Prove distributive laws for unions and intersections of sets

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

Proof ( $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ ). Let $x$ be an arbitrary element of $A \cap (B \cup C)$ . This means that $x \in A$ and $x \in (B \cup C)$ . Further, since $x \in (B \cup C)$ we know $x \in B$ or $x \in C$ . But then, this implies $x \in (A \cap B)$ or $x \in (A \cap C)$ depending on whether $x \in B$ or $x \in C$ , respectively. Then, since $x \in (A \cap B)$ or $x \in (A \cap C)$ , we have $x \in (A \cap B) \cup (A \cap C)$ . Thus, $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$ .
For the opposite inclusion, we let $x$ be an arbitrary element of $(A \cap B) \cup (A \cap C)$ . This means that $x \in (A \cap B)$ or $x \in (A \cap C)$ . If $x \in (A \cap B)$ then $x \in A$ and $x \in B$ , while if $x \in (A \cap C)$ we have $x \in A$ and $x \in C$ . Thus, we have $x \in A$ no matter what, and either $x \in B$ or $x \in C$ . Since $x \in B$ or $x \in C$ , we know $x \in (B \cup C)$ . Since we already had that $x \in A$ no matter what, we now have $x \in A \cap (B \cup C)$ . Therefore, $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$ .
Hence, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .∎

Proof ( $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ ). Let $x$ be an arbitrary element of $A \cup (B \cap C)$ . This means $x \in A$ or $x \in (B \cap C)$ . If $x \in A$ then $x \in A \cup B$ and $x \in A \cup C$ . Hence, $x \in (A \cup B) \cap (A \cup C)$ . Otherwise, if $x \in (B \cap C)$ then $x \in B$ and $c \in C$ . Hence, $x \in A \cup B$ and $x \in A \cup C$ ; therefore, $x \in (A \cup B) \cap (A \cup C)$ . Therefore, $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ .
For the reverse inclusion, let $x$ be an arbitrary element of $(A \cup B) \cap (A \cup C)$ . This means $x \in (A \cup B)$ and $x \in (A \cup C)$ . This implies that $x \in A$ or $x \in B$ and $x \in C$ (since if $x \notin A$ then the fact that $x \in A \cup B$ and $x \in A \cup C$ means $x$ must be in both $B$ and $C$ ). If $x \in A$ , then $x \in A \cup (B \cap C)$ . On the other hand, if $x \in B$ and $x \in C$ , then $x \in B \cap C$ . Hence, $x \in A \cup (B \cap C)$ . Therefore, $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$ .
Therefore, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ .∎

One comment

July 20, 2016 at 2:32 am

David He says:

Hey Rori,

Noticed that on the second proof in the line “then x \in B and c \in C.” should be x \in C

Cheers!

Stumbling Robot

Prove distributive laws for unions and intersections of sets

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