Let $X$ be a compact metric space, $f\colon X\to X$ — isometry. We fix an arbitrary point $x\in X$ and consider sequence $a_n = \underbrace{f(...f}_{n\ \ \mathrm{times}}(x))$. The hypothesis is following: $\{a_n\}$ has a subsequence that converges to $x$. First it seemed unreasonable, and I tried to find counter-example. But in a series of different cases this actually worked. I've failed to find counter-example, but I have no idea how to approach the proof. Does anyone have an idea?
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2Possible duplicate of If $(M,d)$ is a compact metric space and $f:M\rightarrow M$ is an isometry ($d(x,y)=d(f(x),f(y))$ for any $x,y\in M$), then $f$ is a homeomorphism – mathcounterexamples.net Mar 14 '16 at 21:09
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@mathcounterexamples.net, I wouldn't consider it a duplicate. I still don't quite get how these questions are connected. – Glinka Mar 14 '16 at 21:25
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Since $X$ is compact, $a_n$ has a convergent subsequence, say $a_{\phi(n)}$ where $\phi: \mathbb N \to \mathbb N$ is strictly increasing.
Note that $\displaystyle ||a_{\phi(n+1)-\phi(n)}-a_{0}||=||f^{\phi(n)}(a_{\phi(n+1)-\phi(n)})-f^{\phi(n)}(a_{0})||=||a_{\phi(n+1)}-a_{\phi(n)}||\to 0$.
It's possible to choose $\phi_n$ such that $\phi(n+1)-\phi(n)$ is increasing. $a_{\phi(n+1)-\phi(n)}$ is the subsequence you're looking for.
Gabriel Romon
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That tells me that $|a-b|=|f(a-b)|$, but you wrote something like $|a-b|=|f(a)-f(b)|$. Am I missing something? – bartgol Mar 14 '16 at 21:21
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@bartgol an isometry is a mapping that preserves distances, ie. $d(f(x),f(y))=d(x,y)$. Using the distance induced by a norm yields $||f(x)-f(y)||=||x-y||$. – Gabriel Romon Mar 14 '16 at 21:23
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