⇒ ∈ ∪
∈
⇒ ∈ ∪
∪
x A B
x C
x A B
C
(
)
(
)
.
or
Therefore we have
A
B C
A B
C
∪
∪
⊂ ∪
∪
(
) (
)
(1)
(ii) Let us now show that
(
)
(
).
A B
C A
B C
∪
∪ ⊂ ∪
∪
Assume that y is any element of the set (
)
A B
C
∪
∪
y A B
C
y A B
y C
y A
y B
y C
y A
y B
∈ ∪
∪
⇒ ∈ ∪
∈
⇒ ∈
∈
∈
⇒ ∈
∈
(
)
(
)
(
)
(
or
or
or
or
or y C
y A
B C
∈
⇒ ∈ ∪
∪
)
(
)
Therefore we have
(
)
(
)
A B
C A
B C
∪
∪ ⊂ ∪
∪
(2)
From (1) and (2), we have
A
B C
A B
C
∪
∪
=
∪
∪
(
) (
)
Ì Exam ple 0.1.4: Prove that the intersection of sets is associative i.e., if
A, B and C are three sets, then
A
B C
A B
C
∩
∩
=
∩
∩
(
) (
)
.
Solu tion
Let us prove that
A
B C
A B
C
∩
∩
⊂ ∩
∩
(
) (
)
Let x be an element such that
x A
B C
x A
x B C
x A
x B
x C
x A
x
∈ ∩
∩
⇒ ∈
∈ ∩
⇒ ∈
∈
∈
⇒ ∈
(
)
(
)
(
)
(
and
and
and
and ∈
∈
⇒ ∈ ∩
∈
⇒ ∈ ∩
∩
B
x C
x A B
x C
x A B
C
)
(
)
(
)
and
and
There fore
A
B C
A B
C
∩
∩
⊂ ∩
∩
(
) (
)
(1)
Let us prove that
(
)
(
)
A B
C A
B C
∩
∩ ⊂ ∩
∩ .
Let us assume the element y A B
C
∈ ∩
∩
(
)
⇒ ∈ ∩
∈
y A B
y C
(
) and
Introduction
5