Bounds for a covering number of the circle group $\mathbb T$ by some its small subgroups

Question

$\newcommand{\w}{\omega}\newcommand{\A}{\mathcal A}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x}{\mathfrak x}\newcommand{\cov}{\mathrm{cov}}\newcommand{\lac}{\mathrm{lac}}\newcommand{\non}{\mathrm{non}} \newcommand{\IT}{\mathbb T}$Taras Banakh and I proceed a long quest answering a question of ougao at Mathematics.SE.

Recall that a circle $\mathbb T=\{z\in\mathbb C:|z|=1\}$, endowed with the operation of multiplication of complex numbers and the topology inherited from $\mathbb C$ is a topological group. We consider a cardinal $\cov(\A(\IT))$ which is the smallest size of a family $\mathcal U$ of strictly increasing sequences $(u_n)_{n\in\omega}$ of natural numbers such that for each $z\in\IT$ there exists $(u_n)_{n\in\omega}\in\mathcal U$ such that a sequence $(z^{u_n})_{n\in\omega}$ converges to $1$. It would be ideally for us to find a known small cardinal equal to $\cov(\A(\IT))$. While $\cov(\A(\IT))$ remains unknown, we are interested in bounds for it by known small cardinals.

Our try.

Upper bounds.

Let $[\w]^\w$ denote the family of all infinite subsets of $\w$. A subfamily $\mathcal R\subseteq[\w]^\w$ is called reaping if for any set $X\in[\w]^\w$ there is $R\in\mathcal R$ such that one of sets $R\cap X$ and $R\setminus X$ is finite. The reaping number $\mathfrak r$ is the cardinality of the smallest reaping family. By Proposition 9.9 from [1], $\mathfrak r$ is the minimum cardinality of any ultrafilter pseudobase. Recall that a pseudobase for a filter $\F$ on $\omega$ is a family $\mathcal P$ of infinite subsets of $\omega$ such that every set in $\F$ has a subset in $\mathcal P$.

A family $\mathcal R$ of infinite subsets of $\omega$ is called $\sigma$-reaping, if for any countable family $\mathcal X$ of infinite subsets of $\omega$ there is $R\in\mathcal R$ such that for any $X\in \mathcal X$ one of sets $R\cap X$ and $R\setminus X$ is finite. The $\sigma$-reaping number $\mathfrak r_\sigma$ is the cardinality of the smallest $\sigma$-reaping family. Clearly, $\mathfrak r\le\mathfrak r_\sigma$ and there is an old open problem whether $\mathfrak r<\mathfrak r_\sigma$ is consistent, see [4], [3], and [1, 3.6]. By [3], $\mathfrak r_\sigma\le\mathfrak u_p$, where $\mathfrak u_p$ is the smallest base of a $P$-point if a $P$-point exists and $\mathfrak u_p=\mathfrak c$ if no $P$-point exists. It is known that $\mathfrak u_p=\mathfrak u$ if $\mathfrak u<\mathfrak d$. Let us recall that $\mathfrak u$ is the smallest cardinality of a base of a free ultrafilter on $\omega$.

By Theorem 3.7 from [1], $\mathfrak r_\sigma$ is equal to the smallest cardinality of a family $\mathcal R\subseteq[\w]^\w$ such that for any bounded sequence of real numbers $(x_n)_{n\in\w}$ there exists $R\in\mathcal R$ such that the subsequence $(x_n)_{n\in R}$ converges in the real line. It easily follows that $\cov(\A(\IT))\le\mathfrak r_\sigma.$

Problem. Is $\cov(\A(\IT))\le\mathfrak r$?

Lower bounds.

For any family $\I$ of sets with $\bigcup\I\notin\I$ let $\cov(\I)=\min\{|\J|:\J\subseteq\I\;\wedge\;\bigcup\J=\bigcup\I\}$ and $\non(\I)=\min\{|A|:A\subseteq\bigcup\I\;\wedge\;A\notin\I\}$. Let $\M$ and $\N$ be the ideals of meager and Lebesgue null subsets of the real line, respectively.

It is easy to show that $\cov(\A(\IT))\ge\max\{\cov(\M),\cov(\N),\x\}$, where $\x$ is an auxiliary cardinal introduced as follows. An infinite set $R\subseteq\omega$ of natural numbers is called remote if there exists $z\in\IT$ such that $\inf_{n\in R}|z^n-1|>0$. Let $\x$ be the smallest cardinality of a family $\F$ of infinite subsets of $\omega$ such that for any remote set $R$ there exists $F\in\F$ such that $F\cap R$ is finite. So it would be good for us to find a known small cardinal equal to $\x$. While $\x$ remains unknown, we are interested in bounds (especially lower) for it by known small cardinals.

Our try for $\x$.

We can prove that $\cov(\M)\le \x$ and are interested whether this bound can be improved and whether $\cov(\N)\le \x$.

Our bound $\cov(\M)\le \x$ follows from the next

Lemma. For any increasing function $f:\w\to\w$ and any family $\mathcal X\subseteq[\w]^\w$ of cardinality $|\mathcal X|<\cov(\M)$ there exists a set $y\in[\w]^\w$ such that $y\cap x\ne\emptyset$ for every $x\in\mathcal X$ and $y\cap (n,f(n)]=\emptyset$ for any $n\in y$.

Proof. For every $n\in\w$ consider the set $$K_n=\{x\in \mathcal P(\w):n\in x\;\Rightarrow x\cap(n,f(n)]=\emptyset\}$$and observe that it is clopen in the natural compact metrizable topology on $\mathcal P(\w)$. Then the intersection $K=\bigcap_{n\in\w}K_n$ is a compact metrizable space without isolated points.

For each $x\in\mathcal X$ the set $U_x=\{y\in K:x\cap y\ne \emptyset\}$ is open and dense in $K$. Since $|\mathcal X|<\cov(\M)$, the intersection $\bigcap_{x\in\mathcal X}U_x$ is not empty and hence contains some element $y$, which is a set satisfying the required properties. $\square$

Lyubomyr Zdomskyy suggested that it is consistent that $\mathfrak d<\x$, 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 introduced an auxiliary cardinal $\x_{\lac}$, which is the smallest cardinality of a family $\F$ of infinite subsets of $\w$ such that for any lacunary set $L$ there exists $F\in\F$ such that $F\cap L$ is finite. Recall that an infinite set $L$ of natural numbers is called lacunary, if $\inf\{b/a:a,b\in L,\;a<b\}>1$. We have $\x_\lac\le\x$, because Pollington in [2] proved that any lacunary set is remote, as John Griesmer informed us. But it turned out that $\x_\lac$ is rather small. Namely, Will Brian showed that $\x_\lac\le\non(\N)$ and the strict inequality here is consistent.

References

[1] A. Blass, Combinatorial Cardinal Characteristics of the Continuum, in: M. Foreman, A. Kanamori (eds.), Handbook of Set Theory, Springer Science+Business Media B.V. 2010, 395–489.

[2] Andrew D. Pollington, On the density of sequences $\{n_k\xi\}$, Ill. J. Math. 23 (1979) 511–515, ZBL0401.10059.

[3] J. Vaughan, Small uncountable cardinals and topology, Open problems in topology (J. van Mill and G. Reed, eds.), North-Holland, Amsterdam, 1990, 195–218.

[4] P. Vojtáš, Cardinalities of noncentered systems of subsets of $\omega$, Discrete Mathematics 108 (1992) 125–129.

Thanks.

Answer 1

I claim $\mathfrak{x} \leq \mathfrak{r}$.

First, recall the following characterization of $\mathfrak{r}$:

There is a family $\mathcal R$ of infinite subsets of $\mathbb N$, with $|\mathcal R| = \mathfrak{r}$, such that for every bounded countably infinite set $\{ x_n :\, n \in \mathbb N \}$ of real numbers, and every $\varepsilon > 0$, there is some $A \in \mathcal R$ such that the diameter of $\{x_n :\, n \in A\}$ is at most $\varepsilon$.

(This characterization of $\mathfrak{r}$ follows, for example, from Theorem 3.7 in Blass' handbook article you linked to. Directly, this theorem allows us to get an $\mathfrak{r}$-sized family $\mathcal R$ such that any countable subset of $[a,b]$ will be confined to $[a,(a+b)/2]$ or $[(a+b)/2,b]$ on some member of $\mathcal R$. But then, for each $A \in \mathcal R$, we may define, by the same token, an $\mathfrak{r}$-sized family $\mathcal R_A$ of subsets of $A$ such that any $A$-indexed subset of $[a,(a+b)/2]$ or of $[(a+b)/2,b]$ will be confined to just one half of that interval on some member of $\mathcal R_A$. The union of all the $\mathcal R_A$'s is an $\mathfrak{r}$-sized family such that any countable subset of $[a,b]$ will be confined to an interval of length $(a+b)/4$ on some member of the family. We may repeat this finitely many times, to confine our sets to smaller and smaller intervals.)

Using this characterization of $\mathfrak{r}$, we can prove my claim as follows. Let $\mathcal R$ be a family of sets as above. For each $A \in \mathcal R$, let $D_A = \{|a_1-a_2| :\, a_1,a_2 \in A\}$. I claim that $\{D_A :\, A \in \mathcal R\}$ satisfies the definition of $\mathfrak{x}$. This proves the bound we want, since it shows there is an $\mathfrak{r}$-sized family satisfying the definition of $\mathfrak{x}$.

To see that this family works as claimed, let $R \subset \mathbb N$ be any remote set. This is witnessed by some $z \in \mathbb T$ and some $\varepsilon > 0$, which satisfy $|z^n-1| > \varepsilon$ for every $n \in R$. Of course, we may identify $\mathbb T$ with a bounded subset of $\mathbb R$ in the natural way. Thus, by our choice of $\mathcal R$, there is some $A \in \mathcal R$ such that $\{z^n :\, n \in A\}$ has diameter at most $\varepsilon$. If $d \in D_A$, this implies $|z^d - 1| < \varepsilon$, which implies $d \notin R$. Hence $R \cap D_A = \emptyset$. Because $R$ was an arbitrary remote set, this shows the family $\{D_A :\, A \in \mathcal R\}$ really does work as claimed.