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I want to prove the proposition:

Proposition- Let $f:I \to I$ be continuos, and let f have a (2n+1)- periodic orbit {$x_{k}=f^{k}(x_{0})$, $k=0,1,\dots,2n$}, but no (2m+1)-periodic orbit for $1\leqslant m < n$. Suposse $x_{0}$ is in the middle of all $x_{i}$'. Then one of the two permutations

$$ (i) \ x_{2n}<x_{2n-2}<\dots<x_{2}<x_{0}<x_{1}<\dots<x_{2n-3}<x_{2n-1}$$

$$ (i) \ x_{2n-1}<x_{2n-3}<\dots<x_{1}<x_{0}<x_{2}<\dots<x_{2n-2}<x_{2n}$$

is valid.

Remark: I is an interval in the real line.

At the beginning of the proof he makes this statement that I can not justify

Supose n>1. Reorder {$x_{k}=f^{k}(x_{0})$, $k=0,1,\dots,2n$} as {$z_{i}$, $i=1,\dots,2n+1$} such that $$z_{1}<z_{2}<\dots<z_{2n+1}$$

Why i this is correct? What I win with this?

Improve this question
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  • $\begingroup$ I think that this is possible, because the orbit is an invariant set, then $f(\{z_{i}, i=1,2,\dots,2n+1\}) \subset \{x_{k}, i=0,2,\dots,2n\}$. However i don't know if this is answer $\endgroup$
    – user80064
    Commented Sep 9, 2015 at 1:36
  • $\begingroup$ Nothing is happening here. He/she is just taking the set $\{x_0,x_1,\ldots,x_{2n}\}$ and writing it in increasing order as $\{z_1,z_2,\ldots,z_{2n+1}\}$ with $z_1<z_2<\ldots$. The remainder of the proof will show either $z_1=x_{2n}$, $z_2=x_{2n-2}$ etc. or $z_1=x_{2n-1}$, $z_2=x_{2n-3}$ etc. $\endgroup$
    – Anthony Quas
    Commented Sep 9, 2015 at 5:33

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