# Regular open Boolean algebras and homomorphism which does not preserve nearness of sets

I am looking for an example of topological spaces $\langle X_1,\mathscr{O}_1\rangle$ and $\langle X_2,\mathscr{O}_2\rangle$ such that there is a homomorphism $h\colon\mathrm{r}\mathscr{O}_1\longrightarrow\mathrm{r}\mathscr{O}_2$ between the algebras of regular open sets of respective spaces which fails to preserve the information about nearness of sets, in the sense that there are $x,y\in\mathrm{r}\mathscr{O}_1$ such that: $$\mathrm{Cl}\,x\cap \mathrm{Cl}\,y\neq\emptyset\qquad\text{yet}\qquad \mathrm{Cl}\,h(x)\cap \mathrm{Cl}\,h(y)=\emptyset\,.$$ By regular open I mean a set $x$ which is equal to the interior of its closure: $x=\mathrm{Int\,Cl\,}x$.

Similalrly, I am also interested in the dual example in which: $$\mathrm{Cl}\,x\cap \mathrm{Cl}\,y=\emptyset\qquad\text{yet}\qquad \mathrm{Cl}\,h(x)\cap \mathrm{Cl}\,h(y)\neq\emptyset.$$

I will be grateful for help.

## 1 Answer

$\textbf{A simple counterexample}$

The mapping $\phi:Ro(\omega+1)\rightarrow Ro(\omega)$ defined by $\phi(U)=U\cap\omega$ is a Boolean algebra isomorphism, but if $E$ denotes the even positive integers and $O$ denotes the odd positive integers, then $Cl_{\omega+1}(E)\cap Cl_{\omega+1}(O)=\{\omega\}$ while $Cl_{\omega}(\phi(E))\cap Cl_{\omega}(\phi(O))=\emptyset$.

$\textbf{De Vries duality gives many counterexamples}$

A De Vries algebra is a complete Boolean algebra with some additional structure. De Vries duality 1 is a duality between compact Hausdorff spaces and De Vries algebras. The underlying Boolean algebra from the De Vries dual of a compact Hausdorff space is simple the regular open algebra.

If $X,Y$ are compact Hausdorff spaces, and $\phi:Ro(X)\rightarrow Ro(Y)$ is an isomorphism such that $\overline{X}\cap\overline{Y}=\emptyset$ if and only if $\overline{\phi(X)}\cap\overline{\phi(Y)}=\emptyset$, then the mapping $\phi$ is a De Vries isomorphism. Therefore, $\phi$ induces an isomorphisms between the spaces $X$ and $Y$.

It should be said that the morphisms between De Vries algebras are not generally Boolean algebra homomorphisms unless the De Vries algebra morphisms in question are actually De Vries isomorphisms.

Therefore, if $X,Y$ are compact Hausdorff spaces which are not homeomorphic, then for every isomorphism $\phi:Ro(X)\rightarrow Ro(Y)$, we cannot have $\overline{U}\cap\overline{V}=\emptyset\leftrightarrow\overline{\phi(U)}\cap\overline{\phi(V)}=\emptyset$ for each $U,V\in Ro(X)$.

Isomorphisms between regular open algebras are quite common:

For example, if $X$ is a second countable regular space with no isolated points, then

1. $Ro(X)$ has a countable Boolean subalgebra $B$ where $Ro(X)$ is the completion of $B$ and

2. $Ro(X)$ has no atoms.

Since the Cohen algebra is the unique up-to-isomorphism complete atomless Boolean algebra with a countable dense subalgebra, we conclude that $Ro(X)$ is isomorphic to the Cohen algebra.

Stone duality and Gleason covers through de Vries duality Guram Bezhanishvili

• Joseph, only now did I have time to go through your helpful answer. Thanks a lot! Jan 24, 2016 at 21:27