I had a hard time trying to solve exercise 7.24 in Jech's book (3rd edition, 2003) and finally came to the conclusion that the result there, which should be proved might be wrong. The claim goes like this:

Let $A$ be a subalgebra of a Boolean algebra $B$ and suppose that $u \in B-A$. Then there exist ultrafilters $F,G$ on $B$ such that $u \in F$, $-u \in G$ and $F \cap A= G \cap A$.

A (perhaps flawed, as I believe) proof of this can be found here. http://onlinelibrary.wiley.com/doi/10.1002/malq.19690150705/abstract

A counterexample to the claim above is the following:

Let $A$ be the algebra of finite unions of (open, closed, half-open) intervals on $[0,1]$ with rational endpoints, and let $B$ be defined as $A$ but with real endpoints. Each ultrafilter $U$ on $A$ converges to a rational or irrational number $r$ and the elements of $U$ are exactly those sets in $A$ that include $r$. Now if $F$ and $G$ are two ultrafilters on the bigger algebra $B$, both extending $U$ then they converge again towards $r$ and for any $u\in B$ we have that $u\in F$ iff $r \in u$ iff $u\in G$, which makes it impossible to have $u \in B$, yet $-u \in G$.

My questions are now:

  1. Is my counterexample correct?
  2. The claim is used to show that each Boolean algebra of size $\kappa$ has at least $\kappa$ ultrafilters (this is theorem of the paper mentioned above). Does this remain valid ( I suppose not, see the comments)

2 Answers 2


Your counterexample is not correct. Let $r$ be an irrational real number, and let $F$ be the principal ultrafilter in $B$ on the closed interval $[r,r]=\{r\}$, which is an atom in $B$. Note that $F\cap A$ is the ultrafilter of all elements of $A$ in which $r$ is a member. Now consider the complement $-[r,r]=(-\infty,r)\cup (r,\infty)$, which has nonempty intersection with any interval in $A$ containing $r$, since $r$ was irrational. Thus, we add $-[r,r]$ to $F\cap A$ and extend to an ultrafilter $G$. So $F$ and $G$ agree on $A$, but one has the atom $[r,r]$ and the other has the complement.

Meanwhile, the the Balcar-Franek theorem establishes that every [Edit: complete] Boolean algebra of size $\kappa$ has $2^\kappa$ many ultrafilters. This topic also arose in this MO question on density of Boolean algebras.

  • $\begingroup$ Nevertheless, both $F$ and $G$ "converge" to $r$ in the sense of $A$, because they each contain all rational intervals containing $r$, but in $B$ the filter $F$ actually contains the singleton, while $G$ contains the complement of the singleton. $\endgroup$ Aug 25, 2011 at 17:51
  • $\begingroup$ You're welcome; it was my pleasure. $\endgroup$ Aug 25, 2011 at 18:19
  • $\begingroup$ Joel, the Balcar-Franek theorem applies to complete Boolean algebras. The finite-cofinite algebra on the integers has only countably many ultrafilters. $\endgroup$ Aug 25, 2011 at 20:40
  • $\begingroup$ Stefan, yes, I'll correct that. $\endgroup$ Aug 25, 2011 at 21:38

The exercise is correct. Let $u\in B\setminus A$. We say that $u$ splits an ultrafilter $H$ of $A$ if $\{u\}\cup H$ and $\{-u\}\cup H$ both have the finite intersection property. (If $u$ splits $H$, then there are ultrafilters $F$ and $G$ of $B$ such that $F\cap A=H=G\cap A$, $u\in F$, and $-u\in G$.)

Suppose no ultrafilter $H$ is split by $u$.
We say that an ultrafilter $H$ of $A$ is compatible with $b\in B$ iff each $a\in H$ has nonempty intersection (in $B$) with $b$. If no ultrafilter of $A$ is split by $u$, then each ultrafilter is either compatible with $u$ or compatible with $-u$.

Let $C$ be the set of ultrafilters of $A$ compatible with $u$, and let $D$ be the set of ultrafilters of $A$ compatible with $-u$.
Now the set Ult$(A)$ of all ultrafilters of $A$ is the disjoint union of $C$ and $D$. Hence an ultrafilter of $A$ is compatible with $u$ iff it is not compatible with $-u$ and vice versa. So, if $H\in C$, then there is $a\in H$ such that $a$ is disjoint from $-u$. All ultrafilters of $A$ that contain $a$ are incompatible with $-u$ and hence compatible with $u$. This shows that $C$ is open in the Stone space Ult$(A)$ of $A$.
The same is true for $D$. It follows that the two sets are clopen. By the Stone representation theorem, there is $a\in A$ such that $C$ is the set of all ultrafilters $H$ of $A$ that contain $a$. $D$ is the set of all ultrafilters of $A$ that contain $-a$.

In other words, an ultrafilter $H$ of $A$ is compatible with $u$ iff $a\in H$. But this implies that an element $b$ of $A$ has a nonempty intersection with $u$ iff it has a nonempty intersection with $a$. Hence $-a$ is disjoint from $u$. In other words, $u\leq a$. The symmetric argument shows that $-u\leq-a$. It follows that $a=u$ and hence $u\in A$, a contradiction.

And yes, this exercise implies that every infinite Boolean algebra of size $\kappa$ has at least $\kappa$ ultrafilters.


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