Let $f: D \subset \mathbb{R} \rightarrow \mathbb{R}$ continuous with $f(D) \subset D$ and let $x_0$ be a $n$-periodic atractor point of $f$. Show that every point of the cycle $\{ x_0, f(x_0),\dots,f^{n-1}(x_0) \}$ is also an $n$-periodic atractor point of $f$.
I tried using the definition of continuity in $\mathbb{R}$, but I always end up using 3 points of the cycle and I don't know how to solve this.
Can anyone help me?
So essentially, you have per assumptions continuous functions $F,G$, here more precisely $F=f^k$, $G=f^{n-k}$, with a cycle $F(x_*)=y_*$ and $G(y_*)=x_*$ and a neighborhood $U$ of $x_*$ so that $$\lim_{m\to\infty}(G\circ F)^{\circ m}(x)=x_* ~~\text{ for all }~~ x\in U. $$
Now take $V=G^{-1}(U)$ and observe $$(F\circ G)^{\circ m}=F\circ(G\circ F)^{\circ(m-1)}\circ G$$ to find a similar limit property for all $y\in V$ being propagated towards $y_*$.
This redistribution of the compositions means that an $y=y_0\in V$ gets mapped to some $x=x_0\in U$, and the sequences $x_m$, $y_m$ are connected by $x_m=G(x_m)$ and $y_{m+1}=F(x_m)$. Thus convergence of one series is equivalent to the convergence of the other.