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Isometries of $\mathbb{R}^n$
A question about isometry
A isometric map in metric space is surjective?

How to prove that isometry $f:X\to X$ is onto if $X$ is compact??

What about the case when $X$ is non-compact?

Any help would be appreciated.

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marked as duplicate by kahen, rschwieb, David Mitra, Zhen Lin, t.b. Sep 1 '12 at 11:26

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.

See here for your first question. –  David Mitra Sep 1 '12 at 1:08

1 Answer 1

up vote 3 down vote accepted

Let $X$ be a compact metric space and $f\colon X\to X$ an isometry. For $x_0\in X$, consider the sequence $(x_n)$ given by $x_{n+1} = f(x_n)$ and let $x$ be a limit point of it (by compactness). Let $\varepsilon>0$. For some $n,m\in\mathbb N$ with $n<m$ we habe $d(x_n,x)<\frac12\varepsilon$ and $d(x_m,x)<\frac12\varepsilon$, hence $d(x_n,x_m)<\varepsilon$. But then $d(x_0,f(X)) \le d(x_0,x_{m-n})=d(x_n,x_m)<\varepsilon$. Since $\varepsilon>0$ was arbitrary, we conclude $d(x_0,f(X))=0$, i.e. $x_0\in f(X)$.

If $X$ is not compact: For $X=\mathbb{R}_{>0}$, the map $x\mapsto x+1$ is an isometry.

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