# Compact isometries

As seen in this post, any isometry from a compact metric space $K$ to itself is surjective, so the set $I(K)$ of all such isometries is actually a group.

Give $I(K)$ a reasonable metric and prove that it is compact as well. What can one say about the sequence of iterates $K, I(K), I(I(K)), \dots$?

1. A reasonable metric for $I(K)$ is $d(f,g) = sup_{k\in K} d(f(k),g(k))$, which in fact defines a metric on $K^K$. Moreover, the induced topology is the compact open topology.
Now apply (some version of) Arzelà-Ascoli to deduce that $I(K)$ is compact. One must check that the space is complete, which follows from the fact that $K^K$ is complete and $I(K)$ is closed. It is immediate to check the fact that the maps in $I(K)$ are equicontinuous. Finally the maps are pointwise bounded since $K$ is.