I am interested in a "dynamical" modification of the cardinals $\mathfrak r$ and $\mathfrak r_\sigma$, well-known in the theory of cardinal characteristics of the continuum.
For a compact metrizable space $X$, let $\mathfrak r_X$ be the smallest cardinality of a family $\mathcal R$ of infinite subsets of $\omega$ such that for every sequence $(x_n)_{n\in\omega}\in X^\omega$ there exists $R\in\mathcal R$ such that the subsequence $(x_n)_{n\in R}$ converges in $X$.
It is easy to see that $\mathfrak r_2\le \mathfrak r_X\le\mathfrak r_{2^\omega}$ for every compact metric space $X$ containing more than one point. Moreover, it can be shown that
$$\mathfrak r_X=\begin{cases}\mathfrak r_2&\mbox{if $|X|\le\omega$};\\
\mathfrak r_{2^\omega}&\mbox{if $|X|>\omega$}.
\end{cases}
$$
Here $2$ is the doubleton $\{0,1\}$, endowed with the discrete topology.
In the theory of cardinal characteristics of the continuum, the cardinals $\mathfrak r_2$ and $\mathfrak r_{2^\omega}$ are denoted by $\mathfrak r$ and $\mathfrak r_\sigma$, respectively.
As written in the survey of Blass, it is not known whether $\mathfrak r=\mathfrak r_\sigma$ in ZFC.
Now I introduce a dynamical modification of the cardinal $\mathfrak r_X$.
Let us recall that a dynamical system is a pair $(X,f)$ consisting of a compact metric space $X$ and a continuous function $f:X\to X$. Define the iterations $f^n$ of $f$ by the recursive formula: $f^0$ is the identity map of $X$ and $f^{n+1}=f\circ f^n$ for $n\in\omega$.
Definition. For a dynamical system $(X,f)$, let $\mathfrak r_{(X,f)}$ be the smallest cardinality of a family $\mathcal R$ of infinite subsets of $\omega$ such that for every $x\in X$ there exists $R\in\mathcal R$ such that the sequence $(f^n(x))_{n\in R}$ coverges in $X$.
It is clear that $\mathfrak r_{(X,f)}\le\mathfrak r_X$. If $f$ is the identity map of $X$, then $\mathfrak r_{(X,f)}=1$, so the inequality can be strict.
Problem. Is it consistent that $\mathfrak r_{(2^\omega,f)}<\mathfrak r_{2^\omega}$ for the shift map $f:2^\omega\to 2^\omega$, $f:(x_n)_{n\in\omega}\mapsto (x_{n+1})_{n\in\omega}$?
Best Answer
Unfortunately, $\mathfrak r_{(2^\omega,f)}\ge\mathfrak r$. Indeed, let $\mathcal R$ be a family of infinite subsets of $\omega$ such that $|\mathcal R|=\mathfrak r_{(2^\omega,f)}$ and for any $x=(x_n)_{n\in\omega}\in 2^\omega$ there exists $R\in\mathcal R$ such that the sequence $(f^n(x))_{n\in R}$ converges in $2^\omega$. Then the sequence $(x_n)_{n\in R}$ of the first coordinates of the sequence $(f^n(x))_{n\in R}$ stabilizes, so the family $\mathcal R$ witness that $\mathfrak r\le \mathfrak r_{(2^\omega,f)}$.