Download ON DECOMPOSITION OF REGULAR GENERALIZED CONTINUITY

Asian Journal of Current Engineering and Maths1: 5 Sep –Oct(2012) 290 – 295.
Contents lists available at www.innovativejournal.in
ASIAN JOURNAL OF CURRENT ENGINEERING AND MATHS
Journal homepage:http://www.innovativejournal.in/index.php/ajcem
ON DECOMPOSITION OF REGULAR GENERALIZED CONTINUITY IN SUPRA
TOPOLOGICAL SPACES
S. Dayana Mary*, N. Nagaveni
Department of Mathematics, Coimbatore Institute of Technology, Coimbatore – 641 014, India
ARTICLE INFO
ABSTRACT
Corresponding Author
S. Dayana Mary
Department of Mathematics,
Coimbatore Institute of
Technology, Coimbatore – 641 014,
India
[email protected]
Key Words: S-RGLC sets, S-RGL
continuous and S-RGL-irresolute.
The aim of this paper is to introduce and investigate a new
decomposition called supra regular generalized locally closed sets and a new
decomposition of continuous maps called supra regular generalized locally
continuous maps. Furthermore, we obtain several properties of the new
notions.
INTRODUCTION
The concept of generalized closed sets in
topological space was obtained by Levine [8]. Palaniappan
and Rao [10] defined a set called regular generalized
closed. Using the concept of locally closed set due to
Bourbaki [2], Ganster and Reilly [5] introduced and studied
different notions of generalized continuity and gave a
decomposition of continuity. Arockiarani et. al. [1]
introduced regular generalized locally closed sets and
investigate the classes of regular generalized locally
continuous functions in topological spaces.
The notion of supra topological spaces was
introduced by Mashhour et. al. [9] in 1983. Kamaraj et. at
[7] introduced supra regular generalized closed set and
discussed some of their properties. Dayana Mary and
Nagaveni [3] introduced the concept of supra generalized
locally closed sets and study its basic properties. Also they
introduced the concepts of supra generalized locally
continuous functions and investigated.
In this paper we introduce the concept of
supra generalized locally closed sets and study its basic
properties. Also we introduce the concepts of supra
generalized locally continuous functions and investigate
some of the basic properties for this class of functions.
PRELIMINARIES
Throughout this paper, (X, τ), (Y, σ) and (Z, η) (or
simply, X, Y and Z) represent topological space on which no
separation axioms are assumed, unless explicitly stated.
For a subset A if (X, τ), the closure and interior of A in X are
denoted by cl(A) and int (A), respectively. Let P(X) be the
power set of X. The complement of A is denoted by X-A or
Ac .
Now we recall some definitions and results which
are useful in the sequel.
Definition: 2.1 [9,12]
Let X be a non-empty set. The subfamily µ ⊆ P(X)
is said to a supra topology on X if X∈ µ and µ is closed
under arbitrary unions. The pair (X, µ) is called a supra
topological space.
©2012, AJCEM, All Right Reserved.
The elements of µ are said to be supra open in (X,
µ). Complement of supra open sets are called supra closed
sets.
Definition: 2.2 [12]
Let A be a subset (X, µ). Then
(i)
The supra closure of a set A is, denoted by 𝑐𝑙 𝜇 (𝐴),
defined as 𝑐𝑙 𝜇 (𝐴) = ∩ {B : B is a supra closed and A ⊆ B}.
(ii)
The supra interior of a set A is, denoted by 𝑖𝑛𝑡𝜇 (𝐴),
defined as 𝑖𝑛𝑡𝜇 (𝐴) = ∪ {B : B is a supra open and B ⊆ A}.
Definition: 2.3 [9]
Let (X, τ) be a topological space and µ be a supra
topology on X. We call µ is a supra topology associated with
τ if τ ⊆µ
Definition: 2.4 [4]
Let (X, τ) and (Y, σ) be two topological spaces and
τ ⊆ µ. A function f: (X, τ) → (y,σ) is called supra
continuous, if the inverse image of each open set of Y is a
supra open set in X.
Definition: 2.5 [6]
Let (X, τ) and (Y, σ) be two topological spaces and
µ and λ be supra topologies associated with τ and σ
respectively. A function f : (X, τ) → (y,σ) is said to be supra
irresolute , if f-1 (A) is supra open set of X for every supra
open set A in Y.
Definition: 2.6 [11]
Let A and B be subsets of X. Then A and B are said
to be supra separated, if A ∩ 𝑐𝑙 𝜇 (𝐵) = B ∩ 𝑐𝑙 𝜇 (𝐴) = φ.
Definition: 2.7 [11]
Let (X, µ) be a supra topological space. A subset A
of X is called supra g-closed if 𝑐𝑙 𝜇 (𝐴) ⊆U whenever A ⊆ U
and U is supra open in X.
The complement of supra g-closed set is called
supra g-open.
Definition: 2.8 [11]
Let A be a subset (X, µ). Then
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(i)
The supra g-closure of a set A is, denoted by
𝜇
𝜇
𝑐𝑙𝑔 (𝐴), defined as 𝑐𝑙𝑔 (𝐴) = ∩ {B : B is a supra g-closed and
A ⊆ B}.
(ii)
The supra g-interior of a set A is, denoted by
𝜇
𝜇
𝑖𝑛𝑡𝑔 (𝐴), defined as 𝑖𝑛𝑡𝑔 (𝐴) = ∪ {B : B is a supra g-open and
B ⊆ A}.
Remark: 2.9 [11]
(i)
Intersection of two supra g-closed sets is
generally not a supra g-closed set.
(ii)
Union of two supra g-open sets is generally
not a supra g-open set.
Definition: 2.10 [7]
Let (X, µ) be a supra topological space. A subset A
of X is called supra rg-closed if 𝑐𝑙 𝜇 (𝐴) ⊆U whenever A ⊆ U
and U is supra regular open in X.
The complement of supra rg-closed set is called
supra rg-open.
Remark: 2.11 [7]
We have the following implications for properties
of subsets,
Supra
regular
closed→supra
closed→supra
gclosed→supra rg-closed.
None of the above implications is reversible.
Remark: 2.12[7]
(i)
Intersection of two supra rg-closed sets is
generally not a supra rg-closed set.
(ii)
Union of two supra rg-closed sets is
generally not a supra rg-closed set.
Definition: 2.13 [13]
A subset S of a supra topological space X is called
supra locally closed ( briefly 𝐿𝐶𝜇 –closed set) if S =A ∩ B,
where A is supra open and B is supra closed in X.
Definition: 2.14 [13]
A function from a supra topological space X into a
supra topological space Y is called 𝐿𝐶𝜇 –continuous if f1(V)∈𝐿𝐶𝜇 (𝑋) for each supra open set V in Y.
SUPRA REGULAR GENERALIZED LOCALLY CLOSED SETS
In this section, we introduce the notions of supra
regular generalized locally closed sets and discuss some of
their properties.
Definition: 3.1
Let (X, µ) be a supra topological space. A subset A
of (X, µ) is called supra regular generalized locally closed
set (briefly supra rg-locally closed set), if A=U ∩ V, where U
is supra rg-open in (X, µ) and V is supra rg-closed in (X, µ).
The collection of all supra generalized locally
closed sets of X will be denoted by S-RGLC(X).
Remark: 3.2
Every supra rg-closed set (resp. supra rg-open set)
is S-RGLC.
Definition: 3.3
For a subset A of supra topological space (X, µ), A ∈
S-RGLC*(X,µ), if there exist a supra rg-open set U and a
supra closed set V of (X,µ), respectively such that A=U ∩ V.
Definition: 3.4
For a subset A of (X, µ), A ∈ S-RGLC**(X, µ), if there
exist a supra open set U and a supra rg-closed set V of (X,
µ), respectively such that A=U ∩ V.
Theorem: 3.5
Let A be a subset of (X, µ).
(i)
If A ∈ S-RGLC*(X, µ) or A ∈ S-RGLC**(X, µ),
then A is S-RGLC.
If A is supra locally closed, then A ∈ SRGLC*(X, µ) and A ∈ S-RGLC**(X, µ), however
not conversely.
Proof: The proofs are obvious from remark 2.11,
definitions and the following example.
Example: 3.6
Let X = {a, b, c, d} and µ = { φ, X, {a}, {a, b}, {b, c}, {a,
b, c}}.Then the supra locally closed sets are φ, X, {a}, {b}, {c},
{d}, {a, b}, {b, c}, {c, d} {a, d}, {a, b, c}, {b, c, d}} and SRGLC*(X, µ) = S-RGLC**(X, µ) = S-RGLC*(X, µ) = P(X),
because the supra regular generalized closed sets of (X, µ)
are P(X)-{a}. From this, 𝐿𝐶𝜇 is a proper subset of S-RGLC(X,
µ).
Theorem: 3.7
For a subset A of (X, µ), the following are
equivalent:
(i)
A ∈ S-RGLC*(X, µ).
(ii)
A = U ∩ 𝑐𝑙 𝜇 (𝐴), for some supra rg-open set
U.
(iii)
𝑐𝑙 𝜇 (𝐴) – A is supra rg-closed.
(iv)
A∪ [X- 𝑐𝑙 𝜇 (𝐴)], is supra rg-open.
Proof: (i) ⇒ (ii):
Given A ∈ S-RGLC(X, µ)
Then there exist a supra rg-open subset U and a
supra closed subset V such that A=U ∩V. Since A ⊂ U and A
⊂ 𝑐𝑙 𝜇 (𝐴), A ⊂ U ∩ 𝑐𝑙 𝜇 (𝐴).
Conversely, we have 𝑐𝑙 𝜇 (𝐴)⊂ V and hence A = U ∩ V ⊃
U ∩ 𝑐𝑙 𝜇 (𝐴). Therefore A = U ∩ 𝑐𝑙 𝜇 (𝐴)
(ii) ⇒ (i):
Let A = U∩ 𝑐𝑙 𝜇 (𝐴), for some supra
𝜇
rg-open set U. Clearly, 𝑐𝑙 (𝐴) is supra rg-closed and hence
A = U ∩ 𝑐𝑙 𝜇 (𝐴)∈ S − RGLC(X, µ).
(ii) ⇒ (iii):
Let A = U ∩ 𝑐𝑙 𝜇 (𝐴), for some
supra rg-open set U.
Then A ∈ S-RGLC (X, µ). This implies U is supra rg-open and
𝑐𝑙 𝜇 (𝐴) is supra rg-closed. Therefore, 𝑐𝑙 𝜇 (𝐴) − A is supra rgclosed.
(iii) ⇒ (ii):
Let U= X – [𝑐𝑙 𝜇 (𝐴)- A]. By (iii), U is
supra rg-open in X. Then A = U ∩ 𝑐𝑙 𝜇 (𝐴) holds.
(iii) ⇒ (iv):
Let Q =𝑐𝑙 𝜇 (𝐴)–A be supra rgclosed. Then X-Q = X - [𝑐𝑙 𝜇 (𝐴)– A] =A ∪ [(X- 𝑐𝑙 𝜇 (𝐴)]. Since
X-Q is supra rg-open, A ∪ [X- 𝑐𝑙 𝜇 (𝐴)] is supra rg-open.
(vi) ⇒ (iii):
Let U = A ∪ [(X- 𝑐𝑙 𝜇 (𝐴)]. Since X –
U is supra rg-closed and X - U = 𝑐𝑙 𝜇 (𝐴) - A holds, 𝑐𝑙 𝜇 (𝐴) - A
is supra rg-closed.
Definition: 3.8
Let A be a subset (X, µ). Then
(i)
The supra rg-closure of a set A is, denoted by
𝜇
𝜇
𝑐𝑙𝑟𝑔 (𝐴), defined as 𝑐𝑙𝑟𝑔 (𝐴) = ∩ {B : B is a supra rg-closed
and A ⊆ B}.
(ii)
The supra g-interior of a set A is, denoted by
𝜇
𝜇
𝑖𝑛𝑡𝑟𝑔 (𝐴), defined as 𝑖𝑛𝑡𝑟𝑔 (𝐴) = ∪ {B : B is a supra rg-open
and B ⊆ A}.
Theorem: 3.9
For a subset A of (X, µ), the following are
equivalent:
(i)
A ∈ S-RGLC(X, µ).
𝜇
(ii)
A = U ∩ 𝑐𝑙𝑟𝑔 (𝐴), for some supra rg-open
set U.
𝜇
(iii)
𝑐𝑙𝑟𝑔 (𝐴) - A is supra rg-closed.
𝜇
(iv)
A∪ [X - 𝑐𝑙𝑟𝑔 (𝐴)] is supra rg-open.
Proof: (i) ⇒ (ii):
Given A ∈ S-RGLC(X, µ).
Then there exist a supra rg-open subset U and a
supra rg-closed subset V such that A=U ∩V. Since A ⊂ U
𝜇
𝜇
and A ⊂ 𝑐𝑙𝑟𝑔 (𝐴), A ⊂ U ∩ 𝑐𝑙𝑟𝑔 (𝐴).
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𝜇
Conversely by theorem 2.9 (iv), 𝑐𝑙𝑟𝑔 (𝐴) ⊂ V and hence A =
𝜇
𝜇
U ∩ V ⊃ U ∩ 𝑐𝑙𝑟𝑔 (𝐴). Therefore, A = U ∩ 𝑐𝑙𝑟𝑔 (𝐴).
𝜇
(ii) ⇒ (i):
Let A = U∩ 𝑐𝑙𝑟𝑔 (𝐴), for some
𝜇
supra rg-open set U. Then we have, 𝑐𝑙𝑟𝑔 (𝐴) is supra rg𝜇
closed and hence A = U ∩ 𝑐𝑙𝑟𝑔 (𝐴) ∈ S-RGLC(X, µ).
𝜇
(ii) ⇒ (iii):
Let A = U ∩ 𝑐𝑙𝑟𝑔 (𝐴), for some
supra rg-open set U.
Then A ∈ S-RGLC (X, µ). This implies U is supra rg-open and
𝜇
𝜇
𝑐𝑙𝑟𝑔 (𝐴) is supra rg-closed. Therefore, 𝑐𝑙𝑟𝑔 (𝐴) − A is supra
rg-closed.
𝜇
(iii) ⇒ (ii):
Let U= X – [𝑐𝑙𝑟𝑔 (𝐴) - A]. By (iii), U
𝜇
is supra rg-open in X. Then A = U ∩ 𝑐𝑙𝑟𝑔 (𝐴) holds.
𝜇
(iii) ⇒ (iv):
Let Q = 𝑐𝑙𝑟𝑔 (𝐴)–A be supra rg𝜇
𝜇
closed. Then X-Q = X - [𝑐𝑙𝑟𝑔 (𝐴) – A] =A ∪ [(X- 𝑐𝑙𝑟𝑔 (𝐴)].
𝜇
Since X-Q is supra rg-open, A ∪ [X- 𝑐𝑙𝑟𝑔 (𝐴)] is supra rgopen.
𝜇
(vi) ⇒ (iii):
Let U = A ∪ [(X- 𝑐𝑙𝑟𝑔 (𝐴)]. Since X
𝜇
– U is supra rg-closed and X - U = 𝑐𝑙𝑟𝑔 (𝐴) - A is supra rgclosed.
Theorem: 3.10
For a subset A of (X, µ), if A ∈ S-RGLC**(X, µ), then
there exist an supra open set G such that A = G ∩ 𝑐𝑙 𝜇 (𝐴).
Proof: Let A ∈ S-RGLC**(X, µ). Then A=G ∩ V, where G is
supra open set and V is supra rg-closed set. Then A =G ∩ V
⇒ A ⊂ G. Obviously, A ⊂ 𝑐𝑙 𝜇 (𝐴). Thus A ⊂ G ∩ 𝑐𝑙 𝜇 (𝐴) ----(1)
Also we have 𝑐𝑙 𝜇 (𝐴) ⊂ V. This implies A = G ∩ V ⊃ G ∩
𝑐𝑙 𝜇 (𝐴).
Therefore, A ⊃ G ∩ 𝑐𝑙 𝜇 (𝐴) ----- (2). From (1) and (2), we get
A = G ∩ 𝑐𝑙 𝜇 (𝐴).
Theorem: 3.11
For a subset A of (X, µ), if A ∈ S-RGLC**(X, µ), then
𝜇
there exist an supra open set G such that A = G ∩ 𝑐𝑙𝑟𝑔 (𝐴).
Proof: Let A ∈ S-RGLC**(X, µ). By the definition (3.4), A=G
∩ V, where G is supra open set and V is supra rg-closed set.
𝜇
𝜇
Then A =G ∩ V ⇒ A ⊂ G. Since A ⊂ 𝑐𝑙𝑟𝑔 (𝐴), A ⊂ G ∩ 𝑐𝑙𝑟𝑔 (𝐴)
----- (1)
𝜇
𝜇
Also we have 𝑐𝑙𝑟𝑔 (𝐴) ⊂ V. Then A = G ∩ V ⊃ G ∩ 𝑐𝑙𝑟𝑔 (𝐴).
𝜇
Hence A ⊃ G ∩ 𝑐𝑙𝑟𝑔 (𝐴) ----- (2). From (1) and (2), we get A
𝜇
= G ∩ 𝑐𝑙𝑟𝑔 (𝐴).
Theorem: 3.12
Let A be a subset of (X, µ). If A ∈ S-RGLC*(X, µ),
𝜇
𝜇
then 𝑐𝑙𝑟𝑔 (𝐴) - A supra rg-closed and A ∪ [(X- 𝑐𝑙𝑟𝑔 (𝐴)] is
supra rg-open.
Proof :
The proof is obvious from the definitions
and results.
Remark: 3.13
The converse of the above theorem need not be
true as seen the following example.
Example: 3.14
Let X = {a, b, c, d} and µ = {φ, X, {a}, {b, c}, {a, b, c},
{a, b, d}}. Then P(X) – [{a}, {b}and{b, c}] are supra rg-closed
sets in X and S-RGLC*(X, µ) = P(X) – {a, c, d}. If A = {a, c, d},
𝜇
𝜇
then 𝑐𝑙𝑟𝑔 (𝐴) – A is supra rg-closed and A ∪ [(X - 𝑐𝑙𝑟𝑔 (𝐴)] is
supra rg-open but A ∉ S-RGLC*( X, µ).
Theorem: 3.15
Let P ∈ S-RGLC*(X, µ) and Q ∈ S-RGLC*(X, µ). If P
and Q are supra separated, then P ∪ Q ∉ S-RGLC*( X, µ).
Proof :
Let P ∈ S-RGLC*(X, µ) and Q ∈ S-RGLC*(X,
µ). Suppose let us assume P ∪ Q ∈ S-RGLC*(X, µ). By
theorem 3.7, there exist supra rg-open sets U and V of such
that P = U ∩ 𝑐𝑙 𝜇 (𝑃) and Q = V∩ 𝑐𝑙 𝜇 (𝑄). Put G = U ∩ [X 𝑐𝑙 𝜇 (𝑄))] and H = V ∩ [X - 𝑐𝑙 𝜇 (𝑃)]. Then P= G ∩ 𝑐𝑙 𝜇 (𝑃) and
Q =H∩ 𝑐𝑙 𝜇 (𝑄). Also G ∩ 𝑐𝑙 𝜇 (𝑄) =φ and H ∩ 𝑐𝑙 𝜇 (𝑃) =φ.
Hence it follows that G and H are supra rg-open sets of (X,
µ). Therefore P ∪ Q = [G ∩ 𝑐𝑙 𝜇 (𝑃) ] ∪ [H ∩ 𝑐𝑙 𝜇 (𝑄) ] ∈ SRGLC*(X, µ), by our assumption. Then (G ∪ H) ∩ [𝑐𝑙 𝜇 (𝑃) ∪
𝑐𝑙 𝜇 (𝑄)] ∈ S-RGLC*(X, µ). This implies (G ∪ H) is supra rgopen, but it is contradiction to the Remark 2.8(ii). Hence
our assumption is wrong. Thus P ∪ Q ∉ S-RGLC*(X, µ).
Remark: 3.16
The following example is one of the example of the
above theorem.
Example: 3.17
Let X = {a, b, c, d} and µ = {φ, X, {a}, {b, c}, {a, b, c},
{a, b, d}}. Let A= {c} and B = {a, d}. Then A and B are supra
separated, because A ∩ 𝑐𝑙 𝜇 (𝐵) = B ∩ 𝑐𝑙 𝜇 (𝐴) = φ. Then A ∪ B
= {a, c, d}∉ S-RGLC*(X, µ).
Theorem: 3.18
Let A be a subset of (X, µ). If A is supra locally
closed, then A∈S-RGLC.
Proof :
By the definitions and results, the proof is
immediate.
Remark: 3.19
The converse of the above theorem need not be
true as seen from the following example.
Example: 3.20
Let X = {a, b, c, d} and µ = {φ, X, {a}, {a, b}, {b, c}, {a, b,
c}}. Then the supra locally closed sets are {φ, X, {a}, {b}, {c},
{d}, {a, b}, {a, d}, {b, c}, {c, d}, {a, b, c}, {b, c, d}} and S-RGLC
sets are P(X).
Theorem: 3.21
1. Every S-GLC set is S-RGLC.
2. Every S-GLC* set is S-RGLC*.
3. Every S-GLC** set is S-RGLC**.
4. Every S-GLC* set is S-RGLC.
Proof :
By the definitions and results, the proof is
immediate.
Remark: 3.22
The converses of none of the above are need not be
true as seen from the following examples.
Example: 3.23
Let X = {a, b, c, d} and µ = {φ, X, {a}, {b, c}, {a, b, c},
{a, b, d}}.
Then (1) {a, c, d} is S-RGLC but not S-RGLC*.
(2) {a, c, d} is S-RGLC** but not S-RGLC*.
Thus we conclude that S-RGLC* and S-RGLC** are
independent.
Example: 3.24
Let X = {a, b, c}and µ = {φ, X, {a, b}, {a, c}}.
(i)
Then {b, c} is S-RGLC but not S-GLC*.
(ii)
Then {a} is S-RGLC** but not S-GLC**.
(iii)
Then {b, c} is S-RGLC* but not S-GLC*.
Theorem: 3.25
Let A and Z be subset of (X, µ) and let A ⊆ Z. If Z is
supra rg – open in (X, µ) and A ∈ S-RGLC* (Z, µ/Z), then A ∈
S-RGLC* (X, µ).
Proof: Suppose A ∈ S-RGLC*, then there exist a supra rg𝜇
𝜇
open set H of (Z, µ/Z) such that A = H ∩ 𝑐𝑙𝑍 (𝐴). But 𝑐𝑙𝑍 (𝐴)
= Z ∩ 𝑐𝑙 𝜇 (𝐴). Therefore A = H ∩ Z ∩ 𝑐𝑙 𝜇 (𝐴), where H ∩ Z is
supra rg-open. Thus A ∈ S-RGLC* (X, µ).
Remark: 3.26
The following example shows that the assumption
that Z is supra rg-open cannot be removed from the above
theorem.
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Example: 3.27
Let X = {a, b, c, d}, µ = {φ, X, {a}, {b, c,}, {a, b, c}, {a, b,
d}}. Let v be the collection of all supra rg-open sets of (X, µ).
Then V = { φ, X, {a}, {b}, {c}, {d}, {a, b}, {c, d}, {a, c}, {b, c}, {b,
d}, {a, b, c,}, {a, b, d}}. Put Z = A = {a, c, d}. then Z is not supra
rg-open and A∈ S-RGLC* (Z, µ/Z). However A∉ S-RGLC* (X,
µ), since S-RGLC* (X, µ) = P(X) – {a, c, d}.
Theorem: 3.28
Let {𝑍𝑖 : i∈ I}be a finite supra rg-closed cover of (X,
µ) and let A be a subset of (X, µ). If A ∩ 𝑍𝑖 ∈ S-RGLC** (𝑍𝑖 , µ
𝑍𝑖 ) for every i∈ I, then A∈ S-RGLC** (X, µ).
Proof: For every i∈I, there exist an supra open set 𝑣𝑖 ∈µ
and a supra rg-closed set 𝐺𝑖 of such that A ∩ 𝑍𝑖 = 𝑉𝑖 ∩ 𝐺𝑖 ∩
𝑍𝑖 = 𝑉𝑖 ∩ (𝐺𝑖 ∩ 𝑍𝑖 ). Then A = ∪ { A ∩ 𝑍𝑖 / i∈ I} = ∪ {{ 𝑉𝑖 / i∈ I
} ∩ [ ∪{ 𝑍𝑖 ∩ 𝐹𝑖 / i∈I}]}. This shows that A∈ S-RGLC** (X, µ).
Theorem: 3.29
Let (X, µ) and (Y, λ) be the supra topological
spaces.
(1) If M ∈ S-RGLC(X, µ) and N ∈ S-RGLC(Y, λ),
then M × N∈S-RGLC(X × Y, µ × λ).
(2) If M ∈ S-RGLC*(X, µ) and N ∈ S-RGLC*(Y, λ),
then M × N∈S-RGLC*(X × Y, µ × λ).
(3) If M ∈ S-RGLC**(X, µ) and N ∈ S-RGLC**(Y, λ),
then M × N ∈ S-RGLC**(X ×Y, µ × λ).
Proof: Let M ∈ S-RGLC(X, µ) and N ∈ S-RGLC(Y, λ). Then
there exist a supra rg-open sets P and P’ of (X, µ) and (Y, λ)
and supra rg-closed sets Q and Q’ of (X, µ) and (Y, λ)
respectively such that M = P ∩ Q and N = P’ ∩ Q’. Then M ×
N = (P × P’) ∩ (Q ×Q’) holds. Hence M × N∈S-RGLC(X × Y, µ
× λ).
Similarly, the proofs of (2) and (3) follow from the
definitions.
SUPRA REGULAR GENERALIZED LOCALLY CONTINUOUS
FUNCTIONS
In this section the notion of supra regular
generalized locally continuous functions, supra regular
generalized locally irresolute functions are introduced and
their basic properties are investigated and obtained.
Definition: 4.1
Let (X, τ) and (Y, σ) be two topological spaces and
τ ⊆ µ. A function f: (X, τ) → (Y,σ) is called S-RGLcontinuous (resp., S-RGL* - continuous, resp., S-RGL** continuous), if f-1(A) ∈ S-RGLC (X,µ), (resp., f-1 (A) ∈ SRGLC* (X,µ), resp., f-1 (A)∈ S-RGLC** (X,µ)) for each A ∈ σ.
Definition: 4.2
Let (X, τ) and (Y, σ) be two topological spaces and
µ and λ be a supra topologies associated with τ and σ
respectively. A function f : (X, τ) → (Y,σ) is said to be SRGL – irresolute (resp., S-RGL* - irresolute, resp., S-RGL** irresolute) if f-1 (A) ∈ S-RGLC (X,µ), (resp., f-1 (A) ∈ S-RGLC*
(X,µ), resp., f-1 (A) ∈ S-RGLC**(X,µ)) for each A ∈ S-RGLC
(Y, λ) (resp., A∈ S-RGLC* (Y, λ), resp., A ∈ S-RGLC**(Y, λ)).
Theorem: 4.3
Let (X, τ) and (Y,σ) be two topological spaces and µ
be a supra topology associated with τ. Let f: (x, τ) → (y,σ)
be a function. If f is S-RGL*– continuous or S-RGL** continuous, then it is S-RGL - continuous.
Proof: The proof is trivial from the definitions.
Theorem: 4.4
Let (X, τ) and (Y, σ) be two topological spaces and
µ and λ be a supra topologies associated with τ and σ
respectively. Let f: (X, µ) → (Y,σ) be a function. If f is SRGL– irresolute (respectively S-RGL* – irresolute,
respectively S-RGL** – irresolute), then it is S-RGL–
continuous. (respectively S-RGL* – continuous, respectively
S-RGL** – continuous).
Proof :
By the definitions the proof is immediate.
Remark: 4.5
Converse of theorem 4.3 need not be true as seen
from the following example.
Example: 4.6
Let X = Y = {a, b, c }with τ = {φ, X, {a}, {b, c}, {a, b,
c}}, σ = {{φ, Y, {a}, {a, c}, {a, c, d}} and µ = {φ, X, {a}, {b, c},
{a, b, c}, {a, b, d}}. Let f : (X,µ) → (Y,σ) be the identity map.
S-RGLC (X, µ) = P(X) = S-RGLC** (X,µ) and S-RGLC**(X,µ) =
P(X) - {a, c, d}.
Here f is not S-RGL*– continuous, but it is S-RGLcontinuous. Also f is not S-RGL*-continuous, but it is and SRGL** - continuous.
Remark: 4.7
The following example provides a function which
is S –RGL* - continuous function but not S – RGL* irresolute function.
Example: 4.8
Let X = Y = {a, b, c, d }with τ = {φ, X, {a}, {b, c}, {a, b,
c}}, σ = {{φ, Y, {a, b}, {a, b, d}}, µ= {φ, X, {a}, {b, c}, {a, b, c},
{a, b, d}} and λ = {φ, X, {a, b}, {a, b, d}, {b, c, d}}. Let f : (X,µ)
→ (Y,σ) be the identity map.
Here f is not S-RGL*–
irresolute, but it is S-RGL*- continuous.
Theorem: 4.9
Let (X, τ) and (Y,σ) be two topological spaces and µ
be a supra topology associated with τ. Let f: (x, τ) → (y,σ)
be a function. If f S-GL-continuous, then it is S-RGL continuous.
Proof: Consider f is S-GL-continuous. By the definition, f1(V)∈S-GLC(X, µ), for each V ϵ σ. Since every S-GLC set is SRGLC, f-1 (V) ∈ S-RGLC(X, µ), for each V∈(Y,σ). Therefore f
is S-RGL-continuous.
Remark: 4.10
The converse of the above theorem need not be
true as seen from the following example.
Example: 4.11
Let X = Y = {a, b, c, d} with τ = {φ, X, {a, c}, {a, b, c}},
σ = {{φ, Y, {b, c}} and µ= {φ, X, {a, c}, {b, c}, {a, b, c}, {a, b, d},
{b, c, d}}. Define f: (X, µ) → (Y, σ) as f(a) = b, f(b) = c, f(c) = d
and f(d) = a. Then f is S-RGL– continuous, but it is not S-GLcontinuous, because
= {a, b} ∉ S-GLC (X, µ).
𝑓 −1 ({𝑏, 𝑐})
Remark: 4.12
From the above results and examples we have the
following implications.
Theorem: 4.13
If g: X → Y is S-GL- continuous and h: Y → Z is
supra continuous, then hog : X → Z is S-GL – continuous.
Proof: Let g: X → Y is S-GL – continuous and h : Y → Z is
supra continuous. By the definitions, g-1(V) ∈ S-RGLC (X) , V
∈Y and h-1 (W) ∈Y , W∈ Z. Let W ∈ Z. Then (hog)-1(W) = (g-1
h-1)(W)= g-1(h-1(W))=g-1(V), for V ∈ Y. Thus (hog)-1(W) = g-
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∈ S-RGLC (X), W ∈ Z. Therefore, hog is S-GLcontinuous.
Theorem: 4.14
If g: X → Y is S-GL – irresolute and h: Y → Z is SRGLC-continuous, then h o g : X → Z is S-GL – continuous.
Proof: Let g : X → Y is S-GL – irresolute and h : Y → Z is
S-GL-continuous. By the definitions, g-1(V) ∈ S-RGLC (X) ,
for V ∈S-RGLC (Y) and h-1(W) ∈S-RGLC (Y) , for W∈ Z. Let
W ∈ Z. Then (hog)-1(W) = (g-1 h-1) (W) = g-1( h-1(W)) = g1(V), for V ∈ S-RGLC(Y). (hog)-1(W) = g-1(V) ∈ S-RGLC (X),
W ∈ Z. Hence hog is S-GL- continuous.
Theorem: 4.15
If g : X → Y and h : Y → Z are S-GL – irresolute ,
then h o g : X → Z is also S-GL – irresolute.
Proof: By the hypothesis and the definitions, we have g1(V) ∈ S-RGLC(X), for V ∈S-RGLC (Y) and h-1(W) ∈S-RGLC
(Y) , for W∈ S-RGLC(Z). Let W ∈S-RGLC( Z). Then (hog)1(W) = (g-1 h-1) (W) = g-1( h-1(W)) = g-1(V), for V ∈ SRGLC(Y). (hog)-1(W) = g-1(V) ∈ S-RGLC (X), W ∈ S-RGLC(
Z). Thus hog is S-GL – irresolute.
1(V)
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