Weak and Strong Forms of ω-Perfect Mappings

In this paper, we introduce weak and strong forms of ω-perfect mappings, namely the -ω-perfect, weakly -ω-perfect and strongly-ω-perfect mappings. Also, we investigate the fundamental properties of these mappings. Finally, we focused on studying the relationship between weakly -ω-perfect and strongly -ωperfect mappings.


Introduction
In 1943, Formin 1 introduced the concepts of -continuous mappings. In 1966, Bourbaki 2 defined perfect mappings. In 1968, Velicko [3] introduced the concepts of -open and -closed subsets, while in 1968 Singal 4 introduced the notion of almost continuous mappings. In 1981, Long and Herrington 5 introduced the notion of strongly continuous mappings, in 1989, Hdeib 6 introduced the concepts of ω-continuous mappings. In 1991, Chew and Tong 7 introduced the notion of weakly continuous mappings, In this work, (G, τ) and (H ,σ) stand for topological spaces. For a subset K of G, the closure of K and the interior of K will be denoted by cl(K) and int(K), respectively. Let (G , τ ) be a space and K be a subset of G, then a point g  G is called a condensation point of K if, for each S  τ and g  S, the set S ∩ K is uncountable. K is called to be ω-closed [6] if it contains all its condensation points. The complement of ω-closed set is called to be ω-open. It is well known that a subset W of a space (G , τ) is ω-open if and only if, for each g ∈ W , there exists S ∈ τ , such that g ∈ S and S−W is countable. The family of all ω-open sets of a space (G , τ), denoted by τω or ωO(G ), forms a topology on G finer than τ. The ω-closure and ω-interior, that can be known in the same way as cl(K) and int(K), respectively, will be denoted by ωcl(K) and ωint(K), respectively. Several characterizations of ω-closed sets were provided in previous articles [8][9][10][11][12][13][14][15][16]. A point g of G is called θcluster point of K if cl(S) ∩ K  φ, for all open sets S of G containing g. The set of all θ-cluster points ISSN: 0067-2904

Ashaea and Yousif
Iraqi Journal of Science, 2020, Special Issue, pp: 45-55 46 of K is called θ-closure of K and is denoted by cl θ (K). A subset K is called θ-closed if K = cl θ (K) [3]. The complement of θ-closed set is called θ-open. A point g of G is called an ω-θ-cluster point of K if ωcl(S) ∩ K φ for every ω-open set S of G containing g. The set of all ω-θ-cluster points of K is called ω-θ-closure of K and is denoted by ωCl θ (K). A subset K is called ω-θ-closed if K = ωCl θ (K). The complement of ω-θ-closed set is called ω-θ-open. The ω-θ-interior of K is defined by the union of each ω-θ-open sets contained in K and is denoted by ωint θ (K). A mapping  : (G , τ ) → (H ,σ) is called ω-continuous (see 16) (resp., almost weakly ω-continuous(see 11)) if for each g  G and each open set T of H containing (g), there exists an ω-open subset S in G, such that (S)  T (resp., (S)  cl(T) ). A mapping  : (G , τ ) → (H ,σ) is called almost ω-continuous [12] (resp., -ωcontinuous (see 13), strongly -ω-continuous (see [7])) if, for each g  G and for each regular open set T( resp., open ) of H containing (g), there exists an ω-open subset S in G, such that (S)  T ( resp., (ωcl(S))  cl(T) , (ωcl(S)) ⊆ T)). A mapping  : (G , τ ) → (H ,σ) is called θ-continuous (resp., continuous [16] [14] if, for all closed set F and for each point g ∈ G−F, there exist disjoint open sets S and T such that g ∈ S and F ⊆ T. A topological space G is called a semi-regular [15] if, for all point g  G and all open set S containment g, there is an open set T such that g T ⊆ int(cl(T)) ⊆ S. A topological space G is called ω-regular (resp., ω *regular) 12 if, for all ω-closed (resp., closed) set F and for each point g  G −F, there are disjoint ωopen sets S and T such that g ∈ S and F ⊆ T. Also we introduce several results and examples concerning deferent forms of ω-perfect mappings.

Weakly -ω-Perfect Mappings
In this section, we study the weakly  -ω-perfect mappings and several related theorems. Definition 2.1. A mapping  : (G ,τ) → (H ,σ) is said to be weakly -ω-continuous at g  G if, for every open subset T in H containing (g), there exists an -ω-open subset S in G containing g, such that (S)  cl(T). If  is weakly -ω-continuous at every g  G, it is said to be weakly -ωcontinuous.
Then  is almost ω-perfect mapping but it is not ω-perfect mapping. Example 2.7. Let A be the upper half of a plane and B be the X-axis. Let X = A B. If τ hdis be the half disc topology on X and τ r be the relative topology that X inherits by virtue of being a subspace of  2 . Then, the identity of the mapping  :(X, τ r ) → (X, τ hdis ) is that it is an almost weakly ω-perfect mapping but it is not ω-perfect mapping. Then  is almost weakly ω-perfect mapping but it is not almost ω-perfect mapping. Lemma 2.9. [13] A topological space G is ω-regular (resp., ω * -regular) if and only if, for all S ∈ ωO(G) (resp., S ∈ O(G) ) and all point g ∈ S, there is T ∈ ωO(G, g) ; g ∈ T ⊆ ωcl(T) ⊆ S.
be a mapping such that G be an ω-regular space. If  is almost weakly ω-perfect mapping then it is -ω-perfect mapping. Proof: Assume that  is almost weakly ω-perfect mapping. It suffices to be demonstrated that  is ω-continuous, let g  G and T be an open set containment . Also  is almost weakly ω-continuous. Hence,  is almost weakly ω-perfect mapping. Theorem 2.13. Let  : (G , τ)  (H , σ) be a mapping, such that G be an ω-regular space. If  is weakly -ω-perfect mapping then it is ω-perfect mapping. Proof : Let  be a weakly -ω-perfect mapping. It suffices to be demonstrated that  is ω-continuous, Hence is ω-perfect mapping.
be a mapping such that H be an ω-regular space. If  is almost weakly ω-perfect mapping on G, then it is ω-perfect mapping on G.
Proof : Let  be almost weakly ω-perfect mapping. It suffices to be demonstrated that  is ω- almost ω-perfect mapping then it is ω-perfect mapping.
Proof: Let  be an almost ω-perfect mapping. It suffices to be demonstrated that  is ω-continuous, , H is ω-regular space, and there is an ω-open set S1 in G, such that g  S1 and cl(T1 Hence, consider that  is ω-perfect mapping.
almost ω-perfect mapping on G.
Sufficiency. Assume that  is -ω-perfect mapping. It suffices to be demonstrated that  is -ωcontinuous, let g  G and W be an open set of G  H containment  (g). There are the open sets S1 Theorem 2.26. For a mapping  : G → H and H is regular, the following properties are equivalent.

Strongly  -ω-Perfect Mappings
In this section we study the strongly  -ω-perfect mappings and some of their theorems.
Hence  is strongly -ω-perfect mapping.  an ω-open set S containment g in G such that (S) ⊆ T and T ⊆ cl(T), then (S) ⊆ cl(T). Since G is ωregular, there is an ω-open set S1 in G such that g  S1 and cl(S1) ⊆ S, so (cl(S1)) ⊆ (S), (S) ⊆ cl(T) and int(cl((T)) ⊆ cl(T) . It follows that (cl(S1)) ⊆ int(cl((T)), therefore  is almost strongly ωcontinuous. Hence  is almost strongly ω-perfect mapping.  H , σ) be a mapping such that H be an regular space. If  is -ωperfect mapping then it is strongly -ω-perfect mapping. Proof : Let  be an -ω-perfect mapping. It suffices to demonstrate that  is a strongly -ωcontinuous, let g  G also T be an open set containment  (g) in H. Since  is -ω-continuous, there is an ω-open set S containment g in G such that ( ωcl(S)) ⊆ cl(T). Since H is regular, there is an open set W such that  (g) W ⊆ cl(W) ⊆ T, then (ωcl(S)) ⊆ cl(W) ⊆ T, therefore (cl(S)) ⊆ T. So  is strongly -ω-continuous. Hence  is strongly -ω-perfect mapping. Theorem 3.14. Let  : (G , τ)  (H , σ) be a mapping such that H be an ω-regular space. If  is -ωperfect mapping then it is almost strongly ω-perfect mapping.
Proof : Let  be an -ω-perfect mapping. It suffices to demonstrate that  is almost strongly ωcontinuous, let g  G and T be an open set containment  (g) in H. Since  is -ω-continuous, there is an ω-open set S containment g in G such that ( ωcl(S)) ⊆ cl(T). Since H is an ω-regular, there is an ω-open set T1 in H such that  (g) T1, also cl(T1) ⊆ T and int(cl(T1) ⊆ cl(T1). It follows that (cl(S)) ⊆ int(cl(T1), therefore  is almost strongly ω-continuous. So  is almost strongly ω-perfect mapping.
Then  is almost ω-perfect mapping, but not almost strongly ω-perfect mapping. Theorem 3.17. Let  : (G , τ)  (H , σ) be a mapping such that G be an ω-regular space. If  is almost ω-perfect mapping then it is almost strongly ω-perfect mapping.
Proof : Let  be almost ω-perfect mapping. It suffices to demonstrate that  is almost strongly ωcontinuous, let g  G and T be an open set containment  (g) in H. Since  is almost ω-continuous, there is an ω-open set S containment g in G such that (S) ⊆ int(cl(T)). Since G is ω-regular, there is an ω-open set S1 in G such that g  S1, also cl(S1) ⊆ S, so (cl(S1)) ⊆ (S), then (cl(S1)) ⊆ (S) ⊆
Theorem 3.23. Let  : (G , τ)  (H , σ) be a mapping, such that H be an ω-regular space. If  is almost ω-perfect mapping then it is strongly -ω-perfect mapping. Proof: Let  be almost ω-perfect mapping. It suffices to demonstrate that  is strongly-ωcontinuous, let g  G and T be an open set containment  (g) in H. Since  is almost ω-continuous, there is an ω-open set S containment g in G such that (S) ⊆ int(cl(T)). Since G is ω-regular, there is an ω-open set S1 in G such that g  S1 and cl(S1) ⊆ S. So (cl(S1)) ⊆ (S), also int(cl(T)) ⊆ cl(T). It follows that (cl(S1)) ⊆ T, therefore  is strongly -ω-continuous. So  is strongly -ω-perfect mapping. (a)  is almost strongly -ω-perfect.

Relationship between Weak and Strong Forms of ω-Perfect Mappings
In this section, we study the relationship between weakly-ω-perfect mappings and strongly -ωperfect mappings and some theorems concerning them.  H ,σ) is called to be super ω-perfect mapping ( resp., weakly ω-perfect mapping, strongly ω-perfect mapping , almost weakly perfect mapping, almost strongly perfect mapping, weakly θ-perfect mapping) if it is super ω-continuous (resp., weakly ω-continuous Ashaea  The relationships weakly and strongly ω-perfect mappings are given by the following figure: weaklyω-perfect mapping   . Also G is a regular space, there is an open set W such that g  W ⊆ cl(W) ⊆ S, so (cl(W)) ⊆ T . Therefore  is strongly -ω-continuous. Hence consider that  is strongly -ω-perfect mapping.  Proof: Let  be ω-perfect mapping. It suffices to demonstrate that  is super ω-continuous, let g  G and T be an open set containment  (g) in H. Because of  is ω-continuous, there is S  ωO(G, g), such that  (S) ⊆ T. Also, int( cl(S)) ⊆ cl(S), then  (int( cl(S)) ⊆  (cl(S)). Also G is a regular space, there is an open set S1 such that g  S1 ⊆ cl(S1 ) ⊆ S, so (int(cl(S)) ⊆  (cl(S1)) also  (S) ⊆ T. So (int(cl(S)) ⊆ T, then  is super ω-continuous. Hence consider that  is super ω-perfect mapping. Example 4.11. Let  : ( , τ)  ( , τ ) be a mapping, such that (g) = g , and let ( , τ) where τ is the topology with a basis whose members are of the form (a, b) and (a, b) - , such that  = {1\n ; n   + }. Then ( , τ) is a Hausdorff but not ω-regular. Then  is perfect but it is not strongly perfect mapping.
Theorem 4.12. Let  : (G , τ)  (H , σ) be a mapping such that G be an regular space. If  is perfect mapping then it is strongly perfect mapping.
Proof: Let  be perfect mapping. It suffices to demonstrate that  is strongly continuous, let g  G and T be an open set containment  (g) in H. Since  is continuous, there is an open set S containment g in G such that (S) ⊆ T. Since G is regular space, there is an open set S1 in G such that g  S1 and cl(S1) ⊆ S, so (cl(S1)) ⊆ (S). Then (cl(S1)) ⊆ T, therefore  is strongly continuous. So  is strongly perfect mapping.     T)). Because the space G is semi-regular space, there is an open set S1 in G such that g  S1 and T ⊆ int(cl(T)) ⊆ S, so  (T) ⊆  (int(cl(T))) ⊆  (S). Also  (S) ⊆ int(cl( T)). Then  (int(cl(T))) ⊆  (S) ⊆ int(cl(T)). Also the space H is semi-regular space, there is an open set T1 in H such that  (g)  T1 , and S ⊆ int(cl(S)) ⊆ T, so  (S) ⊆  (int(cl(S))) . It follows that  (int(cl(S))) ⊆ T. Then  is super ω-continuous. Hence  is super ωperfect mapping.   H , σ) be a mapping, such that G is a regular space. If  is almost weakly perfect mapping then it is almost strongly perfect mapping.
Proof: Let  be almost weakly perfect mapping. It suffices to demonstrate that  is almost strongly continuous, let g  G and let T be an open set containment  (g) in H. Because of  is almost weakly

Ashaea and Yousif
Iraqi Journal of Science, 2020, Special Issue, pp: 45-55 54 continuous and g  G for each open set T of H containment (g), there is an open set S containment g, such that  (S) ⊆ cl( T). Because the space G is a regular space, there is an open set S1 in G such that g  S1 also cl(S1) ⊆ S, so  (cl(S1)) ⊆  (S). Also  (S) ⊆ cl( T). Then  (cl(S1)) ⊆ cl(T) and int(cl(T1)) ⊆ cl(T1). Then  (cl(S1)) ⊆ int(cl(T1)). It follows that  is almost strongly continuous. Hence  is almost strongly perfect mapping.  where the  defined by  (g) = (g, (g)) for each g  G. If  : (G , τ)  (H , σ) is strongly perfect , then  : G → G  H is strongly perfect. Proof : Assume that  is strongly perfect, let g  G and W be an open set of G  H containment  (g). Yond represents open sets S1 ⊆ G and T ⊆ H such that  (g) = ( g, (g))  S1  T ⊆ W. Since  is strongly continuous and G is a regular space. an open set S containing g in G such that cl(S) ⊆ S1 and  (cl(S) ⊆ T . Therefore  (cl(S)) ⊆ S1  T ⊆ W, then  is strongly continuous. So the mapping  = id x  : G → G  H maps G homeomorphically onto the graph  (g) which is a closed subset of G  H . So  is perfect, and because G is regular, then G  H is regular by theorem 4.22. Hence  : G → G  H is strongly perfect.