 Open Access
 Total Downloads : 241
 Authors : K.Bala Deepa Arasi, S.Navaneetha Krishnan, S.Pious Missier
 Paper ID : IJERTV5IS020169
 Volume & Issue : Volume 05, Issue 02 (February 2016)
 DOI : http://dx.doi.org/10.17577/IJERTV5IS020169
 Published (First Online): 11022016
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
on Contra sbg – Continuous functions in Topological Spaces
K. Bala Deepa Arasi1,
1 Assistant Professor of Mathematics, A.P.C.Mahalaxmi College for Women, Thoothukudi, TN, India

Navaneetha Krishnan2 and S. Pious Missier3
2,3 Associate Professor of Mathematics,

Chidambaram College, Thoothukudi, TN, India
Abstract – In this paper a new class of functions called contra sbcontinuous function is introduced and its properties are studied. Some characterization and several properties concerning Contra sbcontinuity are obtained. Also, Contra sbirresolute function and Perfectly Contra sbirresolute function are introduced.
Keywords: sbclosed sets, sbcontinuous, Contra sb continuous, Contra sbirresolute.
AMS Mathematics Subject Classification : 54C08, 54C10.

INTRODUCTION
In 1996, Dontchev[7] introduced and investigated a new notion of continuity called contra continuity. Follwing this, many authors introduced various types of new generalizations of contra continuity called as contra – continuity, contra semicontinuity[3], contra b continuity[12], contra sgcontinuity[5], contra gs continuity[5], contra gbcontinuity[16], contra g*b continuity[16], contra bcontinuity[15] and so on and they investigated their properties. In 2015, we introduced sb closed sets[3] in Topological spaces.
In this paper, we introduce and investigate some of the properties of contra sbcontinuous, contra sb irresolute functions and we obtain some of its characterization.

PRELIMINARIES

Throughout this paper (X, ) (or simply X) represents topological spaces on which no separation axioms are assumed unless otherwise mentioned. For a subset A of (X,), Cl(A), Int(A) and Ac denote the closure of A, interior of A and the complement of A respectively. We are giving some definitions.
The intersection of all semiclosed (resp.closed, bclosed, regularclosed) sets of X containing A is called the semi closure (resp.closure, bclosure, regular closure) of A and is denoted by sCl(A) (resp.Cl(A), bCl(A), rCl(A)). The family of all semiopen (resp. open, bopen, regular open) subsets of a space X is denoted by SO(X) (resp. O(X), bO(X), rO(X)).
Definition 2.2: A subset A of a topological space (X,) is called a

a sgclosed set[5] if sCl(A) U whenever A
U and U is semiopen in X.

a gsclosed set[5] if sCl(A) U whenever A
U and U is open in X.

a gbclosed set[16] if bCl(A) U whenever A
U and U is open in X.

a g*bclosed set[16] if bCl(A) U whenever A
U and U is gopen in X.

a bclosed set[15] if bCl(A) U whenever A
U and U is open in X.

a sbclosed set[3] if sCl(A) U whenever A

U and U is bopen in X.
The complement of a sgclosed (resp. gsclosed, gbclosed, g*bclosed and bclosed) set is called sgopen (resp. gs open, gbopen, g*bopen and bopen) set.
Definition 2.3: A space (X,) is called a

Tsb space[3] if every sbclosed set in X is closed.
sb
sb

T space[3] if every sbclosed set in X is –
Definition 2.1: A subset A of a topological space (X,) is called

a semiopen set[5] if A Cl(Int(A)).

an open set[8] if A Int(Cl(Int(A))).

a bopen set[1] if A Cl(Int(A)) Int(Cl(A)).

a regular open[14] set if A = Int(Cl(A)).
The complement of a semiopen (resp.open, bopen, regularopen) set is called semiclosed (resp.closed, b closed, regularclosed) set.
closed.
Definition 2.4: A function f: (X,) (Y,) is called a

sb continuous map[4] if 1(V) is sbclosed in (X,) for every closed set V in (Y,).

sb irresolute map[4] if 1(V) is sbclosed in
(X,) for every closed set V in (Y,).
Definition 2.5: A function f: (X,) (Y,) is called a

Contra continuous map[7] if 1(V) is closed in (X,) for every open set V in (Y,).

Contra semicontinuous map[6] if 1(V) is semi closed in (X,) for every open st V in (Y,).

Contra continuous map[8] if 1(V) is closed
in (X,) for every open set V in (Y,).

Contra bcontinuous map[15] if 1(V) is b closed in (X,) for every open set V in (Y,).

Contra sgcontinuous map[5] if 1(V) is sg
closed in (X,) for every open set V in (Y,).

Contra gscontinuous map[5] if 1(V) is gs closed in (X,) for every open set V in (Y,).

Contra gbcontinuous map[16] if 1(V) is gb
closed in (X,) for every open set V in (Y,).

Contra g*bcontinuous map[16] if 1(V) is g*b closed in (X,) for every open set V in (Y,).

Contra bcontinuous map[15] if 1(V) is b
closed in (X,) for every open set V in (Y,).
Definition 2.6:[15] A space (X, ) is said to be locally indiscrete if every open subset of X is closed in X.
Definition 2.7:[5] A topological space (X,) is said to be Urysohn space if for each pair of distinct points x and y in X, there exists two open sets U and V in X such that x U, y V and Cl(U) Cl(V) = .
Definition 2.8:[5] For a map f : X Y, the subset {(x,f(x))
: x X} X Ã— Y is called the graph of f and is denoted by G(f).

CONTRA sbCONTINUOUS FUNCTIONS
We introduce the following definition.
Definition 3.1: A function f: (X,) (Y,) is called Contra sbcontinuous if 1(V) is sbclosed in (X,) for every open set V in (Y,).
Example 3.2: Let X = Y = {a,b,c} with topologies = {X,,{a},{b},{a,b} and = {Y,,{a},{a,b}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = c, f(c)
= b
sb C(X) = {X,,{a},{b},{c},{a,c},{b,c}}
Here, the inverse image of open sets {a} and {a,b} in Y are {a} and {a,c} respectively which are sbclosed sets in X. Hence, f is contra sbcontinuous.
Theorem 3.3:

Every contra continuous function is contra sb continuous function.

Every contra continuous function is contra sb continuous function.

Proof:

Let V be any open set in (Y,). Since f is contra continuous, 1(V) is closed in (X,). By Proposition 3.4 in [3], 1(V) is sbclosed in (X,). Hence, f is contra sbcontinuous function.

Let V be any open set in (Y,). Since f is contra – continuous, 1(V) is closed in (X,). By
Proposition 3.7 in [3], 1(V) is sbclosed in (X,). Hence, f is contra sbcontinuous function.
The following examples show that the converse of the above proposition need not be true.
Example 3.4:

Let X = Y = {a,b,c} with topologies
= {X, , {a},{b},{a,b}} and = {Y, , {a},{b,c}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(X) = {X,,{a},{b},{c},{a,c},{b,c}} C(X) = {X,,{c},{a,c},{b,c}}
Here the inverse image of an open set {a} in (Y,) is {a} which is sbclosed but not closed in (X,). Hence, f is contra sbcontinuous but not contra continuous.

Let X = Y = {a,b,c} with topologies
= {X, , {a},{c},{a,c},{b,c}} and = {Y, , {b}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(X) = {X,,{a},{b},{a,b},{b,c}}
– C(X) = {X,,{a},{a,b},{b,c}}
Here the invrse image of an open set {b} in (Y,) is {b} which is sbclosed but not closed in (X,). Hence, f is contra sbcontinuous but not contra continuous.
Theorem 3.5:

Every contra sbcontinuous function is contra b continuous function

Every contra sbcontinuous function is contra sg continuous function

Every contra sbcontinuous function is contra gs continuous function

Every contra sbcontinuous function is contra gb continuous function

Every contra sbcontinuous function is contra g*bcontinuous function

Every contra sbcontinuous function is contra b continuous function.
Proof:

Let V be any open set in (Y,). Since f is contra sbcontinuous, 1(V) is sbclosed in (X,). By Proposition 3.11 in [3], 1(V) is bclosed in (X,). Hence, f is contra bcontinuous function.

Let V be any open set in (Y,). Since f is contra sbcontinuous, 1(V) is sbclosed in (X,). By Proposition 3.13 in [3], 1(V) is sgclosed in (X,). Hence, f is contra sgcontinuous function.

Let V be any open set in (Y,). Since f is contra sbcontinuous, 1(V) is sbclosed in (X,). By Proposition 3.15 in [3], 1(V) is gsclosed in (X,). Hence, f is contra gscontinuous function.

Let V be any open set in (Y,). Since f is contra sbcontinuous, 1(V) is sbclosed in (X,). By
Proposition 3.17 in [3], 1(V) is gbclosed in (X,). Hence, f is contra gbcontinuous function.

Let V be any open set in (Y,). Since f is contra sbcontinuous, 1(V) is sbclosed in (X,). By Proposition 3.21in [3], 1(V) is g*bclosed in (X,). Hence, f is contra g*bcontinuous function.

Let V be any open set in (Y,). Since f is contra sbcontinuous, 1(V) is sbclosed in (X,). By Proposition 3.23 in [3], 1(V) is bclosed in (X,). Hence, f is contra bcontinuous function.
The converse of the above theorem need not be true as shown in the following example.
Example 3.6:

Let X = Y = {a,b,c} with topologies
= {X, , {a,b},{c}} and = {Y, , {a},{a,b}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(X) = {X,,{c},{a,b}}
b – C(X) = {X,,{a},{b},{c},{a,b},{a,c},{b,c}}
Here the inverse image of an open set {a} in (Y,) is {a} which is bclosed but not sb closed in (X,). Hence, f is contra bcontinuous but not contra sb continuous.

Let X = Y = {a,b,c} with topologies
= {X, , {a},{b,c}} and = {Y, , {a},{a,b},{a,c}} Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(X) = {X,,{a},{b,c}}
sg – C(X) = {X,,{a},{b},{c},{a,b},{a,c},{b,c}}
Here the inverse image of the open sets {a,b} and
{a,c} in (Y,) are {a,b} and {a,c} which are sgclosed but not sb closed in (X,). Hence, f is contra sgcontinuous but not contra sbcontinuous function.

Let X = Y = {a,b,c,d} with topologies
= {X, , {a},{a,c},{a,b,d}} and = {Y, ,
{b},{a,b},{b,c,d}}
Define a function f: (X,) (Y,) by f(a) = d, f(b) = b, f(c)
= c, f(d) = a.
sbC(X)={X,,{b},{c},{d},{b,c},{c,d},{b,d}, {b,c,d}}
gsC(X)={X,,{b},{c},{d},{b,c},{c,d},{b,d},
{a,b,c},{a,c,d},{b,c,d}}
Here the inverse image of an open set {b,c,d} in (Y,) is {a,b,c} which is gsclosed set but not sb closed set in (X,). Hence, f is contra gscontinuous but not contra sbcontinuous function.

Let X = Y = {a,b,c,d} with topologies
= {X, , {a},{a,c},{a,b,d}} and = {Y, ,
{b},{a,b},{b,c,d}}
Define a function f: (X,) (Y,) by f(a) = c, f(b) = b, f(c)
= a, f(d) = d. sbC(X)={X,,{b},{c},{d},{b,c},{c,d},{b,d}, {b,c,d}}
gbC(X)={X,,{b},{c},{d},{b,c},{c,d},{b,d},
{a,b,c},{a,c,d},{b,c,d},{a,b,d}}
Here the inverse image of an open set {b,c,d} in (Y,) is {a,b,d} which is gbclosed but not sb closed in (X,). Hence, f is contra gbcontinuous but not contra sb continuous.

Let X = Y = {a,b,c} with topologies
= {X, ,{a,c}} and = {Y, , {a},{a,b},{a,c}}
Define a function f: (X,) (Y,) by f(a) = b, f(b) = a, f(c)
= c.
sb C(X) = {X,,{b}}
gb – C(X) = {X,,{a},{b},{c},{a,b},{b,c}}
Here the inverse image of the open sets {a,b} and
{a,c} in (Y,) are {a,b} and {b,c} which are g*bclosed but not sb closed in (X,). Hence, f is contra g*b continuous but not contra sbcontinuous.

Let X = Y = {a,b,c,d} with topologies
= {X, , {b},{a,b},{b,c,d}} and = {Y, ,
{a},{a,c},{a,b,d}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c, f(d) = d. sbC(X)={X,,{a},{c},{d},{a,c},{a,d},{c,d}, {a,c,d}}
bC(X)={X,,{a},{c},{d},{c,d},{a,d},{a,c},
{a,b,c},{a,c,d},{a,b,d}}
Here the inverse image of an open set {a,b,,d} in (Y,) is {a,b,d} which is bclosed but not sb closed in (X,). Hence, f is contra bcontinuous but not contra sb continuous.
Theorem 3.7: If f: (X,) (Y,) is contra semicontinuous function if and only if f is contra sbcontinuous function. Proof: Let V be any open set in (Y,). Since f is contra semicontinuous, 1(V) is semiclosed in (X,). Since from Proposition 3.6 in [3], 1(V) is sbclosed in (X,). Hence, f is contra sbcontinuous.
Conversely, Let V be any open set in (Y,). Since f is contra sbcontinuous, 1(V) is sbclosed in (X,). Since from Proposition3.6 in [3], 1(V) is semiclosed in (X,). Hence, f is contra semicontinuous.
Remark 3.8: The following diagram shows the relationships of contra sbcontinuoous function with other known existing functions.
1. Contra sbcontinuous 2. Contra continuous
3. Contra continuous 4. Contra bcontinuous
5. Contra sgcontinuous 6. Contra gscontinuous

Contra gbcontinuous 8. Contra g*bcontinuous 9.Contra bcontinuous
Theorem 3.9: The following are equivalent for a function f: (X,) (Y,),

f is contra sbcontinuous function

For every closed subset F of Y, 1(F) is sb open in X

For each x X and each closed subset F of Y with f(x) F there exists a sbopen set U of X with x U, f(U) F.

Proof:

(b):
Let F be any closed set in Y. Then Fc is an open set in Y. Since f is contra sbcontinuous, 1() is sb closed set in X. Then [1()]c is sbclosed set in X. Therefore 1() is sbopen in X.

(a):
Let F be an open set in Y. Then is closed set in
Y. By (b), 1() is sbopen set in X. Then [1()]c is sbopen set in X. So 1() is sbclosed set in X.
Therefore, f is contra sbcontinuous function

(c):
Let F be any closed subset of Y and let f(x) F where x X. Then by (b), 1(F) is sbopen in X. Also, x
1(F). Take U = 1(F). Then U is a sbopen set containing x and f(U) F.

(b):
Let F be any closed subset of Y. If x 1(F) then f(x) F. By (c), there exists a sbopen set Ux of X with x Ux such that f(Ux) F. Then 1(F) = { Ux : x
1(F)}. Hence, 1(F) is sb open in X.
Theorem 3.10: If X is Tsb space, then for the function f: (X,) (Y,), the following statements are equivalent.

f is contra continuous function

f is contra sbcontinuous function.

Proof:

(ii):
Let V be any open set in Y. Since f is contra continuous, 1(V) is closed in X. From [3] proposition 3.4, 1(V) is sbclosed in X. Therefore, f is contra sb continuous.

(i):
Let V be any open set in Y. Since f is contra sb continuous, 1(V) is sbclosed in X. Also, Since X is Tsb space, 1(V) is closed in X. Therefore, f is contra continuous.
sb
sb
Theorem 3.11: If a function f: (X,) (,) is contra sb continuous and X is T space, then f is contra – continuous.
T 1
T 1
Proof: Let V be any open set in Y. Since f is contra sb continuous, 1(V) is sbclosed in X. Also, Since X is sb space, (V) is closed in X. Therefore, f is
contra continuous.
Theorem 3.12: Let f: (X,) (Y,) be a function then the following statements are equivalent,

f is sbcontinuous function

For each point x X and each open set of Y with f(x) V, there exist a sbopen set U of X such that x U, f(U) V.

Proof:

(ii):
Let f(x) V, then x 1(V). Since f is sb continuous, 1(V) is sbopen in X. Let U = 1(V), then x U and f(U) V.

(i):
Let V be any open set in Y and x 1(V). Then f(x) V. From (ii), there exists a sbopen set Ux of X such that x Ux 1(V) and 1(V) = { Ux}. Then 1(V) is sbopen in X. Hence, f is sbcontinuous.
Theorem 3.13: If a function f: (X,) (Y,) is contra sb continuous and Y is regular, then f is sb continuous.
Proof: Let x X and V be an open set in Y with f(x) V. Since Y is regular, there exists an open set W in Y such that f(x) W and Cl(W) V. Since f is contra sb continuous and Cl(W) is a closed subset of Y with f(x) Cl(W). By theorem 3.9, there exist a sbopen set U of X with x U such that f(U) Cl(W). That is, f(U) V. By theorem 3.12, f is sbcontinuous.
Definintion 3.14: A function f: (X,) (Y,) is said to be strongly sbcontinuous if 1(V) is closed in (X,) for every sbclosed set V in (Y,).
Example 3.15: Let X = Y = {a,b,c} with topologies
= {X, , {a},{b},{a,b},{a,c}} and = {Y, ,
{a},{a,b},{a,c}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(Y) = {Y,,{b},{c},{b,c}}
Here the inverse image of sbclosed sets {b},{c} and {b,c} in (Y,) are {b},{c} and {b,c} respectively which are closed in (X,). Hence, f is strongly sb continuous.
Definition 3.16: A function f: (X,) (Y,) is said to be perfectly sbcontinuous if 1(V) is clopen in (X,) for every sbclosed set V in (Y,).
Example 3.17 : Let X = Y = {a,b,c} with topologies = {X, , {a},{b},{a,b},{a,c}} and = {Y, ,{a,c}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(Y) = {Y,,{b}}
Here the inverse image of sbclosed set {b} in (Y,) is {b} which is both open and closed in (X,). Hence, f is perfectly sbcontinuous.
Definition 3.18: A topological space (X,) is said to be sb Hausdorff (or sbT2 space) if for each pair of distinct points x and y in X, there exists sbopen subsets U and V of X containing x and y respectively such that U V = .
Example 3.19: Let X = {a,b,c} with a topology = {X, ,
{a},{b},{a,b}}
sb O(X) = {X,,{a},{b},{a,b},{a,c},{b,c}} Clearly (X,) is sbHausdorff space.
Theorem 3.20: Let f: (X,) (Y,) be surjective, closed and contra sbcontinuous. If X is Tsb space, then Y is locally indiscrete.
Proof: Let V be any open set in (Y,). Since f is contra sb continuous, 1(V) is sbclosed in (X,). Also, Since X is Tsb space, 1(V) is closed in (X,). By hypothesis, f is closed and surjective, (1(V)) = V is closed in (Y,). Hence, Y is locally indiscrete.
Theorem 3.21: If a function f: (X,) (Y,) is continuous and (X,) is locally indiscrete space, then f is contra sb continuous.
Proof: Let V be any open set in (Y,). Since f is continuous, 1(V) is open in (X,). Since X is locally indiscrete, 1(V) is closed in (X,). By Proposition 3.4 in [3], 1(V) is sbclosed in (X,). Hence, f is contra sb continuous.
Theorem 3.22: If a function f: (X,) (Y,) is contra sb continuous, injective and Y is Urysohn space, then the topological space X is sbHausdorff.
Proof: Let 1 and 2 be two distinct points of X. Suppose
1= f(1 ) and 2= f(2 ). Since f is injective, 1
2 then 1 2. Since Y is Urysohn, there exist open sets
1 and 2 containing 1 and 2 respectively in Y such that Cl(1 ) Cl(2 ) = . Since f is contra sbcontinuous. By theorem 3.12, there exists sbopen sets 1 and 2 containing 1 and 2 respectively in X such that f(1 ) Cl(1 ) and f(2) Cl(2). Since Cl(1 ) Cl(2 ) = , 1 2 = . Hence, X is sbHausdorff space.
Theorem 3.23: Let f: (X,) (Y,) be a function and g: X
XÃ— be a graph function of f defined by g(x) = (x,f(x)) for every x X. If g is contra sbcontinuous, then f is contra sbcontinuous.
Proof: Let V be closed subset of Y. Then XÃ— V is a closed subset of XÃ— . Since g is contra sbcontinuous, 1 (X Ã—
)is sbopen subset of X. Also, 1 (X Ã— ) = 1(V) which is sbopen subset of X. Hence, f is contra sb continuous.
Definition 3.24: A space X is said to be locally sb indiscrete if every sbopen set of X is closed in X.
Example 3.25: Let X = {a,b,c} with topology = {X, ,{c},{a,b}}
sb O(X) = {X,,{c},{a,b}}
Here every sbopen set in X is closed in X. Hence, X is locally sbindiscrete space.
Theorem 3.26: If f: (X,) (Y,) is contra sbcontinuous with X as locally sbindiscrete, then f is continuous.
Proof: Let V be any open set in (Y,). Then 1(V) is sb
closed in X. Since X is locally sbindiscrete space, 1(V) is open in X. Thus, f is continuous.
4. CONTRA sbIRRESOLUTE FUNCTIONS Definition 4.1: A function f: (X,) (Y,) is called contra sbirresolute, if 1(V) is sbclosed in (X,) for every sbopen set V in (Y,).
Example 4.2: Let X = Y = {a,b,c} with topologies = {X, , {a},{b},{a,b}} and = {Y, , {b}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = c, f(c)
= b.
sb C(X) = {X,,{a},{b},{c},{a,c},{b,c}}
sb O(Y) = {Y,,{b},{b,c},{a,b}}
Here the inverse image of sbopen set {b},{b,c} and {a,b} in (Y,) are {c},{b,c} and {a,c} respectively which are sbclosed set in (X,). Hence, f is contra sbirresolute function.
Remark 4.3: The following example shows that the concepts of sbirresolute function and contra sbirresolute are independent of each other.
Example 4.4:

Let X = Y = {a,b,c,d} with topologies
= {X, , {a},{a,c},{a,b,d}} and = {Y, ,
{b},{a,b},{b,c,d}}
Define a function f: (X,) (Y,) by f(a) = b, f(b) = a, f(c)
= c, f(d) =d. sbC(X)={X,,{b},{c},{d},{b,c},{c,d},{b,d}, {b,c,d}}
sbO(Y)={Y,,{a},{c},{d},{a,c},{a,d},{c,d}, {a,c,d}}
sbC(Y)={Y,,{b},{a,b},{b,c},{b,d},{a,b,c},
{a,b,d},{b,c,d}}
Clearly f is contra sbirresolute but not sb closed set {b}, {a,b}, {b,c}, {b,d}, {a,b,c}, {a,b,d} and
{b,c,d} in Y are {a}, {a,b}, {a,c}, {a,d}, {a,b,c}, {a,b,d} and {a,c,d} respectively which are not sbclosed in (X,).

Let X = Y = {a,b,c} with topologies

= {X, , {a},{b},{a,b},{a,c}} and = {Y, , {a},{a,b}} Define a function f: (X,) (Y,) by f(a) = b, f(b) = a, f(c)
= c.
sb C(X) = {X,,{b},{c},{a,c},{b,c}}
sb O(Y) = {Y,,{a},{a,b},{a,c}}
sb C(Y) = { Y,,{b},{c},{b,c}}
Here, the inverse image of sbopen sets {a},{a,b} in (Y,) are {a},{a,b} which are not sbclosed set in (X,). Hence, the f is sbirresolute but not contra sbirresolute.
Theorem 4.5: Every contra sbirresolute function is contra sbcontinuous.
Proof: Let V be any open set in (Y,). By proposition 3.4 in [3], V is sbopen in Y, Since f is contra sb irresolute, V is sbclosed in (X,). Hence, f is contra sb continuous.
The converse of the above theorem need not be true as shown in the following example.
Example 4.6: Let X = Y = {a,b,c} with topologies = {X, , {a},{c},{a,c},{b,c}} and = {Y, , {a}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(X) = {X,,{a},{b},{a,b},{b,c}}
sb OY) = {Y,,{a},{a,b},{a,c}}
Here, the inverse image of sbopen set {a,c} in (Y,) are {a,c} which is not sbclosed set in (X,). Hence, the f is contra sbcontinuous but not contra sbirresolute.
Remark 4.7: The following example shows that the concepts of sbcontinuous and contra sbcontinuous are independent of each other.
Example 4.8:

Let X = Y = {a,b,c,d} with topologies
= {X, , {a},{a,c},{a,b,d}} and = {Y, ,
{b},{a,b},{b,c,d}}
Define a function f: (X,) (Y,) by f(a) = b, f(b) = a, f(c)
= d, f(d) = c. sbC(X)={X,,{b},{c},{d},{b,c},{c,d},{b,d},
{b,c,d}}
Since the inverse image of open sets {b},{a,b} and
{b,c,d} in (Y,) are {a},{a,b} and {a,c,d} respectively which are not sbclosed in (X,), f is not contra sb continuous. Since the inverse image of closed sets
{a},{c,d} and {a,c,d} in (Y,) are {b},{c,d} and {b,c,d} respectively which are sbclosed in (X,), f is sb continuous. Hence, f is sbcontinuous but not contra sb continuous.

Let X = Y = {a,b,c} with topologies = {X, ,{b}} and = {Y, ,{a,c}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb C(X) = {X,,{a},{c},{a,c}}
Since the inverse image of an open set {a,c} in Y is {a,c} which is sbclosed in X, f is contra sb continuous. Also, since the inverse image of a closed set
{b} in Y is {b} which is not sbclosed in X, f is not sb continuous. Hence, f is contra sbcontinuous but not sb continuous.

PERFECTLY CONTRA sbIRRESOLUTE FUNCTION
Definition 5.1: A function f: (X,) (Y,) is called perfectly contra sbirresolute function if 1(V) is sb clopen in (X,) for every sbopen set V in (Y,).
Example 5.2: Let X = Y = {a,b,c} with topologies = {X, , {a},{b,c}} and = {Y, , {a,b},{c}}
Define a function f: (X,) (Y,) by f(a) = c, f(b) = a, f(c)
= b.
sb O(X) = {X,,{a},{b,c}}
sb O(Y) = {Y,,{c},{a,b}}
Since the inverse images of all sbopen sets in (Y,) are sbclopen set in (X,), f is perfectly contra sb irresolute function.
Theroem 5.3:

Every perfectly contra sbirresolute map is contra sbirresolute map.

Every perfectly contra sbirresolute map is sb irresolute map.
Proof:

and (2) directly follows from the definitions 2.4, 4.1 and 5.1.

The converse of the above theorem need not be true as shown in the following example.
Example 5.4:

Let X = Y = {a,b,c} with topologies = {X, , {a},{b},{a,b}} and = {Y, , {b}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = c, f(c)
= b.
sb O(X) = {X,,{a},{b},{a,b},{a,c},{b,c}}
sb O(Y) = {Y,,{b},{a,b},{b,c}}
Since the inverse image of sbopen set {b} in (Y,) is {c} which is sbclosed set in (X,) but not sb open set in X, f is contra sb irresoulute but not perfectly contra sbirresolute function.

Let X = Y = {a,b,c} with topologies
= {X, , {a},{b},{a,b},{a,c}} and = {Y, ,{a},{a,b},{a,c}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c.
sb O(X) = {X,,{a},{b},{a,b},{a,c}}
sb O(Y) = {Y,,{a},{a,b},{a,c}}
Since the inverse image of sbopen sets {a} and
{a,b} in (Y,) are {a} and {a,b} respectively which are sbopen set in (X,) but not sbclosed set in (X,), f is sbirresolute but not perfectly contra sbirresolute function.
Remark 5.5: From the above discussions and known results, we have the following diagram.
In this diagram, A B means A implies B but not conversely. A means A and B are independent of each other.


COMPOSITION OF TWO MAPS
The following example shows that the composition of two contra sbcontinuous function need not be contra sb continuous.
Example 6.1: Let X = Y = Z = {a,b,c} with topologies
= {X, , {a},{c},{a,c},{b,c}}, = {Y, , {b}} and =
{Z, ,{a,c}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = b, f(c)
= c and g: (Y,) (Z,) by g(a) = a, g(b) = b, g(c) = c.
sb C(X) = {X,,{a},{b},{a,b},{b,c}}
sb C(Y) = {Y,,{a},{c},{a,c}}
Clearly f and g are contra sbcontinuous. But their composition is not contra sbcontinuous, since (gf) –
1 of an open set {a,c} in (Z,) is {a,c} which is not sb closed in (X,). Hence, gf is not contra sb continuous.
Theorem 6.2: The composition of two strongly sb continuous function is strongly sbcontinuous function.
Proof: Let f: (X,) (Y,) and g: (Y,) (Z,) be strongly sbcontinuous functions. Let V be sbclosed set in (Z,). Since g is strongly sbcontinuous, 1(V) is closed in (Y,). By Propositon 3.4 in [3], 1(V) is sb closed in (Y,). Since f is strongly sbcontinuous,
1(1()) = (g f)1(V) is closed in (X,). Therefore,
gf is strongly sbcontinuous.
Theorem 6.3: The composition of two perfectly sb continuous function is perfectly sbcontinuous function.
Proof: Let f: (X,) (Y,) and g: (Y,) (Z,) be perfectly sbcontinuous functions. Let V be sbclosed set in (Z,). Since g is perfectly sbcontinuous, 1(V) is clopen in (Y,). By Propositon 3.4 in [3], 1(V) is sb clopen in (Y,). Since f is perfectly sbcontinuous,
1(1()) = (g f)1(V) is clopen in (X,). Therefore,
gf is perfectly sbcontinuous.
The following example shows that the composition of two contra sbirresolute function need not be contra sb irresolute.
Example 6.4: Let X = Y = {a,b,c} with topologies
= {X, , {a},{b},{a,b}}, = {Y, , {b}} and = {Z, ,{a,c}}
Define a function f: (X,) (Y,) by f(a) = a, f(b) = c, f(c)
= b and g: (Y,) (Z,) by g(a) = a, g(b) = b, g(c) = c. sb C(X) = {X,,{a},{b},{c},{a,c},{b,c}}
sb O(Y) = {Y,,{b},{b,c},{a,b}} sb O(Z) = {Z, , {a,c}.
Clearly f and g are contra sbirresolute function. But their composition is not contra sbirresolute, since (gf) 1 of an open set {a,c} in (Z,) is {a,c} which is not sbclosed in (X,). Hence, gf is not contra sb continuous.
Theorem 6.5: The composition of two perfectly contra sb irresolute function is perfectly contra sbirresolute function.
Proof: Let f: (X,) (Y,) and g: (Y,) (Z,) be perfectly contra sbirresolute functions. Let V be any sbopen set in (Z,). Since g is perfectly contra sb irresolute, 1(V) is sbclopen in (Y,). Since f is perfectly contra sbirresolute, 1(1()) = (g f)1(V) is sbclopen in (X,). Therefore, gf is perfectly contra sbirresolute.
Theorem 6.6: If a function f: (X,) (Y,) is strongly sb continuous function and g: (Y,) (Z,) is contra sb continuous function then g f: (X,) (Z,) is contra continuous.
Proof: Let V be any open set in (Z,). Since g is contra sb continuous, 1(V) is sbclosed in (Y,). Since f is strongly sbcontinuous, 1(1()) = (g f)1(V) is closed in (X,). Hence, g f is contra continuous.
Theorem 6.7: If a function f: (X,) (Y,) is contra sb irresolute function and g: (Y,) (Z,) is sb irresolute function then g f: (X,) (Z,) is contra sb irresolute.
Proof: Let V be any sbopen set in (Z,). Since g is sb irresolute, 1(V) is sbopen in (Y,). Since f is contra sbirresolute, 1(1()) = (g f)1(V) is sbclosed in (X,). Hence, g f is contra sbirresolute.
Theorem 6.8: If a function f: (X,) (Y,) is contra sb irresolute function and g: (Y,) (Z,) is sbcontinuous function then g f: (X,) (Z,) is contra sbcontinuous. Proof: Let V be any open set in (Z,). Since g is sb continuous, 1(V) is sbopen in (Y,). Since f is contra sbirresolute, 1(1()) = (g f)1(V) is sbclosed in (X,). Hence, g f is contra sbcontinuous.
Theorem 6.9: If a function f: (X,) (Y,) is sb irresolute function with Y as locally indiscrete space an g: (Y,) (Z,) is contra sbcontinuous function then g f: (X,) (Z,) is sbcontinuous.
Proof: Let V be any open set in (Z,). Since g is contra sb continuous, 1(V) is sbopen in (Y,). But Y is locally sbindiscrete, 1(V) is closed in (Y,). By Proposition
3.4 in [3], 1(V) is sbclosed in (Y,). Since f is sb irresolute, 1(1()) = (g f)1(V) is sbclosed in (X,). Hence, g f is contra sbcontinuous.
Theorem 6.10: If a function f: (X,) (Y,) is sb irresolute function and g: (Y,) (Z,) is contra sb continuous function then g f: (X,) (Z,) is contra sb continuous.
Proof: Let V be any open set in (Z,). Since g is contra sb continuous, 1(V) is sbclosed in (Y,). Since f is sb irresolute, 1(1()) = (g f)1(V) is sbclosed in (X,). Hence, g f is contra sbcontinuous.

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