1. Introduction
In 2012, L. N.T. Nhon and B.Q. Thinh introduced the notations of πgp-normal and
πgp-regular spaces in the Journal of Advanced Studies in Topology, see [4]. From theorem
3.5 in [4], we have the following theorems:
Theorem 1.1. If X is a T1-space and has πgp-normality, then X has πgp-regularity.
Theorem 1.2. If X is finite T1-space and has πgp-regularity, then X has πgp-normality.
A problem being presented by us in [4] was "Is the Theorem 1.2 true for infinite
topological spaces?". In this paper, we will solve the above problem by proving that
Rational Squence Topological space is a T1-space, that has πgp-regularity but also non
πgp-normality.

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JOURNAL OF SCIENCE OF HNUE
Mathematical and Physical Sci., 2013, Vol. 58, No. 7, pp. 59-65
This paper is available online at
gp-REGULARITY OF RATIONNAL SEQUENCE TOPOLOGICAL SPACE
Le Nguyen Thanh Nhon and Bui Quang Thinh
Faculty of Education, Tien Giang University
Abstract. In this paper, we prove that Rational Sequence Topological Space is a
T1-space, which is a πgp-regular and not a πgp-normal space.
Keywords: Regular closed, π-closed, πgp-closed, πgp-normal space, πgp-regular
space.
1. Introduction
In 2012, L. N.T. Nhon and B.Q. Thinh introduced the notations of πgp-normal and
πgp-regular spaces in the Journal of Advanced Studies in Topology, see [4]. From theorem
3.5 in [4], we have the following theorems:
Theorem 1.1. If X is a T1-space and has πgp-normality, then X has πgp-regularity.
Theorem 1.2. If X is finite T1-space and has πgp-regularity, then X has πgp-normality.
A problem being presented by us in [4] was "Is the Theorem 1.2 true for infinite
topological spaces?". In this paper, we will solve the above problem by proving that
Rational Squence Topological space is a T1-space, that has πgp-regularity but also non
πgp-normality.
2. Preliminary Notes
Throughout this paper, space X always means non empty topological spaces on
which no separation axioms are assumed, unless explicitly stated. We will denote the
set of irrational numbers, rational numbers and real numbers by I, Q and R .For a set
A of space X , P (A), X\A, A, int (A) denote to the power set of A, the complement,
the closure and interior of A in X , respectively. Next, we need to recall the following
definitions:
Received September 10, 2013. Accepted October 30, 2013.
Contact Le Nguyen Thanh Nhon, e-mail address: nonh.maths@gmail.com
59
Le Nguyen Thanh Nhon and Bui Quang Thinh
Definition 2.1. [2]. Subset A of space X is said to be regular open or an open domain if
it is the interior of its own closure, or equivalently if it is the interior of some closed set.
SetA is said to be a regular closed or closed domain if its complement is an open domain.
Definition 2.2. [10]. Subset A of space X is called π-open if it is a finite union of open
domain subsets of X . Subset A is called π-closed if its complement is π-open.
Definition 2.3. [3]. Subset A of space X is said to be pre-open (breifly p-open) if A ⊆
int
(
A
)
. The complement of p-open set is called p-closed. The intersection of all p-closed
sets containing A is called p-closure of A and denoted by pcl (A). Dually, the p-interior
of A denoted by pint (A), is defined to be the uinon of all p-open sets contained in A. The
class of all pre-open (resp. pre-closed) sets of X is denoted by PO (X) (resp. PC (X) ).
Definition 2.4. Subset A is said to be a p-neighborhood of x, see [11] if there exists a
p-open set U such that x ∈ U ⊆ A.
Definition 2.5. [5, 6]. Subset A of space X said to be π-generalized closed (briefly
πgp-closed) if pcl (A) ⊆ U whenever A ⊆ U and U is π-open in X . The complement of
πgp-closed set is called πgp-open set. The intersection of all πgp-closed sets containing
A is called π-generalized pre-closure of A and denoted by πgp-cl (A). Dually, the
π-generalized pre-interior of A denoted by πgp-int (A), is defined to be the union of
all πgp-open sets contained in A.The class of all πgp-open (resp. πgp-closed) sets of X
is denoted by πGPO (X) (resp. πGPC (X) ).
Definition 2.6. [1] Subset A of space X is said to be a Gδ (Fσ) set if A is countable
intersections (countable unions) of open (closed) subsets, respectively.
3. Main Results
Definition 3.1. [4] Space X is said to be πgp-normal if for every pair of disjoint
πgp-closed subsets A and B of X , there exist disjointed p-open subsets U and V of X
such that A ⊆ U and B ⊆ V .
Definition 3.2. [4] A space X is said to be (p, πgp) − R0 if pcl ({x}) ⊆ U whenever U
is πgp-open subset of X and x ∈ U .
Definition 3.3. [4] A space X is said to be a πgp-regular if for every πgp-closed subset
F ofX and point x /∈ X , there exist disjoint p-open subset U and V ofX such that x ∈ U
and F ⊆ V .
Theorem 3.1. [4] If X is a (p, πgp)-R0 space and has πgp-normality, then X has
πgp-regularity.
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πgp-Regularity of rationnal sequence topological space
Proof. Let F be a πgp-closed subset ofX and x ∈ X such that x /∈ F . Then, x ∈ (X\F ),
where X\F is πgp-open set in X . Since X is (p, πgp)− R0 space, we have pcl ({x}) ⊆
X\F . Then, F ∩ pcl ({x}) = ∅. Thus, F and pcl ({x}) are disjoint πgp-closed sets in
X . By πgp-normality of X , there exist disjoint p-open subsets U and V of X such that
F ⊆ U and pcl ({x}) ⊆ V . Therefore, there exist p-open subsets U and V of X such that
F ⊆ U and x ∈ V . Hence, X is πgp-regular space.
Corollary 3.1. If X is a T1-space, has πgp-normality, then X has πgp-regularity.
We recall the definition of Rational Sequence Topological Space:
Definition 3.4. Let X = R. For each x ∈ I, fix a sequence {x}n∈N∗ ⊂ Q such that xn
converge to x, that convergency taken in R with its usual topology. Let An (x) denote the
nth tail of the sequence, whereAn (x) = {xi : i ≥ n, n ∈ N∗}. For each x ∈ I, let B (x) =
{Un (x) : n ∈ N}, where Un (x) = An (x)∪ {x}. For each x ∈ Q, let B (x) = {x}. Then
{B (x)}x∈R = {B (x)}x∈I
∪ {B (x)}x∈Q is a neighborhood system. The unique topology
on R generated by this neighborhood system is called Rational Sequence topology on R
and is denoted byRS . Then (R,RS) is called the Rational Sequence topological space.
In this space, we have (R,RS) is T1-sapce, first-countable and separable. Any
singleton {x} is π-closed, Q is an open dense subset of (R,RS). Also, any subset of Q is
an open subset of (R,RS). I is an uncountable closed discrete subspace of (R,RS). I. We
can find some other properties about this space in [8] and [9]. The following Propositions
and lemma help us to show that Rational Sequence topological space is a T1-space and
has πgp-regularity and non πgp-normality.
Lemma 3.1 ( Jone’s Lemma). IfX contains a dense setD and a closed, relatively discrete
subspace S with |S| ≥ 2|D|, then X is not normal.
Proposition 3.1. The Rational Sequence topological space (R,RS) is not normal.
Proof. In this space, we have Q a countable dense subset and I with induced topology on
(R,RS) an uncountable closed discrete subspace of (R,RS). Thus, |P (Q)| = 2|Q| =
2ℵ0 , |I| = c and 2ℵ0 = c. The Rational Sequence topological space (R,RS) satisfies
conditions of the Jone’s Lemma, we have Rational Sequence topological space (R,RS)
is not normal.
Definition 3.5. [7] Space X is said to be sub-maximal if every pre-open subset is open
in X .
Proposition 3.2. If X is a sub-maximal space and has πgp-regularity, then X has
regularity.
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Le Nguyen Thanh Nhon and Bui Quang Thinh
Proposition 3.3. If X is a sub-maximal space and has πgp-normality, then X has
normality.
Lemma 3.2. The Rational Sequence topological space is a submaximal.
Proof. Let G be a non empty pre-open subset of (R,RS), we need to show that G is an
open subset of (R,RS). Since G is a pre-open subset, G ⊆ int (G). We have three cases
about A, which are A ⊆ Q or G ⊆ I or G ∩Q ̸= ∅ and G ∩ I ̸= ∅.
Case 1. Let G ⊆ Q.
Then, observe that G is an open subset of (R,RS).
Case 2. Let G ⊆ I.
Then, we have G ⊆ int (G) = ∅, which is a contradiction with regards to G ̸= ∅.
Therefore, G can not be in I.
Case 3. Let G ∩Q ̸= ∅ and G ∩ I ̸= ∅.
Remark. SinceG∩Q ̸= ∅ is a subset ofQ,G∩Q is an open subset of (R,RS). Therefore,
to show that A is an open subset, we only need to show that:
For each x ∈ G∩ I, there exist an open neighborhood Vx of x in (R,RS) such that
Vx ⊆ G. Then, G = (G ∩Q)
∪( ∪
x∈G∩I
Vx
)
will be an open subset of (R,RS).
In fact, for each x ∈ G ∩ I ⊆ I, there exists a sequence {x}n∈N∗ ⊂ Q such that
xn → x. Let An (x) = {xi : i ≥ n, n ∈ N∗}. Then, Un (x) = An (x) ∪ {x} is an open
neighborhood of x. This implies that {Un (x)}x∈G∩I is a open covering of G ∩ I, let F =∪
x∈G∩I
Un (x). We will prove that for each x ∈ G ∩ I, there exists a number nx ∈ N∗ such
that Unx ⊆ G.
Suppose that for every n ∈ N∗ , there exists a point x0 ∈ G ∩ I such that Un (x0) ̸⊂
G. Since xn converge to x0, every subseteq of {xn}n∈N∗ also converge to x0. Therefore,
without loss of generality, we may assume that Un (x0) ∩G = {x0}. Then, we have:
G = (F\Un (x0)) ∪ {x0} ∪ (G ∩Q)
Since (F\Un (x0)) ∪ (G ∩Q) is the largest open subset contained in G,
int
(
G
)
= (F\Un (x0)) ∪ (G ∩Q)
This implies that x0 /∈ int
(
G
)
, which is a contradiction with regards toa x0 ∈ G ⊆
int
(
G
)
.
Therefore, for each x ∈ G∩ I, there exists a number nx ∈ N∗ such that x ∈ Unx (x) ⊆ G,
let U =
∪
x∈G∩I
Unx (x) Then, we have G = U ∪ (G ∩Q) is an open subset of (R,RS).
Hence, (R,RS) is a submaximal space.
Now, we begin to solve the problem stated in the first section by proving the
following theorem:
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πgp-Regularity of rationnal sequence topological space
Theorem 3.2. The Rational Sequence topological space is a T1-space and it has
πgp-regularity and non πgp-normality.
Proof. Let A be a πgp-closed set and a point x ∈ A in (R,RS). We have three cases
about A, which are A ⊆ Q or A ⊆ I or A ∩ Q ̸= ∅ and A ∩ I ̸= ∅. For each case about
A, there are two subcases about x, which are x ∈ Q or x ∈ I. Now, we show that x and A
can be separated by disjoint pre-open subsets for each case.
Case 1. Let A ⊆ Q.
Subcase 1.1. Let x ∈ Q.
Then, {x} and A are disjoint open subsets in (R,RS). Thus x and A can be
separated by disjoint open subsets.
Subcase 1.2. Let x ∈ I.
Observe that x /∈ A ⊆ Q. We will show that A /∈ A. Since (R,RS) is a
first-countable and T1-space, then any singleton is a Gδ-set. Therefore, {x} is a Gδ-set
in (R,RS). Thus, there exists a decreasing sequence {Vn}n∈N∗ of open sets containing
x. That means: V1 ⊇ V2 ⊇ V3 ⊇ ... ⊇ Vn ⊇ ..., x ∈ Vn for each n ∈ N∗ and
{x} = ∩n∈N∗ Vn.
Claim 1. x /∈ A.
Suppose that x ∈ A. Thus, for each Ux that is an open neighborhood of x, we have
Ux ∩ A ̸= ∅ . Therefore, Vn ∩ A ̸= ∅, for each n ∈ N∗. Since Vn is a decreasing sequence
of open sets, there exists an element y ∈ (A ∩Q) such that y ∈ Vn for each n ∈ N∗. This
implies that y ∈ ∩n∈N∗ Vn = {x}. Thus, we have y = x. Since y ∈ A ∩ Q, y = x, then
x ∈ A ∩Q which is a contradiction with regards to x /∈ A. Therefore, we have x /∈ A.
Since x /∈ A, there there exists an open neighborhood G of x such that G ∩ A = ∅.
Let H = A. Then G and H are disjoint open subsets of (R,RS) such that x ∈ G and
A ⊆ H . Therefore, x and A can be separated by disjoint open subsets of (R,RS).
Case 2. Let A ⊆ I.
Subcase 2.1. Let x ∈ Q.
Then, x /∈ A and {x} is an open subseteq of (R,RS). For each y ∈ A, we have
x ̸= y. Since (R,RS) is a T1-space, there exists an open neighborhood Vy of y such that
x /∈ Vy. Let U = {x} and V =
∪
y∈A
Vy. Then, we have U and V are disjoint open subsets
of (R,RS) such that x ∈ U and A ⊆ V . Therefore, x and A can be separated by disjoint
open subsets of (R,RS).
Subcase 2.2. Let x ∈ I.
Claim 2. There exists an open subset U of (R,RS), such that A ⊆ U ⊆ U and
x ∈ U .
For each y ∈ A, we have y ∈ I and y ̸= x. Thus, there exists a sequence {yn}n∈N∗ ⊂
Q such that yn converge to y. Let An (y) = {yj : j ≥ n, n ∈ N∗}. Then, Un (y) =
63
Le Nguyen Thanh Nhon and Bui Quang Thinh
An (y) ∪ {y} is an open neighborhood of y, for each n ∈ N∗, and we have:
A ⊆
∪
y∈A
Un (y) = A
∪(∪
y∈A
An (y)
)
Let U =
∪
y∈A
Un (y). Observe that A ⊆ U ⊆ U , where:
U = A
∪(∪
y∈A
An (y)
)
= A
∪(∪
y∈A
An (y)
)
Let F =
∪
y∈A
An (y). We need to show that x /∈ F . We have x /∈ An (y) and An (y) ⊆ Q,
for each y ∈ A and n ∈ N∗. Thus, y /∈ F and F ⊆ Q. Then, by Claim 1, we have x /∈ F .
Therefore, x /∈ U .
Since x /∈ U , there exists an open neighborhoodW of x such thatW ∩U = ∅. Therefore,
there exists U , W are disjoint open subsets of (R,RS) such that x ∈ W and A ∈ U .
Hence, x and A can be separated by disjoint open subsets of (R,RS).
Case 3. Let A ⊆ Q.
Subcase 3.1. Let x ∈ Q.
We have {x} is an open subset of (R,RS) and x /∈ A, A ∩ I. Since (R,RS) is a
T1-space, for each y ∈ A∩I, there exists an open neighborhoodGy of y such that x /∈ Gy.
Thus, we have x /∈ ∪
y∈A∩I
Gy. Let U = {x} and V = (A ∩Q)
∪( ∪
y∈A∩I
Gy
)
. Then, U
and V are disjoint open subsets of (R,RS) such that x ∈ U and A ⊆ V . Therefore, x and
A can be separated by disjoint open subsets of (R,RS).
Subcase 3.2. Let x ∈ I.
Since x /∈ A, x /∈ A∩Q and x /∈ A∩I. By Claim 2, there exists an open subsetM of
(R,RS) such that A∩ I ⊆M ⊆M and x /∈M . Then, there exists an open neighborhood
U of x such that U ∩M = ∅. On the other hand, by Claim 1, we have x /∈ A ∩Q. Thus,
there exists an open neighborhood V of x such that V ∩ (A ∩Q) = ∅. LetD = U ∩V and
E = M ∪ (A ∩Q). Therefore, D and E are disjoint open subsets of (R,RS) such that
x ∈ D and A ⊆ E. Hence, x and A can be separated by disjoint open subsets of (R,RS).
We proved that x and A can be separated by disjoint open subsets of (R,RS)
for each case and (R,RS) is a submaximal space. Therefore, by lemma 3.2, we have
x and A can be separated by disjoint pre-open subsets for each case. Hence, (R,RS) is a
πgp-regular space. Next, we prove that the Rational Sequence topological space (R,RS)
is not πgp-normal.
Suppose that (R,RS) is a πgp-normal space. Then, by Proposition 3.3, we have
(R,RS) is a normal space. This is a contradiction with regards to Proposition 3.1.
64
πgp-Regularity of rationnal sequence topological space
Therefore, the Rational Sequence topological space (R,RS) is not πgp-normal.
We have observe that the Rational Sequence Topological Space is a T1-space, which
is a πgp-regular and not a πgp-normal space and we have the following corollary:
Corollary 3.2. The Rational Sequence topological space (R,RS) is a T1-space, and it
has regularity and non normality.
4. Conclusion
We presented some properties of Rational Sequence topological space and we
proved that it is a T1-space, that has πgp-regularity and non πgp-normality. Theorem
3.2 was used as a counterexample to solve the problem stated in the first section of this
paper. We will continue to study properties of πgp-regular and πgp-normal spaces in the
future.
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