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:
- Is my counterexample correct?
- 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)