Katok, Hasselblatt: "Modern theory of dynamical systems", p. 129:
Definition 3.3.3. A point $x\in X$ is nonwandering with respect to the map $f\colon X\to X$ if for any open set $U\ni x$ there is an $N>0$ such that $f^N(U)\cap U\neq\emptyset$. The set of all nonwandering points of $f$ is denoted by $NW(f)$.
Equivalently, one may assume that $N$ is arbitrary big. For, if for every $U$, $f^N(U)\cap U=\emptyset$ for $N\geq N_0$, then $x$ is not periodic. Hence one can find a neighborhood $V\ni x$ such that $f^i(V)\cap V=\emptyset$ for $i=0,1,\ldots,N_0$, and $x$ cannot be nonwandering.
I do not understand this proof of the equivalence.
Let $N_0>0$ be arbitrary.
We suppose that $x$ is nonwandering, i.e. for any open set $U\ni x$ there exists some $N>0$ such that $f^N(U)\cap U\neq\emptyset$.
I guess, we then suppose there exists some open set $V\ni x$ such that for all $N\geq N_0$, we have $f^N(V)\cap V=\emptyset$. But I dont see how to get the contradiction to $x$ is nonwandering.
Laying out the proof in a little more detail, suppose $x$ is nonwandering and that there exists some $N_0$ so that $N > N_0$ implies $f^N(U) \cap U = \emptyset$. This implies $x$ is not periodic. Therefore, we can pick $V \subseteq U$ about $x$ which is small enough so that all of the first $N_0$ iterates $f^i(V)$ do not intersect $V$ (this seems worth verifying instead of taking for granted, but it is necessary for the proof). Now we know that $f^N(V) \subseteq f^N(U)$, so $f^N(U) \cap U = \emptyset$ implies $f^N(V) \cap V = \emptyset$. We already know that for $0 < i \leq N_0$ we have $f^i(V) \cap V = \emptyset$, so putting this together demonstrates that for all $n > 0$ we have $f^n(V) \cap V = \emptyset$, contradicting our claim that $x$ is nonwandering.