

Ratio Mathematica                                                                         Volume 45, 2023 

 
A study on �̂�∗∗𝐬 − 𝑹𝟎 and �̂�
∗∗𝐬 − 𝑹𝟏 spaces  

In Topological spaces 
 

M. Anto* 

S. Andrin Shahila† 

 
Abstract 

In this paper, we introduces the concept of 𝑅0-space, 𝑅1-space, door space, submaximal 
using �̂�∗∗𝑠-closed set and investigate its properties. We have also study their 
relationship with some other higher separation axioms.  We have also, introduced a new 

definition 𝑆�̂�∗-space by using semi-closed and �̂�∗-closed sets and study its relationship 
with other closed sets using �̂�∗∗𝑠-closed set. 
 

Keywords. ĝ∗∗s − R0, ĝ
∗∗s − R1 space, ĝ

∗∗s -door spaces, ĝ∗∗s –submaximal, Sĝ∗-
space. 

 
2010 AMS subject classification: 54D15, 54G05‡ 
 

*Associate Professor, P.G and Research Department of Mathematics Annai Velankanni college, 
Tholayavattam, Kanyakumari District ,629157,Affiliated to Manonmaniam Sundaranar University, 

Abishekapatti, Tirunelveli-627012, Tamilnadu, India; e-mail: andrinshahila@gmail.com. 
†PhD Scholar (Reg.No:19213012092006), P.G and Research Department of Mathematics Annai 

Velankanni college, Tholayavattam, Kanyakumari District, 629157, Affiliated to Manonmaniam 

Sundaranar University, Abishekapatti, Tirunelveli-627012, Tamilnadu, India; e-mail: 

antorbjm@gmail.com. 
‡ Received on July 22, 2022. Accepted on October 15, 2022. Published on January 30, 2023. doi: 

10.23755/rm.v45i0.989. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. This paper is published 

under the CC-BY license agreement. 

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M. Anto and S. Andrin Shahila 

 
1. Introduction 

Topology is a major area of mathematics concerned with spatial properties that 

are preserved under continuous deformations of objects. It emerged through the 

development of concepts from geometry and set theory, such as space, dimension, and 

transformation. Ideas that are now classified as topology were expressed as early as 

1736. By the middle of the 20th century, topology had become an important area of 

study within mathematics. The word topology is used both for the mathematical 

discipline and for a family of sets with certain properties that are used to define a 

topological space, a basic object of topology. Topology includes many subfields namely 

point-set topology, algebraic topology and geometric topology. 

In the literature of General Topology, the concept of semi open sets was 

introduced by Levine in 1963 [11] and g-closed sets in 1970 [12]. M.K.R.S. Veera 

Kumar defined �̂�-closed sets in 2001 [10] and �̂�∗-closed sets in 1996 [9]. In the year 
1995, J. Dontchev [2, 3], defined on door spaces and submaximal spaces. 

The notion of 𝑅0  topological space is introduced by N. A. Shanin [13] in 1943. 
Later, A. S. Davis [13] rediscovered it and studied some properties of this weak 

separation axiom. In 2010, S. Balasubramanian defined separation axioms on 

generalized sets [14]. Several topologists further investigated properties of 

𝑅0topological spaces and many interesting results have been obtained in various 
contexts. In the same paper, A. S. Davis also introduced the notion of 𝑅1 topological 
space which are independent of both 𝑇0 and 𝑇1 but strictly weaker than𝑇2. M. G. 
Murdeshwar and S. A. Naimpally [8] studied some of the fundamental properties of the 

class of 𝑅1 topological spaces. In 1963, N. Levine [11] offered a new notion to the field 
of general topology by introducing semi-open sets. He defined this notion by utilizing 

the known notion of closure of an open set, i.e., a subset of a topological space is semi-

open if it is contained in the closure of its interior. Since the advent of this notion, 

several new notions are defined in terms of semi-open sets of which two are semi-𝑅0 
and semi-𝑅1 introduced by S. N. Maheshwari and R. Prasad [15] and C. Dorsett [1], 
respectively. These two notions are defined as natural generalizations of the separation 

axioms 𝑅0 and 𝑅1 by replacing the closure operator with the semi closure operator and 
openness with semi-openness. Since then, this notion received wide usage in general 

topology.  

In this paper, we continue the study of the above mentioned classes of 

topological spaces satisfying these axioms by introducing two more notions in terms of 

�̂�∗∗𝑠-closed sets [7] called �̂�∗∗𝑠-𝑅0and �̂�
∗∗𝑠-𝑅1. We have also applied the definition in 

[7] to door space and submaximal space.In this paper we introduced Sĝ∗-space using 
semi-closed and �̂�∗-closed sets. 

 
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A study on �̂�∗∗𝑠 − 𝑅0 and �̂�
∗∗𝑠 − 𝑅1 spaces in Topological spaces 

 
2. Preliminaries 

Throughout this paper (𝑋 , 𝜏) represent the non-empty topological spaces on which 
no separation axioms are assumed unless otherwise mentioned. For a subset A of a 

space (𝑋 , 𝜏), 𝑐𝑙(𝐴)and 𝑖𝑛𝑡 (𝐴) denote the closure and interior of A respectively.  
Definition. 2.1 

i) A space (X, τ) is R0 [13] if for each open set U of X, x ∈ U implies cl({x}) ⊆ U. 
ii) A topological space (X, τ) is R1 [13] if for x, y ∈ X such that cl({x}) ≠ cl({y}), 

there are disjoint open sets U and V such that cl({x})⊆ U and cl({y}) ⊆ V. 
iii) A topological space (X, τ) is a door space [3] if every subset of X is either open 

or closed in X. 

iv) A topological space (X, τ) is a submaximal space [4] if every dense subset of X 
is open in X.  

 
3.�̂�∗∗𝐬 − 𝑹𝟎 And �̂�
∗∗𝐬 − 𝑹𝟏 spaces 

Definition. 3.1. A space (𝑋, 𝜏) is said to be ĝ∗∗s − 𝑅0 if for any open set𝑈, 𝑥 ∈ 𝑈, then 
ĝ∗∗scl{x} ⊆ U. 

 
Definition. 3.2. A space (𝑋, 𝜏) is said to be ĝ∗∗s − 𝑅1 if for x and y in X, 
withĝ∗∗scl({x}) ≠ ĝ∗∗scl({y}), there exist two disjoint ĝ∗∗s-open sets U and V such that 
ĝ∗∗scl({x}) ⊆ 𝑈 and ĝ∗∗scl({y}) ⊆ 𝑉 

 
Theorem. 3.3. Every ĝ∗∗s − 𝑅0-space is ĝ
∗∗s − 𝑇0-space. 

Proof. Let (𝑋, 𝜏) be a ĝ∗∗s − 𝑅0-space. Let x and y be two distinct points of X. By 
hypothesis, for any open set U and𝑥 ∈ 𝑈 ⇒ ĝ∗∗scl({x}) ⊆ U. Also, every open set is 
ĝ∗∗s-open, which implies U is ĝ∗∗s-open and𝑥 ∈ 𝑈. Therefore, for any distinct points x 
and y, there exist a ĝ∗∗s-open set U containing x and not y. Hence, (𝑋, 𝜏) is a ĝ∗∗s − 𝑇0-
space. 

Theorem. 3.4. Every ĝ∗∗s − 𝑅0-space is ĝ
∗∗s − 𝑇2-space. 

Proof. Let (𝑋, 𝜏) be a ĝ∗∗s − 𝑅0-space. Let x and y be two distinct points of X. By 
hypothesis, for any open set U and𝑥 ∈ 𝑈 ⇒ ĝ∗∗scl{x} ⊆ U. Ifĝ∗∗scl({x}) ⊈ U, there 
exist another open set V not containing x, so𝑦 ∈ 𝑉. Also,𝑈 ∩ 𝑉 = ∅. Therefore, for any 
distinct points x and y, there exist two disjoint ĝ∗∗s-open sets U and V. Hence, (𝑋, 𝜏) is 
a ĝ∗∗s − 𝑇2-space. 

 
Theorem. 3.5. If (𝑋, 𝜏) isĝ∗∗s − 𝑅0, then for any closed set U and𝑥 ∉ 𝑈, there exist an 
ĝ∗∗s-open set G such that 𝑈 ⊆ 𝐺 and𝑥 ∉ 𝐺.  
Proof. Suppose (𝑋, 𝜏) isĝ∗∗s − 𝑅0. Let U be any closed set and 𝑥 ∉ 𝑈 ⇒ 𝑋\𝑈 is open 
and𝑥 ∈ 𝑋\𝑈. By assumption,ĝ∗∗scl({x}) ⊆ X\U ⇒ U ⊆ X\ĝ∗∗scl({x}). Put𝐺 =

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𝑋\ĝ∗∗scl{x}.Since ĝ∗∗scl{x} is ĝ∗∗s-open. Also, 𝑈 ⊆ 𝐺 and𝑥 ∉ 𝐺. Therefore, for any 
closed set U and 𝑥 ∉ 𝑈, there exist an ĝ∗∗s-open set G such that 𝑈 ⊆ 𝐺 and𝑥 ∉ 𝐺. 
Conversely, suppose for any closed set U and 𝑥 ∉ 𝑈, there exist an ĝ∗∗s-open set G such 
that 𝑈 ⊆ 𝐺 and 𝑥 ∉ 𝐺 

 
Theorem. 3.6. Every ĝ∗∗s − 𝑅0-space is ĝ
∗∗s-regular. 

Proof. Let (𝑋, 𝜏) be a ĝ∗∗s − 𝑅0-space. Let F be a closed set and𝑥 ∈ 𝑋 − 𝐹. By 
theorem: 3.3, there exist an ĝ∗∗s-open set G such that 𝑈 ⊆ 𝐺 and𝑥 ∉ 𝐺. Put𝐻 = {𝑥}. 
Since every singleton set is open, so H is open. Also, we know that, every open set is 

ĝ∗∗s-open. Also, 𝐺 ∩ 𝐻 = 𝐺 ∩ {𝑥} = ∅. Thus, G and H are disjoint ĝ∗∗s-open sets 
containing x and H respectively. Therefore, (𝑋, 𝜏) is ĝ∗∗s-regular.   

 
Theorem. 3.7. A topological space (𝑋, 𝜏) is ĝ∗∗s − 𝑅0 iff for any points x and y in X, 
𝑥 ≠ 𝑦 impliesĝ∗∗scl({x}) ∩ ĝ∗∗scl({y}) = ∅. 
Proof. Let X be ĝ∗∗s − 𝑅0 and 𝑥 ≠ 𝑦 in 𝑋 ⇒ {𝑥} be an open set and𝑦 ∉ {𝑥}. Since,𝑥 ∈
{𝑥}, we have  ĝ∗∗scl({x}) ⊆ {𝑥}. Thusĝ∗∗scl({x}) = {𝑥}. Now,ĝ∗∗scl({x}) ∩
ĝ∗∗scl({y}) = {𝑥} ∩ {𝑦} = ∅. Conversely, Suppose for any points 𝑥 and 𝑦 in𝑋, 𝑥 ≠
𝑦 ⇒ ĝ∗∗scl({x}) ∩ ĝ∗∗scl(y) = ∅. Let V be an open set and 𝑥 ∈ 𝑉 and let 𝑦 ∈
ĝ∗∗scl({x})  (1). Suppose𝑦 ∉ 𝑉. By assumption, ĝ∗∗scl({x}) ∩ ĝ∗∗scl({y}) = ∅ ⇒ 𝑦 ∉
ĝ∗∗scl({x})this is a contradiction to (1). Therefore, 𝑦 ∈ 𝑉 andĝ∗∗scl({x}) ⊆ 𝑉. Thus 
(𝑋, 𝜏) isĝ∗∗s − 𝑅0. 

 
Corollary. 3.8. A topological space (𝑋, 𝜏) is ĝ∗∗s − 𝑅0 iff for any points x and y in X, 
𝑥 ≠ 𝑦 ⇒ 𝑐𝑙({𝑥}) ∩ 𝑐𝑙({𝑦}) = ∅. 
Proof. Directly follows from theorem.3.5  

 
Theorem. 3.9. If (𝑋, 𝜏) isĝ∗∗s − 𝑅0, then it isĝ
∗∗s − 𝑇1. 

Proof. Let (𝑋, 𝜏) be a ĝ∗∗s − 𝑅0-space and 𝑥 ∈ 𝑋 ⇒ {𝑥} is open. By hypothesis, 
ĝ∗∗scl({x}) ⊆ {𝑥} ⇒ ĝ∗∗scl({x}) = {𝑥} ⇒ {𝑥} is ĝ∗∗s-closed ⇒every singleton set is 
ĝ∗∗s-closed⇒  (𝑋, 𝜏) isĝ∗∗s − 𝑇1. 

 
Theorem. 3.10. Every ĝ∗∗s − 𝑅0-space is ĝ
∗∗s − 𝑅1-space. 

Proof. Let (𝑋, 𝜏) isĝ∗∗s − 𝑅0. Let𝑥, 𝑦 ∈ 𝑋, withĝ
∗∗scl({x}) ≠ ĝ∗∗scl({y}). By above 

theorem, ĝ∗∗scl({x}) = {𝑥} and ĝ∗∗scl({y}) = {𝑦} ⇒ {𝑥} and {𝑦} are ĝ∗∗s-open sets and 
{𝑥} ∩ {𝑦} = ∅  ⇒ (𝑋, 𝜏) is ĝ∗∗s − 𝑅1-space.  

 
Theorem. 3.11. For any ĝ∗∗s-closed set H, ĝ∗∗scl({x}) ∩ 𝐻 = ∅, for every𝑥 ∈ 𝑋\𝐻, 
then (𝑋, 𝜏) is ĝ∗∗s − 𝑅0-space. 
Proof. Assume that for any ĝ∗∗s-closed set H,ĝ∗∗scl({x}) ∩= ∅, for every 𝑥 ∈ 𝑋\𝐻. Let 
G be any open set and 𝑥 ∈ 𝐺. Then 𝑥 ∈ 𝐺 ⇒ 𝑥 ∈ 𝑋(𝑋\𝐺) and 𝑋\𝐺 is closed. Therefore, 
by assumption, ĝ∗∗scl({x}) ∩ (𝑋\𝐺) = ∅ ⇒ ĝ∗∗scl({x}) ⊆ 𝐺. Hence proved. 

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∗∗𝑠 − 𝑅1 spaces in Topological spaces 

 
Corollary. 3.12. For any ĝ∗∗s-closed set H,𝑐𝑙({𝑥}) ∩ 𝐻 = ∅, for every 𝑥 ∈ 𝑋\𝐻, then 
(𝑋, 𝜏) is ĝ∗∗s − 𝑅0-space. 
Proof. Assume that for any ĝ∗∗s-closed set H,𝑐𝑙({𝑥}) ∩ 𝐻 = ∅, for every 𝑥 ∈ 𝑋\𝐻. Let 
V be any open set in X and 𝑥 ∈ 𝑉. Then, 𝑥 ∈ 𝑉 = 𝑋\(𝑋\𝑉) and 𝑋\𝑉 is closed. Since 
every closed set is closed. Since, every closed set is ĝ∗∗s-closed which implies that 𝑋\𝑉 
is ĝ∗∗s-closed. By assumption, 𝑐𝑙({𝑥}) ∩ (𝑋\𝑉) = ∅ ⇒ 𝑐𝑙({𝑥}) ⊆ 𝑉 ⇒ ĝ∗∗scl({x}) ⊆
𝑐𝑙({𝑥}) ⊆ 𝑉 ⇒ ĝ∗∗scl({x}) ⊆ V ⇒ (𝑋, 𝜏) is ĝ∗∗s − 𝑅0-space.  
 

Theorem. 3.13. If a topological space, (𝑋, 𝜏) is ĝ∗∗s − 𝑅1 and ĝ
∗∗s − 𝑇0-space.   

Proof. Since every ĝ∗∗s − 𝑇1 is ĝ
∗∗s − 𝑇0. Then, the result is obvious. 

 
Theorem. 3.14. If (𝑋, 𝜏) is ĝ∗∗s − 𝑅0 and ĝ
∗∗s − 𝑇0-space, then it is also a ĝ

∗∗s − 𝑇1-
space. 

Proof. Let 𝑥 ≠ 𝑦 be any two points of X. Since (𝑋, 𝜏) isĝ∗∗s − 𝑅0, there exist an open 
set U such that 𝑥 ∈ 𝑈 andĝ∗∗scl({x}) ⊆ 𝑈 ⇒ 𝑥 ∉ 𝑋\𝑈. Since, 𝑦 ∉ 𝑈, there exist another 
ĝ∗∗s-open set 𝑋\𝑈 = 𝑉(𝑠𝑎𝑦) containing y but not x. Therefore, for any two distinct 
points x and y, there exist two distinct ĝ∗∗s-open sets U and V. Hence, (𝑋, 𝜏) is ĝ∗∗s −
𝑇1-space. 

 
Theorem. 3.15. If a topological space (𝑋, 𝜏) isĝ∗∗s − 𝑅1, then either 𝑐𝑙({𝑥}) = 𝑋, for 
each 𝑥 ∈ 𝑋 or 𝑐𝑙({𝑥}) ≠ 𝑋, for each𝑥 ∈ 𝑋. 
Proof. Assume that (𝑋, 𝜏) isĝ∗∗s − 𝑅1. If 𝑐𝑙({𝑥}) = 𝑋, for each𝑥 ∈ 𝑋, then the theorem 
is obvious. If not, then there exist 𝑦 ∈ 𝑋 such that𝑐𝑙({𝑦}) ≠ 𝑋. Suppose not, there exist 
𝑧 ∈ 𝑋 such that𝑐𝑙({𝑧}) = 𝑋. Now,𝑐𝑙({𝑦}) ≠ 𝑋 = 𝑐𝑙({𝑧}). Since (𝑋, 𝜏) isĝ∗∗s − 𝑅1, 
there exist disjoint ĝ∗∗s-open sets U and V containing 𝑐𝑙({𝑦}) and 𝑐𝑙({𝑧}) respectively. 
Since 𝑐𝑙({𝑧}) = 𝑋, we have 𝑉 = 𝑋 ⇒ 𝑈 ∩ 𝑉 = 𝑈 ≠ ∅ which a contradiction is to𝑈 ∩
𝑉 = ∅. Hence proved. 
 

4. �̂�∗∗𝐬-door space 
Definition. 4.1. A topological space (X, τ) is called a ĝ∗∗s-door space if every subset is 
either ĝ∗∗s-open or ĝ∗∗s-closed. 

 
Definition. 4.2. A topological space (X, τ) is called ĝ∗∗s-submaximal if every dense 
subset of X is ĝ∗∗s-open. 

 
Definition. 4.3. A topological space (X, τ) is called ĝ∗∗s-extremally disconnected space 
in which the ĝ∗∗sclosure of every ĝ∗∗s-open subset is ĝ∗∗s-open. 

 
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M. Anto and S. Andrin Shahila 

 
Definition. 4.4. A topological space is called Sĝ∗-space if the intersection of a semi-
closed set with ĝ∗-closed set is ĝ∗-closed. 

 
Definition. 4.5. A topological space (X, τ) is called ĝ∗∗s-hyperconnected if every non-
empty ĝ∗∗s-open subset of X is ĝ∗∗s-dense. 
 

Theorem. 4.6. Every subspace of ĝ∗∗s-door space is a ĝ∗∗s-door space. 
Proof. Let S be a subspace of X and A ⊆ S is a subset of X. Since X is ĝ∗∗s-door space, 
A is either ĝ∗∗s-open or ĝ∗∗s-closed in X. Hence A is either ĝ∗∗s-open or ĝ∗∗s-closed in 
S. Therefore, S is ĝ∗∗s-door space.  
 

Theorem. 4.7. In a ĝ∗∗s-hyperconnected space, every ĝ∗∗s-submaximal space is a ĝ∗∗s-
door space. 

Proof. Let A ⊂ X and if A is dense in X, so A is ĝ∗∗s-open.                                (1)  
If A is not dense, there exist a non-empty open setB ⊂ Ac. Since every open set is ĝ∗∗s-
open, there exist a non-empty ĝ∗∗s-open setB ⊂ Ac. Since X is ĝ∗∗s-hyperconnected, B 
is dense and Ac is also dense. Again by definition of ĝ∗∗s-submaximal, Ac is ĝ∗∗s-open 
which implies A is ĝ∗∗s-closed.        (2) 
From (1) and (2), A is either ĝ∗∗s-open or ĝ∗∗s-closed. Therefore, (X, τ) is a ĝ∗∗s-door 
space. 

 
Theorem. 4.8. In(X, τ), every door space is ĝ∗∗s-door space. 
Proof. Let A be a door space. Every subset of X is either open or closed. Since every 

closed set is ĝ∗∗s-closed and every open set is ĝ∗∗s-open, every subset of X is either 
ĝ∗∗s-open or ĝ∗∗s-closed. Therefore, A is a ĝ∗∗s-door space. 

 
Remark. 4.9. Converse of above theorem is not true. 

 
Example. 4.10. Let X = {a, b, c}, τ = {∅, X, {a}}, τc = {∅, X, {b, c}},   

ĝ∗∗so(X, τ) = {∅, X, {a}, {a, b}, {a, c}} and ĝ∗∗sc(X, τ) = {∅, X, {b}, {c}, {b, c}}.  

Hence (X, τ) is a ĝ∗∗s-door space but it is not a door space. 

 
Theorem. 4.11. For a subset A of a Sĝ∗-space (X, τ) the following are equivalent: 
i) A is ĝ∗∗s-closed 
ii) ĝ∗cl({x}) ∩ A ≠ ∅, for each x ∈ scl(A) 
iii) scl(A) − A Contains no non-empty ĝ∗-closed set. 
Proof.i) ⇒ ii) Letx ∈ scl(A). Supposeĝ∗cl({x}) ∩ A = ∅. ThenA ⊆ X − ĝ∗cl({x}). 
Since A is ĝ∗∗s-closed, scl(A) ⊆ X − ĝ∗cl({x}) which is a contradiction tox ∈ scl(A). 
Therefore, ĝ∗cl({x}) ∩ A ≠ ∅ 

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ii) ⇒ iii) Let F be a ĝ-closed set such thatF ⊆ scl(A) − A. Supposex ∈ F. 
Thenĝ∗cl({x}) ⊆ F. Therefore ∅ ≠ ĝ∗cl({x}) ∩ A ⊆ F ∩ A ⊆ [scl(A) − A] ∩ A = ∅ 
which is a contradiction. 

iii) ⇒ i) Let A ⊆ G and G be ĝ∗-open in X, Suppose scl (A) not a subset of G. Then 
scl(A) ∩ (X − G) is non-empty ĝ∗-closed subset of scl(A) − A  which is a contradiction. 
Therefore, scl(A) ⊆ G and hence A is ĝ∗∗s-closed. 

 
Theorem. 4.12. If B is a cl open subset of(X, τ), then B is a ĝ∗∗s-closed set. 
Proof. Since B is cl open, B is both open and closed. Let U be a ĝ∗-open set in X 

andB ⊆ U. Since B is clopen, we have int(cl(B)) = B ⇒ int(cl(B)) ⊆ B and B ⊆

int(cl(A)), (ie)int(cl(B)) ⊆ B ⇒ B is semi-closed andscl (B) − B. Therefore, scl(B) ⊆

U. Hence B is ĝ∗∗s-closed in X. 

 
Theorem. 4.13. In(X, τ), every ĝ∗∗s-dense is dense. 
Proof. Let A ⊆ X be ĝ∗∗s-dense in(X, τ) ⇒ ĝ∗∗scl(A) = X. Since, X = ĝ∗∗scl(A) ⊆
cl(A) ⇒ X ⊆ cl(A). Always, cl(A) ⊆ X ⇒ cl(A) = X ⇒ A ⊆ X  be dense in (X, τ). 

 
Remark. 4.14. Converse of above theorem is not true. 

 
Example. 4.15. LetX = {a, b, c}, τ = {∅, X, {a, b}}, τc = {∅, X, {c}}, ĝ∗∗sc(X, τ) =

{∅, X, {c}, {b, c}, {a, c}}  and ĝ∗∗so(X, τ) = {∅, X, {a}, {b}, {a, b}}. Let A = {a} is a 

dense subset of X but it is not ĝ∗∗s-dense.   

Remark. 4.16. Every �̂�∗∗𝑠-door space need not be a �̂�∗∗𝑠-submaximal. 

Example. 4.17. Let 𝑋 = {𝑎, 𝑏, 𝑐}, 𝜏 = {∅, 𝑋, {𝑎, 𝑏}}, �̂�∗∗𝑠𝑐(𝑋, 𝜏) = {∅, 𝑋, {𝑐}, {𝑏, 𝑐},

{𝑎, 𝑐}} and�̂�∗∗𝑠𝑜(𝑋, 𝜏) = {∅, 𝑋, {𝑎}, {𝑏}, {𝑎, 𝑏}}. Here (𝑋, 𝜏) is a �̂�∗∗𝑠-door space but it 

is not �̂�∗∗𝑠-submaximal. 

 
Remark. 4.18. Every �̂�∗∗𝑠-submaximal space need not be �̂�∗∗𝑠-door space. 

 
Example. 4.19. Let 𝑋 = {𝑎, 𝑏, 𝑐, 𝑑}, 𝜏 = {∅, 𝑋, {𝑎}, {𝑑}, {𝑎, 𝑑}, {𝑎, 𝑏, 𝑐}}, 

 𝑠𝑐(𝑋, 𝜏) = {∅, 𝑋, {𝑏}, {𝑐}, {𝑑}, {𝑏, 𝑐}, {𝑐, 𝑑}, {𝑏, 𝑑}, {𝑎, 𝑏, 𝑐}, {𝑏, 𝑐, 𝑑}}, 

�̂�∗𝑐(𝑋, 𝜏)
= {∅, 𝑋, {𝑏}, {𝑐}, {𝑑}, {𝑎, 𝑏}, {𝑏, 𝑐}, {𝑎, 𝑐}, {𝑏, 𝑑}, {𝑐, 𝑑}, {𝑎, 𝑏, 𝑐}, {𝑏, 𝑐, 𝑑}, {𝑎, 𝑐, 𝑑}, {𝑎, 𝑏, 𝑑}},  

�̂�∗∗𝑠𝑐{𝑋, 𝜏} = {∅, 𝑋, {𝑏}, {𝑐}, {𝑑}, {𝑏, 𝑐}, {𝑐, 𝑑}, {𝑏, 𝑑}, {𝑎, 𝑏, 𝑐}, {𝑏, 𝑐, 𝑑}},  

�̂�∗∗𝑠𝑜(𝑋, 𝜏) = {∅, 𝑋, {𝑎}, {𝑑}, {𝑎, 𝑏}, {𝑎, 𝑐}, {𝑎, 𝑑}, {𝑎, 𝑏, 𝑐}, {𝑎, 𝑐, 𝑑}, {𝑎, 𝑏, 𝑑}} . 

 
Remark. 4.20. Every 𝑆�̂�∗-space need not be a door space. 

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Example. 4.21. Let 𝑋 = {𝑎, 𝑏, 𝑐}, 𝜏 = {∅, 𝑋, {𝑎}, {𝑏, 𝑐}}, 𝜏𝑐 = {∅, 𝑋, {𝑎}, {𝑏, 𝑐}},

𝑠𝑐(𝑋, 𝜏) = {∅, 𝑋, {𝑎}, {𝑏, 𝑐}}  𝑎𝑛𝑑 �̂�∗𝑐(𝑋, 𝜏) = {∅, 𝑋, {𝑎}, {𝑏, 𝑐}}. Here 𝑠𝑐 ∩ �̂�∗𝑐 = �̂�∗𝑐 
but it is not a door space.  

 
Theorem. 4.22. In 𝑆�̂�∗-space 
i) (𝑠𝑔)∗-Closed set coincide with �̂�∗∗𝑠-closed set. 
ii) 𝑔𝑠∗∗-Closed set coincide with �̂�∗∗𝑠-closed set. 
iii) �̂�∗-Closed set coincide with �̂�∗∗𝑠-closed set. 
iv) 𝛼𝑔-Closed set coincide with �̂�∗∗𝑠-closed set. 
v) 𝑝𝑠-Closed set coincide with �̂�∗∗𝑠-closed set. 
vi) 𝑔𝑠𝑝-Closed set coincide with �̂�∗∗𝑠-closed set. 
vii) 𝑠𝑝-Closed set coincide with �̂�∗∗𝑠-closed set. 
viii) 𝑠𝑔-Closed set coincide with �̂�∗∗𝑠-closed set. 
ix) 𝑔𝛼-Closed set coincide with �̂�∗∗𝑠-closed set. 
𝑔𝑠-Closed set coincide with �̂�∗∗𝑠-closed set. 

 
Theorem. 4.23. In 𝑆�̂�∗-space, semi-closed and �̂�-closed are independent. 

 
Theorem. 4.24. The property of being �̂�∗∗𝑠-door space is a topological property. 
Proof. Let (𝑋, 𝜏) be a �̂�∗∗𝑠-door space and let 𝑓: 𝑋 → 𝑌 be a homeomorphism. Let𝐴 ⊆
𝑌, consider 𝑓 −1(𝐴) ⊆ 𝑋, since X is a �̂�∗∗𝑠-door space, then 𝑓 −1(𝐴) is either �̂�∗∗𝑠-open 

or �̂�∗∗𝑠-closed in X. Now,𝑓(𝑓 −1(𝐴)) = 𝐴. Then A is either �̂�∗∗𝑠-open or �̂�∗∗𝑠-closed in 

Y. Therefore, Y is a �̂�∗∗𝑠-door space. 

 
Theorem. 4.25. The property of being a �̂�∗∗𝑠-door space is an expensive property. 
Proof. Suppose (𝑋, 𝜏) is a �̂�∗∗𝑠-door space. Let 𝜏 ⊂ 𝜏1 and𝐴 ⊆ 𝑋. Since, (𝑋, 𝜏) is a 
�̂�∗∗𝑠-door space, then A is either �̂�∗∗𝑠 − 𝜏-open or �̂�∗∗𝑠 − 𝜏-closed. Since𝜏1 ⊂ 𝜏, A is 
either �̂�∗∗𝑠 − 𝜏1-open or �̂�

∗∗𝑠 − 𝜏1-closed. Then (𝑋, 𝜏1) is a �̂�
∗∗𝑠-door space. 

 
Theorem. 4.26. Let (𝑋, 𝜏) be a �̂�∗∗𝑠-door space and 𝑌 ⊆ 𝑋 be a clopen subset of Y, 
then (𝑌, 𝜏𝑌) is also a �̂�

∗∗𝑠-door space. 
Proof. Let 𝐴 ⊆ 𝑌 be a subset of Y. Now,𝐴 ⊆ 𝑋. By hypothesis, (𝑋, 𝜏) is a �̂�∗∗𝑠-door 
space which implies that A is either �̂�∗∗𝑠-open or �̂�∗∗𝑠-closed in X. Since Y is both 
open and closed. Then A is either �̂�∗∗𝑠-open or �̂�∗∗𝑠-closed in Y. Therefore, Y is also a 
�̂�∗∗𝑠-door space. 

 
Proposition. 4.27. Let A and B be ĝ∗∗s-closed subsets of X such that 𝑐𝑙(𝐴) = 𝑠𝑐𝑙(𝐴) 
and𝑐𝑙(𝐵) = 𝑠𝑐𝑙(𝐵). Thus 𝐴 ∪ 𝐵 is ĝ∗∗s-closed. 

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A study on �̂�∗∗𝑠 − 𝑅0 and �̂�
∗∗𝑠 − 𝑅1 spaces in Topological spaces 

 
Proof. Let U be ĝ∗-open such that𝐴 ∪ 𝐵 ⊆ 𝑈 ⇒ 𝐴 ⊆ 𝑈 𝑎𝑛𝑑 𝐵 ⊆ 𝑈 ⇒ 𝑠𝑐𝑙(𝐴) ⊆
𝑈 𝑎𝑛𝑑 𝑠𝑐𝑙(𝐵) ⊆ 𝑈 ⇒ 𝑐𝑙(𝐴) ⊆ 𝑈𝑎𝑛𝑑 𝑐𝑙(𝐵) ⊆ 𝑈 ⇒ 𝑐𝑙(𝐴 ∪ 𝐵) = 𝑐𝑙(𝐴) ∪ 𝑐𝑙(𝐵) ⊆ 𝑈 ⇒
𝑐𝑙(𝐴 ∪ 𝐵) ⊆ 𝑈 ⇒ 𝑠𝑐𝑙(𝐴 ∪ 𝐵) ⊆ 𝑐𝑙(𝐴 ∪ 𝐵) ⊆ 𝑈. Therefore, 𝐴 ∪ 𝐵 is ĝ∗∗s-closed.    

 
Proposition. 4.28. If B is a regular open subset of a topological space(𝑋, 𝜏), then B is 
ĝ∗∗s-closed. 
Proof. Let U be ĝ∗-open set such that𝐵 ⊆ 𝑈. Since B is regular open, 𝑖𝑛𝑡(𝑐𝑙(𝐵)) =

𝐵 ⇒ 𝑖𝑛𝑡(𝑐𝑙(𝐵)) ⊆ 𝐵 ⊆ 𝑈 ⇒ 𝐵 is semi − closed ⇒ 𝑠𝑐𝑙(𝐵) = 𝐵 ⊆ 𝑈 ⇒ 𝑠𝑐𝑙(𝐵) ⊆ 𝑈. 

Therefore, B is ĝ∗∗s-closed. 

 
Theorem. 4.29. If A and B are subsets of X such that 𝐴 ⊆ 𝐵 and A is 𝑔∗-closed, then B 
is ĝ∗∗s-closed. 
Proof. Assume 𝐵 ⊆ 𝑈 and U is ĝ∗-open. Therefore, 𝐴 ⊆ 𝐵 ⇒ 𝐴 ⊆ 𝑈. Also, every ĝ∗-
open set is g-open which implies U is g-open in X. Therefore, 𝐴 ⊆ 𝑈 and U is g-open ⇒
𝑐𝑙(𝐴) ⊆ 𝑈 ⇒ 𝑐𝑙(𝐵) ⊆ 𝑈 ⇒ 𝑠𝑐𝑙(𝐵) ⊆ 𝑐𝑙(𝐵) ⊆ 𝑈 ⇒ 𝑠𝑐𝑙(𝐵) ⊆ 𝑈. Therefore B is ĝ∗∗s-
closed. 

 
Theorem. 4.30. If A and B are subsets of X such that 𝐴 ⊆ 𝐵 and A is 𝑔∗𝑠-closed, then 
B is ĝ∗∗s-closed. 
Proof. Assume that 𝐵 ⊆ 𝑈 and U is ĝ∗-open. Since𝐴 ⊆ 𝐵 ⇒ 𝐴 ⊆ 𝑈. Also, every ĝ∗-
open set is gs-open which implies U is gs-open in X. Therefore, 𝐴 ⊆ 𝑈 and U is gs-
open⇒ 𝑠𝑐𝑙(𝐴) ⊆ 𝑈 ⇒ 𝑠𝑐𝑙(𝐴) ⊆ 𝑠𝑐𝑙(𝐵) ⊆ 𝑈 ⇒ 𝑠𝑐𝑙(𝐵) ⊆ 𝑈. Hence proved. 
 

5. Conclusion 

We have studied the concept of 𝑅0-space, 𝑅1-space, door space, and submaximal 
space via ĝ∗∗s-closed set and examples are provided to state the converse doesn’t 
implies. It shows that every ĝ∗∗s − 𝑅0-space is ĝ

∗∗s − 𝑇2-space, ĝ
∗∗s-regular and ĝ∗∗s −

𝑅1-space. In addition we have shown that every �̂�
∗∗𝑠-door space need not be a �̂�∗∗𝑠-

submaximal and suitable examples are given for it. A new concept of 𝑆�̂�∗-space is also 
introduced. Further, we have compared �̂�∗∗𝑠-closed set with other existing closed sets 
and we have also investigated the characterization of these spaces. 

 
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