5
$\begingroup$

$\newcommand{\w}{\omega}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x}{\mathfrak x}\newcommand{\cov}{\mathrm{cov}}\newcommand{\lac}{\mathrm{lac}}$Taras Banakh and I proceed a long quest answering a question 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 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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 18, 2021 at 8:09
  • 1
    $\begingroup$ @AlexRavsky: Thanks -- I'll take a look :) $\endgroup$
    – Will Brian
    Sep 18, 2021 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.