Abstract
An approach is followed here to generate a new topology on a set from an ideal and a family of subsets of . The so-obtained topology is related to other known topologies on . The cases treated here include the one when is taken, then the case when is considered. The approach is open to apply to other choices of . As application, some known results are obtained as corollaries to those results appearing here. In the last part of this work, some ideal-continuity concepts are studied, which originate from some previously known terms and results.
1. Introduction
The interest in the idealized version of many general topological properties has grown drastically in the past 20 years. In this work, no particular paper will be referred to except where it is needed and encountered. However, many symbols, definitions, and concepts used here are as in [1].
2. Open Sets via a Family and an Ideal
Recall that an ideal on a set is a family of subsets of , that is, (the power set of ), such that is closed under finite union, and if and , then (heredity property). An ideal topological space is a triple , where is a set, is a topology on , and is an ideal on . Let be an ideal topological space. The family and forms a base for a topoy on finer than [1].
As a start, this concept will be put in a more general setting as follows.
Definition 2.1. Let be an ideal topological space, and let be a family of subsets of .(a)A set is called an -open set if for each , there exist and such that and , or equivalently . The family of all -open subsets is written .(b)The topology on generated by the subbase is denoted by .
Remark 2.2. (1) The family needs not form, in general, a topology on .
(2) In the case where , it is clear that , and for the case .
Proposition 2.3. Let be an ideal topological space and let . Let and . Let be the topology generated by the subbase , then .
Proof. Note that , and therefore . Now let , then for each , there exists with and such that . This means that where for each . Thus, since .
Proposition 2.4. Let be an ideal topological space. If denotes the topology on generated by the subbase , then .
Proof. To show that , first note that and therefore and . This implies that . Next, consider the base and for . Let , say for some and some . If , then , and so there exist such that , where , that is, , and hence . This shows that .
If is a topological space, we let (resp., ) denote the interior of (resp., the closure of ) in . A subset of is called semiopen if , and is called an set if . The family of all -sets forms a topology on finer than .
Example 2.5. Let be an ideal topological space, where is the ideal of all nowhere dense subsets of . Let , the family of all semiopen subsets of . In this case, it is noted that is the topology studied in [2] and is called the topology of semiopen sets and denoted by . So , the topology of all -sets in the space .
Next, the case where is given and the family is considered. Recall that is a topology on finer than . It is known that . It is then clear that the family and is a base for . In fact, and , see [3].
Definition 2.6. Let be an ideal topological space, , and . is called an --open subset (-set, for short) if for each , there exist and such that , or equivalently . The family of all -sets of is denoted by .
The following is a consequence of [1, Theorem 3.1].
Proposition 2.7. Let be an ideal topological space, then the family and is a base for the topology on .
Proposition 2.8. If is an ideal topological space, then . So .
Proof. It is enough to note that the family and is a base for as well as for .
Let be an ideal topological space. When dealing with the topology , it is found to have a base consisting of elements of the form with and . But as it is well known, the set takes the form for some and . So where . It is now clear that in a more general setting, one needs to deal with a situation where two ideals and are considered on . At this point, let and , where it is easy to see that is itself an ideal on .
Proposition 2.9. Let and be two ideals on a space , then (see [1, Corollaryโ 3.5]) (and recall that is the supremum of the two topologies and which is the topology generated by the subbase ).
Proof. Since and , then (by Theorem 2.3(b) of [1]) it follows that and . Therefore, . Next, if is a base for and , then for some , , and . So . Where is a basic open set in and is a basic open set in . Thus and so which means that .
Corollary 2.10. (see [1, Corollary 3.4]) Let be an ideal topological space, then . In particular, .
3. Sets
Let be a topological space. A subset is called regular open if . The family of all regular open subsets of is denoted by . It is a known fact that is a base for a topology on , finer than , and is called the semiregularization of .
In pursuing the approach used in the first section, it is now the time to consider the case where .
Definition 3.1. Let be an ideal topological space, and let . The set A is called an ideal regular-open set, or set for short, if for each , there exist a regular open set and such that , or equivalently . The family of all sets of is denoted by .
Proposition 3.2. Let be an ideal topological space, then the family is a base for a topology on .
Proof. It is clear that . So it is enough to show that if , then . To this end, let , then there exist , such that , , and . Now, and . So as claimed.
Proposition 3.3. Let be an ideal topological space. If denotes the topology on generated by the base , then (where is the topology constructed in Proposition 3.2).
Proof. We need to appeal to Proposition 2.4, with and so , while .
In general, for an ideal topological space , the two topologies and need not be comparable.
Example 3.4. (a) Let be an ideal topological space where (the set of real numbers), is the left ray topology on , and is the ideal of all finite subsets of . It is easy to see that . Then is the cofinite topology on [1, Example 2.5], and clearly and are incomparable.
(b) Consider the space where , is the cofinite topology on , and is the ideal of all countable subsets of . Again while is the cocountable topology on . Here, .
Definition 3.5 (see [4]). A subset of a space is called -regular open if for each , there exists a regular open set such that and is countable. The family of all -regular open subsets of is denoted by .
The next result is an immediate consequence of definitions.
Proposition 3.6. Let be a topological space, then a subset is an -regular open subset of if and only if is ideal regular open with . Thus, .
Corollary 3.7 (see [4, Theorem 2.1]). Let be a topological space, then is a topological space.
A topological space is called nearly Lindelรถf (see [5]) if every cover of by regular open subsets has a countable subcover.
It is clear that if is nearly Lindelรถf, then is Lindelรถf. In fact, the family forms a base for , and in this case, every basic open cover of will have a countable subcover. On the other hand, by the well-known fact [6, Lemma 1.1] that the two spaces and have the same regular open sets, it follows that is nearly Lindelรถf if and only if is nearly Lindelรถf.
Corollary 3.8 (see [5, Proposition 1.3]). A space is nearly Lindelรถf if and only if is Lindelรถf.
Recall that a subset of a space is called -open (see [7]) if for each , there exists such that and is countable, that is, for some . The family of all -open subsets of a space is denoted by .
The definitions imply directly the following result.
Proposition 3.9. If is a topological space, then .
Corollary 3.10 (see [7, Proposition 2.5]). Let be a topological space. The family is a topology on with .
Proposition 3.11 (see [7, Proposition 4.5]). A space is Lindelรถf if and only if the space is Lindelรถf.
The following result can now be stated.
Proposition 3.12. The following statements are equivalent for a space :(a) is nearly Lindelรถf,(b) is Lindelรถf,(c) is Lindelรถf.
Proof. (a)(b) Follow by Corollary 3.8.
(b)(c) By applying Proposition 3.11 to the space , it follows that is Lindelรถf if and only if is Lindelรถf.
Corollary 3.13 (see [4, Theorem 3.1]). The following statements are equivalent for any space :(a) is nearly Lindelรถf,(b)Every -regular open cover of admits a countable subcover.
Proof. Now, is nearly Lindelรถf if and only if is Lindelรถf (Proposition 3.12). On the other hand, (Proposition 3.6), and therefore is nearly Lindelรถf if and only if is Lindelรถf if and only if every -regular open cover of has a countable subcover.
Corollary 3.14 (see [4, Proposition 3.1]). A space is nearly Lindelรถf if and only if for every family of -regular closed subsets that satisfies the countable intersection property has a nonempty intersection.
Proof. Again is nearly Lindelรถf if and only if is Lindelรถf (note that . Now, the result follows by a well-known fact concerning Lindelรถf spaces and the fact that a subset is -regular closed if it is the complement of an -regular open set.
Let be an ideal topological space. The ideal is called completely codense [8] if , where is the family of all preopen subsets of and is called preopen if .
Proposition 3.15. Let be an ideal topological space, and assume that is completely codense, then is nearly Lindelรถf if and only if is nearly Lindelรถf.
Proof. By a remark on [9, page 3], it follows that . This implies that . Then is nearly Lindelรถf, if and only if is Lindelรถf if and only if is Lindelรถf if and only if is nearly Lindelรถf.
Let be an ideal topological space. The topology is compatible with the ideal , written , [1], if whenever a subset satisfies for each , there exists with and , then .
Proposition 3.16. Let be an ideal topological space, then if and only if .
Proof. Let . Let be satisfying for each , there exists with and . Our assumption and the fact that imply , and so . Conversely, assume that , then and [1, Theorem 4.4]. Let satisfy for each there exists with and . Then with and so . But , so which means that .
Proposition 3.17. A space is hereditarily Lindelรถf if and only if is hereditarily Lindelรถf.
Proof. It is known that is hereditarily Lindelรถf if and only if [1, Theorem 4.10] if and only if if and only if is hereditarily Lindelรถf.
4. Continuity via Ideals
As a start, two known concepts of ideal continuity are stated.
Definition 4.1 (see [10]). Let be an ideal topological space. Let be a given function.(a)The function is called continuous if for every , there exist and such that .(b)The function is called pointwise continuous if for each and for each with , there exists such that and .
It is noted in [10] that every continuous function is pointwise continuous. An example [10, page 327] is provided to show that the converse is not true in general.
Definition 4.2. Let be given. The function is called continuous if is continuous.
Proposition 4.3. A function is continuous if and only if is pointwise continuous.
Proof. Let be continuous, that is, is continuous. Let , , and , then . So there exists a basic open set , for some and , such that . Equivalently, and . Thus, is pointwise continuous. Conversely, let be pointwise continuous. If , then for each , there exists such that , or equivalently, for some . It follows that and so . Thus, is continuous.
The statement of the lemma on [10, page 326] can be formulated as in the next result.
Proposition 4.4. Let be a given function. Assume that is codense (this means that ) and that is regular. If is continuous, then is continuous and hence continuous.
Proposition 4.5. Let be a given function. Assume that , then is continuous if and only if is continuous.
Proof. Let be continuous, then as stated at the beginning of this section, is continuous. Conversely, let be continuous, then by [1, Theorem 4.4], and . So if , then and so for some and . Thus, is continuous.
Definition 4.6 (see [4]). A function is continuous if for each and each regular open set with , there exists an -regular open set of such that and .
Recall that is an -regular open subset of if . So the next result is now clear.
Proposition 4.7. A function is continuous if and only if is continuous (i.e., is continuous).
Proof. Let be continuous. If , then by definition, is a union of -regular open subsets of , that is, is a union of elements of , and therefore . But is a base for , and therefore is continuous. For the converse, let be continuous. Let , , and , then and therefore is an -regular open set containing and . Thus, is continuous.
Corollary 4.8 (see [4, Theorem 4.1]). Let be a continuous surjection. Assume that is nearly Lindelรถf, then is nearly Lindelรถf.
Proof. Assume that is a continuous surjection. This means, by Proposition 4.7, that is an -continuous surjection which means that is a continuous surjection. Assume now that is nearly Lindelรถf, then, by Proposition 3.12, is Lindelรถf. So is Lindelรถf, being the continuous image of a Lindelรถf space. Finally, is nearly Lindelรถf by Corollary 3.8.
Definition 4.9 (see [4]). A function is called continuous if is -regular open in for each open set in .
The following result is an immediate consequence of the definitions involved.
Proposition 4.10. A function is continuous if and only if the function is continuous (i.e., is continuous).
Corollary 4.11 (see [4, Theorem 4.2]). Let be an continuous surjection. If is nearly Lindelรถf, then is Lindelรถf.
Proof. Let be an continuous surjection, then is continuous. If is assumed to be nearly Lindelรถf, then is Lindelรถf, by Proposition 3.12. Therefore , being the continuous image of a Lindelรถf space, is Lindelรถf.
Definition 4.12 (see [4]). A function is called completely continuous if is a regular open set in for each open set in .
Proposition 4.13. If a function is completely continuous, then the function is continuous.
Proof. The easy proof is omitted.
Corollary 4.14. Let be a completely continuous surjection. If is nearly Lindelรถf, then is Lindelรถf.
Proof. The proof is a consequence of Propositions 3.12 and 4.13.
The following example shows that the converse of Proposition 4.13 is not true in general.
Example 4.15. Let be a set and a proper nonempty subset of . Consider the topology on . Let with the topology . Let be the function defined by if and if . Then such that , and so is continuous but not completely continuous.