# (τ1, τ2)α-Continuity for Multifunctions

Authors: Chokchai Viriyapong; Chawalit Boonpok
Journal: Journal of Mathematics
 (2020)

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## Abstract

The main goal of this article is to introduce the concepts of upper and lower (τ1, τ2)α-continuous multifunctions. Characterizations of upper and lower (τ1, τ2)α-continuous multifunctions are investigated. The relationships between upper and lower (τ1, τ2)α-continuous multifunctions and the other types of continuity are discussed.


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## Body

## 1. Introduction

General topology is important in many fields of applied sciences as well as branches of mathematics. Continuity is a basic concept for the study in topological spaces. Generalization of this concept by using weaker forms of open sets such asα-sets [1], semi-open sets [2], preopen sets [3], and semi-preopen sets [4] is one of the main research topics of general topology. In 1965, Njåstad [1] introduced a weak form of open sets called α-sets. Mashhour et al. [5] defined a function to be α-continuous if the inverse image of each open set is an α-set. Noiri [6] called α-continuous functions strong semi-continuous and investigated α-continuous functions in [7] and investigated α-continuous functions. In [8, 9], the authors investigated a class of functions called almost α-continuous or almost feebly continuous. In 1987, Noiri [10] introduced a class of functions called weakly α-continuous functions. Some characterizations of weakly α-continuous functions are investigated in [11–13]. Neubrunn [14] extended these functions to multifunction and introduced the notions of upper and lower α-continuous multifunctions. Popa and Noiri [15] investigated several characterizations of upper and lower α-continuous multifunctions and some basic properties of such multifunctions. In [16], the present authors introduced upper and lower almost α-continuous multifunctions and obtained some characterizations of upper and lower almost α-continuous multifunctions and several properties of such multifunctions. Recently, Popa and Noiri [17] investigated some characterizations and several properties concerning upper and lower weakly α-continuous multifunctions. In this paper, we introduce the notions of upper and lower (τ1, τ2)α-continuous multifunctions and investigate some characterizations of upper and lower (τ1, τ2)α-continuous multifunctions. Section 4 is devoted to introducing and studying upper and lower almost (τ1, τ2)α-continuous multifunctions. In Section 5, several interesting characterizations of upper and lower weakly (τ1, τ2)α-continuous multifunctions are discussed.


## 2. Preliminaries

Throughout the present paper, spaces (X, τ1, τ2) and (Y, σ1, σ2) (or simply X and Y) always mean bitopological spaces on which no separation axioms are assumed unless explicitly stated. Let A be a subset of a bitopological space (X, τ1, τ2). The closure of A and the interior of A with respect to τi are denoted by τi‐Cl(A) and τi‐Int(A), respectively, for i = 1, 2. A subset A of a bitopological space (X, τ1, τ2) is called τ1τ2-semi-open (resp. τ1τ2-regular open [18] and τ1τ2-regular closed [19]) if A ⊆ τ1‐Cl(τ2‐Int(A)) (resp. A = τ1‐Int(τ2‐Cl(A)) and A = τ1‐Cl(τ2‐Int(A))). The complement of a τ1τ2-semi-open set is said to be τ1τ2-semi-closed. The τ1τ2-semi-closure of A is defined by the intersection of τ1τ2-semi-closed sets containing A and is denoted by τ1τ2‐sCl(A). The τ1τ2-semi-interior of A is defined by the union of τ1τ2-semi-open sets contained in A and is denoted by τ1τ2‐sInt(A). A subset A of a bitopological space (X, τ1, τ2) is said to be τ1τ2-closed [20] if A = τ1‐Cl(τ2‐Cl(A)). The complement of a τ1τ2-closed set is said to be τ1τ2-open. The intersection of all τ1τ2-closed sets containing A is called the τ1τ2-closure of A and is denoted by τ1τ2‐Cl(A). The union of all τ1τ2-open sets contained in A is called the τ1τ2-interior of A and is denoted by τ1τ2‐Int(A).By a multifunctionF: X ⟶ Y, we mean a point-to-set correspondence from X into Y, and we always assume that F(x) ≠ ∅ for all x ∈ X. For a multifunction F: X ⟶ Y, following [21], we shall denote the upper and lower inverse of a set B of Y by F +(B) and F−(B), respectively, that is, F+(B) = {x ∈ X∣F(x) ⊆ B} and F−(B) = {x ∈ X ∣ F(x) ∩ B ≠ ∅}. In particular, F−(y) = {x ∈ X∣y ∈ F(x)} for each point y ∈ Y. For each A ⊆ X, F(A) = ∪x∈AF(x). Then, F is said to be surjection if F(X) = Y, or equivalent, if for each y ∈ Y there exists x ∈ X such that y ∈ F(x) and F is called injection if x ≠ y implies F(x) ∩ F(y) = ∅.Definition 1.
(see [22]). A multifunction F: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is said to be:(1)
upper (τ1, τ2)-continuous at a point x ∈ X if for each σ1σ2-open subset V of Y such that F(x) ⊆ V, there exists a τ1τ2-open subset U of X containing x such that F(U) ⊆ V;(2)
lower (τ1, τ2)-continuous at a point x ∈ X if for each σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a τ1τ2-open subset U of X containing x such that F(z) ∩ V ≠ ∅ for every z ∈ U;(3)
upper (resp. lower) (τ1, τ2)-continuous if F has this property at each point of X.Definition 2.
A subsetA of a bitopological space (X, τ1, τ2) is said to be:(i)
(τ1, τ2)α-open if A ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(A)));(ii)
(τ1, τ2)α-closed if its complement is (τ1, τ2)α-open.Definition 3.
LetA be a subset of a bitopological (X, τ1, τ2). Then,(i)
The intersection of all (τ1, τ2)α-closed sets containing A is called the (τ1, τ2)α‐closure of A and is denoted by (τ1, τ2)α- Cl(A).(ii)
The union of all (τ1, τ2)α‐open sets contained in A is called the (τ1, τ2)α –interior of A and is denoted by (τ1, τ2)α- Int(A).Lemma 1.
(see [20]). For a subset A of a bitopological space (X, τ1, τ2), the following properties hold:(1)
τ1τ2‐sCl(A) = τ1‐Int(τ2‐Cl(A)) ∪ A(2)
IfA is τ1τ2-open, then τ1τ2‐sCl(A) = τ1‐Int(τ2‐Cl(A))Lemma 2.
For a subsetA of a bitopological space (X, τ1, τ2), the following properties are equivalent:(1)
A is (τ1, τ2)α-open(2)
U⊆A⊆τ1‐Int(τ2‐Cl(U)) for some τ1τ2-open set U(3)
U⊆A⊆τ1τ2‐sCl(U) for some τ1τ2-open set U(4)
A⊆τ1τ2‐sCl(τ1τ2‐Int(A))Proof.
(1) ⟹ (2): Suppose thatA is a (τ1, τ2)α-open set. Then, we have
(1)
A
⊆
τ
1
‐
Int
τ
2
‐
Cl
τ
1
τ
2
‐
Int
A
.
PutU = τ1τ2‐Int(A); then, U⊆A⊆τ1‐Int(τ2‐Cl(U)).
(2) ⟹ (3): This follows from Lemma1 (2).
(3) ⟹ (4): Suppose thatU⊆A⊆τ1τ2‐sCl(U) for some τ1τ2-open set U. Then, we have U⊆τ1τ2‐Int(A), and hence A⊆τ1τ2‐sCl(τ1τ2‐Int(A)).
(4) ⟹ (1): Suppose thatA⊆τ1τ2‐sCl(τ1τ2‐Int(A)). Since τ1τ2‐Int(A) is τ1τ2-open and by Lemma 1 (2), A⊆τ1‐Int(τ2‐Cl(τ1τ2‐Int(A))). Thus, A is (τ1, τ2)α-open.Lemma 3.
For a subsetA of a bitopological space (X, τ1, τ2), the following properties hold:(1)
A is (τ2, τ2)α-closed if and only if τ1τ2‐sInt(τ1τ2‐Cl(A)) ⊆A(2)
τ1τ2‐sInt(τ1τ2‐Cl(A)) = τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A)))(3)
α(τ1, τ2)‐Cl(A) = A ∪ τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A)))Proof.
(1)
LetA be a (τ1, τ2)α-closed set. Then, X – A is (τ1, τ2)α–open and by Lemma 2, X – A⊆τ1τ2‐sCl(τ1τ2‐Int(X – A)) = X − τ1τ2‐sInt(τ1τ2‐Cl(A)). Thus, τ1τ2‐sInt(τ1τ2‐Cl(A)) ⊆ A.
Conversely, suppose thatτ1τ2‐sInt(τ1τ2‐Cl(A)) ⊆A. Then,
(2)
X
–
A
⊆
X
−
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
A
=
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
X
–
A
.
By Lemma2, we have X − A is (τ1, τ2)α-open, and hence A is (τ1, τ2)α-closed.
(2)
SinceX – τ1τ2‐sInt(τ1τ2‐Cl(A)) = τ1τ2‐sCl(τ1τ2‐Int(X – A)), by Lemma 1 (2),
(3)
X
–
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
A
=
τ
1
‐
Int
τ
2
‐
Cl
τ
1
τ
2
‐
Int
X
–
A
=
X
−
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
A
,
and henceτ1τ2‐sInt(τ1τ2‐Cl(A)) = τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A))).
(3)
We observe that
(4)
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
A
∪
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
A
⊆
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
A
∪
τ
1
τ
2
‐
Cl
A
⊆
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
A
⊆
A
∪
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
A
.
Thus,A ∪ τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A))) is (τ1, τ2)α-closed, and hence (τ1, τ2)α‐Cl(A) ⊆A ∪ τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A))). On the other hand, since (τ1, τ2)α‐Cl(A) is (τ1, τ2)α-closed, we have(5)τ1‐Clτ2‐Intτ1τ2‐ClA⊆τ1‐Clτ2‐Intτ1τ2‐Clτ1,τ2α‐ClA⊆τ1,τ2α‐ClA.
Therefore,A ∪ τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A))) ⊆(τ1, τ2)α‐Cl(A). Consequently, we obtain (τ1, τ2)α‐Cl(A) = A ∪ τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A))).


## 3. Characterizations of Upper and Lowerα(τ1, τ2)-Continuous Multifunctions

In this section, we introduce the notions of upper and lower (τ1, τ2)α-continuous multifunctions. Moreover, several characterizations of upper and lower (τ1, τ2)α-continuous multifunctions are discussed.Definition 4.
A multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is said to be:(1)
upper (τ1, τ2)α-continuous at a point x ∈ X if for each σ1σ2-open subset V of Y such that F(x) ⊆V, there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆V;(2)
lower (τ1, τ2)α-continuous at a point x ∈ X if for each σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a (τ1, τ2)α-open subset U of X containing x such that F(z) ∩ V ≠ ∅ for every z ∈ U;(3)
upper (resp. lower) (τ1, τ2)α-continuous if F has this property at each point of X.Remark 1.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following implication holds:
upper (τ1, τ2)-continuity ⟹ upper (τ1, τ2)α-continuity.
The converse of the implication is not true in general. We give an example for the implication as follows.Example 1.
LetX = {a, b, c, d} with topologies τ1 = {∅, {a}, {b}, {a, b}, X} and τ2 = {∅, {b}, X}. Let Y = {−2, −1, 0, 1, 2} with topologies(6)σ1=∅,−1,−2,2,0,1,Y,and σ2 = {∅, {−1, −2, 2}, Y}. A multifunction F: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is defined as follows:(7)Fx=−1,−2,ifx=a,−1,ifx=b,0,1,ifx=c,−1,0,ifx=d.
Then,F is upper (τ1, τ2)α-continuous but F is not upper (τ1, τ2)-continuous.Theorem 1.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is upper (τ1, τ2)α-continuous at a point x of X(2)
x ∈ τ1τ2‐sCl(τ1τ2‐Int(F+(V))) for each σ1σ2-open subset V of Y containing F(x)(3)
For eachτ1τ2-semiopen subset U of X containing x and each σ1σ2-open subset V of Y containing F(x), there exists a nonempty τ1τ2-open subset G of X such that G ⊆ U and F(G) ⊆ VProof.
(1) ⟹ (2): LetV be any σ1σ2-open subset of Y such that F(x) ⊆V. Then, there exists a (τ1, τ2)α-open set U containing x such that F(U) ⊆V; hence, x ∈ U⊆F+(V). Since U is (τ1, τ2)α-open, by Lemma 2, we have x ∈ U ⊆ τ1τ2‐sCl(τ1τ2‐Int(U)) ⊆ τ1τ2‐sCl(τ1τ2‐IntF+(V))).
(2) ⟹ (3): LetV be any σ1σ2-open subset of Y such that F(x) ⊆ V. Then, we have x ∈ τ1τ2‐sCl (τ1τ2‐ Int(F+(V))). Let U be any τ1τ2-semi-open set containing x. Thus, U ∩ [τ1τ2‐Int(F+(V))] ≠ ∅ and U ∩ [τ1τ2‐Int(F+(V))] is τ1τ2-semi-open. Put G = τ1τ2‐ Int(U ∩ [τ1τ2‐ Int(F+(V))]); then, G is a nonempty τ1τ2-open subset of X such that G ⊆ U and F(G) ⊆ V.
(3) ⟹ (1): Letτ1τ2‐SO(X, x) be the family of all τ1τ2-semi-open subsets of X containing x. Let V be any σ1σ2-open subset of Y containing F(x). For each U ∈ τ1τ2‐SO(X, x), there exists a nonempty τ1τ2-open set HU such that HU ⊆ U and F(HU) ⊆ V. Let W = ∪{HU∣U ∈ τ1τ2‐SO(X, x)}. Then, we have W which is τ1τ2-open in X, x ∈ τ1τ2‐sCl(W), and F(W) ⊆ V. Put G = W  ∪ {x}. By Lemma 1, we have
(8)
W
⊆
G
⊆
τ
1
τ
2
‐
sCl
G
=
τ
1
‐
Int
τ
2
‐
Cl
G
∪
G
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
x
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
τ
2
‐
Cl
x
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
=
τ
1
τ
2
‐
sCl
W
,and by Lemma 2, G is a (τ1, τ2)α-open subset of X containing x such that F(G) ⊆ V. This shows that F is upper (τ1, τ2)α-continuous at x.Theorem 2.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is lower (τ1, τ2)α-continuous at a point x of X(2)
x ∈ τ1τ2‐sCl(τ1τ2‐Int(F‐(V))) for each σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅(3)
For eachτ1τ2-semiopen subset U of X containing x and each σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a nonempty τ1τ2-open subset G of X such that G ⊆ U and F(z) ∩ V ≠ ∅ for every z ∈ GProof.
The proof is similar to that of Theorem1.Definition 5.
A subsetA of a bitopological space (X, τ1, τ2) is said to be a τ1τ2-neighbourhood (resp. (τ1, τ2)α-neighbourhood) of x ∈ X if there exists a τ1τ2-open (resp. (τ1, τ2)α-open) set U such that x ∈ U ⊆ A.Theorem 3.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is upper (τ1, τ2)α-continuous(2)
F+(V) is (τ1, τ2)α-open in X for each σ1σ2-open subset V of Y(3)
F‐(H) is (τ1, τ2)α-closed in X for each σ1σ2-closed subset H of Y(4)
τ1τ2‐sInt(τ1τ2‐Cl(F‐(B))) ⊆ F‐(σ1σ2‐Cl(B)) for each subset B of Y(5)
F+(V) ⊆ τ1τ2‐sCl(τ1τ2‐Int(F+(V))) for each σ1σ2-open subset V of Y(6)
(τ1, τ2)α‐Cl(F‐(B)) ⊆ F‐(σ1σ2‐Cl(B)) for each subset B of Y(7)
For eachx ∈ X and each σ1σ2-neighbourhood V of F(x), F+(V) is a (τ1, τ2)α-neighbourhood of x(8)
For eachx ∈ X and each σ1σ2-neighbourhood V of F(x), there exists a (τ1, τ2)α-neighbourhood U of x such that F(U) ⊆ VProof.
(1) ⟹ (2): LetV be any σ1σ2-open subset of Y and x ∈ F+(V). By Theorem 1, x ∈ τ1τ2‐sCl(τ1τ2‐Int(F+(V))), and hence
(9)
F
+
V
⊆
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
F
+
V
.
It follows from Lemma2 that F+(V) is (τ1, τ2)α-open in X.
(2) ⟹ (3): This follows from the fact thatF+(Y – B) = X – F‐(B) for any subset B of Y.
(3) ⟹ (4): LetB be any subset of Y. Then, σ1σ2‐Cl(B) is σ1σ2-closed in Y and by (3), F‐(σ1σ2‐Cl(B)) is α(τ1, τ2)-closed in X. By Lemma 3 (1),
(10)
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
‐
B
⊆
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
Cl
B
⊆
F
‐
σ
1
σ
2
‐
Cl
B
.
(4) ⟹ (5): LetV be any σ1σ2-open subset of Y. Then, we have Y – V is σ1σ2-closed in Y. By (4),
(11)
X
–
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
F
+
V
=
τ
1
τ
2
‐
sInt
X
–
τ
1
τ
2
‐
Int
F
+
V
=
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
X
–
F
+
V
=
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
–
Y
–
V
⊆
F
–
σ
1
σ
2
‐
Cl
Y
–
V
=
X
–
F
+
V
,
and henceF+(V) ⊆ τ1τ2‐sCl(τ1τ2‐Int(F+(V))).
(5)
⟹ (6): LetB be any subset of Y. Then, Y – σ1σ2‐Cl(B) is σ1σ2-open in Y, and by (5),
(12)
X
–
F
‐
σ
1
σ
2
‐
Cl
B
=
F
+
Y
–
σ
1
σ
2
‐
Cl
B
⊆
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
F
+
Y
–
σ
1
σ
2
‐
Cl
B
=
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
X
–
F
‐
σ
1
σ
2
‐
Cl
B
=
τ
1
τ
2
‐
sCl
X
–
τ
1
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
Cl
B
=
X
–
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
Cl
B
.
Thus,τ1τ2‐sInt(τ1τ2‐Cl(F‐(σ1σ2‐Cl(B)))) ⊆ F‐(σ1σ2‐Cl(B)). By Lemma 3,
(13)
α
τ
1
,
τ
2
‐
Cl
F
‐
B
⊆
α
τ
1
,
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
Cl
B
=
F
‐
σ
1
σ
2
‐
Cl
B
∪
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
Cl
B
=
F
‐
σ
1
σ
2
‐
Cl
B
∪
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
Cl
B
=
F
‐
σ
1
σ
2
‐
Cl
B
.
(6) ⟹ (3): LetH be any σ1σ2-closed subset of Y. By (6), we have
(14)
α
τ
1
,
τ
2
‐
Cl
F
‐
H
⊆
F
‐
σ
1
σ
2
‐
Cl
H
=
F
‐
H
,
and henceF‐(H) is (τ1, τ2)α-closed in X.
(2) ⟹ (7): Letx ∈ X and V be a σ1σ2-neighbourhood of F(x). Then, there exists a σ1σ2-open subset U of Y such that F(x) ⊆ U ⊆ V. Thus, x ∈ F+(U) ⊆ F+(V). By (2), we have F+(U) is (τ1, τ2)α-open in X and hence F+(V) is a (τ1, τ2)α-neighbourhood of x.
(7) ⟹ (8): Letx ∈ X and V be a σ1σ2-neighbourhood of F(x). By (7), F+(V) is a (τ1, τ2)α-neighbourhood of x. Put U = F+(V); then, U is a (τ1, τ2)α-neighbourhood of x and F(U) ⊆ V.
(8) ⟹ (1): Letx ∈ X and V be any σ1σ2-open subset of Y such that F(x) ⊆ V. Then, V is a σ1σ2-neighbourhood of F(x) and by (8), there exists a (τ1, τ2)α-neighbourhood U of x such that F(U) ⊆ V. Since U is (τ1, τ2)α-neighbourhood of x, there exists a (τ1, τ2)α-open set W such that x ∈ W ⊆ U; hence, F(W) ⊆ V. This shows that F is upper (τ1, τ2)α-continuous at x. Consequently, we obtain that F is upper (τ1, τ2)α-continuous.Theorem 4.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is lower (τ1, τ2)α-continuous(2)
F‐(V) is (τ1, τ2)α-open in X for every σ1σ2-open subset V of Y(3)
F+(H) is (τ1, τ2)α-closed in X for every σ1σ2-closed subset H of Y(4)
τ1τ2‐sInt(τ1τ2‐Cl(F+(B))) ⊆ F+(σ1σ2‐Cl(B)) for every subset B of Y(5)
(τ1, τ2)α‐Cl(F+(B)) ⊆ F+(σ1σ2‐Cl(B)) for every subset B of Y(6)
F((τ1, τ2)α‐Cl(A)) ⊆ σ1σ2‐Cl(F(A)) for every subset A of X(7)
F(τ1τ2‐sInt(τ1τ2‐Cl(A))) ⊆ σ1σ2‐Cl(F(A)) for every subset A of X(8)
F(τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A)))) ⊆ σ1σ2‐Cl(F(A)) for every subset A of XProof.
(1) ⟹ (2): LetV be any σ1σ2-open subset of Y and x ∈ F‐(V). By Theorem 2, x ∈ τ1τ2‐sCl(τ1τ2‐Int(F‐(V))), and hence
(15)
F
−
V
⊆
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
F
‐
V
.
It follows from Lemma2 that F‐(V) is (τ1, τ2)α-open in X.
(2) ⟹ (3): This follows from the fact thatF+(Y – B) = X – F‐(B) for any subset B of Y.
(3) ⟹ (4): LetB be any subset of Y. Then, we have σ1σ2‐Cl(B) is σ1σ2-closed in Y and by (3), F+(σ1σ2‐Cl(B)) is α(τ1, τ2)-closed in X. By Lemma 3 (1),
(16)
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
+
B
⊆
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
+
σ
1
σ
2
‐
Cl
B
⊆
F
+
σ
1
σ
2
‐
Cl
B
.
(4) ⟹ (5): LetB be any subset of Y. By (4) and Lemma 3, we have
(17)
τ
1
,
τ
2
α
‐
Cl
F
+
B
⊆
τ
1
,
τ
2
α
‐
Cl
F
+
σ
1
σ
2
‐
Cl
B
=
F
+
σ
1
σ
2
‐
Cl
B
∪
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
F
+
σ
1
σ
2
‐
Cl
B
=
F
+
σ
1
σ
2
‐
Cl
B
.
(5) ⟹ (6): LetA be any subset of X. Since A ⊆ F+(F(A)), we have
(18)
τ
1
,
τ
2
α
‐
Cl
A
⊆
τ
1
,
τ
2
α
‐
Cl
F
+
F
A
⊆
F
+
σ
1
σ
2
‐
Cl
F
A
,
and henceF((τ1, τ2)α‐Cl(A)) ⊆ σ1σ2‐Cl(F(A)).
(6) ⟹ (7): This follows immediately from Lemma3.
(7) ⟹ (8): LetA be any subset of X. By (7) and Lemma 3 (2), we have
(19)
F
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
A
=
F
τ
1
τ
2
‐
sInt
τ
1
τ
2
‐
Cl
A
⊆
σ
1
σ
2
‐
Cl
F
A
.
(8) ⟹ (1): Letx ∈ X and V be any σ1σ2-open subset of Y such that F(x) ∩ V ≠ ∅. Then, we have x ∈ F‐(V). We shall show that F‐(V) is (τ1, τ2)α-open in X. By the hypothesis, F(τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F+(Y – V))))) ⊆ σ1σ2‐Cl(F(F+(Y –V))) ⊆ Y –V. Thus, τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F+(Y –V)))) ⊆ F+(Y –V) = X − F‐(V), and hence F‐(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F‐(V)))). This shows that F‐(V) is α(τ1, τ2)-open in X. Put U = F‐(V); then, U is a (τ1, τ2)α-open set containing x such that F(z) ∩ V ≠ ∅ for every z ∈ U. Consequently, we obtain that F is lower (τ1, τ2)α-continuous.Definition 6.
A functionf: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is said to be (τ1, τ2)α-continuous if for every σ1σ2-open subset V of Y, f−1(V) is (τ1, τ2)α-open in X.Corollary 1.
For a functionf: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
f is (τ1, τ2)α-continuous(2)
f ‐1(F) is (τ1, τ2)α-closed in X for each σ1σ2-closed subset F of Y(3)
τ1τ2‐sInt(τ1τ2‐Cl(f ‐1(B))) ⊆ f ‐1(σ1σ2‐Cl(B)) for each subset B of Y(4)
(τ1, τ2)α‐Cl(f ‐1(B)) ⊆ f ‐1(σ1σ2‐Cl(B)) for each subset B of Y(5)
For eachx ∈ X and each σ1σ2-neighbourhood V of f(x), f ‐1(V) is a (τ1, τ2)α-neighbourhood of x(6)
For eachx ∈ X and each σ1σ2-neighbourhood V of f(x), there exists a (τ1, τ2)α-neighbourhood U of x such that f(U) ⊆ V(7)
f((τ1, τ2)α‐Cl(A)) ⊆ σ1σ2‐Cl(f(A)) for each subset A of X(8)
f(τ1τ2‐sInt(τ1τ2‐Cl(A))) ⊆ σ1σ2‐Cl(f(A)) for each subset A of X(9)
f(τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A)))) ⊆ σ1σ2‐Cl(f(A)) for each subset A of XLemma 4.
(see [20]). For subsets A, B of a bitopological space (X, τ1, τ2), the following properties hold:(1)
A ⊆ τ1τ2‐ker(A)(2)
IfA ⊆ B, then τ1τ2‐ker(A) ⊆ τ1τ2‐ker(B)(3)
IfA is τ1τ2-open, then τ1τ2‐ker(A) = A(4)
x ∈ τ1τ2‐ker(A) if and only if A ∩ H≠∅ for any τ1τ2-closed set H containing xTheorem 5.
LetF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) be a multifunction. If(20)F+σ1σ2‐kerA⊆τ1,τ2α‐IntF+A,for every subset A of X, then F is upper (τ1, τ2)α-continuous.Proof.
LetV be any σ1σ2-open subset of Y. By Lemma 4 (3), we have F+(V) = F+(σ1σ2‐ker(V)) ⊆ (τ1, τ2)α‐Int(F+(V)), and hence (τ1, τ2)α‐Int(F+(V)) = F+(V). Thus, F+(V) is (τ1, τ2)α-open, by Theorem 3, F is upper (τ1, τ2)α-continuous.Theorem 6.
LetF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) be a multifunction. If(21)F−σ1σ2‐kerA⊆τ1,τ2α‐IntF−A,for every subset A of X, then F is lower (τ1, τ2)α-continuous.Proof.
The proof is similar to that of Theorem5.Definition 7.
(see [20]). A collection U of subsets of a bitopological space (X, τ1, τ2) is said to be τ1τ2-locally finite if for each x ∈ X has a τ1τ2-neighbourhood which intersects only finitely many elements of U.Definition 8.
(see [20]). A subset A of a bitopological space (X, τ1, τ2) is said to be(1)
τ1τ2-paracompact if every cover of A by τ1τ2-open sets of X is refined by a cover of A which consists of τ1τ2-open sets of X and is τ1τ2-locally finite in X(2)
τ1τ2-regular if for each x ∈ A and each τ1τ2-open set U of X containing x, there exists a τ1τ2-open set V of X such that x ∈ V ⊆ τ1τ2‐Cl(V) ⊆ ULemma 5.
(see [20]). If A is a τ1τ2-regular τ1τ2-paracompact set of a bitopological space (X, τ1, τ2) and U is a τ1τ2-open neighbourhood of A, then there exists a τ1τ2-open set V of X such that A ⊆ V ⊆ τ1τ2‐Cl(V) ⊆ U.Definition 9.
A multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is called punctually (τ1, τ2)-paracompact (resp. punctually (τ1, τ2)-regular) if for each x ∈ X, F(x) is τ1τ2-paracompact (resp. τ1τ2-regular).
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), by(22)αClF⊛:X,τ1,τ2⟶Y,σ1,σ2,we shall denote a multifunction defined as follows: αClF⊛(x) = (σ1, σ2)α‐Cl(F(x)) for each x ∈ X.Lemma 6.
Let (X, τ1, τ2) be a bitopological space. Then, (τ1, τ2)α‐Cl(A) ⊆ τ1τ2‐Cl(A) for every subset A of X.Lemma 7.
IfF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is punctually (τ1, τ2)-paracompact and punctually (τ1, τ2)-regular, then αClF⊛+V=F+V for every σ1σ2-open subset V of Y.Proof.
LetV be any σ1σ2-open subset V of Y and x∈αClF⊛+V. Then, we have (σ1, σ2)α‐Cl(F(x)) ⊆ V and F(x) ⊆ V. Thus, x ∈ F+(V), and hence αClF⊛+V⊆F+V. On the other hand, let x ∈ F+(V). Then, we have F(x) ⊆ V and by Lemma 5, there exists a σ1σ2-open subset U of Y such that F(x) ⊆ σ1σ2‐Cl(U) ⊆ U ⊆ V. By Lemma 6, (σ1, σ2)α‐Cl(F(x)) ⊆ σ1σ2‐Cl(U) ⊆ V, and hence x∈αClF⊛+V. Thus, F+V⊆αClF⊛+V. Consequently, we obtain αClF⊛+V=F+V.Theorem 7.
LetF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) be punctually (τ1, τ2)-paracompact and punctually (τ1, τ2)-regular. Then, F is upper (τ1, τ2)α-continuous if and only if αClF⊛: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is upper (τ1, τ2)α-continuous.Proof.
Suppose thatF is upper α(τ1, τ2)-continuous. Let x ∈ X and V be any σ1σ2-open subset of Y such that αClF⊛(x) ⊆ V. By Lemma 7, we have x∈αClF⊛+V=F+V. Since F is upper (τ1, τ2)α-continuous, there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ V. Since F(z) is σ1σ2-paracompact and σ1σ2-regular for each z ∈ U, by Lemma 5, there exists a σ1σ2-open set H such that F(z) ⊆ H ⊆ σ1σ2‐Cl(H) ⊆ V. By Lemma 6, we have (σ1, σ2)α‐Cl(F(z)) ⊆ σ1σ2‐Cl(H) ⊆ V for each z ∈ U, and hence αClF⊛(U) ⊆ V. This shows that αClF⊛ is upper (τ1, τ2)α-continuous.
Conversely, suppose thatαClF⊛: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is upper (τ1, τ2)α-continuous. Let x ∈ X and V be any σ1σ2-open subset of Y such that F(x) ⊆ V. By Lemma 7, we have x∈F+V=αClF⊛+V, and hence αClF⊛(x) ⊆ V. Since αClF⊛ is upper (τ1, τ2)α-continuous, there exists a (τ1, τ2)α-open subset U of X containing x such that αClF⊛(U) ⊆ V; hence, F(U) ⊆ V. This shows that F is upper (τ1, τ2)α-continuous.Lemma 8.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), it follows that for each (σ1, σ2)α-open subset V ofYαClF⊛−V=F−V.Proof.
Suppose thatV is any (σ1, σ2)α-open subset of Y. Let x∈αClF⊛−V. Then, we have (σ1, σ2)α‐Cl(F(x)) ∩ V ≠ ∅, and hence F(x) ∩ V ≠ ∅. Thus, x ∈ F−(V). This shows that αClF⊛−V⊆F−V. On the other hand, let x ∈ F−(V). Then, we have ∅ ≠ F(x) ∩ V ⊆ (σ1, σ2)α‐Cl(F(x)) ∩ V, and hence x∈αClF⊛−V. Therefore, F−V⊆αClF⊛−V. Consequently, we obtain αClF⊛−V=F−V.Theorem 8.
A multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is lower (τ1, τ2)α-continuous if and only if αClF⊛: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is lower (τ1, τ2)α-continuous.Proof.
By utilizing Lemma8, this can be proved similarly to that of Theorem 7.Definition 10.
A bitopological space (X, τ1, τ2) is said to be τ1τ2-compact [20] (resp. (τ1, τ2)α-compact) if every cover of X by τ1τ2-open (resp. (τ1, τ2)α-open) sets of X has a finite subcover.Theorem 9.
LetF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) be an upper (τ1, τ2)α-continuous surjective multifunction such that F(x) is σ1σ2-compact for each x ∈ X. If (X, τ1, τ2) is (τ1, τ2)α-compact, then (Y, σ1, σ2) is σ1σ2-compact.Proof.
Suppose that (X, τ1, τ2) is (τ1, τ2)α-compact. Let {Vγ∣γ ∈ Γ} be a σ1σ2-open cover of Y. For each x ∈ X, F(x) is σ1σ2-compact, there exists a finite subset Γ(x) of Γ such that F(x) ⊆ ∪{Vγ∣γ ∈Γ(x)}. Put V(x) = ∪{Vγ∣γ ∈Γ(x)}. Since F is upper (τ1, τ2)α-continuous, there exists a (τ1, τ2)α-open set U(x) containing x such that F(U(x)) ⊆ V(x). The family {U(x) ∣ x ∈ X} is a (τ1, τ2)α-open cover of X and there exists a finite number of points, say, x1, x2, …, xn in X such that X=∪i=1nUxi. Thus, Y=FX=F∪i=1nUxi=∪i=1nFUxi⊆∪i=1nVxi=∪i=1n∪γ∈ΓxiVγ. This shows that (Y, σ1, σ2) is σ1σ2-compact.


## 4. Characterizations of Upper and Lower Almost (τ1, τ2)α-Continuous Multifunctions

In this section, we introduce the notions of upper and lower almost (τ1, τ2)α-continuous multifunctions and investigate some characterizations of these multifunctions.Definition 11.
A multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is said to be:(1)
upper almost (τ1, τ2)α-continuous at a point x ∈ X if for each τ1τ2-semi-open subset U of X containing x and each σ1σ2-open subset V of Y containing F(x), there exists a nonempty (τ1, τ2)α-open subset G of X such that G ⊆ U and F(G) ⊆ σ1σ2‐sCl(V);(2)
lower almost (τ1, τ2)α-continuous at a point x ∈ X if for each τ1τ2-semi-open subset U of X containing x and each σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a nonempty (τ1, τ2)α-open subset G of X such that G ⊆ U and σ1σ2‐sCl(V) ∩ F(z) ≠ ∅ for every z ∈ G;(3)
upper almost (resp. lower almost) (τ1, τ2)α-continuous if F has this property at each point of X.Remark 2.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following implication holds:
upper (τ1, τ2)α-continuity ⟹ upper almost (τ1, τ2)α-continuity.
The converse of the implication is not true in general. We give an example for the implication as follows.Example 2.
LetX = {a, b, c} with topologies τ1 = {∅, {c}, {a, b}, X} and τ2 = {∅, {c}, X}. Let Y = {−2, −1, 0, 1, 2} with topologies σ1 = {∅, {−1, 0, 1}, {−2, 2}, Y} and σ2 = {∅, {−1, 0, 1}, Y}. Define a multifunction F: (X, τ1, τ2) ⟶ (Y, σ1, σ2) as follows:(23)Fx=0,ifx=a,−1,1,ifx=b,−2,2,ifx=c.Then, F is upper almost (τ1, τ2)α-continuous but F is not upper (τ1, τ2)α-continuous.Theorem 10.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is upper almost (τ1, τ2)α-continuous at x ∈ X(2)
For everyσ1σ2-open subset V of Y containing F(x), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐sCl(V)(3)
x ∈ (τ1, τ2)α‐Int(F+(σ1σ1‐sCl(V))) for every σ1σ2-open subset V of Y containing F(x)(4)
x ∈ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ2‐sCl(V))))) for every σ1σ2-open subset V of Y containing F(x)Proof.
(1) ⟹ (2): Letτ1τ2‐SO(X, x) be the family of all τ1τ2-semi-open subsets of X containing x. Let V be any σ1σ2-open subset of Y containing F(x). For each U ∈ τ1τ2‐SO(X, x), there exists a nonempty τ1τ2-open subset GU such that GU ⊆ U and F(GU) ⊆ σ1σ2‐sCl(V). Let W = ∪ {GU ∣ U ∈ τ1τ2‐SO(X, x)}. Then, W is τ1τ2-open, x ∈ τ1τ2‐sCl(W) and F(W) ⊆ σ1σ2‐sCl(V). Put H = W ∪ {x}; by Lemma 1,
(24)
W
⊆
H
⊆
τ
1
τ
2
‐
sCl
H
=
τ
1
‐
Int
τ
2
‐
Cl
H
∪
H
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
x
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
τ
2
‐
Cl
x
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
=
τ
1
τ
2
‐
sCl
W
,
and by Lemma2, H is a (τ1, τ2)α-open subset of X containing x such that F(H) ⊆ σ1σ2‐sCl(V).
(2) ⟹ (3): LetV be any σ1σ2-open subset of Y containing F(x). By (2), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐Cl(V). Thus, x ∈ U ⊆ F+(σ1σ2‐Cl(V)), and hence x ∈ (τ1, τ2)α‐Int(F+(σ1σ1‐sCl(V))).
(3) ⟹ (4):LetV be any σ1σ2-open subset of Y containing F(x). By (3), we have x ∈ (τ1, τ2)α‐Int(F+(σ1σ1‐sCl(V))). Put U = (τ1, τ2)α‐Int(F+(σ1σ1‐sCl(V))); then, U is a (τ1, τ2)α-open subset of X such that x ∈ U ⊆ F+(σ1σ2‐sCl(V)). Consequently, we obtain x ∈ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ2‐sCl(V))))).
(4) ⟹ (1): LetU be any τ1τ2-semi-open subset of X containing x and V be any σ1σ2-open subset of Y containing F(x). By (4) and Lemma 1 (2),
(25)
x
∈
τ
1
‐
Int
τ
2
‐
Cl
τ
1
τ
2
‐
Int
F
+
σ
1
σ
2
‐
sCl
V
=
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
F
+
σ
1
σ
2
‐
sCl
V
.
Thus,τ1τ2‐Int(F+(σ1σ2‐sCl(V))) ∩ U ≠ ∅ and τ1τ2‐Int(F+(σ1σ2‐sCl(V))) ∩ U is τ1τ2-semi-open. Put G = τ1τ2‐Int(τ1τ2‐Int(F+(σ1σ2‐sCl(V))) ∩ U); then, G is a nonempty τ1τ2-open subset of X such that G ⊆ U and F(G) ⊆ σ1σ2‐sCl(V). This shows that F is upper almost (τ1, τ2)α-continuous at x.Theorem 11.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is lower almost (τ1, τ2)α-continuous at x ∈ X(2)
For everyσ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a (τ1, τ2)α-open subset U of X containing x such that σ1σ2‐sCl(V) ∩ F(z) ≠ ∅ for every z ∈ U(3)
x ∈ (τ1, τ2)α‐Int(F−(σ1σ1‐sCl(V))) for every σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅(4)
x ∈ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F−(σ1σ2‐sCl(V))))) for every σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅Proof.
The proof is similar to that of Theorem10.Lemma 9.
Let (X, τ1, τ2) be a bitopological space. If V is τ1τ2-open in X, then τ1τ2‐sCl(V) is τ1τ2-regular open.Proof.
LetV be a τ1τ2-open set. By Lemma 1 (2), τ1τ2‐sCl(V) = τ1‐Int(τ2‐Cl(V)), and hence(26)τ1‐Intτ2‐Clτ1τ2‐sClV=τ1‐Intτ2‐Clτ1‐Intτ2‐ClV=τ1‐Intτ2‐ClV=τ1τ2‐sClV.
Thus,τ1τ2‐sCl(V) is τ1τ2-regular open.Theorem 12.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is upper almost (τ1, τ2)α-continuous(2)
For eachx ∈ X and each σ1σ2-open subset V of Y containing F(x), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐sCl(V)(3)
For eachx ∈ X and each σ1σ2-regular open subset V of Y containing F(x), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ V(4)
F+(V) is (τ1, τ2)α-open in X for every σ1σ2-regular open subset V of Y(5)
F−(K) is (τ1, τ2)α-closed in X for every σ1σ2-regular closed subset K of Y(6)
F+(V) ⊆ (τ1, τ2)α‐Int(F+(σ1σ2‐sCl(V))) for every σ1σ2-open subset V of Y(7)
(τ1, τ2)α‐Cl(F−(σ1σ2‐sInt(K))) ⊆ F−(K) for every σ1σ2-closed subset K of Y(8)
(τ1, τ2)α‐Cl(F−(σ1‐Cl(σ2‐Int(K)))) ⊆ F−(K) for every σ1σ2-closed subset K of Y(9)
(τ1, τ2)α‐Cl(F−(σ1‐Cl(σ2‐Int(σ1σ2‐Cl(B))))) ⊆ F−(σ1σ2‐Cl(B)) for every subset B of Y(10)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F−(σ1‐Cl(σ2‐Int(K)))))) ⊆ F−(K) for every σ1σ2-closed subset K of Y(11)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F−(σ1σ2‐sInt(K))))) ⊆ F−(K) for every σ1σ2-closed subset K of Y(12)
F+(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ2‐sCl(K))))) for every σ1σ2-open subset V of YProof.
(1) ⟹ (2): The proof follows from Theorem10.
(2) ⟹ (3): This is obvious.
(3) ⟹ (4): LetV be any σ1σ2-regular open subset of Y and x ∈ F+(V). Then, we have F(x) ⊆ V. By (3), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ V. Thus, x ∈ τ1‐Int(τ2‐Cl(τ1τ2‐Int(U))) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(V)))), and hence F+(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(V)))). This shows that F+(V) is (τ1, τ2)α-open in X.
(4) ⟹ (5): This follows from the fact thatF+(Y –B) = X –F‐(B) for every subset B of Y.
(5) ⟹ (6): LetV be any σ1σ2-open subset of Y and x ∈ F+(V). Then, we have F(x) ⊆ V ⊆ σ1σ2‐sCl(V), and hence
(27)
x
∈
F
+
σ
1
σ
2
‐
sCl
V
=
X
−
F
‐
Y
‐
σ
1
σ
2
‐
sCl
V
.
By Lemma9, Y– σ1σ2‐sCl(V) is σ1σ2-regular closed in Y, and by (5), F‐(Y– σ1σ2‐sCl(V)) is (τ1, τ2)α-closed in X. This implies that F+(σ1σ2‐sCl(V)) is (τ1, τ2)α-open in X, and hence x ∈ (τ1, τ2)α‐Int(F+(σ1σ2‐sCl(V))). Consequently, we obtain F+(V) ⊆ (τ1, τ2)α‐Int(F+(σ1σ2‐sCl(V))).
(6) ⟹ (7): LetK be any σ1σ2-closed subset of Y. Then, we have Y‐K is σ1σ2-open in Y and by (6),
(28)
X
–
F
‐
K
=
F
+
Y
–
K
⊆
τ
1
,
τ
2
α
‐
Int
F
+
σ
1
σ
2
‐
sCl
Y
–
K
=
τ
1
,
τ
2
α
‐
Int
F
+
Y
–
σ
1
σ
2
‐
sInt
K
=
τ
1
,
τ
2
α
‐
Int
X
–
F
‐
σ
1
σ
2
‐
sInt
K
=
X
–
τ
1
,
τ
2
α
‐
Cl
F
‐
σ
1
σ
2
‐
sInt
K
.
Therefore, (τ1, τ2)α‐Cl(F‐(σ1σ2‐sInt(K))) ⊆ F‐(K).
(7) ⟹ (8): The proof is obvious sinceσ1σ2‐sInt(K) = σ1‐Cl(σ2‐Int(K)) for every σ1σ2-closed subset K of Y.
(8) ⟹ (9): The proof is obvious.
(9) ⟹ (10): LetK be any σ1σ2-closed subset of Y. By (9) and Lemma 3 (3), we obtain
(29)
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
F
‐
σ
1
‐
Cl
σ
2
‐
Int
K
⊆
τ
1
,
τ
2
α
‐
Cl
F
‐
σ
1
‐
Cl
σ
2
‐
Int
K
=
τ
1
,
τ
2
α
‐
Cl
F
‐
σ
1
‐
Cl
σ
2
‐
Int
σ
1
σ
2
‐
Cl
K
⊆
F
‐
σ
1
σ
2
‐
Cl
K
=
F
‐
K
.
(10) ⟹ (11): The proof is obvious sinceσ1σ2‐sInt(K) = σ1‐Cl(σ2‐Int(K)) for every σ1σ2-closed subset K of Y.
(11) ⟹ (12): LetV be any σ1σ2-open subset of Y. Then, we have Y – V is σ1σ2-closed in Y and by (11),
(30)
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
sInt
Y
–
V
⊆
F
‐
Y
–
V
=
X
–
F
+
V
.
Moreover, we have
(31)
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
F
‐
σ
1
σ
2
‐
sInt
Y
–
V
=
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
F
‐
Y
–
σ
1
σ
2
‐
sCl
V
=
τ
1
‐
Cl
τ
2
‐
Int
τ
1
τ
2
‐
Cl
X
–
F
+
σ
1
σ
2
‐
sCl
V
=
X
–
τ
1
‐
Int
τ
2
‐
Cl
τ
1
τ
2
‐
Int
F
+
σ
1
σ
2
‐
sCl
V
,
and henceF+(V) ⊆ τ1 ‐ Int(τ2 ‐ Cl(τ1τ2 ‐ Int(F+(σ1σ2 ‐ sCl(V))))).
(12)
⟹ (1): Letx ∈ X and V be any σ1σ2-open subset of Y containing F(x). By (12), we have x ∈ F+(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ2‐sCl(V))))), and hence F is upper almost (τ1, τ2)α-continuous at x by Theorem 10. Consequently, we obtain F is upper almost (τ1, τ2)α-continuous.Theorem 13.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is lower almost (τ1, τ2)α-continuous(2)
For eachx ∈ X and each σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a (τ1, τ2)α-open subset U of X containing x such that U ⊆ F‐(σ1σ2‐sCl(V))(3)
For eachx ∈ X and each σ1σ2-regular open subset V of Y such that(32)Fx∩V≠∅,there exists a (τ1, τ2)α-open subset U of X containing x such that U ⊆ F‐(V)(4)
F‐(V) is (τ1, τ2)α-open in X for every σ1σ2-regular open subset V of Y(5)
F+(K) is (τ1, τ2)α-closed in X for every σ1σ2-regular closed subset K of Y(6)
F‐(V) ⊆ α(τ1, τ2)‐Int(F‐(σ1σ2‐sCl(V))) for every σ1σ2-open subset V of Y(7)
(τ1, τ2)α‐Cl(F+(σ1σ2‐sInt(K))) ⊆ F+(K) for every σ1σ2-closed subset K of Y(8)
(τ1, τ2)α‐Cl(F+(σ1‐Cl(σ2‐Int(K)))) ⊆ F+(K) for every σ1σ2-closed subset K of Y(9)
(τ1,τ2) α‐Cl(F+(σ1‐Cl(σ2‐Int(σ1σ2‐Cl(B))))) ⊆ F+(σ1σ2‐Cl(B)) for every subset B of Y(10)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F+(σ1‐Cl(σ2‐Int(K)))))) ⊆ F+(K) for every σ1σ2-closed subset K of Y(11)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F+(σ1σ2‐sInt(K))))) ⊆ F+(K) for every σ1σ2-closed subset K of Y(12)
F‐(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F‐(σ1σ2‐sCl(K))))) for every σ1σ2-open subset V of YProof.
The proof is similar to that of Theorem12.Definition 12.
A functionf: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is said to be almost (τ1, τ2)α-continuous if f‐1(V) is (τ1, τ2)α-open in X for every σ1σ2-regular open subset V of Y.Corollary 2.
For a functionf: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
f is almost (τ1, τ2)α-continuous(2)
For eachx ∈ X and each σ1σ2-open subset V of Y containing f(x), there exists a (τ1, τ2)α-open subset U of X containing x such that f(U) ⊆ σ1σ2‐sCl(V)(3)
For eachx ∈ X and each σ1σ2-regular open subset V of Y containing f(x), there exists a (τ1, τ2)α-open subset U of X containing x such that f(U) ⊆ V(4)
For eachx ∈ X and each σ1σ2-open subset V of Y containing f(x), there exists a (τ1, τ2)α-open subset U of X containing x such that f(U) ⊆ σ1‐Int(σ2‐Cl(V))(5)
f−1(F) is (τ1, τ2)α-closed in X for every σ1σ2-regular closed subset F of Y(6)
f−1(V) ⊆ (τ1, τ2)α‐Int(f−1(σ1σ2‐sCl(V))) for every σ1σ2-open subset V of Y(7)
(τ1, τ2)α‐Cl(f−1(σ1σ2‐sInt(F))) ⊆ f−1(F) for every σ1σ2-closed subset F of Y(8)
(τ1, τ2)α‐Cl(f−1(σ1‐Cl(σ2‐Int(F)))) ⊆ f−1(F) for every σ1σ2-closed subset F of Y(9)
(τ1,τ2)α‐Cl(f−1(σ1‐Cl(σ2‐Int(σ1σ2‐Cl(B))))) ⊆ f−1(σ1σ2‐Cl(B)) for every subset B of Y(10)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(f−1(σ1‐Cl(σ2‐Int(F)))))) ⊆ f−1(F) for every σ1σ2-closed subset F of Y(11)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(f−1(σ1σ2‐sInt(F))))) ⊆ f−1(F) for every σ1σ2-closed subset F of Y(12)
f−1(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(f−1(σ1σ2‐sCl(V))))) for every σ1σ2-open subset V of Y
Let {(Xγ, τ1(γ), τ2(γ))∣γ ∈Γ} be a family of bitopological spaces. Let X∗,τ1∗,τ2∗ be the product space, where X∗=∏γ∈ΓXγ and τi∗ denotes the product topology of {τi(γ) ∣ γ ∈ Γ} for i = 1, 2.Lemma 10.
LetAγk be a nonempty subset of Xγk for k = 1, 2, …, n. Then, ∏k=1nAγk×∏γ≠γkXγ is τ1∗,τ2∗α-open in X∗ if and only if Aγkis (τ1(γk), τ2(γk))α-open in Xγk for each k = 1, 2, …, n.Proof.
The proof is similar to that of [23] (Theorem 4.6)Let {(Xγ, τ1(γ), τ2(γ))∣γ ∈Γ} and {(Yγ, σ1(γ), σ2(γ))∣γ ∈ Γ} be two arbitrary families of bitopological spaces with the same set of indices. Let Fγ: (Xγ, τ1(γ), τ2(γ)) ⟶ (Yγ, σ1(γ), σ2(γ)) be a multifunction for each γ ∈ Γ. Let F∗:X∗,τ1∗,τ2∗⟶Y∗,σ1∗,σ2∗ be the product multifunction defined by F∗x∗=∏γ∈ΓFγxγ for each x∗=xγ∈∏γ∈ΓXγ, where τi∗ and σi∗ denote the product topologies for i = 1, 2.Theorem 14.
IfF∗:X∗,τ1∗,τ2∗⟶Y∗,σ1∗,σ2∗ is upper almost τ1∗,τ2∗α-continuous, then(33)Fγ:Xγ,τ1γ,τ2γ⟶Yγ,σ1γ,σ2γ,is upper almost (τ1(γ), τ2(γ))α-continuous for each γ ∈Γ.Proof.
LetVγ be a σ1(γ)σ2(γ)-regular open subset of Yγ. Since F∗ is upper almost τ1∗,τ2∗α-continuous,(34)F∗+Vγ×∏γ≠γ0Yγ0=Fγ+Vγ×∏γ≠γ0Yγ0,is τ1∗,τ2∗α-open in X∗, and by Lemma 10, Fγ+Vγ is (τ1(γ), τ2(γ))α-open in Xγ. Thus, Fγ is upper almost (τ1(γ), τ2(γ))α-continuous.Theorem 15.
IfF∗:X∗,τ1∗,τ2∗⟶Y∗,σ1∗,σ2∗ is lower almost τ1∗,τ2∗α-continuous, then(35)Fγ:Xγ,τ1γ,τ2γ⟶Yγ,σ1γ,σ2γ,is lower almost (τ1(γ), τ2(γ))α-continuous for each γ ∈Γ.Proof.
The proof is similar to that of Theorem14.Theorem 16.
Let (X, τ1, τ2) and (Yγ, σ1(γ), σ2(γ)) be bitopological spaces for each γ ∈Γ. Let Fγ: (X, τ1, τ2) ⟶ (Yγ, σ1(γ), σ2(γ)) be a multifunction for each γ ∈Γ and F:X,τ1,τ2⟶Y∗,σ1∗,σ2∗ a multifunction defined byFx=∏γ∈ΓFγx for each x ∈ X. If F is upper almost (τ1, τ2)α-continuous, then Fγ is upper almost (τ1, τ2)α-continuous for each γ ∈ Γ.Proof.
Letx ∈ X, γ ∈Γ, and Vγ be any σ1(γ)σ2(γ)-regular open subset of Yγ containing Fγ(x). Then, πγ−1Vγ=Vγ×∏γ≠γ0Yγ0 is σ1∗σ2∗-regular open in ∏γ∈ΓYγ containing F(x), where πγ:Y∗,σ1∗,σ2∗⟶Yγ,σ1γ,σ2γ is the projection for each γ ∈Γ. Since F is upper almost (τ1, τ2)α-continuous, there exists a (τ1, τ2)α-open subset U of X containing x such that FU⊆πγ−1Vγ. Therefore, FγU⊆πγFU⊆πγπγ−1Vγ=Vγ. This shows that Fγ is upper almost (τ1, τ2)α-continuous for each γ ∈ Γ.Theorem 17.
Let (X, τ1, τ2) and (Yγ, σ1(γ), σ2(γ)) be bitopological spaces for each γ ∈Γ. Let Fγ: (X, τ1, τ2) ⟶ (Yγ, σ1(γ), σ2(γ)) be a multifunction for each γ ∈Γ and F:X,τ1,τ2⟶Y∗,σ1∗,σ2∗ a multifunction defined by Fx=∏γ∈ΓFγx for each x ∈ X. If F is lower almost (τ1, τ2)α-continuous, then Fγ is lower almost (τ1, τ2)α-continuous for each γ ∈ Γ.Proof.
The proof is similar to that of Theorem16.


## 5. Characterizations of Upper and Lower Weakly (τ1, τ2)α-Continuous Multifunctions

In this section, we introduce the concepts of upper and lower weakly (τ1, τ2)α-continuous multifunctions. Furthermore, some characterizations of upper and lower weakly (τ1, τ2)α-continuous multifunctions are investigated.Definition 13.
A multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is said to be:(1)
upper weakly (τ1, τ2)α-continuous at a point x ∈ X if for each τ1τ2-semi-open subset U of X containing x and each σ1σ2-open subset V of Y containing F(x), there exists a nonempty τ1τ2-open subset G of X such that G ⊆ U and F(G) ⊆ σ1σ2‐Cl(V);(2)
lower weakly (τ1, τ2)α-continuous at a point x ∈ X if for each τ1τ2-semi-open subset U of X containing x and each σ1σ2-open subset V of Y such that F(x) ∩ V≠∅, there exists a nonempty τ1τ2-open subset G of X such that G ⊆ U and F(z) ∩ σ1σ2‐Cl(V) ≠ ∅ for every z ∈ G;(3)
upper weakly (resp. lower weakly) (τ1, τ2)α-continuous if F has this property at each point of X.Remark 3.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following implication holds:
upper almost (τ1, τ2)α-continuity ⟹ upper weakly (τ1, τ2)α-continuity.
The converse of the implication is not true in general. We give an example for the implication as follows.Example 3.
LetX = {a, b, c, d} with topologies τ1 = {∅, {a}, X} and τ2 = {∅, X}. Let Y = {−1, 0, 1, 2} with topologies σ1 = {∅, {−1, 2}, {0, 1}, Y}, and σ2 = {∅, {−1, 2}, Y}. A multifunction F: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is defined as follows:(36)Fx=−1,ifx=a,1,ifx=b,0,1,ifx=c,−1,0,ifx=d.
Then,F is upper weakly (τ1, τ2)α-continuous but F is not upper almost (τ1, τ2)α-continuous.Theorem 18.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is upper weakly (τ1, τ2)α-continuous at x ∈ X(2)
For anyσ1σ2-open subset V of Y containing F(x), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐Cl(V)(3)
x ∈ (τ1, τ2)α‐Int(F+(σ1σ1‐Cl(V))) for every σ1σ2-open subset V of Y containing F(x)(4)
x ∈ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ1‐Cl(V))))) for every σ1σ2-open subset V of Y containing F(x)Proof.
(1) ⟹ (2): Letτ1τ2‐SO(X, x) be the family of all τ1τ2-semi-open subsets of X containing x. Let V be any σ1σ2-open subset of Y containing F(x). For each U ∈ τ1τ2‐SO(X, x), there exists a nonempty τ1τ2-open set HU such that HU ⊆ U and F(HU) ⊆ V. Let W = ∪{HU∣U ∈ τ1τ2‐SO(X, x)}. Then, W is τ1τ2-open in X, x ∈ τ1τ2 ‐ sCl(W), and F(W) ⊆ V. Put G = W ∪ {x}. By Lemma 1, we have
(37)
W
⊆
G
⊆
τ
1
τ
2
‐
sCl
G
=
τ
1
‐
Int
τ
2
‐
Cl
G
∪
G
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
x
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
τ
2
‐
Cl
x
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
∪
W
∪
x
=
τ
1
‐
Int
τ
2
‐
Cl
W
=
τ
1
τ
2
‐
sCl
W
,
and by Lemma2, G is a (τ1, τ2)α-open subset of X containing x such that F(G) ⊆ σ1σ2‐Cl(V).
(2) ⟹ (3): LetV be any σ1σ2-open subset of Y containing F(x). By (2), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐Cl(V). This implies that x ∈ U ⊆ F+(σ1σ2‐Cl(V)), and hence
(38)
x
∈
τ
1
,
τ
2
α
‐
Int
F
+
σ
1
σ
1
‐
Cl
V
.
(3) ⟹ (4): LetV be any σ1σ2-open subset of Y. By (3), we have
(39)
x
∈
τ
1
,
τ
2
α
‐
Int
F
+
σ
1
σ
1
‐
Cl
V
.
Then, there exists a (τ1, τ2)α-open subset U of X such that x ∈ U ⊆ F+(σ1σ1‐Cl(V)). Thus, x ∈ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ1‐Cl(V))))).
(4) ⟹ (1): LetU be any τ1τ2-semi-open set containing x and V be any σ1σ2-open set containing F(x). By (4) and Lemma 1 (2), we have
(40)
x
∈
τ
1
‐
Int
τ
2
‐
Cl
τ
1
τ
2
‐
Int
F
+
σ
1
σ
1
‐
Cl
V
=
τ
1
τ
2
‐
sCl
τ
1
τ
2
‐
Int
F
+
σ
1
σ
1
‐
Cl
V
.
Thus,τ1τ2‐Int(F+(σ1σ1‐Cl(V))) ∩ U ≠ ∅ and τ1τ2‐Int(F+(σ1σ1‐Cl(V))) ∩ U is τ1τ2-semi-open. Put G = τ1τ2‐Int(τ1τ2‐Int(F+(σ1σ1‐Cl(V))) ∩ U); then G is a nonempty σ1σ2-open subset of X such that G ⊆ U and F(G) ⊆ σ1σ2‐Cl(V). This shows that F is upper weakly (τ1, τ2)α-continuous at x.Theorem 19.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is lower weakly (τ1, τ2)α-continuous at x ∈ X(2)
For anyσ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a (τ1, τ2)α-open subset U of X containing x such that σ1σ2‐Cl(V) ∩ F(z)≠∅ for every z ∈ U(3)
x ∈ (τ1, τ2)α‐Int(F−(σ1σ1‐Cl(V))) for every σ1σ2-open subset V of Y such that F(x) ∩ V≠∅(4)
x ∈ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F−(σ1σ1‐Cl(V))))) for every σ1σ2-open subset V of Y such that F(x) ∩ V≠∅Proof.
The proof is similar to that of Theorem18.Definition 14.
LetA be a subset of a bitopological space (X, τ1, τ2). A point x ∈ X is called a (τ1, τ2)θ-cluster point of A if τ1τ2‐Cl(U) ∩ A≠∅ for every τ1τ2-open set U containing x. The set of all (τ1, τ2)θ-cluster point of A is called the (τ1, τ2)θ-closure of A and is denoted by (τ1, τ2)θ‐Cl(A).
A subsetA of a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)θ-closed if A = (τ1, τ2)θ‐Cl(A). The complement of a (τ1, τ2)θ-closed set is said to be (τ1, τ2)θ-open. The union of all (τ1, τ2)θ-open sets contained in A is called the (τ1, τ2)θ-interior of A and is denoted by (τ1, τ2)θ‐Int(A).Lemma 11.
For a subsetA of a bitopological space (X, τ1, τ2), the following properties hold:(1)
IfA is τ1τ2-open in X, then τ1τ2‐Cl(A) = (τ1, τ2)θ‐Cl(A)(2)
(τ1, τ2)θ‐Cl(A) is τ1τ2-closed in XProof.
(1)
In general,τ1τ2‐Cl(A) ⊆ (τ1, τ2)θ‐Cl(A). Suppose that x ∉ τ1τ2‐Cl(A). Then, there exists a τ1τ2-open set U containing x such that U ∩ A =∅; hence, τ1τ2‐Cl(U) ∩ A = ∅. This shows that x ∉ (τ1, τ2)θ‐Cl(A). Thus, (τ1, τ2)θ‐Cl(A) ⊆ τ1τ2‐Cl(A). Consequently, we obtain τ1τ2‐Cl(A) = (τ1, τ2)θ‐Cl(A).
(2)
Letx ∈ X – (τ1, τ2)θ‐Cl(A). Then, we have x∉(τ1, τ2)θ‐Cl(A). There exists a τ1τ2-open set Ux containing x such that τ1τ2‐Cl(Ux) ∩ A =∅. This implies that (τ1, τ2)θ‐Cl(A) ∩ Ux =∅, and hence x ∈ Ux ⊆ X –(τ1, τ2)θ‐Cl(A). Thus, X−τ1,τ2θ‐ClA=∪x∈X−τ1,τ2θ‐ClAUx. This shows that (τ1, τ2)θ‐Cl(A) is τ1τ2-closed.Definition 15.
A subsetA of a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)r-closed (resp. (τ1, τ2)r-open) if A = τ1τ2‐Cl(τ1τ2‐Int(A)) (resp. A = τ1τ2‐Int(τ1τ2‐Cl(A))).Theorem 20.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is upper weakly (τ1, τ2)α-continuous(2)
For eachx ∈ X and each σ1σ2-open subset V of Y containing F(x), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐Cl(V)(3)
F+(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ2‐Cl(V))))) for every σ1σ2-open subset V of Y(4)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F−(σ1σ2‐Int(K))))) ⊆ F−(K) for every σ1σ2-closed subset K of Y(5)
(τ1, τ2)α‐Cl(F−(σ1σ2‐Int(K))) ⊆ F−(K) for every σ1σ2-closed subset K of Y(6)
(τ1, τ2)α‐Cl(F−(σ1σ2‐Int(σ1σ2‐Cl(B)))) ⊆ F−(σ1σ2‐Cl(B)) for every subset B of Y(7)
F+(σ1σ2‐Int(B)) ⊆ (τ1, τ2)α‐Int(F+ (σ1σ2‐Cl(σ1σ2‐Int(B)))) for every subset B of Y(8)
F+(V) ⊆ (τ1, τ2)α‐Int(F+(σ1σ2‐Cl(V))) for every σ1σ2-open subset V of Y(9)
(τ1, τ2)α‐Cl(F−(σ1σ2‐Int(K))) ⊆ F−(K) for every (σ1, σ2)r-closed subset K of Y(10)
(τ1, τ2)α‐Cl(F−(V)) ⊆ F−(σ1σ2‐Cl(V)) for every σ1σ2-open subset V of Y(11)
(τ1, τ2)α‐Cl(F−(σ1σ2‐Int((σ1, σ2)θ‐Cl(B)))) ⊆ F−((σ1, σ2)θ‐Cl(B)) for every subset B of Y(12)
(τ1, τ2)α‐Cl(F−(σ1σ2‐Int(σ1σ2‐Cl(V)))) ⊆ F−(σ1σ2‐Cl(V)) for every σ1σ2-open subset V of YProof.
(1) ⟹ (2): The proof follows immediately from Theorem18.(2) ⟹ (3): LetV be any σ1σ2-open subset of Y and x ∈ F+(V). Then, we have F(x) ⊆ V and by (2), there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐Cl(V). Therefore, we have x ∈ U ⊆ F+(σ1σ2‐Cl(V)), and hence x ∈ U ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ2‐Cl(V))))). Consequently, we obtain F+(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F+(σ1σ2‐Cl(V))))).(3) ⟹ (4): LetK be any σ1σ2-closed subset of Y. Then, we have Y –K is σ1σ2-open in Y and by (3),(41)X–F‐K=F+Y–K⊆τ1‐Intτ2‐Clτ1τ2‐IntF+σ1σ2‐ClY–K=τ1‐Intτ2‐Clτ1τ2‐IntF+Y–σ1σ2‐IntK=τ1‐Intτ2‐Clτ1τ2‐IntX–F‐σ1σ2‐IntK=τ1‐Intτ2‐ClX–τ1τ2‐ClF‐σ1σ2‐IntK=τ1‐IntX–τ2‐Intτ1τ2‐ClF‐σ1σ2‐IntK=X–τ1‐Clτ2‐Intτ1τ2‐ClF‐σ1σ2‐IntK.Thus,τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F‐(σ1σ2‐Int(K))))) ⊆ F‐(K).(4) ⟹ (5): LetK be any σ1σ2-closed subset of Y. By (4), we have(42)τ1‐Clτ2‐Intτ1τ2‐ClF‐σ1σ2‐IntK⊆F‐K,and hence (τ1, τ2)α‐Cl(F‐(σ1σ2‐Int(K))) ⊆ F‐(K) by Lemma 3 (3).(5) ⟹ (6): LetB be any subset of Y. Then, we have σ1σ2‐Cl(B) is σ1σ2-closed in Y, and by (5), (τ1, τ2)α‐Cl(F‐(σ1σ2‐Int(σ1σ2‐Cl(B)))) ⊆ F‐(σ1σ2‐Cl(B)).(6) ⟹ (7): LetB be any subset of Y. By (6), we have(43)F+σ1σ2‐IntB=X–F‐σ1σ2‐ClY–B⊆X−τ1,τ2α‐ClF‐σ1σ2‐Intσ1σ2‐ClY–B=X−τ1,τ2α‐ClF‐Y‐σ1σ2‐Clσ1σ2‐IntB=X–τ1,τ2α‐ClX–F+σ1σ2‐Clσ1σ2‐IntB=τ1,τ2α‐IntF+σ1σ2‐Clσ1σ2‐IntB.(7) ⟹ (8): The proof is obvious.(8) ⟹ (1): Letx ∈ X and V be any σ1σ2-open subset of Y containing F(x). By (8), we have(44)x∈F+V⊆τ1,τ2α‐IntF+σ1σ2‐ClV⊆τ1‐Intτ2‐Clτ1τ2‐IntF+σ1σ2‐ClV,and henceF is upper weakly (τ1, τ2)α-continuous at x by Theorem 18. Consequently, we obtain that F is upper weakly (τ1, τ2)α-continuous.(5) ⟹ (9): The proof is obvious.(9) ⟹ (10): LetV be any σ1σ2-open subset of Y. Then, we have σ1σ2‐Cl(V) is (σ1, σ2)r-closed in Y, and by (9),(45)τ1,τ2α‐ClF‐V⊆τ1,τ2α‐ClF‐σ1σ2‐Intσ1σ2‐ClV⊆F‐σ1σ2‐ClV.(10) ⟹ (8): LetV be any σ1σ2-open subset of Y. By (10), we have(46)X–τ1,τ2α‐IntF+σ1σ2‐ClV=τ1,τ2α‐ClX–F+σ1σ2‐ClV=τ1,τ2α‐ClF‐Y–σ1σ2‐ClV⊆F‐σ1σ2‐ClY–σ1σ2‐ClV=X–F+σ1σ2‐Intσ1σ2‐ClV,and henceF+(V) ⊆ F+(σ1σ2‐Int(σ1σ2‐Cl(V))) ⊆ (τ1, τ2)α‐Int(F+(σ1σ2‐Cl(V))).(10) ⟹ (11): LetB be any subset of Y. Then, σ1σ2‐Int((σ1, σ2)θ‐Cl(B)) is σ1σ2-open in Y. By (10) and Lemma 11 (1),(47)τ1,τ2α‐ClF‐σ1σ2‐Intσ1,σ2θ‐ClB⊆F‐σ1σ2‐Clσ1σ2‐Intσ1,σ2θ‐ClB=F‐σ1,σ2θ‐Clσ1σ2‐Intσ1,σ2θ‐ClB⊆F‐σ1,σ2θ‐ClB.(11) ⟹ (12): LetV be any σ1σ2-open subset of Y. By (11) and Lemma 11 (1), we have (τ1, τ2)α‐Cl(F‐(σ1σ2‐Int(σ1σ2‐Cl(V)))) ⊆ F‐(σ1σ2‐Cl(V)).(12) ⟹ (10): LetV be any σ1σ2-open subset of Y. By (12), we have(48)τ1,τ2α‐ClF‐σ1σ2‐Intσ1σ2‐ClV⊆F‐σ1σ2‐ClV,and hence(49)τ1,τ2α‐ClF‐V⊆τ1,τ2α‐ClF‐σ1σ2‐Intσ1σ2‐ClV⊆F‐σ1σ2‐ClV.Theorem 21.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is lower weakly (τ1, τ2)α-continuous(2)
For eachx ∈ X and each σ1σ2-open subset V of Y such that F(x) ∩ V ≠ ∅, there exists a (τ1, τ2)α-open subset U of X containing x such that U ⊆ F−(σ1σ2‐Cl(V))(3)
F−(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(F−(σ1σ2‐Cl(V))))) for every σ1σ2-open subset V of Y(4)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(F+(σ1σ2‐Int(K))))) ⊆ F+(K) for every σ1σ2-closed subset K of Y(5)
(τ1, τ2)α‐Cl(F+(σ1σ2‐Int(K))) ⊆ F+(K) for every σ1σ2-closed subset K of Y(6)
(τ1, τ2)α‐Cl(F+(σ1σ2‐Int(σ1σ2‐Cl(B)))) ⊆ F+(σ1σ2‐Cl(B)) for every subset B of Y(7)
F−(σ1σ2‐Int(B)) ⊆ (τ1, τ2)α‐Int(F−(σ1σ2‐Cl(σ1σ2‐Int(B)))) for every subset B of Y(8)
F−(V) ⊆ (τ1, τ2)α‐Int(F−(σ1σ2‐Cl(V))) for every σ1σ2-open subset V of Y(9)
(τ1, τ2)α‐Cl(F+(σ1σ2‐Int(K))) ⊆ F+(K) for every (σ1, σ2)r-closed subset K of Y(10)
(τ1, τ2)α‐Cl(F+(V)) ⊆ F+(σ1σ2‐Cl(V)) for every σ1σ2-open subset V of Y(11)
(τ1, τ2)α‐Cl(F+(σ1σ2‐Int((σ1, σ2)θ‐Cl(B)))) ⊆ F+((σ1, σ2)θ‐Cl(B)) for every subset B of Y(12)
(τ1, τ2)α‐Cl(F+(σ1σ2‐Int(σ1σ2‐Cl(V)))) ⊆ F+(σ1σ2‐Cl(V)) for every σ1σ2-open subset V of YProof.
The proof is similar to that of Theorem20.Lemma 12.
IfF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is lower weakly (τ1, τ2)α-continuous, then for each x ∈ X and each subset B of Y such that (τ1, τ2)θ‐Int(B) ∩ F(x) ≠ ∅, there exists a (τ1, τ2)α-open subset U of X containing x such that U ⊆ F−(B).Proof.
Suppose that (τ1, τ2)θ‐Int(B) ∩ F(x) ≠ ∅. Then, there exists a nonempty σ1σ2-open subset V of Y such that V ⊆ σ1σ2‐Cl(V) ⊆ B and F(x) ∩ V ≠ ∅. Since F is lower weakly (τ1, τ2)α-continuous, there exists a (τ1, τ2)α-open subset U of X containing x such that σ1σ2‐Cl(V) ∩ F(z) ≠ ∅ for every z ∈ U, and hence U ⊆ F−(B).Theorem 22.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
F is lower weakly (τ1, τ2)α-continuous(2)
(τ1, τ2)α‐Cl(F+(B)) ⊆ F+((σ1, σ2)θ‐Cl(B)) for every subset B of Y(3)
F((τ1, τ2)α‐Cl(A)) ⊆ (σ1, σ2)θ‐Cl(F(A)) for every subset A of XProof.
(1) ⟹ (2): LetB be any subset of Y. Suppose that x ∉ F+((σ1, σ2)θ‐Cl(B)). Then, we have
(50)
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‐
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σ
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.
Therefore,F(x) ∈ (σ1, σ2)θ‐Int(Y‐B). By Lemma 12, there exists a (τ1, τ2)α-open subset U of X containing x such that U ⊆ F1‐(Y − B) = X − F+(B), and hence U ∩ F+(B) = ∅. Thus, x∉(τ1, τ2)α‐Cl(F+(B)). Consequently, we obtain (τ1, τ2)α‐Cl(F+(B)) ⊆ F+((σ1, σ2)θ‐Cl(B)).
(2) ⟹ (1): LetV be any σ1σ2-open subset of Y. By (2) and Lemma 11 (1), we have (τ1, τ2)α‐Cl(F+(V)) ⊆ F+((σ1, σ2)θ‐Cl(V)) = F+(σ1σ2‐Cl(V)), and by Theorem 21, F is lower weakly (τ1, τ2)α-continuous.
(2) ⟹ (3): LetA be any subset of X. By (2), we have (τ1, τ2)α‐Cl(A) ⊆ (τ1, τ2)α‐Cl(F+(F(A))) ⊆ F+((σ1, σ2)θ‐Cl(F(A))). Thus, F((τ1, τ2)α‐Cl(A)) ⊆ (σ1, σ2)θ‐Cl(F(A)).
(3) ⟹ (2): LetB be any subset of Y. By (3), we have F((τ1, τ2)α‐Cl(F+(B))) ⊆ (σ1, σ2)θ‐Cl(F(F+(B))) ⊆ (σ1, σ2)θ‐Cl(B), and hence
(51)
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τ
2
α
‐
Cl
F
+
B
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F
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.Definition 16.
A functionf: (X, τ1, τ2) ⟶ (Y, σ1, σ2) is said to be weakly (τ1, τ2)α-continuous if for each x ∈ X and each σ1σ2-open subset V containing f(x), there exists a (τ1, τ2)α-open subset U of X containing x such that f(U) ⊆ σ1σ2‐Cl(V).Corollary 3.
For a functionf: (X, τ1, τ2) ⟶ (Y, σ1, σ2), the following properties are equivalent:(1)
f is weakly (τ1, τ2)α-continuous(2)
f‐1(V) ⊆ (τ1, τ2)α‐Int(f‐1(σ1σ2‐Cl(V))) for every σ1σ2-open subset V of Y(3)
(τ1, τ2)α‐Cl(f‐1(σ1σ2‐Int(K))) ⊆ f‐1(K) for every (σ1, σ2)r-closed subset K of Y(4)
(τ1, τ2)α‐Cl(f‐1(V)) ⊆ f‐1(σ1σ2‐Cl(V)) for every σ1σ2-open subset V of Y(5)
(τ1, τ2)α‐Cl(f‐1(σ1σ2‐Int((σ1, σ2)θ‐Cl(B)))) ⊆ f‐1((σ1, σ2)θ‐Cl(B)) for every subset B of Y(6)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(f‐1(V)))) ⊆ f‐1(σ1σ2‐Cl(V)) for every σ1σ2-open subset V of Y(7)
f‐1(V) ⊆ τ1‐Int(τ2‐Cl(τ1τ2‐Int(f‐1(σ1σ2‐Cl(V))))) for every σ1σ2-open subset V of Y(8)
f(τ1‐Cl(τ2‐Int(τ1τ2‐Cl(A)))) ⊆ (σ1, σ2)θ‐Cl(f(A)) for every subset A of X(9)
τ1‐Cl(τ2‐Int(τ1τ2‐Cl(f‐1(B)))) ⊆ f‐1((σ1, σ2)θ‐Cl(B)) for every subset B of YDefinition 17.
A subsetA of a bitopological space (X, τ1, τ2) is said to be τ1τ2-S-closed relative to X if for every cover {Vγ∣γ ∈Γ} of A by τ1τ2-open sets of X, there exists a finite subset Γ0 of Γ such that A ⊆ ∪{τ1τ2‐Cl(Vγ)∣γ ∈ Γ0}. If X is τ1τ2-S-closed relative to X, then the space (X, τ1, τ2) is called τ1τ2-S-closed.Theorem 23.
LetF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) be an upper weakly (τ1, τ2)α-continuous surjective multifunction such that F(x) is τ1τ2-compact for each x ∈ X. If (X, τ1, τ2) is (τ1, τ2)α-compact, then (Y, σ1, σ2) is σ1σ2-S-closed.Proof.
Let {Vγ∣γ ∈Γ} be any σ1σ2-open cover of Y. For each x ∈ X, F(x) is σ1σ2-compact, and hence there exists a finite subset Γ(x) of Γ such that(52)Fx⊆Vγγ∈Γx.
SinceF is upper weakly (τ1, τ2)α-continuous, there exists a (τ1, τ2)α-open subset of X such that F(U(x)) ⊆∪{σ1σ2‐Cl(Vγ)∣γ ∈Γ(x)}. Since (X, τ1, τ2) is (τ1, τ2)α-compact, there exists a finite number of points, say, x1, x2, …, xn in X such that X = ∪{U(xi)∣1 ≤ i ≤ n}. Thus, Y = F(X) = F(∪{U(xi)∣1 ≤ i ≤ n}) ⊆∪{σ1σ2‐Cl(Vγ)∣γ ∈ Γ(xi); 1 ≤ i ≤ n}. This shows that (Y, σ1, σ2) is σ1σ2-S-closed.Theorem 24.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) such that F(x) is σ1σ2-regular σ1σ2-paracompact for each x ∈ X, the following properties are equivalent:(1)
F is upper (τ1, τ2)α-continuous(2)
F is upper almost (τ1, τ2)α-continuous(3)
F is upper weakly (τ1, τ2)α-continuousProof.
We prove only the implication (3) ⟹ (1) since the others are obvious. Suppose thatF is upper weakly (τ1, τ2)α-continuous. Let x ∈ X and V be any σ1σ2-open subset of Y such that F(x) ⊆ V. Since F(x) is σ1σ2-regular σ1σ2-paracompact, there exists a σ1σ2-open subset G of Y such that F(x) ⊆ G ⊆ σ1σ2‐Cl(G) ⊆ V. Since F is upper weakly (τ1, τ2)α-continuous at x and F(x) ⊆ G, there exists a (τ1, τ2)α-open subset U of X containing x such that F(U) ⊆ σ1σ2‐Cl(G), and hence F(U) ⊆ V. This shows that F is upper (τ1, τ2)α-continuous.Lemma 13.
LetA be a subset of a bitopological space (X, τ1, τ2). If A is τ1τ2-regular, then for every τ1τ2-open set G which intersects A, there exists a τ1τ2-open set W such that A ∩ W ≠ ∅ and τ1τ2‐Cl(W) ⊆ G.Proof.
Suppose thatA is a τ1τ2-regular subset of X and G a τ1τ2-open set such that A ∩ G ≠ ∅. Let x ∈ A ∩ G. Then, we have x∉X –G and X – G is τ1τ2-closed. There exist disjoint τ1τ2-open subsets U and V of X such that x ∈ U and X –G ⊆ V, which implies τ1τ2 ‐ Cl(V) ∩ U =∅, and hence x∉τ1τ2-Cl(V). Let W = X– τ1τ2‐Cl(V). Thus, x ∈ W and A ∩ W ≠ ∅. Consequently, we obtain τ1τ2‐Cl(W) = τ1τ2‐Cl(X– τ1τ2‐Cl(V)) ⊆ τ1τ2‐Cl(X –V) = X –V ⊆ G.Theorem 25.
For a multifunctionF: (X, τ1, τ2) ⟶ (Y, σ1, σ2) such that F(x) is σ1σ2-regular for each x ∈ X, the following properties are equivalent:(1)
F is lower (τ1, τ2)α-continuous(2)
F is lower almost (τ1, τ2)α-continuous(3)
F is lower weakly (τ1, τ2)α-continuousProof.
We prove only the implication (3) ⟹ (1) since the others are obvious. Suppose thatF is lower weakly (τ1, τ2)α-continuous. Let x ∈ X and V be any σ1σ2-open subset of Y such that F(x) ∩ V ≠ ∅. Since F(x) is σ1σ2-regular, by Lemma 13, there exists a σ1σ2-open subset G of Y such that F(x) ∩ G ≠ ∅ and σ1σ2‐Cl(G) ⊆ V. Since F is upper weakly (τ1, τ2)α-continuous at x, there exists a (τ1, τ2)α-open subset U of X containing x such that σ1σ2‐Cl(G) ∩ F(z) ≠ ∅ for every z ∈ U and hence V ∩ F(z) ≠ ∅ for every z ∈ U as well, and this shows that F is lower (τ1, τ2)α-continuous at x. This shows that F is lower (τ1, τ2)α-continuous.



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