TITLE: Showing existence of a subsequence
QUESTION [5 upvotes]: The question is one from the previous analysis preliminary exam:

Let $(M, d)$ be a compact metric space and $z ∈ M$. Let
  $T : M → M $ be a function which satisfies 
  $$
d(x, y) ≤ d(T(x), T(y))$$
  for all 
  $x, y \in M,$ i.e. the
  distances are non-decreasing under the mapping T. Define {$x_n$} by
  $x_1 = T(z)$ and $ x_{n+1} = T(x_n)$ for $n ≥ 1.$
  Prove that there exists a sub sequence of {$x_n$} which converges to $z$.

I saw some parallels between this question and to show that an isomtery on a compact set to itself is a surjection 
So I assumed that there is no subsequence which converges to $z$  and therefor there exist $n_o \in \Bbb N $ and $\epsilon$  such that  $d(x_m,z) >\epsilon $ for each  $ m>n_{0}$
And can I say that the sequence therefore has no convergent subsequnece?
and I get $\epsilon< d(x_{m-n},z)=d(z,T^{m-n}(z))≤d(x_n, x_m) $ whenever $m-n>n_{0}$
However i am not totally convinced if I have done everything correctly and also that if the result follows from string of inequalities necessarily .
Any help would be appreciated

REPLY [3 votes]: The sequence $(x_n)$ has a converging subsequence $x_{n_k}\rightarrow p$. The subsequence is then a Cauchy sequence and $$d(x_{n_{k+1}-n_k},z)\leq  d(x_{n_{k+1}},x_{n_k}) \rightarrow 0$$
One may note that if the difference $n_{k+1}-n_k$ stays bounded or, more generally, has a finite accumulation point $m\geq 1$ then $T^m z=z$ is a periodic point, so $x_{km}=z$ for all $k\geq 1$.