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I have recently been interested in some questions which stem from taking subshifts which converge to a limiting subshift in the Hausdorff metric.

More specifically, given an alphabet $\mathcal{A}$, I consider $\mathcal{A}^{\mathbb{Z}^d}$ as a dynamical system with a natural action from $\mathbb{Z}^d$. When I say subshift, I mean a nonempty closed subset $\Omega\subseteq \mathcal{A}^{\mathbb{Z}^d}$ which is invariant under the induced action. I'm interested in conditions of when a sequence of subshifts, $\{ \Omega_n \}_{n=1}^\infty$, converge to a limiting subshift in the Hausdorff metric?

Is anyone familiar with references dealing with this sort of questions? I am wondering if there are pages giving conditions for this or estimates on the rate of convergence for the study of specific subshifts?

Later edit

For example, studying an aperiodic subshift, $\Omega_\infty$, by a sequence of periodic ones converging to it? And learning properties of the limit subshift by the approximating ones.

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  • $\begingroup$ I guess you know that it just means the language converges, and are looking for something more specific? $\endgroup$
    – Ville Salo
    Mar 27, 2023 at 17:37
  • $\begingroup$ This (space of subshifts under Hausdorff metric) is a pretty standard object, here's a generic paper that talks about it arxiv.org/abs/2203.15159 $\endgroup$
    – Ville Salo
    Mar 27, 2023 at 17:39
  • $\begingroup$ @VilleSalo I am aware of that criterion. But I was wondering of examples and methods of approximation. The paper you linked to talks about genericity, which I suspect is not a constructive way to obtain approximations. $\endgroup$ Mar 28, 2023 at 10:19
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    $\begingroup$ Yes it was pretty much just an arbitrary paper where I know this concept appears explicitly by name. It is not clear to me what you are looking for exactly, so I figured it's better than nothing. There is a canonical way to obtain approximations: Take the SFT approximation where you explicitly require that the patterns in a particular domain must come from the said subshift. For any subshift, this converges in Hausdorff metric. $\endgroup$
    – Ville Salo
    Mar 28, 2023 at 10:41
  • $\begingroup$ Ok, I see your edit now. This is a bit more answerable already. $\endgroup$
    – Ville Salo
    Mar 28, 2023 at 10:45

1 Answer 1

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Let $X \subset A^{\mathbb{Z}^d}$ be a subshift. So $A$ a discrete finite set, $A^{\mathbb{Z}^d}$ carries the product topology; $X$ is topologically closed in this topology; for all $\vec v \in \mathbb{Z}^d$ we have $X = \sigma_{\vec v}(X)$ where $\sigma_{\vec v}(x)_{\vec u} = x_{\vec v + \vec u}$.

Given finite $N \subset \mathbb{Z}^d$, and $Z \subset A^{\mathbb{Z}^d}$, write $Z|_N$ for $\{z|_N \;|\; z \in Z\}$. Write $O(x) = \{\sigma_{\vec v}(x) \;|\; \vec v \in \mathbb{Z}^d\}$ for the orbit of a point, and the orbit closure $\overline{O(x)}$ is the closure of the orbit.

Given finite $N \subset \mathbb{Z}^d$, the $N$-SFT approximation $X^N$ is the SFT with forbidden patterns those $P \in A^N$ that do not appear in $X$, i.e. $X^N$ is the subshift of all $y$ such that $P \notin O(y)|_N$ for all $P \notin X|_N$. Note that $X|_N = X^N|_N$, namely the $X \subset X^N$ and by definition any $N$-pattern appearing in $X^N$ appears in $X$ as well.

A subshift is periodic if the subgroup of $\mathbb{Z}^d$ containing those $\vec v \in \mathbb{Z}^d$ such that $\sigma_{\vec v}$ stabilizes $X$ pointwise has full rank. A point $x \in A^{\mathbb{Z}^d}$ is periodic if its orbit closure $\overline{O(x)}$ is periodic. A subshift is aperiodic if it has no periodic points. If $X, Y \subset A^{\mathbb{Z}^d}$ and $N \subset \mathbb{Z}^2$ is finite, we say $X$ is $N$-dense in $Y$ if $Y|_N \subset X|_N$, i.e. $N$-patterns appearing in points of $Y$ can be found also in points of $X$.

Theorem. A subshift $X \subset A^{\mathbb{Z}^d}$ has a sequence of periodic subshifts converging to it in Hausdorff metric iff for all $N$, the $N$th SFT approximation of $X$ has $N$-dense periodic points in $X$ (i.e. periodic points of $X^N$ are $N$-dense in $X$).

Proof. Suppose first that $X$ is the limit of periodic subshifts $X_i$. Let $N \subset \mathbb{Z}^d$ be any finite set and consider the SFT approximation $X^N$. Suppose for a contradiction that $X^N$ does not have $N$-dense periodic points in $X$. This means that some pattern $P \in A^N$ appears in $X$ but not in any periodic point of $X^N$. Take $i$ sufficiently large so that $X_i|_N = X|_N = X^N|_N$. Then $X_i$ contains a point $x$ with $x|_N = P$. Since $X_i|_N = X^N|_N$, we have $X_i \subset X^N$, so $x \in X^N$. But then $x$ is a periodic point in $X^N$ containing the pattern $P$, a contradiction.

Suppose then that the $N$-SFT approximation of $X$ has $N$-dense periodic points for all finite sets $N$. Let $N$ be arbitrary, and for each $P \in X|_N$ pick an arbitrary periodic point $y_P \in X^N$ with $y_P|_N$ = P. Then $\bigcup_{P \in X|_N} \overline{O(y_P)}$ is a finite (thus periodic) subshift which clearly agrees with the language of $X$ in $N$. Square.

For a one-dimensional subshift $X \subset A^{\mathbb{Z}}$, say it is chain-transitive if for any $\epsilon > 0$ and any $x, y \in X$, there exists a sequence of points $x = x_1, x_2, ..., x_k = y$ such that $d(\sigma(x_i), x_{i+1}) < \epsilon$ for all $i$. It is transitive if for any $\epsilon > 0$ and $x, y$ there exist $z \in X$ and $n > 0$ such that $d(z, x) < \epsilon$ and $d(\sigma^n(z), y) < \epsilon$.

Theorem. If $X \subset A^{\mathbb{Z}}$ is chain-transitive, then it is the limit of periodic subshifts in Hausdorff metric.

Proof. It suffices to show that the $N$-SFT approximation of $X$ has $N$-dense periodic points in $X$. For this it suffices to show that the $N$-SFT approximation has dense periodic points. For this, observe that if $X$ is chain-transitive, then the $N$-SFT approximation is transitive for any $N$ (this requires a little proof, but it's kind of well known). It is well known that transitive $\mathbb{Z}$-SFTs have dense periodic points. Square.

Corollary. Every minimal $\mathbb{Z}$-subshift is the limit of periodic subshifts in Hausdorff metric.

Proof. Minimal obviously implies chain-transitive (even transitive). Square.

Let's also show that the corollary fails badly in two dimensions.

Theorem. Two-dimensional SFTs can be aperiodic and minimal.

Proof. This is classical, perhaps first explicitly stated and proved in Mozes, Shahar, Aperiodic tilings, Invent. Math. 128, No. 3, 603-611 (1997). ZBL0879.52011. . Square.

Lemma. Minimal SFTs are isolated point in the space of subshifts under the Hausdorff metric.

Proof. If $X$ is a minimal SFT with forbidden patterns contained in $N$, and $Y$ is sufficiently close to it in Hausdorff metric, then $Y \subset X$ since $Y|_N = X|_N$ and $X$ is equal to its $N$th SFT approximation. Since $X$ is minimal, $Y = X$. Square.

Theorem. There exists a minimal aperiodic two-dimensional subshift which is not a limit of periodic subshifts in Hausdorff metric.

Proof. Let $X$ be a two-dimensional aperiodic minimal SFT. By the previous lemma it is isolated, so any sufficiently good approximation is aperiodic. Square.

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  • $\begingroup$ Thank you for your answer. This is a result I suspected should hold, and was the motive behind some of my past questions. Thanks for clearing this. $\endgroup$ Mar 28, 2023 at 12:24
  • $\begingroup$ Perhaps as an additional question, Is the approximation always 'essentially' by SFT's? I mean the underlying metric is usually the big ball metric which is an ultra-metric, which I think implies the Hausdorff metric is also an ultra-metric. For that reason, can every approximating sequence of subshifts be a replaced by a sequence of SFT's with the same distance from the limit? $\endgroup$ Mar 28, 2023 at 12:26
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    $\begingroup$ Yes in the sense that if $X$ is a subshift, and $X_i \longrightarrow X$ are arbitrary subshifts which are not equal to $X$, then there exist subshifts of finite type $Y_i$ such that $d(X_i, X) = d(Y_i, X)$ (here I assume the Hausdorff metric is defined with one of the usual underlying metrics that only look at positions where configurations differ). Namely take a very large $N$ that witnesses the difference between $X_i$ and $X$ and take the $N$-SFT approximation of $X_i$. $\endgroup$
    – Ville Salo
    Mar 28, 2023 at 12:39
  • $\begingroup$ Okay, thanks again. $\endgroup$ Mar 28, 2023 at 12:40

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