X - B, by (i) there exists an set G such that -open X B G cl (G) X A . That is A X cl (G) X G B . Let U = X cl (G) , then U is -open set and A U cl (U ) B . Theorem 5.4 If f : X Y is continuous bijective, g -open function from normal space X on to a space Y, then Y is g -normal. Proof. Let E and F be disjoint closed sets in Y. Since f is continuous bijective f 1 ( E ) and f 1 ( F ) are disjoint closed sets in X. Now X is normal, there exist disjoint open sets U and V such that f 1 ( E ) U and f 1 ( F ) V . That is, E f (U ) and F f (V ) . Since f is g open function f (U ) and f (V ) are g -open f sets in Y and is injective Imperial Journal of Interdisciplinary Research (IJIR) f (U ) f (V ) = f (U V ) = f ( ) = . Thus, Y is g -normal. Theorem 5.5 If f : X Y is -continuous bijective, g -open function from -normal space X on to a space Y, then Y is g -normal. Proof. Let E and F be disjoint closed sets in Y. Since * f is -continuous bijective f 1 ( E ) and f 1 ( F ) are disjoint -closed sets in X. Now X is -normal, there exist disjoint -open sets U and V ( E ) U and f 1 ( F ) V . That is, E f (U ) and F f (V ) . Since f is g * open function f (U ) and f (V ) are g -open is injective sets in Y and f f (U ) f (V ) = f (U V ) = f ( ) = . Thus, Y is g -normal. such that f 1 Theorem 5.6 If f : X Y is g -continuous, closed, injection and Y is normal, then X is g normal. Proof. Let E and F be disjoint closed sets in Y. Since f is closed injection f (E ) and f (F ) are disjoint closed sets in Y. Now Y is normal, there exist disjoint open sets G and H such that f ( E ) G f ( F ) H . This implies, E f 1 (G) and F f 1 ( H ) . Since f is g -continuous 1 1 function f (G) and f ( H ) are g -open 1 1 sets in X. Further f (G) f ( H ) = . Hence, X is g -normal. and Theorem 5.7 If f : X Y is g -continuous, -closed, injection and Y is -normal, then X is g -normal. Proof. Let E and F be disjoint closed sets in Y. Since f is -closed injection f (E ) and f (F ) are disjoint -closed sets in Y. Now Y is -normal, there exist disjoint -open sets G and H such that f ( E ) G and f ( F ) H . This implies, * E f 1 (G) and F f 1 ( H ) . Since f is g * -continuous f 1 (G) and f 1 ( H ) are sets in X. Further g -open 1 1 f (G) f ( H ) = . Hence, X is g normal. Page 1879 Imperial Journal of Interdisciplinary Research (IJIR) Vol-3, Issue-1, 2017 ISSN: 2454-1362, http://www.onlinejournal.in Theorem 5.8 If f : X Y is g -irresolute, closed, injection and Y is g -normal, then X is g -normal. Proof. Let E and F be disjoint closed sets in Y. Since f is closed injection f (E ) and f (F ) are disjoint closed sets in Y. Now Y is g -normal, there exist disjoint g -open sets G and H such f ( E ) G and f ( F ) H . This implies, E f 1 (G) and F f 1 ( H ) . Since f is g -irresolute f 1 (G) and f 1 ( H ) are sets in X. Further g -open 1 1 f (G) f ( H ) = . Hence, X is g - that normal. Theorem 5.9 If f : X Y is continuous bijective, strongly g -open (resp. quasi g -open) function from a g -normal space X onto a space Y, then Y is g -normal (resp. normal). Proof. Let E and F be disjoint closed sets in Y. Since f is continuous bijective f 1 ( E ) and f 1 ( F ) are disjoint closed sets in X. Now X is g normal, there exist disjoint g -open sets U and V 1 1 such that f ( E ) U and f ( F ) V . That is, E f (U ) and F f (V ) . Since f is strongly g -open (resp. quasi g -open) function f (U ) and f (V ) are g -open (resp. open) sets f in and is injective Y f (U ) f (V ) = f (U V ) = f ( ) = . Thus, X is g -normal (resp. normal). 6. Acknowledgements The authors are grateful to the University Grants Commission, New Delhi, India for its financial support under UGC-SAP-DRS- III(F.510/3/DRS-III / 2016 (SAP-I) dated: 29-02-2016) to the Department of Mathematics, Karnatak University, Dharwad, India. Also this research was supported by the University Grants Commission, New Delhi, India. under No.F.4-1/2006(BSR)/7-101/2007(BSR) dated: 20th June, 2012. 7. References [2] S. S. Benchalli, T. D. Rayanagoudar and P. G. * * Patil, g -pre regular and g -pre normal spaces, Int. Math. Forum. 4(48), 2010, 2399-2408. [3] S. S. Benchalli, P. G. Patil and Pushpa M. Nalwad, Generalized -closed sets in topological spaces, J. New Results in Sci., 7, 2014, 7-14. [4] S. S. Benchalli, P. G. Patil and Pushpa M. Nalwad, Some weaker forms of continuous functions in topological spaces, J. of Advan.Stud.in Top. 7(2), 2016, 101-109. [5] S. N. El-Deeb, I. A. Hasanein, A. S. Mashhour and T. Noiri, On p-regular spaces. Bull. Math. dela Soc. des Sci. Mathe. de Roumanie., 75(27), 1983, 311–315. [6] R. Devi, Studies on generalizations of closed maps and homeomorphisms in topological spaces, Ph.D thesis, Bharathiar University, Coimbatore., 1994. [7] C. Dorsett, Semi regular spaces, Soochow J. Math., 8, 1982, 45-53. [8] C. Dorsett, Semi normal spaces, Kyungpook Math. J., 25 1985, 173-180. [9] Y. Gnanambal, On generalized pre-regular closed sets in topological spaces, Indian. J. pure appl. Math, 28(3), 1997, 351-360. [10] M. S. John, A study on generalizations of closed sets and continuous maps in topological and bitopological spaces, Ph.D. Thesis, Bharathiar University, Coimbatore, India 2002. [11] N. Levine, Semi-open sets and semi-continuity in topological spaces, Amer. Math. Monthly, 70, 1963, 36-41. [12] N. Levine, Generalized closed sets in topology, Rend. Circ. Math. Palermo., 19(2), 1970, 89-96. [13] T. Noiri and V. Popa, On g-regular spaces and some functions, Mem. Fac. Sci. Kochi Univ. Math., 20, 1999, 67-74. [14] S. N Maheshwari and R. Prasad, On s-regular spaces. Glasnik Matematicki Series III., 30(10), 1975, 347-350. [15] S. N. Maheshwar and R. Prasad, On s-normal spaces, Bull. Math. Soc. Sci. Math. R. S. Roumanie., 22, 1978, 27-28. [16] B. M. Munshi, Separation axioms, Acta Ciencia Indica., 12, 1986, 140-146. [17] P. G. Patil, S. S. Benchalli and P. K. Gonnagar -Separation axioms in topological spaces, J. of New Results in Sci. 2014, 96-103 . [18] M. C. Paul and P. Bhattacharyya, On p-normal spaces. Soochow J. of Mathe., 21(3), 1995, 273-289. [19] O. Njastad, On Some Classes of Nearly Open Sets, Pacific. J. Math., 15, 1965, 961-970. [1] S. P. Arya and T. M. Nour, Characterization of s-normal spaces, Indian. J. Pure and Appl. Math., 21(8), 1990, 717-719. Imperial Journal of Interdisciplinary Research (IJIR) Page 1880