$\mathbf{5 5}$ Set $A=\{0,1,2, \cdots, 9\}, B_{1}, B_{2}, \cdots, B_{k}$ are a family of non-empty subsets of $A$, and when $i \neq$ $j$, $B_{i} \cap B_{j}$ has at most two elements. Find the maximum value of $k$.

55 It is easy to know that the number of non-empty subsets containing at most three elements is
$$
C_{10}^{1}+C_{10}^{2}+C_{10}^{3}=175 \text { (sets). }
$$

Let $C$ be the family of subsets satisfying the condition. If $|B| \geqslant 4$, then $B$ and $B-\{b\},(b \in B)$ cannot both be contained in $C$. By replacing $B$ with $B-\{b\}$, the number of elements in $C$ remains unchanged. Similarly, for all sets $B$ in $C$ with more than 4 elements, make the corresponding replacements. After these replacements, each set in the family $C$ is a non-empty set with at most 3 elements. Therefore, the maximum value of $k$ is 175.