#### Abstract

Every topological property can be associated with its relative version in such a way that when smaller space coincides with larger space, then this relative property coincides with the absolute one. This notion of relative topological properties was introduced by Arhangel’skii and Ganedi in 1989. Singal and Arya introduced the concepts of almost regular spaces in 1969 and almost completely regular spaces in 1970. In this paper, we have studied various relative versions of almost regularity, complete regularity, and almost complete regularity. We investigated some of their properties and established relationships of these spaces with each other and with the existing relative properties.

#### 1. Introduction

Arhangel’skii and Ganedi in [1] introduced the concept that with each topological property one can associate a relative version of it formulated in terms of location of in in such a natural way that when coincides with , then this relative property coincides with . The motivation behind this study is to investiagte how the smaller space is located in the larger one. In past few years, a lot of work has been done on relative topology. Grothendieck [2], Tkachuk [3], Chigogidze [4], Ranchin [5], Dow and Varmeer [6], Das and Bhat [7], Das and Raina [8], Just and Tartir [9], and so on are few names who remarkably contributed in this area. An important feature of relative topology is that a whole pack of relative topological properties can be relativised from a single absolute topological property. It depends on the absolute property that how many interesting versions of relative properties can be created out of it.

In general topology, regularity and normality play a very prominent role in studying topological properties which were widely studied by researchers. Like normality, relative version of regularity has also been studied by Arhangel’skii in [10]. In a survey article, Arhangel’skii also studied some versions of relative regularity like relative strong regular, relative superregular, and so on. There are still many important versions of regularity studied by different topologists in general topology which need to be studied in relative sense. Complete regularity is an important classical version of regularity in which a point and a closed set can be functionally separated. Almost regularity and almost complete regularity are some other versions of regularity introduced by Singal and Arya in [11, 12]. In this paper, we have studied these concepts in relative sense. We have also defined relative almost strongly regular and relative almost superregular spaces. Relationships of these notions with each other and with the existing notions have also been studied and counterexamples are provided wherever needed. It is observed that relative almost regularity is the most generalized among all the versions of relative regularity.

#### 2. Basic Definitions and Preliminaries

Let be a topological space and let . A set is said to be regularly open [13] if . The complement of a regularly open set is called regularly closed. A topological space is said to be *almost normal* [11] if for every pair of disjoint sets and one of which is regularly closed and other is closed, there exist disjoint open sets and such that and . A topological space is said to be *completely regular* if for every closed set of and every point such that , there exists a continuous function on into the closed interval such that and . A topological space is said to be *almost regular* [12] if for every regularly closed set and a point , there exist disjoint open sets and such that and . A topological space is said to be *almost completely regular* [11] if for every regularly closed set of and a point such that , there exists a continuous function on into the closed interval such that and .

Let . is said to be in or *relative* [10] if for every , is closed in . is said to be Hausdorff in or *relative Hausdorff* [10] if for every pair of distinct points in there exist disjoint open sets separating them. is said to be regular in [10] or *relative regular* if for every closed set of and a point such that there exist disjoint open subsets and of such that and . is said to be normal in or *relative normal* [10] if for every pair of disjoint closed sets and of , there exist disjoint open subsets and of such that and .

#### 3. Relative Almost Regular Spaces

*Definition 1. *Let be a topological space and . is said to be almost regular in or relative almost regular (almost superregular in or relative almost superregular) if for every regularly closed set in and a point such that , there exist disjoint open sets and in such that and ().

*Definition 2. *Let be a topological space and . is said to be almost strongly regular in or relative almost strongly regular if for every regularly closed set in and a point such that , there exist disjoint open sets and in such that and .

Clearly, if is almost regular, then is almost regular in . If is regular in , then is almost regular in . If is almost superregular in (almost strongly regular in ), then is almost regular in . But none of these implications can be reversed. Also, can be almost superregular in without being almost strongly regular in , and vice versa, can be almost strongly regular in without being almost superregular in .

*Example 1. *Let . Define a topology on by declaring any set open if the set is either or excludes 0. This space is called excluded point topology with 0 as excluded point. Let . Here is not almost regular and is not relative almost superregular because every regularly closed set in includes 0 and there does not exist any open set containing 0. But is relative almost strongly regular in as for every regularly closed set in and a point such that there exist disjoint open sets and in such that and .

*Example 2. *Let be the set of all integers. Define a topology on , where every odd integer is open and a set is open for every even integer , and the successor and the predecessor of also belong to . Let , where ., is not almost strongly regular in because is regularly closed in , , and there do not exist disjoint open sets in separating and 1. But is almost regular in .

*Example 3. *Let and be a topology on . Let . is not almost normal and is not almost strongly regular in because is regularly closed in and for and , there do not exist disjoint open sets in separating them. But is vacuously almost superregular in as are the only regularly closed sets in and there does not exist any which does not belong to these sets.

Arhangel’skii in [10] defined relative superregular and relative strongly regular as follows: is said to be superregular in if for every closed set in and a point such that , there exist disjoint open sets and in such that and . And is said to be strongly regular in if for every closed set in and a point such that , there exist disjoint open sets and in such that and . The idea of defining relative almost superregular and relative almost strongly regular is taken from these concepts. It is clear from the definitions that relative strongly regular implies relative almost strongly regular and relative superregular implies relative almost superregular. But from the following example, it can be easily checked that none can be reversed.

*Example 4. *Let and be a topology on . Let . Since there does not exist any regularly closed set in , is vacuously relative almost superregular and relative almost strongly regular in . But is neither relative superregular nor relative strongly regular because for the closed set in and , there do not exist disjoint open sets in satisfying the conditions of relative superregularity or relative strong regularity.

Theorem 1. *Let . Then, the following statements are equivalent.*(a)* is relative almost superregular.*(b)*For each and for each regularly open set of containing , there exists regularly open set in such that .*(c)*For every regularly closed set of and each such that , there exist open sets and in such that , , and .*

*Proof. * Let and be a regularly open set in containing . Then, is a regularly closed set in and . Since is relative almost superregular, there exist disjoint open sets and in such that and . Since , then , and hence . Thus, . Now . Take . Then, is a regularly open set in and . Let be a regularly closed set in and such that . Since is regularly open set in containing , there exists regularly open set in such that . Again since is a regularly open set in containing , there exists a regularly open set in such that . Now and are two open sets in with disjoint closures containing and , respectively. Let be a regularly closed set in and such that . Then, there exist open sets and in with disjoint closures containing and , respectively. Thus, and are disjoint open sets in containing and , respectively. Hence, is relative almost superregular.

Theorem 2. *Let . Then, for each and for each regularly open set of containing , there exists regularly open set such that implies that for each and for each neighborhood of in , there exists regularly open neighborhood of in such that .*

*Proof. *Let and be a neighborhood of in . Then, there exists an open set in such that implies . is open implies that is regularly open in containing and so there exists a regularly open set containing such that . Thus, .

The proof of the following result is similar to Theorem 1.

Theorem 3. *Let . Then, the following statements are equivalent.*(a)* is relative superregular.*(b)*For each and for each open set of containing , there exists open set in such that .*(c)*For every closed set of and each such that , there exist open sets and in such that , and .*

*Definition 3. *(see [14]). A space is semiregular if the regularly open sets form a base for the topology.

Theorem 4 (see [12]). *A space is semiregular if for every and for each open set containing , there exists an open set such that .*

Theorem 5. *In a semiregular space, every relative almost superregular space is relative superregular.*

*Proof. *Let be a semiregular space and . Let and be any set containing . Since is semiregular, there exists an open set in such that . Again since is open neighborhood of in and is relative almost superregular, there exists an open set such that , which implies . Hence, is relative superregular.

Corollary 1. * is relative superregular iff is relative almost superregular and is semiregular space.**Recall that a space is said to be a Urysohn space if for every pair of distinct points on , there exists two open sets in with disjoint closures separating them, respectively.*

Theorem 6. *Every relative almost superregular, relative Hausdorff space is a Urysohn space.*

*Proof. *Let be a relative almost superregular, relative Hausdorff space. Let and be two distinct points in . Since is relative Hausdorff, there exist and disjoint open sets in such that and . implies . Since is a regularly open set in containing and is relative almost superregular, there exists an open set in such that . This implies that and and .

Theorem 7. *If is relative almost regular, then for each point and each regularly open set containing , there exists a regularly open set in such that .*

*Proof. *Let be relative almost regular. Let and be a regularly open set in . Then, is regularly closed in such that . By relative almost regularity of , there exist disjoint open sets and in such that and . Now implies that , which implies . Thus, .

*Example 5. *Let and be a topology on . Let . . Clearly, is closed and relative almost regular in . But is not almost regular in itself that is with respect to the subspace topology because is regularly closed in and cannot be separated by disjoint open sets in .

*Remark 1. *From the above example, closed relative almost regular space is not almost regular, i.e., if is closed in and is relative almost regular in , then is not necessarily almost regular in itself. Thus, the following result is relevant.

Theorem 8. *Every closed, relative almost regular, relative Hausdorff space is a Urysohn space.*

*Proof. *Let be closed and relative regular, relative Hausdorff space. Let and be two distinct points in . Since is relative Hausdorff, there exist disjoint open sets and in such that and . implies . Since is a regularly closed set in containing and is relative almost regular, there exists an open set in such that which implies . Thus, , and . Using the fact that is closed, it can be easily shown that . Hence, is a Urysohn space.

In above result, the closedness of cannot be dropped (see the following example).

*Example 6. *Consider Example 3. Let . Then, is not closed but relative Hausdorff and relative almost regular. But it can be easily checked that is not a Urysohn space.

*Definition 4. *(see [15]). is said to be almost normal in or relative almost normal if for every pair of disjoint closed sets and of , such that *A* is regularly closed there exist disjoint open subsets and of such that and .

Theorem 9. *If Y is relative almost normal and relative , then Y is relative almost regular.*

*Proof. *Let be a regularly closed set in and be a point in . Since is , is closed in . Since is almost normal in , there exist disjoint open sets and such that and . Hence, is almost regular in .

The following example establishes that relative almost normality does not necessarily imply relative almost regularity.

*Example 7. *Let with same topology as in Example 1. Let , where is any finite subset of . is not relative almost regular because for , which is regularly closed in , and any point in other than 1, there does not exist disjoint open sets separating them. But is vacuously relative almost normal as there does not exist any pair of disjoint regularly closed sets in .

*Definition 5. *(see [16, 17]). is said to be *almost**-normal* if for any two disjoint closed subsets and of such that *A* is regularly closed, there exist open subsets and of such that is dense in and is dense in and .

Theorem 10. *In an almost -normal space, every relative space is relative almost superregular.*

*Proof. *Let be a regularly closed set in and such that . Since is relative , is closed in . Now since is almost -normal, there exist disjoint open sets and in such that and . Since , and are disjoint open sets in containing and , respectively. Hence, is almost superregular in .

In the above theorem, the condition of relative cannot be relaxed as is shown in the following example.

*Example 8. *Consider the space in Example 1. is vacuously normal and thus almost -normal. is not relative because none of the singletons of is closed in . Also, is not relative almost superregular.

*Remark 2. *From the above example, we have also observed that relative Hausdorff does not imply relative . But another relative version of space can be defined which satisfies the condition that for every pair of distinct points and in , there exist disjoint open sets and in such that , and , . In that case, relative Hausdorff implies this relative version of space.

*Definition 6. *A space is said to be seminormal if for every closed set contained in an open set , there exists a regularly open set such that .

Theorem 11. *Every relative subset of a seminormal, almost -normal space is relative superregular.*

*Proof. *Let be a closed set in and such that . Since is relative , is closed in . By seminormality of , there exists regularly open set such that . Here is a regularly closed set containing and . Now since is almost -normal and is relative , by Theorem 10, is relative almost superregular. Thus, there exist disjoint open sets and in such that and . Hence, is relative superregular.

Continuous image of a relative almost superregular (relative almost strongly regular) space is not relative almost superregular (relative almost strongly regular) space. Following is an example.

*Example 9. *(1)Let . Let be the discrete topology on and be a topology on . Let . is relative almost superregular and relative almost strongly regular. Define as , , , and . Then, is continuous, one-one, and onto function. But is not even relative almost regular in because is regularly closed in and , cannot be separated by disjoint open sets in .(2)Also, take with as discrete space and as excluded point topology with 0 as excluded point. Let be the same as Example 7 which is relative almost superregular and relative almost strongly regular in . Define as identity map which is continuous, one-one, and onto. But Example 7 shows that is not even relative almost regular in .In ([16], Theorem 2.25), it is observed that inverse image of a regularly closed set under an onto, continuous, open, and closed mapping is regularly closed.

Theorem 12. *The image of a relative almost regular space is invariant under an onto, continuous, open, and closed mapping.*

*Proof. *Let and be two topological spaces, is relative almost regular in , and is onto, continuous, open, and closed mapping. Let be a regularly closed set in and such that . Then, is regularly closed in , , and . Since is relative almost regular in , there exist disjoint open sets and in such that and . Since , which implies . Also, is closed implies that is closed. Since is a closed set containing , . So, . Thus, there are two disjoint open sets and in such that and . Hence, is relative almost regular in .

#### 4. Relative Almost Completely Regular Spaces

*Definition 7. *Let be a topological space and . Then, is said to be completely regular in if for every closed set of and a point such that , there exists a continuous function on into the closed interval such that and .

*Definition 8. *Let be a topological space and . Then, is said to be almost completely regular in if for every regularly closed set of and a point such that , there exists a continuous function on into the closed interval such that and .

Theorem 13. *If Y is completely regular in , then Y is almost completely regular in X.*

*Proof. *This proof follows from the fact that every regularly closed set is closed.

Theorem 14. *If Y is almost completely regular in X, then Y is almost regular in X.*

*Proof. *Suppose is almost completely regular in . To prove that is almost regular in , let be a regularly closed set in and such that . Since is almost completely regular in , there exists a continuous function such that and , which implies that and are two disjoint open sets in such that and . Hence, is almost regular in .

Theorem 15. *If Y is completely regular in X, then Y is regular in X.*

*Proof. *Suppose is completely regular in . To prove that is regular in , let be a closed set in and such that . Since is completely regular in , there exists a continuous function such that and , which implies that and are two disjoint open sets in such that and . Hence, is regular in .

Theorem 16 (see [15]). *lf is relative almost normal, then for every pair of disjoint regularly closed sets A and B of , there exists a continuous function f on X into closed interval such that and .*

Theorem 17. *In a regular space, every relative almost normal space is relative completely regular.*

*Proof. *Let be a regular space and be relative almost normal. Let be a closed set in and such that . Since is regular, there exist open sets and in such that , , and . Now since is regularly closed in , is closed in , and is relative almost normal, there exists continuous function such that and . Thus, and . Hence, is relative completely regular.

Theorem 18. *In an almost regular space, every relative almost normal space is relative almost completely regular.*

*Proof. *Let be a regularly closed set in and such that . Since is almost regular, there exist open sets and in such that , and . Rest of the proof is same as the proof of Theorem 17.

The proof of the following result is similar to the proof of the previous theorem.

Theorem 19. *Every relative almost superregular, relative almost normal space is relative almost completely regular.**From Example 7, it is clear that in general, relative almost normal does not necessarily imply relative almost regular and hence not imply relative almost completely regular.*

Theorem 20. *In a normal space, every relative space is relative completely regular.*

*Proof. *Let be a closed set in and be a point in . Since is relative , is closed in . By normality of , there exists a continuous function such that and . Hence, is relative completely regular.

Theorem 21. *In an almost normal space, every relative space is relative almost completely regular.*

*Proof. *Proof is similar to Theorem 20.

In above results, the condition of relative cannot be dropped. Consider Example 7 where has no disjoint closed sets, and hence is normal and almost normal as well. But, is neither relative nor relative almost regular. Hence, is neither relative completely regular nor relative almost completely regular.

From the definitions and implications shown in the paper, the interrelation in Figure 1 is obvious. Most of the relations are not reversible and it is natural to ask the following questions for further study on relative versions of variants discussed in the present paper.

*Question 1. *(1)Does there exist a relative almost regular space which is not relative almost completely regular?(2)Does there exist a relative almost completely regular space which is not relative completely regular?(3)Does there exist a relative almost completely regular space such that its larger space is not almost completely regular?(4)Does there exist a relative completely regular space such that its larger space is not completely regular?

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

The first author is thankful to the Department of Science and Technology, Govt. of India, for awarding INSPIRE fellowship (IF160701).