2
$\begingroup$

I'm looking for a simple example in discrete dynamical systems whose periodic points set is not necessary closed.

I've seen some example in websites but they are not that simple and discrete.

Note that :

$(X,f)$ is a Dynamical System if $f:X \to X$ is a homeomorphism and $X$ is a compact metric space. \begin{align} Per(f):=\{x \in X ; f^n(x)=x ,\text{ for some } n \in \mathbb{Z}\} \end{align}

For example I found this example :

Let $X$ be the unit disk $\{z\in\Bbb C: |z|\le 1\}$ and $f:X\to X$, $x\mapsto xe^{|x|i}$. Then $Per(f)=\{x\in X: |x|/\pi\in\Bbb Q\}.$

But It's not discrete and also It's not that simple to me to see why the periodic points is that and why it is not closed.

Could you please help me find a simple example with this property in discrete dynamical systems ?

$\endgroup$
9
  • $\begingroup$ Closed subset of $\Bbb{Z}$? For what topology? $\endgroup$
    – abx
    Jan 21, 2021 at 8:34
  • $\begingroup$ I made a mistake I edited my post I meant the iterates of $f$ for $n \in \mathbb{Z}$. $\endgroup$ Jan 21, 2021 at 8:51
  • 1
    $\begingroup$ Just a guess: What about shift functions on infinite products of finite sets? So you take the space of infinite strings on some finite alphabet and you apply the function shifting the stuff a step to the left or to the right. Then periodic points should be strings with a periodic expression. That's probably something dense in the space. Also the space has a metric where $d(s_1,s_2)$ is say $e^{-|n|}$ where $n$ is the smallest index at which the strings $s_1$ and $s_2$ differ. $\endgroup$ Jan 21, 2021 at 9:07
  • $\begingroup$ The set of periodic points on $\{0,1\}^\mathbf{Z}$ with respect to the shift is dense. $\endgroup$
    – YCor
    Jan 21, 2021 at 9:25
  • $\begingroup$ Any compact space with an action that has infinitely many periodic points and at least 1 transitive (dense) orbit should satisfy the criteria I believe. Of course you could also just ask for compactness, that periodic points are dense and not all points are periodic, for which many examples exist. $\endgroup$
    – Dan Rust
    Jan 21, 2021 at 10:22

2 Answers 2

3
$\begingroup$

Answering based on the comments of mine and YCor. This answer is also a duplicate of the answer to the same question on Math.SE.

Consider the space of infinite strings on a finite alphabet (so $\{1,2,\ldots,n\}^{\mathbb{Z}}$), and let $f$ be a shift function, either to the left or to the right. The space is a compact space when endowed with the product topology, and in fact it's a metric space with $d(s_1,s_2) = e^{-|n|}$ where $n$ is the smallest index at which two strings $s_1$ and $s_2$ differ. The set of periodic points for $f$ is clearly a dense space, consisting of all periodic strings. It's also not closed, since any nonperiodic string can be approximated arbitrarily well by a periodic one.

$\endgroup$
1
$\begingroup$

There are many simple examples in Hyperbolic Dynamical Systems like Solenoid Attractor, Toral Automorphism, The Geometric Horseshoe of Smale, The Henon Map and extra, that have an invariant Hyperbolic set $\Lambda$ for which the periodic points are dense in $\Lambda$. For example, a famous one is Arnold's cat map $f_A : \frac{\mathbb{R}^2}{\mathbb{Z}^2} \to \frac{\mathbb{R}^2}{\mathbb{Z}^2} $ which is defined by $f_A(\bar{x})= A\bar x$ where: $$A=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$ This example can be extended to $\frac{\mathbb{R}^n}{\mathbb{Z}^n}$ for $n\gt 2$ by considering $A\in GL(n, \mathbb{Z})$ with $|\det(A)|=1$ and the condition that no eigenvalue of $A$ lies on $S^1$. Then the peridc points of $f_A$ are dense in $\frac{\mathbb{R}^n}{\mathbb{Z}^n}$.

Actually all the examples that I mentioned above, have a common property that all of them are topologically conjugate with shift space on $n$ symbols for some $n\ge 2$ on some compact invariant set $\Lambda$. So we may ask the following question:

Question

Is there a continuous and compact dynamical system $(X,f)$ with non-compact and non-empty set of periodic points which is not topologically conjugate with some shift map on $n$ symbols for some $n\ge 2$?

$\endgroup$
1
  • 1
    $\begingroup$ Let me answer the question. A dynamical system conjugated to a shift over finitely many symbols has finite topological entropy. Shifts over alphabets with infinitely many symbols (embedded in some compact set) can have infinite entropy. $\endgroup$
    – coudy
    Mar 19, 2023 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.