Ordering periodic orbits 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? 
 A: It's this article, isn't it?: From Intermediate Value Thm to Chaos, by XC Huang, in Math. Mag.
The first question is easy: he can do it, because it's just giving another name. So there are $x_0, \dots, x_{2n-1}$ (distinct points), and we don't know how are they positioned – only that $x_0$ is the one in the middle (I think), but this isn't that important now. Then surely one of them is the smallest – we'll call it $z_1$. Then there's the second smallest, this will be $z_2$. Then the third, and so on, and finally we call the greatest $z_{2n+1}$. Also, $z_{n+1} = x_0$, but I don't see this equality used in proof. If you still have problem with this, see comment.
Now, the second question is more difficult to answer nicely. Even after I saw the article, I can't say much more than 'it makes these points easier to handle'. In the proof the relative position of these points (and intervals $S_{kl}$) is quite important, and we couldn't work well on $x_i$'s – even though we know that $f((x_i, x_j)) = (x_{i+1}, x_{j+1})$, we don't know, where these intervals are, relatively. Now for $z_i$'s we also don't know where $f(z_i)$ is, but we can prove, that there is the greatest $z_i$, such that $f(z_i) > z_i$ (this is $z_m$), which is later important.
Also, there are used the intervals between a point from the orbit and the next closest point from the orbit $(z_i, z_{i+1})$ (so next by their position on the interval, not by the orbit: $x_0, \dots, x_{2n-1}$). I think the author deemed it more inconvenient to write without introducing the $z_i$'s.
EDIT: A remark, which might, alas, mess with your mind: in fact, you could probably write the proof, and nicely, without the $z_i$'s. Instead of $z_m$, we choose $x_m$ as the greatest $x_i$, such that $f(x_i) > x_i$, and similarly with $z_l$. Now instead of writing $z_{i+1}$, that is: next point after $z_i$, we might call $x_{n(i)}$ as the next point after $x_i$. However, they are always used together as a set, so simpler: $N_i = \{x_i, \text{next point after } x_i\}$. Then $S_1 = N_m$, $S_t = N_l$. And also $S_{1m} = \{x_i: x_i\leq x_m\}$, and we got rid of all $z_i$'s. Still, the exposition of the author seems more elegant to me.
