(i, j
)-β
-I
-i-Open Sets and
(i, j
)-β
-I
-i-Almost
Continuous Functions in Ideal Bitopological
Spaces
Hariwan Zikri Ibrahim
Department of Mathematics, Faculty of Science, University of Zakho, Kurdistan-Region, Iraq ∗Corresponding Author: hariwan math@yahoo.com
Copyright c⃝2014 Horizon Research Publishing All rights reserved.
Abstract
In this paper we introduce and study the notion of (i, j)-β-I-i-open sets in ideal bitopological spaces and investigate some of their properties. Also we present and study (i, j)-β-I-i-continuous and (i, j)-β-I-i-almost continuous functions. Furthermore, we obtain basic properties and preservation theorems of (i, j)-β-I
-i-continuous and (i, j)-β-I-i-almost continuous.
Keywords
Ideal bitopological spaces, i-open sets, (i, j)-β-I-open sets, (i, j)-β-I-i-open sets, (i, j)-β-I-i -continuous, (i, j)-β-I-i-almost continuous.1
Introduction
A triple (X, τ1, τ2) whereXis a non-empty set andτ1
andτ2are topologies onXis called a bitopological space.
An idealI on a topological space (X, τ) is a nonempty collection of subsets ofX which satisfies (1)A∈I and
B ⊆A impliesB ∈I and (2) A∈I andB ∈I implies
A∪B ∈ I. An ideal topological space is a topologi-cal space (X, τ) with an ideal I on X and is denoted by (X, τ, I). The subject of ideals in topological spaces has been introduced and studied by Kuratowski [3] and Vaidyanathaswamy [7]. Let (X, τ1, τ2) be a
bitopologi-cal space and letIbe an ideal of subsets ofX. An ideal bitopological space is a bitopological space (X, τ1, τ2)
with an ideal I on X and is denoted by (X, τ1, τ2, I).
A set operator (.)∗i : P(X)→ P(X) is called the local function [7] ofA with respect to τi and I, is defined as follows for A ⊆ X, A∗i(τi, I) = {x ∈ X : U ∩A /∈ I for every U ∈ τi(x)}, where τi(x) = {U ∈ τi|x ∈ U}. For every ideal bitopological space (X, τ1, τ2, I), there
exists a topology τi∗(I), finer than τi. Additionally,
τi−Cl∗(A) =A∪A∗i defines a Kuratowski closure op-erator [3] for τi∗(I). Also, τi−Cl∗(A)⊆τi−Cl(A) for any subsetAofX.
2
Preliminaries
For a subsetAof a bitopological space (X, τ1, τ2), we
denote the closure ofAand the interior ofAwith respect to τi byτi−Cl(A) andτi−Int(A), respectively.
Definition 2.1 [6] A subsetAofX is said to beij-semi open if A⊆τj−Cl(τi−Int(A)).
Definition 2.2 [5] A subsetA of X is said to beij-β -open if A⊆τj−Cl(τi−Int(τj−Cl(A))).
Definition 2.3 [1] A subset A of X is said to be ij -regular open if A=τi−Int(τj−Cl(A)).
Definition 2.4 [4] A point x of X is said to be ij-δ -cluster point of AifA∩U ̸=φfor everyij-reguler open setU containingx, the set of allij-δ-cluster points ofA
is called ij-δ-closure of A and denoted by ij−Clδ(A).
A subset A of X is said to be ij-δ-closed if ij-δ-cluster points ofAis subset ofA, the complement ofij-δ-closed set is ij-δ-open and ij-δ-interior ofA is denoted by ij
-Intδ(A).
Definition 2.5 [2] A subset Aof an ideal bitopological space (X, τ1, τ2, I) is said to be (i, j)-β-I-open if A ⊆
τj −Cl(τi −Int(τj −Cl∗(A))), where i, j = 1,2 and
i̸=j.
Definition 2.6 [2] In an ideal bitopological space
(X, τ1, τ2, I), A ⊆ X is said to be (i, j)-β-I-closed if
X\A is(i, j)-β-I-open in X,i, j= 1,2 andi̸=j.
Theorem 2.7 [2] Arbitrary intersection of (i, j)-β-I -closed sets is always (i, j)-β-I-closed.
Definition 2.8 [2] A function f : (X, τ1, τ2, I) →
(Y, σ1, σ2) is said to be (i, j)-β-I-continuous if the
in-verse image of every i-open set of Y is (i, j)-β-I-open in X, wherei̸=j, i, j= 1,2.
3
(
i, j
)
-
β
-
I
-
i
-Open Sets
The family of all (i, j)-β-I-i-open subsets of ideal bitopological space (X, τ1, τ2, I) is denoted by (i, j)-β
-I-i O(X, τ1, τ2, I) or (i, j)-β-I-i O(X).
Theorem 3.2 A subset A of a space X is (i, j)-β-I
-i-open if and only if A is i-open and it is a union of
(i, j)-β-I-closed sets. That is, A = ∪Fα where A is
i-open andFα is an(i, j)-β-I-closed set for each α.
Proof. Obvious.
Remark 3.3 It is clear from the definition that every
(i, j)-β-I-i-open subset of a space X is i-open but not conversely.
Theorem 3.4 Every (i, j)-β-I-i-open set is (i, j)-β-I -open.
Proof. LetAbe an (i, j)-β-I-i-open subset ofX. Then, we haveA⊆τj−Cl∗(A) and hence A=τi−Int(A)⊆
τi−Int(τj−Cl∗(A))⊆τj−Cl(τi−Int(τj−Cl∗(A))) The following example shows that the converses of Theorem 3.4 is not true in general.
Example 3.5 Let X = {1,2,3,4}, τ1 =
{X, φ,{1},{2},{1,2},{1,2,3}}, τ2 = {X, φ} and
I = {φ,{3},{4},{3,4}}. Then the set {2,3} is
(1,2)-β-I-open but not(1,2)-β-I-1-open.
Theorem 3.6 Let{Aα:α∈∆}be a collection of(i, j)
-β-I-i-open sets in an ideal bitopological space X. Then
∪
{Aα:α∈∆} is also (i, j)-β-I-i-open.
Proof. Since Aα is an (i, j)-β-I-i-open set for each α, thenAα is i-open and
∪
{Aα :α∈∆} is i-open, so for allx ∈Aα, there exists an (i, j)-β-I-closed set F such that x ∈ F ⊆ Aα this implies that x ∈ F ⊆ Aα ⊆
∪
{Aα : α ∈ ∆}, then x∈ F ⊆
∪
{Aα : α ∈ ∆}, and hence∪{Aα:α∈∆}is an (i, j)-β-I-i-open set.
Theorem 3.7 The intersection of two (i, j)-β-I-i-open sets is(i, j)-β-I-i-open.
Proof. LetAandB be two (i, j)-β-I-i-open sets, then
AandB arei-open sets this implies thatA∩B is also an i-open set, we have to prove that A∩B is (i, j)-β
-I-i-open, letx∈A∩B thenx∈A and x∈B, for all
x∈ A there exists an (i, j)-β-I-closed set F such that
x ∈ F ⊆ A and for all x ∈ B there exists an (i, j
)-β-I-closed set E such that x ∈ E ⊆ B, and so that
x∈F∩E⊆A∩B. Since the intersection of two (i, j
)-β-I-closed sets is (i, j)-β-I-closed by Theorem 2.7, this shows thatA∩B is (i, j)-β-I-i-open set.
From Theorems 3.6 and 3.7 we conclude that the fam-ily of all (i, j)-β-I-i-open subsets of a spaceXis a topol-ogy onX.
Theorem 3.8 A subset A of a space (X, τ1, τ2, I) is
(i, j)-β-I-i-open if and only if for each x ∈ A, there exists an(i, j)-β-I-i-open setB such thatx∈B⊆A.
Proof. Assume that A is an (i, j)-β-I-i-open set in (X, τ1, τ2, I), let x∈A. If we putB =A then B is an
(i, j)-β-I-i-open set containingxsuch thatx∈B⊆A. Conversely, suppose that for each x ∈ A, there exists an (i, j)-β-I-i-open set Bx such thatx∈Bx⊆A, thus
A = ∪Bx where Bx ∈ (i, j)-β-I-i O(X) for each x, thereforeAis (i, j)-β-I-i-open.
Definition 3.9 A subsetB of a spaceX is called(i, j)
-β-I-i-closed ifX\B is (i, j)-β-I-i-open.
The family of all (i, j)-β-I-i-closed subsets of ideal bitopological space (X, τ1, τ2, I) is denoted by (i, j)-β
-I-i C(X, τ1, τ2, I) or (i, j)-β-I-i C(X).
Theorem 3.10 A subset B of a space X is(i, j)-β-I
-i-closed if and only if B is an i-closed set and it is an intersection of (i, j)-β-I-open sets.
Proof. Obvious.
Theorem 3.11 Let {Bα : α ∈ ∆} be a collection of (i, j)-β-I-i-closed sets in an ideal bitopological spaceX. Then∩{Bα:α∈∆} is(i, j)-β-I-i-closed set.
Proof. Follows from Theorem 3.6.
Theorem 3.12 The union of two (i, j)-β-I-i-closed sets is (i, j)-β-I-i-closed.
Proof. Follows from Theorem 3.7.
Definition 3.13 Let (X, τ1, τ2, I)be an ideal
bitopolog-ical space and x ∈ X. A subset N of X is said to be
(i, j)-β-I-i-neighborhood of xif there exists an (i, j)-β
-I-i-open set U in X such thatx∈U ⊆N.
Theorem 3.14 A subset A of an ideal bitopological space (X, τ1, τ2, I) is (i, j)-β-I-i-open if and only if it
is an (i, j)-β-I-i-neighborhood of each of its points.
Proof. LetA⊆X be an (i, j)-β-I-i-open set, since for everyx∈A,x∈A⊆AandAis (i, j)-β-I-i-open. This shows that A is (i, j)-β-I-i-neighborhood of each of its points.
Conversely, suppose that A is an (i, j)-β-I-i -neighborhood of each of its points, then for each
x ∈ A, there exists Bx ∈ (i, j)-β-I-i O(X) such that
x ∈ Bx ⊆ A. Therefore A =
∪
{Bx : x ∈ A}. Since each Bx is (i, j)-β-I-i-open, it follows that A is an (i, j)-β-I-i-open set.
Definition 3.15 Let A be a subset of an ideal bitopo-logical space (X, τ1, τ2, I). A point x∈ X is said to be
(i, j)-β-I-i-interior point of A, if there exists an (i, j)
-β-I-i-open set U such that x ∈ U ⊆ A. The set of all (i, j)-β-I-i-interior points of A is called (i, j)-β-I-i -interior of Aand is denoted by(i, j)-β-I-i Int(A).
Theorem 3.16 Let X be a space and A⊆X, x∈X. Thenxis an (i, j)-β-I-i-interior point ofA if and only if Ais an (i, j)-β-I-i-neighborhood of x.
Proof. Obvious.
Theorem 3.17 Let A be any subset of a space X. If a point x is in the (i, j)-β-I-i-interior of A, then there exists an (i, j)-β-I-closed setF of X containingxsuch that F ⊆A.
Proof. Suppose thatx∈(i, j)-β-I-i Int(A), then there exists an (i, j)-β-I-i-open setU ofX containingxsuch thatU ⊆A. SinceU is an (i, j)-β-I-i-open set, so there exists an (i, j)-β-I-closed set F containing xsuch that
x∈F ⊆U ⊆A. Hence,x∈F ⊆A.
Theorem 3.18 For subsetsAandB of a spaceX, the following statements hold:
1. (i, j)-β-I-i Int(A) is the union of all (i, j)-β-I-i -open sets which are contained inA.
2. (i, j)-β-I-i Int(A)is an (i, j)-β-I-i-open set inX. 3. Ais(i, j)-β-I-i-open if and only if A= (i, j)-β-I-i
Int(A).
4. (i, j)-β-I-i Int((i, j)-β-I-i Int(A)) = (i, j)-β-I-i Int(A).
5. (i, j)-β-I-i Int(φ) = φ and (i, j)-β-I-i Int(X) =
X.
6. (i, j)-β-I-i Int(A)⊆A.
7. If A ⊆ B, then (i, j)-β-I-i Int(A) ⊆ (i, j)-β-I-i Int(B).
8. If A∩B=φ, then (i, j)-β-I-i Int(A)∩(i, j)-β-I-i Int(B) =φ.
9. (i, j)-β-I-i Int(A)∪ (i, j)-β-I-i Int(B) ⊆ (i, j)-β
-I-i Int(A∪B).
10. (i, j)-β-I-i Int(A∩B) = (i, j)-β-I-i Int(A)∩(i, j)
-β-I-i Int(B).
11. (i, j)-β-I-i Int(A)is the largest(i, j)-β-I-i-open set contained inA.
Proof. Obvious.
Definition 3.19 Let A be a subset of a space X. A point x∈X is said to be (i, j)-β-I-i-limit point of A if for each (i, j)-β-I-i-open set U containing x, U∩(A\
{x})̸=φ.
The set of all (i, j)-β-I-i-limit points ofAis called the (i, j)-β-I-i-derived set ofAand is denoted by (i, j)-β-I-i D(A).
Some properties of (i, j)-β-I-i-derived set are stated in the following Theorem:
Theorem 3.20 Let A and B be subsets of a space X. Then we have the following properties:
1. (i, j)-β-I-i D(φ) =φ.
2. Ifx∈(i, j)-β-I-i D(A), thenx∈(i, j)-β-I-i D(A\
{x}).
3. If A ⊆ B, then (i, j)-β-I-i D(A) ⊆ (i, j)-β-I-i D(B).
4. (i, j)-β-I-i D(A)∪ (i, j)-β-I-i D(B) (i, j)-β-I-i D(A∪B).
5. (i, j)-β-I-i D(A∩B)⊆(i, j)-β-I-i D(A)∩(i, j)-β
-I-i D(B).
6. (i, j)-β-I-i D((i, j)-β-I-i D(A))\A ⊆ (i, j)-β-I-i D(A).
7. (i, j)-β-I-i D(A∪(i, j)-β-I-i D(A))⊆A∪(i, j)-β-I
-i D(A).
Proof. We will prove (6) and (7) and the other parts can be proved obviously.
(6) x ∈ (i, j)-β-I-i D((i, j)-β-I-i D(A))\A implies that x ∈ (i, j)-β-I-i D((i, j)-β-I-i D(A)) and U is an (i, j)-β-I-i-open set containingx, thenU ∩((i, j)-β-I-i D(A)\ {x})̸=φ. Lety∈U∩((i, j)-β-I-i D(A)\ {x}). Then, since y ∈ (i, j)-β-I-i D(A) andy ∈ U, U ∩(A\
{y})̸=φ. Letz∈U∩(A\ {y}). Then,z̸=xforz∈A
and x /∈A. Hence, U ∩(A\ {x})̸=φ. Therefore,x∈
(i, j)-β-I-i D(A).
(7) Let x ∈ (i, j)-β-I-i D(A∪(i, j)-β-I-i D(A)). If
x ∈ A, the result is obvious. So, let x ∈ (i, j)-β-I-i D(A∪(i, j)-β-I-i D(A))\A, then, for any (i, j)-β-I-i -open set U containing x, U ∩(A∪(i, j)-β-I-i D(A))\
{x} ̸= φ. Thus, U ∩(A\ {x}) ̸= φ or U ∩((i, j)-β
-I-i D(A)\ {x}) ̸= φ. Now, it follows similarly from (6) that U ∩(A\ {x}) ̸= φ. Hence, x ∈ (i, j)-β-I-i D(A). Therefore, in both cases, we obtain that (i, j)-β
-I-i D(A∪(i, j)-β-I-i D(A))⊆A∪(i, j)-β-I-i D(A).
Definition 3.21 For any subset A in a space X, the
(i, j)-β-I-i-closure ofA, denoted by(i, j)-β-I-i Cl(A), is defined by the intersection of all (i, j)-β-I-i-closed sets containingA.
Theorem 3.22 A subset A of an ideal bitopological space X is (i, j)-β-I-i-closed if and only if it contains the set of its (i, j)-β-I-i-limit points.
Proof. Assume thatA is (i, j)-β-I-i-closed and if pos-sible suppose that x is an (i, j)-β-I-i-limit point of A
which belongs to X\A, thenX \Ais (i, j)-β-I-i-open set containing the (i, j)-β-I-i-limit point xof A, there-foreA∩X\A̸=φ, which is a contradiction.
Conversely, assume that A contains the set of its (i, j
)-β-I-i-limit points. For eachx∈X\A, there exists an (i, j)-β-I-i-open setU containingxsuch thatA∩U =φ, that is, x∈ U ⊆ X\A by Theorem 3.8 ,X \A is an (i, j)-β-I-i-open set and hence,Ais an (i, j)-β-I-i-closed set.
Theorem 3.23 Let A be a set in a space X. A point
x ∈ X is in the (i, j)-β-I-i-closure of A if and only if
A∩U ̸=φ for every (i, j)-β-I-i-open set U containing
x.
Proof. Letx /∈(i, j)-β-I-i Cl(A). Thenx /∈∩F, where
F is (i, j)-β-I-i-closed with A ⊆ F. So x ∈ X \∩F
andX\∩F is an (i, j)-β-I-i-open set containingxand hence, (X\∩F)∩A⊆(X\∩F)∩(∩F) =φ. Conversely, suppose that there exists an (i, j)-β-I-i-open set containing xwith A∩U =φ, thenA⊆X\U and
X \U is an (i, j)-β-I-i-closed withx /∈ X\U. Hence,
x /∈(i, j)-β-I-i Cl(A).
Theorem 3.24 Let A be any subset of a space X and
x is a point of X. If A∩F ̸= φ for every (i, j)-β-I -closed set F of X containing x, then the point x is in the (i, j)-β-I-i-closure ofA.
Proof. Suppose thatU is any (i, j)-β-I-i-open set con-tainingx, then by Definition 3.1 , there exists an (i, j
)-β-I-closed setF such thatx∈F⊆U. So by hypothesis
A∩F ̸=φwhich implies thatA∩U ̸=φfor every (i, j
Here, some properties of (i, j)-β-I-i-closure of the sets are introduced.
Theorem 3.25 For subsetsAandB of a spaceX, the following statements are true:
1. The (i, j)-β-I-i-closure of A is the intersection of all(i, j)-β-I-i-closed sets containingA.
2. A⊆(i, j)-β-I-i Cl(A).
3. (i, j)-β-I-i Cl(A)is(i, j)-β-I-i-closed set inX. 4. A is (i, j)-β-I-i-closed set if and only if A=(i, j)
-β-I-i Cl(A).
5. (i, j)-β-I-i Cl((i, j)-β-I-i Cl(A)) =(i, j)-β-I-i Cl(A).
6. (i, j)-β-I-i Cl(φ) =φ and(i, j)-β-I-i Cl(X) =X. 7. If A ⊆ B, then (i, j)-β-I-i Cl(A) ⊆ (i, j)-β-I-i
Cl(B).
8. If (i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(B) = φ, then
A∩B =φ.
9. (i, j)-β-I-i Cl(A)∪(i, j)-β-I-i Cl(B) =(i, j)-β-I-i Cl(A∪B).
10. (i, j)-β-I-i Cl(A∩B)⊆(i, j)-β-I-i Cl(A)∩(i, j)-β
-I-i Cl(B).
11. (i, j)-β-I-i Cl(A) is the smallest(i, j)-β-I-i-closed set containingA.
Proof. Obvious.
Theorem 3.26 For any subsetAof an ideal bitopolog-ical spaceX, the following statements are true:
1. X\(i, j)-β-I-i Cl(A) = (i, j)-β-I-i Int(X\A). 2. X\ (i, j)-β-I-i Int(A) =(i, j)-β-I-i Cl(X\A). 3. (i, j)-β-I-i Cl(A) =X\ (i, j)-β-I-i Int(X\A). 4. (i, j)-β-I-i Int(A) =X\(i, j)-β-I-i Cl(X\A).
Proof. We prove (1) and the other parts can be proved similarly.
For any point x∈X,x∈ X\(i, j)-β-I-i Cl(A) implies thatx /∈(i, j)-β-I-i Cl(A), then there existsG∈(i, j)-β
-I-i O(X) containingx,A∩G=φ, thenx∈G⊆X\A. Thus,x∈(i, j)-β-I-i Int(X\A).
Conversely, the converse can be obtained by reversing the above steps.
Here, the (i, j)-β-I-i-boundary of a subsetAis defined in a spaceX and study its properties.
Definition 3.27 Let A be a subset of a spaceX, then the (i, j)-β-I-i-boundary of A is defined as (i, j)-β-I-i Cl(A)\ (i, j)-β-I-i Int(A)and is denoted by(i, j)-β-I-i Bd(A).
Theorem 3.28 Let (X, τ1, τ2, I) be an ideal
bitopologi-cal space and letA,B be subsets of X. Then
1. (i, j)-β-I-i Bd(A) =(i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(X\A).
2. (i, j)-β-I-i Bd(A)is an(i, j)-β-I-i-closed set. 3. (i, j)-β-I-i Int(A)∩(i, j)-β-I-i Bd(A) =φ.
4. (i, j)-β-I-i Cl(A) = (i, j)-β-I-i Int(A)∪(i, j)-β-I-i Bd(A).
5. (i, j)-β-I-i Bd( (i, j)-β-I-i Int(A)) ⊆ (i, j)-β-I-i Bd(A).
6. (i, j)-β-I-i Bd((i, j)-β-I-i Cl(A)) ⊆ (i, j)-β-I-i Bd(A).
7. (i, j)-β-I-i Bd((i, j)-β-I-i Bd(A)) ⊆ (i, j)-β-I-i Bd(A).
8. (i, j)-β-I-i Bd(A) =(i, j)-β-I-i Bd(X\A). 9. (i, j)-β-I-i Int(A) =A\(i, j)-β-I-i Bd(A).
10. X = (i, j)-β-I-i Int(A)∪ (i, j)-β-I-i Int(X \ A)∪(i, j)-β-I-i Bd(A).
11. X\(i, j)-β-I-i Bd(A) = (i, j)-β-I-i Int(A)∪ (i, j)
-β-I-i Int(X\A).
12. (i, j)-β-I-i Bd(A∪B)⊆ (i, j)-β-I-i Bd(A)∪(i, j)
-β-I-i Bd(B).
13. (i, j)-β-I-i Bd(A∩B)⊆ (i, j)-β-I-i Bd(A)∪(i, j)
-β-I-i Bd(B).
Proof. we have:
1. (i, j)-β-I-i Bd(A) =(i, j)-β-I-i Cl(A)\ (i, j)-β-I-i Int(A) =(i, j)-β-I-i Cl(A)∩X\(i, j)-β-I-i Int(A) = (i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(X\A).
2. (i, j)-β-I-i Cl((i, j)-β-I-i Bd(A)) =(i, j)-β-I-i Cl((i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(X \ A)) ⊆ (i, j)-β-I-i Cl((i, j)-β-I-i Cl(A))∩(i, j)-β-I-i Cl((i, j)-β-I-i Cl(X \ A)) =(i, j)-β-I-i Bd(A). Therefore, (i, j)-β-I-i Bd(A) is (i, j)-β-I-i-closed. 3. (i, j)-β-I-i Int(A)∩(i, j)-β-I-i Bd(A) = (i, j)-β-I-i
Int(A)∩(i, j)-β-I-i Cl(A)\(i, j)-β-I-i Int(A) =φ. 4. (i, j)-β-I-i Int(A)∪(i, j)-β-I-i Bd(A) = (i, j
)-β-I-i Int(A)∪(i, j)-β-I-i Cl(A)\ (i, j)-β-I-i Int(A) =(i, j)-β-I-i Cl(A).
5. (i, j)-β-I-i Bd( (i, j)-β-I-i Int(A)) =(i, j)-β-I-i Cl( (i, j)-β-I-i Int(A))\(i, j)-β-I-i Int( (i, j)-β-I-i Int(A))
= (i, j)-β-I-i Cl( (i, j)-β-I-i Int(A))\ (i, j
)-β-I-i Int(A) ⊆ (i, j)-β-I-i Cl(A)\ (i, j)-β-I-i Int(A) =(i, j)-β-I-i Bd(A).
6. (i, j)-β-I-i Bd((i, j)-β-I-i Cl(A)) =(i, j)-β-I-i Cl((i, j)-β-I-i Cl(A))\ (i, j)-β-I-i Int((i, j)-β-I-i Cl(A)) =(i, j)-β-I-i Cl(A)\ (i, j)-β-I-i Int((i, j
)-β-I-i Cl(A)) ⊆ (i, j)-β-I-i Cl(A)\ (i, j)-β-I-i Int(A) =(i, j)-β-I-i Bd(A).
7. (i, j)-β-I-i Bd((i, j)-β-I-i Bd(A)) =(i, j)-β-I-i Cl((i, j)-β-I-i Bd(A))∩(i, j)-β-I-i Cl(X\(i, j)-β-I
-i Bd(A))
8. (i, j)-β-I-i Bd(A) =(i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(X\A) =(i, j)-β-I-i Bd(X\A).
9. A\(i, j)-β-I-i Bd(A) = A \ ((i, j)-β-I-i Cl(A)\ (i, j)-β-I-i Int(A)) = (i, j)-β-I-i Int(A).
10. Follows from (4) and Theorem 3.26(3). 11. Is a direct consequence of (10).
12. (i, j)-β-I-i Bd(A∪B) =(i, j)-β-I-i Cl(A∪B)∩(i, j
)-β-I-i Cl(X\(A∪B))
= ((i, j)-β-I-i Cl(A)∪(i, j)-β-I-i Cl(B))∩(i, j)-β-I
-i Cl((X\A)∩(X\B))
⊆((i, j)-β-I-i Cl(A)∪(i, j)-β-I-i Cl(B))∩((i, j)-β
-I-i Cl(X\A)∩(i, j)-β-I-i Cl(X\B))
={(i, j)-β-I-i Cl(A)∩((i, j)-β-I-i Cl(X\A)∩(i, j
)-β-I-i Cl(X\B)} ∪ {(i, j)-β-I-i Cl(B)∩((i, j)-β-I-i Cl(X\A)∩(i, j)-β-I-i Cl(X\B)}
= {(i, j)-β-I-i Bd(A)∩(i, j)-β-I-i Cl(X \ B)} ∪
{(i, j)-β-I-i Bd(B)∩(i, j)-β-I-i Cl(X\A)}
⊆(i, j)-β-I-i Bd(A)∪(i, j)-β-I-i Bd(B).
13. (i, j)-β-I-i Bd(A∩B) =(i, j)-β-I-i Cl(A∩B)∩(i, j
)-β-I-i Cl(X\(A∩B))
⊆((i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(B))∩((i, j)-β
-I-i Cl(X\(A∩B)))
= ((i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(B))∩((i, j)-β
-I-i Cl((X\A)∪(X\B)))
= ((i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(B))∩((i, j)-β
-I-i Cl(X\A)∪(i, j)-β-I-i Cl(X\B))
= {((i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(B))∩(i, j)-β
-I-i Cl(X \ A)} ∪ {((i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(B))∩(i, j)-β-I-i Cl(X\B)}
={(i, j)-β-I-i Bd(A)∩(i, j)-β-I-i Cl(B)} ∪ {(i, j
)-β-I-i Bd(B)∩(i, j)-β-I-i Cl(A)}
⊆(i, j)-β-I-i Bd(A)∪(i, j)-β-I-i Bd(B).
Next, (i, j)-β-I-i-open and (i, j)-β-I-i-closed sets in terms of (i, j)-β-I-i-boundary are characterized.
Theorem 3.29 For a subset A of a space X, the fol-lowing statements are true:
1. IfAis(i, j)-β-I-i-closed, then(i, j)-β-I-i Bd(A) =
A\ (i, j)-β-I-i Int(A).
2. If A is (i, j)-β-I-i-open, then (i, j)-β-I-i Bd(A) =(i, j)-β-I-i Cl(A)\A.
3. IfA is both(i, j)-β-I-i-open and(i, j)-β-I-i-closed, then(i, j)-β-I-i Bd(A) =φ.
4. If A is (i, j)-β-I-i-closed and (i, j)-β-I-i Int(A) =
φ, then(i, j)-β-I-i Bd(A) =A.
5. A is (i, j)-β-I-i-open if and only if (i, j)-β-I-i Bd(A) =(i, j)-β-I-i D(A).
6. A is (i, j)-β-I-i-open if and only if (i, j)-β-I-i Bd(A)⊆(X\A). That is,A∩(i, j)-β-I-i Bd(A) =
φ.
7. A is (i, j)-β-I-i-closed if and only if (i, j)-β-I-i Bd(A)⊆A.
Proof. Only the proofs of (6) and (7) are given and the other proofs are obvious.
(6) Assume that A ∈ (i, j)-β-I-i O(X), then by (2) (i, j)-β-I-i Bd(A) =(i, j)-β-I-i Cl(A) \ A. Hence,
A∩(i, j)-β-I-i Bd(A) =φ. That is, (i, j)-β-I-i Bd(A)⊆
X\A.
Conversely, if (i, j)-β-I-i Bd(A) ∩ A = φ, then by Theorem 3.28(1) A∩(i, j)-β-I-i Cl(A) ∩(X\ (i, j)-β
-I-i Int(A)) = φ. This implies that A ∩(X\ (i, j
)-β-I-i Int(A)) = φ by Theorem 3.26(2). Therefore,
A ⊆X\(X\ (i, j)-β-I-i Int(A)) = (i, j)-β-I-i Int(A), but (i, j)-β-I-i Int(A)⊆Aalways. It follows thatA= (i, j)-β-I-i Int(A) which implies that A ∈ (i, j)-β-I-i O(X).
(7) Assume that A ∈ (i, j)-β-I-i C(X), then by (1) (i, j)-β-I-i Bd(A) =A\(i, j)-β-I-i Int(A)⊆A.
Conversely, if (i, j)-β-I-i Bd(A) ⊆ A, then (i, j)-β-I
-i Bd(A)∩X \A = φ which implies that (i, j)-β-I-i Cl(A)∩(X\(i, j)-β-I-i Int(A))∩X\A=φ. Therefore, (i, j)-β-I-i Cl(A)∩(i, j)-β-I-i Cl(X \A))∩X \A =φ
which implies that (i, j)-β-I-i Cl(A)∩X\A =φ, and so X \A ⊆ X\(i, j)-β-I-i Cl(A). Thus, (i, j)-β-I-i Cl(A) ⊆A, but A ⊆ (i, j)-β-I-i Cl(A) always. It fol-lows thatA=(i, j)-β-I-i Cl(A) which implies thatA∈
(i, j)-β-I-i C(X).
The proof of others are simple, therefore omitted.
4
(
i, j
)
-
β
-
I
-
i
-Continuous
and
(
i, j
)
-
β
-
I
-
i
-Almost Continuous
Definition 4.1 A function f : (X, τ1, τ2, I) →
(Y, σ1, σ2) is called (i, j)-β-I-i-continuous at a point
x ∈ X if for each i-open set V of Y containing f(x), there exists an (i, j)-β-I-i-open setU ofX containingx
such thatf(U)⊆V. Iff is(i, j)-β-I-i-continuous at ev-ery pointxofX, then it is called(i, j)-β-I-i-continuous, where i̸=j, i, j= 1,2.
Definition 4.2 A function f : (X, τ1, τ2, I) →
(Y, σ1, σ2) is called (i, j)-β-I-i-almost continuous at a
point x ∈ X if for each i-open set V of Y containing
f(x), there exists an (i, j)-β-I-i-open set U of X con-taining x such that f(U) ⊆ τi−Int(τj −ClV). If f
is (i, j)-β-I-i-almost continuous at every pointx of X, then it is called (i, j)-β-I-i-almost continuous.
It is obvious from the definition that (i, j)-β-I-i -continuous implies (i, j)-β-I-i-almost continuous. How-ever, the converse is not true in general as it is shown in the following example.
Example 4.3 Let X = {1,2,3},τ1 = {X,φ,{1}},τ2
= {X,φ,{1}, {1,2}},σ1 = {X,φ,{1,3}}, σ2 = {X,
φ, {1}} and I = {φ,{1}}. Then the identity function
f : (X, τ1, τ2, I)→(X, σ1, σ2)is(i, j)-β-I-i-almost
con-tinuous but not (i, j)-β-I-i-continuous.
The proof of the following corollary follows directly from thier definitions.
Corollary 4.4
2. Every (i, j)-β-I-i-continuous function is (i, j)-β-I -continuous.
Theorem 4.5 Let X and Y be bitopological spaces. A function f : (X, τ1, τ2, I) → (Y, σ1, σ2) is (i, j)-β-I-i
-continuous if and only if the inverse image under f of everyi-open set in Y is an(i, j)-β-I-i-open in X.
Proof. Assume thatf is (i, j)-β-I-i-continuous and let
V be anyi-open set inY. We have to show thatf−1(V) is (i, j)-β-I-i-open inX. If f−1(V) = φ, there is noth-ing to prove. So let f−1(V) ̸= φ and let x ∈ f−1(V)
so that f(x)∈V. By (i, j)-β-I-i-continuity off, there exists an (i, j)-β-I-i-open setU in X containingxsuch that f(U)⊆V, that is x∈U ⊆f−1(V), so f−1(V) is
an (i, j)-β-I-i-open set.
Conversely, letf−1(V) be (i, j)-β-I-i-open inX for
ev-ery i-open set V in Y. To show that f is (i, j)-β-I
-i-continuous at x ∈ X, let V be any i-open set in Y
such thatf(x)∈V so thatx∈f−1(V). By hypothesis
f−1(V) is (i, j)-β-I-i-open in X. If f−1(V) =U, then
U is an (i, j)-β-I-i-open set inX containingxsuch that
f(U) =f(f−1(V))⊆V
Hencef is an (i, j)-β-I-i-continuous function.
Theorem 4.6 For a function f : (X, τ1, τ2, I) →
(Y, σ1, σ2), the following statements are equivalent:
1. f is(i, j)-β-I-i-continuous.
2. f−1(V) is an (i, j)-β-I-i-open set in X, for each
i-open set V inY.
3. f−1(F) is an (i, j)-β-I-i-closed set in X, for each
i-closed setF inY.
4. f((i, j)-β-I-i Cl(A))⊆τi−Cl(f(A)), for each
sub-setA ofX.
5. (i, j)-β-I-i Cl(f−1(B))⊆f−1(τi−Cl(B)), for each
subsetB of Y.
6. f−1(τ
i −Int(B)) ⊆(i, j)-β-I-i Int(f−1(B)), for
each subsetB of Y.
7. τi −Int(f(A)) ⊆ f((i, j)-β-I-i Int(A)), for each
subsetA ofX.
Proof. (1)⇒(2). Directly from Theorem 4.5.
(2)⇒(3). LetF be anyi-closed set ofY. ThenY \F
is ani-open set ofY. By (2),f−1(Y \F) =X\f−1(F)
is (i, j)-β-I-i-open set in X and hence f−1(F) is (i, j
)-β-I-i-closed set inX.
(3) ⇒ (4). Let A be any subset of X. Then f(A) ⊆
τi−Cl(f(A)) andτi−Cl(f(A)) isi-closed inY. Hence
A ⊆ f−1(τ
i −Cl(f(A))). By (3), we have f−1(τi −
Cl(f(A))) is an (i, j)-β-I-i-closed set in X. Therefore, (i, j)-β-I-i Cl(A)⊆f−1(τi−Cl(f(A))). Hencef((i, j
)-β-I-i Cl(A))⊆τi−Cl(f(A)).
(4) ⇒ (5). Let B be any subset of Y. Then
f−1(B) is a subset of X. By (4), we have f((i, j)-β
-I-i Cl(f−1(B))) ⊆ τ
i −Cl(f(f−1(B))) = τi−Cl(B). Hence (i, j)-β-I-i Cl(f−1(B))⊆f−1(τ
i−Cl(B)). (5) ⇒ (6). Let B be any subset of Y. Then apply (5) to Y \B is obtained (i, j)-β-I-i Cl(f−1(Y \B))⊆
f−1(τ
i −Cl(Y \B)) ⇔(i, j)-β-I-i Cl(X \f−1(B)) ⊆
f−1(Y \τ
i−Int(B))⇔ X\(i, j)-β-I-i Int(f−1(B))⊆
X\f−1(τ
i−Int(B))⇔f−1(τi−Int(B))⊆(i, j)-β-I-i
Int(f−1(B)). Therefore, f−1(τi−Int(B))⊆(i, j)-β-I-i
Int(f−1(B)).
(6) ⇒(7). Let A be any subset ofX. Then f(A) is a subset ofY. By (6), we havef−1(τi−Int(f(A)))⊆(i, j
)-β-I-i Int(f−1(f(A))) =(i, j)-β-I-i Int(A). Therefore,
τi−Int(f(A))⊆f((i, j)-β-I-i Int(A)).
(7)⇒(1). Let x∈X and letV be anyi-open set ofY
containingf(x). Thenx∈f−1(V) andf−1(V) is a
sub-set ofX. By (7), we haveτi−Int(f(f−1(V)))⊆f((i, j
)-β-I-i Int(f−1(V))). Then τ
i−Int(V) ⊆ f((i, j)-β-I-i
Int( f−1(V))). Since V is an i-open set. Then V ⊆
f((i, j)-β-I-i Int(f−1(V))) implies thatf−1(V)⊆(i, j
)-β-I-i Int(f−1(V)). Therefore,f−1(V) is an (i, j)-β-I-i
-open set inXwhich containsxand clearlyf(f−1(V))⊆
V. Hencef is (i, j)-β-I-i-continuous.
Theorem 4.7 For a function f : (X, τ1, τ2, I) →
(Y, σ1, σ2), the following statements are equivalent:
1. f is(i, j)-β-I-i-almost continuous.
2. For each x∈X and each ij-regular open set V of
Y containingf(x), there exists an(i, j)-β-I-i-open
U in X containingxsuch that f(U)⊆V.
3. For each x ∈ X and each ij-δ-open set V of Y
containingf(x), there exists an (i, j)-β-I-i-openU
inX containingxsuch that f(U)⊆V.
Proof. (1) ⇒ (2). Let x ∈ X and let V be any ij -regular open set of Y containing f(x). By (1), there exists an (i, j)-β-I-i-open setU ofX containingxsuch thatf(U)⊆τi−Int(τj−Cl(V)). SinceV isij-regular open, thenτi−Int(τj−Cl(V)) =V. Therefore,f(U)⊆
V.
(2) ⇒ (3). Let x∈ X and letV be any ij-δ-open set of Y containing f(x). Then for each f(x) ∈ V, there exists an i-open set G containing f(x) such that G ⊆ τi−Int(τj−Cl(G))⊆V. Sinceτi−Int(τj−Cl(G)) is
ij-regular open set of Y containingf(x). By (2), there exists an (i, j)-β-I-i-open setU inX containingxsuch that f(U)⊆τi−Int(τj−Cl(G))⊆V. This completes the proof.
(3) ⇒ (1). Let x ∈ X and let V be any i-open set of Y containing f(x). Then τi−Int(τj−Cl(V) is ij
-δ-open set of Y containing f(x). By (3), there exists an (i, j)-β-I-i-open setU in X containing x such that
f(U)⊆τi−Int(τj−Cl(V)). Therefore,f is (i, j)-β-I
-i-almost continuous.
Theorem 4.8 For a function f : (X, τ1, τ2, I) →
(Y, σ1, σ2), the following statements are equivalent:
1. f is(i, j)-β-I-i-almost continuous.
2. f−1(τi−Int(τj−Cl(V)))is an (i, j)-β-I-i-open set
inX, for eachi-open setV inY.
3. f−1(τ
i −Cl(τj −Int(F))) is an (i, j)-β-I-i-closed
set in X, for each i-closed setF inY.
4. f−1(F) is an (i, j)-β-I-i-closed set in X, for each
5. f− (V) is an (i, j)-β-I-i-open set in X, for each
ij-regular open set V ofY.
6. f−1(G) is an (i, j)-β-I-i-open set in X, for each
ij-δ-open setG ofY.
Proof. (1) ⇒(2). Let V be any i-open set in Y. We have to show thatf−1(τi−Int(τj−Cl(V))) is (i, j)-β-I
-i-open set inX. Letx∈f−1(τi−Int(τj−Cl(V))). Then
f(x) ∈ τi−Int(τj−Cl(V)) and τi−Int(τj−Cl(V)) is an ij-regular open set in Y. Since f is (i, j)-β-I-i -almost continuous, then by Theorem 4.7, there exists an (i, j)-β-I-i-open set U of X containing xsuch that
f(U)⊆τi−Int(τj−Cl(V)). Which implies that x∈
U ⊆f−1(τ
i−Int(τj−Cl(V))). Therefore,f−1(τi−Int (τj−Cl(V))) is an (i, j)-β-I-i-open set inX.
(2)⇒(3). LetF be anyi-closed set ofY. ThenY\F is ani-open set ofY. By (2),f−1(τ
i−Int(τj−Cl(Y\F))) is an (i, j)-β-I-i-open set in X and f−1(τ
i−Int(τj −
Cl(Y\F))) =f−1(τ
i−Int(Y\τj−Int(F))) =f−1(Y\
τi−Cl(τj−Int(F))) =X\f−1(τi−Cl(τj−Int(F))) is an (i, j)-β-I-i-open set inX and hencef−1(τi−Cl(τj−
Int(F))) is (i, j)-β-I-i-closed set inX.
(3) ⇒ (4). Let F be any ij-regular closed set of Y. ThenF is ani-closed set ofY. By (3),f−1(τ
i−Cl(τj−
Int(F))) is (i, j)-β-I-i-closed set in X. Since F is ij -regular closed set, then f−1(τ
i −Cl(τj −Int(F))) =
f−1(F). Therefore, f−1(F) is an (i, j)-β-I-i-closed set
inX.
(4)⇒(5). LetV be anyij-regular open set ofY. Then
Y\V is anij-regular closed set ofY and by (4), we have
f−1(Y \V) =X\f−1(V) is an (i, j)-β-I-i-closed set in
X and hencef−1(V) is an (i, j)-β-I-i-open set inX.
(5)⇒(6). LetGbe anyij-δ-open set inY,G=∪{Vα:
α ∈ ∆} where Vα is ij-regular open. Then f−1(G) =
∪
{f−1(Vα)}, from (5) we havef−1(Vα) is an (i, j)-β-I
-i-open set, then f−1(G) =∪{f−1(Vα)} is an (i, j)-β-I
-i-open.
(6) ⇒ (1). Let x∈ X and let V be any ij-δ-open set of Y containing f(x). Then x ∈ f−1(V). By (6), we
havef−1(V) is an (i, j)-β-I-i-open set in X. Therefore,
we obtainf(f−1(V))⊆V. Hence by Theorem 4.7, f is
(i, j)-β-I-i-almost continuous.
Theorem 4.9 For a function f : (X, τ1, τ2, I) →
(Y, σ1, σ2), the following statements are equivalent:
1. f is(i, j)-β-I-i-almost continuous.
2. f((i, j)-β-I-i Cl(A))⊆ij−Clδ(f(A)), for each sub-setA ofX.
3. (i, j)-β-I-i Cl(f−1(B)) ⊆ f−1(ij −Clδ(B)), for
each subsetB of Y.
4. f−1(F) is (i, j)-β-I-i-closed set inX, for eachij
-δ-closed setF of Y.
5. f−1(V)is(i, j)-β-I-i-open set inX, for each ij-δ -open setV ofY.
6. f−1(ijInt
δ(B))⊆(i, j)-β-I-i Int(f−1(B)), for each
subsetB ofY.
7. ijIntδ(f(A))⊆f((i, j)-β-I-i Int(A)), for each
sub-setA ofX.
Proof. (1) ⇒ (2). Let A be a subset of X. Since ij −Clδ(f(A)) is an ij-δ-closed set in Y, then
Y \ ij − Clδ(f(A)) is ij-δ-open, from Theorem 4.8,
f−1(Y \ij−Clδ(f(A))) is (i, j)-β-I-i-open, which im-plies that X \f−1(ij −Clδ(f(A))) is also (i, j)-β-I-i -open, so f−1(ij −Clδ(f(A))) is (i, j)-β-I-i-closed set in X. Since A ⊆ f−1(ij −Clδ(f(A))), so (i, j)-β-I-i
Cl(A)⊆f−1(ij−Cl
δ(f(A))). Therefore, f((i, j)-β-I-i
Cl(A))⊆ij−Clδ(f(A)).
(2) ⇒ (3). Let B be a subset of Y. We have
f−1(B) is a subset of X. By (2), we have f((i, j)-β
-I-i Cl(f−1(B)))⊆ij−Cl
δ(f(f−1(B))) =ij−Clδ(B). Hence (i, j)-β-I-i Cl(f−1(B))⊆f−1(ij−Cl
δ(B)). (3) ⇒ (4). Let F be any ij-δ-closed set of Y. By (3), we have (i, j)-β-I-i Cl(f−1(F))⊆f−1(ij−Cl
δ(F)) =
f−1(F) and hencef−1(F) is an (i, j)-β-I-i-closed set in
X.
(4) ⇒ (5). Let V be any ij-δ-open set of Y. Then
Y \V is an ij-δ-closed set of Y and by (4), we have
f−1(Y \V) =X\f−1(V) is an (i, j)-β-I-i-closed set in
X. Hencef−1(V) is an (i, j)-β-I-i-open set inX. (5) ⇒ (6). For each subset B of Y. We have
ijIntδ(B) ⊆ B. Then f−1(ijIntδ(B))⊆ f−1(B). By (5), f−1(ijInt
δ(B)) is an (i, j)-β-I-i-open set in X. Thenf−1(ijInt
δ(B))⊆(i, j)-β-I-i Int(f−1(B)). (6) ⇒ (7). Let A be any subset of X. Then
f(A) is a subset of Y. By (6), we obtain that
f−1(ijInt
δ(f(A)))⊆(i, j)-β-I-i Int(f−1(f(A))). Hence
f−1(ijInt
δ(f(A))) ⊆(i, j)-β-I-i Int(A), which implies that ijIntδ(f(A))⊆f((i, j)-β-I-i Int(A)).
(7)⇒(1). Letx∈X andV be anyij-reguler open set of Y containingf(x). Thenx∈f−1(V) andf−1(V) is
a subset of X. By (7), we get ijIntδ(f(f−1(V))) ⊆
f((i, j)-β-I-i Int(f−1(V))) implies that ijIntδ(V) ⊆
f((i, j)-β-I-i Int(f−1(V))). Since V is ij-reguler open set and hence ij-δ-open set, then V ⊆ f((i, j)-β-I
-i Int(f−1(V))) this implies that f−1(V) ⊆(i, j)-β-I-i Int(f−1(V)). Therefore, f−1(V) is an (i, j)-β-I-i-open
set inX which containsxand clearlyf(f−1(V))⊆V.
Hence, by Theorem 4.7, f is (i, j)-β-I-i-almost continu-ous.
Theorem 4.10 For a function f : (X, τ1, τ2, I) →
(Y, σ1, σ2), the following statements are equivalent:
1. f is(i, j)-β-I-i-almost continuous. 2. (i, j)-β-I-i Cl(f−1(V))⊆f−1(τ
i−Cl(V)), for each
ji-β-open set V of Y.
3. f−1(τi −Int(F)) ⊆(i, j)-β-I-i Int(f−1(F)), for
each ji-β-closed setF ofY.
4. f−1(τi −Int(F)) ⊆(i, j)-β-I-i Int(f−1(F)), for
each ji-semi closed setF of Y.
5. (i, j)-β-I-i Cl(f−1(V))⊆f−1(τi−Cl(V)), for each
ji-semi open setV ofY.
Proof. (1) ⇒ (2). Let V be any ji-β-open set of Y. SinceV ⊆τi−Cl(τj−Int(τi−Cl(V))) thenτi−Cl(V)⊆
τi−Cl(τj−Int(τi−Cl(V))). Alsoτj−Int(τi−Cl(V))⊆
τi−Cl(V) then τi−Cl(τj −Int(τi−Cl(V))) ⊆ τi −
Cl(V). Then we haveτi−Cl(V) =τi−Cl(τj−Int(τi−
