I am looking how to prove the following fact:
If $ X \subseteq A^\mathbb{Z}$ is an infinite minimal subshift, then for any $N\ge 1$, $X$ is conjugate to a minimal subshift $Y\subseteq B^\mathbb{Z}$ such that for any $y\in Y$, $y(i)\neq y(j)$ if $|i-j|\le N , i\neq j$.
(I have encountered this in a paper in which the author asserts this without a proof.)
Define $X(m)$ as the image of $X$ in $(A^m)^\mathbf{Z}$, mapping $(a_n)_{n\in\mathbf{Z}}$ to $((a_{n+k})_{0\le k<m})_{n\in\mathbf{Z}}$. This is an equivariant embedding.
Fix $N$. We claim that if $X$ has no periodic element then for $m$ large enough, $X(m)$ has the required property: for $0<|i-j|\le N$ and $y\in X(m)$ we have $y(i)\neq y(j)$.
In particular, this applies if $X$ is minimal infinite.
Proof: Otherwise there exists $N$ and an infinite subset $I$ of positive integers such that for every $m\in I$ there exists $y^m\in X(m)$ such that $y^m(0)=y^m(N)$. Write $y^m$ as the image of $z^m\in X$. The condition $y^m(0)=y^m(N)$ means that for all $i\in\{0,\dots,m-1\}$ we have $z(i)=z(i+N)$. Letting $m$ tend to infinity and shifting $z$ by approximatively $-m/2$, and taking a limit point, we obtain an $N$-periodic element in $X$.
