Hello everyone I found the following statement in my lecture but I'm not sure why it's true.
Let $(X,d) $ a metric space (compact). Hence we can say that the isometry group G is a compact in $(C(X,X),d_\infty)$ i.e $G$ is a family of closed and equicontinuous applications.
I thought that using Arzelà-Ascoli theorem would help but it leads nowhere. Thanks in advance for your help.
Hint: Let $f_n$ be a sequence of isometries. Consider $h_n(x)=d(x,f_n(x))$ show that the family $h_n$ is absolutely bounded and equicontinuous.
Ascoli implies the existence of a subsequence $h_p$ which converges towards $h$; $h_p(x)=d(x,f_p(x))$ is a Cauchy sequence implies that $(f_p(x))$ is Cauchy and $f_p(x)$ converges towards $f(x)$. Show that $f$ is an isometry.