The nonwandering set Let $T:X\to X$ be a topological dynamical system. The nonwandering set is defined as $$NW(T)=\{x : \forall\text{ open }U\ni x, \exists N: T^N(U)\cap U\neq\emptyset\}$$
I am looking for an example of a dynamical system showing that in general, $$NW(T)\neq NW(T|_{ NW(T)})$$
 A: Here is an example, taken from a 1978 paper titled Continuous Maps of the Interval with Finite Nonwandering Set. It appears in that paper as Example D (although with a few details missing that I've inserted here). I'm sure that examples exist that require less description, but this is remarkably simple in terms of geometry and visualization.
Let $S$ be a circle in $\mathbb{R}^2$, and let $A$ be an arc joining points $x, y\in S$ such that $A\cap S = \{x, y\}$. We then let $X = S\cup A$ and endow $X$ with the relative topology induced by the usual topology on the plane. Clearly this topological space is pretty nice: it is compact, second-countable, connected, and completely metrizable. We denote $(a, b)$ (resp. $[a, b]$) to mean the open arc (resp. closed arc) on $S$ from $a$ counterclockwise to $b$. We let $T : X\to X$ be a continuous map such that the following hold:


*

*$T(x) = x$

*$T(y') = y$ for some $y'\in (x, y)$

*$T$ maps $[x, y']$ homeomorphically onto $[x, y]$ (with the assumption that $T(p)\in (p, y)$ for all $p\in (x, y']$, i.e. $T$ moves all points in $(x, y']$ strictly closer to $y$)

*$T$ maps $[y', y]$ homeomorphically onto $A$, with $T(y) = x$

*for some $z'\in (y, x)$ such that $T(z') = z\in (x, y')$, $T$ maps $[y, z']$ homeomorphically onto $[x, z]$

*$T$ maps $[z', x]$ homeomorphically onto $[x, z]$

*$T$ maps all of $A$ to the point $x$


To aid in visualizing what's going on here, below is a diagram of $X$ (adapted from the one in the paper):

Now for the dynamics part. We will first show that $NW(T) = \{x, y\}$. Clearly, as $x$ is a fixed point of $T$, $x\in NW(T)$. Then, given a neighborhood $U\ni y$, $T(U)$ contains some interval on $S$ in the form $[x, a)$. As $T$ is expanding on $[x, y']$, $T^N([x, a))$ is eventually close enough to $y'$ so that $T^{N+1}([x, a))\cap U$ is nonempty (this is immediate if $a\geq y'$). Therefore, $T^N(U)\cap U\neq \emptyset$ for some $N\geq 1$, so $y\in NW(T)$.
Next, let's show that $NW(T)$ consists only of those two points. The main thing to recognize is that any point $p\notin \{x, y\}$ has a neighborhood that eventually gets mapped to $\{x\}$, although I will show this in detail. All points in $(y, x)$ are not in $NW(T)$ because $(y, x)$ is not in the image of $T$. Also, all points in $A$ (except for $x$ and $y$) are not in $NW(T)$ because for $p\in A$, given some neighborhood $U\ni p$ such that $U\subset A$ and $U\cap S = \emptyset$, $T^N(U) = \{x\}$ for all $N\geq 1$, so $T^N(U)\cap U = \emptyset$ for all $N\geq 1$. Finally, all points in $(x, y)$ are not in $NW(T)$ because for $p\in (x, y')$, given some neighborhood $U\ni p$ such that $U\subset (x, y')$ and $T(U)\cap U = \emptyset$ (we can do this under the above assumption that $T(p)\in (p, y)$ for all such $p$, as $T$ is continuous and $X$ is Hausdorff, so for a neighborhood of $U'\ni T(p)$ such that $p\notin U'$, we can let $U\ni x$ be a subset of $T^{-1}(U')$ such that $U\cap U' = \emptyset$), we can see that $\partial U = \{p_1, p_2\}$ for some $p_1\in (x, y)$ and $p_2\in (p_1, y)$ (by our previous stipulation, $T(p_1)\in (p_2, y)$ and $T(p_2)\in (T(p_1), y)$). For $N\geq 1$ and $p\in U$, if $T^N(p)\in S$, then $T^N(p)\in (T(p_1), y)$, and if $T^N(p)\in A$, then $T^{N'}(p) = x$ for all $N' > N$. Therefore, as $T^N(p)\notin (x, T(p_1))$ for all $N\geq 1$, we have that $T^N(p)\notin U$ for all $N\geq 1$. This holds for all $p\in U$, so $T^N(U)\cap U = \emptyset$ for all $N\geq 1$. If $p\in (y', y)$, then we have some open neighborhood $U\ni p$ such that $U\subset (y', y)$, and thus $T(U)\subset A$ and $T^N(U) = \{x\}$ for all $N\geq 2$, so $T^N(U)\cap U = \emptyset$ for all $N\geq 1$. If $p = y'$, then we can choose a neighborhood $U\ni p$ such that points in $U\cap (x, y')$ are mapped to $[y', y]\setminus U$ so that $T(U)\cap U = \emptyset$; all points are then mapped to $A$ and eventually to $\{x\}$, so $T^N(U)\cap U = \emptyset$ for all $N\geq 1$. Thus, we have shown that if $p\notin \{x, y\}$, then $p\notin NW(T)$.
Now, as $T(x) = T(y) = x$, we can see pretty clearly that $NW(T|_{NW(T)}) = \{x\}$ ($\{y\}$ is open in $NW(T)$ and $T^N(\{y\})\cap \{y\} = \emptyset$ for all $N\geq 1$). Therefore, $NW(T)\neq NW(T|_{NW(T)})$.
