On Some Types of αrps-Closed Maps

This paper is continues to study of a new type of closed maps which is called αrps-closed map. As well as, we give and study other types of αrps-closed maps which are (αrps-closed maps, strongly αrps-closed maps and almost αrps-closed maps) in topological spaces. Also, we will study the relation between these mappings and discussion some properties of these maps. Citation: Hameed DM, Mushtt IZ, Abdulqader AJ (2018) On Some Types of αrps-Closed Maps. J Appl Computat Math 7: 405. doi: 10.4172/21689679.1000405


Introduction
Mappings play as important role, in the study of modern mathematics, especially in topology and functional analysis [1][2][3][4][5]. Different types of closed and open mappings were studied by various researchers [6]. Generalized closed mappings were introduce and studied. After him different mathematicians worked and studied on different versions of generalized maps [7].
In this paper, we introduce and study new types of closed maps namely αrps-closed map in topological spaces and we use this maps to give other types of αrps-closed map which are (αrps-closed maps [9][10][11][12][13], strongly αrps-closed maps and almost αrps-closed maps) and we discussion the properties of these maps as well as, shows the relationships between some types of these maps [14][15][16][17][18].

rps-Closed Maps
In this section, we introduce a new type of closed sets namely αrpsclosed maps in topological spaces and study some of their properties.
Proposition: Every closed map is αrps-closed map.
Proof : It follows from definition of closed map and fact that every closed set is αrps-closed map.
Remark: The converse of above proposition need not be true as seen from the following example. 6. rgα-Closed map.
The proof of steps 3, 4, 5, and 6 are similar to step 1 and 2.
The following example show the converse of proposition need not be true in general.

Remark:
The concepts of g-closed map and almost closed map are independent to αrps-closed map. As show in the following examples.

Example
Let X={a, b, c} with the topology τ={X, θ, {a}, {a, c}}, where 3. Let A be a closed set in (X, τ), since f is gs-closed map. Thus, f(A) is gs-closed set in (Y, σ), also since Y is a T b -space, then f(A) is a closed set in (Y, σ), and by remark (every closed set is αrps-closed set), hence f(A) is αrps-closed set in (Y, σ) Therefore, f is αrps-closed map.

4.
It is follows from the fact ( every sg-closed map is an gs-closed map) and since Y is a T b − space, then by we get, f is αrps-closed.

5.
Let A be a closed set in (X, τ), since f is rg-closed map. Thus, f(A) is rg-closed set in (Y, σ), also since Y is a T* 1/2 -space, then by definition. We get f(A) is a closed set in (Y,σ), and by remark (every closed set is αrps-closed set), hence f(A) is αrps-closed set in (Y, σ) Therefore, f is αrps-closed map.

6.
It is follows from the fact (every rgα-closed map is an rg-closed map), and since Y is a T *1/2 -space, then by we get, f is αrps-closed.
Similarly, we proof the following proposition.

Remark:
The composition of two αrps-closed maps need not beαrps-closed map in general, the following example to show that.
The following proposition gives the condition to make the composition two αrps-closed maps is also αrps-closed map.
Proof: Let A be a closed set in (X, τ), Thus f(A) is αrps-closed set in (Y, σ), since Y is a T *1/2 -space, then by proposition. we get f(A) is a closed set in (Y, σ), also, since g is αrps-closed map, hence g(f({A}) is aαrps-closed set in (Z,μ). But g(f{A})=g of (A), that is (f{A})=g of is aαrps-closed set in (Z,μ). Therefore g (X, τ) →(Z,μ), is αrps-closed map.
Proof: Let A be a closed set in (X,τ), since f is αrps-closed map. Thus f(A) is αrps-closed set in (Y, σ) and by using remark [every αrpsclosed set is rg-closed set] we get, f(A) is rg-closed set in (Y, σ). Also, since Y is aT* 1/2 -space then, by definition. we get f(A) is a closed set in (Y, σ).
Proof: 1. Let A be a closed set in (X, τ), since f is αg-closed map. Thus f(A) is αgclosed set in (Y, σ), by hypotheses Y is a αT b −space, then by definition, we get, f(A) is a closed set in (Y, σ), and by remark, every closed is αrps-closed set), hence f(A) is αrps-closed set in (Y, σ). Therefore, f is αrps-closed map.

2.
It is follows from the fact (every gα-closed map is an αg-closed map [6]) and since Y is a αT b −space, then by we get, f is αrps-closed.
The following propositions give the condition to make remark true: , that is g of (A)is aαrpsclosed set in (Z, μ). Therefore, g of: (X,τ)→(Z,μ) is αrps-closed map.
2. If g is αrps* -continuous and injective, then f is αrps-closed map.

Some Types of αrps-Closed Maps
Some other types of αrps-closed maps are given in this section such as [α * rps-closed maps, strongly αrps-closed maps and almost αrpsclosed maps) with study the relationships between these types of maps.
Proof: Let f: (X, τ)⟶(Y,σ) be α * rps-closed map and let A be a closed set in (X, τ), by remark ∀ closed set is a αrps-closed set]. Thus, A is aαrps-closed set in (X, τ). Since f is a α * rps-closed map. Then, f(A) is a αrps-closed set in (Y, σ). Therefore, f is αrps-closed map.

Proof
It is follows from proposition.

Remark:
The converse of proposition are not true in general. It is easy to see that in example, f isα*rps-closed map, but is not closed, and in example it is clear that f is (αg-closed map, gαclosed map, sg-closed map, gs-closed map, rg-closed map and rgα-closed map), but is not α*rps-closed map.

1.
The concepts of closed map and almost-closed map are independent to α*rps-closed map. It is clear see that in examples.
The following propositions give the condition to make the proposition, corollary and Remark are true.

Almost-closed map.
Proof (i):-It is follows from proposition, we get f is a closed map.
Proof (ii): It is follows from the fact (∀ closed map is almost-closed map [17]. Proposition: Let f: (X, τ)⟶ (Y, σ) be any map, then f is α*rpsclosed map, if X is T *1/2 -space and f is a 1. αg-closed map and Y is a αT b -space.
2. gα-closed map and Y is a αT b -space.

Proof (i)
Let A be an α*rps-closed set in (X, τ), since X is a T *1/2 -space, then by using proposition. We get, A is a closed set in X. Also, since f is αgclosed map. Thus, f(A) is αg-closed in (Y,σ) and since Y is αT b -space, then f(A) is closed set in (Y, σ), by remark [∀-closed set is an α*rpsclosed set]. Hence, f(A) is α*rps-closed set in (Y, σ). Therefore, f is α*rps-closed map.

Proof (ii)
It is follows from the fact (∀ gα-closed map is αg-closed map [6]), and Similarly, we proof the following proposition.

Proposition
The composition of two α*rps-closed maps is alsoα*rps closed map.
Similarly, we proof the following corollary.
ii. Almost-closed map.

Proof
i. Let A be a closed set in (X, τ), by using remark, step [∀ closed set is an αrps-closed]we get, A is an αrps-closed set in (X, τ). Since f is strongly αrps-closed map. Thus, f(A) is a closed set in (Y, σ). Therefore, f is a closed map.
ii. It is clear that from step [∀ strongly αrps-closed map is a closed and the fact ( closed map is almost closed, [17]. iii. It is clear that from step [∀ strongly αrps-closed map is a closed and the fact (∀ closed map is g-closed, [4]) iv. It is clear that from step and the proposition.
v. Let A be an αrps-closed set in (X, τ). Since f is strongly αrpsclosed map.

Proof
It is clear that from proposition. The following examples show the converse of above proposition and corollary need not be true in general.

Example
Let X={a, b, c} with the topology τ={X, θ, {a}} and let f: (X, τ) →(X, τ) be an identity map. Then, it is clear that f is a closed map, [almostclosed map and g-closed map] but is not strongly αrps-closed, since for closed set A={b}, f(A)=f({b})={b} is not closed set in (X,τ). The following condition to make proposition and corollary are true

Proof
It is follows from proposition and step and proposition.

Proof
Let A be an αrps-closed set in (X, τ). Since f is α * rps-closed map. Thus f(A) is αrps-closed set in (Y, σ). Also, since Y is T *1/2 -space, then f(A) is a closed set in (Y, σ). Therefore, f is strongly αrps-closed map.

Proof
Let A be an αrps -closed set in (X, τ). Since X is a T *1/2 -space. Then, A isa closed set in (X, τ), also since X is a locally-indiscrete, thus A is a regular closed set in (X, τ), hence f(A) is a αrps-closed set in(Y, σ). Therefore, f is a strongly αrps-closed map.
Next, we give some proposition and results about the composition of strongly αrps-closed map.

Proposition
The composition of two strongly αrps-closed maps is also strongly αrps-closed map.
Similarly, we proof the following proposition.

Proof
(i) Let A be αrps-closed set in X, since f is strongly αrps-closed map, then f(A) is a closed set in (Y, σ). Also, since g is αrps-closed map. Thus, g(f({A}) is a αrps-closed set in (Z,μ). That is g(f({A})=g of (A) is a αrpsclosed set in (Z, μ). Therefore, g of: (X,τ)→(Z, μ) is α*rps -closed map.
The proof of steps.

Remark
In the proposition the composition g of: (X,τ)→(Z, μ) need not be in general strongly αrps-closed map. As shows in the following example:
ii. f is a αrps-closed map.

Proof
(i) Let A be a closed set in X, since f is closed map, then f(A) is a closed set in (Y, σ), by remark [∀ closed set is an αrps-closed set, so we get f(A) is an αrps-closed set in(Y, σ). Also, since g is strongly αrpsclosed map. Thus, g(f({A}) is a closed set in (Z, μ). That is g(f({A})=g of (A) is a closed set in (Z,μ) and by remark [∀ closed set is an αrps-closed set], so we get g of (A) is a closed set in (Z, μ). Therefore, g of: (X,τ)→(Z, μ) is α*rps-closed map. The proof of steps.

Remark
In the proposition the composition g of: (X,τ)→(Z, μ) need not be in general strongly αrps-closed map. As shows in the following example. strongly αrps-closed map, but g of: (X,τ)→(Z,μ)) is not strongly αrpsclosed map, since for the closed set A={b}} in {X, τ}, then g of (A)=g of (A)=g of ({b})=g(f({b})=g({b})={c}, which is not closed set in(Z, μ).

Example
The following proposition give the condition to make Remark true:

Proof
Let A be a αrps-closed set in X, then f(A) is a closed set in (Y, σ), by Remark: [∀ closed set is an αrps-closed set], since g is αrps-closed map. Thus g(f({A}) is aαrps-closed set in (Z, μ). That is g(f({A})=g of (A) is an αrps-closed set in (Z, μ). Also, since Z is T *1/2 -space, so we get g of (A) is closed set in (Z, μ).Therefore, g of: (X,τ)→(Z, μ) is stronglyαrpsclosed map.

Proof
(i) Let A be αrps-closed set in X, since X is a T *1/2 -space, then by using proposition we get A is a closed set in X. Thus, f(A) is a closed set in Y, by remark, if (A) is an αrps-closed set inY. Also, since g is strongly αrps-closed map. Thus, g(f({A}) is a closed set in Z. That is g(f({A})=g of (A) is a closed set in Z. Hence, g of (X,τ)→(Z, μ) is strongly αrps-closed map.
(ii) if g of: (X,τ) →(Z, μ) is strongly αrps-closed map and f is strongly αrps-continuous surjective map, then g is strongly αrps-closed map

Proof (i)
Let A be a closed set in Y, since f is continuous, then f −1 (A) is a closed set in (X,τ), by Remark(2-2)[∀ closed set is an αrps-closed set], so we get f -1 (A) is a αrps-closed in X. Also, since g of: (X,τ)→(Z,μ) is a strongly αrps-closed map, thus g of (f -1 (A)is a closed set in (Z,μ) and Remark [∀closed set is anαrps-closed set], hence g of (f -1 (A) is an αrps-closed set in (Z, μ). That is g of (f -1 (A))=g(f(f -1 (A)=g(A), hence g(A) isαrps-closed set in (Z, μ) since, f is subjective. Therefore, g is αrps-closed map.

Proof (ii)
Let A be a αrps-closed set in Y. Since f is strongly αrps-continuous. Then, f -1 (A) is a closed set in X, by Remark [∀ closed set is an αrps-closed set], so we get f -1 (A) is a αrps-closed in X. Also, since g of (X,τ)→(Z, μ) is strongly αrps-closed map, hence map, hence g of (f -1 (A) is a closed set in (Z, μ), since, f is subjective. That is g of (f -1 (A))=g(f(f -1 (A))=g(A), hence g(A) is closed set in Z. Therefore, g is strongly αrps-closed map.
In the following, we give other type of αrps -closed maps which is called almost αrps -closed map.
Definition A map f: (X, τ) ⟶ (Y, σ) is called almostαrps-closed map if f(A) is αrps-closed set in (Y, σ), for every regular closed set A in( X, τ).

Proposition
Every almost closed map is almost αrps-closed map.
Proof: Let f: (X, τ) ⟶ (Y, σ) be a almost closed map and A be a regular closed set in (X, τ). Then, f(A) is a closed set in (Y, σ) and by using remark we get f(A) is an αrps-closed set in (Y, σ). Hence, f is almost αrps-closed map.

Proposition
Everyαrps-closed map is almost αrps-closed map.
Proof: Let f: (X,τ)⟶ (Y,σ) be an αrps-closed map and A be a regular closed set in (X,τ). Since (∀ regular closed [3]). Then, A is a closed set in X.
Thus, f(A) is an αrps-closed set in (Y, σ). Hence, f is almost αrpsclosed map. Corollary 1. Every closed map is almostαrps-closed map.
Proof: The converse of proposition and corollary need not be true in general. Define f:(X, τ)→(Y, σ) by f(a)=b, f(b)=f(c)=c. Then, it is clear that f is almost αrps-closed map but is not almost closed map, since for regular closed set A={b, c} in (X, τ), f(A)=f({b, c})={c} is not closed in (Y, σ).

Example
The following proposition give the condition to make, proposition and corollary are true: