The regular open completion of a Boolean algebra

Put $X=\mathrm{Ult}(B)$. Every clopen $K$ is regular open since $\mathrm{int}\bar{K}=K$. On clopens the operations of F1 give ordinary intersections, complements and finite unions, because these sets are already clopen. Thus F2 gives a Boolean embedding $i:B\to \mathrm{RO}(X)$, whose codomain is complete by F1. If $U\in \mathrm{RO}(X)$ is nonempty, take one $x\in U$ and a basic clopen $[b]$ with $x\in [b]\subseteq U$. This makes $0<i(b)\le U$ and proves order density.

We record two consequences of order density using only Boolean laws. Let $j:B\to C$ be any order-dense Boolean embedding with $C$ complete. For $c\in C$ put $c_0=\bigvee \{j(b):j(b)\le c\}$. Then $c_0\le c$. If strict, density gives $0<j(a)\le c\wedge \mathrm{\neg }{c}_0$. But $j(a)\le c$ puts $j(a)\le c_0$, a contradiction. Thus $c=c_0$. Next suppose $s={\bigvee }_{B}A$ exists. The element $u={\bigvee }_{C}j[A]$ is at most $j(s)$. If $j(s)\wedge \mathrm{\neg }u>0$, density gives $0<j(a)\le j(s)\wedge \mathrm{\neg }u$. Order reflection implies $a\le s$ and $a\wedge b=0$ for every $b\in A$. Then $s\wedge \mathrm{\neg }a$ is an upper bound of $A$ strictly below $s$, contradicting its least-upper-bound property. Therefore $u=j(s)$. Complements convert existing infima to suprema and show their preservation as well. The calculation includes $A=\mathrm{\varnothing }$, since its supremum is zero and embeddings preserve bounds.

Applying step 1.2 to the embedding of step 1.1 proves the claimed preservation of existing bounds. To prove uniqueness of completions, take the stated $j:B\to C$ and $k:B\to D$ and define $\mathrm{\Phi }(c)=\bigvee \{k(b):j(b)\le c\}$. Completeness of $D$ defines this value uniquely for every $c$; monotonicity follows from inclusion of the indexing sets. If $j(b)\le c$, then $k(b)\le \mathrm{\Phi }(c)$. Conversely suppose $j(b)\nleqq c$. Choose by density $0<j(a)\le j(b)\wedge \mathrm{\neg }c$. For every $t$ with $j(t)\le c$, $a\wedge t=0$, so $k(t)\le \mathrm{\neg }k(a)$ and hence $\mathrm{\Phi }(c)\le \mathrm{\neg }k(a)$. But $0<k(a)\le k(b)$, so $k(b)\nleqq \mathrm{\Phi }(c)$. We have proved the exact cut identity $k(b)\le \mathrm{\Phi }(c)$ iff $j(b)\le c$.

Define symmetrically $\mathrm{\Psi }(d)=\bigvee \{j(b):k(b)\le d\}$. The cut identity gives $\mathrm{\Psi }(\mathrm{\Phi }(c))=\bigvee \{j(b):j(b)\le c\}=c$ by step 1.2. Interchanging the roles of the two embeddings gives the reverse cut identity and $\mathrm{\Phi }(\mathrm{\Psi }(d))=d$. Both maps are monotone, so they are inverse order isomorphisms. An order isomorphism preserves bounds and least/greatest bounds: applying its inverse to any competing bound reduces the required inequality to the original least/greatest property. It therefore preserves binary joins and meets; it also preserves complements, since the image of the complement has meet zero and join one with the image element, and complements are unique in a Boolean algebra. Thus $\mathrm{\Phi }$ is a Boolean isomorphism. For $b\in B$, every index $t$ with $j(t)\le j(b)$ satisfies $t\le b$, and $t=b$ itself is included, giving $\mathrm{\Phi }(j(b))=k(b)$.

Any Boolean isomorphism $\mathrm{\Theta }:C\to D$ over $B$ is an order isomorphism and hence preserves every supremum by the bound argument of step 3.1. The dense-supremum equality of step 1.2 forces $\mathrm{\Theta }(c)=\bigvee \{\mathrm{\Theta }(j(b)):j(b)\le c\}=\mathrm{\Phi }(c)$ for all $c$, proving uniqueness. If $B$ is trivial, its Stone space is empty and the regular-open algebra is trivial. Any bound-preserving embedding from this $B$ forces $0_C=1_C$ and likewise for $D$, so the uniqueness statement also holds there. The only selections in the proof are one point and one basic neighborhood or one nonzero dense element at a time; no AC is needed beyond the explicit BPI hypothesis for F2. QED.