
Let x ∈⋃
f ∈F
ðB ∩ A
f
Þ: Then, by definition of union of a family of sets, x is an element of at least one of the
sets that form the union. So, x ∈ B ∩ A
f
for at least one f ∈ F, call tf
1
. Then x ∈ B ∩ A
f
1
. By definition of
intersection, this implies that x ∈ B and x ∈ A
f
1
. The fact that x ∈ A
f
1
implies that x is an element of the
union of A
f
1
with any other set. Thus, x ∈ ð⋃
f ∈F
A
f
Þ: Also x ∈ B . So, x ∈ B ∩ ð⋃
f ∈F
A
f
Þ: This proves that
B ∩ ð⋃
f ∈F
A
f
Þ⊇⋃
f ∈F
ðB ∩ A
f
Þ: The two inclusions prove that the sets are equal.
See the set of exercises below for the second proof.
■
Exercises
18. Consider the family {A
n
for n ∈ ℕ} with {A
n
=[−n, n]={x ∈ ℝ|−n ≤ x ≤ n}. Find ...