Article Citation: Ramesh Bhat.
(2021). ON SUPERCONTINUOUS FUNCTIONS IN
TOPOLOGY. International Journal of Engineering Technologies and Management
Research, 8(4), 1-5. https://doi.org/10.29121/ijetmr.v8.i4.2021.903 Published Date: 12 April 2021 Keywords: Super Continuous
Functions T3 Spaces Neighbourhood Functions ECT The aim of this paper is to introduce and study new classes of continuous functions and its properties in topological spaces comparing with different types of continuous functions.
1. INTRODUCTIONIn this
paper we studed the basic concepts of
super-continuous and their basic results and some other useful results have
been studied. Super-continuous maps were
first introduced and investigated by B. M. Munshi and D. S. Bassan
[1] in 1982.
Later J. L. Reilly and M. K. Vamanamoorthi [2] continued the study of super-continuous
mappings and obtained many useful results in 1983. Super
continuous functions contained in the class of continuous functions. Munshi and
Bassan [1] defined the super-continuous map as follows:
A map f: X ® Y is said to be super-continuous at a point
xÎX if
for every neighbourhood M of f
(x) there is a neighbourhood N of x such that f ()0 Í M. This class is contained in the
class of continuous mappings.
Super-continuous mappings turn out to be the natural tool for studying
nearly compact space of Singal and Mathur [5], almost regular spaces of Singal and Arya [3] and almost completely regular spaces of Singal and Arya [4], M. K.
Signal and A.R. Singal [6] Almost continuous mapping and N. V. Velicko [7],
H-closed topological spaces Various properties of such mappings have been
discussed. 2.
PRILIMINARIES
Throughout
this dissertation work (X, t), (Y, m) and (Z,h) represent non-empty topological
spaces on which no separation axioms are assumed unless explicitly stated, and
they are simply written X, Y and Z respectively. For a subset A of (X, t), the closure of A, the interior
of A with respect to t are denoted by and A0
respectively. The complement of A is denoted by Ac. Now, we recall
some of the following definitions 3. SUPER CONTINUOUS FUNCTIONSDefinition 2.1 : A map f: X ® Y is said to be super-continuous
at a point xÎX if for every neighbourhood
M of f (x) there is a neighbourhood
N of x such that f ()0 Í M. Example 2.2 : Let X={a, b, c, d}, Y={1, 2, 3, 4}, Á1={X, , {a, b}, {a, b, d}} and Á2={ Y, , {1, 3}, {1, 2, 3}}.
Then (X, Á1) and (Y, Á2) are topological spaces. Let f : X
®
Y be a map defined as f (a) = f (b)
=1, f (c) = 2, f (d) = 3. f is continuous: 1)
f is
continuous at aÎX, then for an open set {1, 3}
containing f (a) =1, there exists an
open set {a, b} containing a such that f [{a, b}] = {1} Í {1, 3}. 2)
f
is continuous at bÎX, then for an open set {1, 3}
containing f (b) =1, there exists an
open set {a, b} containing b such that f [{a, b}] = {1} Í
{1, 3}. 3)
f
is continuous at cÎX, then for an open set {1, 2, 3}
in Y containing f (c) = 2,
there exists an open set X containing c such that f [ X ] = {1, 2, 3} Í {1, 2, 3}. 4)
f is continuous at dÎX, then for an open set {1, 2, 3}
in Y containing f (d) = 3, there exists
an open set X containing c such that f [ X ] = {1, 2, 3} Í {1, 2, 3}. Therefore, f is continuous as each point of X. f is not super - continuous
at x = a: For an
open set {1, 3} in Y containing f (a) = 1, there is a neighbourhood {a, b} of a such that f []o = f ( X ) = X Ë {1, 3}. Therefore,
f is not super continuous at x = a. Definition 2.3 : A mapping f: X ® Y is said to be
super-continuous [denoted by SC] if it
is super-continuous at each point of X. Definition 2.4 : A set G is said to be d-open if for each xÎG, there exists a regular open set
H such that xÎ H Í G, or equivalently G can be
expressed as arbitrary union of regular open sets. A set G
is d-closed
if and only if its complement is d-open. Theorem.2.5: Let f: X ® Y be a map. Then the following are equivalent. 1)
f
if super continuous. 2)
Inverse
image of every open subset of Y is a d-open subset of X. 3)
Inverse
image of every closed subset of Y is a d-closed subset of X. 4)
For
each point x of X and for each open neighbourhood M of f(x), there is a d-open neighbourhood
N of x such that f(N) Ì M. Proof: (a) Þ (b): Suppose (a) holds. Let U be any open subset of Y and let xÎ f -1(U). Then f (x) ÎU.
Since f is super continuous, from (a), there exists an open set V in X
such that xÎV and f () Ì U. Thus xÎ Ì f -1(U). Therefore f-1(U) is expressible as
an arbitrary union of regularly open sets.
Hence f -1(U) is d-open. (b) Þ (c): Let U be a closed set in
Y. Then Y – U is open set of Y. Then from (b), f -1(Y – U) is d-open subset of X. Therefore f -1(Y – U) = X – f -1(U)
is d-open
subset of X. Hence f -1(U) is
d-closed
set of X. (c) Þ (d): Let M
be an open set in Y containing f (x),
that is, f (x) Î M.
Since Y – M is closed, by (c), f -1(Y – M) is d-closed subset of X. Therefore f -1(M) is d-open. Also
x Î f-1(M). Let N = f-1(M). Then N is a d-open neighbourhood
of x such that f (N) Í M. (d) Þ (a): Let for each xÎX and for each neighbourhood
M of f (x) there is a neighbourhood
N of f(x), so
N is d-open
neighbourhood N of x such that f (N) Í M, from (d). Then f () Í M. So f is super continuous. Hence the proof. Definition 2.6 : A space X is said to be semi-regular if for
each point x of the space and each open set U containing x there is a open set V such that xÎVÌ Ì U. Theorem 2.7: Let f : X ® Y be a continuous mapping of a
semi-regular space X into Y. Then f is
super - continuous. Proof: Let xÎX and let G be an open set
containing f (x). Since f is continuous, f -1(G) is open in X. Since X is semi-regular space, there is an
open subset M of x such that xÎM Ì Í f -1(G). Therefore f ( x )Î f ( M ) Í f
() Í G.
That is, f () Í G.
Hence f is super continuous. Remark 2.8 : Every open set in a T3 -space can
be written as the union of regular open sets. Corollary 2.9 : Let X be a T3 topological space
and let f : X ® Y be a continuous, then f is
super-continuous. Proof: Proof follows from every regular space ( or
T3 ) space is semi regular. Theorem 2.10: Let X and Y are topological spaces. Then a mapping f : X ® Y is super-continuous if and only
if the inverse image under f of every member of a base ( sub base ) for Y is d-open in X. Proof: Let f be super continuous and B be a subbase
for Y. Since each member of B* is open
in Y, it follows that from the Theorem
that f -1(Y) is d-open for every Y Î B. Conversely,
let f -1(Y) be a d-open in X for every Y Î B and let H be any open set in
Y. Let b be a family of all finite
intersections of members of B* so B is a base for Y. If BÎb, then there exists n1 , n2 ………..nn ( n is finite ) in B* such that
B=n1
Ç n2 Ç……….. Ç nn. Then f -1(B) = f -1(n1) Ç f -1(n2 ) Ç……….. Ç f -1(nn). By
hypothesis, each f -1(ni) is d-open in X and therefore f -1(B) is also d-open in X, since b is a base for X. H = È{B: BÎC Ì b }. Then f -1( H ) = f -1[ È{B:BÎC Ì b }] = È{f -1( B ): BÎC } which is d-open in X, since f -1( B ) is d-open in X. Thus f -1(H) is d-open in X for every open set H in
Y and therefore f is super continuous. Definition 2.11 : A point is said to be a d-adherent point of a set P in a
space X if equivalently, every regular open set containing x has non-empty
intersection with P.the interior of every closed neighbourhood of the point x intersects P or Definition 2.12 : The set (P)d of all d-adherent point of a set P is
called the d-closure of the set P. Theorem 2.13 : A mapping f from a space X into another
space Y is super -continuous if and only if another space Y is super
-continuous if and only if f(A)d Ì for every subset A of
X. Proof: Let f be a super continuous. Since is closed in Y, then
by f -1() is d-closed in X, since f is super
-continuous. Now f (A) Ì implies,
A Ì f -1( ). Therefore (A)d
Ì [f -1( )]d = f -1( ). Therefore f (A)d Ì f [f -1( )] Ì . So
f (A)d Ì . Conversely,
let f (A)d Ì
for every subset A of
X. Let F be any closed set in Y so that =F. Now f -1
() is a subset of X implies that f [f -1( F ) ]d Ì Ì= F implies [f -1( F)]d Ì f -1(F). Therefore [f -1(F)]d = f -1(F), so f -1(F) is d-closed set in X. Hence
f is super -continuous. Theorem 2.14 : A mapping f from a space X into another
space Y is super-continuous if and only if [f -1(B)]d Ì
f -1 ( ) for every B Ì Y. Proof: Let f be super-continuous. Since is closed in Y and
since f is super-continuous, f-1() is d-closed set in X. Therefore f -1() = [f -1()]d.
Now B Ì Ì [()]d implies [f -1(B)]d Ì [f -1( )]d implies [f -1(B)]d Ì
f -1( ). Conversely,
let the condition hold and let F be any closed set in Y. Therefore =F. Now [f -1(F)]d Ì f-1(F) = f -1(F). But f -1(F) Ì Ì [f -1(F)]d .
Hence f -1(F) = [f -1(F)]d . Therefore f -1(F) is
d-closed
set in X. Hence f is super-continuous. Definition 2.15: A point x is called a d-adherent point of a filter base
Ŧ if and only if xÎ Ç{[F]d : F Î Ŧ}. Definition 2.16: A filter base is said to be d-coverage of a point x (written as Ŧd ® x ) if every regular open set
containing x contains F Î Ŧ. Theorem 2.17 : Let f : X ® Y be a mapping. Then f is super-continuous at xÎX if and only if the filter base f
(U
(x)) ® f (x), where U Theorem 2.18 : A mapping
f: X ® Y is super continuous on X if and
only if f (U) ® f (x) for each xÎX and each filter base U
that d-converges
to x. Proof:
Assume that f
is super continuous on X and let U x. Let W be a neighbourhood
of f (x). Then xÎ f -1(W) and f -1(W)
is d-open since f is super- continuous. Therefore xÎH such that f (H) Ì W where H is
regular open and H Ì f -1(W). Therefore,
there exits a uÎ U such that uÎH. Therefore f (u) Ì f (H) Ì
W. Therefore f ( U ) ® f (x). Conversely,
let W be any open subset of Y containing f (x).
Let B be any subset of X. We have to prove that f
([ B ]d ) Ì . Let b Î[ B ]d.
Let U be a filter base on B with U-d-covering
to b so that f (U ) ® f (b). Since f ( U ) is a filter base on f ( B ), therefore f (b)Î [ f (B)] Ì Therefore f is super continuous. Theorem 2.19 ( Restricting the range ) : If
f : X ® Y is super-continuous
and f (X) is taken with the subspace topology then f : X ® f (X) is super-continuous. Proof : Let f
: X ® Y be
super-continuous. Let U be an open
subset of Y then f -1(U)
is d-open in X.
Now f -1(
UÇ f (
X ) ) = f -1( U ) Ç f -1[ f ( X ) )] = f -1( U ) ÇX = f -1( U ) is d-open. Therefore f : X ® Y is super-continuous. Theorem 2.20 ( Expanding the range ) : Let f : X ® Y is super-continuous. If Z is a space having Y as a subspace then the function h : X ® Z obtained by expanding the range
of f
is super-continuous. Proof: We have to show that
h : X ® Z is super-continuous. As Z has Y as a subspace, h is the composite
of the map f : X ® Y which is super-continuous and the
inclusion map g : Y ® Z which is continuous. Thus, h is
super-continuous. Definition 2.21 : A mapping f : X ® Y is said to be almost open if
the image of every regularly open subset of X is an open subset of Y. A mapping f : X ® Y is said to be almost closed if
the image of every regularly closed subset of X is a closed subset of Y. Definition 2.22: A mapping f : X ® Y is said to be almost continuous
at a point xÎX if for every neighbourhood
M of f (x) there is a neighbourhood
N of x such that f (N) Ì . Theorem 2.23 : If f is an almost open, super-continuous
mapping X onto Y and if g is a mapping of Y into Z
then gof is super-continuous if and only if g is
continuous. Proof : Let g be continuous. Let G be an open set in Z. Then g - 1 (G) is open in Y. Also f is super-continuous, f - 1( g – 1(
G ) ) is d-open in X.
Therefore f -1( g
– 1( G ) ) = ( g° f )
– 1(G) is d-open in X.
Hence gof
is super-continuous. Conversely,
let gof be
super-continuous. Let G be an open
subset of Z. Therefore (gof)–1 (G) is d-open subset of X since gof is super-continuous.
That is f -1( g – 1( G ) ) is a d-open subset in X. Also,
f is almost open and onto, f [ f
-1 ( g – 1( G ) ) ] = g – 1( G ) is open
in Y. Hence g is continuous. Theorem 2.24 : Let X, Y, Z be topological spaces and the
mapping f : X ® Y be almost continuous and g
: Y ® Z be super-continuous. Then the composition gof
: X ® Z is
continuous. But if
f : X ® Y is almost continuous and g° f : X ® Z is continuous, then g : Y ® Z need not be super-continuous. Example 2.25 : Let (R, Á1) be the topological space where Á1 is the topology consisting of f, R and complements of countable
subsets of R. Let X = { a, b } and Á2 = {X, f, {a} }. Let f
: R ® X be
defined as follows: Let Y =
{1, 2} and Á3 = {Y, f, {2}}. Let g : X ®Y be defined as g (a) = 2, g (b) = 1.
Then f : R ® X is almost continuous and gof : R ® Y is continuous but g : X ® Y is not super-continuous. SOURCES OF FUNDINGThis research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors. CONFLICT OF INTERESTThe author have declared that no competing interests exist. ACKNOWLEDGMENTNone. REFERENCES
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