= Solution
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 $p\geq1$ 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 \b[there are countably infinitely many <periodic orbits>].
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