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ISSN 2319-8133 (Online) (An International Research Journal), www.compmath-journal.org

A Study on 𝝉 𝟏 𝝉 𝟐 -𝒈 ̂ ∗ 𝒔-closed Sets and 𝝉 𝟏 𝝉 𝟐 -𝒈 ̂ ∗ 𝒔-open Sets in Bitopological Spaces

S. Poius Missier

1

and M. Anto

2

1

Department of Mathematics,

V.O.Chidambaram College of Arts and Science, Thoothukudi, INDIA.

email:[email protected]

2

Associate Professor, Department of Mathematics,

Annai Velankanni College, Tholayavattam, INDIA.

email: [email protected].

(Received on: October 30, 2015) ABSTRACT

In this paper, we introduce 𝜏1𝜏2 -𝑔̂∗𝑠-closed sets. The Properties of 𝜏1𝜏2 - 𝑔̂∗𝑠-closed sets and 𝜏1𝜏2 -𝑔̂∗𝑠-open sets are studied.

Keywords: 𝜏1𝜏2 -𝑔̂∗𝑠-closed sets, 𝜏1𝜏2 -𝑔̂∗𝑠-closure operator and 𝜏1𝜏2 -𝑔̂∗𝑠-open sets.

1. INTRODUCTION

Norman Levine introduced the notion of semi open sets

10

and generalised closed sets (briefly, 𝑔-closed)

8

in 1963 and 1970 respectively. J.C.Kelly

5

introduced the concept of bitopological spaces in the year 1963. Semi open sets and generalised closed sets in a bitopological space were introduced by T.Fukutake in the years 1989

3

and 1985

2

respectively.

After that several authors turned their attention towards generalisations of various concepts of topology in bitopological settings. S. Poius Missier and M. Anto

13

introduced a new class of sets called 𝑔̂

∗

𝑠-closed sets by replacing the closure operator in Levine’s Definition[8] by semi closure operator and replacing openness of the super set with 𝑔̂-openness. In this paper, we introduce 𝑔̂

∗

𝑠-closed sets in a bitopological setting. The aim of this paper is to study the topological properties of 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠-closed sets and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open sets.

2. PRELIMINERIES

Definition 2.1: A subset 𝐴 of a space (𝑋, 𝜏) is called

(2)

(i) semi-open

10

if 𝐴 ⊆ 𝑐𝑙(𝑖𝑛𝑡(𝐴)) and semi-closed if 𝐴

𝑐

is semi open, (ii) pre-open

11

if 𝐴 ⊆ 𝑖𝑛𝑡(𝑐𝑙(𝐴)) and pre-closed if 𝐴

𝑐

is pre open,

(iii) regular-open

7

if 𝐴 = 𝑖𝑛𝑡(𝑐𝑙(𝐴)) and regular closed if 𝐴

𝑐

is regular open.

Definition 2.2:A subset 𝐴 of a space (𝑋, 𝜏) is called

(i) 𝑔-closed

8

if 𝑐𝑙(𝐴) ⊆ 𝑈 whenever 𝐴 ⊆ 𝑈 and 𝑈 is open, (ii) 𝑔̂-closed

17

if 𝑐𝑙(𝐴) ⊆ 𝑈 whenever 𝐴 ⊆ 𝑈 and 𝑈 is semi-open.

Definition 2.3: A subset 𝐴 of a space (𝑋, 𝜏

1

, 𝜏

2

) is called

(i) 𝜏

1

𝜏

2

-semi open

3

if 𝐴 ⊆ 𝜏

2

-𝑐𝑙(𝜏

1

𝑖𝑛𝑡(𝐴)) and 𝜏

1

𝜏

2

-semi-closed if 𝜏

2

-𝑖𝑛𝑡(𝜏

1

-𝑐𝑙(𝐴)) ⊆ 𝐴.

(ii) 𝜏

1

𝜏

2

-pre open

4

if 𝐴 ⊆ 𝜏

1

-𝑖𝑛𝑡(𝜏

2

− 𝑐𝑙(𝐴)) and pre-closed if 𝜏

1

-𝑐𝑙(𝜏

2

-𝑖𝑛𝑡(𝐴)) ⊆ 𝐴 and (iii) 𝜏

1

𝜏

2

- regular open [1] if 𝐴 = 𝜏

1

-𝑖𝑛𝑡(𝜏

2

− 𝑐𝑙(𝐴)) and 𝜏

1

𝜏

2

- regular closed[1] if 𝐴 = 𝜏

1

-

𝑐𝑙(𝜏

2

-𝑖𝑛𝑡(𝐴))

The 𝜏

1

𝜏

2

-semi-closure (respectively 𝜏

1

𝜏

2

-pre closure and 𝜏

1

𝜏

2

-𝛼-closure) of a subset 𝐴 of a space (𝑋, 𝜏

1

, 𝜏

2

) is the intersection of all 𝜏

1

𝜏

2

-semi-closed (respectively 𝜏

1

𝜏

2

-pre closed and 𝜏

1

𝜏

2

-𝛼-closed) sets containing 𝐴 and is denoted by 𝜏

1

𝜏

2

-𝑠𝑐𝑙(𝐴) (resp. 𝜏

1

𝜏

2

-pcl (A) and 𝜏

1

𝜏

2

-𝛼-cl (A)).

Definition 2.4

6

: A subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is called 𝜏

1

𝜏

2

-𝑐𝑙open if A is both 𝜏

1

-closed and 𝜏

2

-open. A is called 𝜏

2

𝜏

1

-𝑐𝑙open if A is both 𝜏

2

-closed and 𝜏

1

-open. A is called pairwise-𝑐𝑙open if it is both 𝜏

1

𝜏

2

-𝑐𝑙open and 𝜏

2

𝜏

1

-𝑐𝑙open.

Lemma 2.5

16

: Arbitrary intersection (union) of 𝑔̂-closed sets is again 𝑔̂-closed.

Lemma 2.6

10

: Arbitrary intersection (union) of semi-closed sets is again semi-closed Lemma 2.7

12

: For a subset 𝐴 of a topological space (𝑋, 𝜏), 𝑠𝑖𝑛𝑡 (𝐴) = 𝑋-𝑠𝑐𝑙(𝐴

𝑐

) Notations Used

 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝐶(𝑋, 𝜏

1

, 𝜏

2

) denotes the class of all 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed subsets of (𝑋, 𝜏

1

, 𝜏

2

)

 𝜏

2

𝜏

1

-𝑔̂

∗

𝑠𝐶(𝑋, 𝜏

1

, 𝜏

2

) denotes the class of all 𝜏

2

𝜏

1

-𝑔̂

∗

𝑠-closed subsets of (𝑋, 𝜏

1

, 𝜏

2

)

 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑂(𝑋, 𝜏

1

, 𝜏

2

) denotes the class of all 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open subsets of (𝑋, 𝜏

1

, 𝜏

2

)

 𝜏

2

-𝑠𝑐𝑙(𝐴) denotes the closure of semi closed subset 𝐴 of 𝑋

 𝑃(𝑋) denotes the class of all subsets of (𝑋, 𝜏

1

, 𝜏

2

)

 Either 𝑋 − 𝐴 or 𝐴

𝑐

denotes the complement of 𝐴 in 𝑋.

3. 𝝉

𝟏

𝝉

𝟐

-𝒈 ̂

∗

𝒔-CLOSED SETS

Definition 3.1

14

: A subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is called a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed set if 𝜏

2

-𝑠𝑐𝑙(𝐴) ⊆ 𝑈 whenever 𝐴 ⊆ 𝑈 and 𝑈 is 𝜏

1

-𝑔̂-open. A subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is called a 𝜏

2

𝜏

1

-𝑔̂

∗

𝑠-closed set if 𝜏

1

-𝑠𝑐𝑙(𝐴) ⊆ 𝑈 whenever 𝐴 ⊆ 𝑈 and 𝑈 is 𝜏

2

-𝑔̂- open.

If a subset 𝐴 is both 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed and 𝜏

2

𝜏

1

-𝑔̂

∗

𝑠-closed in (𝑋, 𝜏

1

, 𝜏

2

), then 𝐴 is

called pairwise 𝑔̂

∗

𝑠-closed.

(3)

S. Poius Missier, et al., J. Comp. & Math. Sci. Vol.6(11), 621-630 (2015) 623

Example 3.2: Let (𝑋, 𝜏

1

, 𝜏

2

) be a bitopological space where 𝑋 = {𝑎, 𝑏, 𝑐, 𝑑} with 𝜏

1

= {∅, 𝑋, {𝑎}, {𝑏}, {𝑎, 𝑏}} and 𝜏

2

= {∅, 𝑋, {𝑎}, {𝑐}, {𝑎, 𝑐}}. Then 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝐶(𝑋, 𝜏

1

, 𝜏

2

) = 𝑃(𝑋) and 𝜏

2

𝜏

1

-𝑔̂

∗

𝑠𝐶(𝑋, 𝜏

1

, 𝜏

2

) = 𝑃(𝑋). All the subsets of (𝑋, 𝜏

1

, 𝜏

2

) are pairwise 𝑔̂

∗

𝑠-closed sets.

Definition 3.3

14

: A subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is called a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open set iff 𝑋 − 𝐴 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

A subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is called a 𝜏

2

𝜏

1

-𝑔̂

∗

𝑠-open set iff 𝑋 − 𝐴 is 𝜏

2

𝜏

1

- 𝑔̂

∗

𝑠-closed.

A subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is called a pairwise 𝑔̂

∗

𝑠-open set iff 𝑋 − 𝐴 is both 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open and 𝜏

2

𝜏

1

-𝑔̂

∗

𝑠-open.

Definition 3.4: All subsets of the bitopological space (𝑋, 𝜏

1

, 𝜏

2

), given in example 3.2, are pairwise 𝑔̂

∗

𝑠-open.

Definition 3.5 : The 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closure of a subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is the intersection of all 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed sets containing 𝐴 and is denoted by 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙(𝐴).

Definition 3.6 : An element x of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is called a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-limit point of a subset A of X, if for each 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open set U containing x, (𝐴 − {𝑥} ∩ 𝑈 ≠ ∅. The set of all 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-limit points of 𝐴, denoted by 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴), is called 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-derived set of 𝐴.

Proposition 3.7 : Let 𝐴 and 𝐵 be subsets of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

). If 𝐴 ⊆ 𝐵, Then (i) 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐵).

(ii) 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐵).

Proof. (i) Follows from definition 3.5 (ii) Follows from definition 3.6

Proposition 3.8 : An element 𝑥 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) iff for every 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open set 𝑈.

Containing 𝑥, 𝐴 ∩ 𝑈 ≠ ∅.

Proof. NECESSITY

Let 𝑥 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) and 𝑈 be a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open set containing 𝑥. Suppose 𝐴 ∩ 𝑈 =

∅. Then 𝐴 ⊆ (𝑋 − 𝑈) and since (𝑋 − 𝑈) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed, we have 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ⊆ 𝑈.

Thus 𝑥 ∈ (𝑋 − 𝑈) which is a contradiction. Therefore 𝐴 ∩ 𝑈 ≠ ∅.

SUFFICIENCY

Suppose for every 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open set 𝑈 containing 𝑥, 𝐴 ∩ 𝑈 ≠ ∅. Let 𝑥 ∉ 𝜏

1

𝜏

2

-

𝑔̂

∗

𝑠𝑐𝑙 (𝐴). Then there exists 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed set 𝐹 in 𝑋 such that 𝐴 ⊆ 𝐹 and 𝑥 ∉ 𝐹. Hence

𝑥 ∈ (𝑋 − 𝐹) where 𝑋 − 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open and (𝑋 − 𝐹) ∩ 𝐴 = ∅ which is a contradiction to

the fact that for every 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open set 𝑈 containing 𝑥, 𝐴 ∩ 𝑈 ≠ ∅. Therefore 𝑥 ∈ 𝜏

1

𝜏

2

-

𝑔̂

∗

𝑠𝑐𝑙 (𝐴). Hence proved.

(4)

Proposition 3.9 : If 𝐴 and 𝐵 are subsets of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

), the following are true.

(i) 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐵) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴 ∪ 𝐵) (ii) 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐵) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴 ∪ 𝐵) (iii) 𝐴 ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴)

(iv) 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) = 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝜏

1

𝜏

2

− 𝑔̂

∗

𝑠𝑐𝑙 (𝐴))

Proof.

(i) Let 𝐴 and 𝐵 be subsets of 𝑋. Since 𝐴 ⊆ 𝐴 ∪ 𝐵 and 𝐵 ⊆ 𝐴 ∪ 𝐵, by 3.7 (ii), 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴 ∪ 𝐵) and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐵) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴 ∪ 𝐵). Hence 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐵) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴 ∪ 𝐵).

(ii) Similar to (i).

(iii) Follows from the Definition 3.5

(iv) By 3.9 (iii) and 3.7 (i), 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙(𝜏

1

𝜏

2

− 𝑔̂

∗

𝑠𝑐𝑙 (𝐴)). Let 𝑥 ∉ 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠𝑐𝑙 (𝐴). Then by 3.8, there exists a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠 open set 𝑈 of 𝑋 containing 𝑥 such that 𝐴 ∩ 𝑈 = ∅. Suppose that 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ∩ 𝑈 ≠ ∅. Then 𝑦 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ∩ 𝑈 and hence 𝑦 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴). Then for every 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠 open set containing 𝑦 intersects 𝐴.

Hence 𝐴 ∩ 𝑈 ≠ ∅ which is a contradiction. Thus 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ∩ 𝑈 = ∅. Hence 𝑥 ∉ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙(𝜏

1

𝜏

2

− 𝑔̂

∗

𝑠𝑐𝑙 (𝐴)). Hence 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙(𝜏

1

𝜏

2

− 𝑔̂

∗

𝑠𝑐𝑙 (𝐴)) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴).

Hence 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) = 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙(𝜏

1

𝜏

2

− 𝑔̂

∗

𝑠𝑐𝑙 (𝐴)).

Proposition 3.10 : If 𝐴 is a subset of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

), then 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) = 𝐴 ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴).

Proof.

First we shall prove that 𝐴 ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴). Let 𝑥 ∈ 𝐴 ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴).

Then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴). Suppose 𝑥 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴). Then every 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open set containing 𝑥 intersects 𝐴 in a point different from 𝑥. Then by 3.8, 𝑥 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴).

Thus 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴). By 3.9 (iii), 𝐴 ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴). Therefore 𝐴 ∪ 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠𝑑 (𝐴) ⊆ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴).

Conversely, let 𝑥 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴). To prove 𝑥 ∈ 𝐴 ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴). If 𝑥 ∈ 𝐴, clearly 𝑥 ∈ 𝐴 ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴). Suppose 𝑥 ∉ 𝐴. We shall prove 𝑥 ∈ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴). Suppose 𝑥 ∉ 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠𝑑 (𝐴). Then there exists a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠 open set 𝑈 containing 𝑥 such that (𝐴 − {𝑥}) ∩ 𝑈 = ∅.

Since 𝑥 ∉ 𝐴, we have 𝐴 ∩ 𝑈 = ∅.

 𝐴 ⊆ (𝑋 − 𝑈) and (𝑋 − 𝑈) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ⊆ (𝑋 − 𝑈)

 𝑥 ∉ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴)

 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ⊆ 𝐴 ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴).

Hence proved.

(5)

S. Poius Missier, et al., J. Comp. & Math. Sci. Vol.6(11), 621-630 (2015) 625 Remark 3.11

The following examples show that, in general,

(i) 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐵) ≠ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴 ∪ 𝐵) (ii) 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐵) ≠ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴 ∪ 𝐵) Example 3.12

Let

(𝑋, 𝜏1, 𝜏2)

be a bitopological space where

𝑋 = {𝑎, 𝑏, 𝑐, 𝑑}

with

𝜏1= {∅, 𝑋, {𝑎}, {𝑎, 𝑏}, {𝑎, 𝑏, 𝑐}}

and 𝜏

2

= {∅, 𝑋, {𝑎}, {𝑏}, {𝑎, 𝑏}}. Let 𝐴 = {𝑎}; 𝐵 = {𝑏, 𝑐} and 𝐴 ∪ 𝐵 = {𝑎, 𝑏, 𝑐}. Then 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠𝑑 (𝐴) = ∅; 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐵) = ∅; 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴 ∪ 𝐵) = {𝑑} and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ∪ 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠𝑑 (𝐵) = ∅. Thus 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴 ∪ 𝐵) ≠ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑑 (𝐵).

Example 3.13

Let

(𝑋, 𝜏1, 𝜏2)

be a bitopological space where 𝑋 = {𝑎, 𝑏, 𝑐, 𝑑} with

𝜏1= {∅, 𝑋, {𝑎}, {𝑎, 𝑏}, {𝑎, 𝑏, 𝑐}}

and 𝜏

2

= {∅, 𝑋, {𝑎}, {𝑏}, {𝑎, 𝑏}}.

Let Let 𝐴 = {𝑎}; 𝐵 = {𝑏, 𝑐} and 𝐴 ∪ 𝐵 = {𝑎, 𝑏, 𝑐}. Then 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) = {𝑎}; 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠𝑐𝑙 (𝐵) = {𝑏}; 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴 ∪ 𝐵) = {𝑎, 𝑏, 𝑐, 𝑑} and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐵) = {𝑎, 𝑏, 𝑐}. Thus 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴 ∪ 𝐵) ≠ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐴) ∪ 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠𝑐𝑙 (𝐵).

Proposition 3.14

Every 𝜏

2

-semi closed set in (𝑋, 𝜏

1

, 𝜏

2

) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

Proof.

Let 𝐴 ⊆ 𝑋 be a 𝜏

2

-semi closed set. Let 𝐴 ⊆ 𝑈 be such that 𝑈 is 𝜏

1

-𝑔̂-open. Since, by assumption, 𝜏

2

-𝑠𝑐𝑙(𝐴) = 𝐴 and hence 𝜏

2

-𝑠𝑐𝑙(𝐴) ⊆ 𝑈. Thus 𝐴 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

Proposition 3.15

If 𝐴 is both 𝜏

1

-𝑔̂-open and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed, then 𝐴 is 𝜏

2

-semi closed.

Proof.

Let 𝐴 ⊆ 𝐴 and 𝐴 𝜏

1

-𝑔̂-open. Since, 𝐴 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed, 𝜏

2

-𝑠𝑐𝑙(𝐴) ⊆ 𝐴. Therefore, 𝜏

2

- 𝑠𝑐𝑙(𝐴) = 𝐴. Therefore, 𝐴 is 𝜏

2

-semi closed. Hence the proof.

Proposition 3.16

Let (𝑋, 𝜏

1

, 𝜏

2

) be a bitopological space and 𝐴 be a subspace of 𝑋 such that 𝑈 ∩ 𝐴 is 𝜏

1

-𝑔̂-open in 𝐴 for every 𝜏

1

-𝑔̂-open set 𝑈 in 𝑋. In addition, let 𝐴 be

𝜏

1

-𝑔̂-open and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋 such that 𝐹 ⊆ 𝐴 ⊆ 𝑋. Then 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴 iff 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋.

Proof. NECESSITY

Let 𝐹 be a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴. Let 𝑈 be a 𝜏

1

-𝑔̂-open subset of 𝑋 such that 𝐹 ⊆ 𝑈. Then, by

assumption, 𝐹 ⊆ 𝑈 ∩ 𝐴 and 𝑈 ∩ 𝐴 is 𝜏

1

-𝑔̂-open in 𝐴. Since 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴, we

have 𝜏

2

-𝑠𝑐𝑙

𝐴

(𝐹) ⊆ 𝑈 ∩ 𝐴. Using the fact that 𝜏

2

-𝑠𝑐𝑙(𝐹) ∩ 𝐴 = 𝜏

2

-𝑠𝑐𝑙

𝐴

(𝐹) [9], 𝜏

2

-𝑠𝑐𝑙(𝐹) ∩

𝐴 ⊆ 𝑈 ∩ 𝐴. Since 𝐴 is 𝜏

1

-𝑔̂-open and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋, by 3.15, 𝐴 is 𝜏

2

-semi closed. So,

we have 𝜏

2

-𝑠𝑐𝑙(𝐹)  𝜏

2

-𝑠𝑐𝑙(𝐴) = 𝐴. Therefore 𝜏

2

-𝑠𝑐𝑙(𝐹) = 𝜏

2

-𝑠𝑐𝑙(𝐹) ∩ 𝐴 ⊆ 𝑈 ∩ 𝐴 ⊆ 𝑈 and

hence 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋.

(6)

SUFFICIENCY

If 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋, then 𝜏

2

-𝑠𝑐𝑙(𝐹)  𝑈 whenever 𝐹 ⊆ 𝑈 and 𝑈 is 𝜏

1

-𝑔̂-open in 𝑋. By Lemma 2.5, 𝑈 ∩ 𝐴 is 𝜏

1

-𝑔̂-open in 𝑋. Since 𝐹 ⊆ 𝐴 ∩ 𝑈 and 𝐴 ∩ 𝑈 is 𝜏

1

-𝑔̂-open in 𝑋, we have 𝜏

2

-𝑠𝑐𝑙(𝐹)  𝐴 ∩ 𝑈 and hence 𝐴 ∩ 𝜏

2

-𝑠𝑐𝑙(𝐹)  𝐴 ∩ 𝑈. But, by assumption, 𝐴 ∩ 𝑈 is 𝜏

1

-𝑔̂-open in 𝐴. Thus we have got 𝐴 ∩ 𝜏

2

-𝑠𝑐𝑙(𝐹)  𝐴 ∩ 𝑈, whenever 𝐹 ⊆ 𝐴 ∩ 𝑈 and 𝐴 ∩ 𝑈 is 𝜏

1

-𝑔̂-open in 𝐴. i.e. 𝜏

2

-𝑠𝑐𝑙

𝐴

(𝐹) ⊆ 𝐴 ∩ 𝑈 whenever 𝐹 ⊆ 𝐴 ∩ 𝑈 and 𝐴 ∩ 𝑈 is 𝜏

1

-𝑔̂-open in 𝐴. Therefore 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴.

Corollary 3.17

Let (𝑋, 𝜏

1

, 𝜏

2

) be a bitopological space and 𝐴 be a subspace of 𝑋 such of 𝑋 such that 𝑈 ∩ 𝐴 is 𝜏

1

-𝑔̂-open in 𝐴 for every 𝜏

1

-𝑔̂-open set 𝑈 in 𝑋. In addition, let 𝐴 be

𝜏

1

-𝑔̂-open and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋. Then 𝐴 ∩ 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴 where 𝐹 is 𝜏

2

-semi closed in 𝑋.

Proof.

We have 𝐴 ∩ 𝐹 ⊆ 𝐴 ⊆ 𝑋 where 𝐴 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋 which is also 𝜏

1

-𝑔̂-open. Then by 3.15, 𝐴 is 𝜏

2

-semi closed in 𝑋. But, by Lemma 2.6, 𝐴 ∩ 𝐹 is 𝜏

2

-semi closed in 𝑋. Therefore by 3.14, 𝐴 ∩ 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋. Hence, by 3.16, 𝐴 ∩ 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴.

Proposition 3.18

Let (𝑋, 𝜏

1

, 𝜏

2

) be a bitopological space and 𝐴 be a subspace of 𝑋 such that 𝑈 ∩ 𝐴 is 𝜏

1

-𝑔̂-open in 𝐴 iff 𝑈 is 𝜏

1

-𝑔̂-open in 𝑋. In addition, let 𝐹 be 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋 such that 𝐹 ⊆ 𝐴 ⊆ 𝑋.

Then 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴.

Proof.

Let 𝐹 ⊆ 𝑈 whenever 𝑈 is 𝜏

1

-𝑔̂-open in 𝑋. Let 𝐹 ⊆ 𝐴 ∩ 𝑈 and 𝐴 ∩ 𝑈 is 𝜏

1

-𝑔̂-open in 𝐴. Then 𝐹 ⊆ 𝑈 and 𝑈 is 𝜏

1

-𝑔̂-open in 𝑋. Since 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋, we have 𝜏

2

-𝑠𝑐𝑙(𝐹) ⊆ 𝑈.

Then 𝐴 ∩ 𝜏

2

-𝑠𝑐𝑙(𝐹) ⊆ 𝐴 ∩ 𝑈 and 𝐴 ∩ 𝑈 is 𝜏

1

-𝑔̂-open in 𝐴. Thus, we have 𝜏

2

-𝑠𝑐𝑙

𝐴

(𝐹) ⊆ 𝐴 ∩ 𝑈 whenever 𝐹 ⊆ 𝐴 ∩ 𝑈 and 𝐴 ∩ 𝑈 is 𝜏

1

-𝑔̂-open in 𝐴. Therefore 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝐴.

Proposition 3.19

If a subset 𝐴 of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed and if 𝐴 ⊆ 𝐵 ⊆ 𝜏

2

-𝑠𝑐𝑙(𝐴), then 𝐵 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋.

Proof.

Let 𝐵 ⊆ 𝑈 and 𝑈 is 𝜏

1

-𝑔̂-open in 𝑋. Since 𝐴 ⊆ 𝐵, it follows that 𝐴 ⊆ 𝑈. Since 𝐴 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠- closed in 𝑋, it follows that 𝜏

2

-𝑠𝑐𝑙(𝐴)  𝑈. By assumption, 𝐵 ⊆ 𝜏

2

-𝑠𝑐𝑙(𝐴) and hence 𝜏

2

- 𝑠𝑐𝑙(𝐵)  𝜏

2

-𝑠𝑐𝑙(𝜏

2

− 𝑠𝑐𝑙(𝐴)) = 𝜏

2

-𝑠𝑐𝑙(𝐴) ⊆ 𝑈. Therefore 𝐵 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋.

Proposition 3.20

Let 𝐴 and 𝐵 be subsets of bitopological space (𝑋, 𝜏

1

, 𝜏

2

) such that 𝜏

2

-𝑠𝑖𝑛𝑡(𝐴) ⊆ 𝐵 ⊆ 𝐴. If 𝐴

is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open, then 𝐵 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open.

(7)

S. Poius Missier, et al., J. Comp. & Math. Sci. Vol.6(11), 621-630 (2015) 627 Proof.

Suppose 𝐴 and 𝐵 are such that 𝜏

2

-𝑠𝑖𝑛𝑡(𝐴) ⊆ 𝐵 ⊆ 𝐴. Let 𝐴 be 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open. We shall show that 𝐵 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open. Let 𝐹 ⊆ 𝐵 and 𝐹 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed in 𝑋. Since 𝐹 ⊆ 𝐵 and 𝐵 ⊆ 𝐴, we have 𝐹 ⊆ 𝐴. Since 𝐴 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open, we have 𝐹 ⊆ 𝜏

2

-𝑠𝑖𝑛𝑡(𝐴). Since 𝜏

2

-𝑠𝑖𝑛𝑡(𝐴) ⊆ 𝐵, we have 𝜏

2

-𝑠𝑖𝑛𝑡(𝜏

2

− 𝑠𝑖𝑛𝑡(𝐴)) ⊆ 𝜏

2

-𝑠𝑖𝑛𝑡(𝐵). i.e., 𝜏

2

-𝑠𝑖𝑛𝑡(𝐴) ⊆ 𝜏

2

-𝑠𝑖𝑛𝑡(𝐵). Therefore 𝐹 ⊆ 𝜏

2

-𝑠𝑖𝑛𝑡(𝐵). Therefore 𝐵 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open.

Proposition 3.21

Let (𝑋, 𝜏

1

, 𝜏

2

) be a bitopological space in which 𝜏

2

-𝑐𝑙(𝐴) = 𝜏

2

-𝑠𝑐𝑙(𝐴) for every subset 𝐴 of 𝑋. If 𝐺 and 𝐻 are two 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed subsets of 𝑋, then 𝐺 ∪ 𝐻 is also 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

Proof.

Let 𝑈 be 𝜏

1

-𝑔̂-open such that 𝐺 ∪ 𝐻 ⊆ 𝑈. Then 𝐺 ⊆ 𝑈 and 𝐻 ⊆ 𝑈. Then 𝜏

2

-𝑠𝑐𝑙(𝐺) ⊆ 𝑈 and 𝜏

2

-𝑠𝑐𝑙(𝐻) ⊆ 𝑈. We know that 𝜏

2

-𝑐𝑙(𝐺 ∪ 𝐻) = 𝜏

2

-𝑐𝑙(𝐺) ∪ 𝜏

2

-𝑐𝑙(𝐻).i.e., 𝜏

2

-𝑐𝑙(𝐺 ∪ 𝐻) = 𝜏

2

- 𝑠𝑐𝑙(𝐺) ∪ 𝜏

2

-𝑠𝑐𝑙(𝐻) ⊆ 𝑈 i.e., 𝜏

2

-𝑐𝑙(𝐺 ∪ 𝐻) ⊆ 𝑈. Therefore, 𝜏

2

-𝑠𝑐𝑙(𝐺 ∪ 𝐻) ⊆ 𝑈. Since 𝜏

2

- 𝑠𝑐𝑙(𝐺 ∪ 𝐻) ⊆ 𝜏

2

-𝑐𝑙(𝐺 ∪ 𝐻). Therefore, 𝐺 ∪ 𝐻 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

Proposition 3.22

Let 𝐵 be a subset of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

). If 𝐵 is 𝜏

1

-𝑔̂-open, 𝜏

1

𝜏

2

-pre open, 𝜏

1

𝜏

2

- 𝑔̂

∗

𝑠-closed and if every 𝜏

2

-semi closed set is 𝜏

2

𝜏

1

-semi closed, then 𝐵 is 𝜏

1

𝜏

2

-regular open.

Proof.

Let 𝐵 be 𝜏

1

-𝑔̂-open and 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed. Then, by 3.15, 𝐵 is 𝜏

2

-semi closed. By assumption, 𝐵 is 𝜏

2

𝜏

1

-semi closed. Therefore 𝜏

1

-𝑖𝑛𝑡(𝜏

2

-𝑐𝑙(𝐵)) ⊆ 𝐵. Since, by assumption, 𝐵 is 𝜏

1

𝜏

2

-pre open, we have 𝐵 ⊆ 𝜏

1

-𝑖𝑛𝑡(𝜏

2

-𝑐𝑙(𝐵)). Therefore 𝐵 = 𝜏

1

-𝑖𝑛𝑡(𝜏

2

-𝑐𝑙(𝐵)) and hence 𝐵 is 𝜏

1

𝜏

2

- regular open.

Proposition 3.23

Let 𝐵 be a subset of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

). If 𝐵 is pairwise-clopen and if every 𝜏

1

𝜏

2

- semi closed set is 𝜏

2

-semi closed, then 𝐵 is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

Proof.

Since 𝐵 pairwise-clopen, 𝜏

2

-𝑖𝑛𝑡(𝜏

1

-𝑐𝑙(𝐵)) = 𝐵. 𝑖. 𝑒. , 𝜏

2

-𝑖𝑛𝑡(𝜏

1

-𝑐𝑙(𝐵)) ⊆ 𝐵. 𝑖. 𝑒. , 𝐵 is 𝜏

1

𝜏

2

- semi closed. By assumption, B is 𝜏

2

- semi closed. i.e., 𝜏

2

-semi 𝑠𝑐𝑙(𝐵) = 𝐵. Therefore B is 𝜏

2

- semi closed and hence, by 3.14, B is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

Lemma 3.24

15

For any subset A of (𝑋, 𝜏

1

, 𝜏

2

), (i) 𝜏

1

𝜏

2

-𝑝𝑖𝑛𝑡(𝐴) = 𝐴 ∩ 𝜏

1

-(𝑖𝑛𝑡 𝜏

2

-𝑐𝑙(𝐴)) (ii) 𝜏

1

𝜏

2

-𝑝𝑐𝑙(𝐴) = 𝐴 ∩ 𝜏

1

-(𝑐𝑙 𝜏

2

-𝑖𝑛𝑡(𝐴))

Proposition 3.25

Let B be a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed subset of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

). If B is 𝜏

1

𝜏

2

- regular

open, 𝜏

1

, 𝜏

2

-𝑝𝑖𝑛𝑡(𝐵) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

(8)

Proof:

Since B is 𝜏

1

𝜏

2

- regular open, we have 𝐵 = 𝜏

1

-(𝑖𝑛𝑡 𝜏

2

-𝑐𝑙(𝐵)). We know that, by Lemma 3.24(i), 𝜏

1

𝜏

2

-𝑝𝑖𝑛𝑡(𝐵) = 𝐵 ∩ 𝜏

1

-(𝑖𝑛𝑡 𝜏

2

-𝑐𝑙(𝐵)) = 𝐵. Therefore 𝜏

1

𝜏

2

-𝑝𝑖𝑛𝑡(𝐵) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠- closed.

Proposition 3.26:

Let B be a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed subset of a topological space (𝑋, 𝜏

1

, 𝜏

2

). If B is 𝜏

1

𝜏

2

- regular closed then, 𝜏

1

𝜏

2

-𝑝𝑐𝑙(𝐵) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed.

Proof:

Since B is 𝜏

1

𝜏

2

- regular closed, we have 𝐵 = 𝜏

1

-𝑐𝑙( 𝜏

2

-𝑖𝑛𝑡(𝐵)). We know that, by Lemma 3.24(ii), 𝜏

1

𝜏

2

-𝑝𝑐𝑙(𝐵) = 𝐵 ∩ 𝜏

1

-(𝑐𝑙 𝜏

2

-𝑖𝑛𝑡(𝐵)) = 𝐵. Therefore 𝜏

1

𝜏

2

-𝑝𝑐𝑙(𝐵) is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠- closed.

Proposition 3.27:

Let A be a 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed subset of (𝑋, 𝜏

1

, 𝜏

2

). Then 𝜏

2

-𝑠𝑐𝑙(𝐴)-A contains no non-empty 𝜏

1

- 𝑔̂ –closed set in X.

Proof:

Let F be a 𝜏

1

-𝑔̂ –closed subset of 𝜏

2

-𝑠𝑐𝑙(𝐴)-A. Since 𝐹 ⊆ 𝜏

2

-𝑠𝑐𝑙(𝐴)-A , we have 𝐴 ⊆ 𝑋 − 𝐹 where A is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed and 𝑋 − 𝐹 is 𝜏

1

-𝑔̂ –open. Thus 𝜏

2

-𝑠𝑐𝑙(𝐴) ⊆ 𝑋 − 𝐹 or equivalently 𝐹 ⊆ 𝜏

2

-𝑠𝑐𝑙(𝐴). But 𝐹 ⊆ 𝑋 − 𝜏

2

-𝑠𝑐𝑙(𝐴). Therefore 𝐹 ⊆ [𝑋 − 𝜏

2

− 𝑠𝑐𝑙(𝐴)] ∩ [𝜏

2

− 𝑠𝑐𝑙(𝐴)]

and hence 𝐹 = ∅. Hence the Proof.

Proposition 3.28:

If a set A is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open in a bitopological space (𝑋, 𝜏

1

, 𝜏

2

), then 𝐺 = 𝑋 whenever G is 𝜏

1

- 𝑔̂ –open and 𝜏

2

-𝑠𝑖𝑛𝑡(𝐴) ∪ 𝐴

𝑐

⊆ 𝐺, where 𝐴

𝑐

denotes 𝑋 − 𝐴.

Proof:

Suppose that A is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-open and 𝐺 is 𝜏

1

-𝑔̂ –open and 𝜏

2

-𝑠𝑖𝑛𝑡(𝐴) ∪ 𝐴

𝑐

⊆ 𝐺.

Then 𝐺

𝑐

⊆ (𝜏

2

− 𝑠𝑖𝑛𝑡(𝐴)) ∪ 𝐴

𝑐𝑐

.

⇒ 𝐺

𝑐

⊆ (𝜏

2

− 𝑠𝑖𝑛𝑡(𝐴)

𝑐

) ∩ 𝐴

⇒ 𝐺

𝑐

⊆ 𝜏

2

− 𝑠𝑐𝑙(𝐴

𝑐

) − 𝐴

𝑐

. (By Lemma 2.7)

Since 𝐴

𝑐

is 𝜏

1

𝜏

2

-𝑔̂

∗

𝑠-closed, by 3.27, we have 𝜏

2

− 𝑠𝑐𝑙(𝐴

𝑐

) − 𝐴

𝑐

contains no non empty.

𝜏

1

-𝑔̂ –closed set in X.

Therefore 𝐺

𝑐

= ∅.

Hence 𝐺 = 𝑋.

Proposition 3.29:

The following are equivalent for any subset A of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

).

(i) 𝐴-𝜏

1

-𝑖𝑛𝑡(𝐴) is 𝜏

2

-𝑔̂

∗

𝑠-open in (𝑋, 𝜏

1

, 𝜏

2

).

(ii) 𝜏

1

-𝑖𝑛𝑡(𝐴) ∪ [𝑋 − 𝐴]is 𝜏

2

-𝑔̂

∗

𝑠-closed in (𝑋, 𝜏

1

, 𝜏

2

)

(9)

S. Poius Missier, et al., J. Comp. & Math. Sci. Vol.6(11), 621-630 (2015) 629 Proof.(i) ⇒ (ii)

Given that 𝐴-𝜏

1

-𝑖𝑛𝑡(𝐴) is 𝜏

2

-𝑔̂

∗

𝑠-open in (𝑋, 𝜏

1

, 𝜏

2

).

Now, 𝑋 − [𝐴 − 𝜏

1

− 𝑖𝑛𝑡(𝐴) ] = 𝑋 ∩ [𝐴 − 𝜏

1

− 𝑖𝑛𝑡(𝐴) ]

𝑐

= [𝐴 − 𝜏

1

− 𝑖𝑛𝑡(𝐴) ]

𝑐

= {𝐴 ∩ [𝜏

1

− 𝑖𝑛𝑡(𝐴)

𝑐

]}

𝑐

= 𝐴

𝑐

∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]

= (𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]

Since 𝐴 − 𝜏

1

− 𝑖𝑛𝑡(𝐴) is 𝜏

2

-𝑔̂

∗

𝑠-open in (𝑋, 𝜏

1

, 𝜏

2

),

we have 𝑋 − [𝐴 − 𝜏

1

− 𝑖𝑛𝑡(𝐴) ] = (𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ] is 𝜏

2

-𝑔̂

∗

𝑠-closed in(𝑋, 𝜏

1

, 𝜏

2

).

(ii) ⇒ (i)

Suppose that (𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ] is 𝜏

2

-𝑔̂

∗

𝑠-closed in(𝑋, 𝜏

1

, 𝜏

2

).

Then 𝑋 − {(𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]} is 𝜏

2

-𝑔̂

∗

𝑠-open in(𝑋, 𝜏

1

, 𝜏

2

).

Now, 𝑋 − {(𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]}

= 𝑋 ∩ {(𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]}

𝑐

= {(𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]}

𝑐

= (𝑋 − 𝐴)

𝑐

∩ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]

𝑐

= 𝐴 ∩ [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]

𝑐

= 𝐴 − [𝜏

1

− 𝑖𝑛𝑡(𝐴) ]

Therefore, 𝐴 − [𝜏

1

− 𝑖𝑛𝑡(𝐴) ] is 𝜏

2

-𝑔̂

∗

𝑠-open in (𝑋, 𝜏

1

, 𝜏

2

).

Proposition 3.30:

Let A be any subset of a bitopological space (𝑋, 𝜏

1

, 𝜏

2

). If 𝜏

1

− 𝑖𝑛𝑡(𝐴) ∪ [𝑋 − 𝐴] is 𝜏

2

-𝑔̂

∗

𝑠- closed in (𝑋, 𝜏

1

, 𝜏

2

),then 𝐺 ∪ 𝜏

1

− 𝑖𝑛𝑡(𝐴) = 𝐴 for some 𝜏

2

-𝑔̂

∗

𝑠-open set G in (𝑋, 𝜏

1

, 𝜏

2

).

Proof:

Suppose that (𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)] is 𝜏

2

-𝑔̂

∗

𝑠-closed in (𝑋, 𝜏

1

, 𝜏

2

). Let 𝑈 = (𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)].

Then 𝑈

𝑐

is 𝜏

2

-𝑔̂

∗

𝑠-open in (𝑋, 𝜏

1

, 𝜏

2

).

Now, 𝑈

𝑐

∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)] = {(𝑋 − 𝐴) ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)]}

𝑐

∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)]

= {𝐴 ∩ [𝜏

1

− 𝑖𝑛𝑡(𝐴)]

𝑐

} ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)]

= {𝐴 ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)]} ∩ {[𝜏

1

− 𝑖𝑛𝑡(𝐴)]

𝑐

∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)]}

= 𝐴 ∩ 𝑋

= 𝐴

Taking 𝐺 = 𝑈

𝑐

, we have 𝐴 = 𝐺 ∪ [𝜏

1

− 𝑖𝑛𝑡(𝐴)] for some 𝜏

2

-𝑔̂

∗

𝑠-open set G in (𝑋, 𝜏

1

, 𝜏

2

).

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s

-spaces, The Journal of Mathematics,

Vol.6, (2), Number 1 : 1-8 (2008).

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Part III, 35, 19-28 (1985).

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16. S.Pious Missier and M.Anto, ĝ

∗

s-closed sets, in topological spaces, International Journal of Modern Engineering Research, Vol 4, issue 11(version 2), 32 – 38 Nov (2014).

17. S.Poius Missier and M.Anto, τ

1

τ

2

− ĝ

∗

s-closed sets, (submitted).

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∗

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References

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