2.2. Generalized Fuzzy Semi-Pre-Open Sets (G_F SPOS)
Definition 2.2.1: Let (X,?) be G_F TS. A generalized fuzzy set G_F S ? in X is called generalized fuzzy semi-open (G_F SOS) if ??cl_? (i_? (?)).
Proposition 2.2.1: G_F OS(X) ? G_F SOS(X)
Proof: Let (X,?) be G_F TS and ? be G_F OS(X). Then i_? (?)=?. Since ??cl_? (?). This implies ?=i_? (?)?(cl_? (i_? (?) )). Hence ? is G_F SOS(X).
Remark 2.2.1: The converse of proposition 2.2.1 is not true as illustrated in Example 2.
2.1
Example 2.2.1: Let X={x_1,x_2}, A={?(x?_1,0.3),?(x?_2,0.7)}, B={?(x?_1,0.7),?(x?_2,0.4)} and C={?(x?_1,0.7),?(x?_2,0.7)}. Clearly ?={0,A,B,C.1} isGFT(X). Thus the set ?={?(x?_1,0.8),?(x?_2,0.7)} is GFSOS(X) but not GFOS(X).
Definition 2.2.2: Let (X,?) be GFTS. A GFS(X) ? is called generalized fuzzy pre-open (GFPOS) if ??i_? (cl_? (?)).
Proof: Let (X,?) be GFTS and ? be GFOS(X). Then cl_? (?)=?. Since ??i_? (?). This implies ?=cl_? (?)?(i_? (cl_? (?) )). Hence ? is GFPOS(X).
Remark 2.2.2: The converse of proposition 2.2.1 is not true as illustrated in Example 2.2.2
Example 2.2.2: In Example 3.1, ?={?(x?_1,0.8),?(x?_2,0.7)} is GFPOS(X)but not GFOS(X).
Definition 2.2.3: Let (X,?) be GFTS. A GFS(X) ? is called generalized fuzzy ?-open (GF?OS) if ??i_? (cl_? (i_? (?) )).
Proof: Let (X,?) be GFTS and ? be GFOS(X). Then i_? (?)=?. Since ??cl_? (?), we have ??cl_? (i_? (?)). This implies ?=i_? (?)?i_? (cl_? (i_? (?) )). Hence ? is GF?OS(X).
Remark 2.2.3: The converse of proposition 2.2.3 is not true as illustrated in Example 2.2.3
Example 2.2.3: In Example 2.
2.1, ?={?(x?_1,0.8),?(x?_2,0.7)} is GF?OS(X)but not GFOS(X).
Definition 2.2.4: Let (X,?) be GFTS. A GFS(X) ? is called generalized fuzzy ?-open (GF?OS) if ??cl_? (i_? (cl_? (?))).
Remark 2.2.4: The converse of proposition 2.2.4 is not true as illustrated in Example 2.2.4
Example 2.2.4: In Example 2.2.1, ?={?(x?_1,0.8),?(x?_2,0.7)} is GF?OS(X) but not GFOS(X).
Proposition 2.3.2: GFSOS(X) ? GFSPOS(X)
Proof: Let X be GFTS and let ?_1be GFSPOS(X) then ?_1?cl_? (i_? (?_1 ). Let us consider GFS, ?_2=i_? (?_1). Since ?_2 isGFOS(X)and a hence GFPOS(X). Now ?_2??_1?cl_? (?_2). This implies ?_1 is GFSPOS(X). Thus each GFSOS(X) is GFSPOS(X).
Remark 2.3.2: The converse of Proposition 2.3.2 is not true in general as shown in Example 2.3.2.
Example 2.3.2: In Example 2.3.1 ?_2 is GFSPOS(X) but not GFSOS(X)
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