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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.)

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    $\begingroup$ No, $\textrm{RO}(X,\tau)$ is complete, while ${\mathcal P}(\omega)/\textrm{fin}$ is not. $\endgroup$ – Keith Kearnes Sep 20 '18 at 12:36
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No, $\mathrm{RO}(X)$ is complete; $\mathcal{P}(\omega)/\mathit{fin}$ is not (no strictly increasing sequence has a supremum).

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  • $\begingroup$ And conversely, every complete Boolean algebra is isomorphic to $\mathrm{RO}(X)$ for a suitable $X$. $\endgroup$ – Emil Jeřábek Sep 20 '18 at 14:57
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The algebra of regular open sets in a completely regular space is complete. As your algebra is not complete, it cannot be isomorphic to such an algebra. So the answer is no, at least on the class of completely regular spaces.

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    $\begingroup$ no separation axioms needed; that algebra is always complete: $\sup\mathcal{O}=\operatorname{int}\operatorname{cl}(\bigcup\mathcal{O})$. $\endgroup$ – KP Hart Sep 20 '18 at 14:45

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