ON SEPARATION AXIOMS IN TOPOLGICAL SPACES

The purpose of this paper is to introduce weak separation axioms via sgp-closed sets in topological spaces and study some of their properties.


INTRODUCTION
General Topology plays very important role in all branches of Mathematics. An important concept in General Topology and Real Analysis concerns the variously modified forms of continuity and separation axioms etc. by utilizing the generalized closed sets.
In 1970, Levine [4] initiated the study of generalized closed(g-closed) sets, that is , a subset A of a topological space X is g-closed if the closure of A included in every open superset of A and defined a T 1/2 space to be one in which the closed sets and g-closed sets coincide. The notion has been studied extensively in recent years by many topologists. The study of g-closed sets has produced some new separation axioms. Some of these have been found useful in computer science and digital topology.
Recently Navalagi and Mahesh Bhat [7] introduced the notion of sgp-closed set utilizing pre closure operator. The notions of sgp-open sets, sgp-contonuity are introduced in [7]. In this paper we continue the study of sgp-closed sets, with introducing and characterizing weak forms of separation axioms.
Proof: Suppose Y is sgp-T 0 -space. Let a and b be two distinct points in X. Since f is an injection, f(a) and f(b) are distinct points in Y. Since Y is sgp-T 0 -space, there exists sgp-open set G in Y such that f(a)  G and f(b)  G. Again since f is sgp-irresolute, f -1 (G) is sgp-open set in X such that a f -1 (G) and b  f -1 (G). Hence X is sgp-T 0 -space.

Theorem 3.5:
If X is sgp-T 0 -space, sgp T c -space and Y is sgp-closed subspace of X, then Y is sgp-T 0 -space. Proof: Let X be sgp-T 0 -space, sgp T c -space and Y is sgp-closed subspace of X. Let a and b be two distinct points of Y. As Y is subspace of X, a and b are two distinct points of X. Since X is sgp- Hence Y is sgp-T 0 -space. Now, we introduce and studysgp-T 1 -space Definition 3.6: A topological space X is said to be sgp-T 1 -space if for any pair of distinct points a and b there exist sgp-open sets G and H such that a  G, b  G and a H, b  H. Theorem 3.10: If f: X  Y is a bijective sgp-open function. If X is a sgp-T 1 -space and sgp T cspace, then Y is a sgp-T 1 -space. Proof: Let y 1 and y 2 be two distinct points of Y. Since f is bijective, there exist distinct points x 1 and x 2 of X such that f(x 1 ) = y 1 and f(x 2 ) = y 2 . Since X is a sgp-T 1 -space, there exist sgp-open sets G and H such that x 1 G and x 2 G and x 1 H and x 2 H. Again since X is sgp T c -space, G and H are open sets in X. As f is sgp-open function, f(G) and f(H) are sgp-open sets such that y 1 Theorem 3.11:If X is sgp-T 1 -space and sgp T c -space and Y is subspace of X, then Y is sgp-T 1space. Proof: Let X be a sgp-T 1 -space and Y be a subspace of X. Let a and b be two distinct points of Y. Since X is a sgp-

Definition4.2:
A space X is said to be i. sgp-D o if for x , y  X containing x but not y or sgp-D-set containing y but not x. ii. sgp-D 1 if for x , y  X such that x ≠ y there exists a sgp-D-set of X containing x but not iii. y and a sgp-D-set containing y but not x. iv.
sgp-D 2 if for x, y X such that x ≠ y there exists a disjoint sgp-Dsets G and E such v. that x G and y  E.

Theorem4.3:
For a space X, the following properties hold: i. If X is sgp-T i , then it is sgp-T i -1 for i =1,2 ii. If X is sgp-T i , then it is sgp-D i for i = 0,1,2 iii. If X is sgp-D i , then it is sgp-D i -1 for i =1,2 Proof: This is obvious from Definition 6.2 Necessity for (ii): Suppose X is sgp-D 1 . It follows from the definition that for any distinct points x and y in X there exists sgp-D-sets G and E such that G containing x but not y and E containing y but not x .Let G = U \ V and E = W \ D, where U, V, W and D aresgp-open. By the fact that x  E, we have two cases, i.e. either x  W or both W and D contain x. If x  W, then from y  G either (i) y  U or (ii) y  U and y  V. If (i) is the case, then it follows from x  U \ V that x  U \ (V  W) and also it follows from y  W \ D that y  W \ (U  D). Thus we have U\ (V  W) and W \ (U D) which are disjoint. If (ii) is the case, it follows from that x  U \ V and y  V since y  U and y  V. Therefore (U\V)  V =  . If x W and x  D, we have y  W \ D and x D. Hence (W \ D)  D =. This shows that X is sgp-D 2 .

Theorem 4.7:
If Y is sgp-D 1 and f  Y is a sgp-irresolute and bijective function, then X is sgp-D 1. Proof: Suppose that Y is sgp-D 1 space. Let x and y be any pair of distinct points in X. Since f is injective and Y is sgp-D 1 , there exist sgp-D-sets S x and S y of S containing f (x) and f (y) respectively, such that f (y) S y . By the Theorem 4.6, f -1 (S x ) and f -1 (S y ) are sgp-D-sets in X containing x and y respectively. This implies that X is a sgp-D 1 space.

Theorem 4.8:
A space X is sgp-D 1 if and only if for each pair of distinct points x and y in X, there exists a sgp-irresolute surjective function f from X onto sgp-D 1 space Y such that f (x) ≠ f (y) Proof: Necessity: For every pair of distinct points of X it suffices to take the identity mapping on X. Sufficiency: Let x and y be any pair of distinct points in X. By hypothesis, there exists a sgpirresolute, surjective function f of a space X onto a sgp-D 1 space Y such that f (x) ≠ f (y). Therefore, there exist disjoint sgp-D-sets S x and S y in Y such that f (x) S x and f (y) S y. Since f is sgp-irresolute and surjective, by Theorem 4.6, f -1 (S x ) and f -1 (S y ) are disjoint sgp-D-sets in X containing x and y respectively. Hence by Theorem 6.4 (ii), X is a sgp-D 1 space.