# Minimal subshift with some $x \in X$ such that $x_{(-\infty,0)}.x_0x_0x_{(0,\infty)} \in X$?

There exists a minimal subshift $$X$$ with a point $$x \in X$$ 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.)
It is well-known that the Chacon substitution $$\tau$$ defined by $$\tau(0) = 0010$$, $$\tau(1) = 1$$ produces a minimal subshift, when you take the legal words to be the words that appear in some $$\tau^n(0)$$. The two-sided fixed point from $$0.0$$ is $$x.y = {...0010001010010.0010001010010...}$$ and the one from $$0. 10$$ is $$x.1y$$. So they are both in $$X$$.
Now apply the additional substitution $$\alpha(0) = 01$$, $$\alpha(1) = 1$$ and you still get a minimal subshift $$Y$$ as image (this is a flow equivalence onto its image). We have $$\alpha(x).\alpha(y) \in Y$$ and $$\alpha(x) . 1 \alpha(y) \in Y$$, and since $$\alpha(x)$$ ends with $$1$$ this is a pair of the kind you are asking for.