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 $\{u\}\notin B$.

  • $\begingroup$ In $\beta \omega$ if $F$ is a countable subset of $C$ and $u\in \cap F$ then $int (\cap F)$ is not empty. I think we can use this to show by transfinite induction that if $u\not \in \omega$ then $\{u\}\not \in B$...Let $B_0$ by the Boolean algebra generated by $C.$ For $ x\in \omega_1 let B_{x+1}$ be the Boolean algebra generated by intersections of countable subsets of $B_x.$ For $0<x=\cup x\in \omega_1$ let $B_x=\cup_{y\in x}B_y.$ Then $B=\cup_{x\in \omega_1}B_x$. Suppose there is a least $x$ such that $\{u\}\in B_x.$ $\endgroup$ – DanielWainfleet Apr 16 at 21:58

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 $A$.

  • $\begingroup$ We need to check that any element of $B$ (or at least its atoms) will tell us only on countably many generators. But somehow I cannot see that. $\endgroup$ – Peter Elias Apr 16 at 10:16
  • $\begingroup$ I have edited my answer to explain the counterexample a bit better (and fixed a mistake in my notation in the countable case). Hopefully this will make things clearer. $\endgroup$ – Mark Kamsma Apr 16 at 10:38
  • $\begingroup$ As far as I can see, you have proved that no countable intersection of clopen sets is a singleton. But what I need is that no element of $B$ is a singleton (or at least not all singletons are there). I am missing why dealing with countable intersections is enough. The elements of $B$ may be more complicated. Remind that $B$ is the $\sigma$-algebra generated by clopen sets. $\endgroup$ – Peter Elias Apr 16 at 10:45
  • $\begingroup$ Hmm, you are right. I oversimplified things a bit here (this is what happens before morning coffee). The trick may still work though, the elements of $B$ are given by countable Boolean combination of clopens. So we would still be able to find some generator that does not appear in that combination. Before I edit my answer, does this sound right to you? $\endgroup$ – Mark Kamsma Apr 16 at 11:10
  • $\begingroup$ Yes, now it is clear. Thank you! It just took me some time to realize that we can define $B$ by an induction of the length $\omega_1$ and that each element of $B$ is obtained at some step $\alpha<\omega_1$ and thus uses only countably many clopens. $\endgroup$ – Peter Elias Apr 16 at 11:28

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.

  1. $\{u\}\in B$,
  2. $\{u\}=\bigcap E$ for some countable set $E\subseteq C$,
  3. $u$ 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.


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