behaviour of kuratowski operators on some new sets in topology
TRANSCRIPT
Global Journal of Pure and Applied Mathematics.
ISSN 0973-1768 Volume 15, Number 6 (2019), pp. 1033-1044
© Research India Publications
http://www.ripublication.com/gjpam.htm
Behaviour of Kuratowski Operators on Some New
Sets in Topology
Dr. G. Hari Siva Annam 1 and N.Raksha Ben 2
1 Assistant Professor, Department of Mathematics Kamaraj College, Tuticorin, India.
2 Research Scholar, Manonmaniam Sundaranar University, Tirunelveli, India.
Abstract
The purpose of this paper is to introduce qk â sets using Kuratowskiâs operator
and to study its basic properties. In Topology, the Kuratowskiâs operator plays
a pivotal role to define a topological structures on a set. As the closure and
interior operator plays a major role in rough set theory, here we introduce qk-
closure and qkâ interior some of its properties are discussed.
AMS Subject Classification(2010) : 54A05
Keywords: qk â sets, qk- closure ,qkâ interior
1. Introduction
In Topology, The Kuratowskiâs operator has a great role to define a topological
structure on a set. Chandrasekhara Rao and P. Thangavelu introduce q-set and studied
its basic properties. As an application of Kuratowski operator on q set, we introduce qk
sets, the basic properties of the qk sets, interior and closure operations are introduced .
We introduce qk interior, qkclosure and its properties are investigated. The concepts of
q sets are studied in [9]. Throughout this paper X is a topological space and A, B are
the subsets of X. The cl A and Int A are the notations which represents closure of A and
Interior of A respectively . The following concepts bring backs the basics which are
used in this paper to our memory. A subset of a topological spaces X is clopen if it is
both open and closed. Let A be the subset of X then ððâðŽ is the intersection of all g-
closed set containing A and then ððð¡âðŽ is the union of all g-open set contained in A.
1034 Dr. G. Hari Siva Annam and N.Raksha Ben
2. qkâ Sets and Its Basic Properties.
Definition 2.1:
A set A of ( ð, ) is said to be qkâ set if ððâ(ððð¡ ðŽ) â ððð¡â(ðð ðŽ).
Definition 2.2 :
A set A of ( ð, ) is said to be pkâ set if ððâ(ððð¡ ðŽ) â ððð¡â(ðð ðŽ).
Result 2.3 :
If A is g- clopen then A is a pk â set.
Result 2.4 :
If A is clopen then A is both qk and pk
Result 2.5 :
Union of qkâ sets need not be qk . The result is described with the help of the
following example. Let ð = {ð, ð, ð, ð} ððð ð = { ð, {ð}, {ð}, {ð, ð}, {ð, ð}, {ð, ð, ð}, ð}
be a topology on ð. Here {a} and {b,c} are qk â sets but their union {a}⪠{b,c} ={a,b,c}
which is not a qk -set.
Result 2.6:
Intersection of qk- sets need not be qk-set . The result is explained in the
following example. Let ð = {ð, ð, ð, ð, ð} ððð ð = { ð, {ð, ð}, {ð, ð}, {ð, ð, ð, ð}, ð} be
a topology on ð. Here {a,b,c,e} and {a,b,d} are two qk â sets but their intersection
{a,b,c,e} â© {a,b,d} = {a,b} which is not a qk- set.
Result 2.7:
Union of pk â sets need not be pk â set. The result is discussed using the example
shown below. Let ð = {ð, ð, ð, ð, ð} ððð ð = { ð, {ð, ð}, { ð, ð}, {ð, ð, ð, ð}, ð} be the
topology on ð. Here {a} and {b} are pk- set but their union {a} ⪠{b} = {a,b} which is
not a pk â set.
Result 2.8:
Intersection of pk- sets need not be pk set. The result is explained by the
following example. Let ð = {ð, ð, ð, ð} ððð ð = { ð, {ð}, {ð}, {ð, ð}, {ð, ð, ð}, ð} be
the topology on ð. Here {a,c} and {a,b} are pk sets but their intersection is {a} which
is not a pk set.
Theorem 2.9 :
Every qkâ sets is a q set .
Behaviour of Kuratowski Operators on Some New Sets in Topology 1035
Proof : Let A be a qkâ set . Clearly, we say that ðð (ððð¡ ðŽ) â ððâ(ððð¡ ðŽ) â ððð¡â(ðð ðŽ) â
ððð¡ (ðð ðŽ) . Since A is a q â set.
Remark 2.10:
The converse of the above theorem is not true that is âEvery q â set need not
be qkâ set . It is shown in the following example. Let ð = {ð, ð, ð, ð} ððð ð =
{ ð, {ð}, {ð}, {ð, ð}, {ð, ð, ð}, ð} be the topology on ð. Here {a,c} and {a,d} are q -sets
but they are not qk â sets .
Theorem 2.11 :
A is a qkâ set if and only if X-A is a qk -set.
Proof : Suppose A is a qk â set .
Then , ððâ(ððð¡ ð â ðŽ) â ð â ððð¡â(ðð ðŽ)
âð â ððâ(ððð¡ ðŽ)
= ððð¡âðð (ð â ðŽ)
â ð â ðŽ is a qk â set .
Conversely, We assume that ð â ðŽ is a qk â set
â ððâ(ððð¡ ðŽ) = ððâ(ððð¡ (ð â (ð â ðŽ))) = ð â ððð¡â ðð(ð â ðŽ)
â ð â ððâððð¡ (ð â ðŽ)
= ððð¡âðð (ð â (ð â ðŽ))
= ððð¡âðð ðŽ
Hence, A is a qk â set .
Theorem 2.12:
Every p set is a pk â set .
Proof :
Suppose A be a p â set.
â ðð (ððð¡ ðŽ) â ððð¡ (ðððŽ)
Now,ððâ(ððð¡ ðŽ) â ðð (ððð¡ ðŽ) â ððð¡(ðð ðŽ) â ððð¡â(ðð ðŽ)
â ððâ(ððð¡ ðŽ) â ððð¡â(ðð ðŽ)
Hence, A is a pk - set.
1036 Dr. G. Hari Siva Annam and N.Raksha Ben
Remark 2.13:
The Converse of the above theorem is not true .It is shown in the following
example:
If ð = {ð, ð, ð, ð} ððð ð = { ð, {ð}, {ð}, {ð, ð}, {ð, ð}, {ð, ð, ð}, ð} be a topology on
ð.Here , every subset of X i.e. P(X) is a pk- set which are behaving like q sets. Hence,
none of them are p sets.
Result 2.14:
The proper sets of X is neither pk setsnor qk sets. It is shown in the following
example:
Let ð = {ð, ð, ð, ð} ððð ð = { ð, {ð}, {ð}, {ð, ð}, {ð, ð}, {ð, ð, ð}, ð} be a topology on ð .
Let A = { a} and B = {b} which are neither pk set nor qkset. Since, ððâ(ððð¡ ðŽ) = {a,d}
and ððð¡â(ðð ðŽ)= {a,c}. Similarly, ððâ(ððð¡ ðµ) = {b,d} and ððð¡â(ðð ðŽ) = {b,c} from
this we conclude that the proper sets of X neither pk sets nor qk sets.
Lemma 2. 15 :
If A is a subset of the topological space X then ðŒð âðð ðŒð â ððð¡ ðŽ = ðð ððð¡ ðŽ
Where ðŒð â interior of A is denoted by ðŒð âððð¡ ðŽ and ðŒð âclosure of A is denoted by
ðŒð âðð ðŽ respectively.
Proof : Suppose x â ðð ððð¡ ðŽ â x â â© ð¹, F is closed set such that ð¹ â ððð¡ ðŽ â
ðŒð â ððð¡ ðŽ. Since every closed set is ðŒð â closed , x belongs to instersection of all ðŒð â
closed set F such that all ðŒð â closed set F contained in ðŒð â ððð¡ ðŽ which implies us
that x â ðŒð âðð ðŒð â ððð¡ ðŽ Hence , ðð ððð¡ ðŽ â ðŒð âðð ðŒð â ððð¡ ðŽ . Now , suppose x â
ðð ððð¡ ðŽ â x â â© ð¹ , F is the closed set such that ð¹ â ððð¡ ðŽ â x â ð¹ , for some
F is ðŒð â-closed such that ð¹ â ððð¡ ðŽ â ðŒð â ððð¡ ðŽ . This implies that x â
ðŒð âðð ðŒð â ððð¡ ðŽ . Hence , ðŒð âðð ðŒð â ððð¡ ðŽ = ðð ððð¡ ðŽ.
Theorem 2.16:
If A is a pk â set if and only if X -A is a pk â set.
Proof :
Suppose A is a pk â set
Then , ððâ(ððð¡ ð â ðŽ) = ð â ððð¡â(ðð ðŽ)
âð â ððâ(ððð¡ ðŽ)
= ððð¡âðð (ð â ðŽ)
â ð â ðŽ is a pk â set .
Behaviour of Kuratowski Operators on Some New Sets in Topology 1037
Conversely, We assume that ð â ðŽ is a pk â set
â ððâ(ððð¡ ðŽ) = ððâ(ððð¡ (ð â (ð â ðŽ) ))= ð â ððð¡â ðð(ð â ðŽ)âð â
ððâððð¡ (ð â ðŽ)
= ððð¡âðð (ð â (ð â ðŽ))
= ððð¡âðð ðŽ
Hence, A is a pk â set .
Proposition 2.17 :
If A is clopen , A and B is a pâ set then Aâ©B is also a pk â set.
3. qkâ Interior and qkâ Closure:
Definition 3.1:
Let A be a subset of a topological space X .The qkâ Interior of A is denoted by qkâ int
and is defined as the union of all qkâ sets contained in A .
Let A be a subset of a topological space X .The qkâ Closure of A is denoted by qkâ cl
and is defined as the intersection of all qkâ sets containing A .
As the collection of qk- sets is not closed under union and intersection it follows that
qkâððð¡ ðŽ need not be a qk â set and also qkâ ðð ðŽ need not be a qk â set. But ,qkâððð¡ ðŽ â
ðŽ â qk â ðð ðŽ is always true for any subset A of a topological space.
Proposition 3.2:
(i) qkâððð¡ ð = ð; qk âðð ð = ð .
(ii) qkâððð¡ ð = ð ; qk âðð ð = ð .
(iii) qkâððð¡ ðŽ â ðŽ â qk â ðð ðŽ
(iv) A â B â qk âððð¡ ðŽ â qk âððð¡ ðµ&A â B â qk âðð ðŽ â qk âðð ðµ
(v) qkâððð¡ (ðŽ â© ðµ)âqkâððð¡ ðŽ â© qk âððð¡ ðµ
(vi) qkâðð(ðŽ â© ðµ)âqkâðð ðŽ â© qk âðð ðµ
(vii) qkâðð(ðŽ ⪠ðµ)âqkâðð ðŽ ⪠qk âðð ðµ
(viii) qkâððð¡(ðŽ ⪠ðµ)âqkâððð¡ ðŽ ⪠qk âððð¡ ðµ
(ix) qkâððð¡ (qk âððð¡ ðŽ)âqkâððð¡ ðŽ
(x) qkâðð (qk âðð ðŽ)âqkâððð¡ ðŽ
(xi) qkâððð¡ (qk âðð ðŽ) â qk âððð¡ ðŽ
(xii) qkâðð (qk âððð¡ ðŽ)âqkâðð ðŽ
1038 Dr. G. Hari Siva Annam and N.Raksha Ben
Proof:
The proof of (i) and (ii) are obvious.In order to prove (iii) we assume ð¥ â ðŽ
whichimplies that ð¥ does not belongs to any qk- sets contained in
ðŽ.Hence, ð¥ â qk âððð¡ ðŽ which gives us qk âððð¡ ðŽâðŽ and suppose , ð¥ â qk â
ðð ðŽ then ð¥ does not belongs to any qk- sets containing ðŽ.Hence we get that
ðŽ â qk âðð ðŽ. Using the above two inclusions we get that qk âððð¡ ðŽ â ðŽ â
qk â ðð ðŽ.(iv) The result is obvious .(v)we have ðŽ â© ðµ â ðŽ and ðŽ â© ðµ â ðµ
using these results we get qkâððð¡ (ðŽ â© ðµ)âqkâððð¡ ðŽ and qk âððð¡ (ðŽ â©
ðµ)âqkâððð¡ ðµ which implies us that
qkâððð¡ (ðŽ â© ðµ)âqkâððð¡ ðŽ â© qk âððð¡ ðµ(vi) Also since we have ðŽ â© ðµ â ðŽ
and ðŽ â© ðµ â ðµ using these results we get qk âðð(ðŽ â© ðµ)âqkâðð ðŽ and qk â
ðð(ðŽ â© ðµ)âqkâðð ðµ which implies us that qk âðð (ðŽ â© ðµ)âqkâðð ðŽ â© qk â
ðð ðµ.(vii) We know that ðŽ ⪠ðµâðŽ and ðŽ ⪠ðµâðµ it follows that qk â
ððð¡(ðŽ ⪠ðµ)âqkâððð¡ ðŽ and qk âððð¡(ðŽ ⪠ðµ)âqkâððð¡ ðµ which implies that qk
âððð¡(ðŽ ⪠ðµ)âqkâððð¡ ðŽ ⪠qk âððð¡ ðµ .(viii) We know that ðŽ ⪠ðµâðŽ and ðŽ âª
ðµâðµ it follows that qk âðð(ðŽ ⪠ðµ)âqkâðð ðŽ and qk âðð(ðŽ ⪠ðµ)âqkâðððµ
which implies that qk âðð(ðŽ ⪠ðµ)âqkâðð ðŽ ⪠qk âðð ðµ . Since we have qk â
ððð¡ ðŽ â ðŽ ðŽ â qk â ðð ðŽ with the help of this inclusions we can easily
establish the proof of (ix)(x) (xi) and (xii). Hence, the proof of the
proposition.
Remark :
However, the reverse inclusions of (v)(vi)(vii)(viii)(ix)(x)(xi) and (xii) of proposition
3.2 are not true in general.
The following results can be established easily.
The inclusions may be strict in some cases they are explained with the help of the
following example.
Lets us see ððð ð¡âð ðððððð€ððð ððððð¢ð ððð A â B â qk âððð¡ ðŽ â qk âððð¡ ðµ
Let X be the topological spaces. Let ð = { ð, {ð, ð}, {ð, ð}, {ð, ð, ð, ð}, ð}. Suppose
A={a} and B={a,b} then qkâððð¡ ðŽ= ð and qkâððð¡ ðµ = {ð} it follows as A â B â qk
âððð¡ ðŽ âqk âððð¡ ðµ. If suppose let A = {b} and B = {b,d} then qkâððð¡ ðŽ= ð and qkâ
ððð¡ ðµ= ð which gives us A â B â qk âððð¡ ðŽ = qk âððð¡ ðµ
Next we check for the inclusion qkâðð(ðŽ ⪠ðµ)âqkâðð ðŽ ⪠qk âðð ðµ For that Let X be
the topological spaces. Let ð = { ð, {ð, ð}, {ð, ð}, {ð, ð, ð, ð}, ð} . Suppose A ={ a,b}
and B = {c,d} then AâªB= { a,b,c,d} here qkâðð(ðŽ ⪠ðµ)= { a,b.c,d} and qkâðð ðŽ ⪠qk
âðð ðµ = {a,b,c,d,e} which follows thatqkâðð(ðŽ ⪠ðµ)âqkâðð ðŽ ⪠qk âðð ðµ . Also if A=
{a} and B = {a,b} , ðŽ ⪠ðµ = {ð, ð}qkâðð(ðŽ ⪠ðµ) = {ð, ð, ð}qkâðð ðŽ ⪠qk âðð ðµ =
{a,b,e} which implies us that qkâðð(ðŽ ⪠ðµ)=qkâðð ðŽ ⪠qk âðð ðµ .
Behaviour of Kuratowski Operators on Some New Sets in Topology 1039
Next we go for the explanation of strict inclusions of qkâððð¡(ðŽ ⪠ðµ)âqkâððð¡ ðŽ ⪠qk â
ððð¡ ðµ.
For this we consider Let X be the topological spaces. Let ð =
{ ð, {ð, ð}, {ð, ð}, {ð, ð, ð, ð}, ð}
Here , Suppose A= { a, d} and B ={ a,b,c} A ⪠B ={a,b,c,d} qkâððð¡(ðŽ ⪠ðµ)=
{a,b,c,d}qkâððð¡ ðŽ ⪠qk âððð¡ ðµ= {a,b} which implies thatqkâððð¡(ðŽ ⪠ðµ)âqkâððð¡ ðŽ ⪠qk
âððð¡ ðµ
Also if suppose A = { c,d,e } B = { b,c,d,e} A ⪠B ={b,c,d,e} then qkâððð¡(ðŽ ⪠ðµ)=
{c,d,e} andqkâððð¡ ðŽ ⪠qk âððð¡ ðµ = {c,d,e} hence we get that qkâððð¡(ðŽ ⪠ðµ)=qkâ
ððð¡ ðŽ ⪠qk âððð¡ ðµ
Next we see the examples for the strict inclusion of qkâððð¡ (qk âððð¡ ðŽ)âqkâððð¡ ðŽ. Let
X be the topological spaces. Let ð = { ð, {ð, ð}, {ð, ð}, {ð, ð, ð, ð}, ð} be the topology
on X .Here , Suppose A ={ a,b} then qkâððð¡ ðŽ = { a,b} and qkâððð¡ (qk âððð¡ ðŽ) = {a,b}
which implies us qkâððð¡ (qk âððð¡ ðŽ)=qkâððð¡ ðŽ . Also if suppose A= { b,d,e} qkâððð¡ ðŽ
= { e} and qkâððð¡ (qk âððð¡ ðŽ)= ð which follows that qkâððð¡ (qk âððð¡ ðŽ)âqkâððð¡ ðŽ.
Next we see the examples of strict inclusions for qkâððð¡ (qk âðð ðŽ) âqkâððð¡ ðŽ. Let X be
the topological spaces. Let ð = { ð, {ð, ð}, {ð, ð}, {ð, ð, ð, ð}, ð} be the topology on X .
Suppose A = {a,c} then qkâððð¡ (qk âðð ðŽ) = {a,b,c,d} and qkâððð¡ ðŽ = ð which implies
us thatqkâððð¡ (qk âðð ðŽ) â qk âððð¡ ðŽ. Also if suppose A = { e} then qkâððð¡ (qk âðð ðŽ)
= {e} and qkâððð¡ ðŽ = {e} which follows that qkâððð¡ (qk âðð ðŽ) = qk âððð¡ ðŽ.
Proposition 3.3:
If A is a qkâ set then qkâððð¡ ðŽ = ðŽ = qk âðð ðŽ.
Proposition 3.4:
If A is a qkâ set then qkâðð (qk âððð¡ ðŽ) = ðŽ = qkâ ððð¡ (qkâ ðð ðŽ) .
Proposition 3.5 :
If A is pre * - open and A is qk- set then A is semi*-open .
Proof :
Suppose A is pre * - open and A is qk- set which implies that ðŽ â ððð¡âðð ðŽ and if
ððâ(ððð¡ ðŽ) â ððð¡â(ðð ðŽ) it follows that ðŽ â ððâððð¡ ðŽ â A is semi * - open .
Proposition 3.6 :
If A is semi * - open and A is pk- set then A is pre*-open .
Result 3.7 :
Every ðŒð â-open is pre*-open
1040 Dr. G. Hari Siva Annam and N.Raksha Ben
Proposition 3.7 :
If A is ðŒð â- open and A is qk- set then A is semi*-open .
Proof : The proof of the proposition follows from the above result and proposition.
Proposition 3.8:
For any subset ð â ð , qk âððð¡ ð = ð â qk âðð (ð â ð).
Proposition 3.9:
Suppose ðŽ â ðµ with ðð ðŽ = ðð ðµ. If A is a qk â set then B is a qkâ set
.
Proof:
Suppose ðŽ â ðµ and A is a qk â set which implies us that ððð¡âðð ðµ = ððð¡âðð ðŽ â
ððâððð¡ ðŽ â ððâððð¡ ðµ . Hence B is a qk â set .
Proposition 3.10:
Suppose ðŽ â ðµ with ðð ðŽ = ðð ðµ. If A and B are pk â set then A is a pkâ set .
4. Algorithm of ðð & ðð âsets
The objectives of the algorithms are to generate ðð ðð ðð âsets if any one of the sets
is clopen
Theorem 4.1:
In a topological space, (ð, ð), where ð = { ð, ð, ð, ðâ², ðⲠ⪠ð, ð } where ðâ² is a
subset of ð and if A is clopen Then A is both ðð & ðð â sets .
Proof:
Case (i) If A is clopen then A is both ðð & ðð â sets.
Case (ii)
Sub case (i)
If A is exact open then A = ðâ² or ðⲠ⪠ð .
Sub case (ii)
If A = ðâ² then int A = ðâ² which implies us that ð= ðð (ðâ²) . Hence ,
ðâ² is not g- closed . Thus we get ðð (ðâ²) = ððâ(ðâ²) = ð Hence, ððâ(ððð¡ ðŽ) =
ððâ(ðâ²) = ð. Also , ððð¡â(ðð ðŽ) = ððð¡â(ð) = ð.
Behaviour of Kuratowski Operators on Some New Sets in Topology 1041
Case (iii)
If A = ðⲠ⪠ð , ðð (ðŽ) = ð , ððð¡ (ðŽ) = ðŽ which implies ððâ(ðŽ) =
ð , ððð¡â ( ðŽ) = ðŽ . Hence A is both ðð & ðð â sets. Thus we get ððâððð¡ (ðŽ) =
ð and ððð¡ âðð (ðŽ) = ð .
Case (iv)
If A is exactly closed ,then A = ð â ðâ² is closed . Hence , ðâ² is open
which implies us ðâ² is both ðð & ðð â sets . Thus , we get ð â ðâ² = ðŽ is
both ðð & ðð â sets.
Case (v)
If A is neither open nor closed
Sub case (i) :
Suppose ðð(ðŽ) is closed and ððð¡(ðŽ) = ð is open ,Hence
ððð¡âðð (ðŽ) = ððð¡â(ðŽ) which implies that ððâððð¡ (ðŽ) = ððâ(ððð¡ðŽ) = ð.Thus
we get ððâððð¡ (ðŽ) â ððð¡â(ðð ðŽ) Hence, The set is a ðð â set .
Sub case (ii) :
Suppose ðð(ðŽ) = X and ððð¡(ðŽ) = ð is open .Then, ððâððð¡ (ðŽ) =
ððâ(ðŽ) and ððð¡âðð (ðŽ) = ð . Thus , we get ððâððð¡ (ðŽ) â ððð¡â(ðð ðŽ) Hence,
The set is a ðð â set .
Theorem 4.2
If (ð, ð) is a topological space and ð = { ð, ð, ð, ðâ², ðⲠ⪠ð, ð } where ð and ð are
clopen and ðâ² â ð then the collection of ðð- set is equal to ð and the collection of
ðð- set is 2ð¥.
Proof of the this theorem is similar to algorithm 4.1
Theorem 4.3:
In a Topological space (X,Ï) where Ï = {Ï,X,U,V} where Uâ©V = ð and UâªV = X
then the member of 2X are pk-set.
Proof:
Let A be any subset of X
Case (i): If A = U or V
Then A is clopen . hence A is pk-set.
Case(ii) : If A â U and V
Then A is not clopen. Then cl(A) and int(A) are clopen.Hence int*cl(A) =cl(A) and
cl*(int(A)) = int (A).Since int(A) â cl(A), cl*(int(A)) â int*cl(A). Hence A is pk-set.
1042 Dr. G. Hari Siva Annam and N.Raksha Ben
Proposition 4.4 :
If (ð, ð) is a topological spaces and any one of the given set is clopen
then the collection of qk â sets are of the form { AâX \ A is either open or closed or
clopen }.
Proposition 4.5 :
If (ð, ð) is a topological spaces and any one of the given set is clopen
then the collection of pk â sets are of the form { AâX \ A is either open nor closed and
ðð(ðŽ) = ð or ððð¡ ðŽ = ð or A and ðð(ðŽ) are open or A and ððð¡ (ðŽ) are closed}
Remark 4.6 :
If (ð, ð) is a topological spaces and any one of the given set is clopen and A is neither
open nor closed and ðð(ðŽ) â ð and ððð¡ ðŽ â ð.Then A is neither pk-set nor ðð- set .
5. Topology Generated by ðð & ðð âsets
Proposition 5.1 :
If A is clopen and B is ðð- set then ðŽ â© ðµ and ðŽ ⪠ðµ are ðð- set
Proposition 5.2:
If A is clopen and B is ðð- set then ðŽ â© ðµ and ðŽ ⪠ðµ are ðð- set
Proposition 5.3:
Let (ð, ð) be a topological space Suppose A is Clopen and B is a ðð- set then ð =
{ð, ðŽ â© ðµ, ðŽ, ðŽ â (ðŽ â© ðµ), ð} is a topology and every member of N is a ðð- set .
Proof:
ð = {ð, ðŽ â© ðµ, ðŽ, ðŽ â (ðŽ â© ðµ), ð} is clearly a topology on X .
Clearly, ð, ðŽ, ð are all ðð- sets. Since A is Clopen and B is a ðð- set, By using the
algorithm ðŽ â© ðµ is a ðð- set. Since the complement of ðð- set is again a ðð- set, ð â(ðŽ â© ðµ) is a ðð- set. This proves that ðŽ â (ðŽ â© ðµ) = ðŽ â© (ð â (ðŽ â© ðµ) is a ðð- set.
Hence every member of N is a ðð- set.
Proposition 5.4:
Let (ð, ð) be a topological space Suppose A is Clopen and B is a ðð- set then ð =
{ð, ðŽ, ðµ, ðŽ â© ðµ, ðŽ ⪠ðµ, ð} is a topology and every member of N is a ðð- set .
Proof:
ð = {ð, ðŽ â© ðµ, ðŽ, ðŽ â (ðŽ â© ðµ), ð} is clearly a topology on X .
Behaviour of Kuratowski Operators on Some New Sets in Topology 1043
Clearly, ð, ðŽ, ð are all ðð- sets. Since A is Clopen and B is a ðð- set, by using algorithm
we get , ðŽ â© ðµ and ðŽ ⪠ðµ are ðð- sets. Hence every member of N is a ðð- set.
Proposition 5.5:
Let (ð, ð) be a topological space Suppose A is Clopen and B is a p- set then ð =
{ð, ðŽ â© ðµ, ðŽ, ðŽ â (ðŽ â© ðµ), ð} is a topology and every member of N is a ðð- set .
Proof:
ð = {ð, ðŽ â© ðµ, ðŽ, ðŽ â (ðŽ â© ðµ), ð} is clearly a topology onX.
Clearly, ð, ðŽ, ð are all ðð- sets. Since A is Clopen and B is a ðð- set, by using the
algorithms we get, ðŽ â© ðµ is a ðð- set. Since the complement of ðð- set is again a ðð-
set, ð â (ðŽ â© ðµ) is a ðð- set. This proves that ðŽ â (ðŽ â© ðµ) = ðŽ â© (ð â (ðŽ â© ðµ) is
a ðð- set. Hence every member of N is a ðð- set.
Conclusion
Thus in this paper we have introduced ðð- sets and ðð- sets . Also some of their
properties were discussed . Henceforth, we are discussing about some more
characteristics of ðð- sets and ðð- sets in the further papers the results will be
established.
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[4] Levine N, Semi-open sets and semi continuity in topological spaces,
Amer.Math.Monthly,70(1963), 36-41.
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1044 Dr. G. Hari Siva Annam and N.Raksha Ben
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