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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 $NW(f)$.

<p>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. </p>

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.

Rhjg
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    $f^N(U) \cap U = \emptyset$ for all open sets $U$ and all $N = N_0, N_0+1, N_0+2, \dots$. So it is enough to find a nbhd $V$ such that $f^i(V) \cap V = \emptyset$ for finitely many $i=1,2,\dots,N_0-1$. And such a $V$ can be found because $f$ is continuous and $f^i(x)$ are distinct. Can you continue now? – user539887 Jul 16 '18 at 11:23
  • For what do we need the continuity? To make sure that preimages of open sets are open? – Rhjg Jul 16 '18 at 16:53
  • I would say that the whole "spirit" of the concept of nonwandering point is topological in nature: a nonwandering point need not return to its original position, but (some) nearby points must. – user539887 Jul 16 '18 at 19:12

1 Answers1

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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.

forget this
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  • By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $U\ni x$ there exists some $N\geq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right? – Rhjg Jul 16 '18 at 11:49
  • It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) \not \in U$ for every $i > 0$. – forget this Jul 16 '18 at 21:59