Some New Separation Axioms In Ideal Topological Spaces

Some New Separation Axioms In Ideal Topological Spaces

Some New Separation Axioms In Ideal Topological Spaces

1. Balaji, N. Rajesh

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   Abstract. In this paper, b- -open sets are used to define some weak separation axioms and to study some of their basic properties. The implications of these axioms among themselves and with the known axioms are investigated.

   1. Introduction

      The notion of R0 topological spaces is introduced by Shanin [13] in 1943. Later, Davis [7] rediscovered it and studied some properties of this weak separation axiom. Several topologists (e.g. [8], [9], [11]) further investigated properties of R0 topological spaces and many in- teresting results have been obtained in various contexts. In the same paper, Davis also introduced the notion of R1 topological space which are independent of both T0 and T1 but strictly weaker than T2. The subject of ideals in topological spaces has been studied by Kuratowski

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      [10] and Vaidyanathasamy [14]. An ideal on a topological space (X, ) is a nonempty collection of subsets of X which satisfies (i) A

      I and B A implies B I and (ii) A I and B I implies A B I. Given a topological space (X, ) with an ideal I on X and if P(X) is the set of all subsets of X, a set operator (.)*: P(X)

      P(X), called the local function [14] of A with respect to and I,

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      is defined as follows: for A X, A*(, I) = {x X|U A / I for

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      every open neighbourhood U of x . A Kuratowski closure operator Cl*(.) for a topology *(, ) called the *-topology, which is finer than

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      is defined by Cl*(A) = A A*(, ), When there is no chance of confusion, A*( ) is denoted by A*. If is an ideal on X, then (X, , ) is called an ideal topological space. By a space, we always mean a topological space (X, ) with no separation properties assumed. If A X, Cl(A) and Int(A) will denote the closure and interior of A in (X, ), respectively. A subset A of a topological space (X, ) is said to be b-open [1] if A Int(Cl(A)) Int(Cl(A)). The notion of b-open sets has been studied extensively in recent years by many topologists [see [4, 5, 12]] because b-open sets are only natural generalization of open sets. More importantly, they also suggest several new properties

      2000 Mathematics Subject Classification. 54D10.

      Key words and phrases. Ideal topological spaces, b-I-R0 space, b-I-R1 space,

      b-I-R2 space.

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      of topological spaces. In this paper, b- -open sets are used to define some weak separation axioms and to study some of their basic proper- ties. The implications of these axioms among themselves and with the known axioms are investigated.

   2. Preliminaries

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      A subset S of an ideal topological space (X, , ) is said to be b- – open [6] if S Int(Cl (S)) Cl (Int(S)). The complement of a b-

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      -open set is called a b- -closed set [6]. The intersection of all b- – closed sets containing S is called the b- -closure of S and is denoted by b Cl(S). The b- -Interior of S is defined by the union of all b- -open sets contained in S and is denoted by b Int(S). The family of all b-

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      -open (resp. b- -closed) sets of (X, , ) containing a point x X is denoted by B O(X, x) (resp. B C(X, x)). A subset U of X is called a b- -neighbourhood of a point x X if there exists a b- -open set V of (X, , ) such that x V U . A function f : (X, , ) (Y, , ) is

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      said to be b- -continuous [6] if f 1(V ) B O(X) for every open set

      V of Y .

      Definition 2.1. A topological space (X, ) is said to be:

      1. b-R0 [4] if every b-open set contains the b-closure of each of its singletons.

      2. b-R1 [4] if for x, y in X with b Cl({x}) b Cl({y}), there ex-

      ist disjoint b-open sets U and V such that b Cl({x}) U and

      b Cl({y}) V .

      R0 and R1 spaces are similarly defined in (see, [7], [13]).

      Definition 2.2. An ideal topological space (X, , I) is said to be:

      1. b-I-T1 [3] if for each pair of distinct points x and y of X, there exist b-I-open sets U and V of X such that x U and y / U , and y V and x / V .

      2. b-I-T2 [3] if for each pair of distinct points x and y in X, there exist disjoint b-I-open sets U and V in X such that x U and y V .

   3. On b-I-R0 spaces

      Definition 3.1. Let (X, , I) be an ideal topological space and A X. Then the b-I-kernel of A, denoted by bI Ker(A) is defined to be the set bI Ker(A) = {G BIO(X) | A G}.

      Lemma 3.2. Let (X, , I) be an ideal topological space and x X. Then, y bI Ker({x}) if and only if x bI Cl({y}).

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      Proof. Suppose that y / b Ker( x ). Then there exists U B O(X, x) such that y / U . Therefore, we have x / b Cl( y ). The proof of the converse case can be done similarly. D

      Lemma 3.3. Let (X, , I) be an ideal topological space and A a subset

      of X. Then, bI Ker(A) = {x X | bI Cl({x}) A }.

      Proof. Let x bI Ker(A) and bI Cl({x}) A = . Hence x / X \

      bI Cl({x}) which is a b-I-open set containing A. This is impossible, since x bI Ker(A). Consequently, bI Cl({x}) A /= . Next, let x

      X such that bI Cl({x}) A and suppose that x / bI Ker(A).

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      Then, there exists a b-I-open set U containing A and x / U . Let y

      b Cl( x ) A. Hence, U is a b- -neighbourhood of y which does

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      not contains x. By this contradiction x b Ker(A) and hence the claim. D

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      Definition 3.4. An ideal topological space (X, , ) is said to be a b-

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      -R0 space if every b- -open set contains the b- -closure of each of its singletons.

      Remark 3.5. Since an ideal topological space (X, , I) is b-I-T1 if and only if the singletones are b-I-closed [3], it is clear that every b-I-T1 space b-I-R0. But the converse is not true in general.

      Example 3.6. Let X = {a, b, c}, = {, {a}, {b, c}, X} and I =

      {, {b}, {c}, {b, c}}. Then (X, , I) is b-I-R0 but none of b-I-T0 and

      b-I-T1.

      Remark 3.7. Th following example and Examlpe 3.6 shows that the notions b-I-T0-ness b-I-R0-ness are independent.

      Example 3.8. Let X = {a, b, c}, = {, {a}, X} and I = {, {a}}. Then (X, , I) is b-I-T0 but not b-I-R0.

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      Proposition 3.9. For an ideal topological space (X, , ), the following properties are equivalent:

      1. (X, , I) is b-I-R0 space;

      2. For any F BIC(X), x /

         some U BIO(X);

         F implies F U and x /

         U for

      3. For any F BIC(X), x / F implies F bI Cl({x}) = ;

      4. For any distinct points x and y of X, either bI Cl({x}) =

      bI Cl({y}) or bI Cl({x}) bI Cl({y}) = .

      Proof. (1)(2): Let F BIC(X) and x / F . Then by (1) bI Cl({x})

      X \ F . Set U = X \ bI Cl({x}), then U BIO(X), F U and x / U . (2)(3): Let F BIC(X) and x / F . There exists U BIO(X)

      such that F U and x / U . Since U BIO(X), U bI Cl({x})

      = and F bI Cl({x}) = . (3)(4): uppose that bI Cl({x}) /= bI Cl({y}) for distinct points x, y X. There exists z bI Cl({x}) such that z / bI Cl({y}) (or z bI Cl({y}) such that z / bI Cl({x})). There exists V BIO(X) such that y / V and z V ; hence x V .

      Therefore, we have x / bI Cl({y}). By (3), we obtain bI Cl({x})

      bI Cl({y}) = . The proof for otherwise is similar. (4)(1): Let V

      BIO(X, x). For each y / V , x /= y and x / bI Cl({y}). This shows

      that bI Cl({x}) bI Cl({y}). By (4), bI Cl({x}) bI Cl({y}) = for

      each y X \ V and hence bI Cl({x}) (yX\V bI Cl({y})) = . On other hand, since V BIO(X) and y X \ V , we have bI Cl({y})

      X \ V and hence X \ V = yX\V bI Cl({y}). Therefore, we obtain (X\V ) bI Cl({x}) = and bI Cl({x}) V . This shows that (X, , I) is a b-I-R0 space. D

      Theorem 3.10. An ideal topological space (X, , I) is a b-I-R0 space

      if and only if for any x and y in X, bI Cl({x}) bI Cl({x}) bI Cl({y}) = .

      bI Cl({y}) implies

      Proof. Suppose that (X, , I) is b-I-R0 and x, y X such that bI Cl({x})

      =/ bI Cl({y}). Then, there exists z bI Cl({x}) such that z / bI Cl({y})

      ( or z / bI Cl({y})) such that z / bI Cl({x}). There exists V

      BIO(X) such that y / V and z V ; hence x V . Therefore, we have

      x / bI Cl({y}). Thus x X \ bI Cl({y}) BIO(X), which implies

      bI Cl({x}) X \ bI Cl({y}) and bI Cl({x}) bI Cl({y}) = . The proof for otherwise is similar. Conversely, let V BIO(X, x). We will

      show that bI Cl({x}) V . Let y X \ V . Then x y and x /

      bI Cl({y}). This shows that bI Cl({x}) /= bI Cl({y}). By assumption,

      bI Cl({x}) bI Cl({y}) = . Hence y / bI Cl({x}) and therefore

      bI Cl({x}) V . D

      x and y in an ideal topological space (X, , I):

      1. bI Ker({x}) /= bI Ker({y});

      2. bI Cl({x}) bI Cl({y}).

      Proof. (1)(2): Suppose that bI Ker({x}) /= bI Ker({y}), then there exists a point z in X such that z bI Ker({x}) and z / bI Ker({y}). It follows from z bI Ker({x}) that {x} bI Cl({z}) /= . This

      implies that x bI Cl({z}). By z / bI Ker({y}), we have {y}

      bI Cl({z}) = . Since x bI Cl({z}), bI Cl({x}) bI Cl({z}) and {y}

      bI Cl({x}) = . Therefore, it follows that bI Cl({x}) bI Cl({y}).

      Now bI Ker({x}) bI Ker({y}) implies that bI Cl({x}) bI Cl({y}).

      (2)(1): Suppose that bI Cl({x}) bI Cl({y}). Then there exists a

      point z in X such that z bI Cl({x}) and z / bI Cl({y}). Then, there exists a b-I-open set containing z and therefore x but not y, namely, y

      / bI Ker({x}) and thus bI Ker({x}) /= bI Ker({y}). D

      Theorem 3.12. An ideal topological space (X, , I) is a b-I-R0 space if and only if for any pair of points x and y in X, bI Ker({x}) /= bI Ker({y}) implies bI Ker({x}) bI Ker({y}) = .

      Proof. Suppose that (X, , I) is a b-I-R0 space. Thus by Lemma

      1. , for any points x and y in X if bI Ker({x}) bI Ker({y}), then

         bI Cl({x}) bI Cl({y}). Now we prove that bI Ker({x}) bI Ker({y})

         = . Assume that z bI Ker({x}) bI Ker({y}). By z bI Ker({x}) and Lemma 3.11, it follows that x bI Cl({z}). Since x bI Cl({x}), by Theorem 3.10 bI Cl({x}) = bI Cl({z}). Similarly, we have bI Cl({y})

         = bI Cl({z}) = bI Cl({x}). This is a contradiction. Therefore, we have bI Ker({x}) bI Ker({y}) = . Conversely, let (X, , I) be an ideal topological space such that for any points x and y in X, bI Ker({x})

         /= bI Ker({y}) implies bI Ker({x}) bI Ker({y}) = . If bI Cl({x})

         =/ bI Cl({y}), then by Lemma 3.2, bI Ker({x}) /= bI Ker({y}). Hence,

         bI Ker({x}) bI Ker({y}) = which implies bI Cl({x}) bI Cl({y})

         = . Because z bI Cl({x}) implies that x bI Ker({z}) and therefore bI Ker({x}) bI Ker({y}) /= . By hypothesis, we have bI Ker({x}) = bI Ker({z}). Then z bI Cl({x}) bI Cl({y}) implies that bI Ker({x}) = bI Ker({z}) =bI Ker({y}). This is a contradiction. Therefore, bI Cl({x}) bI Cl({y}) = and by Theorem 3.10 (X, , I) is a b-I-R0 space. D

         Theorem 3.13. For an ideal topological space (X, , I), the following

         properties are equivalent:

         1. (X, , I) is a b-I-R0 space;

         2. For any nonempty sets A, G BIO(X) such that A G =/

            , there exists F BIC(X) such that A F and F

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            G;

         3. Any G BIO(X), G = {F BIC(X) | F G};

         4. Any F BIC(X), F = {G BIO(X) | F G};

         5. For any x X, bI Cl({x}) bI Ker({x}).

      Proof. (1)(2): Let A be a nonempty set of X and G BIO(X) such that A G /= . There exists x A G. Since x G BIO(X), bI Cl({x}) G. Set F = bI Cl({x}), then F BIC(X), F G and A F /= . (2)(3): Let G BIO(X), then G {F BIC(X)

      | F G}. Let x be any point of G. There exists F BIC(X) such that x F and F G. Therefore, we have x F {F BIC(X)

      | F G} and hence G = {F BIC(X) | F G}. (3)(4): This

      is obvious. (4)(5): Let x be any point of X and y / bI Ker({x}).

      There exists V BIO(X, x) y / V ; hence bI Cl({y}) V = . By

      1. ({G BIO(X) | bI Cl({y}) G}) V = and there exists G

         BIO(X) such that x / G and bI Cl({y}) G. Therefore, bI Cl({x})

         G = and y / bI Cl({x}). Consequently, we obtain bI Cl({x})

         bI Ker({x}). (5)(1): Let G BIO(X, x). Let y bI Ker({x}), then x bI Cl({y}) and y G. This implies that bI Ker({x}) G. Therefore, we obtain x bI Cl({x}) bI Ker({x}) G. This shows that (X, , I) is a b-I-R0 space. D

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         Corollary 3.14. For an ideal topological space (X, , ), the following properties are equivalent:

         1. (X, , I) is a b-I-R0 space;

         2. bI Cl({x}) =bI Ker({x}) for all x X.

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      Proof. (1) (2): Suppose that (X, , ) is a b- -R0 space. By Theorem 3.13, b Cl( x ) b Ker( x ) for each x X. Let y b Ker( x ), then x b Cl( y ) and by Theorem 3.10 b Cl( x ) = b Cl( y ). Therefore, y b Cl( x ) and hence b Ker( x ) b Cl( x ). This shows that b Cl( x ) =b Ker( x ). (2) (1): This is obvious by Theorem 3.13. D

      Corollary 3.15. If for any point x of a b-I-R0 space (X, , I), bI Cl({x})

      bI Ker({x}) = {x}, then bI Ker({x}) = {x}.

      Proof. The proof follows from Theorem 3.13(v). D

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      Theorem 3.16. For an ideal topological space (X, , ), the following properties are equivalent:

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      1. (X, , ) is a b- -R0 space;

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      2. x b Cl( y ) if and only if y b Cl( x ) for any points x

      and y in X.

      Proof. (1)(2): Assume that (X, , I) is b-I-R0. Let x bI Cl({y}) and A BIO(X, y). Now by hypothesis, x A. Therefore, every b-I-open set containing y contains x. Hence y bI Cl({x}). (2)(1):

      Let U BIO(X, x). If y / U , then x / bI Cl({y}) and hence y /

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      bI Cl({x}). This implies that bI Cl({x}) U . Hence (X, , I) is b-I-

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      Theorem 3.17. For an ideal topological space (X, , ), the following properties are equivalent:

      1. (X, , I) is a b-I-R0 space;

      2. If F is a b-I-closed subset of X, then F = bI Ker(F );

      3. If F is a b-I-closed subset of X and x F , then bI Ker({x})

         F ;

      4. If x X, then bI Ker({x}) bI Cl({x}).

      Proof. (1)(2): Let F be b-I-closed subset of X and x / F . Thus

      X\F BIO(X, x). Since (X, , I) is b-I-R0, bI Cl({x}) X \ F . Thus bI Cl({x}) F = and Lemma 3.3 x / bI Ker(F ). Therefore, bI Ker(F ) = F . (2)(3): In general, A B implies bI Ker(A) bI Ker(B). Therefore, it follows from (2) thatbI Ker({x}) bIKer(F )

      = F . (3)(4): Since x bI Cl({x}) and bI Cl({x}) is b-I-closed, by

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      1. bI Ker({x}) bI Cl({x}). (4)(1): We show the implication by using Theorem 3.16. Let x bI Cl({y}). Then by Lemma 3.2 y bI Ker({x}). Since x bI Cl({x}) and bI Cl({x}) is b-I-closed, by (4) we obtain y bI Ker({x}) bI Cl({x}). Therefore, x bI Cl({x}) implies y bI Cl({x}). The converse is obvious and (X, , I) is b-I-

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         Definition 3.18. A filterbase is called b- -convergent to a point x in X, if for any U B O(X, x), there exists B such that B is a subset of U .

         Lemma 3.19. Let (X, , I) be an ideal topological space and let x and y be any two points in X such that every net in X b-I-converging to y b-I-converges to x. Then x bI Cl({y}).

         Proof. Suppose that xn = y for each n N . Then {xn}nN is a net in bI Cl({y}). Since {xn}nN b-I-converges to y, then {xn}nN b-I- converges to x and this implies that x bI Cl({y}). D

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         Theorem 3.20. For an ideal topological space (X, , ), the following statements are equivalent:

         1. (X, , I) is a b-I-R0 space;

         2. If x, y X, then y bI Cl({x}) if and only if every net in X b-I-converging to y b-I-converges to x.

      Proof. (1)(2): Let x, y X such that y bI Cl({x}). Suppose that

      {x}N be a net in X such that {x}N b-I-converges to y. Since y bI Cl({x}), by Theorem 3.10 we have bI Cl({x}) = bI Cl({y}). There- fore x bI Cl({y}). This means that {x}N b-I-converges to x. Con- versely, let x, y X such that every net in X b-I-converging to y b-I- converges to x. Then x bI Cl({y}) by Lemma 3.3. By Theorem 3.10, we have bI Cl({x}) = bI Cl({y}). Therefore y bI Cl({x}). (2)(1): Assume that x and y are any two points of X such that bI Cl({x})

      bI Cl({y}) . Let z bI Cl({x}) bI Cl({y}). So there exists a

      net {x}N in bI Cl({x}) such that {x}N b-I-converges to z. Since z bI Cl({y}), then {x}N b-I-converges to y. It follows that y bI Cl({x}). By the same token we obtain x bI Cl({y}). Therefore bI Cl({x}) = bI Cl({y}) and by Theorem 3.10 (X, , I) is b-I-R0. D

   4. On b-I-R1 spaces

Definition 4.1. An ideal topological space (X, , I) is said to be b-I-

R1 if for x, y in X with bI Cl({x}) bI Cl({y}), there exist disjoint

b-I-open sets U and V such that bI Cl({x}) U and bI Cl({y}) V .

Proposition 4.2. If (X, , I) is b-I-R1, then it is b-I-R0.

Proof. Let U BIO(X, x). If y / U , then since x / bI Cl({y}),

bI Cl({x}) /= bI Cl({y}). Hence there exists a b-I-open Vy such that

bI Cl({y}) Vy and x / Vy , which implies y / bI Cl({x}). Thus

bI Cl({x}) U . Therefore (X, , I) is b-I-R0. D

Theorem 4.3. An ideal topological space (X, , I) is b-I-R1 if and

only if for x, y X, bI Ker({x}) bI Ker({y}), there exist disjoint

b-I-open sets U and V such that bI Cl({x}) U and bI Cl({y}) V .

Proof. It follows from Lemma 3.11 D

1. (X, , I) is b-I-T2,

2. (X, , I) is b-I-R1 and b-I-T1, and

3. (X, , I) is b-I-R1 and b-I-T0.

Proof. (1)(2): Since (X, , I) is b-I-T2, then it is b-I-T1. If x, y X

such that bI Cl({x}) bI Cl({y}), then x y and there exist disjoint

b-I-open sets U and V such that x U and y V and bI Cl({x})

= {x} U and bI Cl({y}) = {y} V . Hence (X, , I) is b-I-R1. (2)(3): Since (X, , I) is b-I-T1, then (X, , I) is b-I-T0. (3)(1):

Since (X, , I) is b-I-R1, and b-I-T1, then (X, , I) is b-I-R0 and b-I-

T0, which implies (X, , I) is b-I-T1. Let x, y X such that x y.

Since bI Cl({x}) = {x} /= {y} = bI Cl({y}), then there exist disjoint b-I-open sets U and V such that x U and y V . Hence, (X, , I) is b-I-T2. D

1. (X, , I) is b-I-R1,

2. for each x, y X one of the following holds:

   1. If U is b-I-open, then x U if and only if y U .

   2. there exist disjoint b-I-open sets U and V such that x U

      and y V , and

3. If x, y X such that bI Cl({x}) /= bI Cl({y}), then there exist

b-I-closed sets F1 and F2 such that x F1, y /

/ F2, and X = F1 F2.

F2, y F1, x

Proof. (1)(2): Let x, y X. Then bI Cl({x}) = bI Cl({y}) or

bI Cl({x}) bI Cl({y}). If bI Cl({x}) = bI Cl({y}) and U is b-I-

open, then x U implies y bI Cl({x}) U and y U implies x bI Cl({y}) U . Thus consider the case that bI Cl({x}) /= bI Cl({y}). Then there exist disjoint b-I-open sets U and V such that x bI Cl({x})

U and y bI Cl({y}) V . (2)(3): Let x, y X such that

bI Cl({x}) /= bI Cl({y}). Then x / bI Cl({y}) or y / bI Cl({x}), say

x / bI Cl({y}). Then there exist a b-I-open set A such that x A

and y / A, which implies there exist disjoint b-I-open sets U and V

such that x U and y V . Then F1 = X \ V and F2 = X \ U are

b-I-closed sets such that x F1, y / F1, y F2, x / F2, X = F1

F2. (3)(1): Let U be b-I-open and let x U . Then bI Cl({x}) U , for suppose not. Let y bI Cl({x}) (X\U ). Then bI Cl({x}) /= bI Cl({y}) and there exist b-I-closed sets F1 and F2 such that x F1,

y / F1, y F2, x / F2, X = F1 F2. Then y F2 \ F1 = X \ F1,

which is b-I-open, and x / X \ F1, which is a contradiction. Hence,

(X, , I) is b-I-R0. Let a, b X such that bI Cl({a}) /= bI Cl({b}). Then there exist b-I-closed sets A1 and A2 such that a A1, b / A1, a / A2, and X = A1 A2. Thus a A1 \ A2 and b A2 \ A1, which are b-I-open, which implies bI Cl({a}) A1 \ A2 and bI Cl({b}) A2

\ A1. Hence, (X, , I) is b-I-R1. D

Theorem 4.6. An ideal topological space (X, , I) is b-I-T2 if and only

if for x, y X such that x y, there exist b-I-closed sets F1 and F2

such that x F1, y / F1, y / F2, x / F2, and X = F1 F2.

Proof. The straightforward proof is omitted. D

Remark 4.7. If {x}A is a net in (X, , I), bI lim({x}A) = {x

X : {x}Ab-I-converges to x}.

1. (X, , I) is b-I-R1;

2. for x, y X, bI Cl({x}) = bI Cl({y}), whenever there exists a net {x}A such that x, y bI lim({x}A);

3. (X, , I) is b-I-R0, and for every b-I-convergent net {x}A

in X, bI lim({x}A) = bI Cl({x}) for some x X.

Proof. (1)(2): Let x, y X such that there exists a net {x}A in X such that x, y bI lim({x}A). Then (a) if U is b-I-open, then x U if and only if y U or (b) there exist disjoint b-I-open sets U and V such that x U and y V . Since x, y bI lim({x}A), then

1. is satisfied, which implies bI Cl({x}) = bI Cl({y}). (2)(3): Let U BIO(X, x). Let y / U . For each n N let xn = x. Then {xn}nN b-I-converges to x and since bI Cl({x}) /= bI Cl({y}), that y A and

   x / A. Thus, y / bI Cl({x}) and bI Cl({y}) U . Hence (X, , I)

   is b-I-R0. Let {x}A be a b-I-convergent net in X. Let x X such that {x}A b-I-converges to x. If y bI Cl({x}), then {x}A b-I-converges to y, which implies bI Cl({x}) bI lim({x}A) and if y bI lim({x}A), then x, y bI lim({x}A), which implies y bI Cl({y}) = bI Cl({x}). Hence bI lim({x}A) = bI Cl({x}). (3)(1): Assume that (X, , I) is not b-I-R1. Then there exists x, y X such that bI Cl({x}) /= bI Cl({y}) and every b-I-open set con- taining bI Cl({x}) intersects every b-I-open set containing bI Cl({y}). Since (X, , I) is b-I-R0, then every b-I-open set containing x contains bI Cl({x}) and every b-I-open set containing y contains bI Cl({y}), which implies that every b-I-open set containing x intersects every b- I-open set containing y. Let Dx = {U X : U BIO(X, x)}. Let

   x be the binary relation on Dx defined by U1 x U2 if and only if U1 U2. Then, clearly (Dx, x) is a directed set. Let Dy = {U X : U BIO(X, y)} and et y be the binary relation on Dy de- fined by U1 y U2 if and only if U1 U2. Then, (Dy , y ) is also a directed set. Let D = {(U1, U2) : U1 Dx and U2 Dy } and let

   be the binary relation on D defined by (U1, U2) (V1, V2) if and only if U1 x V1 and U2 y V2. Then, (D, ) is a directed set. For each (U1, U2) D, let x(U1 ,U2) (U1, U2). Then {x(U1 ,U2)}(U1 ,U2)D is

   a net in X that b-I-converges to both x and y. Thus, there exists z X such that bI lim({x(U1 ,U2)}(U1 ,U2)D) = bI Cl({z}), which implies x, y bI Cl({z}). Since {bI Cl({w}) : w X} is a decomposition of X,

   then bI Cl({x}) = bI Cl({z}) = bI Cl({y}), which is a contradiction. Hence (X, , I) is b-I-R1. D

   Theorem 4.9. An ideal topological space (X, , I) is b-I-T2 if and only if every b-I-convergent net in X b-I-converges to a unique point.

   Proof. The proof follows from Theorems 4.8 4.4. D

   I I

   I I

   (1) (X, , ) is b- -R1,

   I { } / I { }

   I { } / I { }

2. for each pair x, y X, b Cl( x ) = b Cl( y ), there exists a

   I I

   I I

   b- -open, b- -closed set V such and y / V ;

   I I

   I I

   I { } / I { }

   I { } / I { }

3. for each pair x, y X, b Cl( x ) = b Cl( y ), there exists a b- -continuous function f : (X, , ) [0, 1] such that f (x) = 0 and f (y) = 1.

Proof. (1)(2): Let x, y X such that bI Cl({x}) /= bI Cl({y}). Then there exist disjoint b-I-open sets U and W such that bI Cl({x}) U and bI Cl({y}) W and V = bI Cl(U ) is b-I-open, b-I-closed such that x V and y / V . (2)(3): Let x, y X such that bI Cl({x}) =/ bI Cl({y}). Let V be b-I-open, b-I-closed set of X such that x V and y / V . Thus, the function f : (X, , I) [0, 1] defined by f (z) = 0 if z V and f (z) = 1 if z / V satisfies the desired properties. (3)(1): Let x, y X such that bI Cl({x}) /= bI Cl({y}). Let f : (X, , I) [0, 1] such that f is b-I-continuous, f (x) = 0 and f (y) = 1. Then U = f 1([0, 0.5)) and V = f 1((0.5, 1]) are disjoint such that b-I-open, b-I-closed set of X and bI Cl({x}) U and bI Cl({y}) V . D

1. (X, , I) is b-I-R1,

2. for each pair x, y X, x

   set V such and y / V ;

   y, there exists a b-I-open, b-I-closed

3. for each pair x, y X, x /= y, there exists a b-I-continuous function f : (X, , I) [0, 1] such that f (x) = 0 and f (y) = 1.

Proof. The proof is similar to that of Theorem 4.10. D

I

I

{ } I

{ } I

Theorem 4.12. (1). An ideal topological space (X, , ) is b- -R0 (resp. b- -R1) if and only if it is b-R0 (resp. b-R1).

I N

I N

N I I

N I I

1. An ideal topological space (X, , ) is b- -R0 (resp. b- -R1) if and only if it is b- -R0 (resp. b-R1) ( is the ideal of all nowhere dense sets).

   P I I

   P I I

2. An ideal topological space (X, , (X)) is b- -R0 (resp. b- -R1) if and only if it is R0 (resp. R1).

   Proof. It follows from Proposition 2 of [2]. D

   discussed in this paper and examples show that no other implications hold between them:

   R0 R1

   b-I-R0 b-I-R1

   b-R0 b-R1

   Example 4.14. Let X = {a, b, c}, = {, {a}, {b}, {a, b}, X} and

   I = {, {a}}. Then (X, , I) is b-I-R0 but none of R0 and b-I-R1.

   Example 4.15. Let X = {a, b, c}, = {, {a, b}, X} and I = {, {c}}. Then (X, , I) is b-I-R1 but not R1.

   Example 4.16. Let X = {a, b, c}, = {, {a}, {b, c}, X} and I =

   {, {b}}. Then (X, , I) is b-R0 but not b-I-R0.

   References

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      I I

   2. M. Akdag, On b- -open sets and b- -continuous functions, Inter. J. Math. Math. Sci., Volume (2007), ID 75721, 1-13.

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   3. R. Balaji and N. Rajesh, New separation axioms via b- -open sets (submitted).

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   6. A. Caksu Guler and G. Aslim, b- -open sets and decomposition of continuity via idealization, Proc. Inst. Math. Mech. Natl. Acad. Sci. Azerb., 22(2005), 27-32.

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      J. Austr. Math. Soc., 12(1971), 451-455.

   10. K. Kuratowski, Topology, Academic Press, New York, 1966.

   11. S. A. Naimpally, On R0-topological spaces, Ann. Univ. Sci. Budapest Eotvos Sect. Math., 10(1967), 53-54.

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   14. R. Vaidyanathaswamy, The localisation theory in set topology, Proc. Indian Acad. Sci., 20(1945), 51-61.

Department of Mathematics, Agni College of Technology, Kancheep- uram 603103, TamilNadu, India.

E-mail address: balaji 2410@yahoo.co.in

Department of Mathematics, Rajah Serfoji Govt. College, Thanjavur- 613005, TamilNadu, India.

E-mail address: nrajesh topology@yahoo.co.in