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Consider the set $\Sigma=\{0,1\}^\mathbb{Z}$, i.e. the space of bi-infinite sequences of 0's and 1's, and the left-shift $\sigma:\Sigma\to\Sigma$. Define a distance in $\Sigma$ as follows: $$d(s,t)=\sum_{i\in\mathbb{Z}}\dfrac{|s_i-t_i|}{2^{|i|}}.$$

Prove that

(a) there is a countable infinite number of periodic orbits of period arbitrarily large,

(b) there is a non-countable infinite number of non-periodic orbits,

(c) there is a dense orbit.

I think I can prove the first claim by using that for every $k\in\mathbb{N}$ there are exactly $2^k$ periodic orbits of period k, but how can I solve the second claim? Maybe I could say that $\Sigma$ is non-countable and hence (b) has to hold? And how can I prove the last part?

Marc
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1 Answers1

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Extended HINT: For (b) let $\alpha=\langle n_k:k\in\Bbb N\rangle$ be any strictly increasing sequence of positive integers. Define $s^\alpha\in\Sigma$ by

$$s^\alpha_i=\begin{cases} 1,&\text{if }i=0\\ 1,&\text{if }i=\sum_{j=0}^kn_j\text{ for some }k\in\Bbb N\\ 0,&\text{otherwise}\;. \end{cases}$$

Show that each such $s^\alpha$ has a non-periodic orbit.

For (c) you need an element of $\Sigma$ that contains every possible finite string of zeroes and ones infinitely often. There are only countably infinitely many such finite strings, so you can enumerate them as $\{\varphi_n:n\in\Bbb N\}$. For each $\langle m,n\rangle\in\Bbb N\times\Bbb N$ let $\psi_{\langle m,n\rangle}=\varphi_n$. Now use the pairing function $\pi:\Bbb N\times\Bbb N\to\Bbb N$ to let $\rho_n=\psi_{\pi^{-1}(n)}$ for each $n\in\Bbb N$; $\langle\rho_n:n\in\Bbb N\rangle$ is then a sequence of all possible finite sequences of zeroes and ones in which each finite sequence appears infinitely often. Concatenate and pad to the left with zeroes to get the desired member of $\Sigma$.

Brian M. Scott
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