There exists a minimal subshift $X$ with a point $x \in X$ such that $x\_{(-\infty,0)}.x\_0x\_0x\_{(0,\infty)} \in X$?



Minimal subshift with some $x \in X$ such that $x_{(-\infty,0)}.x_0x_0x_{(0,\infty)} \in X$?
We can produce such a subshift by a standard hierarchical construction.
Let $w\_{0,0} = 01$ and $w\_{0,1} = 011$.
For each $k \geq 0$, define $w\_{k+1,0} = w\_{k,0} w\_{k,0} w\_{k,1}$ and $w\_{k+1,1} = w\_{k,0} w\_{k,1} w\_{k,1}$.
Define $X$ by forbidding each word that doesn't occur in any $w\_{k, b}$.
Since each $w\_{k+1,b}$ contains both $w\_{k,0}$ and $w\_{k,1}$, it's easy to show that $X$ is minimal.


For $b \in \{0,1\}$, define $x^b \in X$ as the "limit" of $(w\_{k,b})\_{k \geq 0}$ such that the central $01$ or $011$ is at the origin.
Because of the way we defined $w\_{k,0}$ and $w\_{k,1}$, the only difference between the words, and hence the limiting configurations, is that single extra $1$.
Concretely, $x^0$ and $x^1$ look like
$$
\cdots 0101011\;0101011\;01011011\;0101011\;010.1011\;01011011\;0101011\;01011011\;01011011 \cdots
$$
and
$$
\cdots 0101011\;0101011\;01011011\;0101011\;010.11011\;01011011\;0101011\;01011011\;01011011 \cdots
$$
(Spaces added for clarity.)


