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, $A \cap (A \cup B) = A$ .∎