CHAPTER-!!
rw-CLOSED SETS AND rw-OPEN SETS IN TOPOLOGICAL
SPACES
2.1 Introduction.
N.Levine [50] introduced generalized closed sets in general
topology as a generalization of closed sets. This concept was found to be
useful and many results in general topology were improved. Many
researchers like Balachandran, Sundaram and Maki [13], Bhattacharyya
and Lahiri [15], Arockiarani [3], Dunham [33], Gnanambal [39],
Malghan [57], Palaniappan and Rao [72], Park and Park ([73], [76]),
Arya and Gupta [6] and Devi [23] have worked on generalized closed
sets, their generalizations and related concepts in general topology.
In section 2 of this chapter, a new class of sets, called regular
weakly closed (briefly, rw-closed) sets in topological spaces are
introduced and studied. During this process some of their properties are
obtained. It is found that every closed set is a rw-closed set, which
implies regular generalized closed set.
In section 3 of this chapter, we introduce and study rw-open sets in
topological spaces and some of their properties. Also we introduce rwneighbourhoods in topological spaces by using the notions of rw-open
sets. We prove that every neighbourhood of x in X is a rw-neighbourhood
of x but not conversely.
In section 4 of this chapter, the notion of rw-interior is defined and
some of its basic properties are studied. Also we introduce the concept of
rw-closure in topological spaces by using the notions of rw-closed sets
and obtained some of its basic results. We define xrw and prove that it
17
-
-
forms a topology on X. For any A c X, it is proved that the complement
of rw-interior of A is the rw-closure of the complement of A.
In section 5 of this chapter, a new class of sets called minimal rwopen sets and maximal rw-closed sets in topological spaces are
introduced, which are subclasses of rw-open sets and rw-closed sets
respectively. During this process some of their properties are obtained.
We prove that the complement of a minimal rw-open set is a maximal rwclosed set. Also, we introduce and study maximal rw-open sets and
minimal rw-closed sets in topological spaces.
2.2 rw-closed sets and their basic properties.
In this section, a new class of sets, called regular w-closed
(briefly, rw-closed) sets in topological spaces are introduced and studied.
During this process some of their properties are obtained It is found that
every closed set is a rw-closed set, which implies regular generalized
closed set in X.
2.2.1 Definition: A subset A of a space (X, x) is called a regular w-closed
(briefly, rw-closed) set if cl(A) c U whenever AcU and U is a regular
semiopen in (X, x). We denote the set of all rw-closed sets in (X, x) by
RWC(X).
First we prove that the class of rw-closed sets properly lies between
the class of w-closed sets and the class of regular generalized closed sets.
2.2.2 Theorem: Every w-closed set in X is rw-closed set in X.
Proof: Let A be any w-closed set in X. To prove that A is a rw-closed set
in X. Let U be any regular semiopen in X such that AcU. Since every
-
18
-
regular semiopen set is semiopen in X, by definition of w-closed set in X,
cl(A) c U. Therefore A is a rw-closed set in X.
The converse of the above theorem need not be true as seen from
the following example.
2.2.3 Example: Let X=[a, b, c} with topology x={X, (f), {a}, {b}, {a, b}}.
Then the set A={a, b} is rw-closed, but not w-closed in X.
2.2.4 Theorem: Every rw-closed set is a regular generalized closed set in
X.
Proof: Let A be any rw-closed set in X. Let U be any regular open set in
X such that AcU. Since every regular open set is a regular semiopen set
in X, and since A is rw-closed set in X, it follows that cl(A)cU. Hence A
is a regular generalized closed set in X.
The converse of the above theorem need not be true as seen from
the following example.
2.2.5 Example: Let X={a, b, c, d} and x={X, <|>, {a}, {b}, {a, b}, {a, b,
c}}. Then A={c} is a regular generalized closed set but not a rw-closed
set in X.
2.2.6 Remark: From Sheik John [83] we know that every closed set is a
w-closed set but not conversely. Also from Theorem 2.2.2, every
w-closed set is a rw-closed set but not conversely. Hence every closed
set is a rw-closed set but not conversely.
2.2.7 Remark: From Stone [88] we know that every regular closed set is
closed but not conversely. Also from Remark 2.2.6, every closed set is a
rw-closed set but not conversely. Hence every regular closed set is a rwclosed set but not conversely.
19
-
-
2.2.8 Remark: From [94] we know that every 0-closed set is closed but
not conversely. Also from Remark 2.2.6, every closed set is a rw-closed
set but not conversely and hence every 0-closed set is a rw-closed set but
not conversely.
2.2.9 Remark: From [94] we know that every 5-closed set is closed but
not conversely. Also from Remark 2.2.6, every closed set is a rw-closed
set but not conversely. Hence every 6-closed set is a rw-closed set but not
conversely.
2.2.10 Remark: From Theorem 2.2.4, we know that every rw-closed set
is a rg-closed set but not conversely. Also from G.Gananambal [39], we
that every rg-closed set is a gpr-closed set but not conversely. Hence
every rw-closed set is a gpr-closed set but not conversely.
2.2.11 Remark: From [30] we that every 7i-closed set is closed but not
conversely. Also from Remark 2.2.6, every closed set is a rw-closed set
but not conversely. Hence every 7t-closed set is a rw-closed set but not
conversely.
2.2.12 Theorem: Every rw-closed set is a regular weakly generalized
closed set.
Proof: Let A be any rw-closed set in X. Let U be any regular open set in
X such that AcU. Since every regular open set is a regular semiopen set
in X, U is a regular semiopen set in X. As A is a rw-closed set, we have
cl(A)<=U. Also cl(int(A))ccl(A). Thus cl(int(A))cU. Therefore A is a
regular weakly generalized closed set in X.
The converse of the above theorem need not be true as seen from
the following example.
20
-
-
2.2.13 Example: Let X={a, b, c, d} with the topology x={X, <|>, {a}, {b},
{a, b}, {a, b, c}}. Then the set A= {b, c} is regular weekly generalized
closed, but not a rw-closed set in X.
2.2.14 Remark: The following example shows that rw-closed sets are
independent of g*-closed sets, mildly g-closed sets, g-closed sets, wgclosed sets, semi-closed sets, a-closed sets, ga-elosed sets, ag-closed
sets, sg-closed sets, gs-closed sets, gsp-closed sets, p-elosed sets, pre­
closed sets, gp-closed sets, swg-closed sets, Jtg-closed sets, 0-generalized
closed sets and 6-generalized closed sets.
2.2.15 Example: Let X= {a, b, c, d} with the topology x={X, <j), {a}, {b},
{a, b}, {a, b, c}}. Then
(i)
closed sets in (X, x) are X, <|>, {d}, {c, d}, {a, c, d} {b, c, d},
(ii)
rw-closed sets in (X, x) are X, <|>, {d}, {a, b}, {c, d}, {a, b, c,}, {a,
b,d},{a,c,d}, {b, c, d},
(iii)
g*-closed sets in (X, x) are X, <)>, {d}, {c, d}, {a, d}, {b, d}, {a, b,
d}, {a, c, d}, {b, c, d},
(iv)
mildly g-closed sets in (X, x) are X, 0, {d}, {c, d}, {a, d}, {b, d},
{a, b, d}, {b, c, d}, {a,c,d},
(v)
g-closed sets in (X, x) are X, <{), {d}, {c, d}, {a, d}, {b, d}, {a, b, d},
{a, c. d}, {b, c, d},
(vi)
wg-closed sets in (X, x) are X, (j), {c}, {d}, {c, d}, {a, d}, (b, d},
{a, c}, {a, b, d}, {a, c, d}, {b, c, d},
(vii) semi-closed sets in (X, x) are X, 0, {a}, {b}, fc}, {d}, {c, d}, {a,
d}, {b, c}, {a, c,}, {b, d}, {a, c, d}, fb, c, d},
(viii) a-closed sets in (X,x) are X, (|), {c}, {d}, {c, d}, {a, c. d}, {b, c, d},
-21
-
(ix)
ga-closed sets in (X, t) are X, 0, {c}, {d}, {c, d}, {b, c, d}, {a, c,
d},
(x)
ag-closed sets in (X, x) are X, 0, {c}, {d}, {c, d}, {a, d}, {b, d},
{a, b, d}, {a, c, d}, {b, c,d},
(xi)
sg-closed sets in (X, x) are X, 0, {a}, {b}, {c}, {d}, {c, d}, {b, c},
{a, d}, {b, d}, {a, c}, {a, c, d}, {b, c, d},
(xii) gs-closed sets in (X, x) are X, (j), {a}, {b}, {c}, {d}, {b, c}, {c, d},
{a, d}, {b, d}, {a, c}, {a, c, d}, {a, b, d}, {b, c, d},
(xiii) gsp-closed sets in (X, x) are X, <|>, {c}, {d}, {b, c}, {c, d}, {a, d}
{a, c}, {a, b, d}, {a, c, d}, {b, c, d},
(xiv) (3-closed sets in (X, x) are X, <(), {a}, {b}, {c}, {d}, {c, d}, {a^d},
{b, c}, {b, c}, {a, c},{b, d}, {b, c, d}, {a, c, d},
(xv) pre-closed sets in (X, x) are X, ((), {c}, {d}, {c, d}, {b, c, d}, {a, c,
d},
(xvi) gp-closed sets in (X, x) are X, 0, {c}, {d}, {c, d}, {a, d}, {b, d},
{a, b, d},
(xvii) swg-closed sets in (X, x) are X 0 {c}, {d}, {c, d}, {a, c, d},
{b, c, d},
(xviii)7ig-closed sets in (X, x) are X, 0, {c}, {d}, {a, e), (b, c}, {e, d),
{a, d},{b, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d},
(xix) ©-generalized closed sets in (X, x) are X, 0, {d}, {c, d}, {a, d}, {b,
d}, {a, b, d}, {a, c, d}, {b, c, d},
(xx) 8-generalized closed sets in (X, x) are X, 0, {d}, {c, d}, {a, d},{b,
d}, {a, b, d}, {a, c, d}, (b, c, d},
22
-
-
2.2.16 Remark: From the above discussions and known results we have
the following implications
In the following diagram, by
A ------ ►
A
B we mean A implies B but not conversely and
—► B means A and B are independent of each other.
Figure-2.1
2.2.17 Theorem: Union of two rw-closed subsets of X is also a rw-closed
subset of X.
Proof: Assume that A and B are rw-closed sets in X. Let U be a regular
semiopen set in X such that AuB czU. Then AcU and BcU. Since A and
B
are
rw-closed,
cl(A)c=U
and
cl(B)cU.
Hence
cl(AuB)=cl(A)ucl(B)cU. That is cl(AuB)czU. Therefore AuB is a
rw-closed set in X.
23
-
-
2.2.18 Remark: The intersection of two rw-closed sets in X is generally
not a rw-closed set in X.
2.2.19 Example: Let X={a, b, c, d} and T ={X, (j>, {a}, {b}, {a, b}, {a, b,
c}}. If A={a, b} and B={a, c, d}, then A and B are rw-closed sets In X,
but AnB={a} is not a rw-closed set in X.
2.2.20 Theorem: If a subset A of X is a rw-closed set in X, then cl(A)-A
does not contain any nonempty regular semiopen set in X.
Proof: Suppose that A is a rw-closed subset in X. We prove the result by
contradiction. Let U be a regular semiopen set such that cl(A)-A z>U and
U#|>. Now Uccl(A)-A. Therefore U c X-A which implies A cX-U.
Since U is a regular semiopen set, by Lemma 1.2.5, X-U is also a regular
semiopen set in X. Since A is a rw-closed set in X, by definition, we have
el(A) cX-U.
So U c X-cl(A). Also U a el(A). Therefore U <=
(cl(A)n{X-cl(A)})=()). This shows that U=<j> which is a contradiction.
Hence cl(A)-A dose not contains any non-empty regular semiopen set in
X.
The converse of the above theorem need not be true as seen from
the following example.
2.2.21 Example: If cl(A)-A contains no nonempty regular semiopen
subset in X, then A need not be rw-closed. Consider X={a, b, c} with
topology x={X, (j), {a}, {b}, {a, b}} and A={a}. Then cl(A)-A={a,
e}-{a}={c} does not contain nonempty regular semiopen set, but A is not
a rw-closed set in X.
2.2.22 Corollary: If a subset A of X is a rw-closed set in X, then cl(A)-A
dose not contain any nonempty regular open set in X.
24
-
-
Proof: Suppose that A is a rw-closed set in X. We know that every
regular open set is regular semiopen. By Theorem 2.2.20, cl(A)-A does
not contain any nonempty regular semiopen set in X and hence cl(A)-A
dose not contain any nonempty regular open set in X.
2.2.23 Corollary: If a subset A of X is a rw-closed set in X, then
cl(A)-A does not contain any nonempty regular closed set in X, but not
conversely
Proof: Suppose that A is a rw-closed set in X. By Lemma 1.2.6, every
regular closed set is regular semiopen. By Theorem 2.2.20, cl(A)-A does
not contain any nonempty regular semiopen set in X and hence cl(A)-A
does not contain any nonempty regular closed set in X.
2.2.24 Theorem: For an element xeX, its complement X-{x} is rwclosed or regular semiopen
Proof: Suppose X-{x} is not a regular semiopen set. Then X is the only
regular semiopen set containing X-{x}. This implies cl({X-{x}})czX.
Hence X-{x} is a rw-closed set in X.
2.2.25 Theorem: If A is a regular open and rw-closed set, then A is
regular closed and hence clopen.
Proof: Suppose A is a regular open and rw-closed set. As every regular
open set is a regular semiopen and AcA, we have cl(A)cA. Also
A<zcl(A). Therefore cl(A)=A. That is A is closed. Since A is regular
open, A is open. Now cl{int(A)}=cl{A}=A. Therefore A is regular closed
and clopen.
2.2.26 Theorem: If A is a rw-closed subset of X such that AcBc:cl(A),
then B is a rw-closed set in X.
25
-
-
Proof: Let A be a rw-closed set of X such that AcBccl(A). Let U be a
regular semiopen set of X such that BcU. Then AcU. Since A is rwclosed, we have cl(A) c U. Now cl(B) c cl(cl(A)) = cl(A) c U. That is
cl(B)cU. Therefore B is a rw-closed set in X.
2.2.27 Remark: The converse of the Theorem 2.2.26 need not be true in
general. For consider the topological space (X, x), where X={a, b, c, d}
with the topology x={X, ((», {a}, {b}, {a, b}, {a, b, c}}. Let A= {d} and
B={c, d}. Then A and B are rw-closed sets in (X, x), but A c B & cl (A).
2.2.28 Theorem: Let A be rw-closed in (X, x). Then A is closed if and
only if cl(A)-A is regular semiopen.
Proof: Suppose A is a closed set in X. Then cl(A)=A and so cl(A)-A=(|),
which is regular semiopen in X.
Conversely, suppose cl(A)-A is a regular semiopen set in X. Since
A is rw-closed, by Theorem 2.2.26, cl(A)-A does not contain any
nonempty regular semiopen set in X. Then cl(A)-A=(J). That is cl(A)=A
and hence A is a closed set in X.
2.2.29 Theorem: If A is a regular open and rg-closed set, then A is rwclosed set in X.
Proof: Let A be a regular open and rg-closed set in X. To prove A is a
rw-closed set in X. Let U be any regular semiopen set in X such that
AcU. Since A is regular open and rg-closed, we have cl(A)cA. Then
cl(A)cAcU. That is cl(A)cU. Hence A is a rw-closed set in X.
2.2.30 Theorem: If a subset A of a topological space X is both regular
semiopen and rw-closed, then it is closed.
26
-
-
Proof: Suppose a subset A of a topological space X is both regular
semiopen and rw-closed. Now Ac A. Then cl(A)cA. Also Accl(A) and
so cl(A)=A. Hence A is closed.
2.2.31 Corollary: Let A be a regular semiopen and rw-closed subset in
X. Suppose that F is a closed set in X. Then AnF is a rw-closed set in X.
Proof: Let A be a regular semiopen and rw-closed subset in X and F be
closed. By Theorem 2.2.30, A is closed. So AnF is closed and hence
AnF is a rw-closed set in X.
The next two results are required in the sequel.
2.2.32 Theorem: [21] If A is open and S is semiopen in a topological
space X, then AnS is semiopen in X.
2.2.33 Theorem: [49] Let AcYcX, where X is a topological space and
Y is a subspace of X. If A e SO(X), then Ae SO(Y).
2.2.34 Lemma: Let A cYc X, where X is a topological space and Y is
an open subspace of X. If A e RSO(X), then A e RSO(Y).
Proof: Suppose that AcYcX, Y is an open subspace of X and
AeRSO(X). To prove that A is regular semiopen in Y. That is to prove A
is both semiopen and semi-closed in Y.
Now we show that A is semiopen in Y. Since AcYc X and A is
semiopen in X, by Theorem 2.2.33, A is semiopen in Y.
Now we
show
that A
is
semi-closed in Y.
We have
Y-A=Yn(X-A). Since A is regular semiopen in X, by Lemma 1.2.5,
X-A is also regular semiopen in X and so X-A is semiopen in X. Since
Y is open in X, by Theorem 2.2.32, Yn(X-A) is semiopen in X. That is
27
-
-
Y-A is semiopen in X. Also Y-A c Y c X. By theorem 2.2.33, Y-A is
semiopen in Y and so A is semi-closed in Y. Thus A is both semiopen
and semi-closed in Y. Hence A is a regular semiopen set in Y.
2.2.35 Theorem: Suppose B cA c X, B is rw-closed set relative to A
and that A is both regular open and rw-closed subset of X, then B is a rwclosed set relative to X.
Proof: Let BcG and G be a regular semiopen set in X. But given that
BcAcX, and therefore BcA and BcG. This implies BcAnG.
Now we show that AnG is regular semiopen in A. First we show
that AnG is regular semiopen in X. Since A is an open set and G is
semiopen in X, by Theorem 2.2.32, AnG is semiopen in X. Since A is
regular open and rw-closed by Theorem 2.2.25, A is closed and so A is
semi-closed in X. G is semi-closed in X, as every regular semiopen set is
semi-closed. Then AnG is semi-closed in X. Thus AnG is both
semiopen and semi-closed in X and hence AnG is a regular semiopen in
X. Also AnGcAcX and A is an open subspace of X, by Lemma 2.2.34,
AnG is regular semiopen in A. Since B is rw-closed relative to A,
clA(B)cAnG—(i). But clA(B)=Ancl(B)—(ii). From (i) and (ii) it follows
that Ancl(B)cAnG. Consequently Ancl(B)cG. Since A is regular open
and rw-closed, by Theorem 2.2.25, cl(A)=A and so cl(B)cA. We have
Ancl(B)=cl(B). Thus cl(B)cG and hence B is a rw-closed set relative to
X.
2.2.36 Corollary: [70] Let A be a regularly open set in a space X and B a
subset of A. Then B is a regularly open set in X if and only if B is a
regularly open set in the subspace A.
28
-
-
2.2.37 Lemma: Let Y be a regular open set in a space X and U be a
subset of Y. Then U is a regular semiopen set in X if and only if U is a
regular semiopen set in the subspace Y
Proof: Suppose U is a regular semiopen set in X. Then UcYczX and Y is
an open subspace of X. By Lemma 2.2.34, U is regular semiopen in Y.
Conversely, suppose U is regular semiopen in the subspace Y. To
prove that U is regular semiopen in X. By definition, there exist a regular
open
set V
in Y such that
VcUcclY(V).
That
is
VcUc
clY(V)=Yncl(V)ccl(V). Now VcUc cl(V). Also by Corollary 2.2.36,
V is a regular open set in X. Therefore U is a regular semiopen set in X.
2.2.38 Theorem: Let AcYcX and suppose that A is rw-closed in X.
Then A is rw-closed in Y provided Y is regular open in X.
Proof: Let A be rw-closed in X and Y be regular open subspace in X. Let
U be any regular semiopen set relative to Y such that A c U. Then
AcYc X. By Lemma 2.2.37, U is regular semiopen in X. Since A is rwclosed in X, cl(A)cU. That is Yncl(A)cYnU=U. Thus clY(A)cU.
Hence A is rw-closed in Y.
2.2.39 Theorem: If A is both open and g-closed in X, then it is a rwclosed set in X.
Proof: Let A be an open and a g-closed set in X. Let AcU and U be a
regular semiopen set in X. Now A c A. By hypothesis cl(A)cA. That is
\
el(A)cU. Thus A is a rw-closed set in X.
2.2.40 Remark: If A is both open and g-closed in X, then A need not be
rw-closed, in general, as seen from the following example.
29
-
-
2.2.41 Example: Consider X={a, b, c} with topology x={X, 0, {a}, {b},
{a, b}}. In this topological space the subset {a, b} is an open and a rwclosed set, but not g-closed.
2.2.42 Theorem: If a subset A of a topological space is both open and
wg-closed, then it is rw-closed.
Proof: Suppose a subset A of X is both open and wg-closed. Let AcU
and U be a regular semiopen set in X. Now A z>cl(int(A))=A, as A is an
open set. That is cl(A)c AcU. Thus A is a rw-closed set in X.
2.2.43 Theorem: In a topological space X, if RSO(X)={X, <j)}, then every
subset of X is a rw-closed set.
Proof: Let X be a topological space and RSO(X)={X, 0}. Let A be any
subset of X. Suppose A=<J>.Then
§ is rw-closed set in X.
Suppose A*<j>.
Then X is the only regular semiopen set containing A and so cl(A)cX.
Hence A is a rw-closed set in X.
2.2.44 Remark: The converse of the above Theorem 2.2.43 need not be
true in general as seen from the following example.
2.2.45 Example: Let X={a, b, c, d} with the topology x={X, <j), {a, b},
{c, d}}. Then every subset of (X, x) is a rw-closed set in X, but
RSO(X,x)={X, <)>, {a, b}, {c, d}}.
2.2.46 Theorem: In a topological space (X, x), RSO(X, x)c{FcX: F^x}
if and only if every subset of (X, x) is a rw-closed set.
Proof: Suppose that RSO(X, x)c{Fc X: F^e x}. Let A be a any subset of
(X, x) such that AcU, where U is regular semiopen. Then Ue RSO(X,x)
c{Fc X: I^e x}. That is Ue {Fc X: I^exJ.Thus U is a closed set. Then
cl(U)=U. Also cl(A)ccl (U)=U. Hence A is a rw-closed set in X.
30
-
-
Conversely, suppose that every subset of (X, x) is rw-closed. Let
Ue RSO(X, x). Since UcU and U is rw-closed, we have cl(U)cU. Thus
cl(U)=U and U€ {Fc X: Fcex}. Therefore RSO(X,x)c{Fc X: Fcg x}.
2.2.47 Theorem: Let X be a regular space where “every regular
semiopen subset is open”. If A is compact subset of X, then A is rwclosed.
Proof: Suppose AcU and U is regular semiopen. By hypothesis U is an
open set. But A is a compact subset in the regular space X. Hence there
exists an open set V such that AcVccl(V)cU. Now Accl(V) implies
cl(A)ccl(cl(V))=cl(V)cU. That is cl(A)cU. Hence A is rw-closed in X.
2.2.48 Definition: The intersection of all regular semiopen subsets of
(X, x) containing A is called the regular semi-kernel of A and is denoted
by rsker(A).
2.2.49 Lemma: Let X be a topological space and A be a subset of X. If A
is a regular semiopen set in X, then rsker(A)=A, but not conversely.
Proof: Follows form Definition 2.2.48.
2.2.50 Lemma: For any subset A of (X, x), sker(A)c rsker(A).
Proof: Follows form the implication RSO(X)cSO(X).
2.2.51 Lemma: For any subset A of (X, x), Ac rsker(A).
Proof: Follows form Definition 2.2.48.
2.2.52 Theorem: A subset A of (X, x) is rw-closed if and only if
cl(A)c rsker(A).
Proof: Suppose that A is rw-closed. Then cl(A)cU, whenever AcU and
U is regular semiopen. Let xe cl(A). Suppose x £ rsker(A), then there is a
regular semiopen set U containing A such that xgU. Since A is rw-
-31
-
closed, cl(A)cU. We have xgcl(A), which is a contradiction. Hence
xe rsker(A) and so cl(A)c rsker(A).
Conversely, let cl(A)c rsker(A). If U is any regular semiopen set
containing A, then rsker(A)cU. That is cl(A)cxsker(A)c:U. Hence
cl(A)cU. Therefore A is a rw-closed set in X.
2.2.53 Lemma: [43] Let x be a point of (X, x). Then {x} is either
nowhere dense or preopen.
2.2.54 Remark: [32] In the statement of Lemma 2.2.53, we may consider
the following decomposition of a given topological space (X, x), namely
X=XiuX2 where Xi = {xeX: {x} is nowhere dense] and X2 ={xeX: [x]
is preopen}.
2.2.55 Lemma: For any subset A of (X, x), X2ncl (A)c rsker(A).
Proof: Let x e X2ncl (A) and suppose that x g rsker(A). Then there is a
regular semiopen set U containing A such that xgU. If F=X-U,
then
F
is
regular
semiopen
and
so
F
is
semi-closed.
Now
scl({x})={x}uint(cl({x}))<zF. Since cl({x})ccl(A), we have int(cl({x})cz
Auint(cl(A)). Again since xeX2, we have xgXj and so int(cl({x}>£(|>.
Therefore there has to be some point ye Auint(cl({x})) and hence
ye FnA, which is a contradiction. So xe rsker(A) and hence X2ncl (A)c
rsker(A).
2.2.56 Theorem: For any subset A of (X, x), if Xjncl (A)cA, then A is a
rw-closed set in X.
Proof: Let A be any subset of (X, x). Suppose that Xjncl (A)c A. To
prove A is a rw-closed set in X. Then cl(A)crsker(A), since Acrsker(A),
by Lemma 2.2.51. Now cl(A)=Xncl(A)=(X1uX2)ncl(A). That is
cl(A)=(X]ncl(A))u(X2ncl(A))crsker(A), since X,ncl(A)c rsker(A) and
32
-
-
by Lemma 2.2.55, X2ncl(A) <z rsker(A). That is cl(A) e rsker(A). By
Theorem 2.2.52, A is a rw-closed set in X.
The converse of the above theorem need not be true in general as
seen from the following example.
2.2.57 Example: Let X= {a, b, c, d} be with topology x={X, $, {a}, {b},
{a, b}, {a, b, c}}. Here Xj={c, d} and X2={a, b}. Take A={a, b, c}. Then
A is a rw-closed set in X. Now X!ncl(A)={c, d}nX={c, d}<zA. Thus
Xincl(A)<z A, but A is a rw-closed set in X.
2.3 rw-open sets and rw-neighbourhoods.
In this section, we introduce and study rw-open sets in topological
spaces and obtain some of their properties. Also, we introduce rwneighbourhoods in topological spaces by using the notions of rw-open
sets. We prove that every neighbourhood of x in X is rw-neighbourhood
of x but not conversely.
2.3.1 Definition: A subset A in (X, x) is called regular w-open (briefly,
rw-open) in X if Ac is rw-closed in (X, x).
We denote the family of all rw-open sets in (X, x) by RWO(X).
2.3.2 Theorem: If a subset A of space X is w-open, then it is rw-open but
not conversely.
Proof: Let A be a w-open set in a space X. Then Ac is a w-closed set. By
Theorem 2.2.2, Ac is rw-closed. Therefore A is an rw-open set in X.
The converse of the above theorem need not be true as seen from
the following example.
33
-
-
2.3.3 Example: Let X={a, b, c] be with topology x={X, <j), {a}, {b}, {a,
b}}.
In this topological space the subset {c} is rw-open but not w-open.
2.3.4 Corollary: From Sheik Jhon [83] it is evident that every open set is
w-open set but not conversely. By Theorem 2.3.2, every w-open set
implies rw-open set but not conversely and hence every open set is rwopen set, but not conversely.
2.3.5 Corollary: From Stone [88] it is evident that every regular open set
is open, but not conversely. By corollary 2.3.4, every open set is an rwopen set but not conversely and hence every regular open set is an rwopen set but not conversely.
2.3.6 Corollary: From [94] it is evident that every 0-open set is open but
not conversely. By Corollary 2.3.4, every open set is an rw-open set but
not conversely and hence every 0-open set is an rw-open set but not
conversely.
2.3.7 Corollary: From [94] it is evident that every 8-open set is open but
not conversely. By Corollary 2.3.4, every open set is rw-open, but not
conversely and hence every 8-open set is an rw-open set, but not
conversely.
2.3.8 Theorem: If a subset A of a space X is rw-open, then it is an rgopen set in X.
Proof: Let A be an rw-open set in a space X. Then Ac is a rw-closed set
in X. By Theorem 2.2.4, Ac is a rg-closed set in X. Therefore A is an rgopen set in space X.
The converse of the above theorem need not be true as seen from
the following example.
34
-
-
2.3.9 Example: Let X= {a, b, c, d} be with topology x={X, <)>, {a}, {b},
{a, b}, {a, b, c}}. In this topological space the subset {a, b, d} is an rgopen set but not an rw-open set in X.
2.3.10 Theorem: If a subset A of a space X is rw-open, then it is gpropen but not conversely.
Proof: Let A be an rw-open set in a space X. Then Ac is a rw-closed set
in X. By Remark 2.2.10, Ac is gpr-closed in X. Therefore A is a gpr-open
set in X.
The converse of the above theorem need not be true as seen from
the following example.
2.3.11 Example: Let X= {a, b, c, d} be with topology x={X, 0, {a}, {b},
{a, b}, {a, b, c}}. In this topological space the subset {a, d} is gpr-open,
but not rw-open.
2.3.12 Theorem: If a subset A of a topological space X is rw-open, then
it is rwg-open but not conversely.
Proof: Let A be an rw-open set in a space X. Then Ac is rw-closed in X.
By Theorem 2.2.12, Ac is a rwg-closed set in X. Therefore A is an rwgopen subset in X.
The converse of the above theorem need not be true as seen from
the following example.
2.3.13 Example: Let X= {a, b c, d} be with topology x= {X, <J), {a}, {b},
{a, b}, {a, b, c}}. In this topological space, the subset {b, c} is an rwgopen set in X, but not rw-open.
2.3.14 Theorem: A subset A of a topological space X is rw-open if and
only if Fc int(A) whenever Fc A and F is a regular semiopen in X.
35
-
-
Proof: Suppose that Fez int(A) whenever Fez A and F is regular semiopen
in X. To prove that A is rw-open in X. Let AccU where U is regular
semiopen in X. Then UccA. By Lemma 1.2.5, Uc is also regular
semiopen in X. By hypothesis Uccint(A). Since Ucc int(A), we have
(int (A))c c U. That is cl(Ac) c U, since cl(Ac) = (int(A))c. Thus Ac is
rw-closed. That is A is rw-open.
Conversely, suppose that A is rw-open, FcA and F is regular
semiopen. Then Ac c
By Lemma 1.2.5,
Since Ac is rw-closed, we have cl(Ac) c
is also regular semiopen.
and so Fc int(A), since cl(Ac)
= (int(A))c.
2.3.15 Theorem: If A and B are rw-open sets in a space X, then AnB is
also an rw-open set in X.
Proof: Let A and B be two rw-open sets in X. Then Ac and Bc are rwclosed sets in X. By Theorem 2.2.17, Ac u Bc is also a rw-closed set in X.
That is AcuBc=(An B)c is a rw-closed set in X. Therefore AnB is
an rw-open set in X.
2.3.16 Remark: The union of two rw-open sets in X is generally not an
rw-open set in X.
2.3.17 Example: Let X= {a, b, c} be with topology x={X,
§
{a}, {b},
{a, b}}. If A= {a} and B={c}, then A and B are rw-open sets in X, but
AuB= {a, c} is not an rw-open set in X.
2.3.18 Theorem: If int(A) cBcA and A is an rw-open set of (X, x),
then B is also an rw-open set of (X, x).
Proof: Let int(A) cBcA and A is an rw-open set of X. To prove that B
is an rw-open set of X. Let F be a regular semiopen set of X such that
Fc B. Now F c B c A. That is F c A. Since A is an rw-open set of X, by
36
-
-
Theorem 2.3.14, Fc int(A). By hypothesis int(A)c B. Then int(int (A))c
int(B). That is int(A)c int(B). Then Fc int(B). By Theorem 2.3.14, B is
an rw-open set in X.
2.3.19 Theorem: If A c X is rw-closed, then cl(A)-A is rw-open.
Proof: Let A be rw-closed. Let F be a regular semiopen set such that
Fc el(A)-A. Then by Theorem 2.2.20, F=<|>, so Fc int(cl(A)-A). By
Theorem 2.3.14, cl(A)-A is rw-open.
The reverse implication does not hold.
2.3.20 Example: Let X = {a, b, c} be with topology x={X, <|>, {a}, {b},
{a, b}}. Let A={b}, then cl(A)={b, c} and cl(A)-A={c} which is rwopen in (X, x). But A is not rw-closed in (X, x).
2.3.21 Theorem: Let A and B be subsets of space (X, x). If B is rw-open
and A => int (B), then AnB is rw-open.
Proof: Let B is rw-open in a space X and A 3 int(B). That is int(B) c A.
Then int(B) c AnB. Also int(B) c AnB cB and B is an rw-open set. By
Theorem 2.3.18, AnB is also an rw-open set in X.
2.3.22 Theorem: If a set A is rw-open in a space X, then G=X, whenever
G is regular semiopen and int(A)uAc c G.
Proof: Suppose that A is rw-open in X. Let G be regular semiopen and
int(A) uAc c G. This implies Gcc (int(A) uAc)c= (int(A))c n A. That is
Gcc (int(A))c-Ac, Thus Gcc cl(Ac)-Ac, since (int(A))c=cl(Ac). Now Gc is
also regular semiopen and Ac is rw-closed. By Theorem 2.2.20, it follows
that Gc=€)>. Hence G=X.
The converse of the above theorem need not be true in general as
seen from the following example.
37
-
-
2.3.23 Example: Let X= {a, b, c, d} be with the topology x={X, ((), {a},
{b}, {a, b}, {b, c}, {a, b, c}, {a, b, d}}. Then RWO(X)={ X, (j), {a}, {b},
{c}, {d}, {b, d}, {a, c}, {b, c}, {a, b}, {c, d}, {a, b, c}, {a, b, d}} and
RSO(X)={ X, 0, {a}, {b, c}, {a, d}, {b, c, d}}.
Take
A
=
{b,
c,
d}.
Then
A
is
not
rw-open.
However
int(A)uAc={b, c}u{a}={a, b, c}. So for some regular semiopen G, we
have int(A)uAc= {a, b, c} c G gives G=X, but A is not rw-open.
2.3.24 Theorem: If A c= B cX, where A is rw-open relative to B and B is
an open set relative to X, then A is rw-open relative to X.
Proof: Let F be a regular semiopen set in X and suppose that F c A. By
Lemma 2.2.34, F is regular semiopen relative to B. Since A is rw-open
relative to B, we have Fc intB(A), by Theorem 2.3.14. Hence Fez
(int(A))nB. It follows that Fc int(A). By Theorem 2.3.14, we get A is
rw-open in X.
2.3.25 Theorem: Every singleton point set in a space is either rw-open or
regular semiopen.
Proof: Let X be a topological space. Let xeX. To prove {x} is either rwopen or regular semiopen. That is to prove X-{x} is either rw-closed or
regular semiopen, which follows from Theorem 2.2.24.
2.3.26 Theorem: Let A cYcX and suppose that A is rw-open in X.
Then A is rw-open in Y provided Y is regular open in X.
Proof: Let A be rw-open in X and Y be regular open in X. Let U be any
regular semiopen in Y such that A c U. Then U c A c Y c X. By
Lemma 2.2.37, U is regular semiopen in X. Since A is rw-open in X,
Ucint(A). Also int(A) c intY(A). That is U c intY(A). Hence A is an rwopen set in Y.
38
-
-
Analogous to a neighbourhood in space X, we define rwneighbourhood in a space X as follows:
2.3.27 Definition: Let (X, t) be a topological space and let x e X. A
subset N of X is said to be a rw-neighbourhood of x if and only if there
exists an rw-open set G such that xe GcN.
2.3.28 Definition: A subset N of space X, is called a rw-neighbourhood
of AcX if and only if there exists an rw-open set G such that AcGcN.
We shall use the abbreviated form ‘rw-nhd’ for the word ‘rwneighbourhood’.
2.3.29 Remark: The rw-neighbourhood N of x eX need not be a rwopen in X.
2.3.30 Example: Let X={ 1, 2, 3, 4} be with topology T={X, (|), {1}, {2},
{1, 2}, {1, 2, 3}} Then RWO (X)={X, <>, {1}, {2}, {3}, {4}, {1, 2}, {3,
4}, {1, 2, 3}}. Note that {1, 3} is not an rw-open set, but it is a rw-nhd of
1, since {1} is a rw-open set such that le {1 }c {1, 3}.
2.3.31 Theorem: Every neighbourhood N of xe X is a rw-nieghbourhood
of x.
Proof: Let N be a neighbourhood of point xeX.To prove that N is a rwnhd of x. By definition of neighbourhood, there exists an open set G such
that x e G cN. As every open set is rw-open, G is an rw-open set in X.
Then there exists an rw-open set G such that x e GcN. Hence N is a rwneighbourhood of x.
2.3.32 Remark:
In general, a rw-nhd N of x eX need not be a
neighbourhood of x in X, as seen from the following example.
2.3.33 Example: Let X= {a, b, c, d} be with topology x={X, 0 {a}, {b},
{a, b}, {a, b, c}}. Then RWO(X) ={X,
39
-
-
{a}, {b}, {a, b), {c}, (d}, {c,
d,}, {a, b, c}}. The set {a, c} is a rw-neighbourhood of the point c, since
the rw-open set {c} is such that ce {c}c{a, c}. However, the set {a, c} is
not a neighbourhood of the point c, since no open set G exists such that
ce Gc{a, c}.
2.3.34 Theorem: If a subset N of a space X is rw-open, then N is a rwnhd of each of its points.
Proof: Suppose N is rw-open. Let x e N. We claim that N is a rw-nhd of
x. For N is a rw-open set such that x e N c N. Since x is an arbitrary
point of N, it follows that N is a rw-neighbourhood of each of its points.
2.3.35 Remarks: The converse of the above theorem is not true in
general as seen from the following example.
2.3.36 Example: Let X= {a, b, c, d} be with topology x={X, <j), {a}, {b},
{a, b}, {a, b, c}}. Then RWO(X)={X, <j>, {a}, {b}, {c}, {d}, {a, b}, {c,
d}, {a, b, c}}. The set {a, d} is a rw-nhd of the point a, since the rw-open
set {a}is such that ae {a}c{a, d}. Also the set {a, d} is a rw-nhd of the
point d, since the rw-open set {d} is such that d e {d}c{a, d}. That is {a,
d} is a rw-nhd of each of its points. However the set {a, d} is not an rwopen set in X.
2.3.37 Theorem: Let X be a topological space. If F is a rw-closed subset
of X and xe F0, then there exists a rw-nhd N of x such that NnF=(|>.
Proof: Let F be a rw-closed subset of X and xe F\ Then F° is an rw-open
set of X. So by Theorem 2.3.34, F° contains a rw-nhd of each of its points.
Hence there exists a rw-nhd N of x such that NcF0. That is NnF=<|>.
2.3.38 Definition: Let x be a point in a space (X, x). The set of all rw-nhd
of x is called the rw-neighbourhood system at x, and is denoted by
rw-N(x).
40
-
-
2.3.39 Theorem: Let X be a topological space and for each x e X, let
rw-N(x) be the collection of all rw-nhds of x. Then we have the following
results:
(i)
Vx
(ii)
N
(iii)
Ng rw-N(x), MdN=>M
(iv)
N g rw-N(x), M g rw-N(x) => N n M g rw-N(x).
(v)
N
g
g
g
X, rw-N(x)
rw-N(x) =» x g N.
g
rw-N(x).
rw-N(x) => There exists M
g
rw-N(x) such that M c N and
M g rw-N (y) for every yG M.
Proof: (i) Since X is an rw-open set, it is a rw-nhd of every xgX. Hence
there exists at least one rw-nhd (namely X) for each x
rw-N(x)
g
X. Hence
for every xgX
(ii) If Ng rw-N(x), then N is a rw-nhd of x. So by definition of
rw-nhd, xgN.
(iii) Let Ng rw-N(x) and M z>N. Then there is an rw-open set G such that
xg Gc
N. Since NcM,
xg Gc
M and so M is a rw-nhd of x. Hence M
g
rw- N(x).
(iv) Let N
g
rw-N(x) and M
g
rw-N(x). Then by definition of rw-nhd,
there exist rw-open sets Gi and G2 such that
Hence
xg Gi
xg Gi
c N and
xg G2 c
M.
nG2 c NnM— (1). Since GjnG2 is an rw-open set, (being
the intersection of two rw-open sets) it follows from (1) that NnM is a
rw-nhd of x. Hence NnM
g
rw- N (x).
(v) If Ng rw-N(x), then there exists an rw-open set M such that xg McN.
Since M is an rw-open set, it is a rw-nhd of each of its points. Therefore
MGrw-N(y) V ycM.
-41
-
2.3.40 Theorem: Let X be a nonempty set, and for each
xgX,
let rw-
N(x) be a non-empty collection of subsets of X satisfying following
conditions.
1. N e rw-N(x) => x e N.
2. N € rw-N(x), M e rw-N(x) => N n M e rw-N(x).
Let x consist of the empty set and all those nonempty subsets G of X
having the property that xeG implies that there exists an Ne rw-N(x)
such that xg NcG. Then x is a topology for X.
Proof: (i) $ g x by definition. We now show that Xex. Let x be any
arbitrary element of X. Since rw-N(x) is non-empty, there is an
Ng rw-N(x) and so xgN by (1). Since N is a subset of X, we have
x e N c X. Hence Xg x.
(ii) Let Gj g x and G2g x. If xgGi nG2, then xgGi and xgG2. Since
Gi £ x and G2 g x, there exist Ng rw-N(x) and Mg rw-N(x), such that
xgNcGj
and xgMcG2. Then xg NnMcG]nG2. But NnMGrw-N(x) by
(2). Hence GinG2 g x.
(iii) Let G*g x for every he a . If x
g u{Ga,:Ag a},
then x
g
(3^for some
A/Xg a. Since Q^ex, there exists an Ng rw-N(x) such that xgNc: (3^ and
consequently
xgNc u{G^:
Aga}. Hence u{G*;Xga}gx. It follows that
x is topology for X.
2.4 rw-interior and rw-closure.
In section 4 of this chapter, the notion of rw-interior is defined and
some of its basic properties are studied. Also, we introduce the concept of
rw-closure in topological spaces by using the notations of rw-closed sets
42
-
-
and obtain some of its results. We define Trw and prove that it forms a
topology on X. For any A c X, it is proved that the complement of rwinterior of A is the rw-closure of the complement of A.
2.4.1 Definition: Let A be a subset of (X, x). A point xeA is said to be
rw-interior point of A if and only if A is a rw-neighbourhood of x. The set
of all rw-interior points of A is called the rw-interior of A and is denoted
by rw-int(A).
2.4.2 Theorem: If A is a subset of X, then rw-int(A)= u{G: G is rw-open
GcA}.
Proof: Let A be a subset of (X, x).
xe rw-int(A) <=> x is a rw-interior point of A.
o A is a rw-nhd of point x.
<=> there exists an rw-open set G such that xeGcA
«xe u {G: G is rw-open, GcA}.
Hence rw-int(A) = u {G: G is rw-open, GcA}.
2.4.3 Theorem: Let A and B be subsets of (X, x). Then
(i)
rw-int(X) = X and rw-int((|))
(ii)
rw-int(A) c A
(iii)
If B is any rw-open set contained in A, then B c rw-int(A).
(iv)
If A c B, then rw-int(A) c rw-int(B)
(v)
rw-int(rw-int(A)) = rw-int(A).
Proof: (i) Since X and (]) are rw-open sets, by Theorem 2.4.2,
rw-int(X) = u (G: G is rw-open, G c X} = Xu {all rw-open sets}= X.
That is rw-int(X)=X.
43
-
-
Since <j) is the only rw-open set contained in <(), rw-int(()))=(l).
(ii) Let x e rw-int(A) => x is a rw-interior point of A
=» A is a rw-nhd of x
=> X € A.
Thus x € rw-int(A)=> xe A. Hence rw-int(A) c A.
(iii) Let B be any rw-open set such that B c A. Let xgB. Then since B is
an rw-open set contained in A, x is an rw-interior point of A. That is
xg rw-int(A).
Hence B c rw-int(A).
(iv) Let A and B be subsets of X such that A c B. Let x e rw-int(A).
Then x is an rw-interior point of A and so A is a rw-nhd of x. Since B =)
A, B is also a rw-nhd of x. This implies that x e rw-int(B). Thus we have
shown that x e rw-int(A) => x e rw-int(B). Hence rw-int(A) c rw-int(B).
2.4.4 Theorem: If a subset A of space X is rw-open, then rw-int(A) =A.
Proof: Let A be rw-open subset of X. We know that rw-int(A) c A.
Also, A is rw-open set contained in A. From Theorem 2.4.3. (iii), A <z
rw-int(A). Hence rw-int(A)= A.
The converse of the above theorem need not be true as seen form
the following example.
2.4.5 Example: Let X= {a, b, c} be with topology x = {X, <J>, {a}, {b}, {a,
b}}. Then RWO(X)={X, <|>, {a}, {b}, {c}, {a, b}}. Note that
rw-int({a, c})={a} u{c}u(J) = {a, c}, but {a, c} is not a rw-open set in X.
2.4.6 Theorem: If A and B are subsets of (X, x), then
rw-int(A) u rw-int(B) c rw-int(AuB).
44
-
-
Proof: We know that A c AuB and B c AuB. We have, by Theorem
2.4.3. (iv), rw-int(A) c rw-int(AuB) and rw-int(B) c rw-int(AuB). This
implies that rw-int(A) u rw-int (B) c rw-int(AuB).
2.4.7 Theorem: If A and B are subsets of a space (X, t), then
rw-int(AnB) = rw-int(A) n rw-int(B).
Proof: We know that AnB c A and AnB c B. We have, by Theorem
2.4.3. (iv), rw-int(AnB) c rw-int(A) and rw-int(AnB) c rw-int(B). This
implies that rw-int(AnB) c rw-int(A) n rw-int(B)------(1)
Again, let xe rw-int(A)nrw-int(B). Then xErw-int(A) and XErw-int(B).
Hence x is an rw-interior point of each of sets A and B. It follows that A
and B are rw-nhds of x, so that their intersection AnB is also a rw-nhd of
x. Hence xerw-int(AnB). Thus xe rw-int(A)nrw-int(B) implies that
xe rw-int(AnB).Therefore
rw-int(A)nrw-int(B)c
rw-int(AnB)—(2).
From (1) and (2), we get rw-int(AnB) = rw-int(A) n rw-int(B).
2.4.8 Theorem: If A is a subset of X, then int(A) c rw-int(A).
Proof: Let A be a subset of a space X.
Let xe int(A) => x e u{ G: G is open, GcA}
=> there exists an open set G such that xe GcA.
=» there exists an rw-open set G such that xe GcA, as every
open set is an rw-open set in X.
=> xe u{G: G is rw-open, GcA}
=> xe rw-int(A).
Thus, xe int(A)=> xe rw-int(A). Hence int(A)c rw-int(A)
45
-
-
2.4.9 Remark: Containment relation in the above Theorem 2.4.8 may be
proper as seen from the following example.
2.4.10 Example: Let X= {a, b, c} be with topology x= {X, 0, {a}, {b},
{a, b}}. Then RWO(X)= {X, <|>, {a}, {b}, {c}, {a, b}}. Let A={a, c}.
Now rw-int(A)={a, c} and int(A)={a}. It follows that int(A) c rw-int(A)
and int(A) ^rw-int(A).
2.4.11 Theorem: If A is a subset of X, then w-int(A) c rw-int(A), where
w-int(A) is given by w-int(A) = u{G: G is w-open, GcA}[83].
Proof: Let A be a subset of a space X.
Let xg w-int(A)=> xg u{GcX: G is w-open, GcA}
=> there exists a w-open set G such that xg GcA.
=> there exists a rw-open set G such that xg GcA, as every
w-open set is a rw-open set in X.
=> xg u{GcX: G is a rw-open, GcA}
=>
Thus
xg
rw-int(A).
xg w-int(A)=> xg
rw-int(A).
Hence w-int(A) c rw-int(A).
2.4.12 Remark: Containment relation in the above Theorem 2.4.11 may
be proper as seen from the following example.
2.4.13 Example: Let X={a, b, c} be with topology x = (X, 0, {a}}. Then
RWO(X)={X, 0, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}} and WO(X)={X, <(►,
{a}}. Let A= {a, b}. Then rw-int(A) = {a, b} and w-int(A) = {a}. It
follows that w-int(A) c rw-int(A) and w-int(A) * rw-int(A)
46
-
-
2.4.14 Theorem: If A is a subset of X, then rw-int(A)cgpr-int( A), where
gpr-int(A) is given by gpr-int(A) = u{GcX: G is gpr-open, GcA}[41].
Proof: Let A be a subset of a space X.
Let xg rw-int (A)=> xg u{GcX: G is rw-open, GcA}
=» there exists a rw-open set G such that xg GcA.
=> there exists a gpr-open set G such that xg GcA, as
every rw-open set is a gpr-open set in X.
=> x
=>
g
xg
u{GcX: G is a gpr-open, GcA}
gpr-int(A).
Thus x g rw-int(A) => xg gpr-int(A).
Hence rw-int(A) c gpr-int(A).
Analogous to closure in a space X, we define rw-closure in a space
X as follows
2.4.15 Definition: Let A be a subset of a spce (X, x). We define the rwclosure of A to be the intersection of all rw-closed sets containing A. In
symbols, rw-cl(A) = n {F: A c Fg RWC(X)}
2.4.16 Theorem: If A and B are subsets of a space (X, x), then
(i)
rw-cl(X) = X and rw-cl(<|)) = (j)
(ii)
A c rw-cl(A)
(iii) If B is any rw-closed set containing A, then rw-cl(A) c B
(iv) If Ac B, then rw-cl(A) c rw-cl(B)
(v) rw- cl(A) = rw-cl(rw- cl(A)).
47
-
-
Proof: (i) By the definition of rw-closure, X is the only rw-closed set
containing X. Therefore rw-cl(X)= Intersection of all the rw-closed sets
containing X.
= n{X}=X.
That is rw-cl(X) = X.
By the definition of rw-closure,
rw-cl(<|)) = Intersection of all the rw-closed sets containing (f>
= (f) n any rw-closed set containing (|) = <J>.
That is rw-cl(<|>) = (|).
ii) By the definition of rw-closure of A, it is obvious that A c rw-cl(A).
iii) Let B be any rw-closed set containing A. Since rw-cl(A) is the
intersection of all rw-closed sets containing A, rw-cl(A) is contained in
every rw-closed set containing A. Hence in particular rw-cl(A) c B.
iv) Let A and B be subsets of (X, x) such that A e B. By the definition of
rw-closure, rw-cl(B) = n{F: Be Fe RWC(X)}. IfBcFe RWC(X), then
rw-cl(B) c F. Since AcB,Ac Be Fe RWC(X), we have rw-cl(A)c F.
Therefore rw-cl(A) c n {F: Be F e RWC(X)} = rw-cl(B). That is rwcl(A)c rw-cl(B).
v) Let A be any subset of X. By the definition of rw-closure,
rw-cl(A)=n{F: AcFeRWC(X)}. If AcFeRWC(X), then rw-cl(A)c F.
Since F is a rw-closed set containing rw-cl(A), by (iii) rw-cl(rw-cl(A))c
F. Hence rw-cl(rw-cl(A)) c n {F: Ae Fe RWC(X)}= rw-cl(A). That is
rw-cl{rw-cl(A)}= rw-cl(A).
2.4.17 Theorem: If A e (X, x) is rw-closed, then rw-cl(A) = A.
48
-
-
Proof: Let A be rw-closed subset of (X, x). We know that A c rw-cl(A).
Also AcA and A is rw-closed. By the Theorem 2.4.16 (iii), rw-cl(A)c A.
Hence rw-cl(A) =A.
The converse of the above theorem need not be true as seen from
the following example.
2.4.18 Example: LetX= {a, b, c},andx= {X, <|>, {a}, {b}, {a, b}}.
Here RWC (X) ={X, <>, {c}, {a, b}, {b, c}, {c, a}}
Now rw-cl({a}) = {a, b} n{a, c}n X={a}, but {a} is not a rw-closed
subset in X.
2.4.19 Theorem: If A and B are subsets of a space (X, x), then
rw-cl(AnB) c rw-cl(A) n rw-cl(B).
Proof: Let A and B be subsets of X. Clearly AnBciA and AnBcB. By
Theorem 2.4.16 (iv), rw-cl(AnB)c rw-cl(A) and rw-cl(AnB)ow-cl(B).
Hence rw-cl(AnB) c rw-cl(A) n rw-cl(B).
2.4.20 Remark: In general rw-cl(A) n rw-cl(B) <X rw-cl(AnB), as seen
from the following example.
2.4.21 Example: Consider the either-or topology defined on the interval
X= [-1, 1], The open sets of this topology are <|), [-1,1], (-1,1],[-1,1),
(-1,1), and all the subsets of X that do not contain 0. Let A = (-1,0) and
B=(0,1). Both A and B are not rw-closed sets in X. Now rw-cl(A)= (-1,0]
and rw-cl(B)=[0,l). Then rw-cl(A) n rw-cl(B)={0], but rw-cl(AnB)=
rw-cl{(|)}=(|>. Hence rw-cl(A) n rw-cl(B) <z rw-cl(AnB).
2.4.22 Theorem: If A and B are subsets of a space (X, x), then
rw-cl(AuB) = rw-cl(A) u rw-cl(B).
49
-
-
Proof: Let A and B be subsets of (X, x). Clearly AcAuB and BcAuB.
Hence rw-cl(A)urw-cl(B)c rw-cl(AuB)~(l) Now to prove rw-cl(AuB)
c rw-cl(A) u rw-cl(B). Let x e rw-cl(A uB) and suppose x € rw-cl(A)u
rw-cl(B). Then there exists rw-closed sets A] and Bi with A c Aj B c Bj
and x € AjU Bj. We have AuBcAjuBj and AjUBj is a rw-closed set by
the Theorem 2.2.17, such that xg AjuBj.Thus x £ rw-cl(AuB) which is a
contradiction to x e rw-cl(AuB). Hence rw-cl(AuB)c rw-cl(A)urwcl(B)-(2). From (1) and (2), we have rw-cl(AuB)= rw-cl(A) u rw-cl(B).
2.4.23 Theorem: For an xe X, xerw-cl(A) if and only if VnA^tj) for
every rw-open set V containing x.
Proof: Let x e X and x e rw-cl(A). To prove VnA * <|) for every rw-open
set V containing x. Prove the result by contradiction. Suppose there exists
a rw-open set V containing x such that VnA=<|>. Then Ac X-V and X-V
is rw-closed. We have rw-cl(A)c X-V. This shows that xg rw-cl(A),
which is a contradiction. Hence VnA^(j) for every rw-open set V
containing x.
Conversely, let VnA*<j) for every rw-open set V containing x. To
prove xe rw-cl(A). We prove the result by contradiction. Suppose xg rwcl(A). Then there exists a rw-closed subset F containing A such that xg F.
Then xgX-F and X-F is rw-open. Also (X-F)nA=<|), which is a
contradiction. Hence xe rw-cl (A).
2.4.24 Theorem: If A is a subset of (X, x), then rw-cl(A) c cl(A).
Proof: Let A be a subset of a space (X, x). By the definition of closure,
cl(A)=n{F c X : AcFeC(X)}. If A c Fe C(X), then A c FeRWC(X),
because every closed set is rw-closed. That is rw-cl(A)c F. Therefore rwcl(A) cn{FcX:Ac FeC(X)}=cl(A). Hence rw-cl(A)c cl(A).
50
-
-
2.4.25 Remark: Containment relation in the above Theorem 2.4.24., may
be proper as seen from the following example.
2.4.26 Example: Let X= {a, b, c} be with topology x= {X, (f>, {a},
{a, b}}. Then rw-cl({a})={a) and cl({a})=X .It follows that rw-cl({a})c
cl({a}) and rw-cl({a}) * cl({a}).
2.4.27 Theorem: If A is a subset of a space (X, x), then rw-cl(A)c wcl(A), where
w-cl(A) is given by w-cl(A) = n {FcX: AcF and F is w-
closed set in X}
Proof: Let A be a subset of X. By definition of w-closure
w-cl(A)= n{FcX: AcF and F is w-closed set in X}. If A c F and F is a
w-closed subset of X, then AcFe RWC(X), because every w-closed set is
a rw-closed
subset
in
X.
That is
rw-cl(A)
c
F.
Therefore
rw-cl(A)cn{FcX:AcF and F is w-closed} = w-cl(A). Hence rw-cl(A)c
w-cl(A).
2.4.28 Remark: Containment relation in the above Theorem 2.4.27 may
be proper as seen from following example.
2.4.29 Example: Let X={a, b, c} be with the topology x ={X, <)), {a, b}}.
Let A={a}. Then rw-cl(A) = {a} and w-cl(A) = {a, c}. That is rw-cl(A)c
w-cl(A) and rw-cl(A) ^ w-cl(A).
2.4.30 Theorem: If A is a subset of a space X, then gpr-cl(A) c rw-cl(A)
where gpr-cl(A) is given by gpr-cl(A)=n{FcX: AcFe GPRC(X)}[41],
Proof: Let A be a subset of X. By the definition of rw-closure,
rw-cl(A)=n{ FcX: AcFe RWC(X)}.
If AcFeRWC(X),
then
AcF
eGPRC(X), because every rw-closed set is a gpr-closed set. That is
gpr-cl(A)cF. Therefore gpr-cl(A)cn{FcX: AcFeRWC(X)}= rw-cl<A).
Hence gpr-cl(A) c rw-cl(A).
-51
-
2.4.31 Theorem: rw-closure is a Kuratowski closure operator on a space
X.
Proof: (i) rw-cl((|))=0 (ii) E c rw-cl(E) by Theorem 2.4.16.
(iii) rw-cl(EiuE2) = rw-cl(Ei)urw-cl(E2), by Theorem 2.4.24.
(iv) rw-cl{rw-cl(E)} = rw-cl(E), by Theorem 2.4.16. Hence, rw-closure is
a Kuratowski closure operator on X.
2.4.32 Definition: Let xrw be the topology on X generated by rw-closure
in the usual manner. That is X™ = f UcX: rw-cl (Uc) = 11°}.
2.4.33 Theorem: For any topology xonX,xcxwc x™, where xw={U <=
X: w-cl(Uc)=Uc}[83].
Proof: We know that x c xw from [83]. To prove xw c x™. Let Uexw
which implies w-cl(Uc)=Uc. It follows that Uc is a w-closed set. Now Uc
is rw-closed, as every w-closed set is rw-closed and so rw-cl(Uc)=Uc.
That is Ue xro and so xw c X™. Hence xcxwc X™.
2.4.33 Theorem: Let A be any subset of (X, x). Then
(i)
(rw-int(A))c = rw-cl(Ac)
(ii)
rw-int(A) = (rw-el(Ac))c
(iii)
rw-cl(A) = (rw-int(Ac))c
Proof: Let x e (rw-int(A))c. Then x g rw-int(A). That is every rw-open
set U containing x is such that U <Z A. That is every rw-open set U
containing x is such that U n Ac & 0. By Theorem 2.4.23, x e rw-cl(Ac)
and therefore (rw-int(A))c c rw-cl(Ac).
Conversely, let x e rw-cl(Ac). Then by Theorem 2.4.23, every rwopen set U containing x is such that U n Ac* <|>. That is every rw-open
set U containing x is such that UcA. This implies by Definition of rw-
52
-
-
interior of A, xg rw-int(A). That is xe (rw-int(A))c and rw-cl(Ac) c (rwint(A))c. Thus (rw-int(A))c = rw-cl(Ac).
(ii) Follows by taking complements in (i).
(iii) Follows by replacing A by Ac in (i).
2.5 Minimal rw-open sets and maximal rw-closed sets.
In this section, a new class of sets called minimal rw-open sets and
maximal rw-closed sets in topological spaces are introduced, which are
subclass of rw-open sets and rw-closed sets respectively. During this
process some of their properties are obtained. We prove that the
complement of a minimal rw-open set is a maximal rw-closed set. Also,
we introduce and study maximal rw-open sets and minimal rw-closed sets
in topological spaces.
2.5.1 Definition: A proper nonempty rw-open subset U of X is said to be
a minimal rw-open set if and only if any rw-open set which is contained
in U is <|> or U.
2.5.2 Remark: Minimal open sets and minimal rw-open sets are
independent of each other as seen from the following example.
2.5.3. Example: Let X= {a, b, c} be with the topology x = {X, (|), {a},
{b, c}}. Minimal open sets are {a}, {b, c}. rw-open sets are X, <(>, {a},
{b}, {c}, {a, b}, {b, c}, {a, c}. Minimal rw-open sets are {a}, {b}, {c}.
Here the set {b, c} is a minimal open set but not a minimal rw-open set
and the sets {b} and {c} are minimal rw-open sets but not minimal open
sets.
2.5.4 Remark: From the known the results and by the above Example
2.5.3 we have the following implications:
53
-
-
Minimal rw-open sets
/
*
............. ▼
Minimal open sets ^.......A.........
v
Open sets
Figure- 2.2
2.5.5 Theorem: (i) Let U be a minimal rw-open set and W be a rw-open
set. Then UnW=<|> or Uc W.
(ii) Let U and V be minimal rw-open sets. Then UnV=<|> or U=V.
Proof: (i) Let U be a minimal rw-open set and W be a rw-open set. If
UnW=4, then there is nothing to prove but if UnW*(j) then we have to
prove that UcW. Suppose UnW* $. Then UnWc U and UnW is rwopen, as the finite intersection of rw-open sets is a rw-open set. Since U is
a minimal rw-open set, we have UnW=U. Therefore UcW.
(ii) Let U and V be minimal rw-open sets. Suppose UnV*(|), then we see
that UcV and Vc U by (i). Therefore U=V.
2.5.6 Theorem: Let U be a minimal rw-open set. If x is an element of U,
then UcW for any open neighbourhood W of x.
Proof: Let U be a minimal rw-open set and x be an element of U.
Suppose there exists an open neighbourhood W of x such that UcZ W.
Then UnW is a rw-open set such that UnWcU and UnW*(|). Since U is
a minimal rw-open set, we have UnW=U. That is UcW. This contradicts
our assumption that U<ZW. Therefore UcW for any open neighbourhood
W of x.
2.5.7 Theorem: Let U be a minimal rw-open set. If x is an element of U,
then UcW for any rw-open set W containing x.
54
-
-
Proof: Let U be a minimal rw-open set containing an element x. Suppose
there exists an rw-open set W containing x such that U<xW. Then UnW
is an rw-open set such that UnWcU and UnW*(j). Since U is a minimal
rw-open set, we have UnW=U. That is UcW. This contradicts our
assumption that U<zW. Therefore UczW for any rw-open set W
containing x.
2.5.8 Theorem: Let U be a minimal rw-open set. Then U=n{W: W is
any rw-open set containing x} for any element x of U.
Proof: By Theorem 2.5.7 and from the fact that U is an rw-open set
containing x, we have Ucn{W: Wis any rw-open set containing x}cU.
Therefore we have the result.
2.5.9 Theorem: Let U be a nonempty rw-open set. Then the following
three conditions are equivalent.
(1) U is a minimal rw-open set
(2) U c rw-cl(S) for any nonempty subset S of U
(3) rw-cl(U)=rw-cl(S) for any nonempty subset S of U.
Proof: (1)=> (2) Let U be a minimal rw-open set and S be a nonempty
subset of U. Let xeU. By Theorem 2.5.7, for any rw-open set W
containing x, ScUcW which implies ScW. Now S=SnUc SnW. Since
S is nonempty, therefore SnW* (|). Since W is any rw-open set containing
x, by Theorem 2.4.23, xGrw-cl(S). That is xeU implies xecl(S) which
implies Uc cl(S) for any nonempty subset S of U.
(2) => (3) Let S be a nonempty subset of U. That is S c U which implies
rw-cl(S)c rw-cl(U)------(i). Again from (2) Uc rw-cl(S) for any
nonempty subset S of U which implies rw-cl(U) c rw-cl(rw-cl(S))= rw-
55
-
-
cl(S). That is rw-cl(U)c rw-cl(S)------ (ii). From (i) and (ii), we have
rw-cl(U)=rw-cl(S) for any nonempty subset S of U.
(3)=> (1) From (3) we have rw-cl (U) = rw-cl(S) for any nonempty subset
S of U. Suppose U is not a minimal rw-open set. Then there exists a
nonempty rw-open set V such that VcU and WU. Now there exists an
element ae U such that ag V which implies ae Vc. That is rw-cl({a})c rwcl(Vc)=Vc, as Vc is a rw-closed set in X. It follows that rw-cl({a})£
rw-cl(U). This is a contradiction to fact that rw-cl({a})=rw-cl(U) for any
nonempty subset {a} of U. Therefore U is a minimal rw-open set.
2.5.10 Theorem: Let V be a nonempty finite rw-open set. Then there
exists at least one (finite) minimal rw-open set U such that UcV.
Proof: Let V be a nonempty finite rw-open set. If V is a minimal rw-open
set, we may set U=V. If V is not a minimal rw-open set, then there exists
a (finite) rw-open set V) such that <() * V)cV. If Vi is a minimal rw-open
set, we may set U=VL If V] is not a minimal rw-open set, then there
exists a (finite) rw-open set V2 such that (|#V2 cV). Continuing this
process, we have a sequence of rw-open sets V=>VidV2z>V3 —•
zsVkz>..... . Since V is a finite set, this process repeats only finitely. Then
finally we get a minimal rw-open set U=Vn for some positive integer n.
[A topological space X is said to be locally finite space if each of its
elements is contained in a finite open set.]
2.5.11 Corollary: Let X be a locally finite space and V be a nonempty
rw- open set. Then there exists at least one (finite) minimal rw-open set U
such that UcV.
Proof: Let X be a locally finite space and V be a nonempty rw-open set.
Let
xg V.
Since X is a locally finite space, we have a finite open set Vx
56
-
-
such that xeVx. Then VnVx is a finite rw-open set. By Theorem 2.5.10
there exists at least one (finite) minimal rw-open set U such that
UcVnVx. That is UcVnVxcV. Hence there exists at least one (finite)
minimal
rw-open set U such that UcV.
2.5.12 Corollary: Let V be a finite minimal open set. Then there exists at
least one (finite) minimal rw-open set U such that UcV.
Proof: Let V be a finite minimal open set. Then V is a nonempty finite
rw-open set. By Theorem 2.5.10, there exists at least one (finite) minimal
rw-open set U such that UcV.
2.5.13 Theorem: Let U and Ux be minimal rw-open sets for any element
A, of A. If Uc u Ux, then there exists an element A, of A such that
Ae A
U= UX.
Proof: Let UcUUx. Then Un(LJUx)=U. That is LJ (UnUx) =U.
AeA
AeA
AgA
Also by Theorem 2.5.5 (ii), UnUx= (j> or U=UX for any AeA. It follows
that there exists an element Ae A such that U=UX.
2.5.14 Theorem: Let U and Ux be minimal rw-open sets for any element
A of A. If U=Uxfor any element Ae A, then (Ux) nU=<j).
As A
Proof: Suppose that ( LJ Ux)nU^((). That is LJ (UxnU)^(|). Then there
AeA
yleA
exits an element AeA such that UnU^. By Theorem 2.5.5. (ii), we
have U=UX, which contradicts the fact that U*UX for any AeA. Hence
(U Ux)nU=(|).
57
-
-
2.5.15 Theorem: Let Uybe a minimal rw-open set for any element A of A
and Uj^Un for any elements X and p of A with Xt\i. Assume that | A | >2
Let p be any element of A. Then (
kJ
Ux) nU„ =0.
AeA-{M]
Proof: Put U—Uji in Theorem 2.5.14, then we have the result.
2.5.16 Corollary: Let Ux be a minimal rw-open set for any element X of
for any elements X and p of A with X ^p. If T is a proper
A and
nonempty subset of A, then ( LJlL )n(
MA-r
UY)=(b.
yer
2.5.17 Theorem: Let Ux and Ur be minimal rw-open sets for any element
Ae X and ye V. If there exists an element y of T such that Ux*UY for any
element X of A, then LJ UY <t LJ Uxye r
AeA
Proof: Suppose that an element y1 of T satisfies Ux^UY) for any element X
of A. If U UY c U Ux, then we see Uyic U Ux- By Theorem 2.5.13,
yeT
XeA
XeA
there exists an element A, of A such that UYi= Ux, which is a contradiction.
It follows that LJuy(Z
yeT
Ux.
XeA
2.5.18 Theorem: Let Ux be a minimal rw-open set for any element A of A
and Ux =£ U^ for any elements A and p of A with A ^ p. If T is a proper
nonempty subset of A, then
yef
UY c
XeA
Ux-
Proof: Let k be any element of A-T. Then
Uk n (KJ UY) = U (Uk nUYH> and Uk n( LJ Ux) = LJ (UknUx)= Uk
yeT
A&A
yef
58
-
-
X&A
If LJ IL = CJ Ua., then we have (j)=Uk. This contradicts our assumption
ye r
AeA
that Uk is a minimal rw-open set. Therefore we have the result.
We now introduce maximal rw-closed sets in topological spaces as
follows
2.5.19 Definition: A proper nonempty rw-closed subset F of a
topological space X is said to be maximal rw-closed set if and only if any
rw-closed set which contains F is either X or F.
2.5.20 Remark: Maximal closed sets and maximal rw-closed sets are
independent each other as seen from the following example.
2.5.21 Example: Let X= {a, b, c} be with the topology T={X, (j), {a},
{b, c}}. Closed sets are X, (j), {a }, {b, c}. Maximal closed sets are {a},
{b, c}. rw-closed sets are X, 0, {a}, {b}, {c}, {a, b}> {a, c}, {b, e}.
.Maximal rw-closed sets are {a, b}, {a, c}, {b, c}. Here the set {a} is
maximal closed set but not maximal rw-closed set and the sets {a, b},
{a, c} are maximal rw-closed sets but not maximal closed sets.
2.5.22 Remark: From the known results and by the above Example
2.5.21 we have the following implications:
Figure 2.3
2.5.23 Theorem: A proper nonempty subset F of X is maximal rw-closed
set if and only if X-F is a minimal rw-open set.
59
-
-
Proof: Let F be a maximal rw-closed set. Suppose X-F is not a minimal
rw-open set. Then there exists an rw-open set U ^ X-F such that <|> *Uc
X-F. That is Fc X-U and X-U is a rw-closed set. This contradicts our
assumption that F is a minimal rw-open set.
Conversely let X-F be a minimal rw-open set. Suppose F is not a
maximal rw-closed set. Then there exists a rw-closed set E^F such that
F c E * X. That is 0 * X-E c X-F and X-E is a rw-open set. This
contradicts our assumption that X-F is a minimal rw-open set. Therefore
F is a maximal rw-closed set.
2.5.24 Theorem: (i) Let F be a maximal rw-closed set and W be a rwclosed set. Then FuW=X or Wc F.
(ii) Let F and S be maximal rw-closed sets. Then FuS=X or F=S.
Proof: (i) Let F be a maximal rw-closed set and W be a rw-closed set. If
FuW=X, then there is nothing to prove. But if FuW^X then we have to
prove that Wc F. Suppose FuW^X. Then Fc FuW and FuW is rwclosed, as the finite union of rw-closed sets is a rw-closed set, we have
FuW=X or FuW=F. Therefore FuW=F which implies W c F.
(ii) Let F and S be maximal rw-closed sets. Suppose FuS^X, then we see
that FcS and ScF by (i). Therefore F=S.
2.5.25 Theorem: Let F be a maximal rw-closed set. If x is an element of
F, then for any rw-closed set S containing x, FuS=X or S c F.
Proof: Let F be a maximal rw-closed set and x is an element of F.
Suppose there exists a rw-closed set S containing x such that FuS^X.
Then Fc FuS and FuS is a rw-closed set, as the finite union of rw-closed
sets is a rw-closed set. Since F is a rw-closed set, we have FuS=F.
Therefore ScF.
60
-
-
2.5.26 Theorem: Let Fa, Fp, Fy be maximal rw-closed sets such that
Fa Fp. If FanFp c Fy, then either Fa = Fy or Fp= Fy.
Proof: Given that Fa n Fpc Fy. If Fa= Fy then there is nothing to prove.
But if Fa* Fy then we have to prove Fp= Fy.
Now Fp n Fy= Fp n(Fy nX)
= Fpn(Fyn (Fau Fp) (by Theorem 2.5.24) (ii)
= Fp n ((Fy n Fa) u (Fy n Fp))
= (Fpn Fyn Fa) u (Fpn Fyn Fp)
= (Fa n Fp) u (Fy n Fp) (by Fa n Fp c Fy)
= (FauFy)nFp
=XnFp
(Since
Fa
and
Fy
are
maximal
rw-closed
sets
by
Theorem 2.5.24 (ii), Fa u Fy=X).
= Fp
That is Fp n Fy= Fp which implies Fpc Fy. Since Fp and Fy are maximal
rw-closed sets, we have Fp = Fy. Therefore Fp = Fy.
2.5.27 Theorem: Let Fa, Fp, and Fy be maximal rw-closed sets which are
different from each other. Then (Fan Fp) <£ (Fan Fy).
Proof: Let (Fan Fp) <z (Fan Fy) which implies
(Fa n Fp) u (FynFp) c (Fan Fy) u (FynFp) which implies
(Fa u Fy) nFpcFyn (FauFp). Since by Theorem 2.5.24 (ii), FauFy =X
and FauFp=X which implies X n Fp c Fy nX which implies Fp c= Fy.
From the definition of maximal rw-closed set it follows that Fp = Fy. This
is a contradiction to the fact that Fa, Fp, and Fy are different from each
other. Therefore (Fa nFp)<2 (FanFy).
-61
-
2.5.28 Theorem: Let F be a maximal rw-closed set and x be an element
of F. Then F=u{S: S is a rw-closed set containing x such that FuS^X}.
Proof: By Theorem 2.5.25 and from fact that F is a rw-closed set
containing x, we have Fc u{S: S is a rw-closed set containing x such that
FuS *X}c F. Therefore we have the result.
2.5.29 Theorem: Let F be a proper nonempty cofinite rw-closed subset.
Then there exists (cofinite) maximal rw-closed set E such that FcE.
Proof: If F is a maximal rw-closed set, we may set E=F. If F is not a
maximal rw-closed set, then there exists (cofinite) rw-closed set Fj such
that Fc Fi AX. If Fj is a maximal rw-closed set, we may set E=Fi. If Fi is
not a maximal rw-closed set, then there exists a (cofinite) rw-closed set F2
such that Fc FjC F2AX. Continuing this process, we have a sequence of
rw-closed sets, Fc FjC F2c F3c Fc... c Fk c... . Since F is a cofinite
set, this process repeats only finitely. Then, finally we get a maximal rwclosed set E=E„ for some positive integer n.
2.5.30 Theorem: Let F be a maximal rw-closed set. If x is an element of
X-F, then X-Fc E for any rw-closed set E containing x.
Proof: Let F be a maximal rw-closed set and xe X-F. EcF for any rwclosed set E containing x. Then EuF=X by Theorem 2.5.24 (ii).
Therefore X-Fc E.
We now introduce minimal rw-closed sets and maximal rw-open
sets in topological spaces as follows
2.5.31 Definition: A proper nonempty rw-closed subset F of X is said to
be a minimal rw-closed set if and only if any rw-closed set which is
contained in F is § or F.
62
-
-
2.5.32 Remark: Minimal closed sets and minimal rw-closed sets are
independent of each other as seen from the following example.
2.5.33 Example: Let X= {a, b, c, d} be with the topology x = {X, (j),
{a, b}, {c, d}}. Minimal closed sets are {a, b}, {c, d} and
RWC(X)=P(X). Minimal rw-closed sets are {a}, {b}, {c}, {d}. Here the
sets {a, b} and {c, d} are
minimal closed sets but not minimal rw-
closed sets and the sets {a}, {b}, {c} and {d} are minimal rw-closed sets
but not minimal closed sets.
2.5.34 Definition: A proper nonempty rw-open subset U of a topological
space X is said to be a maximal rw-open set if and only if any rw-open set
which contains U is either X or U.
2.5.35 Remark: Maximal open sets and maximal rw-open sets are
independent of each other as seen from the following example.
2.5.36 Example: Let X= {a, b, c, d} be with the topology z = {X, (f>,
{a, b}, {c, d}}. Minimal open sets are {a, b}, {c, d} and RWO(X)=P(X).
Maximal rw-open sets are {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}. Here the
sets {a, b} and {c, d} are maximal open sets but not maximal rw-open
sets and the sets {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d} are maximal rwopen sets but not maximal open sets.
2.5.37 Theorem: A proper nonempty subset U of X is a maximal rwopen set if and only if X-U is a minimal rw-closed set.
Proof: Let U be a maximal rw-open set. Suppose X-U is not a minimal
rw-closed set. Then there exists a rw-closed set V * X-U such that
(|)*Vc X-U. That is U c X-V and X-V is a rw-open set. This contradicts
our assumption that U is a minimal rw-closed set.
63
-
-
Conversely let X-U be a minimal rw-closed set. Suppose U is not a
maximal rw-open set. Then there exists an rw-open set B*U such that
UcE^X. That is (J^X-EczX-U and X-E is a rw-closed set. This
contradicts our assumption that X-U is a minimal rw-closed set.
Therefore U is a maximal rw-closed set.
64
-
-