Past exam of the nonlinear systems course of the University of Oxford / 2023 / 3 / a

For the left shift, ${\sigma }^p(s)=s$ precisely when the binary sequence is obtained by repeating its first $p$ symbols. There are exactly $2^p$ such sequences, although some have smaller least period. Hence the set of all periodic sequences is a countable union of finite sets and is at most countable. Passing from periodic points to periodic orbits cannot increase this cardinality.

To prove infinitude, for every integer $p\ge 1$ take the repeated word $(0^{p-1}1{)}^{\infty }$. Its least period is $p$, because its ones are separated by exactly $p$ positions; a smaller positive period would carry a one to a zero. Distinct least periods give distinct periodic orbits. Thus there are countably infinitely many periodic orbits.

The space $\Sigma$ of all infinite binary sequences is uncountable by the Cantor diagonal argument: given any list $s^{(1)},s^{(2)},\dots$, the sequence $t_i=1-s_i^{(i)}$ differs from its $i$th entry at position $i$, so it is absent from the list. Part (i) shows that only countably many sequences are periodic. Their complement is therefore uncountable.

Each forward orbit $\{{\sigma }^n(s):n\ge 0\}$ is countable. If there were only countably many nonperiodic forward orbits, their union together with the periodic points would be countable, contradicting the preceding argument. Hence there are uncountably many nonperiodic orbits, rather than merely uncountably many nonperiodic starting points.

One can also exclude eventual periodicity. An eventually periodic sequence is specified by a finite initial word and a finite repeating word, so only countably many exist. Removing all of them still leaves an uncountable set. This proves the assertion even if “nonperiodic” is understood to exclude orbits that eventually enter a periodic orbit.

Construct one infinite binary sequence $u$ by concatenating every binary word of length one, then every word of length two, then every word of length three, and so on, in lexicographic order within each length. Every finite binary word occurs somewhere in $u$.

Given any target sequence $s$ and any $\eta >0$, choose $N$ such that $2^{-N}<\eta$. The word $s_1\cdots s_N$ occurs in $u$, beginning after some $k$ symbols. Thus ${\sigma }^{k}u$ agrees with $s$ in its first $N$ positions, and the prescribed metric givesEvery open open ball about every point therefore meets the forward orbit of $u$. This forward orbit is dense in $\Sigma$. The argument identifies finite-prefix cylinder sets as arbitrarily small neighborhoods in this metric; no binary-expansion convention for real numbers is needed.