Generalized h-Closed Sets in Topological Space

Abstract


I. INTRODUCTION AND PRELIMINARIES
In 1970 Levine [4] first defined and investigated the idea of a generalized closed sets (briefly, g-closed) sets. Dontchev and Maki, in 1999 [2,3], presented the idea of "generalized -(g ), -generalized ( g) respectively" closed sets. Abbas [1] in 2020 introduced the concept of hopen set (h-os). A subset E of is called (h-os) if for every non empty set U in , and , such that . The complement of (h-os) is called h-closed set (h-cs). Our work is divided in to three sections. In the first, gh-closed sets (gh-cs) are defined and provided numerous instances, as well as analyze the link between gh-closed sets and various types of closed sets. The second section is devoted to introduce new class of mappings called gh-continuous mapping. The relationship between gh-continuous and some form of continuous mapping are investigated. In section three we study some classes of separating axioms spaces by explain the relation between them namely .We denoted the topological spaces and simbly by and respectively, open sets(resp. closed sets) by , topological spaces by TS we recall the following definitions and notations. The closure (resp. interior) of a subset of a topological space is denoted by C (E )(resp. (E )).
-Generalized continuous ( g ) [3] suppose that the inverse image of each closed subset of is ( g-cs) in . 3. α -Generalized continuous (αg ) [6] suppose that the inverse image of each closed subset of is (αg-cs) in . 4. Generalized semi-continuous (gs ) [7] suppose that the inverse image of each closed subset of is (gs-cs) in . 5. h-continuous ( ) [1] suppose that the inverse image of each open subset of is (h-os) in . 6. Generalized-continuous (g ) [4] suppose that the inverse image of each closed subset of is (g-cs) in ". Definition.1.5. A TS ( ,τ) is called 1.
space [1] if a, b are to distinct points in there exists (h-os) U such that either and , or and . 2.
space [1] if and , there exists (h-os) U, V containing a, b respectively, such that either and . 3.
space [1] if a, and , there exists disjoint (h-os) U, V containing a, b respectively.

Generalized ( -cs) in TS
This section introduces a new closed set class called generalized -cs) and we investigate the relationship with closed set, (h-cs), (g-cs), (α-cs), ( g-cs), (g -cs), (αg-cs) and (gs-cs). , and E , therefore (E) U. Hence E is (gh-cs) in . ■ As shown in the following example, the converse of the preceding theorem is not true in general.
Assume that E is a subset of the space and let The converse of the above the over is not true in general as shown in the following example". Example 2.8.
The converse is not true in general as shown in the following example". Example 2.14. From Example (2.8)

Each (  -os) is (gh-os) .
Proof. By using the complement of the definition of (ghcs).

III. gh-Continuous Mapping
The gh-continuous map on TS is introduced and studied in this section. Definition 3.1."A mapping is said to be g -continuous (gh-contm), if (F) is (g -cs) in for each (cs) F in " .

Theorem 3.2. If is (contm) then it is (gh-contm)
Proof. Assume that be (contm) and "let F be (cs) in . since is (contm) then is (cs) in . By Corollary (2.15), then is (gh-cs) in . ■ The converse of the above the over is not true in general as shown in the following example".

. Each (g-contm) is (gh-contm).
Proof. Assume that be "(g-contm) and let F be (cs) in . since is (g-contm) and by Theorem (2.13), then is (gh-cs) in . ■ As shown in the following example, the converse of the preceding theorem is not true in general". Example 3.5.
be an identity map. Then is (gh-contm), but is not (g-contm) , because {8, 9} is (cs) in but { } { } is not (g-cs) in . Theorem 3.6. All (g -contm) is (gh-contm). Proof. Suppose that be (g -contm) and F be (cs) in . since is (g -contm) then is (g -cs) in and by Theorem (2.10), then is (gh-cs) in . ■ The converse of the above the over is not true in general as shown in the following example.  Theorem 3.8. Each ( g-contm) is (gh-contm). Proof. Suppose that be ( g-contm) and F be (cs) in . "since is ( g-contm) then is ( gcs) in and by Theorem (2.7), then is (gh-cs) in . ■ As shown in the following example, the converse of the preceding theorem is not true in general". Example 3.9. If = = {3, 4, 5} and in . Remark 3.10. As a result of the above, we have Fig. 2 below.  is said to be girresolute(gh-irrm), if is (g -cs) in for each (gh-cs) F of . Theorem 3.14. Each (gh-irrm) is (gh-contm). Proof. It's obvious. ■ The converse of the above the over is not true in general as shown in the following example. Example 3.15.

A combination of two (gh-irrm) is also (gh-irrm).
Proof. Suppose that and be any two (gh-irrm) . let F be any (gh-cs) in Z. Since H is (gh-irrm), then is (gh-cs) in . Since, is (gh-irrm) then = (F) is (gh-cs) in . Therefore HoF: is (gh-irrm). ■ Definition 3.17. "A mapping is said to be strongly g -continuous , suppose that the inverse image of each (gh-cs) in is closed in " . Theorem 3.18. All strongly (gh-contm) it is (contm). Proof. Assume the following scenario: is strongly (gh-contm). Let F be (cs) in . Since (each (cs) is (gh-cs)), then F is (gh-cs) in . Since is strongly (gh-contm), is (cs) in . Therefore is (contm). ■ As shown in the example (3.15), the converse of the preceding theorem is not true in general. Suppose that be an identity map. Then is continuous , but is not strongly gh-continuous because {3} is (gh-cs) in but { } { } is not (cs) in . Theorem 3.19. Each strongly gh-continuous map it is ghcontinuous. Proof. Assume the following scenario: is strongly (gh-contm). Let F be (cs) in . Since (all (cs) is (gh-cs)), then F is (gh-cs) in . Since is strongly (gh-contm), is (cs) in . Since ( all (cs) is (gh-cs)) , then is (gh-cs) in . ■ As shown in the example (3.15), the converse of the preceding theorem is not true in general. Suppose that be an identity map. Then is (gh-contm) , but is not strongly gh-continuous , since for the (gh-cs) {3} in , { } { } is not closed in .

IV. gh-Closed Sets and Separating Axioms
In this section, we introduce and study a new type of separating axioms spaces for (gh-os) in TS. 2.
-space if a, b and a ≠ b, there exists (ghos) U, V containing a, b respectively, such that either b U and a V. 3.
and a ≠ b, there exists disjoint (gh-os) U, V containing a, b respectively.
Proof: Assume that be -space and a, b be two distinct points in . Since is -space. Then there is one an (os) U in such that a U and b U or b U and a U. Since (each (os) is (gh-os)) proposition 2.23 (2). Then U is (gh-os) in such that a U and b U or b U and a U. Hence is -space. ■ The converse is not true in general as shown in the following example.
Proof: Assume that be -space and a, b be two distinct points in . Since is -space. Then there is one an (h-os) U in such that a U and b U or b U and a U. Since (each (h-os) is (gh-os)) proposition 2.23(1). Then U is (gh-os) in such that a U and b U or b U and a U. Hence is -space. ■ "The converse of the above the over is not true in general as shown in the following example". Example.4.5. Let = {3, 6, 9}, τ = {Ø, , {9}}. Then ( , ho( )) is not -space, but ( , gh ) is -space.

Theorem.4.6"Each -space is -space
Proof: Suppose that be -space and a, b be two distinct points in . Since is -space. Then there exist two (os) U, V in such that a U, b U and b V and a V. Since (each (os) is (gh-os)) proposition 2.23 (2).Then U, V are two (gh-os) in such that a U and b U and b V and a V. Hence is -space. ■ As shown in the following example, the converse of the preceding theorem is not true in general".

Theorem.4.8. Each -space is space
Proof: Suppose that be -space and a, b be two distinct points in . Since is -space. Then there exist two (h-os) U, V in such that a U, b U and b V and a V. Since (each (h-os) is (gh-os)) proposition 2.23(1). Then U, V are two (gh-os) in such that a U and b U and b V and a V.Hence is -space. ■ As shown in the example (4.5), the converse of the preceding theorem is not true in general. ( , ho( )) is not -space, but ( , gh ) is -space.
Proof: Since each -space is space and each space is space. Hence -space is -space. ■ As shown in the example (4.5), the converse of the preceding theorem is not true in general. ( , τ) is not space, but ( , gh o is -space.
Proof: Suppose that be -space and a, b be two distinct points in . Since is -space. Then there exists disjoint (os) U, V containing a, b respectively. From proposition 2.23(2) each (os) is(gh-os). Then U, V are disjoint (gh-os) containing a, b respectively. Hence is -space. ■ As shown in the example (4.5), the converse of the preceding theorem is not true in general. ( , τ) is not space, but ( , gh o is -space.

Theorem.4.12. "Each -space is space.
Proof: Suppose that be -space and a, b be two distinct points in . Since is -space. Then there exists disjoint (h-os) U, V containing a, b respectively. From proposition 2.23(1) each (h-os) is(gh-os). Then U, V are disjoint (gh-os) containing a, b respectively. Hence is -space. ■ As shown in the example (4.5), the converse of the preceding theorem is not true in general. ( , ho( )) is not -space, but ( , gh o is -space".

Conclusion
The generalized h-closed set is not topological space and every closed, h-closed and g-closed sets is generalized hclosed.