Statement of theorem says:
Let $G$ be a countable semi-group admitting a Folner sequence $\{F_N\}$ and let $E \subset G$ be a set with positive upper density. Then there exists a probability space $(X,\hat{B},\mu,T)$ and a measure preserving $G$ action, $T_g$ on $X$ and a set $A \in \hat{B}$ such that $\mu(A)$ is the same as the upper density of $E$. And furthermore, for any $g_1,...,g_k \in G$, one has
$$\mu(A \cap T_{g_1}^{-1}(A) \cap ... \cap T_{g_k}^{-1}(A)) \leq \hat{d}(E \cap (g_1^{-1}(E) \cap ... \cap (g_k^{-1}(E)).$$
So the proof start by setting $X_0 = \{0,1\}^G$ which is compact by Tychonoff, correct? Under the product topology. Then we let $z \in X_0$ be the characteristic function of $E$. I.e., $z_n = 1$ if $n \in E$ and $0$ otherwise. Then consider the $G$ action of $X_0$ given as $$(gx)_n=x_{gn}; \space \text{for any $g,n \in G, x \in X_0$}.$$ Then we set $X$ to being the orbit closure of $z$, i.e.,
$$X= \overline{\{gz:g \in G\}}.$$
If $x \in X$, then for any $g \in G$ we have $gx \in X$. Let $A \subset X$ be the set of all sequences $x_n \in X$ such that $$x_0=1.$$ Then $gz \in A$ if and only if $g \in E$. I do not see why $gz \in A$ iff $g \in E$. Could someone elaborate? I wanna say it has to do with the $G$ action on $X$. From what i gather, i have
$$gz \in A \Leftrightarrow (gz)_0=1 \Leftrightarrow z_{g}=1 \Leftrightarrow g \in E.$$ Does this make sense?
There was an error $A$ is defined as all sequences such that $x_1=1$ then the iff statement holds.