Singletons in the $\sigma$-algebra generated by clopen sets of a Stone space Let $A$ be a Boolean algebra and let $Ult(A)$ be its Stone space, that is, the set of all ultrafilters on $A$. It is well known that $C=\{\{u\in Ult(A)\!:a\in u\}\!:a\in A\}$ is an algebra of sets isomorphic to $A$ and the elements of $C$ are exactly the clopen sets of the topology on $Ult(A)$ generated by $C$.
Questions: Let $B$ be the $\sigma$-algebra of sets generated by $C$. Does $\{\{u\}\!:u\in Ult(A)\}\subseteq B$ necessarily hold true? If not, is there any standard name for the ultrafilters $u$ satisfying $\{u\}\in B$? Is there any characterization of Boolean algebras $A$ for which $\{\{u\}\!:u\in Ult(A)\}\subseteq B$?
My attempt: A candidate counterexample is the Boolean algebra $A=\mathcal{P}(\omega)$. Then $Ult(A)$ is $\beta\omega$, the Stone-Čech compactification of the integers. It is known that $\beta\omega$ is not first countable, see here (statement 7A). More precisely, if $u\in\beta\omega\setminus\omega$ then $u$ does not have a countable local base, hence $\{u\}$ is not a countable intersection of clopen sets. However, it is not clear to me if this implies $\{u\}\notin B$.
 A: If $A$ is countable, then $\{u\} = \bigcap_{a \in u} \{u \in Ult(A) : a \in u\}$. So we should have $\{\{u\} : u \in Ult(A)\} \subseteq B$.
If $A$ is allowed to be uncountable, I think the following would be a counterexample. Let $A$ be freely generated by uncountably many generators $\{a_i\}_{i \in I}$. As you mentioned, the clopens of the Stone space $C$ are given by the elements of $A$ in the sense that for $a \in A$ we have a clopen $[a] = \{u \in Ult(A): a \in u\}$. Any non-empty element $V$ in the $\sigma$-algebra $B$ generated by the Stone space $C$, is a countable Boolean combination of clopens. Every element in $A$ is given by some finite Boolean combination of generators. In particular there will be only countably many generators that are relevant for $V$. Since we had uncountably many generators, we can find some generator $a$ that does not occur anywhere in $V$. So there will be at least two distinct ultrafilters in $V$, one that contains $a$ and one that contains $\neg a$. We can do this for every (non-empty) element of the $\sigma$-algebra, so there will be in fact no singletons in the $\sigma$-algebra generated by the Stone space of $A$.
A: Below we prove that for any Boolean algebra $A$, $\{u\}\in B$ if and only if $u$ has countable character, that is, $u$ is generated by a countable set of generators. As it was shown here or here, no non-principal ultrafilter on $\omega$ is countably generated. Hence, $A=\mathcal{P}(\omega)$ is indeed a counterexample.
Since the character of an ultrafilter $u$ (the minimum size of a set of its generators) is the same as the character of $u$ as a point of the Stone space  (the minimum size of its neighbourhood base), the condition $\{\{u\}\!:u\in Ult(A)\}\subseteq B$ is equivalent to the statement that the Stone space $Ult(A)$ is first countable.
Clearly, the Stone space of a countable Boolean algebra is first countable. On the other side, there are uncountable Boolean algebras having first countable Stone space. One example is the algebra of clopen sets of the Weak Parallel Line Topology, which is known to be compact, Hausdorff, zero-dimensional, first countable, but not second countable (Steen, Seebach: Counterexamples in Topology).


Theorem. Let $A$ be a Boolean algebra, $u\in Ult(A)$, $B$ be the $\sigma$-algebra generated by the family $C=\{\{u\in Ult(A)\!:a\in u\}\!:a\in A\}$. The following conditions are equivalent.
  
  
*
  
*$\{u\}\in B$,
  
*$\{u\}=\bigcap E$ for some countable set $E\subseteq C$,
  
*$u$ is generated by some countable set $D\subseteq A$.

Proof. 1 $\Rightarrow$ 2.
Since $B$ is the $\sigma$-algebra generated by $C$, we have $B=\bigcup_{\alpha<\omega_1}B_\alpha$, where


*

*$B_0=C$,

*$B_{\alpha+1}=\big\{\bigcap E\!:E\subseteq B_\alpha,\ |E|\le\omega\big\}\cup
\big\{\bigcup E\!:E\subseteq B_\alpha,\ |E|\le\omega\big\}$ for all $\alpha<\omega_1$,

*$B_\alpha=\bigcup_{\beta<\alpha}B_\beta$ for all limit $\alpha<\omega_1$.


Let $\{u\}\in B$ and let $\alpha<\omega_1$ be the least ordinal such that $\{u\}\in B_\alpha$. 


*

*If $\alpha=0$ then we are done. 

*If $\alpha=\beta+1$ then $\{u\}=\bigcap E$ for some countable set $E\subseteq B_\beta$, since $\{u\}=\bigcup E$ would imply $\{u\}\in E$. If $\beta=0$ then again we are done. We show that $\beta>0$ is impossible. So assume that $E=\{V_n\!:n\in\omega\}$ and for every $n$ there exists a countable set $F_n\subseteq\bigcup_{\gamma<\beta}B_\gamma$ satisfying $V_n=\bigcap F_n$ or $V_n=\bigcup F_n$. Denote $N=\{n\in\omega\!:V_n=\bigcap F_n\}$. For $n\in\omega\setminus N$ pick some $Z_n\in F_n$ such that $u\in Z_n$. Then
$$\{u\}=\bigcap_{n\in\omega}V_n=\bigcap_{n\in N}\bigcap F_n\cap\bigcap_{n\in\omega\setminus N}\{Z_n\}.$$ Hence, there exists a countable set $F\subseteq\bigcup_{\gamma<\beta}B_\gamma$ such that $\{u\}=\bigcap F$, namely, $F=\bigcup_{n\in N}F_n\cup\{Z_n\!:n\in\omega\setminus N\}$. But this implies $\{u\}\in B_\beta$, which contradicts the minimality of $\alpha$.

*The case of limit $\alpha$ is impossible since it contradicts the minimality of $\alpha$.


2 $\Rightarrow$ 3. If $E\subseteq C$ is countable then there exists a countable set $D\subseteq A$ such that $E=\{\{v\in Ult(A)\!:a\in v\}\!:a\in D\}$. Clearly, $\bigcap E=\{v\in Ult(A)\!:D\subseteq v\}$. If $\{u\}=\bigcap E$ then $u$ must be the only filter containing $D$, that is, $u$ is generated by $D$.
3 $\Rightarrow$ 2 $\Rightarrow$ 1 is clear.
$\ $q.e.d.
