6. Let $n \geqslant 3, b_{n}$ be the number of subsets of the set $\{1, 2, \cdots, n\}$ that have the following property: any two elements in these subsets (which must contain at least two elements) have an absolute difference greater than 1. Then the value of $b_{10}$ is $\qquad$
6. 133

Let $T_{n, k}$ be the number of $k$-element subsets of $\{1,2, \cdots, n\}$ that satisfy the given conditions, denoted as $\mid a_{1} a_{2} \cdots a_{k} \mid$. For such a $k$-element subset, let $b_{1}=a_{1}, b_{2}=a_{2}-1, b_{3}=a_{3}-2, \cdots, b_{k}=a_{k}-(k-1)$. 

We establish a one-to-one mapping $\left\{a_{1} a_{2} \cdots a_{k}\right\} \rightarrow\left\{b_{1} b_{2} \cdots b_{k}\right\}$. Since $\left|a_{1} a_{2} \cdots a_{k}\right|$ has the property that $\left\{b_{1} b_{2} \cdots b_{k} \mid\right.$ is a $k$-element subset of $\{1,2, \cdots, n-k+1\}$, and for every $k$-element subset of $\{1,2, \cdots, n-k+1\}$, there corresponds a subset $\left\{a_{1} a_{2} \cdots a_{k}\right\}$ that meets the requirements, we have $T_{n, k}=C_{n-k+1}^{k}$, and $b_{n}=\sum_{k=2}^{\left[\frac{n+1}{2}\right]} C_{n-k+1}^{k}$.

Therefore, $b_{10}=\sum_{k=2}^{5} C_{11-k}^{k}=C_{9}^{2}+C_{8}^{3}+C_{7}^{4}+C_{6}^{5}=133$.