on Contra sbg – Continuous functions in Topological Spaces

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

1. Navaneetha Krishnan2 and S. Pious Missier3

   2,3 Associate Professor of Mathematics,

   1. Chidambaram College, Thoothukudi, TN, India

      Abstract – In this paper a new class of functions called contra sb-continuous function is introduced and its properties are studied. Some characterization and several properties concerning Contra sb-continuity are obtained. Also, Contra sb-irresolute function and Perfectly Contra sb-irresolute function are introduced.

      Keywords: sb-closed sets, sb-continuous, Contra sb- continuous, Contra sb-irresolute.

      AMS Mathematics Subject Classification : 54C08, 54C10.

      1. 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 semi-continuity[3], contra b- continuity[12], contra sg-continuity[5], contra gs- continuity[5], contra gb-continuity[16], contra g*b- continuity[16], contra b-continuity[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 sb-continuous, contra sb- irresolute functions and we obtain some of its characterization.

      2. 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 semi-closed (resp.-closed, b-closed, regular-closed) sets of X containing A is called the semi- closure (resp.-closure, b-closure, regular closure) of A and is denoted by sCl(A) (resp.Cl(A), bCl(A), rCl(A)). The family of all semi-open (resp. -open, b-open, 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

   1. a sg-closed set[5] if sCl(A) U whenever A

      U and U is semi-open in X.

   2. a gs-closed set[5] if sCl(A) U whenever A

      U and U is open in X.

   3. a gb-closed set[16] if bCl(A) U whenever A

      U and U is open in X.

   4. a g*b-closed set[16] if bCl(A) U whenever A

      U and U is g-open in X.

   5. a b-closed set[15] if bCl(A) U whenever A

      U and U is -open in X.

   6. a sb-closed set[3] if sCl(A) U whenever A

U and U is b-open in X.

The complement of a sg-closed (resp. gs-closed, gb-closed, g*bclosed and b-closed) set is called sg-open (resp. gs- open, gb-open, g*bopen and b-open) set.

Definition 2.3: A space (X,) is called a

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

   sb

   sb

2. T space[3] if every sb-closed set in X is –

Definition 2.1: A subset A of a topological space (X,) is called

1. a semi-open set[5] if A Cl(Int(A)).

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

3. a b-open set[1] if A Cl(Int(A)) Int(Cl(A)).

4. a regular open[14] set if A = Int(Cl(A)).

The complement of a semiopen (resp.open, bopen, regular-open) set is called semi-closed (resp.closed, b closed, regular-closed) set.

closed.

Definition 2.4: A function f: (X,) (Y,) is called a

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

2. sb irresolute map[4] if 1(V) is sb-closed in

(X,) for every closed set V in (Y,).

Definition 2.5: A function f: (X,) (Y,) is called a

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

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

3. Contra -continuous map[8] if 1(V) is -closed

   in (X,) for every open set V in (Y,).

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

5. Contra sg-continuous map[5] if 1(V) is sg-

   closed in (X,) for every open set V in (Y,).

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

7. Contra gb-continuous map[16] if 1(V) is gb-

   closed in (X,) for every open set V in (Y,).

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

9. Contra b-continuous 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 × Y is called the graph of f and is denoted by G(f).

1. CONTRA sb-CONTINUOUS FUNCTIONS

   We introduce the following definition.

   Definition 3.1: A function f: (X,) (Y,) is called Contra sb-continuous if 1(V) is sb-closed 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 sb-closed sets in X. Hence, f is contra sb-continuous.

   Theorem 3.3:

   1. Every contra continuous function is contra sb- continuous function.

   2. Every contra -continuous function is contra sb- continuous function.

Proof:

1. 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 sb-closed in (X,). Hence, f is contra sb-continuous function.

2. 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 sb-closed in (X,). Hence, f is contra sb-continuous function.

The following examples show that the converse of the above proposition need not be true.

Example 3.4:

1. 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 sb-closed but not closed in (X,). Hence, f is contra sb-continuous but not contra continuous.

2. 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 sb-closed but not -closed in (X,). Hence, f is contra sb-continuous but not contra -continuous.

Theorem 3.5:

1. Every contra sb-continuous function is contra b- continuous function

2. Every contra sb-continuous function is contra sg- continuous function

3. Every contra sb-continuous function is contra gs- continuous function

4. Every contra sb-continuous function is contra gb- continuous function

5. Every contra sb-continuous function is contra g*b-continuous function

6. Every contra sb-continuous function is contra b- continuous function.

Proof:

1. Let V be any open set in (Y,). Since f is contra sb-continuous, 1(V) is sb-closed in (X,). By Proposition 3.11 in [3], 1(V) is b-closed in (X,). Hence, f is contra b-continuous function.

2. Let V be any open set in (Y,). Since f is contra sb-continuous, 1(V) is sb-closed in (X,). By Proposition 3.13 in [3], 1(V) is sg-closed in (X,). Hence, f is contra sg-continuous function.

3. Let V be any open set in (Y,). Since f is contra sb-continuous, 1(V) is sb-closed in (X,). By Proposition 3.15 in [3], 1(V) is gs-closed in (X,). Hence, f is contra gs-continuous function.

4. Let V be any open set in (Y,). Since f is contra sb-continuous, 1(V) is sb-closed in (X,). By

   Proposition 3.17 in [3], 1(V) is gb-closed in (X,). Hence, f is contra gb-continuous function.

5. Let V be any open set in (Y,). Since f is contra sb-continuous, 1(V) is sb-closed in (X,). By Proposition 3.21in [3], 1(V) is g*b-closed in (X,). Hence, f is contra g*b-continuous function.

6. Let V be any open set in (Y,). Since f is contra sb-continuous, 1(V) is sb-closed in (X,). By Proposition 3.23 in [3], 1(V) is b-closed in (X,). Hence, f is contra b-continuous function.

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

Example 3.6:

1. 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 b-closed but not sb -closed in (X,). Hence, f is contra b-continuous but not contra sb- continuous.

2. 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 sg-closed but not sb -closed in (X,). Hence, f is contra sg-continuous but not contra sb-continuous function.

3. 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.

   sb-C(X)={X,,{b},{c},{d},{b,c},{c,d},{b,d}, {b,c,d}}

   gs-C(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 gs-closed set but not sb -closed set in (X,). Hence, f is contra gs-continuous but not contra sb-continuous function.

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) = 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}}

   gb-C(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 gb-closed but not sb -closed in (X,). Hence, f is contra gb-continuous but not contra sb- continuous.

5. 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*b-closed but not sb -closed in (X,). Hence, f is contra g*b- continuous but not contra sb-continuous.

6. 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}}

b-C(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 b-closed but not sb -closed in (X,). Hence, f is contra b-continuous but not contra sb- continuous.

Theorem 3.7: If f: (X,) (Y,) is contra semi-continuous function if and only if f is contra sb-continuous function. Proof: Let V be any open set in (Y,). Since f is contra semi-continuous, 1(V) is semi-closed in (X,). Since from Proposition 3.6 in [3], 1(V) is sb-closed in (X,). Hence, f is contra sb-continuous.

Conversely, Let V be any open set in (Y,). Since f is contra sb-continuous, 1(V) is sb-closed in (X,). Since from Proposition3.6 in [3], 1(V) is semi-closed in (X,). Hence, f is contra semi-continuous.

Remark 3.8: The following diagram shows the relationships of contra sb-continuoous function with other known existing functions.

1. Contra sb-continuous 2. Contra continuous

3. Contra -continuous 4. Contra b-continuous

5. Contra sg-continuous 6. Contra gs-continuous

1. Contra gb-continuous 8. Contra g*bcontinuous 9.Contra b-continuous

   Theorem 3.9: The following are equivalent for a function f: (X,) (Y,),

   1. f is contra sb-continuous function

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

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

Proof:

1. (b):

   Let F be any closed set in Y. Then Fc is an open set in Y. Since f is contra sb-continuous, 1() is sb- closed set in X. Then [1()]c is sb-closed set in X. Therefore 1() is sb-open in X.

2. (a):

Let F be an open set in Y. Then is closed set in

Y. By (b), 1() is sb-open set in X. Then [1()]c is sb-open set in X. So 1() is sb-closed set in X.

Therefore, f is contra sb-continuous function

1. (c):

   Let F be any closed subset of Y and let f(x) F where x X. Then by (b), 1(F) is sb-open in X. Also, x

   1(F). Take U = 1(F). Then U is a sb-open set containing x and f(U) F.

2. (b):

   Let F be any closed subset of Y. If x 1(F) then f(x) F. By (c), there exists a sb-open 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.

   1. f is contra continuous function

   2. f is contra sb-continuous function.

Proof:

1. (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 sb-closed in X. Therefore, f is contra sb- continuous.

2. (i):

   Let V be any open set in Y. Since f is contra sb- continuous, 1(V) is sb-closed 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 sb-closed 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,

   1. f is sb-continuous function

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

Proof:

1. (ii):

   Let f(x) V, then x 1(V). Since f is sb- continuous, 1(V) is sb-open in X. Let U = 1(V), then x U and f(U) V.

2. (i):

   Let V be any open set in Y and x 1(V). Then f(x) V. From (ii), there exists a sb-open set Ux of X such that x Ux 1(V) and 1(V) = { Ux}. Then 1(V) is sb-open in X. Hence, f is sb-continuous.

   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 sb-open set U of X with x U such that f(U) Cl(W). That is, f(U) V. By theorem 3.12, f is sb-continuous.

   Definintion 3.14: A function f: (X,) (Y,) is said to be strongly sb-continuous if 1(V) is closed in (X,) for every sb-closed 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 sb-closed 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 sb-continuous if 1(V) is clopen in (X,) for every sb-closed 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 sb-closed set {b} in (Y,) is {b} which is both open and closed in (X,). Hence, f is perfectly sb-continuous.

   Definition 3.18: A topological space (X,) is said to be sb- Hausdorff (or sb-T2 space) if for each pair of distinct points x and y in X, there exists sb-open 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 sb-Hausdorff space.

   Theorem 3.20: Let f: (X,) (Y,) be surjective, closed and contra sb-continuous. 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 sb-closed 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 sb-closed 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 sb-Hausdorff.

   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 sb-continuous. By theorem 3.12, there exists sb-open 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 sb-Hausdorff space.

   Theorem 3.23: Let f: (X,) (Y,) be a function and g: X

   X× be a graph function of f defined by g(x) = (x,f(x)) for every x X. If g is contra sb-continuous, then f is contra sb-continuous.

   Proof: Let V be closed subset of Y. Then X× V is a closed subset of X× . Since g is contra sb-continuous, 1 (X ×

   )is sb-open subset of X. Also, 1 (X × ) = 1(V) which is sb-open subset of X. Hence, f is contra sb- continuous.

   Definition 3.24: A space X is said to be locally sb- indiscrete if every sb-open 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 sb-open set in X is closed in X. Hence, X is locally sb-indiscrete space.

   Theorem 3.26: If f: (X,) (Y,) is contra sb-continuous with X as locally sb-indiscrete, 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 sb-indiscrete space, 1(V) is open in X. Thus, f is continuous.

   4. CONTRA sb-IRRESOLUTE FUNCTIONS Definition 4.1: A function f: (X,) (Y,) is called contra sb-irresolute, if 1(V) is sb-closed in (X,) for every sb-open 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 sb-open set {b},{b,c} and {a,b} in (Y,) are {c},{b,c} and {a,c} respectively which are sb-closed set in (X,). Hence, f is contra sb-irresolute function.

   Remark 4.3: The following example shows that the concepts of sb-irresolute function and contra sb-irresolute are independent of each other.

   Example 4.4:

   1. 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 sb-irresolute 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 sb-closed in (X,).

   2. 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 sb-open sets {a},{a,b} in (Y,) are {a},{a,b} which are not sb-closed set in (X,). Hence, the f is sb-irresolute but not contra sb-irresolute.

Theorem 4.5: Every contra sb-irresolute function is contra sb-continuous.

Proof: Let V be any open set in (Y,). By proposition 3.4 in [3], V is sb-open in Y, Since f is contra sb- irresolute, V is sb-closed 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 sb-open set {a,c} in (Y,) are {a,c} which is not sb-closed set in (X,). Hence, the f is contra sb-continuous but not contra sb-irresolute.

Remark 4.7: The following example shows that the concepts of sb-continuous and contra sb-continuous are independent of each other.

Example 4.8:

1. 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 sb-closed 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 sb-closed in (X,), f is sb- continuous. Hence, f is sb-continuous but not contra sb- continuous.

2. 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 sb-closed in X, f is contra sb- continuous. Also, since the inverse image of a closed set

   {b} in Y is {b} which is not sb-closed in X, f is not sb- continuous. Hence, f is contra sb-continuous but not sb- continuous.

   1. PERFECTLY CONTRA sb-IRRESOLUTE FUNCTION

      Definition 5.1: A function f: (X,) (Y,) is called perfectly contra sb-irresolute function if 1(V) is sb- clopen in (X,) for every sb-open 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 sb-open sets in (Y,) are sb-clopen set in (X,), f is perfectly contra sb- irresolute function.

      Theroem 5.3:

      1. Every perfectly contra sb-irresolute map is contra sb-irresolute map.

      2. Every perfectly contra sb-irresolute map is sb- irresolute map.

         Proof:

         1. 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:

      1. 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 sb-open set {b} in (Y,) is {c} which is sb-closed set in (X,) but not sb- open set in X, f is contra sb irresoulute but not perfectly contra sb-irresolute function.

      2. 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 sb-open sets {a} and

      {a,b} in (Y,) are {a} and {a,b} respectively which are sb-open set in (X,) but not sb-closed set in (X,), f is sbirresolute but not perfectly contra sb-irresolute 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.

   2. COMPOSITION OF TWO MAPS

      The following example shows that the composition of two contra sb-continuous 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 sb-continuous. But their composition is not contra sb-continuous, 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 sb-continuous function.

      Proof: Let f: (X,) (Y,) and g: (Y,) (Z,) be strongly sb-continuous functions. Let V be sb-closed set in (Z,). Since g is strongly sb-continuous, 1(V) is closed in (Y,). By Propositon 3.4 in [3], 1(V) is sb- closed in (Y,). Since f is strongly sb-continuous,

      1(1()) = (g f)-1(V) is closed in (X,). Therefore,

      gf is strongly sb-continuous.

      Theorem 6.3: The composition of two perfectly sb- continuous function is perfectly sb-continuous function.

      Proof: Let f: (X,) (Y,) and g: (Y,) (Z,) be perfectly sb-continuous functions. Let V be sb-closed set in (Z,). Since g is perfectly sb-continuous, 1(V) is clopen in (Y,). By Propositon 3.4 in [3], 1(V) is sb- clopen in (Y,). Since f is perfectly sb-continuous,

      1(1()) = (g f)-1(V) is clopen in (X,). Therefore,

      gf is perfectly sb-continuous.

      The following example shows that the composition of two contra sb-irresolute 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 sb-irresolute function. But their composition is not contra sb-irresolute, 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 6.5: The composition of two perfectly contra sb- irresolute function is perfectly contra sb-irresolute function.

      Proof: Let f: (X,) (Y,) and g: (Y,) (Z,) be perfectly contra sb-irresolute functions. Let V be any sb-open set in (Z,). Since g is perfectly contra sb- irresolute, 1(V) is sb-clopen in (Y,). Since f is perfectly contra sb-irresolute, 1(1()) = (g f)-1(V) is sb-clopen in (X,). Therefore, gf is perfectly contra sb-irresolute.

      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 sb-closed in (Y,). Since f is strongly sb-continuous, 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 sb-open set in (Z,). Since g is sb- irresolute, 1(V) is sb-open in (Y,). Since f is contra sb-irresolute, 1(1()) = (g f)-1(V) is sb-closed in (X,). Hence, g f is contra sb-irresolute.

      Theorem 6.8: If a function f: (X,) (Y,) is contra sb- irresolute function and g: (Y,) (Z,) is sb-continuous function then g f: (X,) (Z,) is contra sb-continuous. Proof: Let V be any open set in (Z,). Since g is sb- continuous, 1(V) is sb-open in (Y,). Since f is contra sb-irresolute, 1(1()) = (g f)-1(V) is sb-closed in (X,). Hence, g f is contra sb-continuous.

      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 sb-continuous function then g f: (X,) (Z,) is sb-continuous.

      Proof: Let V be any open set in (Z,). Since g is contra sb- continuous, 1(V) is sb-open in (Y,). But Y is locally sb-indiscrete, 1(V) is closed in (Y,). By Proposition

      3.4 in [3], 1(V) is sb-closed in (Y,). Since f is sb- irresolute, 1(1()) = (g f)-1(V) is sb-closed in (X,). Hence, g f is contra sb-continuous.

      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 sb-closed in (Y,). Since f is sb- irresolute, 1(1()) = (g f)-1(V) is sb-closed in (X,). Hence, g f is contra sb-continuous.

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