Descending chain in $\mathbb{Z}$ with certain confining property, but not strongly

Question

Call a sequence $A_1 \supseteq A_2 \supseteq \cdots$ of subsets of $\mathbb{Z}$ confining if for all $i$ we have $A_i \supseteq A_{i+1}+A_{i+1}$. (Let us insist that the $A_i$ are symmetric and contain $0$.) Obviously if the $A_i$ are all subgroups of $\mathbb{Z}$ then the sequence is confining. But they don't have to be subgroups, for instance the sequence $B_i := 2^i \mathbb{Z}\setminus\{\pm 2^i\}$ is confining since the "missing" points $\pm 2^i$ are not in $B_{i+1}+B_{i+1}$. Call $(A_i)$ strongly confining if there exists $k\ge 0$ such that for all $i$ we have $A_i\supseteq \langle A_{i+k}\rangle$. For instance, in the above example $(B_i)$ is strongly confining using $k=1$, since the missing points $\pm 2^i$ are not in $\langle B_{i+1}\rangle$.

Question: Does there exist a confining sequence (of symmetric subsets containing $0$) that is not strongly confining?

The motivation here is to try and find an example of a confining subset in $\mathbb{Z}\wr \mathbb{Z}$ that is not equivalent to any confining subgroup, where now "confining" is in the sense of Caprace-Cornulier-Monod-Tessera (in JEMS) and Abbott-Balasubramanya-Rasmussen (in Adv. Math.).

Answer 1

Just the subgroup of $\mathbb Z$ generated by $B$.

Then trivially yes. Just take any very fast increasing sequence $a_k$ of pairwise relatively prime numbers and put $$ A_k=\left\{\sum_i u_ia_{i+k}:|u_i|\le 2^i, \text{ all but finitely many }u_i=0\right\} $$ Then $\langle A_k\rangle=\mathbb Z$ for all $k$ but each $A_k$ is rather sparse by itself.

Answer 2

Here an example, courtesy of Simon Machado, that I might as well record here for the record. First fix an irrational $x$ in $\mathbb{R}$, and embed $\mathbb{Z}$ as a subgroup of $S^1=\mathbb{R}/\mathbb{Z}$ via $\phi\colon n\mapsto nx + \mathbb{Z}$. For each $i\in\mathbb{N}$ let $C_i\subseteq S^1$ be the closed neighborhood of $0+\mathbb{Z}$ that is the image of $[-2^{-i},2^{-i}]$ mod $\mathbb{Z}$, and let $A_i\subseteq \mathbb{Z}$ be $A_i=\phi^{-1}(C_i)$. Since $x$ is irrational, each $A_i$ is infinite. Also note that $A_i$ is symmetric and contains $0$. Clearly $A_i\supseteq A_{i+1}$ and $A_i\supseteq A_{i+1} + A_{i+1}$ by construction. It just remains to show that we never have $A_i \supseteq \langle A_{i+k}\rangle$, and indeed we will show that $A_i$ contains no non-trivial subgroups whatsoever. This is simply because, given any non-trivial $a\in A_i$, again using the fact that $x$ is irrational we can find $n\in\mathbb{Z}$ such that $na\not\in A_i$.

Intuitively, why this worked was, embedding $\mathbb{Z}$ into $S^1$ in this way gives us a natural descending sequence of infinite subsets $A_i$ of $\mathbb{Z}$ coming from a descending sequence of intervals in $S^1$. The intervals shrink fast enough to make the sequence $(A_i)$ confining, but not fast enough for it to be strongly confining (in fact, no sequence of intervals could do make it strongly confining).