Let $w \in \{0,1\}^\mathbb N$ be some word.

Let $\Sigma \subset \{0,1\}^\mathbb N$ be the set of all words obtained from $w$ by deleting an initial segment of $w$. So for every $n$ we have in $\Sigma $ the word obtained by removing the first $n$ characters from $w$. (This is the orbit under a shift.)

What is the closure $\mathrm {Cl}(\Sigma)$ of $\Sigma$ in this product space?

The usual subbase from the closed sets is any set where all coordinates by finitely many can be chosen freely, since each $\{0,1\}$ is discrete, but $\Sigma$ does not seem to be such a set.

By "describing the closure" I mean two goals:

  1. Prove that $\mathrm {Cl} (\Sigma)$ too, like $\Sigma$, has the property that for every word $w \in \mathrm {Cl} (\Sigma)$ we might remove any initial segment and stay in $\mathrm {Cl} (\Sigma)$.

  2. Let $B_n$ be the number of distinct subwords (consecutive substrings) of length $n$ in $\mathrm {Cl} (\Sigma)$ . Let $A_n$ be the corresponding number of $\Sigma$ itself. What I want to prove is that $\lim \frac 1n \log A_n = \lim \frac 1n \log B_n$. (But more might be true.)

  • 1
    $\begingroup$ I don't know about 2. but 1. is essentially obvious: let $v\in \mathrm{Cl}(\Sigma)$, and $n$ be any integer; call $u$ the shift of $v$ by $n$ letters. Let $U$ be a neighbourhood of $u$; we may assume it is of the form $\{l, (l_1,...,l_k) = (u_1,....,u_k)\}$. Then $\{l, (l_1,...,l_{n+k}) = (v_1,...,v_{n+k})\}$ is a neighbourhood of $v$ so it contains a shift of $w$ by $j$ letters, say $f$. But then the shift of $f$ by $n$ letters (which is still a shift of $w$) is in $U$, which shows that $u$ is in the closure $\endgroup$ – Max May 1 '18 at 15:12

For the first question, you only need to observe that shift-by-one (i.e., removing the first symbol) is a continuous map on $\{0,1\}^{\mathbb{N}}$. This is because the pre-image of every cylinder set (i.e., an element of the sub-base) is itself a union of cylinder sets.

For the second question, note that the finite subwords of the elements of $\operatorname{Cl}(\Sigma)$ are precisely the finite subwords of the elements of $\Sigma$. Indeed, if a word $u$ appears in an element $w$ of $\operatorname{Cl}(\Sigma)$, the cylinder set $[u]$ consisting of all infinite words $w'\in\{0,1\}^{\mathbb{N}}$ in which $u$ appears in the same position as in $w$ must intersect $\Sigma$. Hence, $u$ must also appear in an element of $\Sigma$. This means $A_n=B_n$ for each $n$.

In fact, the latter is a characterization of the closure: $\operatorname{Cl}(\Sigma)$ is the set of all words $w\in\{0,1\}^{\mathbb{N}}$ such that every finite subword of $w$ appears also in an element of $\Sigma$.


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