TITLE: Non-isomorphic atomless Boolean algebras
QUESTION [6 upvotes]: All countable atomless algebras are isomorphic. Can one give an example of a pair of mutually non-isomorphic atomless Boolean algebras of cardinaliy continuum?

REPLY [7 votes]: Let $\langle X,\tau\rangle$ be a topological space. A set $U\in\tau$ is a regular open set iff $\operatorname{int}_X\,\operatorname{cl}_XU=U$. Let $\mathrm{RO}(X)$ be the family of regular open subsets of $X$. For $U,V\in\mathrm{RO}(X)$ define
$$\begin{align*}
&U\land V=U\cap V,\\
&U\lor V=\operatorname{int}_X\,\operatorname{cl}_X(U\cup V),\\
&\lnot U=X\setminus\operatorname{cl}_XU,\text{ and}\\
&U\le V\text{ iff }U\subseteq V\;;
\end{align*}$$
it’s well-known that this makes $\mathrm{RO}(X)$ a complete Boolean algebra. Clearly this algebra is atomless if $X$ is $T_2$ and has no isolated points. In particular, $\mathrm{RO}(\Bbb R)$ is atomless. Since $\Bbb R$ is second countable, it’s clear that $|\mathrm{RO}(\Bbb R)|\le 2^\omega$. On the other hand, for any $A\subseteq\Bbb Z$ the set $$\bigcup_{n\in A}\left(n-\frac14,n+\frac14\right)$$ is regular open, and there are clearly $2^\omega$ such sets, so $|\mathrm{RO}(\Bbb R)|=2^\omega$. In short, $\mathrm{RO}(\Bbb R)$ is a complete, atomless Boolean algebra of power $2^\omega$.
Now let $\mathscr{B}=\wp(\omega)/[\omega]^{<\omega}$, the quotient of the power set algebra of $\omega$ by the ideal of finite subsets of $\omega$. Since $|\wp(\omega)|=2^\omega$, and $\left|[\omega]^{<\omega}\right|=\omega$, it’s clear that $|\mathscr{B}|=2^\omega$. It’s also clear that $\mathscr{B}$ is atomless. However, $\mathscr{B}$ is not complete, so it most be distinct from $\mathrm{RO}(\Bbb R)$.
To see that $\mathscr{B}$ is not complete, let $\{A_n:n\in\omega\}$ be a partition of $\omega$ into infinite subsets. For each $A\subseteq\omega$ denote by $\widehat A$ its equivalence class in $\mathscr{B}$. If $\widehat S$ is any upper bound in $\mathscr{B}$ for $\{\widehat{A_n}:n\in\omega\}$, $|A_n\setminus S|<\omega$ for each $n\in\omega$, so for each $n\in\omega$ we may choose $s_n\in S\cap A_n$. Let $T=\{s_n:n\in\omega\}$, and let $S\,'=S\setminus T$; clearly $|A_n\setminus S\,'|<\omega$ for each $n\in\omega$, so $\widehat{S\,'}$ is an upper bound for $\{\widehat{A_n}:n\in\omega\}$, but it’s also clear that $\widehat{S\,'}<_{\mathscr{B}}\widehat{S}$.
(One could also note that $\mathrm{RO}(\Bbb R)$ is ccc, while $\mathscr{B}$ isn’t, but t.b. already used that idea, so I thought that I’d do something different.)