prove that a given subset of $\operatorname{\mathrm{Auttop}}(\Delta)$ is relatively compact, we will generally proceed by contradiction. In this context, the next proposition will be helpful. We introduce a convenient definition to state it.

Let $\Delta$ be a compact polygon. Two convergent sequences $a_n \to a$ and $b_n \to b$ in $\operatorname{\mathrm{Vert}}\Delta$ are said to be **collapsed by** a sequence $\{\varphi_n\} \subseteq \operatorname{\mathrm{Auttop}}(\Delta)$ if $a \neq b$ and if $\varphi_n a_n \to x$ and $\varphi_n b_n \to x$ for some $x \in \operatorname{\mathrm{Vert}}\Delta$.

\[proposition:closer\] Let $\Delta$ be a compact polygon and $S \subseteq \operatorname{\mathrm{Auttop}}(\Delta)$. If $S$ is not relatively compact in $\operatorname{\mathrm{Auttop}}(\Delta)$, then there exist two sequences $a_n \to a$ and $b_n \to b$ in $\operatorname{\mathrm{Vert}}\Delta$ which are collapsed by some sequence $(\varphi_n)$ in $S$ or in $S^{-1}$.

Let $S$ be a subset of $\operatorname{\mathrm{Auttop}}(\Delta)$ which is not relatively compact. By Lemma \[lemma:equicontinuous\], $S$ or $S^{-1}$ is not equicontinuous on $\operatorname{\mathrm{Cham}}\Delta$ and therefore also not equicontinuous on $\operatorname{\mathrm{Vert}}\Delta$. Suppose $S$ is not equicontinuous. Then there exist some $x \in \operatorname{\mathrm{Vert}}\Delta$ and $\varepsilon > 0$ such that for every neighbourhood $U$ of $x$, there exists $\psi \in S$ satisfying $\psi(U) \not \subseteq B(\psi x, \varepsilon)$. In other words, for every $n \in 