# 23 Generalized Binary Topological Space ?G?B TDefinition 231 Let X Essay

2.3. Generalized Binary Topological Space ?(G?_B T)

Definition 2.3.1: Let X and Y are any two non-empty sets. A generalized binary topology ?(G?_B T) from X to Y is a binary structure ???(X)??(Y) that satisfies the following axioms:

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(?,?) and (X,X)??

If {(A_? ,B_? ) ; ???} is a family of members of ?, then (?_(???)?A_? ,?_(???)?B_? )??

If ? is a ?(G?_B T) from X to Y, then the triplet (X,Y,?) is called a generalized binary topological space ?(G?_B TS) and the members of ? are called the generalized binary open subsets ?(G?_B OS) of (X,Y,?).

The elements of X?Y are called the generalized binary points ?(G?_B OP) of (X,Y,?).

Example 2.3.1: Let X={1,2} and Y={a,b,c}. Then ?={(?,?),({1},{a,c} ), ({2},{a,b} ), (X,Y)} is G_B T satisfying the above conditions.

Definition 2.3.2: Let X and Y be any two non-empty sets and (A, B), (C, D) belongs to ?(X)??(Y), we say (A, B) ? (C, D) if A ? C and B ? D.

Proposition 2.3.1: Let {?_? ; ???} be any family of G_B T’S from X to Y. Then ?_(???)??_? is also G_B T from X to Y.

Proof: Let ?_? is G_B T from X to Y. Then (?,?) and (X,X)??_? for all ?? . This implies (?,?)??_(???)??_? and (X,X)??_(???)??_? . Let ?_(???)??_? = ? and (A_? ,B_? )? ? ? ??? , where ? is arbitrary index. Then (A_? ,B_? )? ?_? ? ?? ?. Since ?_? is G_B T from X to Y. It follows (?_(???)?A_? ,?_(???)?B_? )? ?_? ? ?? ? . This implies that (?_(???)?A_? ,?_(???)?B_? )? ?_(???)??_? . Hence ?_(???)??_? is G_B T from X to Y.

Remark 2.3.1: Let {?_? ; ???} be any family of G_B T’S from X to Y. Then ?_(???) ?_? need not be a G_B T as shown in Example 1.3.2.

Example 2.3.2: Let X={a,b,c} and Y={1,2,3}. Then ?_1={(?,?),({a},{1,3} ), ({X},{1,2} ),(X,Y)} and ?_2={(?,?),({b},{1,2} ),({a,c},{1,3} )(X,Y) } are two GBTs from X to Y. Clearly ?_1??_2={(?,?),({b},{1,2} ),({a,b},{X} ),({a,c},{1,3} ), ({X},{1,2} ),(X,Y)} is not G_B T from X to Y.

Definition 2.3.3: Let (X,Y,?) be a G_B TS and A?X,B?Y. Then (A,B) is generalized binary closed (G_B CS) in (X,Y,?) if (XA,YB)??.

Proposition 2.3.2: Let (X,Y,?) be G_B TS. Then

(X,Y) and (?,?) are G_B CS.

If {(A_? ,B_? ): ???} is a family of G_B CS’S , then (?_(???)?A_? ,?_(???)?B_? ) is G_B CS.

Definition 2.3.4: Let (X,Y,?) be G_B TS and (A,B)?(X,Y). Let ??(A,B)?^(1^* )?_?=???{A_?: (A_? ,B_? ) ? is G_B CS and (A,B)?(A_? ,B_? )} and Let ??(A,B)?^(2^* )?_?=???{B_?: (A_? ,B_? ) ? isG_B CS and (A,B)?(A_? ,B_? )}.Then ??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?) isG_B CS and (A,B)???(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?). The ordered pair (??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?)) is called G_B-closure of (A,B) and is denoted Cl_? (A,B) in (X,Y,?) where (A,B)?(X,Y).

Proposition 2.3.3: Let (A,B)?(X,Y). Then (A,B) is G_B CS in (X,Y,?) iff (A,B)=Cl_? (A,B).

Proof: Suppose (A,B) is G_B CS in (X,Y,?). Then by the definition, the order pair (?(A,B)^(1^* )?_? ,?(A,B)^(2^* )?_?)) is G_B-closure of (A,B) and (A,B)?(??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?)) . Since (A,B) is G_B CS containing (A,B) i.e. (??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?))?(A,B). Therefore we get (A,B)=(??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?)).

Conversely, suppose(A,B)=(??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?)). Let (X,Y,?) be G_B T and (A,B)?(X,Y). Then ??(A,B)?^(1^* )?_?=???{A_?: (A_? ,B_? ) ? isG_B C and (A,B)?(A_? ,B_? )} and ??(A,B)?^(2^* )?_?)=???{B_?: (A_? ,B_? ) ? isG_B C and (A,B)?(A_? ,B_? )}. Therefore (??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?)) is G_B C and (A,B)?(??(A,B)?^(1^* )?_? ,??(A,B)?^(2^* )?_?)). Therefore Cl_? (A,B) is G_B C and hence (A, B) is G_B CS.

Proposition 2.3.4: Suppose (A,B)?(C,D)?(X,Y) and (X,Y,?) is G_B TS. Then

### Cl_? (?,?)=(?,?), Cl_? (X,Y)=(X,Y)

(A,B)?Cl_? (A,B)

??(A,B)?^(1^* )?_???(C,D)^(1^* )?_?

??(A,B)?^(2^* )?_?)??(C,D)^(2^* )?_?

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