Pairwise Strongly Θ-Continuous Functions in Bitopological Spaces
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In this paper we introduce the notion of pairwise strongly θcontinuous functions in bitopological spaces and obtain related
concepts. Using these concepts we obtain many results of [5] in this
new setting.
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Before we explain our motivation in clear terms, let us clarify as
prerequisites certain concepts and terminologies to be used in the
sequel. By (X, τ1, τ2) we shall mean a bitopological space [2] where
τ1 and τ2 are topologies on a set X. J. C. Kelley [2] initiated the
systematic study of such spaces by introducing Hausdorff spaces,
pairwise regular spaces and pairwise normal spaces in the theory.
Pairwise compact spaces are defined in [1]. “Recall that a
bitopological space (X, τ1, τ2) is pairwise compact if every pairwise
open cover of X has a finite subcover.”
The problem of defining bitopological compactness has been
considered by several authors in particular, Fletcher, Hoyle and Patty,
Kim. Birsan, Swart, etc. Functions having closed graphs have been
studied extensively by P.E. Long in [3].
______________________________________
Bitopological spaces, pairwise strongly θcontinuous function, pairwise Hausdorff, pairwise Urysohn, pairwise
continuous, etc.
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A.Srivastava and A.Pawar
Later on a stronger condition on the graph was investigated in [4].
In this paper we extend the results established by Paul E. Long
and Larry L. Herrington to functions between bitopological spaces. We
obtain some characterization and several properties of these functions.
Through out a space we mean a bitopological space. Symbols X, Y and
Z are used for spaces and f and g is used for maps between
bitopological spaces.
Recall that a bitopological space (X, τ1, τ2) is
if
for each pair of distinct points x1 and x2 in (X, τ1, τ2), then there is a τ1open set U and τ2-open set V such that x1∈U and x2∈V, U and V are
disjoint.
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Let X be any infinite set, τ1 be the cofinite topology on X
and τ2 be the discrete topology on X. Certainly (X,τ1, τ2) is pairwise
Hausdorff.
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A map f : (X,τ1, τ2) (Y, r1, r2) is said to be
(respectively closed, open homeomorhisms) if f : (X, τ1) (Y, r1) and
if f : (X, τ2) (Y, r2) are continuous (respectively closed, open
homeomor-phisms).
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A bitopological space is
if for every two distinct
points x1, x2 ∈(X, τ1, τ2) there exist U∈τ1 and V∈τ2 such that τ2ClU
∩ τ1ClV = ∅ and one of the points belong to U, the other to V [x1∈U
and x2∈V].
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T. Noiri in [7] defined a function f : X Y for two topological spaces
X and Y to be
θif for each x∈X and each open
set V containing f(x), then there exists an open set U containing x such
that f(Cl(U) ⊂V. Clearly such functions are continuous.
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A function f: (X, τ1, τ2) (Y, r1, r2) is said to be pairwise strongly
θ-continuous if either f : (X, τ1) (Y, r1) is strongly θ-Continuous or
f : (X, τ2) (Y, r2) is strongly θ-continuous (i.e. for each x∈(X, τ1)
240
Pairwise Strongly Θ-Continuous Functions in Bitopological Spaces
and each r1- open V1 containing f(x) there exists an τ1- open set U1,
containing x such that f(τ1Cl(U1)) ⊂ V1 or for each x∈(X, τ2) and each
r2- open set V2 containing f(x) there exists an τ2- open set U2
containing x such that f(τ2Cl(U2)) ⊂ V2.
Let f: (X, τ1, τ2) (Y, r1, r2) be given. Then f is
pairwise strongly θ-continuous if and only if either of the following
two conditions hold:
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(a)
for each x∈(X, τ1) there exists on τ1- open set U1 containing x
and r1- open set V1 containing f(x) such that f(τ1Cl(U1)) ⊂ V1.
(b)
for each x∈(X, τ2) there exists on τ2- open set U2 containing x
and r2- open set V2 containing f(x) such that f(τ2Cl(U2)) ⊂ V2.
: Assume that the condition (b) holds. Now to show that f is
pairwise strongly θ-continuous, take x∈(X, τ1) ; then there exists an τ1open set U1 containing x and an r1-open set V1 containing. f(x) such
that f(τ1Cl(U1) ⊂ V1, which implies that f is pairwise strongly θcontinuous function.
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Conversely, assume that f is pairwise strongly θ-continuous function,
for this we take either f: (X, τ1) (Y, r1) be strongly θ-continuous or
f:(X, τ2) (Y, r2) be strongly θ-continuous. Assume that f : (X, τ2) (Y, r2) is strongly θ-continuous, there exists an τ2-open set U2
containing x and r2-open set V2 containing f(x) such that f(τ2Cl(U2)
⊂ V2.
This gives condition (b).
When we assume that f: (X, τ1) (Y, r1) is strongly θ-continuous then
we get condition (a).
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On the lines of [5] we obtain various results for pairwise strongly
θ − continuous functions in bitopological spaces.
Following [5], if a topological space X is regular, then any continuous
function f: XY is also strongly θ-continuous. Denote by Clθ(A) the θclosure of a set A⊂X. The subset A is θ-closed if Clθ(A) = A.
241
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A.Srivastava and A.Pawar
Similarly, the θ-interior of A is denoted by Intθ(A). A set A is called θopen if Intθ(A) = A.
It is clear that θ-open sets are open and θ-closed sets are closed.
Furthermore, the compliment of a θ-open set is θ-closed and the
complement of a θ-closed set is θ-open.
Since we are working in bitopological space in order to clarify the
topologies concerned we modify some notations in a natural way.
Let (X, τ1, τ2) and (Y, r1, r2) be two bitopological spaces and
f: (X, τ1, τ2) (Y, r1, r2) then –
τ1- clθ(A) is the τ1-θ-closure of a set A⊂ (X, τ1)
r1- clθ(F) is the r1-θ-closure of a set F⊂ (Y, r1)
τ1- Intθ(A) is the τ1-θ-interior of a set A⊂ (X, τ1)
r1- Intθ(F) is the r1-θ-interior of a set F⊂ (Y, r1)
Same as for τ2 and r2 with respect to (X, τ2) and (Y, r2) respectively.
A set A is said to be τ1-θ-closed if τ1-clθ(A) = A.
A set F is said to be r1-θ-closed if r1-clθ(F) = F
A set A is said to be τ1-θ-open if τ1- Intθ(A) = A.
A set F is said to be r1-θ-open if r1- Intθ(F)= F.
Same as for τ2 and r2 with respect to (X, τ2) and (Y, r2) respectively.
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For any f:(X, τ1, τ2) (Y, r1, r2) the following are
equivalent:
(a)
f is pairwise strongly θ-continuous;
(b)
The inverse image of every r1-closed set is τ1–θ -closed and the
inverse image of every r2-closed set is τ2–θ closed;
(c)
The inverse image of every r1-open set is τ1–θ–open and the
inverse image of every r2-open set is τ2–θ–open;
(d)
For each x∈(X, τ1, τ2) and each net xα 
→ x, we have
θ
f(xα) f(x).
242
Pairwise Strongly Θ-Continuous Functions in Bitopological Spaces
Let F1 and F2 be subsets of (Y, r1) and (Y, r2)
respectively, such that F1 is r1-closed and F2 is r2 -closed. Suppose
that f -1 (F1) is not τ1–θ–closed in (X, τ1) and F2 is not
τ2–θ–closed in (X, τ2). Then there is a point x∉f -1 (F1) ∪ f -1 (F2) such
that for every τ1–open set U1 and every τ2-open set U2 both
containing x we have τ1Cl(U1) ∩ f -1 (F1) ≠ ∅ and τ2Cl(U2) ∩ f -1 (F2)
≠ ∅.
Since f(x) ∉F1 ∪ F2 , Y \ F1 is r1-open and Y \ F2 is r2-open, both
containing f(x), having the property that no closed neighborhood of x
will map into Y \ F1 and Y \ F2 under f. Consequently, f is not pairwise
strongly θ-continuous at x. This contradiction implies that f -1(F1) is
τ1-θ-closed in (X, τ1) and f-1(F2) is τ2-θ-closed in (X, τ2) respectively.
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Let V1 be r1 -open in (Y, r1) and V2 be r2 -open in (Y,
r2). Then Y \ F1 is r1-closed and Y \ F2 is r2 -closed and by (b) f -1 (Y \
V1) is r1-θ-closed and f -1 (Y \ F2) is r2-θ-closed. But
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X \ f -1(Y \V1) = f -1(V1) is τ1–θ-open and
X \ f -1(Y \ V2) = f -1(V2) is τ2–θ-open.
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Let x∈(X, τ1, τ2) and let xα 
→ x. Let V1 be r1-
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open and V2 be r2-open set, both containing f(x). Then by (c), f -1 (V1)
is τ1–θ-open and f -1 (V2) is τ2–θ-open both containing x. Thus there
exists an τ1–open set U1 and τ2–open set U2 such that
x∈U1 ⊂ τ1 Cl (U1) ⊂ f -1 (V1) and
x∈U2 ⊂ τ2 Cl (U2) ⊂ f -1 (V2).
The τ1–θ convergence and τ2–θ convergence of xα is eventually
in τ1 Cl(U1) and τ1 Cl(U1) respectively. So that f(xα) is eventually in V1
and V2. This shows that f(xα) f(x).
suppose f is not pairwaise strongly θ-continuous for
some x∈(X, τ1, τ2). Then there is an r1-open set V1 and r2-open set
V2 both containing f(x) such that for every τ1–open U1 and τ2–open set
U2 both containing x, such that f{τ1 Cl (U1)} ⊄ V1 and f{τ2 Cl (U2)}
⊄ V2.
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A.Srivastava and A.Pawar
Now consider the directed sets D1 = {xα, τ1Cl(U1α)} and D2 = {xα,
τ2Cl(U2α)} using by reverse inclusion where U1α and U2α both
contains x and x2∈ τ1 Cl (U1α) ∪ τ2 Cl (U2α) such that f(xα)
⊄ V1 ∪ V2. Then the net g: D1 (X, τ1) and g: D2 (X, τ2)
defined by g(xα, U1α) = xα , τ1 –θ converges to x and g(xα, U2α) = xα
τ2 –θ converges to x, but the net fg does not converge to f(x). The
contradiction we obtained implies that f is pairwise strongly θcontinuous function.
Let f (X, τ1, τ2) (Y, r1, r2) be a pairwise
strongly θ-continuous injective function and let (Y, r1, r2) be pairwise
Hausdorff. Then (X, τ1, τ2) is pairwise Urysohn.
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:- Let x1≠ x2 where x1,x2 ∈ X . Then f(x1) ≠ f(x2). By hypothesis
(Y, r1, r2) is pairwise Hausdorff; then there exist disjoint sets r1-open s
V1 and r2-open V2 containing f(x1) and f(x2) respectively. Thus, there
exist τ1–open set U1 and τ2–open set U2 containing x1 and x2,
respectively, such that f(τ2Cl(U1) ⊂V1 and f(τ1Cl(U2) ⊂V2, because f is
pairwise strongly θ-continuous.
It follows that f -1 (f(τ2Cl(U1)) ⊂ f -1 V1 and f -1 (f(τ1Cl(U2)) ⊂ f -1 V2
therefore τ2Cl(U1) ⊂ f -1 V1 and τ1Cl(U2) ⊂ f -1 V2. Then
τ2Cl(U1) ∩ τ1Cl(U2) = ∅ , from which we conclude that (X, τ1, τ2) is
pairwise Urysohn.
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: Let f: (X, τ1, τ2) (Y, r1, r2) be injective and pairwise
strongly θ-continuous. If (Y, r1, r2) is a pairwise T1-space, then (X, τ1,
τ2) is pairwise Hausdorff.
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Let x1, x2 ∈ X such that x1≠ x2. Then f(x1) ≠ f(x2) . So there
exist an r1-open set V1 and an r2-open set V2 containing f(x1) and f(x2)
respectively, such that f(x2) ∉ V1 and f(x1) ∉ V2. Since f is pairwise
strongly θ-continuous, there exist an τ1-open set U1 and an τ2-open set
U2 containing x1 and x2 respectively, such that f(τ1Cl(U1) ⊂ V1 and
f (τ2Cl(U2) ⊂ V2. Thus x2 ∉ τ1Cl(U1) and x1 ∉ τ2Cl(U2). Therefore,
U1 and X \ τ1Cl(U1) , respectively U2 and X \ τ2Cl(U2), are disjoint τ1open sets and τ2-open sets separating x1 and x2 respectively.
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Pairwise Strongly Θ-Continuous Functions in Bitopological Spaces
Let f:(X, τ1, τ2) (Y, r1, r2) be pairwise strongly
θ-continuous and g:(Y, r1, r2) (Z, α1, α2) be pairwise continuous.
Then the composition gf: (X, τ1, τ2) (Z, α1, α2) is pairwise strongly
θ-continuous.
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Let V1 be α1-open set in (Z, α1) and V2 be α2-open set in
(Z, α2). Then g-1(V1) is r1-open set in (Y, r1) and g-1(V2) is r2-open set
in (Y, r2), hence f -1(g-1V1) = (gf)-1 V1 and f -1(g-1V2) = (gf)-1 V2 are
τ1-θ-open and τ2-θ-open respectively, by Theorem 3.1(c). Thus gf is
pairwise strongly θ-continuous by Theorem 3.1. It is clear that the
composition of two pairwise strongly θ-continuous is pairwise strongly
θ-continuous.
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The function f:(X, τ1, τ2) (Y, r1, r2) is pairwise strongly
θ-continuous if and only if for each pairwise sub-basis r1-open set V
and r2-open set S, subsets of (Y, r1, r2), f -1(V) and f-1(S) are τ1-open
and τ2-open in (X, τ1) and (X, τ2) respectively.
The necessity follows from Theorem 3.1. Conversely, let {Vα,
Sα : α∈∆ } be a pairwise sub basis for (Y, r1, r2) and assume that
f-1(Vα) and f-1(Sα) are τ1-θ-open and τ2-θ-open respectively, for all
α∈∆. Every r1-open V and r2-open S, subsets of (Y, r1, r2), can be
written as
V = U {Vα 1 ∩ Vα 2 ∩… ∩ Vα n : {α1, α2 ,…,αn} ⊂ ∆ }
and
S = U {Sα 1 ∩ Sα 2 ∩ … ∩Sα n :{α1, α2 ,…,αn} ⊂ ∆ }.
Then
f -1(V) = U {f -1 (Vα 1 )∩ f -1 (Vα 2 )∩ … ∩ f -1 (Vα n )}
and
f -1(S) = U { f -1 (Sα 1 ) ∩ f -1 (Sα 2 ) ∩ … ∩ f -1 (Sα n )}.
The finite intersection of τ1-θ-open sets is τ1-θ-open and the finite
intersection of τ2-θ-open sets is τ2-θ-open and also the union of τ1-θopen sets is τ1-θ-open and the union of τ2-θ-open sets is τ2-θ-open.
Therefore f -1(V) is τ1-θ-open and f-1(S) is τ2-θ-open and hence f is
pairwise strongly θ-continuous, by Theorem 3.1.
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A.Srivastava and A.Pawar
Let f:(X, τ1, τ2) ( Π Xα, r1, r2) be given. Then f is
pairwise strongly θ-continuous if and only if the composition with each
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projection ∏α is pairwise strongly θ-continuous.
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If f is pairwise strongly θ-continuous, then ∏α f is pairwise
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strongly θ-continuous by the continuity of ∏α and by Theorem 3.4.
Conversely, let V1 and V2 be a pairwise sub-basic r1-open set in
( ∏ Xα, r1) and r2-open set in ( ∏ Xα, r2) respectively for all α∈∆. Then
V1 = ∏α−1 (S) for some r1-open S in (Xα, r1) and
V2 = ∏α−1 (T) for some r2-open T in (Xα, r2)
Thus
f -1(V1) = f -1 ( ∏α−1 (S)) = ( ∏α f)-1 (S) is r1-θ-open
and
f -1(V2) = f -1 ( ∏α−1 (T)) = ( ∏α f)-1 (T) is r2-θ-open
due to ( ∏α f) being pairwise strongly θ-continuous, by Lemma 3.5.
Let f: (X, τ1, τ2) (Y, r1, r2) be a
function, let g: (X, τ1, τ2) (X, τ1, τ2) × (Y, r1, r2), given by g(x) =
(x, f(x)) be its pairwise graph map. Then f is pairwise strongly θcontinuous if and only if g is pairwise strongly θ-continuous.
Furthermore, if g: (X, τ1, τ2) (X, τ1, τ2) × (Y, r1, r2) is pairwise
strongly θ-continuous, then (X, τ1, τ2) is pairwise regular.
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Only the last statement needs verification. Let g be pairwise
strongly θ-continuous , x∈(X, τ1, τ2). Then for any τ1-open U and τ2open V containing x, where U × (Y, r2) is τ1-r2-open in (X × Y, τ1 × r2 )
and V × (Y, r1) is τ2r1-open in (X × Y, τ2 × r1 ) (by pairwise graph map).
Thus there exists an τ1-open set U0 and τ2-open V0 both containing x
such that
g(τ1Cl(U0) ⊂ U × (Y, r2) and
g(τ2Cl(V0) ⊂ V × (Y, r1)
Consequently,
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Pairwise Strongly Θ-Continuous Functions in Bitopological Spaces
x ∈ τ1(U0) ⊂ τ1 Cl(U0) ⊂ U and
x ∈ τ2(V0) ⊂ τ2 Cl(V0) ⊂ V
which implies that (X, τ1, τ2) is pairwise regular.
Let (Xα i , τ1, τ2) be a bitopological space and let
Uα i and Sα i be subsets of (Xα i , τ1) and (Xα i , τ2) respectively, for
each i = 1, 2, …….., n. Then
Uα 1 × Uα 2 × ………… × Uα n × ∏ × (Xα, τ1) ⊂ ∏ (Xα, τ1)
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α ≠αi
Sα 1 × Sα 2 × ……………. × Sα n ×
α∈∆
∏ × (Xα, τ2) ⊂ ∏ (Xα, τ2)
α ≠α i
α∈∆
are τ1–θ-open and τ2–θ-open respectively, if and only if Uα i is τ1–θ-open
and Sα i is τ2–θ-open in (Xα i , τ1) and (Xα i , τ2) respectively, for each i = 1,
2, ….., n.
Suppose that Uα i ⊂ (Xα i , τ1) is τ1-θ-open in (Xα i , τ1) and
Sα i ⊂ (Xα i , τ2) is τ2-θ-open in (Xα i , τ2) for each i = 1, 2, ……, n.
Then for each i and each xi∈Uα i ∩ Sα i then there exists an τ1-open
Vα i and τ2-open Tα i both containing xi such that
xi ∈ Vαi ⊂ τ1 Cl (Vαi) ⊂ Uαi
xi ∈ Tαi ⊂ τ2 Cl (Tαi) ⊂ Sαi
Thus, for each
{xα} ∈ Uα 1 × Uα 2 × ………… × Uα n × ∏
i
α ≠ αi
{x} ∈ Vα 1 × Vα 2 × ………..… × Vα n × ∏
α ≠ αi
(Xα, τ1),
(Xα, τ1) ⊂
τ1Cl (Vα 1 ) × τ1Cl (Vα 2 ) × ……… × τ1Cl(Vα n ) × ∏ (Xα, τ1) ⊂
α ≠ αi
Uα 1 × Uα 2 × ………………… × Uα n × ∏
α ≠ αi
(Xα, τ1).
This shows that Uα 1 × Uα 2 × ………………… × Uα n × ∏
α ≠ αi
(Xα, τ1) is τ1-θ-open. Same as for Sα i , we get
Sα 1 × Sα 2 × ………………… × Sα n × ∏ (Xα, τ2) is τ2-θ-open.
α ≠ αi
247
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A.Srivastava and A.Pawar
The converse is left for reader.
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: Define
∏ fα
: ∏ (Xα, τ1, τ2) ∏ (Yα, r1, r2)
α
α
by {xα} {fα (xα)}. Then
α
∏ fα is pairwise strongly
θ-continuous if
α
and only if each fα : (Xα, τ1, τ2) (Yα, r1, r2) is pairwise strongly θcontinuous.
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: Let V = Vα 1 × Vα 2 × ………..… × Vα n × ∏
α ≠ αi
(Xα, τ1)
be a basic r1-open set in ∏ (Yα, r1) and
α
S = Sα 1 × Sα 2 × ………..… × Sα n × ∏
α ≠ αi
be a basic r2-open set in ∏
(Xα, τ2)
(Yα, τ2).
α
Then if fα−1 (Vα i ) is τ1-θ-open in (Xα i , τ1) and fα−1 (Sα i ) is τ2i
i
θ- open in (Xα i , τ2), for each α i , where i = 1, 2, ……………..,n. Then we
( ∏ f α ) -1 (V) = f
have
(Vα n )
×
-1
(Vα 1 ) × f
-1
(Vα 2 ) × ……….. × f
-1
α
∏ (X α , τ1) and ( ∏ f α
α ≠ αi
……….. × f
-1
−1
) (S) = f
-1
(Sα 1 ) × f
-1
(Sα 2 ) ×
α
(Sα n ) × ∏ (X α , τ2) are
α ≠ αi
τ1-θ-open in ( ∏ Xα, τ1) and
α
τ2-θ-open in ( ∏ Xα, τ2) respectively by Lemma 3.8. This shows that
α
is pairwise strongly θ-continuous.
Conversely, suppose that
∏ fα
α
∏ fα
is pairwise strongly θ-continuous.
α
Let Vα i ⊂ (Yα i , r1) be r1-open and Sα i ⊂ (Yα i , r2) be r2-open. Then V
= Vα i × ∏ (Yα, r1) and S = Sα i × ∏ (Yα, r2) are pairwise sub basic
α ≠ αi
α ≠ αi


r1-open in ( Π Yα, r1) and r2-open in ( Π Yα, r2) respectively. And  ∏ fα 


α

248
−1
Pairwise Strongly Θ-Continuous Functions in Bitopological Spaces
−1


(V) = fα (Vα i ) × ∏ (Xα, τ1) is τ1-θ-open and  ∏ fα  (S) = fα−1


i
i
α ≠ αi
α

(Sα i ) × ∏ (Xα, τ2) is τ2-θ-open. Thus f -1 (Vα i ) is τ1-θ-open in (Xα i ,
−1
α ≠ αi
τ1) and f
(Sα i ) is τ2-θ-open in (Xα i , τ2) which implies that fα i is pairwise
strongly θ-continuous, by Theorem 3.1.
-1
1.
I. E. Cooke, E. Ian and I. L. Reilly;
London Math. Soc. (2), 9 (1975), 518-522.
2.
J. C. Kelley;
(1963), 71-89.
3.
P. E. Long,
Monthly 76(8), October, 1969.
4.
P. E. Long and L. L. Herringon,
, Boll. Un. Mat. Ital. (4), 12 (1975), 381-384.
, J.
, Proc. London Math. Soc. (3) 13
, The American Math.
5.
θ
P. E. Long and L. L. Herrington,
J. Korean Math. Soc. (18), 1 (1981), 21-28.
6.
M. N. Mukherjee and G. K. Banerjee
Indian J. Pure, Appl. Math. (21) 7 (1990), 639-648.
7.
T. Noiri,
θ
(16), 2 (1980), 161-166.
8.
I. L. Reilly,
Mathamatica (to appear).
"
249
"
;
s,
, J. of the Korean Math. Soc.
"
, Nanta
…………..
A.Srivastava and A.Pawar
School of Studies in Mathematics
Vikram University,
Ujjain- 456 010 (M.P.) INDIA.
E-mail : [email protected]
250