[proofplan] The proof uses the coding map as a conjugacy between the dynamics of $F$ on $\Lambda$ and the full two-shift on $\Sigma_2$. A periodic symbolic sequence is pulled back by the bijection $\pi$ to a unique point of $\Lambda$, and the conjugacy relation transfers the equation $\sigma^p(s)=s$ into $F^p(x)=x$. Exact periods are preserved because $\pi$ is injective. Finally, periodic sequences form a countable union of finite sets, and quotienting by finite orbits preserves countability. [/proofplan] [step:Transfer a periodic symbolic sequence to a periodic point in $\Lambda$] Let $s \in \Sigma_2$ and let $p \in \mathbb N$ satisfy $\sigma^p(s)=s$. Since $\pi: \Lambda \to \Sigma_2$ is bijective by hypothesis, there exists a unique point $x \in \Lambda$ such that $\pi(x)=s$. For each $m \in \mathbb N$, let $F^m: \Lambda \to \Lambda$ and $\sigma^m: \Sigma_2 \to \Sigma_2$ denote the $m$-fold iterates of $F|_\Lambda$ and $\sigma$, respectively. We extend the conjugacy relation from one iterate to $m$ iterates. For every $m \in \mathbb N$ and every $y \in \Lambda$, \begin{align*} \pi(F^m(y)) = \sigma^m(\pi(y)). \end{align*} This follows by induction on $m$: the case $m=1$ is the stated conjugacy relation, and the induction step uses $F(\Lambda) \subset \Lambda$ so that the relation may be applied to $F^m(y)$. Applying this identity with $m=p$ and $y=x$ gives \begin{align*} \pi(F^p(x)) = \sigma^p(\pi(x)). \end{align*} Since $\pi(x)=s$ and $\sigma^p(s)=s$, we obtain \begin{align*} \pi(F^p(x)) = s = \pi(x). \end{align*} Because $\pi$ is injective, this implies $F^p(x)=x$. Thus every symbolic point fixed by $\sigma^p$ gives a point of $\Lambda$ fixed by $F^p$. [guided] Let us start with a symbolic orbit. We are given a sequence $s \in \Sigma_2$ and a positive integer $p \in \mathbb N$ such that $\sigma^p(s)=s$. The assumed coding bijection $\pi: \Lambda \to \Sigma_2$ is bijective, so there is exactly one point $x \in \Lambda$ whose symbolic code is $s$; that is, $\pi(x)=s$. The conjugacy relation is stated for one application of the map: \begin{align*} \pi(F(y)) = \sigma(\pi(y)) \end{align*} for every $y \in \Lambda$. To use the period equation $\sigma^p(s)=s$, we need the corresponding relation for $p$ iterates. Define $F^m: \Lambda \to \Lambda$ to be the $m$-fold iterate of $F|_\Lambda$, and define $\sigma^m: \Sigma_2 \to \Sigma_2$ to be the $m$-fold iterate of $\sigma$. Since $\Lambda$ is $F$-invariant, each point $F^m(y)$ remains in $\Lambda$, so the conjugacy relation may be applied repeatedly. Induction on $m$ gives \begin{align*} \pi(F^m(y)) = \sigma^m(\pi(y)) \end{align*} for every $m \in \mathbb N$ and every $y \in \Lambda$. Now take $m=p$ and $y=x$. Then \begin{align*} \pi(F^p(x)) = \sigma^p(\pi(x)). \end{align*} Since $\pi(x)=s$, the right-hand side becomes $\sigma^p(s)$. The sequence $s$ is fixed by $\sigma^p$, so \begin{align*} \pi(F^p(x)) = \sigma^p(s) = s = \pi(x). \end{align*} The final step uses injectivity of $\pi$. If two points of $\Lambda$ have the same code, they are the same point. Therefore $F^p(x)=x$. This proves that every periodic symbolic sequence produces a periodic point of the horseshoe dynamics. [/guided] [/step] [step:Show exact symbolic period gives exact dynamical period] Assume now that $s \in \Sigma_2$ has exact period $p$ under $\sigma$, meaning $\sigma^p(s)=s$ and $\sigma^k(s)\ne s$ for every integer $k$ with $1 \le k < p$. Let $x \in \Lambda$ be the unique point satisfying $\pi(x)=s$. The previous step gives $F^p(x)=x$. Suppose, for contradiction, that $x$ has period $k$ under $F|_\Lambda$ for some integer $k$ with $1 \le k < p$. Then $F^k(x)=x$. Applying $\pi$ and using the iterate form of the conjugacy relation gives \begin{align*} \sigma^k(s) = \sigma^k(\pi(x)) = \pi(F^k(x)) = \pi(x) = s. \end{align*} This contradicts the exact-period assumption on $s$. Hence $x$ has exact period $p$ under $F|_\Lambda$. [/step] [step:Produce a symbolic sequence of exact period $p$ for every positive integer $p$] Fix $p \in \mathbb N$. Define $s_p \in \Sigma_2$ by setting $(s_p)_k = 1$ when $k \equiv 0 \pmod p$ and $(s_p)_k = 0$ otherwise, for every $k \in \mathbb Z$. Here $(s_p)_k$ denotes the $k$-th coordinate of the bi-infinite sequence $s_p$. The definition gives $\sigma^p(s_p)=s_p$. If $1 \le k < p$, then the coordinate $0$ of $\sigma^k(s_p)$ equals $(s_p)_k=0$, while the coordinate $0$ of $s_p$ equals $1$. Therefore $\sigma^k(s_p)\ne s_p$ for every $1 \le k < p$. Thus $s_p$ has exact period $p$ under $\sigma$. By the preceding step, the unique point $x_p \in \Lambda$ satisfying $\pi(x_p)=s_p$ has exact period $p$ under $F|_\Lambda$. Since $p \in \mathbb N$ was arbitrary, $F|_\Lambda$ has periodic points of exact period $p$ for every positive integer $p$. [/step] [step:Count the periodic orbits by counting periodic sequences] For each $p \in \mathbb N$, define \begin{align*} \operatorname{Fix}(\sigma^p) := \{s \in \Sigma_2 : \sigma^p(s)=s\}. \end{align*} A sequence in $\operatorname{Fix}(\sigma^p)$ is determined by its coordinates $s_0,s_1,\dots,s_{p-1}$, so \begin{align*} |\operatorname{Fix}(\sigma^p)| \le 2^p. \end{align*} Thus $\operatorname{Fix}(\sigma^p)$ is finite for every $p \in \mathbb N$. Let $\operatorname{Per}(\sigma) \subset \Sigma_2$ denote the set of all periodic sequences: \begin{align*} \operatorname{Per}(\sigma) := \bigcup_{p=1}^{\infty} \operatorname{Fix}(\sigma^p). \end{align*} This is a countable union of finite sets, hence countable. Since $\pi: \Lambda \to \Sigma_2$ is bijective and sends periodic points of $F|_\Lambda$ exactly to periodic sequences of $\sigma$, the set of periodic points of $F|_\Lambda$ is countable. Each periodic orbit of $F|_\Lambda$ contains at least one periodic point, and distinct periodic orbits are disjoint subsets of the [countable set](/page/Countable%20Set) of periodic points. Therefore the collection of periodic orbits contained in $\Lambda$ is countable. This completes the proof. [/step]