Abstract:
In
this paper, we introduce the concept of
l(m.
n) -
J- closed sets in bigeneralized topological spaces and study some of
their properties. The notion of Jg(m, n)- continuous
functions and strongly l(m.
n) -
J- closed set is also defined on bigeneralized topological spaces and
investigate some of their characterizations.
Mathematics
Subject Classification:
54A05, 54A10
Keywords:
bigeneralized topological spaces,
l(m.n)-J-closed
sets, Jg(m,n)- continuous
functions, l(m,
n) � J
s � closed set.
1. Introduction
�.Cs�sz�r [3]
introduced the concepts of generalized neighborhood systems and he
also introduced the concepts of continuous functions and associated
interior and closure operators on generalized neighborhood systems and
generalized topological spaces. In particular, he investigated
characterizations for generalized continuous functions (= (g,g�)-
continuous functions). In [4], he introduced and studied the notions
of g- a
- open sets, g - semi-open sets, g � pre open sets and g -
b
open sets in generalized topological spaces.
After the
introduction of the concept of generalized closed set by Levine[9] in
a topological spaces. Several other authors gave their ideas to the
generalizations of various concepts in topology. Kelly[8] introduced
the concept of bitopological spaces .Since many mathematicians
generalized the topological concepts into bitopological setting. In
1986, Fukutake [7] generalized the notion to bitopological spaces and
he defined a set ij-generalized closed set. Some new types of
generalized closed sets in bitopological spaces were introduced. In
this paper, we introduce the notion of
lm,
n- J -
closed sets in bigeneralized topological spaces and study some of
their properties. We also introduce the concept of strongly
lm,
n- J -
closed sets and strongly lm,
n- J
-continuous functions and investigate some of their characterizations.
2. Preliminaries
Let X be a non �
empty set and l
be a collection of subsets of X. Then
l
is called a generalized topology (briefly GT) on X iff
f
�
l
and Gi �
l
for i �
I �
f
implies G = .We
call the pair (X,l),
a generalized topological space(briefly GTS)on X. The elements of
l
are called l-
open sets and the complements are called
l-
closed sets. The generalized closure of a subset S of X, denoted by cl(S),
is the intersection of generalized closed sets including S and the
interior of S, denoted by il(S),
is the union of generalized open sets contained in S.
Definitions 2.1.
Let (X,
l)
be a generalized topological space and A
�
X, then A is said to be
(1)
l
- semi open if A �
cl
(il
(A))
(2)
l
- pre open if A �
il(
cl(A))
(3)
l
- a-
open if A �
il(
cl
(il
(A)))
(4)
l
- b
open if A �
cl
(il
(cl(A)))
The complement of
l
- semi open (resp l
- pre open , l
- a-
open, l
- b
open) is said to be l
- semi closed (resp l
- pre closed , l
- a-
closed, l
- b
closed). The class of all l
- semi open sets on X is denoted by
s(lx)(briefly
sx
or
s).
The class of l
- pre open (l
- a-
open and l
- b
open) sets on X as [p(gx),
a(gx),
b(gx)
] or briefly [[p,
a,
b].
Definition 2.3.
[2] Let
X be non-empty set and let l1
and l2
be generalized topologies on X. A triple (X,
l1,
l2)
is said to be a bigeneralized topological space (briefly BGTS).
Let (X,
l1,
l2)
be a bigeneralized topological space and A be a subset of X. The
closure of A and the interior of A with respect to
lm
are denoted by clm
(A)
and ilm
(A),
respectively, for m = 1,2.
Definition 2.4.
[10]Let
(X, l)
be a generalized topological space and A
�
X, then A is said to be l
- J- open
if A �
il
(cπ
(A)). The complement of
l
- J- open is said to be l
- J- closed.
The class of all
l
- J - open sets on X is denoted by λ � JO(X). The class of all
l
- J - closed sets on X is denoted by λ � JC(X).The
l
- J- closure of a subset S of X, denoted by cJ(S) is the
intersection of l
- J- closed sets including S. The
l
- J- interior of a subset S of X, denoted by iJ (S) is the
union of l
- J- open sets contained in S.
The class of all
l
- J - open sets is properly placed between
l
- open sets and l
- pre � open sets.
3. l(m,
n)
- J- closed sets
Definition 3.1.
A Subset A of a bigeneralized topological space (X,
l1,
l2)
is said to be l(m,
n) � J
- closed set
if cln
(A)
�
U, whenever A �
U and U is lm
� J- open, where m, n = 1,2 and m�n.
The complement of l(m,
n) � J
- closed set is said to be l(m,
n) �
J-open set.
Remark 3.2.
Every
(m, n) - closed set is l(m,
n) � J
- closed.
The converse is not
true as can be seen from the following example.
Example 3.3.
Let X =
{a, b, c}. Consider two generalized topologies
l1
= {f,{a},{a,
b}, X}, and l2
= {f,{c},{b,
c}},.Let A = {a, b}, then A is
l(m,
n) �
J- closed but not (m, n) - closed.
Proposition 3.4.
Every l(m,
n) �
closed set is l(m,
n) �
J- closed.
Proof:
Let U be a lm
� J- open set such that A �
U. Since A is l(m,
n) �
closed set, cln
(A)
�
U. Therefore A is l(m,
n) �
J- closed.
Proposition 3.5.
Let (X,
l1,
l2)
be a bigeneralized topological space and A be a subset of X. If A is
ln
� closed, then A is l(m,
n) �
J- closed, where m, n = 1, 2 and m
�
n.
Remark 3.6.(i)
The
union of two l(m,
n) � J
- closed need not be l(m,
n) �
J- closed.
(ii)
The
intersection of two l(m,
n) �
J- closed sets need not be
l(m, n)
� J � closed.
Proposition
3.7.
Let (X, l1,
l2)
be a bigeneralized topological space. If A is
l(m,
n) � J
- closed and F is (m, n) � J- closed, then A
�
F is l(m,
n) � J
� closed, where m, n = 1, 2 and m
�
n.
Proof:
Let U be a lm
� J- open set such that A �
F �
U. Then A �
U �
(X - F) and so cln
(A)
�
U �
(X - F).Therefore cln
(A)
�
F �
U .Since F is (m, n) � J- closed, c
ln
(A �
F) �
c ln
(A) �
c ln
(F) �
U . Hence A �
F is l(m,
n) � J
� closed.
Proposition
3.8.
For each element x of a bigeneralized topological space (X,
l1,
l2),
{x} is lm
- J- closed or X � {x} is l(m,
n) - J
- closed, where m, n = 1, 2 and m
�
n.
Proof:
Let x �
X and the singleton {x} be not
lm
� J- closed. Then X � {x} is not
lm
�J- open, if X �
lm,
then X is only lm
� J- open set which contains X � {x}, hence X � {x} is
l(m,
n) �
J- closed and if X �
lm,
then X � {x} is l(m,
n) �
J� closed.
Proposition
3.9.
Let (X, l1,
l2)
be a bigeneralized topological space. Let A
�
X be a l(m,
n) � J
- closed subset of X, then cln
(A) \
A does not contain any non- empty
lm
- J-
closed set, where m, n = 1, 2 and m
�
n.
Proof:
Let A be a l(m,
n) � J
- closed set and F �
f
is lm
� J- closed such that F �
c ln
(A) \ A. Then F �
X \ A and hence A �
X \ F. Since A is l(m,
n) �
J- closed, c
ln
(A) �
X \ F and hence F �
X \ c ln
(A) .So, F �
cln
(A)
�
(X \ cln
(A)) =
f.Therefore
c ln
(A) \ A does not contain any non- empty
lm
- J-
closed set.
Proposition
3.10.
Let l1
and l2
be
generalized topologies on X. If A is
l(m,
n) -
J- closed set, then clm
({x})
�
A �
f
holds for each x �
cln
(A),
where m, n = 1, 2 and m �
n.
Proof:
Let x �
cln
(A).
Suppose that clm
({x})
�
A = f.
Then A �
X - clm
({x}).
Since A is l(m,
n) �
J� closed and X - clm
({x})
is lm
� J- open. Thus cln
(A)
�
X - clm
({x}).
Hence cln
(A)
�
clm
({x})
= f.
This is a contradiction.
Proposition
3.11. If A is a l(m,
n) � J
- closed set of (X, l1,
l2)
such that A �
B �
cln
(A)
then B is l(m,
n) � J
- closed set, where m, n = 1, 2 and m
�
n.
Proof:
Let A be a l(m,
n) �
J- closed set and A �
B �
cln
(A).
Let B �
U and U is lm
� J-open. Then A �
U. Since A is l(m,
n) � J
� closed, we have cln
(A)
�
U. Since B �
cln
(A),
then cln
(B)
�
cln
(A)
�
U. Hence B is l(m,
n) � J
� closed.
Remark.3.12.
l(1,2)
� JC(X) is generally not equal to
l(2,1)
� JC(X) as can be seen from the following example.
Example 3.13.
Let X =
{a, b, c}. Consider two generalized topologies
l1
= {f,
{a}, {a, b}, X} and l2
= {f,
{c}, {b, c}}.Then l(1,2)
� JC(X) = {{a}, {b},{a, b}, {b, c},X} and
l(2,1)
� JC(X) = {{b}, {c},{a, b}, {b, c},f,
X }. Thus l(1,2)
� JC(X) �
l(2,1)
� JC(X).
Proposition
3.14.
Let l1
and l2
be generalized topologies on X. if
l1
�
l2,
then l(2,1)
� JC(X) �
l(1,2)
� JC(X)
Proposition
3.15.
For a bigeneralized topological space (X,
l1,
l2),
lm
- JO(X)
�
ln
- JC(X)
if and only if
every subset of X is l(m,
n) - J
- closed set, where m, n = 1, 2 and m
�
n.
Proof:
Suppose that lm
- JO(X)
�
ln
- JC(X).
Let A be a subset of X such that A
�
U, where U �
lm
- JO(X).
Then cln
(U)
�
cln
(A) =
U and hence A is l(m,
n) � J
� closed set.
Conversely, suppose
that every subset of X is l(m,
n) � J
� closed. Let U �
lm
-
JO(X).Since U is l(m,
n) � J
� closed, we have cln
(U)
�
U. Therefore, U �
ln
- JC(X)
and hence lm
- jO(X)
�
ln
- JC(X).
Proposition
3.16.
Let (X, l1,
l2)
be a bigeneralized topological space. Let A
�
X be a l(m,
n) � J
- closed subset of (X, l1,
l2).
Then A is lm
- J-
closed if and only if cln
(A) \
A is lm
- J-
closed, where m, n = 1, 2 and m
�
n.
Proof:
Let A be a l(m,
n) �
J- closed set. If A is lm
� J- closed, then cln
(A) \
A = f,
but f
is lm
� J- closed . Therefore c
ln
(A) \ A is lm
- J-
closed.
Conversely, Suppose
that c ln
(A) \ A is lm
- J-
closed, As A is l(m,
n) �
J� closed, cln
(A) \
A = f,
Consequently cln
(A) =
A.
Proposition
3.17.
Let (X, l1,
l2)
be a bigeneralized topological space. If A is
l(m,
n) � J
- closed and A �
B �
cln
(A),
then cln
(B) \ B has no non empty lm
- J-
closed subset.
Proof:
A
�
B implies X � B �
X � A and B �
cln
(B)
implies cln
((cln
(A)) =
cln
(A).Thus
cln
(B)
�
(X - B) �
cln
(A)
�
(X -A) which yields cln
(B) /
B �
cln
(A) /
A. As A is l(m,
n) � J
� closed, c
ln
(A) \ A has no non empty lm
� J-
closed subset. Therefore c
ln
(B) \ B has no non empty lm
- J-
closed subset.
Remark 3.18.(i)
The
intersection of two l(m,
n) �
J- open sets need not be l(m,
n) �
J- open
(ii) The union of two
l(m,
n) �
J- open sets need not be l(m,
n) � J
� open.
Proposition
3.19.
A subset of a bigeneralized topological space (X,
l1,
l2)
is l(m,
n) - J
- open iff every subset of F of X, F
�
iln
(A) whenever F is lm
� J- closed and F �
A, where m, n = 1, 2 and m �
n.
Proof:
Let A be a lm,
n � J-
open. Let F �
A and F is lm
� J-closed. Then X � A �
X � F and X � F is lm
� J- open, we have X � A is l(m,
n) �
J� closed, then cln
(X -
A) �
X - F. Thus X - iln
(A) �
X � F and hence F �
iln
(A).
Conversely, let F
�
iln
(A) where F is a lm
� J- closed set such that F �
A. Let X - A �
U where U is a lm
� J-
open. Then X - U �
A and X - U is lm
� J- closed. By the assumption, X - U
�
iln
(A), then X - iln
(A) �
U. Therefore, cln
(X -
A) �
U. Hence X - A is l(m,
n) - J
- closed and hence A is l(m,
n) -J
- open.
Proposition
3.20.
Let A and B be subsets of a bigeneralized topological space(X,
l1,
l2)
such that iln
(A) �
B �
A. If A is l(m,
n) - J
- open then B is also l(m,
n) -
J- open, where m, n = 1, 2 and m
�
n.
Proof: Suppose
that iln
(A) �
B �
A. Let F be a lm
� J- closed set such that F �
B. Since A is l(m,
n) - J
- open, F �
iln.
Since iln
�
B, we have iln
(iln(A))
�
iln(B).
Consequently iln(A)
�
iln(B).
Hence F �
iln(B).
Therefore, B is l(m,
n) - J
-open.
Proposition
3.21.
If A be a subset of a bigeneralized topological space (X,
l1,
l2)
is l(m
,n) -
J- closed , then cln
(A) -
A is lm,
n � J
� open, where m, n = 1, 2 and m
�
n.
Proof:
Suppose
that A is lm,
n � J
� closed. Let X � (cln
(A) -
A) �
U and U is lm
� J-
open. Then X � U �
X � (cln
(A) -
A) and X � U is lm
� J-
closed. Thus we have cln
(A) -
A does not contain non � empty
lm
� J-
closed by Proposition 3.10 . Consequently, X � U =
f,
then U = X. Therefore, cln
(X- (
cln
(A) �
A)) �
U, so we obtain X � (cln
(A) �
A) is lm,
n � J-
closed. Hence cln
(A) -
A is lm,
n - J
- open.
4. Jg(m, n)- continuous functions
Definition 4.1.
Let (X,lX1,
lX2)
and (Y,lY1,
lY2)
be generalized topological spaces. A function f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is said to be (m, n) - J- generalized continuous (briefly Jg(m,
n)- continuous) if f-1(F) is
l(m,
n)-
J-closed in X for every ln-
closed F of Y, where m, n = 1,2 and m�
n.
A
function f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is said to be pairwise J-generalized continuous(briefly pairwise Jg-continuous)
if f is Jg(1,2)-continuous and Jg(2,1)-continuous.
Proposition
4.2.
For an injective function f: (X,lX1,
lX2)
�
(Y,lY1,
lY2),
the following properties are equivalent:
(1)
f is
Jg(m, n)- continuous,
(2)
For
each x �
X and for every l
n-
open set V containing f(x), there exists a
l(m,
n)-J-open
set U containing x such that f(U)
�
V:
(3)
f( (A))
�
(f(A))
for every subset A of X;
(4)
(f-1(B))
�
f-1( (B)
for every subset B of Y.
Proof:
(1)
�(2):
Let x �
X and V be a ln-open
subset of Y containing f(x). Then by (1), f -1(V)
is l(m,
n)-J-open
of X containing x. If U = f -1(V), then f (U)
�
V.
(2)�(3):
Let A be a subset of X and f(x)
�
( (f(A)).
Then, there exists a l
n �
open subset V of Y containing f(x) such that V
�
f(A) = f.Then
by(2), there exist a l(m,
n)-J-open
set such that f(x) �
f(U) �
V. Hence, f(U) �
f(A) = f
implies U �
A = f.
Consequently, x �
and
f(x) �
f ((A)).
(3)
�(4):
Let B be a subset of Y. By (3) we obtain f((f
-1(B)) �
(f(f
-1(B)). Thus
(f
-1(B)) �
f -1 (
(B)).
(4)
�(1):
Let F be a ln
� closed subset of Y. Let U be a
lm
� J- open subset of X such that f-1 (F)
�
U. Since (F)
= F and by (4), (f
-1(F)) �
U. Hence f is Jg (m, n) continuous.
Definition 4.3.
Let (X,lX1,
lX2)
�
(Y,lY1,
lY2)
be generalized topological spaces. A function f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is said to be Jgm- continuous if f -1(F) is
lm
�J-closed in X for every lm
� closed F of Y, for m = 1,2.
Definition 4.4.
Let (X,lX1,
lX2)
�
(Y,lY1,
lY2)
be generalized topological spaces. A function f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is said to be Jgm- closed(resp. Jgm- open) if
f(F) is lm
� J-closed(resp lm
�J- open) of Y for every lm
� closed (resp. lm
� open) F of X, for m = 1,2.
Proposition 4.5.
If f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is Jgm - continuous and Jgm - closed , then f(A)
is l
(m, n)- J-closed subset of Y for every
l
(m, n)- J-closed subset A of X, where m, n = 1,2 and m�
n.
Proof:
Let U be lm
� J- open subset of Y such that f (A)
�
U. Then A �
f-1(U) and f-1(U) is
lm
� J-open subset of X. Since A is
l
(m, n)- J-closed, (f(A))
�
f-1(U) and hence f( (A))
�
U. Therefore we have (f(A))
�
(f((A)))
= f((A))
�
U. Therefore, f(A) is l
(m, n)- J-closed subset of Y.
Lemma 4.6.
If f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is Jgm � closed, then for each subset S of Y and each
lm
� J- open subset U of X containing f -1(S), there exists a
lm
� J- open subset V of Y such that f-1(V)
�
U.
Proposition 4.7.
If f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is injective, Jgm-closed and Jg(m ,n)-continuous,
then f-1(B) is l
(m, n)- J-closed subset of X for every
l
(m, n)- J-closed subset of B of Y, where m, n = 1,2 and m�
n.
Proof:
Let B be a l
(m, n)- J-closed subset of Y. Let U be a
lm
� J- open subset of X such that f-1(B)
�
U. Since f is Jgm-closed and by lemma 4.6, there exists a
lm
� J- open subset of Y such that B
�
V and f-1(V) �
U. Since B is l
(m, n)- J-closed set and B
�
V, then (B)
�
V. Consequently, f-1((B))
�
f-1(V) �
U. By theorem 4.2, (f-1(B))
�
f-1((B)
�U
and hence f -1(B) is
l
(m, n)- J-closed subset of X.
Definition 4.8.
A bigeneralized topological space (X,lX1,
lX2)
is said to be l
(m, n) � JT1/2 � space if, for every
l
(m, n)- J-closed set is
ln
� closed, where m, n = 1,2 and m�
n.
Definition 4.9.
A bigeneralized topological space (X,lX1,
lX2)
is said to be pairwise l
� JT1/2 � space if it is both
l
(1, 2) �JT1/2 � space and
l
(2, 1) � J1/2 � space.
Proposition
4.10. A
bigeneralized topological space is a
l
(m, n)- JT1/2-space if and only if {x} is
ln
- open or lm
�J-closed for each x �
X, where m, n = 1,2 and m�
n.
Proof:
Suppose that {x} is not lm
�J-
closed. Then X � {x} is l(m,
n)
-J-closed by proposition 3.9. Since X is
l
(1, 2) -T1/2 -space, X � {x} is
ln
-closed. Hence, {x} is ln
- open.
Conversely, let F
be a l
(m, n) - J-closed set. By assumption , {x} is
ln
- open or lm
�J- closed for any x �
cln(F).Case
(i) Suppose that {x}is ln
- open. Since {x} �
F�
f,
we have x �
F. Case (ii) Suppose that {x} is
lm
� J-closed. If x �
F, then {x} �
cln(F)
� F, which is a contradiction to proposition 3.9, Therefore, x
�
F. Thus in both case, we conclude that F is
ln
- closed. Hence, (X,lX1,
lX2)
is a l
(m, n) -JT1/2 -space.
Definition 4.11.
Let (X,lX1,
lX2)
and (Y,lY1,
lY2)
be generalized topological spaces. A function f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is said to be Jg (m, n) - irresolute if f -1(F)
is l(m,
n)-J-
closed in X for every
l(m,
n)
�J-closed F of Y, where m, n = 1,2 and m�
n.
Proposition 4.12.
Let f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
and g: (Y,lY1,
lY2)
�
(Z,lZ1,
lZ2)
be functions, the following properties hold:
(i)
If f is Jg (m, n) - irresolute and Jg (m, n)
-continuous, then g
�
f is Jg (m, n) -continuous.
(ii)
If f and g are Jg (m, n) - irresolute , then g
�
f is Jg (m, n) - irresolute;
(iii)
Let (Y,lY1,
lY2)
be a l
(m, n) -JT1/2 -space. If f and g are Jg (m,
n) -continuous, then g
�
f is Jg (m, n) -continuous.
Proof:
(i) Let F be a
ln
-closed subset of Z. Since g is Jg (m, n) - continuous,
then g -1(F) is
l(m,
n)
-J-closed subset of Y. Since f is Jg (m, n) - irresolute,
then (g �
f)-1(F) = f -1(g -1(F)) is
l(m,
n)
-J-closed subset of X. Hence g
�
f is Jg (m, n) -continuous.
(ii) Let F be a
l(m,
n)
-J- closed subset of Z. Since g is Jg (m, n) - irresolute,
then g -1(F) is
l(m,
n)
-J-closed subset of Y. Since f is Jg (m, n) - irresolute,
then (g �
f)-1(F) = f -1(g -1(F)) is
l(m,
n)
�J- closed subset of X. Hence, g
�
f is Jg (m, n) - irresolute.
(iii) Let F be a
ln
-closed subset of Z. Since g is Jg (m, n) -continuous, then
g -1(F) is
l(m,
n)
-J- closed subset of Y. Since (Y,lY1,
lY2)
is a l
(m, n) -JT1/2 -space, then g -1(F) is
ln
-closed subset of Y. Since f is Jg (m, n) -continuous, then
(g �
f)-1(F) = f -1(g -1(F)) is
l(m,
n)
�J-closed subset of X. Hence then g
�
f is Jg (m, n) -continuous.
Proposition 4.13.
Let: (X,lX1,
lX2)
be a l
(m, n) -JT1/2 -space. If f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is surjective, gn -closed and Jg (m, n)
-irresolute, then (Y,lY1,
lY2)
is a l
(m, n) -JT1/2 -space, where m, n = 1,2 and m�
n.
Proof:
Let F be a
l
(m, n) -J-closed subset of Y. Since f is Jg (m, n)
- irresolute, we have f -1(F) is a
l
( m, n) -J-closed subset of X. Since (X,lX1,
lX2)
is a l
(m, n) -JT1/2 -space, f -1(F) is a
l
n
-closed subset of X. It follows by assumption that F is a
l
n
-closed subset of Y. Hence (Y,lY1,
lY2)
is a l
(m, n) -JT1/2 space.
Proposition 4.14.
Let: (X,lX1,
lX2)
be a l
(m, n) - JT1/2 space. If f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
is bijective, gn -open and Jg (m, n) �
irresolute, then (Y,lY1,
lY2)
is a l
(m, n) -JT1/2space, where m, n = 1,2 and
m�
n.
5. Strongly-λ(m, n) - J - closed sets
Definition 5.1.
A subset A of a bigeneralized topological space (X,
l1,
l2)
is said to be strongly
l(m,
n)
� J s - closed set
(briefly
l(m,
n)
� J s - closed)
if cln
(A) �
U, whenever A
�
U and U is l(m,
n)
� J - open, where m, n = 1,2 and m�n.
Definition 5.2.
A subset A of a bigeneralized topological space (X,
l1,
l2)
is said to be pairwise strongly
l(m,
n)
� P- J s
closed (briefly
l(m,
n)
� P- J s
closed)if A is
l(1,2)
-Js � closed and
l(2,1)
- Js � closed..
Proposition 5.3.
Every λn - closed set is
l(m,
n)
� J s � closed.
Proof:
Let U be a
l(m,
n)
� J � open set such that A
�
U and A be λn - closed set. Then cln
(A) = A and so cln
(A) �
U. Thus A is
l(m,
n)
� J s � closed.
Proposition 5.4.
Let (X, l1,
l2)
be a bigeneralized topological space. Let A
�
X be a l(m,
n)
� Js - closed if and only if , then cln
(A) \ A does not contain any non- empty
l
(m,n) - J- closed set, where m, n = 1, 2 and m
�
n.
Proof:
Let F be a
l(m,
n)
� J - closed subset of cln
(A) \ A . Now, F
�
c
ln
(A) \ A and A
�
X \ F where A is
l(m,
n)
� J s � closed and X \ F is
l(m,
n)
� J � open. Thus c
ln
(A) �
X \ F and hence F
�
X \ c
ln
(A) .So, F
�
cln
(A) �
(X \ cln
(A)) = f.Therefore
F = j.
Conversely, assume cln
(A) \ A contains no non- empty
l
(m,n) - J- closed set. Let U be
l
(m,n) - J- open such that A
�
U. Suppose that cln
(A) is not contained in U, then cln
(A) �
Uc is a nonempty
l
(m, n) - J- closed set of cln
(A) \ A, which is a contradiction. Therefore cln
(A) �
U and hence A is
l(m,
n)
� J s � closed.
Proposition 5.5.
Let (X, l1,
l2)
be a bigeneralized topological space. Let A
�
X be a l(m,
n)
� Js - closed and A
�
B �
cln
(A) then B is
l(m,
n)
� Js - closed set, where m, n = 1, 2 and m
�
n.
Proof:
Let A be a
l(m,
n)
� Js- closed set and A
�
B �
cln
(A). Let B
�
U and U is
l(m,
n)
� J - open. Then A
�
U. Since A is
l(m,
n)
� Js � closed, we have cln
(A) �
U. Since B
�
cln
(A), then cln
(B) �
cln
(A) �
U. Hence B is
l(m,
n)
� Js � closed.
Definition 5.6.
A subset A of a bigeneralized space X is called strongly
l(m,
n)
� J s � open if Ac is
l(m,
n)
� J s � closed.
Proposition 5.7.
Let (X, l1,
l2)
be a bigeneralized topological space. Let A
�
X be a l(m,
n)
� Js � open if and only if F
�
iln
(A) whenever F
�
A and F is
l(m,
n)
� J � closed.
Proof:
Let A be l(m,
n)
� Js- open and suppose F
�
A where F is a
l(m,
n)
� J � closed. Then X \ A is
l(m,
n)
� J s � closed and X \ A
�
X \ F, where X \ F is
l(m,
n)
� J � open set. This implies that cln
(X \ A) �
X \ F. Now cln
(X \ A) = X \ iln
(A). Hence X \ iln
(A) �
X \ F and F
�
iln
(A).
Conversely. If F is an
l(m,
n)
� J � closed with F
�
iln
(A) whenever F
�
A. Then X \ A
�
X \F and X \ iln
(A) �
X \ F. Thus cln
(X \ A) �
X \ F. Hence X \ A is
l(m,
n)
� Js- closed and A is
l(m,
n)
� Js- open.
Proposition 5.8.
Let (X, l1,
l2)
be a bigeneralized topological space. For each x
�
X.{x} is
l(m,
n)
� J � closed or {x}c is
l(m,
n)
� Js- closed.
Proof:
If {x} is not
l(m,
n)
� J � closed, then the only
l(m,
n)
� J � open containing {x}c is X. Thus cln
({x}c)
�
X and {x}c is
l(m,
n)
� Js- closed.
6. On Strongly - λ(m, n) - J - continuous and irresolute
functions
Definition 6.1.
Let (X,lX1,
lX2)
and (Y,lY1,
lY2)
be generalized topological spaces. A function f: (X,lX1,
lX2)
�(Y,lY1,
lY2)
is said to be
l(m,
n)
-strongly J - continuous(briefly λ(m, n)-Js-
continuous) if f-1(F) is
l(m,
n)-
Js-closed in X for every
ln-
closed F of Y, where m, n = 1,2 and m
�
n.
Definition 6.2.
Let (X,lX1,
lX2)
and (Y,lY1,
lY2)
be generalized topological spaces. A function f: (X,lX1,
lX2)
�(Y,lY1,
lY2)
is said to be
l(m,
n)
-strongly J � irresolute (briefly λ(m, n)-Js-
irresolute) if f-1(F) is
l(m,
n)-
Js-closed in X for every
l(m,
n)-
Js-closed F of Y, where m, n = 1,2 and m
�
n.
Proposition 6.3.
Let f: (X,lX1,
lX2)
�
(Y,lY1,
lY2),
then, (1)
�(2):
(2)�(3):
(3) �(4)
(1)
f is λ(m, n)-Js- continuous,
(2)
For each x
�
X and for every
l
n-
open set V containing f(x), there exists a
l(m,
n)-Js-open
set U containing x such that f(U)
�
V:
(3)
f (λ(m, n)-Js cλ(A))
�
λn- cλ(f(A)) for every subset A of
X;
(4)
λ(m, n)-Js cλ (f-1(B))
�
f-1(λn- cλ (B)) for every
subset B of Y.
Proof:
(1) �(2):
Let x �
X and V be a
ln-open
subset of Y containing f(x). Then by (1), f -1(V)
is l(m,
n)-Js-open
of X containing x. If U = f -1(V), then f (U)
�
V.
(2)�(3):
Let A be a subset of X and f(x)
�(λn-
cλ (f(A)). Then, there exists a
l
n
� open subset V of Y containing f(x) such that V
�
f(A) = f.Then
by(2), there exist a
l(m,
n)-Js-open
set such that f(x)
�
f(U) �
V. Hence, f(U)
�
f(A) = f
and U �
A = f.
Consequently, x
�
λ(m, n)-Js cλ(A) and f(x)
�
f (λ(m, n)-Js cλ(A)).
(3) �(4):
Let B be a subset of Y and A = f-1(B). By (3) we obtain f(
λ(m, n) - Js cλ(f -1(B)))
�
l
n
- cλ (f(f -1(B)))
�
l
n
- cλ (B).Thus λ(m, n)-Js cλ
(f-1(B))
�
f-1(λn- cλ (B)).
Proposition 6.4.
If f: (X,lX1,
lX2)
�
(Y,lY1,
lY2)
and g: (Y,lY1,
lY2)
�
(Z,lZ1,
lZ2)
be two functions, then the following will hold
(i)
If g is ln
� continuous and f is λ(m, n)-Js � continuous,
then g�f
is λ(m, n)-Js � continuous.
(ii)
If g is λ(m, n) - Js �irresolute and f is λ(m,
n)-Js �irresolute, then g�f
is λ(m, n)-Js �irresolute.
(iii)
If g is λ(m, n) - Js �continuous and f is λ(m,
n)-Js�irresolute, then g�f
is λ(m, n)-Js �continuous.
Proof:
obvious
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