Let $f \colon \omega^\alpha + 1 \to \omega^\alpha + 1$, where $\alpha \geq 1$ a continous funciton such that, exists $w \in \omega^\alpha + 1$ with $\mathcal{O}_f(w)$ dense in $\omega^\alpha + 1$.
The set $\mathcal{O}_f(w)$ is the orbit of $w$, ie, $\mathcal{O}_f(w) = \{ w, f(w), f^2(w), \dots \}$.
With this conditions, we have that: If $y \in \omega^\alpha + 1$ is limit point and $f(y)$ is isolated point, then:
- the set $\{ z \in \omega^\alpha + 1 \colon f(z) = f(y) \}$ is open and infinite.
- There are $k < l$ such that $f(f^k(w)) = f(f^l(w)) = f(y)$.
I am not entirely clear why the above is true, I would appreciate a suggestion regarding the two previous statements.
For any successor ordinal $\beta + 1$, the set $\{z \in \omega^\alpha + 1 : f(z) = \beta + 1\}$ is open. Notice it is just $f^{-1}[(\beta,\beta+2)]$, so it is open by continuity of $f$.
Then $f^{-1}[\{f(y)\}]$ is open ($f(y)$ is a successor ordinal) and contains $y$, so for some $z < y$ we see $(z,y+1) \subseteq f^{-1}[\{f(y)\}]$. Since $y$ is a limit ordinal, $(z,y+1)$ is infinite. In fact, $(z,y+1)$ contains infinitely many (we only need 2 of them) isolated points. By the density of the orbit of $w$, there exists $k < l$ such that $f^k(w)$ and $f^l(w)$ are distinct elements of $f^{-1}[\{f(y)\}]$, in which case $f(f^k(w)) = f(f^l(w)) = f(y)$.