I'm looking for an example in discrete dynamics which its periodic set of periodic points is not necessary closed.
$f:X \to X$ is a homeomorphism from compact metric $X$ to itself, and $\text{Per(f)}=\{x \in X: \exists k \in \mathbb{Z}, f^k(x)=x\}$.
An example I was thinking about is the rotation map for $\alpha \in \mathbb{R}$ : \begin{align} &f:S^1 \to S^1\\ &f(x)= x + \alpha, \text{mod} 1 \end{align} I do know for $\alpha \in \mathbb{Q}$ The set $\text{Per}(f)= S^1$ and for $\alpha \in \mathbb{Q}^c$ for all $x \in S^1$ the orbit of $x$ under $f$ is dense in $S^1$, so the periodic set of periodic points for $\alpha \in \mathbb{R}$ is points which is rotated by rational $\alpha$'s. I don't know how to show if this set is not closed in $S^1$?
If we consider doubling map $T:[0,1) \to [0,1)$ where $T(x)=2x,\text{mod}1$ then: \begin{align} T^n(x)=x (\text{mod}1) \Rightarrow 2^nx=x(\text{mod}1) \end{align} so $2^n(x)-x=p$ for $p\in \mathbb{Z}^+$ so $x= \frac{p}{2^n-1}$ for $p=0 ,1 ,\cdots, 2^n-2$ so the set of periodic points of $T$ is: \begin{align} \text{Per}(T)=\{\frac{p}{2^n-1}, p=0,1,\cdots,2^n-2\} \end{align} It is not necessary closed since when $p=2^n-2$ and when $n \to \infty$ then we could have $1$ as a limit point which does not belong to $\text{Per}(T)$.