Exercises in Set Theory: Ultrafilters and Boolean algebras
Since my previous blog post, I have been working on exercises from chapter 7 of Thomas Jech’s book "Set Theory". I’ve published my solutions but I’m not yet done with exercises 7.22 and 7.28 which seem to be generalizations of theorems 4.3 and 3.2. I guess that before coming back to these two exercises, I’ll consider the content and exercises of subsequent chapters…
I had the opportunity to read chapter 7 again and to study more precisely the correctness of some claims (you know, those that are "clearly seen to be obvious via an easy proof based on a straightforward argument that makes them trivial" but actually take time before really being understood) I describe below the two statements for which the ratio between the time I spent on them and the length of the proof provided was the largest. I indicate the arguments I used to convince myself. I also had trouble to understand the proof of theorem 7.15 but most of the required arguments can actually be found in the exercices.
The first part of the chapter essentially deals with ultrafilters on an infinite set. In particular, a nonprincipal ultrafilter on is a Ramsey ultrafilter if for every partition of into pieces such that , there exists such that . We have:
The continuum hypothesis implies the existence of a Ramsey ultrafilter.
The proof given in the book is to consider an enumeration of all partitions of and to construct by induction a sequence of infinite subsets. is chosen to be included in some element of or to intersect each of them at, at most, one element. For limit, is chosen such that is finite for all . Such a is claimed to exist because is countable. Finally, is claimed to be a Ramsey ultrafilter.
There were several points that were not obvious to me. First, it’s good to count the number of partitions of . On the one hand, given any which is neither empty nor cofinite we can define a partition of by and so there are at least such partitions (cofinite subsets are in bijection with finite subsets and ). On the other hand, a partition is determined by the family of subsets and so there are at most such partitions. Since we assume the continuum hypothesis we get the enumeration .
Next, I’m not even sure that as it is defined is a filter. Because is infinite and but if and , it does not seem really clear which . However, if is a nonprincipal ultrafilter a remark from chapter 10 suggests that does not contain any finite set. By exercise 7.3, this is equivalent to the fact that does not contain any singleton . This latter point is easy to verify: if we assume the contrary, because is nonprincipal, there exists such that and so would be in . Hence any ultrafilter containing should also contain . is a filter: the arguments above still hold and and so we only need to verify that is finite if . This is already what is claimed for a limit ordinal. In general if is the greatest limit ordinal such that then by construction, and so is empty if and is finite if . So we can replace by a "ultrafilter extending ". It’s not too difficult to check that a ultrafilter containing must be Ramsey. For any partition then by construction either for some and so or . In the latter case we can find (so ) such that (we pick one element from each such that and put these elements into ).
Finally, the most difficult part was to understand how the sequence can be built. We can choose arbitrarily as this set is not involved in the arguments above. For any , if there exists some such that is infinite, we just set . Otherwise, since is infinite, is nonempty for infinitely many . Pick one element from each nonempty set to form the set . By definition, . Now we consider the case limit. Let be a cofinal -sequence in . Note that it is the only place where we use the continuum hypothesis! For each , we define . We have infinite and finite so is infinite. Similarly, and are finite so is infinite etc and finally is infinite. Thus we can use a classical technique from Cantor (I can’t find a reference) and construct the set by picking each so that is finite. Moreover for each consider some : is finite.
The second part of the chapter is about Boolean algebras. A Boolean algebra is a completion of a Boolean algebra if it is complete (every subset has a least upper bound ) and if is a dense subalgebra of (for any there exists such that ). I got stuck on the apparently easy proof of the following lemma:
The completion of a Boolean algebra is unique up to isomorphism
Let are two completions of . It is indicated in the book that the density of in and is used to prove that defines an isomorphism between and . The example of is given: indeed if we can find an element such that and so . ∎
Two or three pages before, it is mentioned that "if is a one-to-one mapping such that then is an isomorphism". That seems to be the most natural method to prove the lemma above. However, this statement is false: consider the morphism of Boolean algebra which is one-to-one but not surjective. Actually, the property implies the fact that is one-to-one, which becomes vacuous. So I suspect the author rather means "surjective" instead of "one-to-one". Indeed, in that case is a bijection that preserves the order in both directions and since all the operations of the Boolean algebra can be described in terms of this order, is an isomorphism.
So first, given , we want to find such that . By symmetry, the natural candidate is . For all , we have and so . If then by density we can find such that . Hence and . The latter implies that for all we have an so . Combining this with the former, we would have . Thus is surjective.
It remains to prove . If then . Let’s consider the contrapositive of the converse implication: suppose . Then and (we use the hint given in the book). If we are able to prove then we would have and so as desired. It suffices to show two inequalities on the and operations. Note that . Let . Let . Then for all , so . Similarly, we have for all , . So .
As a conclusion, I notice a similarity between these too cases: both have a sketchy proof based on a statement that seems incorrect (" is a filter", "any one-to-one mapping between Boolean algebra that preserves the order in both directions is an isomorphism"). Sometimes, I’m wondering if these kinds of inaccuracy are not intentional in order to force readers to try and find the proof themselves :-)