Chapter 6.2, Problem 6ES

Let $\cap$ and $\cup$ stand for the words “intersection” and “union,” respectively. Fill in the blanks in the following proof that for all sets A, B, and C, $A\cap \left(B\cup C\right)=\left(A\cap B\right)\cup \left(A\cap C\right)$.

Proof: Suppose A, B and C are any sets.

(1) Proof that $A\cap \left(B\cap C\right)\subseteq \left(A\cap B\right)\cap \left(A\cap C\right):$

Let $x\in A\cap \left(B\cup C\right)$ . [We must show that $x\in$(a) ].

By definition of $\cap$ , $x\in$(b) and $x\in B\cup C$ .

Thus $x\in A$ and, by definition of $\cup ,x\in B$ or (c) .

Case 1 ( $x\in A$ and $x\in B$ ): In this case, $x\in A\cap B$ by definition of $\cup$ .

Case 2 ( $x\in A$ and $x\in C$ ): In this case, $x\in A\cap C$ by definition of $\cup$ .

By cases 1 and 2, $x\in A\cap B$ or $x\in A\cap C$ , and so, by definition of $\cup$ . (d) .
[So $A\cap \left(B\cup C\right)\subseteq \left(A\cap B\right)\cup \left(A\cap C\right)$by definition of subset.]
(2) Proof that $\left(A\cap B\right)\cup \left(A\cap C\right)\subseteq A\cap \left(B\cup C\right):$

Let $x\in \left(A\cap B\right)\cup \left(A\cap C\right)$ . [We must show that $x\in A\cap \left(B\cup C\right)$ .]
By definition of $\cup$ , $x\in A\cap B$ (a) $x\in A\cap C$ .

Case 1 $\left(x\in A\cap B\right):$ In this case, by definition of $\cup$ , $x\in A$ and $x\in B$ .

Since $x\in B$ , then $x\in B\cup C$ by definition of $\cup$ .

Case 2 $\left(x\in A\cap C\right):$ In this case, by definition of $\cup$ , $x\in A$(b) $x\in C$ .

Since $x\in C$ , then $x\in B\cup C$ by definition of $\cup$ .

In both cases $x\in A$ and $x\in B\cup C$ , and so, by definition of $\cup$ , (c) .

[So $\left(A\cap B\right)\cup \left(A\cap C\right)\subseteq A\cap \left(B\cap C\right)$by definition of (d) .]
(3) Conclusion: [Since both subset relations have been proved, it follows, by definition of set equality, that (a).]