If $x \in X$ is a non-wandering point of $f:X \to X$, then $(\forall U \ni x)$, there exist infinitely many $n$ s.t. $f^n(U) \cap U \neq \emptyset$. Let $X$ be a metric space and $f:X \to X$ a continuous map.
$x \in X$ is called a non-wandering point of $f$ if for every open set $U \ni x$, there exists $n \in \mathbb{N}$ such that $f^n(U) \cap U \neq \emptyset$. 
I want to show that if there exists one such $n$, that there exist infinitely many such $n$. 
I haven't gotten far, but here's my progress: suppose for the sake of contradiction that there exist only finitely many such $n$, and that they are $n_{1},...,n_{k}$ for some open set $U$. Now I want to find an open set containing $x$ which will never intersect itself, no matter how many times we apply $f$ to it, but I have no idea on how to arrive at such a set. Is there a good hint or maybe a different way to do this?
 A: Let $d$ be a metric for $X$ and define $B_n := B(x,1/n)$ for all $n \geq 1$. Because $U \ni x$ is open, it is clear that $B_n$ will be contained in $U$ for all $n$ sufficiently large. Therefore, by omitting finitely many $B_n$ if necessary, we obtain a family $\{B_n\}_{n}$ of open balls entered at $x$ such that


*

*$B_n \supseteq B_{n+1}$ for all $n$;

*$B_n \subseteq U$ for each $n$.


Arguing by contradiction, let us assume that $f^n(U) \cap U = \varnothing $ for all $n \geq K$. Clearly, this means that $f^{-n}(U) \cap U = \varnothing$ for all $n \geq K$. For each $n \in \mathbb{N}$ let us now define
$$
a_n := \max\left\{ m \in \mathbb{N} : f^{-m}(B_n) \cap B_n \neq\varnothing \right\}. 
$$
We now check that $a_n$ is well defined. Indeed, if $f^{-m}(B_n) \cap B_n \neq \varnothing$, then $f^{-m}(U) \cap U$ must also be non-empty. Thus, $m \leq K$. Furthermore, since every $B_n$ is open and $x$ is a non-wandering point, the set described above is non-empty. From this, we see that $a_n \leq K$ is well defined.
Because $B_n \supseteq B_{n+1}$, the $(a_n)$ will be a decreasing sequence of natural numbers. Hence, $a_n = a \in \mathbb{N}$ for all $n$ sufficiently large. For large $n$, pick $x_n \in f^{-a}(B_n) \cap B_n$. Clearly, $x_n \in B_n$ implies that $d(x_n,x) < 1/n$ and so $x_n \to x$ as $n \to \infty$.
On the other hand, $f^a(x_n) \in B_n$ also asserts that $f^a(x_n) \to x$ as $n \to \infty$. From this, we infer that $f^a(x) = x$. But this is a problem because then $f^{a^m}(x) = x$ for all $m \in \mathbb{N}$. Namely, $f^n(U) \cap U \neq \varnothing$ for infinitely many $n$.
