7.1.21

Let $X$ be a nonempty set and let $\mathcal{P}(X)$ be the set of all subsets of $X$ (the power set of $X$). Define addition and multiplication on $\mathcal{P}(X)$ by: $$A+B \equiv (A-B) \cup (B-A)$$ and $$A \cdot B \equiv A \cap B$$ i.e. addition is the symmetric difference of sets and multiplication is intersection of sets.
(a) Prove that $\mathcal{P}(X)$ is a ring under these operations. Note: $\mathcal{P}(X)$ and its subrings are often referred to as rings of sets.
(b) Prove that this ring is commutative, has an identity and is a Boolean ring.

$\textit{Proof}$ : To show that $\mathcal{P}(X)$ is a ring, it suffices to show that ($\mathcal{P}(X),+,\cdot)$ satisfies the 3 axions of rings. Using basic set theory, we see that
(i) $(\mathcal{P}(X),+)$ is an Abelian group since $$A+B=(A-B)\cup (B-A)=(B-A) \cup (A-B)=B+A$$ (ii) $\cdot$ is associative since $$A\cdot(B\cdot C)=A \cap (B \cap C)=(A \cap B) \cap C= (A \cdot B) \cdot C$$ (iii) ($\mathcal{P}(X),+,\cdot$) also satisfies distributive laws since $$(A+B) \cdot C=((A-B) \cup (B-A)) \cap C$$ $$=((A\cap C)-(B\cap C)) \cup ((B \cap C)-(A \cap C))$$ $$=(A \cap C) + (B \cap C) = A \cdot C + B \cdot C$$ Right distribution is shown similarly.
So, ($\mathcal{P}(X),+,\cdot$) is a ring.

$\cdot$ is a commutative operation since $$A \cdot B=A \cap B = B \cap A = B \cdot A$$ The universe or entire set $X$ is s.t. $$A \cdot X= A \cap X= X \cap A = A \quad \forall A \in{\mathcal{P}(X)}$$ So, $X$ is the (multiplicative) identity. As a note, $\emptyset$ is the additive identity since $$A + \emptyset = (A-\emptyset) \cup (\emptyset - A) = A \cup \emptyset = \emptyset + A=A$$ Finally, ($\mathcal{P}(X),+,\cdot$) is a Boolean ring since $$A^2=A \cdot A = A \cap A = A$$ $ \therefore$ ($\mathcal{P}(X),+,\cdot$) is a Boolean ring (automatically commutative also by exercise 7.1.15) with identity. $\blacksquare$