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Stone duality for a finite Boolean algebra

Example

For the power-set algebra $B = P ({1, 2, 3})$ the ultrafilter space $Ult (B)$ has exactly three points, the principal ultrafilters $↑ {1}, ↑ {2}, ↑ {3}$ , with the discrete topology; the map $S \mapsto [S]$ is the identity identification of $P ({1, 2, 3})$ with $Clop (Ult (B))$ under the correspondence $i \leftrightarrow ↑ {i}$ . More generally, every finite Boolean algebra $B$ is isomorphic to the power set of its finite set of atoms, and its Stone space is the finite discrete space on those atoms; no choice principle is used.

Facts & Assumptions

Given: The Boolean algebra $B = P ({1, 2, 3})$ and its ultrafilter space (Boolean algebra and Boolean ultrafilter, Stone space and clopen algebra).

[L1]

A Boolean ultrafilter is a maximal proper filter (Boolean algebra and Boolean ultrafilter). Such a filter decides every element: if $b \notin U$ , maximality makes the filter generated by $U \cup {b}$ improper, so some $u \in U$ has $u \land b = 0$ and hence $u \leq \neg b$ , giving $\neg b \in U$ ; both cannot lie in a proper filter. The principal filter $↑ ⁣ {i} = {S : i \in S}$ is an ultrafilter of $P ({1, 2, 3})$ by this criterion.

[L2]

An atom of a Boolean algebra $B$ is a minimal nonzero element. Every nontrivial finite Boolean algebra has atoms, and every element is the join of the atoms below it. In the trivial algebra the atom set is empty and its sole element $0 = 1$ is the empty join. [algebra]

Verification

technique · direct

1.1

Every ultrafilter $U$ of $P ({1, 2, 3})$ is principal: the join ${1} \cup {2} \cup {3} = {1, 2, 3}$ belongs to $U$ , so one of the singletons belongs to $U$ by the dichotomy [L1], and then $U = ↑ ⁣ {i}$ for that $i$ ; conversely each $↑ ⁣ {i}$ is an ultrafilter by [L1].

L1algebra

1.2

Consequently $Ult (B) = {↑ ⁣ {1}, ↑ ⁣ {2}, ↑ ⁣ {3}}$ has three points, and the basic open sets $[S] = {U : S \in U}$ are in bijection with the subsets $S \subseteq {1, 2, 3}$ through $S \leftrightarrow {i : i \in S}$ ; in particular every subset of the three-point space is basic open, so the topology is discrete and $Clop (Ult (B)) = P (Ult (B))$ has eight elements, matching $∣ B ∣ = 8$ .

1.1algebra

2.1

Let $B$ be a general finite Boolean algebra. If $B$ is trivial, then $A t o m s (B) = \emptyset$ , the unique map $B \to P (\emptyset)$ is an isomorphism, and both $B$ and $P (\emptyset)$ have no ultrafilters; their Stone space is the empty discrete space. If $B$ is nontrivial, [L2] says that distinct atoms have meet $0$ and every $b \in B$ is the join of the atoms below it. Thus $b \mapsto {a atom : a \leq b}$ is a bijection onto the power set of the finite atom set and preserves joins, meets and complements. Hence $B ≅ P (A t o m s (B))$ ; the argument of [step 1.1], with the finite nonempty atom set in place of ${1, 2, 3}$ , says its ultrafilters are the principal ones at atoms, so its Stone space is finite and discrete.

step 1.1L2algebra∎

Remarks

Finiteness makes choice unnecessary: the atoms are found by descending chains in a finite poset, and no extension of filters is needed because every ultrafilter is principal.
The example is the degenerate case of Stone duality in which the Stone space is finite and the functors are the identity identifications on finite power sets.

Depends on

Used by

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Dependency tree · two levels

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Sources

Marcus Tressl, Stone Duality for Boolean Algebras — Theorem 2.3.4, pp. 9–10, and Theorem 3.1.5, pp. 11–12; the finite computation is supplied locally (standard reference, not scraped)