Download ON REGULAR PRE-SEMIOPEN SETS IN TOPOLOGICAL SPACES

International Journal of General Topology
Vol. 4, Nos. 1-2 January-December 2011, pp. 17– 26; ISSN: 0973-6751
ON REGULAR PRE-SEMIOPEN SETS
IN TOPOLOGICAL SPACES
T. Shyla Isac Mary*1 and P. Thangavelu2
ABSTRACT
The generalized open sets in point set topology have been found considerable interest among general
topologists. Veerakumar introduced and investigated pre-semi-open sets and Anitha introduced pgpropen sets in topological spaces. In this paper the concept of regular pre-semiopen sets is introduced and
its relationships with other generalized sets are investigated.
Keywords: pre-semiopen, pgpr-open.
1. INTRODUCTION
Levine[9] introduced generalized open (briefly g-open) sets in topology. Researchers in
topology studied several versions of generalized open sets. In this paper the concept of
regular pre-semiopen (briefly rps-open) set is introduced and their properties are
investigated. This class of sets is properly placed between the class of semi-preopen sets[1]
and the class of pre-semiopen sets[20]. The preliminary concepts are given in the section
2 and the concept of rps-open sets is studied in section 3.
2. PRELIMINARIES
Throughout this paper X and Y represent the topological spaces on which no separation
axioms are assumed unless otherwise mentioned. For a subset A of a topological space X,
clA and intA denote the closure of A and the interior of A respectively. X\A denotes the
complement of A in X. We recall the following definitions.
Definition 2.1: A subset A of a space X is called
(i) pre-open [12] if A ⊆ int clA and pre-closed if cl intA ⊆ A;
(ii) semi-open [8] if A ⊆ cl intA and semi-closed if int clA ⊆ A;
(iii) semi-pre-open [1] if A ⊆ cl int clA and semi-pre-closed if int cl intA ⊆ A;
(iv) α-open [14] if A ⊆ int cl intA and α-closed if cl int clA ⊆ A;
(v) regular open [18] if A = int clA and regular closed if A = cl intA.
(vi) Π -open [21] if A is a finite union of regular open sets.
* Corresponding Author: [email protected]
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The semi-pre-closure (resp. semi-closure, resp. pre-closure, resp. α-closure) of a subset
A of X is the intersection of all semi-pre-closed (resp. semi-closed, resp. pre-closed, resp.
α-closed) sets containing A and is denoted by spclA (resp. sclA, resp. pclA, resp. clA).
Definition 2.2: A subset A of a space X is called
(i) generalized closed[9] (briefly g-closed) if clA ⊆ U whenever A ⊆ U and U is open.
(ii) regular generalized closed[15] (briefly rg-closed) if clA ⊆ U whenever A ⊆ U and U is
regular open.
(iii) α-generalized closed[10] (briefly αg-closed ) if aclA ⊆ U whenever A ⊆ U and U is
open.
(iv) generalized-semi pre-regular-closed [16] (briefly gspr-closed) if spclA ⊆ U whenever
A ⊆ U and U is regular-open.
(v) generalized semi-closed [3] (briefly gs-closed) if sclA ⊆ U whenever A ⊆ U and U is
open.
(vi) Π-generalized closed [5](briefly Π g-closed) if clA ⊆ U whenever A ⊆ U and U is
Π -open.
(vii) generalized pre-closed [11] (briefly gp-closed) if pclA ⊆ U whenever A ⊆ U and U
is open.
(viii) generalized semi-pre-closed [4] (briefly gsp-closed) if spclA ⊆ U whenever A ⊆ U and
U is open.
(ix) Π -generalized pre-closed [7] (briefly Π gp-closed) if pclA ⊆ U whenever A ⊆ U and U
is Π -open.
(x) generalized pre-regular closed[6] (briefly gpr-closed) if pclA ⊆ U whenever A ⊆ U
and U is regular open.
(xi) weakly generalized closed [13] (briefly wg-closed) if cl intA ⊆ U whenever A ⊆ U
and U is open.
(xii) Π -generalized semi-pre-closed[16] (briefly Π gsp-closed) if spclA ⊆ U whenever A ⊆
U and U is Π -open.
(xiii) regular weakly generalized closed[19] (briefly rwg-closed) if cl intA ⊆ U whenever A
⊆ U and U is regular open.
(xiv) pre-semiclosed [20] if spclA ⊆ U whenever A ⊆ U and U is g-open.
(xv) pre-generalized pre-regular-closed [2] (briefly pgpr-closed) if pclA ⊆ U whenever A ⊆
U and U is rg-open.
A subset B of a space X is generalized open (briefly g-open) if X\B is g-closed. The
concepts of rg-open, αg-open, gspr-open, gs-open, πg-open, gp-open, gsp-open, πgp-open,
gpr-open, wg-open, πgsp-open, rwg-open, pre-semi-open, pgpr-open sets can be analogously
defined.
The authors introduced and studied regular pre-semiclosed sets[17].
On Regular Pre-semiopen Sets in Topological Spaces
19
Definition 2.3: A subset A of a space X is called regular pre-semi closed[17] (briefly
rps-closed) if spclA ⊆ U whenever A ⊆ U and U is rg-open.
We use the following notations.
RPSC(X, τ) - The collection of all rps-closed sets in (X, τ)
RPSO(X, τ) - The collection of all rps-open sets in (X, τ)
SPO(X, τ) - The collection of all semi-pre-open sets in (X, τ)
3. rps-OPEN SETS
In this section, we introduce and study rps-open sets in topological spaces and obtain
some of their basic properties. Also, we introduce rps-neighbourhood (shortly rps-nbhd) in
topological spaces by using the notion of rps-open sets.
Definition 3.1: A subset A of X is called regular pre-semiopen (briefly rps-open) if its
complement is rps-closed.
Proposition 3.2:
(i) Every semi-pre-open set is rps-open.
(ii) Every pgpr-open set is rps-open.
(iii) Every pre-open set is rps-open.
Proof:
(i) Let A be a semi-pre-open set in a space X. Then X \ A is semi-pre-closed . Since every
semi-pre-closed set is rps-closed, X \ A is rps-closed. Therefore A is rps-open set in
X.
(ii) Let A be a pgpr-open set in X. Then X\A is pgpr-closed. Since every pgpr-closed set is
rps-closed, X\A is rps-closed. Therefore A is rps-open Set in X.
(iii) Let A be a pre-open set in X. Then X\A is pre-closed. Since every pre-closed set is rpsclosed, X\A is rps-closed. Therefore A is rps-open.
The reverse implications are not true as shown in Example 3.4
Proposition 3.3:
(i) Every rps-open set is pre-semi-open.
(ii) Every rps-open set is gspr-open.
(iii) Every rps-open set is gsp-open.
Proof:
(i) Let A be a rps-open set in X. Then X\A is rps-closed. Since every rps-closed set is
pre-semi-closed, X\A is pre-semi-closed. Therefore A is pre-semi-open set in X.
(ii) Let A be a rps-open set in X. Then X\A is rps-closed. Since every rps-closed set is gsprclosed , X\A is gspr-closed. Therefore A is gspr-open set in X.
(iii) Let A be a rps-open set in X. Then X\A is rps-closed. Since every rps-closed set is gspclosed, X\A is gsp-closed. Therefore A is gsp-open.
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The reverse implications are not true as shown in Example 3.4
Example 3.4: Let X = {a, b, c, d} with topology τ = {φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}. Then
{c} is rps-open but it is neither semi-pre-open nor pre-open. Also {a, c} is both pre-semiopen and gsp-open but it is not rps-open. {b, d} is rps-open but not pgpr-open set. {a, c, d} is
gspr-open but not rps-open.
The concepts of rwg-open, wg-open, gpr-open, πg-open, πgp-open, gp-open, rg-open,
αg-open sets are independent with the concept of rps-open as shown in the following
example.
Example 3.5: Let X = {a, b, c, d} with topology τ = {φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}. Then
the set {b, c, d} is rps-open. However it can be verified that it is not rwg-open, not wg-open,
not gpr-open, not πg-open, not πgp-open, not gp-open, not rg-open and not αg-open. Also
{c, d} is both rwg-open and rg-open but not rps-open. {a, c} is wg-open, πg-open, πgp-open,
gp-open and αg-open but not rps-open. {a, c, d} is gpr-open but not rps-open.
The concepts of g-open and rps-open sets are independent as shown in the following
example.
Example 3.6: Let X = {a, b, c, d} with topology τ = {φ, {a}, {a, b}, X}. Then {a, c, d} is rpsopen but not g-open and {b, d} is g-open but not rps-open.
The concepts of gs-open and rps-open sets are independent as shown in the following example
Example 3.7: Let X = {a, b, c} with topology τ = {φ, {a, b}, X}. Then {b, c} is rps-open but
not gs-open.
From Example 3.5, we see that {a, c} is gs-open but not rps-open.
Thus the above discussions lead to the following implication diagram. In this diagram,
by “A → B” we mean A implies B but not conversely and “A ↔
/ B” means A and B are
independent of each other.
Diagram
On Regular Pre-semiopen Sets in Topological Spaces
21
Union and intersection of two rps-open sets need not be rps-open set as shown in the
following example.
Example 3.8: Let X = {a, b, c, d} with topology τ = {φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}.
Let A = {a, B = {c}, C = {a, d} and D = {b, d}. Here A and B are rps-open but A ∪ B = {a, c}
is not rps-open. Also C and D are rps-open but C ∩ D = {d} is not rps-open.
The intersection of two rps-open sets is rps-open if atleast one of them is semi-open.
Theorem 3.9: If A and B are rps-open sets and let A be semi-open. Then A ∩ B is also
rps-open set in X.
Proof: If A and B are rps-open sets in a space X, then X\A and X\B are rps-closed sets
in a space X. Also given that A is semi-open. X\A is semi-closed and by Theorem 3.12 of
[17]. (X\A) ∪ (X\B) is rps-closed set in X. That is (X\A) ∪ (X\B) = X\(A ∩ B) is rps-closed
set in X. Therefore A ∩ B is rps-open set in X.
Theorem 3.10: Every singleton point set in a space X is either rps-open or rg-open.
Proof: Let X is a topological space. Let x ∈ X. To prove {x} is either rps-open or rg-open.
That is to prove X\{x} is either rps-closed or rg-open, which follows from Theorem 3.16
of [17].
Theorem 3.11: A set A ⊆ X is rps-open if and only if F ⊆ spint A whenever F ⊆ A, F is
rg-closed.
Proof: Let A ⊆ X be rps-open. Let F be rg-closed and F ⊆ A. Then X\A ⊆ X\F where
X\F is rg-open. Since X\A is rps-closed, spcl (X\A) ⊆ X\F and hence X\spint A ⊆ X\F that
implies F ⊆ spint A.
Conversely, assume that F ⊆ spint A whenever F ⊆ A, F is rg-closed. Suppose X\A ⊆ U
where U is rg-open. Then X\U ⊆ A where X\U is rg-closed. By assumption, X\U ⊆ spint
A that implies spcl (X\A) ⊆ U. This proves that X\A is rps-closed and hence A is rps-open.
The next theorem shows that all the sets between spint A and A are rps-open whenever
A is rps-open.
Theorem 3.12: If spint A ⊆ B ⊆ A and A is rps-open, then B is rps-open.
Proof: Let A be rps-open and spint A ⊆ B ⊆ A. Then X\A ⊆ X\B ⊆ X\spint A that
implies X\A ⊆ X\B spcl (X\A). Since X\A is rps-closed, by Theorem 3.15 (i) of [17] X\B is
rps-closed. This proves that B is rps-open.
Theorem 3.13: If A ⊆ X is rps-closed, then spcl A\A is rps-open.
Proof: Let A ⊆ X is rps-closed and let F be a rg-closed set such that F ⊆ spcl A\A. Then
by Theorem 3.13 of [17], F = ∅ that implies F ⊆ spint (spcl A\A). This proves that spcl A\
A is rps-open.
The converse of the above theorem does not hold as shown in the following example.
Example 3.14: Let X = {a, b, c, d} with topology τ = {φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}.
Let A = {b, d}. Then spcl A = {b, c, d} and spcl A\A = {c} which is rps-open. But A is not
rps-closed.
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Theorem 3.15: If a set A is rps-open in X and if G is rg-open in X with spint A ∪ (X\A)
⊆ G then G = X.
Proof: Suppose that G is rg-open and spint A ∪ (X\A) ⊆ G. Now (X\G) ⊆ spcl (X\A) ∩
A = spcl (X\A) \ (X\A). Suppose A is rps-open. Since X\G is rg-closed and since X\A is
rps-closed, by Theorem 3.13 of [17], X\G = ∅ and hence G = X.
Theorem 3.16: Let X be a topological space and A, B ⊆ X. If B is rps-open and spint B ⊆
A, then A ∩ B is rps-open.
Proof: Since B is rps-open and spint B ⊆ A, spint B ⊆ A ∩ B ⊆ B. By Theorem 3.12, A ∩
B is rps-open.
Definition 3.17: Let X be a topological space and let x ∈ X. A subset N of X is said to be
a rps-neighbourhood (briefly rps-nbhd) of x if there exists a rps-open set G such that x ∈ G
⊆ N.
Definition 3.18: A subset N of a space X is called a rps-nbhd of A ⊆ X if there exists a
rps-open set G such that A ⊆ G ⊆ N.
Remark 3.19: The rps-nbhd N of x ∈ X need not be rps-open in X.
Example 3.20: Let X = {a, b, c, d} with topology τ = {φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}.
Then RPSO(X, τ) = {φ, {b, c, d}, {a, b, d}, {a, b, c}, {a, d}, {a, b}, {b, c}, {b, d}, {b}, {c}, {a}, X}.
Note that {a, c} is not a rps-open set, but it is a rps-nbhd of {a}, since {a} is rps-open set such
that a ∈ {a} ⊆ {a, c}.
The next two theorems follows easily from the definition.
Theorem 3.21: Every open nbhd N of x ∈ X is a rps-nbhd of x.
Theorem 3.22: If a subset N of a space X is rps-open, then N is a rps-nbhd of each of its
points.
Theorem 3.23: Let X be a topological space. If F is a rps-closed subset of X and x ∈ X\F,
then there exist a rps-nbhd N of x such that N ∪ F = ∅ .
Proof: Let F be a rps-closed subset of X and x ∈ X\F. Then X\F is rps-open set of X.
Then using Theorem 3.22, X\F contains a rps-nbhd of each of its points. Hence there exists
a rps-nbhd N of x such that N ⊆ X\F. That is N ∩ F = ∅ .
4. rps-CLOSURE
The intersection of rps-closed sets is not rps-closed. However we define the rps-closure of
A as the intersection of all rps-closed sets containing A and study its basic properties.
Definition 4.1: For a subset A of a space X, rps-cl A = ∩ {F: A ⊆ F and F is rps-closed in
X} is called the rps-closure of A.
Remark 4.2:
(i) rps-closure of a set A is not always rps-closed.
(ii) If A is rps-closed then rps-cl A = A
On Regular Pre-semiopen Sets in Topological Spaces
23
However if rps-cl A = A then it is not true that A is rps-closed as seen in the following
example.
Example 4.3: Let X = {a, b, c, d} and τ = {φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}.
Let A = {b, d}. rps-cl A = {b, d} but A is not rps-closed.
The following lemma follows immediately from Definition 4.1.
Lemma 4.4: Let A and B be subsets of (X, τ). Then
(i) rps-cl φ = φ and rps-cl X = X.
(ii) If A ⊆ B, rps-cl A ⊆ rps-cl B.
(iii) A ⊆ rps-cl A.
Lemma 4.5: Let x ∈ X. Then x rps-cl A if and only if V ∩ A ≠ ∅ for every rps-open set V
containing x.
Proof: Let x ∈ rps-cl A. Suppose there exists a rps-open set V containing x such that V ∩
A = ∅ . Since A ⊆ X\V and by Lemma 4.4 (ii), rps-cl A ⊆ X\V. This implies x ∈ rps-cl A
which is a contradiction. Conversely, we assume that V ∩ A ≠ ∅ for every rps-open set V
containing x. Suppose x ∉ rps-cl A.
Then by Definition 4.1, there exists a rps-closed subset F containing A such that x ∉ F.
Therefore x ∈ X\F and X\F is rps-open. Since A ⊆ F, (X\F) ∩ A = ∅ which is impossible
as x ∈ X\F and x ∈ A. This proves the lemma.
Lemma 4.6: Let A and B be subsets of (X, τ). Then
(i) rps-cl A = rps-cl(rps-cl A)
(ii) (rps-cl A) ∪ (rps-cl B) ⊆ rps-cl (A ∪ B)
(iii) rps-cl (A ∩ B) ⊆ rps-cl A ∩ rps-cl B
Proof: Since A rps-cl A, by Lemma 4.4 (ii), rps-cl A ⊆ rps-cl (rps-cl A). Suppose x ∈ rps-cl
(rps-cl A). Let V be a rps-open set containing x. Then using Lemma 4.5, rps-cl A ∩ V ≠ ∅ .
Let y ∈ rps-cl A ∩ V. Then y ∈ rps-cl A and y ∈ V. Again by Lemma 4.5, A ∩ V ≠ ∅ so that
x ∈ rps-cl A. Therefore rps-cl (rps-cl A) ⊆ rps-cl A. This implies rps-cl A = rps-cl (rps-cl A)
This proves (i).
Now using Lemma 4.4 (ii), rps-cl A ⊆ rps-cl (A ∪ B) and rps-cl B ⊆ rps-cl (A ∪ B). This
implies that rps-cl A ∪ rps-cl B ⊆ rps-cl (A ∪ B). This proves (ii).
Again using Lemma 4.4 (ii) rps-cl (A ∩ B) ⊆ rps-cl A and rps-cl (A ∩ B) ⊆ rps-cl B. Thus
rps-cl (A ∩ B) ⊆ rps-cl A ∩ rps-cl B. This proves (iii).
The following example shows that the equality need not hold in Lemma 4.6 (ii)
and (iii).
Example 4.7: Let X = {a, b, c, d} and τ = {φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}. Then RPSC(X,
τ) = {φ, {a}, {c}, {d}, {b, c}, {c, d}, {a, d}, {a, c}, {a, c, d}, {a, b, d}, {b, c, d}, X}
Let A = {a}, B = {b}, C = {a, b} and D = {b, c}. Then rps-cl A = {a} and rps-cl B = {b, c}.
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International Journal of General Topology
Therefore rps-cl A ∪ rps-cl B = {a, b, c} and rps-cl (A ∪ B) = {a, b, d}.
rps-cl C = {a, b, d} and rps-cl D = {b, c}.rps-cl C ∩ rps-cl D = {b}
rps-cl (C ∩ D) = {b, c}.
Proposition 4.8: Let RPSC(X, τ) be closed under finite union. Then rps-cl (A ∪ B) = (rpscl A) ∪ (rps-cl B) for every A, B ∈ RPSC(X, τ)
Proof: Let A and B be rps-closed sets in (X, τ). Since RPSC(X, τ) is closed under finite
unions, A ∪ B is rps-closed. Using Remark 4.2, rps-cl (A ∪ B) = A ∪ B = (rps-cl A) ∪ (rps-clB).
Theorem 4.9: Let SPC(X, τ) be closed under finite union. Then RPSC(X, τ) is closed
under finite union.
Proof: Let A, B RPSC(X, τ) and let A ∪ B ⊆ U where U is rg-open in X. Then A ⊆ U and
B ⊆ U. By Definition 2.3, spcl A ⊆ U and spcl B ⊆ U implies spcl A ∪ spcl B ⊆ U. Since SPC
(X, τ) is closed under finite union, spcl (spcl A ∪ spcl B) = spcl A ∪ spcl B.
Using Theorem 3.11 of [17], spcl (A ∪ B) = spcl A ∪ spcl B. Therefore spcl (A ∪ B) ⊆ U.
Again using Definition 2.3 A ∪ B ∈ RPSC(X, τ).
The next corollary follows immediately from Theorem 4.9.
Corollary 4.10: Let SPO(X, τ) be closed under finite intersection. Then RPSC(X, τ) is
closed under finite intersection.
Definition 4.11: Let (X, τ) be a topological space τrps = {V ⊆ X: rps-cl(X\V) = X\V}
Theorem 4.12: For a space (X, τ), Every rps-closed set is semi-preclosed if and only if
τrps = SPO(X, τ) holds.
Proof:
Necessity: Let A ∈τrps. Then rps-cl(X\A) = X\A.
By hypothesis, spcl(X\A) = rps-cl(X\A) = X\A. This implies A ∈ SPO(X, τ).
Sufficiency: Suppose τrps = SPO(X, τ). Let A be a rps-closed set. Then rps-cl(A) = A.
This implies X\A ∈τrps = SPO(X, τ). So A is semi-preclosed.
Theorem 4.13: Let RPSO(X, τ) be a topology. Then τrps is a topology.
Proof: Clearly ∅ , X ∈rps. Let {Aα: α ∈Ω} ⊆ rps.
Then rps-cl(X\(∪ Aα)) = rps-cl (∩ (X\Aα))
⊆

α∈Ω
rps-cl(X\Aα), by Lemma 4.6 (iii).
= X\(  α∈Ω Aα) follows from Definition 4.1.
Hence

α∈Ω
Aα ∈τ rps .
Let A, B ∈ τrps.
Now rps-cl(X\(A ∩ B)) = rps-cl ((X\A) ∪ (X\B))
= rps-cl(X\A) ∪ rps-cl(X\B)
On Regular Pre-semiopen Sets in Topological Spaces
25
= (X\A) ∪ (X\B)
= X\(A ∩ B)
This implies that A ∩ B ∈ τrps. Thus τrps is a topology.
Remark 4.14: If A is rps-closed in (X, τ), then A is closed in (X, τrps) provided τrps is a
topology.
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*T. Shyla Isac Mary1
Department of Mathematics,
Nesamony Memorial Christian College,
Martandam-629165, India.
E-mail: [email protected]
P. Thangavelu2
Department of Mathematics,
Aditanar College, Tiruchendur-628216, India.
E-mail: [email protected], [email protected]