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

Question

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. — Sahiba Arora, May 22 '20 at 20:33
As $f$ is bijective, it trivially follows from the compactness of $M$. — Ishan Deo, May 22 '20 at 20:36

score 2 · Accepted Answer · answered May 22 '20 at 20:37

Oh, that's a nice one.

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.

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

1 Answers1