Example 3 Let $A_{1}, A_{2}, \cdots, A_{k}$ be a family of subsets of $X=\{1,2, \cdots, n\}$. If for any pair of elements $i, j$ in $X$, there is always one subset among $A_{1}, A_{2}, \cdots, A_{k}$ that contains exactly one of $i, j$, then this family of subsets is called separable. Find the smallest $k$ such that there is a family of subsets $A_{1}, A_{2}, \cdots, A_{k}$ that is both a cover and separable.
Consider the subordination table mentioned in Section 1.2. When $A_{1}, A_{2}, \cdots, A_{k}$ form a cover, each column has at least one 1. When $A_{1}, A_{2}, \cdots, A_{k}$ are separable, no two columns are identical.

Since the table has $k$ rows and each element in the table is either 0 or 1, at most $2^{k}-1$ distinct columns can be formed, with no column consisting entirely of 0s. Therefore,
$$
2^{k}-1 \geqslant n,
$$

which implies
$$
k \geqslant\left[\log _{2} n\right]+1 .
$$

On the other hand, take $k$ such that
$$
2^{k}-1 \geqslant n \geqslant 2^{k-1} .
$$

Construct $n$ distinct columns of length $k$ composed of 0s and 1s, with no column consisting entirely of 0s (this is possible because $2^{k}-1 \geqslant n$). The $k$ rows of this table represent $k$ sets that are both a cover and separable. Therefore, the minimum value of $k$ is $\left[\log _{2} n\right]+1$.