I am reading from the paper here. In Lemma 5, the authors make the following assertion:
Let $f : [0, 1] \to [0, 1]$ be a continuous function with a dense set of periodic points. If $f$ has no proper invariant nondegenerate subinterval, then either:
- $f$ is transitive; or
- $f$ permutes two subintervals $[0, a]$ and $[a, 1]$ for some $0 < a < 1$.
($f$ is said to be transitive if for any two open nonempty sets $U$ and $V$,
there is some $n \in N$ such that $f^n(U) \cap V$ is nonempty)
I will repeat the initial part of the proof given in the paper verbatim:
Suppose now that $f$ has no proper invariant subinterval. Certainly it has no
interval of fixed points. Suppose that $f$ is not transitive. Then there are two nonempty
open subintervals $U$ and $V$ of $[0, 1]$ such that $U′ = \bigcup_{i \ge 0}f^i(U)$ and $V$ are disjoint. $U'$ can be written as a union of at most countably many disjoint, nondegenerate intervals. If one of these intervals is, say $J = \bigcup_{i∈N} f^i(U)$, for some subset $N$ of $\mathbb{N}$, then its image under $f$ , $f(J) = \bigcup_{i∈N} f^{i+1}(U)$, is a subset of another of these intervals. But
since $J$ must contain a periodic point of period $k$, then $f^k(J)$ must be a subset of $J$.
Moreover, $f^k(J)$ must be dense in $J$, since otherwise $J \setminus f^k(J)$ contains some open
subinterval that contains no periodic points.
I was able to follow the reasoning upto this part. But I am not able to understand the next line.
We may, therefore, express the closure of $U′$ as a finite union of closed, nondegenerate, intervals $I_i = [r_i, s_i]$, with $r_i < s_i ≤ r_{i+1}$,
for $1 ≤ i ≤ n$, that are permuted by $f$.
Why are we able to to express $\overline{U'}$ as finite union of closed intervals? I am able to see that it should be a countable union of closed intervals.