$2 \cdot 111$ two-digit number set $\{00,01, \cdots, 98,99\}$ has a subset $X$ with the following property: in any infinite sequence of digits, there are two adjacent digits that form an element of $X$. How many elements does $X$ have at minimum?
[Solution] For any $i, j \in\{0,1, \cdots, 9\}, X$ should contain either $ij$ or $ji$, otherwise the sequence $ijijij \cdots$ would not contain any element of $X$. There are $10+C_{10}^{2}=55$ such unordered pairs $(i, j)$, so $|X| \geqslant 55$.
On the other hand, let
$$
X=\{ij \mid 0 \leqslant i \leqslant j \leqslant 9\},
$$

then $|X|=55$, and for any infinite sequence, let $i$ be the smallest digit in it, and $j$ be the digit following $i$, then $ij \in X$.
In summary, $X$ must contain at least 55 elements.