Infinite $\omega$-limit set for map implies no isolated points for $\omega$-limit set Let $X$ be a metric space and $f:X\to X$ be any map. Assume that for $x\in X$ the $\omega$-limit set  $\omega(x)$, which is the set of points $y$ such that there is some sequence $(n_k)$ such that $d(f^{n_k}(x),y)\to0$ as $k\to\infty$, is infinite.  I want to prove that $\omega(x)$ contains no isolated points.
Now if $y\in\omega(x)$ would be isolated, then we cannot have $f^n(x)=y$: since $\omega(x)$ is positively invariant, we have for some $m>n$ that $f^m(x)=y$ and then $\omega(x)$ consists of at most $m-n$ points. However, I do not see how to proceed from here, especially without continuity assumptions on the map $f$. Why can't we have that for any neighbourhood $U\ni y$ disjoint from $\omega(x)\setminus\{y\}$ we have that $f^n(x)\in U$ infinitely many times?
Any help is much appreciated.
 A: This is false, even if $f$ is assumed to be continuous.
Let's build a counterexample for a possibly discontinuous $f$ first.
Consider $X = [0,1] \times (\{0\} \cup \{1/n : n \in \mathbb{N} \})$ (here I use the convention that $0 \notin \mathbb{N}$).
Let $C \subset [0,1]$ be any closed set.
For $k \in \mathbb{N}$, let $A_k = \{a/k : 0 \leq a \leq k, d(a/k, C) \leq 1/k\}$ where $d(x, C) = \inf \{|x-c| : c \in C\}$.
Let $(q_n)_{n \in \mathbb{N}}$ be the sequence of rational numbers defined by listing all elements of $A_1$, then $A_2$, then $A_3$, etc.
Define $f : X \to X$ by $f(q_n, 1/n) = (q_{n+1}, 1/(n+1))$ and arbitrarily for other points of $X$.
Then $\omega((q_1, 1)) = C \times \{0\}$.
In particular, if we choose $C = [0,1/2] \cup \{1\}$ then $C$ is uncountable and contains an isolated point.
If you choose $C = \{0\} \cup \{1/k : k \in \mathbb{N}\}$ and restrict to $Y = (C \times \{0\}) \cup \{(q_n,1/n) : n \in \mathbb{N}\}$, then you get a counterexample with continuous $f$ as long as you choose its values on $C \times \{0\}$ in the right way.
