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Yet more proofs on intersections and unions of sets

Prove that A \cup (A \cap B) = A and A \cap (A \cup B) = A.


Proof. First, it is clear that A \subseteq A \cup (A \cap B) (see Exercise 12 of Section I.2.5, or simply note that x \in A implies x \in A \cup (A \cap B) since x is in A).
For the reverse inclusion, if x is any element of A \cup (A \cap B) then x \in A or x \in (A \cap B). But x \in A \cap B implies x \in A (and x \in B). So, in either case x \in A; hence, A \cup (A \cap B) \subseteq A.
Therefore, A \cup (A \cap B) = A.∎

Proof. From Exericse 12 (Section I.2.5) we know that A \cap C \subseteq A for any set C; hence, A \cap (A \cup B) \subseteq A.
For the reverse inclusion, if x is any element in A, then x \in A and x \in A \cup B; hence, x \in A \cap (A \cup B). Therefore, A \cap (A \cup B) \subseteq A.
Hence, A \cap (A \cup B) = A.∎

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