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DECOMPOSITION OF β -CONTINUITY IN IDEAL TOPOLOGICAL
SPACES
S. JAFARI β , K. VISWANATHAN, J.JAYASUDHA
* College of Vestsjaelland South, Herrestrade 11,
4200 Slagelse,Denmark
e-mail : πππ πππππππ ππ@πππππ.πππ
Post-Graduate and Research Department of Mathematics, N G M College,
Pollachi - 642 001, Tamil Nadu, INDIA
e-mail : π£ππ π’πππ@π¦πβππ.πππ
Abstract
In this paper, we introduce and investigate the notions of βπ -continuous maps
and βπ -irresolute maps in ideal topological spaces. Also we introduce the notion of
π ππβ β πΌ sets and π ππβ β πΌ continuous maps to obtain a decomposition of β -continuity.
1
Introduction and Preliminaries
The concept of ideals in topological spaces is treated in the classic text by Kuratowski
[7] and Vaidyanathaswamy [13]. The notion of β -open sets in topological spaces was
introduced by Jankovic and Hamlett [6]. Dontchev et al. [2] introduced and studied
the notion of βπ -closed sets. Recently, Navaneethakrishnan and Paulraj Joseph [10]
have studied further the properties of βπ -closed sets and βπ -open sets. An ideal β
on a topological space (π, π ) is a non-empty collection of subsets of π satisfying the
following properties: (1) π΄ β β and π΅ β π΄ imply π΅ β β (heredity); (2) π΄ β β and
π΅ β β imply π΄ βͺ π΅ β β (ο¬nite additivity).A topological space (π, π ) with an ideal β on
π is called an ideal topological space and is denoted by (π, π, β). For a subset π΄ β π,
π΄β (β) = {π₯ β π : π β© π΄ β
/ β for every π β π (π₯)}, is called the local function [7] of π΄
with respect to β and π . We simply write π΄β in case there is no chance for confusion.
A Kuratowski closure operator πΆπβ (.) for a topology π β (β) called the β -topology ο¬ner
than π is deο¬ned by πΆπβ (π΄) = π΄ βͺ π΄β [13]. Let (π, π ) denote a topological space on
which no separation axioms are assumed unless explicitly stated. In a topological space
0
0
2000 Mathematics subject classiο¬cation: Primary 54A05; Secondary 54D15, 54D30.
keywords : βπ -closed set, βπ -open set, π ππβ β πΌ set
2
S. Jafari, K. Viswanathan, J.Jayasudha
(π, π ), the closure and the interior of any subset π΄ of π will be denoted by πΆπ(π΄) and
πΌππ‘(π΄), respectively. A subset π΄ of a space is said to be semi-open [9] if π΄ β ππ(πππ‘π΄) .
A subset π΄ of a topological space (π, π ) is said to be g-closed [8] (resp. π -closed [12])
if πΆπ(π΄) β π whenever π΄ β π and π is open(resp.semi-open) in π; A subset π΄ of a
topological space (π, π ) is locally closed (brieο¬y LC) if π΄ = π β© π where U is open and
V is closed. A subset π΄ of a topological space (π, π ) is semi-locally closed (brieο¬y slc)
if π΄ = π β© π where U is semi-open and V is semi-closed. A subset π΄ of a topological
space (π, π ) is called π ππβ set if π΄ = π β© π where U is semi-open and V is closed.
A function π : (π, π ) β (π, π) is g-continuous[8] (resp. π -continuous) if π β1 (π ) is
g-closed(resp. π -closed) in (π, π ) for every closed set V of (π, π) . A subset π΄ of an
ideal space (π, π, β) is β -closed [6] if π΄β β π΄ . A subset π΄ of an ideal space (π, π, β)
is βπ -closed [2] (resp. βπ -closed [11]) if π΄β β π whenever π΄ β π and π is open(resp.
semi-open). An ideal space (π, π, β) is called a πβ space [2] (resp. πβπ [11]) if every
βπ -closed (resp. βπ -closed) subset of X is β -closed.
Lemma 1.1. [6] Let (π, π, β) be a space with an ideal β on π, and π΄ is a subset of
π. Then,
1. Every β -closed set is βπ -closed,
2. Every βπ -closed set is βπ -closed.
We denote the family of all βπ -closed (resp. βπ - open) subsets of an ideal space (π, π, β)
by βπ(π) (resp. βππ(π) ) .
Deο¬nition 1.2. [5] A function f: (π, π, β) β (π, π) is said to be β -continuous if π β1 (π΄)
is β -closed in (π, π, β) for every closed set A in (π, π) .
2
βπ -continuity and βπ -irresoluteness
Deο¬nition 2.1. A function f: (π, π, β) β (π, π) is said to be βπ -continuous if π β1 (π΄)
is βπ -closed in (π, π, β) for every closed set A in (π, π) .
Remark 2.2.
If β = {β
} in the above deο¬nition then the notion of βπ -continuity
coincides with the notion of π -continuity.
Decomposition of β -continuity in ideal topological spaces
3
Deο¬nition 2.3. A function f: (π, π, β) β (π, π, π₯ ) is said to be βπ -irresolute if π β1 (π΄)
is βπ -closed in (π, π, β) for every βπ -closed set A in (π, π, π₯ ) .
Theorem 2.4. For a function f: (π, π, β) β (π, π) , the following hold:
1. Every continuous function is βπ -continuous;
2. Every β -continuous function is βπ -continuous;
3. Every π -continuous function is βπ -continuous;
4. Every βπ -continuous function is βπ -continuous.
Proof. (1) Let f be a continuous function and V be a closed set in (π, π) . Then π β1 (π )
is closed in (π, π, β) . Since every closed set is β -closed and hence βπ -closed, π β1 (π ) is
βπ -closed in (π, π, β) .Therefore, f is βπ -continuous.
(2) Let f be a β -continuous function and V be a closed set in (π, π) . Then π β1 (π )
is β -closed in (π, π, β) . Since every β -closed set is βπ -closed, π β1 (π ) is βπ -closed in
(π, π, β) .Therefore, f is βπ -continuous.
(3) Let f be an π -continuous function and V be a closed set in (π, π) . Then π β1 (π ) is
π -closed in (π, π, β) . Since every π -closed set is βπ -closed [11], π β1 (π ) is βπ -closed
in (π, π, β) .Therefore, f is βπ -continuous.
(4) Let f be a βπ -continuous function and V be a closed set in (π, π) . Then π β1 (π ) is
βπ -closed in (π, π, β) . Since every βπ -closed set is βπ -closed, π β1 (π ) is βπ -closed in
(π, π, β) .Therefore, f is βπ -continuous.
β‘
Remark 2.5. The relationships deο¬ned above, are shown in the following diagram:
continuity β π -continuity β π -continuity
β
β
β
β -continuity β βπ -continuity β βπ -continuity
None of these implications is reversible as shown by the following examples.
Example 2.6. Let π = {π, π, π}, π = {β
, {π}, {π, π}, π} , β = {β
, {π}} and π =
{β
, {π}, π} . Deο¬ne f: (π, π, β) β (π, π) as f(a)=b, f(b)=a, f(c)=c. Then for the set
4
S. Jafari, K. Viswanathan, J.Jayasudha
π = {π, π} , π β1 (π ) = {π, π} is βπ -closed but not closed. Hence f is βπ -continuous but
not continuous.
Example 2.7. Let π = {π, π, π}, π = {β
, {π}, {π, π}, π} , β = {β
, {π}} and π =
{β
, {π, π}, π} . Deο¬ne f: (π, π, β) β (π, π) to be the identity map. Then for the set
π = {π} , π β1 (π ) = {π} is βπ -closed but not β -closed. Hence f is βπ -continuous but
not β -continuous.
Example 2.8. Let π = {π, π, π, π}, π = {β
, {π}, {π, π, π}, π} , β = {β
, {π}} and
π = {β
, {π, π}, π} . Deο¬ne f: (π, π, β) β (π, π) to be the identity map. Then for the set
π = {π, π} , π β1 (π ) = {π, π} is βπ -closed but not βπ -closed. Hence f is βπ -continuous
but not βπ -continuous.
Theorem 2.9. A function f: (π, π, β) β (π, π) is βπ -continuous if and only if π β1 (π )
is βπ -open in (π, π, β) for every open set V in (π, π) .
Proof. Let f: (π, π, β) β (π, π) be βπ - continuous and V be an open set in (π, π) . Then
π π is closed in (π, π) and π β1 (π π ) is βπ -closed in (π, π, β) . But π β1 (π π ) = (π β1 (π ))π
and so π β1 (π ) is βπ -open in (π, π, β) .
Conversely, Suppose that π β1 (π ) is βπ -open in (π, π, β) for each open set V in (π, π) .
Let F be a closed set in (π, π) . Then πΉ π is open in (π, π) and by hypothesis π β1 (πΉ π )
is βπ -open in (π, π, β) . Since π β1 (πΉ π ) = (π β1 (πΉ ))π , we have π β1 (πΉ ) is βπ -closed in
(π, π, β) and so f is βπ -continuous.
Deο¬nition 2.10. For A function f: (π, π, β) β (π, π), the subset {(π₯, π (π₯)) : π₯ β π} β
π × π is called the graph of f and is denoted by G(f).
Lemma 2.11. The following properties are equivalent for a graph G(f ) of a function f:
1. G(f ) is βπ -continuous;
2. For each (π₯, π (π₯)) β π × π βπΊ(π ), there exists an βπ -open set ππ of X containing
π₯ and an open set V containing π (π₯) such that π(ππ ) β© π β= π.
Theorem 2.12. Let f: (π, π, β) β (π, π) be a function and g: π β π × π the graph
function of f. If g is βπ -continuous then f is βπ -continuous.
Decomposition of β -continuity in ideal topological spaces
5
Proof. Suppose that g is βπ -continuous. Let π₯ β π and V be any open set of Y
containing f(x). Then π × π is open in π × π and by βπ -continuity of g, there exists
π β βππ(π, π ) containing π₯ such that π(π ) β π × π . Therefore we obtain π (π ) β π.
This shows that f is βπ -continuous.
Theorem 2.13. Let (π, π, β) be an ideal topological space.
If π΄ β βπ(π) and
π΄ β π0 β βπ(π) then π΄ β βπ(π0 ).
Proof. Let π΄ β βπ(π) and π΄ β π0 β βπ(π) then ππβ (π΄) β© π0 β π whenever π΄ β π
β (π΄) = ππβ (π΄) β© π (Lemma 3.2 [1]).Hence π΄ β βπ(π ).
and U is semi-open. Also πππ
0
0
0
Theorem 2.14. If f: (π, π, β) β (π, π) is an βπ -continuous function and π0 β βππ(π) .
Then the restriction π /π0 : (π0 , π /π0 , β/π0 ) β (π, π) is βπ -continuous.
Proof. Let V be any open set of (π, π) . Since f is βπ -continuous, π β1 (π ) is βπ -open
in (π, π, β) and π β1 (π ) β© π0 = (π /π0 )β1 (π ) β βππ(π) . Moreover by Theorem 2.11
we have (π /π0 )β1 (π ) β βππ(π0 ) . This shows that π /π0 is βπ -continuous.
Remark 2.15. The composition of two βπ -continuous maps need not be βπ -continuous
as seen from the following example:
Example 2.16. Let π = π = π = {π, π, π, π}, π = {β
, {π}, {π, π, π}, π} and
π = {β
, {π}, π }, π = {β
, {π, π}, π}, β = {β
, {π}, {π}, {π, π}}, π₯ = {β
, {π}}. Deο¬ne
f: (π, π, β) β (π, π, π₯ ) and g: (π, π, π₯ ) β (π, π) to be the identity map. Then the map f
is βπ -continuous, g is βπ -continuous but π β π is not βπ -continuous.
Theorem 2.17. Let f: (π, π, β) β (π, π, π₯ ) be βπ -continuous and g: (π, π, π₯ ) β (π, π)
be continuous. Then π β π : (π, π, β) β (π, π) is βπ -continuous.
Proof. Let g be a continuous function and V be any open set in (π, π) . Then π β1 (π ) is
open in (π, π, π₯ ) . Since f is βπ -continuous, π β1 (π β1 (π )) = (π β π )β1 (π ) is βπ -open in
(π, π, β) . Hence π β π is βπ -continuous.
Theorem 2.18. A function f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute if and only if the
inverse image of every βπ -open set in (π, π, π₯ ) is βπ -open in (π, π, β) .
Proof. This is a consequence of Theorem 2.9.
6
S. Jafari, K. Viswanathan, J.Jayasudha
Theorem 2.19. Let f: (π, π, β) β (π, π, π₯ ) and g: (π, π, π₯ ) β (π, π, π¦) be βπ -irresolute.
Then (π β π ) : (π, π, β) β (π, π, π¦) is βπ -irresolute.
Proof. Let g be a βπ -irresolute function and V be any π¦π -open set in (π, π, π¦) . Then
π β1 (π ) is π₯π -open in (π, π, π₯ ) . Since f is βπ -irresolute, π β1 (π β1 (π )) = (π β π )β1 (π )
is βπ -open in (π, π, β) . Hence π β π is βπ -irresolute.
Theorem 2.20. If f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute and g: (π, π, π₯ ) β (π, π) is
β - continuous then π β π : (π, π, β) β (π, π) is βπ - continuous.
Proof. Let V be any closed set of (π, π) . Then π β1 (π ) is β -closed in (π, π, π₯ ) .
Therefore, π β1 (π β1 (π )) = (π β π )β1 (π ) is βπ -closed in (π, π, β) , since every β -closed set
is βπ - closed. Hence π β π is βπ -continuous.
Theorem 2.21. Let (π, π, β) be an ideal topological space, (π, π, π₯ ) be a πβπ -space
and (π, π) be a topological space.
If f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute and
g: (π, π, π₯ ) β (π, π) is βπ - continuous then π β π is βπ - continuous.
Proof. Analogous to the Proof of Theorem 2.20.
Theorem 2.22. Let (π, π, β) be an ideal topological space, (π, π, π₯ ) be a πβ -space
and (π, π) be a topological space.
If f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute and
g: (π, π, π₯ ) β (π, π) is βπ - continuous then π β π is βπ - continuous.
Proof. Analogous to the Proof of Theorem 2.20.
3
Decomposition of β -continuity
In this section, we obtain a decomposition of β -continuity in ideal topological spaces.
In order to obtain the decomposition of β -continuity we introduce the notion of π ππβ β πΌ
set and π ππβ β πΌ continuous mapping in ideal topological spaces and prove that a map is
β -continuous if and only if it is both βπ -continuous and π ππβ β πΌ continuous.
Deο¬nition 3.1. A subset π΄ of an ideal topological space (π, π, β) is called π ππβ β πΌ set
if π΄ = π β© πΉ where U is semi-open and F is β -closed.
Remark 3.2. Every β -closed set is π ππβ β πΌ set but not conversely.
7
Decomposition of β -continuity in ideal topological spaces
Example
3.3. Let
π
=
{π, π, π, π},
π
=
{β
, {π}, {π, π, π}, π}
and
β
=
{β
, {π}, {π}, {π, π}}. Then the set π΄ = {π} is a π ππβ β πΌ set but not a β -closed set.
Remark 3.4. The notions of βπ -closed sets and π ππβ β πΌ sets are independent.
Example 3.5. Let π = {π, π, π}, π = {β
, {π}, {π, π}, π} and β = {β
, {π}}. Then the set
π΄ = {π} is βπ -closed but not π ππβ β πΌ set.
Example
3.6. Let
π
=
{π, π, π, π},
π
=
{β
, {π}, {π, π, π}, π}
and
β
=
{β
, {π}, {π}, {π, π}}. Then the set π΄ = {π} is π ππβ β πΌ set but not βπ -closed.
Theorem 3.7. A subset of an ideal topological space (π, π, β) is β -closed if and only if
it is both βπ -closed and an π ππβ β πΌ set.
Proof. Necessity is trivial. To prove the suο¬ciency, assume that A is both βπ -closed
and an π ππβ β πΌ -set. Then π΄ = π β© πΉ wher U is semi-open and F is β -closed. Therefore
π΄ β π and π΄ β πΉ and so by hypothesis, π΄β β π and π΄β β πΉ . Thus π΄β β π β© πΉ = π΄ .
Hence A is β -closed.
Deο¬nition 3.8. A function f: (π, π, β) β (π, π) is said to be π ππβ β πΌ continuous if
π β1 (π ) is an π ππβ β πΌ set in (π, π, β) for every closed set V in (π, π).
Example 3.9. Let π = {π, π, π, π}, π = {β
, {π}, {π, π, π}, π} , β = {β
, {π}, {π}, {π, π}}
and π = {β
, {π}, π}. Deο¬ne the map f: (π, π, β) β (π, π) to be the identity map. Then f
is a π ππβ β πΌ continuous map.
Remark 3.10. From the deο¬nitions it is clear that every β -continuous map is π ππβ β πΌ
continuous. But the converse is not true.
Example 3.11. Let π = {π, π, π, π}, π = {β
, {π}, {π, π, π}, π} , β = {β
, {π}, {π}, {π, π}}
and π = {β
, {π, π, π}, π}. Deο¬ne the map f: (π, π, β) β (π, π) to be the identity map.
Then f is π ππβ β πΌ continuous but not β -continuous, as for the closed set π = {π} in
(π, π) π β1 (π ) = {π} is not β -closed.
Remark 3.12. The concepts of βπ -continuity and π ππβ β πΌ continuity are independent
as seen from the following examples.
Example 3.13. The map f deο¬ned in Example 3.11 is π ππβ β πΌ continuous but not βπ continuous, as for the closed set π = {π} in (π, π) π β1 (π ) = {π} is not βπ -closed.
8
S. Jafari, K. Viswanathan, J.Jayasudha
Example 3.14. Let π = {π, π, π}, π = {β
, {π}, {π, π}, π} , β = {β
, {π}} and π =
{β
, {π, π}, π}. Deο¬ne the map f: (π, π, β) β (π, π) to be the identity map.Then f is
βπ -continuous but not π ππβ β πΌ continuous, as for the closed set π = {π} in (π, π)
π β1 (π ) = {π} is not an π ππβ β πΌ set.
Theorem 3.15. A function f: (π, π, β) β (π, π) is β -continuous if and only if it is both
βπ -continuous and π ππβ β πΌ continuous.
Proof. This is an immediate consequence of Theorem 3.7.
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Decomposition of β -continuity in ideal topological spaces
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