§ Boolean algebras

• Has meet, join, complement, 1, 0 with usual laws

§ Atomic boolean algebras

• Consider $2^S$ where $S$ is finite. Then the elements of the form {s} ∈ 2^S are said to be atoms because if x ⊂ {s} then x = 0 or x = {s}.

§ Atomless boolean algebras

• Let $S$ be an infinite set, and let $I$ be a collection of its finite subsets. Then $I$ is an ideal (downward closed subset which has all joins), because the union of two finite sets is finite, and the subset of any finite set is finite.
• The quotient $T = 2^S/I$ will be an atomless boolean algebra.
• Note that the quotient kills all finite subsets.
• So for any non-zero $x \in T$, then it must be an equivalence class with some infinite subset. If we take $k, k'$ to be non-empty disjoint subsets of $x$, then neither is equivalent to $x$ or to $\emptyset$, because they differ at infinitely many locations from each. Thus, $x$ is not an atom.
• Furthermore, the boolean algebra is not complete, because, if we have $k_1, k_2, \dots$ be a countable collection of countably infinite subsets of $S$ (for example, if $S \equiv \mathbb N$, then we could take $k_i$ to be the set of numbers with $i$ bits as 1 in their binary representation), then this collection has no least upper bound.
• Suppose $u$ is an upper bound. Then $u$ differs from each $k_i$ in only finitely many locations.
• Now build $e_i \in u \cap k_i$, and consider the set $c \equiv u / \{ e_i \}$. That is, we remove one element from $u$from the intersection with each $k_i$. This new $c \subseteq u$, and $c$ is still an upper bound, since it differs from each of the $k_i$ at finitely many locations. Thus, this algebra is not complete.

§ Or, how to embed a poset into a boolean algebra.

• Every poset $P$ can be embedded into a complete atomic boolean algebra $2^P$by sending $p \mapsto \{ x : x \leq p \}$ (the ideal of $p$).
• Alternatively, that's just $Hom(-, p)$. God bless yoneda embedding.
• We can thus consider a ring map from $2^p \to 2$, which gives us a maximal ideal of $2^P$ (ideal is maximal because quotient is field).
• This assigns to us consistent truth values of $p$.
• In this way, maximal ideals of posets completed to rings correspond to truth values.
• Dualize the story via Grothendieck/Geometry to talk about filters :)