## Set theory seminar -Stefan Geschke: Cofinalities of algebraic structures

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 $A$ be a (first order) structure in a countable language. Write ${rm cf}(A)$ for the smallest $delta$ such that $A=bigcup_{alpha for a strictly increasing union of proper substructures.

Since the structures $A_alpha$ need to be proper, ${rm cf}(A)$ is not defined if $A$ is finite. It may also fail to exist if $A$ is countable, but it is defined if $A$ is uncountable. Moreover, if ${rm cf}(A)$ exists, then

1. ${rm cf}(A)le|A|$, and
2. ${rm cf}(A)$ is a regular cardinal.

Example 1. Groups can have arbitrarily large cofinality. This is not entirely trivial, as the sets $A_alpha$ may have size $|A|$

Question 1. Is every regular cardinal realized this way?

Example 2. (Shelah-Thomas) ${rm cf}(S_omega)$ can vary between $aleph_1$ and $2^{aleph_0}$.

Example 3. Boolean algebras. The cofinality of any  (infinite) Boolean algebra is defined and at  most $2^{aleph_0}$. If the algebra is complete, its cofinality is precisely $aleph_1$.

Here is a sketch of the argument giving the upper bound:

1. Suppose $f:Ato B$ is an epimorphism. Then ${rm cf}(A)le{rm cf}(B)$. This is because if $B=bigcup_{alpha, then we can set $A_alpha=f^{-1}[B_alpha]$ and $A=bigcup_{alpha. Moreover, since $f$ is onto, no $A_alpha$ is $A$.
2. Now we argue that if $A$ is a Boolean algebra, then it admits a quotient of size at most $2^{aleph_0}$. We use Stone duality. Let $X={rm ult}(A)$, the Stone space of $A$, so quotients of $A$ correspond to closed subspaces of $X$. Under this correspondence, a homomorphism $f:Ato B$ corresponds to $f^*:{rm ult}(B)to{rm ult}(A)$. Now let ${x_n}_nsubseteq X$ and set $Y$ be the closure of ${x_n}_n$, so $Y$ is compact, 0-dimensional and has weight at most $2^{aleph_0}$ (i.e., $Y$ admits a basis of size at most $2^{aleph_0}$). Let $Y^*$ be the Boolean algebra of clopen subsets of $Y$. Then $Y^*$ is a quotient of $A$, as desired. To see the claim about the weight of $Y$, simply notice that for any $Csubseteq X$, any clopen set in the closure of $C$ is coded by its intersection with $C$.

Open question 1. Is it consistent to have a Boolean algebra of cofinality strictly larger than $aleph_1$?

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 $B$ is a Banach space, then ${rm cf}(B)$ is the smallest lenght of a strictly increasing chain of closed proper subspaces of $B$ whose union is dense in $B$. Tis is well defined as long as $B$ is infinite dimensional, and an upper bound is the density of $B$.

Theorem. ${rm cf}(B)=aleph_0$ iff $B$ has a separable infinite dimensional quotient.

The problem here is the following: If $f:Bto C$ 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 $B$ is uncountable, then the union of a chain witnessing this is automatically closed, and therefore equals all of $B$. 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 $C(K)$ where $K$ is compact: Either $K$ is scattered, in which case the conclusion is easy, or else $K$ 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 $2^{aleph_0}$, because every Banach space has an infinite dimensional quotient isomorphic to a subspace of $ell^infty$.

Why is open problem 1 difficult? It is a bit more convenient to switch the discussion to the topological setting, via Stone duality. Let $X$ be compact. The coinitiality of $X$ is the minimal length of a nontrivial inverse system with limit $X$,

$Xtodotsto X_{alpha+1}to X_alphatodotsto X_0.$

Here, each $X_{alpha+1}$ is a quotient of $X_alpha$. Intuitively, the “resolution” with which we see $X$ increases as we go back.

If $X$ is compact and 0-dimensional, then its coinitiality is precisely the cofinality of the Boolean algebra of clopen subsets of $X$. For general spaces $X$, we look at the $C^*$-algebra $C(X)$. The coinitiality of $X$ is always defined for $X$ compact infinite; the argument is a bit more delicate than before.

One can characterize countable coinitiality: Let $(x_n)_n$ be a sequence in compact $X$ and let $pinbetaomega$. The $p$-limit of the sequence $x_n$ is the unique $xin X$ such that

$bigcap_{Ain p}{rm cl}{x_n:nin A}={x}.$

Another description is the following: Let $f:nmapsto x_n$. Then $f$ has a unique continuous extension $hat f:betaomegato X$, and $pmbox{-}lim x_n=hat f(p).$

Definition. $(x_n)_n$ is a double sequence iff it is discrete and for each $pinbetaomegasetminusomega$

$pmbox{-}lim x_{2n}=pmbox{-}lim x_{2n+1}.$

A typical example of a double sequence is a convergent sequence, but being a double sequence is a more general notion.

Theorem. $X$ is of countable coinitiality iff $X$ has a double sequence.

Proof. Assume first there is a double sequence $(x_n)_n$. Get quotients of $X$ by identifying $x_{2n}$ and $x_{2n+1}$ for all $n$ from some $n_0$ on. The inverse limit of this sequence of spaces is $X$.

Assume now that $X$ has countable coinitiality, and consider a witnessing sequence $Xtodotsto X_{n+1}to X_ntodotsto X_0$. In each $X_{n+1}$ choose two distinct points $t,t'$ that were previously identified in $X_n$, and let $p_{2n},p_{2n+1}$ be preimages in $X$ of $t,t'$, respectively. Then $(p_n)_n$ is a double sequence. ${sf QED}$

Definition. A subset $B$ of a Boolean algbera is independent iff whenever $C_1,C_2$ are finite subsets of $B$, $bigwedge C_1landbigwedge_{pin C_2}-p>0.$

Lemma. If $A$ is a Boolean algebra and $A$ has an uncountable independent subset, then ${rm cf}(A)lealeph_1.$

Proof. Let ${rm Fr}(kappa)$ denote the free Boolean algebra on $kappa$ generators. Then ${rm Fr}(omega_1)$ embeds into $A$ (by hypothesis) and is dense in ${rm RO}({rm Fr}(omega_1))={mathbb C}(omega_1)$ so, from general theory of Boolean algebras, we get a homomorphic map $f:Ato{mathbb C}(omega_1)$.

Consider now a filtration $(B_alpha)_{alpha of ${rm Fr}(omega_1)$, say $B_alpha={rm Fr}(alpha)$. Also, let $(C_alpha)_{alpha be a filtration of ${mathbb C}(omega_1)$, say $C_alpha={mathbb C}(alpha)$. We can now let $A_alpha=f^{-1}[{mathbb C}(alpha)]$, and we verify at once that the $A_alpha$ are increasing substructures with union $A$. ${sf QED}$

Now we can prove that ${rm cf}(A)>aleph_0$ for $A$ complete: Or we would have a double sequence in the Stone space, but we can separate $(x_{2n})_n$ and $(x_{2n+1})_n$ in the Stone space by “separating them” in the complete Boolean algebra.

If the coinitiality of $X$ is larger than $aleph_1$, then $X$ contains no double sequence and no uncountable dyadic family (the dual of an independent family). In particular, $X$ cannot contain a copy of $betaomega$. We say that such an $X$ is an Efimov space.

Definition. An Efimov space is a compact space without convergent sequences nor copies of $betaomega$.

[For $X$ 0-dimensional, it is enough to see that it admits no countable quotient nor a quotient isomorphic to ${mathcal P}(omega)$.]

Open question 3. Assume $X$ 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 ${sf ZFC}$; it is known that it has a positive answer under ${sf CH}$ 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 ${mathbb P}$ be the corresponding forcing notion), and look at ${mathcal P}(omega)^V$; take its Stone space in $V^{{mathbb P}}$. This is an Efimov space in the extension!

To see this, argue as follows:

1. No convergent sequences are added (this follows from general results about random forcing).
2. ${mathcal P}(omega)^V$ is too small, so its Stone space cannot contain $betaomega$.

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 ${sf PFA}$ refutes the existence of Efimov spaces.

Here is a brief sketch of a construction of an Efimov space under $diamondsuit$. The construction is inductive and at limit stages we simply take unions; given $B_alpha$, we get $B_{alpha+1}=B_alpha(a_alpha)$ such that any subalgebra containing $B_alpha$ is either $B_alpha$ or $B_{alpha+1}$.—A minimal extension.

Then $bigcup_{alpha cannot contain an uncountable independent subset.

Finally, $diamondsuit$ allows us to kill convergent sequences by “predicting” them.

Theorem. (Geschke) Under $diamondsuit$ (or even ${sf CH}$), there are Efimov spaces of coinitialities $aleph_0$ and $aleph_1$.

[For $aleph_1$, the argument from $diamondsuit$ sketched above also allows us to kill double sequences. For $aleph_0$, one keeps one double sequence, while killing all convergent sequences.]