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The double-negation nucleus on a frame gives a dense sublocale whose frame is a complete Boolean algebra. Its elements are regular open elements, with Boolean joins $\neg \neg {\bigvee }_i{a}_i$.

Use the order on locale maps in which $f\le g$ means $f^{\ast }(a)\le g^{\ast }(a)$ for every open $a$. A Hausdorff locale has a closed diagonal in its localic square. A totally unordered locale, or $T_U$ locale, has discretely ordered hom-sets: every comparable pair of maps with any common domain is equal.

In the localic square of $Y$, letThis is the largest open disjoint from the diagonal: inverse image along the diagonal sends the rectangle $a\times b$ to $a\wedge b$. Hence $N$ is the complement of its closure. If $f\le g$, thenfor every summand of $N$. Thus $(f,g)$ factors through the closure of the diagonal. For a Hausdorff locale that closure is the diagonal itself, so $f=g$. This proof of Hausdorff implies $T_U$ does not require classical logic.

Here is an explicit classical counterexample to the converse implication from spatial Hausdorffness. Give $\mathbb{R}$ the topology generated by the usual opens and by $\mathbb{Q}$. Every open has the form $U=V\cup (W\cap \mathbb{Q})$, where $V,W$ are usual open sets and we may arrange $V\subseteq W$. The topology is finer than the usual one, so the space is Hausdorff. Put $B=\mathrm{RO}(\mathbb{R})$, the complete Boolean algebra of ordinary regular open elements, and write $\rho (V)=\mathrm{int}\overline{V}$. Restriction from this space to its rational and irrational subspaces, followed by Booleanization, gives frame homomorphismsTo justify these formulas, each subspace has its ordinary subspace topology and is dense in the ordinary real line. Regular-open algebras of a space and a dense subspace are isomorphic: restrict a regular open and take its interior-closure, with inverse obtained by taking interior-closure in the ambient space. Restriction is a frame homomorphism, and double negation is a frame homomorphism into its fixed-element frame. These observations also show independence of the chosen representation of $U$. Since $V\subseteq W$, we have $f^{\ast }\le g^{\ast }$, but $f^{\ast }(\mathbb{Q})=0$ and $g^{\ast }(\mathbb{Q})=1$. Thus this Hausdorff space fails $T_U$ as a locale.

For the remaining arguments about connected locales, use the classical-logic option: positive opens of a locale are exactly the nonzero opens. There is a necessary positivity qualification. The empty locale has no positive opens, so the printed condition on those opens holds vacuously, although no point $1\to \varnothing$ exists. The point assertions therefore require $X$ to be positive. This qualification is automatic if “totally connected” includes nonemptiness; otherwise it must be added explicitly.

Let $X$ be positive and satisfy the condition on its positive opens of a locale. Put $F=\{a\in \mathcal{O}(X):a\ne 0\}$. It contains $1$, excludes $0$, and is upward closed. If $a,b\in F$ had $a\wedge b=0$, their union would be a positive open of a locale with a decomposition into two disjoint nonzero opens, contradicting its being a connected locale. Thus $a\wedge b\in F$. If ${\bigvee }_i{a}_i\ne 0$, not all $a_i$ can be zero; classical logic supplies a nonzero summand. Hence $F$ is a completely prime filter, and its characteristic frame homomorphism defines a point $p:1\to X$.

The point is dense. Indeed $p^{\ast }(a)=0$ implies $a=0$ by the definition of $F$. In nucleus language the image of $p$ has nucleus $p_{\ast }{p}^{\ast }$; its value at $0$ is $p_{\ast }(0)=0$, so the image is a dense sublocale.

Conversely suppose an open locale has a dense point $p$. Density means $p^{\ast }(a)=0$ only if $a=0$, and classically this implies $p^{\ast }(a)=1$ for every positive open of a locale. If a positive open $u$ split as $a\vee b=u$ with $a\wedge b=0$ and both summands nonzero, then $p^{\ast }(a)=p^{\ast }(b)=1$, contradicting preservation of their meet. Thus every positive open of a locale is a connected locale, proving the converse under the declared logical convention.

Let $x:X\to 1$ be the terminal map. For a positive totally connected locale $X$, the point above satisfiesbecause the right side is $1$ when $a\ne 0$ and both sides are $0$ otherwise. Hence $1_X\le px$. If $f:X\to Y$ and $Y$ is a totally unordered locale, composition gives $f\le fpx$, so $f=fpx$. Thereforewhich is the required constant factorization. For the empty source the factorization need not exist: the unique map $\varnothing \to \varnothing$ does not factor through $1$. Thus the positivity qualification matters for this assertion too.

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 $\gamma X\to X$.

Here “constructive ZF” is understood as full intuitionistic Zermelo–Fraenkel set theory, with power set, full separation, collection and set induction. It is not the predicative theory CZF with restricted separation and subset collection. We construct a negative inner interpretation of classical ZF. Its classical logic belongs to the interpreted universe, not to the ambient intuitionistic universe. A bare double-negation of the ordinary set axioms does not by itself solve the extensionality and power-set difficulties.

Work inside IZF. Let $1=\{0\}$ and regard subsets of $1$ as truth values. For $p\subseteq 1$, put $j(p)=\{0\in 1:\neg \neg (0\in p)\}$. Define the double-negation Boolean algebraMeets are intersections, joins are $j$ of unions, and complement is $\neg p=\{0\in 1:0\notin p\}$. Arbitrary intersections of stable predicates are stable. The laws $\neg \neg \neg P\leftrightarrow \neg P$ and $\neg \neg (P\vee \neg P)$ give complementation, and distributing a stable conjunction through a regularized disjunction gives distributivity. Thus $B$ is a complete Boolean algebra, constructively. In particular $p{\vee }_B\neg p=1$, even when their ordinary union is not known to be $1$.

Construct a class $W$ of Boolean-valued names. A name has a set of immediate subnames and a coefficient in $B$ on each; the subname relation is inductively well-founded. More concretely, take set-sized rooted-tree codes with Boolean labels on their edges, requiring that every progressive subset of their nodes contains every node. A subset is progressive when containing all children of a node implies containing that node. This set-quantified condition is definable using power set and supplies induction for each formula by separation. On these trees recursively code each root by the function assigning its children's coded names their edge coefficients; repeated equal codes can be combined by joining their coefficients. The class of these codes is definable by the existence of such a well-founded construction tree. Every set-indexed family of names with coefficients produces another name by adjoining a new root. Collection gathers a set of construction trees witnessing the given subnames, and gluing all these witnesses under the new root avoids choosing one witness tree for each subname; repetitions have identical codes and coefficients. Well-founded recursion on these trees collects compatible partial evaluations by collection; uniqueness identifies all overlapping evaluations, so no selection of witnesses is required. Double induction gives the corresponding recursion on pairs of trees. This avoids assuming a classical linear ordering of all constructive ordinals.

For names $a,b$, define valued membership and equality simultaneously:Here $p\Rightarrow q=\neg p{\vee }_{B}q$. Each recursive call replaces at least one argument by an immediate subname, so well-founded double recursion applies. Induction proves reflexivity, symmetry and transitivity of valued equality and its substitution laws. For example the diagonal term in the membership join gives $a(u)\le [[u\in a]]$, establishing reflexivity; expanding joins and using distributivity gives transitivity and substitution at the next step.

Interpret logical connectives by the Boolean operations, and quantifiers by meets and joins over names. Although $W$ is a class, the values being joined form a subset of the set $B$: for each fixed formula, full separation forms $\{p\in B:\exists a\in W(p=[[\phi (a)]])\}$. This defines each fixed formula's truth value inductively. It does not give a uniform truth predicate for all formulas of the ambient universe. Substitution and the complete Boolean algebra laws verify the classical logical rules, including excluded middle. We now verify the set axioms with explicit names.

The empty-domain name is empty. The name with domain $\{a,b\}$ and coefficient $1$ on both is a pair, since its membership value is $[[x=a]]{\vee }_B[[x=b]]$. For union of $a$, use domain ${\bigcup }_{u\in \mathrm{dom}a}\mathrm{dom}u$, with coefficient at $v$ equal to ${\bigvee }_{u}a(u)\wedge u(v)$, taking missing coefficients to be zero. The membership definition distributes to the required existential membership condition. Extensionality follows from the definition of valued equality: agreeing on the membership value of every name in particular implies both displayed inclusion conditions.

For separation, keep the domain of $a$ and replace its coefficient at $u$ by $a(u)\wedge [[\phi (u)]]$. Substitution makes membership in the resulting name equal to $[[x\in a]]\wedge [[\phi (x)]]$. For power set, write $d=\mathrm{dom}a$. Every function $h:d\to B$ gives a name $a_h$ with coefficients $a(u)\wedge h(u)$. The functions form the set $B^d$, so collect all $a_h$ as a domain with coefficient $1$. Each is a valued subset of $a$. Conversely, for an arbitrary name $c$, put $h(u)=[[u\in c]]$. If $t=[[c\subseteq a]]$, the membership and substitution formulas giveThus, to the full extent that $c$ is a subset of $a$, it belongs to the constructed power-set name. This proves the full power-set axiom without treating every external subset of a name as an internal subset.

The subtle point is collection for Boolean-valued names. For a fixed formula $\phi (u,v)$ form by full separation the setEvery pair in $S$ has a witness name. The ambient axiom schema of collection supplies a set $C$ of witness names covering all these pairs; separation discards any non-name elements. Make a name $b$ with domain $C$ and all coefficients $1$. For each $u\in d$, every value $a(u)\wedge [[\phi (u,v)]]$ occurring with an arbitrary name $v$ is itself a permissible $p$ in $S$. Its collected witness $w\in C$ has value at least $p$. HenceIf $t=[[\forall x\in a\exists y\phi (x,y)]]$, this inequality says $t\wedge a(u)$ is covered by the witnesses in $b$ for every $u$. The bounded universal truth formula therefore gives $t\le [[\forall x\in a\exists y\in b\phi (x,y)]]$. This is exactly the collection axiom's implication at value $1$. It also yields replacement under a uniqueness hypothesis. We collected all Boolean portions of the existential truth; no choice of a single top-valued witness was assumed.

For infinity, form the canonical names $\check{n}$ recursively, with domain $\{\check{m}:m<n\}$ and coefficients $1$, and the name $\check{\omega }$ with all $\check{n}$ in its domain. It is inductive. For set induction, let $p$ be the value of $\forall x[(\forall y\in x\psi (y))\to \psi (x)]$. Induction on the construction of a name $a$, using the bounded universal formula, gives $p\le [[\psi (a)]]$ from the same inequalities for all its subnames. Consequently $p\le [[\forall x\psi (x)]]$. Thus set induction has value $1$, and classical logic in the interpreted universe converts it to the usual foundation axiom.

We have proved all axioms of classical ZF in the Boolean-valued inner universe over intuitionistic set theory, with bottom distinct from top. Any finite classical proof of contradiction would translate, instance by instance, to an intuitionistic proof that $0=1$ in $B$, an ambient contradiction. ThereforeConversely classical ZF proves the intuitionistic set axioms and allows their logical rules, so the consistency strengths agree. This is a proof-translation argument, not an assertion that IZF proves its own consistency. The valued construction is an inner interpretation of the same negative-model kind as the stable-set approach. Closing arbitrary sets under the unbounded predicate $\neg \neg (x\in a)$ would require an additional double-complement axiom; the bounded truth-value construction above avoids that assumption. The distinction is discussed in the study of double complement; the treatment of intuitionistic valued models gives the broader semantic setting.