All spaces considered below are compact Hausdorff.

If $K$ is a space, then $w(K)$ is its weight. For a Boolean algebra $\mathcal{A}$, $K_\mathcal{A}$ denotes its Stone space. I am interested in possible cardinalities of algebras such that their Stone spaces do not have non-trivial convergent sequences. Let me thus define the following cardinal number called (by me) the convergence number:

$\mathfrak{z}=\min\{|\mathcal{A}|:\ K_\mathcal{A}\text{ does not have non-trivial convergent sequences}\}$

($\mathfrak{z}$ from the Polish word "zbieżność" meaning "convergence")

Of course, $\mathfrak{z}$ is not greater than the continuum $\mathfrak{c}$ (consider $\mathcal{A}=\wp(\omega)$).

On the other hand, it is well-known that the splitting number $\mathfrak{s}$ is not greater than $\mathfrak{z}$ -- it follows from the following equivalent definition of $\mathfrak{s}$ due to Booth '74:

$\mathfrak{s}=\min\{w(K):\ K\text{ is not sequentially compact}\}.$

An example of a space $K$ from this definition is $2^\mathfrak{s}$ (which is the Stone space of an algebra).

Also, one can prove (see Geschke '06) that if a space $K$ has weight less than the covering number of category $\text{cov}(\mathcal{M})$, then $K$ must contain a non-trivial convergent sequence, thus $\text{cov}(\mathcal{M})\le\mathfrak{z}$.

It can be shown that the inequalities $\mathfrak{s}<\text{cov}(\mathcal{M})$ and $\text{cov}(\mathcal{M})<\mathfrak{s}$ are relatively consistent (see here). Under Martin's Axiom, all those numbers are equal (to the continuum $\mathfrak{c}$). A natural ZFC simultaneous upper bound of $\mathfrak{s}$ and $\text{cov}(\mathcal{M})$ is the dominating number $\mathfrak{d}$. My question is thus about relations between $\mathfrak{z}$ and $\mathfrak{d}$, especially I am interested in the following:

Question: Is it consistent that $\mathfrak{d}<\mathfrak{z}$ ($<\mathfrak{c}$)?

Recall that the cofinality of measure $\text{cof}(\mathcal{N})$ is not less than $\mathfrak{d}$. If $\kappa$ is a cardinal number such that $\text{cof}([\kappa]^\omega)=\kappa<\mathfrak{c}$, then assuming $\text{cof}(\mathcal{N})=\kappa$ I can construct an example of a Boolean algebra without non-trivial convergent sequences and of cardinality $\kappa$; hence, it is consistent that $\mathfrak{z}\le\text{cof}(\mathcal{N})<\mathfrak{c}$.

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    $\begingroup$ Am I right that in the formula s=min{w(K): K is not sequentially compact} you assume that K is compact? $\endgroup$ – Lajos Soukup Mar 25 '16 at 14:38
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    $\begingroup$ Nice question. I believe that if you state your question in a more combinatorial/elementary manner, it may attract more efforts. Even if you just talk directly on the BA rather than its stone space (I know, that is straightforward, but those not used to this language my appreciate). $\endgroup$ – Boaz Tsaban Mar 28 '16 at 20:23
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    $\begingroup$ @LajosSoukup: yes, I always think only about compact Hausdorff spaces. (As I wrote in the first line ;)) $\endgroup$ – Damian Sobota Apr 3 '16 at 19:47
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    $\begingroup$ I meant, could you replace "the Stone space of $A$ does not have a convergent subsequence" by a direct, combinatorial assertion on the BA $A$? My experience is that people (including some prominent mathematicians), prefer such questions to be presented to them in a purely set theoretic/combinatorial language, if possible. $\endgroup$ – Boaz Tsaban Apr 3 '16 at 21:54
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    $\begingroup$ @BoazTsaban: Oh, that's (among others) the problem! I do not know any characterizations of BAs whose Stone spaces do (not) have non-trivial convergent sequences. (Btw, this is closely related to Efimov's problem, which is now open for 50 years, asking whether there exists a space without any copies of $\omega+1$ and $\beta\omega$.) $\endgroup$ – Damian Sobota Apr 3 '16 at 23:43

First of all, let me say that I love this question.

Alan Dow and I have been thinking about this question and its relatives quite a lot recently. We completed a paper on the topic last week (available on arXiv), and I'll summarize the results here.

As you mention in the comments, the number $\mathfrak{z}$ does not seem to admit a simple combinatorial description, and can be very difficult to work with. Our paper introduces a new cardinal characteristic of the continuum that is closely related to $\mathfrak{z}$ and is "almost" an upper bound for it (in a sense I'll explain below). But the new characteristic has a simple description and is much easier to work with. This allows us to analyze $\mathfrak{z}$ indirectly, by working with a more manageable proxy instead.

Definition: If $U$ and $A$ are infinite sets, we say that $U$ splits $A$ if both $A \cap U$ and $A \setminus U$ are infinite. The splitting number of the reals, denoted $\mathfrak{s}(\mathbb R)$, is the smallest possible cardinality of a collection $\mathcal U$ of open subsets of $\mathbb R$ such that every infinite $A \subseteq \mathbb R$ is split by some $U \in \mathcal U$.

The classical splitting number $\mathfrak{s}$ is the smallest possible cardinality of a collection $\mathcal S$ of subsets of $\mathbb N$ such that every infinite subset of $\mathbb N$ is split by some member of $\mathcal S$. The new number $\mathfrak{s}(\mathbb R)$ is just a topological variant of $\mathfrak{s}$, where instead of splitting subsets of $\mathbb N$ with subsets of $\mathbb N$, we're splitting subsets of $\mathbb R$ with open sets.

I should mention that the value of $\mathfrak{s}(\mathbb R)$ does not change if one replaces $\mathbb R$ with any other uncountable Polish space in the above definition.

Our main theorem relating $\mathfrak{s}(\mathbb R)$ to $\mathfrak{z}$ is the following:

Theorem: If $\mathfrak{s}(\mathbb R) < \aleph_\omega$, then $\mathfrak{z} \leq \mathfrak{s}(\mathbb R)$.

(Actually we have a slightly stronger theorem: if there is a cardinal $\kappa$ such that $\mathfrak{s}(\mathbb{R}) \leq \kappa = \mathrm{cof}(\kappa^{\aleph_0},\subseteq)$, then $\mathfrak{z} \leq \kappa$. It follows that $\mathfrak{z} \leq \mathfrak{s}(\mathbb R)$ whenever $\mathfrak{s}(\mathbb R) < \aleph_\omega$, as stated above, and more: if $\mathfrak{z} > \mathfrak{s}(\mathbb R)$, then either $\mathfrak{s}(\mathbb R)$ has countable cofinality, or there are inner models containing measurable cardinals.)

After proving this theorem relating $\mathfrak{z}$ and $\mathfrak{s}(\mathbb R)$ near the beginning of our paper, we go on to analyze $\mathfrak{s}(\mathbb R)$ in detail. We prove three lower bounds and one upper bound from $\mathsf{ZFC}$:

$\bullet$ $\mathfrak{s},\,\mathrm{cov}(\mathcal M),\,\mathfrak{b} \ \leq \ \mathfrak{s}(\mathbb R)$.

$\bullet$ $\max\{\mathfrak{b},\mathrm{non}(\mathcal N)\} \, \geq \, \mathfrak{s}(\mathbb R)$.

The second bullet point is particularly significant to your question, because it gives an upper bound for $\mathfrak{z}$ also (at least assuming $\mathfrak{s}(\mathbb R) < \aleph_\omega$). In addition to these inequalities, we prove two consistency results via forcing showing that it is possible to have either of

$\bullet$ $\mathfrak{s}(\mathbb R) \,<\, \mathrm{non}(\mathcal N)$

$\bullet$ $\mathfrak{s}(\mathbb R) \,>\, \mathrm{cof}(\mathcal M) = \mathfrak{d}$.

Taken together, these results completely determine the place of $\mathfrak{s}(\mathbb R)$ in Cichoń's diagram:

enter image description here

In this picture, the green cardinals are (consistently strict) lower bounds for $\mathfrak{s}(\mathbb R)$, the red cardinals are (consistently strict) upper bounds, and a carindal $\kappa$ is yellow we know both that both $\kappa < \mathfrak{s}(\mathbb R)$ and $\mathfrak{s}(\mathbb R) < \kappa$ are consistent.

Returning to your question, I am sad to admit that we still do not know whether $\mathfrak{d} < \mathfrak{z}$ is consistent. In our forcing model with $\mathfrak{d} < \mathfrak{s}(\mathbb R)$, we do not know the value of $\mathfrak{z}$, but this model is of course a good candidate for getting $\mathfrak{d} < \mathfrak{z}$. (It is possible though that $\mathfrak{z} < \mathfrak{s}(\mathbb R)$ in this model; we already know that the inequality $\mathfrak{z} < \mathfrak{s}(\mathbb R)$ is consistent because it holds in the Laver model, although a proof of this isn't available yet -- this result will go into a future paper.) Another good candidate we've thought of is a model obtained by adding $\aleph_1$ random reals to a model of $\mathsf{MA}+\neg \mathsf{CH}$, but once again we are not yet able to compute $\mathfrak{z}$ in such a model. I will note that $\mathfrak{d}$ is not a lower bound for $\mathfrak{z}$, because $$\aleph_1 = \mathfrak{z} = \mathfrak{s}(\mathbb R) = \max\{\mathfrak{b},\mathrm{non}(\mathcal N)\} < \mathfrak{d} = \aleph_2$$ in the Miller model.

In closing I'll include one more image: it's a picture like the one above for $\mathfrak{s}(\mathbb R)$, but showing what we currently know about $\mathfrak{z}$ instead. The cardinals in the striped region are those that we know are consistently $>\! \mathfrak{z}$, but we don't know yet whether they're consistently $<\! \mathfrak{z}$ as well.

enter image description here


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