2.4. Generalized Binary Continuous Maps (G_B CM)

Definition 2.4.1: Let (X,Y,?) be G_B TS and (Z,T_G ) be G_T S. Then the map F:Z?X?Y is called generalized binary continuous (G_B CM) at z?Z if for any G_B OS (A,B)?(X,Y,?) with f(z)?(A,B) then there exists a T_G-open set G in (Z,T_G ) such that z?G and F(G)?(A,B). The map F is called G_B CM if it is G_B C at each z?Z.

Proposition 2.4.1: Let (X,Y,?) be G_B TS and (Z,T_G ) be G_T S. Let F:Z?X?Y be a mapping. Then F is called G_B CM if F^(-1) (A,B) is T_G-open in (Z,T_G) for every G_B OS (A, B) in (X,Y,?).

Proof: Let F be G_B CM and (A, B) be G_B OS in (X,Y,?).

If F^(-1) (A,B)=?. Then ? is T_G-open set. If F^(-1) (A,B)?? and let z?F^(-1) (A,B). Then F(z)?(A,B). Since F is G_B C at z, there exists a T_G-open set G in (Z,?_g ) such that z?G and F(G)?(A,B). Hence z?G?F^(-1) (A,B). Therefore F^(-1) (A,B) is T_G-open in (Z,T_G ).

Conversely to show that the mapping F:Z?X?Y is G_B C, let z?Z and (A, B) be a G_B OS in (X,Y,?) with F(z)?(A,B). Then z?F^(-1) (A,B), where F^(-1) (A,B) is T_G-open. Also ?F(F?^(-1) (A,B))?(A,B). Hence F is G_B C at z. Therefore F is G_B CM.

Example 2.4.1: LetZ={1,2,3}, X={a_1,a_2 } and Y={b_1,b_2 }. Then T_G={?,{3},{1,2},{2,3},Z} and ?={(?,?),({a_1 },{b_1 } ),({a_2 },{Y} ),(X,Y) }. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(1)=(a_1,b_1 )=F(2) and F(3)=(a_2,Y). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={1,2}, F^(-1) ({a_2 },{Y})={3}, and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-open in (Z,T_G ). Hence F is G_B CM.

Definition 2.4.2: Let (X,Y,?) be G_B TS. Let (A,B)? ?(X)??(Y). Then (A,B) is called

Generalized binary semi-open ?(G?_B SOS) if (A,B)?Cl_? (I_? (A,B)).

Generalized binary pre-open?(G?_B POS) if (A,B)?I_? (Cl_? (A,B)).

Generalized binary ?-open ?(G?_B ?OS) if (A,B)?I_? (Cl_? (I_? (A,B))).

Generalized binary ?-open ?(G?_B ?OS) if (A,B)?Cl_? (I_? (Cl_? (A,B))).

Generalized binary regular open?(G?_B ROS) if (A,B)=I_? (Cl_? (A,B)).

Definition 2.4.2: Let (Z,T_G ) be G_T S and (X,Y,?) be G_B TS. Then the mapping F:Z?X?Y is called generalized binary semi-continuous (G_B SCM) if F^(-1) (A,B) is T_G-semi-open in (Z,T_G ) for every G_B OS (A, B) in (X,Y,?).

Example 2.4.2: Let Z={1,2,3}, X={a_1,a_2 } and Y={b_1,b_2 }. Then T_G={?,{1},{1,2},{2,3},Z} and?={(?,?),({a_1 },{b_1 } ),({a_1 },{Y} ),({a_2 },{Y} ),(X,Y) }. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(1)=(a_1,b_1 ),F(2)=(a_2,b_2 )=F(3). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={1}, F^(-1) ({a_1 },{Y})={1}, F^(-1) ({a_2 },{Y})={2,3} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-semi-open in (Z,?_g ). Hence F is G_B SCM.

### Proposition 2.4.2: G_B CM ?? G_B SCM

Proof: Let (A,B) be G_B OS in (X,Y,?). Since F is G_B CM, we have F^(-1) (A,B) is T_G-open in (Z,T_G ). We know that every T_G-open set in G_T S is T_G-semi-open. Hence F^(-1) (A,B) is T_G-semi-open in (Z,T_G ). Thus F is G_B SCM.

Remark 2.4.1: The converse of Proposition 2.4.2 is illustrated in Example 2.4.3

Example 2.4.3: Let Z={1,2,3,4}, X={a_1,a_2 } and Y={b_1,b_2 }. Then T_G={?,{3},{3,4},{1,2,4},Z} and ?={(?,?),({a_1 },{b_1 } ),({a_1 },{Y} ),({a_2 },{Y} ),(X,Y)}. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(1)=(a_1,b_1 ),F(2)=F(3)=F(4)=(a_2,b_2 ). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={1}, F^(-1) ({a_1 },{Y})={1}, F^(-1) ({a_2 },{Y})={2,3,4} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-semi-open in (Z,T_G ). Hence F is G_B SCM but not G_B CM because {2,3,4} is T_G-semi-open but not T_G-open in (Z,T_G ).

### Proposition 2.4.3: G_B CM ?? G_B ?CM

Proof: Let (A,B) be G_B OS in (X,Y,?). Since F is G_B CM, we have F^(-1) (A,B) is T_G-open in (Z,T_G ). We know that every T_G-open set in G_T S is T_G-?-open. Hence F^(-1) (A,B) is T_G-?-open in (Z,T_G ). Thus F is G_B ?CM.

Remark 2.4.2: The converse of the Proposition 2.4.3 is illustrated in Example 2.4.4.

Example 2.4.4: Let Z={1,2,3,4}, X={a_1,a_2 } and Y={b_1,b_2 }. Then T_G={?,{1,2} {2,3,4},Z} and ?={(?,?),({a_1 },{b_1 } ),({a_1 },{b_2 } ),({a_1 },{Y} ),(X,Y) }. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(1)=(a_1,b_1 )= F(2)=F(3). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={1,2,3} , F^(-1) ({a_1 },{b_2 })={?}, F^(-1) ({a_1 },{Y})={1,2,3} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-?-open in (Z,T_G ). Hence F is G_B ?CM but not G_B CM because {1,2,3} is T_G-?-open but not T_G-open in (Z,T_G ).

## Proposition 2.4.4:

## G_B ?CM ?? G_B SCM

## G_B ?CM ?? G_B PCM

## G_B RCM ?? G_B CM

Proof: Let (A,B) be G_B OS in (X,Y,?). Since F is G_B ?CM, we have F^(-1) (A,B) is T_G-?-open in (Z,T_G ). We know that every T_G-?-open set in G_T S is T_G-semi-open (T_G-pre-open). Hence F^(-1) (A,B) is T_G-semi-open (T_G-pre-open) in (Z,T_G ). Thus F is G_B SCM (G_B PCM).

Similarly let (A,B) be G_B OS in (X,Y,?).. Suppose F is G_B RCM , we have F^(-1) (A,B) is T_G-r-open in (Z,T_G ). We know that every T_G-r-open set is T_G-open. Hence F^(-1) (A,B) is T_G-open in (Z,T_G ). Thus F is G_B CM, which proves (iii).

Remark 2.4.3: The converse of the Proposition 2.4.4 is illustrated in Example 2.4.5, Example 2.4.6 and Example 2.4.7.

Example 2.4.5: Let Z={1,2,3,4}, X={a_1,a_2 }, Y={b_1,b_2 }. Then T_G={?,{2},{2,3},{3,4},{2,3,4},Z} and ?={(?,?),({a_1 },{b_1 } ),({a_1 },{Y} ),({a_2 },{Y} ),(X,Y) }. Clearly T_G is G_Ton Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(1)=F(2)=(a_1,b_1 ), F(3)=F(4)=(b_1,b_2 ). NowF^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={1,2}, F^(-1) ({a_1 },{Y})={1,2}, F^(-1) ({a_2 },{Y})={3,4} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-semi-open in (Z,T_G ). Hence F is G_B SCM but not G_B ?CM because {1,2} is T_G-semi-open but not T_G-?-open in (Z,T_G ).

Example 2.4.6: Let Z={1,2,3,4}, X={a_1,a_2 }, Y={b_1,b_2 }. Then T_G={?,{1,2},{2,3},{3,4},{1,2,3},{2,3,4},Z} and?={(?,?),({a_1 },{b_1 } ),({a_1 },{Y} ),({a_2 },{Y} ),(X,Y) }. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(1)=F(3)=F(4)=(a_1,b_1 ),F(2)=(b_1,b_2 ). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={1,3,4}, F^(-1) ({a_1 },{Y})={1,3,4}, F^(-1) ({a_2 },{Y})={?} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-pre-open in (Z,T_G ). Hence F is G_B PCM but not G_B ?CM because {1,3,4} is T_G-pre-open but not T_G-?-open in (Z,T_G ).

Example 2.4.7: Let Z={1,2,3,4}, X={a_1,a_2 }, Y={b_1,b_2 }. Then T_G={?,{2},{2,3},{3,4},{2,3,4},Z} and ?={(?,?),({a_1 },{b_1 } ),({a_1 },{Y} ),({a_2 },{Y} ),(X,Y) }. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(2)=F(3)=(a_1,b_1 ),F(1)=f(4)=(b_1,b_2 ). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={2,3}, F^(-1) ({a_1 },{Y})={2,3}, F^(-1) ({a_2 },{Y})={?} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-open in (Z,T_G ). Hence F is G_B CM but not G_B RCM because {2,3} is T_G-open but not T_G-regular open in (Z,T_G ).

### Remark 2.4.4: G_B SCM ? G_B PCM

This can be illustrated in Example 2.4.8 and Example 2.4.9.

Example 2.4.8: Let Z={1,2,3,4}, X={a_1,a_2 } and Y={b_1,b_2 }. Then T_G={?,{1},{2} {1,2},{1,3},{2,3},{1,2,3},Z}and?={(?,?),({a_1 },{b_1 } ),({a_1 },{Y} ),({a_2 },{Y} ),(X,Y) }. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(2)=F(3)=F(4)=(a_2,b_2 ). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={?}, F^(-1) ({a_1 },{Y})={?}, F^(-1) ({a_2 },{Y})={2,3,4} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-semi-open in (Z,T_G ). Hence F is G_B SCM but is not G_B PCM, because {2,3,4} is not T_G-pre-open in (Z,T_G ).

Example 2.4.9: Let Z={1,2,3}, X={a_1,a_2 } and Y={b_1,b_2 }. Then T_G={?,{1,2},{2,3},Z} and ?={(?,?),({a_1 },{b_1 } ),({a_1 },{Y} ),({a_2 },{Y} ),(X,Y) }. Clearly T_G is G_T on Z and ? is G_B T from X to Y. Define F:Z?X?Y by F(1)=(a_1,b_1 )=F(3),F(2)=(?,b_2 ). Now F^(-1) (?,?)=?, F^(-1) ({a_1 },{b_1 })={1,3}, F^(-1) ({a_1 },{Y})={1,3}, F^(-1) ({a_2 },{Y})={?} and F^(-1) (X,Y)=Z. This shows that the inverse image of every G_B OS in (X,Y,?) is T_G-pre-open in (Z,T_G ). Hence F is G_B PCM but is not G_B SCM because {1,3} is not T_G-semi-open in (Z,T_G ).