Let $f:X\to X$ be a homeomorphism on compact metric space $(X, d)$. A point $x\in X$ is called a non-wandering point, $x\in\Omega(f)$, if for every open set $U$ of $x$ there is $n\in\mathbb{N}$ such that $f^n(U)\cap U\neq \emptyset$.
We say that a point $x\in X$ is strong non-wandering, $x\in\Omega_s(f)$, if for every open set $U$ of $x$, there is $n\in\mathbb{N}$ such that $f^{nk}(U)\cap U\neq\emptyset$ for all $k\in\mathbb{N}$.
In the following we show that $\Omega_s(f)$ is a closed set.
Let $x_n\to x$ and $x_n\in\Omega_s(f)$ for all $n$. Let $U$ be an pen set of $x$, hence there is $x_n\in U$, this implies that there is $m\in\mathbb{N}$ such that $f^{mk}(U)\cap U\neq \emptyset$ for all $k\in\mathbb{N}$, i.e. $x\in\Omega_s(f)$.
It is clear that $\Omega_s(f)\subseteq \Omega(f)$.
What can say about reverse inclusion?
Please help me.
Reverse is not true. Take irrational rotation $f$ on circle, we can rotate by some small angle, for example $\alpha=\frac{2\pi}{100}$. Then take some small neighbourhood $U$ of any point $x$ in the circle. There is $n\in\mathbb{N}$ such that $f^n(U)\cap U\neq\emptyset$, but there also is $k\in\mathbb{N}$ such that $f^{nk}(U)\cap U=\emptyset$, since $f^k$ is also irrational rotation and is transitive, so $f^{nk}(U)$ raches point $-x$ (on the opposite site of circle) and, since $U$ is small enough, $f^{nk}(U)\cap U=\emptyset$.