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International Journal of Mathematical Archive-4(10), 2013, 239-243
Available online through www.ijma.info ISSN 2229 – 5046
ON REGULAR GENERALISED β – CONTINUOUS FUNCTIONS
Y. Palaniappan1* and V. Senthilkumaran2
1Associate
Professor of Mathematics (Retired), Arignar Anna Government Arts College,
Musiri, Tamilnadu, India.
2Associate
Professor of Mathematics, Arignar Anna Government Arts College,
Musiri, Tamilnadu, India.
(Received on: 12-09-13; Revised & Accepted on: 11-10-13)
ABSTRACT
In this paper rgβ-continuous function, rgβ-open functions, rgβ-closed functions are defined and their properties are
investigated.
MSC: 54D99.
Key words: rgβ-open sets, rgβ-closed sets, rgβ-continuous functions.
1. INTRODUCTION
Levine [5] introduced generalised closed sets as a generalisation of closed sets. Andrijevic [1] introduced semi preopen
sets(β-open sets) and semi preclosed sets(β-closed sets). Recently Palaniappan[7] has introduced and studied the
properties of rgβ- closed sets. In this paper rgβ-continuous function is defined and its properties are investigated. rgβopen maps, rgβ- closed maps and rgβ- homeomorphism are introduced and their properties are studied.
Throughout this paper (X, τ) and (Y, σ) represent non empty topological spaces on which no separation axioms are
assumed unless otherwise stated.cl A and int A denote the closure and the interior of A respectively.(X, τ) and (Y, σ)
will be denoted by X and Y respectively when there is no confusion.
2. PRELIMINARIES
Definition 2.1: A subset A of a topological space X is called
1. A pre open set [6] if A⊂int cl A and a pre closed set if cl int A⊂A
2. A regular open set if A=int cl A and a regular closed set if A= cl int A
3. A β –open set [1] (semi preopen set) if A⊂ cl int cl A and a β –closed set (semi preclosed set) if Int cl int A⊂ A.
The intersection of all β-closed subsets of X containing A is denoted by βcl A. βcl A is a β closed subset of X.
Definition 2.2: A subset A of a topological space X is called a regular generalised β-closed set [7] (rgβ- closed set) if
βcl A ⊂ U whenever A ⊂ U and U is regular open in X.
The complement of rgβ-closed set in X is rgβ-open in X.
The intersection of all rgβ- closed subsets of X containing A is called rgβ- closure of A and is denoted by rgβ-cl A. In
general rgβ- cl A is not rgβ- closed. If we assume that arbitrary intersection of rgβ closed sets of X is rgβ closed in X
then, rg clA is rgβ closed and rgβ int A is rgβ open.
3. rgβ - CONTINUOUS FUNCTIONS.
Definition 3.1: A function f:X→Y is called rgβ- continuous if the inverse image of every open set in Y is rgβ- open in
X.
Corresponding author: Y. Palaniappan1*
1Associate
Professor of Mathematics (Retired), Arignar Anna Government Arts College,
Musiri, Tamilnadu, India.
International Journal of Mathematical Archive- 4(10), Oct. – 2013
239
1*
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Y. Palaniappan and V. Senthilkumaran / ON REGULAR GENERALISED β – CONTINUOUS FUNCTIONS/ IJMA- 4(10), Oct.-2013.
Remark 3.1: f: X→ Y is rgβ- continuous implies the inverse image of every closed set of Y is rgβ- closed in X.
Theorem3.1: Let arbitrary intersections of rgβ- closed sets of X be rgβ- closed in X. f: X→Y is rgβ- continuous
implies rgβ- cl f-1(B) ⊂ f-1(cl B) , for every B⊂ Y.
Proof: Let f be rgβ- continuous. Cl B is closed in Y. So f-1(cl B) is rgβ- closed in X.
Hence rgβ-cl f-1(cl B) =f-1(cl B).
Therefore rgβ-clf-1(B)⊂rgβ -cl f-1(cl B)=f-1(cl B)
Theorem 3.2: Let arbitrary intersections of rgβ- closed sets of X be rgβ- closed in X. A mapping f:X→Y is rgβcontinuous if and only if f(rgβ- clA)⊂clf(A), for every A ⊂ X.
Proof: Let f be rgβ- continuous and A ⊂ X. Let f(A)=B. By Theorem 3.1, rgβ- cl f-1(B) ⊂f-1(clB). That is,
rgβ-cl f-1(f(A))⊂ f-1(cl f(A)). But A⊂ f-1(f(A)). rgβ-cl A⊂ rgβ-cl f-1(f(A))⊂f-1 (clf(A)).
f(rgβ-cl A) ⊂cl f(A).
Conversely, let f: X→Y be such that f(rgβ-cl A)⊂cl f(A), for every A ⊂X.
Let B be an arbitrary closed set of Y. Then f-1(B)⊂X. f (rgβ-cl f-1(B))⊂ cl f(f-1(B))=cl B=B, as B is closed in Y
f-1(f(rgβ- cl f-1(B)))⊂ f-1(B). That is,rgβ- cl f-1(B)⊂ f-1 (B). But f-1(B)⊂ rgβ-cl f-1(B).
Hence f-1(B)=rgβ- cl f-1(B), which is rgβ- closed in X. So, f is rgβ- continuous.
Theorem 3.3: Let arbitrary intersections of rgβ- closed sets of X be rgβ- closed in X. f: X→Y is rgβ- continuous if and
only if f-1(int B)⊂ rgβ-int(f-1(B)), for every B⊂Y.
Proof: Let f be rgβ- continuous. Int B is open in Y. So, f-1(int B) is rgβ- open in X.
Hence rgβ- int(f-1(intB))=f-1(int B). Therefore f-1(int B) = rgβ- int f-1(int B) ⊂rgβ- int f-1(B).
Conversely, let f-1(intB) ⊂rgβ-int f-1(B). Let G be open in Y. So, int G=G.
f-1(G)=f-1(intG)⊂rgβ- int(f-1(G)). But rgβ-int f-1(G)⊂ f-1(G). Hence f-1(G) =rgβ- int f-1(G).
So f-1(G) is rgβ- open. Hence f is rgβ- continuous.
Theorem 3.4: Let X be any discrete topological space and let Y be any topological space. Every mapping of X into Y
is rgβ- continuous.
Proof: Let f: X→Y be any mapping. Let G be open in Y. Then f-1(G) is open in X, since X is discrete space. Every
open set is rgβ- open. So f-1(G) is rgβ- open and hence f is rgβ- continuous. Theorem 3.5 Let X be any topological
space and let Y be any indiscrete space. Every mapping of X into Y is rgβ- continuous.
Theorem 3.5: Let X be any topological space and let Y be any indiscrete space. Every mapping of X into Y is rgβcontinuous.
Proof: Let f: X→Y be any mapping. φ and Y are the only open sets of Y. Then inverse image in X under f are φ and
X which are open in X and hence rgβ open in X. Hence f is rgβ- continuous.
Theorem 3.6: The charactertistic function of a subset E of a topological space X is rgβ- continuous if and only if E is
both rgβ- open and rgβ- closed in X.
Proof: KE(X) =
1, if x∈E
0, if x∈X-E
Let E be both rgβ - open and rgβ- closed.
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Y. Palaniappan and V. Senthilkumaran / ON REGULAR GENERALISED β – CONTINUOUS FUNCTIONS/ IJMA- 4(10), Oct.-2013.
Let us prove KE is rgβ - continuous mapping of X into R, with usual topology.
Let G be any open subset of R
KE-1(G) =
E if 1 ∈G but 0∉ G
X-E
if 0∈ G but 1∉ G
X,
if both 0∈ G and 1∈G
φ,
if neither 0∈G nor 1∈G
E is both rgβ - open and rgβ- closed.
Hence X-E is both rgβ- open and rgβ- closed. X is both rgβ- open and rgβ-closed.
φ is both rgβ- open and rgβ- closed. Hence KE-1(G) is rgβ- open in X. So KE is rgβ- continuous.
Conversely, let KE be rgβ - continuous.
Let G be an open subset of R containing 1 but not 0. So KE-1(G) =E, rgβ – open
Let H be an open subset of R containing 0 but not 1.
KE (H) =X-E, rgβ - open. So E is rgβ - closed. Hence E is both rgβ - open and rgβ - closed.
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4.rgβ - OPEN and rgβ - CLOSED FUNCTION
Definition 4.1: Let X and Y be topological spaces. A function f: X→ Y is said to be rgβ - open if the image of every
rgβ - open set in X is open in Y. Similarly a function f: X→ Y is said to be rgβ - closed if the image of every rgβ closed set in X is closed in Y.
Example 4.1: Let (X, τ) be any topologogical space. Let (Y, σ) be such that Y= {p, q, r} and σ = {φ, {p}, {p, r}, Y}.
Let f: X→ Y be such that f(x) =p, for every x∈X. Let us see f is rgβ- continuous and rgβ- open but not rgβ- closed.
f-1(φ )=φ , f-1({p}) = X, f-1({p, r})=X, f-1(Y)=X. Hence f is rgβ - continuons.
Let G be rgβ - open in X. f(G)= φ , if G=φ. f(G)={p}, if G≠φ. Hence f is rgβ- open.
Let V be any nonempty rgβ - closed set in X. f(V)={p}, which is not closed in Y.
Hence f is not rgβ - closed.
Let g: X → Y be such that g(x)=q, for every x∈ X.
Let us see g is rgβ - continuous and rgβ - closed but not rgβ - open.
g-1(φ )=φ ,g-1({p})= φ, g-1({p,r})= φ, g-1(Y)=X. Hence g is rgβ - continuous.
g(X)={q}, which is not open in Y. So g is not rgβ - open
Let H be any rgβ - closed set in X. g(H)=φ, if H=φ . g(H)={q}, if H≠φ.
As φ and {q} are closed in Y, g is rβg - closed.
Theorem 4.1: Let arbitrary intersections of rgβ- closed sets of X be rgβ- closed in X.
Let X and Y be topological spaces. Then a mapping f: X → Y is rgβ - open if and only if f(rgβ -intA)⊂ int f (A), for
every A⊂ X.
Proof: Let f be rgβ - open and A⊂ X. rgβ-int A is rgβ - open in X. As f is rgβ-open, f(rgβ - int A) is open in Y.
f(rgβ - int A)⊂ f(A). f(rgβ - int A) is open in Y. So int f(rgβ - int A) = f(rgβ - int A).
f(rgβ - int A)⊂ f(A) implies f(rgβ - int A)⊂ int f( A).
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Y. Palaniappan and V. Senthilkumaran / ON REGULAR GENERALISED β – CONTINUOUS FUNCTIONS/ IJMA- 4(10), Oct.-2013.
Conversely, let f(rgβ - int A)⊂ int f(A), for every A ⊂X.
Let G be rgβ - open in X. Hence rgβ-int G=G. f(G)= f(rgβ - int G)⊂ int f(G).
But int f(G)⊂ f(G). So, f(G)=int f(G), which is open inY. Hence f is rgβ - open.
Theorem 4.2: Let X and Y be topological spaces. Let arbitrary intersection of rgβ - closed sets in X be rgβ - closed in
X. Then a mapping f : X→ Y is rgβ- closed if and only if cl f(A)⊂ f(rgβ-cl A), for every A⊂ X.
Proof: Let f be rgβ - closed and A⊂ X. rgβ - cl A is rgβ - closed in X. So, f(rgβ -clA) is closed in Y. cl f(rgβ - clA) =
f(rgβ -cl A). A⊂rgβ - cl A implies f(A)⊂ f(rgβ - cl A).
This implies cl f(A)⊂ cl f(rgβ - cl A).
Hence cl f (A)⊂ f(rgβ - cl A).
Let cl f(A)⊂ f(rgβ- cl A). Let H be rgβ- closed in X. So, rgβ- cl H = H.
Then f(rgβ- cl H) = f(H). Given that cl f(H) ⊂f(rgβ - cl H) =f (H). But f(H)⊂ cl f(H).
Hence f(H)= cl f(H), which is closed in Y. Therefore f is rgβ - closed.
Definition 4.2: Let X and Y be topological spaces. A mapping f: X→Y is said to be rgβ - homeomorphism if f is one to
one, onto, rgβ - continuous and rgβ - closed.
Theorem 4.3: Let X and Y be topological spaces. Let arbitrary intersections of rgβ - closed sets in X be rgβ - closed
in X. Let the mapping f: X→Y be one to one and onto. Then f is a rgβ homeomorphism if and only if
f(rgβ -cl A)= cl f( A), for every A ⊂X.
Proof: Let f be a rgβ homeomorphism. f is one to one and onto and rgβ- continuous. Let A⊂ X. Then by theorem 3.2,
f(rgβ - cl A)⊂ cl f(A). A⊂rgβ – cl A implies f(A)⊂ f(rgβ-clA). This implies cl f(A)⊂ cl f(rgβ -clA). f is rgβ -closed.
rgβ- cl A is rgβ - closed in X. f(rgβ - cl A) is closed in Y.
Hence cl f(rgβ- clA)=f(rgβ - cl A).
So, f(rgβ-cl A)=cl f(A).
Conversely, let f(rgβ - cl A)=cl f(A), for every A⊂ X. f(rgβ - cl A)⊂cl f(A). Therefore, f is rgβ - continuous, by
theorem 3.2. Let H be rgβ- closed in X. So, rgβ - cl H=H. f(rgβ -cl H)=f(H).
Hence f(H)=cl f(H),which is closed in Y. So, f is rgβ - closed map. f is one to one, onto, rgβ- continuous and rgβclosed.
Hence f is a rgβ - homeomorphism.
Remark4.1: In the definition of rgβ - homeomorphism, we can take f to be rgβ- open instead of f is rgβ - closed. Both
are equivalent.
Theorem 4.4: Let arbitrary intersections of rgβ- closed sets of X be rgβ- closed in X. If f is an one to one mapping of a
space X onto a space Y,then f is a rgβ- homeomorphism ,if f(rgβ- int A)=int f(A),for every A⊂ X.
Proof: Let f(rgβ - int A)=int f(A),for every A ⊂X. Hence f(rgβ - int A)⊂ int f(A),for every A⊂ X. So,by theorem 4.1,f
is rgβ - open.
Let B be open in Y. Let f-1(B)=A.f(A)=B. Int f(A)=int B= B. Hence f(rgβ - int A)=B.
Therefore, f-1(B)=rgβ -intA, which is rgβ - open in X. So, f is rgβ - continuous. This completes the proof.
Theorem 4.5: Let X and Y be topological spaces. Let arbitrary intersections of rgβ - closed sets in X be rgβ- closed in
X.A mapping f:X→Y is rgβ - continuous implies for x∈ X and for every open set V of Y containing f(x), there exists
an rgβ - open set U of X containing x such that f(U)⊂ V.
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Y. Palaniappan and V. Senthilkumaran / ON REGULAR GENERALISED β – CONTINUOUS FUNCTIONS/ IJMA- 4(10), Oct.-2013.
Proof: Let f be rgβ - continuous. Let x ∈X and V be an open set of Y containing f(x). f-1(V) is an rgβ- open set
containing x. Let f-1(V) =U. f(U)=V⊂ V.
Let the given condition hold. Let V be open in Y. Let x∈ f-1(V). So f(x)∈ V.
Therefore, there exists an rgβ - open set Ux containing x such that f(Ux)⊂ V or Ux ⊂f-1(V). So, f-1(V)=∪ Ux, where
x∈f-1(V), which is rgβ - open in X, since arbitrary intersection of rgβ- closed sets in X is rgβ - closed in X. So, f is rgβ
- continuous.
Remark 4.2: f is rgβ - continuous implies, for x∈ X and for every open set V containing f(x), there exists an rgβ- open
set U of X, containing x such that f(U)⊂ V. This condition is said to be f is rgβ - continuous at x∈ X .
Theorem 4.6: Let X be a topological space and let a singleton {a} be rgβ - open in X. Let Y be any topological space.
Then every function f: X →Y is rgβ - continuous at a.
Proof: Let V be open in Y containing f(a). a∈ f-1(V). That is, {a}⊂ f-1(V). So f({a}) ⊂V.
Therefore f is rgβ - continuous at a.
CONCLUSION
rgβ- continuous functions ideas can be extended to study rgβ-contra continuous functions and rgβ- semi continuous
functions
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Source of support: Nil, Conflict of interest: None Declared
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