Topological Dynamics: closure of forward orbit vs. $\omega$ limit set Let $f: X \rightarrow X$ be continuous, where $X$ is a topological space. This forms a topological dynamical system. For $x \in X$, define $\omega (x) = \cap_{n \in \mathbb{N}} \overline{\cup_{i \geq n}f^{i}(x)}$. Define the forward orbit of $x$ to be $\mathcal{O}^{+}(x) = \cup_{n \in \mathbb{N}} f^{n}(x)$. Clearly $\omega (x) \subseteq \overline{\mathcal{O}^{+}(x)}$. My intuition is that $\overline{\mathcal{O}^{+}(x)} - \mathcal{O}^{+}(x) \subseteq \omega (x)$ also holds. Is this true? The closures of two sets that only differ in a finite number of points should be the same, but this reasoning can't be used because we're dealing with an infinite intersection..
What I've tried:
Say $z \in \overline{\mathcal{O}^{+}(x)} - \mathcal{O}^{+}(x)$. Then every nbhd $U$ of $z$ contains some $f^{j}(x)$. Fixing $n$, it suffices to show that $z \in \overline{\cup_{i \geq n}f^{i}(x)}$, i.e. that every nbhd $U$ of $z$ also contains some $f^{i}(x)$ for $i \geq n$. Could there be some funky nbhd of $z$ that contains some iterate $f^{k}(x)$ for $k < n$ but never any for $k \geq n$?
For some context, my larger question is: "Is $\overline{\mathcal{O}^{+}(x)}$ f-invariant?" which I think will follow from the answer to the above, since $\omega$-limit sets are $f$-invariant.
Thank you for any help/advice!
 A: The accepted answer is still not true given your (lack of) hypotheses. You need your space to be at least $T_1$. Heuristically, $\omega(x)$ is the set of points for which the orbit of $x$ under $f$ approaches infinitely often and $\overline{O^{+}(x)}-O^{+}(x)$ is the set of points for which the orbit approaches (but does not meet). Your intuition is right, that the set of points that the $x$ visits often under $f$ should be visited infinitely often, but that relies on the $T_1$ characterization of a limit point $p$ of a set $S$: that every neighborhood of $p$ contains infinitely many members of $S$. Regrettably, in a general topological space, a limit point $p$ of a set $S$ only implies that every neighborhood of $p$ contains a member of $S$. 
With that in mind, consider the natural numbers with two zeros. That is, let $\tilde{\mathbb{N}}=\mathbb{N}\cup\{*\}$, and let $\{0,*\}$ and the singletons be a basis for your topology. So $0$ and $*$ are topologically indistinguishable. 
Define a function $f:\mathbb{N}\to \mathbb{N}$ by letting $f(n)=n+1$ if $n\in \mathbb{N}$ and $f(*)=1$. The pre-image of every basic open set is open, and so $f$ is continuous.  With that said, $O^{+}(0)=\mathbb{N}$ and $\overline{O^{+}(0)}\setminus O^{+}(0)=\tilde{\mathbb{N}}\setminus \mathbb{N}=\{*\}$. Moreover, by the same argument in my other answer above/below, $\omega(x)=\emptyset$, the key in this part being the fact that $*$ lies in the closure of the orbit of $0$ but not $f(0)=1$. Since $\{*\}\not\subseteq \emptyset$, this completes the counterexample.
However, since every space you might care about is most likely $T_1$, this is probably not an issue for you. 
A: Take $X=\mathbb{R}$ and $f(x)=x+1$. Then $\overline{O^+(x)}=O^+(x)=\{x+n: n\in \mathbb{N}\}$. For a given $n$, we have $\overline{\cup_{i\ge n}f^i(x)}=\cup_{i\ge n}f^i(x)=\{x+i: i\ge n\}$. So $\cap_{n=1}^m\{x+i: i\ge n\}=\{x+i: i\ge m\}$ and so $\omega(x)=\emptyset$. 
So $\omega(x)=\emptyset\subseteq \{x+n: n\in \mathbb{N}\}=\overline{\mathcal{O}^+(X)}$ for every $x\in \mathbb{R}$, the reverse inclusion is not true. 
A: The closure of $\mathcal O^+(x)$ contains, in particular, all the points in $\mathcal O^+(x)$ itself, and there is no reason that these should be in $\omega(x)$.
A: The first question is false, as pointed out. The larger question is true however. Indeed, let $y \in \overline{\mathcal{O}^{+}(x)}$ : then there exists integers $n_k$ such that $f^{n_k}(x)$ converges to $y$, by definition. But then $f^{n_k +1}(x)$ converges to $f(y)$ (by continuity of $f$) so $f(y) \in \overline{\mathcal{O}^{+}(x)}$ as well.
It does not imply the first question, because you can perfectly well have two $f$-invariant closed sets $A \subset B$ without having $A=B$ (take $f(x)=x/2$, $A=\{0 \}$, $B=\{1/2^n \} \cup  \{0 \}$). 
The property "$B$ is a close $f$-invariant set that does not contain any other $f$-invariant set" is called irreducibility. 
