# Prove that the sequence $x_0 \operatorname{:= }x, x_n \operatorname{:= } f(x_{n-1})$ in compact spaces has a convergent subsequence converging to $x$

Let $$(M,d)$$ be a compact metric space, and $$f:M\to M$$ be bijective and satisfy $$d(f(x),f(y))\le d(x,y)\ \ \forall x,y\in M$$.

Now, if $$x_0 \operatorname{:= }x, x_n \operatorname{:= } f(x_{n-1})$$, then show that there exists a convergent subsequence of $$(x_n)$$ converging to $$x$$.

I was able to get a convergent subsequence above, but I wasn't able to show that it converged to $$x$$.

• Please show your work about how you got a convergent subsequence. May 22, 2020 at 20:33
• As $f$ is bijective, it trivially follows from the compactness of $M$. May 22, 2020 at 20:36

First, go back to the past. Since $$f$$ is bijective and $$M$$ compact, there exists a positive increasing sequence $$(n_k)_{k \geq 0}$$ and $$y \in M$$ such that
$$\lim_{k \to + \infty} x_{-n_k} = y$$
Now, for all $$k$$, $$\ell \geq 0$$,
$$d(x, f^{n_\ell-n_k} (x)) \leq d(f^{-n_\ell} (x), f^{-n_k} (x)),$$
and $$f^{-n_k} (x)$$ and $$f^{-n_\ell} (x)$$ are both close to $$y$$ if $$k$$ and $$\ell$$ are large enough.