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

$$\newcommand{\w}{\omega}\newcommand{\A}{\mathcal A}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x}{\mathfrak x}\newcommand{\cov}{\mathrm{cov}}\newcommand{\lac}{\mathrm{lac}}\newcommand{\non}{\mathrm{non}} \newcommand{\IT}{\mathbb T}$$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 , $$\mathfrak r$$ 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 , , and [1, 3.6]. By , $$\mathfrak r_\sigma\le\mathfrak u_p$$, 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 , $$\mathfrak r_\sigma$$ 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 $$\inf\{b/a:a,b\in L,\;a1$$. We have $$\x_\lac\le\x$$, because Pollington in  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

 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.

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

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

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

Thanks.

• My first thought is to consider what a "generic" $z$ does. For each $z \in \mathbb T$, define $R_z = \{n :\, |z^n-1| \geq 1/2 \}$. (Of course the "1/2" could be made smaller, but I doubt it matters.) I don't have a proof right now, but it seems like a Cohen-generic $z$ should have the property that $R_z \cap A$ is infinite for any infinite $A$ in the ground model. In other words: given $A$, there are only a meager set of $z$'s with $R_z \cap A$ finite. This means you need at least $\mathrm{cov}(\mathcal M)$ $A$'s to get a sufficiently large collection $\mathcal F$. . . . Sep 18, 2021 at 10:53
• This shows (modulo my claim about Cohen-generic $z$'s) that $\mathfrak{x} \geq \mathrm{cov}(\mathcal M)$. Similarly, if you can show that a "random" $z$ has the same property, then this would show $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}$ also. Sep 18, 2021 at 10:54
• Perhaps I'm missing something, but I don't see why $\mathrm{cov}(\mathcal{A}(\mathbb T)) \leq \mathfrak{r}_\sigma$. The characterization of $\mathfrak{r}_\sigma$ you mention gives you an increasing sequence $(u_n)_{n \in \mathbb N}$ such that $(z^{u_n})_{n \in \mathbb N}$ converges, but you have no control over what this sequence converges to. Your definition of $\mathrm{cov}(\mathcal{A}(\mathbb T))$ requires that it converge to $1$. How do you do this? Sep 18, 2021 at 12:24
• @WillBrian The key idea is for every $R\in\mathcal R$ pick a sequence $(r^R_n)_{n\in\omega}\in R^\omega$ such that $u^R_n=r^R_{n+1}-r^R_{n}<r^R_{n+2}-r^R_{n+1}$ for each $n\in\omega$. Sep 18, 2021 at 13:28
• Aha! That's a nice idea. Sep 18, 2021 at 13:34

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.

• Thanks for your answer. We introduce the cardinal $\mathfrak x$ as an auxilliary bound for $\mathrm{cov}(\mathcal A(\mathbb T))$. We hoped that $\mathfrak x$ can be helpful to obtain a new bound for $\mathrm{cov}(\mathcal A(\mathbb T))$, where $\mathfrak x$ is not used explicitly. Taras Banakh suggested to dismiss $\mathfrak x$ if it yields us no such bound. So, in particular, a suggested bound $\mathrm{cov}(\mathcal N)\le \mathfrak x$ is desirable for us, but not crucial. Sep 18, 2021 at 14:50
• @AlexRavsky: OK. I think $\mathrm{cov}(\mathcal N)$ is true (and can be proved as I outlined in my comments on the question), but I don't think I'll have any time to write the details down today. Sep 18, 2021 at 15:36