Boolean algebra and Boolean ultrafilter

A Boolean algebra is a set $B$ with distinguished elements $0,1\in B$, binary operations $\wedge ,\vee :B\times B\to B$ and a unary operation $\mathrm{\neg }:B\to B$, such that for all $a,b,c\in B$:

• $(B,\wedge ,1)$ and $(B,\vee ,0)$ are commutative monoids (associativity and the identities $a\wedge 1=a$, $a\vee 0=a$);
• the absorption laws hold: $a\wedge (a\vee b)=a$ and $a\vee (a\wedge b)=a$;
• the distributive laws hold: $a\wedge (b\vee c)=(a\wedge b)\vee (a\wedge c)$ and $a\vee (b\wedge c)=(a\vee b)\wedge (a\vee c)$;
• the complement laws hold: $a\wedge \mathrm{\neg }a=0$ and $a\vee \mathrm{\neg }a=1$.

The trivial Boolean algebra is the one-element algebra $\{0\}$, in which $0=1$; it is allowed here, and it corresponds to the empty Stone space in Stone space and clopen algebra.

A Boolean homomorphism $\phi :B\to B^{\mathrm{\prime }}$ is a map with $\phi (0)=0$, $\phi (1)=1$, $\phi (a\wedge b)=\phi (a)\wedge \phi (b)$, $\phi (a\vee b)=\phi (a)\vee \phi (b)$ and $\phi (\mathrm{\neg }a)=\mathrm{\neg }\phi (a)$ for all $a,b$.

A proper filter in $B$ is a subset $F\subseteq B$ with

1. $1\in F$ and $0\mathrm{\notin }F$;
2. $a,b\in F$ implies $a\wedge b\in F$;
3. $a\in F$ and $a\le b$ (meaning $a\wedge b=a$) imply $b\in F$.

A Boolean ultrafilter is a proper filter that is maximal with respect to inclusion among proper filters.

The following two facts are used repeatedly below, and are proved here rather than assumed. Let $U\subseteq B$ be a proper filter.

Dichotomy. $U$ is an ultrafilter if and only if for every $a\in B$ exactly one of $a\in U$ and $\mathrm{\neg }a\in U$ holds.

If $U$ is an ultrafilter and $a\mathrm{\notin }U$, then $\mathrm{\neg }a\in U$: if also $\mathrm{\neg }a\mathrm{\notin }U$, the family $U^{\mathrm{\prime }}:=\{b:b\ge a\wedge u$ for some $u\in U\}$ is a proper filter strictly containing $U$ — it is a filter by construction, it contains $a$ and hence is strictly larger, and it is proper because $a\wedge u\mathrm{\ne }0$ for every $u\in U$ (were $a\wedge u=0$ then $u\le \mathrm{\neg }a$ and $\mathrm{\neg }a\in U$ by upward closure, contrary to assumption), so $0\mathrm{\notin }{U}^{\mathrm{\prime }}$; this contradicts maximality. The two alternatives are exclusive because $a\wedge \mathrm{\neg }a=0\mathrm{\notin }U$.

Conversely, suppose $U$ decides every element. If $U\subseteq V$ is a proper filter and $a\in V$, then $\mathrm{\neg }a\mathrm{\notin }V$ (else $0=a\wedge \mathrm{\neg }a\in V$), so $\mathrm{\neg }a\mathrm{\notin }U$ and hence $a\in U$ by the dichotomy applied to $U$; thus $V\subseteq U$ and $V=U$, so $U$ is maximal.

Two-valued homomorphisms. The assignments $U\mapsto {\chi }_U$, where ${\chi }_U(a)=1$ for $a\in U$ and ${\chi }_U(a)=0$ otherwise, and $\chi \mapsto {\chi }^{-1}(\{1\})$, are mutually inverse bijections between Boolean ultrafilters on $B$ and Boolean homomorphisms $B\to \{0,1\}$ with the two-element Boolean algebra as codomain.

That ${\chi }_U$ is a homomorphism uses the dichotomy: ${\chi }_U(\mathrm{\neg }a)=1-{\chi }_U(a)$ by exclusivity, ${\chi }_U(a\wedge b)={\chi }_U(a){\chi }_U(b)$ because $U$ is closed under $\wedge$ and upward closed, and the identity for $\vee$ follows from de Morgan and the other two, or directly from the fact that $a\vee b\in U$ if and only if $a\in U$ or $b\in U$ (if $a\vee b\in U$ and both $a\mathrm{\notin }U$ and $b\mathrm{\notin }U$, then $\mathrm{\neg }a,\mathrm{\neg }b\in U$, so $\mathrm{\neg }(a\vee b)=\mathrm{\neg }a\wedge \mathrm{\neg }b\in U$, contradicting $(a\vee b)\wedge \mathrm{\neg }(a\vee b)=0\mathrm{\notin }U$; the converse is upward closure). That ${\chi }^{-1}(\{1\})$ is an ultrafilter is immediate from the homomorphism identities: it is a proper filter, and it decides each element because $\chi (a)\in \{0,1\}$ forces exactly one of $\chi (a)=1$, $\chi (\mathrm{\neg }a)=1$.

• Filters are proper by convention, as for filters on a set; the improper family $B$ itself is not a filter here, so "ultrafilter" means a maximal proper filter.
• The trivial algebra has no ultrafilters and no two-valued homomorphisms. In the one-element algebra $0=1$, a proper filter would have to contain $1$ and omit $0=1$, which is impossible; and a Boolean homomorphism to $\{0,1\}$ would have to send $1$ to $1$ and $0=1$ to $0$, which is also impossible. This matches the empty Stone space under the convention of Stone space and clopen algebra.