1. (i) Prove that A UB=BU A [Hints. Show that A UB CBU A and B UA SAU B] (ii) Prove that An B=BnA
Answers
Answer:
Step-by-step explanation:
Solution. (a) Prove that A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C). Suppose x ∈
A∩(B ∪C), then x ∈ A and x ∈ B ∪C. The latter means that either x ∈ B,
or x ∈ C. If x ∈ B, then x ∈ A ∩ B, and if x ∈ C, then x ∈ A ∩ C. Thus,
either x ∈ A ∩ B, or x ∈ A ∩ C, i. e. x ∈ (A ∩ B) ∪ (A ∩ C).
Prove the converse inclusion. Take x ∈ (A ∩ B) ∪ (A ∩ C). So, either
x ∈ A ∩ B, or x ∈ A ∩ C. In both cases x ∈ A. If x ∈ A ∩ B, then x ∈ B,
and if x ∈ A ∩ C, then x ∈ C. So, either x ∈ B, or x ∈ C, i. e. x ∈ B ∪ C.
Thus, we proved x ∈ A and x ∈ B ∪ C, hence x ∈ A ∪ (B ∩ C).
(b) Prove the inclusion A∪(B∩C) ⊂ (A∪B)∩(A∪C). Let x ∈ A∪(B∩C).
Then either x ∈ A, or x ∈ B ∩C. In the first case x ∈ A, which is a subset of
both A∪B and A∪C. So, x ∈ (A∪B)∩(A∪C). In the second case x ∈ B∩C,
which is a subset of B ⊂ A∪B and C ⊂ A∪ C. Thus, x ∈ (A∪B)∩(A∪ C)
in this case.
Now prove the converse inclusion. Choose x ∈ (A ∪ B) ∩ (A ∪ C). So
x ∈ A∪B and x ∈ A∪ C. If x ∈ A, then obviously x ∈ A∪(B ∩ C), because
A ⊂ A∪(B ∩C). Otherwise x ∈ B and x ∈ C, i. e. x ∈ B ∩C ⊂ A∪(B ∩C).