516 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

1. Introduction

In 1996, Dontchev introduced the notion of contra-continuous functions [6].Baker introduced almost contra-continuous and contra almost β-continuousfunctions [4]. Dontchev, Ganster and Reilly introduced a new class of func-tions called regular set connected functions [7] and Jafari and Noiri intro-duced contra-precontinu-ous functions [11]. Ekici introduced and studiedabout almost contra-precontinuous functions [8]. Joseph and Kwack intro-duced the notion of (θ, s)-continuous functions [12]. Noiri and Popa [17]pointed out that almost contra-continuity is equivalent to (θ, s)-continuity.Recently, the new class of functions called contra-Pβ-continuous functions[30] was studied also (see [[26]-[30]]). In this paper, we introduced thenotion of almost contra-Pβ-continuity which is a strong form of almostcontra-precontinuity and weaker then the notion of regular set connectedfunctions, perfectly continuous functions, and contra-Pβ-continuous func-tions. Also, it is shown that almost contra-Pβ-continuity is independent to(θ, s)-continuity and contra-precontinuity.

This paper is organized as follows. Section-2 develops the necessarypreliminaries. In Section-3, the concept of almost contra-Pβ-continuity isintroduced, and we obtain characterization and basic properties of thisnotion. In Section-4 axioms are studied in relation to almost contra-Pβ-continuity. In section-5, Pβ-closed, Pβ-regular and contra-Pβ-closed graphsof an almost contra-Pβ-continuous functions are defined. Sections-6 givessufficient conditions for a space to be connected and hyperconnected.

2. Preliminaries

Let (X, τ) or X (resp. (Y, σ) or Y , (Z, υ) or Z) be a topological spaceor space. We will denote by Cl(A) and Int(A), the closure of A and theinterior of A, respectively for a subset A of a space.

Definition 2.1. A subset A of a space is said to be

1. α-open [15] if A ⊆ Int(Cl (Int(A)));

2. pre-open [19] if A ⊆ Int(Cl(A));

3. semi-open [14] if A ⊆ Cl(Int(A));

4. β-open [1] if A ⊆ Cl(Int(Cl(A)));

5. regular open [23] if A = Int(Cl(A)) ;

Independent form of (θ, s)-continuous functions in topological... 517

The complement of pre-open (resp. semi-open, α-open, β-open, regularopen) set is said to be pre− closed (resp. semi− closed, α− closed, β −closed, regular closed).

Definition 2.2. [26] A pre-open subset A of a topological space X is saidto be Pβ-open, if for each x ∈ A there exists a β-closed set F such thatx ∈ F ⊆ A.

That is, the pre-open sets is expressed as a union of β-closed sets.

The family of all pre-open (resp. semi-open, α-open, β-open, regularopen, Pβ-open) subsets of a topological space (X, τ) is denoted by PO(X)(resp. SO(X), αO(X), βO(X), RO(X) PβO(X)). The collections SO(X),PO(X), βO(X) and PβO(X) are closed under arbitrary unions. The fam-ily of all closed (resp. pre-closed, semi-closed, α-closed, β-closed, reg-ular closed, Pβ-closed) subsets of a topological space (X, τ) is denotedby C(X) (resp. PC(X), SC(X), αC(X), βC(X), RC(X), PβC(X)).Set C(X,x) = V ∈ C(X)|x ∈ V for x ∈ X and similarly PO(X,x),SO(X,x), αO(X,x), βO(X,x), PβO(X,x).

Definition 2.3. A point x ∈ X is said to be θ-semi-cluster point [12] of asubset A of x if Cl(U) ∩ A 6= φ for every U ∈ SO(X,x). The set of all θ-semi-cluster points of A is called the θ-semi-closure of A and is denoted byθsCl(A). A subset is called θ-semi-closed if A = θsCl(A). The complementof a θ-semi-closed set is called θ-semi-open.

Definition 2.4. A function f : X → Y is called

1. contra-Pβ-continuous [30] if f−1(V ) is Pβ-closed for each open set V

of Y.

2. almost-Pβ-continuous [30] if f−1(V ) is Pβ-open for each regular open

set V of Y.

3. Pβ-continuous [26] if for each x ∈ X and each open set V of Y con-taining f(x), there exists a U ∈ PβO(X,x) such that f(U) ⊂ V .

4. precontinuous [19] if for each x ∈ X and each open set V of Y con-taining f(x), there exists a U ∈ PO(X,x) such that f(U) ⊂ V .

5. regular set connected [7] if f−1(V ) is clopen for each regular open setV of Y.

518 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

6. almost contra-continuous [4] ((θ, s)-continuous [12]) if f−1(V ) is closedfor each regular open set V of Y.

7. almost contra-precontinuous [8] if f−1(V ) is preclosed for each regularopen set V of Y.

8. almost contra-β-continuous [4] if f−1(V ) is β-closed for each regularopen set V of Y.

Lemma 2.5. [16] For a subset A of a topological space Y, the followingstatements hold:

1. αCl(A) = Cl(A) for every A ∈ βO(Y );

2. pCl(A) = Cl(A) for every A ∈ SO(Y );

3. sCl(A) = Int(Cl(A)) for every A ∈ PO(Y ).

Definition 2.6. Let A be a subset of a space (X, τ).

1. The set ∩U ∈ RO(X)| A ⊂ U is called the r-kernel [9] of A and isdenoted by rker(A).

2. The set ∩F ∈ PβC(X)| A ⊂ F is called the Pβ-closure [26] of Aand is denoted by PβCl(A).

3. The set ∪U ∈ PβO(X)| U ⊂ A is called the Pβ-Interior [26] of Aand is denoted by PβInt(A).

Lemma 2.7. [9] The following statements hold for subsets A and B of aspace X:

1. x ∈ rker(A) if and only if A ∩ F 6= φ for any F ∈ RC(X,x).

2. A ⊂ ker(A) and A = ker(A) if A is regular open in X.

3. if A ⊂ B, then rker(A) ⊂ rker(B).

3. Almost contra-Pβ-continuity

Definition 3.1. A function f : X → Y is called almost contra-Pβ-continuousif f−1(V ) is Pβ-closed in X for each regular open set V of Y.

Theorem 3.2. Every regular set connected function is almost contra-Pβ-continuous.

Independent form of (θ, s)-continuous functions in topological... 519

Proof. Suppose that f : X → Y is a regular set connected function.Then for each regular open set V of Y , f−1(V ) is clopen, which impliesf−1(V ) is Pβ-closed. 2

Theorem 3.3. Every almost contra-Pβ-continuous function is almost contra-precontinuous.

Proof. Suppose that f : X → Y is a almost contra-Pβ-continuousfunction. Then for each regular open set V of Y , f−1(V ) is Pβ-closed andhence preclosed. 2

Theorem 3.4. Every contra-Pβ-continuous function is almost contra-Pβ-continuous.

Proof. Suppose that f : X → Y is a contra-Pβ-continuous function.Then for each regular open set V of Y , f−1(V ) is Pβ-closed. 2The following diagram summarizes the obvious implications, A → B de-notes A implies B but not conversely:

Example 3.5. Let X = a, b, c, τ = φ, a, b,X andσ = φ, a, b, a, b,X. Define a function f : (X, τ) → (X,σ) byf(a) = b, f(b) = a and f(c) = c. Then f is almost contra-Pβ-continuousbut not regular set-connected, not contra-Pβ-continuous not contra pre-continuous and not almost contra-continuous.

Example 3.6. Let X = a, b, c, d, τ = φ, a, c, a, c, a, b, a, b, c,a, c, d,X and σ = φ, a, b, a, b, a, b, c,X. Define a function f :(X, τ)→ (X,σ) by f(d) = b and f(b) = a. Then f is contra-precontinuous,almost contra-pre-continuous and almost contra-continuous but not almostcontra-Pβ-continuous.

520 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

The other implications not reversible are shown in [4, 8, 30, 6]

Remark 3.7. From Example 3.5 and Example 3.6 note that almost contra-Pβ-continuity is independent to both almost contra-continuity and contra-precontinuity. Hence almost contra-Pβ-continuity is independent to both(θ, s)-continuity and contra-precontinuity

Theorem 3.8. The following statements are equivalent for a function f :X → Y :

1. f is almost contra-Pβ-continuous;

2. the inverse image of each regular closed set in Y is Pβ-open in X;

3. the inverse image of a θ-semi-open set of Y is Pβ-open;

4. the inverse image of a θ-semi-closed set of Y is Pβ-closed;

5. for each point x in X and each V ∈ RC(Y, f(x)), there is U ∈PβO(X,x) such that f(U) ⊂ V ;

6. for each point x in X and each regular open set V in Y non-containingf(x), there exist a Pβ-closed set K in X not containing x such thatf−1(V ) ⊂ K;

7. for each point x in X and each V ∈ SO(Y, f(x)), there is U ∈PβO(X,x) such that f(U) ⊂ Cl(V );

8. f−1(V ) ⊂ PβInt(f−1(Cl(V ))) for every V ∈ SO(Y );

9. f(PβCl(A)) ⊂ rker(f(A)) for every subset A of X;

10. PβCl(f−1(B)) ⊂ f−1(rker(B)) for every subset B of Y;

11. f−1(Cl(V )) is Pβ-open for every V ∈ βO(Y );

12. f−1(Cl(V )) is Pβ-open for every V ∈ SO(Y );

13. f−1(Int(Cl(V ))) is Pβ-closed for every subset V of Y;

14. f−1(Cl(Int(V ))) is Pβ-open for every subset V of Y;

15. f(PβCl(A)) ⊂ θsCl(f(A)) for every subset A of X;

16. PβCl(f−1(B)) ⊂ f−1(θsCl(B)) for every subset B of Y;

Independent form of (θ, s)-continuous functions in topological... 521

17. PβCl(f−1(V )) ⊂ f−1(θsCl(V )) for every open subset V of Y;

18. PβCl(f−1(V )) ⊂ f−1(sCl(V )) for every open subset V of Y;

19. PβCl(f−1(V )) ⊂ f−1(Int(Cl(V ))) for every open subset V of Y.

Proof. (1) ⇔ (2). Obvious.(2) ⇒ (3). Since any θ-semi-open set is a union of regular closed sets, by(2), (3) holds.(3) ⇒ (7). Let x ∈ X and V ∈ SO(Y, f(x)). Since Cl(V) is θ-semi-openin Y , there exist a U ∈ PβO(X,x) such that x ∈ U ⊂ f−1(Cl(V )). Hencef(U) ⊂ Cl(V ).(7) ⇒ (2). Let V be any regular closed set of Y and x ∈ f−1(V ). ThenV is a semiopen set containing f(x) and there exists a U ∈ PβO(X,x)such that U ⊂ f−1(Cl(V )) = f−1(V ). Therefore, x ∈ U ⊂ f−1(V ) andhence x ∈ U ⊂ PβInt(f

−1(V )) and f−1(V ) ⊂ PβInt(f−1(V )). Therefore,

f−1(V ) = PβInt(f−1(V )). Hence, f−1(V ) is Pβ-open.

(7) ⇒ (8). Let V ∈ SO(Y ) and x ∈ f−1(V ). Then f(x) ∈ V . By(7), there exists a U ∈ PβO(X,x) such that f(U) ⊂ Cl(V ). It fol-lows that x ∈ U ⊂ f−1(Cl(V )). Hence, x ∈ PβInt(f

−1(Cl(V ))). Thus,f−1(V ) ⊂ PβInt(f

−1(Cl(V ))).(8) ⇒ (1). Let F be any regular closed of Y. Since F ∈ SO(Y ), by (8),f−1(F ) ⊂ PβInt(f

−1(F )). This shows that f−1(F ) is Pβ-open.(7) ⇒ (16). Let A be any subset of X. Suppose that x ∈ PβCl(A) andG is any semiopen set of Y containing f(x). By (7), there exists a U ∈PβO(X,x) such that f(U) ⊂ Cl(G). Since x ∈ PβCl(A), U ∩ A 6= φ andhence φ 6= f(U) ∩ f(A) ⊂ Cl(G) ∩ f(A). Therefore, f(x) ∈ θsCl(f(A)).(16)⇒ (17). Let B be any subset of Y. Then f(PβCl(f

−1(B))) ⊂ θsCl(f(f−1(B)))⊂ θsCl(B).(17)⇒ (7). Let V be any semiopen set of Y containing f(x). Since Cl(V )∩(Y −Cl(V )) = φ, f(x) /∈ θsCl((Y −Cl(V ))) and x /∈ f−1(θsCl(Y −Cl(V ))).By (17) x /∈ PβCl(f

−1(Y − Cl(V )). Hence, there exists a U ∈ PβO(X,x)such that U ∩ f−1(Y − Cl(V )) = φ so that f(U) ∩ (Y − Cl(V )) = φ.(3) ⇔ (4). Obvious.(2) ⇒ (5). Let x ∈ X and V be a regular closed set of Y containingf(x). Then x ∈ f−1(V ). By (2), f−1(V ) is Pβ-ope. So there exists aU ∈ PβO(X,x) such that U ⊂ f−1(V ).(5) ⇒ (9). Let A be any subset of X. Suppose that y /∈ rker(f(A)).Then, by Lemma 2.7, there exists a V ∈ RC(Y, y) such that f(A)∩V = φ.For any x ∈ f−1(V ), by (5), there exists a Ux ∈ PβO(X,x) such that

522 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

f(Ux) ⊂ V . Hence, f(A ∩ Ux) ⊂ f(A) ∩ f(Ux) ⊂ f(A) ∩ V = φ so thatA ∩ Ux = φ. This shows that x /∈ PβCl(A) for any x ∈ f−1(V ). There-fore, f−1(V ) ∩ PβCl(A) = φ and therefore, V ∩ f(PβCl(A)) = φ. Thus,y /∈ f(PβCl(A)).(9)⇔ (10). LetB be any subset of Y . By (9) and Lemma 2.7, f(PβCl(f

−1(B))) ⊂rker(f(f−1(B))) ⊂ rker(B) and PβCl(f

−1(B)) ⊂ f−1(rker(A)). Con-versely, let B = f(A), where A is a subset of X. Then PβCl(A) ⊂PβCl(f

−1(B)) ⊂ f−1(rker(f(A))). Therefore, f(PβCl(A)) ⊂ rker(f(A)).(10)⇒ (1). Let V be any regular open set of Y . Then, by (10) and Lemma2.7, PβCl(f

−1(V )) ⊂ f−1(rker(V )) = f−1(V ) and PβCl(f−1(V )) = f−1(V ).(5) ⇔ (6). Let V be any regular open set in Y not containing f(x). ThenY −V is a regular closed set containing f(x). By (5), there exists a Pβ-openset U in X containing x such that f(U) ⊂ Y −U . Hence, U ⊂ X − f−1(V )and then f−1(V ) ⊂ X − U , converse follows similarly.(2)⇒ (11). Let V be any β-open set of Y. It follows from ([2], Theorem.2.4)that Cl(V ) is regular closed. By (2), f−1(Cl(V )) is Pβ-open in X.(11) ⇒ (12). This is clear since every semiopen set is β-open.(12) ⇒ (13). Let V be any subset of Y . Then Int(Cl(V )) is regular open.Therefore, Y − Int(Cl(V )) is regular closed and hence it is semiopen. By(12), X−f−1(Int(Cl(V ))) = f−1(Y − Int(Cl(V ))) = f−1(Cl(Y −Cl(V )))is Pβ-open. Hence, f

−1(Int(Cl(V ))) is Pβ-closed.(13) ⇒ (1). Let V be any regular open set of Y . By (13), f−1(V ) =f−1(Int(Cl(V ))) is Pβ-closed. It show that f is almost contra-Pβ-continuous.(2) ⇔ (14). Let V be any subset of Y . Since Cl(Int(V )) is regular closed,f−1(Cl(Int(V )) is Pβ-open. The converse follows similarly.(16) ⇒ (17). Obvious.(17) ⇒ (18). It follows from [12] the fact that θsCl(V ) = sCl(V ) for anopen set V.(18) ⇒ (19). Obvious from Lemma 2.5.(19)⇒ (1). Let V ∈ RO(Y ). Then, by (20), PβCl(f

−1(V )) ⊂ f−1(Int(Cl(V )) =f−1(V ). 2

The statement (5) is the pointwise definition of a almost contra Pβ-continuous function.

Corollary 3.9. For a function f : X → Y , the following are equivalent:

1. f is almost contra Pβ-continuous.

2. f−1(αCl(A)) is Pβ-open for every A ∈ βO(Y );

3. f−1(pCl(A)) is Pβ-open for every A ∈ SO(Y );

Independent form of (θ, s)-continuous functions in topological... 523

4. f−1(sCl(A)) is Pβ-open for every A ∈ PO(Y );

5. PβCl(f−1(B)) ⊂ f−1(θsCl(B)) for every subset B ∈ SO(Y );

6. PβCl(f−1(B)) ⊂ f−1(θsCl(B)) for every subset B ∈ PO(Y );

7. PβCl(f−1(B)) ⊂ f−1(θsCl(B)) for every subset B ∈ βO(Y ).

Proof. The equivalences of (1), (2), (3) and (4) follow from Lemma 2.5,and the equivalences of (1), (5), (6) and (7) follow from Theorem 3.8 ((1)⇔ (16)). 2

Theorem 3.10. If a function f : X → Y is an almost contra-Pβ-continuousfunction which satisfies the property PβInt(f

−1(Cl(V ))) ⊂ f−1(V ) for eachopen set V of Y , then f is Pβ-continuous.

Proof. Let V be any open set of Y . Then Cl(V ) is a semi-open set inY . Since f is almost contra-Pβ-continuous, by Theorem 3.8(8), f−1(V ) ⊂f−1(Cl(V )) = PβInt(PβInt(f

−1(Cl(V )))) ⊂ (PβInt(f−1(V )) ⊂ f−1(V ).

2

Definition 3.11. [31] A topological space is said to be PΣ if for any openset V of X and each x ∈ V , there exists a regular closed set F of X containingx such that x ∈ F ⊂ V .

Theorem 3.12. If f : X → Y is an almost contra-Pβ-continuous functionand Y is PΣ, then f is Pβ-continuous.

Proof. Soppose that V is any open set of Y . By the fact that Yis PΣ, there exists a subfamily Ω of regular closed sets of Y such thatV = ∪F | F ∈ Ω. Since f is almost contra-Pβ-continuous, then f−1(F )is Pβ-open in X for each F ∈ Ω. Therefore, f−1(V ) is Pβ-open in X. 2

Definition 3.13. [13] A topological space (X, τ) is said to be extremallydisconnected if the closure of every open set of X is open in X.

Theorem 3.14. If (Y, σ) is said to be extremally disconnected, then thefollowing properties are equivalent for a function f : X → Y :

1. f is almost contra Pβ-continuous;

2. f is almost Pβ-continuous.

524 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

Proof. (1) ⇒ (2). Let x ∈ X and U be any regular open set of Ycontaining f(x). Since Y is extremally disconnected, U is clopen and henceU is regular closed. Thus, f−1(U) is Pβ-open in X. (2)⇒ (1). Let B be anyregular closed set of Y . since Y is extremally disconnected, B is regularopen and f−1(B) is Pβ-open in X. 2We cannot relax the condition extremally disconnected in Theorem 3.14.

Example 3.15. The set of the real numberR with usual topology τ , is nota extremally disconnected space. Define the identity function f : (R, τ)→(R, τ) is almost Pβ-continuous but not almost contra-Pβ-continuous.

Definition 3.16. [30] The Pβ-frontier of a set A in a space X4, denotedPβFr(A), is defined as PβFr(A) = PβCl(A) ∩ PβCl(X −A).

Theorem 3.17. The set of all points x of X at which f : X → Y is notalmost contra-Pβ-continuous is identical with the union of the Pβ-frontierpoints of the inverse images of regular closed sets of Y containing f(x).

Proof. Suppose that f is not almost contra-Pβ-continuous at a x ∈ X.Then there exists a F ∈ RC(Y, f(x)) such that f(U)∩(Y −F ) 6= φ for everyU ∈ PβO(X,x). Then U ∩ f−1(Y −F ) 6= φ. Therefore, x ∈ PβCl(f

−1(Y −F )) = PβCl(X − f−1(F )). Since x ∈ f−1(F ), x ∈ PβCl(f

−1(F )). There-fore, x ∈ PβFr(f

−1(F )). Now suppose x ∈ PβFr(f−1(F )) for some

F ∈ RC(Y, f(x)). Assume that f is almost contra-Pβ-continuous at x ∈ X.Then there exists a U ∈ PβO(X,x) such that f(U) ⊂ F . Therefore,x ∈ U ⊂ f−1(F ), so that x ∈ PβInt(f

−1(F )) ⊂ X − PβFr(f−1(F )), a

contradiction. This means that f is not almost contra-Pβ-continuous at x.2

Definition 3.18. A function f : X → Y is said to be:

1. R-map [5] if f−1(V ) is regular closed in X for every regular closed setV of Y.

2. contra R-map [9] if f−1(V ) is regular closed in X for every regularopen set V of Y.

3. Pβ-resolute [26] if f(V ) is Pβ-closed in Y for every Pβ-closed set V ofX.

4. Pβ-irresolute [26] if f−1(V ) is Pβ-open in X for every Pβ-open set V

of Y.

Independent form of (θ, s)-continuous functions in topological... 525

Theorem 3.19. Let Xi| i ∈ ∆ be a family of topological spaces. Ifa function f : X → Q

Xi is almost contra Pβ-continuous, then Pri f :X → Xi is almost contra Pβ-continuous for each i ∈ ∆, where Pri is theprojection of

QXi onto Xi.

Proof. For a fixed i ∈ ∆, let Vi be any regular open sets of Xi. SincePri is continuous, P

−1ri (Vi) is open in

QXi, it is an R-map and hence

P−1ri (Vi) is regular open inQXi. Since f is almost contra Pβ-continuous

f−1(P−1ri (Vi)) = (Pri f)−1(Vi) is Pβ-closed in X. Therefore, Pri f isalmost contra Pβ-continuous for each i ∈ ∆. 2

Theorem 3.20. Let f : X → Y is a Pβ-resolute surjection and g : Y → Zis a function such that gf is almost contra Pβ-continuous then, g is almostcontra Pβ-continuous.

Proof. Let V be any regular open set in Z. Since g f is almost contraPβ-continuous, f

−1(g−1(V )) = (g f)−1(V ) is Pβ-closed. Since f is Pβ-resolute surjection, f(f−1(g−1(V ))) = g−1(V ) is Pβ-closed. 2

Theorem 3.21. Let f : X → Y and g : Y → Z be functions. Then thefollowing statements hold:

1. If f is almost contra Pβ-continuous and g is R-map, then gf is almostcontra-Pβ-continuous.

2. If f is almost Pβ-continuous and g is contra R-map, then gf is almostcontra-Pβ-continuous.

3. If f is Pβ-irresolute and g is almost contra Pβ-continuous., then g fis almost contra-Pβ-continuous.

Proof. The proof follows from the fact f−1(g−1(V )) = (g f)−1(V ) andthe definitions. 2

Definition 3.22. A topological space is said to be

1. Pβc-closed [30] (resp. nearly compact [20]) if every Pβ-closed (resp.regular open) cover of X has a finite subcover.

2. countably Pβc-closed [30] (resp. nearly countably compact [10, 21])if every countable cover of X by Pβ-closed (resp. regular open) setshas a finite subcover.

526 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

3. Pβc-Lindelof [30] (resp. nearly Lindelof [8]) if every cover of X byPβ-closed (resp. regular open) sets has a countable subcover.

Theorem 3.23. Let f : X → Y be an almost contra Pβ-continuous sur-jection. Then, the following statements hold:

1. If X is Pβc-closed, then Y is nearly compact;

2. If X is countably Pβc-closed, then Y is nearly countably compact;

3. If X is Pβc-Lindelof, then Y is nearly Lindelof.

Proof. (1). Let Vα : α ∈ I be any regular open cover of Y. Since f isalmost contra Pβ-continuous, then f−1(Vα) : α ∈ I is a Pβ-closed coverof X. Since X is Pβc-closed, there exists a finite subset I0 of I such thatX = ∪f−1(Vα) : α ∈ I0. Thus, we have Y = ∪Vα : α ∈ I0 and Y isnearly compact. The proofs of (2) and (3) are similar to the proof of (1).2

Definition 3.24. A topological space is said to be

1. Pβ-compact [30] (resp. S-closed [25]) if every Pβ-open (resp. regularclosed) cover of X has a finite subcover.

2. countably Pβ-compact [30] (resp. countably S-closed [1]) if everycountable cover of X by Pβ-open (resp. regular closed) sets has afinite subcover.

3. Pβ-Lindelof [30] (resp. S-Lindelof [18]) if every cover of X by Pβ-open(resp. regular closed) sets has a countable subcover.

Theorem 3.25. Let f : X → Y be an almost contra Pβ-continuous sur-jection. Then the following statements hold:

1. If X is Pβ-compact, then Y is S-closed;

2. If X is countably Pβ-compact, then Y is countably S-closed;

3. If X is Pβ-Lindelof, then Y is S-Lindelof.

Proof. (1). Let Vα : α ∈ I be any regular closed cover of Y . Since fis almost contra Pβ-continuous, then f−1(Vα) : α ∈ I is a Pβ-open coverof X. Since X is Pβ-compact, there exists a finite subset I0 of I such thatX = ∪f−1(Vα) : α ∈ I0. Thus, we have Y = ∪Vα : α ∈ I0 and Y isS-closed. 2

Independent form of (θ, s)-continuous functions in topological... 527

4. Separation Axioms

Definition 4.1. A topological space (X, τ) is said to be:

1. Pβ − T0 [30] (resp. Pβ − T1 [30]) if for any distinct points x and y inX, there is a Pβ-open U in X containing x but not y or (resp. and)a Pβ-open set V in X containing y but not x.

2. Pβ − T2 [30] if for every pair of distinct points x and y, there existtwo Pβ-open sets U and V such that x ∈ U, y ∈ V and U ∩ V = φ.

From above definitions, we have following implications:

Pβ − T2 ⇒ Pβ − T1 ⇒ Pβ − T0.

Example 4.2. LetX = a, b, c, τ = φ, a, b,X. Then (X, τ) is Pβ−T1but not Pβ − T2 space.

Example 4.3. LetR be the set of real numbers with the right order topol-ogy [23] generated by the basis Sa = x : x > a, a ∈ Q, where Q is theset of rational numbers. Then (X, τ) is Pβ − T0 but not Pβ − T1 space .

Theorem 4.4. Let X be a topological space. Suppose that for each pairof distinct points x1 and x2 in X, there exists a function f of X intoa Uryshon space Y 4suchthatf(x1) 6= f(x2). Moreover, let f be almostcontra-Pβ-continuous at x1 and x2. Then X is Pβ − T2.

Proof. Let x1 and x2 be any distinct points in X. Suppose that thereexist a Uryshon space Y and a function f : X → Y such that f(x1) 6= f(x2)and f is almost contra-Pβ-continuous at x1 and x2. Let w1 = f(x1) andw2 = f(x2). Then w1 6= w2, since Y is Uryshon, there exist open sets Uand V containing w1 and w2, respectively such that Cl(U) ∩ Cl(V ) = φ.Since f is almost contra-Pβ-continuous at x1 and x2, there exist Pβ-opensets A and B containing x1 and x2 respectively, such that f(A) ⊂ Cl(U)and f(B) ⊂ Cl(V ). So A ∩B = φ since Cl(U) ∩ Cl(V ) = φ. 2

Corollary 4.5. If f is almost contra-Pβ-continuous injection of a topolog-ical space X into a Uryshon space Y , then X is Pβ − T2.

Definition 4.6. A topological space (X, τ) is said to be:

528 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

1. Pβ-normal [30] if for each pair of nonempty disjoint closed sets F1and F2, there exist disjoint Pβ-open sets U and V containing F1 andF2, respectively.

2. Weakly Hausdorff [22] (briefly weakly-T2) if every point of X is anintersection of regular closed sets of X.

3. s-Urysohn [3] if for each pair of distinct points x and y in X, thereexist U ∈ SO(X,x) and V ∈ SO(X, y) such that Cl(U)∩Cl(V ) = φ.

4. Ultra Hausdorff [24] if for each pair of distinct points x and y in X,there exist disjoint clopen sets U and V containing x and y, respec-tively.

Theorem 4.7. If f is an almost contra-Pβ-continuous injection of a topo-logical space X into a s-Urysohn space Y, then X is Pβ − T2.

Proof. Since f is injective, we have f(x) 6= f(y) for any distinct pointsx and y in X. Since Y is s-Urysohn, there exist U1 ∈ SO(X,x) andU2 ∈ SO(X, y) such that Cl(U1) ∩ Cl(U2) = φ. Since f is almost contra-Pβ-continuous, there exist Pβ-open setsV1 and V2 in X containing x andy, respectively such that f(V1) ⊂ Cl(U1) and f(V2) ⊂ Cl(U2). Therefore,V1 ∩ V2 = φ. 2

Theorem 4.8. If f is an almost contra-Pβ-continuous injection of a topo-logical space X into a Ultra Hausdorff space Y , then X is Pβ − T2.

Proof. Let x and y be distinct points in X. Since f is injective andY is Ultra Hausdorff, there exist disjoint clopen sets U1 and U2 containingf(x) and f(y), respectively. Then U1 and U2 are regular closed set in Y .Since f is almost contra-Pβ-continuous, there exist Pβ-open sets V1 andV2 in X containing x and y respectively, such that f(V1) ⊂ Cl(U1) andf(V2) ⊂ Cl(U2). Therefore, V1 ∩ V2 = φ. 2

Theorem 4.9. If f is an almost contra-Pβ-continuous injection of a topo-logical space X into a weakly Hausdorff space Y , then X is Pβ − T1.

Independent form of (θ, s)-continuous functions in topological... 529

Proof. Since Y is weakly Hausdorff and f is injective, for any distinctpoints x1 and x2 of X, there exist V1, V2 ∈ RC(X) such that f(x1) ∈ V1,f(x2) /∈ V1, f(x2) ∈ V2, and f(x1) /∈ V2. Since f is almost contra-Pβ-continuous, f−1(V1) and f−1(V2) are Pβ-open sets and x1 ∈ f−1(V1), x2 /∈f−1(V1), x2 ∈ f−1(V2), and x1 /∈ f−1(V2). 2

Theorem 4.10. If f : X → Y is a almost contra-Pβ-continuous closedinjection and Y is normal, then X is Pβ-normal.

Proof. Let F1 and F2 be disjoint closed subsets of X. Since f is closedand injective, f(F1) and f(F2) are disjoint closed subsets of Y . Since Yis normal, there exist disjoint open sets V1 and V2 containing f(F1) andf(F2), respectively. Then there exist open sets Ui such that f(Fi) ⊂ Ui ⊂Cl(Ui) ⊂ Vi. Hence, Fi ⊂ f−1(Cl(Ui)), f

−1(Cl(Ui)) ∈ PβO(X) for i=1,2and f−1(Cl(U1)) ∩ f−1(Cl(U2)) = φ. 2

Definition 4.11. A topological space is called a Pβc-space if the union ofany two Pβ-closed sets is Pβ-closed set.

From Definition 4.11, it follows that in a Pβc-space, Pβ-open sets form atopology.

Example 4.12. 1. Let X = [−1, 1] ⊂ R, define either-or topology onX[23] by τ = U : 0 /∈ U or (−1, 1) ⊂ U. Here, PβO(X) =(−1, 1), (−1, 1], [−1, 1), [−1, 1] ∪ A : 0 /∈ A. Then (X, τ) is aPβc-space.

2. Let X be the set of positive integers and P = 2k − 1, 2k : k ∈ Nbe a partition of X. Then the partition P together with the emptyset φ is a basis for a topology τ on X, known as partition topologyon X[23]. Every subset in this space is pre-open and hence Pβ-open.Thus, (X, τ) is a Pβc-space.

3. In Example 4.3, let U = 1, 2, 3, 4, 6, 8, 10, ... and V = 1, 2, 3, 5, 7, 9, 11, ....Then U and V are Pβ-open sets but U ∩ V = 1, 2, 3 is not Pβ-openset. Thus, (X, τ) is not a Pβc-space.

Theorem 4.13. If f, g : X → Y are almost contra-Pβ-continuous func-tions, X is a Pβc-space and Y is s-Urysohn, then E = x ∈ X| f(x) = g(x)is Pβ-closed in X.

530 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

Proof. If x ∈ X−E, then f(x) /∈ g(x). Since Y is s-Urysohn, there existV1 ∈ SO(Y, f(x)) and a V2 ∈ SO(Y, g(x)) such that Cl(V1) ∩ Cl(V2) = φ.By the fact that f and g are almost contra Pβ-continuous functions, thereexist Pβ-open sets U1 and U2 in X containing x such that f(U1) ⊂ Cl(V1)and g(U2) ⊂ Cl(V2). We put U = U1 ∩ U2. Then U is Pβ-open in X.Thus, f(U) ∩ g(U) = φ. It follows that x /∈ PβCl(E). This shows that Eis Pβ-closed in X. 2

Theorem 4.14. Let f : X → Y be a function and g : X → Y the graphfunction, given by g(x) = (x, f(x)) for every x ∈ X. Then f is almostcontra-Pβ-continuous if g is almost contra-Pβ-continuous.

Proof. Let x ∈ X and V be a regular open subset of Y containingf(x). Then X×V is regular open. Since g is almost contra-Pβ-continuous,g−1(X × V ) = f−1(V ) is Pβ-closed. 2

Theorem 4.15. [26] If X is T1-space, then PO(X) = PβO(X).

Theorem 4.16. Let f : X → Y be function such that X is T1-space,then f is almost contra-Pβ-continuous if and only if f is almost contraprecontinuous.

Proof. Proof follows from Theorem 4.15. 2

5. Pβ-closed, Pβ-regular, strongly contra-Pβ-closed

Recall that for a function f : X → Y , the subset G(f) = (x, f(x)) : x ∈X ⊂ X × Y is said to be the graph of f .

Definition 5.1. The graph G(f) of a function f : X → Y is said to bePβ-closed (resp. Pβ-regular, strongly contra-Pβ-closed) if for each (x, y) ∈(X × Y ) − G(f), there exists a U ∈ PβO(X,x) (resp. U ∈ PβC(X,x),U ∈ PβO(X,x)) and a V ∈ O(Y, y) (resp. V ∈ RO(Y, y), V ∈ RC(Y, y))such that (U × V ) ∩G(f) = φ.

Lemma 5.2. The graph G(f) of a function f : X → Y , the followingstatements are equivalent:

1. G(f) is Pβ-closed (resp. Pβ-regular, strongly contra-Pβ-closed);

2. for each (x, y) ∈ (X × Y ) − G(f), there exists a U ∈ PβO(X,x)(resp. U ∈ PβC(X,x), U ∈ PβO(X,x)) and a V ∈ O(Y, y) (resp.V ∈ RO(Y, y), V ∈ RC(Y, y)) such that f(U) ∩ V = φ.

Independent form of (θ, s)-continuous functions in topological... 531

Proof. The proof follow from the fact that for any subsets A ⊂ X andB ⊂ Y , (A×B) ∩G(f) = φ if and only if f(A) ∩B = φ. 2

Theorem 5.3. If f : X → Y is a function with a Pβ-closed graph, thenfor each x ∈ X, f(x) = ∩Cl(f(U)) | U ∈ PβO(X,x).

Proof. Suppose there exists an element y 6= f(x) such thaty ∈ ∩Cl(f(U)) | U ∈ PβO(X,x). Then y ∈ Cl(f(U)) for every U ∈PβO(X,x). So V ∩ f(U) 6= φ, for every V ∈ O(Y, y), which contradicts thehypothesis that f is a function with a Pβ-closed graph. 2

Theorem 5.4. If f : X → Y is almost contra-Pβ-continuous and Y isHausdroff, then G(f) is Pβ-closed in X × Y .

Proof. Let (x, y) ∈ (X × Y ) − G(f), then y 6= f(x) and there existdisjoint open sets V and W such that y ∈ V and f(x) ∈ W . Then f(x) /∈Y −Cl(W ), since Y −Cl(W ) is a regular open set containing V , it followsthat f(x) /∈ rker(V ) and hence x /∈ f−1(rker(V )). Then, by Theorem3.8(9), x /∈ PβCl(f

−1(V )). Therefore, (x, y) ∈ (X − PβCl(f−1(V ))) × V .

2

Theorem 5.5. If a function f : X → Y has Pβ-closed graph and X isPβc-space, then f−1(K) is Pβ-closed for every compact subset K of Y .

Proof. Let K be a compact subset of Y and x ∈ X− f−1(K). Then foreach y ∈ K, (x, y) ∈ (X × Y ) − G(f). So there exists a Uy ∈ PβO(X,x)and an open set Vy of Y containing y such that f(Uy)∩Vy = φ. The familyVy| y ∈ K is an open cover of K and hence there is a finite subcoverVyi | i = 1, ...., n. Let U = ∩Uyi | i = 1, ...., n. Then U ∈ PβO(X,x)and f(U) ∩K = φ. Hence, U ∩ f−1(K) = φ, which proves that f−1(K) isPβ-closed in X. 2

Theorem 5.6. If an injective mapping f : X → Y has a Pβ-regular graphG(f), then X is Pβ − T0.

beginproof Let x1 and x2 be any two distinct points of X. Then, (x1, f(x2)) ∈(X × Y ) −G(f). Since G(f) is Pβ-regular, there exists a Pβ-closed set Uin X containing x1 and V ∈ RO(Y, f(x2)) such that f(U) ∩ V = φ. Thus,U ∩ f−1(V ) = φ. Therefore, x2 /∈ U . 2

Theorem 5.7. If f : X → Y has a Pβ-regular graph G(f) and f is surjec-tive, then Y is weakly Hausdroff.

532 Brij K. Tyagi, Sumit Singh and Manoj Bhardwaj

Proof. Let y1 and y2 be any two distinct points of Y. Since Y is sur-jective, there exists a point x ∈ X such that f(x) = y1 for some x ∈ Xand (x, y2) ∈ (X × Y )−G(f). Since f has a Pβ-regular graph, there exista U ∈ PβC(X,x) and a V ∈ RO(Y, y2) such that f(U) ∩ V = φ. Thus,y2 /∈ (Y − V ) ∈ RC(Y ) and y1 ∈ (Y − V ). 2

Theorem 5.8. Let f : X → Y be a mapping with a Pβ-closed graph.

1. if f is injective, then X is Pβ − T1;

2. if f is surjective, then Y is T1.

Proof. (1). Let x1 and x2 be any two distinct points of X. Then,(x1, f(x2)) ∈ (X × Y ) − G(f). Then there exists a Pβ-open set U in Xcontaining x1 and V ∈ O(Y, f(x2)) such that f(U) ∩ V = φ. Since x2 ∈f−1(V ), x2 /∈ U . Therefore, U is a Pβ-open set containing x1 but not x2.(2). Let y1 and y2 be any two distinct points of Y. Since Y is surjective,there exists a x ∈ X such that f(x) = y1. Then (x, y2) ∈ (X × Y )−G(f).Since f has a Pβ-closed graph, there exist a U ∈ PβO(X,x) and an openset V of Y containing y2 such that f(U) ∩ V = φ. Since y1 = f(x) andx ∈ U , y1 /∈ V . 2

Theorem 5.9. Let f : X → Y be a mapping with strongly contra-Pβ-closed graph G(f). If f is an almost contra-Pβ-continuous injection, then Xis Pβ − T2.

Proof. Let x and y be any distinct points of X. Since f is injective, wehave f(x) 6= f(y). Then, we have (x, f(y)) ∈ X × Y − G(f). Since G(f)is strongly contra-Pβ-closed, there exist a U ∈ PβO(X,x) and a regularclosed set V containing f(y) such that f(U) ∩ V = φ. Since f is almostcontra-Pβ-continuous, f

−1(V ) ∈ PβO(X, y). 2

6. Connectedness

Definition 6.1. A space X is said to be

1. hyperconnected [13] if every nonempty open subset of X is dense inX.

2. Pβ-ultra-connected [30] if every two nonempty Pβ-closed subsets of Xintersect.