# Some Results ofτ1τ2-δ Semiconnectedness and Compactness in Bitopological Spaces

Authors: M. Arunmaran; K. Kannan
Journal: Journal of Mathematics
 (2018)

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## Abstract

We are going to establish some results ofτ1τ2-δ semiconnectedness and compactness in a bitopological space. Besides, we will investigate several results in τ1τ2-δ semiconnectedness for subsets in bitopological spaces. In particular, we will discuss the relationship related to semiconnectedness between the topological spaces and bitopological space. That is, if a bitopological space (X,τ1,τ2) is τ1τ2-δ semiconnected, then the topological spaces (X,τ1) and (X,τ2) are δ-semiconnected. In addition, we introduce the result which states that a bitopological space (X,τ1,τ2) is τ1τ2-δ semiconnected if and only if X and ϕ are the only subsets of X which are τ1τ2-δ semiclopen sets. Moreover, we have proved some results in compactness also. Altogether, several results of τ1τ2-δ semiconnectedness and compactness in a bitopological space have been discussed.


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## Body

## 1. Introduction

The concept “bitopological space” was established by Kelly in [1]. He introduced this concept in his journal of London Mathematical Society in 1963. He initiated his study about bitopological space by using quasimetric and its conjugate. A quasimetric on a set X is a nonnegative real valued function p(,) on the Cartesian product of X(X×X) that satisfies the following three axioms:(1)
p(x,x)=0,  ∀x∈X.(2)
p(x,z)≤p(x,y)+p(y,z),  ∀x,y,z∈X.(3)
p(x,y)=0 if and only if x=y,  ∀x,y∈X.However, the quasimetric cannot be a metric. Because the symmetric property does not hold for quasimetric. Moreover, every metric space is a topological space. But this is not true for bitopological space in general. Anyhow, bitopological spaces exist for quasimetric spaces. Maheshwari and Prasad [2] introduced semiopen sets in bitopological spaces in 1977. In 1987, the notion δ-open sets in bitopological spaces was introduced by Banerjee [3]. After that, Khedr [4] introduced and studied about τ1τ2- open sets. Later, Fukutake [5] defined one kind of semiopen sets in 1989. Recently, Edward Samuel and Balan [6] established the concept τ1τ2-δ semiopen sets in bitopological spaces. We have already published some properties of τ1τ2-δ semiopen/closed sets in bitopological spaces in [7]. Moreover, we have presented some results of τ1τ2-δ semiconnectedness in bitopological spaces in [8]. In this paper, we are going to discuss the following results:(1)
LetAα:α∈I be a family of τ1τ2-δ semiconnected subsets of a bitopological space (X,τ1,τ2) with ⋂Aα≠ϕ. Then ⋃α∈IAα is also τ1τ2-δ semiconnected.(2)
If a bitopological space(X,τ1s,τ2s) is τ1sτ2s-δ semiconnected, then (X,τ1,τ2) is τ1τ2-δ semiconnected.(3)
If a bitopological space(X,τ1,τ2) is τ1τ2-δ semiconnected, then the topological spaces (X,τ1) and (X,τ2) are δ-semiconnected.(4)
A bitopological space(X,τ1,τ2) is τ1τ2-δ semiconnected if and only if X and ϕ are the only subsets of X which are τ1τ2-δ semiclopen sets.(5)
In a bitopological space(X,τ1,τ2), if X is τ1τ2-δ semiconnected then X cannot be expressed as the union of two sets A(≠ϕ) and B(≠ϕ) with A∩B=ϕ such that A is τ1-δ semiopen and B is τ2-δ semiopen.(6)
In a bitopological space(X,τ1,τ2), if X cannot be expressed as the union of two nonempty sets A and B with A∩B=ϕ such that A is τ1-δ semiopen and B is τ2-δ semiopen, then X does not contain any nonempty proper subset which is both τ1-δ semiopen and τ2-δ semiclosed.(7)
The union of any family ofτ1τ2-δ semiconnected sets with a nonempty intersection is τ1τ2-δ semiconnected.(8)
Everyτ1τ2-δ semicompact space is τ1τ2-δ compact.(9)
IfY is τ1-δ semiclosed subset of a τ1τ2-δ semicompact space (X,τ1,τ2), then Y is τ2-δ semicompact.(10)
IfY is τ1-δ closed subset of a τ1τ2-δ semicompact space (X,τ1,τ2), then Y is τ2-δ semicompact.


## 2. Materials and Methods

Letτi-int(A), τi-cl(A), τi-δint(A), τi-δcl(A), and τi-δscl(A) be the interior, closure, δ-interior, δ-closure, and δ-semiclosure of A with respect to the topology τi, respectively, i=1,2. Let τj-δint(A) and τj-δcl(A) are the δ-interior and δ-closure of A with respect to the topology τj, respectively, j=1s,2s, where τ1s and τ2s are semiregularization of τ1 and τ2, respectively.Definition 1.
For a nonempty setX, we define two topologies τ1 and τ2  (τ1 and τ2 may be the same or distinct). Then the triple (X,τ1,τ2) is called a bitopological space.Definition 2 (see [1]).
LetA be subset of a bitopological space (X,τ1,τ2). Then A is called τ1τ2-open, if A∈τ1∪τ2. Complement of τ1τ2-open set is called τ1τ2-closed set.Definition 3 (see [9]).
LetA be subset of a bitopological space (X,τ1,τ2). Then A is called(1)
τ12-regular open, if A=τ1-int(τ2-cl(A));(2)
τ21-regular open, if A=τ2-int(τ1-cl(A));(3)
τ1τ2-semiopen, if A⊆τ2-cl(τ1-int(A));(4)
τ1τ2-semiclosed, if A⊇τ2-int(τ1-cl(A)).Definition 4 (see [9]).
LetA be subset of bitopological space (X,τ1,τ2). Then,(1)
A is said to be τ1-δ open set, if, for x∈A, there exists τ12-regular open set G such that x∈G⊂A. Complement of τ1-δ open set is called τ1-δ closed set;(2)
A is said to be τ2-δ open set, if for x∈A, there exists τ21-regular open set G such that x∈G⊂A. Complement of τ2-δ open set is called τ2-δ closed set;(3)
Collection of allτ1-δ open sets and τ2-δ open sets are denoted by τ1s and τ2s respectively. And also τ1s⊂τ1 and τ2s⊂τ2.Definition 5 (see [6]).
LetA be subset of bitopological space (X,τ1,τ2). Then, A is called τ1τ2-δ semiopen set, if there exists an τ1-δ open set U such that U⊆A⊆τ2-cl(U).
Complement ofτ1-δ open set is called τ1-δ closed set.Definition 6 (see [6]).
A subsetY is called a τ1τ2-δ semidisconnected subset of a bitopological space (X,τ1,τ2), if ∃τ1τ2-δ semiopen sets U,V such that U∩Y≠ϕ≠V∩Y, U∩V∩Y=ϕ and Y⊆U∪V. Otherwise Y is called a τ1τ2-δ semiconnected subset.Definition 7 (see [6]).
A bitopological space(X,τ1,τ2) is called τ1τ2-δ semiconnected space, if X cannot be expressed as the union of two disjoint sets A(≠ϕ) and B(≠ϕ) such that (A∩τ1-δscl(B))∪(B∩τ2-δscl(A))=ϕ. Suppose X can be so expressed, then X is called τ1τ2-δ semidisconnected space and we write X=A\B and it is called τ1τ2-δ semiseparation of X.Definition 8.
A nonempty collectionC=Ai:i∈I is called a τ1τ2-δ semiopen cover of a bitopological space (X,τ1,τ2), if X=⋃i∈IAi and C⊆τ1-δSO(X,τ1,τ2)∪τ2-δSO(X,τ1,τ2) and C contains at least one member of τ1-δSO(X,τ1,τ2) and one member of τ2-δSO(X,τ1,τ2).Definition 9.
A coverC=Ai:i∈I of a bitopological space is called τ1τ2-δ open cover of X, if C⊆τ1s∪τ2s and C∩τ1s≠ϕ, C∩τ2s≠ϕ and X=⋃C.Definition 10.
A bitopological space(X,τ1,τ2) is called τ1τ2-δ compact, if every τ1τ2-δ open cover of X has a finite subcover.Definition 11.
A bitopological space(X,τ1,τ2) is called τ1τ2-δ semicompact, if every τ1τ2-δ semiopen cover of X has a finite subcover.


## 3. Results

### 3.1. Semiconnectedness

Proposition 12 (see [8]).
LetAα:α∈I be family of τ1τ2-δ semiconnected subsets of a bitopological space (X,τ1,τ2) such that ⋂α∈IAα≠ϕ; then ⋃α∈IAα is also τ1τ2-δ semiconnected.Proof.
Letx0∈⋂α∈IAα. Now, assume that Y=⋃α∈IAα is not τ1τ2-δ semiconnected. Then there exist two τ1τ2-δ semiopen sets U and V if U∩Y≠ϕ≠V∩Y, U∩V∩Y=ϕ and Y⊆U∪V, then let x0∈U (other case is similar). Now there exist α∈I such that Aα∩V≠ϕ also Aα∩U≠ϕ (since x0∈Aα) and also U∩V∩Aα=ϕ and Aα⊂U∪V (since Aα⊂Y), which shows that Aα is τ1τ2-δ semidisconnected subset and it is a contradiction. So, ⋃α∈IAα is τ1τ2-δ semiconnected.Proposition 13.
If a bitopological space(X,τ1s,τ2s) is τ1sτ2s-δ semiconnected, then (X,τ1,τ2) is τ1τ2-δ semiconnected.Proof.
Suppose(X,τ1s,τ2s) is τ1sτ2s-δ semiconnected. Then X cannot be expressed as the union of two nonempty disjoint sets A and B such that (A∩τ1s-δscl(B))∪(B∩τ2s-δscl(A))=ϕ. Also A is τ1s-δ semiopen and B is τ2s-δ semiopen. Since τ1s⊆τ1 and τ2s⊆τ2, we have every τ1s-δ semiopen and τ2s-δ semiopen are τ1-δ semiopen and τ2-δ semiopen, respectively. Therefore, X cannot be expressed as the union of two nonempty disjoint sets A and B such that A is τ1-δ semiopen and B is τ2-δ semiopen, respectively. Hence, (X,τ1,τ2) is τ1τ2-δ semiconnected.Proposition 14 (see [8]).
If a bitopological space(X,τ1,τ2) is τ1τ2-δ semiconnected, then the topological spaces (X,τ1) and (X,τ2) are δ-semiconnected.Proof.
Since everyτ1-δ open set and τ2-δ open set are τ1-δ semiopen set and τ2-δ semiopen set, respectively, So if (X,τ1) and (X,τ2) are δ-semidisconnected spaces then the bitopological space (X,τ1,τ2) becomes δ-semidisconnected. But this is impossible. So (X,τ1) and (X,τ2) are δ-semiconnected spaces.Proposition 15 (see [8]).
A bitopological space(X,τ1,τ2) is τ1τ2-δ semiconnected if and only if X and ϕ are the only subsets of X which are τ1τ2-δ semiclopen sets (simultaneously semiopen and semiclosed.)Proof.
Consider aτ1τ2-δ semiconnected space (X,τ1,τ2); let ϕ≠A≠X and A is τ1τ2-δ semiclopen set; then X=A∪(X\A) is τ1τ2-δ semidisconnected in the bitopological space, which is contradiction. So X and ϕ are the only subsets of X which are both τ1τ2-δ semiclopen sets. Conversely, let X and ϕ be the only subsets of X which are both τ1τ2-δ semiclopen sets. If the bitopological space is τ1τ2-δ semidisconnected, so there exists a τ1τ2-δ semidisconnection X=A∪B of the bitopological space. So A=X\B and B=X\A; then A and B both are τ1τ2-δ semiclopen sets and each of them is neither X nor ϕ. This is a contradiction. Therefore, (X,τ1,τ2) is τ1τ2-δ semiconnected.Proposition 16.
IfX is τ1τ2-δ semiconnected then X cannot be expressed as the union of two nonempty sets A,B with A∩B=ϕ such that A is τ1-δ semiopen and B is τ2-δ semiopen.Proof.
Assume thatX can be expressed as the union of two nonempty disjoint sets A and B such that A is τ1-δ semiopen and B is τ2-δ semiopen, respectively. Since A∩B=ϕ, we have A⊆Bc. Then τ2-δscl(A)⊆τ2-δscl(Bc)=Bc. Therefore, τ2-δscl(A)∩B=ϕ. Similarly, we can prove A∩τ1-δscl(B)=ϕ. Hence, (A∩τ1-δscl(B))∪(τ2-δscl(A)∩B)=ϕ. This contradicts our supposition. Thus, X cannot be expressed as the union of two nonempty disjoint sets A and B such that A is τ1-δ semiopen and B is τ2-δ semiopen.Proposition 17.
IfX cannot be expressed as the union of two disjoint sets A(≠ϕ) and B(≠ϕ) such that A is τ1-δ semiopen and B is τ2-δ semiopen, then X does not contain any nonempty proper subset which is both τ1-δ semiopen and τ2-δ semiclosed.Proof.
LetX cannot be expressed as the union of two nonempty sets A,B with A∩B=ϕ such that A is τ1-δ semiopen and B is τ2-δ semiopen. If X contains a nonempty proper subset A which is both τ1-δ semiopen and τ2-δ semiclosed. Then X=A∪Ac, where A is τ1-δ semiopen, Ac is τ2-δ semiopen, and A,Ac are disjoint. This is a contradiction. Thus, X does not contain any nonempty proper subset which is both τ1-δ semiopen and τ2-δ semiclosed.Proposition 18.
The union of any family ofτ1τ2-δ semiconnected sets with a nonempty intersection is τ1τ2-δ semiconnected.Proof.
TakeA=⋃i∈IAi, where each Ai is τ1τ2-δ semiconnected with ⋂Ai≠ϕ. Suppose that A is not τ1τ2-δ semiconnected. Then A=C∪D, where C and D are two nonempty disjoint sets such that (C∩τ1-δscl(D))∪(τ2-δscl(C)∩D)=ϕ. Since Ai is τ1τ2-δ semiconnected and Ai⊆A, we have Ai⊆C or Ai⊆D. Therefore, ⋃Ai⊆C or ⋃Ai⊆D as A=⋃Ai=C∪D. Hence, A⊆C or A⊆D. Since ⋂Ai≠ϕ, we have x∈⋂Ai. Therefore, x∈Ai,  ∀i. So, x∈A. Therefore, x∈C or x∈D. Suppose x∈C. Since C∩D=ϕ, we have x∉D. Therefore, A⊈D. Thus, A⊆C. This contradicts A=C∪D. Thus, A is τ1τ2-δ semiconnected.


### 3.2. Semicompactness

Proposition 19.
Everyτ1τ2-δ semicompact space is τ1τ2-δ compact.Proof.
Take(X,τ1,τ2) to be τ1τ2-δ semicompact. Let C=Ai:i∈I be a pairwise open cover of X. Then X=⋃i∈IAi and C⊆τ1s∪τ2s and C contains at least one member of τ1s and one member of τ2s. Since every τ1-δ open set is τ1-δ semiopen, we have X=⋃i∈IAi and C⊆τ1-δSO(X,τ1,τ2)∪τ2-δSO(X,τ1,τ2) and C contains at least one member of τ1-δSO(X,τ1,τ2) and one member of τ2-δSO(X,τ1,τ2). Therefore, C is τ1τ2-δ semiopen cover of X. Since X is τ1τ2-δ semicompact, C has a finite subcover. Thus, (X,τ1,τ2) is τ1τ2-δ compact.Proposition 20.
IfY is τ1-δ semiclosed subset of a τ1τ2-δ semicompact space (X,τ1,τ2) then Y is τ2-δ semicompact.Proof.
Let(X,τ1,τ2) be a τ1τ2-δ semicompact space. Let Aα:α∈I be τ2-δ semiopen cover of Y. Since Y is τ1-δ semiclosed, Yc is τ1-δ semiopen. And Yc∪Aα:α∈I be a τ1τ2-δ semiopen cover of X. Since X is τ1τ2-δ semicompact, X=Yc∪⋃i=1nAi. Hence, Y=⋃i=1nAi. Thus, Y is τ2-δ semicompact.Proposition 21.
IfY is τ1-δ closed subset of a τ1τ2-δ semicompact space (X,τ1,τ2) then Y is τ2-δ semicompact.Proof.
Since everyτ1-δ closed set is τ1-δ semiclosed, Y is τ1-δ semiclosed. Then by applying Proposition 20, Y is τ2-δ semicompact.


## 3.1. Semiconnectedness

Proposition 12 (see [8]).
LetAα:α∈I be family of τ1τ2-δ semiconnected subsets of a bitopological space (X,τ1,τ2) such that ⋂α∈IAα≠ϕ; then ⋃α∈IAα is also τ1τ2-δ semiconnected.Proof.
Letx0∈⋂α∈IAα. Now, assume that Y=⋃α∈IAα is not τ1τ2-δ semiconnected. Then there exist two τ1τ2-δ semiopen sets U and V if U∩Y≠ϕ≠V∩Y, U∩V∩Y=ϕ and Y⊆U∪V, then let x0∈U (other case is similar). Now there exist α∈I such that Aα∩V≠ϕ also Aα∩U≠ϕ (since x0∈Aα) and also U∩V∩Aα=ϕ and Aα⊂U∪V (since Aα⊂Y), which shows that Aα is τ1τ2-δ semidisconnected subset and it is a contradiction. So, ⋃α∈IAα is τ1τ2-δ semiconnected.Proposition 13.
If a bitopological space(X,τ1s,τ2s) is τ1sτ2s-δ semiconnected, then (X,τ1,τ2) is τ1τ2-δ semiconnected.Proof.
Suppose(X,τ1s,τ2s) is τ1sτ2s-δ semiconnected. Then X cannot be expressed as the union of two nonempty disjoint sets A and B such that (A∩τ1s-δscl(B))∪(B∩τ2s-δscl(A))=ϕ. Also A is τ1s-δ semiopen and B is τ2s-δ semiopen. Since τ1s⊆τ1 and τ2s⊆τ2, we have every τ1s-δ semiopen and τ2s-δ semiopen are τ1-δ semiopen and τ2-δ semiopen, respectively. Therefore, X cannot be expressed as the union of two nonempty disjoint sets A and B such that A is τ1-δ semiopen and B is τ2-δ semiopen, respectively. Hence, (X,τ1,τ2) is τ1τ2-δ semiconnected.Proposition 14 (see [8]).
If a bitopological space(X,τ1,τ2) is τ1τ2-δ semiconnected, then the topological spaces (X,τ1) and (X,τ2) are δ-semiconnected.Proof.
Since everyτ1-δ open set and τ2-δ open set are τ1-δ semiopen set and τ2-δ semiopen set, respectively, So if (X,τ1) and (X,τ2) are δ-semidisconnected spaces then the bitopological space (X,τ1,τ2) becomes δ-semidisconnected. But this is impossible. So (X,τ1) and (X,τ2) are δ-semiconnected spaces.Proposition 15 (see [8]).
A bitopological space(X,τ1,τ2) is τ1τ2-δ semiconnected if and only if X and ϕ are the only subsets of X which are τ1τ2-δ semiclopen sets (simultaneously semiopen and semiclosed.)Proof.
Consider aτ1τ2-δ semiconnected space (X,τ1,τ2); let ϕ≠A≠X and A is τ1τ2-δ semiclopen set; then X=A∪(X\A) is τ1τ2-δ semidisconnected in the bitopological space, which is contradiction. So X and ϕ are the only subsets of X which are both τ1τ2-δ semiclopen sets. Conversely, let X and ϕ be the only subsets of X which are both τ1τ2-δ semiclopen sets. If the bitopological space is τ1τ2-δ semidisconnected, so there exists a τ1τ2-δ semidisconnection X=A∪B of the bitopological space. So A=X\B and B=X\A; then A and B both are τ1τ2-δ semiclopen sets and each of them is neither X nor ϕ. This is a contradiction. Therefore, (X,τ1,τ2) is τ1τ2-δ semiconnected.Proposition 16.
IfX is τ1τ2-δ semiconnected then X cannot be expressed as the union of two nonempty sets A,B with A∩B=ϕ such that A is τ1-δ semiopen and B is τ2-δ semiopen.Proof.
Assume thatX can be expressed as the union of two nonempty disjoint sets A and B such that A is τ1-δ semiopen and B is τ2-δ semiopen, respectively. Since A∩B=ϕ, we have A⊆Bc. Then τ2-δscl(A)⊆τ2-δscl(Bc)=Bc. Therefore, τ2-δscl(A)∩B=ϕ. Similarly, we can prove A∩τ1-δscl(B)=ϕ. Hence, (A∩τ1-δscl(B))∪(τ2-δscl(A)∩B)=ϕ. This contradicts our supposition. Thus, X cannot be expressed as the union of two nonempty disjoint sets A and B such that A is τ1-δ semiopen and B is τ2-δ semiopen.Proposition 17.
IfX cannot be expressed as the union of two disjoint sets A(≠ϕ) and B(≠ϕ) such that A is τ1-δ semiopen and B is τ2-δ semiopen, then X does not contain any nonempty proper subset which is both τ1-δ semiopen and τ2-δ semiclosed.Proof.
LetX cannot be expressed as the union of two nonempty sets A,B with A∩B=ϕ such that A is τ1-δ semiopen and B is τ2-δ semiopen. If X contains a nonempty proper subset A which is both τ1-δ semiopen and τ2-δ semiclosed. Then X=A∪Ac, where A is τ1-δ semiopen, Ac is τ2-δ semiopen, and A,Ac are disjoint. This is a contradiction. Thus, X does not contain any nonempty proper subset which is both τ1-δ semiopen and τ2-δ semiclosed.Proposition 18.
The union of any family ofτ1τ2-δ semiconnected sets with a nonempty intersection is τ1τ2-δ semiconnected.Proof.
TakeA=⋃i∈IAi, where each Ai is τ1τ2-δ semiconnected with ⋂Ai≠ϕ. Suppose that A is not τ1τ2-δ semiconnected. Then A=C∪D, where C and D are two nonempty disjoint sets such that (C∩τ1-δscl(D))∪(τ2-δscl(C)∩D)=ϕ. Since Ai is τ1τ2-δ semiconnected and Ai⊆A, we have Ai⊆C or Ai⊆D. Therefore, ⋃Ai⊆C or ⋃Ai⊆D as A=⋃Ai=C∪D. Hence, A⊆C or A⊆D. Since ⋂Ai≠ϕ, we have x∈⋂Ai. Therefore, x∈Ai,  ∀i. So, x∈A. Therefore, x∈C or x∈D. Suppose x∈C. Since C∩D=ϕ, we have x∉D. Therefore, A⊈D. Thus, A⊆C. This contradicts A=C∪D. Thus, A is τ1τ2-δ semiconnected.


## 3.2. Semicompactness

Proposition 19.
Everyτ1τ2-δ semicompact space is τ1τ2-δ compact.Proof.
Take(X,τ1,τ2) to be τ1τ2-δ semicompact. Let C=Ai:i∈I be a pairwise open cover of X. Then X=⋃i∈IAi and C⊆τ1s∪τ2s and C contains at least one member of τ1s and one member of τ2s. Since every τ1-δ open set is τ1-δ semiopen, we have X=⋃i∈IAi and C⊆τ1-δSO(X,τ1,τ2)∪τ2-δSO(X,τ1,τ2) and C contains at least one member of τ1-δSO(X,τ1,τ2) and one member of τ2-δSO(X,τ1,τ2). Therefore, C is τ1τ2-δ semiopen cover of X. Since X is τ1τ2-δ semicompact, C has a finite subcover. Thus, (X,τ1,τ2) is τ1τ2-δ compact.Proposition 20.
IfY is τ1-δ semiclosed subset of a τ1τ2-δ semicompact space (X,τ1,τ2) then Y is τ2-δ semicompact.Proof.
Let(X,τ1,τ2) be a τ1τ2-δ semicompact space. Let Aα:α∈I be τ2-δ semiopen cover of Y. Since Y is τ1-δ semiclosed, Yc is τ1-δ semiopen. And Yc∪Aα:α∈I be a τ1τ2-δ semiopen cover of X. Since X is τ1τ2-δ semicompact, X=Yc∪⋃i=1nAi. Hence, Y=⋃i=1nAi. Thus, Y is τ2-δ semicompact.Proposition 21.
IfY is τ1-δ closed subset of a τ1τ2-δ semicompact space (X,τ1,τ2) then Y is τ2-δ semicompact.Proof.
Since everyτ1-δ closed set is τ1-δ semiclosed, Y is τ1-δ semiclosed. Then by applying Proposition 20, Y is τ2-δ semicompact.


## 4. Discussion

In this paper, we have used the result that everyτ1-δ closed set is τ1-δ semiclosed. Moreover, if A and B are two τ1τ2-δ semicompact subsets of X, then A∪B is also τ1τ2-δ semicompact. Besides, every τ1τ2-δ semicompact space is τ1τ2-δ compact. The concept semiconnectedness and compactness is used in various parts of Mathematics. Simultaneously, the bitopological spaces have several applications in analysis, general topology, and theory of ordered topological spaces.


## 5. Concluding Remarks

In this paper, some results ofτ1τ2-δ semiconnectedness and compactness in bitopological spaces have been discussed. Simultaneously, we have some important results which are related to connectedness and compactness. So we are interested to check whether those results will work for bitopological space or not. If the results fail to hold for bitopology, we are going to illustrate by examples. Further, we want to find how uniform continuity will work in bitopological spaces.



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