Infinite closed shift-invariant set of binary sequences with zero entropy?

Question

Does there exist an infinite closed shift-invariant $X \subset \{0,1\}^\Bbb Z$ with zero topological entropy? How to think of an example? Will periodic points of shift$|_X$ have zero entropy?

Answer 1

The set $$ X_k := \{x: \text{there are at most $k$ occurrences of $1$ in $x$}\} $$ (for any $k\geq 1$) is shift-invariant and closed, and has topological entropy $0$. It is clearly infinite, though only countably infinite.

If you prefer an uncountably infinite set, recall that the entropy of a shift space $X$ is really "average entropy per position". So, you can still have infinite amount of information in your sequence spread enough in such a way that the average amount of information per position is $0$.

To construct an example, for a non-decreasing function $g:\mathbb{Z}^+\to\mathbb{Z}^+$, let $$ X_{g(\cdot)}:= \{x: \text{for each $i\in\mathbb{Z}$, $n\in\mathbb{Z}^+$, the number of $1$s in $x|_{[i,i+n)}$ is at most $g(n)$}\} \;. $$ Now, choose $g(\cdot)$ such that

• $g(n)\to\infty$ as $n\to\infty$ (so that $X_{g(\cdot)}$ is uncountable: exercise!), but
• $\log\sum_{k\leq g(n)}\binom{n}{k}=o(n)$ as $n\to\infty$. (This means, the number of possible words of length $n$ appearing on a sequence $x\in X_{g(\cdot)}$ is asymptotically negligible compared to $2^n$.)

Choosing $g(n):=\lfloor\log (n+1)\rfloor$ or $g(n):=\lfloor\sqrt{n}\rfloor$ would certainly do.