If $\omega_f(y) \subseteq \omega_f(x)$ and $int(\omega_f(y)) \neq \varnothing$ then $\omega_f(y) = \omega_f(x)$

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We let $X$ be a compact metric space and $f:X\rightarrow X$ a continuous function. For any point $x$ we define the orbit of $x$ under $f$ as $orb_f(x) = \{f^n(x): n \in \mathbb{N}\}$, where $f^n$ is used to denote $f$ composed with itself $n$ times. We define $\omega_f(x) = \cap_{n \in \mathbb{N}} \overline{orb_{f}(f^n(x))}$. This set is the omega limit set of $f$ on point $x$ which basically consists of the accumulation points of the orbit of $x$. Now, I'm not sure how to tackle this problem because it's enough to prove that for any point $p \in \omega_f(x)$ there's an increasing sequence of natural numbers $(n_k)_{k \in \mathbb{N}}$ such that $\lim_{k \rightarrow \infty}f^{n_k}(y) = p$ but I'm not really sure of how to prove this or how to use the fact that $int(\omega_f(y)) \neq \varnothing$

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First observe that $f^n(x)$ will frequently lie in every neighborhood intersecting $\omega_f(x)$. In particular, the nonempty interior condition on $\omega_f(y)\subseteq \omega_f(x)$ implies that $f^n(x)$ is frequently in $\omega_f(y)$.

Now let $x_0\in \omega_f(x)$, and let $U\ni x_0$ be a neighborhood. Then $f^n(x)$ frequently lies in $U$.

In particular, we have some $m,k$ such $f^m(x)\in \omega_f(y)$ and $f^{m+k}(x)\in U$. But since $f$ is continuous, $f^{-k}(U)$ is open and contains $f^m(x)$, so $f^{-k}(U)\cap \omega_f(y)\neq \emptyset$, and so there are infinitely many iterates $f^l(y)$ in $f^{-k}(U)$, hence infinitely many $f^{l+k}(y)$ in $U$.

Since $U$ was an arbitrary neighborhood of $x_0$, this implies $x_0\in \omega_f(y)$, and since $x_0\in \omega_f(x)$ was arbitrary, we then have $\omega_f(x)\subseteq \omega_f(y)$, and the reverse inclusion was given.

Remark

As you can see from the proof this statement in fact applies to an arbitrary continuous self map $f$ on a topological space $X$. No other properties are needed (other than the stated assumptions $\omega_f(y)\subseteq \omega_f(x)$ and $\text{int}(\omega_f(y))\neq \emptyset$.)