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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.

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    In particular, [Qiaochu's answer](http://math.stackexchange.com/a/36511/9754) answers this as well.2012-07-15

1 Answers 1

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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 $X$.

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    Interesting argument!2014-12-11