International Journal of Scientific & Engineering Research, Volume 4, Issue 9, September-2013 2035
ISSN 2229-5518
The aim of this paper is to introduce the concept of fuzzy γ-boundary and fuzzy γ- semi boundary of a fuzzy topological space. Some characterizations are discussed, examples are given and properties are established.
The concepts of fuzzy set operations were first introduced by L.A.Zadeh[8] in his paper. After that Chang[3]defined and studied the notion of fuzzy topological space. Pu and Liu[6] defined the notion of fuzzy boundary in fuzzy topological spaces in 1980. Following this, Ahmad and Athar[1] studied their properties. They are also defined the
concept of fuzzy semi boundary and discussed their properties. In 2011, Swidi and Oon[5] introduce fuzzy γ-open
set and fuzzy γ-closed set and discussed their properties. Usha et al[7] defined the concept of fuzzy γ-semi open sets
and fuzzy γ-semi closed sets in fuzzy topological spaces.
Using this, we have introduce fuzzy γ-boundary and fuzzy γ-semi boundary and present several properties of fuzzy
γ-boundary and fuzzy γ-semi boundary with proper examples.
For the basic concepts and notations one can refer Chang. The following definitions and lemmas are useful in studying the properties of fuzzy γ-boundary and fuzzy γ-semi boundary.
(a) Φ, X ϵ τ,
(b) If a, b ϵ τ, then A ˄ B ϵ τ,
(c) If Ai ϵ τ for each i ϵ I, then ˅I Ai ϵ τ.
τ is called fuzzy topology for X and the pair (X, τ) is a fuzzy topological space.
A ≤ ( int( cl A)) ∨ cl( int(A)). The complement of a fuzzy γ-open set is called fuzzy γ-closed.
γ-cl(A)=˄{B: B is fuzzy γ-closed and B≥A} and γ-int(A)=˅{B: B is fuzzy γ-open and B≤A}.
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i. (γ-int (A)) c = γ-cl (Ac) and ii. (γ-cl (A))c = γ-int (Ac).
(ii) γ-int(A) is fuzzy γ-open in X,
(iii) γ-cl(A) ≤ γ-cl(B) if A≤ B,
(iv) γ-int(γ-int A) = γ-int A,
(v) γ-cl(γ-cl A) = γ-cl A,
(vi) γ-int (A ∧ B) = (γ-int A) ∧ (γ-int B),
(vii) γ-int (A ∨ B) ≥ (γ-int A) ∨ (γ-int B),
(viii) γ-cl (A ∨ B) = γ-cl (A) ∨ γ-cl (B) and
(ix) γ-cl (A ∧ B) ≤ γ-cl (A) ∧ γ-cl (B).
of A (briefly γ-sint (A)) is the union of all fuzzy γ-semi open sets of X contained in A. That is, γ-sint (A) = ∨ {B: B
≤ A, B is fuzzy γ-semi open in X}.
of A (briefly γ-scl (A)) is the intersection of all fuzzy γ-semi closed sets contained in A. That is, γ-scl (A) = ∧{ B: B
≥ A, B is fuzzy γ-semi closed}.
(i) γ-sint (A ∧ B) = (γ-sint A) ∧ (γ-sint B), (ii) γ-sint (A ∨ B) ≥ (γ-sint A) ∨ (γ-sint B), (iii) γ-scl γ-scl(A) = γ-scl(A),
(iv) γ-sint γ-sint(A)= γ-sint(A).
(i) γ-scl (A ∨ B) = γ-scl (A) ∨ γ-scl (B) and
(ii) γ-scl (A ∧ B) ≤ γ-scl (A) ∧ γ-scl (B).
Lemma 2.12[7]: Let (X, τ) be a fuzzy topological space. Then for any fuzzy subsets A of X, we have i. (γ-sint (A)) c = γ-scl (Ac) and
ii. (γ-scl (A))c = γ-sint (Ac)
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Lemma 2.13[7]: Let (X, τ) and (Y, σ) be any two fuzzy topological spaces such that X is product related to Y. Then the product A1 × A2 of fuzzy γ-open set A1 of X and a fuzzy γ-open set A2 of Y is a fuzzy γ-open set of the fuzzy product space X × Y.
In this section, we introduce the concept of fuzzy γ-Boundary and their properties are analysed.
γ-boundary of A is defined as γ-Bd(A)= γ-cl(A) ˄ γ-cl(Ac). Obviously γ-Bd(A) is a fuzzy γ-closed set.
Example 3.3: Let X = {a, b} and τ = {0, 1, {a.3 , b.4 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.7, b.6 }}. Let A = {a.6 , b.4 }.
Then γ-cl(A) = {a.7, b.4 } and γ-Bd(A)= {a.5 , b.4 }. It follows that γ-cl(A) ≠ {a.6 , b.4 }= A˅ γ-Bd(A).
(1) γ-Bd(A)= γ-Bd(AC).
(2) If A is fuzzy γ-closed , then γ-Bd(A) ≤ A.
(3) If A is fuzzy γ-open , then γ-Bd(A) ≤ AC.
(4) Let A ≤ B and B ϵ FγC(X) (resp., B ϵ FγO(X)). Then γ-Bd(A) ≤ B (resp.,γ-Bd(A) ≤ BC), where FγC(X)
(resp., FγO(X)) denotes the class of fuzzy γ-closed (resp., fuzzy γ-open) sets in X.
(5) (γ-Bd A)C = γ-int(A) ˅ γ-int(AC).
(6) γ-Bd(A) ≤ Bd(A).
(7) γ-cl(γ-Bd(A)) ≤ Bd(A).
By Definition 3.1, γ-Bd(A) = γ-cl(A) ˄ γ-cl(AC) and γ-Bd(AC)= γ-cl(AC) ˄ γ-cl(A). Therefore γ-Bd(A) = γ-Bd(AC). Hence (1).
Let A be fuzzy γ-closed. By Lemma 2.5, γ-cl(A)= A.
γ-Bd(A) ≤ γ-cl(A) ˄ γ-cl(AC) ≤ γ-cl(A) = A. Hence (2).
Let A be fuzzy γ-open. By Lemma 2.5, γ-int(A) = A. It follows that
γ-Bd(A) ≤ γ-cl(AC) = [ γ-int(A)]C = AC. Hence (3).
Let A ≤ B. Then by Properties [5]2.7, γ-cl(A) ≤ γ-cl(B).
Since B ϵ F γC(X), we have γ-cl(B) = B.
This implies that , γ-Bd(A) = γ-cl(A) ˄ γ-cl(AC) ≤ γ-cl(B) ˄ γ-cl(BC) ≤ γ-cl(B) = B.
That is γ-Bd(A) ≤ B.
Let B ϵ FγO(X). Then Bc ϵ FγC(X). Using the above, γ-Bd(A) ≤ BC. Hence (4).
By Definition 3.1, γ-Bd(A)= γ-cl(A) ˄ γ-cl(AC). Taking complement on both sides, we get
[γ-Bd(A)]C = [ γ-cl(A) ˄ γ-cl(AC)]C = [γ-cl(A)]C ˅ [ γ-cl(AC)]C =γ-int(AC) ˅ γ-int(A).
Hence (5).
Since γ-cl(A) ≤ cl(A) and γ-cl(AC) ≤ cl(AC), we have
γ-Bd(A)= γ-cl(A)˄ γ-cl(AC) ≤ cl(A) ˄ cl(AC) = Bd(A). Hence (6).
γ-cl( γ-Bd(A)) = γ-cl(γ-cl(A) ˄ γ-cl(AC)) ≤ γ-cl(γ-cl(A)) ˄ γ-cl(γ-cl(AC))
= γ-cl(A) ˄ γ-cl(AC) = γ-Bd(A) ≤ Bd(A). Thus γ-cl(γ-Bd(A)) ≤ Bd(A). Hence (7).
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The converse of (2) and (3) and reverse inequalities of (6) and (7) in the Proposition 3.4 are in general, not true as is shown by the following example .
Example 3.5:(i) Let X = {a, b} and τ = {0, 1, {a.2 , b.1 },{a.8, b.9 }, {a.7 , b.2 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.8 , b.9 },{a.2 , b.1 }, {a.3 , b.8 }}. Let A = {a.8 , b.7 }. Then γ-cl(A) = {a.8 , b.8 }}and γ-Bd(A)= {a.3 , b.3 }.
Therefore γ-Bd(A) = {a.3 , b.3 } ≤ A, but A is not fuzzy γ-closed.
(ii) Let X = {a, b} and τ = {0, 1, {a.2 , b.2 },{a.8 , b.8 }, {a.7 , b.7 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.8 , b.8 },{a.2 , b.2 }, {a.3 , b.3 }}. Let A = {a.2 , b.3 }.
Then γ-cl(A) = {a.3, b.3 }} and γ-Bd(A)= {a.3 , b.3 }.
Therefore γ-Bd(A)= {a.3 , b.3 } ≤ AC = {a.8 , b.7 }, but A is not fuzzy γ-open.
(iii) Let X = {a, b} and τ = {0, 1, {a.3, b.4 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.7 , b.6 }}. Let A = {a.6 , b.4 }. Then γ-cl(A) = {a.7 , b.4 } and γ-Bd(A) = {a.5 , b.4 }.
Now Bd(A) = cl(A) ˄ cl(AC) = {a.7 , b.6 } ˄ {a.7 , b.6 } = {a.7, b.6 }.Thus γ-Bd(A) ≥/ Bd(A). Now γ-cl γ-Bd(A)= {a.5 , b.5 }. This implies that γ-cl γ-Bd(A) ≥/ Bd(A).
(1) γ-Bd(A) = γ-cl(A) ˄ (γ-int A)C, (2) γ-Bd(γ-int(A)) ≤ γ-Bd(A),
(3) γ-Bd(γ-cl(A)) ≤ γ-Bd(A),
(4) γ-int(A) ≤ A ˄ ( γ-Bd(A))C.
Proof:
Since γ-cl(AC) = (γ-int A)C, we have γ-Bd(A)= γ-cl(A) ˄ γ-cl(AC) = γ-cl(A)˄ (γ-int A)C. This proves (1). By Definition 3.1, γ-Bd(γ-int(A)) = γ-cl(γ-int(A)) ˄ γ-cl(γ-int A)C = γ-cl(γ-int(A)) ˄ γ-cl(γ-cl(AC))
= γ-cl(γ-sint(A)) ˄ γ-cl(AC) = γ-cl(γ-int(A)) ˄ (γ-int A)C ≤ γ-cl(A) ˄ (γ-int A)C
= γ-Bd(A). Hence (2).
γ-Bd(γ-cl(A)) = γ-cl(γ-cl(A))˄ γ-cl(γ-cl A)C = γ-cl(γ-cl(A)) ˄ [γ-int( γ-cl A)]C
≤ γ-cl(A)˄ (γ-int A)C = γ-Bd(A). Thus proves (3).
A ˄ (γ-BdA)C = A ˄ (γ-cl A ˄ γ-cl AC)C = A ˄ (γ-int AC ˅ γ-int A)
= (A ˄ γ-int AC) ˅ (A ˄ γ-int A) = (A ˄ γ-int AC)˅ γ-int(A) ≥ γ-int(A). Hence (4).
To show that the inequalities (2), (3) and (4) of Proposition 3.6 are in general irreversible, we have the following example.
Example 3.7: (i) Let X = {a, b} and τ = {0, 1,{a.2 , b.1 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.8, b.9 }}. Let A = {a.3 , b.2 }.
Then γ-cl(A)= {a.6 , b.5 } , γ-int(A)= {a.2 , b.2 } and calculations give γ-Bd(A) = {a.6 , b.5 }. This shows that γ-Bd(A) ≰ γ-Bd(γ-int(A)) = {a.3 , b.2 }.
Now we calculate γ-Bd γ-cl(A)= {a.3, b.4 }.This shows that γ-Bd(A) ≰ γ-Bd(γ-cl(A))
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and γ-int(A) = {a.2 , b.2 } ≥/ A ˄ (γ-Bd(A))C = {a.3 , b.2 } ˄ {a.4, b.5 } = {a.3 , b.2 }.
γ-Bd(A ˅ B) ≤ γ-Bd(A) ˅ γ-Bd(B).
We use Lemma 2.5 (viii), (ix) to prove this.
γ-Bd(A˅B) = γ-cl(A˅B) ˄ γ-cl(A˅B)C = γ-cl(A˅B) ˄ γ-cl(AC ˄ BC)
≤ (γ-cl(A) ˅ γ-cl(B)) ˄ (γ-cl (A C) ˄ γ-cl (BC)) ≤ (γ-cl(A) ˄ γ-cl (AC)) ˅ ( γ-cl(B) ˄ γ-cl(BC))
= γ-Bd(A)˅ γ-Bd(B). Hence the Proof.
The reverse in equality in Theorem 3.8 is in general not true as shown by the following example.
Example 3.9: Let X = {a, b} and τ = {0, 1, {a.3 , b.4 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.7, b.6 }}. Let A = {a.3 , b.6 } and
B = {a.4 , b.3 }. Then calculations give γ-Bd(A)= {a.4 , b.5 } and γ-Bd(B)={a.6 , b.5 }. Now A ˅ B = {a.4 , b.6 } and γ-Bd(A˅B) = {a.5 , b.5 }.
This gives that γ-Bd(A) ˅ γ-Bd(B) = {a.6 , b.5 } ≰ γ-Bd(A ˅ B) = {a.5 , b.5 }.
The following example shows that γ-Bd(A ˄ B) ≰ γ-Bd(A) ˄ γ-Bd(B) and
γ-Bd(A) ˄ γ-Bd(B) ≰ γ-Bd(A ˄ B).
Example 3.10: Using the Example 3.9, A = {a.3 , b.6 } and B = {a.4 , b.3 }. Then calculations give γ-Bd(A)= {a.4 ,
b.5 } and γ-Bd(B) = {a.6 , b.5 }. Now A ˄ B = {a.3 , b.3 } and γ-Bd(A ˄ B) = {a.5 , b.4 }.
This gives that γ-Bd(A) ˄ γ-Bd(B) = {a.4 , b.5 } ≰ γ-Bd(A˄B)= {a.5 , b.4 } and γ-Bd(A ˄ B) ≰ γ-Bd(A) ˄ γ-Bd(B).
γ-Bd(A ˄ B) ≤ ( γ-Bd(A) ˄ γ-cl(B)) ˅ (γ-Bd(B) ˄ γ-cl(A)).
We use Lemma 2.5 (viii), (ix) to prove this.
γ-Bd(A ˄ B) = γ-cl(A ˄ B) ˄ γ-cl(A ˄ B)C = γ-cl(A ˄ B) ˄ γ-cl(AC ˅ BC)
≤ (γ-cl(A) ˄ γ-cl(B)) ˄ (γ-cl(AC) ˅ γ-cl(BC))
= (γ-cl(A) ˄ γ-cl(B) ˄ γ-cl(AC)) ˅ (γ-cl(A) ˄γ-cl(B) ˄ γ-cl(BC))
= (γ-Bd(A) ˄ γ-cl(B)) ˅ ( γ-Bd(B) ˄ γ-cl(A)). Hence proved.
γ-Bd(A ˄ B) ≤ γ-Bd(A) ˅ γ-Bd(B).
The reverse in equality in Theorem 3.11 is in general not true as shown by the following example.
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Example 3.13: Using the Example 3.9, A = {a.3 , b.6 }, γ-cl(A)= {a.4 , b.6 }, γ-Bd(A)= {a.4 , b.5 } and B = {a.4 , b.3 },
γ-cl(B)= {a.6 , b.5 }, γ-Bd(B)= {a.6 , b.5 }. Now A ˄ B ={a.3, b.3 } and γ-Bd(A ˄ B) = {a.4 , b.4 }. This gives that
( γ-Bd(A) ˄ γ-cl(B)) ˅ (γ-Bd(B) ˄ γ-cl(A))= {a.4 , b.5 } ≰ γ-Bd(A ˄ B)= {a.4, b.4 }.
(1) γ-Bd( γ-Bd(A)) ≤ γ-Bd(A).
(2) γ-Bd( γ-Bd( γ-Bd(A))) ≤ γ-Bd( γ-Bd(A)).
We use Lemma 2.5 (v) and Definition 3.1 to prove this.
γ-Bd( γ-Bd(A)) = γ-cl( γ-Bd(A)) ˄ γ-cl( γ-Bd(A))C ≤ γ-cl( γ-Bd(A))
= γ-cl (γ-cl(A) ˄ γ-cl(AC)) = γ-cl( γ-cl( A)) ˄ γ-cl( γ-cl(AC))
= γ-cl(A) ˄ γ-cl(AC) = γ-Bd(A). This proves (1).
γ-Bd( γ-Bd( γ-Bd(A))) = γ-cl( γ-Bd(γ-Bd(A))) ˄ γ-cl(γ-Bd( γ-Bd(A))C)
= γ-Bd( γ-Bd(A)) ˄ γ-cl( γ-Bd( γ-Bd(A))C) ≤ γ-Bd( γ-Bd( A)). Hence the proof.
The reverse inequality in Theorem 3.14 is in general not true as shown by the following example.
Example 3.15: Let X = {a, b} and τ = {0, 1, {a.2 , b.2 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.8, b.8 }}. Let A = {a.3 , b.4 }.
Then calculations give γ-Bd(A)= {a.7 , b.7 } ≰ γ-Bd(γ-Bd(A)) = {a.6 , b.7 }.
Again we calculate γ-Bd( γ-Bd( γ-Bd( A))) ={a.4 , b.4 }. This shows that
γ-Bd( γ-Bd(A)) = {a.6 , b.7 } ≰ γ-Bd( γ-Bd( γ-Bd(A))) ={a.4 , b.4 }.
(λ × μ)(x, y) = min { λ(x), μ(y)}, for each X × Y.
Definition 3.17[2]: An fuzzy topological space (X, τ1 ) is a product related to an fuzzy topological space (Y, τ2 ) if for fuzzy sets A of X and B of Y whenever C c ≥/ A and D c ≥/ B implies Cc × 1 ∨ 1 × D c ≥ A × B, where C ∈τ1 and D∈τ2 , there exist C1 ∈ τ1 and D1 ∈ τ2 such that C1 c ≥ A or D1 c ≥ B and C1 c × 1∨1 × D1 c = Cc × 1∨1 × D
c.
(λ ∧ μ) × (υ ∧ ω) = (λ × ω) ∧ (μ × υ).
(1) γ-cl A × γ-cl B ≥ γ-cl(A × B).
(2) γ-int A × γ-int B ≤ γ-int(A × B).
By using Definition 3.16, (γ-cl A × γ-cl B)(x, y) = min{γ-cl A(x), γ-cl B(y)}
≥ min{A(x), B(y)} = (A × B)(x, y). This shows that γ-cl A × γ-cl B ≥ (x, y).
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Thus By Lemma 2.5, γ-cl(A × B) ≤ γ-cl (γ-cl A × γ-cl B) = γ-cl A × γ-cl B.
By using Definition 3.16, (γ-int A × γ-int B)(x, y)= min{γ-int A(x), γ-int B(y)}
≤ min{A(x), B(y)} = (A × B)(x, y). This shows that γ-int A × γ-int B ≤ (x, y).
Thus By Lemma 2.5, γ-int(A × B) ≥ γ-int (γ-int A × γ-int B)=γ-int A × γ-int B.
For fuzzy sets A i’s of X and B j’s of Y, we first note that
(i) inf{ A i , B j}= min{inf A i , inf B j}, (ii) inf{ A i × 1}=(inf A i)× 1,
(iii) inf{1× B j }= 1× (inf B j ) .
In view of above theorem it is sufficient to show that γ-cl(A × B) ≥ γ-cl A × γ-cl B.
Let A I ϵ FγO(X) and B j ϵ FγO(Y). Then, γ-cl(A × B) = inf{(A i ×B j)C/ (A i ×B j)C ≥ A ×B }
= inf {A iC ×1 ˅ 1 ×B jC / A iC ×1 ˅ 1 ×B jC ≥ A ×B }
= inf{ A iC ×1 ˅ 1 ×B jC /A i C ≥ A or B jC ≥ B }
= min ( inf{ A iC ×1 ˅ 1 ×B jC /A i C ≥ A }, inf{A iC ×1 ˅ 1 ×B j C /B j C ≥ B }).
Since inf{A iC ×1 ˅ 1 ×B jC /A i C ≥ A } ≥ inf{A iC ×1 / A i C ≥ A } = inf{A iC /A i C ≥ A } × 1
= γ-cl(A) ×1, inf{A iC ×1 ˅ 1 ×B jC /A i C ≥ A } ≥ inf{ B jC ×1 / B j C ≥ B }
=1 × inf{ B j C / B j C ≥ B } = 1 × γ-cl(B).
Thus we have γ-cl(A × B) ≥ min(γ-cl A × 1, 1 × γ-cl B) = γ-cl A × γ-cl B.
Theorem 3.21: Let Xi, i=1, 2, …n be a family of product related fuzzy topological spaces. If each Ai is a fuzzy set in Xi, then
γ-Bd∏𝑛
𝐴𝑖 = [γ-Bd A1 × γ-cl A2 × …. × γ-cl An ] ˅ [γ-cl A1 × γ-Bd A2 × γ-cl A3 × ….×
γ-cl An ] ˅ ….˅ [γ-cl A1 × γ-cl A2 × ….× γ-Bd An ].
It suffices to prove this for n=2, consider γ-Bd(A1 ˅A2 ) = γ-cl(A1 × A2 ) ˄ [ γ-int(A1 × A2 )]C
= (γ-cl A1 × γ-cl A2 ) ˄ [ γ-int A1 × γ-int A2 ]C = ( γ-cl A1 × γ-cl A2 ) ˄ [( γ-int A1 ˄ γ-cl A1 ) × (γ-int A2 ˄ γ-cl A2 )]C =
( γ-cl A1 × γ-cl A2 ) ˄[ (γ-int A1 × γ-cl A2 ) ˄ ( γ-cl A1 × γ-int A2 ]C
= [( γ-cl A1 × γ-cl A2 ) ˄ ( γ-int A1 × γ-cl A2 )C] ˅ [( ( γ-cl A1 × γ-cl A2 ) ˄ ( γ-cl A1 × γ-int A2 )C]
= [( γ-cl A1 ˄ γ-int A1 ) × γ-cl A2 ] ˅ [ γ-cl A1 × (γ-cl A2 ˄ γ-int A2 )]
=( γ-Bd A1 × γ-clA2 ) ˅ ( γ-cl A1 × γ-Bd A2 ).
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Then γ-Bd( f -1( B)) ≤ f -1(γ-Bd( B)), for any fuzzy subset B in Y.
Let f be a fuzzy continuous function and B be a fuzzy subset in Y. By using Definition 3.1, we have γ-Bd (f -1( B)) =
γ-cl(f -1( B)) ˄ γ-cl(f -1( B))C= γ-cl(f -1( B))˄ γ-cl(f -1( BC)).
Since f is fuzzy continuous and f -1( B) ≤ f -1(γ-cl(B)) it follow that γ-cl(f -1( B)) ≤ f -1(γ-cl(B)). This together with the above imply that γ-Bd( f -1( B)) ≤ f -1(γ-cl(B)) ˄ f -1(γ-cl(BC))
= f -1(γ-cl(B) ˄ γ-cl(BC)). That is γ-Bd( f -1( B)) ≤ f -1(γ-Bd( B)).
In this section we define γ-semi boundary and discuss their properties with examples. Definition 4.1: Let A be a fuzzy set in an fuzzy topological space (X, τ). Then the fuzzy γ-semi boundary of A is defined as γ-sBd(A)= γ-scl(A)˄ γ-scl(AC).
Obviously γ-sBd(A) is a fuzzy γ-semi closed set.
Example 4.3: Let X = {a, b} and τ = {0, 1,{a.8 , b.8 }, {a.2 , b.2 }, {a.3, b.7 } }. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.2 , b.2 }, {a.8 , b.8 }, {a.7 , b.3 }}. Let A = {a.6 , b.9 }. Then γ-scl(A) = {a.8 , b.9 } and γ-sBd(A)= {a. 4 , b.3 }. It follows that γ-scl(A) ≠ {a.6 , b.9 } = A ˅ γ-sBd(A).
(1) γ-sBd( A) = γ-sBd( AC).
(2) If A is fuzzy γ-semi closed, then γ-sBd( A) ≤ A.
(3) If A is fuzzy γ-semi open, then γ-sBd( A) ≤ AC.
(4) Let A ≤ B and B ϵ FγSC(X) (resp., B ϵ FγSO(X)). Then γ-sBd( A) ≤ B (resp.,γ-sBd( A) ≤ BC), where
FγSC(X) (resp., FγSO(X)) denotes the class of fuzzy γ-semi closed (resp., fuzzy γ-semi open) sets in X.
(5) (γ-sBd( A))C = γ-sint(A) ˅ γ-sint( AC).
(6) γ-sBd( A) ≤ Bd( A).
(7) γ-scl(γ-sBd( A)) ≤ Bd( A).
By Definition 4.1, γ-sBd( A)= γ-scl( A) ˄ γ-scl(AC) and
γ-sBd( AC)= γ-scl(AC) ˄ γ-scl(A). Therefore γ-sBd(A) = γ-sBd(AC). Hence (1).
Let A be fuzzy γ-semi closed. By Proposition[7] 6.3(ii), γ-scl(A)= A.
γ-sBd(A)= γ-scl(A)˄ γ-scl(AC) ≤ γ-scl(A) = A. Hence (2).
Let A be fuzzy γ-semi open. By Proposition[7] 5.2(ii), γ-sint(A)= A.
γ-sBd(A) = γ-scl(A) ˄ γ-scl(AC) ≤ γ-scl(AC) = [ γ-sint(A)]C = AC. Hence (3).
Let A ≤ B. Then by Proposition[7] 6.3(iv), γ-scl(A) ≤ γ-scl(B). Since B ϵ F γSC( X), we have γ-scl( B) = B.
This implies that , γ-sBd γ-sBd(A) ≤ γ-scl(A) ≤ γ-scl( B)=B.
That is γ-sBd( A) ≤ B. Let B ϵ FγSO( X). Then BC ϵ FγSC( X). Then by the above,
γ-sBd( A) ≤ BC. Hence (4).
By Definition 4.1, γ-sBd(A)= γ-scl(A)˄ γ-scl(AC). Taking complement on both sides, we get
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[γ-sBd(A)]C = [ γ-scl(A)˄ γ-scl(AC)]C = [γ-scl(A)]C ˅ [ γ-scl(AC)]C
= γ-sint(AC) ˅ γ-sint(A). Hence (5).
Since γ-scl(A) ≤ cl(A) and γ-scl(AC) ≤ cl(AC), then we have
γ-sBd(A) = γ-scl( A) ˄ γ-scl(AC) ≤ cl(A) ˄ cl(AC) = Bd(A). Hence (6).
γ-scl( γ-sBd(A)) = γ-scl( γ-scl(A) ˄ γ-scl(AC)) ≤ γ-scl(γ-scl( A)) ˄ γ-scl(γ-scl( AC))
= γ-scl( A) ˄ γ-scl( AC) = γ-sBd( A) ≤ Bd( A). Thus γ-scl( γ-sBd( A)) ≤ Bd(A).
Hence (7).
The converse of (2) and (3) and reverse inequalities of (6) and (7) in the Proposition 4.4 are in general, not true as is shown by the following example .
Example 4.5: Let A = {a.6 , b.9 } and B = {a.4 , b.1 } in the fuzzy topological space (X, τ) be defined in Example 4.3. Then γ-sBd(A) = {a.4 , b.3 } ≤ A, but A is not fuzzy γ-semi closed.
γ-sBd( B) ={a.4, b.3 } ≤ BC, but B is not fuzzy γ-semi open.
Bd( B)= {a.7 , b.3 } ≰ γ-sBd( B) = {a.4 , b.3 } and Bd( B) ≰ γ-scl(γ-sBd( A)) = {a.5 , b.3 }.
(1) γ-sBd( A) = γ-scl(A) ˄ (γ-sint(A))C, (2) γ-sBd( γ-sint( A)) ≤ γ-sBd(A),
(3) γ-sBd(γ-scl(A)) ≤ γ-sBd(A),
(4) γ-sint(A) ≤ A˄( γ-sBd(A))C.
Proof:
Since γ-scl(AC) = (γ-sint(A))C, we have γ-sBd(A)= γ-scl(A)˄ γ-scl(AC)
= γ-scl(A) ˄ (γ-sint(A))C. This proves (1).
γ-sBd( γ-sint( A)) = γ-scl( γ-sint( A)) ˄ γ-scl( γ-sint(A))C
= γ-scl( γ-sint( A)) ˄ γ-scl( γ-scl(AC)) = γ-scl( γ-sint( A)) ˄ γ-scl( AC)
= γ-scl(γ-sint( A)) ˄ (γ-sint( A))C ≤ γ-scl( A) ˄ (γ-sint( A))C = γ-sBd( A). Hence (2).
γ-sBd(γ-scl( A)) = γ-scl(γ-scl( A)) ˄ γ-scl(γ-scl(A))C
= γ-scl(γ-scl( A)) ˄ [γ-sint( γ-scl(A))]C ≤ γ-scl( A) ˄ (γ-sint( A))C = γ-sBd( A).
Thus proves (3).
A ˄ (γ-sBd( A))C = A ˄ (γ-scl( A) ˄γ-scl( AC))C = A ˄ (γ-sint( AC) ˅ γ-sint( A))
= ( A ˄ γ-sint(AC)) ˅ (A ˄ γ-sint(A)) = (A ˄ γ-sint(AC)) ˅ γ-sint(A) ≥ γ-sint(A).
Hence (4).
To show that the inequalities (2), (3) and (4) of Proposition 4.6 are in general irreversible, we have the following example.
Example 4.7: Choose A= {a.4 , b.1 } in the fuzzy topological space X defined in Example 4.3. Then calculations give
γ-sint(A) = {a.1 , b.1 } and
γ-sBd(A) = {a.4 , b.3 } ≰ γ-sBd(γ-sint(A)) = {a.2 , b.1 }.
The following example shows that γ-sBd(A) ≰ γ-sBd(γ-scl(A)).
Example 4.8: Let X = {a, b} and τ = {0, 1,{a.2 , b.1 }, {a.8 , b.8 }}. Then (X, τ) is a fuzzy topological space. The
family of all fuzzy closed sets of τ is τc = {0, 1, {a.8 , b.9 }, {a.2 , b.2 }}. Let A = {a.6, b.3 }, then calculations give γ-
scl(A) = {a.6 , b.5 } and
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γ-sBd(A) = {a.6 , b.5 } ≰ γ-sBd(γ-scl(A)) = {a.4 , b.5 }.
The following example shows that A ˄ ( γ-sBd(A))C ≰ γ-sint(A).
Example 4.9: Let X = {a, b} and τ = {0, 1,{a.8 , b.8 }, {a.2 , b.2 }, {a.3, b.7 } }. Then (X, τ) is a fuzzy topological
space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.2 , b.2 }, {a.8 , b.8 }, {a.7 , b.3 }}. Let A = {a.4 , b.1 }. Then
γ-sint(A) = {a.2 , b0 } and γ-sBd(A)= {a.4 , b.3 }. It follows that A ˄ [γ-sBd(A)]C = {a.4 , b.1 } ≰ γ-sint(A).
γ-sBd( A ˅ B) ≤ γ-sBd( A) ˅ γ-sBd( B). Proof:
We use Lemma 2.11 to prove this.
γ-sBd(A ˅ B) = γ-scl(A ˅ B) ˄ γ-scl( A ˅ B )C = γ-scl( A ˅ B) ˄ γ-scl( AC ˄ BC)
≤ (γ-scl( A) ˅ γ-scl( B)) ˄ (γ-scl(AC) ˄ γ-scl (BC))
≤ (γ-scl( A) ˄ γ-scl(AC)) ˅ ( γ-scl( B) ˄ γ-scl(BC)) = γ-sBd( A)˅ γ-sBd( B). Hence the Proof.
The reverse in equality in Theorem 4.10 is in general not true as shown by the following example.
Example 4.11: Let X = {a, b} and τ = {0, 1, {a.8 , b.8 }, {a.2 , b.1 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.2, b.2 }, {a.8 , b.9 }}. Let A = {a.4 , b.1 } and B = {a.3 , b.4 }. Then calculations give γ-sBd(A) = {a.4 , b.3 } and
γ-sBd(B) = {a.3, b.6 }. Now A ˅ B = {a.4 , b.4 } and γ-sBd(A ˅ B) = {a.4 , b.5 }. This gives that
γ-sBd( A) ˅ γ-sBd( B) = {a.4 , b.6 } ≰ γ-sBd( A ˅ B) = {a.4 , b.5 }.
The following example shows that γ-sBd( A ˄ B) ≰ γ-sBd( A) ˄ γ-sBd( B) and
γ-sBd( A) ˄ γ-sBd( B) ≰ γ-sBd( A ˄ B).
Example 4.12: Let X = {a, b} and τ = {0, 1, {a.8 , b.6 }, {a.2 , b.3 }}. Then (X, τ) is a fuzzy topological space. The
family of all fuzzy closed sets of τ is τc = {0, 1, {a.2, b.4 }, {a.8 , b.7 }}. Let A = {a.3 , b.4 } and B = {a.6 , b.2 }. Then
calculations give γ-sBd(A)= {a.3 , b.5 } and
γ-sBd( B) ={a.5, b.4 }. Now A ˄ B = {a.3 , b.2 } and γ-sBd( A ˄ B)= {a.4 , b.3 }. This gives that
γ-sBd( A) ˄ γ-sBd( B) = {a.3 , b.4 } ≰ γ-sBd( A ˄ B) = {a.4 , b.3 } and
γ-sBd( A ˄ B) ≰ γ-sBd( A) ˄ γ-sBd( B).
γ-sBd( A ˄ B) ≤ ( γ-sBd( A) ˄ γ-scl( B)) ˅ (γ-sBd( B) ˄ γ-scl(A)).
We use Lemma 2.11 to prove this.
γ-sBd( A ˄ B) = γ-scl( A ˄ B) ˄ γ-scl( A ˄ B)C = γ-scl( A ˄ B) ˄ γ-scl( AC ˅ BC)
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≤ (γ-scl( A) ˄ γ-scl( B)) ˄ (γ-scl(AC) ˅ γ-scl(BC))
= (γ-scl( A) ˄ γ-scl( B) ˄ γ-scl(AC)) ˅ (γ-scl( A) ˄ γ-scl( B) ˄γ-scl(BC))
= (γ-sBd( A) ˄ γ-scl(B)) ˅ ( γ-sBd( B) ˄ γ-scl(A)). Hence proved.
γ-sBd( A ˄ B) ≤ γ-sBd( A) ˅ γ-sBd( B).
The reverse in equality in Theorem 4.13 is in general not true as shown by the following example.
Example 4.15: Let X = {a, b} and τ = {0, 1, {a.3 , b.2 }, {a.6 , b.8 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.7, b.8 }, {a.4 , b.2 }}. Let A = {a.4 , b.6 } and B = {a.5 , b.3 }. Then calculations give γ-sBd(A)= {a.5 , b.4 },
γ-scl(A) = {a.5 , b.6 }, γ-scl( B) = {a.5 , b.4 } and γ-sBd( B) ={a.5 , b.4 }. Now A ˄ B = {a.4 , b.3 } and γ-sBd( A ˄ B)= {a.4 , b.4 }. This gives that
( γ-sBd( A) ˄ γ-scl( B)) ˅ (γ-sBd( B) ˄ γ-scl( A)) = {a.5 , b.4 } ≰ γ-sBd( A ˄ B)= {a.4 , b.4 }.
(1) γ-sBd( γ-sBd( A)) ≤ γ-sBd( A).
(2) γ-sBd( γ-sBd( γ-sBd( A))) ≤ γ-sBd( γ-sBd( A)).
γ-sBd( γ-sBd( A)) = γ-scl( γ-sBd( A)) ˄ γ-scl( γ-sBd( A))C
≤ γ-scl( γ-sBd(A)) = γ-scl( γ-scl( A) ˄ γ-scl(AC)) = γ-scl(γ-scl(A)) ˄ γ-scl(γ-scl(AC))
= γ-scl(A) ˄ γ-scl(AC) = γ-sBd(A). This proves (1).
γ-sBd( γ-sBd( γ-sBd( A))) = γ-scl( γ-sBd( γ-sBd( A))) ˄ γ-scl( γ-sBd( γ-sBd( A))C)
= γ-sBd( γ-sBd( A)) ˄ γ-scl( γ-sBd( γ-sBd( A))C) ≤ γ-sBd( γ-sBd( A)). Hence the proof.
The reverse inequality in Theorem 4.16 is in general not true as shown by the following example.
Example 4.17: Let X = {a, b} and τ = {0, 1, {a.3 , b.2 }, {a.6 , b.8 }}. Then (X, τ) is a fuzzy topological space. The family of all fuzzy closed sets of τ is τc = {0, 1, {a.7 , b.8 }, {a.4 , b.2 }}. Let A = {a.4, b.6 }.Then calculations give
γ-sBd(A) = {a.5 , b.7 } ≰ γ-sBd(γ-sBd(A))={a. 5 , b.6 }.
It follows that γ-sBd( γ-sBd( A)) = {a.5 , b.6 } ≰ γ-sBd( γ-sBd( γ-sBd( A))) ={a.5 , b.5 }.
is a fuzzy subset of a fuzzy topological space Y. Then
(1) γ-scl A × γ-scl B ≥ γ-scl(A × B)
(2) γ-sint A × γ-sint B ≤ γ-sint (A × B). Proof:
By using Definition 3.16, (γ-scl A × γ-scl B)(x, y) = min{γ-scl A(x), γ-scl B(y)}
≥ min{A(x), B(y)} = (A × B) (x, y). This shows that γ-scl A × γ-scl B ≥ (x, y).
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Thus By Lemma 2.10, γ-scl( A × B) ≤ γ-scl (γ-scl A × γ-scl B) = γ-scl A × γ-scl B.
By using Definition 3.16, (γ-sint A × γ-sint B)(x, y) = min{γ-sint A(x), γ-sint B(y)}
≤ min{A(x), B(y)} = (A × B)(x, y). This shows that γ-sint A × γ-sint B ≤ (x, y).
Thus By Lemma 2.10, γ-sint(A × B) ≥ γ-sint (γ-sint A × γ-sint B) = γ-sint A × γ-sint B.
of X and a fuzzy set B of Y γ-scl (A × B) = γ-scl A × γ-scl B.
Proof:
For fuzzy sets A i’s of X and B j’s of Y, we first note that
(i) inf{ A i , B j}= min{inf A i , inf B j}, (ii) inf{ A i × 1}= (inf A i)× 1,
(iii) inf{1× B j }= 1× (inf B j ) .
In view of above theorem it is sufficient to show that γ-scl(A × B) ≥ γ-scl A × γ-scl B. Let A i ϵ FγSO(X) and B j ϵ FγSO(Y). Then, γ-scl(A × B) = inf{( A I × B j )C /(A i × B j )C
≥ A ×B }= inf{A iC ×1 ˅ 1 ×B jC / A iC ×1 ˅ 1 ×B jC ≥ A ×B }
= inf{ A iC ×1 ˅ 1 ×B jC /A i C ≥ A or B jC ≥ B }
= min (inf { A iC ×1 ˅ 1 ×B jC /A i C ≥ A }, inf {A iC ×1 ˅ 1 ×B j C /B j C ≥ B }). Since inf{A iC ×1 ˅ 1 ×B jC /A i C ≥ A } ≥ inf{A iC ×1 /A i C ≥ A }
= inf{ A iC /A i C ≥ A } × 1 = γ-cl(A) × 1, inf {A iC × 1 ˅ 1 × B jC / A i C ≥ A }
≥ inf {B jC ×1 /B j C ≥ B } = 1 × inf{ B j C / B j C ≥ B }= 1 × γ-cl(B).
Thus we have γ-scl( A × B) ≥ min( γ-scl A × 1, 1 × γ-scl B) = γ-scl A × γ-scl B.
Theorem 4.20: Let Xi, i=1, 2, …n be a family of product related fuzzy topological spaces. If each Ai is a fuzzy set in Xi, then
γ-sBd∏𝑛
𝐴𝑖 = [γ-sBd(A1 ) × γ-scl(A2 ) × ….× γ-scl(An )] ˅ [γ-scl(A1 ) × γ-sBd(A2 ) × γ-scl(A3 ) × ….× γ-scl(An ) ] ˅
….˅ [γ-scl(A1 ) × γ-scl(A2 ) × ….× γ-sBd(An )].
It suffices to prove this for n=2, consider γ-sBd( A1 ˅ A2 ) = γ-scl( A1 × A2 ) ˄ [ γ-sint( A1 × A2 )]C = (γ-scl(A1 ) × γ- scl (A2 )) ˄ [ γ-sint(A1 ) × γ-sint(A2 )]C
= ( γ-scl(A1 ) × γ-scl (A2 )) ˄ [( γ-sint(A1 ) ˄ γ-scl(A1 )) × (γ-sint(A2 ) ˄ γ-scl(A2 ))]C
= ( γ-scl(A1 ) × γ-scl (A2 )) ˄ [ (γ-sint(A1 ) × γ-scl(A2 )) ˄ ( γ-scl(A1 ) × γ-sint(A2 ) ]C
= [( γ-scl(A1 ) × γ-scl (A2 )) ˄ ( γ-sint(A1 ) × γ-scl(A2 ))C] ˅ [( ( γ-scl(A1 ) × γ-scl (A2 )) ˄( γ-scl(A1 ) × γ-sint(A2 ))C] = [( γ-scl(A1 ) ˄ γ-sint(A1 )) × γ-scl(A2 )] ˅ [ γ-scl(A1 ) × (γ-scl(A2 ) ˄ γ-sint(A2 ))]
= ( γ-sBd(A1 ) × γ-scl(A2 )) ˅ ( γ-scl(A1 ) × γ-sBd(A2 )).
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Then γ-sBd( f -1( B)) ≤ f -1(γ-sBd( B)), for any fuzzy subset B in Y.
Let f be a fuzzy continuous function and B be a fuzzy subset in Y. By using Definition 4.1, we have γ-sBd (f -1( B))
= γ-scl(f -1( B)) ˄ γ-scl(f -1( B))C= γ-scl(f -1( B))˄ γ-scl(f -1( BC)).
Since f is fuzzy continuous and f -1( B) ≤ f -1(γ-scl(B)) it follow that γ-scl(f -1( B)) ≤ f -1(γ-scl(B)). This together with the above imply that γ-sBd( f -1( B)) ≤ f -1(γ-scl(B)) ˄ f -1(γ-scl(BC)) = f -1(γ-scl(B) ˄ γ-scl(BC)).
That is γ-sBd( f -1( B)) ≤ f -1(γ-sBd( B)).
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