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X
,itis
Notice that when
μ
A
∈{
0
,
1
}
•[
μ
(
0
)
A
]=
X
, since for all
x
∈
X
it is 0
μ(
x
)
[
μ
(
r
)
A
•
If
r
>
0,
]=
A
, since for all
x
∈
A
it is 0
<
r
1
=
μ
A
(
x
),
then,
the only
r-cuts of a crisp subset
A
of
X
are
X
and
A
.
If
r
[
μ
(
s
)
]ↂ[
μ
(
r
)
]
μ(
)
μ(
)
s
, since
s
x
implies
r
x
,itresults
: r-cuts are
decreasing when their indices increase.
Let us show an example with
X
={
1
,
2
,
3
,
4
,
5
}
and
μ
=
0
.
8
/
1
+
0
.
6
/
2
+
0
.
7
/
3
+
1
5, where the only significative values for the
r
-cuts are 0.6, 0.7, 0.8, and 1:
•[
μ
(
0
.
6
)
]={
/
4
+
1
/
1
,
2
,
3
,
4
,
5
}=
X
•[
μ
(
0
.
7
)
]={
1
,
3
,
4
,
5
}
•[
μ
(
0
.
8
)
]={
1
,
4
,
5
}
•[
μ
(
1
)
]={
.
It is clear that 0
4
,
5
}
[
μ
(
1
)
]ↂ[
μ
(
0
.
8
)
]ↂ[
μ
(
0
.
7
)
]ↂ[
μ
(
0
.
6
)
]
.
6
0
.
7
0
.
8
1, and
.
X
,is
Theorem 2.1.1
(Theorem of resolution)
Fo r a l l
μ
∈[
0
,
1
]
μ(
x
)
=
max
{
r
∈
), μ
(
r
)
(
[
0
,
1
];
min
(μ
r
(
x
x
)
}
, for all x
∈
X.
Proof
), μ
(
r
)
(
, μ
(
r
)
(
0
r
1
min
max
(μ
r
(
x
x
))
=
0
r
1
min
max
(
r
x
))
0
r
1
min
r
1
r
,
if
r
μ(
x
),
,
if
r
μ(
x
),
=
max
,
=
max
0
r
1
=
μ(
x
).
0
,
otherwise,
0
,
otherwise,
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