On Some Continuous and Irresolute Maps In Ideal Topological Spaces

IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 11, Issue 6 Ver. 1 (Nov. - Dec. 2015), PP 01-08
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DOI: 10.9790/5728-11610108 www.iosrjournals.org 1 | Page
On Some Continuous and Irresolute Maps In Ideal Topological
Spaces
M. Navaneethakrishnan1
, P. Periyasamy2
, S. Pious Missier3
1
Department of Mathematics, Kamaraj College, Thoothukudi, Tamilnadu, India.
2
Department of Mathematics, Kamaraj College, Thoothukudi, Tamilnadu, India.
3
Department of Mathematics, V.O.Chidambaram College, Thoothukudi, Tamilnadu, India.
Abstract: In this paper we introduce some continuous and irresolute maps called δˆ -continuity, δˆ -irresolute,
δˆ s-continuity and δˆ s-irresolute maps in ideal topological spaces and study some of their properties.
Keywords: δˆ -continuity, δˆ -irresolute, δˆ s- continuity, δˆ s-irresolute.
I. Introduction
Ideals in topological space (X, ) is a non-empty collection of subsets of X which satisfies the
properties (i) AI and BA  BI (ii) AI and BI  ABI. An ideal topological space is denoted by the
triplet (X, , I). In an ideal space (X, , I), if P(X) is the collection of all subsets of X, a set operator (.)* : P(X)
 P(X) called a local function [4] with respect to the topology  and ideal I is defined as follows: for AX, A*
= {xX / UAI, for every open set U containing x}. A Kuratowski closure operator cl*(.) of a subset A of X
is defined by cl*(A) = AA* [12]. Yuksel, Acikgoz and Noiri [14] introduced the concept of -I-closed sets in
ideal topological space. M. Navaneethakrishnan, P.Periyasamy, S.Pious missier introduced the concept of δˆ -
closed set [9] and δˆ s-closed set [8] in ideal topological spaces. K. Balachandran, P. Sundaram and H.Maki [1],
B.M. Munshi and D.S. Bassan [7], T.Noiri [10], Julian Dontchev and Maximilian Ganster [3], N.Levine [5]
introduced the concept of g-continuity, supercontinuity, -continuity, g-continuity, w-continuity, respectively.
The purpose of this paper is to introduce the concept of δˆ -continuity, δˆ -irresolute, δˆ s-continuity, δˆ s-irresolute
maps. Also, we study some of the characterization and basic properties of these maps.
II. Preliminaries
Definition 2.1 A subset A of a topological space (X, ) is called a
(i) Semi-open set [5] if A  cl(int(A))
(ii) g-closed set [6] if cl(A)  U whenever A  U and U is open in (X, )
(iii) w-closed set [11] cl(A)  U whenever A  U and U is semi-open.
(iv) -closed set [13] if cl(A) = A, where cl(A) = {xX : int(cl(U))A, for each U(x)}.
(v) g-closed set [3] if cl(A)  U whenever AU and U is open.
The complement of semi-open (resp. g-closed, w-closed, -closed, g-closed) set is called semi-closed
(resp. g-open, w-open, -open, g-open) set.
Definition 2.2 [14] Let (X, , I) be an ideal topological space, A a subset of X and x a point of X.
(i) x is called a -I-cluster point of A if Aint cl*(U)   for each open neighbourhood of x.
(ii) The family of all -I-cluster points of A is called the -I-closure of A and is denoted by [A]-I and
(iii) A subset A is said to be -I-closed if [A]-I = A. The complement of -I-closed set of X is said to be -I-
open.
Remark 2.3 [9] From the Definition 2.2 it is clear that [A]-I = {xX: int(cl*(U))  A  , for each U(x)}.
Notation 2.4 [9]. Throughout this paper [A]-I is denotd by cl(A).
Definition 2.5 A subset A of an ideal topological space (X, , I) is called
(i) Ig-closed set [2] if A*  U whenever A  U and U is open.
(ii) δˆ -closed set [9] if cl(A)  U whenever A  U and U is open.
(iii) δˆ s-closed set [8] if cl(A)  U whenever A  U and U is semi-open.
The complement of Ig-closed (resp. δˆ -closed, δˆ s-closed) set is called Ig-open (resp. δˆ -open; δˆ s–open) set.

On Some Continuous and Irresolute Maps In Ideal Topological Spaces
Definition 2.6 Let A be a subset of an ideal space (X, , I) then the δˆ s–closure of A is defined to be the
intersection of all δˆ s–closed sets containing A and is denoted by δˆ scl(A). That is δˆ scl(A) = {F: AF and F is
δˆ s–closed}.
Definition 2.7 A map f : (X, )  (Y, ) is said to be
[1]. g-continuous [1] if f-1
(F) is g-closed in X for every closed set F of Y.
[2]. g-irresolute [1] if f-1
(F) is g-closed in X for every g-closed set F of Y.
[3]. w-continuous [5] if f-1
(F) is w-closed in X for every closed set F of Y.
[4]. w-irresolute [5] if f-1
(F) is w-closed in X for every w-closed set F of Y.
[5]. g-continuous [3] if f-1
(F) is g-closed in X for every closed set F of Y.
[6]. g-irresolute [3] if f-1
(F) is g-closed in X for every g-closed set F of Y.
[7]. -continuous [10] if f-1
(U) is -open in X for every -open set U in Y.
[8]. Supercontinuous [7] if f-1
(U) is -open in X for every open set U in Y.
Definition 2.8 A map f: (X, , I1)  (Y, , I2) is said to be
(i) Ig-continuous if f-1
(F) is Ig-closed in X for every closed set F of Y.
(ii) Ig-irresolute if f-1
(F) is Ig-closed in X for every Ig-closed set F of Y.
Definition 2.9 A topological space (X, ) is called
(i) T½-space [6] if for every g-closed subset of X is closed.
(ii) T¾-space [3] if for every g-closed subset of X is -closed.
Definition 2.10 [2] An ideal space (X, , I) is called TI-space if for every Ig-closed subset is *-closed.
III. δˆ -Continuous And δˆ -Irresolute Maps
Definition 3.1 A function f from an ideal space (X, , I1) into an ideal space (Y, , I2) is called δˆ -continuous if
f-1
(F) is δˆ -closed in (X, , I1) for every closed set F of (Y, , I2).
Definition 3.2 A function f from an ideal space (X, , I1) into an ideal space (Y, , I2) is called δˆ -irresolute if f-1
(F) is δˆ -closed in (X, , I1) for every δˆ -closed set F in (Y, , I2).
Theorem 3.3 If a map f: (X, , I1)  (Y, , I2) from an ideal space (X, , I1) into an ideal space (Y, , I2) is δˆ -
continuous, then it is g-continuous.
Proof. Let f: (X, , I1)  (Y, , I2) be δˆ -continuous and F be any closed set in Y. Then the inverse image f-1
(F)
is δˆ -closed in (X, , I1). Since every δˆ -closed set is g–closed, f-1
(F) is g-closed in (X, , I1). Therefore f is g-
continuous.
Remark 3.4 The converse of Theorem 3.3 need not be true as seen in the following Example.
Example 3.5 Let X=Y=a,b,c,dwith topologies  = X, ,b,a,b,b,c,a,b,c, {a,b,d}},  = Y, ,b,
c,d, b,c,d and ideals I1={,{c}, {d}, {c,d}}, I2= ,b, c,b,c}. Let f: (X, , I1)  (Y, , I2) be a map
defined by f(a)=b, f(b)=c, f(c)=d and f(d)=a, then f is g–continuous but not δˆ - continuous, since for the closed
set F=a,b in (Y, , I2), f-1
(F) = a,d is not δˆ - closed set in (X, , I1).
Theorem 3.6 If a map f: (X, , I1)  (Y, , I2) from an ideal space (X, , I1) into an ideal space (Y, , I2) is δˆ -
continuous, then it is Ig-continuous.
Proof. Let f: (X, , I1)  (Y, , I2) be δˆ -continuous and F be any closed set in (Y, , I2). Then the inverse
image f-1
(F) is δˆ -closed in (X, , I1). Since every δˆ -closed set is Ig –closed, f-1
(F) is Ig-closed. Therefore f is Ig-
continuous.
Remark 3.7 The converse of the above Theorem is not always true as shown in the following Example.
Example 3.8 Let X=Y=a,b,c,dwith topologies  = X, ,b,d,b,d,b,c,d},  = Y, ,b, a,b,
b,c,a,b,c, a,b,d} and ideals I1={,{d}}, I2= ,c, d,c,d}. Let f: (X, , I1)  (Y, , I2) be a map
defined by f(a)=d, f(b)=a, f(c)=b and f(d)=c, then f is Ig –continuous but not δˆ - continuous, since for the closed
set F=c in (Y, , I2), f-1
(F) = d is not δˆ - closed set in (X,,I1).
Theorem 3.9 A map f: (X, , I1)  (Y, , I2) from an ideal space (X, , I1) into an ideal space (Y, , I2) is
supercontinuous, then f is δˆ - continuous.
Proof. Let f : (X, , I1)  (Y, , I2) is supercontinuous and U be an open set in (Y, , I2). Then f-1
(U) is -open
in (X, , I1). Since f-1
(Uc
) = [f-1
(U)]c
, f-1
(Uc
) is -closed in (X, , I1) for every closed set Uc
in (Y, , I2). Also,
since every -closed set is δˆ -closed f-1
(Uc
) is δˆ -closed for every closed set Uc
in (Y, , I2). Hence f is δˆ -
continuous.
Remark 3.10 The converse of the above Theorem is not always true as shown in the following Example.

Example 3.11 Let X = Y = {a, b, c, d} with topologies  = {X, , {b}, {c, d}, {b, c, d}},  = {Y, , {c}, {a, d},
{a, c, d}} and ideals I1 = {, {c}}, I2 = {, {a}, {d}, {a, d}}. Let f: (X, , I1)  (Y, , I2) be a map defined by
f(a) = b, f(b) = a, f(c) = c and f(d) = d, then f is δˆ -continuous but not supercontinuous because, for the open set
U = {a, d} in (Y, , I2), f-1
(U) = {b, d} is not -open in (X, , I1).
Theorem 3.12 A function f: (X, , I1)  (Y, , I2) is δˆ -continuous if and only if f-1
(U) is δˆ - open in (X,,I1)
for every open set U in (Y,,I2).
Proof. Necessity - Let f: (X, , I1)  (Y, , I2) be δˆ -continuous and U any open set in (Y, , I2). Then f-1
(Uc
) is
δˆ -closed in (X, , I1). But f-1
(Uc
) = [f-1
(U)]c
and so f-1
(U) is δˆ -open in (X, , I1).
Sufficiency - Suppose f-1
(U) is δˆ -open in (X, , I1) for every open set U in (Y,,I2). Again since f-1
(Uc
) = [f-
1
(U)]c
, f-1
(Uc
) is δˆ -closed in X, for every closed set Uc
in (Y,,I2). Therefore f is δˆ -continuous.
Definition 3.13 A map f: (X,,I1)(Y,,I2) is called -I-closed if the image of -I-closed set under f is -I-
closed.
Theorem 3.14 Let f: (X, , I1)  (Y, , I2) be continuous and -I-closed, then for every δˆ -closed subset A of
(X, , I1), f(A) is δˆ -closed in (Y, , I2).
Proof. Let A be δˆ -closed in (X, , I1). Let f(A)  U where U is open in (Y, , I2). Since A  f–1
(U) and A is
δˆ -closed and since f-1
(U) is open in (X,,I1), then cl(A)f-1
(U). Thus f(cl(A))U. Hence
cl(f(A))cl(f(cl(A)))= f(cl(A))U. Since f is -I-closed. Hence f(A) is δˆ -closed in (Y, , I2).
Remark 3.15 Let f: (X, , I1)  (Y, , I2) be δˆ -continuous function then it is clear that f(cl(A))  cl(f(A)) for
every -I-closed subset A of X. But the converse is not true. For instant, let X = Y = {a, b, c, d}, with topologies
 = {X, , {b}, {c, d}, {b, c, d}},  = {Y, , {a}, {b}, {a, b}, {c, d}, {a, c, d}, {b, c, d}} and ideals I1 = {,
{c}}, I2 = {, {a}}. Let f: (X, , I1)  (Y, , I2) be identity map. Then f(cl(A))  cl(f(A)) for every -I-closed
subset A of X. But for the closed set {b} of Y, f-1
({b}) = {b} is not δˆ -closed set in X. Therefore f is not δˆ -
continuous.
Remark 3.16 Let f: (X, , I1)  (Y, , I2) be a continuous function then it is clear that, f(cl(A))cl(f(A)) for
every -I-closed subset A of X. But the converse is not true. For instant, let X, Y, , , I1, I2, f be as Example
given in Remark 3.15. Then f(cl(A))  cl(f(A)) for every subset A of X. But for the closed set B = {b} in Y, f-
1
(B) = {b} is not closed in X. Hence f is not continuous.
Theorem 3.17 A map f: (X, , I1)  (Y, , I2) from an ideal space (X, , I1) into an ideal space (Y, , I2) is δˆ -
irresolute if and only if the inverse image of every δˆ -open set in (Y, , I2) is δˆ -open in (X, , I1).
Proof. Necessity - Assume that f is δˆ -irresolute. Let U be any δˆ -open set in (Y, , I2). Then X–U is δˆ -closed
in (Y, , I2). Since f is δˆ -irresolute f-1
(X–U) is δˆ -closed in (X, , I1). But f-1
(Uc
) = [f-1
(U)]c
and so f-1
(U) is δˆ -
open in X. Hence the inverse image of every δˆ -open set in (Y, , I2) is δˆ -open in X.
Sufficiency - Assume that the inverse image of every δˆ -open set in (Y, , I2) is δˆ -open in (X, , I1). Let V be
any δˆ -closed set in (Y, , I2). Then X–V is δˆ -open in (Y, , I2). By assumption, f-1
(X–V) is δˆ -open in (X, ,
I1). But f-1
(Vc
) = [f-1
(V)]c
and so f-1
(V) is δˆ -closed in (X, , I1). Therefore f is δˆ -irresolute.
Theorem 3.18 Let every δˆ -closed set is -closed in (X, , I1) and f: (X, , I1)  (Y, , I2) be δˆ -irresolute.
Then f is -continuous.
Proof. Let F be a -closed subset of (Y,,I2). By Theorem 3.3 [9], F is δˆ -closed. Since f is δˆ -irresolute, f-1
(F)
is δˆ -closed in (X, , I1). By hypothesis f-1
(F) is -closed. Then f is -continuous.
Remark 3.19 The converse of Theorem 3.18 is need not be true as shown in the following Example.
Example 3.20 Let X = Y = {a, b, c, d} with topologies  = {X, , {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}},
 = (Y, , {c}, {c, d}, {a, c, d}} and ideals I1 = {, {a}, {b}, {a, b}}, I2 = {, {b}, {d}, {b, d}}. Here every δˆ -
closed set is -closed in (X, , I1). Let f: (X, , I1)  (Y, , I2) be identity map. Then f is -continuous, but not
δˆ -irresolute because for the δˆ -closed set A = {a, b} in (Y, , I2), f-1
(A) = {a, b} is not δˆ -closed in (X, , I1).
Theorem 3.21 If f : (X, , I1)  (Y, , I2) is bijective, open and δˆ -continuous then f is δˆ -irresolute.
Proof. Let F be any δˆ -closed set in Y and f-1
(F)U, where U. Then it is clear that cl(F)f(U) and therefore
f-1
(cl(F))U. Since f is δˆ -continuous and cl(F) is a closed subset of (Y, , I2), cl(f-1
(cl(F)))U and hence
cl(f-1
(F))  U. Thus f-1
(F) is δˆ -closed in (X, , I1). This shows that f is δˆ -irresolute.

Theorem 3.22 Let f: (X, , I1)  (Y, , I2) be surjective, δˆ -irresolute and -closed. If every δˆ -closed set is -
closed in (X, , I1) then the same in (Y, , I2).
Proof. Let F be a δˆ -closed set in (Y, , I2). Since f is δˆ -irresolute, f-1
(F) is δˆ -closed in (X, , I1). Then by
hypothesis, f-1
(F) is -closed in (X, , I1). Since f is surjective and F is -closed in (Y, , I2).
Theorem 3.23 Let f: (X, , I1)  (Y, , I2) and g: (Y, , I2)  (Z, , I3) be any two functions. Then the
following hold.
(i) gof is δˆ -continuous if f is δˆ -irresolute and g is δˆ -continuous.
(ii) gof is δˆ -irresolute if f is δˆ -irresolute and g is δˆ -irresolute.
(iii) gof is g-continuous if f is g-irresolute and g is δˆ -continuous.
(iv) gof is Ig-continuous if f is Ig-irresolute and g is δˆ -continuous
Proof. (i) Let F be a closed set in (Z, , I3). Since g is δˆ -continuous, g-1
(F) is δˆ -closed in (Y, , I2). Since f is
δˆ -irresolute, f-1
(g-1
(F)) is δˆ -closed in (X,, I1). Thus (gof)-1
(F) = f-1
(g-1
(F)) is a δˆ -closed set in (X, , I1) and
hence gof is δˆ -continuous.
(ii) Let V be a δˆ -closed set in (Z, , I3). Since g is δˆ -irresolute, g-1
(V) is δˆ -closed in (Y, , I2). Also, since f is
δˆ -irresolute, f-1
(g-1
(V)) is δˆ -closed in (X, , I1). Thus (gof)-1
(V) = f-1
(g-1
(V)) is a δˆ -closed set in (X, , I1) and
hence gof is δˆ -irresolute.
(iii) Let F be a closed set in (Z, , I3). Since g is δˆ -continuous, g-1
(F) is δˆ -closed in (Y, , I2). Since every δˆ -
closed set is g-closed g-1
(F) is g-closed in (Y, , I2). Since f is g-irresolute, f-1
(g-1
(F)) is g-closed in (X, , I1).
Thus (gof)-1
(F) = f-1
(g-1
(F)) is g-closed in (X, , I1) and hence gof is g-continuous.
(iv) Let F be a closed set in (Z, , I3). Since g is δˆ -continuous g-1
(F) is δˆ -closed in (Y, , I2). Since every δˆ -
closed set is Ig-closed g-1
(F) is Ig-closed in (Y, , I2). Since f is Ig-irresolute, f-1
(g-1
(F)) is Ig-closed in (X, , I1).
Thus (gof)-1
(F) = f-1
(g-1
(F)) is Ig-closed in (X, , I1) and hence gof is Ig-continuous.
Remark 3.24 Composition of two δˆ -continuous functions need not be δˆ -continuous as shown in the following
Example.
Example 3.25 Let X = Y = Z = {a, b, c, d} with topologies  = {X, , {a}, {c, d}, {a, c, d}, {b, c, d}},  = {Y,
, {a, b, c}},  = {Z, , {b}, {c, d}, {b, c, d}} and ideals I1 = {, {a}}, I2 = {, {a}} and I3 = {, {d}}. Let f: (X,
, I1)  (Y, , I2) and g: (Y, , I2)  (Z,,I3) defined by f(a) = g(a) = d, f(b) = g(b) = b, f(c) = g(c) = c, and f(d)
= g(d) = a. Then f and g are δˆ -continuous but their composition gof: (X, , I1)  (Z,,I3) is not δˆ -continuous,
because for the closed set A = {a, c, d} in (Z,,I3), (gof)-1
(A) = {a, c, d} is not δˆ -closed in (X, , I1).
IV. δˆ S -Continuous And δˆ S –Irresolute Maps
Definition 4.1 A function f from an ideal space (X, , I1) into an ideal space (Y,,I2) is called δˆ s–continuous if
f-1
(F) is δˆ s – closed in (X,, I1) for every closed set F of (Y,,I2).
Definition 4.2 A function f from an ideal space (X,, I1) into an ideal space (Y,,I2) is called δˆ s–irresolute if f-
1
(F) is δˆ s – closed in (X,, I1) for every δˆ s -closed set F of (Y,,I2).
Theorem 4.3 If a map f: (X,, I1) (Y,,I2) from an ideal space (X,, I1) into an ideal space (Y,,I2) is δˆ s –
continuous then it is δˆ –continuous.
Proof : Let f: (X,, I1) (Y,,I2) be δˆ s –continuous and F be any closed set in (Y,,I2). Then the inverse
image f-1
(F) is δˆ s-closed. Since every δˆ s – closed set is δˆ -closed, f-1
(F) is δˆ -closed in (X,, I1). Therefore f is
δˆ -continuous.
Remark 4.4 The converse of Theorem 4.3 is not always true as shown in the following Example.
Example 4.5 Let X=Y={a,b,c,d} with topologies ={X, ,{c}, {c,d}, {a,c,d}}, ={Y, , {a}, {c,d}, {a,c,d},
{b,c,d}} and ideals I1={,{b},{d},{b,d}}, I2={,{a}}. Let f: (X,, I1) (Y,,I2) be an identity map, then F is
δˆ -continuous but not δˆ s–continuous since for the closed set F={a,b} in (Y,,I2), f-1
(F)={a,b} is not δˆ –closed
in (X,, I1).
continuous then it is g-continuous.

Proof : Let f: (X,, I1) (Y,,I2) be δˆ s –continuous and F be any closed set in (Y,,I2) . Then the inverse
image f-1
(F) is δˆ s – closed in (X,, I1). Since every δˆ s – closed set is g-closed, f-1
(F) is g-closed in (X,, I1).
Therefore f is g-continuous.
Remark 4.7 The reversible implication of Theorem 4.6 is not true as shown in the following Example.
Example 4.8 Let X=Y={a,b,c,d} with topologies  ={X, , {a}, {b}, {a,b}},  = {X, ,
{b},{a,b},{b,c},{a,b,c,},{a,b,d}} and ideals I1={{d}}, I2 = {, {a}, {b}, {a,b}}. Let f: (X,, I1) (Y,,I2) be
an identify map, then f is g- continuous but not δˆ s–continuous, since for the closed set F={c} in (Y,,I2), f-1
(F)={c} is not δˆ s –closed in (X,, I1) .
continuous then it is w-continuous.
Proof : Let f: (X,, I1) (Y,,I2) be δˆ s –continuous and F be any closed set in (Y,,I2). Then the inverse image
f-1
(F) is δˆ s –closed in (X,, I1). Since every δˆ s –closed set is w-closed f-1
(F) is w-closed in X. Therefore F is w-
continuous.
Remark 4.10 The converse of Theorem 4.9 need not be true as seen in the following Example.
Example 4.11 Let X=Y={a,b,c,d} with topologies ={ X, , {c}, {a,c},{b,c}, {a,b,c,},{a,c,d}},  =
{Y,,{b},{b,c}} and ideals I1={,{c}}, I2={}. Let f: (X,, I1) (Y,,I2) be a map defined by f(a)=c, f(b)=a,
f(c)=b and f(d)=d, then f is w-continuous but not δˆ s –continuous, because, for the closed set F={a,d} in
(Y,,I2), f-1
(F)={b,d} is not δˆ s–closed in (X,, I1).
continuous, then it is Ig-continuous.
Proof : Let f: (X,, I1) (Y,,I2) be δˆ s –continuous and F be any closed set in (Y,,I2). Then the inverse
image f-1
(F) is δˆ s –closed. Since every δˆ s –closed set is Ig-closed, f-1
(F) is Ig-closed in (X,, I1). Therefore f is
Ig-continuous.
Remark 4.13 The reversible direction of Theorem 4.12 is not always true as shown in the following Example.
Example 4.14 Let X=Y={a,b,c,d} with topologies  = {X,,{b},{d},{b,d},{b,c,d}},  = {Y,,{b}, {a,d},
{a,b,d},{a,c,d}} and ideals I1={,{d}}, I2={,{b}}. Let f:(X,, I1) (Y,,I2) be a map defined by f(a)=c, f(b)=a,
f(c)=d, and f(d)=b, then f is Ig-continuous, but not δˆ s –continuous, because for the closed set F={b,c} in
(Y,,I2), f-1
(F)={a,d} is not δˆ s –closed in (X,, I1)
Theorem 4.15 If f: (X,, I1) (Y,,I2) is supercontinuous then f is δˆ s –continuous.
Proof: Let f: (X,, I1) (Y,,I2) is supercontinuous and U be any open set in (Y,,I2). Then f-1
(U) is -open in
(X,, I1). Since f-1
(Uc
)=[f-1
(U)]c
, f-1
(Uc
) is -closed, in (X,, I1) for every closed Set Uc
in (Y,,I2). Also since
every -closed set is δˆ s –closed f-1
(Uc
) is δˆ s –closed for every closed Uc
in (Y,,I2). Hence, f is δˆ s –
continuous.
Remark 4.16 The following Example shows that the converse of Theorem 4.15 is not true.
Example 4.17 Let X,Y,Z,,I1,I2 and f be as in Example 3.11. Then f is δˆ s–continuous but it is not
supercontinuous because, for the open set U={a,d} in (Y,,I2), f-1
(U) ={b,d} is not –open in (X,, I1).
Theorem 4.18 Let f:(X,,I1)  (Y,,I2) be a map from an ideal space (X,,I1) into an ideal space (Y,,I2), then
the following are equivalent.
(i) f is δˆ s - continuous
(ii) The inverse image of each open set in Y is δˆ s open in X.
Proof: (i)  (ii) Assume that f:(X,,I1)  (Y,,I2) be a δˆ s – continuous. Let U be open in (Y,,I2). Then Uc
is
closed in (Y,,I2). Since f is δˆ s – continuous, f-1
(Uc
) is a δˆ s – closed in (X,,I1) But f-1
(Uc
) = [f-1
(U)]c
. Thus
[f-1
(U)]c
is δˆ s –closed in (X,,I1) and so f-1
(U) is δˆ s –open in (X,,I1)
(ii)  (i) Assume that the inverse image of each open set is δˆ s – open in (X,,I1). Let F be any closed set in
(Y,,I2). Then Fc
is open in (Y,,I2). By assumption, f-1
(Fc
) is δˆ s – open in (X,,I1). But f-1
(Fc
) = [f-1
(F)]c
. Thus
[f-1
(F)]c
is δˆ s – open in (X,,I1) and so f-1
(F) is δˆ s – closed in (X,,I1). Therefore f is δˆ s – continuous.
Theorem 4.19 Let f:(X,,I1)(Y,,I2) be a δˆ s-continuous function. Then f( δˆ scl(A))cl(f(A)) for every subset
A of X.

Proof: Since f(A)cl(f(A)), we have Af-1
(cl(f(A))). Also since cl(f(A)) is a closed set in (Y,,I2) and hence f-
1
(cl(f(A)) is a δˆ s-closed set containing A. Consequently δˆ scl(A) f-1
(cl(f(A))). Therefore f( δˆ scl(A))  f(f-
1
(cl(f(A)))) cl(f(A)).
Remark 4.20 The following Example shows that the converse of Theorem 4.19 is not true.
Example 4.21 Let X,Y,, ,I1,I2 and f be f as Example given in Remark 3.15. Then f(δˆ scl(A))cl(f(A)) for
every subset A of X. But for the closed set A={b}, f-1
{A}={b} is not δˆ s – closed in X. Hence f is not δˆ s–
continuous.
Remark 4.22 Let f:(X,,I1)  (Y,,I2) be a δˆ s–continuous function then it is clear that f(cl(A)) cl(f(A)) for
every -I-closed subset A of X. The following Example shows that the converse is not true. Let X,Y,, ,I1,I2,
and f be as Example given in Remark 3.15. Then f(cl(A))cl(f(A)) for every -I-closed subset A of X. But for
the closed set B={b}, f-1
{B}={b} is not δˆ s – closed in X. Therefore f is not δˆ s–continuous.
Theorem 4.23 Let :(X,,I1)  (Y,,I2) and g: (Y,,I2)  (Z,,I3) be any two functions. Then the following
hold.
(i) gof is δˆ s – continuous if f is δˆ s – continuous and g is continuous.
(ii) gof is g– continuous if f is g – irresolute and g is δˆ s – continuous.
(iii) gof is Ig – continuous if f is Ig – irresolute and g is δˆ s – continuous.
(iv) gof is w– continuous if f is w– irresolute and g is δˆ s – continuous.
(v) Let (Y,,I2) be T¾ - space. Then gof is δˆ s – continuous if f is δˆ s – continuous and g is g – continuous.
(vi) Let every δˆ s – closed set is -I-closed in (Y,,I2). Then gof is δˆ s – continuous if both f and g are δˆ s –
continuous.
(vii) Let (Y,,I2) be T½ – Space. Then gof is δˆ s – continuous if f is δˆ s – continuous and g is g – continuous
(viii) Let every Ig-closed set is closed in (Y,,I2). Then gof is δˆ s-continuous if f is δˆ s – continuous and g is Ig –
continuous
(ix) Let (Y,, I2) be T½ – Space. Then gof is g- irresolute if f is δˆ s – continuous and g is g– irresolute
(x) Let every Ig-closed set is closed in (Y,,I2). Then gof is Ig –irresolute if f is δˆ s – continuous and g is Ig –
irresolute
Proof: (i) Let F be a closed set in (Z,,I3). Since g is continuous g-1
(F) is also closed in (Y,, I2). Since f is δˆ s-
continuous f-1
(g-1
(F)) is δˆ s closed in (X,,I1). Thus (gof)-1
= f-1
(g-1
(F)) is δˆ s closed in (X,,I1). Therefore gof is
δˆ s-continuous.
(ii) Let F be a closed set in (Z, , I3). Since g is δˆ s –continuous, g-1
(F) is δˆ s closed in (Y,, I2). Since every δˆ s-
closed set is g–closed, g-1
(F) is g - closed in (Y,,I2). Also, since f is g-irresolute f-1
(g-1
(F)) is g- closed in
(X,,I1) Thus (gof)-1
= f-1
(g-1
(F)) is g-closed in (X,,I1). Therefore gof is g-continuous.
(iii). Since g is δˆ s-continuous, for any closed set F in (Z,,I3), g-1
(F) is δˆ s - closed in (Y,, I2). Since every δˆ s
- closed set is Ig-closed and f is Ig-irresolute, f-1
(g-1
(F)) is Ig – closed in (X,,I1). Hence gof is Ig-continuous.
(iv) Since g is δˆ s-continuous for any closed set F in (Z, , I3), g-1
(F) is δˆ s - closed in (Y, , I2). Since every δˆ s-
closed set is w–closed and f is w–irresolute , f-1
(g-1
(F)) is w-closed in (X, , I1). Hence gof is w-continuous.
(v) Since g is g – continuous, for every closed set F in (Z, , I3), g-1
(F) is g-closed in (Y,,I2). Since by
hypothesis and f is δˆ s-continuous, f-1
(g-1
(F)) is δˆ s-closed in (X, , I1). Hence gof is δˆ s-continuous.
(vi) Since g is δˆ s-continuous, for every closed set F in (Z, , I3), g-1
(F) is δˆ s - closed in (Y,,I2). By
hypothesis g-1
(F) is -I-closed. Since every -I –closed set is closed, g-1
(F) is closed in (Y, ,I2). Also since f is
δˆ s-continuous f-1
(g-1
(F) is δˆ s-closed in (X, , I1). Therefore gof is δˆ s-continuous.
(vii) Since g is g-continuous and by the assumption, for every closed set F in (Z, , I3), g-1
(F) is closed. Also
since f is δˆ s–continuous. f-1
(g-1
(F)) is δˆ s-closed in (X, ,I1). Therefore gof is δˆ s –continuous.
(viii) Since g is Ig-continuous and by the assumption, for every closed set F in (Z, , I3), g-1
(F) is *-closed in (Y,
, I2). Also, since f is δˆ s-continuous, f-1
(g-1
(F)) is δˆ s-closed in (X, , I1) and hence gof is δˆ s-continuous.
(ix) Since g is g-irresolute and by the assumption, for every g-closed set in (Z, , I3), g-1
(F) is closed in (Y,,I2).
Also, since f is δˆ s-continuous and every δˆ s-closed set is g–closed, f-1
(g-1
(F)) is g–closed in (X,,I1). Therefore
gof is g- irresolute.

(x) Let F be an Ig-closed set in (Z, , I3), Since g is Ig-irresolute and by the assumption, g-1
(F) is closed in (Y, ,
I2). Again, since f is δˆ s –continuous and every δˆ s-closed set is Ig-closed, f-1
(g-1
(F)) is Ig-closed in (X,,I1).
Therefore gof is Ig- irresolute.
Remark 4.24 Composition of two δˆ s-continuous functions need not be δˆ s-continuous as shown in the
following Example.
Example 4.25 Let X,Y,Z, , , , I1, I2, I3, f, g and A be as in Example 3.25. Then f and g are δˆ s-continuous but
their composition gof is not δˆ s– continuous because, (gof)-1
(A) =a,c,d is not δˆ s– closed in (X, , I1).
Definition 4.26 Let (X, , I1) be an ideal space and δˆ s =UX : δˆ scl(X–U) = X–U, δˆ s is a topology in (X,
, I1).
Theorem 4.27 Let f : (X, , I1) → (Y, , I2) be a function from an ideal space (X, , I1) into an ideal space (Y, ,
I2) such that δˆ s is a topology on (X, , I1). Then the following are equivalent.
(i) For every subset A of X, f( δˆ s cl(A))  cl (f(A)) holds.
(ii) f : (X,  δˆ s)  (Y,) is continuous.
Proof: (i) (ii) Let A be a closed subset in Y. By hypothesis f ( δˆ s cl (f-1
(A))  cl (f(f-1
(A)  cl(A) = A.
Therefore δˆ s cl (f-1
(A)  f-1
(A). Also f-1
(A)  δˆ s cl (f-1
(A). Hence δˆ s cl (f-1
(A)) = f-1
(A). Thus f-1
(A) is
closed in (X,δˆ s) and so f is continuous.
(ii) (i) Let AX, then cl(f(A)) is closed in (Y, , I2). Since f: (X,δˆ s, I1)  (Y, , I2) is continuous, f-1
(cl(f(A))) is closed in (X, δˆ s, I1) and hence δˆ scl (f-1
(cl(f(A))))= f-1
(cl(f(A))). Since A  f-1
(f(A))  f-1
(cl(f(A))), δˆ s cl(A)  δˆ s cl (f-1
(cl(f(A))) = f-1
(cl(f(A))). Therefore f( δˆ s cl(A))  cl (f(A)).
Theorem 4.28 Let f : (X, , I1)  (Y,, I2) be a function from an ideal space (X, , I1) into an ideal space (Y, ,
I2). Then the following are equivalent.
(i) For each point x in X and each open set V in Y with f(x)V, there is a δˆ s-open set U in X such that xU
and f(U)V.
(ii) For each subset A of X, f (δˆ scl (A))  cl(f(A))
(iii) For each subset G of Y, δˆ scl (f-1
(G)  f-1
(cl (G)
(iv) For each subset G of Y, f-1
(int(G)  δˆ sint (f-1
(G))
Proof : (i) (ii) Let yf ( δˆ scl (A)) and V be any open set of Y containing y. Since y  f ( δˆ scl (A)), there
exists xδˆ scl(A) such that f(x)=y. Since f(x)V, then by hypothesis there exists a δˆ s- open set U in X such that
xU and f(U)V. Since xδˆ scl(A), then by Theorem 5.7 [8] UA ≠. Then =f(UA)f(U)f(A)  Vf(A)
and hence Vf(A). Therefore we have y=f(x)cl(f(A).
(ii) (i) Let x  X and V be any open set in Y containing f(x). Let A = f-1
(Vc
). Then xA. Since f( δˆ s
cl(A))cl(f(A)Vc
, δˆ s cl(A)f-1
(Vc
)=A. Since xA, x δˆ s cl (A). By Theorem 5.7 [8] there exists a δˆ s -open
set U containing x such that UA = , and so UAc
and hence f(U)f(Ac
)V.
(ii) (iii) Let G be any subset of Y. Replacing A by f-1
(G) in (ii), we get f( δˆ scl (f-1
(G)))  cl(f(f-1
(G)))  cl
(G).
(iii) (ii) Put G=f(A) in (iii) we get, δˆ scl (f-1
(f(A)))  f-1
(cl(f(A))) and hence f( δˆ scl(A))  cl (f(A)).
(iii) (iv) Let G be any subset in Y. Then Y–GY. By (iii), δˆ scl(f-1
(Y–G))  f-1
(cl(Y–G)). Therefore X– δˆ s int
(f-1
(G))  X – f-1
(int (G)) and so f-1
(int (G))  δˆ s int (f-1
(G).
iv) (iii) Let G be any subset in Y. Then Y–GY. By (iv), f-1
(int (Y–G)) δˆ s int(f-1
(Y–G)). Therefore X–f-1
(cl(G))  X–δˆ s cl (f-1
(G)) and hence δˆ scl(f-1
(G))  f-1
(cl(G)).
Theorem 4.29 A map f: (X, , I1)  (Y, , I2) is δˆ s – irresolute if and only if the inverse image of every δˆ s-
open set in (Y, , I2) is δˆ s –open in (X, , I1).
Proof : Necessity - Assume that f is δˆ s - irresolute. Let U be any δˆ s - open set in (Y,, I2). Then Uc
is δˆ s -
closed in (Y, , I2). Since f is δˆ s – irresolute, f-1
(Uc
) is δˆ s – closed in (X, , I1). But f-1
(Uc
)=[f-1
(U)]c
and so f-1
(U) is δˆ s-open in (X, , I1). Hence the inverse image of every δˆ s-open set in (Y, , I2) is δˆ s-open in (X, , I1).
Sufficiency – Assume that the inverse image of every δˆ s - open set in (Y,, I2) is δˆ s-open in (X, , I1). Let V be
any δˆ s- closed set in (Y, , I2). Then Vc
is δˆ s-open in (Y,, I2). By assumption, f-1
(Vc
) is δˆ s-open in (X, , I1).
But f-1
(Vc
)=[f-1
(V)]c
and so f-1
(V) is δˆ s-closed in (X, , I1). Therefore f is δˆ s-irresolute.

Theorem 4.30 Let every δˆ s - closed set is -closed in (X, , I1). If f: (X, , I1)  (Y, , I2) be δˆ s-irresolute.
Then f is -continuous.
Proof : Let F be a –closed subset of (Y, , I2). By Theorem 3.2 [8], F is δˆ s-closed. Since f is δˆ s-irresolute, f-
1
(F) is δˆ s-closed in (X, , I1). By hypothesis f-1
(F) is –closed. Then f is –continuous.
Theorem 4.31 Let f: (X, , I1)  (Y, , I2) and g: (Y, , I2)  (Z, , I3) be any two functions. Then the
following hold.
(i) gof is δˆ s – continuous if f is δˆ s – irresolute and g is δˆ s – continuous.
(ii) gof is δˆ s – irresolute if f is δˆ s – irresolute and g is δˆ s – irresolute.
(iii) hof is g – continuous if f is g – irresolute and h is δˆ s – continuous.
(iv) gof is w – continuous if f is w – irresolute and g is δˆ s – continuous.
(v) gof is Ig- continuous if f is Ig – irresolute and g is δˆ s – continuous.
(vi) gof is δˆ s – continuous if f is δˆ s – irresolute and g is δˆ s – continuous.
Proof : (i) Since g is δˆ s – continuous, for every closed set F in (Z, , I3), g-1
(F) is δˆ s – closed in (Y, , I2).
Since f is δˆ s – irresolute, f-1
(g-1
(F) is δˆ s – closed in (X, , I1).
(ii) Since g is δˆ s – irresolute, for every δˆ s – closed set F in (Z, , I3), g-1
(F) is δˆ s – closed in (Y, , I2). Since f
is δˆ s – irresolute, f-1
(g-1
(F)) is δˆ s – closed in (X, , I1).
(iii) Since h is δˆ s –continuous, for every closed set F in (Z, , I3), h-1
(F) is δˆ s – closed set in (Y,, I2). Since f is
g-irresolute and every δˆ s - closed set is g–closed, f-1
(h-1
(F)) is g – closed in (X, , I1).
(iv) Since g is δˆ s – continuous, f is w – irresolute and every δˆ s – closed set is w – closed, f-1
(g-1
(F)) is w –
closed in (X, , I1) for every closed set F in (Z,,I3).
(v) Since g is δˆ s – continuous, f is Ig – irresolute and every δˆ s – closed set is Ig – closed, f-1
(g-1
(F)) is Ig –
closed in (X, , I1) for every closed set F in (Z, , I3).
(vi) Since g is δˆ s – continuous, f is δˆ – irresolute and every δˆ s – closed set is δˆ – closed, f-1
(g-1
(F)) is δˆ –
closed in (X, , I1) for every closed set F in (Z, , I3).
Theorem 4.32 (i) f:( X, , I1) (Y, , I2) is a δˆ s – continuous, surjection and X is δˆ s – connected then Y is
connected.
(ii) If f:( X, , I1) (Y, , I2) is δˆ s – irresolute, surjection and X is δˆ s – connected then Y is δˆ s – connected.
Proof: (i) Suppose Y is not connected. Then Y = AB where AB = , A, B and A, B are open in Y.
Since f is δˆ s – continuous and onto X = f-1
(A) f-1
(B) where f-1
(A) and f-1
(B) are disjoint non-empty δˆ s – open
sets in X. This contradicts the fact that X is δˆ s – connected. Hence Y is connected.
(ii) The proof is similar to the proof of (i).
References
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