EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 1, 2022, 15-29
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Supra b limit points and supra b separation axioms

Tareq M. Al-shami1, Abdelwaheb Mhemdi2,∗, Mohammed Jameel1,3,
Mohamed Abouhawwash 3

1 Department of Mathematics, Sana’a University, Sana’a, Yemen
2 Department of Mathematics, College of Sciences and Humanities in Aflaj,
Prince Sattam bin Abdulaziz University, Riyadh, Saudi Arabia
3 Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, Egypt

Abstract. In recent years, some generalized structures of topology were introduced. Supra topol-
ogy was one the most important of those generalizations. To contribute in this orientation, we
devoted this work to studying limit points and separation axioms on supra topological spaces by
using supra b-open sets. We define them in a similar way of their counterparts on topological
spaces. In general, we demonstrate their main properties and investigate the sufficient conditions
for some equivalent relations between them. Some novel and interesting examples are provided.

2020 Mathematics Subject Classifications: 54A05, 54C08, 54D99

Key Words and Phrases: Supra b-open set, supra b limit point, SbTk-space (k = 0, 1, 2, 3, 4)

1. Introduction and preliminaries

The term “Topology” on a nonempty set U is used to describe a subfamily of the power
set U which is closed under arbitrary union and is closed under finite intersection. To model
some real-life issues problems and keep some topological properties under conditions fewer
than topology’s conditions, various types of topology’s extensions have been defined and
discussed. One of the celebrated extensions of a topology is a supra topology defined by
Mashhour et al. [24] in 1983. Then, Maki et al. [23] initiated the concept of minimal
structures. Császár [17], in 2002, presented the concepts generalized topology and weak
structure. Al-Odhari [1], in 2015, familiarized another extension of a topology called an
infra topology.

In the pioneering work of Mashhour et al. [24] the concepts of supra continuity and
supra separation axioms were studied. Following this work, many researchers have ex-
plored the topological concepts and notions in the frame of supra topology. For exam-
ple, some classes of generalizations of open sets were furnished to supra topologies in

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i1.4260

Email addresses: tareqalshami83@gmail.com (T.M. Al-shami), mhemdiabd@gmail.com (A. Mhemdi),
mohjameel555@gmail.com (M. Jameel), saleh1284@mans.edu.eg (M. Abouhawwash)

http://www.ejpam.com 15 © 2022 EJPAM All rights reserved.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 16

[3, 18, 19, 21, 27, 29]. The class of supra R-open sets was studied in a topological spaces
under the name of somewhere dense sets [4, 16]. Mustafa and Qoqazeh [26] applied supra
D-sets to investigate new families of separation axioms. Mustafa [25] defined supra b-
compact and supra b-Lindelof spaces. Then, Al-shami established the concepts of supra
paracompactness [7], supra complete Hausdorffness and supra complete regularity [8]. In
[9–11, 20], Al-shami with his coauthors scrutinized the main properties of limit points
using the famous generalizations of supra open sets. Also, Al-shami with his coauthors
[2, 5, 6, 12] introduced different kinds of supra compactness. The connected spaces were
also initiated in supra topologies by the authors of [22, 28]. As evidence of the importance
of supra topologies, they were explored and discussed in the soft frames; see, [13–15].

It should be noted that some topology’s properties are evaporated in supra topologies
like the distribution property for the interior and closure operators with respect to the
operations of intersection and union. Another example of these missing properties is the
relationship between compact and closed subsets of Hausdorff space.

We organize the rest of this manuscript as follows. In Section (2), we recall the concepts
and findings that make this work readable. In Section (3), we apply the class of supra b-
open sets to display novel kinds of limit points of sets. We devoted Section (4) to introduce
new families of spaces, namely SbTk-spaces (k = 0, 1, 2, 3, 4). Finally, we provide some
conclusions and suggest some future work in Section (5).

2. Preliminaries

Herein, we mention some concepts and findings given in the literature of supra topolo-
gies that are necessary to understand this article.

Definition 1. [24] We call a subfamily Ω of the power set of U ̸= ∅ a supra topology (in
short, ST) provided that U ∈ Ω and the arbitrary union of members of Ω is also a member
of Ω.

We call a pair (U ,Ω) a supra topological space (in short, STS). The members of Ω are
said to be supra open sets and the complement of each member of Ω is said to be a supra
closed set.

Remark 1. (i) We call Ω an associated ST with a topology τ if τ ⊆ Ω.

(ii) Henceforth, we consider (U ,Ω) and (V,Ψ) as associated STSs with the topological
spaces (U , τ) and (V, θ), respectively.

Definition 2. [29] A subset O of (U ,Ω) is said to be supra b-open if O ⊆ int(cl(O))
⋃
cl(int(O)).

Definition 3. [18, 24] Let O ⊆ U .

(i) The union of all supra open (respectively, supra b-open) subsets of an STS (U ,Ω)
contained in O is denoted by intΩ(O) (respectively, bintΩ(O)).



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 17

(ii) The intersection of all supra closed (respectively, supra b-closed) subsets of an STS
(U ,Ω) including O is denoted by clΩ(O) (respectively, bclΩ(O)).

For simplicity, we sometimes write int(O), cl(O), bint(O) and bcl(O) instead of intΩ(O),
clΩ(O), bintΩ(O) and bclΩ(O), respectively.

Definition 4. [18] Let g : (U ,Ω) → (Y,Ψ) be a mapping.

(i) If the inverse image of every open set is supra b-open, then we call g a supra b-
continuous mapping.

(ii) If the image of every open (respectively, closed) set is supra b-open (respectively,
supra b-closed), then we call g a supra b-open (respectively, supra b-closed) mapping.

Definition 5. [2] Let O be a subset of an STS (U ,Ω). We call a class ΩA = {A
⋂
Θ :

Θ ∈ Ω} a supra relative topology on O, call (O,ΩA) a supra subspace of (U ,Ω).

Definition 6. [12] We call β a basis for an STS (U ,Ω) if every element of Ω can be
expressed as a union of elements of β.

Definition 7. [12] Let {(Uk,Ωk) : k = 1, 2, ..., n} be a family of STSs. Then, {Θ1 ×Θ2 ×
... × Θn : Θk ∈ Ωk} constitutes a basis for an ST C on U =

∏n
k=1 Uk. We call (U , C) a

finite product of STSs.

Proposition 1. [12] Let O ⊆ (U ,Ω) and P ⊆ (V,Ψ). Then, int(O×P ) = int(O)×int(P )
and cl(O × P ) = cl(O)× cl(P ).

3. Supra b-limit points of a set

We devote this section to defining a new type of supra limit points of a set by using
supra b-open sets. With the help of examples, we probe their fundamental properties and
describe their behaviors in some spaces.

Definition 8. We call O a supra b neighbourhood of ξ in an STS (U ,Ω) provided that
there is a supra b-open set F including ξ such that ξ ∈ F ⊆ O.

Definition 9. We call ξ ∈ U a supra b-limit point of a subset O of an STS (U ,Ω) if every
supra b neighborhood W of ξ satisfying that W \ {ξ}

⋂
A ̸= ∅.

A supra b derived set of O, symbolized by Ob′, is all supra b-limit points of O.
One can prove the two results below easily, so we omit their proofs.

Proposition 2. For every O,P ⊆ (U ,Ω), we have O ⊆ P implies Ob′ ⊆ P b′.

Corollary 1. Let O and P be subsets of (U ,Ω). Then,

(i) Ob′⋃P b′ ⊆ (O
⋃
P )b′.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 18

(ii) (O
⋂
P )b′ ⊆ Ob′⋂P b′.

The converse of the above proposition and corollary are in general false as the following
example shows.

Example 1. Let Ω = {∅,U , {ξ1, ξ2}, {ξ2, ξ3}, {ξ1, ξ2, ξ3}} be a supra topology on U =
{ξ1, ξ2, ξ3, ξ4}. Then {∅,U , {ξ2}, {ξ1, ξ2}, {ξ2, ξ3}, {ξ1, ξ3}, {ξ2, ξ4}, {ξ1, ξ2, ξ3}, {ξ1, ξ2, ξ4},
{ξ1, ξ3, ξ4}, {ξ2, ξ3, ξ4}} is the collection of all supra b-open subsets of (U ,Ω). If O =
{ξ1, ξ4}, P = {ξ2, ξ3}, C = {ξ1, ξ2, ξ4} and D = {ξ2, ξ3, ξ4}, then Ob′ = ∅, P b′ = {ξ1},
Cb′ = {ξ3, ξ4} and Db′ = {ξ1, ξ2, ξ4}. Now, we have the following cases:

(i) Ob′ ⊆ P b′, but P ̸⊆ O.

(ii) Ob′⋃P b′ = {ξ1} and (O
⋃
P )b′ = {ξ1, ξ3, ξ4}. Then (O

⋃
P )b′ ̸⊆ Ob′⋃P b′.

(iii) Cb′⋂Db′ = {ξ4} and (C
⋂
D)b′ = ∅. Then Cb′⋂Db′ ̸⊆ (C

⋂
D)b′.

Proposition 3. Let O be a subset of (U ,Ω) and ξ ∈ U . Then ξ ∈ Ob′ iff ξ ∈ (O \ {ξ})b′.

Proof. ⇒: Consider ξ ∈ Ob′. Then, (Θ \ {ξ})
⋂
O ̸= ∅ for any supra b-open set Θ

including ξ. Obviously, we obtain (Θ\{ξ})
⋂
(O\{ξ}) ̸= ∅. This means that ξ ∈ (O\{ξ})b′.

Proposition (2) proves the sufficient part.

Theorem 1. Let O be a subset of (U ,Ω). Then.

(i) O is a supra b-closed set iff Ob′ ⊆ O.

(ii) O
⋃

Ob′ is a supra b-closed set.

(iii) bcl(O) = O
⋃
Ob′.

Proof.

(i) Let O be a supra b-closed set and ξ ̸∈ O. Since Oc
⋂
O = ∅ and Oc is a supra b-open

set including ξ, we obtain ξ ̸∈ Ob′. Therefore, Ob′ ⊆ O. Conversely, suppose that
ξ ∈ Oc. Since Ob′ ⊆ O, ξ ̸∈ Ob′. Accordingly, we find a supra b-open set Θξ satisfying
Θξ \ {ξ}

⋂
O = ∅. Now, Θξ

⋂
O = ∅ because ξ ∈ Oc. Therefore, Θξ ⊆ Oc, which

means that Oc =
⋃
{Θξ : ξ ∈ Oc}. This ends the proof that O is supra b-closed.

(ii) Suppose that ξ ̸∈ (O
⋃
Ob′). Then, ξ ̸∈ O and ξ ̸∈ Ob′. This means that there exists

a supra b-open set Θ satisfying the following equality.

Θ
⋂

O = ∅ (1)

Now, for every ξ ∈ Θ, we obtain ξ ̸∈ Ob′. So we obtain the following equality.

Θ
⋂

Ob′ = ∅ (2)

It comes from (1) and (2) that Θ
⋂
(O

⋃
Ob′) = ∅. Thus, ξ ̸∈ (O

⋃
Ob′)b′. Hence,

(O
⋃
Ob′)b′ ⊆ (O

⋃
Ob′). According to (i), O

⋃
Ob′ is supra b-closed.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 19

(iii) It is clear that O
⋃
Ob′ ⊆ bcl(O). Conversely, it well known that bcl(O) is the smallest

supra b-closed set including O. Since O
⋃
Ob′ is a supra b-closed set including O,

bcl(O) ⊆ O
⋃
Ob′. Hence, bcl(O) = O

⋃
Ob′.

Corollary 2. If O is a supra b-closed subset of (U ,Ω), then Ob′, (Ob′)b′, ((Ob′)b′)b′, ...
are supra b-closed sets.

Definition 10. We call a mapping g : (U ,Ω) → (V,Ψ):

(i) supra b⋆-continuous if g−1(H) is supra b-open for any supra b-open set H.

(ii) supra b⋆-open (respectively, supra b⋆-closed) if g(H) is supra b-open (respectively,
supra b-closed) for any supra b-open (respectively, supra b-closed) set H.

We call a bijective, supra b⋆-continuous and supra b⋆-open mapping a supra b⋆-homeomorphism.

Theorem 2. If f : (U ,Ω) → (V,Ψ) is a supra b⋆-homeomorphism mapping, then f(Ob′) =
(g(O))b′ for each O ⊆ U .

Proof. Consider ξ ̸∈ (f(O))b′. Then, (H\{ξ})
⋂
f(O) = ∅ for some supra b-open setsH

including ξ. Directly, f−1[(H \ {ξ})
⋂
f(O)] = f−1(∅). Now, (f−1(H) \ f−1(ξ))

⋂
O = ∅.

This means that f−1(ξ) ̸∈ Ob′. The bijectiveness of f implies that ξ ̸∈ f(Ob′). Thus,
f(Ob′) ⊆ (f(O))b′. Following similar technique, we obtain (f(O))b′ ⊆ f(Ob′).

Definition 11. A sub-collection Λ of the power set of U ̸= ∅ is said to have the difference
property if Θ ∈ Λ implies that Θ \ {ξ} ∈ Λ.

To illustrate the difference property, we supply the next two examples.

Example 2. Consider Ω = {∅,Θ ⊆ N : Θ is infinite } as an ST on the natural numbers
set N . Obviously, Θ ∈ Ω implies Θ \ {ξ} ∈ Ω. Then, (N ,Ω) has the difference property.
Note that the families of supra open and supra b-open subsets of (N ,Ω) are identical.
Hence, (N ,Ω) has the difference property for the family of supra b-open sets.

Example 3. Consider Ω = {∅,N \ {1, 3},N \ {1, 4},N \ {2, 3},N \ {2, 4}}
⋃
{Θ ⊆ N : Θ

such that {1, 2} ⊆ Θ or {3, 4} ⊆ Θ} as an ST on the natural numbers set N . Then
{1, 2} ∈ Ω, but {1, 2}\{2} = {1} ̸∈ Ω. Thus, (U ,Ω) does not have the difference property..
Hence, it does not have the difference property for the family of supra b-open sets.

Theorem 3. Let O be a subset of (U ,Ω) which has the difference property for the family
of supra b-open sets. Then.

(i) (Ob′)b′ ⊆ Ob′.

(ii) bcl(Ob′) = Ob′ = (bcl(O))b′.

(iii) Ob′ = ∅ if O is finite.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 20

Proof.

(i) Consider ξ ̸∈ Ob′. Then, Θ \ {ξ}
⋂
O = ∅ for some supra b-open sets Θ including

ξ. By the difference property for the family of supra b-open subsets of (U ,Ω), we
obtain Θ\{ξ} is a supra b-open set. So that, Θ\{ξ}

⋂
Ob′ = ∅. Now, Θ

⋂
(Ob′)b′ = ∅

because ξ ̸∈ Ob′. Therefore, ξ ̸∈ (Ob′)b′. Hence, (Ob′)b′ ⊆ Ob′.

(ii) Since (Ob′)b′ ⊆ Ob′, it comes from Theorem (1) that Ob′ is supra b-closed. Thus,

bcl(Ob′) = Ob′ (3)

Also, (O)b′ ⊆ (bcl(O))b′ because O ⊆ bcl(O). Conversely, let ξ ̸∈ (O)b′. Then, it
comes from 1 above that Θ \ {ξ}

⋂
O = ∅ and Θ \ {ξ}

⋂
Ob′ = ∅. This means that

Θ \ {ξ}
⋂
bcl(O) = ∅. Therefore, ξ ̸∈ (bcl(O))b′. Thus, (bcl(O))b′ ⊆ (O)b′. Hence

(bcl(O))b′ = (O)b′ (4)

Equalities (3) and (4) end the proof.

(iii) Consider O is a finite subset of U . Suppose that ξ ∈ U such that ξ ∈ Ob′. Then,
Θ\{ξ}

⋂
O ̸= ∅ for every supra b-open set Θ including ξ. Now, for every ζ ∈ O such

that ζ ̸= ξ, we have Θ \ {ξ, ζ} is a supra b-open set. Thus, Θ \ [O
⋃
{ξ}] is a supra

b-open set such that Θ \ [O
⋃
{ξ}]

⋂
O = ∅. This implies that ξ ̸∈ Ob′. But this is a

contradiction. Hence, Ob′ = ∅.

To show that the converse of Theorem 3 is not true if (U ,Ω) does not have the difference
property for the family of supra b-open sets, consider O = {2, 4, 5} as a subset of STS
furnished in Example (3). Note that the families of supra open and supra b-open sets are
identical. So that, we obtain Ob′ = N \{2, 4}, (Ob′)b′ = N \{1, 3} and cl(Ob′) = N , which
means that

(i) (Ob′)b′ ̸⊆ Ob′.

(ii) bcl(Ob′) ̸= Ob′.

(iii) Ob′ ̸= ∅ in spite of O is finite.

4. Supra b separation axioms

In this section, we familiarize the concepts of regularity, normality and Tk-spaces using
supra b-open sets. We give them some descriptions and reveal the interrelations between
them with the assistant of examples and counterexamples.

Definition 12. An STS (U ,Ω) is said to be:

(i) SbT0 if for every ξ ̸= ζ ∈ U , there is a supra b-open set O such that ξ ∈ O or ζ ∈ O.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 21

(ii) SbT1 if for every ξ ̸= ζ ∈ U , there are supra b-open sets O,Θ such that ξ ∈ O \ Θ
and ζ ∈ Θ \O.

(iii) SbT2 (or, supra b Hausdorff) if for every ξ ̸= ζ ∈ U , there are disjoint supra b-open
sets O,Θ such that ξ ∈ O and ζ ∈ Θ.

(iv) supra b regular if for every supra b-closed set F and each ξ ̸∈ F , there are disjoint
supra b-open sets O,Θ such that ξ ∈ O and F ⊆ Θ.

(v) supra b normal if for every two disjoint supra b-closed sets are separated by two
disjoint supra b-open sets.

(vi) SbT3 (respectively, SbT4) if it is supra b regular (respectively, supra b normal) and
SbT1.

Theorem 4. The next three properties are identical.

(i) (U ,Ω) is SbT0;

(ii) bcl({ξ}) ̸= bcl({ζ}) for every ξ ̸= ζ ∈ U ;

(iii) {ξ}b′ is a union of supra b-closed sets for every ξ ∈ U .

Proof. 1 → 2: Let ξ ̸= ζ ∈ U . Then there is a supra b-open set O such that ξ ∈ Ω
or ζ ∈ Ω. Say, ξ ∈ Θ and ζ ̸∈ Θ. Since Θ is a supra b-open set including ξ such that
Θ
⋂
{ζ} = ∅, we obtain ξ ̸∈ bcl({ζ}). Obviously, ξ ∈ bcl({ξ}); hence, bcl({ξ}) ̸= bcl({ζ}).
2 → 3: Suppose that ζ ∈ {ξ}b′. Then, ζ ∈ bcl({ξ}). Consequently, bcl(ζ) ⊆ bcl({ξ}).

This means that ζ ∈ bcl(ζ) ⊆ {ξ}b′. So that, {ξ}b′ =
⋃
{bcl(ζ): for every ζ ∈ {ξ}b′}.

3 → 1: Consider ξ ̸= ζ. Then

(i) Either ζ ∈ {ξ}b′. Then there is a supra b-closed set F such that ζ ∈ F ⊆ {ξ}b′. This
means that ξ ̸∈ F . Thus, F c is a supra b-open set including ξ such that ζ ̸∈ F c.

(ii) Or ζ ̸∈ {ξ}b′. Then there is a supra b-open set Θ including ζ such that ξ ̸∈ Θ.

It follows from the above two cases that (U ,Ω) is SbT0.

Corollary 3. In an SbT0-space (U ,Ω), there exists at most a singleton set which is a
supra b dense ({ξ} is a supra b dense set if bcl{ξ} = U).

Proof. Suppose that {ξ} and {ζ} are two distinct singleton subset of an SbT0-space
(U ,Ω) such that bcl({ξ}) = bcl({ζ}) = U . This leads to a contradiction because (U ,Ω) is
SbT0. Hence, we obtain the desired result.

Theorem 5. The next there properties are identical.

(i) (U ,Ω) is SbT1;



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 22

(ii) All singleton subsets of (U ,Ω) are supra b-closed;

(iii) The intersection of all supra b-open sets including a set O is exactly O;

(iv) {ξ}b′ = ∅ for every ξ ∈ U .

Proof. 1 → 2: Let (U ,Ω) be SbT1 such that {ξ} ⊆ U . For each ζ ̸= ξ ∈ U , there is
a supra b-open set Θ including ζ satisfies that Θ

⋂
{ξ} = ∅. Then, ζ ̸∈ bcl({ξ}). Thus,

bcl({ξ}) = {ξ}. Hence, we obtain the required result.
2 → 3: Let O ⊆ U and ξ ∈ Oc. Then {ξ}c is a supra b-open set including O. Therefore,

O ⊆ {Θ : Θ is a supra b-open set including O} ⊆ {{ξ}c : ξ ∈ Oc} ⊆ O. Hence, O = {Θ : Θ
is a supra b-open set including O}.

3 → 4: Let ξ ∈ U such that {ξ}b′ ̸= ∅. Then there is ζ ̸= ξ such that ζ ∈ {ξ}b′. So that,
Θ \ {ζ}

⋂
{ξ} ≠ ∅ for every supra b-open set Θ including ζ. This contradicts 3 because the

intersection of all supra b-open sets including ζ is not equal {ζ}. Hence, {ξ}b′ = ∅.
4 → 1: Let ξ ̸= ζ. Then, the singleton sets {ξ} and {ζ} are supra b-closed because

{ξ}b′ = ∅ and {ζ}b′ = ∅. This implies that {ξ}c and {ζ}c are supra b-open sets including
{ζ} and {ξ}, respectively. Hence, (U ,Ω) is SbT1.

Proposition 4. If (U ,Ω) satisfies the difference property for the family of supra b-open
sets, then it is SbT1.

Proof. Since U is a supra b-open set and (U ,Ω) satisfies the difference property for
the family of supra b-open sets, we find that U \ {ξ} is a supra b-open set for each ξ ∈ U .
Hence, we obtain the desired result.

To clarify that the converse of the aforementioned proposition is in general false, we
furnish the example below.

Example 4. Every singleton subset of (U ,Ω), given in Example (1), is supra b-closed;
therefore, (U ,Ω) is SbT1. In contrast, a set {ξ1, ξ3} is supra b-open, whereas {ξ1, ξ3}\{ξ1}
is not supra b-open. Consequently, (U ,Ω) does not have the difference property for the
class of supra b-open sets.

The next definition will help to prove that SbT0- and SbT1-spaces are equivalent.

Definition 13. We call (U ,Ω) a supra b symmetric space if ξ ∈ bcl{ζ} implies that
ζ ∈ bcl{ξ} for ξ ̸= ζ ∈ U .

Theorem 6. Let (U ,Ω) be a supra b symmetric space. Then it is SbT1 iff it is SbT0.

Proof. ⇒: It is straightforward.
⇐: For two distinct points ξ and ζ, there is a supra b-open set Θ including only one of
them. Assume that ξ ∈ Θ and ζ ̸∈ Θ. Then, ξ ̸∈ bcl{ζ}. Now, ζ ̸∈ bcl{ξ} because (U ,Ω)
is supra b symmetric. Thus, (bcl{ξ})c is a supra b-open set including b. Hence, (U ,Ω) is
SbT1.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 23

Theorem 7. The next three properties are identical.

(i) (U ,Ω) is SbT2;

(ii) {ξ} =
⋂
{Fk : Fk is a supra b-closed neighborhood of ξ} for every ξ ∈ U ;

(iii) △ = {(ξ, ξ) : ξ ∈ U} forms a supra b-closed set in the product of supra spaces U ×U .

Proof. 1 → 2: Since (U ,Ω) is SbT2, for ξ ̸= ζ there are disjoint supra b-open sets Θk

and Ok such that ξ ∈ Θk and ζ ∈ Ok. Then, ξ ∈ bcl(Θk) ⊆ Oc
k = Fk. Therefore, Fk is

a supra b-closed neighborhood of ξ such that ζ ̸∈ Fk. Thus, {ξ} =
⋂
{Fk : Fk is a supra

b-closed neighborhood of ξ}.
2 → 1: Consider ξ ̸= ζ. Since {ξ} =

⋂
{Fk : Fk is a supra b-closed neighborhood of

ξ}, there is a supra b-closed neighborhood Fk0 of ξ such that ζ ̸∈ Fk0 . So that, there is a
supra b-open set Θ including ξ such that ξ ∈ bcl(Θ) ⊆ Fk0 . Obviously, (bcl(Θ))c is a supra
b-open set including ζ and Θ

⋂
(bcl(Θ))c = ∅. Hence, (U ,Ω) is SbT2.

1 → 3: Suppose that (ξ, ζ) ∈ U ×U −△. Then ξ ̸= ζ. By hypothesis, there are disjoint
supra b-open sets Θ and O respectively including ξ and ζ. Now, (ξ, ζ) ∈ Θ×O ⊆ U×U−△,
which proves that U × U −△ is a supra b neighbourhood of any of its points. Thus, △ is
supra b-closed.

3 → 1: Consider △ as a supra b-closed subset of U × U . For ξ ̸= ζ ∈ U , we have
U × U −△ is a supra b-open set including (ξ, ζ). So that, there are supra b-open subsets
Θ and O of (U ,Ω) such that (ξ, ζ) ∈ Θ × O ⊆ U × U − △. Thus, Θ and O are disjoint
supra b-open sets respectively including ξ and ζ. This ends the proof that (U ,Ω) is SbT2.

Theorem 8. The next three properties are identical.

(i) (U ,Ω) is a supra b regular space;

(ii) For every supra b-open subset O of (U ,Ω) including ξ, there is a supra b-open subset
V of (U ,Ω) such that ξ ∈ V ⊆ bcl(V ) ⊆ O;

(iii) Every supra b-open subset O of (U ,Ω) is written: O =
⋃
{H : H is a supra b-open

subset of (U ,Ω) and bcl(H) ⊆ O}.

Proof. 1 → 2: Let O be a supra b-open set such that ξ ∈ U . By hypothesis, there it
disjoint supra b-open sets V and W respectively including ξ and Oc. So that, ξ ∈ V ⊆
W c ⊆ O. Hence, ξ ∈ V ⊆ bcl(V ) ⊆ O.

2 → 3: Let O be a supra b-open set. By hypothesise, for each ξ ∈ O, there exists a
supra b-open set H such that ξ ∈ H ⊆ bcl(H) ⊆ O. Thus, O =

⋃
{H : H is supra b-open

and bcl(H) ⊆ O}.
3 → 1: Suppose that F is a supra b-closed set such that ξ ̸∈ F . Then F c =

⋃
{H : H

is supra b-open and bcl(H) ⊆ F c}. Since ξ ∈ F c, there is a supra b-open set Hξ including
ξ such that bcl(Hξ) ⊆ F c. Put V = (bcl(Hξ))

c; this means that V is a supra b-open set
including F . Thus, V

⋂
Hξ = ∅, which finishes the proof.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 24

Theorem 9. The concepts of SbT2, SbT1 and SbT0 are equivalent under a supra b regular
space.

Proof. The directions SbT2 ⇒ SbT1 ⇒ SbT0 are clear.
To obtain the desired result, we prove that SbT0 ⇒ SbT2. To this end, let ξ ̸= ζ ∈ U . It
follows from Theorem (4) that bcl{ξ} ≠ bcl{ζ}. So that, ξ ̸∈ bcl{ζ} or ζ ̸∈ bcl{ξ}. Consider
ξ ̸∈ bcl{ζ}. By the condition of supra b regularity, there are disjoint supra b-open sets Θ
and O respectively including ξ and bcl{ζ}. Hence, the proof is complete.

Theorem 10. The next properties are identical.

(i) (U ,Ω) is a supra b normal space;

(ii) For every supra b-closed set F and supra b-open set O including F , there is a supra
b-open set V such that F ⊆ V ⊆ bcl(V ) ⊆ O;

(iii) For every supra b-open sets O and V such that O
⋃
V = U , there are two supra

b-closed sets F and H respectively included in O and V such that F
⋃
H = U .

Proof. 1 → 2: Let (U ,Ω) be supra b normal and F be a supra b-closed subset of
a supra b-open set O. Then Oc and F are disjoint supra b-closed sets. So that, there
are two disjoint supra b-open sets W and V respectively including Oc and F . Now,
F ⊆ V ⊆ W c = bcl(W c) ⊆ O, which means that F ⊆ V ⊆ bcl(V ) ⊆ O.

2 → 3: Let O and V be supra b-open sets such that O
⋃
V = U . Then, Oc is a

supra b-closed sets such that Oc ⊆ V . By 2, there is a supra b-open set Θ such that
Oc ⊆ Θ ⊆ bcl(Θ) ⊆ V . Hence, Θc ⊆ O and bcl(Θ) ⊆ V are supra b-closed sets such that
Θc

⋃
bcl(Θ) = U .

3 → 1: Let F and H be disjoint supra b-closed sets. Since F c and Hc are supra open
sets such that F c

⋃
Hc = U , then there are two supra b-closed sets M and N such that

M ⊆ F c, N ⊆ Hc and M
⋃
N = U . Thus, M c and N c are two disjoint supra b-open sets

respectively including F and H. This ends the proof that (U ,Ω) is supra b normal.

Theorem 11. Every SbTk-space is SbTk−1 for k = 1, 2, 3, 4.

The four examples below elucidate that the converse of Theorem (11) is in general
false.

Example 5. Let Ω = {∅,U , {ξ4}, {ξ1, ξ3, ξ4}, {ξ2, ξ3, ξ4}, {ξ1, ξ2, ξ4}} be an ST on U =
{ξ1, ξ2, ξ3, ξ4}. Then the class of all supra b-open subsets of (U ,Ω) is {∅,U , {ξ4}, {ξ1, ξ4},
{ξ2, ξ4}, {ξ3, ξ4}, {ξ1, ξ3, ξ4}, {ξ2, ξ3, ξ4}, {ξ1, ξ2, ξ4}}. Therefore, (U ,Ω) is not an SbT1-
space because ξ1 ̸= ξ4 and all supra b-open sets including ξ1 contain ξ4 as well. On
the other hand, it can be checked that (U ,Ω) is SbT0.

Example 6. Since all singleton subsets of (U ,Ω), given in Example (1), are supra b-
closed, (U ,Ω) is SbT1. In contrast, (U ,Ω) is not SbT2 because ξ1 ̸= ξ3 and there do not
exist disjoint supra b-open sets such that one of them contains ξ1 and the other contains
ξ3.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 25

Example 7. Let Ω = {∅,U , {ξ1, ξ2}, {ξ3, ξ4}, {ξ1, ξ3}, {ξ2, ξ4}, {ξ2, ξ3}, {ξ1, ξ2, ξ3}, {ξ1, ξ2, ξ4},
{ξ1, ξ3, ξ4}, {ξ2, ξ3, ξ4}} be an ST on U = {ξ1, ξ2, ξ3, ξ4}. In (U ,Ω), a set is supra open iff
it is supra b-open. Now, {ξ1, ξ4} is a supra b-closed set and ξ2 ̸∈ {ξ1, ξ4}. Since there do
not exist two disjoint supra b-open sets such that one of them contains ξ2 and the other
contains {ξ1, ξ4}, we obtain (U ,Ω) is not SbT3. In contrast, one can check that (U ,Ω) is
SbT2.

Example 8. Let Ω = {∅,U , {ξ2}, {ξ4}, {ξ2, ξ4}, {ξ1, ξ3}, {ξ1, ξ4}, {ξ1, ξ5}, {ξ2, ξ3}, {ξ2, ξ5},
{ξ3, ξ5}, {ξ4, ξ5}, {ξ1, ξ2, ξ3}, {ξ1, ξ2, ξ4}, {ξ1, ξ2, ξ5}, {ξ1, ξ3, ξ4}, {ξ1, ξ3, ξ5}, {ξ1, ξ4, ξ5}, {ξ2,
ξ3, ξ4}, {ξ2, ξ3, ξ5}, {ξ2, ξ4, ξ5}, {ξ3, ξ4, ξ5}, {ξ1, ξ2, ξ3, ξ4}, {ξ1, ξ2, ξ3, ξ5}, {ξ1, ξ2, ξ4, ξ5}, {ξ1,
ξ3, ξ4, ξ5}, {ξ2, ξ3, ξ4, ξ5}} be an ST on U = {ξ1, ξ2, ξ3, ξ4, ξ5}. In (U ,Ω), a set is supra
open iff it is supra b-open. Now, {ξ1, ξ2} and {ξ3, ξ4} are disjoint supra b-closed subsets
of (U ,Ω). Since there do not exist two disjoint supra b-open sets such that one of them
contains {ξ1, ξ2} and the other contains {ξ3, ξ4}, we obtain (U ,Ω) is not supra b normal.
Hence, it is not SbT4. In contrast, one can check that (U ,Ω) is SbT3.

Theorem 12. Every STk-space (U ,Ω) is SbTk for k = 0, 1, 2.

Proof. Straightforward.

The converse of the above theorem is in general false as the next example shows.

Example 9. Let Ω = {∅,U , {ξ1}, {ξ2, ξ3}} be an ST on U = {ξ1, ξ2, ξ3}. One can check
that (U ,Ω) is SbT4 but is not ST4.

Definition 14. Let O ̸= ∅ be a subset of (U ,Ω). We call a class ΩO = {O
⋂
Θ : Θ is a

supra b-open subset of (U ,Ω)} a relative b-topology on O, and call (O,ΩA) a b-subspace of
(U ,Ω).

Definition 15. We call a property a relative b-hereditary if the it passes from an STS to
every relative b-subspace.

One can prove the next two propositions easily; so we omit their proofs.

Proposition 5. Let (Y,ΩY ) be an b-subspace of (U ,Ω). A subset H of Y is supra b-closed
in (Y,ΩY ) iff there exists a supra b-closed subset F of (U ,Ω) such that H = Y

⋂
F .

Proposition 6. A property of being an SbTk-space is a relative b-hereditary for k =
0, 1, 2, 3.

Proposition 7. Let g : (U ,Ω) → (V, θ) be an injective supra b-continuous mapping. If
(V, θ) is Tk, then (U ,Ω) is SbTk for k = 0, 1, 2.

Proof. We give a proof when k = 2.
Let ξ ̸= ζ ∈ U . Since g is injective, there are x ̸= y ∈ V such that x = f(ξ) and

y = f(ζ). Since (V, θ) is T2, there are two disjoint open subsets M and N of (V, θ)
respectively including x and y. Thus, g−1(M) and g−1(N) are disjoint supra b-open
subsets of (U ,Ω) respectively including ξ and ζ. This proves that (U ,Ω) is SbT2.

Similarly, one can prove the next findings.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 26

Proposition 8. Let g : (U , τ) → (V,Ψ) be a bijective supra b-open mapping. If (U , τ) is
Tk, then (V,Ψ) is SbTk for k = 0, 1, 2.

Proposition 9. Let g : (U , τ) → (V,Ψ) be an injective supra b⋆-continuous mapping. If
(U , τ) is SbTk, then (V,Ψ) is SbTk for k = 0, 1, 2.

Proposition 10. Let g : (U , τ) → (V,Ψ) be a bijective supra b⋆-open mapping. If (U , τ)
is SbTk, then (V,Ψ) is SbTk for k = 0, 1, 2.

Proposition 11. Let g : (U , τ) → (V,Ψ) be a supra b⋆-homeomorphism mapping. Then
(U , τ) is SbTk iff (V,Ψ) is SbTk for k = 0, 1, 2, 3, 4.

Recall that: (O×P )
⋃
(C×D) ⊆ (O

⋃
C)×(P

⋃
D) for every O,C ⊆ U and P,D ⊆ Y .

Theorem 13. The product of two supra b-open sets O and P is supra b-open.

Proof. For two supra b-open sets O and P , we have O ⊆ int(cl(O))
⋃

cl(int(O)) and
P ⊆ int(cl(int(P )))

⋃
int(cl(int(P ))). Therefore

O × P ⊆ [int(cl(O))
⋃
cl(int(O))]× [int(cl(P ))

⋃
cl(int(P ))]

⊆ [int(cl(O))× int(cl(P ))]
⋃
[cl(int(O))× cl(int(P ))]

= [int(cl(O × P ))]
⋃
[cl(int(O × P ))].

Thus, O × P is supra b-open.

Theorem 14. The finite product of SbTk-spaces is SbTk for k = 0, 1, 2.

Proof. We give a proof for two STSs (U ,Ω) and (V,Ψ) when k = 2.
Consider (U ×V, T ) as the product of STSs (U ,Ω) and (V,Ψ). Suppose that (ξ1, ζ1) ̸=

(ξ2, ζ2). Then, ξ1 ̸= ξ2 or ζ1 ̸= ζ2. Suppose, without loss of generality, that ξ1 ̸= ξ2.
By hypothesis, there are disjoint supra b-open subsets M and N of (U ,Ω) respectively
including ξ1 and ξ2. It follows from Theorem (13) that M × V and N × V are two
supra b-open subsets of (U × V, T ) respectively including (ξ1, ζ1) and (ξ2, ζ2) such that

(M × V)
⋂̃
(N × V) = ∅. This proves that (U × V, T ) is SbT2.

Definition 16. Let (U × V, T ) be the product of STSs (U ,Ω) and (V,Ψ) such that C1
and C2 are respectively the classes of all supra b-open subsets of (U ,Ω) and (V,Ψ). Then
β = {Θ×H : Θ ∈ C1 and H ∈ C2} forms a basis for an ST C on U ×V. We call (U ×V, C)
a b-finite product of supra spaces.

Lemma 1. Let (U × V, C) be the b-product of STSs (U ,Ω) and (V,Ψ). If E is a supra
closed subset of (U × V, C), then E =

⋂
k∈I

[(Fk × V)
⋃
(U × Hk)], where Fk and Hk are

respectively supra b-closed subsets of (U ,Ω) and (V,Ψ).

Theorem 15. The b-finite product of SbTk-spaces is STk for k = 0, 1, 2, 3.



A. Mhemdi et al. / Eur. J. Pure Appl. Math, 15 (1) (2022), 15-29 27

Proof. We give a proof for two STSs (U ,Ω) and (V,Ψ) when k = 3.
Let (U×V, C) be the b-product supra space of (U ,Ω) and (V,Ψ). First, we demonstrate

that (U×V, C) is ST1. To do this, let (ξ1, ζ1) ̸= (ξ2, ζ2). Then, ξ1 ̸= ξ2 or ζ1 ̸= ζ2. Suppose,
without loss of generality, that ξ1 ̸= ξ2. So that, there are supra b-open subsets M and N
of (U ,Ω) respectively including ξ1 and ξ2. It comes from Definition(16) that M × V and
N ×V are two supra open subsets of (U ×V, C) respectively including (ξ1, ζ1) and (ξ2, ζ2)
such that (ξ1, ζ1) ̸∈ N × V and (ξ2, ζ2) ̸∈ M × V. Thus, (U × V, C) is ST1. Second, we
demonstrate that (U × V, C) is supra regular. Suppose that (x, y) ∈ U × V and E is a
supra closed subset of (U ×V, C) such that (ξ, ζ) ̸∈ E =

⋂
k∈I

[(Fk×V)
⋃
(U ×Hk)], where Fk

and Hk are respectively supra b-closed subsets of (U ,Ω) and (V,Ψ). Then there is j ∈ I
such that (ξ, ζ) ̸∈ [(Fj ×V)

⋃
(U ×Hj)]. This means that ξ ̸∈ Fj and ζ ̸∈ Hj . Since (U ,Ω)

and (V,Ψ) are supra b regular, there are disjoint supra b-open subsets O and P of (U ,Ω)
respectively including ξ and Fj , and there are disjoint supra b-open subsets M and N of
(V,Ψ) respectively including ζ and Hj . Now, O × M and [(P × V)

⋃
(U × N)] are two

supra open subsets of (U × V, C) respectively including (ξ, ζ) and [(Fj × V)
⋃
(U × Hj)].

Obviously, E ⊆ [(Fj × V)
⋃
(U × Hj)] and (O × M)

⋂
[(P × V)

⋃
(U × N)] = ∅. Hence,

(U × V, T ) is supra regular.

5. Conclusion

Supra topology is one of the famous extensions of topological spaces. We have devoted
this article to present novel types of limit points and separation axioms in the frame
of supra topology. We have formulated these types using supra b-open sets. We have
scrutinized their basic features and revealed some relationships between them. To validate
the introduced findings, we have displayed some illustrative examples. To complete this
path of study, we plan to investigate the following topics:

(i) Formulate other kinds of regular and normal spaces.

(ii) Explore strong types of separation axioms induced from supra b-open set like SbTk-
spaces for (k = 1

2 , 2
1
2 , 3

1
2).

Conflict of interest

The authors declare that there is no conflict of interest regarding the publication of
this paper.

Acknowledgements

This publication was supported by the Deanship of scientific research at Prince Sattam
bin Abdulaziz University, Alkharj, Saudi Arabia.



REFERENCES 28

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