Example 3 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 two elements, and the absolute difference between any two elements in each subset is greater than 1.
Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ that satisfy the given property. Then, the subsets of $\{1,2, \cdots, k, k+1, k+2\}$ that satisfy the given property can be divided into two categories: the first category does not contain $k+2$, and there are $a_{k+1}$ such subsets; the second category contains $k+2$, and these subsets are either the union of a corresponding subset of $\{1,2, \cdots, k\}$ and $\{k+2\}$, or the union of a single-element subset of $\{1,2, \cdots, k\}$ and $\{k+2\}$, totaling $a_{k}+k$ subsets.
Thus, $a_{k+2}=a_{k+1}+a_{k}+k$.
Clearly, $a_{3}=1, a_{4}=3$.
Therefore, $a_{5}=7, a_{6}=14, a_{7}=26, a_{8}=46, a_{9}=79, a_{10}=133$.
Comment: Establishing a recurrence relation is an important method in combinatorial counting. We can also solve the problem by classifying and using the gap-filling method: $C_{9}^{2}+C_{8}^{3}+C_{7}^{4}+C_{6}^{5}=133$. Then, how can we solve the general case $\{1,2, \cdots, n\}$ using classification and the gap-filling method? The two methods have different forms of expression, but what is the intrinsic relationship between them? Please think about it yourself.