Help clarifying this property of $\omega$ limit sets. Let $X$ be a metric space and $f:X \to X$ be continuous. For $x \in X$, the $\omega$-limit set $\omega(x)$ is defined as all cluster points of $(x(n))$, defined as $x(n+1)=f(x(n))$ with $x(0)=x$.
Wikipedia claims that the limit set is always $f$-invariant, provided $f$ is continuous. However, no constraint was placed on $X$. $X$ could be non-complete then. I want to make sure this is true because a set might not contain it's limit points. I could have a sequence of rationals converge to an irrational. I tried finding a proof of this online yet only found ones for the continuous case (not the discrete case as I described).
Could someone confirm whether this really is true?
 A: This is just an "if--then" statement: if $p \in \omega(x)$ then $f(p) \in \omega(x)$.
And that's pretty easy to prove: assuming $p \in \omega(x)$ it follows that $p = \lim_{i \to \infty} x(n_i)$ for some subsequence $n_i$. It follows that $f(p) = \lim_{i \to \infty} f(x(n_i)) = x(n_{i+1})$ (this is where continuity of $f$ is used), and therefore $f(p) \in \omega(x)$.
Added: Regarding your comment below, let's consider the case of a metric space $Y$ and a subspace $X \subset Y$ such that $X$ is complete with respect to the metric that is obtained by restriction of the metric on $Y$. I'll leave it as an exercise in topology to verify an important conclusion in this case: $X$ is a closed subset of $Y$, and therefore if a sequence in $X$ converges to a limit in $Y$ that limit is actually contained in $X$.
So suppose now that $f : Y \to Y$ is continuous, and that $f(X)=X$, so we can restrict to $f : X \to X$. In this situation, if $x \in X$, if $p \in Y$, and if $p \in \omega(x)$ then we can conclude that $p$ is a limit of a sequence in $X$, namely some sequence of the form $(x(n_i))$ with $1 \le n_1 < n_2 < \cdots$. But as said above $X$ is closed in $Y$, and therefore $p \in X$.
