# Must every infinite subshift contain a non-periodic point?

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.

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.