Separation axioms in ideal bitopological spaces

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https://doi.org/10.26637/MJM0801/0017

Separation axioms in ideal bitopological spaces

P. Maragatha Meenakshi

1

* and A. Vanitha

2

Abstract

In this paper, we introduce and study (i,j)-semi-I-R0 and (i,j)-semi-I-R1 spaces. Also we obtain several characterizations of this axioms.

Keywords

Ideal bitopological spaces,(i,j)-semi-I-closed set,(i,j)-semi-I-open set,(i,j)-semi-I-closure,(i,j)-semi-I -kernal.

AMS Subject Classification 54D10.

1_{Department of Mathematics, Periyar E.V.R. College, Affiliated to Bharathidasan University, Tiruchirappalli-620023, Tamil Nadu, India.} 2_{Department of Mathematics, Valluvar College of Science and Management, Affiliated to Bharathidasan University, Karur-639003, Tamil Nadu,}

India.

*Corresponding author:[email protected]; [email protected]

Article History: Received11October2019; Accepted25December2019 c2020 MJM.

1. Introduction

The concept of ideals in topological spaces has been intro-duced and studied by Kuratowski [6] and Vaidyanathasamy [10]. An idealI on a topological space(X,τ)is a nonempty

collection of subsets ofX which satisfies (i)A∈I andB

⊂AimpliesB∈I and (ii)A∈I andB∈I impliesA

∪B ∈I. Given a bitopological space(X,τ1,τ2)with an idealI onX and ifP(X)is the set of all subsets ofX, a set operator(.)?_i:P(X)→P(X), called the local function [10] of A with respect toτi and I, is defined as follows: forA ⊂X, A?_i(τi,I)={x∈X|U∩A∈/ I for everyU ∈

τi(x)}, whereτi(x)={U∈τi|x∈U}. Observe additionally thatτi-Cl?(A)=A∪A?i(τi,I)defines a Kuratowski closure operator forτ?(I), when there is no chance of confusion,

A?_i(I)is denoted byA?_i andτi-Int?(A)denotes the interior of Ainτ_i?(I). In this paper, we introduce and study(i, j)-semi-I-R0and(i,j)-semi-I-R1spaces. Also we obtain several characterizations of this axioms.

2. Preliminaries

LetAbe a subset of a bitopological space(X,τ1,τ2). We denote the closure ofAand the interior ofAwith respect toτi byτi-Cl(A)andτi-Int(A), respectively.

Definition 2.1. [1] A subset A of an ideal bitopological space (X,τ1,τ2,I)is said to be(i,j)-semi-I-open [1] if A⊂τj -Cl?(τi-Int(A)).

The complement of an(i,j)-semi-I-open set is called an (i,j)-semi-I-closed set.

Definition 2.2. [1] The intersection (resp. union) of all(i, j)-semi-I-closed (resp.(i,j)-semi-I-open) sets of X contain-ing (resp. contained in) A⊂X is called the(i,j)-semi-I -closure (resp.(i,j)-semi-I-interior) of A and is denoted by (i,j)-sICl(A)(resp.(i,j)-sIInt(A)). The the intersection of all(i,j)-semi-I-open sets of X containing A is called the (i,j)-semi-I-kernal of A and is denoted by(i,j)-sIKer(A).

Definition 2.3. [7] An ideal bitopological space(X,τ1,τ2,I) is said to be

1. (i,j)-semi-I-T0if for every pair of distinct points in X , there exists an(i,j)-semi-I-open set of X containing one of the points but not the other.

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3. (i,j)-semi-I-T2if for every pair of distinct points x, y of X , there exists a pair of disjoint(i,j)-semi-I-open sets, one containing x and the other containing y.

3. On

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Definition 3.1. An ideal bitopological space(X,τ1,τ2,I)is said to be(i,j)-semi-I-R0if for every(i,j)-semi-I-open set of X contains the(i,j)-semi-I-closure of each of its sin-gletons.

Definition 3.2. An ideal bitopological space(X,τ1,τ2,I) is said to be (i,j)-semi-I-symmetric if for each x,y∈X , x∈(i,j)-sICl({y})implies y∈(i,j)-sICl({x}).

Theorem 3.3. An ideal bitopological space(X,τ1,τ2,I)is (i,j)-semi-I-R0if, and only if it is(i,j)-semi-I-symmetric.

Proof. Assume that(X,τ1,τ2,I)is(i,j)-semi-I-R0. Let x∈(i,j)-sICl({y})andU be any(i,j)-semi-I-open set such thaty∈U. Then by hypothesis,x∈U. Therefore, every (i,j)-semi-I-open set which containsycontainsx. Hence, y∈(i,j)-sICl({x}). Conversely, letUbe an(i,j)-semi-I -open set andx∈U. Ify∈/U, thenx∈/(i,j)-sICl({y}), and thus by assumption,y∈/(i,j)-sICl({x}). Therefore,(i, j)-sICl({x})⊂U, and hence,(X,τ1,τ2,I)is(i,j)-semi-I -R0.

Theorem 3.4. An ideal bitopological space(X,τ1,τ2,I)is (i,j)-semi-I-T1if, and only if(X,τ1,τ2,I)is(i,j)-semi-I -T0and(i,j)-semi-I-R0

Proof. Letx,y∈X andx6=y. Since (X,τ1,τ2,I)is(i, j)-semi-I-T0, we may assume without loss of generality that x∈G⊂X\{y}for some(i,j)-semi-I-open setG. Thus,x∈/

(i,j)-sICl({y}), and by Theorem3.3,y∈/(i,j)-sICl({x}). Therefore,X\(i,j)-sICl({x})is an(i,j)-semi-I-open set containingybut notx. Hence,(X,τ1,τ2,I)is(i,j)-semi-I -T1. The converse is clear.

Proposition 3.5. The following statements are equivalent for an ideal bitopological space(X,τ1,τ2,I):

1. (X,τ1,τ2,I)is(i,j)-semi-I-R0space;

2. If for any F∈(i,j)-SIC(X), x∈/F, then F ⊂U and x∈/U for some U∈(i,j)-SIO(X);

3. If for any F∈(i,j)-SIC(X)such that x∈/F, then F

∩(i,j)-sICl({x})=/0;

4. If for any two distinct points x,y∈X , then either(i, j)-sICl({x})=(i,j)-sICl({y})or(i,j)-sICl({x})∩

(i,j)-sICl({y})=/0.

Proof. (1)⇒(2): LetF∈(i,j)-SIC(X)andx∈/F. Then by (1)(i,j)-sICl({x})⊂X\F. SetU =X\(i,j)-sICl({x}), thenU∈(i,j)-SIO(X)withF⊂Uandx∈/U.

(2)⇒(3): LetF∈(i,j)-SIC(X)such thatx∈/ F. Then by (2), there existsU∈(i,j)-SIO(X)such thatF⊂Uandx∈/

U. SinceU∈(i,j)-SIO(X),U∩(i,j)-sICl({x})= /0 and F∩(i,j)-sICl({x})= /0.

(3)⇒(4): Suppose that(i,j)-sICl({x})6= (i,j)-sICl({y}) for the distinct pointsx,y∈X. Then there existsz∈(i, j)-sICl({x}) such that z ∈/ (i,j)-sICl({y}) (or z ∈ (i, j)-sICl({y}) such that z ∈/ (i,j)-sICl({x})). Then there existsV ∈(i,j)-SIC(X,z)such thaty∈/ V, hencex∈V. Therefore, x∈/ (i,j)-sICl({y}). By (3), we obtain(i, j)-sICl({x})∩(i,j)-sICl({y})= /0. The proof for the other case is similar.

(4)⇒(1): LetV ∈ (i,j)-SIO(X,x). For each y∈/ V, we havex6=yandx∈/(i,j)-sICl({y}). This shows that(i, j)-sICl({x})6= (i,j)-sICl({y}). Hence(i,j)-sICl({x})∩

(i,j)-sICl({y})= /0 for eachy∈X\Vand(i,j)-sICl({x})

∩( ∪

y∈X\V

(i,j)-sICl({y})) = /0. SinceV ∈(i,j)-SIO(X)

andy∈X \V,(i,j)-sICl({y})⊂X \V and henceX \

V = ∪

y∈X\V

(i,j)-sICl({y}). Therefore, we obtain(X\V)∩

(i,j)-sICl({x})= /0 and hence(i,j)-sICl({x})⊂V. Then (X,τ1,τ2,I)is(i,j)-semi-I-R0.

Theorem 3.6. An ideal bitopological space (X,τ1,τ2,I) is (i,j)-semi-I-R0 if, and only if for any x,y∈X , (i, j)-sICl({x})6= (i,j)-sICl({y})implies(i,j)-sICl({x})∩

(i,j)-sICl({y})=/0.

Proof. Suppose that(X,τ1,τ2,I)is(i,j)-semi-I-R0andx, y∈Xsuch that(i,j)-sICl({x})6= (i,j)-sICl({y}). There existsz∈(i,j)-sICl({x})such thatz∈/ (i,j)-sICl({y}) (or z ∈ (i,j)-sICl({y}) such that z ∈/ (i,j)-sICl({x})). Sincez∈/(i,j)-sICl({y}), there existsV ∈(i,j)-SIO(X,z) such thaty∈/V. Butz∈(i,j)-sICl({x})sox∈V. Then x∈/ (i,j)-sICl({y}). Hence x∈X \ (i,j)-sICl({y})∈

(i,j)-SIO(X). Since(X,τ1,τ2,I)is(i,j)-semi-I-R0, we have(i,j)-sICl({x})⊂X\(i,j)-sICl({y}). Hence(i, j)-sICl({x})∩ (i,j)-sICl({y}) = /0. The proof for other-wise is similar. Conversely, letV ∈(i,j)-SIO(X,x). We will show that (i,j)-sICl({x}) ⊂V. Lety∈/ V, that is, y∈X \V. Then x6= yand x ∈/ (i,j)-sICl({y}). This shows that(i,j)-sICl({x})6= (i,j)-sICl({y}). By assump-tion,(i,j)-sICl({x})∩(i,j)-sICl({y})= /0. Hencey∈/

(i,j)-sICl({x})and therefore(i,j)-sICl({x})⊂V. Hence (X,τ1,τ2,I)is(i,j)-semi-I-R0.

Theorem 3.7. An ideal bitopological space (X,τ1,τ2,I) is (i,j)-semi-I-R0 if, and only if for any points x and y in X ,(i,j)-sIKer({x})6= (i,j)-sIKer({y})implies(i, j)-sIKer({x})∩(i,j)-sIKer({y}) =/0.

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As-sume thatz∈(i,j)-sIKer({x})∩(i,j)-sIKer({y}). Byz∈

(i,j)-sIKer({x}),x∈(i,j)-sICl({z}). Thus by Theorem 3.6,(i,j)-sICl({x})=(i,j)-sICl({z}). Similarly, we have (i,j)-sICl({y})=(i,j)-sICl({z})=((i,j)-sICl({x}), a contradiction. Hence(i,j)-sIKer({x})∩(i,j)-sIKer({y}) = /0. Conversely, let(i,j)-sIKer({x})6= (i,j)-sIKer({y}) implies (i,j)-sIKer({x}) ∩ (i,j)-sIKer({y}) = /0. As-sume that(i,j)-sICl({x})= (i6 ,j)-sICl({y}). Then(i, j)-sIKer({x})= (i6 ,j)-sIKer({y}), and therefore by assump-tion, (i,j)-sIKer({x}) ∩(i,j)-sIKer({y}) = /0. Now if z∈(i,j)-sICl({x}), thenx∈(i,j)-sIKer({z}), and there-fore,(i,j)-sIKer({x})∩(i,j)-sIKer({z})6=/0. By hypoth-esis, (i,j)-sI({x}) = (i,j)-sIKer({z}). Thus z∈(i, j)-sICl({x}) ∩ (i,j)-sICl({y}) implies (i,j)-sIKer({x}) = (i,j)-sIKer({z}) = (i,j)-sIKer({y}), a contradiction. Therefore(i,j)-sICl({x})6= (i,j)-sICl({y})implies that (i,j)-sICl({x})∩(i,j)-sICl({y})= /0, and Theorem3.6, (X,τ1,τ2,I)is(i,j)-semi-I-R0.

Theorem 3.8. For an ideal bitopological space(X,τ1,τ2,I), the following statements are equivalent:

1. (X,τ1,τ2,I)is(i,j)-semi-I-R0.

2. For any nonempty subset A of X and G∈(i,j)-SIO(X) such that A∩G6= /0, there exists F∈(i,j)-SIC(X) such that A∩F6=/0and F⊂G.

3. For any G ∈ (i,j)-SIO(X), G = ∪{F :F ∈ (i, j)-SIC(X), F⊂G}.

4. For any F ∈ (i,j)-SIC(X), F = ∩{G:G∈(i, j)-SIO(X), F⊂G}.

5. For any x∈X ,(i,j)-sICl({x})⊂(i,j)-sIKer({x}).

Proof. (1)⇒(2): Let A be a nonempty set ofXandG∈(i, j)-SIO(X)such thatA∩G6= /0. Then there existsx∈A∩G. Sincex∈G∈(i,j)-SIO(X),(i,j)-sICl({x})⊂G. SetF =(i,j)-sICl({x}). ThenF∈(i,j)-SIC(X),F⊂GandA

∩F6=/0.

(2)⇒(3): LetG∈(i,j)-SIO(X), thenG⊃ ∪{F:F∈(i, j)-SIC(X),F⊂G}. Letxbe any point ofG. Then there exists F ∈(i,j)-SIC(X)such thatx∈FandF ⊂G. Therefore, x∈F⊂ ∪{F:F∈(i,j)-SIC(X),F ⊂G}, and henceG=

∪{F:F∈(i,j)-SIC(X),F ⊂G}. (3)⇒(4): This is obvious.

(4)⇒(5): Letxbe any point ofX andy∈/(i,j)-sIKer({x}). Then there existsV ∈ (i,j)-SIO(X,x) anyy∈/ V; hence (i,j)-sICl({y})∩V= /0. By (4),∩{G:G∈(i,j)-SIO(X), (i,j)-sICl({y})⊂G}, and there existsG∈(i,j)-SIO(X) such thatx∈/Gand(i,j)-sICl({y})⊂G. Therefore,(i, j)-sICl({x})∩G= /0 andy∈/(i,j)-sICl((i,j)-sICl({x})) = (i,j)-sICl({x}). Consequently, we obtain(i,j)-sICl({x})

⊂(i,j)-sIKer({x}).

(5)⇒(1): LetG∈(i,j)-SIO(X,x). Ify∈(i,j)-sIKer({x}), thenx∈(i,j)-sICl({y})and soy∈G. This implies that (i,j)-sIKer({x})⊂G. Therefore,x∈(i,j)-sICl({x})⊂

(i,j)-sIKer({x})⊂G. This shows that(X,τ1,τ2,I)is an (i,j)-semi-I-R0space.

Corollary 3.9. For an ideal bitopological space(X,τ1,τ2,I) is(i,j)-semi-I-R0if, and only if(i,j)-sICl({x})=(i, j)-sIKer({x})for each x∈X .

Proof. Suppose(X,τ1,τ2,I)is an(i,j)-semi-I-R0space. By Theorem3.8,(i,j)-sICl({x})⊂(i,j)-sIKer({x})for eachx∈ X. Let y∈(i,j)-sIKer({x}). Then we havex

∈(i,j)-sICl({y})and by Theorem3.6(i,j)-sICl({x})= (i,j)-sICl({y}). Therefore,y∈(i,j)-sICl({x})and hence (i,j)-sIKer({x})⊂(i,j)-sICl({x}). This shows that(i, j)-sICl({x})=(i,j)-sIKer({x}). The converse follows from Theorem3.8.

Theorem 3.10. The following statements are equivalent for an ideal bitopological space(X,τ1,τ2,I):

2. x∈(i,j)-sICl({y})⇔y∈(i,j)-sICl({x})for any points x and y in X .

Proof. (1)⇒(2): Assume that(X,τ1,τ2,I)is(i,j)-semi-I -R0andx∈(i,j)-sICl({y}). Then(i,j)-sICl({x})=(i, j)-sICl({y}). Hencey∈(i,j)-sICl({x}). The other part is similar.

(2)⇒(1): Letx∈U∈(i,j)-SIO(X,x). Ify∈/U, thenx∈/

(i,j)-sICl({y}) and hence y∈/ (i,j)-sICl({x}) (by (2)). Thus (i,j)-sICl({x})⊂U. Hence (X,τ1,τ2,I)is (i, j)-semi-I-R0.

2. If F is an(i,j)-semi-I-closed subset of X , then F = (i,j)-sIKer(F).

3. If F is an(i,j)-semi-I-closed subset of X and x∈F, then(i,j)-sIKer({x})⊂F.

4. If x∈X , then(i,j)-sIKer({x})⊂(i,j)-sICl({x}).

Proof. (1)⇒(2): LetFbe an(i,j)-semi-I-closed subset ofX andx∈/F. ThusX\F∈(i,j)-SIO(X,x). Since(X,τ1,τ2,I) is(i,j)-semi-I-R0,(i,j)-sICl({x})⊂X \F. Thus(i, j)-sICl({x})∩F = /0 andx∈/ (i,j)-sIKer(F). Therefore, (i,j)-sIKer(F)=F.

(2)⇒(3): IfA⊂B, then(i,j)-sIKer(A)⊂(i,j)-sIKer(B). Then(i,j)-sIKer({x})⊂(i,j)-sIKer(F)=F.

(3)⇒(4): Sincex∈(i,j)-sICl({x})and(i,j)-sICl({x})is (i,j)-semi-I-closed,(i,j)-sIKer({x})⊂(i,j)-sICl({x}). (4)⇒(1): Ifx∈(i,j)-sICl({y}), theny∈(i,j)-sIKer({x}). Since x∈ (i,j)-sICl({x}) and (i,j)-sICl({x}) is (i, j)-semi-I-closed, by (4), we obtainy∈(i,j)-sIKer({x})⊂

(i,j)-sICl({x}). Thenx∈(i,j)-sICl({y})implies thaty∈

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Definition 3.12. A net{xα}α∈Λ in an ideal bitopological

space(X,τ1,τ2,I)is called(i,j)-semi-I-convergent to a point x in X if for every U∈(i,j)-SIO(X,x), there exists

α0∈Λsuch that xα∈U for eachα≥α0.

Lemma 3.13. Let (X,τ1,τ2,I)be an ideal bitopological space and let x and y any two points in X such that every net in X (i,j)-semi-I-converging to y(i,j)-semi-I-converges to x. Then x∈(i,j)-sICl({y}).

Proof. Supposexn=yfor eachn∈N. Then{xn}n∈Nis a net inXthat(i,j)-semi-I-convergence toy. Thus by assumption, (i,j)-semi-I-converges tox. Sox∈(i,j)-sICl({y}).

2. If x, y∈X , then y∈(i,j)-sICl({x})if, and only if every net in X (i,j)-semi-I-converging to y also(i, j)-semi-I-converges to x.

Proof. (1)⇒(2): Letx,y∈Xsuch thaty∈(i,j)-sICl({x}). Suppose that{xα}α∈Λbe a net inXsuch that{xα}α∈Λ(i,

j)-semi-I-converges toy. Sincey∈(i,j)-sICl({x}), by The-orem3.3,x∈(i,j)-sICl({y}). Conversely, letx,y∈Xsuch that every net inX(i,j)-semi-I-converging toy(i,

j)-semi-I-converges tox. Thenx∈(i,j)-sICl({y}). By Theorem 3.10,y∈(i,j)-sICl({x}).

(2)⇒(1): Assume thatxandyare any two points ofXsuch that(i,j)-sICl({x})∩(i,j)-sICl({y})6=/0. Letz∈(i, j)-sICl({x})∩(i,j)-sICl({y}). There exists a net{xα}α∈Λ

in(i,j)-sICl({x}) (i,j)-semi-I-converges to z. Sincez

∈(i,j)-sICl({y}),{xα}α∈Λalso(i,j)-semi-I-converges

toy. Hence by (2)z∈(i,j)-sICl({y}). Therefore(i, j)-sICl({z})⊂(i,j)-sICl({y})(?). Soy∈(i,j)-sICl({z}) gives(i,j)-sICl({y})⊂(i,j)-sICl({z})(??). Hence from (?) and (??),(i,j)-sICl({y})=(i,j)-sICl({z}). Similarly it can be shown that(i,j)-sICl({x})=(i,j)-sICl({z})by taking the net in(i,j)-sICl({y}). So(i,j)-sICl({x})= (i,j)-sICl({y}). Then (X,τ1,τ2,I) is (i,j)-semi-I-R0.

Definition 3.15. An ideal bitopological space(X,τ1,τ2,I) is said to be(i,j)-semi-I-R1if for each points x and y of X such that(i,j)-sICl({x})6= (i,j)-sICl({y}), there exist disjoint(i,j)-semi-I-open subsets of X , say, U and V such that(i,j)-sICl({x})⊂U and(i,j)-sICl({y})⊂V .

Proposition 3.16. Every(i,j)-semi-I-R1space is(i,

j)-semi-I-R0.

Proof. LetU ∈(i,j)-SIO(X,x). Ify∈/U, thenx∈/(i, j)-sICl({y}). So(i,j)-sICl({x})6= (i,j)-sICl({y}). Since (X,τ1,τ2,I)is(i,j)-semi-I-R1, there exists an(i,

j)-semi-I-open setVysuch that(i,j)-sICl({y})⊂Vyandx∈/Vy, implies thaty∈/ (i,j)-sICl({x}). Hence(i,j)-sICl({x})

⊂U. Then(X,τ1,τ2,I)is(i,j)-semi-I-R0.

1. (X,τ1,τ2,I)is(i,j)-semi-I-T2,

2. (X,τ1,τ2,I) is (i,j)-semi-I-R1 and (i,j)-semi-I -T1,

3. (X,τ1,τ2,I) is (i,j)-semi-I-R1 and (i,j)-semi-I -T0.

Proof. (1)⇒(2): Since(X,τ1,τ2,I)is(i,j)-semi-I-T2, then it is(i,j)-semi-I-T1. Ifx,y∈X such that(i,j)-sICl({x})

6

= (i,j)-sICl({y}), thenx6=yand there exist disjoint(i, j)-semi-I-open setsU andV such that x ∈ U and y ∈ V. Hence by Theorem3.4, (i,j)-sICl({x}) ={x} ⊂U and (i,j)-sICl({y})={y} ⊂V. Hence (X,τ1,τ2,I)is(i, j)-semi-I-R1.

(2)⇒(3): Since(X,τ1,τ2,I)is(i,j)-semi-I-T1, then it is (i,j)-semi-I-T0.

(3)⇒(1): Since(X,τ₁,τ2,I)is(i,j)-semi-I-R1and(i, j)-semi-I-T0, then by Proposition3.16,(X,τ1,τ2,I)is(i, j)-semi-I-R0and (i,j)-semi-I-T0. Hence by Theorem 3.4, (X,τ1,τ2,I)is(i,j)-semi-I-T1. Letx,y∈Xsuch thatx6= y. Then(i,j)-sICl({x})={x} 6={y}=(i,j)-sICl({y}). Since(X,τ1,τ2,I)is(i,j)-semi-I-R1, there exist disjoint (i,j)-semi-I-open setsUandVsuch that(i,j)-sICl({x}) =

{x} ⊂U and(i,j)-sICl({y}) ={y} ⊂V. Hence we have (X,τ1,τ2,I)is(i,j)-semi-I-T2and thus by Theorem3.3 (X,τ)is an(i,j)-semi-I-R0space.

Corollary 3.18. For an(i,j)-semi-I-R1space(X,τ1,τ2,I), the following statements are equivalent:

1. (X,τ1,τ2,I)is(i,j)-semi-I-T2.

Theorem 3.19. For an ideal bitopological space(X,τ1,τ2,I) is(i,j)-semi-I-R1if, and only if x∈X\(i,j)-sICl({y}) im-plies that x and y have disjoint(i,j)-semi-I-neighbourhoods.

Proof. Letx∈X\(i,j)-sICl({y}). Then(i,j)-sICl({x})6= (i,j)-sICl({y}) and x and y have disjoint (i,j)-semi-I -neighbourhoods. Conversely, first we show that(X,τ1,τ2,I) is(i,j)-semi-I-R0. LetUbe an(i,j)-semi-I-open set and x∈U. Suppose thaty∈/U. Then,(i,j)-sICl({y})∩U=/0 and x∈/ (i,j)-sICl({y}). Then there exist disjoint (i, j)-semi-I-open setsUxandUysuch thatx∈Uxandy∈Uyand Ux∩Uy=/0. Hence,(i,j)-sICl({x})⊂(i,j)-sICl(Ux)and (i,j)-sICl(x)∩Uy⊂(i,j)-sICl({Ux})∩Uy= /0. There-fore,y∈/(i,j)-sICl({x}). Consequently,(i,j)-sICl({x})⊂

(5)

thatz∈Vz,y∈Vy. Sincez∈(i,j)-sICl({x}),x∈Vz. Since (X,τ1,τ2,I)is(i,j)-semi-I-R0,(i,j)-sICl({x})⊂Vz, (i,j)-sICl({y})⊂Vy andVz∩Vy= /0. Then(X,τ1,τ2)is (i,j)-semi-I-R1and thus by Theorem3.3(X,τ)is an(i,

j)-semi-I-R0space.

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

(a) If U is(i,j)-semi-I-open, then x∈U if, and only if y∈U .

(b) there exist disjoint(i,j)-semi-I-open sets U and V such that x∈U and y∈V .

3. If x, y∈X and(i,j)-sICl({x})6= (i,j)-sICl({y}), then there exist(i,j)-semi-I-closed sets F1and F2such that x∈F1, y∈/F1, y∈F2, x∈/F2, and X = F1∪F2.

Proof. (1)⇒(2): Letx,y∈X. Then(i,j)-sICl({x})=(i, j)-sICl({y})or(i,j)-sICl({x})6= (i,j)-sICl({y}). If(i, j)-sICl({x})=(i,j)-sICl({y})andUis(i,j)-semi-I-open, then x∈U impliesy∈ (i,j)-sICl({x})⊂U andy ∈U impliesx∈(i,j)-sICl({y})⊂U. Thus consider the case that(i,j)-sICl({x})6= (i,j)-sICl({y}). Then there exist disjoint(i,j)-semi-I-open setsUandV such thatx∈(i, j)-sICl({x})⊂Uandy∈(i,j)-sICl({y})⊂V.

(2)⇒(3): Letx,y∈X,(i,j)-sICl({x})6= (i,j)-sICl({y}). Thenx∈/(i,j)-sICl({y})ory∈/(i,j)-sICl({x}), sayx∈/

(i,j)-sICl({y}). Then there exists an(i,j)-semi-I-open set Asuch thatx∈A andy∈/A. Then by (2) there exist disjoint (i,j)-semi-I-open setsUandV such thatx∈Uandy∈V. ThenF1=X\VandF2=X\Uare(i,j)-semi-I-closed sets such thatx∈F1,y∈/F1,y∈F2,x∈/F2andX=F1∪F2. (3)⇒(1): We shall first show that(X,τ1,τ2,I)is(i,

j)-semi-I-R0space. LetU be an(i,j)-semi-I-open set such that x∈U. We claim that(i,j)-sICl({x})⊂U. For supposey∈

(i,j)-sICl({x})∩(X\U). Then(i,j)-sICl({x})6= (i, j)-sICl({y})(if(i,j)-sICl({x}) = (i,j)-sICl({y}), theny∈

U ) and hence by (3), there exist(i,j)-semi-I-closed sets F1andF2such thatx∈F1,y∈/F1,y∈F2,x∈/F2andX = F1∪F2. Theny∈F2\F1=X\F1∈(i,j)-SIO(X)andx∈/ X\F1, a contradicts the fact thaty∈(i,j)-sICl({x}). Hence (X,τ1,τ2,I)is(i,j)-semi-I-R0space. Let p,q∈Xbe such that(i,j)-sICl({p})6= (i,j)-sICl({q}). Then by the given condition there exist(i,j)-semi-I-closed setsH1andH2such thatp∈H1,q∈/H1,q∈H2,p∈/H2andX=H1∪H2. Thus p∈H1\H2 and q∈H2\H1, where H1\H2 and H2\H1 are disjoint(i,j)-semi-I-open sets. Hence(i,j)-sICl({p})⊂

H1\H2and(i,j)-sICl({q})⊂H2\H1. Hence(X,τ1,τ2,I) is(i,j)-semi-I-R1space. and thus by Theorem3.3(X,τ)is

an(i,j)-semi-I-R0space.

In view of Theorems3.17and3.20, it now follows that

Theorem 3.21. A bitopological space(X,τ1,τ2,I)is(i, j)-semi-I-T2if, and only if for each x, y∈X such that x6=y, there exist(i,j)-semi-I-closed sets F1and F2such that x∈ F1, y∈/F1, y∈F2, x∈/F2and X = F1∪F2.

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Separation axioms in ideal bitopological spaces