Functionally Separation Axioms on General Topology
In this paper, we define a new family of separation axioms in the classical topology called functionally spaces for . With the assistant of illustrative examples, we reveal the relationships between them as well as their relationship with spaces for . We demonstrate that functionally spaces are preserved under product spaces, and they are topological and hereditary properties. Moreover, we show that the class of each one of them represents a transitive relation and obtain some interesting results under some conditions such as discrete and Sierpinski spaces.
1. Introduction and Preliminaries
Topology is a branch of mathematics that researches properties of spaces that are invariant under any continuous deformation. That is, it forms a novel kind of geometry that relies on nearness or neighborhood of elements instead of measuring distance between them. Recently, topological spaces have been applied to model practical problems, especially those related to information system; see, for example [1–3].
Separation axioms are one among the most common, important, and interesting ideas in topology. They can be used to initiate different classes of topological spaces and determine the type of some subsets. Some separation axioms were defined using continuous maps (see, [4–6]) on one hand. On the other hand, some separation axioms were introduced using generalized open sets (see  and the references mentioned therein). Also, separation axioms have been generalized to other spaces in general topology like pretopological spaces, ordered topological spaces, and generalized topological spaces (see [8–11]). Another generalization of separation axioms was given using functions in the category of topological spaces (see ). We draw attention to some separation axioms have been very useful in the study of certain objects in digital topology (see [13, 14]).
This work is organized as follows: after this introduction, we recall some definitions and results which are required to make this work self-contained. In Section 2, we introduce the concepts of functionally space for . We study their main properties, especially those are related to product spaces and topological and hereditary properties. In Section 3, we explore some findings that associated these concepts with the discrete and Sierpinski spaces. We write some conclusions and propose some future works in Section 4.
The following definitions are mentioned in .
Definition 1. A subcollection of is called a topology on provided that it is closed under arbitrary union and finite intersection. A pair is called a topological space. Every subset of belong to is called an open set, and its complement is called a closed set.
Definition 2. A bijective map is said to be a homeomorphism if it is continuous (i.e., the inverse image of every open (resp., closed) set is open (resp., closed)) and open (i.e., the image of every open set is open).
A property which is preserved by every homeomorphism map is called a topological property.
Definition 3. Let be a set in a topological space . The family is called a relative topology on . A pair is called a subspace of .
A property which is preserved by every subspace is called a hereditary property.
Definition 4. is said to be as follows:(1) provided that, for every , there exists an open set containing only one of them(2) provided that, for every , there exist two open sets such that one containing but not and the other containing but not (3) (or Hausdorff) provided that there exist two disjoint open sets and for every such that and
Proposition 1. For any two sets and , we have .
Definition 5. Let be the collection of topological spaces. Then, defines a base for a topology on . We call a finite product space.
Definition 6. A binary relation on (i.e., ) is said to be transitive if whenever and .
2. Relative Separation Axioms
In this portion, we formulate the concepts of functionally space using continuous maps. We explore their basic properties and provide some interesting examples to clarify the presented results.
Definition 7. Let be two topological spaces. is said to be as follows:(1)A functionally space relatively to or, for short, is if for any distinct points , there exist a continuous map from to and such that (2)A functionally space relatively to or, for short, is if for any distinct points , there exist a continuous map from to and such that and (3)A functionally space relatively to or, for short, is if for any distinct points , there exist a continuous map from to and such that , and
Remark 1. (1)If is a singleton, then for any topological space , is for each (2)If is the indiscrete topology and , then is not for each topology on
Proposition 2. is a-space if and only if it isfor each.
Proof. It is sufficient to take the identity map from to itself. It is clear that is a continuous map satisfying the desired result.
It is easy to prove the following implications.
Proposition 3. Everyspace isfor each.
We give the following example to clarify the fail of the converse of Proposition 3.
Example 1. Let be a topology on and be the cofinite topology on the set of natural numbers . It is clear that and are, respectively, and spaces. On the other hand, is not a space and is not a space. Without loss of generality, consider . Then, it follows from Proposition 3 that is but not . Also, consider . Then, is but not .
Proposition 4. For all , ifis, thenis a-space.
Proof. When . Suppose that in an -space. Then, there exist a continuous map from to and such that . Now, is an open subset of containing only one of the elements and but not both. Hence, is a -space.
By analogous it is easy to prove the proposition for .
Corollary 1. Ifis a finitespace, thenis the discrete topology.
The converse of the proposition above fails as shown in the following example.
Example 2. Let be the discrete topology on and be a topology on . It is clear that is a space. On the other hand, let be any continuous map. While , and are three distinct points in , a topology contains only one proper open subset of ; hence, is not .
We summarize the previous relationships in Figure 1.
Theorem 1. Ifandarespace such thatis finite, thenis an-space, where.
Proof. We only prove the theorem in case of .
Since a finite space, then every map from to is continuous. Therefore, for every , we define a continuous map such that . Since is a space, there exist open subsets of such that and . Hence, the proof is complete.
In the following result, we investigate under which conditions the converse of Proposition 4 is true.
Proposition 5. Letbe an injective bicontinuous map. Then,is aspace if and only ifis aspace for each.
Proof. We only prove the proposition incase of . The other cases are proved similarly.
The necessary part follows from Proposition 4. To prove the sufficient part, let . By hypothesis, there exist two disjoint open sets such that and . Since is injective, we obtain , and since is open, then and are disjoint open sets in containing and , respectively. It comes from the continuity of that is a . Hence, we obtain the desired result.
Theorem 2. The finite product ofspaces isfor each.
Proof. We only prove the proposition incase of .
Let and be two spaces. Then, for all , there exists a continuous map from to and such that and , and for all , there exists a continuous map from to and such that and . Suppose that . Then, or . Without loss of generality, consider . Taking a map defined by . It can be checked that is a continuous map. Now, we have is an open subset of containing and is an open subset of containing such that . Hence, is a .
By analogous one can prove the other two cases.
Theorem 3. The property of being anspace is a topological property for each.
Proof. We only prove the proposition incase of .
Suppose that is a homomorphism map. To prove that is a space, let . Then, . Since is a space, then there exists a continuous map from to and such that , and . Since is a homomorphism and is continuous, then is a continuous map satisfying that is a space. Hence, the proof is complete.
By analogous one can prove the other two cases.
Theorem 4. The property of being an space is a hereditary property for each.
Proof. The proof follows from the fact a continuous map implies that the restriction map is also continuous for every nonempty subset of .
Theorem 5. The binary relation defined on the set of topological spaces by if and only ifisis transitive.
Proof. Let , and be three topological spaces such that and . If , then there exists a continuous map from to satisfying the conditions given in Definition 2.1 for each . Now, since , then there exists also a continuous map from to satisfying the same conditions, so that the composition is a continuous map from to verifying the conditions in the separation for the points and . Thus, ; hence, is transitive.
3. Properties and Particular Cases
In this section, we present some properties of spaces under some particular of topological spaces such as discrete and Sierpinski spaces.
Proposition 6. Letandbe two topological spaces such thatand. Ifis, thenandare discrete topologies.
Proof. Let . Then, there exists a continuous map from to and such that and , then . Similarly, we prove that . Thus, . Also, it follows from Corollary 1 that is the discrete topology on .
In the rest of this paper we will denote by the topological space with the discrete topology.
Theorem 6. The next statements are identical:(1) is (2)For all , there exists a clopen (open and closed) containing and not containing
Proof. : suppose be . Let , then there exists such that and . So that is a clopen containing but not containing . : let and a open set containing and not containing . Now, the map from to defined by and is a continuous map. Now, we have is an open neighborhood of such that , as required.Using Theorem 6, we can deduce the next corollary.
Corollary 2. We have the following equivalences:Now, we denote by the Sierpinski space with the topology .
Theorem 7. Letbe a topological space such thatis not a singleton. Then, the next statements are identical:(1) is an -space(2) is a space
Proof. : by Proposition 4. : suppose is a -space. Let . Then, there is open set containing or but not both. Say, . Now, the map from to defined by and is continuous and is an open subset containing and not containing . This fact completes the proof.
Remark 2. There is no topological space which is , so that there is no one which is .
We denote by the set of all open neighborhoods of in a topological space .
Theorem 8. The next statements are identical:(1) is (2) such that strictly
Proof. : let be a continuous map from to and such that and . Then, strictly. : let such that strictly. We define the map from to by and . On the one hand, is continuous because every open set containing contain also . On the other hand, since the inclusion is strict, there exists an open neighborhood of which does not contain ; hence, we deduce that is .
Corollary 3. If is, thenis not a-space.
Let be a Functionally Hausdorff space. Then, can be seen as . What about a topological space such that is . The following proposition gives a sufficient condition for such topological space.
Proposition 7. Letbe a topological space. Ifsatisfies the following conditions:
Then, is .
Proof. Let and be the map from to defined by , and . Then, satisfies the third separation axioms in Definition 7.
The following example shows that the condition in the previous proposition is sufficient but not necessary for a topological space such that is .
Example 3. Let with the topology . Then, is . Note that does not satisfy the condition given in the above proposition because for any three elements of, we have , , and . On the other hand, is .
It is natural now to pose the question about the necessary and sufficient condition for a topological space such that is . The following theorem answers this question.
Theorem 9. Letbe a topological space. Then, the next properties are identical:(1) is (2) and a continuous map from to such that , and
Proof. The first implication is trivial by the definition of .
Conversely, if and are two distinct real numbers, then there exists an isomorphism from to itself satisfying and . So that the map is continuous such that and . This completes the proof.
It is well known that any map from a discrete topological space to any topological space is continuous. We benefit from this fact to establish the following results.
Proposition 8. Let be a topological space and . Then, the following properties are equivalent:(1) is (2)
Proof. : this direction is obvious. : let and . We define a map (continuous) from to by and for all . Hence, is , as required.
Proposition 9. Letbe a topological space and. Then, the following properties are equivalent:(1) is (2)There exist two distinct points such that the topology of the subspace is the discrete topology
Proof. Suppose that is and let . Then, there exists such that and . It is clear that the subspace is equipped by the discrete topology.
Conversely, it is sufficient, for every two distinct points in , a map such that and .
Theorem 10. Letbe a topological space and. Then, the following properties are equivalent:(1) is (2)There exist such that
Proof. : this direction follows from 3 of Definition 7. : let . We define a (continuous) map from to such that and for all . Hence, we obtain that is , as required.
Corollary 4. If is disconnected topological space, then is for each .
This manuscript contributes to the area of separation axioms. We have applied continuous maps to introduce the concepts of functionally spaces for . We have elucidated the relationships between them and clarified that they are weaker than spaces, see Proposition 3 and Proposition 4. Also, we have investigated some results associated them with product spaces and categories. Moreover, their behaviours with some special spaces such as discrete and Sierpinski spaces have been studied.
In the upcoming work, we plan to explore functionally regular and functionally normal spaces. Also, we generalize the presented concepts presented in this manuscript using somewhere dense sets and SD-continuous maps given in [7, 16, 17].
No data were used to support this study.
Conflicts of Interest
The authors declare no conflicts of interest.
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