Proof of Stone's representation theorem for finite Boolean algebras

Proof

Let $B$ be a finite Boolean algebra, with order $a \le b$ meaning $a \wedge b = a$. Write $\varphi(b) = \set{x \in \operatorname{At}(B) : x \le b}$ for $b \in B$. An isomorphism here is a bijection $B \to \mathcal{P}(\operatorname{At}(B))$ that sends $\wedge, \vee, \neg, 0, 1$ to $\cap, \cup$, complement in $\operatorname{At}(B)$, $\varnothing$ and $\operatorname{At}(B)$. We first derive basic facts on the order, then on atoms.

1. For $u \in B$, absorption gives $u \wedge u = u \wedge (u \vee (u \wedge u)) = u$, since $u \vee (u \wedge u) = u$. In the same way $u \vee u = u$.
2. For $u \in B$, $0 \wedge u = 0$: absorption gives $0 \wedge (0 \vee u) = 0$, and $0 \vee u = u \vee 0 = u$. Also $u \wedge 1 = u$, so $u \le 1$.
3. The relation $\le$ is a partial order. Reflexivity is step 1. If $a \le b$ and $b \le a$, then $a = a \wedge b = b \wedge a = b$.
4. Transitivity: if $a \le b$ and $b \le c$, then $a \wedge c = (a \wedge b) \wedge c = a \wedge (b \wedge c) = a \wedge b = a$.
5. If $a \le b$, then $a \vee b = (b \wedge a) \vee b = b \vee (b \wedge a) = b$ by absorption.
6. For $a, b \in B$, $a \wedge b \le a$, since $(a \wedge b) \wedge a = (a \wedge a) \wedge b = a \wedge b$ by step 1. Also $a \le a \vee b$, since $a \wedge (a \vee b) = a$ by absorption.
7. If $c \le a$ and $c \le b$, then $c \le a \wedge b$, since $c \wedge (a \wedge b) = (c \wedge a) \wedge b = c \wedge b = c$.
8. By steps 4, 6 and 7, for all $c, a, b \in B$: $c \le a \wedge b$ if and only if $c \le a$ and $c \le b$.
9. If $c \le 0$, then $c = c \wedge 0 = 0 \wedge c = 0$ by commutativity and step 2. So no atom lies below $0$, and $\varphi(0) = \varnothing$. By step 2, $\varphi(1) = \operatorname{At}(B)$.
10. Let $x$ be an atom and $b \in B$. We claim that exactly one of $x \le b$ and $x \le \neg b$ holds.
11. By step 6, $x \wedge b \le x$. If $x \wedge b \ne 0$, then $x \wedge b = x$ because $x$ is minimal in $B \setminus \set{0}$. So $x \le b$.
12. If $x \wedge b = 0$, then $x = x \wedge (b \vee \neg b) = (x \wedge b) \vee (x \wedge \neg b) = 0 \vee (x \wedge \neg b) = x \wedge \neg b$. So $x \le \neg b$.
13. If both held, step 8 would give $x \le b \wedge \neg b = 0$, so $x = 0$ by step 9. This is false for an atom, which proves the claim of step 10.
14. By step 10, $\varphi(\neg b) = \operatorname{At}(B) \setminus \varphi(b)$. By step 8, $\varphi(a \wedge b) = \varphi(a) \cap \varphi(b)$.
15. Let $x$ be an atom and $a, b \in B$. If $x \le a$ or $x \le b$, then $x \le a \vee b$ by steps 6 and 4.
16. Conversely, let $x \not\le a$ and $x \not\le b$. By step 10, $x \le \neg a$ and $x \le \neg b$. By step 8 and De Morgan's law, $x \le \neg a \wedge \neg b = \neg(a \vee b)$.
17. Then step 10 gives $x \not\le a \vee b$. With step 15, $\varphi(a \vee b) = \varphi(a) \cup \varphi(b)$.
18. Every nonzero $b \in B$ has an atom below it. Let $S = \set{y \in B \setminus \set{0} : y \le b}$. It contains $b$ and is finite.
19. Choose $y \in S$ for which the set $D(y) = \set{z \in B : z \le y}$ has the fewest elements. Let $z \ne 0$ with $z \le y$.
20. Then $z \le b$ by step 4, so $z \in S$, and $D(z) \subseteq D(y)$ by step 4. If $z \ne y$, then $y \notin D(z)$ by step 3, while $y \in D(y)$.
21. So $z \ne y$ would give $\abs{D(z)} < \abs{D(y)}$, against the choice of $y$. Hence $z = y$, so $y$ is an atom, and $y \le b$.
22. The map $\varphi$ is injective. Let $\varphi(a) = \varphi(b)$, and suppose $a \not\le b$. Put $c = a \wedge \neg b$.
23. If $c = 0$, then $a = a \wedge (b \vee \neg b) = (a \wedge b) \vee (a \wedge \neg b) = (a \wedge b) \vee 0 = a \wedge b$, so $a \le b$. Hence $c \ne 0$.
24. By step 18, some atom $x$ satisfies $x \le c$. By step 8, $x \le a$ and $x \le \neg b$.
25. So $x \in \varphi(a)$, and $x \notin \varphi(b)$ by step 10. This contradicts $\varphi(a) = \varphi(b)$, so $a \le b$.
26. By symmetry $b \le a$, so $a = b$ by step 3.
27. The map $\varphi$ is surjective. Let $T \subseteq \operatorname{At}(B)$, and list its elements as $t_1, \dots, t_k$. Put $s_0 = 0$ and $s_i = s_{i-1} \vee t_i$ for $1 \le i \le k$.
28. By step 9 and step 17, $\varphi(s_k) = \varphi(t_1) \cup \dots \cup \varphi(t_k)$, by induction on $k$.
29. For atoms $x$ and $t$, $x \le t$ holds if and only if $x = t$. Indeed $x \ne 0$ and $x \le t$ give $x = t$, by minimality of $t$.
30. So $\varphi(t_i) = \set{t_i}$ for each $i$, and $\varphi(s_k) = T$ by step 28.
31. By steps 22 to 30, $\varphi$ is a bijection from $B$ to $\mathcal{P}(\operatorname{At}(B))$. By steps 9, 14 and 17, it preserves $0$, $1$, $\neg$, $\wedge$ and $\vee$.
32. Hence $\varphi$ is an isomorphism from $B$ to the power set Boolean algebra of $\operatorname{At}(B)$.