Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In the book "Recurrence in Ergodic Theory and Combinatorial Number Theory", 1981, Furstenberg introduced the notion of central sets.

He proved in Theorem 8.8 that in each finite partition of $\mathbb{N}$, $\mathbb{N}=B_1 \cup B_2 \cup \dots \cup B_q$ one of the sets $B_j$ contains a central set. Then he states without proof that the same is true for partitions of central sets and that one can prove this "along the same lines" as Theorem 8.8 (p. 163).

I know that this fact follows immediately from the characterization of central sets by minimal idempotent ultrafilters. However, this is obviously not the proof that Furstenberg had in mind 1981 as this characterization was not available then. Thus my question:

How can one prove in the fashion of Theorem 8.8, i.e. using only tools recurrence in dynamical systems, that for each finite partition of a central set $S$, $S=B_1 \cup B_2 \cup \dots \cup B_q$ one of the $B_j$ contains a set that is again central?


For the sake of reference I include the necessary definitions and the proof of Theorem 8.8 below.

Let $(X,T)$ be a dynamical system. $x_1,x_2\in X$ are proximal if for some sequence $n_k\to \infty$, $d(T^{n_k}x_1,T^{n_k}x_2) \to 0$. A point $x\in X$ is called uniformly recurrent if for each neighborhood $U$of $x$, the sequence of values $n_1 < n_2 < \dots$, for which $T^{n_k}x\in U$, satisfies that $\{ n_{k+1}-n_k \}$ is bounded.

As set $S$ is called central if there is a $x,y\in X$ and a neighborhood $U$ of $y$ such that $y$ is uniformly recurrent and proximal to $x$ and $S=\{ n : T^nx\in U \}$.

The only tools we need for the proof Theorem 8.8 is then the Auslander-Ellis Theorem (Theorem 8.7) which reads as follows. If $(X,T)$ is a dynamical system an $X$ is compact. Then every point $x$ in $X$ is proximal to a uniformly recurrent point.

Proof of Theorem 8.8:

Consider the dynamical system with $(\Omega,T)$ with $\Omega =\{ 1, \dots, q \}^\mathbb{Z}$ and $T$ the shift $T\omega(n)=\omega(n+1)$. Let $\xi\in\Omega$ be a point with $\xi(n)=i$ implies $n\in B_i$ for $n\in\mathbb{N}$.

Let now $\eta\in\Omega$ be a uniformly recurrent point proximal to $\xi$. This point exists by the Auslander-Ellis Theorem. Let $j=\eta(0)$ and set $U=\{ \omega : \omega(0)=j \}$ and let $S=\{n: T^n\xi\in U \}$. Clearly $S$ is central and if $n\in S$ then $\xi(n)=T^n\xi(0)=j$, so that $n\in B_j$ and thus $S\subseteq B_j$.

share|improve this question
    
How does the definition of "proximal" make sense when the space $X$ is compact? The metric has to be bounded then. Or do you mean 0 instead of infinity in that definition? –  Gabor Szabo Feb 1 '13 at 11:13
    
@Gabor: your right that should be 0, thank you. It is now corrected. –  alexod Feb 1 '13 at 12:21
    
Maybe I'm missing something here, but why not to look in the system $\{1,\ldots,q\}^{S}$, $S=\{s_n\}$ and the shift will correspond to advancing to the successor the number inside $S$, $s_{n}\mapsto s_{n+1}$. This will produce a central set $S′ \subset \mathbb{N}$ which will correspond to $\{s_{n'}\mid n'\in S'\}$ just as in Hillel's proof –  Asaf Feb 1 '13 at 13:38
    
Asaf: The sets $S,S′$ are clearly central. But I do not see why this implies that $\{ s_{n′} \mid n′\in S′ \}$ is central. However, this might be what Hillel was thinking about. Maybe there is just an argument missing –  alexod Feb 1 '13 at 15:09
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.