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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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marked as duplicate by Zhen Lin, Jonas Meyer, Jason DeVito, Qiaochu Yuan Jul 15 '12 at 5:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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What have you tried? –  Jonas Meyer Jul 15 '12 at 4:46
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See math.stackexchange.com/q/74326 and math.stackexchange.com/q/36502. (These were found with the Google search site:math.stackexchange.com isometric compact surjective metric.) –  Jonas Meyer Jul 15 '12 at 4:50
    
In particular, Qiaochu's answer answers this as well. –  mixedmath Jul 15 '12 at 4:53

1 Answer 1

up vote 11 down vote accepted

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! –  Groups Dec 11 at 10:56

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