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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 $\neg \neg {\bigvee }_i{a}_i$.

A sublocale is flat when its nucleus on a frame preserves finite joins, including $0$. Equivalently its fixed elements are closed under ambient finite joins. In particular every flat sublocale is a dense sublocale.

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.