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I am dealing with discrete dynamical systems in $\mathbb{R}^n$, and the sequences I refer to are orbits of such systems, i.e. $(x_n)_{n\geq 0}=(f^n(x_0))_{n=0}^\infty$. (However, most theory here applies to sequences in general.)

I have the following two definitions of limit points of sequences and sets respectively ('(.)' will always denote a sequence, while '{.}' will denote a set):

$\textbf{Def 1.}$ A point $p$ is called a $\textbf{limit point}$ of $(x_n)_{n\geq 0}$, if there exists a subsequence $(x_{n_k})_{k\geq 0}$ in $(x_n)_{n\geq 0}$, such that $x_{n_k}\to p$, as $k\to\infty$. Further, we call the set of all limit points of $(x_n)_{n\geq 0}$ the $\textbf{limit set}$ of the orbit of $x_0$, and we denote it $L(x_0)$.

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$\textbf{Def 2.}$ A point $p\in A\subset \mathbb{R}^n$ is called a $\textbf{limit point}$ of $A$ if every neighborhood of $p$ contains a point of $A$ other than $p$.

Ex 1). If an orbit is $k$-periodic, i.e. $(x_n)_{n\geq 0}=(x_0,..., x_k,x_0,...,x_k,x_0,...)$, then $L(x_0)=\{x_0,...x_k\}$.

I also have the following definition of an aperiodic orbit:

Def 3. We call the orbit $(x_n)_{n\geq 0}$ $\textbf{aperiodic}$ if $L(x_0)$ is not finite.

Since no finite set can have a limit point in the sense of Def 2, if the orbit is periodic, the points of $L(x_0)$ are not limit points of the set of points in the orbit.

Questions:

1.) If the orbit is only asymptotically periodic, meaning $L(x_0)$ is finite but the orbit only approaches a periodic orbit, are the points of $L(x_0)$ limit points of $\{x_n\}_{n\in\mathbb{N}}$, by Def 2?

<p><strong>2.)</strong>, If an orbit is aperiodic, are the points of $L(x_0)$ limit points of $\{x_n\}_{n\in\mathbb{N}}$, by Def 2?</p>

Thoughts: I really want to be able to draw the conclusion that an aperiodic orbit visits every open neighborhood of every point in $L(x_0)$, which my intuition tells me it does, but I want to make sure and be able to state more rigorously why and under what circumstances. I have looked at the following posts for some clarity:

Are the limit points of a sequence and of a set defined differently? - I am already familiar with the difference between the definitions so this did not add much.

Limit point of sequence vs limit point of the set containing all point of the sequence - This also mostly talked about the difference between the definitions, but I also realized the following: If an orbit is aperiodic, all points must be distinct, so if we have a convergent subsequence, shouldn't every nbhd. of the limit of that subseq. contain infinitely many different points of $\{x_n\}_{n\in\mathbb{N}}$?

Christopher.L
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1 Answers1

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The answer (to both questions) is yes. You gave the reason yourself:

If an orbit is aperiodic, all points must be distinct, so if we have a convergent subsequence, shouldn't every nbhd. of the limit of that subseq. contain infinitely many different points of $\{x_n\}_{n \in \mathbb{N}}$?

Exactly. By the definition of a limit point of a sequence, for every neighbourhood $U$ of the limit point (call it $\lambda$), the set $\{n \in \mathbb{N} : x_n \in U\}$ is infinite. Since for an aperiodic (that includes asymptotically periodic) orbit, all the points $x_n$ are different, this means that the set

$$U \cap \{ x_n : n \in \mathbb{N}\} = \bigl\{ x_m : m \in \{ n \in \mathbb{N} : x_n \in U\}\bigr\}$$

is infinite, in particular it contains points other than $\lambda$.

Daniel Fischer
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  • Right! Just to clarify, when you wrote "Since for an aperiodic (that includes asymptotically periodic) orbit...", did you mean that an asymptotically periodic orbit is aperiodic, or did you mean to write "...for aperiodic orbits, as well as for asymptotically periodic ones, ..."? If you meant the first, are you using another definition of aperiodic or have I misunderstood something? For example, let $f=x(1-x)$, then $(f^n(1/2))_{n=1}^\infty$ is asymptotically periodic (of period 1, since 0 is a fix point) but not aperiodic since $L(1/2)=0$. – Christopher.L Jun 06 '17 at 16:19
  • Not being familiar with the standard terminology in the field of discrete dynamic systems, I took "aperiodic" to mean "not periodic". And since you mentioned asymptotically periodic orbits separately, I added the parenthetical remark to make clear that the same reasoning also applies to that case. All that matters is that we don't have a truly periodic orbit, so all $x_n$ are different. – Daniel Fischer Jun 06 '17 at 17:30
  • Ok, I defined 'aperiodic' in Def. 3 above, but I can definitely see how one could make such an assumption. I think we usually define it like that because we are trying to exclude even asymptotically periodic behaviour, and focus on the "truly" non-periodical. Thank you for clarifying this. – Christopher.L Jun 06 '17 at 17:42
  • So you did. Selective reading struck again :( – Daniel Fischer Jun 06 '17 at 17:44
  • Definitely no worries, aperiodic sounds like it speaks for itself, I agree. In retrospect, I also think I just somewhat needed to verify what I already knew, but also making everything perfectly clear both to you and myself, so that there were no ambiguities left. – Christopher.L Jun 06 '17 at 17:53