MAT 125a, HW1 Solutions
5
(a) (i) First assume
x
∈
B
∩ {∪
i
∈
I
X
i
}
. Since
x
∈
B
∩ {∪
i
∈
I
X
i
}
we have that
x
∈
B
and
∃
j
∈
I
such that
x
∈
X
j
. Since
x
∈
B
and
x
∈
X
j
we have
x
∈
B
∩
X
j
⊂ ∪
i
∈
I
(
B
∩
X
i
).
So
x
∈
B
∩ {∪
i
∈
I
X
i
}
implies
x
∈ ∪
i
∈
I
(
B
∩
X
i
) hence
B
∩ {∪
i
∈
I
X
i
} ⊂ ∪
i
∈
I
(
B
∩
X
i
).
(ii) Next assume
x
∈ ∪
i
∈
I
(
B
∩
X
i
). Since
x
∈ ∪
i
∈
I
(
B
∩
X
i
)
∃
j
∈
I
such that
x
∈
B
∩
X
j
.
Since
x
∈
B
∩
X
j
we have that
x
∈
B
and
x
∈
X
j
hence
x
∈
B
and
x
∈ ∪
i
∈
I
X
i
which
implies that
x
∈
B
∩ {∪
i
∈
I
X
i
}
. Therefore
∪
i
∈
I
(
B
∩
X
i
)
⊂
B
∩ {∪
i
∈
I
X
i
}
.
Therefore by (i) and (ii) we have
∪
i
∈
I
(
B
∩
X
i
) =
B
∩ {∪
i
∈
I
X
i
}
.
(b) (i) First assume
x
∈
(
∪
i
∈
I
X
i
)
{
. Since
x
∈
(
∪
i
∈
I
X
i
)
{
we have that
x
∈
S
and
x
6∈ ∪
i
∈
I
X
i
which implies that
x
∈
S
and
x
6∈
X
i
∀
i
∈
I
. Since
x
∈
S
and
∀
i
∈
I s
6∈
X
i
we have
that
x
∈
X
{
i
∀
i
∈
I
and hence
x
∈ ∩
i
∈
I
X
{
i
. Therefore
x
∈
(
∪
i
∈
I
X
i
)
{
⊂ ∩
i
∈
I
X
{
i
.
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 Spring '10
 anne
 Number Theory, Empty set, ∪i∈I Xi

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