7・118 Let $S=\{1,2, \cdots n\}, A$ be a set composed of several subsets of set $S$, such that any two elements in $A$ as two subsets of $S$ do not contain each other. Find the maximum value of the number of elements $|A|$ in set $A$.
[Solution] For any such set $A$, let $A$ contain $f_{k}$ $k$-element subsets of $S$, for $k=1,2, \cdots, n$. Thus, $f_{k}$ are non-negative integers and
$$
|A|=\sum_{k=1}^{n} f_{k} \text {. }
$$

When $f_{k}$ is a positive integer, for each $k$-element subset $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ of $S$ that belongs to $A$, the number of all permutations of $S$ with $a_{1}, a_{2}, \cdots, a_{k}$ as the first $k$ elements is $k!(n-k)!$. Since any two elements in $A$ as subsets of $S$ do not contain each other, the permutations derived from the subsets in $A$ are all distinct. Therefore, we have
$$
\sum_{k=1}^{n} f_{k} k!(n-k)!\leqslant n!\text {. }
$$

Furthermore, for a fixed positive integer $n$, the binomial coefficients $\left\{C_{n}^{0}, C_{n}^{1}, \cdots, C_{n}^{n}\right\}$ are maximized at $C_{n}^{\left[\frac{n}{2}\right]}$. Hence, from (1) and (2), we have
$$
\begin{aligned}
|A| & =\sum_{k=1}^{n} f_{k} \leqslant C{ }_{n}^{\left[\frac{n}{2}\right]} \sum_{k=1}^{n} \frac{f_{k}}{C_{n}^{k}} \\
& =C_{n}^{\left[\frac{n}{2}\right]} \frac{1}{n!} \sum_{k=1}^{n} f_{k} k!(n-k)!\leqslant C_{n}^{\left[\frac{n}{2}\right]} .
\end{aligned}
$$

On the other hand, when $A_{0}$ is the set of all $\left[\frac{n}{2}\right]$-element subsets of $S$, we have $\left|A_{0}\right|=C_{n}^{\left[\frac{n}{2}\right]}$.
In summary, the maximum number of elements in a set $A$ that meets the requirements of the problem is $C_{n}^{\left[\frac{n}{2}\right]}$.