WEAK SEPARATION AXIOMS VIA g ~ -OPEN SETS

Far East Journal of Mathematical Sciences (FJMS) Volume 30, Issue 3, 2008, Pages 457-471 Published online: October 17, 2008 This paper is available online at http://www.pphmj.com  2008 Pushpa Publishing House

~ -OPEN SETS WEAK SEPARATION AXIOMS VIA g MOHAMMAD SARSAK and NEELAMEGARAJAN RAJESH Department of Mathematics Faculty of Science Hashemite University P. O. Box 150459, Zarqa 13115, Jordan e-mail: [email protected] Department of Mathematics Kongu Engineering College Perundurai, Erode-638 052, Tamilnadu, India e-mail: [email protected] Abstract ~ -open sets are used to define some weak separation In this paper, g axioms and to study some of their basic properties. The implications of these axioms among themselves and with the known axioms Ti ,

(i = 0, 1 2, 1, 2) are investigated.

1. Introduction In 1970, Levine [5] initiated the study of the so-called g-closed sets. The notion has been studied extensively in recent years by many topologists. In the same paper Levine also introduced the notion of T1 2 spaces which properly lie between T1 spaces and T0 spaces. In a recent 2000 Mathematics Subject Classification: 54D10. ~ -open set, g ~ -kernel, g ~ -closure, g ~−T , Keywords and phrases: g 0

~ −T , g 2

~−R. ~ − R , weak g ~−R , g g 0 0 1

Submitted by Takashi Noiri Received May 6, 2008

~−T , g 12

~−T , g 1

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MOHAMMAD SARSAK and NEELAMEGARAJAN RAJESH

~ -closed sets were introduced and year, a generalization of closed sets, g studied by Jafari et al. [3]. This notion was further studied by Rajesh and Ekici [7-10]. In this paper, we continue the study of related spaces with ~ -open sets (i.e., complements of g ~ -closed sets). We introduce and g ~ − T , (i = 0, 1 2, 1, 2). characterize four new separation axioms called g i ~ − T , i = 0, 1 2, 1, 2 is weaker than T , i = 0, 1 2, 1, 2, We show that g i i respectively. Throughout this paper, a space stands for a topological space and a function f : X → Y denotes a function from a space X into a space Y. For a subset A of a space X, the closure and the interior of A in X are denoted by cl( A ) and int( A ) , respectively.

2. Preliminaries Before entering our work we recall the following definitions and results which are used in this paper.

Definition 2.1. A subset A of a space X is said to be semi-open [6] if A ⊂ cl(int( A )) . The complement of a semi-open set is called semi-closed. The intersection of all semi-closed subsets of X that contains A, or equivalently, the smallest semi-closed subset of X that contains A, is called the semi-closure of A [2] and is denoted by scl( A ) .

Definition 2.2. Let A be a subset of a space X. Then (i) A is generalized closed (briefly g-closed [5]) if cl( A ) ⊂ U whenever A ⊂ U and U is open in X. (ii) A is gˆ -closed [13] if cl( A ) ⊂ U whenever A ⊂ U and U is semi-open in X. The complement of a gˆ -closed set is called gˆ -open. (iii) A is ∗ g -closed [12] if cl( A ) ⊂ U whenever A ⊂ U and U is

gˆ -open in X. The complement of a ∗ g -closed set is called ∗ g -open. (iv) A is ‡ g -semi-closed [14] if scl( A ) ⊂ U whenever A ⊂ U and

U is ∗ g -open. The complement of a ‡ g -semi-closed set is called ‡ g semi-open.

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~ -closed [3] if cl( A ) ⊂ U whenever A ⊂ U and U is (v) A is g ~ -closed set is called g ~ -open. The g - semi-open. The complement of a g ~ -open (resp. g ~ -closed) subsets of X is denoted by g ~( X ) class of all g (resp. g~C ( X )). ‡

~ -closed (resp. g ~ -open) sets Definition 2.3. The intersection of all g ~ -closure (resp. g ~ -kernel) of A [10] and is containing A is called the g

~ − cl( A ) (resp. g ~ − ker ( A )). denoted by g Definition 2.4. A space X is called a T1 2 -space [5] if every g-closed subset of X is closed in X, or equivalently, if every singleton subset of X is open or closed.

Theorem 2.5 [3]. In any space X, the following hold:

~ -closed sets is g ~ -closed. (i) An arbitrary intersection of g ~ -closed sets is g ~ -closed. (ii) The finite union of g ~ -closed if and only if it coincides with its Remark 2.6. A subset is g ~ -closure. g Definition 2.7 [4]. A subset U x of a space X is said to be a ~ -neighborhood of a point x ∈ X if there exists a g ~ -open set G in X g such that x ∈ G ⊂ U x .

~ -open in X if and only if Lemma 2.8 [4]. A subset A of a space X is g ~ -neighborhood of each of its points. it is a g ~ is said to be g ~ -open in X. continuous if the inverse image of every open set in Y is g Definition 2.9 [8]. A function

f : X →Y

~ Definition 2.10 [7]. A function f : X → Y is said to be g ~ -open set in Y is g ~ -open in X. irresolute if the inverse image of every g

~∗ -closed Definition 2.11 [1]. A function f : X → Y is said to be g ~ -closed) if the image of every g ~ -closed (resp. closed) set in X is (resp. g ~ g -closed in Y.

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~∗ -open Definition 2.12 [1]. A function f : X → Y is said to be g ~ -open) if the image of every g ~ -open (resp. open) set in X is (resp. g ~ -open in Y. g ~ -regular if for each Definition 2.13 [9]. A space X is said to be g ~ -open closed subset F of X and each point x ∈ F c , there exist disjoint g sets U and V such that F ⊂ U and x ∈ V .

~ -irresolute if and only if Theorem 2.14. A function f : X → Y is g ~ -open subset W of Y and for each x ∈ X such that f (x ) ∈ W , for each g ~ -open subset U of X such that x ∈ U and then there exists a g f (U ) ⊂ W . ~ − T Spaces 3. g 0 ~ − T if to each pair of Definition 3.1. A space X is said to be g 0 ~ distinct points x, y of X there exists a g -open set A containing x but not y

~ -open set B containing y but not x. or a g Theorem 3.2. For a space X, the following are equivalent:

~−T . (i) X is g 0 (ii) For each x ∈ X , {x} =

∩ {F ∈ g~(X ) ∪ g~C (X ) : x ∈ F } = g~ − cl({x})

~ − ker ({x}). ∩g

Proof. The proof follows from the definitions.

~ − T if and only if for each pair of Theorem 3.3. A space X is g 0 ~ ~ distinct points x, y of X, g − cl({x}) ≠ g − cl({y}) . ~ − T space and x, y be any two Proof. Necessity. Let X be a g 0 ~ distinct points of X. There exists a g -open set G containing x but not y or ~ -closed set containing y but not x, say, x but not y. Thus X − G is a g ~ − cl({y}) is the smallest which does not contain x but contains y. Since g ~ -closed set containing y, g ~ − cl({y}) ⊂ X − G , and so x ∉ g ~ − cl({y}) . g ~ − cl({x}) ≠ g ~ − cl({y}) . Consequently, g

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~ − cl({x}) Sufficiency. Let x , y ∈ X , x ≠ y. Then by assumption, g ~ − cl({y}). Thus there exists a point z ∈ X such that z belongs to ≠ g ~ − cl({x}) but not to g ~ − cl({y}) or z belongs to g ~ − cl({y}) but not to g ~ − cl({x}) , say, g ~ − cl({x}) but not to g ~ − cl({y}) . If we suppose that g ~ − cl({y}), then z ∈ g ~ − cl({x}) ⊂ g ~ − cl({y}), which is a contradiction. x∈g

~ -open and does ~ − cl({y})) is g ~ − cl({y})), but X − ( g Thus x ∈ X − ( g ~−T . not contain y, hence X is g 0

~ -closure Definition 3.4. A function f : X → Y is said to be point g ~ − cl({x}) ≠ g ~ − cl({y}) , then one-to-one if for each x , y ∈ X such that g ~ − cl({f (x )}) ≠ g ~ − cl({f ( y )}) . g

~ -closure one-to-one function Theorem 3.5. If f : X → Y is a point g ~ − T space, then f is one-to-one. and X is g 0

~ − T , by Theorem Proof. Let x , y ∈ X with x ≠ y. Since X is g 0 ~ ~ ~ 3.3, g − cl({x}) ≠ g − cl({y}) . But f is point g -closure one-to-one, so ~ − cl({f (x )}) ≠ g ~ − cl({f ( y )}) . Hence f (x ) ≠ f ( y ) . Thus, f is one-to-one. g ~ − T space X Theorem 3.6. Let f : X → Y be a function from a g 0 ~ -closure one-to-one if and only if ~ − T space Y. Then f is point g into a g 0 f is one-to-one.

Proof. Follows from Theorem 3.5 and from the definitions.

~ -irresolute function. Theorem 3.7. Let f : X → Y be an injective g ~−T . ~ − T , then X is g If Y is g 0 0 Proof. Let x , y ∈ X with x ≠ y. Since f is injective, f (x ) ≠ f ( y ) ,

~ -open set V in Y such that ~ − T , so there exists a g but Y is g x 0 ~ -open set V in Y such f (x ) ∈ Vx and f ( y ) ∉ Vx or there exists a g y ~ -irresoluteness of f, f −1 (V ) is that f ( y ) ∈ V y and f (x ) ∉ V y . By g x ~ -open in X such that x ∈ f −1 (V ) and y ∉ f −1 (V ) or f −1 (V ) is g x x y

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MOHAMMAD SARSAK and NEELAMEGARAJAN RAJESH

~ -open in X such that y ∈ f −1 (V ) and x ∉ f −1 (V ) . This shows that g y y ~−T . X is g 0 Theorem 3.8. Let

f : X →Y

be an injective

~ - continuous g

~−T . function. If Y is T0 , then X is g 0 Proof. The proof is similar to that of Theorem 3.7. ~ − T Spaces 4. g 1 ~ − T if to each pair of Definition 4.1. A space X is said to be g 1

~ -open sets, one containing x but distinct points x, y of X, there exist two g not y and the other containing y but not x. ~ − T . However, the next It is evident that every T1 space is g 1 question asks about the converse. ~ − T space that is not T ? Question 1. Is there an example of a g 1 1

Theorem 4.2. For a space X, the following statements are equivalent:

~−T . (i) X is g 1 ~ -closed in X. (ii) Each singleton subset of X is g ~ − ker ( A ) , or equivalently, every (iii) For every subset A of X, A = g ~ -open sets. subset of X is the intersection of g (iv) For each

x ∈ X,

{x} = g~ − ker({x}), or equivalently, every

~ -open sets. singleton subset of X is the intersection of g Proof. (i ) ⇒ (ii ) : Let x ∈ X . Then by (i), for any y ∈ X , y ≠ x , there exists a

~ - open set V containing y but not x. Hence g y

y ∈ V y ⊂ {x}c . Now varying y over {x}c we get {x}c =

∪ {V y : y ∈ {x}c }.

~ -open sets. Since an arbitrary union of g ~ -open So {x}c is the union of g ~ -open, {x} is g ~ -closed. sets is g

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(ii ) ⇒ (iii ): If A ⊂ X , then for each point y ∉ A, {y}c is g~ -open by (ii). Hence A =

∩ {{y}c : y ∈ A c }

~ -open sets. is the intersection of g

(iii ) ⇒ (iv ): Obvious. (iv ) ⇒ (i ): Let x , y ∈ X and x ≠ y. Then by (iv), there exists a ~ -open set U such that x ∈ U and y ∉ U . Similarly, there exists g x x x ~ ~−T . a g -open set U y such that y ∈ U y and x ∉ U y . Hence X is g 1 ~ -closed Theorem 4.3. Let X be a T1 space and f : X → Y be a g ~−T . surjective function. Then Y is g 1 Proof. Suppose y ∈ Y . Since f is surjective, there exists a point

x ∈ X such that y = f (x ) . Since X is T1 , {x} is closed in X. Since f is ~ -closed, f ({x}) = {y} is g ~ -closed in Y. Hence by Theorem 4.2, Y is g ~−T . g 1 ~∗ -closed function ~ − T space and f be a g Theorem 4.4. Let X be a g 1 ~−T . from X onto a space Y. Then Y is g 1 Proof. Similar to that of Theorem 4.3. Definition 4.5. Let A be a subset of a space X and x ∈ X . Then x is ~ -limit point of A if for each U ∈ g ~( X ), x ∈ U , then said to be a g

~ -limit points of A is called the U ∩ ( A \ {x}) ≠ ∅ and the set of all g ~ -derived set of A and is denoted by g ~d( A ) . g ~d( A ) for some A ⊂ X , ~ − T and x ∈ g Theorem 4.6. If X is g 1 ~ -neighborhood of x contains infinitely many points of A. then every g ~ -neighborhood of x such that U ∩ A is Proof. Suppose U is a g ~ -closed set. finite. Let U ∩ A = {x , x , ..., x } = B. Clearly B is a g 1

2

n

~ -neighborhood of x and V ∩ ( A − {x}) Hence V = U − (B − {x}) is a g ~d( A ) , a contradiction. = ∅, which implies that x ∉ g

464

MOHAMMAD SARSAK and NEELAMEGARAJAN RAJESH The proof of the following theorem is straightforward and thus

omitted. ~ − T space X, then g ~d( A ) is Theorem 4.7. If A is a subset of a g 1

~ -closed. g ~ -irresolute function. Theorem 4.8. Let f : X → Y be an injective g ~ − T , then X is g ~−T . If Y is g 1 1 Proof. Similar to the proof of Theorem 3.7

~ -open ~ − R [4] if every g Definition 4.9. A space X is said to be g 0 ~ -closure of each of its singletons. subset of X contains the g ~ − T if and only if it is g ~ − T and Theorem 4.10. A space X is g 1 0 ~−R . g 0

~ − T . It ~ − T space. Then by definitions, X is g Proof. Let X be a g 1 0 ~−R . follows also by Theorem 4.2 that X is g 0

~ − R . We want to ~ − T and g Conversely, suppose that X is both g 0 0 ~ show that X is g − T1 . Let x, y be any distinct points of X. Since X is ~ -open set G such that x ∈ G and y ∉ G or ~ − T , there exists a g g 0

~ -open set H such that y ∈ H and x ∉ H . Without loss there exists a g ~ -open set G such that of generality, we may assume that there exists a g

~−R , ~cl({x}) ⊂ G. As y ∉ G , x ∈ G and y ∉ G. Since X is g g 0 ~cl({x}). Hence y ∈ H = X − g ~cl({x}) and it is clear that x ∉ H . y∉g ~ -open sets G and H containing x and y, Thus it follows that there exist g ~−T . respectively, such that y ∉ G and x ∉ H . Hence X is g 1

Definition 4.11. A subset A of a space X is called g -closed if ~ ~ - open in X, or g − cl( A ) ⊂ U whenever A ⊂ U and U is g ~ − cl( A ) ⊂ g ~ − ker ( A ) . equivalently, if g

~ -closed set is It is clear from the above definition that every g g -closed.

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465

~−T Definition 4.12. A space X is said to be g 1 2 if every g -closed ~ -closed. subset of X is g The following two theorems are immediate consequences of the definitions.

Theorem 4.13. For a space X, the following statements are equivalent:

~−T . (i) X is g 12

~ -open or g ~ -closed. (ii) Every singleton subset of X is g Theorem 4.14. For a space X, the following statements are equivalent:

~−T . (i) X is g 12 (ii) For each subset A of X , A =

∩ {F ∈ g~(X ) ∪ g~C (X ) : A ⊂

F} =

~ − cl( A ) ∩ g ~ − ker ( A ). g ~−T ~ − T , every g ~ − T space is g Clearly, every g 12 1 2 space is 1

~ ~ − T and every T g 1 2 space is g − T1 2 . However, the converses are 0 not true as shown by the following examples.

Example 4.15. Let X = {a, b, c} with the topology τ = {∅, {a}, {a, b},

~ − T . Observe that the ~ − T but not g X }. Then the space X is g 12 0 ~ -open subsets of X are the open sets. g Example 4.16. Let X = {a, b, c, d} with the topology τ = {∅, {a},

{b, c}, {a, b, c}, X }. Then the space X is g~ − T1 2 but not T1 2 as every ~ -open or g ~ -closed. Observe that every semisingleton subset of X is g ~ -closed sets are the closed sets open subset of X is open and thus the g ~−T together with {b, d}, {c, d}, {a, c, d}, {a, b, d}. Also X is g 1 2 but not ~ − T space but not T . ~ − T . It is also an example of a g g 1 0 0

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MOHAMMAD SARSAK and NEELAMEGARAJAN RAJESH

~ − R if for each x ∈ X Definition 4.17. A space X is called weak g 0 ~ − cl({x}) ∩ g ~ − ker ({x}) , then {x} = g ~ − ker({x}) . such that {x} = g ~ − R . However, ~ − R space is weak g It is easy to see that every g 0 0 the converse is not true as shown by the following example.

Example 4.18. Let X = {a, b, c} with the topology τ = {∅, {a}, X }. ~ − R but not g ~ − R . Observe that the Then the space X is weak g 0 0

~ -open subsets of X are the open sets. g It is easy to verify now the following improvement of Theorem 4.10.

Theorem 4.19. For a space X, the following are equivalent:

~−T . (i) X is g 1 ~−R . ~ − T and g (ii) X is g 0 0 ~ − T and weak g ~−R . (iii) X is g 0 0

Definition 4.20. Let f be a function from a space X into a space Y. Then the graph G ( f ) = {(x , f (x )) : x ∈ X } of f is said to be strongly

~ -closed if for each (x , y ) ∈ ( X × Y ) − G ( f ) , there exist a g ~ -open g subset U of X and an open subset V of Y containing x and y, respectively, such that (U × V ) ∩ G ( f ) = ∅.

Lemma 4.21. Let f be a function from a space X into a space Y. Then ~ -closed if and only if for each point (x , y ) ∈ its graph G ( f ) is strongly g

( X × Y ) − G ( f ), there exist a g~ -open subset U of X and an open subset V of Y containing x and y, respectively, such that f (U ) ∩ V = ∅.

Proof. Follows immediately from the above definition. Theorem 4.22. If f : X → Y is an injective function with a strongly ~ -closed graph, then X is g ~−T . g 1 Proof. Suppose that x and y are distinct points of X. Since f is injective,

f (x ) ≠ f ( y ) .

Thus

(x , f ( y )) ∈ ( X × Y ) − G ( f ), but G ( f ) is

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467

~ -closed, so there exist a g ~ -open set U and an open set V strongly g containing x and f ( y ) , respectively, such that f (U ) ∩ V = ∅. Hence

~ -open set M and an open set N y ∉ U . Similarly there exist a g containing y and f (x ) , respectively, such that f ( M ) ∩ N = ∅. Hence ~−T . x ∉ M . Thus it follows that X is g 1

Theorem 4.23. If f : X → Y is a surjective function with a strongly

~ -closed graph, then Y is T . g 1 Proof. Let y1 and y2 be two distinct points of Y. Since f is such that f (x ) = y2 . Hence (x , y1 ) ∉ ~ -open set U and an open G ( f ) and thus by Lemma 4.21 there exist a g

surjective, there exists x ∈ X

set V containing x and y1 , respectively, such that f (U ) ∩ V = ∅. Hence

y2 ∉ V . Similarly there exists x 0 ∈ X such that f (x 0 ) = y1 . Hence (x 0 , y2 ) ∉ G ( f ) and thus there exist a g~ -open set M and an open set N containing x 0 and y2 , respectively, such that f ( M ) ∩ N = ∅. Hence

y1 ∉ N . Thus it follows that Y is T1 . Remark 4.24. In Definition 4.20, if we consider U and V both are ~ -open, then Theorem 4.23 yields that Y is g ~−T . g 1 ~ − T Spaces 5. g 2 ~ − T if to each pair of Definition 5.1. A space X is said to be g 2

~ -open sets, one distinct points x, y of X, there exist two disjoint g containing x and the other containing y.

~ − T . However, the next It is clear that every T2 -space is g 2 question asks about the converse. ~ − T space that is not T ? Question 2. Is there an example of a g 2 2

Remark 5.2. We observe that every

~−T g 2

space is

~−T . g 1

However, the converse is not true as shown by the following example.

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MOHAMMAD SARSAK and NEELAMEGARAJAN RAJESH

Remark 5.3. An infinite set X with the finite complement topology is ~ ~ − T since any two non-empty open g − T1 . It is, however, not g 2 ~ -open subsets of X intersect. subsets of X and hence any two non-empty g

~ -open subset of X is open. Observe that a g Theorem 5.4. For a space X, the following statements are equivalent: ~−T . (i) X is g 2

~ -neighborhood of x} ~ − cl(U ) : U is a g (ii) For each x ∈ X , ∩{g x x = {x} or equivalently, every singleton subset of X is the intersection of

~ -closed neighborhoods of x. g ~ − T space and x ∈ X . Then to Proof. (i ) ⇒ (ii ) : Let X be a g 2 each y ∈ X ,

~ -open sets G and H such that y ≠ x , there exist g

x ∈ G , y ∈ H and G ∩ H = ∅. Since x ∈ G ⊂ X − H , X − H is a ~ -closed g ~ -neighborhood of x to which y does not belong. Consequently, g ~ -closed g ~ -neighborhoods of x is reduced to {x}. the intersection of all g

(ii ) ⇒ (i ): Suppose that x , y ∈ X and x ≠ y. Then by hypothesis ~ -closed g ~ -neighborhood U of x such that y ∉ U . Now there exists a g ~ -open set G such that x ∈ G ⊂ U . Thus G and X − U are there is a g ~ -open sets containing x and y, respectively. Hence X is disjoint g ~−T . g 2 The proof of the following theorem is straightforward and thus omitted.

~ − T if and only if for each x , y ∈ X Theorem 5.5. A space X is g 2 ~ -closed sets F and F such that such that x ≠ y, there exist g 1 2 x ∈ F1 , y ∉ F1 , y ∈ F2 , x ∉ F2 and X = F1 ∪ F2 . Recall that a subset A of a space X is called sg-closed if whenever A ⊂ U , where U is semi-open in X, then scl( A ) ⊂ U .

~ -open sets need not be g ~ -open as Remark 5.6. The product of two g the following example tells.

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Example 5.7. Let X = {a, b, c} and τ = {∅, {a, b}, X }. Then A =

{b, c} is g~ -closed. Now A × X is not sg-closed because if U = ( X × X ) −{(a, c )}, then U is semi-open in X × X and A × X ⊂ U . However,

~ -closed set is sg-closed, it follows X × X = scl( A × X ) { U . Since every g ~ -closed. From this we conclude that the product of that A × X is not g ~ -closed sets need not be g ~ -closed. Since the union of g ~ -open sets two g ~ -open, it follows that the product of two g ~ -open sets need not be is g ~ -open. g ~ -regular T space is g ~−T . Theorem 5.8. Every g 0 2 ~ -regular T space and let x , y ∈ X be such Proof. Let X be a g 0 that x ≠ y. Since X is T0 , there exists an open set V containing x but not y or y but not x, say x but not y. Then y ∈ X − V , X − V is closed ~ -regularity of X, there exist g ~ -open sets G and H and x ∉ X − V . By g ~−T . such that x ∈ G , y ∈ X − V ⊂ H and G ∩ H = ∅. Hence X is g 2

~ -irresolute (resp. Theorem 5.9. If f : X → Y is an injective g ~ -continuous) function and Y is g ~−T . ~ − T (resp. T ) , then X is g g 2 2 2 Proof. We show the first case, the other case is similar. Suppose that ~ − T , so x , y ∈ X , x ≠ y. Since f is injective, f (x ) ≠ f ( y ) , but Y is g 2

~ -open sets G, H in Y such that f (x ) ∈ G , f ( y ) ∈ H and there exist g G ∩ H = ∅. Let U = f −1 (G ) and V = f −1 (H ) . Then by hypothesis, U

~ -open sets in X. Also x ∈ f −1 (G ) = U , y ∈ f −1 (H ) = V and and V are g ~−T . U ∩ V = ∅. Hence X is g 2 The following three theorems have easy proofs and thus omitted:

~∗ -open) ~ -open (resp. g Theorem 5.10. If f : X → Y is a bijective g ~ − T ) , then Y is g ~−T . function and X is T2 (resp. g 2 2

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MOHAMMAD SARSAK and NEELAMEGARAJAN RAJESH

~ -open function from a space X onto a space Theorem 5.11. If f is a g Y and the set {(x1 , x 2 ) : f (x1 ) = f (x 2 )} is closed in X × X , then Y is ~−T . g 2

~∗ -open function from a space X onto a space Theorem 5.12. If f is a g ~ -closed graph, then Y is g ~−T . Y and f has a strongly g 2

Remark 5.13. The above theorem is still true if we consider in the ~ -closed graph U and V to be both g ~ -open. definition of a strongly g ~ − R [4] if for each Definition 5.14. A space X is said to be g 1 ~ -open sets ~ ~ x , y ∈ X with g − cl({x}) ≠ g − cl({y}), there exist disjoint g ~ − cl({x}) ⊂ U and g ~ − cl({y}) ⊂ V . U and V such that g ~ − T if and only if it is g ~ − R and Theorem 5.15. A space X is g 2 1 ~−T . g 0

Proof. Similar to that of Theorem 4.10. Remark 5.16. In the following diagram we denote by arrows the implications between the separation axioms which we have introduced and discussed in this paper. However, none of these implications is reversible.

⇒

~−T g 2

⇓ T1

⇒

⇓ ~ g − T1

⇓ T1 2

⇒

⇓ T0

⇒

T2

⇓ ~−T g 12 ⇓ ~−T g 0

~ − R space is Remark 5.17. It is not difficult to see that every g 1 ~ − R . However, it follows from Theorem 4.19 and Theorem 5.15 that g 0 ~ − T but not g ~ − T is an example of a g ~−R any space which is g 1 2 0

~−R. space that is not g 1

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S. G. Crossley and S. K. Hildebrand, Semi-closure, Texas J. Sci. 22 (1971), 99-112. S. Jafari, T. Noiri, N. Rajesh and M. L. Thivagar, Another generalized closed sets, Kochi J. Math. 3 (2008), to appear.

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