when $X$ is a topological space and $Y$ is a metric space, a subset $\mathcal{F}$ of $C(X,Y)$ is said to be **equicontinuous at $x_0 \in X$** if, for all $\varepsilon > 0$, there exists a neighbourhood $U$ of $x_0$ such that $$f(U) \subseteq B(f(x_0), \varepsilon) \quad\text{for all } f \in \mathcal{F},$$ where $B(x, r)$ denotes the open ball of radius $r$ centered at $x$. The set $\mathcal{F}$ is then called **equicontinuous on $X$** if it is equicontinuous at each point of $X$.

Let $X$ be a topological space and $Y$ be a metric space. Let $\mathcal{F}$ be a subset of $C(X,Y)$. If $\mathcal{F}$ is equicontinuous on $X$ and if $\{f(x) \mid f \in \mathcal{F}\}$ is relatively compact in $Y$ for all $x \in X$, then $\mathcal{F}$ is relatively compact in $C(X,Y)$ equipped with the compact-open topology. The converse holds if $X$ is locally compact.

See [@Munkres]\*[Theorem 47.1]{}.

We can now already give some characterization of relative compactness in $\operatorname{\mathrm{Auttop}}(\Delta)$.

\[lemma:equicontinuous\] Let $\Delta$ be a compact spherical building. A set $S \subseteq \operatorname{\mathrm{Auttop}}(\Delta)$ is relatively compact in $\operatorname{\mathrm{Auttop}}(\Delta)$ if and only if $S$ and $S^{-1}$ are equicontinuous on $\operatorname{\mathrm{Cham}}\Delta$.

First assume that $S$ is relatively compact in $\operatorname{\mathrm{Auttop}}(\Delta)$. The set $S$ 