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On Some Types of αrps- Closed Maps
ISSN: 2168-9679

Journal of Applied & Computational Mathematics
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On Some Types of αrps- Closed Maps

Hameed DM*, Mushtt IZ and Abdulqader AJ
Department of Mathematics, College of Education, Al-Mustansiriyah University, Iraq
*Corresponding Author: Hameed DM, Department of Mathematics, College of Education, Al-Mustansiriyah University, Baghdad, Iraq, Tel: +9647709281272, Email: [email protected]

Received Date: May 21, 2018 / Accepted Date: May 29, 2018 / Published Date: Jun 09, 2018

Keywords: αrps-closed; Topology; Mappings; Subset

Introduction

Mappings play as important role, in the study of modern mathematics, especially in topology and functional analysis [1-5]. 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].

Hammed introduced and studied αrps-closed sets and also introduce the notion (αrps-continuous, αrps-irresolute and strongly αrps-continuous) functions [8].

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 [9-13], 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 [14-18].

Throughout this paper (X,τ), (Y,σ) and (Z,μ) (or simply X,Y and Z) represent non-empty topological spaces [19-22]. For a sub set A of a space X.cl (A), int (A) and Ac denoted the closure of A, the interior of A and the complement of A in X respectively [23].

Preliminaries

Some definition and basic concepts have been given in this section.

Definition

A subset A of a space X is said to be:

1. Semi-open [9] If A ⊆ cl(int t(A)) and semi-closed set if int( cL(A)) ⊆ A.

2. α-Open set [16] If A ⊆ int(cl(int(A))) and α-closed set if cl(int(cl(A)) ⊆ A.

3. Preopen set [15] If A ⊆ int(cl(A)) and preclosed if cl(int(A) ⊆ A.

4. Semi-preopen set [1] If A ⊆ cl(int(cl(A)) and semi-preclosed if int (Cl(int(A)) ⊆ A).

5. Regular open [20] If A=int(cl(A) and regular closed if A=cl(int(A)).

6. Regular α-open [21] if there is a regular open set U such that U ⊆ A ⊆ αcl(U).

The semi-closure (resp. α-closure, semi-pre closure), of a sub set A of X is the intersection of all semi-closed (resp. α-closed, semi-pre closed) sets containing A and denoted by scl(A) (resp. αcl(A), resp. spcl(A).

Remark: It has been proved that:

1. Every regular closed set and closed set in a space X is an αrpsclosed set.

2. Every αrps-closed set is (sg-closed, gs-closed, αg-closed, gα- closed, rg-closed and rgα-closed) set.

Definition

A sub set A of a space X is said to be a:

1. Generalized closed set (briefly, g-closed) [10] if cl(A) ⊆ U whenever A ⊆ U and U is an open set in X.

2. Generalized semi-closed set (briefly, gs-closed) [3] if scl(A) ⊆ U whenever A ⊆ U and U is an open set in X.

3. Semi-generalized closed set (briefly, sg-closed) [5] if scl(A) ⊆ U whenever A ⊆ U and U is a semi-open set in X.

4. Generalized α-closed set (briefly, gα-closed) [13] if αcl(A) ⊆ U whenever A ⊆ U and U is an α-open set in X.

5. α-Generalized closed set (briefly, α g-closed) [12] if αcl(A) ⊆ U whenever A ⊆ U and U is an open set in X.

6. Regular generalized closed set (briefly, rg-closed) [18] if cl(A) ⊆ U whenever A ⊆ U and U is a regular open set X.

7. Regular generalized α-closed set (briefly, rgα-closed) [21] if cl(int(A)) ⊆ U whenever A ⊆ U and U is a regular α-open set.

8. Pre-semi closed] if spcl(A) ⊆ U whenever A ⊆ U and U is a g-open.

9. Regular pre-semi closed (briefly, rps-closed) [19] if spcl (A) ⊆ U whenever A ⊆ U and U is an rg-open set in X.

10. αrps-Closed set [8] if αcl(A) ⊆ U whenever A ⊆ U and U is arps-open set in X.

The complement

g-Closed (resp. gs-closed, sg-closed, gα-closed, αg-closed, rgclosed, rgα-closed, rps-closed and αrps-closed) sets is called a g-open (resp. gs-open, sg-open, gα-open, αg-open, rg-open, rgα-open, rpsopen and αrps-open) sets,

The class: The regular closed (resp. g-closed, sg-closed, gs-closed, αg-closed, gα-closed, rg-closed, rgα-closed, rps-closed and αrpsclosed) subsets of X is denoted by RC(X,τ)[resp. GC(X,τ), SGC(X,τ), GSC(X,τ), αGC(X,τ), GαC (X,τ),RGC(X,τ), RGαC(X,τ), RPSC(X,τ) and αRPSC(X,τ)].

Definition

A map f: (X, τ) ⟶ (Y, σ) is called

1. Closed map: if f(A) is a closed set in (Y,σ), for every closed set A in (X, τ) [ 20].

2. g-Closed map: if f(A) is g-closed set in (Y,σ), for every closed set A in (X, τ) [11].

3. sg-Closed map: if f(A) is sg-closed set in (Y,σ), for every closed set A in ( X, τ) [7].

4. gs-Closed map: if f(A) is gs-closed set in (Y, σ), for every closed set A in( X, τ) [7].

5. gα-Closed map: if f(A) is gα-closed set in (Y,σ), for every closed set A in (X, τ) [6].

6. αg-Closed map: if f(A) is αg-closed set in (Y, σ), for every closed set A in (X, τ) [6].

7. rg-closed map: if f(A) is rg-closed set in (Y, σ), for every closed set A in (X, τ) [4].

8. rgα-Closedmap: if f(A) is rgα-closed in (Y, σ), for every closed set A in (X, τ) [22].

9. Almost-Closed map: if f(A) is a closed set in (Y, σ), for every regular closed set A in (X, τ) [17].

Definition : A topological space X is said to be:

1. T* 1/2 - spaces: If every rg-closed sets is closed [13].

2. Tb-space: If every gs-closed sets is closed [6].

3. αTb-space: If every αg- closed sets is closed [6].

4. T1/2-space: If every g- closed sets is closed [10].

5. Locally Indiscrete space: if every closed set is a regular closed [3].

Definition: A function f: (X, τ) →(Y, σ) is said to be:

1. Continuous function: If the inverse image of each open (closed) set in Y is an open(closed) set in X [9].

2. αrps-Continuous: If f −1(A) is αrps- closed set in X for every closed set A in Y [8].

3. αrps-Irresolute continuous: If f −1 (A) is an αrps-closed set in X for every αrps-closed set A in Y [8].

4. Stronglyαrps-continuous continuous: If f −1 (A) is closed set in X for every αrps-closed set A in Y [8].

Proposition

1. If X is a T1/2-space, then every g-closed set in X is αrps-closed.

2. If X is a Tb-space, then every αrps-closed set in X is g-closed.

3. If X is a T*1/2-space, then every αrps-closed set in X is closed.

αrps-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.

Definition : A map f: (X, τ)⟶(Y, σ) is called αrps-closed map if f(A) is αrps-closed set in (Y, σ), for every closed set A in (X, τ).

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.

Example: Let X={a, b, c} with the topology τ={X, θ, {a}, {a, c}}, where αRPSC(X, τ)={X, θ, {b}, {c}, {b, c}} and define f: (X, τ)→(X, τ) by f(a)=a, f(b)=c and f(c)=b, then f is αrps-closed map, but f is not closed map, since for the closed set A={b} in {X, τ}, but f(A)=f({b})={c} is αrps-closed set in (X, τ), but is not closed set in (X, τ).

Proposition : Every αrps-closed map f : (X, τ) →(Y, σ) is

1. αg-Closed map.

2. gα-Closed map.

3. sg-Closed map.

4. gs-Closed map.

5. rg- Closed map.

6. rgα-Closed map.

Proof: 1: 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 αg-closed set] we get f(A) is αg-closed set in (Y, σ). Hence, f: (X, τ) →(Y, σ) is αg-closed map. 2: 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 αrps-closed set is gα-closed set] we get f(A) is gα-closed set in (Y, σ). Hence, f: (X, τ) →(Y, σ) is gα-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.

Example : Let X={a, b, c}, with the topology τ={X, θ, {a}, {b, c}}, where αRPSC(X, τ)={X, θ, {a}, {b, c}}, GαC(X,τ) =SGC(X, τ)=GSC(X, τ)=RGC(X, τ) =RGαC(X, τ)={X, θ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}}, define f: (X, τ) →(X, τ) by f(a)=b, f(b)=a and f(c)=c, then it is clear that f is (αg-closed, gα-closed, sg-closed, gsclosed, rg-closed and rgα-closed ) map, but f is not αrps-closed, since for the closed set A={b, c} in {X, τ}, but f(A)=f({b, c})={a,c} is (αg-closed, gα-closed, sg-closed, gs-closed, rg-closed and rgα-closed) set in (X, τ), but is not αrps-closed set in (X, τ).

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 αRPSC(X, τ)={X, θ {b},{c}},{b, c}}, GC(X,τ)={X,θ,{b},{a, b},{b, c} and define f: (X, τ) →(X, τ) by f(a)=a, f(b)=c and f(c)=b, then f is αrps closed map, but f is not g-closed map, since for the closed set A={b} in {X, τ}, but f(A)=f({b})={c} is αrps-closed set in (X, τ), but is not g-closed set in (X, τ).

Example

Let X={a, b, c}, with the topology τ={X, θ, {a}, {b, c}}, where αRPSC(X, τ)={ X, θ, {a}, {b, c} }, GC(X, τ)={X, θ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c} }. Define f: (X, τ) →(X, τ) by f(a)=b, f(b)=a and f(c)=c, then it is clear that f is a g-closed map, but f is not αrps−closed map, since for the closed set A={a} in {X, τ}, but f(A)=f({a})={b} g-closed set in (X, τ), but is not αrps-closed set in (X, τ).

Example

Let X={a, b, c} with the topology τ={X, θ, {a}}, where αRPSC(X, τ)={X, θ {b},{c},{b,c}}, and the set of regular closed set={X,θ} define f: (X, τ) →(X, τ) by f(a)=b, f(b)=a and f(c)=b, then f is almost -closed map, but f is not αrps-closed map, since for the closed set A={b, c} in {X, τ}, but f(A)=f({b, c})={a, c} is not αrps-closed set in (X, τ).

Example

Let X =Y={a, b, c} with the topologies τ={X, θ, {a}, {b}, {a, b} } and σ={Y,θ,{a},{a,c}}, where αRPSC(Y,σ)={Y, θ, {b}, {c}, {b, c}}, and RC(X, τ)={X,θ}, {a, c}, {b, c}}, let f: (X, τ)→(Y, σ) be a map defined by f(a)=f(c)=c and f(b)=b, then f is αrps-closed map, but f is not almost -closed map, since for the regular closed set A={a, c} in {X, τ}, but f(A)=f({a, c})={c} is not closed set in (Y, σ).

The following propositions give the condition to make the propositions and remark are true:

Proposition

If f: (X, τ) →(Y, σ) is αrps-closed map and Y is aT*1/2 - space, then f is a 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).

Hence, f is an αrps-closed map

Proposition: A map f : (X, τ) →(Y, σ)is αrps-closed map, if f is a

1. αg-Closed map and Y is αTb-space.

2. gα-Closed map and Y is αTb-space.

3. gs-Closed map and Y is Tb-space.

4. sg-Closed map and Y is Tb-space.

5. rg-Closed map and Y isT*1/2-space.

6. rgα-Closed map and Y isT*1/2-space.

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 αTb −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 αTb −space, then by we get, f is αrps-closed.

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 Tb-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 Tb − 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.

Proposition

1. If f: (X, τ) →(Y, σ) is g-closed map Y is T1⁄2-space, then f is a αrps closed map.

2. If f: (X, τ) →(Y, σ) is αrps-closed map Y isT*1/2-space, then f is a g closed map.

3. If f: (X, τ) →(Y, σ) is αrps-closed map Y is T*1/2-space, then f is almost-closed map.

4. If f: (X, τ) →(Y, σ) is almost-closed map and (X, τ) is a locally indiscrete then f is αrps-closed map.

Remark: The composition of two αrps-closed maps need not beαrps-closed map in general, the following example to show that.

Example

Let X=Y=Z={a, b, c} with the topologies τ={X, θ, {a}, {a, c}}, σ={y, θ,{a}},μ={Z,θ,{a},{b},{a,b}} where αRPSC(Y,σ)={Y, θ, {b},{c},{b, c}} and αRPSC(Z, μ)={Z, θ, {c}{a, c}, {b, c}}. let f: (X, τ)→(Y, σ) be the identity map, and g: (Y,σ)→(Z, μ) be a map defined by g(a)=b and g(b)=a and g(c)=c, then it is easy to see that f and g are αrps-closed map, but g of: (X,τ)→(Z,μ) is not αrps-closed map,since for the closed set A={b} in {X, τ}.

g of (A)=g of ({b})=g(f({b})=g({b})={a}, which is not αrps-closed set in (Z, μ). Hence, f is not αrps-closed map.

The following proposition gives the condition to make the composition two αrps-closed maps is also αrps-closed map.

Proposition

Let f: (X, τ) →(Y, σ) and g: ( Y, σ)→(Z,μ) be two αrps-closed maps and Y is a T*1/2-space, then g of (X,τ) →(Z,μ) is 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.

Proposition: If f: (X, τ) →(Y, σ) is a closed map and g: (Y, σ) →(Z, μ) is an αrps- closed map, then g of (X,τ)→(Z,μ) is αrps-closed map.

Proof: Let A be a closed set in (X, τ), Thus f(A) is a closed set in (Y,σ), since g is αrps-closedmap, hence g(f({A}) is an αrps-closed set in (Z,μ). That is g(f({A})=g of (A) is aαrps-closed set in (Z,μ). Therefore, g of: (X,τ)→(Z,μ) is αrps-closed map.

Remark

If f: (X, τ) →(Y, σ) is αrps-closed map and g: (Y, σ) →(Z, μ) is a closed map, then g of (X,τ)→(Z,μ) need not be αrps-closed map, and this is shown by the following example:

Example

Let X=Y=Z={a, b, c} with the topologies τ={X, θ, {a}, {a, c}}, σ={Y, θ,{a}}, μ={Z ,θ,{a},{b, c}} where αRPSC (Y,σ)={Y, θ, {b},{c}, {b, c}} and αRPSC(Z, μ)={Z, θ, {a}, {a, c}.let f:(X,τ)→(Y, σ), and g:(Y,σ)→(Z,μ) be two the identity maps then it is easy to see that f isαrps-closed map and g is a closed map, but g of: (X,τ)→(Z,μ)) is not αrps-closed map, since for the closed set A={b} in{X, τ}, then g of (A)=g of ({b})=g(f ({b})=g({b})={b}, which is not αrps-closed set in (Z,μ). Therefore, g of (X,τ)→(Z,μ) is not αrps-closed map.

The following propositions give the condition to make remark true:

Proposition

If f(X,τ)→(Y, σ) is an αrps- closed map, g:(Y, σ)→(Z, μ) is a closed map and let Y be T*1/2-space, then g of: (X,τ) →(Z, μ) is 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*1/2space, then by proposition. we get f(A) is a closed set in (Y,σ), also, since g is a closed map, hence g(f({A}) is a closed set in (Z, μ), by remark. [∀ closed set is] αrps-closed set]. Then, g(f({A}) is an αrps-closed setin (Z,μ). But, g(f({A})=g of (A), that is g of (A)is aαrpsclosed set in (Z, μ). Therefore, g of: (X,τ)→(Z,μ) is αrps-closed map.

Proposition

Let f: (X, τ) →(Y, σ) and g: (Y,σ)→(Z,μ) be two maps, such that their composition g of (X,τ)→(Z,μ) is αrps-closed map,

1. If f is a continuous and subjective, then g is αrps-closed map.

2. If g is αrps* - continuous and injective, then f is αrps-closed map.

Proof: (i):- Let A be a closed set of (Y, σ), since f is a continuous, then f-1(A)) is a closed set in (X,τ). Also, since g of: (X,τ)→(Z,μ) is an αrps-closed map, thus g of (f−1(A)) is 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 surjective. Therefore, g is αrps-closed map.

Proof: (ii):- Let E be a closed set in (X,τ). Since g of (X,τ)→(Z,μ) is an αrps-closed map, thus g of (E)is an αrps-closed set in (Z,μ), since g is αrps*-continuous. Then g-1(g of (E))=f(E) is an αrps-closed setin (Y, σ). Also, since f is injective. Therefore, 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.

Definition

A map f: (X, τ) ⟶ (Y, σ) is called α*rps-closed map if f(A) is αrpsclosed set in (Y, σ), for every αrps-closed set A in (X, τ).

Proposition

Every α*rps-closed map is αrps-closed map.

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.

Corollary : Everyα*rps-closed map is

1. αg-Closed map.

2. gα-Closed map.

3. sg-Closed map.

4. gs-Closed map.

5. rg-Closed map.

6. rgα-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.

Proposition: Let f: (X,τ)⟶(Y, σ) be an α*rps-closed map and Y is T*1/2 space then f is a

1. Closed map.

2. 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 αTb-space.

2. gα-closed map and Y is a αTb-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 αTb-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

Let f: (X, τ)⟶ (Y, σ) be any map, then f is α*rps-closed map, if X is T*1/2-space and f is a

1. gs-Closed map and Y is a Tb-space.

2. sg-Closed map and Y is a Tb -space.

3. rg-Closed map and Y is T*1/2-space.

4. rgα-closed map and Y is T*1/2-space

5. Closed map.

6. αrps-Closed map.

7. g-Closed map.

Proposition

If f: (X, τ) ⟶ (Y, σ) is a almost - closed map and X is a T*1/2-space and locally indiscrete, then f is α*rps-closed map.

Proof

Let A is a αrps-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 X is a locally indiscrete, then by definition of locally indiscrete we have, A is a regular closed set in X. since f is a almost-closed map. Thus, f(A) is a closed set in Y and by remark [∀ closed set in X]. Hence, f(A) is a αrps -closed set in Y. Therefore, f is α*rps- closed map.

Proposition

The composition of two α*rps-closed maps is alsoα*rps closed map.

Proof: Let f: (X, τ) ⟶ (Y, σ) and g: (Y,σ)→(Z,μ) be two α*rps -closed map, and A be αrps-closed set in X, since f is α*rps-closed map, then f(A) is an αrps - closed set in (Y,σ). Also, since g is an α*rps-closed map. Thus, g(f({A}) is aαrps-closed set in (Z,μ). That is g(f ({A})=g of (A)is a αrps- closed set in (Z, μ).

Therefore, g of (X,τ)→(Z,μ) is α*rps-closed map.

Proposition

If f:(X,τ)⟶(Y,σ) is αrps-closed map and g:(Y,σ) →(Z, μ) is α*rpsclosed map, then g of: (X,τ)→(Z,μ) is αrps-closed map.

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

Similarly, we proof the following corollary.

Corollary

If f:(X, τ)⟶(Y,σ) is a closed map and g:(Y, σ →(Z, μ) is α*rps -closed map, then g of (X,τ)→(Z,μ) is αrps-closed map.

Now, we give another type of αrps-closed map is called strongly αrps-closed map.

Definition

A map f: (X, τ) ⟶ (Y, σ) is called strongly αrps-closed map if f(A) is closed set in (Y, σ), for every αrps-closed set A in (X, τ).

Proposition

Every strongly αrps- closed map f: (X, τ)⟶(Y, σ) is

i. Closed map.

ii. Almost-closed map.

iii. g-Closed map.

iv. αrps-Closed map.

v. α*rps-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.

Thus, f(A) is a closed set in (Y, σ), by using remark, [∀ closed set is an αrps-closed set], then A is an αrps-closed set in (Y, σ). Therefore f is a α*rps-closed

Corollary: Every strongly αrps-closed map f: (X, τ) ⟶ (Y, σ) is

1. αg-Closed map.

2. gα-Closed map.

3. sg-Closed map.

4. gs-Closed map.

5. rg-Closed map.

6. rgα-Closed 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.

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).

Example

Let X=Y={a, b, c} with the topologies τ={X, θ, {a}}, and σ={ Y, θ, {a}, {a, c}}, where αRPSC (X, τ)={X, θ, {b},{c}, {b, c}} and let f: (X, τ) →(Y, σ) be an identity map. Then, it is clear that f is αrps-closed map and α*rps -closed map but is not strongly αrps-closed map, since for closed set A={c}, then f(A)=f({c})={c} is not closed set in (Y, σ).

Example

LetX=Y={a, b, c} with the topologies τ={X, θ, {a}}, and σ={Y, θ, {a}, {b, c} }. Define f: (X, τ) →(Y, σ) by f(a)=b, f(b)=a and f(c)=c. Then, it is clear that f is αg-closed map (gα-closed map, sg-closed map, gsclosed map, rg-closed map and rgα-closed map), but is not strongly αrps-closed map, since for closed set A={c}, then f(A)=f({c})={c} is not closed set in (Y,σ).

The following condition to make proposition and corollary are true

Proposition

Let f : (X, τ) →(Y, σ) be any map, then f is a strongly αrps- closed map if X is a T*1/2-space and

1. Closed map

2. Almost- closed map.

3. g-Closed map.

4. gα-Closed map.

5. αg-Closed map.

6. rg-Closed map.

7. rgα-Closed map.

8. αrps-closed map.

Proof

It is follows from proposition and step and proposition.

Proposition

If f: (X, τ) →(Y, σ)is an α*rps-closed map and Yis a T*1/2-space, then f is a strongly αrps-closed map.

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.

Proposition

If f: (X, τ) →(Y, σ) is almost-closed map and Xis a locally - indiscrete and T*1/2-space, then f is a 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.

Proof

Let f: (X, τ) ⟶ (Y, σ) and g: (Y, σ) →(Z, μ) be two strongly αrpsclosed map, and A be αrps-closed set in X, since f is strongly αrpsclosed map, then f(A) is a closed set in (Y, σ), by remark [∀ closed set is an αrps-closed set] Thus, f(A) is an αrps-closed set in (Y,σ). Also, since g is an strongly αrps closed map. Thus, g(f({A}) is a closed set in (Z,μ). That is g(f({A})=g of (A)is aαrps-closed set in (Z, μ). Therefore, g of: (X,τ)→(z,μ) is strongly αrps-closed map.

Similarly, we proof the following proposition.

Proposition

1. If f: (X, τ) ⟶ (Y, σ) is a strongly αrps-closed map and g: (Y, σ) →(Z, μ) is closed map, then g of (X,τ)→(Z, μ) is a strongly αrps-closed map.

2. If f: (X, τ) ⟶ (Y, σ) is a α*rps - closed map and g: (Y, σ) →(Z, μ) is strongly αrps-closed map, then g of (X,τ) →(Z, μ) is a strongly αrps-closed map.

Proposition

Let f: (X, τ) ⟶ (Y, σ) and g: (Y, σ) →(Z, μ) be two any maps, then g of(X,τ) →(Z, μ) is a α*rps -closed map, if f is strongly αrps-closed map and

i. g- is αrps-closed map.

ii. g- is α*rps -closed map.

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:

Example

Let X=Y=Z={a, b, c} with the topologies τ={X, θ, {a}, {b, c}}, σ ={Y,θ,{a}, {a, c}},μ={z,θ,{a}}. Let f: (X, τ) →(Y, σ) by f(a)=f(b)=b and f(c)=c and g: (Y,σ)→(Z,μ) be an identity map, then it is easy to see that f is strongly αrps-closed map and g is α*rps- closed map, but g of (X,τ)→(Z,μ).

Is not strongly αrps-closed map,since for the closed set A={a}in {X, τ}, then g of (A)=g of ({a})=g(f({a})=g({b})={b}, which is not closed set in (Z,μ).

Proposition

Let f: (X, τ) ⟶ (Y, σ) and g: (Y, σ) →(Z, μ) be two any maps then g of: (X,τ)→(Z, μ) is αrps-closed map, if g is strongly αrps-closed and

i. f is a closed map.

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.

Example

Let X=Y=Z={a, b, c} with the topologies τ={X,θ,{a}}, σ={Y, θ,{a}, {b},{a,b}},μ={Z,θ,{a},{a,c}} and let f: (X, τ)→(Y, σ) be an identity map and g: (Y,σ)→(Z,μ) be a mapping defined by g(a)=g(b)=c and g(c)=b, then it is easy to see that f is a closed map and αrps-closed map and g is 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).

The following proposition give the condition to make Remark true:

Proposition

Let f:(X, τ) ⟶ (Y, σ) and g: (Y, σ) →(Z, μ) be two any maps, then g of (X,τ) →(Z, μ) is a strongly αrps-closed map, if f is strongly αrps-closed map and (Z, μ ) is a T*1/2-space

1. g is αrps-closed map.

2. g is α*rps-closed map.

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.

Proposition

Let f:(X, τ) ⟶ (Y, σ) and g: (Y, σ) →(Z, μ) be two any maps, then g of :(X,τ) →(Z, μ) is strongly αrps-closed map, if g is strongly αrps-closed map, X is a T*1/2-space and

1. f is a closed map

2. f is a αrps-closed 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.

Proposition: Let f:(X, τ) ⟶(Y, σ) and g: (Y, σ) →(Z, μ) be two any maps

(i) if g of: (X,τ) →(Z, μ) is strongly αrps-closed map and f is continuous surjective map, then g is a 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(2-2)[∀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.

2. Everyα*rps-closed map is almost αrps-closed map.

3. Every strongly αrps-closed map is almostαrps-closed map.

Proof: The converse of proposition and corollary need not be true in general.

Example

Let X=Y={a, b, c} with the topologies τ={X, θ, {a}} and σ={Y, θ, {a}, {b}, {a, b}}, where αRPSC(X,τ)={X, θ, {b}, {c}, {b, c}}RC(X,τ)={X,θ} and αRPSC(Y, σ)={Y, θ, {c}, {a, c}, {b, c}}. Define f:(X, τ)→(Y, σ)by f(a)=c f(b)=a and f(c)=b. Then, it is clear that f is almost αrps-closed map but is not closed map (αrps-closed, α*rps-closed and strongly αrps-closed) map, since for closed set A={b, c} in (X, τ), f(A)=f({b, c})={a, b} is not closed and (αrps-closed )set in Y.

Example

Let X=Y={a, b, c} with the topologies τ={X, θ, {a}} and σ={Y, θ, {a}, {a, c}}, where RC(X, τ)={X, θ, {b, c}} and αRPSC(Y, σ)={Y, θ, {b}, {c}, {b, c}}.

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).

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

Proposition

If f: (X, τ) →(Y, σ) is almost αrps-closed map and (Y, σ) is a T*1/2- space, then f is a almost-closed set.

Proof: Let A be a regular closed set in (X, τ), since f is a almost αrps-closed map Then, f(A) is an αrps-closed set in Y. Also, since Y is a T*1/2space, then by proposition. we get, f(A) is a closed set in Y. Hence, f is almost-closed map.

Proposition

If f: (X,τ)→(Y, σ) is almost αrps-closed map and X is a locally indiscrete space, then f is αrps-closed set.

Proof

Let A be a closed set in (X, τ), since X is a locally indiscrete, then by definition. We get, A is a regular closed set in X. Also, since f is a almost αrps-closed map. Then, f(A) is an αrps-closed set in (Y, σ). Therefore, f is αrps-closed map.

Proposition: Let f: (X, τ) →(Y, σ)be almost αrps- closed map and X be a locally indiscrete space and Y be a T*1/2space, then f is a closed set.

Proof: Let A be a closed set in (X, τ), since X is a locally indiscrete, then by definition we get, A is a regular closed set in X. Also, since f is a almost αrps- closed map. Then, f(A) is an αrps-closed set in (Y, σ) and since Y is a T*1/2space, then by proposition we get f(A) is a closed set in.

Proposition

Let f: (X,τ) →(Y, σ) be almost αrps-closed map and X be a locally indiscrete space and T*1/2-space, then

1. f is aα*rps-closed set.

2. f is a strongly αrps-closed set if Y is a T*1/2space.

Proof: (i) Let A be a αrps-closed set in (X, τ), since X is a T*1/2space, then by proposition. we have, A is a closed set in X and since X is a locally indiscrete, then by definition. we get, A is a regular closed set in X. Also, since f is a almost αrps-closed map. Then, f(A) is an αrpsclosed set in (Y, σ). Therefore, f is an α*rps -closed set.

Proof: (ii) Let A be a αrps-closed set in (X, τ), since X is a T*1/2space, then by proposition. we have, A is a closed set in X and since X is a locally indiscrete, then by definition, we get, A is a regular closed set in X. Also, Thus, f(A) is an αrps-closed set in (Y, σ). Also, since Y is a T*1/2space Hence, f(A) is a closed set in Y. Therefore, f is a strongly αrps-closed set.

Remark: The composition of two strongly αrps-closed maps need not be strongly αrps-closed map in general, the following example to show that.

Example: Let X={a, b, c, d }, Y=Z={a, b, c} with the topologies τ={X, θ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, { a, b, d }},σ={Y,θ,{a}},μ={Z,θ,{a},{b,c}} where RC(X, τ)={X,θ,{a,d},{b,c,d}}}, RC(Y, σ)={Y, θ},RC(Z,θ){a},{b, c}}, αRPSC(Y, σ)={Y, θ, {b},{c}, {b, c}} and αRPSC(Z, μ)={Z, θ, {a}, {b, c}} Define f:(X, τ)→(Y, σ) by f(a)=f(d)=b, f(b)=f(c)=c and g: (Y,σ)→(Z,μ) be an identity map, then it is easy to see that f and g are almostαrpsclosed map, but g of: (X,τ)→(z,μ)) is not almostαrps-closed map, since for the regular closed set A={a,d} in {X, τ} g of (A)=g of ({a,d})==g(f({a, d}))=g({b})={b}, which is not αrps-closed set in (Z,μ). Hence, g of is not strongly αrps-closed map the following proposition give the condition to make remark is true:

Proposition

If f:(X, τ)⟶(Y, σ) and g:(Y, σ)→(Z, μ) are two almost αrps-closed maps and Y is locally indiscrete and T*1/2-space, then g of: (X,τ)→(Z, μ) is a almost αrps-closed map.

Proof: Let A be a regular closed set in X, thus f(A) is an αrps-closed set in Y. Since, Y is a T*1/2space, then by proposition. we get f(A) is a closed set in Y. Also, since Y is a locally indiscrete, hence f(A) is a regular closed set in Y, since g is almost αrps-closed map. then g(f(A)) is a αrps-closed set in Z. But g(f(A))=g of (A). Therefore, g of: (X,τ) →(Z, μ) is a almost αrps-closed map. The proof of the following proposition it is easy.

Proposition

Let f:(X, τ)⟶(Y, σ) and g:(Y, σ)→(Z, μ) be two maps, then g of (X,τ)→(Z,μ) is a almost αrps-closed map, if f is almost αrps-closed and g is Figure 1.

(1) α*rps -closed map.

(2) Strongly αrps-closed map.

computational-mathematics-mapping-types

Figure 1: Illustrates the relation between the αrps- closed mapping types.

Remark: Here in the following diagram illustrates the relation between the αrps-closed mapping types (without using condition), where the converse is not necessarily true.

References

Citation: Hameed DM, Mushtt IZ, Abdulqader AJ (2018) On Some Types of αrps- Closed Maps. J Appl Computat Math 7: 405. DOI: 10.4172/2168-9679.1000405

Copyright: © 2018 Hameed DM, et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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