Proof. To each $\alpha = \omega ^{\zeta }$ or $\alpha = \omega ^{\zeta }+1$ for $\zeta \geq 1$ , we associate a countable $X \subset [0,1]$ , such that for $\alpha = \omega ^{\lambda + n}$ , the close-up rank of E is $\lambda + 2n$ , and for $\alpha = \omega ^{\lambda + n} + 1$ , it is $\lambda + 2n + 1$ .

Let $\alpha $ be as above and let $A = \{0,1,\ldots ,\alpha \}$ with the order topology. As a Von Neumann ordinal, $A = \alpha +1$ , but the intended reading of $\phi (A)$ is $\phi (A) = \{\phi (a) \;|\; a \in A\}$ . Take an injective order homomorphism $\phi : A \to [0,1]$ , so that $\phi (0) = 0$ , $\phi (\alpha ) = 1$ . Identify $0 \equiv 1$ to obtain a continuous function $\phi : A \to \phi (A) \subset \mathbb {S}^1 = [0,1]/{\equiv }$ , and let $\phi (A) = X$ . The reason for arranging the ordinal on a sphere is simply to compactify the non-compact ordinals, and the reason for wanting non-compact ordinals is to allow the last step of the close-up process to be a transitive closure step.

While we do the construction on $\mathbb {S}^1 = [0,1]/{\equiv }$ rather than $[0,1]$ , we note that any non-full closed subset of $\mathbb {S}^1$ is homeomorphic to a closed subset of $[0,1]$ .

Let $E \subset (\mathbb {S}^1)^2$ be the union of the diagonal relation $\Delta $ and the successor and predecessor relations, that is,

$$ \begin{align*} E = & \{(\phi(\beta), \phi(\beta)), (\phi(\beta), \phi(\beta+1)), (\phi(\beta+1), \phi(\beta)) \;|\; \beta+1 < \alpha \}. \end{align*} $$

Observe that if $\alpha $ is a successor, we do not join $\phi (\alpha ) = \phi (0)$ with its ‘predecessor’ $\phi (\alpha -1)$ . It is easy to see that X and E are closed. Consider the close-up process for this relation, starting from $E^{(1)} = E$ .

Note that at all times, $E^{(\lambda )}$ for $\lambda \geq 1$ is symmetric and reflexive, so we simply have to figure out which pairs $(\phi (\beta ), \phi (\beta + \gamma ))$ are contained in $E^{(\lambda )}$ for each ordinal $\lambda \geq 1$ . What happens is that only the length of the jumps $\gamma $ matters. Alternately, we are allowed to first take less than $\omega $ steps with the current relation (a transitive closure step), and then to close up these jumps into a single superjump consisting of $\omega $ many previous jumps (a topological closure step).

Concretely, we prove an exact characterization of $E^{(\lambda )}$ by transfinite induction: every ordinal can be written in a unique way as $\lambda + 2n$ (even ordinals) or $\lambda + 2n + 1$ (odd ordinals) where $n \in \mathbb {N}$ and $\lambda $ is a limit ordinal (interpreting $0$ as a limit ordinal). For all $\lambda , n$ , we prove by induction for all ordinals $\lambda +2n+b$ greater than $0$ ( $n \in \mathbb {N}, b \in \{0,1\}$ ) that

$$\begin{align*}E^{(\lambda + 2n)} = \{ (\phi(\beta), \phi(\beta + \gamma)), (\phi(\beta + \gamma), \phi(\beta)) \;|\; \gamma < \omega^{\lambda + n}, \beta + \gamma < \alpha \}, \end{align*}$$

$$\begin{align*}E^{(\lambda + 2n + 1)} = \{ (\phi(\beta), \phi(\beta + \gamma)), (\phi(\beta + \gamma), \phi(\beta)) \;|\; \gamma \leq \omega^{\lambda + n}, \beta + \gamma < \alpha \}. \end{align*}$$

The claim for $E^{(1)}$ is (setting $\lambda = 0, n = 0, b = 1$ ) equivalent to having (up to symmetry) $E^{(1)} \ni (\beta , \beta +\gamma )$ if and only if $\gamma < 2$ , $\beta + \gamma < \alpha $ . This is precisely the definition of $E^{(1)} =~E$ .

For the induction steps, suppose $E^{(\lambda +2n)}$ is of the claimed form, and consider $E^{(\lambda +2n+1)}$ . This is a topological closure step, that is, $E^{(\lambda +2n+1)} = \overline {E^{(\lambda +2n)}}$ . Taking the closure of the pairs given by the inductive assumption, we clearly have

$$\begin{align*}\{ (\phi(\beta), \phi(\beta + \gamma)) \;|\; \gamma \leq \omega^{\lambda + n}, \beta + \gamma < \alpha \} \subset E^{(\lambda+2n+1)}. \end{align*}$$

If $\beta> 0$ , then $(\phi (\beta ), \phi (\beta + \gamma )) \notin E^{(\lambda +2n+1)}$ for $\gamma \geq \omega ^{\lambda + n}+1$ , $\beta + \gamma < \alpha $ since otherwise $(\phi (\theta ), \phi (\beta +\gamma )) \in E^{(\lambda +2n)}$ for $\theta \leq \beta $ arbitrarily close to $\beta $ and $\gamma \geq \omega ^{\lambda + n}+1$ (note that $\beta + \omega ^{\lambda + n}+1$ is isolated), contradicting the inductive assumption.

If $\beta = 0$ , then there may be additional limits on the right end of $[0,1]/{\equiv }$ . In this case, suppose we have $(\phi (\beta ), \phi (\beta + \gamma )) = (0, \phi (\gamma )) \in E^{(\lambda +2n+1)}$ for $\omega ^{\lambda + n}+1 \leq \gamma < \alpha $ . To obtain such pairs, we must have $(\phi (\delta ), \phi (\theta )) \in E^{(\lambda +2n)}$ for $\theta < \alpha $ arbitrarily close to $\alpha $ and $\delta \leq \gamma $ arbitrarily close to $\gamma $ . If $\alpha = \omega ^{\zeta }+1$ , then $\alpha $ cannot be approximated from below, so this is impossible.

If $\alpha = \omega ^{\zeta }$ , then observe that since $\gamma < \alpha $ , we have $\omega ^{\lambda + n}+1 \leq \gamma < \alpha $ , so since $\alpha = \omega ^{\zeta }$ , we have $\omega ^{\lambda + n}+1 \leq \gamma \leq \omega ^{\zeta '} \cdot m$ for some $\zeta ' < \zeta $ , $m \in \mathbb {N}$ (here we simplify the successor and limit case of $\zeta $ into one). Observe that we even have $\omega ^{\zeta '} \cdot \omega \leq \alpha $ . Now, for any $\delta \leq \gamma $ and $\omega ^{\zeta '} \cdot 2m \leq \theta < \alpha $ , we have

$$\begin{align*}\delta + \omega^{\lambda + n} + 1 \leq \omega^{\zeta'} \cdot 2m \leq \theta \end{align*}$$

by left distributivity and monotonicity of ordinal multiplication, which means $(\phi (\delta ), \phi (\theta )) \notin E^{(\lambda +2n)}$ , and thus $(\phi (\delta ), \phi (\theta )) \notin \overline {E^{(\lambda +2n)}}$ . This concludes the analysis of the step from $\lambda +2n$ to $\lambda +2n+1$ .

Suppose then that $E^{(\lambda +2n+1)}$ is of the claimed form and consider $E^{(\lambda +2n+2)}$ . Since $(\phi (\beta ), \phi (\beta + \gamma )) \in E^{(\lambda +2n+1)}$ for all $\beta $ , $\beta + \gamma < \alpha $ with $\gamma \leq \omega ^{\lambda + n}$ , taking the transitive closure, we have $(\phi (\beta ), \phi (\beta + \gamma )) \in E^{(\lambda +2n+2)}$ for any $m \in \mathbb {N}$ and $\gamma \leq \omega ^{\lambda + n} \cdot m$ with $\beta + \gamma < \alpha $ . This is indeed equivalent to having $(\phi (\beta ), \phi (\beta + \gamma )) \in E^{(\lambda +2n+2)}$ for any $\gamma < \omega ^{\lambda + n} \cdot \omega = \omega ^{\lambda + n + 1}$ as required.

To see that $E^{(\lambda +2n+2)}$ does not contain any pairs $(\phi (\beta ), \phi (\beta + \gamma ))$ with ${\gamma> \omega ^{\lambda + n} \cdot \omega} $ , $\beta + \gamma < \alpha $ , consider any tuple $(\phi (\beta _1), \phi (\beta _2), \ldots , \phi (\beta _k))$ with $(\phi (\beta _i), \phi (\beta _{i+1})) \in E^{(\lambda +2n+1)}$ , $\beta _i \neq \beta _{i+1}$ for all i, and $\beta _1 < \beta _k$ . If $\beta _j$ is minimal among of the $\beta _i$ , then it easily follows from the inductive assumption that

$$\begin{align*}\beta_k \leq \beta_j + \omega^{\lambda + n} \cdot k \leq \beta_1 + \omega^{\lambda + n} \cdot k < \beta_1 + \omega^{\lambda + n} \cdot \omega. \end{align*}$$

This concludes the analysis of the step from $\lambda +2n+1$ to $\lambda +2n+2$ .

Finally, we look at the case where $\lambda $ is an infinite limit ordinal. By definition, $E^{(\lambda )} = \bigcup _{\delta < \lambda } E^{(\delta )}$ . We need to show that $E^{(\lambda )}$ contains $(\phi (\beta ), \phi (\beta +\gamma ))$ for $\beta +\gamma < \alpha $ if and only if $\gamma < \omega ^{\lambda }$ . Since $\gamma < \omega ^{\lambda }$ is equivalent to $\text { there exists } \delta < \lambda : \gamma < \omega ^{\delta }$ , this follows from the inductive assumption on the relations $E^{(\delta )}$ . This concludes the analysis of the limit steps.

Now observe that by the above characterization, $E^{(\lambda +2n+b)} = X^2$ if and only if $(0,\gamma ) \in E^{(\lambda +2n+b)}$ for all $\gamma < \alpha $ . This happens if and only if $b = 0$ and $\gamma < \alpha \! \implies \! \gamma < \omega ^{\lambda + n}$ , that is, $\alpha \leq \omega ^{\lambda + n}$ ; or $b = 1$ and $\gamma < \alpha \!\implies \! \gamma \leq \omega ^{\lambda + n}$ , that is, $\alpha \leq \omega ^{\lambda + n} + 1$ .

Now, to prove the original claim, for $\lambda = 1$ , we can realize close-up rank $1$ by setting $E = X^2$ for any choice of X. For larger countable ordinals, in the above construction, we obtained for the choice $\alpha = \omega ^{\lambda + n}$ that the close-up rank of E is $\lambda + 2n$ , and that for the choice $\alpha = \omega ^{\lambda + n}+1$ , it is $\lambda + 2n + 1$ .

For the last sentence, observe that in the case of odd $\lambda + 2n + 1$ , the pairs added at the last step are just those joining some $\phi (\beta )$ for $\beta < \alpha -1$ with $\phi (\alpha -1)$ . We can realize any such limit through pairs $(\phi (\beta ), \phi (\beta _i)) \in E^{(\lambda +2n)}$ , where $\beta _i \nearrow \alpha -1$ .

Now consider any $(a_1,\ldots ,a_k,b_1,\ldots ,b_k) \in (E^{[k]})^2$ , where we can take the $a_i$ to be sorted in increasing order of $\phi ^{-1}(a_i)$ , and similarly for the $b_i$ . If $(a_1,b_1) \in E^{(\lambda +2n)}$ , then

$$\begin{align*}(a_1,\ldots,a_k,b_1,\ldots,b_k) \in (E^{(\lambda+2n)})^{[2k]} \cap (E^{[k]})^2 \subset \overline{(E^{(\lambda+2n)})^{[2k]} \cap (E^{[k]})^2} \end{align*}$$

follows from transitivity of $E^{(\lambda +2n)}$ . Otherwise, by the above paragraph and up to symmetry, we can assume $(a_1, b_1)$ is the limit of $(\phi (\beta ), \phi (\beta _i))$ . We have $b_i = b_1$ for all i since $\phi (\alpha -1)$ is not in E-relation with any other point. We have $a_i \in \{ a_1, \phi (\phi ^{-1}(a_1) + 1) \}$ for all i (recall that the original relation E is just the successor relation, symmetrized and reflexivized), and by the characterization of $E^{(\lambda +2n)}$ , we then have $(a_i, \phi (\beta _i)) \in E^{(\lambda +2n)}$ for any $a_i$ , as required.

Proof. We need to show that the orbit-closure of x contains a Toeplitz point with the same language as x. Let $w_1 = x|_{[-1,1]}$ and let $j_1$ be such that an arithmetic progression of $w_1$ s with difference $n_1$ begins at $j_1$ . Replace x by $x^1 = \sigma ^{j_1}(x)$ , so that we still have $w_1 = x^1|_{[-1,1]}$ , but now the central three coordinates are in an arithmetic progression. From now on, restrict to shifting by multiples of $n_1$ .

By the original assumption, the central word $x^1|_{[-n_1-|j_1|-2, n_1+|j_1|+2]}$ is in an arithmetic progression beginning at some $j_2$ with difference $n_2$ , we may take $n_1 \;|\; n_2$ . Let $x^2 = \sigma ^{h_2 n_1}(x^1)$ , where and where denotes integer division discarding remainder. Since we shifted by a multiple of $n_1$ , in $x^2$ , the word at $[-1,1]$ still lies in an arithmetic progression with difference $n_1$ . However, since we shifted by a number with distance at most $n_1$ from $j_2$ , now the subword $x^2|_{[-2-|j_1|, 2+|j_1|]}$ lies in an arithmetic progression with difference $n_2$ . Observe that this contains the word $x|_{[-2, 2]}$ .

In $x^2$ , we have inserted Toeplitz periods for more coordinates, without modifying the periods we already introduced for $x^1$ , and also introduced larger subwords of x into the periodic area. An obvious induction gives a Toeplitz point with the same language as that of x in the limit.

Proof. Define the subshift $X' \subset \{0,1,\#\}^{\mathbb {Z}}$ by requiring that in every point, there is an arithmetic progression $\{3k+i \;|\; k \in \mathbb {Z}\}$ containing only $\#$ , such that $\{3k+i+1 \;|\; k \in \mathbb {Z}\}$ is a constant sequence containing only $0$ or only $1$ , and recursively $\{3k+i+2 \;|\; k \in \mathbb {Z}\}$ is a point of $X'$ .

This indeed describes a unique subshift whose typical points look like

$$\begin{align*}\ldots \#z_0\#\#z_0z_1\#z_0\#\#z_0\#\#z_0z_1\#z_0z_2\#z_0\#\#z_0z_1\#z_0\#\#\ldots, \end{align*}$$

where $z \in \{0,1\}^{\mathbb {N}}$ , and we associate a sequence of bits to every point in $X'$ by locating the unique infinite $3$ -progression of $\#$ s (which can be deduced from any word of length $3$ ), outputting the bit to the right of one (thus any) of them, and recursively extracting the rest of the bits from the third $3$ -progression. Let $\phi : X' \to \{0,1\}^{\mathbb {N}}$ extract this sequence of bits, so $\phi $ is continuous, and $\phi (x) = \phi (\sigma (x))$ for all $x \in X'$ .

Since Z is compact, metrizable, and zero-dimensional, we can take (up to homeomorphism) $Z \subset \{0,1\}^{\mathbb {N}}$ a closed set such that no point in Z is eventually constant, equivalently, $01$ appears infinitely many times in every $z \in Z$ . Define $X = \phi ^{-1}(Z)$ , which is clearly a subshift. A simple computation shows that the number of words of length n in X (even in $X'$ ) is $O(n^{\log _3 6})$ , so X has zero entropy.

The map $\phi $ is surjective and continuous by definition, and it is closed since X is compact and Z Hausdorff, so it is a quotient map from X to Z.

Let us now analyze the dynamical structure of Z. Call the unique infinite $3$ -progression of $\#$ s in a point of X the $0$ -skeleton. Recursively, we refer to the $0$ -skeleton of the point of $X'$ obtained after i iterations of the recursive extraction process as the i-skeleton. The skeleton refers collectively to the family of all the i-skeletons, more precisely, the skeleton of $x \in X$ is $\pi (x)$ , where $\pi (\#) = \#$ , $\pi (0) = \pi (1) = 0$ .

Dynamically, extraction of the skeletons gives an almost-one-to-one subshift cover of the $3$ -odometer $(\mathbb {Z}_3, (a \mapsto a+1))$ (where $\mathbb {Z}_3$ is the space of $3$ -adic integers), more precisely, $\pi (X)$ is a two-to-one, and almost one-to-one, cover of the $3$ -odometer, by the obvious map $\psi : \pi (X) \to \mathbb {Z}_3$ that records the offsets of the $\#$ -progressions in an intertwining way. What happens is that a typical $\psi (\pi (x))$ determines the skeleton $\pi (x)$ entirely. If it does not, then there is a single ‘hole’ left after filling in the $\#$ s. If there is a hole left, then $|\psi ^{-1}(\psi (\pi (x)))| = 2$ .

It is easy to see that if $\psi ^{-1}(\psi (\pi (x)))$ is a singleton, then for any $z \in Z$ , there is exactly one point $x' \in X$ with $\phi (x') = z$ and $\pi (x') = \pi (x)$ , since after filling in the skeleton and the bits to the right of it, there is no hole left to insert a $\#$ , thus no hole to insert a bit either. If there is a hole left, then we can insert any of $0,1,\#$ , but all other cells are determined by $\psi (\pi (x))$ and $\phi (x)$ .

It follows that the subshift X itself is a three-to-one and almost-one-to-one subshift cover of the system $\mathbb {Z}_3 \times Z$ under $(\alpha , z) \mapsto (\alpha +1, z)$ , where $\mathbb {Z}_3$ denotes the $3$ -odometer.

We now show the openness of $\phi $ . Suppose $x \in [u]$ and $\phi (x) = z$ . Pick $x' \in [u]$ such that $\phi (x') = z$ and

$$\begin{align*}\text{ for all } y, y': \pi(y') = \pi(y) = \pi(x') \wedge \phi(y) = \phi(y')\! \implies\! y = y'. \end{align*}$$

This is possible because $[u]$ determines only part of the skeleton of x, and we may modify the rest to find $x'$ such that $\pi (x')$ is a singleton $\psi $ -fiber. The bits left in u after determining part of the skeleton do not constrain this since there are no eventually constant points in Z; that is, if there is a hole left in x after determining the skeleton and filling in the bits of z, and this hole is filled with $a \in \{0,1,\#\}$ , then we can pick $x'$ so that this hole is filled with a, at the point of the process where a suitable bit emerges.

Now pick $w \in \{0,1\}^{*}$ , with $z \in [w]$ , long enough so the bits in u are determined by those in w (when using the skeleton of $x'$ ), so that for all $z' \in [w]$ , there exists a (unique) point $y \in [u]$ with $\pi (y) = \pi (x') \wedge \phi (y) = z'$ . This means $\phi ([u]) \supset [w] \ni z$ . Thus $\phi $ is open.

To see that every point generates a Toeplitz subshift, let $x \in X$ be arbitrary and suppose $x \in [u]$ for some $u \in A^{**}$ , $u : [a,b] \to A$ . If $\psi ^{-1}(\psi (\pi (x)))$ is a singleton, that is, $\pi (x)$ and $\phi (x)$ determine x uniquely, then clearly there exists a period $3^n> 0$ such that $\sigma ^{i3^n}(x) \in [u]$ for all $i \in \mathbb {Z}$ , that is, x is already Toeplitz. In general, it suffices to show the condition from the previous lemma, so we show that the word u appears in some arithmetic progression in x.

For this, note that since $\pi (X)$ is an almost-one-to-one cover of $\mathbb {Z}_3$ and $\mathbb {Z}_3$ is minimal, we can find j such that the set $\psi ^{-1}(\psi (\pi (\sigma ^j(x))))$ has very small diameter. In particular, we can ensure that the symbols in $\pi (x)_{[a,b]}$ are uniquely determined by $\psi (\pi (\sigma ^j(x)))$ . It follows (as discussed in the proof) that the skeleton $\psi (\pi (\sigma ^j(x)))$ and finitely many bits in $\phi (x)$ actually determine the word $\sigma ^j(x)_{[a,b]}$ uniquely. Since $\phi (x)$ is not eventually constant, we can pick j so that together with the bits of z, the forced word is $\sigma ^j(x)_{[a, b]} = u$ . The subword $[j+a, j+b]$ of x then appears in an arithmetic progression because of the equicontinuity properties of the $3$ -odometer, as required.

Observe now that (since every Toeplitz subshift is minimal) every point generates a minimal subshift, and this subshift maps to a single point in $\phi $ . Conversely, the language of a minimal subshift is determined by the bits in its image in a continuous way (again use that no point in Z is eventually constant). Thus the fibers of $\phi $ are precisely the minimal subshifts of X and $\phi $ induces a homeomorphism between them and Z.

Proof. Let $\hat A = A \cup \{\#\}$ , fix a sequence $(x^n, y^n) \in E$ such that each pair $(x, y) \in E$ is a limit point of the sequence, and define

$$\begin{align*}Y_n = {\{ \ldots \# u_i \# u_{i+1} \# u_{i+2} \# \ldots \;|\; \text{for all } i: u_i \in \{x^n_{[-n,n]}, y^n_{[-n,n]}\} \}} \end{align*}$$

(that is, $Y_n$ is the shift-invariant set of all points of the stated form). Let $Y = \overline {\bigcup _{n} Y_n}$ .

If $(x,y) \in E \setminus \Delta $ , then obviously by picking larger and larger n such that $(x^n, y^n)$ is very close to $(x,y)$ , $Y_n \subset Y$ implies that $(x,y) \in \mathrm {EP}(Y)$ . (This is obvious from the intuition stated in the introduction: even if we only obtain information when the orbit is very close to x and y, we obtain an exponential amount of information from points of $Y_n$ if $d((x_n, y_n), (x, y))$ is very small. The concrete computation is also straightforward.)

Suppose then that $(x,y) \in X^2 \setminus E$ . Since E is closed, for some n, we have $([x_{[-k,k]}] \times [y_{[-k,k]}]) \cap E = \emptyset $ . We claim that for $U = [x_{[-k,k]}], V = [y_{[-k,k]}]$ , the cover $(C_0, C_1) = (U^c, V^c)$ has zero topological entropy. By definition, to show this, we must give for each m, a set of binary words $W \subset \{0,1\}^m$ , whose size grows subexponentially in m, such that for any $z \in Y$ , there exists $w \in W$ such that $\sigma ^i(z) \in C_{w_i}$ for all $i \in \{0,1, 2,\ldots , m-1\}$ . In other words, we must cover all partial orbits of length m with a subexponential (in m) number of sets of the form $\bigcap _{i = 0}^{m-1} \sigma ^{-i}(C_{w_i})$ .

Since the sets $C_i$ are clopen, we can equivalently consider a covering problem for words, that is, it is sufficient to find a set of binary words $W \subset \{0,1\}^m$ such that for each $t \in \mathcal {L}_m(Y)$ , there exists $w \in W$ such that $(U^c, V^c)^w$ intersects $[t]$ , where $(U^c, V^c)^u$ is defined inductively on word length by

$$\begin{align*}(U^c, V^c)^0 = U^c, \;(U^c, V^c)^1 = V^c, \,(U^c, V^c)^{a \cdot w} = (U^c, V^c)^a \cap \sigma^{-1}((U^c, V^c)^w) \end{align*}$$

for $a \in \{0,1\}, w \in \{0,1\}^m$ . In fact, we need at most $(|A|+1)^{2k}$ times more sets to cover the partial orbits than for this covering problem for words, and this constant factor is subsumed by subexponential growth. Let us say $w \in \{0,1\}^m$ covers $t \sqsubset \mathcal {L}_m(Y)$ if $(U^c, V^c)^w \cap [t] \neq \emptyset $ .

We first note that the words t containing at most one $\#$ are easy to cover. For such $t \sqsubset Y$ , we always have $t \sqsubset X$ or $t = u\#v$ , where u and v are words in X. All such words appear in a morphic image of the zero entropy subshift $X^2 \times \overline {\mathcal {O}(\ldots 000111\ldots )}$ , so they cannot generate entropy with respect to any cover. More concretely, letting $f(m)$ be the number of words of length m in X, the number of words of length m of the forms $t \sqsubset X$ or $u \# v$ , where $u,v \sqsubset X$ , is at most $f(m) + \sum _{i = 0}^m f(i) f(m-i-1)$ . If f grows subexponentially, its self-convolution grows subexponentially as well. Thus for each word of these two forms, we can simply include an arbitrary covering word in W.

To cover words with at least two $\#$ -symbols, observe that in Y, all such words have an arithmetic progression of $\#$ s. Consider a progression $\alpha = \{i(2n+2) + c \;|\; i \in \mathbb {Z}\} \cap \{0,1, 2,\ldots , m-1\}$ , where $n \geq 0, c \in [0,2n+1]$ and $|\alpha | \geq 2$ . Then the set of words where $\#$ s appear exactly in this arithmetic progression consists precisely of words obtained by inserting any combination of $x^n_{[-n,n]}$ or $y^n_{[-n,n]}$ between $i(2n+2) + c$ and $(i+1)$ $(2n+2) + c$ for each i (and we have up to four choices in total for the prefix and suffix).

Since $(x^n,y^n) \in E$ , by shift-invariance also $(\sigma ^\ell (x^n), \sigma ^\ell (y^n)) \in E$ for all $\ell \in \mathbb {Z}$ and thus from $([x_{[-k,k]}] \times [y_{[-k,k]}]) \cap E = \emptyset $ , we obtain that for all $\ell $ ,

$$\begin{align*}\{x^n_{[\ell-k,\ell+k]}, y^n_{[\ell-k,\ell+k]}\} \neq \{x_{[-k,k]}, y_{[-k,k]} \}, \end{align*}$$

and thus see that there is a single word $w_{n,c}$ of length m that covers every word t with progression $\alpha $ . This gives a polynomial bound on the number of words w needed to cover those t with at least two $\#$ s, all in all, we have obtained a subexponential bound.

Proof. Define $w_0 = 1$ and inductively $w_{i+1} = w_i 0^{i+1} w_i$ . Define

$$\begin{align*}w_\omega = \lim_i w_i \in \{0,1\}^{\mathbb{N}} = 101001010001010010100001010010100010100101\ldots \end{align*}$$

Another description is that $w_\omega $ is obtained from the ruler sequence A007814 in OEIS [13] or universal counterexample [Reference Ferenczi6]

$$\begin{align*}01020103010201040102010301020105\ldots \end{align*}$$

by applying the substitution $n \mapsto 10^{n+1}$ (or $0 \mapsto 1$ , $n \mapsto 0^{n}$ ). We refer to the words $w_i$ and their obvious positioned variants as full words and refer to the process of extending a positioned $w_i \in A^{**}$ to $w_{i+1} \in A^{**}$ (which can be done in exactly two ways) as the left or right full extension.

By induction $n_i = |w_i| = 3 \cdot 2^i - i - 2$ and $\sum _j (w_i)_j = 2^i$ . A short computation shows that for every $k \geq 1$ , there exists a subword of $w_\omega $ of length k (for example its prefix), where the symbol $1$ appears at least $k/3$ times (any $k/c$ works just as well for what follows).

We now perform some substitutions to obtain a subshift. Let $\tau (0) = \{0\}, \tau (1) = \{1, 11\}$ be a non-deterministic substitution and define $Y''' = \tau (w_\omega ) \subset \{0,1\}^{\mathbb {N}}$ , that is, $Y'''$ is the set of all possible points obtained by replacing $0$ s in $w_\omega $ by $0$ s, and $1$ s by either $1$ or $11$ . Let $\hat A = A \sqcup \{\#\}$ where $\# \notin A$ , and let $\tau ' : \hat A \to \{0,1\}$ be the deterministic substitution mapping $\# \mapsto 1$ and $a \mapsto 0$ for $a \in A$ . Let $Y'' = (\tau ')^{-1}(Y''')$ . Let $Y'$ be the $\mathbb {Z}$ -subshift whose language consists of the left-extendable words in the language of $Y''$ . Let Y be the subshift of $Y'$ with additional forbidden words

$$\begin{align*} & \big\{ u_0\#u_1\#\ldots\#u_k \;|\;\text{there exists } x_0, x_1, \ldots, x_k \in X:\text{for all } i, j: x_i \in [u_i] \kern1.2pt{\wedge}\kern1.2pt (x_i, x_j) \kern1.2pt{\in}\kern1.2pt E\big \} \end{align*}$$

(where the $u_i$ are quantified over $A^*$ ). The double occurrences of $\#$ fit this pattern by picking empty words $u_i$ .

Say that a set of words $U \subset A^*$ is a friendship if for any finite subset $V \subset U$ there exist points $(b_v)_v \in X^V$ such that $\text { for all } u, v \in V: b_u \in [u] \wedge (b_u, b_v) \in E$ , so since E is doubly shift-invariant, the forbidden words of Y precisely require that the set of words $U \subset A^*$ that appear in y is a friendship. (We remark that even for a closed doubly shift-invariant equivalence relation, for a finite set U, in general, being a friendship is stronger than all pairs $\{u, v\}$ being friendships for $u, v \in U$ .)

For a concrete picture of points in $Y'$ , consider the sequence $w_\omega \in \{0,1\}^{\mathbb {N}}$ . Its two-sided orbit closure contains points of the form

$$\begin{align*}\ldots101001010001010010100001010010100010100101\ldots \end{align*}$$

We allow any of the $1$ s to be duplicated, e.g.

$$\begin{align*}\ldots101100101000110110010110000101001101000101001011\ldots, \end{align*}$$

and then replace the $1$ s by $\#$ , that is,

$$\begin{align*}\ldots\#0\#\#00\#0\#000\#\#0\#\#00\#0\#\#0000\#0\#00\#\#0\#000\#0\#00\#0\#\#\ldots, \end{align*}$$

finally replace maximal $0$ -sequences by words over A of the same length, e.g.

$$\begin{align*}\ldots\#0\#\#01\#0\#011\#\#1\#\#10\#1\#\#0100\#1\#01\#\#1\#101\#1\#11\#1\#\#\ldots, \end{align*}$$

if $A = \{0,1\}$ . Such are the points of $Y'$ . In points of Y, the finite words should additionally be from the language of X, and should form a friendship.

If $U \subset A^*$ is a friendship, then by compactness to every word $u \in U$ , we can associate a point $b_u \in [u]$ , so that $(b_u, b_v) \in E$ for all $u, v \in U$ . Fix such points for each set U and call $b_u$ the (friendship) bracelet of u (with respect to U). Of particular importance in what follows are (sets of) bracelets for sets of words $U \subset A^*$ that appear in some point $y \in Y$ , or the union of such sets for two points $y, z$ . (It is worth noting that in such a situation, the bracelets are not enforced to be in the orbit closure of y or z.)

Call the binary word $\tau '(w)$ recording positions of $\#$ by $1$ in a word w the skeleton, similarly for points $x \in Y$ , and the word where each maximal occurrence of $11$ is further replaced by $1$ the preskeleton. We call a (positioned or not) word $w \in \hat A^* \cup \hat A^{**}$ full if it is equal to (a shift of) $(\tau ')^{-1}(w_n)$ for some n, or if (more generally) its preskeleton is full.

We now explain the technical trick for characterizing the entropy pairs $\mathrm {EP}(Y)$ , and show that $(y,z) \notin \Delta _Y$ is an entropy pair if and only if the set

$$\begin{align*}U = \{ u \in A^* \;|\; u \sqsubset y \vee u \sqsubset z \}, \end{align*}$$

containing all non- $\#$ subwords from y and z, is a friendship.

Suppose $u', v' \in \hat A^{**}$ are any two positioned words that appear in some points of Y, with $\#$ as their first and last symbols (note that words beginning and ending with $\#$ are dense in Y in the natural topology of $\hat A^{**} \cup \hat A^{\mathbb {Z}}$ ), and let U be the set of words containing $u \in A^*$ if and only if $\#u\# \sqsubset u'$ or $\#u\# \sqsubset v'$ . Suppose U is a friendship. Extend $u'$ and $v'$ to full positioned words whose preskeletons are the same length, in an arbitrary way, though so that their finite subwords over $A^*$ remain a friendship (safe continuations for words can always be found in the bracelets proving the friendship). The left and right ends of the resulting words $u''$ and $v''$ have some left and right offsets $n_1, n_2$ , respectively, in the sense that $u'' \in A^{[a,b]}$ and $v'' \in A^{[a+n_1,b+n_2]}$ .

Now, continue extending the words by alternately turning them into full positioned words by adding a left and a right full extension, so that no new occurrence of $\#\#$ appears, and all finite subwords over $A^*$ remain a friendship. Note that the left and right ends continue to differ by exactly $n_1$ and $n_2$ , since the length of the left or right continuation is always the same if no occurrence of $\#\#$ appears. Extend the words so many times that at least $\max (|n_1|, |n_2|)$ undoubled $\#$ s are obtained on both sides. Now finally double some of the $\#$ to get two positioned full words $u, v$ which have $u'$ and $v'$ at the center, have the same preskeleton, and have the same left and right length, that is, $u, v \in A^{[c,d]}$ for some $c, d \in \mathbb {Z}$ .

Now suppose u and v have preskeleton $w_i$ . By the analysis of the density of $1$ s in the ruler sequence, for each k, we can find a word $w \in (\hat A \cup \{\text {\textvisiblespace }\})^*$ of length at most $2k$ such that:

• • the $\text {\textvisiblespace }$ -symbols appear in segments of length exactly $|u| = |v|$ ;

• • there are at least $k/(3 \cdot 2^i) - 2$ such segments; and

• • replacing the $\text {\textvisiblespace }^{|u|}$ -subwords individually by u or v always leads to a valid word of Y.

To see this, recall that for all k, we can find a word of length k in the ruler sequence with at least $k/3$ many $1$ -symbols. Observing that all $1$ s appear in disjoint $w_i$ -words, we can have at least $k/(3 \cdot 2^i) - 2$ occurrences of $w_i$ in a word $w'$ of length k in the ruler sequence. Replacing these occurrences of $w_i$ by $\text {\textvisiblespace }^{|u|}$ in $w'$ , and then filling the remaining gaps with arbitrary words from the bracelets of words in U, we obtain a word w with the desired properties.

The interchangeability, together with the lower bound on maximal density, shows that the cover $([u]^c, [v]^c)$ generates entropy in Y, and thus we have shown that any points whose non- $\#$ subwords form a friendship are entropy pairs. However, if the union $U \subset A^*$ of the sets of subwords of two words $u, v$ is not a friendship, then these words do not even appear together in any point of Y, by definition, so neither can u and v, and it follows that the cover $([u]^c, [v]^c)$ has zero entropy. This concludes the characterization of $\mathrm {EP}(Y)$ .

An important corollary of this characterization is that $\mathrm {EP}(Y)$ is doubly shift-invariant, and therefore so is any $\mathrm {EP}(Y)^{(\lambda )}$ by an obvious transfinite induction. If $(x, y) \in F$ and F is a doubly shift-invariant closed relation, then $(x', y) \in F$ for any $x'$ in $\overline {\mathcal {O}(x)}$ . Let us call a doubly shift-invariant equivalence relation F with this property hereditary. It is easy to show by transfinite induction that if a doubly shift-invariant equivalence relation F is hereditary, then so is $F^{(\lambda )}$ , for any ordinal $\lambda $ . In particular, the relations $\mathrm {EP}(Y)^{(\lambda )}$ are all hereditary since $\mathrm {EP}(Y)$ is doubly shift-invariant and closed.

It is clear from the characterization that $\mathrm {EP}(Y) \cap X^2 = E$ as required. We need to show $\mathrm {EP}(Y)^{(\lambda )} \cap X^2 = E^{(\lambda )}$ for all $\lambda $ , and we do this by transfinite induction. On limit ordinal steps, this obviously continues to hold. On even successor steps, that is, transitive closure steps, we use the fact that points of Y can be replaced by any point in their orbit closure, and the orbit-closure of every point contains a point of X. This allows turning any transitivity sequence $(x = y_0, y_1, \ldots , y_k = x')$ , where $x, x' \in X$ , $y_i \in Y$ , $(y_i, y_{i+1}) \in E^{(\lambda -1)}$ , into one where $y_i \in X$ for all i. This shows $(\mathrm {EP}(Y)^{(\lambda -1)})^* \cap X^2 = (\mathrm {EP}(Y)^{(\lambda -1)} \cap X^2)^*$ as required.

Now, note that if $\mathrm {EP}(Y)^{(\lambda )}$ is transitively closed, $y \in Y \setminus X$ and b is any bracelet of y, or more generally, any point in X whose set of $A^*$ -subwords together with $A^*$ -subwords of Y forms a friendship, then we have $(y, z) \in \mathrm {EP}(Y)^{(\lambda )}$ if and only if $(b, z) \in \mathrm {EP}(Y)^{(\lambda )}$ . This is because $(b, y) \in \mathrm {EP}(Y)$ by the characterization of entropy pairs of Y. In other words, on every transitive step, every $y \in Y$ satisfies the same relations as any of its bracelets or any point that could be chosen as a bracelet for it.

On a topological closure step, that is, for an odd ordinal $\lambda $ , suppose $\lim _i (x_i, y_i) = (x, y) \in X^2$ where $(x_i, y_i) \in \mathrm {EP}(Y)^{(\lambda -1)}$ for all i. It is enough to show that the $x_i$ can be replaced by points from X, as then (also applying this idea to the right component) we have $\overline {\mathrm {EP}(Y)^{(\lambda -1)}} \cap X^2 = \overline {\mathrm {EP}(Y)^{\lambda -1} \cap X^2}$ . If there is a subsequence of the $x_i$ in X, then we can directly restrict to that. Otherwise, restrict to a subsequence such that $x_i \notin X$ for all i.

In $x_i$ , take a maximal central positioned word $u_i : A^{[-a, a]}$ not containing $\#$ (where necessarily $a \rightarrow \infty $ as $i \rightarrow \infty $ ) and replace $x_i$ by the bracelet $b_{u_i}$ with respect to the language of $x_i$ , centered to have $u_i$ in the same place as in $x_i$ . We have $(b_{u_i}, y_i) \in \mathrm {EP}(Y)^{(\lambda -1)}$ since $\mathrm {EP}(Y)^{(\lambda -1)}$ is transitively closed and $(b_{u_i}, x_i) \in \mathrm {EP}(Y)$ , and clearly $(b_{u_i}, y_i) \rightarrow (x,y)$ , as required.

Finally, we verify the last claim. Suppose thus that $E^{(\lambda )} = X^2$ for some $\lambda $ . If $\lambda $ is even, then $\mathrm {EP}(Y)^{(\lambda )}$ is transitively closed so since any $y \in Y \setminus X$ appears in the same relations as any bracelet of its, $\mathrm {EP}(Y)^{(\lambda )} = Y^2$ .

Suppose then that $\lambda $ is odd. Let $(y, z) \in Y^2$ , and without loss of generality suppose $y, z \in Y \setminus X$ (this suffices since $\mathrm {EP}(Y)^{(\lambda )}$ is closed and $X \subset \overline {Y \setminus X}$ ). Let $(b_u)_u$ and $(c_v)_v$ be the respective bracelets of their subwords with respect to their languages. If any two of these bracelets, one from y and one from z, are in relation on step $\lambda -1$ , then already $(y, z) \in \mathrm {EP}(Y)^{(\lambda -1)}$ since $\mathrm {EP}(Y)^{(\lambda -1)}$ is transitive.

Otherwise, let m be arbitrary and let (up to swapping y and z) $n \geq m$ and k be such that $y_{[-m,m]} = u_1\#u_2\#\ldots \#u_k$ , $U = (u_1, \ldots , u_k) \in (A^*)^k$ and $z_{[-m,n]} = v_1\#v_2\#\ldots \#v_k$ , $V = (v_1, \ldots , v_k) \in (A^*)^k$ . Let $B = (b_u)_{u \in U}$ and $C = (c_v)_{v \in V}$ be bracelets for these words with respect to the subwords of y and z, respectively.

Now, recall that the bracelets of a point are always (by definition) in the E-relation together, and apply the odd condition to $(B, C) \in (E^{(k)})^2$ . Since $(E^{[k]})^2 \subset \overline {(E^{(\lambda -1)})^{[2k]} \cap (E^{[k]})^2}$ , we can find in $(E^{(\lambda -1)})^{[2k]} \cap (E^{[k]})^2$ arbitrarily good approximations $B' = (b_u')_{u \in U}$ and $C' = (c_v')_{v \in V}$ to B and C, respectively. In particular, we can find good enough approximations to ensure that $b_u'$ contains u and $c^{\prime }_v$ contains v, for each $u \in U, v \in V$ . Then since $\mathrm {EP}(Y)^{(\lambda -1)}$ is transitive and $(b_u', c^{\prime }_v) \in E^{(\lambda -1)}$ , any pair of points $(y_k, z_k)$ in Y, for which we can take $b_u'$ and $c^{\prime }_v$ as one of their bracelets, satisfies $(y_k, z_k) \in \mathrm {EP}(Y)^{(\lambda -1)}$ .

We construct the pair $y_k, z_k$ so that $(y_k, z_k)_{[-k,k]} = (y,z)_{[-k,k]}$ . For this, directly let $(y_k)_{[-m,m]} = u_1\#u_2\#\ldots \#u_k$ and $(z_k)_{[-m,n]} = v_1\#v_2\#\ldots \#v_k$ . Extend $y_k$ to a point of Y by copying the skeleton from y and filling all the remaining gaps arbitrarily so that all words between two $\#$ s (and possible infinite tails without an occurrence of $\#$ ) are subwords of the words in $B'$ . Extend $z_k$ similarly. These are indeed valid points: the skeletons are correct since we copied them from valid points, and the set of $A^*$ -subwords of each point is a friendship, since $B', C' \in E^{[k]}$ . Clearly we can take any $b_u'$ and $c^{\prime }_v$ as one of their bracelets, so $(y_k, z_k) \in \mathrm {EP}(Y)^{(\lambda -1)}$ gives an approximation to $(y,z)$ that is correct in the interval $[-m,m]$ .

Proof. Suppose f is open and $f(x) = y$ . Then every neighborhood $U \ni x$ maps to a neighborhood $f(U)$ of y, in particular, if $\lim _i y_i = y$ , then for all large enough i, $y_i$ has a preimage $x_i$ in U. Picking open sets U from a decreasing countable neighborhood basis of x, and always picking preimages for $y_i$ from U until they start having preimages in the next neighborhood of x, the claim is proved.

For the other direction, suppose f is not open, and let U be open such that for some $y \in f(U)$ , every neighborhood of y contains a point without a preimage in U. Pick $x \in U$ such that $f(x) = y$ . By first-countability, there is a sequence $y_i \rightarrow y$ such that the points $y_i$ have no preimage in U. The second condition fails, since we cannot pick $x_i \in U$ .

Proof. Since X is first-countable, so is $X^2$ , and thus the closure of a set $C \subset X^2$ is just the set of limits of sequences in C. (In other words, $X^2$ has the Fréchet–Urysohn property. For general topological spaces, even if X is Frechét–Urysohn, $X^2$ need not be [Reference Gruenhage8].)

Write $g = f \times f : X^2 \to Y^2$ , and observe this is also an open quotient map. We have $F^{(\zeta )} = g^{-1}(E^{(\zeta )})$ where $\zeta = 0$ if E is not closed, and $\zeta = 1$ otherwise, since E is closed if and only if F is closed since g is a quotient map. We proceed by induction. Suppose $\lambda $ is an odd ordinal, and $g(x,x') = (y,y')$ . If $(x,x') \in F^{(\lambda )}$ , then there exist $(x_i,x_i') \in F^{(\lambda -1)}$ with $(x_i,x_i') \rightarrow (x,x')$ and then $g(x_i,x_i') \rightarrow (y,y')$ by continuity showing $E^{(\lambda )} \supset g(F^{(\lambda )})$ , which implies $g^{-1}(E^{(\lambda )}) \supset F^{(\lambda )}$ .

For the other inclusion, suppose $(y, y') \in E^{(\lambda )}$ , so $(y, y') = \lim _i (y_i, y_i')$ for some $(y_i, y_i') \in E^{(\lambda -1)}$ . By the previous lemma, there exist $(x_i,x_i') \in X^2$ such that $g(x_i,x_i') = (y,y')$ and $(x_i, x_i') \rightarrow (x, x')$ . Then by induction, we have $(x_i,x_i') \in F^{(\lambda -1)}$ for all i, so $(x,x') \in F^{(\lambda )}$ , showing $g^{-1}(E^{(\lambda )}) \subset F^{(\lambda )}$ .

If $\lambda $ is an even successor ordinal, suppose $g(x,x') = (y,y')$ . If $(x,x') \in F^{(\lambda )}$ , there exist $x = x_0, x_1, \ldots , x_k = x'$ such that $(x_i, x_{i+1}) \in F^{(\lambda -1)}$ for all i, and then by induction, $g(x_i, x_{i+1}) \in E^{(\lambda -1)}$ and since $y = f(x_0), y' = f(x_k)$ , we have $(y,y') \in E^{(\lambda )}$ .

If however $(y, y') \in E^{(\lambda )}$ , then there exist $y = y_0, y_1,\ldots , y_k = y'$ with $(y_i, y_{i+1}) \in E^{(\lambda -1)}$ . Pick any preimages $f(x_i) = y_i$ with $x_0 = x, x_k = x'$ by induction $(x_i, x_{i+1}) \in F^{(\lambda -1)}$ for all i, and thus $(x,x') \in F^{(\lambda )}$ .

For the limit ordinal case, observe that union commutes with preimage.

Proof. Suppose $(E^{[k]})^2 \subset \overline {(E^{(\lambda -1)})^{[2k]} \cap (E^{[k]})^2}$ and consider $(A', B') \in (F^{[k]})^2$ . Consider $f(A')$ and $f(B')$ (where f is applied diagonally to all elements of the tuple), which are by definition k-tuples $A, B \in E^{[k]}$ . In $(E^{(\lambda -1)})^{[2k]} \cap (E^{[k]})^2$ , we find arbitrarily good approximations $(A_i, B_i)$ to $(A,B)$ . Take any f-preimages, $f(A_i') = A_i$ , $f(B_i') = B_i$ which are close to $A', B'$ elementwise, using Lemma 9, obtaining by definition $A^{\prime }_i, B_i' \in F^{[k]}$ . By the previous lemma, we have $(F^{(\lambda -1)})^{[2k]} = (f \times f)^{-1}(E^{(\lambda -1)})^{[2k]}$ so we have $(A_i', B_i') \in (F^{(\lambda -1)})^{[2k]}$ . This implies $(F^{[k]})^2 \subset \overline {(F^{(\lambda -1)})^{[2k]} \cap (F^{[k]})^2}$ .

The other direction is similar, but using continuity instead of openness.

Proof. If $\alpha = 1$ , $X = A^{\mathbb {Z}}$ is an example.

Let $\lambda \geq 2$ be a countable ordinal. By Lemma 7, there exists a closed countable set $Z \subset [0,1]$ and a closed symmetric reflexive relation $E \subset Z^2$ such that E is equalizing with close-up rank $\lambda $ . Furthermore, if $\lambda $ is an odd ordinal, then $(E^{[k]})^2 \subset \overline {(E^{(\lambda -1)})^{[2k]} \cap (E^{[k]})^2}$ .

By Lemma 4, since every countable compact subspace of $[0,1]$ is zero-dimensional, there exists a pointwise zero-entropy Toeplitz subshift X and a doubly shift-invariant closed equivalence relation $F \subset X^2$ such that Z is the open quotient of X, by $\phi : X \to Z$ , and $(\phi (x), \phi (y)) \in E\! \iff (x,y) \in F$ .

Since $\phi $ is an open quotient map, Lemma 10 shows that $E^{(\lambda )} = Z^2$ if and only if $(\phi ^{-1} \times \phi ^{-1})(E) = F$ satisfies $F^{(\lambda )} = X^2$ . If $\lambda $ is an odd ordinal, then $(E^{[k]})^2 \subset \overline {(E^{(\lambda -1)})^{[2k]} \cap (E^{[k]})^2}$ , and by the previous lemma, we have $(F^{[k]})^2 \subset \overline {(F^{(\lambda -1)})^{[2k]} \cap (F^{[k]})^2}$ .

Now, apply Lemma 5 to obtain $Y \supset X$ such that $\lambda $ is the minimal ordinal satisfying $\mathrm {EP}(Y)^{(\lambda )} = Y^2$ . This Y is a subshift in CPE class $\lambda $ .