EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 2, 2022, 403-414
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On Nowhere Dense Sets
Preecha Yupapin1,2, Vadakasi Subramanian3, Yasser Farhat4,∗

1 Computational Optics Research Group, Science and Technology Advanced Institue,
Van Lang University, Ho Chi Minh City, Vietnam
2 Faculty of Technology, Van Lang University, Ho Chi Minh City, Vietnam
3 Department of Mathematics, A.K.D.Dharma Raja Women’s College, Rajapalayam
4 Academic Support Department, Abu Dhabi Polytechnic, P. O. Box 111499, Abu Dhabi,
UAE

Abstract. We introduce two types of strongly nowhere dense sets, namely (s, v)-strongly nowhere
dense set, (s, v)?-strongly nowhere dense set and analyze their characteristics in a bigeneralized
topological space (BGTS). Further, it is also given some relations between these two types of
strongly nowhere dense sets along with its various properties for (s, v)?-strongly nowhere dense
set. Finally, the necessary and sufficient condition is found between µ-strongly nowhere dense set
and (s, v)?-strongly nowhere dense set in a BGTS.
2020 Mathematics Subject Classifications: 54A05, 54A10
Key Words and Phrases: Bigeneralized topological spaces, µ(s,v)-open, (s, v)-open, (s, v)-
nowhere dense

1. Introduction

The concept of a generalized topological space was introduced by Császár in [4]. Let
X be any non-null set. A collection µ of subsets of X is a generalized topology [8] in X if it
contains the empty set and it closed under arbitrary union. Then the pair (X,µ) is called
as a generalized topological space (GTS) [8]. The pair (X,µ) is called a strong generalized
topological space (sGTS) [8] if X ∈ µ.

If Q ∈ µ, then Q is called a µ-open set and if X − Q ∈ µ, then Q is said to be a
µ-closed set. Let D be a subset of a GTS (X,µ). The interior of D [8] denoted by iD,
is the union of all µ-open sets contained in D and the closure of D [8] denoted by cD,
is the intersection of all µ-closed sets containing D when no confusion can arise. Denote
{D ∈ µ | D 6= ∅} by µ̃ [7] and denote {D ∈ µ | x ∈ D} by µ(x) [7].

Define a generalized topology µ? as follows; µ? = {
⋃

t(U
t
1 ∩ U t

2 ∩ U t
3 ∩ ... ∩ U t

nt
) |

U t
1, U

t
2, ..., U

t
nt

∈ µ} [7]. Then µ ⊂ µ? and µ? is closed under finite intersection [7].

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i2.4283

Email addresses: preecha.yupapin@vlu.edu.vn (P. Yupapin), farhat.yasser.1@gmail.com (Y. Farhat)

https://www.ejpam.com 403 © 2022 EJPAM All rights reserved.



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 404

2. Preliminaries

Let (X,µ) be a GTS and Q ⊂ X. Then Q is called a µ-nowhere dense [6] (resp.
µ-dense [6, 7], µ-codense [7]) set if icQ = ∅ (resp. cQ = X ; c(X −Q) = X).

Let µ1 and µ2 be two generalized topologies on a non-null set X. Then (X,µ1, µ2) is
called as a bigeneralized topological space (briefly, BGTS) [2].

Let (X,µ1, µ2) be a BGTS and D ⊂ X. Then cs(D) denote the closure of D and is(D)
denote the interior of D with respect to µs, respectively, for s = 1, 2 [2].

A subset Q of a BGTS (X,µ1, µ2) is called (s, v)-closed if cs(cv(Q)) = Q, where s, v = 1
or 2 ; s 6= v. If X −Q is (s, v)-closed, then Q is called as (s, v)-open [2] set.

In [2], let Q be a subset of a BGTS (X,µ1, µ2) is called
(1) (s, v)-g-preopen if Q ⊆ is(cv(Q)) where s, v = 1 or 2 ; s 6= v.
(2) (s, v)-g-α-open if Q ⊆ is(cv(is(Q))) where s, v = 1 or 2 ; s 6= v.

Lemma 1. [3] Let Q be a subset of a generalized topological space (X,µ). Then y ∈ c(Q)
if and only if M ∩Q 6= ∅ for any M ∈ µ(y).

Lemma 2. [8, Lemma 3.2] Let (X,µ) be a generalized topological space and D,B ⊂ X.
If B ∈ µ̃;B ∩D = ∅, then B ∩ cD = ∅.

3. Nowhere dense sets

In this section, we define a set namely, (s, v)?-nowhere dense and give some of their
properties in a BGTS.

Let Q be a subset of a generalized topological space (X,µ). Then Q is called µ-semi-
open if Q ⊂ cµ(iµ(Q)) [5]. If X − Q is a µ-semi-open set, then Q is called µ-semi-closed
[5].

Moreover, σ(µ) or σ(µ(X)) = {Q ⊂ X | Q is µ-semi-open set in X} [8]. Also, iσ(Q)
denote the µ-semi-interior of Q ⊂ X is defined by the union of all µ-semi-open subsets of
(X,µ) contained in Q [8].

Let Q be a subset of a BGTS (X,µ1, µ2) is called (s, v)-nowhere dense [1] set in X if
is(cv(Q)) = ∅ where s, v = 1, 2 ; s 6= v.

Definition 1. Let (X,µ1, µ2) be a bigeneralized topological space and K be a non-null
subset of X. Then K is called to be a (s, v)?-nowhere dense set if iσv(cs(K)) = ∅ where
s, v = 1, 2 ; s 6= v;σv = σµv .

Moreover, (s, v)? − N (X) = {Q ⊂ X | Q is (s, v)?-nowhere dense set in X} where
s, v = 1, 2 ; s 6= v.

Example 2. Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};
µ1 = {∅, {p, q}, {q, r}, {p, q, r}} and µ2 = {∅, {p, s}, {q, s}, {p, q, s}}. Then σ1 = {∅, {s}, {p,
q}, {q, r}, {p, q, r}, {p, q, s}, {q, r, s}, X} and σ2 = {∅, {r}, {p, s}, {q, s}, {p, q, s}, {p, r, s}, {q,
r, s}, X}.
1. Take E = {s}. Then iσ2(c1(E)) = iσ2(E) = ∅. Thus, E is a (1, 2)?-nowhere dense set in
X.



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 405

2. Choose F = {p, r}. Then iσ1(c2(F )) = iσ1({p, r}) = ∅. Then F is a (2, 1)?-nowhere
dense in X.

In a bigeneralized topological space, if K ∈ (s, v)? − N (X) and L ⊂ K, then L ∈
(s, v)? − N (X) where s, v = 1, 2 and s 6= v. Also, every (s, v)?-nowhere dense set where
s, v = 1, 2 and s 6= v, is a µv-codense set for v = 1, 2 in X.

Moreover, any (s, v)?-nowhere dense set is a (v, s)-nowhere dense set in a bigeneralized
topological space (X,µ1, µ2) where s, v = 1, 2 and s 6= v, since µ ⊂ σ [3].

Example 3. Consider the BGTS (X,µ1, µ2) where X = {p, q, r, s} and µ1, µ2 are defined
in Example 2.
Take P = {s}. Then P is (1, 2)?-nowhere dense set, by Example 2. Now i2(c1(P )) =
i2(P ) = ∅. Therefore, P is (2, 1)-nowhere dense set in X.
Choose D = {p, r}. In Example 2, D is (2, 1)?-nowhere dense set in X. Here i1(c2(D)) =
i1(D) = ∅. Thus, D is (1, 2)-nowhere dense set in X.

Theorem 4. Let (X,µ1, µ2) be a bigeneralized topological space. Then the followings are
true.
(a) If (X,µ1) is a sGTS and Q ⊂ X is a (1, 2)-nowhere dense set, then Q ∈ (2, 1)?−N (X).
(b) If (X,µ2) is a sGTS and J ⊂ X is a (2, 1)-nowhere dense set, then J ∈ (1, 2)?−N (X).

Proof. (a). Assume that, (X,µ1) is a sGTS and Q is a (1, 2)-nowhere dense set. Then
i1(c2(Q)) = ∅. Suppose iσ1(c2(Q)) 6= ∅. Then there exist G ∈ σ̃1 such that G ⊂ c2(Q).
Since G ∈ σ̃1 we have G ⊂ c1(i1(G)) which implies c1(i1(G)) 6= ∅ which turn implies that
i1(G) 6= ∅, by assumption. Thus, i1(G) ∈ µ̃1 and i1(G) ⊂ c2(Q). Then i1(c2(Q)) 6= ∅
which is not possible. Therefore, iσ1(c2(Q)) = ∅.
(b). Follows from the similar arguments in (a).

In Theorem 4, the condition “µ1 is a sGT” is necessary as shown by the below Example
5. The condition “µ2 is a sGT” in Theorem 4 is necessary as shown by Example 6.

Example 5. Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s,
t};µ1 = {∅, {p, q}, {p, s}, {p, q, s}};µ2 = {∅, {p, q}, {q, r}, {p, q, r}}. Here µ1 is not a sGT.
Then σ1 = {∅, {r}, {t}, {r, t}, {p, q}, {p, s}, {p, q, r}, {p, q, s}, {p, q, t}, {p, r, s}, {p, s, t}, {p, q,
r, s}, {p, q, r, t}, {p, q, s, t}, {p, r, s, t}, X}.
Take D = {r}. Then i1(c2(D)) = i1({r, s, t}) = ∅. Thus, D is a (1, 2)-nowhere dense set in
X. But iσ1(c2(D)) = iσ1({r, s, t}) = {r, t} 6= ∅. Thus, D /∈ (2, 1)? −N (X).

Example 6. Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s,
t};µ1 = {∅, {p, q, r}, {p, q, s}, {q, r, s}, {p, q, r, s}} and µ2 = {∅, {p, q}, {q, r}, {p, q, r}}. Here
µ2 is not a sGT. Then σ2 = {∅, {s}, {t}, {s, t}, {p, q}, {q, r}, {p, q, r}, {p, q, s}, {p, q, t}, {q, r,
s}, {q, r, t}, {p, q, r, s}, {p, q, r, t}, {p, q, s, t}, {q, r, s, t}, X}.
Choose D = {s}. Then i2(c1(D)) = i2({s, t}) = ∅. Thus, D is a (2, 1)-nowhere dense set
in X. But iσ2(c1(D)) = iσ2({s, t}) = {s} 6= ∅. Thus, D /∈ (1, 2)? −N (X).

Theorem 7. Let (X,µ1, µ2) be a BGTS and E ⊂ X. Then the followings are true.
(a) If (X,µ2) is a sGTS and if c1(E) does not contain a non-null µ2-open set, then E ∈



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 406

(1, 2)? −N (X).
(b) If (X,µ1) is a sGTS and if c2(E) does not contain a non-null µ1-open set, then E ∈
(2, 1)? −N (X).

Proof. (a). Assume that, (X,µ2) is a sGTS. Suppose iσ2(c1(E)) 6= ∅. Then there
is a non-null σ2-open set M such that M ⊂ c1(E). Since M is a non-null σ2-open set
we have M ⊂ c2(i2(M)). This implies that c2(i2(M)) 6= ∅ which implies i2(M) 6= ∅, by
assumption. Thus, c1(E) contain a non-null µ2-open set which is not possible. Therefore,
E ∈ (1, 2)? −N (X).
(b). By similar arguments in (a), we get the proof.

Theorem 8. Let (X,µ1, µ2) be a bigeneralized topological space. If µs ⊂ µv and Q ∈
(s, v)? −N (X), then Q is a µv-nowhere dense set in X where s, v = 1, 2 ; s 6= v.

Proof. Take s = 1 and v = 2. Suppose µ1 ⊂ µ2 and Q ∈ (1, 2)? − N (X). Then
iσ2(c1(Q)) = ∅. This implies that iµ2(cµ1(Q)) = ∅ which implies iµ2(cµ2(Q)) = ∅, by
hypothesis. Hence Q is a µ2-nowhere dense set in X.
Similarly, we can prove the result for s = 2 and v = 1.

In Theorem 8, the conditions “µ1 ⊂ µ2” and “µ2 ⊂ µ1” are can not be dropped as
shown by the below Example 9.

Example 9. Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};
µ1 = {∅, {p, q}, {q, r}, {r, s}, {p, q, r}, {p, q, s}, {q, r, s}, X} and µ2 = {∅, {p, s}, {q, s}, {p, q,
s}}. Here µ1 * µ2. Now σ2 = {∅, {r}, {p, s}, {q, s}, {p, q, s}, {p, r, s}, {q, r, s}, X}. Take
Q = {p, q}. Then iσ2(c1(Q)) = iσ2({p, q}) = ∅. Thus, Q is a (1, 2)?-nowhere dense set in
X. Here i2(c2(Q)) = i2(X) = {p, q, s} 6= ∅. Thus, Q is not a µ2-nowhere dense set in X.
(b) Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {p, r}, {q, r}, {p, q, r}} and µ2 = {∅, {p, q}, {q, s}, {p, q, s}}. Here µ2 * µ1. Now σ1 =
{∅, {s}, {p, r}, {q, r}, {p, q, r}, {p, r, s}, {q, r, s}, X}. Choose H = {r}. Then iσ1(c2(H)) =
iσ1({r}) = ∅. Thus, H is a (2, 1)?-nowhere dense set in X. But i1(c1(H)) = i1(X) =
{p, q, r} 6= ∅. Thus, H is not a µ1-nowhere dense set in X.

Theorem 10. Let (X,µ1, µ2) be a bigeneralized topological space. If µv ⊂ µs and if µv

is a strong generalized topology, then any µv-nowhere dense set in X is a (s, v)?-nowhere
dense set in X where s, v = 1, 2 and s 6= v.

Proof. Take s = 1 and v = 2. Assume that, µ2 ⊂ µ1 and Q is a µ2-nowhere dense
set in X. Then iµ2(cµ2(Q)) = ∅. Suppose iσ2(c1(Q)) 6= ∅. Then there exists M ∈ µ̃σ2 such
that M ⊂ c1(Q). Since M ∈ µ̃σ2 we have c2(i2(M)) 6= ∅. Then by hypothesis, i2(M) 6= ∅
and so i2(c1(Q)) 6= ∅. By hypothesis, i2(c2(Q)) 6= ∅, which is not possible. Therefore,
iσ2(c1(Q)) = ∅. Hence Q ∈ (1, 2)? −N (X).
Similarly, we can prove the result for s = 2 and v = 1.

The following Example 11 shows that the hypothesis of Theorem 10 can not be
dropped.



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 407

Example 11. (a). Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s};µ1 = {∅, {p, s}, {q, s}, {p, q, s}} and µ2 = {∅, {p, r}, {p, s}, {q, r}, {p, r, s}, {p, q,
r}, X}. Here µ2 * µ1 but µ2 is a sGT. Now σ2 = {∅, {p, r}, {p, s}, {q, r}, {p, r, s}, {p, q, r}, X}.
Take H = {s}. Then i2(c2(H)) = i2(H) = ∅ and so H is µ2-nowhere dense set in X. But
iσ2(c1(H)) = iσ2(X) = X 6= ∅. Thus, H is not a (1, 2)?-nowhere dense set in X.
(b). Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {p, q}, {p, s}, {q, s}, {p, q, s}} and µ2 = {∅, {p, s}, {q, s}, {p, q, s}}. Here µ2 ⊂ µ1 but
µ2 is not a sGT. Now σ2 = {∅, {r}, {p, s}, {q, s}, {p, q, s}, {p, r, s}, {q, r, s}, X}. Choose
P = {r}. Then i2(c2(P )) = i2(P ) = ∅ so that P is a µ2-nowhere dense set in X. But
iσ2(c1(P )) = iσ2(P ) = P 6= ∅. Thus, P is not a (1, 2)?-nowhere dense set in X.
(c). Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {p, q}, {q, r}, {p, q, r}, X} and µ2 = {∅, {p, r}, {q, r}, {p, q, r}}. Clearly, µ1 * µ2 but
µ1 is a sGT. Here σ1 = {∅, {p, q}, {q, r}, {p, q, r}, {p, q, s}, {q, r, s}, X}. Take Q = {r, s}.
Then i1(c1(Q)) = i1(Q) = ∅ so that Q is a µ1-nowhere dense set in X. But iσ1(c2(Q)) =
iσ1(X) = X 6= ∅. Hence Q is not a (2, 1)?-nowhere dense set in X.
(d). Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {q, r}, {q, s}, {q, r, s}} and µ2 = {∅, {p, q}, {q, r}, {q, s}, {p, q, r}, {p, q, s}, {q, r, s}, X}.
Here µ1 ⊂ µ2 but µ1 is not a sGT. Now σ1 = {∅, {p}, {q, r}, {q, s}, {q, r, s}, {p, q, r}, {p, q, s},
X}. Let K = {p, r}. Then i1(c1(K)) = i1(K) = ∅ so that K is a µ1-nowhere dense set in
X. But iσ1(c2(K)) = iσ1(K) = {p} 6= ∅. Hence K is not a (2, 1)?-nowhere dense set in X.

4. (s, v)-strongly nowhere dense sets

In this section, we define a set namely, (s, v)-strongly nowhere dense set and give
some of its properties in a BGTS (X,µ1, µ2).

Let Q be a subset of a GTS (X,µ). Then Q is called µ-strongly nowhere dense [7] set
if for every K ∈ µ̃, there is P ∈ µ̃ such that P ⊂ K and P ∩Q = ∅.

A generalized topology µ on X is said to satisfy the I-property [9] whenever W1,W2, . . . ,
Wn ∈ µ with W1 ∩W2 ∩ · · · ∩Wn 6= ∅, iµ(W1 ∩W2 ∩ · · · ∩Wn) 6= ∅.

A GTS (X,µ) is called as a hyperconnected space [6] if cµ(Q) = X for each Q ∈ µ̃.

Definition 12. Let B be a non-null subset of a bigeneralized topological space (X,µ1, µ2).
Then B is said to be (s, v)-strongly nowhere dense if for every P ∈ µ̃v there is Q ∈ µ̃s such
that Q ⊂ P and Q ∩B = ∅ where s, v = 1, 2 ; s 6= v.

Moreover, (s, v) − S(X) = {Q ⊂ X | Q is a (s, v)-strongly nowhere dense set in X}
where s, v = 1, 2 ; s 6= v.

In a bigeneralized topological space, if P ∈ (s, v) − S(X) and Q ⊂ P, then Q ∈
(s, v) − S(X) where s, v = 1, 2 and s 6= v. Moreover, every non-null µv-open set is need
not be a (s, v)-strongly nowhere dense set in X where s, v = 1, 2 ; s 6= v.

Example 13. (a). Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s};µ1 = {∅, {p, q}, {q, r}, {p, q, r}} and µ2 = {∅, {p, q, r}, {p, q, s}, {q, r, s}, X}. Let
P = {s}. Then P ∈ (1, 2)−S(X).



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 408

(b). Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s}; µ1 =
{∅, {p, q}, {p, r}, {p, q, r}} and µ2 = {∅, {p}, {q, r}, {q, s}, {p, q, r}, {p, q, s}, {q, r, s}, X}. Let
J = {q, r}. Then J ∈ (2, 1)−S(X).

Proposition 14. Let (X,µ1, µ2) be a bigeneralized topological space and D ⊂ X. Then
D ∈ (s, v)−S(X) if and only if cs(D) ∈ (s, v)−S(X) where s, v = 1, 2 ; s 6= v.

Example 15 shows that the collection (s, v)−S(X) is need not be closed under finite
union in a BGTS (X,µ1, µ2) where s, v = 1, 2 and s 6= v.

Example 15. (a). Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s, t};µ1 = {∅, {s}, {p, q}, {p, r}, {p, q, r}, {p, q, s}, {p, r, s}, {p, q, r, s}} and µ2 = {∅,
{p, q, r}, {p, r, s}, {p, q, r, s}}. Take K = {q, s}, L = {r, t}. Then K,L ∈ (1, 2)−S(X). Now
K ∪ L = {q, r, s, t}. But K ∪ L /∈ (1, 2)−S(X). Because, Here, for every G ∈ µ̃2 there is
no J ∈ µ̃1 such that J ⊂ G and J ∩ (K ∪ L) = ∅.
(b). Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s, t};µ1 =
{∅, {p, q, s}, {p, r, s}, {p, q, r, s}} and µ2 = {∅, {r}, {p, q}, {p, s}, {p, q, r}, {p, q, s}, {p, r, s},
{p, q, r, s}}. Take L = {q, r},M = {s, t}. Then L,M ∈ (2, 1) − S(X). Now L ∪ M =
{q, r, s, t}. But L ∪M /∈ (2, 1) − S(X). Here, for every H ∈ µ̃1 there is no K ∈ µ̃2 such
that K ⊂ H and K ∩ (L ∪M) = ∅.

Theorem 16. Let (X,µ1, µ2) be a BGTS where µ2 = µ?
1. Then the family (s, v)−S(X)

is closed under finite union where s, v = 1, 2 ; s 6= v.

Theorem 17. Let (X,µ1, µ2) be a bigeneralized topological space. If (X,µs) is hy-
perconnected and µs satisfy the I-property, then A1 ∪ A2 ∈ (s, v) − S(X) whenever
A1, A2 ∈ (s, v)−S(X) where s, v = 1, 2 ; s 6= v.

Proof. Take s = 1 and v = 2. Assume that, (X,µ1) is hyperconnected and µ1 satisfy
the I-property. Suppose that, A1 and A2 are (1, 2)-strongly nowhere dense sets in X. Take
D = A1 ∪ A2. Let G ∈ µ̃2. Then there exists Hi ∈ µ̃1 such that Hi ⊂ G and Hi ∩ Ai = ∅
for i = 1, 2. By our assumption, iµ1(H1 ∩H2) 6= ∅. Take J = iµ1(H1 ∩H2). Then J ∈ µ̃1.
Thus, there is J ∈ µ̃1 such that J ⊂ G and J ∩D = ∅. Hence D ∈ (1, 2)−S(X).
Similarly, we can prove the result for s = 2 and v = 1.

Corollary 18. Let (X,µ1, µ2) be a bigeneralized topological space. If (X,µs) is a hyper-
connected space and µs satisfy the I-property, then the family (s, v)−S(X) is closed under
finite union where s, v = 1, 2 ; s 6= v.

The following Example 19 shows that (s, v)-strongly nowhere dense and (s, v)-nowhere
dense sets are not comparable in a BGTS.

Example 19. (a). Consider the bigeneralized topological space (X,µ1, µ2) where X =
[0, 3];µ1 = {∅, [0, 1), {3

2}, [1, 2], [0, 1) ∪ {3
2}, [0, 2]} and µ2 = {∅, [0, 32), [1, 3], [0, 3]}. Let

G = (2, 3]. Then G ∈ (1, 2)−S(X). But G is not a (1, 2)-nowhere dense set in X.
(b). Consider the bigeneralized topological space (X,µ1, µ2) where X = [0, 3]; µ1 =
{∅, [0, 2), (1, 3], [0, 3]} and µ2 = {∅, [0, 1), (1, 2), {2}, [0, 1) ∪ {2}, (1, 2], [0, 1) ∪(1, 2), [0, 1) ∪



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 409

(1, 2]}. Let H = (2, 3]. Then H ∈ (2, 1) −S(X). But H is not a (2, 1)-nowhere dense set
in X.
(c). Consider the bigeneralized topological space (X,µ1, µ2) where X = [0, 3]; µ1 =
{∅, [0, 2), (1, 3], [0, 3]} and µ2 = {∅, [0, 1), [1, 2), [0, 2)}. Let K = [2, 3]. Then K is a (s, v)-
nowhere dense set in X where s, v = 1, 2 and s 6= v. But K /∈ (s, v)−S(X) where s, v = 1, 2
and s 6= v.

Theorem 20. Let (X,µ1, µ2) be a BGTS and Q ⊂ X. If Q ∈ (s, v)−S(X), then Q is a
(v, s)-nowhere dense set in X where s, v = 1, 2 ; s 6= v.

Proof. Suppose Q ∈ (s, v)−S(X) where s, v = 1, 2 ; s 6= v. Assume that, iv(cs(Q)) 6= ∅
where s, v = 1, 2 ; s 6= v. Then there is a set J ∈ µ̃v such that J ⊂ cs(Q) where s, v = 1, 2
and s 6= v which implies that Q /∈ (s, v) − S(X) where s, v = 1, 2 ; s 6= v which is not
possible. Therefore, Q is a (v, s)-nowhere dense set in X where s, v = 1, 2; s 6= v.

Definition 21. Let Q be a non-null subset of a BGTS (X,µ1, µ2). Then for s, v = 1, 2
and s 6= v,
(a) Q is called (s, v)-meager if Q =

⋃
m∈NDm where each Dm is a (s, v)-nowhere dense

set in X.
(b) Q is called (s, v)-residual if X −Q is a (s, v)-meager set in X.
(c) Q is of (s, v)-second category set if Q is not a (s, v)-meager set in X.

Definition 22. Let B be a non-null subset of a BGTS (X,µ1, µ2). Then for s, v = 1, 2
and s 6= v,
(a) B is said to be a (s, v)-s-meager set if B =

⋃
m∈NBm for each Bm ∈ (s, v)−S(X).

(b) B is called as a (s, v)-s-residual set if X −B is a (s, v)-s-meager set in X.
(c) B is of (s, v)-s-second category set if B is not a (s, v)-s-meager set in X.

Corollary 23. Let (X,µ1, µ2) be a BGTS and D ⊂ X. For s, v = 1, 2 and s 6= v, the
followings are true.
(a) If D is (s, v)-s-meager, then it is a (v, s)-meager set.
(b) If D is (s, v)-s-residual, then it is a (v, s)-residual set.
(c) If D is of (s, v)-second category set, then it is of (v, s)-s-second category set.

Corollary 24. Let (X,µ1, µ2) be a BGTS. Then the followings are true.
(a) If µ2 is a strong generalized topology, then (1, 2)−S(X) ⊂ (1, 2)? −N (X).
(b) If µ1 is a strong generalized topology, then (2, 1)−S(X) ⊂ (2, 1)? −N (X).

Proof. (a). Assume that, µ2 is a strong generalized topology. Let Q ∈ (1, 2)−S(X).
By Theorem 20, Q is a (2, 1)-nowhere dense set in X. By our assumption and Theorem 4
(b), Q is a (1, 2)?-nowhere dense set in X.
(b). Suppose that, µ1 is a strong generalized topology. Let D ∈ (2, 1)−S(X). By Theorem
20 and Theorem 4 (a), D ∈ (2, 1)? −N (X).

Definition 25. Let (X,µ1, µ2) satisfy the condition;

if B1 ∈ µ̃s, B2 ∈ µ̃v and B1 ∩B2 6= ∅, then is(B1 ∩B2) 6= ∅



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 410

where s, v = 1, 2 and s 6= v. Then the BGTS (X,µ1, µ2) is said to satisfy the IS-property.

Theorem 26. Let (X,µ1, µ2) be a BGTS which has the IS-property and K,L,Q ⊂ X.
Then
(a) If K ∈ (s, v) − S(X) and L ∈ (v, s) − S(X), then K ∪ L ∈ (s, v) − S(X) where
s, v = 1, 2 ; s 6= v.
(b) If L is a (v, s)-nowhere dense set, then L ∈ (s, v)−S(X) where s, v = 1, 2 ; s 6= v.
(c) If Q ∈ (s, v)? −N (X), then Q ∈ (s, v)−S(X) where s, v = 1, 2 ; s 6= v.

Proof. (a). Let G ∈ µ̃v for v = 1, 2. Then there is a set J ∈ µ̃s such that J ⊂ G and
J ∩ K = ∅ for s = 1, 2. By hypothesis, there is a set M1 ∈ µ̃v such that M1 ⊂ J and
M1 ∩L = ∅ for v = 1, 2. Take P = J ∩M1. Then P ⊂ G and is(P ) 6= ∅, by hypothesis for
s = 1, 2. Also, is(P )∩(K∪L) = ∅ for s = 1, 2. Thus, there is is(P ) ∈ µ̃s such that is(P ) ⊂ G
and is(P )∩ (K ∪L) = ∅ where s, v = 1, 2 ; s 6= v. Therefore, K ∪L ∈ (s, v)−S(X) where
s, v = 1, 2 and s 6= v.
(b). Suppose L is a (v, s)-nowhere dense set where s, v = 1, 2 and s 6= v. Then X − cs(L)
is µv-dense and also µs-open set where s, v = 1, 2 and s 6= v. Let V ∈ µ̃v for v = 1, 2.
Then V ∩ (X − cs(L)) 6= ∅ for s = 1, 2. By hypothesis, is(V ∩ (X − cs(L)) 6= ∅ for s = 1, 2.
Take P = is(V ∩ (X − cs(L)) for s = 1, 2. Then P ⊂ V and P ∩ L = ∅. Therefore, L is a
(s, v)-strongly nowhere dense set in X where s, v = 1, 2 ; s 6= v.
(c). It follows from (b) and the fact that every (s, v)?-nowhere dense set is a (v, s)-nowhere
dense set where s, v = 1, 2 ; s 6= v.

Theorem 27. Let (X,µ1, µ2) be a BGTS. If µs ⊂ µv and Q ∈ (s, v)−S(X), then Q is a
µs-strongly nowhere dense set in X where s, v = 1, 2 and s 6= v.

Definition 28. Let (X,µ1, µ2) satisfy the condition;

if B1 ∈ µ̃s, B2 ∈ µ̃v and B1 ∩B2 6= ∅, then iv(B1 ∩B2) 6= ∅

where s, v = 1, 2 and s 6= v. Then the BGTS (X,µ1, µ2) is said to satisfy the IV -property.

Theorem 29. Let (X,µ1, µ2) be a BGTS which has the IV -property. If D ∈ (s, v)−S(X),
then D is a µv-strongly nowhere dense set in X where s, v = 1, 2 and s 6= v.

Proof. Take s = 1 and v = 2. Assume that, the bigeneralized topological space
(X,µ1, µ2) satisfy the IV -property. Let D ∈ (1, 2) − S(X) and G ∈ µ̃2. Then there is
a set J ∈ µ̃1 such that J ⊂ G and J ∩D = ∅. Here G ∈ µ̃2, J ∈ µ̃1 and J ∩G 6= ∅. By our
assumption, iµ2(G ∩ J) 6= ∅. Take K = iµ2(G ∩ J). Then K ∈ µ̃2. Thus, there is K ∈ µ̃2

such that K ⊂ G and K ∩D = ∅. Therefore, D is a µ2-strongly nowhere dense set in X.
Similarly, we can prove that the result is true for the case s = 2 and v = 1.

Theorem 30. Let (X,µ1, µ2) be a bigeneralized topological space. If µv ⊂ µs where
s, v = 1, 2 and s 6= v, then the following hold.
(a) If Q is a µv-strongly nowhere dense set, then Q ∈ (s, v)−S(X) where s, v = 1, 2 and
s 6= v.



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 411

(b) If J is a µs-strongly nowhere dense set, then J ∈ (s, v)−S(X) where s, v = 1, 2 and
s 6= v.

Proof. Assume that, µv ⊂ µs where s, v = 1, 2 and s 6= v.
(a). Suppose that, Q is a µv-strongly nowhere dense set where v = 1, 2. Take s = 1 and
v = 2. Then Q is a µ2-strongly nowhere dense set and µ2 ⊂ µ1. Let G ∈ µ̃2. Then there
is H ∈ µ̃2 such that H ⊂ G and H ∩Q = ∅. By hypothesis, H ∈ µ̃1. Thus, there is a set
H ∈ µ̃1 such that H ⊂ G and H ∩Q = ∅. Therefore, Q ∈ (1, 2)−S(X).
Similarly, we can prove that the result is true for the case s = 2 and v = 1.
(b). Let J be a µs-strongly nowhere dense set for s = 1, 2. Choose s = 1 and v = 2. Then
J is a µ1-strongly nowhere dense set and µ2 ⊂ µ1. Let H ∈ µ̃2. Then H ∈ µ̃1 and so there
is a set K ∈ µ̃1 such that K ⊂ H and K ∩ J = ∅. Thus, there is a set K ∈ µ̃1 such that
K ⊂ H and K ∩ J = ∅. Hence J ∈ (1, 2)−S(X).
By Similar arguments, we can prove that the result is true for the case s = 2 and v = 1.

5. (s, v)?-strongly nowhere dense sets

In this section, we introduce (s, v)?-strongly nowhere dense set and analzye its nature
in a BGTS (X,µ1, µ2).

Definition 31. Let (X,µ1, µ2) be a BGTS and B be a non-null subset of X. Then B is
called (s, v)?-strongly nowhere dense if for every K ∈ µ̃s there is M ∈ σ̃v such that M ⊂ K
and M ∩B = ∅ where s, v = 1, 2 ; s 6= v.

Moreover, (s, v)? −S(X) = {Q ⊂ X | Q is a (s, v)?-strongly nowhere dense set in X}
where s, v = 1, 2 ; s 6= v.

Moreover, every non-null µs-open set is need not be an element of (s, v)?−S(X) where
s, v = 1, 2 and s 6= v.

Definition 32. Let D be a non-null subset of a BGTS (X,µ1, µ2). Then for s, v = 1, 2
and s 6= v,
(a) D is said to be a (s, v)?-s-meager set if D =

⋃
m∈NDm, for each Dm ∈ (s, v)? −S(X).

(b) D is called (s, v)?-s-residual if X −D is a (s, v)?-s-meager set in X.
(c) D is of a (s, v)?-s-second category set if D is not a (s, v)?-s-meager set in X.

In a bigeneralized topological space, if P ∈ (s, v)? − S(X) and Q ⊂ P, then Q ∈
(s, v)? −S(X) where s, v = 1, 2 and s 6= v.

Moreover, (s, v)−S(X) ⊂ (v, s)? −S(X) where s, v = 1, 2 and s 6= v.

Theorem 33. Let (X,µ1, µ2) be a bigeneralized topological space. Then the following
hold.
(a) If µ2 is a strong generalized topology, then (1, 2)? −S(X) ⊂ (2, 1)−S(X).
(b) If µ1 is a strong generalized topology, then (2, 1)? −S(X) ⊂ (1, 2)−S(X).

Proof. (a). Suppose µ2 is a strong generalized topology and Q ∈ (1, 2)? −S(X). Let
G ∈ µ̃1. Then there is a set P ∈ σ̃2 such that P ⊂ G and P ∩Q = ∅. Since P ∈ σ̃2 we have
i2(P ) ∈ µ̃2, by assumption. Take J = i2(P ). Then J ∈ µ̃2 and J ⊂ G. Also, J ∩ Q = ∅.



P. Yupapin, V. Subramanian, Y. Farhat / Eur. J. Pure Appl. Math, 15 (2) (2022), 403-414 412

Thus, there is a set J ∈ µ̃2 such that J ⊂ G and J ∩Q = ∅. Therefore, Q ∈ (2, 1)−S(X).
By Similar arguments, we get the proof for (b).

Moreover, the family (s, v)? − S(X) is need not be closed under finite union where
s, v = 1, 2 and s 6= v as shown by Example 34.

Example 34. (a). Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s};µ1 = {∅, {p, q}, {q, r}, {p, q, r}};µ2 = {∅, {p}, {q}, {p, q}, {p, s}, {q, s}, {p, q, s}}.
Then σ2 = {∅, {p}, {q}, {r}, {p, q}, {p, r}, {q, r}, {p, s}, {q, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q,
r, s}, X}. Take P = {q, s} and Q = {r, s}. Then P and Q are (1, 2)?-strongly nowhere dense
sets in X. But P ∪Q = {q, r, s} /∈ (1, 2)? −S(X).
(b). Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {r}, {p, s}, {q, s}, {p, q, s}, {p, r, s}, {q, r, s}, X};µ2 = {∅, {q, r}, {r, s}, {q, r, s}, {p, q, s},
X}. Then σ1 = {∅, {r}, {p, s}, {q, s}, {p, q, s}, {p, r, s}, {q, r, s}, X}. Take K = {p} and
L = {q}. Then K and L are (2, 1)?-strongly nowhere dense sets in X. But K ∪ L =
{p, q} /∈ (2, 1)? −S(X).

The following Example 35 shows that
a. P ∪ Q /∈ (s, v)? − S(X) even if P ∈ (s, v)? − S(X) and Q ∈ (v, s)? − S(X) where
s, v = 1, 2 and s 6= v.
b. P ∪ Q /∈ (v, s)? − S(X) even if P ∈ (s, v)? − S(X) and Q ∈ (v, s)? − S(X) where
s, v = 1, 2 and s 6= v.

Example 35. Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r,
s};µ1 = {∅, {p}, {r}, {p, r}, {p, q}, {q, r}, {p, q, r}} and µ2 = {∅, {p}, {p, q}, {p, s}, {q, s},
{p, q, s}}. Then σ1 = {∅, {p}, {r}, {s}, {p, r}, {p, q}, {p, s}, {q, r}, {r, s}, {p, q, r}, {p, q, s},
{p, r, s}, {q, r, s}, X} and σ2 = {∅, {p}, {r}, {p, q}, {p, r}, {p, s}, {q, s}, {p, q, r}, {p, q, s}, {p,
r, s}, {q, r, s}, X}.
Take P = {q, s} and Q = {q, r}. Then P ∈ (1, 2)? −S(X) and Q ∈ (2, 1)? −S(X). Here
P ∪Q = {q, r, s}. But P ∪Q /∈ (1, 2)? −S(X). Also, P ∪Q /∈ (2, 1)? −S(X).

Theorem 36. Let (X,µ1, µ2) be a BGTS. If µ1 and µ2 are strong generalized topologies,
then the followings are true.
(a) (1, 2)? −S(X) ⊂ (2, 1)? −N (X).
(b) (2, 1)? −S(X) ⊂ (1, 2)? −N (X).

Proof. It is enough to prove (a) only. Let E ∈ (1, 2)?−S(X). Suppose iσ1(c2(E)) 6= ∅.
Then there exist G ∈ σ̃1 such that G ⊂ c2(E). Since G ∈ σ̃1 we have i1(G) 6= ∅, by
assumption. Thus, i1(G) ∈ µ̃1. Since G ⊂ c2(E) we have H ∩ E 6= ∅ for every H ∈ σ̃2
such that H ⊂ i1(G) which is a contradiction to hypothesis. For, H ∈ σ̃2 which implies
H ⊂ c2(i2(H)). Since µ2 is a strong generalized topology, i2(H) ∈ µ̃2. Here i2(H) ⊂
i1(G) ⊂ c2(E). This implies i2(H) ∩ c2(E) 6= ∅ which implies that i2(H) ∩ E 6= ∅, by
Lemma 2. Thus, H ∩ E 6= ∅. Therefore, E ∈ (2, 1)? −N (X).

Theorem 37. Let (X,µ1, µ2) be a BGTS which has the IS-property. Then (s, v)? −
N (X) ⊂ (v, s)? −S(X) where s, v = 1, 2 and s 6= v.



REFERENCES 413

Proof. Assume that, (X,µ1, µ2) satisfy the IS-property. Let Q ∈ (s, v)?−N (X) where
s, v = 1, 2 and s 6= v. By hypothesis and Theorem 26, Q ∈ (s, v)−S(X) where s, v = 1, 2
and s 6= v. Also, (s, v) −S(X) ⊂ (v, s)? −S(X) where s, v = 1, 2 and s 6= v. Therefore,
Q ∈ (v, s)? −S(X) where s, v = 1, 2 and s 6= v.

Theorem 38. Let (X,µ1, µ2) be a BGTS. If µv ⊂ µs, µv is a sGT and Q ∈ (s, v)?−S(X),
then Q is a µv-strongly nowhere dense set where s, v = 1, 2 and s 6= v.

Proof. Assume that, µv ⊂ µs, µv is a sGT and Q ∈ (s, v)? − S(X) where s, v = 1, 2
and s 6= v. Take s = 1 and v = 2. Then Q ∈ (1, 2)? −S(X);µ2 ⊂ µ1 and µ2 is a sGT. Let
H ∈ µ̃2. Then H ∈ µ̃1. By assumption, there is K ∈ σ̃2 such that K ⊂ H and K ∩Q = ∅.
Since K ∈ σ̃2 we have K ⊂ c2(i2(K)). This implies i2(K) 6= ∅, since µ2 is a sGT which
implies that i2(K) ∈ µ̃2. Take B = i2(K). Thus, there is B ∈ µ̃2 such that B ⊂ H and
B ∩Q = ∅. Hence Q is a µ2-strongly nowhere dense set in X.
By similar arguments, we can prove the result for the case s = 2 and v = 1.

Theorem 39. Let (X,µ1, µ2) be a BGTS which has the IS-property. If µv is a sGT and
D ∈ (s, v)? −S(X), then D is a µs-strongly nowhere dense set where s, v = 1, 2 ; s 6= v.

Proof. We give the detailed proof only for s = 2 and v = 1. Assume that, the bigen-
eralized topological space (X,µ1, µ2) satisfy the IS-property and µ1 is a strong generalized
topology. Let D be (2, 1)?-strongly nowhere dense set and G ∈ µ̃2. Then there is a set
P ∈ σ̃1 such that P ⊂ G and P ∩ D = ∅. Since P ∈ σ̃1 we have iµ1(P ) 6= ∅, by our
assumption. Take J = iµ1(P ). Then J ∈ µ̃1. Here G ∈ µ̃2, J ∈ µ̃1 and J ∩ G 6= ∅. By
our assumption, iµ2(J ∩ G) 6= ∅. Take E = iµ2(J ∩ G). Then E ∈ µ̃2. Thus, there exists
E ∈ µ̃2 such that E ⊂ G and E ∩D = ∅. Hence D is µ2-strongly nowhere dense in X.

References

[1] Santanu Acharjee, Binod Chandra Tripathy, and Kyriakos Papadopoulos. Two forms
of pairwise lindelöfness and some results related to hereditary class in a bigeneralized
topological space. New Mathematics and Natural Computation, 13(02):181–193, 2017.

[2] Chawalit Boonpok. Weakly open functions on bigeneralized topological spaces. Int.
Journal of Math. Analysis, 4(18):891–897, 2010.

[3] A Csaszar. Extremally disconnected generalized topologies. In Annales Univ. Sci.
Budapest, volume 47, pages 151–161, 2004.

[4] Akos Császár. Generalized open sets. Acta mathematica hungarica, 75, 1997.

[5] Akos Császár. Generalized open sets in generalized topologies. Acta mathematica
hungarica, 106, 2005.

[6] Erdal Ekici. Generalized hyperconnectedness. Acta Mathematica Hungarica, 133, 2011.



REFERENCES 414

[7] Ewa Korczak-Kubiak, Anna Loranty, and Ryszard J Pawlak. Baire generalized topolog-
ical spaces, generalized metric spaces and infinite games. Acta Mathematica Hungarica,
140(3):203–231, 2013.

[8] Zhaowen Li and Funing Lin. Baireness on generalized topological spaces. Acta Math-
ematica Hungarica, 139(4), 2013.

[9] V Renukadevi and S Vadakasi. On lower and upper semi-continuous functions. Acta
Mathematica Hungarica, 160(1):1–12, 2020.