Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 20 / 4

Put $A=\mathcal{O}(X)$ and $\neg a=a\to 0$. The closure of the open $a$ is the closed sublocale with closed nucleus $c(\neg a)$; its interior is the open $\neg \neg a$. The closed sublocale complementary to $\neg a$ is also open precisely when $\neg a$ is complemented, in which case its open representative is $\neg \neg a$. Thus the closure is clopen precisely when the regular open element $\neg \neg a$ is complemented. Every regular open element has this form, so (i) and (ii) are equivalent.

Suppose every regular open element is complemented. In a Heyting algebra, a complemented element is regular: if $a\vee b=1$ and $a\wedge b=0$, then $b=\neg a$ and $a=\neg b$. Finite joins of complemented elements are complemented. Hence $\neg \neg a\vee \neg \neg b$ is regular, and it is the least regular upper bound of $a\vee b$. ThereforeThis proves (ii) implies (iii). The identification of the regular open elements with interiors of closures was established in the preceding part.

Conversely assume binary-join preservation by double negation. Since $\neg (a\vee \neg a)=0$, applying the assumption givesAlso $\neg \neg a\wedge \neg a=0$. Thus $\neg \neg a$ is complemented, and every regular open element is clopen. This proves (iii) implies (ii). Double negation already preserves $0$ and all finite meets, so binary-join preservation here is equivalent to preservation of all finite joins.

The Booleanization of a locale is the sublocale with double-negation nucleus. A flat sublocale has a nucleus preserving finite joins, equivalently its fixed-element set is closed under ambient finite joins. For completeness, if a nucleus preserves finite joins then its fixed points are closed under them. Conversely, if fixed points are closed under joins, $j(a)\vee j(b)$ is fixed and bounds $a\vee b$, giving $j(a\vee b)\le j(a)\vee j(b)$; monotonicity gives equality, and preservation of $0$ is precisely $j(0)=0$. Since $\neg \neg 0=0$, condition (iv) is exactly (iii). Together the preceding implications prove all four characterizations of an extremally disconnected locale.

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 $\gamma X\to X$.