Let $f\colon X \rightarrow X$ is continuous and $X$ a topological space which is locally compact, metrizable and second countable.
We say that a set $A\subset X$ is $f-$invariant iff for all $n \in \mathbb{Z}$ we have that $f^n(A)\subset A$.
A point $y\in X$ is a $\omega-$limit point of $x$ if there exist a sequence $n_k \rightarrow \infty$ such that $f^{n_k}(x)\rightarrow y$.
A point $y\in X$ is a $\alpha-$limit point of $x$ if there exist a sequence $n_k \rightarrow \infty$ such that $f^{-n_k}(x)\rightarrow y$.
The $\omega-$limit set of $x$ is the set of all $\omega-$limit points of $x$. The $\alpha-$limit set is defined analogously, if $f$ is invertible.
An exercise says that $\omega-$limit set of $x$ and $\alpha-$limit set of $x$ are $f-$invariant.
I have done the following:
If $y$ is $\omega-$limit point of $x$, then exist a sequence $n_k \rightarrow \infty$ such that $f^{n_k}(x)\rightarrow y$. As $f$ is continuous, then for all $n \in \mathbb{N}$ we have that $f^n$ is also continuous. As $X$ is metrizable, this implies that $f^n(f^{n_k}(x))\rightarrow f^n(y)$. Then $f^n(y)$ is in $\omega-$limit set of $x$, so $f^n(\omega(x))\subset \omega(x)$ for all $n\in \mathbb{N}$.
But my question is, if I don't know if $f$ has continuous inverse, how can I show that $f^n(\omega(x))\subset \omega(x)$ for all $n\in \mathbb{Z}$, I mean, in the definition it doesn't assume that $f$ must be an homeomorphism, just assumes that $f$ is invertible.
Any help would be appreciated.