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Given an countable amenable group $G$, let $\{T_n\}_{n \in \mathbb{N}}$ be a Folner sequence for $G$, i.e., $\lim_{n \to +\infty} \frac{|gT_n \Delta T_n|}{|T_n|} = 0$, for every $g \in G$. Now, for each $n \in \mathbb{N}$, consider $S_n = \bigcup_{k=1}^n T_k$. My question is whether or not $\{S_n\}_{n \in \mathbb{N}}$ is a Folner sequence.

Obviously, if $\{T_n\}$ is increasing, the answer is yes, but I am not being able to prove the case when $\{T_n\}$ is not increasing and I don't even know if this is true.

What I did so far is: given $g \in G$ and $n \in \mathbb{N}$,

\begin{align*} \frac{|gS_n \Delta S_n|}{|S_n|} &\leq \frac{|\bigcup_{k=1}^n(gT_k \Delta T_k)|}{|S_n|}\\ &\leq \frac{\sum_{k=1}^{n}|gT_k \Delta T_K|}{|S_n|}\\ &= \sum_{k=1}^{n} \frac{|gT_k \Delta T_K|}{|S_n|}\\ &= \sum_{k=1}^{n} \frac{|gT_k \Delta T_K|}{|T_k|} \end{align*} and I know that what is inside the sum goes to zero, but this does not help me (or at least I don't see how it could help me).

Does someone know how to prove it or have a counterexample?

Luísa Borsato
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  • You need $G$ to be countable, otherwise it has no Følner sequence. – YCor Feb 09 '20 at 13:50
  • Ok. But assuming $G$ countable and assuming that ${T_n}$ is a Folner sequence, can you decide whether or not the sequence ${S_n}$ is Folner? – Luísa Borsato Feb 09 '20 at 13:56

1 Answers1

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No, there are counterexamples.

Define in $\mathbf{Z}$, $F_n=[0,\sqrt{n}]\cup\{n^2\}$ (in the segment I implicitly intersect with $\mathbf{Z}$). It's Følner, because adding the singleton is negligible. But $F'_n=\bigcup_{k\le n}F_n=[0,\sqrt{n}]\cup\{k^2:k\le n\}$ is not Følner because for $k>n^{1/4}$ we have $k^2\in F'_n$ but $k^2+1\notin F'_n$, so $\frac{|F'_n|\Delta|F'_n+1|}{|F'_n|}$ even tends to 1.

On the other hand, it's true (in every group) if the $F_n$ are pairwise disjoint (use that if $(a_n)$ and $(b_n)$ are real sequences with $a_n\ge 0$ and $b_n\ge 1$ and $a_n/b_n\to 0$, then $\sum a_n/\sum b_n\to 0$).

YCor
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