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Asian Journal of Current Engineering and Maths 2: 3 May – June (2013) 201 - 204.
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ASIAN JOURNAL OF CURRENT ENGINEERING AND MATHS
Journal homepage: http://www.innovativejournal.in/index.php/ajcem
A NOTE ON VARIOUS MAPPINGS IN SIMPLE EXTENSION TOPOLOGY
*F. Nirmala Irudayam, I. Arockiarani**
*Department of Mathematics with Computer Applications, Nirmala College for Women, Coimbatore, India
**Department of Mathematics, Nirmala College for women, Coimbatore, India
ARTICLE INFO
ABSTRACT
Corresponding Author
F. Nirmala Irudayam,
Department of Mathematics with
Computer Applications,
Nirmala College for Women,
Coimbatore, India
[email protected]
Key Words: τ+p g regular , totally
τ+p g continuous, τ+p g connected
space , Strongly τ+p g continuous
functions and Totally τ+p g
irresolute functions
In this paper we intend to introduce new concepts of continuity in terms of
generalizations of pre-open sets in simple extension topology as totally τ+p g
continuous, strongly τ+p g continuous functions and totally τ+p g irresolute
functions. We also attempt to elucidate some of the decompositions using
these mappings.
INTRODUCTION
The concept of pre open set was introduced by
Corson and Micheal [4] who used the term “locally dense”.
This set defined by Corson was redefined by the name “preopen set” by A. S. Mashhour[11] and extensive study was
done by various topologists.
Levine[10] in 1963 ,initiated the concept of simple
extension topology, defined as τ+(B)= { O ∪ (O’ ∩ B) / O,O’
∈τ}, where B ∉ τ. By the definition of simple expansion we
infer that all topologies are simple expansion topologies.
Another significant contribution in the field of general
topology was due to Levine[9]in 1970,who also introduced
the notion of g closed sets.
This concept was found to be useful and many
results in general topology were improved by researches
like Balachandran, Sundaram and Maki[2], Arockiarani[1],
Devi[5],Benchalli and Wali[3] .Jain [7] introduced totally
continuous functions in general topology. Nour[16]
introduced totally semi continuous and a strongly semi
continuous functions as stronger forms of totally
continuous functions and strongly continuous functions
respectively.
Veerakumar [18] introduced totally pre-continuous and
strongly pre-continuous functions as alternative stronger
forms of totally continuous functions and strongly
continuous functions respectively.
2 . Preliminaries:
Throughout this paper, (X, τ+) or (Y, σ+) will denote
simple extension topological spaces on which no
separation axioms are assumed unless otherwise stated.
For a subset A of a space (X, τ+), cl(A) (resp. int(A),C(A))
will denote the closure (resp. interior, complement) of A in
(X, τ+).
Let us recall the following definitions and notations, which
are useful in the sequel.
©2013, AJCEM, All Right Reserved.
Definition 2.1:
A subset A of a space (X, τ) is said to be
(i) semi-open [9] if A ⊆ cl(int(A))
(ii) preopen [11] if A ⊆ int(cl(A)
(iii) α-open [15] if A ⊆ int(cl(int(A))) .
(iv) A is called semi-closed if C(A) is semi-open or
equivalently int(cl(A)) ⊆ A.
(v) A is called a preclosed set if C(A) is preopen or
equivalently cl(int(A)) ⊆ A.
(vi) A is called an α-closed set if C(A) is an α-open set or
equivalently cl(int(cl(A))) ⊆ A.
Definition 2.2:
A function f : (X, τ) → (Y, σ) is said to be semicontinuous[9] (resp. α-continuous [12], pre-continuous
[11], totally continuous [7], totally semi-continuous [16]) if
the inverse image of every open subset of (Y, σ) is a
semi-open (resp. α-open, preopen, clopen, semi-clopen)
subset of (X,τ).
Definition 2.3 :
A function f : (X, τ) → (Y, σ) is said to be strongly
continuous [8] (resp. strongly semi-continuous [16]) if the
inverse image of every subset of Y is clopen (resp. semiclopen) subset of (X, τ).
Definition 2.4 :[6]
A space X is said to be a door space if every subset of X is
either open or closed in X.
Definition 2.4:[14]
A subset A of a topological space (X, τ) is said to be a τp+
generalized closed if τ+ cl(A)⊆ U whenever A ⊆ U and U is
pre-open in (X, τ).
Also τ+ cl(A) = ∩ {S ⊆ X / A ⊆ S and S is closed in τ+(B)}
The complement of τp+ generalized closed is known as τp+
generalized open in (X, τ).
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Totally τ+p g continuous functions:
Definition 3.1 :
A subset A of a space X is called τ+p g regular if it is
both τ+p g open and τ+p g closed.
Definition 3.2:
A function f : X→ Y is τ+p g continuous if the
inverse image of each closed subset of Y is a τ+p g
closed in X.
Definition 3.3:
A function f : X→ Y is totally τ+p g continuous
if the inverse image of each open subset of Y is a τ+p g
regular subset of X.
Theorem 3. 4:
(i) Every totally τ+p g continuous function is τ+p g
continuous.
(ii) Every totally continuous function is totally τ+p g
continuous.
Proof : Obvious.
But the converses need not be true.
EXAMPLE 3.5 :
Let X = {a, b, c} and τ = {φ, X, {a, b}}; B={a} ; τ+(B) = {φ, X,
{a}{a, b}}. Let
g : (X, τ+) → (X, σ+) be the identity map. g is not totally τ+p
g continuous since {a, b} is open in (X, τ) but g -1({a, b}) =
{a, b} is not τ+p g regular in (X, τ+). However g is a τ+p g
continuous map.
Lemma 3.6 [7]
If V ∈ PO(X) and U ∈ SO(X), then U ∩ V ∈ PO(U).
Theorem 3.7 :
Suppose f : (X, τ+) → (Y, σ+) is totally τ+p g continuous
and A is a semi-open set in (X, τ+). Then the restriction fA
: A → Y is also totally τ+p g continuous.
Proof:
Let V be an open set in (Y, σ+). Since f is totally τ+p g
continuous, then f -1(V) is τ+p g regular.
Since A is semi-open, then by the above Lemma, f -1(V)
∩ A is τ+p g regular in A.
Therefore fA is totally τ+p g continuous.
Definition 3.8 :
A topological space (X, τ+) is said to be τ+p g connected if
it cannot be written as the union of two non-empty disjoint
τ+p g open sets.
Theorem 3.9 :
If f is a totally τ+p g -continuous map from a τ+p g
connected space (X, τ+) onto another space (Y, σ+), then (Y,
σ+) is an indiscrete space.
Proof:
Let us assume that (Y, σ+) is not an indiscrete space. Let
A be a proper non-empty open subset of (Y, σ+). Since f is a
totally τ+p g continuous function, then f -1(A) is a
proper non-empty τ+p g regular subset of X. Then X = f -1(A)
∪ C(f -1(A)). Thus X is a union of
two non-empty disjoint τ+p g -open sets which is a
contradiction. Therefore X must be an indiscrete space.
Hence the proof.
Theorem 3.10:
connected if every totally τ+p g
A space X is τ+p g
continuous function from a space X into any To space Y
is a constant map.
Proof:
Assume that f : X → Y is totally τ+p g
continuous and Y is a To space .
Suppose that f is not a constant map, then let us
consider two points x and y in X such that f(x) ≠ f(y). As
Y is a To space and f(x), f(y) are distinct points of Y
then there exists an open set H say in Y containing
f(x) but not f(y).
Now f -1 (H) is τ+p g regular subset of X, as f is totally
τ+p g continuous function.
Thus x ∈ f -1 (H) and y ∉ f -1 (H)
Now X = f -1 (H) U C (f -1 (H) ) which is the union of two
non- empty τ+p g open subsets of X. Hence X is not τ+p g
connected, which is a contradiction to our fact. Thus f
is a constant map.
Theorem 3.11:
If f is a totally τ+p g continuous map from a
+
τ p g connected space X onto another space Y then Y
is an indiscrete space.
Proof:
Let us assume that Y is not an indiscrete space.
Suppose that N is a proper
non – empty open subset of Y. Since f is totally τ+p g
is a proper non – empty τ+p g
continuous, f -1 ( N)
regular subset of X. Then X = f -1 ( N) U C(f -1 ( N)).
Hence X is a union of two non – empty disjoint τ+p g
open sets, a contradiction .Thus Y is an indiscrete space.
Theorem 3.12:
Let f : (X, τ+) → (Y, σ+) be a totally τ+p g continuous
function and Y is a T1 space. If A is a non-empty τ+p g
connected subset of X, then f(A) is singleton.
Proof:
On the contrary let us assume that f(A) is not singleton.
Let f(x1) = y1 ∈ A and f(x2) = y2 ∈ A. Since y1, y2 ∈ Y and Y
is a T1 space, then there exists an open set G in Y containing
y1 but not y2. Since f is totally τ+p g continuous, then f 1(G) is a τ+ g regular set containing x but not x . Now X =
p
1
2
f -1(G) ∪ C(f -1(G)). Thus we have expressed X as a union
of two non-empty τ+p g open sets which is a contradiction.
Hence the proof.
Definition 3.13:
A subset A of topological space X is called τ+p g
T2 if, for each pair, of distinct points x, y of X, there
exists τ+p g open sets U,V in X such that x ∈ U, y ∈ V and
U∩V =φ
Theorem 3.14:
continuous
Let f : X → Y be a totally τ+p g
injection. If Y is To, then X is
τ+p g T2.
Proof :
Assume that x and y be any two distinct
points of X. As f is an injection, f(x) ≠ f(y).Since Y is To,
there exists an open subset V of Y containing f(x) but
not f(y). Then x ∈ f -1 (V), y ∉ f -1 (V), and f -1 (V), is a
τ+p g regular subset of X. Assume that A = f -1 (V), and
B = C(f -1 (V)) here A and B are two disjoint τ+p g
open subsets of X containing x and y respectively. Then
X is τ+p g T2-space.
Theorem 3.15 :
Let f : X → Y be totally τ+p g continuous and Y
be a T1 space. If A is a non – empty τ+p g connected
subset of X, then f (A) is a singleton set.
Proof :
Let us assume the contrary that f (A) is not a
singleton set.
Suppose f( x1) = y1 ∈A, f (x2) = y2 ∈A, As y1 , y2 ∈ Y and Y
is a T1 space, there exist an open set H in y (say)
τ +p g
containing y1 but not y2. As f is totally
continuous ,
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f -1 (H) is τ+p g regular set containing x1 but not x2.
Now X = f -1 (H) U C(f -1 (H)). Hence X is the union of
two non – empty τ+p g open sets, a contradiction. Thus
f(A) is singleton set.
Strongly τ+p g continuous functions
Definition 4.1:
A function f : (X, τ+) → (Y, σ+), is said to be strongly τ+p g
continuous if the inverse image of every subset of Y is a τ+p
g regular subset of (X, τ+).
Remark 4.2 :
Every strongly τ+p g continuous function is totally τ+p g
continuous.
Remark 4.3 :
Every strongly continuous function is strongly τ+p g
continuous but the following example supports that the
converse is not true in general.
Example 4.4:
Let X = {a, b, c} and τ = {φ, X, {a}}B= {b, c}}τ+(B) = {φ, X, {a},
{b, c}}.
Define h : (X, τ+) → (X, τ+) by h(a) = a, h(b) = c and h(c) =b.
It is easy to see that h is a strongly τ+p g continuous map
but not a strongly continuous map.
Theorem 4.5 :
Let f : (X, τ+) → (Y, σ+), be a totally τ+p g continuous
injection. If (Y,σ+) is To, then (X, τ+) is τ+p g -T2.
Proof:
Let x and y be any two distinct points of X. f(x) ≠ f(y) since f
is an injection. Since
(Y, σ+) is To, then there exists an open subset U(say) of (Y,
σ+), containing f(x) but not f(y).
Then x ∈ f -1(U) and y ∉ f -1(U). f -1(U) is a τ+p g regular
subset of (X, τ) since f is totally τ+p g
continuous and U is an open subset of (Y, σ+). Take G = f 1(U) and H = C(f -1(U)). Clearly G
and H are two disjoint τ+p g open subsets of (X, τ+)
containing x and y respectively. Thus (X, τ+) is a pre-T2
space.
Lemma 4.6:
For a topological space (X, τ+), a subset A of X is clopen if
and only if A is α+-clopen in X.
The following theorem gives a characterization for totally
continuous functions.
Theorem 4.7:
A function f : (X, τ+) → (Y, σ+) is totally continuous if and
only if the
inverse image of every open subset of (Y, σ+) is an α+clopen subset of (X,τ +) .
Proof: Follows from the above Lemma.
The following theorem gives a decomposition for totally
continuous functions.
Theorem 4.8:
A function f : (X,τ +) → (Y,σ +) is totally continuous if and
only if f is both totally semi-continuous and totally τ+p g
continuous.
Proof: Obvious.
Theorem 4.9 :
(i) A function f : (X, τ+) → (Y, σ+) is strongly continuous if
and only if the inverse image of every subset of Y is an α+clopen subset of (X, τ+).
(ii) A function f : (X, τ+) → (Y, σ+) is strongly continuous if
and only if f is both strongly semi-continuous and strongly
τ+p g continuous.
5. Totally τ+p g irresolute functions
Definition 5.1:
(i) A function f : (X, τ+) → (Y, σ+) is τ+p g irresolute if
the inverse image of every τ+p g
closed set of Y is τ+p g closed in X.
(ii) A function f : (X, τ+) → (Y, σ+) is totally τ+p g
irresolute if the inverse image of
every τ+p g open subset of Y is τ+p g regular in X.
Theorem 5.2 :
(i) Every totally τ+p g irresolute function is τ+p g
irresolute
(ii) Every totally τ+p g irresolute function is totally τ+p g
continuous.
Proof : The proof is immediate.
Definition 5.3:
τ +p g
A function
f : (X, τ+) → (Y, σ+) is quasi
irresolute if the inverse image of every τ+p g regular
subset of Y is τ+p g regular in X.
Theorem 5.4 :
Every totally τ+p g irresolute function is quasi τ+p g
irresolute function.
Proof : The proof is immediate from the definition.
Theorem 5.5 :
If Y is a τ+p g T2 space and f : (X, τ+) → (Y,
σ+) is a totally τ+p g irresolute injection, then X is τ+p g
T2.
Proof :
Assume x, y to be distinct points of X. As f is
an injection, then f(x) ≠ f(y). Now there exist disjoint
τ+p g open sets U and V such that U ∩V = φ, Since Y is a
τ+p g T2 space. As f is totally τ+p g irresolute, f -1 (U) and
f -1 (V) are τ+p g regular in X such that x ∈ f -1 (U) , y ∈ f
-1 (V) and f -1 (U) ∩ f -1 (V) = φ. Thus X is τ+ g T .
p
2
Theorem 5.6:
Let f : X → Y and g : Y → Z be two functions then the
following hold.
(i) If f is totally τ+p g irresolute and g is τ+p g
continuous then (g ° f) is totally τ+p g continuous.
(ii) If f is totally τ+p g irresolute and g is τ+p g
irresolute , then (g ° f) is totally τ+p g irresolute.
(ii) If f is quasi - τ+p g irresolute and g is totally τ+p g
irresolute, then (g ° f) is totally τ+p g irresolute.
Proof : Obvious.
Theorem 5.7:
The composition of two totally
τ+p g irresolute
+
functions is totally τ p g irresolute.
Proof :
Follows from theorem 5.6(ii) and theorem 3.4
Theorem 5.8 :
Let f: X → Y be a function and g : X → X xY, be
the graph of f. If g is to τ+p g irresolute , then f is
totally τ+p g irresolute
Proof :
Assume that U ∈ τ+p g O(Y). Then X x U∈ τ+p gO(X). As g
is totally τ+p g irresolute ,
g -1 ( X x U) is τ+p g regular in X. Thus X∩f-1(U)= f-1(U) is
τ+p g regular in X. Hence f is totally τ+p g irresolute .
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