KM-Fuzzy Approach Space - EUSFLAT

FA5. F(x,A,t + s) ≥ F(x,A (r) ,t), ifs> r ≥ 0,
F(x,A,+∞) ≥ F(x,A (+∞) ,t),
where A (r) = {y ∈ X : F(y,A,t)=1,∀t > r},
and A (+∞) = {y ∈ X : F(y,A,+∞)=1};
FA6. F(x,A,·): [0,+∞) → [0,1] is left-continuous for all x ∈
X and A ⊂ X.
The map F is symmetric (or the FA-space (X,F) is symmetric)
if
F(x,{y},t)=F(y,{x},t) for all x,y ∈ X, t > 0.
Firstly, we will prove that there are two families of spaces
satisfying the above six properties.
Lemma 2.2 (A-space ↩→ FA-space) Let (X,δ) be an Aspace.
If we define a fuzzy set Fδ on X × P (X) × [0,+∞] as
follows:
Fδ (x,A,t)=0 if t ≤ δ(x,A),
Fδ (x,A,t)=1 if t > δ(x,A),
then (X,F δ) is a FA-space.
Proof. FA1, FA2 and FA3 are immediate. In order to
prove FA4, if A,B ⊆ X we can suppose min(δ(x,A),δ(x,B)) =
δ(x,A) ≤ δ(x,B) and, then max(F(x,A,t),F(x,B,t)) =
F(x,A,t). Thus Fδ (x,A ∪ B,t) =0 for t ≤ δ(x,A); and
Fδ (x,A ∪ B,t) =1 for t > δ(x,A). This completes the proof
of FA4.
For every r ∈ [0,+∞), the set A (r) is the same in A-space
and in FA-space, since
A (r)
A−space = x ∈ X δ(x,A) ≤ r
= x ∈ X Fδ (x,A,t)=1,∀t > r = A (r)
FA−space .
Let us suppose that s > r. Since δ(x,A) ≤ δ(x,A (r) )+r, thus
for t ≤ δ(x,A) − s, wehave
t + r < t + s ≤ δ(x,A) ≤ δ(x,A (r) )+r;
i.e., t < δ(x,A (r) ) and F δ(x,A,t + s)=F δ(x,A (r) ,t)=0 for t +
s ≤ δ(x,A).
Otherwise, if t + s > δ(x,A), then F δ(x,A,t + s) =1 ≥
F δ(x,A (r) ,t). This completes the proof.
Theorem 2.3 (FM-space ↩→ FA-space) Let (X,M,∗) be a
FM-space. If we define a fuzzy set F on X × P (X) × [0,+∞]
for every x ∈ X, t ∈ [0,∞] and A ⊆ X as follows:
FM (x,A,t)=sup M (x,a,t),
a∈A
then (X,FM) is a FA-space.
Proof. By FM1 and FM2, conditions FA1 and FA3 hold.
The proof of the second condition FA2 follows from the
definition of supA for empty sets, sup/0 = 0. If x ∈ X and
A,B ⊆ X are subsets of X, then
Max
sup M (x,a,t), sup M (x,b,t)
a∈A
b∈B
ISBN: 978-989-95079-6-8
IFSA-EUSFLAT 2009
= sup M (x,a,t).
a∈A∪B
Consequently FA4 is proved.
We will prove now FA5. Let x ∈ X, A ∈ P (X) and r ∈ [0,∞].
If A = /0, the condition FA5 is trivial. It is enough to prove that
the property FA5 holds true for every A = /0. Since, for every
x,y,z ∈ X, t,s > 0
in particular,
M(x,z,t + s) ≥ M(x,y,t) ∗ M(y,z,s);
M(x,y,t + s) ≥ M(x,a,t) ∗ M(a,y,s)
for every x ∈ X, a ∈ A (r) , y ∈ A, t > 0,s > r ≥ 0.
Taking supremum over y ∈ A, then
FM (x,A,t + s)=sup M (x,y,t + s)
y∈A
≥ M(x,a,t) ∗ sup M(a,y,s)
y∈A
= M(x,a,t) ∗ FM(a,A,s).
Since, a ∈ A (r) , then FM(a,A,s)=1 for s > r. Thus
FM (x,A,t + s) ≥ M(x,a,t).
Taking supremum over a ∈ A (r) ,wehave
FM (x,A,t + s) ≥ FM(x,A (r) ,t).
Note that additionally the following condition is satisfied:
FA7 Let x,y ∈ X be such that FM (x,{y},t)=1 for every t > 0,
then x = y.
3 Characterization
Now we ask about the maps F : X × P (X) × [0,+∞] → [0,1]
that could provide to X of a FA-space structure. In this cases,
it could be interesting to weaken axioms FA4 and FA5.
Proposition 3.1 Let X be a non-empty set, F : X × P (X) ×
[0,+∞] → [0,1] be a map satisfying axiom FA2 and such that
F (x,A,t)=1 for every t > 0 and x ∈ A ⊆ X. Then:
The axiom FA4 is equivalent to
FA4 ′ . F (x,A ∪ B,t) ≥ Max(F (x,A,t),F (x,B,t)) for t > 0,
for every non-empty subsets A,B ⊆ X such that A = B
and A ∪ B ⊂ X, and for each x ∈ X\(A ∪ B).
The axiom FA5 is equivalent to
FA5 ′ . F(x,A,t +s) ≥ F(x,A (r) ,t) for t > 0, for every s > r ≥
0, /0 = A ⊂ X and for each x ∈ X\A.
Proof. We prove that there exists some cases in which FA4
and FA5 are always true.
Property FA4 ′ .IfA is empty, then F (x,A ∪ B,t)=F (x,B,t)
and, by the axiom FA2
Max(F (x,A,t),F (x,B,t)) = Max(F (x, /0,t),F (x,B,t))
= Max(0,F (x,B,t)) = F (x,B,t).
A similar conclusion is valid for B = /0. We will suppose that
A and B are non-empty sets. If x ∈ A ∪ B, then, by hypothesis,
F (x,A ∪ B,t)=1 ≥ Max(F (x,A,t),F (x,B,t)).
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