EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 1, 2023, 386-403
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Submaximality on bigeneralized topological spaces

Yasser Farhat1, Muthumari Krishnan2,∗, Vadakasi Subramanian2,
M. R. Ahmadi Zand3

1 Academic Support Department, Abu Dhabi Polytechnic, P. O. Box 111499, Abu Dhabi,
UAE
2 Department of Mathematics, A.K.D.Dharma Raja Women’s College, Rajapalayam
3 Department of Mathematics, Yazd University, Yazd, Iran

Abstract. In this article, in a bigeneralized topological space, we introduce a new space namely,
(s, v)-bigeneralized submaximal space, and analyze its nature. Also, the characterization theorem
for a (s, v)-bigeneralized submaximal space, image and preimage of (s, v)-bigeneralized submaximal
is a (s, v)-bigeneralized submaximal space under (µ, η)-open, (µ, η)-continuous map, respectively
are proved. Further, the relationship between hyperconnected space and submaximal space in a
pairwise bigeneralized submaximal space is given.

2020 Mathematics Subject Classifications: 54A05, 54A10

Key Words and Phrases: Bigeneralized topological spaces, (s, v)-bigeneralized submaximal
space, (s, v)⋆-bigeneralized submaximal space, (s, v)-open, (s, v)-dense.

1. Introduction

The new most interesting tool namely, generalized topological space was founded
by Császár in [2]. Most of them studied the nature of this space and some, researchers
have defined some new tools in this space and examined their significance in generalized
topological space. Particularly, submaximal space was introduced by Ekici in a generalized
topological space. In generalized topological space, he launched some characterization
theorems for submaximal space. Based on this, some mathematicians established some
new results for generalized submaximal space e.g. [9, 19]. In 2016, Ahmadi Zand et.al gave
few results for submaximal space and defined a space namely, generalized Gδ-submaximal
space, and studied the nature of this space [19].

In [11], J.C. Kelly introduced the notion of bitopological space. Motivated by this,
Boonpok defined the concept of bigeneralized topological space in 2010 [5]. In bigeneralized
topological space, he proved some results for (m,n)-closed sets. In [13, 14, 17, 18], some

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4655

Email addresses: farhat.yasser.1@gmail.com (Y. Farhat),
muthumarikrishnan060196@gmail.com (K.Muthumari)

https://www.ejpam.com 386 © 2023 EJPAM All rights reserved.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 387

new properties of different types of sets are proved. Based on this, here we prove some
interesting results for nowhere-dense sets in bigeneralized topological space.

From the previous observations, it has been motivated, in section 3, (s, v)-bigenralized
submaximal space is defined and the equivalent conditions for this space are given and some
properties for nowhere dense sets are proved, also, the significance of an image and pre-
image of (s, v)-bigeneralized submaximal space are analyzed. In a pairwise bigeneralized
submaximal space, the relationship between generalized submaximal space and generalized
hyperconnected space is discussed.

In section 4, we define the notion (s, v)⋆-bigeneralized submaximal space and the re-
lationship between (s, v) and (s, v)⋆ bigeneralized submaximal space is proven. In the
last section, the necessary conditions for a bigeneralized topological space is a (s, v)⋆⋆-
bigeneralized topological space are given. In this space, few results for closed sets are
proven.

With the results given in the mentioned sections, it is possible to easily check whether
a given space is (s, v) or (s, v)⋆ or (s, v)⋆⋆ bigeneralized submaximal space or not, is obtain
some tricks to check in a (s, v) ((s, v)⋆ and (s, v)⋆⋆) bigeneralized submaximal space, a given
set is (v, s)-nowhere dense or not, are given the necessary conditions for check whether a
given space is submaximal space or not.

2. Preliminaries

Let X be any non-null set. A collection µ of subsets of X is a generalized topology
[2] in X if it contains the empty set and is closed under arbitrary union. The pair (X,µ) is
called a generalized topological space (GTS). The pair (X,µ) is called a strong generalized
topological space (sGTS) [20] if X ∈ µ.

In [3], if Q ∈ µ, then Q is called a µ-open set ; 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 [20] denoted by iD,
is the union of all µ-open sets contained in D and the closure of D [20] denoted by cD, is
the intersection of all µ-closed sets containing D. For simplicity of notation, we will write
i(D) and c(D) when no confusion can arise.

Next, we present some definitions and lemmas that are useful for the development of
the following sections.

In [10], notated by;

µ̃ = {D ∈ µ | D ̸= ∅},
µ(x) = {D ∈ µ | x ∈ D}.

A subset Q of a GTS (X,µ) is said to be;
• µ-nowhere dense [8] if icQ = ∅.
• µ-dense [8] if cQ = X.
• µ-codense [9] if c(X −Q) = X.

A GTS (X,µ) is called as a ;
• hyperconnected space [8] if cµ(Q) = X whenever Q ∈ µ.
• generalized submaximal [9] if Q ∈ µ̃ whenever cµ(Q) = X.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 388

Let µ1 and µ2 be two generalized topologies defined on a non-null set X. Then the
triple (X,µ1, µ2) is called as bigeneralized topological space (briefly, BGTS) [5].

Let Q be a subset of a BGTS (X,µ1, µ2). Then the closure of D and the interior of D
with respect to µs are denoted by cs(D) and is(D), respectively, for s = 1, 2 [5].

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 ̸= v. If X − J is (s, v)-closed, then J is called as (s, v)-open where s, v = 1 or 2 ;
s ̸= v [5].

Let (X,µ) and (Y, η) be two GTSs. A function h : (X,µ) → (Y, η) is called as ;
• (µ, η)-continuous [6] if h−1(Q) ∈ µ for each Q ∈ η.
• (µ, η)-open [12] if h(Q) ∈ η for each Q ∈ µ.

Lemma 1. [19, Lemma 2.6] Let (X,µ) be a GTS and Q ⊂ X. Then iµ(cµ(Q))−Q = ∅.

Lemma 2. [7] A mapping h : (X,µ) → (Y, η) is (µ, η)-continuous if and only if c(h−1(Q)) ⊂
h−1(cQ) for any Q ⊂ Y.

Lemma 3. [20, Lemma 7.3] A mapping h : (X,µ) → (Y, η) is (µ, η)-open if and only if
h−1(cQ) ⊂ ch−1(Q) for any Q ⊂ Y.

3. (s, v)-bigeneralized submaximal Space

Previously, few debilitated forms of open sets are studied. Császár was introduced
and studied some new open sets, namely, semi-open, pre-open, etc.... in [2, 3, 6]. Using
these tools, the necessary condition for a BGTS is a (s, v)-bigeneralized submaximal space
is proved.

In this section, some shortcuts for examining whether a bigeneralized topological space
is (s, v)-bigeneralized submaximal space or not are given. In a (s, v)-bigeneralized sub-
maximal space, we prove some easier way to check whether the given set is nowhere dense
or not.

We begin by remembering some needed definitions.
In [3], let (X,µ) be a GTS and Q ⊂ X is called as;

i). µ-semi-open if Q ⊂ cµ(iµ(Q)).
ii). µ-pre-open if Q ⊂ iµ(cµ(Q)).
iii). µ-α-open if Q ⊂ iµ(cµ(iµ(Q))).
iv). µ-β-open if Q ⊂ cµ(iµ(cµ(Q))).
v). µ-b-open [4] if Q ⊂ cµ(iµ(Q)) ∪ iµ(cµ(Q)).

Let Q be a subset of a GTS (X,µ). Then frontier of Q [15] is denoted by Frµ(Q) and
defined by Frµ(Q) = cµQ ∩ cµ(X −Q).

A subset Q of a BGTS (X,µ1, µ2) is said to be (s, v)-nowhere dense [1] set in X if
is(cv(Q)) = ∅ where s, v = 1, 2 and s ̸= v.

We denoted by, (s, v)−N (X) = {Q ⊂ X | Q is a (s, v)-nowhere dense set in X} where
s, v = 1, 2 ; s ̸= v [16].



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 389

A subset Q of a BGTS (X,µ1, µ2) is called (s, v)-dense [16] if cs(cv(Q)) = X, where
s, v = 1, 2 ; s ̸= v. Notated by, (s, v) − D(X) = {Q ⊂ X | Q is a (s, v)-dense set in X}
where s, v = 1, 2 ; s ̸= v [16].

Definition 1. Let (X,µ1, µ2) be a bigeneralized topological space. A space X is called
(µs, µv)-bigeneralized submaximal (briefly, (s, v)-bigeneralized submaximal) if Q ∈ µs

whenever cµv(Q) = X where s, v = 1, 2 ; s ̸= v.

Example 2. (a). Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s};µ1 = {∅, {p, q}, {p, s}, {q, r}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X} and
µ2 = {∅, {p, r}, {q, s}, {p, r, s}, X}. Then {p, q}, {p, s}, {q, r}, {r, s}, {p, q, r}, {p, q, s}, {p, r,
s}, {q, r, s} and X are µ2-dense subsets of X. Also, every µ2-dense subset of X is a µ1-open
set in X so that X is a (1, 2)-bigeneralized submaximal space.

(b). Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {p, s}, {q, r}, {q, r, s}, X} and µ2 = {∅, {p, q}, {p, r}, {p, s}, {q, s}, {r, s}, {p, q, r}, {p, q, s},
{p, r, s}, {q, r, s}, X}. For that, {p, q}, {p, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}
and X are µ1-dense subsets of X. Also, every µ1-dense subset of X is a µ2-open set in X.
Thus, X is a (2, 1)-bigeneralized submaximal space.

Definition 3. Let (X,µ1, µ2) be a bigeneralized topological space. IfX is (1, 2)-bigeneralized
submaximal and (2, 1)-bigeneralized submaximal, then X is called pairwise bigeneralized
submaximal space.

Example 4. Consider the bigeneralized topological space (X,µ1, µ2) whereX = {p, q, r, s};
µ1 = {∅, {p, q}, {p, r}, {p, s}, {p, q, r}, {p, q, s}, {p, r, s}, X} and µ2 = {∅, {p}, {p, q}, {p, r},
{p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X}.

(a). So that {p, q, r}, {p, q, s}, {p, r, s} and X are µ2-dense subsets of X, also, every µ2-
dense subset of X is a µ1-open set of X for that X is (1, 2)-bigeneralized submaximal.

(b). Since {p}, {p, q}, {p, r}, {p, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s} and X are µ1-dense
subset of X and also, every µ1-dense subset of X is a µ2-open set in X we have X is
(2, 1)-bigeneralized submaximal.
Therefore, X is pairwise bigeneralized submaximal.

The following Theorem 5 is the characterization of (s, v)-bigeneralized submaximality
for a BGTS.

Theorem 5. Let (X,µ1, µ2) be a BGTS. Then the following are equivalent.
(a) (s, v)-bigeneralized submaximal space
(b) Every Q ⊂ X with iv(Q) = ∅, is a µs-closed set.
(c) Q ⊂ X, cv(Q)−Q is a µs-closed set where s, v = 1, 2 and s ̸= v.

Proof. We give the detailed proof only for s = 1, v = 2.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 390

(a) ⇒ (b) Assume that, X is a (1, 2)-bigeneralized submaximal space and take Q ⊂ X with
i2(Q) = ∅ so that cµ2(X −Q) = X which implies X −Q ∈ µ1, by hypothesis. Therefore,
Q is a µ1-closed set.

(b) ⇒ (c) By Lemma 1, i2(c2(Q)−Q) = ∅ whereby by (b) c2(Q)−Q is a µ1-closed set.
(c) ⇒ (a) Consider, cµ2(Q) = X for that c2(Q) − Q = X − Q so that X − Q is a µ1-
closed set, by assumption which implies Q ∈ µ1. Therefore, X is a (1, 2)-bigeneralized
submaximal space.

Proposition 6. Let (X,µ1, µ2) be a BGTS. If X is a (s, v)-bigeneralized submaximal
space, then (X,µs) is a sGTS where s, v = 1, 2 ; s ̸= v.

Proof. If s = 1, v = 2 and X is a (1, 2)-bigeneralized submaximal space, then every µ2-
dense subset of X is µ1-open and so X is µ1-open whereby (X,µ1) is a sGTS. Similarly,
we can prove the result for s = 2; v = 1.

Proposition 7. Let (X,µ1, µ2) be a (s, v)-bigeneralized submaximal space. If Q ∈ (v, s)−
N (X), then it is a µs-closed set where s, v = 1, 2 and s ̸= v.

Proof. Fix s = 2, v = 1, by hypothesis, X is a (2, 1)-bigeneralized submaximal space
and let Q ∈ (1, 2)−N (X) so i1(c2(Q)) = ∅ for that i1(Q) = ∅ whereby by Theorem 5, Q
is a µ2-closed set in X. By similar considerations, we get the proof for s = 1; v = 2.

The following Example 8 shows that the reverse implication of Proposition 7 need not
be true.

Example 8. Consider the bigeneralized topological space (X,µ1, µ2) whereX = {p, q, r, s};
µ1 = {∅, {p, q}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X} and µ2 =
{∅, {q, r}, {q, s}, {r, s}, {q, r, s}}.Fix s = 1 and v = 2. Here every µ2-dense is µ1-open so
that X is (1, 2)-bigeneralized submaximal space. Choose L = {r, s} we get L is µ1-closed
but not in (2, 1)−N (X). Because, i2(c1(L)) = i2(L) = L ̸= ∅.

Choose s = 2 and v = 1. Take µ1 = {∅, {r}, {q, r}, {q, s}, {q, r, s}};µ2 = {∅, {q, r}, {q, s}, {r,
s}, {p, q, r}, {p, r, s}, {q, r, s}, X}. Since every µ1-dense set is µ2-open we have X is (2, 1)-
bigeneralized submaximal space. Consider K = {p, r}. Then K is µ2-closed but not in
(1, 2)−N (X), because, i1(c2(K)) = i1(K) = {r} ≠ ∅.

Proposition 9. Let (X,µ1, µ2) be a (s, v)-bigeneralized submaximal space. If µs ⊂ µv,
then (X,µv) is a generalized submaximal space where s, v = 1, 2 ; s ̸= v.

Proof. Assume that, X is a (s, v)-bigeneralized submaximal space. Take s = 2, v = 1
then we get X is (2, 1)-bigeneralized submaximal and µ2 ⊂ µ1. Let K be a µ1-dense set
in X we have iµ1(X − K) = ∅ so that X − K is µ2-closed, by hypothesis and Theorem
5. Thus, K is µ2-open for that K is µ1-open, since µ2 ⊂ µ1. Hence (X,µ1) is generalized
submaximal. By similar arguments, we can prove this result for the case s = 1, v = 2.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 391

In a BGTS (X,µ1, µ2), if µ1 = µ2 = µ, then every generalized submaximal space is a
pairwise bigeneralized submaximal space, and conversely.

Theorem 10. Let (X,µ1, µ2) and (X, η1, η2) be two BGTSs. If µi ⊂ ηi where i = 1, 2,
and if X is a (µs, µv)-bigeneralized submaximal space, then X is a (ηs, ηv)-bigeneralized
submaximal space for s, v = 1, 2 and s ̸= v.

Theorem 11. Let (X,µ1, µ2) be a BGTS and Q be a non-null subset of X. If X is a
(s, v)-bigeneralized submaximal space, then cv(Q)−Q ∈ (v, s)−N (X) where s, v = 1, 2 ;
s ̸= v.

Proof. It is enough to the case only for s = 1, v = 2. Assume that,

X is a (1, 2)− bigeneralized submaximal space (1)

and Q is a non-null subset of X, consider, G = c2(Q)−Q. By Lemma 1, i2(G) = ∅ and so
G is a µ1-closed set, by (1). Thus, i2(c1(G)) = ∅. Therefore, c2(Q)−Q ∈ (2, 1)−N (X).

The below Theorem 12 easily follows from similar considerations in the above Theorem
11.

Theorem 12. Let (X,µ1, µ2) be a BGTS and H be a non-null subset of X. If X is a
(v, s)-bigeneralized submaximal space, then csH − H ∈ (s, v) − N (X) where s, v = 1, 2 ;
s ̸= v.

Theorem 13 provides tricks to effortlessly check, in a (s, v)-bigeneralized submaximal
space, whether the frontier of a given set is (s, v)-nowhere dense or not.

Theorem 13. Let (X,µ1, µ2) be a (s, v)-bigeneralized submaximal space and Q ∈ µ̃v.
Then Frv(Q) ∈ (v, s)−N (X) where s, v = 1, 2 ; s ̸= v.

Proof. We give the detailed proof only for s = 1, v = 2. Assume that, X is a (1, 2)-
bigeneralized submaximal space and

Q ∈ µ̃2. (2)

Take E = Fr2(Q). Then E = c2Q ∩ c2(X − Q). From (2), we have E = c2Q − Q. By
Lemma 1, i2(E) = ∅ and so E is a µ1-closed set, by our assumption. Thus, i2(c1(E)) = ∅.
Therefore, Fr2(Q) ∈ (2, 1)−N (X).

Example 14 shows that the condition “Q ∈ µ̃v” can not dropped in Theorem 13.
Example 15 shows that the condition “X is a (s, v)-bigeneralized submaximal space” is
necessary for Theorem 13.

Example 14. (a) Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s};µ1 = {∅, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s},
{q, r, s}, X} and µ2 = {∅, {q, s}, {r, s}, {q, r, s}}. Here {s}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q,
r}, {p, q, s}, {p, r, s}, {q, r, s} and X are µ2-dense subsets of X. Also, every µ2-dense is
µ1-open. Therefore, X is a (1, 2)-bigeneralized topological space. Let J = {q, r}. Then



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 392

J /∈ µ̃2. Here Fr2(J) = c2(J) ∩ c2(X − J). This implies Fr2(J) = X which implies that
i2(c1(Fr2(J))) = i2(X) = {q, r, s}. Thus, i2(c1(Fr2(J))) ̸= ∅.

(b) Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {p, s}, {q, s}, {p, q, s}} and µ2 = {∅, {s}, {p, q}, {p, r}, {p, s}, {q, s}, {r, s}, {p, q, r}, {p, q,
s}, {p, r, s}, {q, r, s}, X}.Here {s}, {p, q}, {p, s}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q,
r, s} and X are µ1-dense subsets of X. Also, every µ1-dense set is µ2-open. There-
fore, X is a (2, 1)-bigeneralized topological space. Let K = {p, q}. Then K /∈ µ̃1.
Here Fr1(K) = c1(K) ∩ c1(X − K). This implies Fr1(K) = X which implies that
i1(c2(Fr1(K))) = i1(X) = {p, q, s}. Thus, i1(c2(Fr1(K))) ̸= ∅.

Example 15. (a) Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s};µ1 = {∅, {p, s}, {r, s}, {p, r, s}} and µ2 = {∅, {q, r}, {r, s}, {q, r, s}}. Take P =
{q, s}. Then cµ2(P ) = X. But P /∈ µ1. Thus, X is not a (1, 2)-bigeneralized topolog-
ical space. Let K = {q, r}. Then K ∈ µ̃2. Here Fr2(K) = c2(K) ∩ c2(X − K). This
implies Fr2(K) = {p, s} which implies that i2(c1(Fr2(K))) = i2(X) = {q, r, s}. Thus,
i2(c1(Fr2(K))) ̸= ∅.

(b) Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s, t};µ1 =
{∅, {p, q, s}, {p, q, t}, {q, s, t}, {p, q, s, t}} and µ2 = {∅, {q, r, s}, {q, r, t}, {r, s, t}, {q, r, s, t}}.
Take M = {q, r}. Then cµ1(M) = X. But M /∈ µ̃2. Thus, X is not a (2, 1)-bigeneralized
topological space. Let J = {p, q, t}. Then J ∈ µ̃1. Here Fr1(J) = c1(J) ∩ c1(X − J). This
implies Fr1(J) = {r, s} which implies that i1(c2(Fr1(J))) = i1(X) = {p, q, s, t}. Thus,
i1(c2(Fr1(J))) ̸= ∅.

The below two theorems (Theorem 16 and Theorem 17) reduce the complexity of
checking whether the frontier of a given set is in (s, v)−N (X) or not.

Theorem 16. Let (X,µ1, µ2) be a BGTS. If Q ∈ (s, v)−N (X), then Frv(Q) ∈ (s, v)−
N (X) where s, v = 1, 2 ; s ̸= v.

Proof. Fix s = 1, v = 2 and let Q ∈ (1, 2) − N (X) for that i1(c2(Q)) = ∅, also,
Fr2(Q) ⊆ c2(Q) which implies that i1(Fr2(Q)) = ∅ which turn implies that Fr2(Q) ∈
(1, 2)−N (X), because, Fr2(Q) is µ2-closed. Similarly, we can prove that the result is true
for s = 2, v = 1.

Theorem 17. Let (X,µ1, µ2) be a BGTS. If cµvQ = X and if Q is a (s, v)-open set, then
Frv(Q) ∈ (v, s)−N (X) where s, v = 1, 2 ; s ̸= v.

Proof. Consider, s = 1, v = 2 and given

c2(Q) = X. (3)

Assume that, Q is a (1, 2)-open set for that, i1(i2(Q)) = Q, also, Fr2(Q) = c2(Q) ∩
c2(X −Q) by which Fr2(Q) = X − i2(Q) which implies that c1(Fr2(Q)) = c1(X − i2(Q))
so c1(Fr2(Q)) = X − i1(i2(Q)), thus, c1(Fr2(Q)) = X − Q, whereby by (3), we get
i2(c1(Fr2(Q))) = ∅ and hence Fr2(Q) ∈ (2, 1) − N (X). Similarly, we can prove that the
result is true for s = 2, v = 1.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 393

The below Corollary 18 directly follows from Theorem 5 and Theorem 17 so the trivial
proof is neglected.

Corollary 18. Let (X,µ1, µ2) be a (s, v)-bigeneralized submaximal space and cµvQ = X.
If Q is a (s, v)-open set, then Frv(Q) is µs-closed and hence Frv(Q) is a (s, v)-closed in
X where s, v = 1, 2 ; s ̸= v.

The following two theorems (Theorem 19 and Theorem 20) are gives the necessary
condition for a BGTS is a (s, v)-bigeneralized submaximal space.

Theorem 19. Let (X,µ1, µ2) be a BGTS and Q ⊂ X. If cs(X −Q) ⊂ cvQ −Q, then X
is a (s, v)-bigeneralized submaximal space for s, v = 1, 2 ; s ̸= v.

(a) X is a (s, v)− bigeneralized submaximal space (b)

(c)

where, (a) Every µv-pre-open is µs-open.
(b) Every µv-β-open is µs-open.
(c) Every µv-b-open is µs-open.

The following Theorem 20 describes the above diagram.

Theorem 20. Let (X,µ1, µ2) be a BGTS and µv is a sGT. Then X is a (s, v)-bigeneralized
submaximal space if any one of the following hold;
(a) Every µv-pre-open is µs-open.
(b) Every µv-β-open is µs-open.
(c) Every µv-b-open is µs-open where s, v = 1, 2 and s ̸= v.

Proof. We give the detailed proof for only s = 1, v = 2.

(a) Assume that, µ2 is a sGT, every µ2-pre-open is µ1-open and let c2(Q) = X by which
i2(c2(Q)) = X, thus, Q ⊂ i2(c2(Q)) so for Q is µ2-pre-open, whereby by hypothesis, Q is
a µ1-open set and hence X is a (1, 2)-bigeneralized submaximal space.

(b) Consider, µ2 is a sGT, every µ2-β-open is µ1-open, we take c2(P ) = X so for
i2(c2(P )) = X, by hypothesis, this implies P ⊂ c2(i2(c2(P ))) which implies P is µ2-β-
open which turn implies that P is a µ1-open set by which X is a (1, 2)-bigeneralized
submaximal space.

(c) Given µ2 is a sGT, every µ2-b-open is µ1-open and consider, c2(J) = X implies
i2(c2(J)) = X, by our assumption so for J ⊂ c2(i2(J)) ∪ i2(c2(J)), this implies J is
µ2-b-open which implies that J is a µ1-open set, by hypothesis. Therefore, X is a (1, 2)-
bigeneralized submaximal space.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 394

Next, in the rest of this section with the series of theorems in a BGTS, the significance
of (s, v)-bigeneralized submaximality is analyzed via functions.

Theorem 21. Let (X,µ1, µ2) and (Y, η1, η2) be two BGTS and h : (X,µi) → (Y, ηi) be
a (µi, ηi)-open map for i = 1, 2. If h is surjective, then image of a (s, v)-bigeneralized
submaximal space is (s, v)-bigeneralized submaximal space where s, v = 1, 2 ; s ̸= v.

Proof. It is enough to prove the case only for s = 1, v = 2. Assume that, X is a
(1, 2)-bigeneralized submaximal space. Given h : (X,µi) → (Y, ηi) be a (µi, ηi)-open map
for i = 1, 2. Then

h is a (µ1, η1)− open map (4)

h is a (µ2, η2)− open map (5)

Let Q be a η2-dense subset of Y. From (5) and Lemma 3, cµ2(h
−1(Q)) = X. By our

assumption, h−1(Q) ∈ µ1. By (4), h(h−1(Q)) ∈ η1. Thus, Q ∈ η1, by hypothesis. Hence Y
is a (1, 2)-bigeneralized submaximal space

Example 22 shows that the hypothesis cannot be dropped in Theorem 21.

Example 22. Consider the bigeneralized topological spaces (X,µ1, µ2) and (Y, η1, η2)
where X = {p, q, r, s};Y = {p1, q1, c1, s1}. Define a map h : (X,µi) → (Y, ηi) for i = 1, 2
as follows h(p) = q1;h(q) = p1;h(r) = s1;h(s) = r1. Clearly, h is a surjective map.

(a) Let µ1 = {∅, {p, q}, {p, s}, {q, r}, {p, q, r}, {p, q, s}, {p, r, s}, X};µ2 = {∅, {p}, {s}, {p, s},
{p, q, r}, {q, r, s}, X}; η1 = {∅, {q1}, {p1, q1}, {p1, s1}, {q1, r1}, {p1, q1, r1}, {p1, q1, s1}, {q1, r1,
s1}, Y } and η2 = {∅, {p1, s1}, {q1, s1}, {p1, q1, s1}}. Here h(P ) ∈ η1 whenever P ∈ µ1.
Therefore, h is a (µ1, η1)-open map. Let J = {p}. Then J ∈ µ2. But h(J) /∈ η2. Thus,
h is not a (µ2, η2)-open map. Here {p, s}, {p, q, s}, {p, r, s} and X are µ2-dense subsets of
X. Also, every µ2-dense set is µ1-open. Therefore, X is a (1, 2)-bigeneralized submaximal
space. Let K = {s1}. Then cη2(K) = Y. But K /∈ η1. Thus, Y is not a (1, 2)-bigeneralized
submaximal space.

(b) Let µ1, η1 and η2 are generalized topologies defined as in (a). Take µ2 = {∅, {p}, {s}, {p,
s}, {p, q}, {p, r}, {q, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X}. Here h(P ) ∈ η1 whenever
P ∈ µ1. Therefore, h is a (µ1, η1)-open map. Let Q = {p}. Then Q ∈ µ2. But h(Q) /∈ η2.
Thus, h is not a (µ2, η2)-open map. Here {p, q}, {p, r}, {q, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q,
r, s} and X are µ1-dense subsets of X. Also, every µ1-dense set is µ2-open. Therefore, X is
a (2, 1)-bigeneralized submaximal space. Let J = {p1, q1}. Then cη1(J) = Y. But J /∈ η2.
Thus, Y is not a (2, 1)-bigeneralized submaximal space.

(c) Let µ1 = {∅, {p}, {s}, {p, q}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r,
s}, X};µ2 = {∅, {p, s}, {q, s}, {p, q, s}}; η1 = {∅, {p1, r1}, {r1, s1}, {p1, r1, s1}} and η2 =
{∅, {p1, r1}, {q1, r1}, {p1, q1, r1}}. Here h(P ) ∈ η2 whenever P ∈ µ2. Therefore, h is a
(µ2, η2)-open map. Let J = {p}. Then J ∈ µ1. But h(J) /∈ η1. Thus, h is not a (µ1, η1)-
open map. Here {s}, {p, q}, {p, s}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s} and X



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 395

are µ2-dense subsets of X. Also, every µ2-dense set is µ1-open. Therefore, X is a (1, 2)-
bigeneralized submaximal space. Let M = {p1, q1, r1}. Then cη2(M) = Y. But M /∈ η1.
Thus, Y is not a (1, 2)-bigeneralized submaximal space.

(d) Let µ1 and η1 are generalized topologies defined as in (c). Take µ2 = {∅, {p, r}, {p, q, r},
{p, q, s}, {p, r, s}, {q, r, s}, X} and η2 = {∅, {q1, s1}, {p1, q1, r1}, {p1, q1, s1}, {p1, r1, s1}, {q1,
r1, s1}, Y }. Here h(P ) ∈ η2 whenever P ∈ µ2. Therefore, h is a (µ2, η2)-open map. Let
K = {p}. Then K ∈ µ1. But h(K) /∈ η1. Thus, h is not a (µ1, η1)-open map. Here
{p, q, r}, {p, q, s}, {p, r, s}, {q, r, s} and X are µ1-dense subsets of X. Also, every µ1-dense
set is µ2-open. Therefore, X is a (2, 1)-bigeneralized submaximal space. Let K = {r1}.
Then cη1(K) = Y. But K /∈ η2. Thus, Y is not a (2, 1)-bigeneralized submaximal space.

Consider the bigeneralized topological spaces (X,µ1, µ2) and (Y, η1, η2) where X =
{p, q, r, s};Y = {p1, q1, r1, s1, t1}. Define a map h : (X,µi) → (Y, ηi) for i = 1, 2 as follows
h(p) = p1;h(q) = q1;h(r) = r1;h(s) = s1. Clearly, h is not a surjective map.

(e) Let µ1 = {∅, {q}, {p, q}, {p, r}, {q, r}, {q, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X};µ2 =
{∅, {p, q}, {q, r}, {p, q, r}}; η1 = {∅, {q1}, {p1, q1}, {p1, r1}, {q1, r1}, {q1, s1}, {p1, q1, r1}, {p1,
q1, s1}, {p1, r1, s1}, {q1, r1, s1}, {p1, q1, r1, s1}} and η2 = {∅, {p1, q1}, {q1, r1}, {p1, q1, r1}}.
Here h(P ) ∈ η1 whenever P ∈ µ1. Therefore, f is a (µ1, η1)-open map. Also, h(M) ∈ η2
wheneverM ∈ µ2. Hence h is a (µ2, η2)-open map. Here {q}, {p, q}, {p, r}, {q, r}, {q, s}, {p,
q, r}, {p, q, s}, {p, r, s}, {q, r, s} andX are µ2-dense subsets ofX. Also, every µ2-dense set is
µ1-open. Therefore, X is a (1, 2)-bigeneralized submaximal space. Let J = {p1, q1, r1, t1}.
Then cη2(J) = Y. But J /∈ η1. Thus, Y is not a (1, 2)-bigeneralized submaximal space.

(f) Let µ1 and η1 are generalized topologies defined as in (e). Take µ2 = {∅, {p, q}, {q, r}, {p,
q, r}, {p, q, s}, {q, r, s}, X}; η2 = {∅, {p1, q1}, {q1, r1}, {p1, q1, r1}, {p1, q1, s1}, {q1, r1, s1}, {p1,
q1, r1, s1}}. Here h(K) ∈ η1 whenever K ∈ µ1. Therefore, h is a (µ1, η1)-open map. Also,
h(Q) ∈ η2 wheneverQ ∈ µ2.Hence h is a (µ2, η2)-open map. Here {p, q}, {q, r}, {p, q, r}, {p,
q, s}, {q, r, s} and X are µ1-dense subsets of X. Also, every µ1-dense set is µ2-open.
Therefore, X is a (2, 1)-bigeneralized submaximal space. Let P = {q1, r1, s1, t1}. Then
cη1(P ) = Y. But P /∈ η2. Thus, Y is not a (2, 1)-bigeneralized submaximal space.

Theorem 23. Let (X,µ1, µ2) and (Y, η1, η2) be two BGTSs and h : (X,µi) → (Y, ηi)
be a (µi, ηi)-continuous map for i = 1, 2. Then inverse image of a (s, v)-bigeneralized
submaximal space is (s, v)-bigeneralized submaximal space, s, v = 1, 2 ; s ̸= v.

Proof. It follows from Lemma 2 and the similar arguments in Theorem 21.

The hypothesis of Theorem 23 is necessary as shown by Example 24.

Example 24. Consider the bigeneralized topological spaces (X,µ1, µ2) and (Y, η1, η2)
where X = {p, q, r, s};Y = {p1, q1, r1, s1}. Define a map h : (X,µi) → (Y, ηi) for i = 1, 2
as follows h(p) = p1;h(q) = q1;h(r) = r1;h(s) = s1. Clearly, h is a surjective map.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 396

(a) Let µ1 = {∅, {p}, {r}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s},
{q, r, s}, X};µ2 = {∅, {p, q}, {q, r}, {p, q, r}}; η1 = {∅, {p1}, {r1}, {p1, q1}, {p1, r1}, {p1, s1},
{q1, r1}, {q1, s1}, {r1, s1}, {p1, q1, r1}, {p1, q1, s1}, {p1, r1, s1}, {q1, r1, s1}, Y } and η2 = {∅,
{p1, q1, r1}, {p1, q1, s1}, {p1, q1, r1, s1}}. Here h−1(P ) ∈ µ1 whenever P ∈ η1. Therefore, h
is a (µ1, η1)-continuous map. Let J = {p1, q1, s1}. Then J ∈ η2. But h

−1(J) /∈ µ2. Thus, h
is not a (µ2, η2)-continuous map. Here {p1}, {r1}, {p1, q1}, {p1, r1}, {p1, s1}, {q1, r1}, {q1,
s1}, {r1, s1}, {p1, q1, r1}, {p1, q1, s1}, {p1, r1, s1}, {q1, r1, s1} and Y are η2-dense subsets of
Y. Also, every η2-dense set is η1-open. Therefore, Y is a (1, 2)-bigeneralized submaximal
space. Let K = {q}. Then cµ2(K) = X. But K /∈ µ1. Thus, X is not a (1, 2)-bigeneralized
submaximal space.

(b) Let µ1, η1 and µ2 are generalized topologies defined as in (a). Take η2 = {∅, {p1, q1, r1},
{p1, r1, s1}, Y }. Here h−1(M) ∈ µ1 whenever M ∈ η1. Therefore, h is a (µ1, η1)-continuous
map. Let Q = {p1, r1, s1}. Then Q ∈ η2. But h−1(Q) /∈ µ2. Thus, h is not a (µ2, η2)-
continuous map. Here {p1, q1, r1}, {p1, r1, s1} and Y are η1-dense subsets of Y. Also, every
η1-dense set is η2-open. Therefore, Y is a (2, 1)-bigeneralized submaximal space. Let
J = {p, r, s}. Then cµ1(J) = X. But J /∈ µ2. Thus, X is not a (2, 1)-bigeneralized sub-
maximal space.

(c) Let µ1 = {∅, {q, s}, {r, s}, {q, r, s}};µ2 = {∅, {p}, {p, s}, {q, s}, {p, q, s}}; η1 = {∅, {p1, q1},
{p1, s1}, {p1, q1, r1}, {p1, q1, s1}, {p1, r1, s1}, Y } and η2 = {∅, {p1}, {p1, s1}, {q1, s1}, {p1, q1,
s1}}. Here h−1(P ) ∈ µ2 whenever P ∈ η2. Therefore, h is a (µ2, η2)-continuous map. Let
K = {p1, q1}. Then K ∈ η1. But h

−1(K) /∈ µ1. Thus, h is not a (µ1, η1)-continuous map.
Here {p1, q1}, {p1, s1}, {p1, q1, r1}, {p1, q1, s1}, {p1, r1, s1} and Y are η2-dense subsets of Y.
Also, every η2-dense set is η1-open. Therefore, Y is a (1, 2)-bigeneralized submaximal
space. Let M = {p, s}. Then cµ2(M) = X. But M /∈ µ1. Thus, X is not a (1, 2)-
bigeneralized submaximal space.

(d) Let µ1 and η1 are generalized topologies defined as in (c). Take µ2 = {∅, {p}, {p, q}, {p, r},
{p, s}, {q, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X} and η2 = {∅, {p1}, {p1, q1}, {p1, r1}, {p1,
s1}, {q1, s1}, {p1, q1, r1}, {p1, q1, s1}, {p1, r1, s1}, {q1, r1, s1}, Y } Here h−1(P ) ∈ µ2 when-
ever P ∈ η2. Therefore, h is a (µ2, η2)-continuous map. Let J = {p1, s1}. Then J ∈ η1. But
h−1(J) /∈ µ1. Thus, h is not a (µ1, η1)-continuous map. Here {p1}, {p1, q1}, {p1, r1}, {p1, s1},
{q1, s1}, {p1, q1, r1}, {p1, q1, s1}, {p1, r1, s1}, {q1, r1, s1} and Y are η1-dense subsets of X.
Also, every η1-dense set is η2-open. Therefore, Y is a (2, 1)-bigeneralized submaximal
space. Let K = {q, r, s}. Then cµ1K = X. But K /∈ µ2. Thus, X is not a (2, 1)-
bigeneralized submaximal space.

An interesting result has been proved that, in a pairwise bigeneralized submaximal
space, every hyperconnected space is submaximal. The proof is a direct consequence of
the definitions so the proof is neglected.

Theorem 25. Let (X,µ1, µ2) be a BGTS. If X is pairwise bigeneralized submaximal space,
then the followings are true.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 397

(a) If (X,µ1) is hyperconnected, then (X,µ2) is a generalized submaximal space.
(b) If (X,µ2) is hyperconnected, then (X,µ1) is a generalized submaximal space.

Example 26 shows that the condition “(X,µi) is hyperconnected” for i = 1, 2 is neces-
sary in Theorem 25.

Example 26. (a) Let (X,µ1, µ2) be a BGTS defined as in 4. Then X is a pairwise
bigeneralized topological space. Here {p} ∈ µ2 but cµ2({p}) ̸= X. Therefore, (X,µ2) is
not a hyperconnected space. Also, cµ1({p}) = X but {p} /∈ µ1. Thus, (X,µ1) is not a
generalized submaximal space.

(b) Let (X,µ1, µ2) be a BGTS whereX = {p, q, r, s};µ1 = {∅, {p}, {q}, {p, q}, {p, r}, {p, s},
{q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X} and µ2 = {∅, {p, q, r}, {p, q, s},
{p, r, s}, {q, r, s}, X}. Then {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r,
s}, {q, r, s} and X are µ2-dense subsets of X. Also, every µ2-dense set is µ2-open. There-
fore, X is (1, 2)-bigeneralized submaximal space. Here {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}
and X are µ1-dense subsets of X. Also, every µ1-dense set is µ2-open. Therefore, X
is (2, 1)-bigeneralized submaximal space. Hence X is pairwise bigeneralized submaximal
space. Let Q = {p}. Then Q ∈ µ̃1. But cµ1(Q) ̸= X. Therefore, (X,µ1) is not a hyper-
connected space. Also, cµ2({p, q}) = X but {p, q} /∈ µ2. Thus, (X,µ2) is not a generalized
submaximal space.

4. (s, v)⋆-bigeneralized submaximal space

Here, we define a space namely, (s, v)⋆-bigeneralized submaximal space and give few
results about this space which is helpful to reduce the complexity to check whether the
given BGTS is (s, v)⋆-bigeneralized submaximal space or not.

We begin with a definition of (s, v)⋆-bigeneralized submaximal space.

Definition 27. Let (X,µ1, µ2) be a BGTS. A space X is said to be (µs, µv)
⋆-bigeneralized

submaximal (briefly, (s, v)⋆-bigeneralized submaximal) if Q ∈ µs whenever Q ∈ (s, v) −
D(X) where s, v = 1, 2 ; s ̸= v.

Example 28. (a) Consider the bigeneralized topological space (X,µ1, µ2) where X =
{p, q, r, s};µ1 = {∅, {q}, {s}, {p, q}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q,
r, s}, X} and µ2 = {∅, {p, s}, {q, s}, {p, q, s}}. Here {s}, {p, q}, {p, s}, {q, s}, {r, s}, {p, q, r},
{p, q, s}, {p, r, s}, {q, r, s} and X are (1, 2)-dense subsets of X. Thus, (1, 2) − D(X) ⊂ µ1.
Hence X is a (1, 2)⋆-bigeneralized submaximal space.

(b) Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {p, q}, {q, r}, {p, q, r}} and µ2 = {∅, {q}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r},
{p, q, s}, {p, r, s}, {q, r, s}, X}.Here {q}, {p, q}, {p, r}, {q, r}, {q, s}, {p, q, r}, {p, q, s}, {p, r, s},
{q, r, s} and X are (2, 1)-dense subsets of X. Thus, (2, 1) − D(X) ⊂ µ2. Hence X is a
(2, 1)⋆-bigeneralized submaximal space.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 398

Theorem 29. Let (X,µ1, µ2) be a BGTS. If cµs(Frv(J)) = X, then J ∈ (s, v) − D(X)
where s, v = 1, 2 and s ̸= v.

Proof. Fix s = 1, v = 2; assume that, cµ1(Fr2(J)) = X by which c1(c2J ∩c2(X−J)) =
X and so c1(c2(J)) = X, thus, J ∈ (1, 2)−D(X). Similarly, we can prove that the result
is true for s = 2, v = 1.

The below Corollary 30 is the direct consequence of the above Theorem 29 and Defi-
nition 27, so the easy proof is omitted.

Corollary 30. Let (X,µ1, µ2) be a (s, v)⋆-bigeneralized submaximal space. If cµs(Frv(Q)) =
X, then Q ∈ µs where s, v = 1, 2 ; s ̸= v.

Theorem 31. Let (X,µ1, µ2) be a BGTS. If X is a (s, v)⋆-bigeneralized submaximal space,
then X is a (s, v)-bigeneralized submaximal space where s, v = 1, 2 ; s ̸= v.

Proof. Consider, s = 1, v = 2; assume that, X is a (1, 2)⋆-bigeneralized submaximal
space and let c2(J) = X for that reason c1(c2(J)) = X it turns out J ∈ (1, 2) − D(X)
whereby by our assumption, J ∈ µ1 so it result that X is a (1, 2)-bigeneralized submaximal
space. By the same considerations in the above case, we can prove that this result is true
for s = 2, v = 1.

Theorem 32 describes the below diagram.

(a) X is a (s, v)⋆ − bigeneralized submaximal space (b)

(c)

where, (a) Every µv-pre-open is µs-open.
(b) Every µv-β-open is µs-open.
(c) Every µv-b-open is µs-open.

The following Theorem 32 gives a shortcut for finding the relationship between (v, s)-
bigeneralized submaximal space and (s, v)⋆-bigeneralized submaximal space in a bigener-
alized topological space.

Theorem 32. Let (X,µ1, µ2) be a (v, s)-bigeneralized submaximal space. Then X is a
(s, v)⋆-bigeneralized submaximal space if any one of the following is true;
(a) Every µv-pre-open is µs-open.
(b) Every µv-β-open is µs-open.
(c) Every µv-b-open is µs-open where s, v = 1, 2 and s ̸= v.

Proof. We give the detailed proof for only s = 1, v = 2. Suppose that X is (2, 1)-
bigeneralized submaximal space.

(a) Assume that, every µ2-pre-open is µ1-open. LetQ ∈ (1, 2)−D(X). Then c1(c2(Q)) = X
and so c2(Q) is a µ1-dense set in X. By our assumption, c2(Q) ∈ µ2. Thus, Q ⊂ i2(c2(Q)).
Therefore, Q is µ2-pre-open. By hypothesis, Q ∈ µ1. Hence X is a (1, 2)⋆-bigeneralized



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 399

submaximal space.

(b) Suppose every µ2-β-open is µ1-open. Let K ∈ (1, 2) − D(X). Then c2(K) ∈ µ2, by
similar arguments in (a). This implies K ⊂ c2(i2(c2(K))) which implies that K is µ2-β-
open. By our assumption, K ∈ µ1. Therefore, X is a (1, 2)⋆-bigeneralized submaximal
space.

(c) Assume that, every µ2-b-open is µ1-open. Let P ∈ (1, 2) − D(X). Then c2(P ) ∈ µ2,
by similar arguments in (a). Thus, P ⊂ c2(i2(P )) ∪ i2(c2(P )). This implies P is µ2-b-
open which implies that P ∈ µ1, by hypothesis. Therefore, X is a (1, 2)⋆-bigeneralized
submaximal space.

The below Theorem 33 gives a characterization of the (s, v)⋆-bigeneralized submaximal
space in terms of a closed set. This theorem is a direct implication of Definition 27 so the
proof is skipped.

Theorem 33. Let (X,µ1, µ2) be a BGTS. Then the following are equivalent.
(a) (s, v)⋆-bigeneralized submaximal space.
(b) Every Q ⊂ X with is(iv(Q)) = ∅, is a µs-closed set where s, v = 1, 2 ; s ̸= v.

Corollary 34. Let (X,µ1, µ2) be a (s, v)⋆-bigeneralized submaximal space. Then cv(Q)−Q
is µs-closed and hence (v, s)-nowhere dense where Q ⊂ X and s, v = 1, 2 ; s ̸= v.

Proof. This proof is directly follows from above Theorem 33.

Theorem 35. Let (X,µ1, µ2) be a pairwise bigeneralized submaximal space. Then the
followings are true.
(a) If (X,µ2) is hyperconnected and µ1 is sGT, then X is a (1, 2)⋆-bigeneralized submax-
imal space.
(b) If (X,µ1) is hyperconnected and µ2 is sGT, then X is a (2, 1)⋆-bigeneralized submax-
imal space.

Proof. We will present the detailed proof only for (a). Suppose (X,µ2) is hyper-
connected and µ1 is a sGT. Let P ∈ (1, 2) − D(X). Then c2(P ) is a µ1-dense set in X
and so c2(P ) ∈ µ2, since X is (2, 1)-bigeneralized submaximal space. Thus, c2(P ) ∈ µ̃2.
By our assumption, c2(P ) is a µ2-dense set. This implies P is a µ2-dense set which im-
plies that P ∈ µ1, since X is a (1, 2)-bigeneralized submaximal space. Therefore, X is a
(1, 2)⋆-bigeneralized submaximal space.

5. (s, v)⋆⋆-bigeneralized submaximal space

In this part, we introduce the notion namely, (s, v)⋆-bigeneralized submaximal space.
Some conditions are proven to examine the given space as either (s, v)⋆-bigeneralized
submaximal or not.



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 400

Definition 36. Let (X,µ1, µ2) be a bigeneralized topological space. A space X is said
to be (µs, µv)

⋆⋆-bigeneralized submaximal (briefly, (s, v)⋆⋆-bigeneralized submaximal) if
Q ∈ µv whenever Q ∈ (s, v)−D(X) where s, v = 1, 2 ; s ̸= v.

Example 37. (a) 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}, {s}, {p, q}, {p, r}, {p, s}, {q, r},
{q, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X}.Here {p}, {p, q}, {p, r}, {p, s}, {q, r}, {p, q, r},
{p, q, s}, {p, r, s}, {q, r, s} and X are (1, 2)-dense subsets of X. Also, (1, 2) − D(X) ⊂ µ2.
Hence X is a (1, 2)⋆⋆-bigeneralized submaximal space.

(b) Consider the bigeneralized topological space (X,µ1, µ2) where X = {p, q, r, s};µ1 =
{∅, {r}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, X} and
µ2 = {∅, {p, r}, {q, r}, {p, q, r}}.Here {r}, {p, q}, {p, r}, {q, r}, {r, s}, {p, q, r}, {p, q, s}, {p, r,
s}, {q, r, s} and X are (2, 1)-dense subsets of X. Also, (2, 1) − D(X) ⊂ µ1. Hence X is a
(2, 1)⋆⋆-bigeneralized submaximal space.

The following Theorem 38 describes the below diagram.

(a) X is a (s, v)⋆⋆ − bigeneralized submaximal space (b)

(c)

where, (a) Every µs-pre-open is µv-open.
(b) Every µs-β-open is µv-open.
(c) Every µs-b-open is µv-open.

The following Theorem 38 provides an easier way to check the significance of bigener-
alized topological space using different types of open sets.

Theorem 38. Let (X,µ1, µ2) be a BGTS and µs is sGT. Then X is a (s, v)⋆⋆-bigeneralized
submaximal space if any one of the following is true;
(a) Every µs-pre-open is µv-open.
(b) Every µs-β-open is µv-open.
(c) Every µs-b-open is µv-open where s, v = 1, 2 and s ̸= v.

Proof. It is enough to prove the case for s = 1, v = 2.

(a) Assume that, µ1 is sGT, every µ1-pre-open is µ2-open and let Q ∈ (1, 2)−D(X), then
c1(c2(Q)) = X whereby by hypothesis, c1(Q) ⊃ c2(Q), this implies c1(Q) ⊃ c1(c2(Q))
which implies that c1(Q) = X. Also, i1(c1(Q)) = X, by our assumption. Thus, Q ⊂
i1(c1(Q)) so it result that Q is µ1-pre-open it turns out Q ∈ µ2 and hence X is a (1, 2)⋆⋆-
bigeneralized submaximal space.

(b) Suppose µ1 is sGT and every µ1-β-open is µ2-open. Consider, K ∈ (1, 2) − D(X) so
cµ1(K) = X, by similar arguments in (a). Thus, i1(c1(K)) = X, by hypothesis. By which
K ⊂ c1(i1(c1(K))) implies that K is µ1-β-open. By our assumption, K ∈ µ2. Therefore,



Y. Farhat et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 386-403 401

X is a (1, 2)⋆⋆-bigeneralized submaximal space.

(c) Assume that, µ1 is sGT and every µ1-b-open is µ2-open. Take J ∈ (1, 2) − D(X) so
for cµ1(J) = X, by similar arguments in (a). By our assumption, i1(c1(J)) = X. Thus,
J ⊂ c1(i1(J)) ∪ i1(c1(J)). This implies J is µ1-b-open which implies that J ∈ µ2, by
hypothesis. Hence X is a (1, 2)⋆⋆-bigeneralized submaximal space.

The following Example 39 shows that the hypothesis of Theorem 38 can not be
dropped.

Example 39. Consider the bigeneralized topological space (X,µ1, µ2) whereX = {p, q, r, s};
µ1 = {∅, {p, r}, {q, r}, {p, q, r}, X} and µ2 = {∅, {p, r}, {q, r}, {p, s}, {q, s}, {p, q, r}, {p, q, s},
{p, r, s}, {q, r, s}, X}. Here, µ1 ⊂ µ2 and µ1 is a strong generalized topology.
Fix s = 1 and v = 2.
• Take K = {r}. Then K is µ1-pre-open but not µ2-open.
• Choose L = {p, q}. We get L is µ1-β-open but L /∈ µ2.
• Let W = {r, s}. Then W is µ1-b-open but not in µ2.
Since {r} is (1, 2)-dense but not in µ2 we have X is not a (1, 2)⋆⋆-bigeneralized submaximal
space.
Fix s = 2 and v = 1.
Take µ1 = {∅, {p, q}, {p, s}, {r, s}, {p, q, s}, {p, r, s}, X} and µ2 = {∅, {p, s}, {r, s}, {p, r, s}, X}.
Thus, µ2 ⊂ µ1 and µ2 is a sGT.
Here {s} is µ2-pre-open, µ2-β-open and µ2-b-open but not µ1-open.
Clearly, X is not a (2, 1)⋆⋆-bigeneralized submaximal space. Because, Choose K = {p, r}.
then K is (2, 1)-dense but not µ1-open.

Theorem 40. Let (X,µ1, µ2) be a BGTS. Then the following are equivalent.
(a) (s, v)⋆⋆-bigeneralized submaximal space.
(b) Every Q ⊂ X with is(iv(Q)) = ∅, is a µv-closed set where s, v = 1, 2 ; s ̸= v.

Proof. This proof is directly follows from Definition 36 so the easy proof is neglected.

Corollary 41. Let (X,µ1, µ2) be a (s, v)⋆⋆-bigeneralized submaximal space. Then cv(Q)−
Q is a µv-closed set where Q ⊂ X and s, v = 1, 2 ; s ̸= v.

Theorem 42. Let (X,µ1, µ2) be a (s, v)⋆⋆-bigeneralized submaximal space. If Q ∈ (v, s)−
N (X), then csQ is µv-closed, s, v = 1, 2 ; s ̸= v.

Proof. We give the detailed proof only for s = 1, v = 2. Assume that, X is a (1, 2)⋆⋆-
bigeneralized submaximal space. Let Q ∈ (2, 1) − N (X). Then i2(c1(Q)) = ∅ and so
i1(i2(c1(Q))) = ∅. By hypothesis, c1(Q) is a µ2-closed set in X.

(a) X is a (v, s)⋆⋆ − bigeneralized submaximal space (b)

(c)



REFERENCES 402

where, (a) Every µs-pre-open is µs-open.
(b) Every µs-β-open is µs-open.
(c) Every µs-b-open is µs-open.

The following Theorem 43 describes the above diagram.

Theorem 43. Let (X,µ1, µ2) be a (s, v)-bigeneralized submaximal space. Then X is a
(v, s)⋆⋆-bigeneralized submaximal space if any one of the following is true.
(a) Every µs-pre-open set is µs-open.
(a) Every µs-β-open set is µs-open.
(a) Every µs-b-open set is µs-open where s, v = 1, 2 and s ̸= v.

Proof. It is enough to prove (a) only and the case s = 1, v = 2. Assume that, X is
a (1, 2)-bigeneralized submaximal space. Let J ∈ (2, 1) − D(X). Then c1(J) is µ2-dense
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hypothesis. Hence X is a (2, 1)⋆⋆-bigeneralized submaximal space.

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