A isometric map in metric space is surjective?

Question

Possible Duplicate:
Isometries of $\mathbb{R}^n$

Let $X$ be a compact metric space and $f$ be an isometric map from $X$ to $X$. Prove $f$ is a surjective map.

Answer 1

Here is an alternative to the proof linked to in the comments:

Suppose there existed $x \in X\setminus f(X)$. Then $x$ has positive distance $d$ from the compact set $f(X)$. Now consider the recursively defined sequence $$x_0 := x, \qquad x_n := f(x_{n-1}) \quad \forall \, n>0$$ We have $d(x_0, x_n)\ge d$ for all $n>0$, by assumption on $x$. This implies that we also have $d(x_k, x_{k+n}) = d(x_0, x_n) \ge d$ for all $k,n>0$ (here we use that $f$ is an isometry). Therefore $d(x_n, x_m) \ge d$ for all $m\ne n$, which is in contradiction to sequential compactness of $X$.