# [disjunction] and [conjunction]-sets of generalized topologies.

1. Introduction

In 199[gamma], Professor A. Csaszar [1] nicely presented the open sets and all weak forms of open sets in a topological space X in terms of monotonic functions defined on [??](X), the collection of all subsets of X. For each such function [gamma], he defined a collection [mu] of subsets of X, called the collection of [gamma]-open sets. A is said to be [gamma]-open if A [subset] [gamma](A). B is said to be [gamma]-closed if its complement is [gamma]-open. With respect to this collection of subsets of X, for A [subset] X, the [gamma]-interior of A, denoted by [i.sub.[gamma]](A), is defined as the largest [gamma]-open set contained in A and the [gamma]-closure of A, denoted by [i.sub.[gamma]](A), is the smallest [gamma]-closed set containing A. It is established that [mu] is a generalized topology [3]. In [5], [gamma]-semiopen sets are defined and discussed. In [7], [gamma][alpha]-open sets, [gamma]-preopen sets and [gamma]-open sets are defined and discussed. [gamma][beta]-open sets are defined in [6]. If [alpha] is the family of [gamma][alpha]-open sets, is the family of all [gamma]-semiopen sets, is the family of all [gamma]-preopen sets, [pi] is the family of all [gamma][beta]-open sets and [beta] is the family of all [gamma]-open sets, then each collection is a generalized topology. Since every topological space is a generalized topological space, we prove that some of the results established for topological spaces are also true for the generalized topologies [Omega] = {[mu], [alpha], [sigma], [pi], b, [beta]}. In section 2, we list all the required definitions and results. In section 3, we define the [[conjunction].sub.k] and [[disjunction].sub.k], operators for each k [member of] [Omega] and discuss its properties. Then, we define [[conjunction].sub.k]-sets, [[disjunction].sub.k]-sets, g. [[conjunction].sub.k]-sets and g. [[disjunction].sub.k]-sets and characterize these sets. In section 4, for each k [member of] [Omega], we define and characterize the separation axioms k - [T.sub.i], i = 0, 1, 2 and k - [R.sub.i], i = 0, 1.

2. Preliminaries

Let X be a nonempty set and [Gamma] = {[gamma]: [??](X) [right arrow] [??](X)|[gamma](A) [subset] [gamma](B) whenever A [subset] B}. For [gamma] [member of] [Gamma], a subset A [subset] X is said to be [gamma]-open [1] if A [subset] [gamma](A). The complement of a [gamma]-open set is said to be a [gamma]-closed set. A family [xi] [subset] [??](X) is said to be a generalized topology [3] if 0 [member of] [xi] and [xi] is closed under arbitrary union. The family of all [gamma]-open sets, denoted by [mu], is a generalized topology [4]. A [subset] X, is said to be [gamma]-semiopen [5] if there is a [gamma]-open set G such that G [subset] A [subset] [c.sub.[gamma]](G) or equivalently, A [subset] [c.sub.[gamma]][i.sub.[gamma]]. (A) [8, Theorem 2.4]. A is said to be [gamma]-preopen [7] if A [subset] [i.sub.[gamma]][c.sub.[gamma]] (A). A [subset] X, is said to be [gamma][alpha]-open [7] if A [subset] [i.sub.[gamma]][c.sub.[gamma]][i.sub.[gamma]] (A). A is said to be [gamma][beta]-open [7] if A [subset] [c.sub.[gamma]][i.sub.[gamma]][c.sub.[gamma]] (A). A is said to be [gamma][beta]-open [6] if A [subset] [i.sub.[gamma]][c.sub.[gamma]](A) [union] [c.sub.[gamma]][i.sub.[gamma]] (A). In [4], [5], [6] and [7], it is established that each [member of] [Omega] is a generalized topology and so [c.sub.k] and [i.sub.k], can be defined, similar to the the definition of [c.sub.[gamma]] and [i.sub.[gamma]]. In this paper, for k [member of] [Omega], the pair (X, k) is called a generalized topological space or simply a space. For each [gamma] [member of] [Gamma], a mapping [gamma]*: [??](X) [right arrow] [??](X) [1] is defined by [gamma]*(A) = X - [gamma](X -A). Clearly, [gamma]* [member of] [Gamma]. The following lemmas will be useful in the sequel. Moreover, one can easily prove the following already established results of Lemma 2.1.

Lemma 2.1. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following hold.

(a) A [subset] [c.sub.k] (A) for every subset A of X.

(b) If A [subset] B, then [c.sub.k](A) [subset] [c.sub.k](B).

(c) For every subset A of X, x [member of] [c.sub.k] (A) if and only if there exists a k-open set G such that A [intersection] G [not equal to] 0 (For k = [mu], it is established in Lemma 2.1 of [2]).

(d) A is k-closed if and only if A = [c.sub.k](A).

(e) [c.sub.k] (A) is the intersection of all k-closed sets containing A.

Lemma 2.2. Let X be a nonempty set and [gamma] [member of] [Gamma](X). Then X is [gamma]-semiopen [5, Proposition 1.2].

3. U-sets and n-sets

In this section, for the space (X, k), k [member of] [Omega], we define [disjunction]-sets, [conjunction]-sets, g. [disjunction]-sets and g. [conjunction]-sets. For [sigma] [member of] [Omega], [disjunction]-sets and [conjunction]-sets are defined in [5]. For A [subset] X, we define [[conjunction].sub.k] (A) = [intersection] {U [subset] X| A [subset] U and U [member of] k} and [[disjunction].sub.k] (A) = [union] {U [subset] X|U [subset] A and U is k-closed}. The following Theorem 3.1 gives the properties of the operator [[conjunction].sub.k]. Example 3.2 below shows that the two sets in 3.1(e) are not equal.

Theorem 3.1. Let A, B and {[C.sub.[??]]|[??] [member of] [Delta]} be subsets of X, [gamma] [member of] [Gamma] and [member of] [Omega]. Then the following hold.

(a) If A [subset] B, then [[conjunction].sub.k](A) [subset] [[conjunction].sub.k](B).

(b) A [subset] [[conjunction].sub.k](A).

(c) [[conjunction].sub.k] ([[conjunction].sub.k](A)) = [[conjunction].sub.k](A).

(d) [[conjunction].sub.k] ([union]{[C.sub.[??]]|[??][member of][Delta]}) = [union]{[[conjunction].sub.k]([C.sub.[??]])|[??][member of][Delta]}

(e) [[conjunction].sub.k] ([intersection]{[C.sub.[??]]|[??][member of][Delta]}) [subset] [intersection]{[[conjunction].sub.k]([C.sub.[??]])|[??][member of][Delta]}.

(f) If A [member of] k, then [[conjunction].sub.k](A) = A.

(g) [[conjunction].sub.k](A) = {x|[c.sub.k]({x})[intersection] A [not equal to] 0}.

(h) y [member of] [[conjunction].sub.k]({x}) if and only if x [member of] [c.sub.k]({y}).

(i) [[conjunction].sub.k]({x}) [not equal to] [[conjunction].sub.k]({y}) if and only if [c.sub.k]({x}) [not equal to] [c.sub.k]({y}).

Proof.

(a). Suppose x [??] [[conjunction].sub.k](B). Then there exists G [member of] k such that B [subset] G and x [??] G. Since A [subset] B, there exists G [member of] k such that A [subset] G and x [??] G and so x [??] [[conjunction].sub.k](A) which proves (a).

(b). The proof follows from the definition of [[conjunction].sub.k].

(c). By (b), A [subset] [[conjunction].sub.k](A) and so by (a), [[conjunction].sub.k](A) [subset] [[conjunction].sub.k]([[conjunction].sub.k](A)). Let x [??][[conjunction].sub.k](A). then there exists G [member of] such that A [subset] G and x [??] G which implies that [[conjunction].sub.k](A) [subset] G and x [??] G. Therefore, x [??] [[conjunction].sub.k] ([[conjunction].sub.k](A)) which implies that [[conjunction].sub.k]([[conjunction].sub.k](A)) [subset] [[conjunction].sub.k](A). This completes the proof.

(d) Clearly, by (a), [union]{[[conjunction].sub.k] ([C.sub.[??]]) [??] [member of] [Delta]} [subset] [[conjunction].sub.k]([union]{[C.sub.[??]]|[??] [member of] [Delta]}). Conversely, suppose x [??] [union] {[[conjunction].sub.k]([C.sub.[??]])|[??] [member of] [Delta]}. Then x [??][[conjunction].sub.k]([C.sub.[??]]) for every [??] [member of] [Delta]. Therefore, for every [??] [member of] [Delta], there exists [G.sub.[??]] [member of] k such that [C.sub.[??]] [subset] [G.sub.[??]] and x [??][G.sub.[??]]. Let G = [union]{[G.sub.[??]]|[??] [member of] [Delta]}. Then x [??] G and [union]{C [member of] [Delta]} [subset] G which implies that x [??] [[conjunction].sub.k]([union]{[C.sub.[??]]|[??] [member of] [Delta]}). This completes the proof.

(e) The proof follows from (a).

(f) The proof follows from the definition of n.

(g) Let x [member of] [[conjunction].sub.k](A). If [c.sub.k]({x}) [intersection] A = 0, then X - [c.sub.k]({x}) is a k-open set such that A [subset] X - [c.sub.k]({x}) and x [??] X - [c.sub.k]({x}). Therefore, x [??] [[conjunction].sub.k](A), a contradiction to the assumption and so [c.sub.k]({x}) [intersection] A [not equal to] 0. Hence [[conjunction].sub.k](A) [subset] {x|[c.sub.k]({x})[intersection] A [not equal to] 0}. Conversely, suppose for x [member of] X, [c.sub.k]({x}) [intersection] A [not equal to] 0. If x [??][[conjunction].sub.k](A), then there exists a k-open set G such that A [subset] G and x [??] G. Therefore, x [member of] X - G which implies that [c.sub.k]({x}) [subset] [c.sub.k] (X - G) = X - G [subset] X - A and so [c.sub.k]({x}) [intersection] A = 0, a contradiction. Therefore, {x| [c.sub.k] ({x}) [intersection] A [not equal to] 0} [subset] [[conjunction].sub.k](A). This completes the proof.

(h) Suppose y [member of] [[conjunction].sub.k]({x}). Then y [member of] G whenever G is a k-open set containing x. Suppose x [??] [c.sub.k] ({y}), then there is a k-closed set F such that {y} [subset] F and x [??] F. Since X - F is a k-open set containing x, y [member of] F and so [c.sub.k]({y}) [subset] [c.sub.k](F) = F which implies that [c.sub.k]({y})[intersection]{x} = 0. By (g), y [??][[conjunction].sub.k]({x}), a contradiction. Hence x [member of] [c.sub.k]({y}).

Conversely, suppose x [member of] [c.sub.k]({y}). If y [??] [[conjunction].sub.k]({x}), there exists a k-open set G containing x such that y [??] G. Now y [member of] X - G implies that [c.sub.k]({y}) [subset] [c.sub.k](X - G) = X - G [subset] X - {x} and so [c.sub.k]({y}) [intersection] {x} = 0 which implies that x [??] [c.sub.k]({y}), a contradiction to the hypothesis. Therefore, y [member of] [[conjunction].sub.k]({x}). This completes the proof.

(i) Suppose [[conjunction].sub.k]({x}) [not equal to] [[conjunction].sub.k]({y}). Assume that z [member of] [[conjunction].sub.k]({x}) and z [??] [[conjunction].sub.k]({y}). Then by (h) and (g), x [member of] [c.sub.k]({z}) and {y} [intersection] [c.sub.k]({z}) = 0 and so [c.sub.k]({x}) [subset] [c.sub.k]({z}) and {y} [intersection] [c.sub.k]({z}) = 0. Therefore, [c.sub.k]({x}) [intersection] {y} = 0 which implies that [c.sub.k]({x}) [c.sub.k]({y}).

Conversely, suppose [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Assume that z [member of] [c.sub.k]({x}) and z [??] [c.sub.k]({y}). By (h), x [member of] [[conjunction].sub.k]({z}) and y [??] [[conjunction].sub.k]({z}) and so [[conjunction].sub.k]({x}) [subset] [[conjunction].sub.k]({z}) and {y} [intersection] [[conjunction].sub.k]({z}) = 0 which implies that {y} [intersection] n({x}) = 0. Therefore, [[conjunction].sub.k]({y}) [[conjunction].sub.k]({x}) which completes the proof.

Example 3.2. Let X = {a, b} and [gamma]: [??](X) [right arrow] [??](X) be defined by [gamma](0) = 0, [gamma]({a}) = {b}, [gamma]({b}) = {b}, [gamma](X) = X. Then [mu] = {0, {b}, X}. If A = {a}, B = {b}, then [[conjunction].sub.[mu](A) = X, [[conjunction].sub.[mu]](B) = B and [[conjunction].sub.[mu] (A [intersection] B) = [[conjunction].sub.[mu]] (0) = 0. Since [[conjunction].sub.[mu]](A) [intersection] [[conjunction].sub.[mu]](B) = B, [[conjunction].sub.[mu]](A) [intersection] [[conjunction].sub.[mu]](B) (A [intersection] B).

The proof of the following Theorem 3.3 is similar to that of Theorem 3.1 and hence the proof is omitted. Example 3.4 shows that the two sets in 3.3(d) are not equal. Theorem 3.5 below gives the relation between the operators [[conjunction].sub.k] and [[disjunction].sub.k].

Theorem 3.3. Let A, B and {[C.sub.[??]]|[??] [subset] [member of] [Delta]} be subsets of X, [gamma] [member of] [Gamma] and k [member of] [Omega]. Then the following hold.

(a) If A [subset] B, then [[disjunction].sub.k](A) [subset] [[disjunction].sub.k](B).

(b) [[disjunction].sub.k] (A) [subset] A.

(c) [[disjunction].sub.k] ([[disjunction].sub.k](A)) = [[disjunction].sub.k](A).

(d) [[disjunction].sub.k] ([union]{C.sub.[??]|[??] [member of] [Delta]}) [contains] [union] {[[disjunction].sub.k](C.sub.[??])|[??] [member of] [Delta]}.

(e) [[disjunction].sub.k] ([intersection]{C.sub.[??]|[??] [member of] [Delta]}) = [intersection] {[[disjunction].sub.k](C.sub.[??])|[??] [member of] [Delta]}.

(f) If A is k-closed, then [[disjunction].sub.k] (A) = A.

Example 3.4. Let X = {a, b, c, d} and [gamma]: [??](X) [right arrow] [??](X) be defined by [gamma](0) [gamma]({a}) = {b}, [gamma]({b}) = {c}, [gamma]({c}) = {b}, [gamma]({d}) = {d}, [gamma]({a, b}) = {b, c}, [gamma]({a, c}) = {b}, [gamma]({a, d}) = {b, d}, [gamma]({b, c}) = {b, c}, [gamma]({b, d}) = {[c, d}, [gamma]({[c, d}) = {b, d}, [gamma]({a, b, c}) = {b, c}, [gamma]({b, c, d}) = {b, c, d}, [gamma]({a, c, d}) = X, [gamma]({a, b, d}) {b, c, d} and y(X) = X. Then [mu] = {0, {d}, {b, c}, {a, c, d}, {b, c, d}, X}. If A = {a}, B = {d}, then [V.sub.[mu]] (A) {a}, [V.sub.[mu]] (B) = 0 and [V.sub.[mu]] (AU B) = [V.sub.[mu]] ({a, d}) {a, d}. Since [V.sub.[mu]] (A) [union] [V.sub.[mu]] (B) = A, [V.sub.[mu]] (A) [union] [V.sub.[mu]] (B) [V.sub.[mu]] (A [union] B). If A = {a, c, d}, then [V.sub.[mu]] (A) = A but A is not [mu]-closed. This shows that the reverse direction of Theorem 3.3 (f) is not true.

If A = {[c, d}, then [[conjunction].sub.k](A) = A but A is not [mu]-open. This shows that the reverse direction of Theorem 3.1(f) is not true.

Theorem 3.5. Let A be a subset of X, [gamma] [member of] [Gamma] and [member of] [Omega]. Then the following hold.

(a) [[conjunction].sub.k](X - A) = X - [[conjunction].sub.k](A).

(b) [[disjunction].sub.k](X - A) = X - [[disjunction].sub.k](A).

(c) [[conjunction].sub.k]* = [[disjunction].sub.k].

(d) [[disjunction].sub.k]* = [[conjunction].sub.k].

Proof. (a) and (b) follow from the definitions of [[conjunction].sub.k] and [[disjunction].sub.k].

(c) If A [subset] X, then ([[conjunction].sub.k])* (A) = X - [[conjunction].sub.k] (X -A) = X - (X - [[disjunction].sub.k](A)) = [[disjunction].sub.k](A) and so ([[conjunction].sub.k])* = [[disjunction].sub.k].

(d) The proof is similar to the proof of (c).

If X is a nonempty set, [gamma] [member of] [Gamma] (X) and k [member of] [Omega], a subset A of X is said to be a [[disjunction].sub.k]-set if A = [[disjunction].sub.k](A) and A is said to be a [[conjunction].sub.k]-set if A = [[conjunction].sub.k](A). In any space (X, k), the following Theorem 3.6 lists out the [[disjunction].sub.k]-sets and the [[conjunction].sub.k]-sets.

Theorem 3.6. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then the following hold.

(a) 0 is a [[conjunction].sub.k]-set.

(b) A is a [[conjunction].sub.k]-set if and only if X - A is a [[disjunction].sub.k]-set.

(c) X is a [[disjunction].sub.k]-set.

(d) The union of [[conjunction].sub.k]-sets is again a [[conjunction].sub.k]-set.

(e) The union of [[disjunction].sub.k]-sets is again a [[disjunction].sub.k]-set.

(f) The intersection of [[conjunction].sub.k]-sets is again a [[conjunction].sub.k]-set.

(g) The intersection of [[disjunction].sub.k]-sets is again a [[disjunction].sub.k]-set.

(h) If k [member of] [[Omega].sub.1] = [Omega] - {[mu], [alpha], [pi]}, then X is a [[conjunction].sub.k]-set and so 0 is a [[disjunction].sub.k]-set.

Proof.

(a) follows from Theorem 3.1(f) since 0 [member of] k for every [member of] [Omega].

(b) Suppose A is a [[conjunction].sub.k]-set. Then A = [[conjunction].sub.k](A). Now X - A = X - [[conjunction].sub.k](A) = [[disjunction].sub.k](X - A), by Theorem 3.5(b). Therefore, X - A is a [[disjunction].sub.k]-set. The proof of the converse is similar with follows from Theorem 3.5(a).

(c) follows from (a) and (b).

(d) Let {[A.sub.[??]]|[??] [member of] [Delta]} be a family of [[conjunction].sub.k]-sets. Therefore, [A.sub.[??]] = [[conjunction].sub.k]([A.sub.[??]]) for every [member of] [Delta]. Now [[conjunction].sub.k]([union]{[A.sub.[??]]|[??] [member of] [Delta]}) = [union] {[[conjunction].sub.k] ([A.sub.[??]])|[??] [member of] [Delta]}, by Theorem 3.1(d) and so [[conjunction].sub.k]([union]{[A.sub.[??]]|[??] [member of] [Delta]}) = [union]{[A.sub.[??]]|[??] [member of] [Delta]}.

(e) Let {[A.sub.[??]]|[??] [member of] [Delta]} be a family of [[disjunction].sub.k]-sets. Therefore, [A.sub.[??]] = [[disjunction].sub.k]([A.sub.[??]]) for every [member of] [Delta]. Now [union]{[A.sub.[??]]|[??] [member of] [Delta]} = [union] {[[disjunction].sub.k] ([A.sub.[??]])|[??] [member of] [Delta]} [subset] [[disjunction].sub.k]([union]{[A.sub.[??]]|[??] [member of] [Delta]}) by Theorem 3.3(d) and so [[disjunction].sub.k]([union]{[A.sub.[??]]|[??] [member of] [Delta]}) = ([union]{[A.sub.[??]]|[??] [member of] [Delta]}) = U{A [member of] O} by Theorem 3.3(b).

(f) Let {{[A.sub.[??]]|[??] [member of] [Delta]} be a family of [[conjunction].sub.k]-sets. Therefore, [A.sub.[??]] = [[conjunction].sub.k]([A.sub.[??]]) for every [??] [member of] [Delta]. Now [intersection]{[A.sub.[??]]|[??] [member of] [Delta]} = [intersection]{[[conjunction].sub.k] [A.sub.[??]]|[??] [member of] [Delta]} [contains] [[conjunction].sub.k] [intersection]{[A.sub.[??]]|[??] [member of] [Delta]}) by Theorem 3.1 (e) and so [[conjunction].sub.k] [intersection]{[A.sub.[??]]|[??] [member of] [Delta]}) = [intersection]{[A.sub.[??]]|[??] [member of] [Delta]} by Theorem 3.1(b).

(g) Let {[A.sub.[??]]|[??] [member of] [Delta]} be a family of [[disjunction].sub.k]-sets. Therefore, [A.sub.[??]] = [[disjunction].sub.k]([A.sub.[??]]) for every [??] [member of] [Delta]. Now [[disjunction].sub.k] [intersection]{[A.sub.[??]]|[??] [member of] [Delta]}) = [intersection] {[[disjunction].sub.k], ([A.sub.[??]])|[??] [member of] [Delta]}, by Theorem 3.3(e) and so [[disjunction].sub.k] [intersection]{[A.sub.[??]]|[??] [member of] [Delta]}) = [intersection] {[A.sub.[??]]|[??] [member of] [Delta]}.

(h) Since X [member of] [sigma] by Lemma 2.2, X [member of] k for every [member of] [[Omega].sub.1] and so the proof follows from (a) and (b).

Remark 3.7. Let [[tau].sub.k] = {A [subset] X|A = [[conjunction].sub.k](A)} and [[tau].sup.k] = {A [subset] X|A = [[disjunction].sub.k] (A)}. Then [[tau].sub.k] and [[tau].sup.k] are topologies by Theorem 3.6, such that arbitrary intersection of [[tau].sub.k]-open sets is a [[tau].sub.k]-open set and an arbitrary intersection of [[tau].sup.k]-open sets is a [[tau].sup.k]-open set.

Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. A subset A of X is called a generalized [[conjunction].sub.k]-set (in short, g. [[conjunction].sub.k]-set) if [[conjunction].sub.k](A) [subset] F whenever A [subset] F and F is k-closed. B is called a generalized [[disjunction].sub.k]-set (in short, g. [[disjunction].sub.k]-set) if X - B is a g. [[conjunction].sub.k]-set. We will denote the family of all g. [[conjunction].sub.k]-sets by [D.sup.[[conjunction].sub.k]] and the family of all g. [[disjunction].sub.k]-sets by [D.sup.[[disjunction].sub.k]]. The following Theorem 3.8 shows that [D.sup.[[conjunction].sub.k]] is closed under arbitrary union and D is closed under arbitrary intersection. Theorem 3.9 below gives a characterization of g. [[disjunction].sub.k]-sets.

Theorem 3.8. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then the following hold.

(a) If [B.sub.[??]] [member of] [D.sup.[[conjunction].sub.k]] for every [??] [member of] [Delta], then [union]{[B.sub.[??]]|[??] [member of] [Delta]} [member of] [D.sup.[[conjunction].sub.k]].

(b) If [B.sub.[??]] [member of] [D.sup.[[disjunction].sub.k]] for every [??] [member of] [Delta], then [intersection]{[B.sub.[??]]|[??] [member of] [Delta]} [member of] [D.sup.[[disjunction].sub.k]].

Proof. (a) Let [B.sub.[??]] [member of] [D.sup.[[conjunction].sub.k]] for every [??] [member of] [Delta]. Then each [B.sub.[??]] is a g. [[conjunction].sub.k]-set. Suppose F is k-closed and [union] {[B.sub.[??]]|[??] [member of] [Delta]} [subset] F. Then for every [??] [subset] [member of] [Delta], [B.sub.[??]] [subset] F and F is k-closed. By hypothesis, for every [??] [member of] [Delta], [[conjunction].sub.k]([B.sub.[??]]) [subset] F and so [union]{[[conjunction].sub.k] ([B.sub.[??]]) [??] [member of] [Delta]} [subset] F. By Theorem 3.1(d), [[conjunction].sub.k] ([union]{[B.sub.[??]]|[??] [member of] [Delta]}) [subset] F and so [union]{[B.sub.[??]]|[??] [member of] [Delta]} [member of] [D.sup.[[conjunction].sub.k]].

(b) Let [B.sub.[??]] [member of] [D.sup.[[disjunction].sub.k]] for every [??] [member of] [Delta]. Then each [B.sub.[??]] is a g. [[disjunction].sub.k]-set and so X - [B.sub.[??]] [member of] [D.sup.[[conjunction].sub.k]] for every [??] [member of] [Delta]. Now X - ([intersection]{[B.sub.[??]] |[??][member of] [Delta]}) = [union]({X - [B.sub.[??]]|[??][member of] [Delta]) [member of] [D.sup.[[conjunction].sub.k]], by (a). Therefore, [intersection]{[B.sub.[??]] |[??][member of] [Delta]} [member of] [D.sup.[[disjunction].sub.k]].

Theorem 3.9. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then a subset A of X is a g. [[disjunction].sub.k]-set if and only if U [subset] [[disjunction].sub.k](A) whenever U [subset] A and U is k-open.

Proof. Suppose A is a g. [[disjunction].sub.k]-set. Let U be a k-open set such that U [subset] A. Then X - U is a k-closed set such that X - U [contains] X - A and so [[conjunction].sub.k](X - U) [contains] [[conjunction].sub.k](X - A). Therefore, X - U [contains] [[conjunction].sub.k] (X - A) = X - [[disjunction].sub.k](A) and so U [subset] [[disjunction].sub.k](A). Conversely, suppose the condition holds. Let A be a subset of X. Let F be a k-closed subset of X such that X - A [subset] F. Then X - F [subset] A and so by hypothesis, X - F [subset] [[disjunction].sub.k](A). Then X - [[disjunction].sub.k](A) [subset] F and so [[conjunction].sub.k](X - A) [subset] F which implies that X - A is a g. [[conjunction].sub.k]-set. Therefore, A is a g. [[disjunction].sub.k]-set.

The remaining theorems in this section give some properties of g. [[disjunction].sub.k]-sets and g. [[conjunction].sub.k]-sets.

Theorem 3.10. Let x [member of] X, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then the following hold.

(a) {x} is either a k-open set or X - {x} is a g. [[conjunction].sub.k]-set.

(b) {x} is either a k-open set or a g. [[disjunction].sub.k]-set.

Proof. (a) Suppose {x} is not a k-open set. Then X is the only k-closed set containing X - {x} and so [[conjunction].sub.k](X - {x}) [subset] X. Therefore, X - {x} is a g. [[disjunction].sub.k]-set.

(b) Suppose {x} is not a k-open set. By (a), X - {x} is a g. [[conjunction].sub.k]-set and so {x} is a g. [[disjunction].sub.k]-set.

Theorem 3.11. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. If B is a g. [[disjunction].sub.k]-set and F is a k-closed set such that [[disjunction].sub.k](B) [union] (X - B) [subset] F, then F = X.

Proof. Since B is a g. [[disjunction].sub.k]-set, X - B is a g. [[conjunction].sub.k]-set such that X - B [subset] F. Therefore, [[conjunction].sub.k](X - B) [subset] F which implies that X - F [subset] [[disjunction].sub.k] (B). Also, [[disjunction].sub.k](B) [subset] F and so X - F [subset] X - [[disjunction].sub.k](B). Hence X - F [subset] [[disjunction].sub.k](B) [intersection] (X - [[disjunction].sub.k](B)) . Therefore, F = X.

Corollary 3.12. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. If B is a g. [[disjunction].sub.k]-set such that [[disjunction].sub.k](B) U (X - B) is k-closed, then B is a [[disjunction].sub.k]-set.

Proof. By Theorem 3.11, [[disjunction].sub.k](B) U (X - B) = X and so X - ([[disjunction].sub.k](B) [union] (X - B)) which implies that (X - [[disjunction].sub.k](B)) [intersection] B = 0. Therefore, B [subset] [[disjunction].sub.k](B) and so by Theorem 3.3(b), B = [[disjunction].sub.k](B).

Theorem 3.13. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. If A and B are subsets of X such that A [subset] B [subset] [[conjunction].sub.k](A) and A is a g. [[conjunction].sub.k]-set, then B is a g. [[conjunction].sub.k]-set. In particular, if A is a g. [[conjunction].sub.k]-set, then [[conjunction].sub.k](A) is a g. [[conjunction].sub.k]-set.

Proof. Since A [subset] B [subset] [[conjunction].sub.k](A), [[conjunction].sub.k](A) [subset] [[conjunction].sub.k](B) [subset] [[conjunction].sub.k]([[conjunction].sub.k] (A)) = [[conjunction].sub.k](A) and so [[conjunction].sub.k](A) [[conjunction].sub.k](B). If F is any k-closed set such that B [subset] F, then A [subset] F and so [[conjunction].sub.k](B) = [[conjunction].sub.k](A) [subset] F. Therefore, B is a g. [[conjunction].sub.k]-set.

Corollary 3.14. Let X be a nonempty set, [gamma][member of] [Gamma](X) and [member of] [Omega]. If A and B are subsets of X such that [[disjunction].sub.k](A) [subset] B [subset] A and A is a g. [[disjunction].sub.k]-set, then B is a g. [[disjunction].sub.k]-set. In particular, if A is a g. [[disjunction].sub.k]-set, then [[disjunction].sub.k](A) is a g. [[disjunction].sub.k]-set.

4. Some separation axioms in generalized topological spaces

Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then (X, k) is called a - [T.sub.1] space if for all x, y [member of] X, x [not equal to] y, there exists a k-open set G such that x [member of] G and y [??] G and there exists a k-open set H such that x [??] H and y [member of] H. [sigma]- [T.sub.1] space is defined in [5]. The following Theorem 4.1 gives a characterization of k - [T.sub.1] spaces in terms of k-closed sets and Theorem 4.2 gives a characterization of k - [T.sub.1] spaces in terms of [[conjunction].sub.k]-sets.

Theorem 4.1. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k - [T.sub.1] space if and only if every singleton set is a k-closed set.

Proof. Suppose (X, k) is a k - [T.sub.1] space. Let x [member of] X. If g [member of] X - {x}, then x g. By hypothesis, there exists a k-open set [H.sub.y] such that y [member of] [H.sub.y] and x [??][H.sub.y] and so y [member of] [H.sub.y] [subset] X -{x}. Hence X - {x} = [union] {[H.sub.y]|y [member of] X - {x}} is k-open and so {x} is -closed. Conversely, suppose each singleton set is a k-closed set. Let x, y [member of] X such that x [not equal to] y. Then X - {x} and X - {y} are k-open sets such that y [member of] X - {x} and x [member of] X - {y}. Therefore, (X, k) is a k - [T.sub.1] space.

Theorem 4.2. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a - [T.sub.1] space if and only if every subset of X is a [[conjunction].sub.k]-set.

Proof. Suppose (X, k) is a k - [T.sub.1] space. Let A be subset of X. By Theorem 3.1(b), A [subset] [[conjunction].sub.k](A). Suppose x [??] A. Then X - {x} is a k-open set such that A [subset] X - {x} and so [[conjunction].sub.k](A) [subset] X - {x}. Hence every subset of X is a [[conjunction].sub.k]-set. Conversely, suppose every subset of X is a [[conjunction].sub.k]-set and so [[conjunction].sub.k]({x}) = {x} for every x [member of] X. Let x, y [member of] X such that x [not equal to] y. Then y [??] [[conjunction].sub.k]({x}) and x [??][[conjunction].sub.k]({y}). Since y [??] [[conjunction].sub.k]({x}), there is a k-open set U such that x [member of] U and y [??] U. Similarly, since x [??] [[conjunction].sub.k]({y}), there is a k-open set V such that y [member of] V and x [??] V. Therefore, (X, k) is a k - [T.sub.1] space.

Corollary 4.3. Let X be a nonempty set, [member of] [Gamma](X) and k [member of] [Omega]. Then the following are equivalent.

(a) (X, k) is a - [T.sub.1] space.

(b) Every subset of X is a [[conjunction].sub.k]-set.

(c) Every subset of X is a [[disjunction].sub.k]-set.

Proof. (a) and (b) are equivalent by Theorem 4.2. (b) and (c) are equivalent by Theorem 3.5(b).

Theorem 4.4. Let X be a nonempty set, [gamma] [member of] [Gamma](X), [member of] [Omega] and A [subset] X. Then the following hold.

(a) If A is a [[conjunction].sub.k]-set, then A is a g. [[conjunction].sub.k]-set.

(b) If A is a [[disjunction].sub.k]-set, then A is a g. [[disjunction].sub.k]-set. The reverse directions are true if (X, k) is a k - [T.sub.1] space.

Proof.

(a) Suppose A is a [[conjunction].sub.k]-set. Then A = [[conjunction].sub.k](A). If A [subset] F where F is k-closed, then A = [[conjunction].sub.k](A) [subset] F. Therefore, A is a g. [[conjunction].sub.k]-set.

(b) The proof is similar to the proof of (a).

The reverse directions follow from Corollary 4.3.

Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. (X, k) is said to be a k - [T.sub.0] space if for distinct points x and y in X, there exists a k-open set G containing one but not the other. Clearly, every - [T.sub.1] space is a - [T.sub.0] space. Since every topology is a generalized topology and in a topological space, [T.sub.0] spaces need not be [T.sub.1] spaces, - [T.sub.0] spaces need not be - [T.sub.0] spaces. Theorem 4.5 gives a characterization of - [T.sub.0] spaces. The easy proof of Corollary 4.6 is omitted.

Theorem 4.5. Let (X, k) be a k-space where [gamma][member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k - [T.sub.0] space if and only if distinct points of X have distinct k-closures.

Proof. Suppose (X, k) is a k - [T.sub.0] space. Let x and y be points of X such that x [not equal to] y. Then there exists a k-open set G containing one but not the other, say x [member of] G and y [??] G. Then y [??] [c.sub.k]({x}) and so [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Conversely, suppose distinct points of X have distinct k-closures. Let x and y be points of X such that x [not equal to] y. Then [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Suppose z [member of] [c.sub.k]({x}) and z [??] [c.sub.k]({y}). If x [subset] [c.sub.k]({y}), then [c.sub.k]({x}) [subset] [c.sub.k]({y}) and so z [member of] [c.sub.k]({y}), a contradiction. Therefore, x [??] [c.sub.k]({y}) which implies that x [member of] X - [c.sub.k]({y}) and X - [c.sub.k]({y}) is k-open. Hence (X, k) is a k - [T.sub.0] space.

Corollary 4.6. Let (X, k) be a k-space where [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k - [T.sub.0] space if and only if for distinct points x and y of X, either x [??] [c.sub.k]({y}) or y [??] [c.sub.k]({x}).

Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. (X, k) is said to be a k - [T.sub.2] space if for distinct points x and y in X, there exist disjoint k-open sets G and H such that x [member of] G and y [member of] H. Clearly, every k - [T.sub.2] space is a k - [T.sub.1] space and the converse is not true. Theorem 4.7 below gives characterizations of k - [T.sub.2] spaces.

Theorem 4.7. In a k-space (X, k), where [gamma] [member of] [Gamma](X) and [member of] [Omega], the following statements are equivalent.

(a) (X, k) is a k - [T.sub.2] space.

(b) For each x [member of] X and y [not equal to] x, there exists a k-open set U such that x [member of] U and y [??] [c.sub.k](U).

(c) For every x [member of] X, {x} = [intersection]{[c.sub.k](U) x [member of] U and U is k - open}.

Proof. (a)[??](b). Let x, y [member of] X such that y [not equal to] x. Then, there exists disjoint k-open sets U and H such that x [member of] U and y [member of] H. Then X - His a k-closed set such that U [subset] X - H and so [c.sub.k](U) [subset] X - H. U is the required k-open set such that x [member of] U and y [??] [c.sub.k](U).

(b)[??](c). Let x [member of] X. If y [member of] X such that x [not equal to] y, by (b), there exists a k-open set U such that x [member of] U and y [c.sub.k](U). Clearly, {x} = [intersection]{[c.sub.k](U) x [member of] U and U is k - open}.

(c)[??](a). Let x, y [member of] X such that y [not equal to] x. Then y [??] {x} = [intersection]{[c.sub.k](U) x [member of] U and U is k-open}, by (c). Therefore, y [??] [c.sub.k](U) for some k-open set containing x. U and X - [c.sub.k](U) are the required disjoint k-open sets containing x and y respectively.

Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is said to be a k - [R.sub.0] space if every k-open subset of X contains the k-closure of its singletons. (X, k) is said to be a k - [R.sub.1] space if for x, y [member of] X with [c.sub.k]({x}) [c.sub.k]({y}), there exist disjoint k-open sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H. Clearly, every k - [R.sub.1] space is a k - [R.sub.0] space but the converse is not true. The following Theorem 4.8 follows from Theorem 4.1. Theorem 4.9 below gives a characterization of k - [R.sub.0] spaces.

Theorem 4.8. Let X be a nonempty set, [gamma] [member of] [Gamma] (X) and [member of] [Omega]. Then every - [T.sub.1] space is a - [R.sub.0] space.

Theorem 4.9. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k - [R.sub.0] space if and only if every k-open subset of X is the union of k-closed sets.

Proof. Suppose (X, k) is a k - [R.sub.0] space. If A is k-open, then for each x [member of] A, [c.sub.k]({x}) [subset] A and so [union]{[cl.sub.k],{x} x [member of] A} [subset] A. It follows that A = [union]{[cl.sub.k],{x} x [member of] A}. Conversely, suppose A is k-open and x [member of] A. Then by hypothesis, A = [union]{[B.sub.[??]]|[??] [subset] [member of] [Delta]} where each [B.sub.[??]] is k-closed. Now x [member of] A, implies that x [member of] [B.sub.[??]] for some [??] [member of] [Delta]. Therefore, [c.sub.k]({x}) [subset] [B.sub.[??]] [subset] A and so (X, k) is a k - [R.sub.0] space.

Theorem 4.10. Let X be a nonempty set, [gamma] [member of] [gamma] (X) and [member of] [Omega]. Then (X, k) is a k - [R.sub.0] space if and only if for x, y [member of] X, [c.sub.k]({x}) [c.sub.k]({y}) implies that [c.sub.k]({x}) [intersection] [c.sub.k]({y}) = 0.

Proof. Suppose (X, k) is a k - [R.sub.0] space. Let x,y [member of] X, such that [c.sub.k]({x}) [??][c.sub.k]({y}). Suppose z [member of] [c.sub.k]({x}) and z [c.sub.k]({y}). Since z [c.sub.k]({y}), there exists a k-open set G containing z such that y G. Since z [member of] [c.sub.k]({x}), x [member of] G. Since y G, it follows that x [c.sub.k]({y}) and so x [member of] X-[c.sub.k]({y}). By hypothesis, [c.sub.k]({x}) [subset] X-[c.sub.k]({y}) and so [c.sub.k]({x}) [intersection][c.sub.k]({y}) = 0. Conversely, suppose the condition holds. Let G be a k-open set such that x [member of] G. If y [??] G, then x [not equal to]y and so x [c.sub.k]({y}) which implies that [c.sub.k]({x}) [c.sub.k]({y}). By hypothesis, [c.sub.k]({x}) [intersection] [c.sub.k]({y}) and so y [c.sub.k]({x}). Hence [c.sub.k]({x}) [subset] G which implies that (X, k) is a k - [R.sub.0] space.

Theorem 4.11. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k - [R.sub.0] space if and only if for x, y [member of] X, [[conjunction].sub.k] ({x}) [not equal to] [[conjunction].sub.k]({y}) implies that [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({y}) = 0.

Proof. Suppose (X, k) is a k - [R.sub.0] space. Let x, y [member of] X such that [[conjunction].sub.k]({x}) [not equal to] [[conjunction].sub.k]({y}). Suppose that z [member of] [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({y}). Then z [member of] [[conjunction].sub.k]({x}) and z [member of] [[conjunction].sub.k]({y}). By Theorem 3.1(h), x [member of] [c.sub.k]({z}) and y [member of] [c.sub.k]({z}) and so [c.sub.k]({x}) [intersection][c.sub.k]({z}) [not equal to] 0 and [c.sub.k]({y}) [intersection][c.sub.k]({z}) [not equal to] 0. By Theorem 4.10, [c.sub.k]({x}) = [c.sub.k]({z}) and [c.sub.k]({y}) = [c.sub.k]({z}) and so [c.sub.k]({x}) = [c.sub.k]({y}). By Theorem 3.1(i), [[conjunction].sub.k]({x}) = [[conjunction].sub.k]({y}), a contradiction. Therefore, [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({y}) = 0. Conversely, suppose the condition holds. Let x, y [member of] X such that [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Suppose that z [member of] [c.sub.k]({x}) [intersection] [c.sub.k]({y}). Then z [member of] [c.sub.k]({x}) and z [member of] [c.sub.k]({y}). By Theorem 3.1(h), x [member of] [[conjunction].sub.k]({z}) and y [member of] [[conjunction].sub.k]({z}) and so [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({z}) [not equal to] 0 and [[conjunction].sub.k]({y}) [intersection] [[conjunction].sub.k]({z}) [not equal to] 0. By hypothesis, [[conjunction].sub.k]({x}) = [[conjunction].sub.k]({z}) and [[conjunction].sub.k]({x}) = [conjunction].sub.k]({z}) and so [[conjunction].sub.k] ({x}) = [[conjunction].sub.k],({y}) By Theorem 3.1(i), [c.sub.k]({x}) = [c.sub.k]({y}), a contradiction. Therefore, [c.sub.k]({x}) [intersection] [c.sub.k]({y}) = 0. By Theorem 4.10, (X, k) is a k - [R.sub.0] space.

Theorem 4.12. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equal.

(a) (X, k) is a k - [R.sub.0] space.

(b) For any nonempty set A and a k-open set G such that A [intersection] G [not equal to] 0, there exists a k-closed set F such that A [intersection] F [not equal to] 0 and F [subset] G.

(c) If G is k-open, then G = [union]{F|F [subset] G and F is k-closed}.

(d) If F is k-closed, then F = [intersection]{G|F [subset] G and G is k-open}.

(e) For every x [member of] X, [c.sub.k]({x}) [subset] [[conjunction].sub.k]({x}).

Proof. (a)[??](b). Suppose (X, k) is a k - [R.sub.0] space. Let A be a nonempty set and G be a k-open set such that A [intersection] G [not equal to] 0. If x [member of] A [intersection] G, then x [member of] G and so by hypothesis, [c.sub.k]({x}) [subset] G.

If F = [c.sub.k]({x}), then F is the required k-closed set such that A [intersection] F [not equal to] 0 and F [subset] G.

(b)[??](c). Let G be k-open. Clearly, G [contains][intersection] [union]{F|F [subset] G and F is k-closed}. If x [member of] G, then {x} [intersection] G [not equal to] 0 and so by (b), there is a k-closed set F such that {x} [intersection] F [not equal to] 0 and F [subset] G which implies that x [member of] {F|F [subset] G and F is k-closed}. Therefore, G [subset] {F|F [subset] G and F is k-closed}. This completes the proof.

(c)[??](d). Let F be k-closed. By (c), X - F = [union]{K|K [subset] X - F and K is k-closed} and so F = [intersection]{X - K|F [subset] X - K and X - K is k-open}.

(d)[??](e). Let x [member of] X. If y [??] [[conjunction].sub.k]({x}), then by Theorem 3.1(g), {x} [intersection] [c.sub.k]({y}) = 0. By (d), [c.sub.k]({y}) = [intersection]{G [c.sub.k]({y}) [subset] G and G is k-open}. Therefore, there is a k-open G such that [c.sub.k]({y}) [subset] G and x [??] G which implies that y [??] [c.sub.k]({x}). Therefore, [c.sub.k]({x}) [subset] [[conjunction].sub.k]({x}).

(e)[??](a). Let G be a k-open set such that x [member of] G. If y [member of] [c.sub.k]({x}), then by (e), y [member of] n({x}). Since [[conjunction].sub.k]({x}) [subset] [[conjunction].sub.k](G) = G, y [member of] G. Hence (X, k) is a k - [R.sub.0] space.

Corollary 4.13. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent.

(a) (X, k) is a k - [R.sub.0] space.

(b) For every x [member of] X, [c.sub.k]({x}) = [[conjunction].sub.k]({x}).

Proof. (a)(b). Let x [member of] X. By Theorem 4.12, [c.sub.k]({x}) [subset] [[conjunction].sub.k]({x}). To prove the direction, assume that y [member of] [[conjunction].sub.k]({x}). By Theorem 3.1(h), x [member of] [c.sub.k]({y}) and so [c.sub.k]({x}) [subset] [c.sub.k]({y}) which implies that [c.sub.k]({x}) [intersection] [c.sub.k]({y}) [not equal to] 0. By Theorem 4.10, [c.sub.k]({x}) = [c.sub.k]({y}) and so y [member of] [c.sub.k]({x}). Hence [c.sub.k]({x}) [[conjunction].sub.k]({x}).

(b)(a). The proof follows from Theorem 4.12.

Theorem 4.14. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent.

(a) (X, k) is a k - [R.sub.0] space.

(b) For all x, y [member of] X, x [member of] [c.sub.k]({y}) if and only if y [member of] [c.sub.k]({x}).

Proof. (a)[??](b). Let x, y [member of] X such that x [member of] [c.sub.k]({y}). By Corollary 4.13, x [member of] [[conjunction].sub.k]({y}) and so by Theorem 3.1(h), y [member of] [c.sub.k]({x}). Thus x [member of] [c.sub.k]({y}) implies that y [member of] [c.sub.k]({x}). Similarly, we can prove that y [member of] [c.sub.k]({x}) implies that x [member of] [c.sub.k]({y}).

(b)[??](a). Conversely, suppose the condition holds. Let x [member of] X. If y [member of] [c.sub.k]({x}), then by hypothesis, x [member of] [c.sub.k]({y}) and so by Theorem 3.1(h), y [member of] n({x}) which implies that [c.sub.k]({x}) [subset] n({x}). By Theorem 4.12(e), (X, k) is a k - [R.sub.0] space.

The following Theorem 4.15 gives a characterization of k - [R.sub.1] space and Theorem 4.16 gives a characterization of k - [T.sub.2] space.

Theorem 4.15. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent.

(a) (X, k) is a k - [R.sub.1] space.

(b) For x, y [member of] X such that [[conjunction].sub.k]({x}) [[conjunction].sub.k]({y}), there exist disjoint k-open sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H.

Proof. (a)[??](b). Let x, y [member of] X such that [[conjunction].sub.k]({x}) [[conjunction].sub.k]({y}). Then by Corollary 4.13, since (X, k) is a k - [R.sub.0] space, [c.sub.k]({x}) [c.sub.k]({y}) and so there exists disjoint k-open sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H.

(b)[??](a). Let x, y [member of] X such that [c.sub.k]({x}) [c.sub.k]({y}). By Theorem 3.1(i), [[conjunction].sub.k]({x}) [[conjunction].sub.k]({y}). By hypothesis, there exist disjoint k-open sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H and so (X, k) is a k - [R.sub.1] space.

Theorem 4.16. Let X be a nonempty set, y [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent.

(a)(X, k) is a k - [T.sub.2] space.

(b) (X, k) is both a - [R.sub.1] space and a k - [T.sub.1] space.

(c)(X, k) is both a - [R.sub.1] space and a k - [T.sub.0] space.

Proof. (a)[??](b). Suppose (X, k) is a k - [T.sub.2] space. Clearly, (X, k) is a k - [T.sub.1] space and so singletons are -closed sets, by Theorem 4.1. If x, y [member of] X with [c.sub.k]({x}) [c.sub.k]({y}), then x [not equal to] y and so there exist disjoint k-open sets G and H such that x [member of] G and y [member of] H. Therefore, [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H which implies that (X, k) is a k - [R.sub.1] space.

(b)[??](c). The proof is clear.

(c)[??](a). Let x, y [member of] X such that x [not equal to] y. Since (X, r) is a k - [T.sub.0] space, by Theorem 4.5, [c.sub.k]({x}) [c.sub.k]({y}). Since (X, k) is a k - [R.sub.1] space, there exist disjoint -open sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H. Therefore, (X, k) is a k - [T.sub.2] space.

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P. Sivagami ([dagger]) and D. Sivaraj ([double dagger])

([dagger]) Department of Mathematics, Kamaraj College, Thoothukudi-628 003, Tamil Nadu, India

([double dagger]) Department of Computer Science, D.J. Academy for Managerial Excellence, Coimbatore-641 032, Tamil Nadu, India

E-mail: ttn_sivagamiyahoo.co.in