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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.

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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.

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  • $\begingroup$ Thanks. So the naive approach is at least the first part, and then one just needs to shift the elements in the sequence so that to one side, it is periodic, with the same periodic block as the original limit point, and to the other side, you know that it is definitely not periodic with that block. It makes sense. $\endgroup$
    – Dan Rust
    Commented Nov 22, 2019 at 15:15

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