Must every infinite subshift contain a non-periodic point?

Question

Let $\mathcal{A}$ be a finite alphabet and let $(X,\sigma)$ be some subshift over $\mathcal{A}$, a non-empty, closed, $\sigma$-invariant subspace of the full-shift $(\mathcal{A}^\mathbb{Z},\sigma)$, where $\sigma \colon \mathcal{A}^\mathbb{Z} \to \mathcal{A}^\mathbb{Z}$ is the shift map given by $\sigma(x)_i = x_{i+1}$. Note that, as $\mathcal{A}^\mathbb{Z}$ is compact (with the product topology), and $X$ is a closed subspace, then $X$ is also compact.

If $X$ is infinite, is it true that there exists a non-periodic element of $X$? That is, does there exist an $x \in X$ such that $\sigma^n(x) \neq x$ for all $\sigma \in \mathbb{Z}\setminus\{0\}$?

The naive argument using compactness and a sequence of points with increasing periods doesn't work, because of the example $x_n = \cdots ba^{2n}ba^n.a^nba^{2n}b\cdots$, which each have period $2n+1$, but whose limit $x = \cdots aaaa.aaaa\cdots$ has period $1$. Of course, this doesn't consitute a counterexample, as the point $ a^\infty .b a^\infty$ is also in the subshift, so this just says that the argument is not careful enough.

I feel like a counterexample is unlikely, so either I'm missing something obvious here, or there is some obscure counterexample, as the question seems a natural one.

Answer 1

Yes, in every infinite subshift $X$ one finds a non-periodic element. Take any sequence of distinct elements of $X$, then extract a converging subsequence $x^i \to x \in X$. If $x$ is not periodic, then we are done; otherwise it has some period $p$, and we may assume (by removing a finite number of elements from the sequence) that none of the $x^i$ are $p$-periodic. Then in every $x^i$, there is a coordinate $j_i \in \mathbb{Z}$ such that $x^i_{j_i} \neq x^i_{j_i + p}$. Taking a subsequence and noting the symmetry of the situation, we may assume the $j_i$ are all positive, and we choose the minimal positive $j_i$ for each $i$. Then the region $[0, j_i]$ is $p$-periodic in each $x^i$, and $j_i \to \infty$. The shifted sequence $\sigma^{j_i}(x^i)$ has a limit point $y$ which is $p$-periodic in $(-\infty, 0]$ but $y_0 \neq y_p$, which means it's not periodic.

The same proof (taken from this article) works more generally for $d$-dimensional subshifts.