$T$ contains the diagonal set $\Delta_X$. Let $n$ be a large natural number. If $m$ is large enough, then we have $$\begin{aligned}
\inf \left\{ \frac{ \sharp(T[Y]) }{ \sharp(Y) } \ \colon\ 
\emptyset \neq Y \subseteq X_m, Y \textrm{\ is\ a } 
T^{\circ n} \textrm{-bounded\ set} \right\} > 1.\end{aligned}$$ Take an element $x \in X_m$. By the inequality, we have $$\begin{aligned}
1 
= \sharp(\{x\})
< \sharp(T[x])
< \sharp(T \circ T [x])
< \cdots
< \sharp(T^{\circ n} [x])\end{aligned}$$ and $n < \sharp(T^{\circ n} [x]) \le \sharp(X_m)$. It follows that $\sharp(X_m)$ is grater than $n$ for large $m$.

For a box space of a residually finite group $G$, being a weak expander sequence means non-amenability of $G$.

Let $G$ be a finitely generated residually finite group. Let $H_1 \supseteq H_2 \supseteq \cdots$ be a sequence of finite index normal subgroups of $G$ which satisfies $\bigcap_{m = 1}^\infty H_m = \{1_G\}$. Then the following conditions are equivalent:

1.  \[LemmaBoxConditionNonAme\] The group $G$ is not amenable,

2.  \[LemmaBoxPropertyNotA\] The box space $\square G = \bigsqcup_{m = 1}^\infty G / H_m$ does not have property A,

3.  \[LemmaBoxPropertyWES\] The box space $\square G$ is a sequence of weak expander spaces.

By [@RoeLectureNote Proposition 11.39], conditions (\[LemmaBoxConditionNonAme\]) 