# Is the boundary of an open set in a $\sigma$-space empty?

Recall that a Boolean space is a $\sigma$-space in case the closure of every open Borel set is open.

Let $\{B_i\}$ be a denumerable family of open-closed sets in a $\sigma$-space $X$. Then $\bigcup_i B_i$ is open but not closed (hence, not open-closed). Since $X$ is a $\sigma$-space, however, the closure of $\bigcup_i B_i$ is open and, consequently, the closure of $\bigcup_i B_i$ is open-closed.

Now, $\bigcup_i B_i$ and ${\rm cl}\ \bigcup_i B_i$ symmetrically differ by a meagre set, namely, the boundary of $\bigcup_i B_i$. This boundary is meager, but is it empty? (I think not, because the boundary of set is empty if and only if the set is open-closed, and $\bigcup_i B_i$ is open. But I am not sure...)

• It seems that each open set is Borel (by definition), so your $\sigma$-space is the same and an extremally disconnected space. If the boundary of any open set is empty, then each open set is closed. But this property does not necessarily hold in extremally disconnected spaces. For example, $\beta\mathbb N$ is extremaly disconnected, and $\mathbb N$ is an open set in $\beta\mathbb N$, which is not closed. – Taras Banakh Sep 12 '18 at 18:01
• In the definition of $\sigma$-space, "Borel" should be replaced by "Baire", I think (or equivalently $F_\sigma$, or equivalently cozero, because Boolean spaces (i.e. Stone spaces) are compact Hausdorff spaces). – Robert Furber Sep 13 '18 at 1:52
• Robert, aren't Baire and Borel sets essentially the same in Stone spaces? (Since Stone spaces are second-countable.) – puzzled Sep 13 '18 at 8:54
• open Borel set is redundant. $\sigma$-space already has another meaning; your notion is just called extremally disconnected. – Henno Brandsma Sep 13 '18 at 21:53
• @puzzled Not a bit. Stone spaces are second countable iff there are only countably many clopen sets (i.e. the Boolean algebra is countable). An infinite $\sigma$-complete Boolean algebra contains a subalgebra isomorphic to $\mathcal{P}(\omega)$ so is never countable. Therefore a Stone space that is a $\sigma$-space is second countable iff it is finite. (Also, write @ before the user name if you want to reply to a comment -- then the user you are replying to will be notified of it). – Robert Furber Sep 14 '18 at 10:03

In effect, you are asking if $\bigvee\limits_{i \in \omega}B_i = \bigcup\limits_{i \in \omega}B_i$, where the left hand side is the closure of the union, which is the join/supremum of the family $\{B_i\}_{i \in \omega}$ in the Boolean algebra of clopen sets of $X$, which we will write as $\mathrm{Clopen}(X)$. It is possible to characterize exactly when this happens.
We first consider the case when the family is not "genuinely infinite", in that there exists a finite set $I \subset \omega$ such that $\bigcup\limits_{i \in \omega}B_i = \bigcup\limits_{i \in I} B_i$. It follows that $\bigcup\limits_{i \in I} B_i$ is clopen, so the boundary of $\bigcup\limits_{i \in \omega}B_i$ is empty.
In the other case, suppose the family is "genuinely infinite", which is to say that for any finite set $I \subset \omega$, $\bigcup_\limits{i \in I}B_i$ is a proper subset of $\bigcup\limits_{i \in \omega}B_i$. We will show that the boundary of $\bigcup\limits_{i \in \omega}B_i$ is not empty.
We do this using Stone duality and the Boolean prime ideal theorem. Let's define $B = \mathrm{cl}\left(\bigcup_{i \in \omega}B_i\right) = \bigvee_{i \in \omega}B_i$. We are looking for a point $x \in X$ such that $x \in B$ but $x \not\in \bigcup_{i \in \omega}B_i$. By Stone duality, this is equivalent to finding an ultrafilter $u$ on $\mathrm{Clopen}(X)$ such that $B \in u$ but $B_i \not\in u$ for all $i \in I$. Let $\mathcal{F}$ be the principal filter on $B$ in $\mathrm{Clopen}(X)$, i.e. the set of clopen sets containing $B$, and let $\mathcal{I}$ be the ideal generated by the set $\{B_i\}_{i \in \omega}$ in $\mathrm{Clopen}(X)$, i.e. the set of clopen sets $G$ such that there exists a finite set $I \subset \omega$ such that $G \subseteq \bigcup_{i \in I}G_i$. Our assumption that the join is "genuinely infinite" ensures that $\mathcal{F} \cap \mathcal{I} = \emptyset$. So by the Boolean prime ideal theorem there exists an ultrafilter $u$ such that $\mathcal{F} \subseteq u$ and $u \cap \mathcal{I} = \emptyset$. Therefore $B \in u$ and $B_i \not\in u$ for all $i \in \omega$, as required.