Countable topological dynamical system

Question

Definitions

For the purpose of this post, a (topological) dynamical system is a compact metric space $X$ equipped with a homemomorphism $T:X\to X$.

We say that a subset $S$ of $\mathbb Z$ is relatively dense in $\mathbb Z$ if there is a positive integer $N$ such that for all $a\in \mathbb Z$ the set $\{a+1, a+2, \ldots, a+N\}$ has non-empty intersection with $S$.

Let $x$ be a point in a dynamical system $(X, T)$.

$\bullet$ The orbit of $x$ is defined as $O_x=\{T^nx:\ n\in\mathbb Z\}$.

$\bullet$ We say that a point $x\in X$ is almost periodic if for all neighborhoods $U$ of $x$ in $X$, the set $\{n\in \mathbb Z:\ T^nx\in U\}$ is relatively dense in $\mathbb Z$.

$\bullet$ We say $x$ periodic if the orbit of $x$ is finite.

Clearly, any periodic point is almost periodic.

Question 1

Assuming $(X, T)$ is a dynamical system with $|X|=|\mathbb N|$, is it necessary that every almost periodic point is also periodic?

I do not know the answer to the above question. In fact, I do not know any "good" examples of a countable dynamical system. If you are aware of good examples then please feel free to share.

Question 2

Assuming $(X, T)$ is a dynamical system with $|X|=|\mathbb N|$, is it necessary that $X$ has a periodic point?

The answer to this question is in the affirmative. This is because we know that there is a $T$-invariant probability measure $\mu$ on $X$. Since $X$ is countable, there is a point $x$ in $X$ such that $\mu(x)>0$. Now the orbit of $x$ must be finite, for otherwise, by the $T$-invariance of $\mu$, we would have that $\mu(X)=\infty$.

Can we give an argument with does not go via measure theory and is purely topological in nature?

Answer 1

Here is the answer rewritten to avoid the transfinite induction.

Given a topological space $X$, let $X'$ denote the set of non-isolated points of $X$.

Definition. Given a homeomorphism $T: X\to X$, a point $x\in X$ is called recurrent (with respect to $(X,T)$) if for each neighborhood $U$ of $x$ there exists $n\ge 1$ such that $T^n(x)\in U$.

This condition is weaker than almost periodic.

Lemma 1. Let $X$ be a countable compact metrizable space, $T: X\to X$ a homeomorphism. Then every recurrent point $x\in X$ is periodic.

Proof. Consider the collection ${\mathcal I}_x$ of all compact $T$-invariant subsets of $X$ containing $x$. Let $C_x$ denote the intersection of all members of ${\mathcal I}_x$. Clearly, $C_x\in {\mathcal I}_x$. I claim that $x$ is an isolated point of $C_x$. Indeed, since $C_x$ is countable and compact metrizable, it has some isolated points, $C'_x\ne C_x$. If $x\in C'_x$ then $C'_x\in {\mathcal I}_x$ and $C'_x$ is strictly smaller than $C_x$, which is a contradiction. Hence, $x$ is isolated in $C_x$. The point $x$ is still recurrent with respect to $(C_x,T)$. Since $x$ is isolated in $C_x$, $\{x\}$ is a neighborhood of $x$ in $C_x$. Hence, by recurrence, there exists $n\ge 1$ such that $T^n(x)\in \{x\}$, i.e. $T^n(x)=x$, i.e. $x$ is $T$-periodic. qed

This answers Question 1. To answer Question 2, I will prove a stronger result:

Lemma 2. Let $X$ be a nonempty compact metrizable topological space, $T: X\to X$ is a homeomorphism. Then $X$ contains recurrent points. Equivalently, $X$ contains a $T$-invariant compact nonempty subset $X_0$ such that every $T$-orbit in $X_0$ is dense in $X_0$.

Proof. Consider the poset ${\mathcal I}$ of all nonempty $T$-invariant compact subsets of $X$ (with the partial order given by inclusion). Clearly, the intersection of members of each totally ordered (nonempty) subset in ${\mathcal I}$ belongs to ${\mathcal I}$. Hence, by Zorn's Lemma, ${\mathcal I}$ contains a minimal element $X_0$. By the minimality, every $T$-orbit in $X_0$ is dense (otherwise, take the closure of a non-dense $T$-orbit in $X_0$). qed

Combining the two lemmata, we see that if $X$ is countable, compact, metrizable, nonempty, then for each homeomorphism $T: X\to X$, there exists a $T$-periodic point.