Hyperbolic sets

Question

I recently started reading about hyperbolic dynamics in the notes of L. Wen,

http://www6.cityu.edu.hk/rcms/publications/ln5.pdf

and in this (page 8) there is the following statement: If the definition of hyperbolicity we allow $E^{s}= \{0\}$ or $E^u=\{0\}$ in this case the hyperbolic set $\Lambda$ must be finite. He says as something trivial (perhaps) but I can not see.

Answer 1

Let's suppose $E^u=0$ on $\Lambda$ and $\lambda\in(0,1)$ be the contracting constant of $Df$ on $\Lambda$. Pick an open neighborhood $U\supset \Lambda$ (small enough) such that $\|Df|_{U}\|<\lambda<1$. Then pick $N$ large with $\lambda^N<1/6$.

First step is to show that, every nonwandering point $x\in\Omega(f,\Lambda)$ is a periodic point.

• Pick a neighborhood $B(x,\delta)\subset U$ small enough such that we can show inductively, $f^nB(x,\delta)\subset U$ for all $n\ge1$. By the nonwandering assumption, there exists $y\in B(x,\delta/3)\cap\Lambda$ such that $f^ny\in B(x,\delta/3)$ for some $n\ge N$. In particular $$f^nB(x,2\delta/3)\subset f^nB(y,\delta)\subset B(f^ny,\delta/6)\subset B(x,\delta/2).$$ As a contracting self-map, there exists a unique point $p\in B(x,\delta/2)$ fixed by $f^n$ and every point in $B(x,\delta/2)$ is attracted to $p$ under $f^n$. Hence every point in $B(x,\delta/2)$ is wandering (unless the point is $p$ itself). Since $x$ is assumed to be nonwandering, we must have $x=p$ be a periodic point.

Secondly, we show every point $x\in\Lambda$ is a periodic point.

• Pick a general point $x\in \Lambda$. Note that every point $y\in\alpha(f,x)$ is nonwandering and hence periodic. So every point lying close enough to $y$ will stay close and approximate the orbit $y$ under forward iterates (by the assumption $E^u=0$ on $\Lambda$). Since we assume $y\in\alpha(f,x)$, we can pick $x_k=f^{-n_k}x\to y$. So $d(x,y)= d(f^{n_k}(x_k),y)\le d(x_k,y)\to 0$, which implies $x=y$ is also a periodic point.

Finally, the finiteness: this also follows from $E^u=0$ on $\Lambda$, since every periodic point $x\in\Lambda$ will admits an open attracting neighborhood. If $x_k\in\Lambda\to x$, then $x\in\Lambda$ and $x_k$ will fall into the attracting neighborhood of $x$ for $k\ge K$ , which will force $x_k=x$ for all $k\ge K$.