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Every proper lattice ideal of a Boolean algebra extends to a prime lattice ideal. More generally, an ideal disjoint from a lattice filter extends to a prime ideal still disjoint from that filter. This Boolean case of the Stone prime filter theorem is equivalent to the ultrafilter lemma over Zermelo–Fraenkel set theory and is weaker than the axiom of choice.

A chain-complete partially ordered set has a supremum for every chain, with the empty-chain condition requiring a least element if empty chains are included. This implies the weaker upper-bound hypothesis of Zorn lemma, which requires only that each chain have some upper bound, not necessarily a least one. Some treatments use “chain-complete” for that weaker condition, so the convention should be specified. In the inclusion order on lattice ideals avoiding a specified element, the union of a nonempty chain is an ideal and is its supremum. This union observation is the step needed to apply Zorn in prime-ideal separation arguments.

A homomorphism between lattices preserves binary joins and meets: $f(x\vee y)=f(x)\vee f(y)$ and $f(x\wedge y)=f(x)\wedge f(y)$. It is order preserving, since $x\le y$ implies $x=x\wedge y$ and hence $f(x)=f(x)\wedge f(y)$. For a map into the two-element lattice, the inverse image of zero is a lattice ideal, and meet preservation says this ideal is prime: if a meet belongs to it, at least one factor does. Conversely the indicator of the complement of a prime lattice ideal preserves both operations. Preservation of distinguished top and bottom elements is an additional requirement for homomorphisms of bounded lattices.

The lattice ideals of a Boolean algebra form a frame under inclusion. Meets are intersections and joins are ideals generated by unions. A complemented ideal is principal: if $I\vee J=B$ and $I\cap J=\{0\}$, choose $b\in I,c\in J$ with $b\vee c=1$; then $c=\neg b$ and $I=\downarrow b$. This ideal frame defines an extremally disconnected locale exactly when $B$ is complete.

For a Boolean algebra $B$, write $\downarrow b$ for its principal lattice ideal. The ideal frame of a Boolean algebra has intersections as meets and ideals generated by unions as joins. The pseudocomplement of an ideal $I$ isIn particular $(\downarrow b{)}^{\ast }=\downarrow \neg b$, so every principal ideal is complemented and regular.

If $B$ is complete, put $s={\bigvee }_{B}I$. Then $I^{\ast }=\downarrow \neg s$ and $I^{\ast \ast }=\downarrow s$. Thus all regular ideals are principal and complemented, so the corresponding locale is extremally disconnected.

Conversely, if the ideal frame of a Boolean algebra is extremally disconnected, each $I^{\ast \ast }$ is complemented. Any complemented ideal is principal: if $J\vee K=B$ and $J\cap K=\{0\}$, the fact that their generated join contains $1$ supplies $b\in J,c\in K$ with $b\vee c=1$. Then $b\wedge c=0$, so $c=\neg b$; every $j\in J$ is disjoint from $c$ and therefore $j\le b$. Thus $J=\downarrow b$. Write $I^{\ast \ast }=\downarrow s$. Since $I\subseteq I^{\ast \ast }$, $s$ bounds $I$. If $t$ is another upper bound, $I\subseteq \downarrow t$ implies $I^{\ast \ast }\subseteq (\downarrow t{)}^{\ast \ast }=\downarrow t$, so $s\le t$. An arbitrary subset has the same upper bounds as the lattice ideal it generates, so every subset of $B$ has a supremum. Hence

Now put $A=\mathcal{O}(X)$ and $B=A_{\neg \neg }$. This is a complete Boolean algebra, with arbitrary Boolean joins $\neg \neg {\bigvee }_A{b}_i$, so the preceding result makes $\gamma X$ extremally disconnected. We construct its map to $X$ using the permitted Joyal extension lemma.

The map $\sigma :\mathrm{Idl}(B)\to B$, $I\mapsto {\bigvee }_{B}I$, is a frame homomorphism, with right adjoint $b\mapsto \downarrow b$. Thus $k(I)=\downarrow {\bigvee }_{B}I$ is a nucleus on a frame whose fixed frame is $B$. It preserves finite joins: both $k(I\vee J)$ and $k(I)\vee k(J)$ equal $\downarrow ({\bigvee }_{B}I{\vee }_B{\bigvee }_{B}J)$, and $k(0)=0$. Consequently the locale with frame $B$ is a flat sublocale of $\gamma X$. It is also the Booleanization of a locale $X$, so it has its inclusion map to $X$.

The Joyal extension lemma says that a map from a flat sublocale to a compact locale that is a regular locale extends uniquely to the ambient locale. Apply it to this inclusion. Its frame formula giveswhere $u\prec a$ is the well-inside relation $\neg u\vee a=1$. This also proves directly that the map is surjective. Indeed $u\prec a$ implies $\neg \neg u\le a$: meet $\neg u\vee a=1$ with $\neg \neg u$. Every member of the ideal $g^{\ast }(a)$ is therefore at most $a$ in $A$, since a finite Boolean join of its generators is below $\neg \neg (u_1\vee \cdots \vee u_n)\le a$; the last inequality follows because $u_1\vee \cdots \vee u_n\prec a$. Regularity givesThus $a$ can be recovered from $g^{\ast }(a)$, so $g^{\ast }$ is injective. A locale map with injective inverse-image frame homomorphism is a surjection. We have obtained an extremally disconnected $\gamma X$ and a surjection $\gamma X\to X$.

A chain-complete partially ordered set has a supremum for each chain (and a least element if the empty chain is included). The weaker condition needed for Zorn lemma is that every chain have an upper bound; some terminology also calls this chain-complete. Zorn's lemma says that a nonempty partially ordered set with this upper-bound property has a maximal element. We verify that exact hypothesis below.

Consider the lattice ideals containing $a$ but not $b$, ordered by inclusion. The principal ideal $\downarrow a$ belongs to this family because $b\nleqq a$. The union of a nonempty chain of such ideals is an ideal: two of its elements lie together in one member of the chain, which contains their join and everything below it. The union still avoids $b$. The empty chain has the initial principal ideal as an upper bound. Zorn supplies a maximal such ideal $I$.

We prove that $I$ is a prime lattice ideal. If $x,y\notin I$, maximality says that the ideals generated by adjoining $x$ and by adjoining $y$ both contain $b$. Thus $b\le i\vee x$ and $b\le j\vee y$ for some $i,j\in I$: the first generated ideal consists of elements below $i\vee x$, and similarly for $y$. If $x\wedge y\in I$, distributivity givesa contradiction. Consequently $x\wedge y\in I$ implies $x\in I$ or $y\in I$.

Define $f(x)=0$ on $I$ and one elsewhere. An ideal contains $x\vee y$ exactly when it contains both $x$ and $y$, and its primeness says that it contains $x\wedge y$ exactly when it contains at least one of them. Thus $f$ preserves both joins and meets, and $f(a)=0,f(b)=1$ as required.

For completeness, take the Lindenbaum algebra of classical propositional formulae modulo provable equivalence, including any given premises as axioms. It is a Boolean, hence distributive, lattice; provable implication is its order. If a formula $P$ is not provable from the premises, then $1\nleqq [P]$. The separation just proved, with $a=[P],b=1$, gives a homomorphism of lattices sending $P$ to zero and one to one. It also sends zero to zero, since $0\le [P]$, and preserves complements because their meet is zero and join is one. Its values on atomic propositions form a Boolean valuation satisfying all premises and falsifying $P$. Thus semantic consequence implies provability, the completeness theorem for propositional logic.

The Stone prime filter theorem says that if a lattice filter $F$ and a lattice ideal $I$ of a distributive lattice are disjoint, then there is a prime filter of a distributive lattice $P$ such thatEquivalently, whenever $a\nleqq b$, there is a prime filter containing $a$ and omitting $b$.

For a prime filter $P\in \hat{H}$, first suppose $a\Rightarrow b\in P$. If $Q\supseteq P$ and $a\in Q$, then $a\Rightarrow b\in Q$ andso $b\in Q$. Thus no prime filter above $P$ belongs to $a^{\ast }\setminus b^{\ast }$, and

Conversely, suppose $a\Rightarrow b\notin P$. The lattice filter generated by $P\cup \{a\}$ is disjoint from the principal lattice ideal $\downarrow b$. Indeed, an intersection would give some $p\in P$ with $p\wedge a\le b$, whence $p\le a\Rightarrow b$ and then $a\Rightarrow b\in P$, a contradiction. The Stone prime filter theorem therefore extends this filter to a prime filter $Q\supseteq P$ that omits $b$. Then $Q\in a^{\ast }\setminus b^{\ast }$, so $P\in \uparrow (a^{\ast }\setminus b^{\ast })$.

We have proved, for every $P$,which is the required identity

A proper lattice ideal $I$ is prime if $x\wedge y\in I$ implies $x\in I$ or $y\in I$. Its complement is a prime filter of a distributive lattice when the lattice is distributive: downward closure of the ideal makes the complement upward closed, primeness makes it meet-closed, and join closure makes that filter prime. Equivalently, its complement indicator is a homomorphism of lattices to the two-element lattice. Given $b\nleqq a$ in a distributive lattice, maximize an ideal containing $a$ and avoiding $b$ using Zorn lemma. If $x,y$ lie outside it, the two enlarged ideals contain $b$, so $b\le i\vee x$ and $b\le j\vee y$ for some $i,j\in I$. Distributivity shows that $x\wedge y\in I$ would force $b\in I$, a contradiction. Thus this maximal avoiding ideal is prime and separates the two specified elements.

If a lattice filter $F$ and lattice ideal $I$ in a distributive lattice are disjoint, there is a prime filter of a distributive lattice containing $F$ and disjoint from $I$.

The Boolean prime ideal theorem is the Boolean-algebra version; distributivity gives this corresponding filter-extension statement.