Download alpha-closed sets in topological spaces

International Mathematical Forum, 5, 2010, no. 24, 1167 - 1178
Remarks on gα -Closed Sets
in Topological spaces
1
Saeid Jafari, 2 M. Lellis Thivagar and 3 Nirmala Rebecca Paul
1
College of Vestsjaelland South, Herrestraede 11
4200 Slagelse,Denmark
jafari@stofanet.dk
2
Department of Mathematics, Arul Anandar College
Karumathur, Madurai (Dt.)-625514, Tamilnadu, India
mlthivagar@yahoo.co.in
3
Department of Mathematics, Lady Doak College
Madurai - 625 002, Tamilnadu, India
nimmi rebecca@yahoo.com
Abstract
In this paper we introduce a new class of sets called gα - closed sets
in topological space. We prove that this class lies between α- closed sets
and gα-closed sets.We discuss some basic properties of gα - closed sets.
Applying this we introduce a new space called Tgα - space.
Mathematics Subject Classification: 54C08
Keywords: Topological spaces,generalized closed sets,# gs-closed sets,
#
gs-open sets and α- Closure
1
Introduction
The concept of generalised closed sets play a significant role in topology.There
are many research papers which deals with different types of generalised closed
sets.Devi et al[5] and Maki et al.[6] introduced semi-generalised closed sets(briefly
sg-closed) ,generalised semi-closed sets(briefly gs-closed),generalised α - closed
(briefly gα- closed) sets and α-generalized closed (briefly α g - closed) sets
respectively.Veera kumar introduced and studied # gs - closed sets [11].Rajesh
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S. Jafari, M. Lellis Thivagar and N. Rebecca Paul
and Lellis Thivagar [9] introduced g - closed sets and proved that they form a
topology. In this paper we introduce a new class of sets namely gα - closed sets
for topological spaces and study their basic properties.
2
preliminaries
We list some definitions in a topological space (X, τ ) which are useful in the
following sections. The interior and the closure of a subset A of (X, τ ) are
denoted by Int(A) and Cl(A), respectively. Throughout the present paper
(X, τ ) and (Y, σ)(or X and Y ) represent non-empty topological spaces on which
no separation axiom is defined, unless otherwise mentioned.
Definition 2.1 A subset A of a space X is called
1. a semi-open set [2] if A ⊆ Cl(Int(A))
2. a pre-open set [8] if A ⊆ Int(Cl(A))
3. an α -open set [3] if A ⊆ Int(Cl(Int(A)))
The complement of a semi-open (resp. pre open and α- open) set is called a
semi-closed (resp. pre closed and α-closed) set. The α-closure[9] (resp semiclosure[7]) of a subset A of X denoted by αcl(A) (resp scl (A)) is defined to be
the intersection of all α -closed sets (resp. semi-closed sets) containing A. If
A ⊆ B ⊆ X then αclB (A) and αintB (A) denote the α- closure of A relative to
B and α- interior of A relative to B, respectively.A subset A of a topological
space (X, τ ) is α- closed (resp. semi-closed) if and only if αcl(A) = A(resp
sCl(A) = A).
Definition 2.2 A subset A of a space X is called
1. an α- generalized closed set [6] (briefly αg- closed) if αcl(A) ⊆ U whenever A ⊆ U and U is open in X.
2. an ω- closed set [9] if Cl(A) ⊆ U whenever A ⊆ U and U is semi-open
in X.
3. a ∗ g -closed set [10 ] if Cl(A) ⊆ U whenever A ⊆ U and U is ω-open in
X.
4. a # g- semi-closed set [11] (briefly
A ⊆ U and U is ∗ g-open in X.
#
gs-closed) if scl(A) ⊆ U whenever
Remarks on gα - closed sets in topological spaces
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5. a g- closed set of [9] if Cl(A) ⊆ U whenever A ⊆ U and U is # gs- open
in X.
6. a g-semi-closed set [9] (briefly g s- closed) if scl(A) ⊆ U whenever A ⊆ U
and U is # gs-open in X.
The complements of the above sets are called their respective open sets.
Definition 2.3 A topological space X is called a
1. α Tb - space [1] if every αg-closed set in it is closed.
2. Tb - space [6] if every gs-closed set in it is closed.
g -closed set in it is closed.
3. Tg -space [9] if every 4. Tgs -space [9] if every g s-closed set in it is closed.
3
gα -closed and open sets
After investigating several closed sets through extensive study we introduce
the following definition.
Definition 3.1 A subset A of a space X is called gα - closed if αCl(A) ⊆ U
#
whenever A ⊆ U and U is gs-open in X.
Proposition 3.2 Every closed set is gα - closed set but not conversely.
Proof.Let A be a closed set and U be any # gs -open set containing A. Since
A is closed we have αCl(A) ⊆ Cl(A) = A ⊆ U. Hence A is gα - closed.
Example 3.3 Let X = {a, b, c} and τ = {φ, {a}, X} The set {b} is gα - closed
but not closed.
Proposition 3.4 Every α- closed set is gα - closed but not conversely.
Proof.Let A be an α- closed set and U be any # gs - open set containing A.
Since A is α- closed, we have αCl(A) = A ⊆ U Hence A is gα - closed.
Example 3.5 Let X = {a, b, c} and τ = {φ, {a, b}, X} The set {a, c} is gα closed but not α- closed.
Proposition 3.6 Every g -closed set is gα - closed but not conversely.
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Proof.Let A be a g - closed set and U be any # gs - open set containing A. We
have αCl(A) ⊆ Cl(A) ⊆ U. Hence A is gα - closed.
Example 3.7 Let X = {a, b, c} and τ = {φ, {a}, X} The set {c} is gα - closed
but not g- closed.
Proposition 3.8 Every gα - closed set is gα- closed but not conversely.
Proof.Let A be an gα - closed set and U be any α- open set containing A.
Since any α- open set is semi open and any semi open set is # gs - open[9], we
have αCl(A) ⊆ U.Hence A is gα- closed.
Example 3.9 Let X = {a, b, c} and τ = {φ, {a}, {b, c}, X} The set {c} is
gα-closed but not gα - closed .
Proposition 3.10 Every gα - closed set is αg - closed but not conversely.
Proof.Let A be an gα - closed set and U be any open set containing A. Since
#
any open set is gs - open we have αCl(A) ⊆ U.Hence A is αg- closed.
Example 3.11 Let X = {a, b, c} and τ = {φ, {a}, {b, c}, X} The set {c} is
αg-closed but not gα - closed .
Proposition 3.12 Every gα - closed set is gs - closed (sg-closed) closed but
not conversely.
Proof.Let A be any gα -closed set and U be any open set (semi-open) containing
A.Since any open set (semi-open) is # gs - open,we have sCl(A) ⊆ αCl(A) ⊆ U.
Hence A is gs-closed (sg - closed)
Example 3.13 Let X = {a, b, c} and τ = {φ, {a}, {b, c}, X} The set {c} is
sg-closed and gs-closed but not gα - closed .
g s - closed set but not conversely.
Proposition 3.14 Every gα - closed set is Proof.Let A be an gα - closed set and U be any # gs - open set containing A.
Since sCl(A) ⊆ αCl(A) ⊆ U. Hence A is gs - closed.
Example 3.15 Let X = {a, b, c} and τ = {φ, X, {a}, {b}, {a, b}},The set {a}
is g s - closed but not gα - closed.
Remark 3.16 The following examples show that gα - closedness is independent
of g-closedness and ω- closedness.
Remarks on gα - closed sets in topological spaces
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Example 3.17 Let X ={a, b, c} and τ = {φ, X, {a}, {b, c}} ,The set {a, c} is
g -closed,ω-closed but not gα -closed.
Example 3.18 Let X = {a, b, c, d} and τ = {φ, X, {a}, {a, b, d}},The set {b}
is gα - closed sets but not g-closed .
Example 3.19 Let X = {a, b, c, d} and τ = {φ, X, {a}, {b}, {a, b}},The set
{c} is gα - closed but not ω- closed.
Remark 3.20 From the above discussion and known results we have the
following implications A → B represents A implies B but not conversely .
Remark 3.21 1.
gα -closed. 2.closed.
3.
g -closed. 4.ω-closed.
4.g-closed 5.αg-closed.
6.gs-closed. 7.gs-closed.
8.sg-closed. 9.gα-closed.
10.α-closed
Definition 3.22 A subset A of X is called gα - open if and only if Ac is gα closed in X.
Theorem 3.23 i) Every open set is gα - open but not conversely.
ii) Every α- open set is gα - open but not conversely.
iii)Every g - open set is gα - open but not conversely.
iv) Every gα -open set is gα(αg)-open but not conversely.
v) Every gα - open set is sg(gs-open) - open but not conversely .
4
Characterization of gα -closed and open sets
Theorem 4.1 If A and B are gα - closed in X then, A ∪ B is gα - closed in X.
Proof.Let A and B be any two gα - closed sets in X and U be any # gs open set containing A and B.We have αCl(A) ⊆ U and αCl(B) ⊆ U. Thus
αCl(A ∪ B) = αCl(A) ∪ αCl(B) ⊆ U[1].Hence A ∪ B is gα - closed in X.
Theorem 4.2 If a set A is gα - closed, then αCl(A)−A contains no non empty
closed set.
Proof.Suppose that A is gα - closed. Let F be a closed subset of αCl(A) − A.
Then A ⊆ F c . F c is open and hence # gs - open. Since A is gα - closed
,αCl(A) ⊆ F c .Consequently F ⊆ (αCl(A))c . Every closed set is α- closed and
hence F is α-closed.Therefore F ⊆ αCl(A).Hence F ⊆ αCl(A) ∩ (αCl(A))c
and hence F is empty. Observe that the converse of the above theorem is not
true.
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Example 4.3 Let X = {a, b, c, d} and τ = {φ, X, {b, c}, {b, c, d}, {a, b, c, }} If
A = {a, b} then αCl(A) − A = X − {a, b} = {c, d} does not contain non empty
closed set.However A is not an gα - closed subset of X.
Theorem 4.4 A set A is gα - closed if and only if αCl(A) − A contains no
#
non empty gs - closed set.
Proof.Necessary.Suppose that A is gα - closed. Let F be a # gs - closed set
contained in αCl(A) − A. Since F c is # gs - open with A ⊆ F c and A is gα c
c
closed αCl(A) ⊆ F .Then F ⊆ (α(Cl(A)) .Also F ⊆ αCl(A) − A.Therefore
F ⊆ (α(Cl(A))c ∩ (αCl(A)) = φ.Hence F = φ.
Sufficiency. Suppose that αCl(A)−A contains no nonempty # gs-closed set. Let
A ⊆ G and G be # gs-open. If αCl(A) is not a subset of G then αCl(A)∩Gc is a
non empty # gs-closed subset of αCl(A)−A, which is a contradiction.Therefore
αCl(A) ⊆ G and hence A is gα - closed.
Theorem 4.5 If A is
α-closed sub set of X.
#
gs-open and gα - closed subset of (X, τ ) then A is an
Proof.Since A is # gs-open and gα -closed,αCl(A) ⊆ A.Hence A is α- closed.
Theorem 4.6 Let A be an gα - closed subset of X.If A ⊆ B ⊆ αCl(A) then B
is also an gα - closed subset of X.
Proof.Let U be a # gs-open set of X such that B ⊆ U.Then A ⊆ U.Since A is an
gα - closed set αCl(A) ⊆ U.Also B ⊆ αCl(A), αCl(B) ⊆ αCl(A) ⊆ U.Hence
B is also an gα - closed subset of X.
Theorem 4.7 Let A ⊆ Y ⊆ X and Y ∈ τ .If A is gα - closed in X then A is
αg - closed relative to Y.
Proof.Since every gα - closed set is αg - closed the proof, follows from [9,
Theorem 3.22].
Proposition 4.8 For each a ∈ X either {a} is # gs-closed or {a}c is gα - closed
in X.
Proof.Suppose that {a} is not # gs-closed in X.Then {a}c is not # gs-open and
the only # gs-open set containing {a}c is the space X itself.That is {a}c ⊆
X.Therefore αCl({a}c ) ⊆ X and so {a}c is gα -closed.
Theorem 4.9 Let A be an gα -closed set in X.Then A is α- closed if and only
if αCl(A) − A is closed.
Remarks on gα - closed sets in topological spaces
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Proof.Necessity. Let A be an gα -closed subset of X. Then αCl(A) = A and
so αCl(A) − A = φ which is closed.
Sufficiency. Since A is gα - closed by Theorem 4.2 αCl(A) − A contains no
nonempty closed set.But αCl(A) − A is closed.This implies αCl(A) − A = φ.
That is αCl(A) = A.Hence A is α-closed.
Definition 4.10 The intersection of all # g-semi-open subsets of (X, τ ) containing A is called the # g-semi-kernel of A and is denoted by # gs-ker(A).
Lemma 4.11 A subset A of (X, τ ) is gα -closed if and only if αCl(A) ⊆ # gs −
ker(A).
Proof.Suppose that A is gα -closed in X. Then αCl(A) ⊆ U whenever A ⊆ U
#
and U is gs-open in (X, τ ). Let x ∈ αCl(A). If x ∈
/ # gs − ker(A) then there
/ U .Since U is a # gs-open set containing A,
is a # gs-open set U such that x ∈
we have x ∈
/ αCl(A),a contradiction.
Conversely let αCl(A) ⊆ # gs − ker(A). If U is any # gs-open set containing
gα -closed.
A, then αCl(A) ⊆ # gs − ker(A) ⊆ U.Therefore A is Lemma 4.12 (4) Let x be a point of (X, τ ).Then {x} is either nowhere dense
or pre-open.
Remark 4.13 In the notation of lemma 4.12,we consider the following decomposition of a given topological space (X, τ ),namely X = X1 ∪ X2 where
X1 = {x ∈ X : {X}is no where dense} and X2 = {x ∈ X : {x}is pre-open}
Proposition 4.14 For any subset A of X, X2 ∩ αCl(A) ⊆ # gs − ker(A).
/ # gs − ker(A).Then there is a sgProof.Let x ∈ X2 ∩ αCl(A) and if x ∈
open set U containing A such that x ∈
/ U.If F = X − U,then F is # gs-closed.
Since αCl({x}) ⊆ αCl(A), we have Int(αCl({x})) ⊆ A ∩ Int(αCl(A)) ⊆ A ∩
/ X1 and so Int(Cl({x}) = φ.Therefore
Int(Cl(A)).Since x ∈ X2 ,we have x ∈
there has to be some point y ∈ A ∩ Int(Cl({x}) and hence y ∈ F ∩ A.A
contradiction.
Theorem 4.15 A subset A of (X, τ ) is gα -closed if and only if X1 ∩αCl(A) ⊆
A.
Proof.Suppose that A is gα -closed. Let x ∈ X1 ∩ αCl(A).Then x ∈ X1 and
x ∈ αCl(A).Since x ∈ X1 , Int(Cl({x}) = φ.Therefore, {x} is semi-closed,since
Int(Cl({x}) ⊆ {x}. Since every semi-closed set is # gs-closed {x} is # gs-closed.
If x ∈
/ A and U = X − {x},then U is a # gs-open set containing A and so
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αCl(A) ⊆ U,a contradiction.
Conversely ,let X1 ∩ αCl(A) ⊆ A.Then X1 ∩ αCl(A) ⊆ # gs − ker(A).Since
A ⊆ # gs − ker(A).Now αcl(A) = X ∩ αCl(A) = X1 ∪ X2 ∩ αCl(A) = (X1 ∩
αCl(A)) ∪ (X2 ∩ αCl(A)) ⊆ # gs − ker(A).Since X1 ∩ αCl(A) ⊆ # gs − ker(A)
and by Proposition 4.14 and by Lemma 4.11 A is gα -closed .
gα -closed.
Theorem 4.16 Arbitrary intersection of gα -closed sets is Proof.Let F = {Ai : i ∈ Λ} be a family of gα -closed sets and let A = i∈Λ Ai .
Since A ⊆ Ai for each i,X1 ∩ αCl(A) ⊆ X1 ∩ αCl(Ai ) for each i.By Theorem
4.15 for each gα -closed set Ai , we have X1 ∩ αCl(Ai ) ⊆ Ai for each i and so
X1 ∩ αCl(Ai ) ⊆ A for each i.Thus X1 ∩ αCl(A) ⊆ X1 ∩ αCl(Ai ) ⊆ A for each
gα -closed by Theorem 4.15.
i ∈ Λ.That is X1 ∩ αCl(A) ⊆ A and so A is Remark 4.17 Thus from Theorem 4.1 and Theorem 4.16 leads us into another class of closed sets namely gα -closed sets which are closed under finite
union and arbitrary intersection.Note that that while no other class of closed
sets form a topology this class forms a topology.
Remark 4.18 For a subset A of X, αCl(Ac ) = (αInt(A))c .
Theorem 4.19 A sub set A of X is gα - open if and only if F ⊆ αInt(A)
#
whenever F is gs - closed and F ⊆ A.
Proof.Necessity Let A be an gα - open set in X. Let F be a # gs-closed such
that F ⊆ A.Then Ac ⊆ F c where F c is # gs-open.Ac is gα -closed implies that
c
c
c
c
αCl(A ) ⊆ F i.e(αInt(A)) ⊆ F .That is F ⊆ αInt(A)
Sufficiency. Suppose F is # gs-closed and F ⊆ A.Also F ⊆ αInt(A).Let Ac ⊆
U where U is # gs-open.Then U c ⊆ A where U c is # gs-closed.By hypothesis
U c ⊆ αInt(A).That is (αInt(A))c ⊆ U.i.e.αCl(Ac ) ⊆ U. This implies that
Ac is gα -closed .Hence A is gα -open.
Proposition 4.20 If αInt(A) ⊆ B ⊆ A and A is gα -open then B is gα -open.
Proof.αInt(A) ⊆ B ⊆ A implies Ac ⊆ B c ⊆ (αintA)c i.e.Ac ⊆ B c ⊆ (αCl(Ac )
and Ac is gα -closed. By theorem 4.7 B c is gα -closed. Hence B is gα - open.
Theorem 4.21 If A and B are gα - open sets in X then A ∩ B is gα -open in
X.
Theorem 4.22 A set A is gα - open in X if and only if G = X whenever G is
#
c
gs-open and αInt(A) ∪ A ⊆ G.
Remarks on gα - closed sets in topological spaces
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Proof.Necessity. Let A be gα - open and G be # gs- open and αInt(A) ∪ Ac ⊆
G.This gives Gc ⊆ (αInt(A) ∪ Ac )C = (αInt(A))c ∩ A = (αInt(A))c − Ac =
gα -closed and Gc is # gs - closed by theorem 4.5 it
αCl(Ac ) − Ac Since Ac is follow that Gc = φ Therefore G = X
Sufficiency. Suppose that F is # gs - closed and F ⊆ A.Then αInt(A) ∪ Ac ⊆
αInt(A) ∪ F c .It follows by hypothesis that αIntA ∪ F c = X and hence F ⊆
αInt(A).Therefore by Theorem 4.18 A is gα - open in X.
5
gα -Closure
Definition 5.1 Let (X, τ ) be a topological space and B ⊆ X. We define
gα -closed
the gα -closure of B (briefly gα − Cl(B) to be the intersection of all sets containing B which is denoted by gα − Cl(B) = {A : B ⊆ A andA ∈
gα C(X, τ )} where gα C(X, τ ) is set of all gα -closed subsets of X.
Lemma 5.2 For any B ⊆ X, B ⊆ gα − Cl(B) ⊆ Cl(B)
Proof.It follows from Propsition 3.2.
Remark 5.3 The relations in Lemma 5.2 may be proper as seen from the
following example.
Example 5.4 Let X = {a, b, c} and τ = {φ, X, {a, b}} Let B = {b}.
gα − Cl(B) ⊆ Cl(B)
Then gα − Cl(B) = {b, c} and so B ⊆ Theorem 5.5 The gα -closure is a Kuratowski closure operator on X.
Proof.(i)
gα − Cl(φ) = φ and ii)B ⊆ gα − Cl(B) by Lemma 5.2
gα C(X, τ ),then Bi ⊆ A and by definition 5.1
iii)Let B1 ∪ B2 ⊆ A and A ∈ gα − Cl(Bi ) ⊆ A for i=1,2. Therefore gα − Cl(B1 ) ∪ gα − Cl(B2 ) ⊆ {A : B1 ∪
gα C(X, τ )} = gα − Cl(B1 ∪ B2 ).For the reverse inclusion,let
B2 ⊆ A andA ∈ x ∈ gα −Cl(B1 ∪B2 ) and suppose that x ∈
/
gα −Cl(B1 )∪
gα −Cl(B2 ).Then there
exist gα -closed sets A1 and A2 with B1 ⊆ A1 , B2 ⊆ A2 and x ∈
/ A1 ∪ A2 .We
gα -closed set by theorem 4.1 such
have B1 ∪ B2 ⊆ A1 ∪ A2 and A1 ∪ A2 is a that x ∈
/ A1 ∪ A2 .Thus x ∈
/ gα − Cl(B1 ∪ B2 ) which is a contradiction to
gα − Cl(B1 ) ∪ gα − Cl(B2 ) = gα − Cl(B1 ∪ B2 ).
x ∈ gα − Cl(B1 ∪ B2 ).Hence iv) Let B ⊆ A and A ∈ gα C(X, τ ).Then by definition 5.1 gα − Cl(B) ⊆
A and gα − Cl(
gα − Cl(B)) ⊆ A.We have gα − Cl(
gα − Cl(B)) ⊆ {A :
gα − Cl(B).By Lemma 5.2 gα − cl(B) ⊆
B ⊆ AandA ∈ gα − C(X, τ )} = gα − Cl(
gα − Cc(B)) and therefore gα − Cl(B) = g
α − Cl(
gα − Cl(B)). Hence
gα -closure is a Kuratowski closure operator on X.
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Proposition 5.6 Let (X, τ ) be a topological space and B ⊆ A. The following
properties hold. i) gα − Cl(B) is the smallest gα -closed set containing B.
gα − Cl(B) = B
ii) B is gα -closed if and only if Proposition 5.7 For any two subsets A and B of (X, τ ) i) If A ⊆ B, then
gα − Cl(B).
gα − Cl(A) ⊆ ii)
gα − Cl(A ∩ B) ⊆ gα − Cl(A) ∩ gα − Cl(B)
6
Applications
Definition 6.1 A space X is called a Tgα -space if every gα - closed set in it is
α- closed.
Theorem 6.2 For a space X the following conditions are equivalent. i) X is
a Tgα - space.
ii) Every singleton of X is either # gs-closed or α- open.
Proof.i) ⇒ ii) Let x ∈ X suppose that {x} is not a # gs-closed set of X.Then
X − {x} is not a # gs - open set.So X is the only # gs-open set containing
X − {x}.Then X − {x} is an gα - closed set of X. Since X is a Tgα - space
X − {x} is an α- closed set of X and hence {x} is an α-open set of X.
ii) ⇒ i) Let A be an gα - closed set of X. A ⊆ αCl(A) Let x ∈ αCl(A) by ii)
#
{x} is either gs- closed or α- open. Case i) suppose that {x} is # gs- closed.If
x ∈
/ A, αCl(A) − A contains a non empty # gs-closed set {x}.By Theorem
4.4 we arrive at a contradiction. Thus x ∈ A.Case ii) suppose that {x} is
α-open.Since x ∈ αCl(A), {x} ∩ A = φ.This implies x ∈ A.Thus in any case
x ∈ A,So αCl(A) ⊆ A.Therefore αCl(A) = A or equivalently A is α- closed.
Hence X is a Tgα - space.
Theorem 6.3 i)Every Tb - space is a Tgα -space.
ii) Every α Tb - space is a Tgα - space.
gα -closed set in X.By Proposition
Proof.Let X be a Tb space and A be an 3.12 A is a gs - closed set. Since it is a Tb - space A is closed in X and so it is
α- closed in X. Hence X is a Tgα -space.
gα - closed set in X.By Proposition 3.10
ii) Let X be a α Tb - space and A be an A is αg-closed.Since X is a α Tb -space it is closed in X. So it is α-closed.Hence
X is a Tgα -space.
Definition 6.4 A topological space X is called a # Tgα -space if every gα - closed
set in it is closed.
Remarks on gα - closed sets in topological spaces
1177
Theorem 6.5 Every # Tgα -space is a Tgα -space but not conversely.
Proof.Since every closed set is an α- closed set the result follows.
Example 6.6 Let X = {a, b, c, d} and τ = {φ, X, {a}, {a, b}} . (X, τ ) is a
Tgα -space but not a # Tgα - space.
Theorem 6.7 Every # Tgα -space is a Tg -space but not conversely.
g - closed sub set of X.Any g - closed
Proof.Let X be a # Tgα - space and A be a #
set is gα -closed. Since X is a Tgα -space,A is closed in X.Hence X is a Tg space.
Example 6.8 Let X = {a, b, c, d} and τ = {φ, X, {a}, {a, b}} . (X, τ ) is a Tg
but not a # Tgα - space.
Theorem 6.9 Every Tgs -space is a # Tgα - space.
gα - closed subset of X.By ProposiProof.Let X be a Tgs - space and A be any tion 3.14. A is g s -closed.Since X is a Tgs - space, A is closed in X.Hence X is
a # Tgα -space.
Example 6.10 Let X = {a, b, c, d} and τ = {φ, X, {a}, {c}, {d}, {a, c}, {a, d}
{c, d}, {a, c, d}} . (X, τ ) is a # Tgα -space but not a Tgs -space.
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Received: August, 2009