Some Applications of Supra Preopen Sets

Abstract

The aim of this work is to define some concepts on supra topological spaces using supra preopen sets and investigate main properties. We started this paper by correcting some results obtained in previous study and presenting further properties of supra preopen sets. Then, we introduce a concept of supra prehomeomorphism maps and discuss its main properties. After that we explore the concepts of supra limit and supra boundary points of a set with respect to supra preopen sets and examine their behaviours on the spaces that possess the difference property. Finally, we formulate the concepts of supra pre--spaces and give completely descriptions for each one of them. In general, we study their main properties in detail and show the implications of these separation axioms among themselves as well as with -space with the help of some interesting examples.

1. Introduction and Preliminaries

A set with a family of its subsets is called a supra topological space [1], denoted by , if and the arbitrary union of members of is in . Mashhour et al. [1] generalized some topological notions such as interior and closure operators and continuity and separation axioms. Al-Shami [2] has studied the classical topological notions such as limit points of a set, compactness, and separation axioms on the supra topological spaces.

Some results via topology are 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 -space is closed and is invalid on the supra topologies. To extend a class of supra open sets, the notions of supra -open [3], supra preopen [4], supra -open [5], supra -open [6], supra -open [7], and supra semiopen sets [8] have been introduced and their main properties have been discussed. These generalizations 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 utilized to define new versions of compactness and connectedness, see, for example [9–14]. Mustafa and Qoqazeh [15] took advantage of supra -sets to define separation axioms on supra topological spaces. Recently, Al-Shami and El-Shafei [16] have studied separation axioms on supra soft 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 [17]. Also, the possibility of applying semiopen sets to deal with some problems on digital topology has been demonstrated in [18].

The layout of the paper is as follows. In Section 2, we correct some results of [4] and investigate further properties of supra preopen sets. Also, it presents the concept of supra prehomeomorphism maps and explores main properties. The concepts of supra prelimit and supra preboundary points of a set are studied in Section 3. Section 4 introduces new types of separation axioms using supra preopen sets and elucidates the relationships between them with the help of examples. Section 5 concludes the paper with summary and further works.

In what follows, we collect the relevant definitions and results from supra topology and supra preopen sets to make this paper self-contained and easy to read.

Definition 1. (see [1]). A family of subsets of a nonempty set is called a supra topology provided that the following two conditions hold:(1) and (2) is closed under arbitrary unionThen, the pair is called a supra topological space. Every element of is called a supra open set and its complement is called a supra closed set.

Remark 1. (1)Since , then some authors remove the empty set from the first condition of a supra topology(2) is called an associated supra topology with a topology if (3)Through this paper, we consider and are associated supra topological spaces with the topological spaces and , respectively

Definition 2. (see [1]). Let be a subset of . Then, is the union of all supra open sets contained in and is the intersection of all supra closed sets containing .
If there is no confusion, we write and in the places of and , respectively.

Definition 3. A subset of is said to be(1)Supra -open [3] if (2)Supra preopen [4] if (3)Supra -open [5] if

Definition 4. (see [1]). For a subset of , is the union of all supra preopen sets contained in and is the intersection of all supra preclosed sets containing .
If there is no confusion, we write and in the places of and , respectively.

Definition 5. (see [4]). A map is said to be(1)Supra precontinuous if is a supra preopen set in for each open set in (2)Supra preopen (resp. supra preclosed) if is a supra preopen (supra preclosed) set in for each open (resp. closed) set in

Theorem 1 (see [4]). For a map , we have the following results for every (1) is supra precontinuous if and only if (2) is supra preopen if and only if (3) is supra preclosed if and only if

Definition 6. (see [1]). A map is said to be(1)-continuous if is a supra open set in for every supra open set in (2)-open (resp. -closed) if is a supra open (resp. supra closed) set in for every supra open set in (3)-homeomorphism if it is bijective, -continuous, and -open

Theorem 2. For a map , we have the following results for every :(1) is -continuous if and only if (2) is -open if and only if (3) is -closed if and only if

Definition 7. (see [1]). Let be a subset of . The family is called a supra relative topology on . A pair is called a supra subspace of .

Definition 8. (see [11]). is called a basis for a supra topology if every member of can be expressed as a union of elements of .

Definition 9. (see [11]). Let be the collection of supra topological spaces. Then, defines a basis for a supra topology on . The pair is called a finite product supra spaces.

Theorem 3 (see [11]). and are supra preopen sets iff the product of them is supra preopen.

2. Some Corrections and New Results of Supra Preopen Sets

We begin this section with Proposition 1 below, originally proposed as Proposition 2.1 in [4].

Proposition 1. (1)The intersection of supra open and supra preopen is supra preopen(2)The intersection of supra -open and supra preopen is supra preopen

We give the below example to demonstrate that the above proposition need not be true in general.

Example 1. Let be a supra topology on . Since , then is a supra preopen set and since , then is a supra -open set. Now, and . Since and , then is neither a supra preopen set nor a supra -open set. This emphasizes that the above proposition is erroneous.

The above proposition is true in the case of is a topological space because the following two properties are satisfied on topological spaces, but they do not valid in supra topological spaces:(1) for every subsets of a topological space (2)If is an open set, then for every subset of a topological space

Proposition 2. Let be a supra b-open subset of such that . Then, and are supra preopen.

Proof. Since is a supra b-open set, then . Since , then . Therefore, . Thus, is a supra preopen set. Also, . This implies that . Therefore, is supra preopen set.

In general, there does not exist a relationship between supra preopen sets in and its subspaces as the next two example show.

Example 2. Let be a supra topology on . If , then is a relative supra topology . In , a set is not supra preopen. On the contrary, it can be seen that is a supra preopen set in .

Theorem 4. The property of being a supra preopen set is preserved under -continuous and -open map.

Proof. Let a map be -continuous and -open and let be a supra preopen subset of . Obviously, . Since is -open, then it follows from Theorem 2 that . Since is -continuous, then it follows from Theorem 2 that . Thus, . Hence, is a supra preopen set.

Definition 10. For a nonempty subset of , the family is a supra preopen subset of is called a relative pretopology on . A pair is called a presubspace of .

One can easily prove that a presubspace of is a supra topological space.

Proposition 3. Let be a presubspace of . A subset of is supra preclosed in iff there exists a supra preclosed subset of such that .

Proof.  Necessity: let be a supra preclosed subset of . Then, there exists a supra preopen subset of such that . Now, there exists a supra preopen subset of such that . Therefore, . By taking , the proof of the necessary part is complete. Sufficiency: let such that is a supra preclosed subset of . Then, . Since is a supra preopen subset of , then is a supra preopen subset of . Thus, is a supra preclosed subset of .

In the rest part of this section, we present a concept of supra prehomeomorphism and supra pre-homeomorphism maps and discuss some basic properties (denotes another type of homeomorphism maps).

Definition 11. A bijective map is said to be a supra prehomeomorphism if it is supra precontinuous and supra preopen.

Since every supra open set is supra preopen, then every supra homeomorphism map is a supra prehomeomorphism. However, the converse is not always true as it is illustrated in the following example.

Example 3. Assume that and are two topologies on the real numbers set . Let or and be two associated supra topologies with and , respectively. Then, the identity map is a supra prehomeomorphism, but it is not a supra homeomorphism because the image of a supra open set is on a supra open set.

Theorem 5. The equivalence of the following properties hold if a map is bijective and supra precontinuous:(1) is a supra prehomeomorphism(2) is supra precontinuous(3) is supra preclosed

Proof. Straightforward.

Theorem 6. A bijective map is a supra prehomeomorphism if and only if and for every .

Proof. If is a supra prehomeomorphism map. Then, is supra precontinuous and supra preclosed. It follows from Theorem 1 that and .
If is a bijective map such that and , then is supra precontinuous and supra preclosed. It follows from Theorem 1 that is a supra prehomeomorphism map.

Definition 12. A map is said to be(1)Supra pre-continuous if is a supra preopen set in for every supra preopen set in (2)Supra pre-open (resp. supra pre-closed) if is a supra preopen (resp. supra preclosed) set in for every supra preopen (resp. supra preclosed) set in (3)Supra pre-homeomorphism if it is bijective, supra pre-continuous, and supra pre-open

3. Limit and Boundary Points of a Set with Respect to Supra Preopen Sets

This section defines the concepts of supra prelimit and supra preboundary points of a set and studies the interrelations between them. It provides some examples to show the obtained results and examines some properties of the supra prederived set on the spaces that possess the difference property.

Definition 13. A subset of is said to be a supra preneighbourhood of provided that there is a supra preopen set containing such that .

Definition 14. A point is said to be a supra prelimit point of a subset of provided that every supra preneighborhood of contains at least one point of other than itself.

All supra prelimit points of is said to be a supra prederived set of and is denoted by .

Proposition 4. If , then for every subsets and of .

Proof. Straightforward.

Corollary 1. We have the following results for any two subsets and of :(1)(2)

The following example illustrates that the converse of the above proposition and corollary fails.

Example 4. Let be a supra topology on . Then, , is the collection of all supra preopen subsets of . If and , then and . Obviously, . Now, we have the following cases:(i), but (ii) and (iii) and

Proposition 5. Let be a subset of and . Then, if and only if .

Proof.  Necessity: let . Then, for every supra preopen set containing , we have . Therefore, . Thus, . Sufficiency: it follows from Proposition 4.

Theorem 7. Let be a subset of . Then, the following results hold.(i) is a supra preclosed set iff (ii) is a supra preclosed set(iii)

Proof. (i)Suppose that is a supra preclosed set and . Then, is a supra preopen set containing . In this case, leads to . Therefore, . Conversely, let and let . Then, . Therefore, there is a supra preopen set such that . Since , then . Now, . Therefore, . Thus, is supra preclosed.(ii)Let . Then, and . Therefore, there is a supra preopen set such that Now, for each , we have . This means that From (1) and (2), we obtain . This implies that . Hence, . By , is a supra preclosed set, as required.(iii)Since and , then . Since is a supra preclosed set containing and is the smallest supra preclosed set containing , then . Therefore, .

Corollary 2. If is a supra preclosed subset of , then , , , …, are supra preclosed sets.

Theorem 8. If is a supra pre-homeomorphism map, then for each .

Proof. Let . Then, there is a supra preopen set containing such that . So, . This implies that . Thus, . Since is bijective, then . Therefore, . By reversing the preceding steps, we find that . Hence, the proof is complete.

Definition 15. For a nonempty set , a subcollection of is said to have the difference property provided that implies that .

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

Example 5. Let be a supra topology on the set of natural numbers . It is clear that the infinity of implies the infinity of . That is, implies . Then, has the difference property. Also, it can be seen that the collection of supra preopen subsets of coincides with the collection of supra open sets. Hence, has the difference property for the collection of supra preopen sets.

Example 6. Let such that or be a supra topology on the set of natural numbers . Then, , but . Therefore, does not have the difference property for the collection of supra open sets. Hence, it does not have the difference property for the collection of supra preopen sets.

Theorem 9. If has the difference property for the collection of supra preopen sets, then the following properties hold for :(1)(2)(3) if is finite

Proof. (1)Let . Then, there is a supra preopen set containing such that . Since has the difference property for the collection of supra preopen sets, then is a supra preopen set. Therefore, . Since , then . Thus, . Hence, .(2)Since , then it follows from Theorem 7 that is a supra preclosed set. Therefore, Also, because . On the contrary, let . Then, it follows from 1 above that and . This means that . Therefore, . Thus, . Hence, From (3) and (4), the desired result is proved.(3)Let be a finite subset of . Suppose that there exists an element such that . Then, for every supra preopen set containing , we have . Therefore, for every such that , we have is a supra preopen set. Thus, is a supra preopen set such that . This implies that . However, this is a contradiction. Hence, it must be that .

We explain that the three properties mentioned in the above theorem need not be true if does not have the difference property for the collection of supra preopen sets. Let be a subset of supra topological space given in Example 6. Note that the collection of supra open sets coincides with the collection of supra preopen sets. By calculating, we find that , , and . This leads to the following three properties:(1)(2)(3) in spite of is finite

Definition 16. Let be a subset of . The supra preboundary of (denoted by ) is the set of all elements which belongs to .

Lemma 1. Let be a subset of . Then,(1)(2)

Proof. We prove (1) and (2) is proved in a similar way.
Since , then . Therefore, . Conversely, . Then, . Therefore, . Thus, . Hence, the desired result is proved.

Proposition 6. Let be a subset of . Then, .

Proof. and   and   and .

Corollary 3. Let be a subset of . Then,(1)(2) is a supra preclosed set(3)(4)(5)(6)(7)

Proof. (1)(2)It follows from noting that is the intersection of two supra preclosed sets which is supra preclosed(3)(4)(5)(6)(7); Since , then the result holds

Proposition 7. Let be a subset of . Then,(1) is a supra preopen set iff (2) is a supra preclosed set iff

Proof. (1)Necessity: since is supra preopen, then . Sufficiency: let . Then, or . Since , then . Therefore, . Thus, is supra preopen.(2)Necessity: since is supra preclosed, then . Therefore, . Sufficiency: suppose that and let . Then, . Since , then so that for each supra preopen set containing , we obtain . This implies that . Thus, is a union of supra preopen sets. Hence, is supra preopen.

Corollary 4. We have the following properties for a subset of :(1) is both supra preopen and supra preclosed iff (2)(3)

4. Separation Axioms with Respect to Supra Preopen Sets

In this section, we use a class of supra preopen sets to introduce new types of separation axioms, namely, -spaces . We establish some characterizations of each one of these spaces and study them in terms of hereditary and topological properties and product space. Illustrative examples are provided to show the obtained results.

Definition 17. A supra topological space is said to be(1) if for every , there exists a supra preopen set containing only one of them.(2) if for every , there exist two supra preopen sets one of them contains but not and the other contains but not (3)Supra pre-Hausdorff (or ) if for every , there exist two disjoint supra preopen sets and containing and , respectively(4)Supra preregular if for every supra preclosed set and each , there exist disjoint supra preopen sets and containing and , respectively(5)Supra prenormal if for every disjoint supra preclosed sets and , there exist disjoint supra preopen sets and containing and , respectively(6) (resp. ) if it is both supra preregular (resp. supra prenormal) and

Theorem 10. The following three statements are equivalent:(1) is an -space(2) for each (3)For each , we have is a union of supra preclosed sets

Proof. : for each , there exists a supra preopen set containing but not or containing but not . Say and . Then, because is a supra preopen set containing such that . Since , then .
: let . Then, and . Therefore, . Thus, . Hence, : for each .
: let . Then, we have two cases:(i)Either . Then, there is a supra preclosed set such that . Since , then . Therefore, is a supra preopen set containing such that .(ii)Or . Then, there is a supra preopen set containing such that .In the both cases above, we infer that is an -space.

Corollary 5. An -space contains at most a supra predense singleton set is a supra predense set if .

Proof. Let be an -space. Suppose that there are two distinct singleton set and such that . Then, is not an -space, a contradiction. Hence, contains at most a supra predense singleton set.

Theorem 11. The following there statements are equivalent:(1) is an -space(2)Every singleton subset of is supra preclosed(3)The intersection of all supra preopen sets containing a set is exactly (4) for each

Proof. : consider is an -space and let . For all such that , there exists a supra preopen set containing such that . Then, . Therefore, . Thus, is a supra preclosed set.
: let be a subset of . Then, for each , we have is a supra preopen set containing . Now, is a supra preopen set containing . Thus, is a supra preopen set containing , as required.
: suppose that there exists such that . Then, there exists such that . Therefore, for every supra preopen set containing . This implies that any supra preopen set containing contains as well. Thus, the intersection of all supra preopen sets containing is not equal . However, this contradicts 3. Hence, it must be .
: let . Since and , then and are supra preclosed sets. Therefore, and are supra preopen sets containing and , respectively. Thus, is an -space.

Proposition 8. Every satisfying the difference property for the collection of supra preopen sets is an -space.

Proof. Let . Since is a supra preopen set and satisfies the difference property for the collection of supra preopen sets, then and are supra preopen sets containing and , respectively, such that and . Hence, is an -space.

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

Example 7. Let be a supra topology on . Then, the collection of all supra preopen subsets of is . Therefore, is an -space. On the contrary, does not satisfy the difference property for the collection of supra preopen sets because is a supra preopen set, but is not supra preopen.

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

Definition 18. is called a supra presymmetric space if implies that for .

Theorem 12. Let be a supra presymmetric space. Then, it is iff it is .

Proof. The necessary condition is obvious.
To prove the sufficient condition, let . Then, there exists a supra preopen set containing only one of them. Say, and . Therefore, . By the supra presymmetry of , we have . Thus, is a supra preopen set containing . Hence, is .

Theorem 13. The following three statements are equivalent:(1) is an -space(2) is a supra preclosed neighborhood of for each (3)The diagonal is supra preclosed in the product supra space

Proof. : consider is an -space. Then, for , there exist two disjoint supra preopen sets and such that and . Obviously, . Therefore, . Thus, is a supra preclosed neighborhood of such that . Hence, is a supra preclosed neighborhood of .
: to prove that is an -space, let . Since is a supra preclosed neighborhood of , then there exists a supra preclosed neighborhood of such that . Therefore, there exists a supra preopen set containing such that . It is clear that is a supra preopen set containing and . Hence, is an -space.
: suppose that is and let . Then, . Therefore, there exist two disjoint supra preopen sets and containing and , respectively. Thus, , proving that is a supra preneighbourhood of any of its points. Hence, is supra preclosed.
: suppose that is a supra preclosed subset of and let . Then, is a supra preopen set containing . Therefore, there exist two supra preopen subsets and of such that . This implies that and are two disjoint supra preopen sets containing and , respectively. Hence, is .

Theorem 14. The following three statements are equivalent:(1) is supra preregular(2)For each supra preopen subset of containing , there exists a supra preopen subset of such that (3)Every supra preopen subset of can be represented as follows: is a supra preopen subset of and

Proof. : let be a supra preregular space and be a supra preopen set such that . Then, there exist disjoint supra preopen sets and containing and , respectively. Therefore, . Thus, .
: suppose that is a supra preopen set. By hypothesis, for each , there exists a supra preopen set such that . Then, is supra preopen and .
: let be a supra preclosed set such that . Then, is supra preopen and . Since , then there exists a supra preopen set containing such that . Take . Then, is a supra preopen set containing and . This completes the proof.

Theorem 15. Consider is a supra preregular space. Then, the following concepts are equivalent:(1) is an -space(2) is an -space(3) is an -space

Proof. The implications are obvious.
: let such that . Since is an -space, then from Theorem 10, we obtain . Therefore, or . Say, . Since is supra preregular, then there exist disjoint supra preopen sets and containing and , respectively. Thus, is an -space.

Theorem 16. The following statements are equivalent:(1) is supra prenormal(2)For each supra preclosed set and each supra preopen set containing , there exists a supra preopen set such that (3)For every supra preopen sets and such that , there are two supra preclosed sets and contained in and , respectively, such that

Proof. : consider is supra prenormal and is a supra preclosed subset of a supra preopen set . Then, and are disjoint supra preclosed sets. Therefore, there exist two disjoint supra preopen sets and containing and , respectively. Thus, . Hence, .
: consider and are supra preopen sets such that . Then, is a supra preclosed sets such that . By 2, there is a supra preopen set such that . Thus, and are supra preclosed sets such that .
: consider and are disjoint supra preclosed sets. Since and are supra open sets such that , then there are two supra preclosed sets and such that , , and . Thus, and are two disjoint supra preopen sets containing and , respectively. Hence, is supra prenormal.

Theorem 17. If all members of are supra clopen, then is an -space.

Proof. Let such that . By hypothesis, is a supra clopen set. Then, . Therefore, is a supra preopen set. The arbitrary selection of implies that every singleton subset of is supra preopen. Therefore, every subset of is supra preopen. Hence, the desired result is proved.

To explain that the converse of the above theorem is not always true, we provide the following example.

Example 8. Let be a supra topology on . Then, the collection of all supra preopen subsets of is . It easily checks that is an -space. On the contrary, a set is not a supra clopen set.

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

It should be noted that the concepts of -space which were defined by replacing “supra preopen” by “supra open” in Definition 4.1, see [1, 2].

Theorem 18. Every -space is for .

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

Example 9. Let be a supra topology on . Then, the collection of all supra preopen subsets of is . Therefore, is not an -space because and every supra preopen set containing 1 contains 4 as well. On the contrary, it can be checked that is .

Example 10. Assume that is the same as in Example 7. Then, is not an -space because and there do not exist disjoint supra preopen sets such that one of them contains 2 and the other contains 3. On the contrary, it can be checked that is .

Example 11. Let be a supra topology on . In , every set is supra open iff it is supra preopen. Now, is a supra preclosed set and . Since there do not exist two disjoint supra preopen sets such that one of them contains 2 and the other contains , then is not supra . On the contrary, it can be checked that is .

Example 12. Let , , be a supra topology on . In , every set is supra open iff it is supra preopen. Now, and are disjoint supra preclosed subsets of . Since there do not exist two disjoint supra preopen sets such that one of them contains and the other contains , then is not supra prenormal. Therefore, it is not . On the contrary, one can check that is .

Theorem 19. Every -space is for .

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

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

Example 13. Let be a supra topology on . Then, is not an -space. On the contrary, it follows from Theorem 17 that is .

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

Definition 19. A property is said to be a relative prehereditary property if the property passes from a supra topological space to every relative presubspace.

Theorem 20. A property of being an -space is a relative prehereditary for .

Proof. We prove the theorem in the case of and the other cases follow similar lines.
Suppose that is a relative presubspace of an -space . We first show that is an -space. Let . Then, there are two supra preopen subsets and of containing and , respectively, such that and . Now, and are two supra preopen subsets of containing and , respectively, such that and . Thus, is . Second, we show that is supra preregular. Let be a supra preclosed subset of and such that . It follows from Proposition 3 that there is a supra preclosed subset of such that . Since , then there exist disjoint supra preopen subsets and of containing and , respectively. Now, and are disjoint supra preopen subsets of containing and , respectively. Thus, is supra preregular. Hence, the proof is complete.

Proposition 9. Let be an injective supra precontinuous map. If is , then is for .

Proof. We only prove the proposition in the case of and the other cases can be made similarly.
Let . Then, it follows from the injectivity of , that there are such that and . Since is , then there are two disjoint open subsets and of containing and , respectively. Now, and are disjoint supra preopen subsets of containing and , respectively. Hence, is , as required.
In a similar way, one can prove the following results.

Proposition 10. Let be a bijective supra preopen map. If is , then is for .

Proposition 11. Let be an injective supra pre-continuous map. If is , then is for .

Proposition 12. Let be a bijective supra pre-open map. If is , then is for .

Proposition 13. Let be a supra pre-homeomorphism map. Then, is iff is for .

Theorem 21. The finite product of -spaces is for .

Proof. We prove the theorem for two supra topological spaces and in the case of . One can prove the other cases similarly.
Let be the product supra space of and . Suppose that . Then, either or . Without loss of generality, suppose that . Therefore, there exist two disjoint supra preopen subsets and of containing and , respectively. It follows from Theorem 3 that and are two supra preopen subsets of containing and , respectively, such that . Hence, is .

Definition 20. Let and be two supra topological spaces and be their product supra space such that and are the collections of all supra preopen subsets of and , respectively. Then, and defines a basis for a supra topology on . We called a prefinite product supra space.

Lemma 2. Let and be two supra topological spaces and be their preproduct supra space. If is a supra closed subset of , then , where and are supra preclosed subsets of and , respectively.

Theorem 22. The prefinite product of -spaces is for .

Proof. We prove the theorem for two supra topological spaces and in the case of . One can prove the other cases similarly.
Let be the preproduct supra space of and . We first prove that is . Suppose that . Then, either or . Without loss of generality, suppose that . Therefore, there exist two supra preopen subsets and of containing and , respectively. According to Definition 20, and are two supra open subsets of containing and such that and . Hence, is . Second, we prove that is supra regular. Suppose that and is a supra closed subset of such that , where and are supra preclosed subsets of and , respectively. Then, there exists such that . This means that and . Since and are supra pre regular, then there exist disjoint supra preopen subsets and of containing and , respectively, and there exist disjoint supra preopen subsets and of containing and , respectively. Therefore, and are two supra open subsets of containing and , respectively. Obviously, and . Thus, is supra regular. Hence, the proof is complete.

5. Conclusion

We began this work by correcting some results of [4]. Then, we have presented the concept of supra pre homeomorphism and investigated main properties. Also, we have introduced and studied the concepts of supra limits and supra boundary points with respect to preopen sets. Finally, we have defined the concepts of supra preregular, supra prenormal, and -spaces and discussed their basic properties. From the concrete thoughts given in this work, more investigations can be carried out on the theoretical parts of these generalized ideas which are valuable by studying the following themes:(1)Define weak types of supra preregular and supra prenormal spaces(2)Study -spaces for 1/2, 2 1/2, 3 1/2, 5(3)Explore the concepts introduced herein using the classes of supra -open sets, supra semiopen sets, supra -open sets, and supra -open sets(4)Investigate of the possibility of applying these concepts on information system, especially, separation axioms

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.