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## The powerset of a set is a Boolean ring under intersection and symmetric difference

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.21

Let $X$ be a nonempty set and let $R = \mathcal{P}(X)$ denote the power set of $X$. Define operators $+$ and $\cdot$ on $R$ as follows: $$A + B = (A \setminus B) \cup (B \setminus A)$$ and $A \cdot B = A \cap B$.

(1) Prove that $R$ is a ring under these operators.
(2) Prove that this ring is commutative, has an identity, and is Boolean.

Solution: Before we begin, note that for all sets $X$, $Y$, and $Z$, the following hold. $$(X \cup Y) \setminus Z = (X \setminus Z) \cup (Y \setminus Z)$$ $$X \setminus (Y \setminus Z) = (X \setminus Y) \cup (X \cap Z)$$ $$X \setminus (Y \cup Z) = (X \setminus Y) \cap (X \setminus Z)$$ $$X \cap (Y \setminus Z) = (X \cap Y) \setminus (X \cap Z)$$ (1) Now let $A,B,C \subseteq X$. We have the following.\begin{align*}&\ (A + B) + C\\ =&\ ((A \setminus B) \cup (B \setminus A)) + C\\
=&\ (((A \setminus B) \cup (B \setminus A)) \setminus C) \cup (C \setminus ((A \setminus B) \cup (B \setminus A)))\\
=&\  ((A \setminus B) \setminus C) \cup ((B \setminus A) \setminus C) \cup ((C \setminus(A \setminus B)) \cap (C \setminus (B \setminus A)))\\
=&\ (A \setminus (B \cup C)) \cup (B \setminus (A \cup C)) \cup (((C \setminus A) \cup (C \cap B)) \cap ((C \setminus B) \cup (C \cap A)))\\
=&\  [(A \setminus B) \cap (A \setminus C)] \cup [(B \setminus A) \cap (B \setminus C)] \cup [(C \setminus A) \cap (C \setminus B)] \cup [(C \setminus A) \cap C \cap A] \cup [(C \setminus B) \cap C \cap B] \cup [C \cap B \cap C \cap A]\\
=&\  [(A \setminus B) \cap (A \setminus C)] \cup [(B \setminus A) \cap (B \setminus C)] \cup [(C \setminus A) \cap (C \setminus B)] \cup [A \cap B \cap C]\\
=&\  [(A \setminus B) \cap (A \setminus C)] \cup [(B \setminus A) \cap (B \setminus C)] \cup [(C \setminus A) \cap (C \setminus B)] \cup [(A \setminus B) \cap (A \cap B)] \cup [(A \setminus C) \cap (A \cap C)] \cup [A \cap B \cap A \cap C]\\
=&\ [((A \setminus B) \cup (A \cap C)) \cap ((A \setminus C) \cup (A \cap B)] \cup [(B \setminus A) \cap (B \setminus C)] \cup [(C \setminus A) \cap (C \setminus B)]\\
=&\ [(A \setminus (B \setminus C)) \cap (A \setminus (C \setminus B))] \cup [((B \setminus A) \cap (B \setminus C)) \cup ((C \setminus A) \cap (C \setminus B))]\\
=&\ (A \setminus ((B \setminus C) \cup (C \setminus B))) \cup [(B \setminus (C \cup A)) \cup (C \setminus (B \cup A))]\\
=&\ (A \setminus (B+C)) \cup [((B \setminus C) \setminus A) \cup ((C \setminus B) \setminus A)]\\
=&\ (A \setminus (B + C)) \cup (((B \setminus C) \cup (C \setminus B)) \setminus A)\\
=&\ (A \setminus (B+C)) \cup ((B+C)\setminus A)\\
=&\  A + (B+C)\end{align*}So $+$ is associative. Moreover,$$A + \emptyset = (A \setminus \emptyset) \cup (\emptyset \setminus A) = A \cup \emptyset = A$$ and similarly $\emptyset + A = A$, so that $\emptyset$ is an additive identity.$$A+A = (A \setminus A) \cup (A \setminus A) = \emptyset \cup \emptyset = \emptyset,$$ so that every element of $R$ is its own additive inverse. Thus $(R,+)$ is a group.$$A+B = (A \setminus B) \cup (B \setminus A) = (B \setminus A) \cup (A \setminus B) = B + A,$$ so that $R$ is an abelian group.$$A \cdot (B \cdot C) = A \cdot (B \cap C) = A \cap (B \cap C) = (A \cap B) \cap C = (A \cdot B) \cdot C,$$ so that multiplication is associative.\begin{align*}A \cdot (B + C) =&\ A \cap ((B \setminus C) \cup (C \setminus B))\\ =&\ (A \cap (B \setminus C)) \cup (A \cap (C \setminus B))\\ =&\ ((A \cap B) \setminus (A \cap C)) \cup ((A \cap B) \setminus (A \cap C))\\ =&\ (A \cap B) + (A \cap C) \\=&\ A \cdot B + A \cdot C;\end{align*} distributivity on the right is similar, since $\cap$ is commutative.

Thus $R$ is a ring.

(2) Since $$A \cdot B = A \cap B = B \cap A = B \cdot A,$$ $R$ is commutative. Since $$A \cdot X = A \cap X = A$$ and $$X \cdot A = X \cap A = A,$$ $R$ has an identity. For all $A \in R$, $$A \cdot A = A \cap A = A,$$ so that $R$ is boolean.