EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 3, 2020, 427-443
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Limit points and separation axioms with respect to
supra semi-open sets

T. M. Al-shami1,∗, E. A. Abo-Tabl2,3, B. A. Asaad4,5, M. A. Arahet6

1 Department of Mathematics, Sana’a University, Sana’a, Yemen
2 Department of Mathematics, College of Arts and Sciences,Methnab, Qassim University,
P. O. Box 931, Buridah 51931, Methnab, Saudia Arabia
3 Department of Mathematics, Faculty of Science, Assiut University, Assiut, Egypt
4 Department of Computer Science, College of Science, Cihan University-Duhok, Kurdistan
Region, Iraq
5 Department of Mathematics, Faculty of Science, University of Zakho, Kurdistan Region,
Iraq
6 Department of Mathematics, Amran University, Amran, Yemen

Abstract. Sometimes we need to minimize the conditions of topology for different reasons such
as obtaining more convenient structures to describe some real-life problems, or constructing some
counterexamples whom show the interrelations between certain topological concepts, or preserving
some properties under fewer conditions of those on topology. To contribute this research area,
in this paper, we establish some new concepts on supra topological spaces using supra semi-open
sets and give some characterizations of them. First, we introduce a concept of supra semi limit
points of a set and study main properties, in particular, on the spaces that possess the difference
property. Second, we define and investigate new separation axioms, namely supra semi Ti-spaces
(i = 0, 1, 2, 3, 4) and give complete descriptions for each one of them. We provide some examples
to show the relationships between them as well as with STi-space.

2020 Mathematics Subject Classifications: 54A05, 54C10, 54D10, 54D15

Key Words and Phrases: Supra semi-open set, supra semi limit point, SsTi-space (i =
0, 1, 2, 3, 4)

1. Introduction and preliminaries

A structure on a nonempty set X is a subset of its power set P (X), topological spaces is
an example of structure satisfying three conditions. In fact, topological spaces have been
generalized in many ways. Alexendroff [12], in 1940, developed abstract spaces where

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v13i3.3743

Email addresses: tareqalshami83@gmail.com (T. M. Al-shami),
a.adotabl@qu.edu.sa (E. A. Abo-Tabl), baravan.asaad@uoz.edu.krd (B. A. Asaad),
malroheet@yahoo.com (M. A. Arahet)

https://www.ejpam.com 427 c© 2020 EJPAM All rights reserved.



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 428

he strengthened the intersection condition. Mashhour et al. [21], in 1983, considered
supra topological spaces by neglecting the intersection condition. Maki et al. [20], in
1996, presented a minimal structure as a collection contains the empty and universal sets.
Császár [13], in 2002, introduced generalized topological spaces if it contains the empty
set and is closed under a nonempty union; and in 2011, he [14] studied a weak structure
as a collection contains the empty set. Al-Odhari [1], in 2015, defined infra topological
spaces by dropping only the union condition.

Many aspects of such spaces have already been studied. Supra topological spaces is
one of the most important developments of general topology in recent years. A family µ
of subsets of a nonempty set X is called a supra topology if it satisfies two conditions:
The first one is the empty and universal sets belong to µ; and the second one is it is
closed under arbitrary union. Some authors remove the empty set from the first condition
of a supra topology because it is obtained from the second condition as the union of an
empty collection of sets. Mashhour et al. [21] studied some basic operators, continuity
and separation axioms on supra topological spaces. Al-shami [2] investigated the classical
topological notions such as limit points of a set, compactness, and separation axioms on
the supra topological spaces. He [7] also studied paracompactness on supra topological
spaces. It should be noted that the supra topological frame can be more convenient to
solve some practical problems and to model some phenomena as pointed out in [19].

Some results via topology do not still valid via supra topology such as the distribution
of the closure operator between the union of two sets and the distribution of the interior
operator between the intersection of two sets. Also, the property of a compact subset of
a T2-space is closed is invalid on the supra topologies.

To extend a class of supra open sets, the notion of supra α-open [15], supra pre-open
[24], supra b-open [26], supra β-open [18], supra R-open [16] and supra semi-open sets [3]
have been introduced and their main properties have been discussed. These generaliza-
tions of supra open sets were defined in a similar way of defining them on general topology.
In other words, their definitions were formulated using supra interior and supra closure
operators instead of interior and closure operators. These generalizations have been uti-
lized to define new versions of compactness, connectedness and separation axioms, see,
for example [5, 6, 17, 22, 25]. The class of supra R-open sets has been studied in [4, 11]
under the name of somewhere dense sets. Mustafa and Qoqazeh [23] took advantage of
supra D-sets to define separation axioms on supra topological spaces. Recently, Al-shami
and El-Shafei [9, 10] have studied separation axioms on supra soft topological spaces and
supra soft topological ordered spaces.

The layout of the paper is as following: In Section (2), we explore a concept of supra
semi limit points of a set. In Section (3), we initiate new types of separation axioms
using supra semi-open sets and illustrate the relationships between them with the help of
examples. Section (4) concludes the paper with summary and further works.

In the rest of this section, we mention some definitions and results of supra topology
and supra semi-open sets that make this study self-contained and easy to read.



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 429

Definition 1. [21] A family µ of subsets of a nonempty set X is called a supra topology
provided that the following two conditions hold.

(i) X and ∅ ∈ µ.

(ii) µ is closed under arbitrary union.

Then the pair (X,µ) is called a supra topological space. Every element of µ is called a
supra open set and its complement is called a supra closed set.

Remark 1. (i) µ is called an associated supra topology with a topology τ if τ ⊆ µ.

(ii) Through this paper, we consider (X,µ) and (Y, ν) are associated supra topological
spaces with the topological spaces (X, τ) and (Y, θ), respectively.

Definition 2. [21] Let A be a subset of (X,µ). Then intµ(A) is the union of all supra
open sets contained in A and clµ(A) is the intersection of all supra closed sets containing
A.

If there is no confusion, we write int(A) and cl(A) in the places of intµ(A) and clµ(A),
respectively.

Definition 3. [3] A subset A of (X,µ) is said to be supra semi-open if A ⊆ cl(int(A)).

Definition 4. [3] For a subset A of (X,µ), sintµ(A) is the union of all supra semi-open
sets contained in A and sclµ(A) is the intersection of all supra semi-closed sets containing
A.

If there is no confusion, we write sint(A) and scl(A) in the places of sintµ(A) and
sclµ(A), respectively.

Definition 5. [3] A map g : (X,µ)→ (Y, ν) is said to be:

(i) supra semi-continuous if the inverse image of each open subset of Y is a supra semi-
open subset of X.

(ii) supra semi-open (resp. supra semi-closed) if the image of each open (resp. closed)
subset of X is a supra semi-open (resp. supra semi-closed) subset of Y .

Definition 6. [2] Let A be a subset of (X,µ). The family µA = {A
⋂
G : G ∈ µ} is called

a supra relative topology on A. A pair (A,µA) is called a supra subspace of (X,µ).

Definition 7. [8] β is called a basis for a supra topology (X,µ) if every member of µ can
be expressed as a union of elements of β.

Definition 8. [8] Let {(Xi, µi) : i = 1, 2, ..., n} be the collection of supra topological
spaces. Then β =

∏n
i=1 µi = {

∏n
i=1Gi : Gi ∈ µi} defines a basis for a supra topology T

on X =
∏n
i=1Xi. The pair (X,T ) is called a finite product supra spaces.

Proposition 1. [8] Let A and B be two subsets of (X,µ) and (Y, ν), respectively. Then:

(i) cl(A)× cl(B) = cl(A×B).

(ii) int(A)× int(B) = int(A×B).



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 430

2. Limit points of a set with respect to supra semi-open sets

In this section, we introduce and study supra semi limit points of a set. We explore
many properties of them and discuss their behaviour on the spaces that possess the differ-
ence property. For more illustration of the presented findings, some interesting examples
are given.

Definition 9. A subset A of (X,µ) is said to be a supra semi neighbourhood of x ∈ X
provided that there is a supra semi open set F containing x such that x ∈ F ⊆ A.

Definition 10. A point x ∈ X is said to be a supra semi limit point of a subset A of
(X,µ) provided that every supra semi neighborhood of x contains at least one point of A
other than x itself.

All supra semi limit points of A is said to be a supra semi derived set of A and is
denoted by As′.

Proposition 2. If A ⊆ B, then As′ ⊆ Bs′ for every A,B ⊆ (X,µ).

Proof. Straightforward.

Corollary 1. We have the following results for every A,B ⊆ (X,µ).

(i) As′
⋃
Bs′ ⊆ (A

⋃
B)s′.

(ii) (A
⋂
B)s′ ⊆ As′

⋂
Bs′.

The following example illustrates that the converse of the above proposition and corol-
lary fails.

Example 1. Let µ = {∅, X, {1, 2}, {2, 3}, {1, 2, 3}} be a supra topology on X = {1, 2, 3, 4}.
Then {∅, X, {1, 2}, {2, 3}, {1, 2, 3}, {1, 2, 4}, {2, 3, 4}} is the collection of all supra semi-
open subsets of (X,µ). If A = {2}, B = {2, 4}, C = {1, 3, 4} and D = {2, 3, 4}, then
As′ = {1, 3, 4}, Bs′ = {1, 3, 4}, Cs′ = {2, 4} and Ds′ = {1, 3, 4}. Now, one readily checks
the following:

(i) Bs′ ⊆ As′, but B 6⊆ A.

(ii) Bs′⋃Ds′ = {1, 3, 4} and (B
⋃
D)s′ = Xs′ = X.

(iii) Bs′⋂Cs′ = {4} and (B
⋂
C)s′ = {4}s′ = ∅.

Proposition 3. Let A be a subset of (X,µ) and x ∈ X. Then x ∈ As′ if and only if
x ∈ (A \ {x})s′.

Proof. Necessity : Let x ∈ As′. Then for every supra semi open set G containing x, we
have (G \ {x})

⋂
A 6= ∅. Therefore (G \ {x})

⋂
(A \ {x}) 6= ∅. Thus x ∈ (A \ {x})s′.

Sufficiency : It follows from Proposition (2).



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 431

Theorem 1. Let A be a subset of (X,µ). Then the following results hold.

(i) A is a supra semi-closed set iff As′ ⊆ A.

(ii) A
⋃
As′ is a supra semi-closed set.

(iii) scl(A) = A
⋃
As′.

Proof.

(i) Suppose that A is a supra semi-closed set and x 6∈ A. Then Ac is a supra semi-open
set containing x. In this case Ac

⋂
A = ∅ leads to x 6∈ As′. Therefore As′ ⊆ A.

Conversely, let x ∈ Ac and let As′ ⊆ A. Then x 6∈ As′. Therefore there is a supra
semi-open set Gx such that Gx \{x}

⋂
A = ∅. Since x ∈ Ac, then Gx

⋂
A = ∅. Now,

Gx ⊆ Ac. Therefore Ac =
⋃
{Gx : x ∈ Ac}. Thus A is supra semi-closed.

(ii) Let x 6∈ (A
⋃
As′). Then x 6∈ A and x 6∈ As′. Therefore there is a supra semi-open set

G such that
G
⋂
A = ∅ (1)

Now, for each x ∈ G, we have x 6∈ As′. This means that

G
⋂
As′ = ∅ (2)

From (1) and (2), we obtain G
⋂

(A
⋃
As′) = ∅. This implies that x 6∈ (A

⋃
As′)s′.

Hence (A
⋃
As′)s′ ⊆ (A

⋃
As′). By (i), A

⋃
As′ is a supra semi-closed set, as required.

(iii) Since A ⊆ scl(A) and As′ ⊆ (scl(A))s′ ⊆ scl(A), then A
⋃
As′ ⊆ scl(A). Since

A
⋃
As′ is a supra semi-closed set containing A and scl(A) is the smallest supra

semi-closed set containing A, then scl(A) ⊆ A
⋃
As′. Therefore scl(A) = A

⋃
As′.

Corollary 2. If A is a supra semi-closed subset of (X,µ), then As′, (As′)s′, ((As′)s′)s′, ...
are supra semi-closed sets.

Definition 11. A map g : (X,µ)→ (Y, ν) is said to be:

(i) supra semi?-continuous if g−1(H) is a supra semi-open set in X for every supra
semi-open set in Y .

(ii) supra semi?-open (resp. supra semi?-closed) if g(H) is a supra semi-open (resp.
supra semi-closed) set in Y for every supra semi-open (resp. supra semi-closed) set
in X.

(iii) supra semi?-homeomorphism if it is bijective, supra semi?-continuous and supra
semi?-open.

Theorem 2. If g : (X,µ)→ (Y, ν) is a supra semi?-homeomorphism map, then g(As′) =
(g(A))s′ for each A ⊆ X.



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 432

Proof. Let a 6∈ (g(A))s′. Then there is a supra semi-open set H containing a such that
(H \ {a})

⋂
g(A) = ∅. So g−1[(H \ {a})

⋂
g(A)] = g−1(∅). This implies that (g−1(H) \

g−1(a))
⋂
A = ∅. Thus g−1(a) 6∈ As′. Since g is bijective, then a 6∈ g(As′). Therefore

g(As′) ⊆ (g(A))s′. By reversing the preceding steps, we find that (g(A))s′ ⊆ g(As′).
Hence, the proof is complete.

Definition 12. For a nonempty set X, a sub collection Λ of 2X is said to have the
difference property provided that G ∈ Λ implies that G \ {x} ∈ Λ.

The following two examples illustrate the existence and uniqueness of the difference
property.

Example 2. Let µ = {∅, G ⊆ N : G is infinite } be a supra topology on the set of natural
numbers N . It is clear that the infinity of G implies the infinity of G\{x}. That is, G ∈ µ
implies G \ {x} ∈ µ. Then (N , µ) has the difference property. Also, it can be seen that the
collection of supra semi-open subsets of (N , µ) coincides with the collection of supra open
sets. Hence, (N , µ) has the difference property for the collection of supra semi-open sets.

Example 3. Let µ = {∅, G ⊆ N : G such that {1, 2} ⊆ G or {2, 3} ⊆ G} be a supra
topology on the set of natural numbers N . Then {1, 2} ∈ µ, but {1, 2} \ {2} = {1} 6∈ µ.
Therefore (X,µ) does not have the difference property.

Theorem 3. If (X,µ) has the difference property for the collection of supra semi-open
sets, then the following properties hold for A ⊆ X.

(i) (As′)s′ ⊆ As′.

(ii) scl(As′) = As′ = (scl(A))s′.

(iii) As′ = ∅ if A is finite.

Proof.

(i) Let x 6∈ As′. Then there is a supra semi-open set G containing x such that G \
{x}

⋂
A = ∅. Since (X,µ) has the difference property for the collection of supra

semi-open sets, then G \ {x} is a supra semi-open set. Therefore G \ {x}
⋂
As′ = ∅.

Since x 6∈ As′, then G
⋂

(As′)s′ = ∅. Thus x 6∈ (As′)s′. Hence, (As′)s′ ⊆ As′.

(ii) Since (As′)s′ ⊆ As′, then it follows from Theorem (1) that As′ is a supra semi-closed
set. Therefore

scl(As′) = As′ (3)

Also, (A)s′ ⊆ (scl(A))s′ because A ⊆ scl(A). On the other hand, let x 6∈ (A)s′. Then
it follows from 1 above that G \ {x}

⋂
A = ∅ and G \ {x}

⋂
As′ = ∅. This means

that G \ {x}
⋂
scl(A) = ∅. Therefore x 6∈ (scl(A))s′. Thus (scl(A))s′ ⊆ (A)s′. Hence

(scl(A))s′ = (A)s′ (4)

From (3) and (4), the desired result is proved.



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 433

(iii) Let A be a finite subset of X. Suppose that there exists an element x ∈ X such that
x ∈ As′. Then for every supra semi-open set G containing x, we have G\{x}

⋂
A 6= ∅.

Therefore for every y ∈ A such that y 6= x, we have G \ {x, y} is a supra semi-open
set. Thus G\[A

⋃
{x}] is a supra semi-open set such that G\[A

⋃
{x}]

⋂
A = ∅. This

implies that x 6∈ As′. But this is a contradiction. Hence, it must be that As′ = ∅.

We explain that the three properties mentioned in the above theorem need not be
true if (X,µ) does not have the difference property for the collection of supra semi-open
sets. Let A = {1, 3} be a subset of supra topological space given in Example (3). Then
As′ = N \ {1, 3}, (As′)s′ = N and scl(As′) = N . This leads to the following three
properties.

(i) (As′)s′ 6⊆ As′.

(ii) scl(As′) 6= As′.

(iii) As′ 6= ∅ in spite of A is finite.

3. Separation axioms with respect to supra semi-open sets

In this section, we utilize supra semi-open sets to introduce the concepts of supra semi
regular, supra semi normal and SsTi-spaces (i = 0, 1, 2, 3, 4). We give some characteri-
zations for each one of them and elucidate the relationships among themselves as well as
with STi-space.

Definition 13. A supra topological space (X,µ) is said to be:

(i) SsT0 if for every a 6= b ∈ X, there exists a supra semi-open set containing only one
of them.

(ii) SsT1 if for every a 6= b ∈ X, there exist two supra semi-open sets one of them
contains a but not b and the other contains b but not a.

(iii) supra semi Hausdorff (or SsT2) if for every a 6= b ∈ X, there exist two disjoint supra
semi-open sets U and V containing a and b, respectively.

(iv) supra semi regular if for every supra semi-closed set F and each a 6∈ F , there exist
disjoint supra semi-open sets U and V containing F and a, respectively.

(v) supra semi normal if for every disjoint supra semi-closed sets F and H, there exist
disjoint supra semi-open sets U and V containing F and H, respectively.

(vi) SsT3 (resp. SsT4) if it is both supra semi regular (resp. supra semi normal) and
SsT1.

Theorem 4. The following three statements are equivalent:

(i) (X,µ) is an SsT0-space;



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 434

(ii) scl({a}) 6= scl({b}) for each a 6= b ∈ X;

(iii) For each a ∈ X, we have {a}s′ is a union of supra semi-closed sets.

Proof. 1→ 2: For each a 6= b ∈ X, there exists a supra semi-open set G containing a
but not b, or containing b but not a. Say a ∈ G and b 6∈ G. Then a 6∈ scl({b}) because
G is a supra semi-open set containing a such that G

⋂
{b} = ∅. Since a ∈ scl({a}), then

scl({a}) 6= scl({b}).
2 → 3: Let b ∈ {a}s′. Then b 6= a and b ∈ {a}

⋃
{a}s′ = scl({a}). Therefore

scl(b) ⊆ scl({a}). Thus b ∈ scl(b) ⊆ {a}s′. Hence, {a}s′ =
⋃
{scl(b): for each b ∈ {a}s′}.

3→ 1: Let a 6= b. Then we have two cases:

(i) Either b ∈ {a}s′. Then there is a supra semi-closed set F such that b ∈ F ⊆ {a}s′.
Since a 6∈ {a}s′, then a 6∈ F . Therefore F c is a supra semi-open set containing a
such that b 6∈ F c.

(ii) Or b 6∈ {a}s′. Then there is a supra semi-open set G containing b such that a 6∈ G.

In the both cases above, we infer that (X,µ) is an SsT0-space.

Corollary 3. An SsT0-space (X,µ) contains at most a supra semi dense singleton set
({a} is a supra semi dense set if scl{a} = X).

Proof. Let (X,µ) be an SsT0-space. Suppose that there are two distinct singleton
set {a} and {b} such that scl({a}) = scl({b}) = X. Then (X,µ) is not an SsT0-space, a
contradiction. Hence, (X,µ) contains at most a supra semi dense singleton set.

Theorem 5. The following four statements are equivalent:

(i) (X,µ) is an SsT1-space;

(ii) Every singleton subset of (X,µ) is supra semi-closed;

(iii) The intersection of all supra semi-open sets containing a set A is exactly A;

(iv) {a}s′ = ∅ for each a ∈ X.

Proof. 1 → 2: Consider (X,µ) is an SsT1-space and let {a} ⊆ X. For all b ∈ X such
that a 6= b, there exists a supra semi-open set G containing b such that G

⋂
{a} = ∅. Then

b 6∈ scl({a}). Therefore, scl({a}) = {a}. Thus, {a} is a supra semi-closed set.
2 → 3: Let A be a subset of (X,µ). Then for each a ∈ Ac, we have {a}c is a

supra semi-open set containing A. Now, A ⊆ {G : G is a supra semi-open set containing
A} ⊆ {{a}c : a ∈ Ac} ⊆ A. Thus, A = {G : G is a supra semi-open set containing A}, as
required.

3 → 4: Suppose that there exists a ∈ X such that {a}s′ 6= ∅. Then there exists b 6= a
such that b ∈ {a}s′. Therefore G\{b}

⋂
{a} 6= ∅ for every supra semi-open set G containing

b. This implies that any supra semi-open set containing b contains a as well. Thus the



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 435

intersection of all supra semi-open sets containing b is not equal {b}. But this contradicts
3. Hence, it must be {a}s′ = ∅.

4 → 1: Let a 6= b. Since {a}s′ = ∅ and {b}s′ = ∅, then {a} and {b} are supra semi-
closed sets. Therefore {a}c and {b}c are supra semi-open sets containing {b} and {a},
respectively. Thus, (X,µ) is an SsT1-space.

Proposition 4. Every (X,µ) satisfying the difference property for the collection of supra
semi-open sets is an SsT1-space.

Proof. Let a 6= b ∈ X. Since X is a supra semi-open set and (X,µ) satisfies the
difference property for the collection of supra semi-open sets, then X \ {a} and X \ {b}
are supra semi-open sets containing b and a, respectively, such that a 6∈ X \ {a} and
b 6∈ X \ {b}. Hence, (X,µ) is an SsT1-space.

We show by the following example that the converse of the above proposition is not
always true.

Example 4. Let µ = {∅, G ⊆ N : 1 ∈ G, or 1 6∈ G and Gc is finite} be a supra topology on
the natural numbers set N . Note that (N , µ) is not a topological space. Then the collection
of all supra semi-open subsets of (N , µ) coincides with the collection of supra open sets. It
is clear that (N , µ) is not an SsT1-space. On the other hand, (N , µ) does not satisfy the
difference property for the collection of supra semi-open sets because {1, 2, 3} is a supra
semi-open set, but {1, 2, 3} \ {1} is not supra semi-open.

We need the following definition to obtain the equivalence between SsT0 and SsT1.

Definition 14. (X,µ) is is called a supra semi symmetric space if a ∈ scl{b} implies that
b ∈ scl{a} for a 6= b ∈ X.

Theorem 6. Let (X,µ) be a supra semi symmetric space. Then it is SsT1 iff it is SsT0.

Proof. The necessary condition is obvious.
To prove the sufficient condition, let a 6= b. Then there exist a supra semi-open set G

containing only one of them. Say, a ∈ G and b 6∈ G. Therefore a 6∈ scl{b}. By the supra
semi symmetry of (X,µ), we have b 6∈ scl{a}. Thus (scl{a})c is a supra semi-open set
containing b. Hence, (X,µ) is SsT1.

Theorem 7. The following three statements are equivalent:

(i) (X,µ) is an SsT2-space;

(ii) {a} =
⋂
{Fi : Fi is a supra semi-closed neighborhood of a} for each a ∈ X;

(iii) The diagonal 4 = {(a, a) : a ∈ X} is supra semi-closed in the product supra space
X ×X.



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 436

Proof. 1 → 2: Consider (X,µ) is an SsT2-space. Then for a 6= b, there exist two
disjoint supra semi-open sets Gi and Hi such that a ∈ Gi and b ∈ Hi. Obviously, Gi ⊆ Hc

i .
Therefore a ∈ scl(Gi) ⊆ Hc

i = Fi. Thus, Fi is a supra semi-closed neighborhood of a such
that b 6∈ Fi. Hence, {a} =

⋂
{Fi : Fi is a supra semi-closed neighborhood of a}.

2 → 1: To prove that (X,µ) is an SsT2-space, let a 6= b. Since {a} =
⋂
{Fi : Fi is a

supra semi-closed neighborhood of a}, then there exists a supra semi-closed neighborhood
Fi0 of a such that b 6∈ Fi0 . Therefore there exists a supra semi-open set G containing a
such that a ∈ scl(G) ⊆ Fi0 . It is clear that (scl(G))c is a supra semi-open set containing
b and G

⋂
(scl(G))c = ∅. Hence, (X,µ) is an SsT2-space.

1→ 3: Suppose that (X,µ) is SsT2 and let (a, b) ∈ X×X−4. Then a 6= b. Therefore
there exist two disjoint supra semi-open sets G and H containing a and b, respectively.
Thus, (a, b) ∈ G×H ⊆ X×X−4, proving that X×X−4 is a supra semi neighbourhood
of any of its points. Hence, 4 is supra semi-closed.

3 → 1: Suppose that 4 is a supra semi-closed subset of X × X and let a 6= b ∈ X.
Then X × X − 4 is a supra semi-open set containing (a, b). Therefore there exist two
supra semi-open subsets G and H of (X,µ) such that (a, b) ∈ G × H ⊆ X × X − 4.
This implies that G and H are two disjoint supra semi-open sets containing a and b,
respectively. Hence, (X,µ) is SsT2.

Theorem 8. The the following three statements are equivalent:

(i) (X,µ) is supra semi regular;

(ii) For each supra semi-open subset U of (X,µ) containing a, there exists a supra semi-
open subset V of (X,µ) such that a ∈ V ⊆ scl(V ) ⊆ U ;

(iii) Every supra semi-open subset U of (X,µ) can be represented as follows: U =
⋃
{H :

H is a supra semi-open subset of (X,µ) and scl(H) ⊆ U}.

Proof. 1 → 2: Let (X,µ) be a supra semi regular space and U be a supra semi-open
set such that a ∈ U . Then there exist disjoint supra semi-open sets V and W containing
a and U c, respectively. Therefore a ∈ V ⊆W c ⊆ U . Thus a ∈ V ⊆ scl(V ) ⊆ U .

2→ 3: Suppose that U is a supra semi-open set. By hypothesise, for each a ∈ U , there
exists a supra semi-open set H such that a ∈ H ⊆ scl(H) ⊆ U . Then U =

⋃
{H : H is

supra semi-open and scl(H) ⊆ U}.
3 → 1: Let F be a supra semi-closed set such that a 6∈ F . Then F c =

⋃
{H : H is

supra semi-open and scl(H) ⊆ F c}. Since a ∈ F c, then there exists a supra semi-open
set Ha containing a such that scl(Ha) ⊆ F c. Take V = (scl(Ha))

c. Then V is a supra
semi-open set containing F and V

⋂
Ha = ∅. This completes the proof.

Theorem 9. Consider (X,µ) is a supra semi regular space. Then the following concepts
are equivalent:

(i) (X,µ) is an SsT2-space;



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 437

(ii) (X,µ) is an SsT1-space;

(iii) (X,µ) is an SsT0-space.

Proof. The implications 1→ 2→ 3 are obvious.
3 → 1: Let a, b ∈ X such that a 6= b. Since (X,µ) is an SsT0-space, then from Theorem
(4), we get scl{a} 6= scl{b}. Therefore a 6∈ scl{b} or b 6∈ scl{a}. Say, a 6∈ scl{b}. Since
(X,µ) is supra semi regular, then there exist disjoint supra semi-open sets G and H
containing a and scl{b}, respectively. Thus (X,µ) is an SsT2-space.

Theorem 10. The following statements are equivalent:

(i) (X,µ) is supra semi normal;

(ii) For each supra semi-closed set F and each supra semi-open set U containing F ,
there exists a supra semi-open set V such that F ⊆ V ⊆ scl(V ) ⊆ U ;

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

semi-closed sets F and H contained in U and V , respectively, such that F
⋃
H = X.

Proof. 1 → 2: Consider (X,µ) is supra semi normal and F is a supra semi-closed
subset of a supra semi-open set U . Then U c and F are disjoint supra semi-closed sets.
Therefor there exist two disjoint supra semi-open sets W and V containing U c and F ,
respectively. Thus F ⊆ V ⊆W c = scl(W c) ⊆ U . Hence, F ⊆ V ⊆ scl(V ) ⊆ U .

2→ 3: Consider U and V are supra semi-open sets such that U
⋃
V = X. Then U c is

a supra semi-closed sets such that U c ⊆ V . By 2, there is a supra semi-open set G such
that U c ⊆ G ⊆ scl(G) ⊆ V . Thus Gc ⊆ U and scl(G) ⊆ V are supra semi-closed sets such
that Gc

⋃
scl(G) = X.

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

⋃
Hc = X, then there are two supra semi-closed sets M and

N such that M ⊆ F c, N ⊆ Hc and M
⋃
N = X. Thus M c and N c are two disjoint supra

semi-open sets containing F and H, respectively. Hence, (X,µ) is supra semi normal.

Now, we show the implications of these separation axioms among themselves as well
as with STi-space.

It should be noted that the concepts of STi-space which were defined by replacing
’supra semi-open’ by ’supra open’ in Definition (13), see, [2, 21].

Theorem 11. Every SsTi-space is SsTi−1 for i = 1, 2, 3, 4.

Converse of this theorem is not necessary true as it is seen in the following examples.

Example 5. Let µ = {∅, X, {1}} be a supra topology on X = {1, 2, 3}. Then the collection
of all supra semi-open subsets of (X,µ) is {∅, X, {1}, {1, 2}, {1, 3}}. Therefore (X,µ) is
not an SsT1-space, because 1 6= 2 and every supra semi-open set containing 2 contains 1
as well. On the other hand, it can be checked that (X,µ) is SsT0.



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 438

Example 6. Let µ = {∅, X, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} be a supra topology on X =
{1, 2, 3, 4}. Then the collection of all supra semi-open subsets of (X,µ) is {∅, X, {1, 2}, {1, 3},
{2, 3}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}}. Therefore (X,µ) is not an SsT2-space, because
3 6= 4 and there do not exist disjoint supra semi-open sets such that one of them contains
3 and the other contains 4. On the other hand, it can be checked that (X,µ) is SsT1.

Example 7. Let µ = {∅, X, {1, 2}, {3, 4}, {1, 3}, {2, 4}, {2, 3}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4},
{2, 3, 4}} be a supra topology on X = {1, 2, 3, 4}. In (X,µ), every set is supra open iff it
is supra semi-open. Now, {1, 4} is a supra semi-closed set and 2 6∈ {1, 4}. Since there do
not exist two disjoint supra semi-open sets such that one of them contains 2 and the other
contains {1, 4}, then (X,µ) is not supra SsT3. On the other hand, it can be checked that
(X,µ) is SsT2.

Example 8. Assume that (N , µ) is the same as in Example (4). Then {2n : n ∈ N} and
{2n+ 3 : n ∈ N} are disjoint supra semi-closed subsets of (N , µ). Since there do not exist
two disjoint supra semi-open sets such that one of them contains {2n : n ∈ N} and the
other contains {2n + 3 : n ∈ N}, then (N , µ) is not supra semi normal. Therefore it is
not SsT4. On the other hand, it can be checked that (N , µ) is SsT3.

Theorem 12. Every STi-space (X,µ) is SsTi for i = 0, 1, 2.

Proof. It follows from the fact that every supra open set is supra semi-open.

Converse of this theorem is not necessary true as it is seen in the following examples.

Example 9. Assume that (X,µ) is the same as in Example (5). Then (X,µ) is not an
ST0-space. On the other hand, the collection of all supra semi-open subsets of (X,µ) is
{∅, X, {1}, {1, 2}, {1, 3}}. Hence, (X,µ) is SsT0.

Example 10. Let µ = {∅, X, {1}, {2}, {1, 2}} be a supra topology on X = {1, 2, 3}. Then
(X,µ) is not an ST1-space. On the other hand, the collection of all supra semi-open subsets
of (X,µ) is {∅, X, {1}, {2}, {1, 2}, {1, 3}, {2, 3}}. Now, it can be checked that (X,µ) is
SsT4.

We complete this section by discussing these separation axioms in terms of hereditary
and topological properties and finite product space.

Definition 15. For a nonempty subset A of (X,µ), the family µA = {A
⋂
G : G is a

supra semi-open subset of (X,µ)} is called a relative semi-topology on A. A pair (A,µA)
is called a semi-subspace of (X,µ).

One can easily prove that a semi-subspace (A,µA) of (X,µ) is a supra topological
space.

Proposition 5. Let (Y, µY ) be a semi-subspace of (X,µ). A subset H of Y is supra
semi-closed in (Y, µY ) iff there exists a supra semi-closed subset F of (X,µ) such that
H = Y

⋂
F .



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 439

Proof. Necessity : Let H be a supra semi-closed subset of (Y, µY ). Then there exists a
supra semi-open subset W of (Y, µY ) such that H = Y \W . Now, there exists a supra semi-
open subset V of (X,µ) such that W = Y

⋂
V . Therefore H = Y \ (Y

⋂
V ) = Y

⋂
V c.

By taking F = V c, the proof of the necessary part is complete.
Sufficiency : Let H = Y

⋂
F such that F is a supra semi-closed subset of (X,µ). Then

Y \H = Y \(Y
⋂
F ) = (Y

⋂
X)\(Y

⋂
F ) = Y

⋂
(X \F ). Since X \F is a supra semi-open

subset of (X,µ), then Y \ H is a supra semi-open subset of (Y, µY ). Thus H is a supra
semi-closed subset of (Y, µY ).

Definition 16. A property is said to be a relative semi-hereditary property if the property
passes from a supra topological space to every relative semi-subspace.

Theorem 13. A property of being an SsTi-space is a relative semi-hereditary for i =
0, 1, 2, 3.

Proof. We shall suffice with proof of case i = 3 which directly contains the case i = 1.
In a similar way, one can prove the cases i = 0, 2.

Suppose that (A,µA) is a relative semi-subspace of an SsT3-space (X,µ). We first
show that (A,µA) is an SsT1-space. Let x 6= y ∈ A ⊆ X. Then there are two supra
semi-open subsets U and V of (X,µ) containing x and y, respectively, such that x 6∈ V
and y 6∈ U . Now, G = A

⋂
U and H = A

⋂
V are two supra semi-open subsets of (A,µA)

containing x and y, respectively, such that x 6∈ H and y 6∈ G. Thus, (A,µA) is SsT1.
Second, we show that (A,µA) is supra semi regular. Let H be a supra semi-closed subset
of (A,µA) and a ∈ A such that a 6∈ H. It follows from Proposition (5) that there is a
supra semi-closed subset F of (X,µ) such that H = F

⋂
A. Since a 6∈ F , then there exist

disjoint supra semi-open subsets U and V of (X,µ) containing F and a, respectively. Now,
M = U

⋂
A and N = V

⋂
A are disjoint supra semi-open subsets of (A,µA) containing H

and a, respectively. Thus (A,µA) is supra semi regular. Hence, the proof is complete.

Proposition 6. Let g : (X,µ) → (Y, θ) be an injective supra semi-continuous map. If
(Y, θ) is Ti, then (X,µ) is SsTi for i = 0, 1, 2.

Proof. We only prove the proposition in the case of i = 2 and the other cases can be
made similarly.

Let a 6= b ∈ X. Then, it follows from the injectivity of g, that there are x 6= y ∈ Y
such that x = f(a) and y = f(b). Since (Y, θ) is T2, then there are two disjoint open
subsets U and V of (Y, θ) containing a and b, respectively. Now, g−1(U) and g−1(V ) are
disjoint supra semi-open subsets of (X,µ) containing a and b, respectively. Hence, (X,µ)
is SsT2, as required.

In a similar way, one can prove the following results.

Proposition 7. Let g : (X, τ) → (Y, ν) be a bijective supra semi-open map. If (X, τ) is
Ti, then (Y, ν) is SsTi for i = 0, 1, 2.



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 440

Proposition 8. Let g : (X, τ) → (Y, ν) be an injective supra semi?-continuous map. If
(X, τ) is SsTi, then (Y, ν) is SsTi for i = 0, 1, 2.

Proposition 9. Let g : (X, τ)→ (Y, ν) be a bijective supra semi?-open map. If (X, τ) is
SsTi, then (Y, ν) is SsTi for i = 0, 1, 2.

Proposition 10. Let g : (X, τ) → (Y, ν) be a supra semi?-homeomorphism map. Then
(X, τ) is SsTi iff (Y, ν) is SsTi for i = 0, 1, 2, 3, 4.

Theorem 14. A and B are supra semi-open sets iff the product of them is supra semi-
open.

Proof. Let A and B be two supra semi-open sets. Then A ⊆ cl(int(A)) and B ⊆
cl(int(B)). So A × B ⊆ cl(int(A)) × cl(int(B)) = cl(int(A × B)). Thus A × B is a
supra semi-open set. Conversely, let A × B be a supra semi-open set. Then A × B ⊆
cl(int(A× B)) = cl(int(A))× cl(int(B)). So A ⊆ cl(int(A)) and B ⊆ cl(int(B)). Hence,
the proof is complete.

Theorem 15. The finite product of SsTi-spaces is SsTi for i = 0, 1, 2.

Proof. We shall suffice with proof of case i = 3 which directly contains the case i = 1.
In a similar way, one can prove the cases i = 0, 2. For simplicity, the proof is given for
two supra topological spaces (X,µ) and (Y, ν).

Let (X×Y, T ) be the product supra space of (X,µ) and (Y, ν). Suppose that (x1, y1) 6=
(x2, y2). Then either x1 6= x2 or y1 6= y2. Without loss of generality, suppose that x1 6= x2.
Therefore there exist two disjoint supra semi-open subsets U and V of (X,µ) containing
x1 and x2, respectively. It follows from Theorem (14) that U × Y and V × Y are two
supra semi-open subsets of (X × Y, T ) containing (x1, y1) and (x2, y2), respectively, such

that (U × Y )
⋂̃

(V × Y ) = ∅. Hence, (X × Y, T ) is SsT2.

Definition 17. Let (X,µ) and (Y, ν) be two supra topological spaces and (X × Y, T ) be
their product supra space such that C1 and C2 are the collections of all supra semi-open
subsets of (X,µ) and (Y, ν), respectively. Then β = {G×H : G ∈ C1 and H ∈ C2} defines
a basis for a supra topology C on X ×Y . We called (X ×Y, C) a semi-finite product supra
space.

Lemma 1. Let (X,µ) and (Y, ν) be two supra topological spaces and (X × Y, C) be
their semi-product supra space. If E is a supra closed subset of (X × Y, C), then E =⋂
i∈I

[(Fi × Y )
⋃

(X × Hi)], where Fi and Hi are supra semi-closed subsets of (X,µ) and

(Y, ν), respectively.

Theorem 16. The semi-finite product of SsTi-spaces is STi for i = 0, 1, 2, 3.

Proof. We shall suffice with proof of case i = 3 which directly contains the case i = 1.
In a similar way, one can prove the cases i = 0, 2. For simplicity, the proof is given for
two supra topological spaces (X,µ) and (Y, ν).



T. M. Al-shami et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 427-443 441

Let (X × Y, C) be the semi-product supra space of (X,µ) and (Y, ν). We first prove
that (X × Y, C) is ST1. Suppose that (x1, y1) 6= (x2, y2). Then either x1 6= x2 or y1 6=
y2. Without loss of generality, suppose that x1 6= x2. Therefore there exist two supra
semi-open subsets U and V of (X,µ) containing x1 and x2, respectively. According to
Definition(17), U × Y and V × Y are two supra open subsets of (X × Y, C) containing
(x1, y1) and (x2, y2) such that (x1, y1) 6∈ V ×Y and (x2, y2) 6∈ U ×Y . Hence, (X ×Y, C) is
ST1. Second, we prove that (X × Y, C) is supra regular. Suppose that (x, y) ∈ X × Y and
E is a supra closed subset of (X × Y, C) such that (x, y) 6∈ E =

⋂
i∈I

[(Fi × Y )
⋃

(X ×Hi)],

where Fi and Hi are supra semi-closed subsets of (X,µ) and (Y, ν), respectively. Then
there exists j ∈ I such that (x, y) 6∈ [(Fj × Y )

⋃
(X ×Hj)]. This means that x 6∈ Fj and

y 6∈ Hj . Since (X,µ) and (Y, ν) are supra semi regular, then there exist disjoint supra semi-
open subsets U and V of (X,µ) containing x and Fj , respectively, and there exist disjoint
supra semi-open subsets M and N of (Y, ν) containing y and Hj , respectively. Therefore
U ×M and [(V × Y )

⋃
(X × N)] are two supra open subsets of (X × Y, C) containing

(x, y) and [(Fj × Y )
⋃

(X ×Hj)], respectively. Obviously, E ⊆ [(Fj × Y )
⋃

(X ×Hj)] and
(U ×M)

⋂
[(V × Y )

⋃
(X ×N)] = ∅. Thus, (X × Y, T ) is supra regular. Hence, the proof

is complete.

4. Conclusion

There are many generalizations of topological spaces which help us to picture and
satisfy many topological properties under fewer conditions such as supra topology, minimal
structure and generalized topology. This work is devoted to introducing and discussing
the concepts of limits points of a set and separation axioms with respect to semi-open
sets. We have established their main properties and provided some examples to show
the obtained results. From the concrete thoughts given in this work, it can be done
more investigations on the theoretical parts of these generalized ideas which is valuable
by studying the following themes:

(i) Define weak types of supra semi regular and supra semi normal spaces.

(ii) Study SsTi-spaces for (i = 1
2 , 2

1
2 , 3

1
2 , 5)

(iii) Explore the concepts introduced herein using the classes of supra α-open sets, supra
b-open sets and supra β-open sets.

(iv) Investigate of the possibility of applying these concepts on information system, es-
pecially, separation axioms.

Conflict of interest

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



REFERENCES 442

Acknowledgements

The authors would like to thank the reviewers for their valuable comments which
improved the presentation of this paper.

References

[1] A M Al-Odhari. On infra topological spaces. International Journal of Mathematical
Archive, 6(11):179–184, 2015.

[2] T M Al-shami. Some results related to supra topological spaces. Journal of Advanced
Studies in Topology, 4(7):283–294, 2016.

[3] T M Al-shami. On supra semi open sets and some applications on topological spaces.
Journal of Advanced Studies in Topology, 8(2):144–153, 2017.

[4] T M Al-shami. Somewhere dense sets and ST1-spaces. Punjab University Journal of
Mathematics, 49(2):101–111, 2017.

[5] T M Al-shami. Utilizing supra α-open sets to generate new types of supra compact
and supra lindelöf spaces. Facta Universitatis, Series: Mathematics and Informatics,
32(1):151–162, 2017.

[6] T M Al-shami. Supra semi-compactness via supra topological spaces. Journal of
Taibah University for Science, 12(3):338–343, 2018.

[7] T M Al-shami. Paracompactness on supra topological spaces. Journal of Linear and
Topological Algebra, 9(2):121–127, 2020.

[8] T M Al-shami, B A Asaad, and M A El-Gayar. Various types of supra pre-compact
and supra pre-lindelöf spaces. Missouri Journal of Mathematical Science, 32(1):1–20,
2020.

[9] T M Al-shami and M E El-Shafei. On supra soft topological ordered spaces. Arab
Journal of Basic and Applied Sciences, 26(1):433–445, 2019.

[10] T M Al-shami and M E El-Shafei. Two types of separation axioms on supra soft
topological spaces. Demonstratio Mathematica, 52(1):147–165, 2019.

[11] T M Al-shami and T Noiri. More notions and mappings via somewhere dense sets.
Afrika Matematika, 30(7):1011–1024, 2019.

[12] P Alexandroff. Diskrete räume. Rec. Math. [Mat. Sbornik] N.S., 2(44):501–519, 1937.

[13] Á Császár. Generalized topology, generalized continuity. Acta Mathematica Hungar-
ica, 96:351–357, 2002.

[14] Á Császár. Weak structure. Acta Mathematica Hungarica, 131(11):193–195, 2011.



REFERENCES 443

[15] R Devi, S Sampathkumar, and M Caldas. On α-open sets and sα-continuous maps.
General Mathematics, 16:77–84, 2008.

[16] M E El-Shafei, M Abo-Elhamayel, and T M Al-shami. On supra R-open sets and some
applications on topological spaces. Journal of Progressive Research in Mathematics,
8(2):1237–1248, 2016.

[17] M E El-Shafei, A H Zakari, and T M Al-shami. Some applications of supra preopen
sets. Journal of Mathematics, Volume 2020, Article ID 9634206:11 pages, 2020.

[18] S Jafari and S Tahiliani. Supra β-open sets and supra β-continuity on topological
spaces. Annales Univ. Sci. Budapest., 56:1–9, 2013.

[19] A M Kozae, M Shokry, and M Zidan. Supra topologies for digital plane. AASCIT
Communications, 3(1):1–10, 2016.

[20] H Maki, J Umehara, and T Noiri. Every topological space is pret 1
2
. Mem. Fac. Sci.

Kochi. Univ. Ser. A Math., 17:33–42, 1996.

[21] A S Mashhour, A A Allam, F S Mahmoud, and F H Kheder. On supra topological
spaces. Indian Journal of Pure and Applied Mathematics, 14(4):502–510, 1983.

[22] J M Mustafa. Supra b-compact and supra b-lindelof spaces. Journal of Mathematics
and Applictions, 36:79–83, 2013.

[23] J M Mustafa and H A Qoqazeh. Supra d-sets and associated separation axioms.
International Journal of Pure and Applied Mathematics, 80(5):657–663, 2012.

[24] O R Sayed. Supra pre-open sets and supra pre-continuous on topological spaces.
Series Mathematics and Information, 20(2):79–88, 2010.

[25] O R Sayed. Supra β-connectedness on topological spaces. Proceedings of the Pakistan
Academy of Sciences, 49(1):19–23, 2012.

[26] O R Sayed and T Noiri. On supra b-open sets and supra b-continuity on topological
spaces. European Journal of Pure and Applied Mathematics, 3:295–302, 2010.