/compile/output.dvi


New Generalized Classes of τω

Otchana Thevar Ravi1,∗, Ilangovan Rajasekaran1, Soundararajan Satheesh Kanna2

and Malliharjunaiah Paranjothi3

1 Department of Mathematics, P. M. Thevar College, Usilampatti, Madurai District,

Tamil Nadu, India.
2 Department of Mathematics, Research Scholar, Bharathidasan University, Tiruchirapalli, Tamil

Nadu, India.
3 Department of Mathematics, Sree Sowdambiga College of Engineering, Aruppukottai,

Virudhunagar District, Tamil Nadu, India.

Abstract. The purpose of this paper is to introduce a new class of sets called semi-ω-open which lies

between the class of α − ω-open sets and the class of β − ω-open sets and to investigate the basic

properties of such sets. This apart, some new generalized classes of τω are introduced and investigated

on the line of research.

2010 Mathematics Subject Classifications: 54C05, 54C08, 54C10

Key Words and Phrases: ω-open set, α−ω-open set, pre-ω-open set, β −ω-open set, b−ω-open set,

ω− t-set, δ−ω-open set, semi⋆ −ω-closed set.

1. Introduction

In 1982, the notions of ω-closed sets and ω-open sets were introduced and studied by

Hdeib [7]. In 2009, Noiri et al. [10] introduced some generalizations of ω-open sets and

investigated some properties of the sets. Moreover, they used them to obtain decompositions

of continuity.

In this paper, we introduce and investigate the new notion called semi-ω-open sets which

is weaker than α −ω-open sets and stronger than β −ω-open sets. Also we introduce and

investigate some new generalized classes of τω.

∗Corresponding author.

Email addresses: siingam@yahoo.com (O. Ravi), rajasekarani@yahoo.com (I. Rajasekaran),

satheesh−kanna@yahoo.co.in (S. Kanna) and jothimp123@gmail.com (M. Paranjothi)

http://www.ejpam.com 152 c© 2016 EJPAM All rights reserved.

EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 9, No. 2, 2016, 152-164

ISSN 1307-5543 – www.ejpam.com



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 153

2. Preliminaries

Throughout this paper, R (resp. Q, Q⋆) denotes the set of all real numbers (resp. the set

of all rational numbers, the set of all irrational numbers).

By a space (X ,τ), we always mean a topological space (X ,τ)with no separation properties

assumed. If H ⊂ X , cl(H) and int(H) will, respectively, denote the closure and interior of H

in (X ,τ). τH denotes the relative topology on H and τu denotes the usual topology on R.

Definition 1. A subset H of a space (X ,τ) is said to be semi-open [9] if H ⊂ cl(int(H)).

Definition 2 ([11]). Let H be a subset of a space (X ,τ), a point p in X is called a condensation

point of H if for each open set U containing p, U ∩ H is uncountable.

Definition 3 ([7]). A subset H of a space (X ,τ) is calledω-closed if it contains all its condensation

points.

The complement of an ω-closed set is called ω-open.

It is well known that a subset W of a space (X ,τ) is ω-open if and only if for each x ∈W ,

there exists U ∈ τ such that x ∈ U and U −W is countable. The family of all ω-open sets,

denoted by τω, is a topology on X , which is finer than τ. The interior and closure operator in

(X ,τω) are denoted by intω and clω respectively.

Lemma 1 ([7]). Let H be a subset of a space (X ,τ). Then

(i) H is ω-closed in X if and only if H = clω(H).

(ii) clω(X\H) = X\intω(H).

(iii) clω(H) is ω-closed in X .

(iv) x ∈ clω(H) if and only if H ∩ G 6= φ for each ω-open set G containing x.

(v) clω(H) ⊂ cl(H).

(vi) int(H) ⊂ intω(H).

Remark 1. For a subset of a space (X ,τ), the following property holds:

Every closed set is ω-closed but not conversely [2, 7].

Definition 4. [1] A space (X ,τ) is called anti-locally countable if each non-empty open set is

uncountable.

Lemma 2 ([8]). Let (H,τH) be an anti-locally countable subspace of a space (X ,τ). Then

cl(H) = clω(H).

Lemma 3 ([6]). If U is an open set, then cl(U ∩ H) = cl(U ∩ cl(H)) and hence U ∩ cl(H) ⊂
cl(U ∩ H) for any subset H.

Lemma 4 ([1, 4]). If (X ,τ) is an anti-locally countable space, then intω(H) = int(H) for every

ω-closed set H of X and clω(H) = cl(H) for every ω-open set H of X.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 154

Definition 5 ([10]). A subset H of a space (X ,τ) is called

(i) α−ω-open if H ⊂ intω(cl(intω(H)));

(ii) pre-ω-open if H ⊂ intω(cl(H));

(iii) β −ω-open if H ⊂ cl(intω(cl(H)));

(iv) b−ω-open if H ⊂ intω(cl(H))∪ cl(intω(H)).

Definition 6 ([10]). A subset H of a space (X ,τ) is called an ω− t-set if int(H) = intω(cl(H)).

Definition 7. A space (X ,τ) is called submaximal [5] if every dense subset is open.

Definition 8. A subset H of a space (X ,τ) is called ω-dense [3] if clω(H) = X .

3. Properties of Semi-ω-Open Sets

Definition 9. A subset H of a space (X ,τ) is said to be

(i) semi-ω-open if H ⊂ cl(intω(H)).

(ii) semi-ω-closed if int(clω(H)) ⊂ H.

The complement of semi-ω-open set is called semi-ω-closed.

Example 1. Let X = {a, b, c} with the topology τ= {φ, X , {a}, {a, b}}. Then {a} is semi-ω-open.

Example 2. Let X = Rwith the usual topology τu. Let H = (0,1)∩Q. Then H is not semi-ω-open,

since cl(intω(H)) = cl(φ) = φ.

Proposition 1. In a space (X ,τ), every semi-open subset is semi-ω-open.

Proof. Let H be semi-open in (X ,τ). Then H ⊂ cl(int(H)) ⊂ cl(intω(H)). This proves that

H is semi-ω-open.

Remark 2. The converse of Proposition 1 is not true.

Example 3. Let X = R with the usual topology τu. Then H = Q⋆ is semi-ω-open for

cl(intω(H)) = cl(H) = R and H ⊂ cl(intω(H)). But H is not semi-open for

cl(int(H)) = cl(φ) = φ and H 6⊆ cl(int(H)).

From the above Example, we observe that the converse fails in an anti-locally countable

space also.

Theorem 1. In an anti-locally countable space, an ω-closed and a semi-ω-open subset is semi-

open.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 155

Proof. Let (X ,τ) be an anti-locally countable space and H be an ω-closed and a semi-ω-

open subset.

Since H is semi-ω-open, H ⊂ cl(intω(H)). Since (X ,τ) is anti-locally countable and H is

ω-closed, intω(H) = int(H) by Lemma 4. Hence H ⊂ cl(intω(H)) = cl(int(H)) and thus H

is semi-open.

Theorem 2. For a subset of space (X ,τ), the following properties hold:

(i) Every ω-open set is semi-ω-open.

(ii) Every α−ω-open set is semi-ω-open.

(iii) Every semi-ω-open set is β −ω-open.

(iv) Every semi-ω-open set is b−ω-open.

Proof. (i). If H is an ω-open set, then H = intω(H) ⊂ cl(intω(H)). Therefore H is semi-

ω-open.

(ii). If H is an α−ω-open set, then H ⊂ intω(cl(intω(H))) ⊂ cl(intω(H)). Therefore H is

semi-ω-open.

(iii). If H is an semi-ω-open set, then H ⊂ cl(intω(H)) ⊂ cl(intω(cl(H))). Therefore H is

β −ω-open.

(iv). If H is an semi-ω-open set, then H ⊂ cl(intω(H)) ⊂ intω(cl(H)) ∪ cl(intω(H)).

Therefore H is b−ω-open.

The following Examples support that the separate converses of Theorem 2 are not true in

general.

Example 4. Let X = R with the usual topology τu.

(i) Let H = (0,1]. Then H is semi-ω-open set but not ω-open, since

H = (0,1] 6= (0,1) = intω(H).

(ii) Let H = (0,1]. Then H is semi-ω-open set but not α−ω-open, since

intω(cl(intω(H))) = intω(cl(0,1)) = intω([0,1]) = (0,1).

(iii) Let H = [0,1]∩Q. Then H is β −ω-open set but not semi-ω-open, since

cl(intω(H)) = cl(φ) = φ.

(iv) Let H = Q. Then H is b−ω-open set but not semi-ω-open, since cl(intω(H)) = cl(φ) = φ.

Theorem 3. Let H be a subset of a space (X ,τ). Then H is α−ω-open if and only if it is semi-ω-

open and pre-ω-open.

Proof. Let H be an α−ω-open. Then H ⊂ intω(cl(intω(H))). It implies that

H ⊂ intω(cl(intω(H))) ⊂ cl(intω(H)) and H ⊂ intω(cl(intω(H))) ⊂ intω(cl(H)). Thus H is

semi-ω-open and pre-ω-open.

Conversely, let H be semi-ω-open and pre-ω-open. Then we have H ⊂ cl(intω(H)) and

H ⊂ intω(cl(H)). Hence H ⊂ intω(cl(H)) ⊂ intω(cl(intω(H)))which implies that H is α−ω-

open.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 156

Remark 3. The concepts of semi-ω-openness and pre-ω-openness are independent.

Example 5. Let X = R with the usual topology τu. The interval H = (0,1] is semi-ω-open but

not pre-ω-open, since intω(cl(H)) = intω([0,1]) = (0,1).

Example 6. Let X = R with the usual topology τu. Let H = Q. Then H is pre-ω-open but not

semi-ω-open, since cl(intω(H)) = cl(φ) = φ.

Proposition 2. The intersection of a semi-ω-open set and an open set is semi-ω-open.

Proof. Let H be a semi-ω-open and U be an open set in X. Then H ⊂ cl(intω(H)) and

int(U) = U . By Lemma 3, we have

U ∩ H ⊂U ∩ cl(intω(H)) ⊂ cl(U ∩ intω(H))

=cl(int(U)∩ intω(H)) ⊂ cl(intω(U)∩ intω(H))

=cl(intω(U ∩ H)).

Therefore U ∩ H is semi-ω-open.

Remark 4. The intersection of two semi-ω-open sets need not be semi-ω-open. This can be seen

from the following Example.

Example 7. Let X = R with the usual topology τu. Let A= (0,1] and B = [1,2), then A and B

are semi-ω-open, but A∩ B = {1} which is not semi-ω-open, since cl(intω(A∩ B)) = cl(φ) = φ.

Theorem 4. Let H be a subset of a space (X ,τ). If H is both closed and β −ω-open, then H is

semi-ω-open.

Proof. Since H is a β −ω-open set, H ⊂ cl(intω(cl(H))) = cl(intω(H)), H being closed.

Therefore H is semi-ω-open.

Theorem 5. Let H be a subset of a space (X ,τ). If H is both β −ω-open and ω− t-set, then H

is semi-ω-open.

Proof. Since H is a ω− t-set, int(H) = intω(cl(H)). Since H is β −ω-open also,

H ⊂ cl(intω(cl(H))) ⊂ cl(int(H)) ⊂ cl(intω(H)).

Therefore H is semi-ω-open.

Theorem 6. Let H be a subset of a space (X ,τ). If H is both b−ω-open and ω− t-set, then H is

semi-ω-open.

Proof. Since H is ω − t-set, intω(cl(H)) = int(H) ⊂ intω(H). Since H is b − ω-open

also, H ⊂ intω(cl(H))∪ cl(intω(H)) ⊂ intω(H)∪ cl(intω(H)) = cl(intω(H)). Therefore H is

semi-ω-open.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 157

Proposition 3. Let H be a subset of a space (X ,τ). Then H is semi-ω-open if and only if

cl(H) = cl(intω(H)).

Proof. Let H be semi-ω-open. Then H ⊂ cl(intω(H)) and cl(H) ⊂ cl(intω(H)). But always

cl(intω(H)) ⊂ cl(H). Thus, we obtain that cl(H) = cl(intω(H)).

Conversely, let the condition hold. We have H ⊂ cl(H) = cl(intω(H)), by the given condi-

tion. Thus H ⊂ cl(intω(H)) and hence H is semi-ω-open.

Proposition 4. Let H ⊂ (X ,τ) be a b−ω-open set such that cl(H) = φ. Then H is semi-ω-open.

Theorem 7. For a subset H of a submaximal space (X ,τ), the following properties are equivalent.

(i) H is semi-ω-open,

(ii) H is β −ω-open.

Proof. (i)⇒ (ii): It follows from the fact that every semi-ω-open set is β −ω-open.

(ii)⇒ (i): Let H be a β −ω-open set in X. Then H ⊂ cl(intω(cl(H))) and

cl(H) ⊂ cl(intω(cl(H))). Thus, cl(H) is semi-ω-open. Put A= cl(H) and K = H ∪ (X\cl(H)).

We have H = cl(H) ∩ K and cl(K) = X . This implies that H = A∩ K , where A is semi-ω-

open and K is dense. Since X is submaximal, then K is open. By Proposition 2, H = A∩ K is

semi-ω-open.

Theorem 8. A subset H of a space (X ,τ) is semi-ω-open if and only if there exists U ∈ τω such

that U ⊂ H ⊂ cl(U).

Proof. Let H be semi-ω-open. Then H ⊂ cl(intω(H)). Take intω(H) = U . Then, we have

U ⊂ H ⊂ cl(U).

Conversely, let U ⊂ H ⊂ cl(U) for some U ∈ τω. Since U ⊂ H, we have U ⊂ intω(H) and

hence cl(U) ⊂ cl(intω(H)). Thus we obtain H ⊂ cl(intω(H)) and H is semi-ω-open.

Corollary 1. If A is a semi-ω-open set in a space (X ,τ) and A⊂ B ⊂ cl(A), then B is semi-ω-open

in X.

Proof. Since A is semi-ω-open, A⊂ cl(intω(A)) ⊂ cl(intω(B)) for A⊂ B. So

cl(A) ⊂ cl(intω(B)). Since B ⊂ cl(A), B ⊂ cl(intω(B)). Thus B is semi-ω-open.

4. Properties of δ−ω-Open Sets

Definition 10. A subset H of a space (X ,τ) is said to be

(i) δ−ω-open if intω(cl(H)) ⊂ cl(intω(H)).

(ii) δ−ω-closed if int(clω(H)) ⊂ clω(int(H)).

The complement of δ−ω-open set is called δ−ω-closed.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 158

Example 8. Let X = R with the usual topology τu. Let H = Q. Then H is not δ−ω-open, since

intω(cl(Q)) = intω(R) = R and cl(intω(Q)) = cl(φ) = φ.

Example 9. Let X = R with the usual topology τu. Let H = (0,1]. Then H is δ−ω-open, since

intω(cl((0,1])) = intω([0,1]) = (0,1) and cl(intω(H)) = cl(0,1) = [0,1].

Proposition 5. For a subset of a space (X ,τ), the following properties hold:

(i) Every α−ω-open set is δ−ω-open.

(ii) Every ω− t-set is δ−ω-open.

Proof. (i) Since H is an α−ω-open set, H ⊂ intω(cl(intω(H))) ⊂ cl(intω(H)). Then we

obtain cl(H) ⊂ cl(intω(H)) and intω(cl(H)) ⊂ cl(H) ⊂ cl(intω(H)). Therefore H is δ −ω-

open.

(ii) Since H is an ω− t-set, intω(cl(H)) = int(H) ⊂ H. Then we obtain

intω(cl(H)) ⊂ intω(H) ⊂ cl(intω(H)).

Therefore H is δ−ω-open.

Example 10. Let X = R with the usual topology τu.

(i) Let H = (0,1]. Then H is δ−ω-open but not α−ω-open, since intω(cl(H)) = (0,1) and

cl(intω(H)) = [0,1].

(ii) Let H = Q⋆. Then H is δ−ω-open but notω− t-set, since int(Q⋆) = φ, intω(cl(Q⋆)) = R

and cl(intω(Q
⋆)) = cl(Q⋆) = R.

Definition 11. A subset H of a space (X ,τ) is said to be β −ω-closed if int(clω(int(H))) ⊂ H.

The complement of β −ω-open set is called β −ω-closed.

Proposition 6. Let H be a subset of a space (X ,τ). Then H is β −ω-closed if and only if

int(clω(int(H))) = int(H).

Proof. Since H is β −ω-closed set, int(clω(int(H))) ⊂ H and then we obtain

int(clω(int(H))) ⊂ int(H). But int(H) ⊂ int(clω(int(H))). Thus we have

int(H) = int(clω(int(H))).

Conversely, let the condition hold. We have int(clω(int(H))) = int(H) ⊂ H. Therefore H

is β −ω-closed.

Theorem 9. For a subset H of a space (X ,τ), the following properties are equivalent:

(i) H is semi-ω-closed.

(ii) H is β −ω-closed and δ−ω-closed.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 159

Proof. (i)⇒ (ii): Let H be semi-ω-closed. By Theorem 2(iii), H is β −ω-closed. Since H

is semi-ω-closed, int(clω(H)) ⊂ H and int(clω(H)) ⊂ int(H). It gives that

clω(int(clω(H))) ⊂ clω(int(H)). Thus int(clω(H)) ⊂ clω(int(clω(H))) ⊂ clω(int(H)) and so

H is δ−ω-closed.

(ii)⇒ (i): Since H is δ−ω-closed, int(clω(H)) ⊂ clω(int(H)) and

int(clω(H)) ⊂ int(clω(int(H))). Since H is β − ω-closed, int(clω(int(H))) ⊂ H. Then

int(clω(H)) ⊂ H and so H is semi-ω-closed.

Remark 5. The concepts of β −ω-closedness and δ−ω-closedness are independent.

Example 11.

(i) Let X = R with the topology τ = {φ, X ,Q⋆}. Let H = Q⋆. Then H is δ−ω-closed but not

β −ω-closed, since Q is not β −ω-open.

(ii) Let X = R with the usual topology τu. Let H = Q⋆. Then H is β − ω-closed but not

δ−ω-closed, since Q is not δ−ω-open.

Theorem 10. Let (X ,τ) be a space. Then a subset of X is α −ω-open if and only if it is both

δ−ω-open and pre-ω-open.

Proof. Necessity: Let H be an α −ω-open set. Then H ⊂ intω(cl(intω(H))). It implies

that cl(H) ⊂ cl(intω(H)) and intω(cl(H)) ⊂ intω(cl(intω(H))) ⊂ cl(intω(H)). Hence, H is a

δ−ω-open set. On the other hand, since H is an α−ω-open set, H is a pre-ω-open set.

Sufficiency: Let H be both δ−ω-open and pre-ω-open. Since H is δ−ω-open, we have

intω(cl(H)) ⊂ cl(intω(H)) and hence intω(cl(H)) ⊂ intω(cl(intω(H))). Since H is pre-ω-

open, we have H ⊂ intω(cl(H)). Therefore we obtain that H ⊂ intω(cl(intω(H))) which

proves that H is an α−ω-open set.

Remark 6. The concepts of δ−ω-openness and pre-ω-openness are independent.

Example 12. Let X = R with the usual topology τu.

(i) H = (0,1] is δ−ω-open but not pre-ω-open.

(ii) H = Q is pre-ω-open but not δ−ω-open.

Proposition 7. Let A and B be subsets of a space (X ,τ). If A⊂ B ⊂ cl(A) and A is δ−ω-open in

X, then B is δ−ω-open in X.

Proof. Suppose that A⊂ B ⊂ cl(A) and A is δ−ω-open in X. Then, we have

intω(cl(A)) ⊂ cl(intω(A)). Since A⊂ B, cl(intω(A)) ⊂ cl(intω(B)) and

intω(cl(A)) ⊂ cl(intω(B)). Since B ⊂ cl(A), we have cl(B) ⊂ cl(cl(A)) = cl(A) and

intω(cl(B)) ⊂ intω(cl(A)). Therefore we obtain that intω(cl(B)) ⊂ cl(intω(B)). This shows

that B is a δ−ω-open set.

Corollary 2. Let (X ,τ) be a space. If A⊂ X is δ−ω-open and dense in (X ,τ), then every subset

of X containing A is δ−ω-open.

Proof. It is obvious by Proposition 7.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 160

5. Properties of Semi ⋆ −ω-Open Sets

Definition 12. A subset H of a space (X ,τ) is said to be

(i) semi⋆ −ω-open if H ⊂ clω(int(H)).

(ii) semi⋆ −ω-closed if intω(cl(H)) ⊂ H.

The complement of a semi⋆ −ω-open set is called semi⋆ −ω-closed.

Example 13. Let X = {a, b, c} with the topology τ= {φ, X , {a}, {a, b}}.

(i) Let H = {a}. Then H is semi⋆ −ω-open, since int(H) = {a} and clω(int(H)) = {a}.

(ii) Let H = {c}. Then H is not semi⋆ −ω-open, since int(H) = φ and clω(int(H)) = φ.

Proposition 8. For a subset of a space (X ,τ), every semi⋆ −ω-open set is semi-ω-open.

Proof. If H is semi⋆ −ω-open set, then H ⊂ clω(int(H)) ⊂ cl(intω(H)). Therefore H is

semi-ω-open.

Example 14. Let X = R with the usual topology τu. Let H = Q⋆. Then H is semi-ω-open but

not semi⋆ −ω-open, since cl(intω(H)) = cl(Q⋆) = R and clω(int(H)) = clω(φ) = φ.

Proposition 9. A subset H of a space (X ,τ) is semi⋆−ω-open if and only if clω(H) = clω(int(H)).

Proof. If H is semi⋆ −ω-open set, then H ⊂ clω(int(H)) and clω(H) ⊂ clω(int(H)). But

clω(int(H)) ⊂ clω(H). Hence clω(H) = clω(int(H)).

Conversely, let the condition hold. We have H ⊂ clω(H) and clω(H) = clω(int(H)). There-

fore H is semi⋆ −ω-open.

Definition 13. A subset H of a space (X ,τ) is said to be ω⋆ − t-set if intω(cl(H)) = intω(H).

Example 15. Let X = R with the usual topology τu.

(i) Let H = (0,1]. Then H is a ω⋆ − t-set.

(ii) Let H = Q⋆. Then H is not a ω⋆ − t-set.

Proposition 10. In a space (X ,τ), every closed set is a ω⋆ − t-set.

Proof. Let H be a closed set. Then H = cl(H) and we have intω(cl(H)) = intω(H) which

proves that H is a ω⋆ − t-set.

The converse of Proposition 10 is not true as can be seen from the following Example.

Example 16. Let X = R with the usual topology τu. Let H = (0,1]. Then H isω⋆− t-set but not

closed.

Proposition 11. In a space (X ,τ), every ω− t-set is a ω⋆ − t-set.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 161

Proof. If H is a ω− t-set, then intω(cl(H)) = int(H) ⊂ intω(H) ⊂ intω(cl(H)). Thus we

have intω(cl(H)) = intω(H) and hence H is a ω⋆ − t-set.

Example 17. Let X = {a, b, c} with the topology τ = {φ, X , {a}, {a, b}}. Then H = {c} is a

ω⋆− t-set but not aω− t-set. Since intω(H) = H, int(H) = φ and intω(cl(H)) = intω(H) = H,

we have intω(cl(H)) = intω(H) and intω(cl(H)) 6= int(H). This proves that H is a ω⋆ − t-set

but not a ω− t-set.

Theorem 11. A subset H of a space (X ,τ) is semi⋆ −ω-closed if and only if H is a ω⋆ − t-set.

Proof. Let H be a semi⋆−ω-closed set in X. Then X\H is semi⋆−ω-open. By Proposition 9,

we have clω(X\H) = clω(int(X\H)). It follows that

X\intω(H) = clω(X\cl(H)) = X\intω(cl(H)).

Thus, intω(cl(H)) = intω(H) and hence H is a ω⋆ − t-set in X.

Conversely, let H be a ω⋆ − t-set. Then intω(cl(H)) = intω(H) ⊂ H. Therefore H is

semi⋆ −ω-closed.

Proposition 12. If A and B are ω⋆ − t-sets of a space (X ,τ), then A∩ B is a ω⋆ − t-set.

Proof. Let A and B be ω⋆ − t-sets. Then we have

intω(A∩ B) ⊂intω(cl(A∩ B)) ⊂ intω(cl(A)∩ cl(B))

=intω(cl(A))∩ intω(cl(B)) = intω(A)∩ intω(B) = intω(A∩ B).

Then intω(A∩ B) = intω(cl(A∩ B)) and hence A∩ B is an ω⋆ − t-set.

Definition 14. A subset H of a space (X ,τ) is said to be semi-ω-regular if H is semi-ω-open and

a ω⋆ − t-set.

Example 18. Let X = R with the usual topology τu.

(i) Let H = (0,1]. Then H is semi-ω-regular.

(ii) Let H = R\Q. Then H is not semi-ω-regular, since H is not ω⋆ − t-set.

Theorem 12. Let H be a subset of a space (X ,τ). Then H is semi-ω-regular if and only if H is

both β −ω-open and semi⋆ −ω-closed.

Proof. If H is semi-ω-regular, then H is both semi-ω-open and a ω⋆ − t-set. Since every

semi-ω-open set is β −ω-open, H is both β −ω-open and a ω⋆ − t-set. By Theorem 11, we

obtain the result.

Conversely, let H be semi⋆ −ω-closed and β −ω-open. Since H is a semi⋆ −ω-closed, by

Theorem 11 H is a ω⋆ − t-set. Since H is β −ω-open, H ⊂ cl(intω(cl(H))) = cl(intω(H)).

Therefore H is semi-ω-open. Since H is both semi-ω-open and a ω⋆ − t-set, H is semi-ω-

regular.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 162

Remark 7. The concepts of β −ω-openness and semi⋆ −ω-closedness are independent.

Example 19. (i) Let X = Rwith the topology τ= {φ, X ,Q⋆}. Then H = Q is semi⋆−ω-closed

but not β −ω-open. Since intω(cl(H)) = intω(H) = φ ⊂ H, H is semi⋆−ω-closed. Again

since H 6⊆ cl(intω(cl(H))) = φ, H is not β −ω-open.

(ii) Let X = R with the usual topology τu. Let H = Q. Then H is β−ω-open but not semi⋆−ω-

closed, since intω(cl(H)) = intω(R) = R.

6. Properties of ω−R-Closed Sets

Definition 15. A subset H of a space (X ,τ) is called ω−R-closed if H = cl(intω(H)).

Theorem 13. Let (X ,τ) be a space and H a subset of X. Then the following properties are equiv-

alent.

(i) H 6= φ is ω−R-closed.

(ii) There exists a non-empty ω-open set G such that G ⊂ H = cl(G).

(iii) There exists a non-empty ω-open set G such that H = G ∪ (cl(G)− G).

Proof. (i)⇒ (ii): Suppose H 6= φ is an ω−R-closed set. Then H = cl(intω(H)). Let

G = intω(H). G is the required ω-open set such that G ⊂ H = cl(G).

(ii)⇒ (iii): Since H = cl(G) = G ∪ (cl(G)− G) where G is a nonempty ω-open set, (iii)

follows.

(iii)⇒ (i): H = G∪(cl(G)−G) implies that H = cl(G) = cl(intω(G)) ⊂ cl(intω(H)), since

G is ω-open and G ⊂ H. Again intω(H) ⊂ H implies that cl(intω(H)) ⊂ cl(H) = cl(G) = H.

Therefore H = cl(intω(H)) which implies that H is ω−R-closed.

Theorem 14. Let H be a subset of a space (X ,τ). If H is β −ω-open, then cl(H) is ω−R-closed.

Proof. Suppose H is β −ω-open. Then H ⊂ cl(intω(cl(H))) and so

cl(H) ⊂ cl(intω(cl(H))) ⊂ cl(H)which implies that cl(H) = cl(intω(cl(H))). Therefore cl(H)

is ω−R-closed.

Theorem 15. Let H be a subset of a space (X ,τ). Then the following properties are equivalent.

(i) H is ω−R-closed.

(ii) H is semi-ω-open and closed.

(iii) H is β −ω-open and closed.



O. Ravi, I. Rajasekaran, S. Kanna and M. Paranjothi / Eur. J. Pure Appl. Math, 9 (2016), 152-164 163

Proof. (i)⇒ (ii): If H is ω−R-closed, then H = cl(intω(H)) and cl(H) = cl(intω(H)).

Since H ⊂ cl(intω(H)), H is semi-ω-open. Also, H = cl(H) and so H is closed.

(ii)⇒ (iii): It follows from the fact that every semi-ω-open set is a β −ω-open.

(iii) ⇒ (i): Suppose H is β −ω-open and closed. Then H ⊂ cl(intω(cl(H))) and H =

cl(H). Now cl(intω(H)) ⊂ cl(H) = H. Also, H ⊂ cl(intω(H)). Therefore H = cl(intω(H))

which implies that H is ω−R-closed.

Remark 8. (i) The concepts of semi-ω-openness and closedness are independent.

(ii) The concepts of β −ω-openness and closedness are independent.

Example 20. (i) Let X = R with the usual topology τ. Let H = (0,1]. Then H is semi-ω-open

but not closed.

(ii) Let X = R with the topology τ = {φ,R,Q⋆}. Let H = Q. Then H is closed but not semi-ω-

open.

Example 21. (i) Let X = R with the usual topology τu. Let H = (0,1]. Then H is β−ω-open

but not closed.

(ii) Let X = R with the topology τu = {φ,R,Q⋆}. Let H = Q. Then H is closed but not

β −ω-open.

7. Further Properties

Definition 16. A space (X ,τ) is called ω-submaximal if every ω-dense subset of X is ω-open.

Proposition 13. Every submaximal space is ω-submaximal.

Proof. Let H ⊂ X beω-dense. Then X = clω(H) ⊂ cl(H) and X = cl(H). Thus H is dense in

X. Since X is submaximal, H is open and henceω-open in X. Therefore, X isω-submaximal.

Example 22. Let X = {a, b, c} with the topology τ = {φ, X , {c}, {b, c}}. Set H = {a, c}. Then

cl(H) = X and H /∈ τ. Hence X is not submaximal but it isω-submaximal, since the onlyω-dense

set is X.

Definition 17. A subset H of a space (X ,τ) is called ω-codense if X\H is ω-dense.

Theorem 16. For a space (X ,τ), the following are equivalent.

(i) X is ω-submaximal,

(ii) Every ω-codense subset H of X is ω-closed.

Proof. (i)⇒ (ii): Let H be a ω-codense subset of X. Then X\H is ω-dense and therefore

X\H is ω-open, X being ω-submaximal by assumption. Thus H is ω-closed.

(ii)⇒ (i): Let H be aω-dense subset of X. Then X\H isω-codense in X and by assumption

X\H is ω-closed. Hence H is ω-open and thus X is ω-submaximal.



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