An isometry

Let $K$ be a compact metric space, and $\phi :K\to K$ 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 $x\in K\backslash f\left(K\right)$ . The image $f\left(K\right)$ of $f$ is a compact, and since a metric space is Hausdorff it is also closed. Then, there is an $\epsilon$ such that $𝒝(x,\epsilon )\cap f\left(K\right)=\varnothing$ . Now consider the sequence $(x_n{)}_{n\in 𝕅}$ , where $x_0=x$ , $x_1=f\left(x\right)$ , $x_2=f\left(f\right(x\left)\right)$ and so on. Now, since the distance of $x$ from $f\left(K\right)$ is at least $\epsilon$ , $d(x_0,x_n)\ge \epsilon$ , $n\le 1$ . But since $f$ is an isometry, it is also true that $d(x_m,x_n)\ge \epsilon$ , 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 $\left(x_n\right)$ cannot have any convergent subsequence. Hence, $K$ cannot be compact.