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Let $f:X\to X$ be a homeomorphism on compact metric space $(X, d)$ with two following property:

1) Every minimal set of $X$ is a fixed point i.e. if $K\subseteq X$ is a closed $f$-invariant set with $\overline{O_f(a)}=K$ for all $a\in K$, then $K=\{p\}$.

2) $f$ has finitely fixed point.

It is known that every closed $f$-invariant set $A$ contains a minimal set, hence I think that (1) implies that $\omega_f(x)=\{p\}$. Take $\omega_f(B)=\bigcup_{b\in B}\omega_f(b)$, I think that by (2), we have $\omega_f(B)$ is a finite set.

What can say about $f:X\to X$. Can we say that $X$ is countable or $f$ has finite orbit? Would you please your idea about them.

user479859
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  • I think you only get that $ω_f(B)$ is a finite set of fixed points of $f$, not that it reduces to a single one. What exactly is the claim of the task, and what is your question about it? Why do you think that $X$ should be countable? It has a countable dense subset by compactness. What indications do you have that $f^n=id_X$ for some $n$? – Lutz Lehmann Jan 17 '20 at 12:04
  • @LutzLehmann, Thanks. Yes I only get that $\omega_f(B)$ is a finite set. Indeed I am studying a paper entitle "Hyper-expansive Homeomorphisms". In this paper $f:X\to X$ is called hyper-expansive whenever there is $c>0$ such that for every closed sets $A\neq B$, there is $n\in \mathbb{Z}$ with $d_H(f^n(A), f^n(B))>c$. – user479859 Jan 17 '20 at 18:17
  • Author claim that if $f$ is hyper-expansive, then $X$ is countable. In this direction, he gave its proof. It is not clear for me the following statement in its proof: – user479859 Jan 17 '20 at 18:22
  • He said that for every $\epsilon>0$ there is $n\in \mathbb{N}$ such that for every $x\notin B_\epsilon(\Omega(f))$, we have $f^j(x)\in B_\epsilon(\Omega(f))$ if $|j|>n$.
    • – user479859 Jan 17 '20 at 18:24
  • He showed that if $f$ is hyper-expansive, then it has finite number of orbits, because if it has infinite number orbits and $\epsilon<c$, then $X-B_\epsilon(\Omega(f))$ is infinite(???) and there are $x, y\notin B_\epsilon(\Omega(f))$ and $p, q\in \Omega(f)$ with $\omega_f(x)=\omega_f(y)={p}$ and $\alpha_f(x)=\alpha_f(y)={q}$. Thus if $d(x, y)$is small, this two points contradicts the expansiveness of $f$(???) – user479859 Jan 17 '20 at 18:36
  • @LutzLehmann, Note that by hyper-expansiveness of $f$, we have every minimal set is a fixed point hence $\omega_f(x)={p}$ for every $x\in X$. Also hyper-expansivity implies expansivity, hence $f$ has finite number of fixed points. – user479859 Jan 17 '20 at 18:40
  • You should put those details not in comments, but in the question text. That these are questions about a paper changes the context of the question. The topological properties of $f$ are assumptions without which your questions do not make sense. – Lutz Lehmann Jan 17 '20 at 19:17
  • @LutzLehmann, I asked a new question with the following link. Please help me to know it. – user479859 Jan 17 '20 at 19:43
  • https://math.stackexchange.com/questions/3512815/a-question-about-finite-number-of-orbits-fx-to-x – user479859 Jan 17 '20 at 19:43