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From the answer of the question isometry $f:X\to X$ is onto if $X$ is compact , I found out that the answer is valid only if the range of the sequence $\{x_n:n \in\mathbb N\}$ is infinite.

Now I'd like to know if the range is finite. Then $x_{n+k}=x_n$ for some $k\in \mathbb N$, so the sequence is periodic. By intuition, I guess the isometry $f$ satisfies $f^k=id.$, but I have no idea to prove this. I only know that the isometry is injective. Does anyone have any hints?

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bellcircle
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  • Why is the answer valid only when the sequence's range is infinite? I suppose it would be better if the answer said "cluster point" or "accumulation point," rather than "limit point," but it seems like it works fine with that adjustment. Am I missing something? – Cameron Buie Dec 30 '16 at 02:50
  • The three terms has the same meaning, and the Bolzano-Weierstrass property states that every "infinite" subset of a compact space has an accumulation point. If the range of sequence is finite, then the finite number of points cannot have accumulation point. – bellcircle Dec 30 '16 at 03:19
  • The terms have the same meaning for sets, but for sequences, "limit point" implies uniqueness, while the other two do not. For example, the sequence $n\mapsto (-1)^n$ has no limit point. However, both $-1$ and $1$ are cluster/accumulation points for the sequence. – Cameron Buie Dec 30 '16 at 03:23
  • Then your argument in the previous question is under assumption that $x$ is 'limit point' of the sequence, so I guess the argument should be changed by 'accumulation point' or 'cluster point' argument. – bellcircle Dec 30 '16 at 03:35
  • I read your answer again, and I found that just changing 'limit point' to 'accumulation point' makes your answer valid even if the range is finite. Thank you for your explanation. – bellcircle Dec 30 '16 at 03:39
  • You're very welcome! (P.S.: It isn't my answer. I just edited a typo while I was looking at it. ;-)) – Cameron Buie Dec 30 '16 at 03:49

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