 Open Access
 Total Downloads : 1612
 Authors : R. Balaji, N. Rajesh
 Paper ID : IJERTV2IS4079
 Volume & Issue : Volume 02, Issue 04 (April 2013)
 Published (First Online): 04042013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Some New Separation Axioms In Ideal Topological Spaces
Some New Separation Axioms In Ideal Topological Spaces

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.

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,
}
}
is defined as follows: for A X, A*(, I) = {x XU 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 bopen [1] if A Int(Cl(A)) Int(Cl(A)). The notion of bopen sets has been studied extensively in recent years by many topologists [see [4, 5, 12]] because bopen 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, bIR0 space, bIR1 space,
bIR2 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.

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:

bR0 [4] if every bopen set contains the bclosure of each of its singletons.

bR1 [4] if for x, y in X with b Cl({x}) b Cl({y}), there ex
ist disjoint bopen 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:

bIT1 [3] if for each pair of distinct points x and y of X, there exist bIopen sets U and V of X such that x U and y / U , and y V and x / V .

bIT2 [3] if for each pair of distinct points x and y in X, there exist disjoint bIopen sets U and V in X such that x U and y V .


On bIR0 spaces
Definition 3.1. Let (X, , I) be an ideal topological space and A X. Then the bIkernel 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}).
I { }
I { }
I { } I
I { } I
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 bIopen 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).
I { } I
I { } I
Then, there exists a bIopen 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 bIT1 if and only if the singletones are bIclosed [3], it is clear that every bIT1 space bIR0. 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 bIR0 but none of bIT0 and
bIT1.
Remark 3.7. Th following example and Examlpe 3.6 shows that the notions bIT0ness bIR0ness are independent.
Example 3.8. Let X = {a, b, c}, = {, {a}, X} and I = {, {a}}. Then (X, , I) is bIT0 but not bIR0.
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Proposition 3.9. For an ideal topological space (X, , ), the following properties are equivalent:

(X, , I) is bIR0 space;

For any F BIC(X), x /
some U BIO(X);
F implies F U and x /
U for

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

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 bIR0 space. D
Theorem 3.10. An ideal topological space (X, , I) is a bIR0 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 bIR0 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):

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

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 bIopen 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 bIR0 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 bIR0 space. Thus by Lemma

, 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 bIR0 space. D
Theorem 3.13. For an ideal topological space (X, , I), the following
properties are equivalent:

(X, , I) is a bIR0 space;

For any nonempty sets A, G BIO(X) such that A G =/
, there exists F BIC(X) such that A F and F
G;
G;

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

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

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

({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 bIR0 space. D
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Corollary 3.14. For an ideal topological space (X, , ), the following properties are equivalent:

(X, , I) is a bIR0 space;

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

I { } I { }
I { } I { }
I { } I { } I { }
I { } I { } I { }
I { } I { } I { }
I { } I { } I { }
I { } I { } I { }
I { } I { } I { }
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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 bIR0 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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(X, , ) is a b R0 space;
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I { } I { }

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 bIR0. Let x bI Cl({y}) and A BIO(X, y). Now by hypothesis, x A. Therefore, every bIopen 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 /
R . D
R . D
bI Cl({x}). This implies that bI Cl({x}) U . Hence (X, , I) is bI
0
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Theorem 3.17. For an ideal topological space (X, , ), the following properties are equivalent:

(X, , I) is a bIR0 space;

If F is a bIclosed subset of X, then F = bI Ker(F );

If F is a bIclosed subset of X and x F , then bI Ker({x})
F ;

If x X, then bI Ker({x}) bI Cl({x}).
Proof. (1)(2): Let F be bIclosed subset of X and x / F . Thus
X\F BIO(X, x). Since (X, , I) is bIR0, 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 bIclosed, by
R . D
R . D

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 bIclosed, 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 bI
0
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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 bIconverging to y bIconverges 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 bIconverges to y, then {xn}nN bI 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:

(X, , I) is a bIR0 space;

If x, y X, then y bI Cl({x}) if and only if every net in X bIconverging to y bIconverges 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 bIconverges 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 bIconverges to x. Con versely, let x, y X such that every net in X bIconverging to y bI 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 bIconverges to z. Since z bI Cl({y}), then {x}N bIconverges 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 bIR0. D


On bIR1 spaces

Definition 4.1. An ideal topological space (X, , I) is said to be bI
R1 if for x, y in X with bI Cl({x}) bI Cl({y}), there exist disjoint
bIopen sets U and V such that bI Cl({x}) U and bI Cl({y}) V .
Proposition 4.2. If (X, , I) is bIR1, then it is bIR0.
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 bIopen 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 bIR0. D
Theorem 4.3. An ideal topological space (X, , I) is bIR1 if and
only if for x, y X, bI Ker({x}) bI Ker({y}), there exist disjoint
bIopen sets U and V such that bI Cl({x}) U and bI Cl({y}) V .
Proof. It follows from Lemma 3.11 D

(X, , I) is bIT2,

(X, , I) is bIR1 and bIT1, and

(X, , I) is bIR1 and bIT0.
Proof. (1)(2): Since (X, , I) is bIT2, then it is bIT1. If x, y X
such that bI Cl({x}) bI Cl({y}), then x y and there exist disjoint
bIopen 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 bIR1. (2)(3): Since (X, , I) is bIT1, then (X, , I) is bIT0. (3)(1):
Since (X, , I) is bIR1, and bIT1, then (X, , I) is bIR0 and bI
T0, which implies (X, , I) is bIT1. Let x, y X such that x y.
Since bI Cl({x}) = {x} /= {y} = bI Cl({y}), then there exist disjoint bIopen sets U and V such that x U and y V . Hence, (X, , I) is bIT2. D

(X, , I) is bIR1,

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

If U is bIopen, then x U if and only if y U .

there exist disjoint bIopen sets U and V such that x U
and y V , and


If x, y X such that bI Cl({x}) /= bI Cl({y}), then there exist
bIclosed 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 bI
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 bIopen 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 bIopen set A such that x A
and y / A, which implies there exist disjoint bIopen sets U and V
such that x U and y V . Then F1 = X \ V and F2 = X \ U are
bIclosed sets such that x F1, y / F1, y F2, x / F2, X = F1
F2. (3)(1): Let U be bIopen 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 bIclosed 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 bIopen, and x / X \ F1, which is a contradiction. Hence,
(X, , I) is bIR0. Let a, b X such that bI Cl({a}) /= bI Cl({b}). Then there exist bIclosed 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 bIopen, which implies bI Cl({a}) A1 \ A2 and bI Cl({b}) A2
\ A1. Hence, (X, , I) is bIR1. D
Theorem 4.6. An ideal topological space (X, , I) is bIT2 if and only
if for x, y X such that x y, there exist bIclosed 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}AbIconverges to x}.

(X, , I) is bIR1;

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);

(X, , I) is bIR0, and for every bIconvergent 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 bIopen, then x U if and only if y U or (b) there exist disjoint bIopen sets U and V such that x U and y V . Since x, y bI lim({x}A), then

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 bIconverges 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 bIR0. Let {x}A be a bIconvergent net in X. Let x X such that {x}A bIconverges to x. If y bI Cl({x}), then {x}A bIconverges 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 bIR1. Then there exists x, y X such that bI Cl({x}) /= bI Cl({y}) and every bIopen set con taining bI Cl({x}) intersects every bIopen set containing bI Cl({y}). Since (X, , I) is bIR0, then every bIopen set containing x contains bI Cl({x}) and every bIopen set containing y contains bI Cl({y}), which implies that every bIopen set containing x intersects every b Iopen 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 bIconverges 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 bIR1. D
Theorem 4.9. An ideal topological space (X, , I) is bIT2 if and only if every bIconvergent net in X bIconverges 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 { }

for each pair x, y X, b Cl( x ) = b Cl( y ), there exists a
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I I
b open, b closed set V such and y / V ;
I I
I I
I { } / I { }
I { } / I { }

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 bIopen sets U and W such that bI Cl({x}) U and bI Cl({y}) W and V = bI Cl(U ) is bIopen, bIclosed such that x V and y / V . (2)(3): Let x, y X such that bI Cl({x}) =/ bI Cl({y}). Let V be bIopen, bIclosed 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 bIcontinuous, 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 bIopen, bIclosed set of X and bI Cl({x}) U and bI Cl({y}) V . D

(X, , I) is bIR1,

for each pair x, y X, x
set V such and y / V ;
y, there exists a bIopen, bIclosed

for each pair x, y X, x /= y, there exists a bIcontinuous 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 bR0 (resp. bR1).
I N
I N
N I I
N I I

An ideal topological space (X, , ) is b R0 (resp. b R1) if and only if it is b R0 (resp. bR1) ( is the ideal of all nowhere dense sets).
P I I
P I I

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
bIR0 bIR1
bR0 bR1
Example 4.14. Let X = {a, b, c}, = {, {a}, {b}, {a, b}, X} and
I = {, {a}}. Then (X, , I) is bIR0 but none of R0 and bIR1.
Example 4.15. Let X = {a, b, c}, = {, {a, b}, X} and I = {, {c}}. Then (X, , I) is bIR1 but not R1.
Example 4.16. Let X = {a, b, c}, = {, {a}, {b, c}, X} and I =
{, {b}}. Then (X, , I) is bR0 but not bIR0.
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Department of Mathematics, Agni College of Technology, Kancheep uram 603103, TamilNadu, India.
Email address: balaji 2410@yahoo.co.in
Department of Mathematics, Rajah Serfoji Govt. College, Thanjavur 613005, TamilNadu, India.
Email address: nrajesh topology@yahoo.co.in