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 ()
## 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 A ## 3. SUPER CONTINUOUS FUNCTIONS
Let f : X
®
Y be a map defined as f (a) = f (b)
=1, f (c) = 2, f (d) = 3.
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.
For an
open set {1, 3} in Y containing f (a) = 1, there is a neighbourhood {a, b} of a such that f [] Therefore,
f is not super continuous at x = a.
A set G
is d-closed
if and only if its complement is d-open.
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.
Let U be any open subset of Y and let xÎ f (b) Þ (c): Let U be a closed set in
Y. Then Y – U is open set of Y. Then from (b), f (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 (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.
Conversely,
let f
Conversely,
let f (A)
Conversely,
let the condition hold and let F be any closed set in Y. Therefore =F. Now [f
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 ] Let b Î[ B ]
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.
Conversely,
let gof be
super-continuous. Let G be an open
subset of Z. Therefore (gof)
But if
f : X ® Y is almost continuous and g° f : X ® Z is continuous, then g : Y ® Z need not be super-continuous.
Let Y =
{1, 2} and Á ## 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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K. Singal and S. P. Arya, On almost regular spaces. Math. Vesni
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