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I need to prove this:

Show that if X has no isolated points and $O^{+}(x)$ is dense, then $\omega (x)$ is dense.

The set $O^{+}(x)=\bigcup_{n\in\mathbb{N_0}}f^{n}(x)$ where $f^n$ is the $n$-th composition, and $\omega (x)= \bigcap_{n\in \mathbb{N}}\overline{\bigcup_{i\geq n} f^{i}(x)}$, i tried to prove $\bigcap_{n\in \mathbb{N}}\overline{\bigcup_{i\geq n} f^{i}(x)}$ $\bigcap U \neq \emptyset$ where $U$ is a neighborhood of $x$ but i can't, any help would be appreciated.

P3peM4th.
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    $\omega(x)$ is the intersection of the closures of subsets of $O^+(x)$ that are obtained by removing finitely many points. Since $X$ has no isolated points, removing finitely many points from $O^+(x)$ doesn't change that the resulting set is dense. – EEE Oct 12 '17 at 12:58
  • you are right, thanks i think i can solve it whit this, greetings. – P3peM4th. Oct 12 '17 at 13:02

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