Download (1, 2) extremally disconnectedness via bitopological open sets

International Journal of General Topology
© Serials Publications
Vol. 4, Nos. 1-2, January-December 2011, pp. 9– 15;
ISSN:0973-6751
(1, 2) EXTREMALLY DISCONNECTEDNESS VIA
BITOPOLOGICAL OPEN SETS
M. Lellis Thivagar1, B. Meera Devi2 and G. Navalagi3
ABSTRACT
In this paper, we introduce and study of some new bitopological open sets namely (1, 2) semi open sets,
(1 ,2)* semi open sets and (1, 2) b-open sets. The notions are vital in facilitating us to expand the some of
the concepts and results of Classical Topological Spaces into the field of Bitopology.
AMS Subject Classification : 54E55
Key words: (1, 2) extremally disconnected, (1, 2) semi-open, (1, 2) semi T2, (1, 2)* Semi-open, (1, 2)
b-open.
1. INTRODUCTION
Levin [4], Njastad [8] and Monsef [7] introduced semi-open sets, pre-open sets, α-open
sets and β-open sets respectively which are called nearly open sets. J C Kelly [3] introduced
the concept of bitopological spaces in 1963. The notion of pairwise Hausdorff spaces was
the innovative idea of Reilly [10] and Kelly [3]. The purpose of this paper is to define some
properties by using (1, 2) extremally disconnected, (1, 2) semi-open and (1, 2) b-open sets
in bitopological spaces and investigate the relationships between them.
2. PRELIMINARIES
Definition 2.1: A subset S of (X, τ) is called
(i) A semi-open set [4] if S ⊆ cl(int S)
(ii) A pre-open set [4] if S ⊆ int(cl S)
(iii) A b-open set [1] if S ⊆ int(cl S) ∪ cl(int S)
Remark 2.2: Njastad [8] has shown that SO(X) is a topology iff X is extremally
disconnected.
Definition 2.3: [3] A topological space (X, τ) is said to be semi-Hausdorff space if
given any two distinct points x and y, there is an semi-open set U containing x and an
semi-open set V containing y such that U ∩ V = φ
Definition 2.4: [2] A space X is called an extremally disconnected space if closure of
each open set is open.
Remark 2.5: Any result in the bitopological space (X, τ1, τ2) for any topology τ is true
for the topological space (X, τ)
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Definition 2.6: [3] A bitopological space (X, τ1, τ2) will be said to be bi-℘ whenever (X,
τ1) and (X, t2) both posses the property.
Definition 2.7: A bitopological space satisfying a bitopological property ℘ will be
called pairwise ℘ .
Theorem 2.8: In a topological space (X, τ) is extremally disconnected, X is Hausdorff
iff X is semi - Hausdorff.
Proof: Assume X is Haudorff. Let x ≠ y. Then there exist open sets U and V such that x
∈ U and y ∈ V and U ∩ V = φ. Since every open set is an semi-open set, U, V ∈ SO(X).
Hence U ∩ V = φ, where U and V are semi-open sets containing x and y respectively. Thus
X is semi-Hausdorff. Conversely, suppose X is semi-Hausdorff. Let x ≠ y, then there exist
semi-open sets A and B such that x ∈ A, y ∈ B and A ∩ B = φ. We know that cl (int A) ∩ cl
(int B) = φ. Let U = cl (int A) and V = cl (int B). Then A ⊆ U and B ⊆ V and U, V ∈ τ. x ∈ A ⇒
x ∈ U and y ∈ B ⇒ y ∈ V. Thus U and V are disjoint open sets containing x and y respectively.
Hence, X is Hausdorff.
Definition 2.9: [3] A bitopological space (X, τ1, τ2) is said to be pairwise Hausdorff if
given two distinct points x, y of X, there is a t1 - open set U and a t2 - open set V such that
x ∈ U, y ∈ V, and U ∩ V = φ.
Definition 2.10: [3] A bitopological space (X, τ1, τ2) is said to be pairwise semi-Hausdorff
if given two distinct points x, y of X, there is a τ1 semi open- set U and a τ2 semi open- set
V such that x ∈ U, y ∈ V, and U ∩ V = φ.
The above definition leads to the fact that the proof of theorem 2.8 does not
automatically extend to bitopological space since there is a difficulty that τ1 int A ∩ τ2 cl(τ2
int B) need not be empty set. The following example substantiates this claim.
Example 2.11: Let X ={a, b, c}; τ1 = {φ, X, {a}, {b}, {a, b}} and τ2 = {φ, X, {b}, {c}, {b, c} }. Let
U = {a} and V = {c}. Hence, τ1 int U ∩ τ2 cl(τ2 int V) = {a} ≠ φ.
Therefore, in order to overcome the above difficulty, we apply the following
bitopological (1, 2) semi- open sets.
Definition 2.12: [6] A subset A of a bitopological space (X, τ1, τ2) is called
(i) τ1τ2 open if A ∈ τ1 or A ∈ τ2
(ii) τ1τ2 closed if X – A is τ1τ2 open
(iii) τ1τ2 closure of A if τ1τ2 cl (A) = ∩ {F/A ⊆ F and F is τ1τ2 closed}
Note 2.13: τ1τ2 cl A ⊆ τ1 cl A and τ1τ2 cl A ⊆ τ2 cl A
Definition 2.14: [5] A subset of (X, τ1, τ2) is said to be
(i) (1, 2) semi open-set if S ⊆ τ1τ2 cl(τ1 int(S))
(ii) (1, 2) pre open-set if S ⊆ τ1 int(τ1τ2 cl(S))
Definition 2.15: [6] Let S be a subset of (X, τ1, τ2). Then S is said to be τ1,2 open if S = A
∪ B where A ∈ τ1 and B ∈ τ2.
(1, 2) Extremally Disconnectedness via Bitopological Open Sets
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Definition 2.16: [6] Let S be a subset of X. Then S is said to be (1, 2)* semi -open set if
S ⊆ τ1τ2 cl (τ1τ2 int S)
3. (1, 2) EXTREMAL DISCONNECTED SPACES
We introduce the following definition.
Definition 3.1: A space (X, τ1, τ2) is called an (1, 2) extremally disconnected space if
τ1τ2 closure of each τ1 open set is τ1 open similarly τ1τ2 closure of each τ2 open set is
τ2 open.
Example 3.2: Let X = {a, b, c}, τ1 = {φ, X, {a}, {b}, {a, b}}
t2 = {φ, X, {a}, {c}, {a, c}}; τ1τ2 open = {φ, X, {a}, {b}, {c}, {a, b}, {a, c}}
τ1τ2 closed = {φ, X, {b}, {c}, {a, b}, {a, c}, {b, c}}.
Hence every τ1τ2 closure of τ1 open is τ1 open and also every τ1τ2 closure of τ2 open is τ2
open.
Definition 3.3: A subset A of (X, τ1, τ2) is said to be (1, 2) b-open set if A ⊆ τ1τ2 cl(τ1
int(S)) ∪ τ1int(τ1τ2 cl(S))
Definition 3.4: A subset A of (X, τ1, τ2) is called pairwise semi - open set in X if A is (1,
2) semi - open set and (2, 1) semi - open set.
Lemma 3.5: In a bitopological space (X, τ1, τ2), let A be τ1 semi - open set and B be τ2
semi - open set such that A ∩ B = φ. Then τ1 int A ∩ τ1τ2 cl(τ2 int B) = φ.
Proof: We know that τ1 int A ∩ τ2 int B = φ where A is τ1 semi - open set and B is τ2 semi
- open set. Suppose τ1int A ∩ τ1τ2 cl(τ2 int B) ≠ φ. Then there is an element x such that x ∈
τ1int A and x ∈ τ1τ2 cl(τ2 int B). Since x ∈ τ1τ2 cl (τ2 int B), if C is any τ1 closed set containing
τ2 int B, then x ∈ C. Similarly, if D is any τ2 closed set containing τ2 int B then x ∈ D. In
particular, if we take C to be τ1cl (τ2 int B), then x ∈ C. Similarly, if we take D to be τ2 cl (τ2
int B) then, x ∈ D. (ie) if V is any τ1 open set containing x then V ∩ τ2 int B ≠ φ. Also, if V is
any τ2 open set containing x then V ∩ τ2 int B ≠ φ. Thus τ1 int A ∩ τ2 int B ≠ φ which is a
contradiction. Hence the claim.
Proposition 3.6: In a bitopological space (X, τ1, τ2), any open set in (X, τ1) is (1, 2) semi
- open set and any open set in (X, τ2) is (2, 1) semi - open set.
Proof: Let A be any open set in (X, τ1). (ie) A = τ1 int A.
Now A ⊆ τ1τ2 cl A = τ1τ2 cl (τ1 int A). Therefore A is (1, 2) semi - open set. Similarly, any
open set in (X, τ2) is (2, 1) semi - open set.
Remark 3.7: Converse of the above proposition need not be true.
Example 3.8: Let X = {a, b, c}, τ1 = {φ, X, {a}} and τ2 = {φ, X, {a, c}}. Here, {a, b} is (1, 2) semi
- open set but not open in τ1.
Theorem 3.9: Every (1, 2) semi-open set is (1, 2) b-open.
Proof: Let A be a (1, 2) semi-open set in (X, τ1, τ2). Then A ⊆ τ1τ2 cl(τ1 int(A)) ⊆ τ1τ2 cl(τ1
int(A)) ∪ τ1 int(τ1τ2 cl(A)). Hence A is (1, 2) b-open.
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Remark 3.10: The converse of the above theorem need not be true as shown in the
following example.
Example 3.11: Let X = {a, b, c, d}, τ1 = {φ, X, {a}, {a, b}, {a, c, d}}
τ2 = {φ, X, {a}, {d}, {a, d}}; τ1τ2 open = {φ, X, {a}, {d}, {a, b}, {a, d}, {a, c, d}}
τ1τ2 closed = {φ, X, {b}, {b, c}, {c, d}, {a, b, c}, {b, c, d}}
(1, 2) SO(X) = {φ, X, {a}, {a, b}, {a, c}, {a, b, c}, {a, c, d}}
(1, 2) bO(X) = {φ, X, {a}, {a, b}, {a, c}, {a, d},{a, b, c}, {a, b, d},{a, c, d}}.
Here {a, d} is (1, 2) b-open set but it is not (1, 2) semi-open.
Using Lemma 3.5 and Proposition 3.6, we extend Theorem 2.8 to bitopological spaces.
Theorem 3.12: In a bitopological space (X, τ1, τ2) is (1, 2) extremally disconnected, X is
pairwise Hausdorff iff X is (1, 2) semi -Hausdorff.
Proof: Let X be pairwise Hausdorff. Let x ≠ y. Then there is a τ1 open set U and a τ2
open set V such that U ∩ V = φ, x ∈ U and y ∈ V. Since every open set is (1, 2) semi- open
set, by proposition 2.12, U ∩ V = φ where U is (1, 2) semi - open set and V is (2, 1) semiopen set containing x and y respectively. Hence X is (1, 2) semi -Hausdorff. Conversely, let
X be (1, 2) semi - Hausdorff. Let x ≠ y. Then there is a (1, 2) semi - open set A and
(2, 1) semi-open set B such that x ∈ A and y ∈ B and A ∩ B = φ. Hence τ1 int A ∩ τ2 int
B = φ.
⇒ τ1 int A ∩ τ1τ2 cl(τ2 int B) = φ, by lemma 2.11. ⇒ τ1τ2 cl(τ1 int A) ∩ τ1τ2 cl(τ2 int B)
= φ.
Let U = τ1τ2 cl(τ1 int A) and V = τ1τ2 cl(τ2 int B). Then A ⊆ U and B ⊆ V where U is τ1 open
and V is τ2 open. x ∈ A ⇒ x ∈ U and y ∈ B ⇒ y ∈ V. Then, there exist τ1 open set U and τ2
open set V containing x and y respectively such that U ∩ V = φ. Hence X is pairwise
Hausdorff. This is analogous to Theorem 2.8
Theorem 3.13: Every (1, 2) semi - open set is τ1 semi - open set.
Proof: Let A be a (1, 2) semi - open set. (i. e) A ⊆ τ1τ2 cl (τ1 int A) ⊆ τ1 cl (τ1 int A)
since τ1τ2 cl A ⊆ τ1 cl A. Hence (1, 2) semi - open set implies τ1 semi - open set.
Remark 3.14: The converse of the above theorem need not be true.
Example 3.15: Let X = {a, b, c, d}, τ1 = {φ, X, {a}, {a, b}, {a, c, d}}
τ1 SO(X) = {φ, X, {a}, {a, b}, {a, c}, {a, d}, {a, b, c}, {a, b, d}, {a, c, d}}
τ2 = {φ, X, {a}, {d}, {a, d}}; τ1τ2 open = {φ, X, {a}, {d}, {a, b}, {a, d}, {a, c, d}}
τ1τ2 closed = {φ, X, {b}, {b, c}, {c, d}, {a, b, c}, {b, c, d}}
(1, 2) SO(X) = {φ, X, {a}, {a, b}, {a, c}, {a, b, c}, {a, c, d}}.
It is obvious that {a, d} ∈ τ1 semi-open but {a, d} ∉ (1, 2) semi - open set.
Remark 3.16: (1, 2) semi-open sets need not form a topology, because the intersection
of (1, 2) semi-open sets need not be a (1, 2) semi - open set as seen from the following
example.
(1, 2) Extremally Disconnectedness via Bitopological Open Sets
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Example 3.17: Let X = {a, b, c, d}, τ1 = {φ, X, {a}, {c}, {a, c}, {a, b, c}}
τ2 = {φ, X, {d}} ; τ1τ2 open = {φ, X, {a}, {c}, {d}, {a, c}, {a, b, c}}
τ1τ2 closed = {φ, X, {d}, {b, d}, {a, b, d}, {a, b, c}, {b, c, d}};
(1, 2) SO(X) = {φ, X, {a}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}.
Clearly, {b, c} ∩ {a, b} = {b} ∉ (1, 2) semi- open set.
Theorem 3.18: If (X, τ1, τ2) is (1, 2) extremally disconnected then (1, 2) semi-open sets
form a topology.
Proof: Let A and B are two (1, 2) semi-open sets. Then A ⊆ τ1τ2 cl(τ1 int(A)) and B ⊆ τ1τ2
cl(τ1 int(B)). Now A ∪ B = [τ1τ2 cl(τ1 int(A))] ∪ [τ1τ2 cl(τ1 int(B))] ⊆ τ1τ2 cl[τ1 int(A ∪ B)].
Hence A ∪ B is (1, 2) semi-open. Also A ∩ B = [τ1τ2 cl(τ1 int(A))] ∩ [τ1τ2 cl(τ1 int(B))] ⊆ τ1
int(A) ∩ τ1 int(B) [since (X, τ1, τ2) is (1, 2) extremally disconnected] = τ1 int(A ∩ B) ⊆ τ1τ2
cl(τ1 int(A ∩ B)). Hence (1, 2) semi-open sets form a topology.
Theorem 3.19: In a bitopological space (X, τ1, τ2), every τ1-open is (1, 2) pre-open set
and every τ2-open is (2, 1) pre-open.
Proof: Let A be any τ1-open set. That is A = τ1 int(A). Now A ⊆ τ1τ2 cl(A) = τ1τ2 cl
(τ1 int A). This implies A ⊆ τ1 int(τ1τ2 cl(A)). Hence A is (1, 2) pre- open set. Similarly, every
τ2- open set is (2, 1) pre - open set.
Remark 3.20: The converse of the above theorem need not be true as shown in the
following example.
Example 3.21: Let X = {a, b, c}, τ1 = {φ, X, {c}, {a, b}}
τ2 = {φ, X, {a}} ; τ1τ2 open = {φ, X, {a}, {c}, {a, b}}
τ1τ2 closed = {φ, X, {c}, {a, b}, {b, c}} ;
(1, 2) PO(X) = {φ, X, {a}, {c}, {a, b}, {a, c}, {a, c}}. Here {a, c} is (1, 2) pre-open but not
τ1-open.
Remark 3.22: The concept of (1, 2) semi-open and (1, 2) pre-open are independent
notions as seen from the following examples.
Example 3.23: Let X = {a, b, c}, τ1 = {φ, X, {a}, {b}, {a, b}}
τ2 = {φ, X, {b}} ; τ1τ2 open = {φ, X, {a}, {b}, {a, b}}
τ1τ2 closed = {φ, X, {c}, {a, c}, {b, c}} ;
(1, 2) PO(X) = {φ, X, {a}, {b}, {a, b}}
(1, 2) SO(X) = {φ, X, {a}, {b}, {a, b}, {a, c}, {b, c}}. Here {a, c} is (1, 2) semi-open but not (1,
2) pre-open.
Example 3.24: Let X = {a, b, c}, τ1 = {φ, X, {a}, {b, c}}
τ2 = {φ, X, {a}} ; τ1τ2 open = {φ, X, {a}, {b, c}}
τ1τ2 closed = {φ, X, {a}, {b, c}} ;
(1, 2) PO(X) = {φ, X, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}}
(1, 2) SO(X) = {φ, X, {a}, {b, c}}. Here {a, b} is (1, 2) pre-open but not (1, 2) semi-open.
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Theorem 3.25: Every (1, 2) pre-open set is (1, 2) b-open.
Proof: Let A be a (1, 2) pre-open set in (X, τ1, τ2). Then A ⊆ τ1int(τ1τ2 cl(A)) ⊆ τ1τ2 cl(τ1
int(A)) ∪ τ1int(τ1τ2 cl(A)). Hence A is (1, 2) b-open.
Remark 3.26: The converse of the above theorem need not be true. In Example 3.11, {a,
c} is (1, 2) b-open but not (1, 2) pre-open.
Remark 3.27: From the above discussions we have the following diagram. Here
A → B represents A implies B. A →
| B represents A does not implies B.
1. τ1-open 2. (1,2) pre-open 3. (1,2) semi-open 4. (1,2) b-open
Theorem 3.28: If (X, τ1, τ2) is (1, 2) extremally disconnected then every (1, 2) b-open is
(1, 2) preopen.
Proof: Let A is (1, 2) b-open. Then A ⊆ τ1τ2 cl(τ1 int(A)) ∪ τ1int(τ1τ2 cl(A))
⊆ τ1int(τ1τ2cl(A)) τ1int(τ1τ2 cl(A)) [since (X, τ1, τ2) is (1, 2) extremally disconnected]
= τ1int(τ1τ2cl(A)). Hence A is (1, 2) pre-open.
Remark 3.29: The semi-open sets in (X, τ2) and the (1, 2)* semi-open sets in (X, τ1, τ2)
are independent.
Proposition 3.30: Every (1, 2) semi- open set is (1, 2)* semi- open set.
Proof: Let S be (1, 2) semi- open set. Then S ⊆ τ1τ2 cl (τ1 int (S)). Since τ1 int(S) ⊆ τ1τ2(int
S), S ⊆ τ1τ2 cl (τ1τ2 int(S)). Hence, S is (1, 2)* semi- open set.
Remark 3.31: The Converse of the above proposition need not be true as shown below.
Example 3.32: Let X = {a, b, c, d}, τ1 = {φ, X, {a}, {a, b}, {a, c, d}}
τ2 = {φ, X, {a}, {d}, {a, d}}; τ1τ2 open = {φ, X, {a}, {d}, {a, b}, {a, d}, {a, c, d}}
τ1τ2 closed = {φ, X, {b}, {b, c}, {c, d}, {a, b, c}, {b, c, d}}
(1, 2) SO(X) = {φ, X, {a}, {a, b}, {a, c}, {a, b, c}, {a, c, d}}.
(1, 2) Extremally Disconnectedness via Bitopological Open Sets
15
(1, 2)* SO(X) = {φ, X, {a}, {d}, {a, b}, {a, c}, {a, d}, {c, d}, {a, b, c}, {a, c, d}, {a, b, d}}
It is obvious that {c, d} ∈ (1, 2)* semi -open but {c, d} ∉ (1, 2) semi -open.
Theorem 3.33: Every pairwise semi- open sets are (1, 2)* semi- open sets.
Proof: Let S be pairwise semi-open set. Then S is (1, 2) semi-open and (2, 1) semi-open.
That is S ⊆ τ1τ2 cl (τ1 int (S)) ∩ τ1τ2 cl (τ2 int (S)). Since τ1int(S) ⊆ τ1τ2 int (S) and τ2int(S) ⊆ τ1τ2
int (S), S ⊆ τ1τ2 cl (τ1τ2 int (S)) ∩ τ1τ2cl(τ1τ2 int (S)) = τ1τ2 cl (τ1τ2 int (S)). Hence S is (1, 2)*
semi-open set.
Remark 3.34: The converse of the above theorem need not be true. In example 3.34,
pairwise semi-open = {φ, X, {a}, {a, b}, {a, c}, {a, b, c}},
(1, 2)* SO(X) = {φ, X, {a}, {d}, {a, b}, {a, c}, {a, d}, {c, d}, {a, b, c}, {a, c, d}, {a, b, d}}.
Clearly {a, d} is (1, 2)* semi- open but not pairwise semi-open.
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1
Department of Mathematics,
Arul Anandar College, Karumathur-625514,
Madurai Dt., India. E-mail : [email protected]
2
Department of Mathematics,
Sri S.R.N.M College, Sattur-626203, India.
E-mail: [email protected]
3
Department of Mathematics,
KLE,Society’s, G.H. College,
Haveri-581110, Karnataka, INDIA,
E-mail: [email protected]$