$2 \cdot 41$ Given the set $\{1,2,3,4,5,6,7,8,9,10\}$. Find the number of subsets of this set that have the following property: each subset contains at least 2 elements, and the absolute difference between any two elements in each subset is greater than 1.
[Solution] Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ that satisfy the given property.
The subsets of the set $\{1,2, \cdots, n, n+1, n+2\}$ that satisfy the given property can be divided into two categories: the first category of subsets includes the element $n+2$, and there are $a_{n}+n$ such subsets (i.e., the union of each subset of $\{1,2, \cdots, n\}$ with $\{n+2\}$, as well as $\{1, n+2\},\{2, n+2\}, \cdots, \{n, n+2\}$). The second category of subsets does not include $n+2$, and there are $a_{n+1}$ such subsets.
Thus, we have
$$
a_{n+2}=a_{n}+a_{n+1}+n .
$$

Clearly, $a_{3}=1, a_{4}=3$ (i.e., $\{1,3\},\{2,4\},\{1,4\}$ ).
Therefore, $a_{5}=7, a_{6}=14, a_{7}=26, a_{8}=46, a_{9}=79, a_{10}=133$.
The number of subsets we are looking for is 133.