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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!

2 Answers 2

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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?

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    We have the following property: $\omega \in \Sigma_2$ is positively recurrent iff every block in $\omega$ occurs infinitely often. Hence your word, $00...0 0010 0010 ... 0010$, is positively recurrent. But I don't understand why this word is not negatively recurrent. Can you explain me? Thank you! – 2017-03-30
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    I think you might be mistaking the definition of positive recurrent with just recurrent. Otherwise wouldn't the result you stated imply that left recurrence and right recurrence are equivalent? That's what you're trying to prove is false! – 2017-03-30
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    The word isn't left recurrent because the subword $0010$ does not appear anywhere to the left of the origin. – 2017-03-30
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    So, the word is positively recurrent because the subword 0010 occurs infinitely often to the right of the origin? – 2017-03-30
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    Not just the word $0010$ but every word that appears in $\omega$, because of how we have constructed $\omega$. – 2017-03-30
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    In general, how can be described positively recurrent but not negatively recurrent points? I have to show that such points are dense in $\Sigma_2$. So I have to prove that in every symmetric cylinder there exists such a point, but I don't know how to describe them in general. Thank you! – 2017-03-30
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    All you have to do is show that for every finite word, there exists a right recurrent bi-infinite word which is not left recurrent that contains the finite word as a subword. You can use the exact same construction as in this answer to show that such a bi-infinite word exists. – 2017-03-30
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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...