3
$\begingroup$

$\newcommand{\w}{\omega}\newcommand{\A}{\mathcal A}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x}{\mathfrak x}\newcommand{\cov}{\mathrm{cov}}\newcommand{\lac}{\mathrm{lac}}\newcommand{\non}{\mathrm{non}} \newcommand{\IT}{\mathbb T}$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.

$\endgroup$
7
  • 1
    $\begingroup$ 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 $\mathcal F$. . . . $\endgroup$
    – Will Brian
    Sep 18, 2021 at 10:53
  • 1
    $\begingroup$ 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. $\endgroup$
    – Will Brian
    Sep 18, 2021 at 10:54
  • 1
    $\begingroup$ 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? $\endgroup$
    – Will Brian
    Sep 18, 2021 at 12:24
  • 1
    $\begingroup$ @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 $n\in\omega$. $\endgroup$ Sep 18, 2021 at 13:28
  • 1
    $\begingroup$ Aha! That's a nice idea. $\endgroup$
    – Will Brian
    Sep 18, 2021 at 13:34

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Sep 18, 2021 at 14:50
  • 1
    $\begingroup$ @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. $\endgroup$
    – Will Brian
    Sep 18, 2021 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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