# Bounds for a small cardinal


Recently we encountered a small cardinal $$\x_{\lac}$$, which is the smallest cardinality of a family $$\F$$ of infinite subsets of $$\w$$ such that for any lacunary set $$R$$ there exists $$F\in\F$$ such that $$F\cap R$$ is finite. Recall that a set $$R\subset\w\setminus\{0\}$$ is called lacunary, if $$\inf\{b/a:a,b\in R,\;a1$$.

It would be ideally for us to find a known small cardinal equal to $$\x_{\lac}$$. While $$\x_{\lac}$$ remains unknown, we are interested in bounds (especially lower) for it by known small cardinals.

Our try. For any family $$\I$$ of sets let $$\cov(\I)=\min\{|\J|:\J\subseteq\I\;\wedge\;\bigcup\J=\bigcup\I\}$$. Let $$\M$$ and $$\N$$ be the ideals of meager and Lebesgue null subsets of the real line, respectively. We can prove that $$\cov(\M)\le \x_{\lac}$$ and are interested whether this bound can be improved and whether $$\cov(\N)\le \x_{\lac}$$.

Lyubomyr Zdomskyy suggested that it is consistent that $$\mathfrak d<\x_{\lac}$$, where $$\mathfrak d$$ is the cofinality of $$\w^\w$$ endowed with the natural partial order: $$(x_n)_{n\in\w}\le (y_n)_{n\in\w}$$ iff $$x_n\le y_n$$ for all $$i$$. We are interested whether $$\x_{\lac}\le \mathfrak a$$, where $$\mathfrak a$$ is the minimum size of a maximal (with respect to inclusion) pairwise almost disjoint family of infinite subsets of $$\omega$$.

Thanks.

EDIT: In my original post, I showed that $$\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$$ in the random model. Upon further reflection, I think we can prove a stronger result, with an arguably easier (but completely different) proof:

Theorem: $$\mathfrak{x}_{lac} \leq \mathrm{non}(\mathcal N)$$.

Note that this implies $$\mathfrak{x}_{lac} < \mathrm{cov}(\mathcal N)$$ in the random model, and it also implies the consistency of $$\mathfrak{x}_{lac} < \mathfrak{d}$$.

In addition to this theorem, let me also point out that it is consistent to have $$\mathfrak{a} < \mathrm{cov}(\mathcal M)$$. (See Corollary 2.6 in this paper of Brendle.) Therefore the lower bound $$\mathrm{cov}(\mathcal M) \leq \mathfrak{x}_{lac}$$ mentioned in the post already implies $$\mathfrak{a}$$ is not an upper bound for $$\mathfrak{x}_{lac}$$.

Proof of the theorem: Suppose we form an infinite set $$B$$ by choosing from each interval of the form $$[2^k,2^{k+1})$$ exactly one integer $$b_k$$ at random, and then taking $$B = \{b_k :\, k \in \omega \}$$. (By "at random" I mean that we choose with the uniform distribution, so each integer in $$[2^k,2^{k+1})$$ has probability $$1/2^k$$ of being selected.) I claim that if $$A$$ is lacunary, then it is almost surely true that $$A \cap B$$ is finite.

To see this, fix some $$c > 1$$ and some $$n_0 \in \omega$$ such that if $$a$$ and $$b$$ are consecutive members of $$A$$ above $$n_0$$, then $$b/a > c$$. If $$k$$ is large enough that $$n_0 < 2^k$$, then this implies there are at most $$log_c(2)$$ members of $$[2^k,2^{k+1})$$ in $$A$$. This implies that the probability of choosing $$b_k \in A$$ is $$\log_c(2)/2^k$$ when $$k$$ is large enough. It follows that the probability of there being $$>K$$ members of $$A$$ in $$B$$ (when $$K$$ is large) is $$\sum_{k > K} \log_c(2)/2^k = \log_c(2)/2^K$$. Since this goes to $$0$$ as $$K$$ goes to infinity, the probability of $$A \cap B$$ being infinite is $$0$$.

The idea of choosing $$B$$ randomly, one point at a time, like this can be formalized by defining a probability measure on a Polish space, where points of the space correspond to possible choices of the sequence of $$b_k$$'s. What the previous paragraph shows is that in this probability space, the set of all $$B$$'s with $$A \cap B$$ infinite form a null set, for any given lacunary set $$A$$. Hence any non-null subset of this probability space will contain a $$B$$ with $$A \cap B$$ finite. Since this holds for any $$A$$, we see that any non-null subset $$X$$ of this probability space contains, for any lacunary set $$A$$, some infinite $$B$$ with $$A \cap B$$ finite. The smallest possible size of such a set $$X$$ is $$\mathrm{non}(\mathcal N)$$.

QED

One more observation: The strict inequality $$\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$$ is also consistent, so this upper bound cannot be improved to an equality. To see this, begin with a model of Martin's Axiom $$+ \, \neg \mathsf{CH}$$, and then do a legnth-$$\omega_1$$, finite support iteration of the eventually different reals forcing. It is not difficult to see that this forcing will make $$\mathfrak{x}_{lac} = \aleph_1$$ in the extension. But the iteration is $$\sigma$$-centered, and forcing with a $$\sigma$$-centered poset over a model of $$\mathsf{MA}$$ does not change the value of $$\mathrm{non}(\mathcal N)$$. Thus we get $$\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$$ in the extension.

Original post:

It is not provable that $$\mathrm{cov}(\mathcal N) \leq \mathfrak{x}_{lac}$$, because $$\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$$ in the random model.

To see this, let me sketch an argument that after forcing to add any number of random reals (in the usual way, via a measure algebra), the collection $$[\omega]^\omega \cap V$$ of infinite subsets of $$\omega$$ in the ground model has the property described in the definition of $$\mathfrak{x}_{lac}$$. That is: for every lacunary set $$A \subseteq \omega$$ in the extension, there is an infinite $$B \subseteq \omega$$ in the ground model such that $$A \cap B$$ is finite.

We work in the ground model. Suppose $$\dot A$$ is a name for an infinite lacunary set in the extension. There is some fixed $$c > 1$$ and $$n_0 \in \omega$$, and some forcing condition $$p$$, such that $$p \Vdash$$ if $$a$$ and $$b$$ are consecutive elements of $$\dot A$$ above $$n_0$$, then $$b/a > c$$.

I claim that for every $$\varepsilon > 0$$, there are arbitrarily large $$n \in \omega$$ such that $$m(p \wedge [n \in \dot A]) < \varepsilon$$.

(Note: Here I'm using the standard notation for forcing with measure algebras. If $$\varphi$$ is any well-formed formula in the forcing language, then we write $$[\varphi]$$ to mean the supremum of all the conditions forcing $$\varphi$$, and $$m([\varphi])$$ denotes the measure of this supremum. Roughly, you may think of $$m([\varphi])$$ as the probability that $$\varphi$$ ends up being true in the forcing extension.)

To prove my claim, suppose, aiming for a contradiction, that it is false. Then there is some $$\varepsilon > 0$$ and some $$N \in \omega$$ such that $$m([n \in \dot A]) \geq \varepsilon$$ for all $$n \geq N$$. But this is just another way of saying that the "expected size" of $$A \cap \{n\}$$ is $$\geq\varepsilon$$ for all $$n \geq N$$. By the linearity of expectation, this means the expected size of $$A \cap \{k,k+1,\dots,\ell-2,\ell-1\}$$ is $$\geq (\ell-k)\varepsilon$$ whenever $$\ell > k > N$$. But by our choice of $$p$$, if $$k \geq N$$ and $$\ell < ck$$, then $$p \Vdash |A \cap \{k,k+1,\dots,\ell-2,\ell-1\}| \leq 1$$. Since $$c > 1$$, this yields a contradiction for sufficiently large $$k$$, namely $$k > N,2/c\varepsilon$$.

Using this claim, we can now find an infinite ground model set almost disjoint from the set named by $$\dot A$$ in the extension. Using the claim, we may find for each $$k \in \omega$$ some $$n_k > n_{k-1}$$ such that $$m([n_k \in \dot A]) < m(p)/2^{k+2}$$. Now let $$p' = p - \bigvee_{k \in \omega}[n_k \in \dot A]$$. Then $$m(p') \geq m(p) - \sum_{k \in \omega}m([n_k \in \dot A]) > m(p)/2 > 0$$, so $$p'$$ is a condition in our measure algebra, and $$p'$$ forces $$\dot A$$ to be disjoint from $$\{n_k :\, k \in \omega \}$$ (because it extends the complement of each $$[n_k \in \dot A]$$).

This shows that it is impossible to have a name $$\dot A$$ for a lacunary set such that $$\dot A$$ is forced to have infinite intersection with every infinite subset of $$\omega$$ from the ground model. Therefore there is no such set.

• Thanks a lot for your answer. This is our question on cardinals $\mathrm{cov}(\mathcal A(\mathbb T))$ and $\mathfrak x$, for which we hoped $\mathfrak x_{\mathrm{lac}}$ is a good bound. Sep 18, 2021 at 8:09
• @AlexRavsky: Thanks -- I'll take a look :) Sep 18, 2021 at 10:42