1. Introduction and Preliminaries
Defining a new type of generalized open sets and utilizing it to define new topological concepts is now a very hot research topic [
1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
11]. As a new type of generalized open sets, Al-Ghour and Samarah in [
12] defined coc-open sets as follows: A subset
A of a topological space
is called coc-open set if
A is a union of sets of the form
, where
and
C is a compact subset of
X. Authors in [
12] proved that the family of all coc-open sets of a topological space
forms a topology on
X finer than
, and via this class of sets they obtained a decomposition theorem of continuity. Research via coc-open sets was and still a hot area of research, indeed authors in [
13] introduced and coc-closed, coc-open functions and coccompact spaces, authors in [
14] defined new types of connectedness, in [
15] authors have studied s-coc-separation axioms, coc-convergence as a new type of convergence for nets and filters was introduced in [
16], new types of dimension theory were introduced in [
17], in [
18] new classes of functions were introduced, and in [
19] the suthors generalized co-compact open sets. In this paper, we use coc-open sets to define and investigate new separation axioms and new class of functions.
The material of this paper lies in four chapters. In
Section 2, we define and investigate co-
as a new topological property and as a generalization of
topological property. In
Section 3, we introduce and investigate co-regularity as a weaker form of regularity, s-regularity as a stronger form of regularity and co-normality as a weaker form of normality. In
Section 4, we introduce and investigate slightly coc-continuous functions.
In this paper, for a subset A of a topological space , will denote the closure of A is and will denote the relative topology of on A.
The following definition and theorem will be used in this sequel:
Definition 1. [12] A subset A of a topological space is called co-compact open set (notation: coc-open) if for every , there exists an open set and a compact subset K of such that . The complement of a coc-open subset is called coc-closed. The family of all coc-open subsets of a topological space will be denoted by . Theorem 1. [12] Let be a topological space. Then - (a)
The collection forms a topology on X with .
- (b)
and is compact in forms a base for .
2. Co- Topological Spaces
Let us start by the following main definition:
Definition 2. A topological spaceis called co-if and only if for all ∈ X with there exist ∈ such that , and .
Theorem 2. Every topological space is co-.
The following example shows that the converse of Theorem 2 is not true in general:
Example 1. Let , ∪, and τ is the topology on X that having as a base. Then
- (a)
is co-.
- (b)
is not .
Proof. - (a)
Let ∈ with .
Case 1. . Let U= and V=. Then ∈, and .
Case 2.. Let U= and V=. Then ∈, and .
Case 3. and . Let and . Then ∈, , and .
- (b)
Suppose to the contrary that is . Then there are ∈ such that and .Since , then U= and . Thus, , a contradiction.
□
Theorem 3. [12] If is a hereditarily compact topological space, then is the discrete topology on X. Theorem 4. If is a hereditarily compact topological space, then is .
Proof. Let be a hereditarily compact topological space. Then by Theorem 3, is the discrete topology on X. Thus, is . □
Theorem 5. If is hereditarily compact and topological space, then is co-.
Proof. Let with . Take and . Since is hereditarily compact, then by Theorem 4, . Since is , then . Thus, we have ∈, and . □
Theorem 6. If X is any non-empty set and τ is the cofinite topology on X, then is hereditarily compact.
Proof. If A is a non-empty subset of X, then is the cofinite topology on A and hence A is a compact subset of . Therefore, is hereditarily compact space. □
Corollary 1. If X is any non-empty set and τ is the cofinite topology on X, then is the discrete topology on X.
Proof. Theorems 3 and 6. □
Corollary 2. If X is any non-empty set and τ is the cofinite topology on X, then is co-.
Proof. Theorems 3 and 6 and the fact that is . □
Corollary 3. If X is any infinite set and τ is the cofinite topology on X, then is an example on a co- topological space that is not .
Theorem 7. If is a co- topological space, then is .
Proof. Suppose that is co- and let with . Since is co-, there exist ∈ such that and . Therefore, is . □
The following example shows that Theorem 7 is not reversible:
Example 2. Let and τ be the topology on having as a base. It is proved in example 2.16 of [2] that is the discrete topology on and so is . If is co-, then there are ∈ such that , and . Without loss of generality, we can assume that . Choose a basic element B such that . Then and so . Therefore, is not co-. Theorem 8. A topological space is co- if and only if for all with there exist and a compact set K of such that ( and ) or ( and ).
Proof. - ⟹)
Assume that is co- and let with . Then there exist ∈ such that , and . Without loss of generality we may assume that . Since , there are W and a compact set of such that . Since , then .
- ⟸)
Let with . Then there exist and a compact set K of such that ( and ) or ( and ). If ( and ), then we have , , and . If and , then and . Therefore, is co-.
□
Theorem 9. A topological space is co- if and only if for all with there exist and a compact set K of such that ( and ) or ( and ).
Proof. - ⟹)
Suppose that is co- and let with . Since is co-, by Theorem 8 there are and a compact set K of such that ( and ) or ( and ).
Case 1. and
. We have
and so
Since , then .
Case 2. and
. We have
and so
Since , then .
- ⟸)
Let with . By assumption there exist and a compact set K of such that ( and ) or ( and ).
Case 1. and . Set . Then and .
Case 2. and . Set . Then and .
Therefore, by Theorem 8 we conclude that is co-. □
Lemma 1. [20] Let be a closed function in which its fiber is a compact subset of for all . If B is a compact subset of , then is a compact subset of . Theorem 10. Let be continuous, closed and injective function where is a co- topological space, then is co-.
Proof. Let
with
. Since
f is injective,
. Since
is a co-
, then by Theorem 8 there exist
and compact set
K of
such that (
and
) or (
and
). Without loss of generality we can assume that
and
. Since
f is continuous,
. Since
f is injective, then its fibers are compact. Hence, by Lemma 1,
is compact in
. We have
and
This ends the proof according to Theorem 8. □
Corollary 4. Being’co-’ is a topological property.
Lemma 2. [12] Let be a topological space and be a closed subset of . Then . Theorem 11. If is a co- topological space and is a closed subset of , then is co-.
Proof. Let
with
. Then
with
. Since
is
-
, then there exist
∈
such that
,
and
. Set
and
. Then
and by Lemma 2
. Since
,
. Moreover,
and
□
3. Co-Regulerity and Co-Normality
We start by the following main definition:
Definition 3. A topological space is said to be co-regular if for each closed set F of and each there exist such that and .
Theorem 12. If is a topological space such that is the discrete topology on X, then is co-regular.
Proof. Let F be a closed set in and let . Let and . Then , and . Therefore, is co-regular. □
Corollary 5. If is a hereditarily compact topological space, then is co-regular.
Proof. Theorems 3 and 12 □
Corollary 6. If X is any non-empty set and τ is the cofinite topology on X, then is co-regular.
Proof. Corollary 1 and Theorem 12 □
Theorem 13. Every regular topological space is co-reguler.
Proof. Let be a regular topological space. Let F be a closed set in and let . Since is regular, there exist such that and . Since , it follows that is co-reguler. □
Remark 1. The converse of Theorem 13 is not true in general: By Corollary 6, where τ is the cofinite topology on is co-regular. On the other hand, it is well known that is not regular.
Theorem 14. Let be a topological space. If is regular, then is co-reguler.
Proof. Let F be a closed set in and let . Since F is closed in and , then F is closed in . Since is regular, there exist such that and . Therefore, is co-reguler. □
Example 3. Let , . Then
- (a)
is co-reguler.
- (b)
is regular.
- (c)
is not regular.
- (d)
is co-.
Proof. - (a)
Let F be closed in and let .
Case 1. . Take and . Then U and .
Case 2.. Then . Take and . Then and . It follows that is co-reguler.
- (b)
Let F be closed in and let . By (a), we only need to discuss the case when F is not closed in . It is not difficult to see that . Thus, we must have and F is finite. Take and . Then , , , and . Therefore, is regular.
- (c)
Suppose to the contrary that is regular. Take and . Then F is closed in with and by regularity there are such that and . Since , then and so . Thus, , a contradiction.
- (d)
Let with . Without loss of generality we may assume that there are only two cases:
Case 1. and . Take and . Then , , and .
Case 2. and . Take and . Then , , and .
It follows that is co-. □
Theorem 15. A topological space is co-regular if and only if for every with , there exists such that .
Proof. - ⟹)
Suppose that
is co-regular and let
with
. Then we have
is closed in
with
. So by co-regularity of
there are
, such that
and
. Now
with
- ⟸)
Let F be closed in and . Then and by assumption there exists such that . Let . Then we have and .
□
Theorem 16. If is a co-regular topological space, then for every closed set F in we have Proof. Suppose that is co-regular. To see that , let . Then by co-regularity of , there are such that and . Thus we have and which implies that . Conversely, it is obvious that . □
Theorem 17. A closed subspace of a co-regular topological space is co-reguler.
Proof. Let
be a co-regular topological space and let
A be a non-empty closed set of
. Let
and let
B be a closed set in
with
. Since
A is closed in
, then
B is closed set in
. Since
is co-regular, then there are
such that
and
. We have
and by Lemma 2,
. Now we have
,
and
It follows that is co-reguler. □
Definition 4. A topological space , is called s-regular if for each closed set F in and , there exist such that and .
Theorem 18. Every s-regular topological space is regular.
The following example shows that the converse of Theorem 18 is not true in general:
Example 4. Let X be a set which contains at least three points and let be a partition on X in which for some , B has at least two points. Let τ be the topology on X having is a base. Since is zero-dimensional, it is regular. Choose such that and let . Then F is closed in with . If such that and , then . Therefore, , is not s-regular.
Lemma 3. [12]If is a topological space, then . Theorem 19. Let be a topological space. Then is s-regular if and only if is regular.
Proof. Theorem 18 and Lemma 3 □
Corollary 7. Every topological space is s-regular.
Definition 5. A topological space is said to be co-normal if for each disjoint closed sets of there exist such that and .
Theorem 20. Every normal topological space is co-normal.
Proof. Let be a normal topological space. Let A and B be two disjoint closed sets in . Since , is normal space then there exist such that and . It follows that is co-normal. □
Theorem 21. Let be a topological space. If is normal, then is co-normal.
Proof. Let be two disjoint closed sets in . Since , then are closed sets in . Since , is normal, then there exist such that and . Therefore, is co-normal. □
Corollary 8. If is a topological space such that is the discrete topology on X, then is co-normal.
The following example shows that converse of Theorem 20 is not true in general:
Example 5. Let and let τ be the cofinite topology. Then
- (a)
is co-normal.
- (b)
is not normal.
Proof. - (a)
By Corollary 2, is the discrete topology on X. So by Corollary 8, is co-normal.
- (b)
This is well known in general topology.
□
Theorem 22. A topological space is co-normal if and only if for every open set U and any closed set A in with , there exists V such that
Proof. - ⟹)
Suppose that
is a co-normal topological space. Let
and let
A be closed set in
with
. We have
A and
are disjoint closed sets in
. Thus, since
is co-normal, then there exist
such that
and
. Now
- ⟸)
Let
A and
B be two disjoint closed sets in
. Put
. Then
with
. By assumption there exists
V such that
. Put
. Then
W,
and
Therefore, is co-normal. □
Corollary 9. A topological space is co-normal if for any pair of disjoint closed sets , there exists such that and .
Proof. Let
A and
B be two disjoint closed sets of
. Put
. Then
with
. By Theorem 22, there exists
V such that
. Thus,
□
Theorem 23. A closed subspace of a co-normal topological space is co-normal.
Proof. Let
be a co-normal topological space and let
A be a non-empty closed set of
. Let
be two disjoint closed sets in
. Since
A is closed in (
, then
B and
C are closed in
. Since
is co-normal, then there exist
such that
with
. Let
and
. Then
∈
. Since
A is closed, then by Lemma 2,
. Also, we have
and
□
This shows that is co-normal.
Theorem 24. A topological space is co-normal if and only if for every pair of open sets U and V in such that , there exist coc-closed sets A and B in such that and .
Proof. - ⟹)
Suppose that
is a co-normaltopological space. Let
U and
V be two open sets in
with
. Then
and
are disjoint closed sets of
, and since
is co-normal, then there exist
such that
and
. Let
and
. Then
A and
B are coc-closed sets of
with
,
. Moreover,
- ⟸)
Let
A and
B be two disjoint closed sets in
. Set
and
. Then
U and
V are open sets in
such that
. Thus by assumption, there exist coc-closed sets
and
in
such that
and
. Let
and
. Then
and
are coc-open sets in
,
and
□
Theorem 25. A co-normal topological space is co-reguler.
Proof. Let be and co-normal. Let , and F be a closed set in such that . Since is , then is closed. Since is co-normal and , then there exist such that and . □
4. Slightly Coc-Continuous Functions
Definition 6. [21] A subset A of a topological space is said to be coc-clopen if both A and are coc-open. Example 6. Let and . It is known that (see [12]). Then the set of coc-clopen sets is . Definition 7. [22] A function is said to be slightly continuous if for each , and for each clopen set V in containing , there exists an open set U in containing x such that . Definition 8. A function is said to be slightly coc-continuous if for each , and for each clopen set V in containing , there exists a coc-open set U of containing x such that .
Theorem 26. Every slightly continuous function is slightly coc-continuous.
Proof. Let be a slightly continuous function. Let V be a clopen set in such that . Since f is slightly continuous, then there exists containing x such that . This shows that f is slightly coc-continuous. □
Theorem 27. [22] Let and be topological spaces. The following statements are equivalent for a function : - (a)
f is slightly continuous.
- (b)
For every clopen set , is open.
- (c)
For every clopen set , is closed.
- (d)
For every clopen set , is clopen.
Theorem 28. Let and be topological spaces. The following statements are equivalent for a function :
- (a)
f is slightly coc-continuous.
- (b)
For every clopen set , is coc-open.
- (c)
For every clopen set , is coc-closed.
- (d)
For every clopen set , is coc-clopen.
Proof. - (a)⟹(b):
Let V be a clopen set in . Since f is slightly coc-continuous, then for every there exists containing x, such that . It is easy to check that . Hence is coc-open.
- (b)⟹(c):
Let
V be a clopen set in
. Then
is clopen and by (b)
is coc-open. Moreover, we have
It follows that is coc-closed.
- (c)⟹(d):
Let V be a clopen set in . Then is also clopen and by (c), and are coc-closed. It follows that is coc-clopen.
- (d)⟹(a):
Let and V be a clopen set in containing . By (d), is coc-clopen. Take . Then U is coc-open with . It follows that f is slightly coc-continuous.
□
The converse of Theorem 26 is not true in general as the following two examples show:
Example 7. Let , and . Let be the identity function. Then
- (a)
f is slightly coc-continuous.
- (b)
f is not slightly continuous.
Proof. - (a)
Let U be a clopen set in . Since clearly is the discrete topology, then is coc-open in . So f is slightly coc-continuous.
- (b)
Let . Then U is clopen in but which is not open in . Therefore, f is not slightly continuous.
□
Example 8. Let be the real numbers. Take two topologies on , τ and σ where τ is cofinite topology and σ is discrete topology. Let be an identity function. By Corollary 2, is the discrete topology and hence, f is slightly coc-continuous. On the other hand, since is clopen in but is not open in , f is not slightly continuous.
Theorem 29. If is a slightly coc-continuous function and A is a closed set in , then the restriction is slightly coc-continuous.
Proof. Let V be a clopen set in . By Lemma 2, is coc-open in . Thus is slightly coc-continuous. □
Definition 9. A function is said to be
- (a)
coc-irresolute [13], if for every coc-open subset U of , is coc-open in . - (b)
coc-open [13], if for every coc-open subset A of , is coc-open in . - (c)
coc-continuous [12], if for each point and each open subset V in containing , there exists an coc-open subset U in containing x such that . - (d)
weakly coc-continuous, if for each point and each open subset V in containing , there exists an coc-open subset U in containing x such that ,
- (e)
contra coc-continuous, if is coc-open in for each closed set F in .
Theorem 30. Let and be functions. Then the following properties hold:
- (a)
If f is coc-irresolute and g is slightly coc-continuous, then is slightly coc-continuous.
- (b)
If f is slightly coc-continuous and g is slightly continuous, then is slightly coc-continuous.
Proof. - (a)
Let V be any clopen set in . Since g is slightly coc-continuous, is coc-open set in . Since f is coc-irresolute, is coc-open. Therefore, is slightly coc-continuous.
- (b)
Let V be any clopen set in . Since g is slightly continuous, then is clopen. Since f is slightly coc-continuous, is coc-open. Therefore, is slightly coc-continuous.
□
Corollary 10. Let and be functions. Then, the following properties hold:
- (a)
If f is coc-irresolute and g is coc-continuous, then is slightly coc-continuous.
- (b)
If f is coc-irresolute and g is slightly continuous, then is slightly coc-continuous.
- (c)
If f is coc-continuous and g is slightly continuous, then is slightly coc-continuous.
- (d)
If f is slightly coc-continuous and g is continuous, then is slightly coc-continuous.
Theorem 31. Let be surjective, coc-irresolute and coc-open and be a function. Then is slightly coc-continuous if and only if g is slightly coc-continuous.
Proof. - ⟹)
Let be slightly coc-continuous and V be clopen set in . Then is coc-open in . Since f is coc-open, then is coc-open in . Since f is surjective, then . It follows that g is slightly coc-continuous.
- ⟸)
If g is slightly coc-continuous, then by Theorem 30, is slightly coc-continuous.
□
Definition 10. [21] A topological space is said to be coc-connected if X can not be written as a union of two disjoint non-empty coc-open sets. A topological space is said to be coc-disconnected if it is not coc-connected. It is clear that a topological space is coc-connected if and only if is connected. Therefore, where is the co-countable topology on is an example of a coc-connected topological space.
Also, it is clear that coc-connected topological spaces are connected, however where is the cofinite topology on is an example of a connected topological space that is coc-disconnected.
Theorem 32. Let be slightly coc-continuous and surjective function. If is coc-connected, then is connected.
Proof. Suppose to the contrary that
is disconnected. Then there exist non-empty disjoint open sets
U and
V such that
. It is clear that
U and
V are clopen sets in
. Since
f is slightly coc-continuous, then
and
are coc-open in
. Also,
and
□
Since f is surjective, and are non-empty. Therefore, is coc-disconnected space. This is a contradiction.
Corollary 11. The inverse image of a disconnected topological space under a surjective slightly coc-continuous function is coc-disconnected.
Recall that a topological space is called extremally disconnected if the closure of each open set is open.
Theorem 33. If is slightly coc-continuous such that is extremally disconnected, then f is weakly coc-continuous.
Proof. Let and let V be an open subset of Y containing . Since is extremally disconnected, then is open and hence clopen. Since f is slightly coc-continuous, then there exists a coc-open set such that and . It follows that f is weakly coc-continuous. □
Recall that a topological space is called locally indiscrete if every open set is closed.
Theorem 34. If is slightly coc-continuous such that is locally indiscrete, then f is coc-continuous and contra coc-continuous.
Proof. Let V be an open set of Y. Since is locally indiscrete, then V is clopen. Since f is slightly coc-continuous, then is coc-clopen set in X. Therefore, f is coc-continuous and contra coc-continuous. □
Recall that a topological space is called zero-dimensional if has a base consists of clopen sets.
Theorem 35. If is slightly coc-continuous such that is zero-dimensional, then f is coc-continuous.
Proof. Let and let such that . Since is zero-dimensional, there exists a clopen set B such that . Since f is slightly coc-continuous, then there exists a coc-open set W in such that . Therefore, f is coc-continuous. □
Definition 11. [23] A topological space is said to be: - (a)
Clopen (Clopen Hausdorff or Ultra-Hausdorff ) if for each pair of distinct points x and y in X, there exist disjoint clopen sets U and V in X such that and .
- (b)
Clopen regular, if for each clopen set F and each point , there exist disjoint open sets U and V such that and .
Theorem 36. If is slightly coc-continuous and injective function and is clopen , then is .
Proof. Let , with . Since f is injective, then . Since is clopen , then there exist two disjoint clopen sets U and V in such that and . Since f is slightly coc-continuous, . Now and . Therefore, is a topological space. □
Theorem 37. If is slightly coc-continuous injective open function from an s-regular topological space onto a topological space , then is clopen regular.
Proof. Let
F be a clopen set in
Y and let
such that
. Since
f is onto then there is
such that
. Since
f is slightly coc-continuous, then by Theorem 28 (c),
is a coc-closed set. Since
is s-regular, and
, there exist two disjoint open sets
U and
V such that
and
. Since
f is onto,
. Also,
. Since
f is injective
Therefore, is clopen regular. □
Theorem 38. If is slightly continuous function and is slightly coc-continuous function and is clopen Hausdorff, then is coc-closed in .
Proof. Let . Then . Since is clopen Hausdorff, then there exists two clopen sets such that , and . Since f is slightly continuous and g is slightly coc-continuous, then is open and is coc-open in with and . Set . Then O is coc-open. with . It follows that is coc-open and E is coc-closed. □
Theorem 39. Let be slightly coc-continuous and injective function. If is zero-dimensional and , then is .
Proof. Let with . Since f is injective, then . Since is , then there exist two open sets U and V in Y such that and . Since is zero-dimensional, then there exist two clopen sets such that and . Since f is slightly coc-continuous, then and are coc-open sets. We have and . This shows that is . □
Theorem 40. Let be slightly coc-continuous, open and injective function. If is zero-dimensional, then is co-regular.
Proof. Let and U be an open set containing x. Since f is an open function, then is open in with . Since is zero-dimensional, then there exist a clopen set V such that . Since f is slightly coc-continuous, then is coc-clopen set in X with . This implies that is co-regular. □
Theorem 41. Let be slightly coc-continuous, closed and injective function. If is zero-dimensional, then is co-regular.
Proof. Let and F be a closed set of X such that . Since f is a closed function, then is closed in . Since f is injective, then and so . Since is zero-dimensional, then there exists a clopen set V in Y such that . Since f is slightly coc-continuous, then is a coc-clopen set in . Moreover, and which is also coc-clopen set of . Therefore is co-regular. □
Theorem 42. Let be slightly coc-continuous, closed and injective function. If is normal, then is co-normal.
Proof. Let A and B be any two non-empty disjoint closed sets in . Since f is closed and injective, we have and are two disjoint closed sets in Y. Since is normal, then there exist two open sets U and V in Y such that , and . For each , and since is zero-dimensional there exists a clopen set such that . Thus, . Put . Then . Since f is slightly coc-continuous, is coc-open for each and so is coc-open in X. Similarly, there exists a coc-open set H in X such that and . This shows that is co-normal. □