Download Applications of a Conditional Preopen Set in Bitopological Spaces

Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 44, 2173 - 2187
Applications of a Conditional Preopen Set
in Bitopological Spaces
Alias B. Khalaf
Department of Mathematics, College of Science
University of Duhok, Kurdistan-Region, Iraq
[email protected]
Hardi N. Aziz
Department of Mathematics, College of Science
University of Sulaimani, Kurdistan-Region, Iraq
[email protected]
Abstract. The aim of this paper is to introduce a new type of sets in
bitopological spaces called (i, j)-Ps -open sets and study its basic properties.
Also, by utilizing (i, j)-Ps -open sets, we introduce some strong types of precontiuity and almost precontinuity in bitopological spaces. Moreover a type of
contra-precontinuity is defined and characterized.
Mathematics Subject Classification: Primary: 54A05, 54A10; Secondary: 54E55
Keywords: (i, j)-Ps -open, semi-open, preopen, precontinuous functions,
(i, j)-Ps -continuous functions, almost (i, j)-Ps -continuous functions, contra (i, j)Ps -continuous functions
1. Introduction
In 1963 Kelley J. C. [15] was first introduced the concept of bitopological
spaces, where X is a nonempty set and τ1 , τ2 are topologies on X. In 1982
Mashhour et al [18] introduced the concept of preopen sets in topological
spaces. By using this concept, several authers defined and studied stronger or
weaker types of topological concept such as closedness [10], continuity [18] and
compactness [19].
In this paper, we introduce the concept of a conditional preopen set in a
bitopological space, and we find basic properties and relationships with other
concepts of sets. Throughout this paper, (X, τ1 , τ2 ) is a bitopological space
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and if A ⊆ Y ⊆ X, then i-Int(A) and i-Cl(A) denote respectively the interior
and closure of A with respect to the topology τi on X and i-IntY (A), i-ClY (A)
denote respectively the interior and the closure of A with respect to the induced
toplogy on Y.
2. Preliminaries
We recall the following definitions and results:
Definition 2.1. A subset A of a space (X, τ ) is called:
1. preopen [18], if A ⊆ Int(Cl(A)).
2. semi-open [16], if A ⊆ Cl(Int(A)),
3. α-open [22], if A ⊆ Int(Cl(Int(A))).
4. regular open [26], if A = Int(Cl(A)).
The family of all preopen (resp., semi-open, α-open, regular open) sets in X
is denoted by P O(X) (resp., SO(X), αO(X), RO(X)). The complement of
a preopen (resp., semi-open, α-open, regular open) set is said to be preclosed
(resp., semi-closed, α-closed, regular closed) and is denoted by P C(X) (resp.,
SC(X), αC(X), RC(X)). The intersection of all preclosed (resp., semi-closed,
α-closed) sets of X containing A is called preclosure (resp., semi-closure, αclosure) of A and is denoted by pClA (resp., sClA, αClA). The union of all
preopen (resp., semi-open, α-open) sets of X contained in A called preinterior
(resp., semi-interior, α-interior) of A and is denoted by pIntA (resp., sIntA,
αIntA). A subset A of a space X is called regular semi-open [3], if A =
sInt(sCl(A)). A subset A of a space X is called δ-open [28], if for each x ∈ A,
there exists an open set G such that x ∈ G ⊆ Int(Cl(G)) ⊆ A). A subset
A of a space X is called θ-semi-open [14] (resp., semi-θ-open [5]) if for each
x ∈ A, there exists a semi-open set G such that x ∈ G ⊆ Cl(G) ⊆ A (resp.,
x ∈ G ⊆ sCl(G) ⊆ A). The family of all regular semi-open (resp., δ-open,
θ-semi-open, semi-θ-open) sets in X is denoted by RSO(X) (resp., δO(X),
θSO(X), SθO(X)). The complemet of regular semi-open (resp., δ-open, θsemi-open, semi-θ-open) sets in X is called regular semi-closed (resp., δ-closed,
θ-semi-closed, semi-θ-closed). A subset A of a topological space (X, τ ) is called
η-open [25], if A is a union of δ-closed sets. The complement of η-open sets
is called η-closed. A subset A of a space (X, τ ) is said to be preregular [21],
if A is both preopen and preclosed. A subclass τ ∗ of subsets of X is called
a supratopology on X [12], if X, φ ∈ τ ∗ and τ ∗ is closed under arbitrary
unions. In a bitopological space (X, τ1 , τ2 ), a point x of A is said to be in
the ij-θ-interior of A, denoted by ij-Intθ (A), if there exists U ∈ τi such that
x ∈ U ⊆ j-Cl(U) ⊆ A. A is said to be ij-θ-open [11], if A = ij-Intθ (A),
the complement of an ij-θ-open set is called ij-θ-closed. A subset A of X
is said to be (τi , τj )-generalized pre-open) set (briefly,(τi , τj )-gp-open[29], if
F ⊆ pIntj (A)), whenever F ⊆ A and F is τi -closed set. The family of all
(τi , τj )-gp-open set of X denoted by (τi , τj )-GP O(X).
We recall the following definitions on a space X.
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Definition 2.2. A topological space X is called,
1. Submaximal [13], if every dense set of X is open in X. Equivalently if
every preopen set is open.
2. Extermally disconnected [5], if Cl(U) ∈ τ for every U ∈ τ .
3. Hyperconnected [8], if every nonempty open subset of X is dense.
4. Locally indiscrete [8], if every open subset of X is closed.
Lemma 2.3. [6] A space (X, τ ) is hyperconnected if and only if RO(X) =
{φ, X}.
Theorem 2.4. [8] For a topological space (X, τ ) the following conditions are
equivalent:
1. X is locally indiscrete.
2. Every subset of X is preopen.
Remark 2.5. From the above theorem it is obvious that if X is locally indiscrete space, then every semi-open subset of X is closed and hence every
semi-closed subset of X is open.
Definition 2.6. [23] A space X is said to be semi-regular if for any open set
U of X and each point x ∈ U, there exists a regular open set V of X such that
x ∈ V ⊆ U.
Theorem 2.7. [1] Let Y be any subspace of the topological space (X, τ ). Then
the following statements are true:
1. If A ∈ P O(X) and A ⊆ Y , then A ∈ P O(Y ).
2. If F ∈ SC(X) and F ⊆ Y , then F ∈ SC(Y ).
3. If F ∈ SC(Y ) and Y ∈ SC(X) then F ∈ SC(X).
4. If G ∈ τ , and Y ∈ P O(X) then G ∩ Y ∈ P O(X).
From the above theorem it is clear that, if (Y, τY ) is a subspace of a space
(X, τ ), F ∈ P C(X) and F ⊆ Y then F ∈ P C(Y ).
Lemma 2.8. [10] For any topological space (X, τ ) the following statements
are true:
1. If A ∈ P O(X) and B ∈ SO(X), then A ∩ B ∈ P O(B).
2. If Y is an open subspace of the space X and if F ∈ SC(X), then F ∩ Y ∈
SC(Y ).
Lemma 2.9. [8] If R ∈ RO(X) and P ∈ P O(X), then R ∩ P ∈ RO(P )
Theorem 2.10. [29] In a bitopological space (X, τ1 , τ2 ). If A ∈ j-P O(X),
then A ∈ (i, j)-GP O(X).
3. (i, j)-Ps -open sets
In this section, we introduce and define a new type of sets in bitopological
spaces and we find some of its properties.
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Definition 3.1. A subset A of a bitopological space (X, τ1 , τ2 ) is said to be
(i, j)-Ps -open, if A is a j-pre-open set and for all x in A, there exist an i-semiclosed set F such that x ∈ F ⊆ A. A subset B of X is called (i, j)-Ps - closed if
and only if B c is (i, j)-Ps - open. The family of (i, j)-Ps -open (resp., (i, j)-Ps closed) subset of x is denoted by (i, j)-Ps O(X) (resp.,(i, j)-Ps C(X)).
Corollary 3.2. A subset A of a bitopological space X is (i, j)-Ps -open, if A
is j-pre open set and it is a union of i-semi-closed sets. This means that
A = ∪Fγ , where A is a j-preopen and Fγ is an i-semi-closed set for each γ.
Corollary 3.3. A subset B of a bitopological space X is (i, j)-Ps - closed if
and only if B is j-pre closed and it is an intersection of i-semi-open sets.
Proof. Follows from Corollary 3.2 taking A = B c .
It is clear that the union of any family of (i, j)-Ps -open sets in a space X is also
(i, j)-Ps -open. The intersection of two (i, j)-Ps -open sets is not (i, j)-Ps -open
set in general, as shown in the following example:
Example 3.4. Let X = {a, b, c} , τ1 = {φ, {a}, {c}, {a, c}, X}, τ2 = {φ, {c}, {a, b}, X},
then (i, j)-PS O(X) = {φ, {a}, {c}, {a, b}, {b, c}, {a, c}, X}, if A = {a, b} and
B = {b, c} then A ∩ B = {b} ∈
/ (i, j)-PS O(X)
From the above example we notice that the family of all (i, j)-Ps -open sets
is a supratopology and it is not a topology in general.
A space X is said to be semi-T1 [17] if for every two distinct points of X there
exist two semi-open sets each one contains one of the points but not the other.
Lemma 3.5. [17] A space X is semi-T1 if and only if every singleton {x} is
semi-closed in X.
Proposition 3.6. Let (X, τ1 , τ2 ) be a bitopological space if (X, τi ) is a semiT1 -space , then (i, j)-Ps O(X) = j-P O(X).
Proof. Let A be any subset of a space X and A is j-pre-open set, if A = φ,
then A ∈ (i, j)-Ps O(X), if A = φ, now let x ∈ A, since (X, τ1 ) is semi-T1 space, then by Theorem 3.5 every singleton is i-semi-closed set , and hence
x ∈ {x} ⊆ A, therefore A ∈ (i, j)-Ps O(X), hence j-P O(X) ⊆ (i, j)-Ps O(X)
but (i, j)-Ps O(X) ⊆ j-P O(X) generally, thus (i, j)-Ps O(X) = j-P O(X)
Theorem 3.7. In a bitopological space (X, τ1 , τ2 ). If the space (X, τi ) is hyperconnected, then (i, j)-Ps O(X) = {φ, X}.
Proof. Suppose that a subset A of a bitopological space X is an (i, j)-Ps −open
set. If A = φ, then for each x ∈ A there exist i-semi-closed set F such
that x ∈ F ⊆ A. Since F is i-semi-closed, then i-int(i-Cl(F )) ⊆ F , so
i-sCl(i-int(i-Cl(F )) ⊆ F . Since (X, τi ) is hyperconnected, then every nonempty preopen subset of X is semi-dense [16]. This implies that i−sCl(i-int(iCl(F ))) = X ⊆ F , so F = X. Hence (i, j)-Ps O(X) = {φ, X}.
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Theorem 3.8. In a bitopological space (X, τ1 , τ2 ) if a space (X, τi ) is locally
indiscrete then (i, j)-Ps O(X) ⊆ τi .
Proof. Let V ∈ (i, j)-Ps O(X) then V ∈ j-P O(X) and for each x ∈ V , there
exist i-semi-closed F in X such that x ∈ F ⊆ V , by Remark 2.5, F is i-open,
it follows that V ∈ τi , and hence (i, j)-Ps O(X) ⊆ τi .
The converse of Theorem 3.8 is not true in general, as shown in the following
example:
Example 3.9. Let X = {a, b, c}, τ1 = {φ, {a}, {b, c}, X} and τ2 = {φ, {b}, X},
then (i, j)-Ps O(X) = {φ, {b, c}, X} and it is clear that (X, τ1 ) is locally indiscrete but tau1 is not a subset of (i, j)-Ps O(X)
Remark 3.10. In any bitopological space (X, τ1 , τ2 ) we have:
1. If τi (resp. j-P O(X)) is indiscrete, then (i, j)-Ps O(X) is also indiscrete .
2. If (i, j)-Ps O(X) is discrete, then j-P O(X) is also discrete.
Lemma 3.11. [1] For any two topological spaces X and Y . If A ⊆ X and
B ⊆ Y then,
1. pIntX×Y (A × B) = pIntX (A) × pIntY (B)
2. sClX×Y (A × B) = sClX (A) × sClY (B)
Proposition 3.12. Let X1 , X2 be two bitopological space and X1 × X2 be the
bitopological product , let A1 ∈ (i, j)-Ps O(X1) and A2 ∈ (i, j)-Ps O(X2) then
A1 × A2 ∈ (i, j)-Ps O(X1 × X2 )
Proof. Let (x1 , x2 ) ∈ A1 × A2 , then x1 ∈ A1 and x2 ∈ A2 . Since A1 ∈ (i, j)Ps O(X1) and A2 ∈ (i, j)-Ps O(X2 ), then A1 ∈ j-P O(X1) and A2 ∈ j-P O(X2)
also, there exist F1 ∈ i-SC(X1 ) and F2 ∈ i-SC(X2 ) such that x1 ∈ F1 ⊆ A1
and x2 ∈ F2 ⊆ A2 . Therefore (x1 , x2 ) ∈ F1 × F2 ⊆ A1 × A2 . Since A1 ∈ jP O(X1) and A2 ∈ j-P O(X2), then by Theorem 3.11 part(1), A1 × A2 =
j-pIntx1 (A1 ) × j-pIntx2 (A2 ) = j-pIntx1 ×x2 (A1 × A2 ), hence A1 × A2 ∈ jP O(X1 × X2 ).Since F1 ∈ i-SC(X1 ) and F2 ∈ i-SC(X2 ), then by Theorem 3.11
part (2), we get F1 × F2 = i-sClx1 (F1 ) × i-sClx2 (F2 ) = i-sClx1 ×x2 (F1 × F2 ).
Hence F1 × F2 ∈ i-SC(X1 × X2 ), so A1 × A2 ∈ (i, j)-Ps O(X).
Theorem 3.13. For any bitopological space (X, τ1 , τ2 ), if A ∈ j-P O(X) and
either A ∈ i-ηO(X) or A ∈ i-SθO(X), then A ∈ (i, j)-Ps O(X)
Proof. Let A ∈ i-ηO(X) and A ∈ j-P O(X), if A = φ , then A ∈ (i, j)-Ps O(X).
If A = φ, since A ∈ i-ηO(X) , then A = ∪Fγ , where Fγ ∈ i-δC(X) for each γ.
Since i-δC(X) ⊆ i-SC(X), so Fγ ∈ i-SC(X) for each γ, and A ∈ j-P O(X).
Hence, by Corollary 3.2, A ∈ (i, j)-Ps O(X)
On the other hand suppose that A ∈ i-SθO(X) and A ∈ j-P O(X). If A = φ,
then A ∈ (i, j)-Ps O(X). If A = φ, since A ∈ i-SθO(X), then for each x ∈ A,
there exists an i-semi-open set U such that x ∈ U ⊆ i-sCl(U) ⊆ A. This
implies that x ∈ i-sCl(U) ⊆ A and A ∈ j-P O(X). Therefore, by Corollary
3.2, A ∈ (i, j)-Ps O(X)
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Theorem 3.14. Let Y be a subspace of a bitopological space (X, τ1 , τ2 ), if
A ∈ (i, j)-Ps O(X) and A ⊆ Y , then A ∈ (i, j)-Ps O(Y )
Proof. Let A ∈ (i, j)-Ps O(X), then A ∈ j-P O(X) and for each x ∈ A, there
exists an i-semi-closed set F in X such that x ∈ F ⊆ A. Since A ∈ j-P O(X)
and A ⊆ Y , then by Theorem 2.7, A ∈ j-P O(Y ), and since F ∈ i-SC(X) and
F ⊆ Y , then by Theorem 2.7, F ∈ i-SC(Y ). Hence, A ∈ (i, j)-Ps O(Y ).
Corollary 3.15. Let (X, τ1 , τ2 ) be a bitopological space , A and Y be two subsets of X such that A ⊆ Y ⊆ X, Y ∈ RO(X, τj ) and Y ∈ RO(X, τi ). Then
A ∈ (i, j)-Ps O(Y ) if and only if A ∈ (i, j)-Ps O(X)
Proposition 3.16. Let Y be a subspace of a bitopological space (X, τ1 , τ2 ). If
A ∈ (i, j)-Ps O(Y ) and Y ∈ i-SC(X), then for each x ∈ A, there exists an
i-semi-closed set F in X such that x ∈ F ⊆ A.
Proof. Let A ∈ (i, j)-PS O(Y ), then A ∈ j-P O(Y ) and for each x ∈ A there
exists an i-semi-closed set F in Y such that x ∈ F ⊆ A. Since Y ∈ i-SC(X),
so by Theorem 2.7, F ∈ i-SC(X), which completes the proof.
Proposition 3.17. Let A and Y be any subsets of a bitopological space X. If
A ∈ (i, j)-Ps O(X), Y ∈ RO(X, τj ) and Y ∈ RO(X, τi ), then A ∩ Y ∈ (i, j)Ps O(X).
Proof. Let A ∈ (i, j)-PS O(X), then A ∈ j-P O(X) and A = ∪Fγ ,where Fγ ∈ iSC(X) for each γ, then A ∩ Y = ∪Fγ ∩ Y = ∪(Fγ ∩ Y ). Since Y ∈ RO(X, τj ),
then Y is j−open, by Theorem 2.7, A ∩ Y ∈ j-pO(X) . Since Y ∈ RO(X, τi ),
then Y ∈ i-SC(X) and hence Fγ ∩ Y ∈ i-SC(X), for each γ. Therefore by
Corollary 3.2, A ∩ Y ∈ (i, j)-Ps O(X).
Proposition 3.18. Let A and Y be any subsets of a bitopological space X. If
A ∈ (i, j)-PS O(X), Y ∈ RO(X, τi ) and Y ∈ RO(X, τj ), then A ∩ Y ∈ (i, j)PS O(Y ).
Proof. Let A ∈ (i, j)-PS O(X), then A ∈ j-P O(X) and A = ∪Fγ where Fγ ∈ iSC(X) for each γ, then A∩Y = ∪Fγ ∩Y = ∪(Fγ ∩Y ). Since Y ∈ RSO(X, τj ),
then Y ∈ j-SO(X) [27] and by Theorem 2.8, A ∩ Y ∈ j-P O(Y ). Since
Y ∈ RSO(X, τi ) then Y ∈ i-SC(X) [27]. Hence Fγ ∩ Y ∈ i-SC(X) for each
γ. Since Fγ ∩ Y ⊆ Y and Fγ ∩ Y ∈ i-SC(X) for each γ, then by Theorem 2.7,
Fγ ∩ Y ∈ i-SC(Y ). Therefore, by Corollary 3.2, A ∩ Y ∈ (i, j)-Ps O(Y ).
Proposition 3.19. If Y is an i-open and j-open subspace of a bitopological
space X and A ∈ (i, j)-Ps O(X), then A ∩ Y ∈ (i, j)-Ps O(Y ).
Proof. Let A ∈ (i, j)-Ps O(X), then A ∈ j-P O(X) and A = ∪Fγ where Fγ ∈ iSC(X) for each γ, then A ∩ Y = ∪Fγ ∩ Y = ∪(Fγ ∩ Y ). Since Y is j-open
subspace of X, then Y ∈ j-SO(X) and hence by Theorem 2.8 Part (1), A∩Y ∈
j-P O(Y ). Since Y is an i-open subspace of X, then by Theorem 2.8 Part(2),
Fγ ∩ Y ∈ i-SC(Y ) for each γ. Therefore, by Corollary 3.2, A ∩ Y ∈ (i, j)Ps O(Y ).
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Corollary 3.20. If either Y ∈ RSO(X, τj ) and Y ∈ RSO(X, τi) or Y is an
i-open and j-open subspace of a bitopological space X, and A ∈ (i, j)-Ps O(X),
then A ∩ Y ∈ (i, j)-Ps O(Y ).
In any bitopological space (X, τ1 , τ2 ) , (1, 2)-Ps O(X) is incomparable with
(2, 1)-Ps O(X), as it is shown in the following example:
Example 3.21. Let X = {a, b, c}, τ1 = {φ, {c}, {a, b}, X}, τ2 = {φ, {a}, {c}, {a, c}, X},
then (1, 2)-PsO(X) = {φ, {c}, X} and (2, 1)-PsO(X) = {φ, {a}, {c}, {a, b}, {b, c}, {a, c}, X}.
It is clear that {a} ∈ (2, 1)-PsO(X), but {a} ∈
/ (1, 2)-Ps O(X)
Proposition 3.22. Let (X, τ1 , τ2 ) be a bitopological space and A be a subset
of X. If A is both i-regular semi-open and j-preregular set, then A is both
(i, j)-Ps -open and (i, j)-Ps - closed set.
Proof. Suppose that A is both i-regular semi-open and j-preregular set, then
A is both i-semi-closed [27] and j-preopen set. Hence A is (i, j)-Ps - open.
Again since A is both i-regular semi-open and j-preregular set, then A is both
i-semi-open [27] and j-pre closed set, hence A is (i, j)-Ps - closed set.
Proposition 3.23. If Y is j-pre regular and i-pre regular subspace of a bitopological space X and B ∈ (i, j)-Ps C(X), then B ∩ Y ∈ (i, j)-Ps C(Y )
Proof. Let B ∈ (i, j)-Ps C(X), then B is j-pre closed and B = ∩Vγ , where Vγ
is an i-semi-open set in X for each γ. So B ∩ Y = ∩Vγ ∩ Y = ∩(Vγ ∩ Y ),
since Y is i-pre regular subspace of X, then Y is i-preopen. By Theorem 2.8,
Vγ ∩ Y ∈ i-SO(Y ) for each γ, and since Y is j-pre regular subspace of X, then
Y is j-pre closed. Hence B ∩ Y ∈ j-P C(X), but B ∩ Y ⊆ Y , by Theorem ??,
B ∩ Y ∈ j-P C(Y ). Therefore, by Proposition 3.3, B ∩ Y ∈ (i, j)-Ps C(Y ).
Proposition 3.24. If Y is i-preregular and j-preregular subspace of a bitopological space X, and B ∈ (i, j)-Ps O(X), then B ∩ Y ∈ (i, j)-Ps O(Y ).
Definition 3.25. If A is a subset of a bitopological space (X, τ1 , τ2 ), then the
(i, j)-Ps -interior ((i, j)-Ps Int(A)), the (i, j)-Ps -closure ((i, j)-Ps Cl(A)) and the
(i, j)-Ps -boundary ((i, j)-Ps Bd(A)) of A are defined as follows:
1. (i, j)-Ps Cl(A) = ∩{F : A ⊆ F , X − F ∈ (i, j)-Ps O(X)}.
2. (i, j)-Ps Int(A) =∪{U : U ⊆ A, U ∈ (i, j)-Ps O(X)}.
3. (i, j)-Ps Bd(A)=(i, j)-Ps Cl(A) − (i, j)-Ps Int(A).
Proposition 3.26. For any two subsets A and B of a bitopological space X,
the following statements are true:
1. (i, j)-Ps Cl(X − A)=X − ((i, j)-Ps Int(A)) and (i, j)-Ps Int(X − A)=X −
((i, j)-Ps Cl(A)),
2. (i, j)-Ps Cl(X)=X, (i, j)-Ps Cl(φ)=φ, (i, j)-Ps Int(X)=X and (i, j)-Ps Int(φ)=φ,
3. If A ⊆ B, then (i, j)-Ps Cl(A) ⊆ (i, j)-Ps Cl(B) and (i, j)-Ps Int(A) ⊆
(i, j)-Ps Int(B),
4. A ⊆ (i, j)-Ps Cl(A) and (i, j)-Ps Int(A) ⊆ A,
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5. (i, j)-Ps Cl((i, j)-Ps Cl(A))= (i, j)-Ps Cl(A) and (i, j)-Ps Int((i, j)-Ps Int(A))=
(i, j)-Ps Int(A).
In general (i, j)-Ps Int(A) ∪ (i, j)-Ps Int(B) = (i, j)-Ps Int(A ∪ B) and (i, j)Ps Int(A ∩ B) = (i, j)-Ps Int(A) ∩ (i, j)-Ps Int(B), as it shown in the following
examples:
Example 3.27. Let X = {a, b, c}, τ1 = {φ, {a}, {a, c}, X} and τ2 = {φ, {a, c}, X},
then (i, j)-Ps O(X) = {φ, {c}, {b, c}, X}. If we take A = {a, b} and B =
{a, c}, then (i, j)-Ps Int(A) = φ and (i, j)-Ps Int(B) = {c}. Therefore, (i, j)Ps Int(A) ∪ (i, j)-Ps Int(B) = {c} but (i, j)-Ps Int(A ∪ B) = (i, j)-Ps Int(X) =
X.
Example 3.28. Let X = {a, b, c}, τ1 = {φ, {a}, {c}, {a, c}, X} and τ2 =
{φ, {c}, {a, b}, X}, then (i, j)-Ps O(X) = {φ, {a}, {c}, {a, b}, {b, c}, {a, c}, X}.
If A = {a, b} and B = {b, c}, then (i, j)-Ps Int(A) = {a, b}, (i, j)-Ps Int(B) =
{b, c} and so, (i, j)-Ps Int(A) ∩ (i, j)-Ps Int(B) = {b}. On the other hand,
(i, j)-Ps Int(A ∩ B) = (i, j)-Ps Int({b}) = φ, so (i, j)-Ps Int(A ∩ B) = (i, j)Ps Int(A) ∩ (i, j)-Ps Int(B).
The proof of the following result is obvious.
Proposition 3.29. If A is any subset of a bitopological space X, then (i, j)Ps Int(A) ⊆ j-pInt(A) ⊆ A ⊆ j-pCl(A) ⊆ (i, j)-Ps Cl(A).
In general, (i, j)-Ps Int(A) = j-pInt(A) and (i, j)-Ps Cl(A) = j-pCl(A) as it
is shown in the following example:
Example 3.30. Considering the space X in Example 3.28. If we take A =
{a}, then j-pInt(A) = A and (i, j)-Ps Int(A) = φ. This shows that (i, j)Ps Int(A) = j-pInt(A).
Example 3.31. Considering the space X as defned in Example 3.28. If we
take F = {b}, then j-pCl(F ) = {b} and (i, j)-Ps Cl(F ) = {a, b}. Hence (i, j)Ps Cl(F ) is not a subset of j-pCl(F ).
The following results can be proved by straightforward statments:
Proposition 3.32. Let A be a subset of a bitopological space X, then (i, j)Ps Bd(A) = φ if and only if A is both (i, j)-Ps - open and (i, j)-Ps -closed set.
Theorem 3.33. For any subset A of a bitopological space X, the following
statements are true:
1. (i, j)-Ps Bd(A) = (i, j)-Ps Bd(X − A)
2. A ∈ (i, j)-Ps O(X) if and only if (i, j)-Ps Bd(A) ⊆ X − A, i.e., A ∩ (i, j)Ps Bd(A) = φ.
3. A ∈ (i, j)-Ps C(X) if and only if (i, j)-Ps Bd(A) ⊆ A.
4. (i, j)-Ps Bd((i, j)-Ps Bd(A)) ⊆ (i, j)-Ps Bd(A)
5. (i, j)-Ps Bd((i, j)-Ps Int(A)) ⊆ (i, j)-Ps Bd(A)
6. (i, j)-Ps Bd((i, j)-Ps Cl(A)) ⊆ (i, j)-Ps Bd(A)
7. (i, j)-Ps Int(A) = A-((i, j)-Ps Bd(A))
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4. On (i, j)-Ps -Continuity
Recall the fopllowing definitions:
Definition 4.1. A function f : X → Y is called,
1. precontinuous, [18] (resp., almost precontinuous, [20]) if the inverse image
of each open (resp., regular open) subset of Y is preopen in X.
2. Contra continuous [7], if the inverse image of every open subset of Y is
closed in X.
3. R-map [4], if the inverse image of each regular open subset of Y is regular
open in X.
Definition 4.2. [9] A function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is said to be ijperfectly continuous if f −1 (V ) is both j-closed and j-open in X for each i-open
subset V of Y .
Definition 4.3. A function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is called (i, j)-Ps continuous (resp. almost (i, j)-Ps -continuous ) at a point x ∈ X, if for each
i-open set V of Y containing f (x), there exist (i, j)-Ps -open U of X containing
x such that f (U) ⊆ V (resp., f (U) ⊆ i-Int(i-Cl(V )) ). if f is (i, j)-Ps continuous (resp. almost (i, j)-Ps -continuous) at every point x of X, then it is
called (i, j)-Ps -continuous (resp. almost (i, j)-Ps -continuous ).
It is obvious from definition that (i, j)-Ps -continuity implies almost (i, j)Ps -continuity however the converse is not true in general , as shown in the
following example:
Example 4.4. Let X = {a, b, c}, τ1 = {φ, {a}, {a, c}, X} and τ2 = {φ, {a, c}, X},
then the family of (i, j)-Ps -open subsets of X is (i, j)-Ps O(X) = {φ, {c}, {b, c}, X}.
Let f : (X, τ1 , τ2 ) → (X, τ1 , τ2 ) be the identity function, then it is almost (i, j)Ps -continuous. Since {a} is an i-open set in X containing f (a) = a but there
is no (i, j)-Ps -open set U of X containing a such that f (U) ⊆ {a}, so it is not
(i, j)-Ps - continuous.
Proposition 4.5. A function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is (i, j)-Ps -continuous
if and only if the inverse image of every i-open set in Y is (i, j)-Ps -open in X.
It is clear that Every (i, j)-Ps -continuous function is j-pre continuous but
the converse is not true in general, as shown in the following example:
Example 4.6. In Example 4.4, f is j-pre continuous, but it is not (i, j)-Ps continuous.
Theorem 4.7. For a function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ), the following statements are equivalent:
1. f is (i, j)-Ps -continuous,
2. f −1 (V ) is (i, j)-Ps -open set in X, for each i-open set V in Y ,
3. f −1 (F ) is (i, j)-Ps -closed set in X, for each i-closed set F in Y ,
4. f ((i, j)-Ps Cl(A)) ⊆ i-Cl(f(A)), for each subset A of X,
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5. (i, j)-Ps Cl(f −1 (B)) ⊂ f −1 (i-Cl(B)), for each subset B of Y ,
6. f −1 (i-Int(B)) ⊆ (i, j)-Ps Int(f −1 (B)), for each subset B of Y ,
7. i-Int(f (A)) ⊆ f ((i, j)-Ps Int(A)), for each subset A of X.
Proof. (1) ⇒ (2) : Follows from definition.
(2) ⇒ (3) : Let F be any i-closed set of Y , then Y − F is an i-open set of Y .
By (2), we have f −1 (Y − F ) = X − f −1 (F ) is (i, j)-ps -open set in X. Hence
f −1 (F ) is (i, j)-ps - closed set in X.
(3) ⇒ (4) : Let A be any subset of X, then f (A) ⊆ i − Cl(f (A)) and i −
Cl(f (A)) is i-closed set in Y . Hence A ⊆ f −1 (i − Cl(f (A))), by (3), f −1 (i −
Cl(f (A))) is (i, j)-ps - closed set in X. Therefore, (i, j)-ps Cl(A) ⊆ f −1 (i −
Cl(f (A))), so f ((i, j)-ps Cl(A)) ⊆ i − Cl(f (A)).
(4) ⇒ (5) : Let B be any subset of Y , then f −1 (B) is a subset of X, by (4)
we have f ((i, j)-psCl(f −1 (B))) ⊆ i − Cl(f (f −1(B))) = i − Cl(B), hence (i, j)ps Cl(f −1 (B)) ⊂ f −1 (i − Cl(B)).
(5) ⇒ (6) : Let B be any subset of Y , then by applying conditiomn (5) on
Y − B we obtain,
(i, j)-Ps Cl(f −1 (Y − B)) ⊂ f −1 (i-Cl(Y − B))
⇔ (i, j)-Ps Cl(X − f −1 (B)) ⊂ f −1 (Y − i-Int(B))
⇔ X − (i, j)-Ps Int(f −1 (B)) ⊂ X − f −1 (i-Int(B))
⇔ f −1 (i-Int(B)) ⊂ (i, j)-Ps Int(f −1 (B)). This completes the proof.
(6) ⇒ (7) : Let A be any subset of X, then f (A) is a subset of Y . By (6),
we have f −1 (i-Int(f (A))) ⊆ (i, j)-Ps Int(f −1 (f (A))) = (i, j)-Ps Int(A). Therefore, i-Int(f (A)) ⊆ ((i, j)-Ps Int(A)).
(7) ⇒ (1) : Let x ∈ X and let V be any i-open set of Y containing f (x),
then x ∈ f −1 (V ) and f −1 (V ) is a subset of X. Hence, by (7), we have
i − Int(f (f −1 (V ))) ⊂ f ((i, j)-ps Int(f −1 (V ))). So, i − Int(V ) ⊆ f ((i, j)ps Int(f −1 (V ))). Since V is an i-open set in Y , then V ⊆ f ((i, j)-ps Int(f −1 (V )))
which implies that f −1 (V ) ⊆ (i, j)-ps Int(f −1 (V )). Therefore, f −1 (V ) is (i, j)ps -open set in X contains x and clearly f (f −1 (V )) ⊆ V . Hence, f is (i, j)-ps continuous.
Theorem 4.8. For a function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ), the following statements are equivalent:
1. f is almost (i, j)-Ps -continuous.
2. for each x ∈ X and each i-open set V of Y containing f (x), there exist
an (i, j)-Ps -open set U of X containing x such that f (U) ⊆ i-sCl(V )
3. for each x ∈ X and each i-regular open set V of Y containing f (x), there
exist an (i, j)-Ps -open set U of X containing x such that f (U) ⊆ V .
4. for each x ∈ X and each i-δ-open set V of Y containing f (x), there exist
an (i, j)-Ps -open set U of X containing x such that f (U) ⊆ V .
Proof. (1) ⇒ (2): Let x ∈ X and let V be any i-open set of Y containing
f (x), by (1) there exists an (i, j)-ps - open set U of X containing x such that
f (U) ⊆ i − Int(i − Cl(V )). Since V is i-open and hence V is i-preopen set, so
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i − Int(i − Cl(V )) = i − sCl(V ). Therefore, f (U) ⊆ i − sCl(V ).
(2) ⇒ (3): Let x ∈ X and let V be any i-regular open set of Y containing f (x),
then V is i-open set of Y containing f (x). So by (2), there exists an (i, j)-ps open set U of X containing x such that f (U) ⊆ i−sCl(V ). Since V is i-regular
open and hence V is i-open, so i − sCl(V ) = i − Int(i − Cl(V )). Therefore,
f (U) ⊆ i − Int(i − Cl(V )) and since V is i-regular open, then f (U) ⊆ V .
(3) ⇒ (4): Let x ∈ X, and let V be any i − δ-open set of Y containing f (x),
then for each f (x) ∈ V there exists an i-open set G containing f (x) such that
G ⊆ i − Int(i − Cl(G)) ⊆ V . Since i − Int(i − Cl(G)) is i-regular open set of Y
containing f (x), so by (3) there exists an (i, j)-ps - open set U of X containing
x such that f (U) ⊆ i − Int(i − Cl(G)) ⊆ V . This implies that f (U) ⊆ V .
(4) ⇒ (1): Let x ∈ X, and let V be any i-open set of Y containing f (x), then iInt(i−Cl(V )) is i−δ-open set of Y containing f (x). Hence, by (4), there exists
an (i, j)-ps -open set U of X containing x such that f (U) ⊆ i − Int(i − Cl(V )).
Therefore, f is almost (i, j)-ps - continuous.
Theorem 4.9. For a function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ), the following statements are equivalent:
1. f is almost (i, j)-Ps -continuous.
2. f −1 (i-Int(i-Cl(V ))) is (i, j)-Ps -open set in X, for each i-open set V in
Y.
3. f −1 (i-Cl(i-Int(F ))) is (i, j)-Ps -closed set in X, for each i-closed set F in
Y.
4. f −1 (F ) is (i, j)-Ps -closed set in X, for each i-regular closed set F in Y .
5. f −1 (V ) is (i, j)-Ps -open set in X, for each i-regular open set V in Y .
Proof. Similar to the proof of Theorem4.7
Proposition 4.10. A function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is (i, j)-Ps -continuous
if and only if f is j-pre continuous and for each x ∈ X and each i-open set
V of Y containing f (x) there exist an i-semi-closed set F of X containing x
such that f (F ) ⊆ V .
Proof. Necessity. Let f be an (i, j)-ps - continuous function. Let x ∈ X and
V be any i-open set of Y containing f (x), then by hypothesis there exists an
(i, j)-ps -open set U of X containing x such that f (U) ⊆ V . Since U is (i, j)ps - open, then for each x ∈ U there exists an i-semi-closed set F of X such
that x ∈ F ⊆ U. Therefore we have f (F ) ⊆ V and (i, j)-ps -continuity always
implies j-pre continuity.
Sufficiency. Let V be any i-open set of Y . To show that f −1 (V ) is an (i, j)ps -open set in X. Since f is j-pre continuous, then f −1 (V ) is an j-pre open
set in X. Let x ∈ f −1 (V ), then f (x) ∈ V . So, by hypothesis, there exists an
i-semi-closed set F of X containing x such that f (F ) ∈ V . This implies that
x ∈ F ⊆ f −1 (V ), so f −1 (V ) is an (i, j)-ps - open set in X. Therefore, so by
Proposition ??, f is (i, j)-ps - continuous.
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Theorem 4.11. Let f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) be a function and let β be any
basis for σi in Y , then f is (i, j)-Ps -continuous if and only if for each B ∈ β,
f −1 (B) is an (i, j)-Ps -open subset of X.
Proof. Obvious.
Theorem 4.12. Let f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) be ij-perfectly continuous and
i-contra continuous, then f is (i, j)-Ps -continuous.
Proof. Let V be any i-open suset of Y . Since f is ij-perfectly continuous, so
f −1 (V ) is j-open in X. Since f is i-contra continuous, so f −1 (V ) is i-closed
in X. Therefore, f −1 (V ) is (i, j)-ps - open set in X, so by Proposition ??, f is
(i, j)-ps - continuous.
Theorem 4.13. Let f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) be (i, j)-Ps -continuous, if Y
is an i-open subset of a bitopological space (Z, η1 , η2 ), then f : (X, τ1 , τ2 ) →
(Z, η1 , η2 ) is (i, j)-Ps -continuous.
Proof. Let V be any i-open set in Z, then V ∩ Y is an i-open set in Y . Since
f is (i, j)-Ps -continuous, so by Theorem 4.7, f −1 (V ∩ Y ) is (i, j)-Ps -open set
in X but f (x) ∈ Y for each x ∈ X. Thus f −1 (V ) = f −1 (V ∩ Y ) is (i, j)Ps -open set in X. Therefore, by Theorem 4.7, f : (X, τ1 , τ2 ) → (Z, η1, η2 ) is
(i, j)-Ps -continuous.
Theorem 4.14. Let f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) be almost (i, j)-Ps -continuous
, if Y is an i-preopen subset of a bitopological space (Z, η1 , η2 ), then f :
(X, τ1 , τ2 ) → (Z, η1 , η2 ) is almost (i, j)-Ps -continuous.
Proof. Let V be any i-regular open set of Z. Since Y is i-preopen then by
Theorem 2.9, V ∩Y is i-regular open set in Y . Since f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 )
is almost (i, j)-Ps -continuous, so by Theorem 4.9, f −1 (V ∩ Y ) is (i, j)-Ps -open
set in X but f (x) ∈ Y for each x ∈ X, hence f −1 (V ) = f −1 (V ∩ Y ) is (i, j)Ps -open set in X. Therefore, by Theorem 4.9, f : (X, τ1 , τ2 ) → (Z, η1, η2 ) is
almost (i, j)-Ps -continuous.
Theorem 4.15. If f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is almost (i, j)-Ps -continuous
and g : (Y, σ1 , σ2 ) → (Z, η1 , η2 ) is i-continuous and i-open, then the composition function g ◦ f : (X, τ1 , τ2 ) → (Z, η1 , η2 ) is almost (i, j)-Ps -continuous.
Proof. Let x ∈ X and V be an i-open set of Z. Since g is i-continuous,
so g −1 (V ) is i-open set of Y containing f (x). Since f is almost (i, j)-Ps continuous, then there exists an (i, j)-Ps -open set U of X containing x such
that f (U) ⊆ i-Int(i-Cl(g −1(V ))), also g is i-continuous, then (g ◦ f )(U) ⊆ g(iInt(g −1 (i-Cl(V )))). Since g is i-open we obtain that (g ◦ f )(U) ⊆ i-Int(iCl(V )). Therefore, g ◦ f is almost (i, j)-Ps -continuous.
Theorem 4.16. If f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is almost (i, j)-Ps -continuous
and Y is i-semi-regular, then f is (i, j)-Ps -continuous.
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Proof. Let x ∈ X and let V be any i-open set of Y containing f (x), by the
semi-regularity of Y , there exists an i-regular open set G of Y such that f (x) ∈
G ⊆ V . Since f is almost (i, j)-Ps -continuous, so by Theorem 4.8, there exists
an (i, j)-Ps -open set U of X containing x such that f (U) ⊆ G ⊆ V . Therefore,
f is (i, j)-Ps -continuous.
Definition 4.17. A function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is called contra (i, j)Ps -continuous if f −1 (V ) is (i, j)-Ps -closed in X for every j-open set V of Y .
The following result can be proved easily.
Theorem 4.18. The function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is contra (i, j)-Ps continuous if and only if for each j-closed set A containing f (x), there exist
an (i, j)-Ps -open set G containing x such that f (G) ⊆ A.
Theorem 4.19. If a function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is contra (i, j)-Ps continuous at x, then for each j-closed set A containing f (x), there exist an
i-semi-closed set F such that f (F ) ⊆ A.
Proof. Let A be any j-closed set containing f (x). Since f is contra (i, j)-Ps continuous, so by Theorem 4.18, there exists an (i, j)-Ps - open set G containing
x such that f (G) ⊆ A. Since G is (i, j)-Ps - open set, so for all x ∈ G there
exists an i-semi-closed set F such that x ∈ F ⊆ G. This implies that f (F ) ⊆
f (G) ⊆ A which completes the proof.
Theorem 4.20. The following statments are equivalent for a function f :
(X, τ1 , τ2 ) → (Y, σ1 , σ2 ):
1. f is contra (i, j)-Ps -continuous.
2. The inverse image of every j-closed set of Y is (i, j)-Ps -open in X.
3. For each x ∈ X and each j-closed set B in Y with f (x) ∈ B, there exists
an (i, j)-Ps -open set A in X such that x ∈ A and f (A) ⊆ B.
4. f ((i, j)-Ps Cl(A)) ⊆ j-ker(f (A)), for every subset A of X.
5. (i, j)-Ps Cl(f −1 (A)) ⊆ f −1 (j-ker((A))), for every subset A of Y .
Proof. (1) ⇒ (3): Let x ∈ X and B be j-closed set in Y with f (x) ∈ B. By
(1), it follows that f −1 (Y − B) = X − f −1 (B) is (i, j)-Ps -closed in X and
so f −1 (B) is (i, j)-Ps -open. Take A = f −1 (B), we obtain that x ∈ A and
f (A) ⊆ B.
(3) ⇒ (2): Let B be j-closed set in Y with x ∈ f −1 (B). Since f (x) ∈ B, so by
(3), there exists an (i, j)-Ps -open set A in X containing x such that f (A) ⊆ B.
It follows that x ∈ A ⊂ f −1 (B), hence f −1 (B) is (i, j)-Ps -open set in X.
(2) ⇒ (1): Follows from Theorem ??.
(2) ⇒ (4): Let A be any subset of X and let y ∈ j-ker(f (A)), then there
exist j-closed set F containing y such that f (A) ∩ F = φ. Hence we have
A ∩ f −1 (F ) = φ and (i, j)-Ps Cl(A) ∩ f −1 (F ) = φ, so f ((i, j)-PsCl(A)) ∩ F = φ
and y ∈
/ f ((i, j)-PsCl(A)). Thus f ((i, j)-Ps Cl(A)) ⊂ j-ker(f (A)).
(4) ⇒ (5): let B be any subset of Y , so by (4), f ((i, j)-PsCl(f −1 (B))) ⊂ jker(B). Hence, (i, j)-Ps Cl(f −1 (B)) ⊂ f −1 (j-ker(B)).
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(5) ⇒ (1): Let B be any j-open set of Y , by (5), (i, j)-Ps Cl(f −1 (B)) ⊂ f −1 (jker(B)) = f −1 (B) and hence (i, j)-Ps Cl(f −1 (B)) = f −1 (B). Therefore, we
obtain that f −1 (B) is (i, j)-Ps -closed in X, so f is contra (i, j)-Ps −continuous.
Proposition 4.21. If a function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is both j-contra
continuous and ji-perfectly continuous, then f is contra (i, j)-Ps -continuous.
Proof. Let V be any j-open set in Y . Since f is ji-perfectly continuous so
f −1 (V ) is both i-open and i-closed in X and since f is j-contra continuous, so
f −1 (V ) is j-closed in X. Hence f −1 (V ) is (i, j)-Ps − closed set in X, so f is
contra (i, j)-Ps -continuous.
Theorem 4.22. If a function f : (X, τ1 , τ2 ) → (Y, σ1 , σ2 ) is contra (i, j)-Ps continuous and g : (Y, σ1 , σ2 ) → (Z, η1 , η2 ) is j-R-map, then g ◦ f is contra
(i, j)-Ps -continuous.
Proof. Let V be any j-regular open set of Z. Since g is j-R-map, so g −1 (V )
is j-regular open set in Y and since every regular open is open set, so g −1 (V )
is j-open set in Y . Since f is contra (i, j)-Ps -continuous, so (g ◦ f )−1 (V ) =
f −1 (g −1 (V )) is (i, j)-Ps -closed set in X. Therefore, g ◦ f is contra (i, j)-Ps continuous.
Theorem 4.23. If a function f : (X, τ1 , τ2 ) → (ΠYi , σ1 , σ2 ) is contra (i, j)-Ps continuous, then gi ◦ f : (X, τ1 , τ2 ) → (Y i, σ1 , σ
2 ) is contra (i, j)-Ps -continuous
for each i ∈ I, where gi is the j-projection of Yi onto Yi.
−1
(V i) is jProof. Let
Vi be any j-open set of Yi . Since gi is j-continuous, so−1gi −1
open in
Yi and since f is contra (i, j)-Ps -continuous, so f (g (Vi )) =
(gi ◦ f )−1 (Vi ) ∈ (i, j)-Ps C(X). This shows that gi ◦ f is contra (i, j)-Ps continuous for each i ∈ I.
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Received: May, 2010