Amenability of the integers (with the Følner condition)

Question

I only know the following definition of amenability of a discrete group (but I think the definition is suitable for non-dicrete groups as well):

Def: A discrete group $G$ satisfies the Følner condition if for every finite subset $E\subseteq G$ and for every $\epsilon >0$ there exists a non-empty finite set $F\subseteq G$ such that $$\max\limits_{s\in E}\frac{|F\triangle sF|}{|F|}<\epsilon,$$ where $F\triangle sF$ denotes the symmetric difference of the sets $F$ and $sF$.

I want to know how to prove that the discrete group $(\mathbb{Z}, +)$ is amenable using this definition (instead of using Følner sequences or the definition with left invariant means as I have seen on many other websites).

Idea: I guess that the proof can use the idea with Følner sequences, for example $F_n:=[-n,n]\cap \mathbb{Z}$ for $n\in\mathbb{N}$ is such a sequence. Write $\mathbb{Z}=\bigcup\limits_{n\in\mathbb{N}} F_n$, let $E$ be a finite subset of $G$ and let $\epsilon >0$. Then there exists a $n_0\in\mathbb{N}$ so that $E\subseteq F_{n_0}$, where $F_{n_0}$ is non-empty. I guess one can take $F=F_{n_0}$, but I'm not sure: Is $$\max\limits_{s\in E}\frac{|F_{n_0}\triangle sF_{n_0}|}{|F_{n_0}|}<\epsilon ?$$ If not, how to do it?

Answer 1

Here the operation of the group is addition, so it is more proper to write $F+s$ instead of $sF$.

For the symmetric difference to be small, you need the intersection of the sets to be big. We can replace $E=[-m,m]\cap\mathbb Z$ for some $m$ with $m\geq\max E$ (this would only increase the max, so if we are below $\epsilon$ for this new $E$, we will also be for the original one).

Now fix $\epsilon>0$. We need a set that is not "changed much" by a translation by $m$. Let us take $$ F=[-m^2,m^2]\cap\mathbb Z.$$ Then, for $k\in E$, $F+k$ is the set $F$ shifted $k$ units (one way or the other, depending on the sign of $k$). The important thing is that $$ |F\Delta (F + k)|=2|k|. $$ Thus, for any $k\in E$, $$ \frac{|F\Delta (F + k)|}{|F|}=\frac{2|k|}{m^2}\leq\frac{2m}{m^2}=\frac2m. $$ Now it is enough to choose $m$ with $m>2/\epsilon$.