$\newcommand{\w}{\omega}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x}{\mathfrak x}\newcommand{\cov}{\mathrm{cov}}\newcommand{\lac}{\mathrm{lac}}$[Taras Banakh](https://mathoverflow.net/users/61536/taras-banakh) and I proceed a long quest answering a [question](https://math.stackexchange.com/questions/1203722/approximate-vanishing-in-pontryagin-dual) of ougao at Mathematics.SE.


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 $\inf\{b/a:a,b\in R,\;a<b\}>1$.


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](http://www.logic.univie.ac.at/%7Elzdomsky) 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.



Bounds for a small cardinal
**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](https://arxiv.org/pdf/math/9309205.pdf) 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.


