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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

infinity
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1 Answers1

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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 $k\geq 1$.

H. H. Rugh
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  • Thank you, I think what I tried to do was to show that if there isn't any subsequence there is no Cauchy subsequence. Do you see any thing wrong in what I did – infinity Oct 09 '16 at 12:21
  • It's close, but you should extract a further subsequence. Suppose you define (using your constants) $y_k=x_{k(n_0+1)}$. Then the sequence $(y_k)$ has no convergent subsequence (by the argument you correctly give). This contradicts compactness. – H. H. Rugh Oct 09 '16 at 12:59
  • Or, you may in fact note that having found your $n_0$ then $\delta=\min {d(z,x_1),...,d(z,x_{n_0}),\epsilon}>0$. The sequence $(x_n)$ then verifies $d(x_m,x_n)\geq \delta$ for all $n\neq m$ and can not have any convergent subsequence. – H. H. Rugh Oct 09 '16 at 13:09
  • Best one-line proof I've seen in quite a while. – DanielWainfleet Oct 09 '16 at 16:43
  • You need to also choose your subsequence so that $n_{k+1}-n_k\to\infty$. – Eric Wofsey Oct 09 '16 at 18:29
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    @EricWofsey Not really. If it stays bounded but the distance goes to zero then it must be zero for some positive iterate, whence a periodic point and this then gives a subsequence. – H. H. Rugh Oct 09 '16 at 18:54
  • Oh, true. But you do need to state that separately for the proof to be complete. – Eric Wofsey Oct 09 '16 at 18:57