What the statement asserts

The objects involved

The unit interval and its order automorphisms. Let $I = [0,1] \subset \mathbb{R}$, with the order inherited from $\mathbb{R}$. Consider the set $\operatorname{Aut}_{\le}(I)$ of all order isomorphisms $f : I \to I$, i.e. bijections with $s \le t \iff f(s) \le f(t)$ (increasing bijections of $[0,1]$ onto itself; such a map necessarily fixes $0$ and $1$). This set is a group under

$$(f g)(t) = f\bigl(g(t)\bigr), \qquad 1 = \mathrm{id}_I, \qquad f^{-1} = \text{the inverse map}.$$

So a product $fg$ means "apply $g$ first, then $f$".

Dyadic rationals. A real number $x$ is called dyadic if $x = m / 2^k$ for some integer $m \in \mathbb{Z}$ and some natural number $k \in \{0,1,2,\dots\}$.

The generating maps. Call an order automorphism $f$ of $I$ Thompson-type if there exists a finite set $B \subset \mathbb{R}$, every element of which is dyadic (no requirement that $B \subseteq [0,1]$; $B$ may be empty), such that: for all $x, y \in I$ with $x < y$ and with the open interval $(x,y)$ containing no point of $B$, there exist an integer $n \in \mathbb{Z}$ and a real number $c$ with

$$f(z) = 2^{n} z + c \qquad \text{for every } z \in I \text{ with } x \le z \le y .$$

(Here $2^n$ is the real power with integer exponent, so $n$ may be negative. The pair $(n,c)$ may depend on $x$ and $y$.)

The group $F$. $F$ is the subgroup of $\operatorname{Aut}_{\le}(I)$ generated by all Thompson-type maps: the smallest subgroup containing every Thompson-type $f$ (its elements are finite products of Thompson-type maps and their inverses). $F$ is regarded as a group in its own right, with the composition product, identity and inverse described above.

Følner-type inequality. For finite subsets $\Gamma, A \subseteq F$ and a real number $\varepsilon$, say that $A$ is $(\Gamma,\varepsilon)$-Følner if

$$\sum_{\gamma \in \Gamma} \bigl|\, \gamma A \;\triangle\; A \,\bigr| \;<\; \varepsilon \cdot |A| ,$$

where $\gamma A = \{\gamma a : a \in A\}$ is the left translate of $A$ by $\gamma$ (product in $F$, i.e. $\gamma a = \gamma \circ a$), $\triangle$ is symmetric difference $(X \setminus Y) \cup (Y \setminus X)$, $|\cdot|$ is the number of elements, and the sum and inequality are computed in $\mathbb{R}$. The inequality is strict. The sum is over the elements of the set $\Gamma$ (each counted once).

Note: if $A = \varnothing$ the left side is $0$ and the right side is $0$, so the strict inequality fails; hence a $(\Gamma,\varepsilon)$-Følner set is always non-empty. Likewise no set is $(\Gamma,\varepsilon)$-Følner when $\varepsilon \le 0$.

Iterated exponential. Define $T : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ by

$$T(0, m) = m, \qquad T(p+1, m) = 2^{\,T(p,m)} .$$

Only $T(n, 0)$ occurs below: $T(0,0) = 0$, $T(1,0) = 1$, $T(2,0) = 2$, $T(3,0) = 4$, $T(4,0) = 16$, $T(5,0) = 65536$, and in general $T(n,0)$ is a tower of $n-1$ twos stacked as exponents over $2^0 = 1$ (for $n \ge 1$).

The statement

Let $\Gamma$ be a finite subset of $F$ such that

1. (symmetric) for every $\gamma \in \Gamma$, also $\gamma^{-1} \in \Gamma$; and
2. (generating) the subgroup of $F$ generated by $\Gamma$ is all of $F$.

Then there exists a real number $C$ with $C > 1$ such that for every natural number $n \ge 0$ and every finite subset $A \subseteq F$:

$$\text{if}\quad \sum_{\gamma \in \Gamma} \bigl|\gamma A \,\triangle\, A\bigr| \;<\; C^{-n}\,|A| \quad\text{then}\quad T(n, 0) \;\le\; |A| .$$

Equivalently: every finite $A \subseteq F$ that is $(\Gamma, C^{-n})$-Følner has at least $T(n,0)$ elements.

Remarks on quantifiers and edge cases

• The constant $C$ is chosen once, depending (at most) on $\Gamma$; the same $C$ must work for all $n$ and all $A$ simultaneously.
• $C^{-n} = 1/C^n$ is a power with integer exponent $-n$; since $C > 1$, it is a positive number $\le 1$, equal to $1$ when $n = 0$.
• $n = 0$: the conclusion is $0 \le |A|$, which holds for every $A$, so this case imposes nothing.
• $n = 1$: the conclusion is $|A| \ge 1$, i.e. $A \ne \varnothing$; by the note above this follows from the hypothesis itself.
• $A$ ranges over all finite subsets of $F$; no condition relates $A$ to $\Gamma$ beyond the displayed inequality.
• $\Gamma$ being empty is not excluded by the form of the hypotheses, but then the generating hypothesis says that the trivial subgroup equals $F$, i.e. that $F$ is the trivial group.
• The symmetry hypothesis is about the set $\Gamma$ only: it is not required that $\Gamma$ contains the identity, nor that $\gamma \ne \gamma^{-1}$.
• The conclusion is a lower bound on the cardinality $|A|$ (a natural-number inequality, non-strict), not on any other size of $A$.