Let $(X, T)$ be a minimal subshift, i.e. $X$ is a closed $T$-invariant subset of $A^\mathbb{Z}$, where $T$ is the shift. A pair $x,y\in X$ is asymptotic if $d(T^nx, T^ny)$ goes to zero as $n\to\infty$. Always exists such a pair when $X$ is infinite: for every $n\geq1$ there exists $x^{(n)}, y^{(n)} \in X$ such that $x^{(n)}_0\not= y^{(n)}_0$ and $x^{(n)}_{[1,N]}= y^{(n)}_{[1,N]}$ (if not, for some $n$, $x_{[1,n]} = y_{[1,n]}$ implies $x_0=y_0$, i.e., $x_{[1,n]}$ determines $x_0$, and this forces $X$ to be periodic), and any pair of convergent subsequences of $x^{(n)}$ and $y^{(n)}$ will do the trick.

My question is: do exist $k$-tuples of asymptotic points, for every $k\geq1$? More precisely, is it true that for every $k\geq1$ there exists $x_1,\dots,x_k\in X$ such that $$\lim_{n\to\infty}d(T^nx_i, T^nx_j) = 0\ \forall i,j$$

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    $\begingroup$ No. Consider the Sturmian shift. There asymptotic pairs, but no triples. $\endgroup$ – Anthony Quas May 16 '19 at 6:23

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