Positive recurrent point, but not negative recurrent in the shift space

Question

Consider the shift space $\Sigma_2 = \{\omega = (...\omega_{-1} \; \omega_0 \; \omega_1, ...) : \omega_i \in \{0, 1\}, \; i \in \mathbb{Z}\}$ and the shift map $\sigma : \Sigma_2 \to \Sigma_2, \; \sigma(\omega)_i = \omega_{i+1}$.

$\omega \in \Sigma_2$ is $\textit{positive recurrent}$ if $\; \exists n_k \nearrow \infty$ such that $\omega = \underset{n_k \to \infty}{lim} \sigma^{n_k}(\omega)$.

$\omega \in \Sigma_2$ is $\textit{negative recurrent}$ if $\; \exists n_k \searrow -\infty$ such that $\omega = \underset{n_k \to -\infty}{lim} \sigma^{n_k}(\omega)$.

The metric we consider is the following: $$d_\lambda(\omega, \eta) = \sum_{i=-\infty}^{\infty} \frac{|\omega_i-\eta_i|}{\lambda^{|i|}}.$$

Can someon give me a hint on how to find a point in $\Sigma_2$ that is positive recurrent, but is not negative recurrent?

Thank you!

Answer 1

I'll try to give hints only.

Suppose we put some arbitrary word at the origin. Say $u_0 = 00.10$. If our word is right recurrent, then certainly that length-4 word at the origin should appear infinitely often to the right. So let's extend our word to the right by that word to get $u_1 = 00.100010$. Ok, but then how do we extend to the left? Well we don't want the word to be left-recurrent so we should probably avoid the word that is at the origin. The word $0000$ doesn't appear at the origin so let's extend by that to the left to get $u_2 = 000000.100010$.

Can you see what the next step $u_3$ would be? How about $u_4$?

Write a rule for $u_{2k}$ in terms of $u_i$ where $i\leq 2k$. Then do the same for $u_{2k+1}$.

Why does this word have the properties that you asked for?

Answer 2

Hint: To the negative side choose 1's and let $\omega_0=0$. Then it is not negative recurrent. On the positive side construct recursively the sequence so that it contains arbitray long finite sequences of the form $(\omega_{-N},...,\omega_{N-1})$, i.e. given the construction for a given $N$, add that word from $N$ to $3N-1$ and redo the recursive step for $N'=3N$, etc...