Let $(X,\rho)$ be a metric compact space and $F$ is the set of isometries from $X$ to itself. For every pair $f,g\in F$, define $$d(f,g)=\sup_{x\in X} \rho(f(x),g(x)).$$ Prove that $d$ is a metric in $F$ and with that distance, prove $F$ is a metric compact space.
Let just pay attention to the part $F$ is compact (since it's easy to check $d$ is a metric in $F$).
My purpose is to apply Arzela Ascoli theorem
Let $(X,\rho)$ be a compact metric space, $(Y,d)$ is a complete metric space and $F$ is a subset of $C(X,Y)$ ($C(X,Y)$ is set of continuous mappings from $X$ to $Y$). Then $F$ is relatively compact $\iff\begin{cases}\forall x\in X,\{f(x)|f\in F\}\text{ is totally bounded in } Y\\ F\text{ is equicontinuous on }X \end{cases}$
Consider $x_0\in X$.
$\forall\epsilon>0$, let choose $\delta=\epsilon$. Then $\forall x\in X: \rho(x,x_0)<\delta,\forall f\in F$, we have $$\sup_{x\in X}\rho(f(x),f(x_0))=\sup_{x\in X}\rho(x,x_0)\leq\epsilon$$
$\Rightarrow F$ is equicontinuous on $X$.
Assume $\{f_n\}\subset F$ converges to $f$. Then $\rho(f_n(x),f_n(y))=\rho(x,y)\forall x,y\in X$.
Fixing $x,y$ and let $n\to\infty$, we have $\rho(f(x),f(y))=\rho(x,y).$
$\Rightarrow f\in F\Rightarrow F$ is closed.
I just need to prove $F$ is totally bounded in $C[X,X]$ to finish the proof. But I don't know how to do it :( Could anyone help me? Thanks in advance
Here is different approach: Let $(x_i)$ be a countable dense set in $X$ and $(f_n)$ be a sequence in $F$. By a diagonal procedure show that there is subsequnce $(n_i)$ such that $\lim_{i \to \infty} f_{n_i}(x_k)$ exists for every $k$. A simple argument using the isometry property of $f_n$'s and triangle inequality shows that $f_{n_i}(x)$ is Cauchy, and hence convegent, for every $x\in X$. Let $f(x)=\lim_i f_{n_i}(x)$ and note that $d(f(x),f(y))=\lim_i d(f_{n_i}(x),f_{n_i}(y))=d(x,y)$. Thus, $f \in F$ and $f_{n_i} \to f$.