Let $K$ be a compact metric space, and $\phi : K \rightarrow K$ an isometry. Prove that $\phi$ is surjective.

Update: Changed formulas to LaTeX. Nobody has solved this yet? come on... i know you can do it!

Update (11/02/2008): Solution posted! (show solution)

Suppose that $f$ is not surjective, and that $x\in K\f(K)$. The image $f(K)$ of $f$ is a compact, and since a metric space is Hausdorff it is also closed. Then, there is an $\epsilon$ such that $\mathcal{B}(x,\epsilon) \cap f(K) = \emptyset$. Now consider the sequence $(x_n)_{n\in \mathbb{N}}$, where $x_0 = x$, $x_1 = f(x)$, $x_2 = f(f(x))$ and so on. Now, since the distance of $x$ from $f(K)$ is at least $\epsilon$, $d(x_0, x_n) \geq \epsilon$, $n \leq 1$. But since $f$ is an isometry, it is also true that $d(x_m, x_n) \geq \epsilon$, for each $n > m$ (in facts, this is true for the pair $x_o, x_{n-m}$, and applying $f$ ($m$ times) preserves the distances), and so the $(x_n)$ cannot have any convergent subsequence. Hence, $K$ cannot be compact.