Travel of the point via isometry: does it get closer to home?

Question

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?

Answer 1

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.