1

I understand what is meant by a limit set but I don't understand what it would mean for this set to be empty. Could someone provide an example?

usainlightning
  • 1,217

2 Answers2

3

Well, in order for the limit set to be empty, you want $\{f^n(x): n\in\mathbb{N}\}$ to have no cluster points for any $x\in X$. So basically, you want $f$ to keep pushing points further and further away. Can you think of any nice continuous function $f$ on $\mathbb{R}$ which "keeps moving to the right?"

Noah Schweber
  • 245,398
  • Do you mean a function like $y=$log$x$, so the limit set can only contain real numbers?( not $\infty$) – usainlightning Aug 28 '15 at 09:59
  • 1
    But $\log$ isn't defined on all of $\mathbb{R}$. Also, if the space we're living on is $\mathbb{R}$, then the limit set by definition can only contain real numbers. Think about the function $f(x)=x+1$ . . . – Noah Schweber Sep 03 '15 at 18:09
  • 1
    So because $f(x)$ tends to infinity it's limit is not a real number, so it has an empty limit set? – usainlightning Sep 04 '15 at 18:35
  • Bingo! (But let's be a bit careful - what we need to say is "For each $x$, the sequence $(x, f(x), f(f(x)), . . .)$ tends to $\infty$.") Note how crucial not having $\infty$ is. Exercise: if your underlying space is compact, then any $f$ at all - continuous or not - has a nonempty limit set. – Noah Schweber Sep 04 '15 at 18:50
  • @NoahSchweber This is an interesting question and answer (+1). Makes me wonder what happens if we define the function over the extended reals? Wouldn't $\pm \infty$ be legitimate members and then the limit set would not be empty? (not sure about this). If so, would one need to resort to fractional-dimensional objects (e.g., strange attractors) to ensure that no limit set exists while at the same time remaining bounded? –  Sep 09 '15 at 10:40
  • @Bey The extended reals are compact, so any $f$ at all would have a nonempty limit set. – Noah Schweber Sep 09 '15 at 14:15
0

Let us remember the definition of flow:

Let $E \subseteq \mathbb{R}^{n}$ open and $f \in C^{1}\left(E\right)$. Let >$D\left(x\right)$ maximal interval corresponding to $x$. We define

$\Omega=\left\{\left(t,x\right)\in \mathbb{R}\times E \: : \: t \in D\left(x\right) \right\}.$

Then, we define the flow associated to $f$ as the funtion $\Phi:\Omega\rightarrow E$ such that

$\Phi\left(0,x\right)=x$

$\frac{d\Phi\left(t,x\right)}{dt}=f\left(\Phi\left(t,x\right)\right)$

In the case of continuous dynamical systems, especially for flows, the $\omega$**-limit set** for a given $x$ is defined by:

$\omega \left(\Phi\right)=\left\{y \in E \: : \: \mbox{There is } \left(t_{n}\right)_{n \in \mathbb{N}} \mbox{ such that } \lim_{n\rightarrow\infty}t_{n}=\infty \mbox{ and } \lim_{n\rightarrow\infty}\Phi\left(t_{n},x\right)=y\right\}$

This set can be empty, for example, we consider the flow defined by $\Phi\left(t,x\right)=x+t$ associated to the constant function $f\left(x\right)=1$.

Diego Fonseca
  • 3,072