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.