Prove that for a class of sets we have,
Proof. Let





















For the reverse inclusion, let












Hence,

Proof. Let be any element of
. This means that
and
. Further,
not in the intersection of the sets
means that there is at least one
, say
, such that
. Since
and
, we know
. Then we can conclude
. Thus,
.
For the reverse inclusion, we let be any element in
. This means there is at least one
, say
, such that
, which means
and
. Since
, we know
; therefore,
. Hence,
.
Therefore, ∎