Proposition 5.8.
Assuming that:
(
A
j
)
j
∈
J
,
U
,
∼
as usual
n
∈
ℕ
for each
j
we have
B
j
,
C
j
⊆
A
j
n
Then
(1)
[
(
B
j
)
j
∈
J
]
∩
[
(
C
j
)
j
∈
J
]
=
[
(
B
j
∩
C
j
)
j
∈
J
]
(2)
[
(
B
j
)
j
∈
J
]
∪
[
(
C
j
)
j
∈
J
]
=
[
(
B
j
∪
C
j
)
j
∈
J
]
(3)
[
(
B
j
)
j
∈
J
]
∖
[
(
C
j
)
j
∈
J
]
=
[
(
B
j
∖
C
j
)
j
∈
J
]
(4)
If
n
>
0
,
k
∈
{
1
,
…
,
n
}
then
π
k
(
[
(
B
j
)
j
∈
J
]
)
=
[
(
π
k
(
B
j
)
)
j
∈
J
]
.