# Can the Boolean Algebra of regular open sets be isomorphic to ${\cal P}(\omega)/(\text{fin})$?

Let $(X,\tau)$ be a topological space. $A\subseteq X$ is said to be regular open if $A = \text{int}(\text{cl}(A))$ and let $\text{RO}(X,\tau)$ denote the collection of regular open sets of $X$. A standard exercise exercise shows that $(\text{RO}(X,\tau),\subseteq)$ is not only a lattice, but even a Boolean algebra.

Question. Is there a topological space $(X,\tau)$ such that $$\text{RO}(X,\tau) \cong {\cal P}(\omega)/(\text{fin})?$$

(The Boolean algebra ${\cal P}(\omega)/(\text{fin})$ is defined here.)

• No, $\textrm{RO}(X,\tau)$ is complete, while ${\mathcal P}(\omega)/\textrm{fin}$ is not. – Keith Kearnes Sep 20 '18 at 12:36

No, $\mathrm{RO}(X)$ is complete; $\mathcal{P}(\omega)/\mathit{fin}$ is not (no strictly increasing sequence has a supremum).
• And conversely, every complete Boolean algebra is isomorphic to $\mathrm{RO}(X)$ for a suitable $X$. – Emil Jeřábek Sep 20 '18 at 14:57
• no separation axioms needed; that algebra is always complete: $\sup\mathcal{O}=\operatorname{int}\operatorname{cl}(\bigcup\mathcal{O})$. – KP Hart Sep 20 '18 at 14:45