This is a short overview of a talk given by Stefan Geschke on November 21, 2008. Stefan’s topic, Cofinalities of algebraic structures and coinitialities of topological spaces, very quickly connects set theory with other areas, and leads to well-known open problems. In what follows, compact always includes Hausdorff. Most of the arguments I show below are really only quick sketches rather than complete proofs. Any mistakes or inaccuracies are of course my doing rather than Stefan’s, and I would be grateful for comments, corrections, etc.
Definition. Let be a (first order) structure in a countable language. Write for the smallest such that for a strictly increasing union of proper substructures.
Since the structures need to be proper, is not defined if is finite. It may also fail to exist if is countable, but it is defined if is uncountable. Moreover, if exists, then
- , and
- is a regular cardinal.
Example 1. Groups can have arbitrarily large cofinality. This is not entirely trivial, as the sets may have size .
Question 1. Is every regular cardinal realized this way?
Example 2. (Shelah-Thomas) can vary between and .
Example 3. Boolean algebras. The cofinality of any (infinite) Boolean algebra is defined and at most . If the algebra is complete, its cofinality is precisely .
Here is a sketch of the argument giving the upper bound:
- Suppose is an epimorphism. Then . This is because if , then we can set and . Moreover, since is onto, no is .
- Now we argue that if is a Boolean algebra, then it admits a quotient of size at most . We use Stone duality. Let , the Stone space of , so quotients of correspond to closed subspaces of . Under this correspondence, a homomorphism corresponds to . Now let and set be the closure of , so is compact, 0-dimensional and has weight at most (i.e., admits a basis of size at most ). Let be the Boolean algebra of clopen subsets of . Then is a quotient of , as desired. To see the claim about the weight of , simply notice that for any , any clopen set in the closure of is coded by its intersection with .
Open question 1. Is it consistent to have a Boolean algebra of cofinality strictly larger than ?
Example 4. Banach spaces. Here we want the cofinality to consider the metric structure as well, so we modify the definition, by requiring that if is a Banach space, then is the smallest lenght of a strictly increasing chain of closed proper subspaces of whose union is dense in . Tis is well defined as long as is infinite dimensional, and an upper bound is the density of .
Theorem. iff has a separable infinite dimensional quotient.
The problem here is the following: If is onto, a preimage of a dense set is not necessarily dense, so the argument for the case of Boolean algebras does not apply here. However, if the cofinality of is uncountable, then the union of a chain witnessing this is automatically closed, and therefore equals all of . Hence, realizing a countable cofinality seems to be the issue here. However:
Open question 2. Is it consistent to have a Banach space of uncountable cofinality?
This is equivalent to the separable quotient problem, a famous question in functional analysis.
All concrete examples whose cofinality is known have countable cofinality. For example, this holds for all where is compact: Either is scattered, in which case the conclusion is easy, or else admits a perfect subset, and by Riesz duality we can build a measure on it, and use the measure to find the desired quotient.
Remark. For Banach spaces, we also have that the cofinality is at most , because every Banach space has an infinite dimensional quotient isomorphic to a subspace of .
Why is open problem 1 difficult? It is a bit more convenient to switch the discussion to the topological setting, via Stone duality. Let be compact. The coinitiality of is the minimal length of a nontrivial inverse system with limit ,
Here, each is a quotient of . Intuitively, the “resolution” with which we see increases as we go back.
If is compact and 0-dimensional, then its coinitiality is precisely the cofinality of the Boolean algebra of clopen subsets of . For general spaces , we look at the -algebra . The coinitiality of is always defined for compact infinite; the argument is a bit more delicate than before.
One can characterize countable coinitiality: Let be a sequence in compact and let . The -limit of the sequence is the unique such that
Another description is the following: Let . Then has a unique continuous extension , and
Definition. is a double sequence iff it is discrete and for each ,
A typical example of a double sequence is a convergent sequence, but being a double sequence is a more general notion.
Theorem. is of countable coinitiality iff has a double sequence.
Proof. Assume first there is a double sequence . Get quotients of by identifying and for all from some on. The inverse limit of this sequence of spaces is .
Assume now that has countable coinitiality, and consider a witnessing sequence . In each choose two distinct points that were previously identified in , and let be preimages in of , respectively. Then is a double sequence.
Definition. A subset of a Boolean algbera is independent iff whenever are finite subsets of ,
Lemma. If is a Boolean algebra and has an uncountable independent subset, then
Proof. Let denote the free Boolean algebra on generators. Then embeds into (by hypothesis) and is dense in so, from general theory of Boolean algebras, we get a homomorphic map .
Consider now a filtration of , say . Also, let be a filtration of , say . We can now let , and we verify at once that the are increasing substructures with union .
Now we can prove that for complete: Or we would have a double sequence in the Stone space, but we can separate and in the Stone space by “separating them” in the complete Boolean algebra.
If the coinitiality of is larger than , then contains no double sequence and no uncountable dyadic family (the dual of an independent family). In particular, cannot contain a copy of . We say that such an is an Efimov space.
Definition. An Efimov space is a compact space without convergent sequences nor copies of .
[For 0-dimensional, it is enough to see that it admits no countable quotient nor a quotient isomorphic to .]
Open question 3. Assume is compact and homogeneous. Does it have a convergent sequence?
Open question 4. (Efimov problem) Is there an Efimov space?
Problem 4 asks for examples in ; it is known that it has a positive answer under and under certain inequalities among cardinal invariants. Another example can be obtained as follows: Start with a model with small continuum, add many random reals (let be the corresponding forcing notion), and look at ; take its Stone space in . This is an Efimov space in the extension!
To see this, argue as follows:
- No convergent sequences are added (this follows from general results about random forcing).
- is too small, so its Stone space cannot contain .
A related argument shows that the existence of real valued measurable cardinals also implies a positive answer to question 4.
On the other hand, it is expected that refutes the existence of Efimov spaces.
Here is a brief sketch of a construction of an Efimov space under . The construction is inductive and at limit stages we simply take unions; given , we get such that any subalgebra containing is either or .—A minimal extension.
Then cannot contain an uncountable independent subset.
Finally, allows us to kill convergent sequences by “predicting” them.
Theorem. (Geschke) Under (or even ), there are Efimov spaces of coinitialities and .
[For , the argument from sketched above also allows us to kill double sequences. For , one keeps one double sequence, while killing all convergent sequences.]