Example 5 Let $A=A_{1} \cup A_{2} \cup \cdots \cup A_{k}, S=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$. If (i) the intersection of any $r$ elements in $S$ is non-empty; (ii) the intersection of any $r+1$ elements is an empty set. Questions: (1) What is the minimum size of $|A|$? (2) When $|A|$ is minimal, what is the size of $|A_{i}|$?
(1) Consider any $r$-element subset $\left\{i_{1}, i_{2}, \cdots, i_{r}\right\}$ of the index set $\{1,2, \cdots, k\}$. Take any element $a$ in $A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i}$. Define a mapping $f:\left\{i_{1}, i_{2}, \cdots, i_{r}\right\} \rightarrow a$.

Thus, $f$ is an injection from the set of $r$-element subsets of the index set to $A$. In fact, if $\left\{j_{1}, j_{2}, \cdots, j_{r}\right\}$ also corresponds to $a$, it would imply that the intersection of $r+1$ subsets is non-empty, which is a contradiction.
Therefore, $|A|$ is no less than the number of $r$-element subsets of the index set, i.e., $|A| \geqslant C_{k}^{r}$.
(2) Consider any $A_{i}$, and take any other $r$ sets in $S$. Their intersection contains at least one element (is non-empty), and this intersection is disjoint from $A_{i}$. Since there are $C_{k-1}^{r}$ different ways to choose these sets, $\left|A_{i}\right| \leq|A|-C_{k-1}^{r}$. When $|A|=C_{k}^{r}$, $\left|A_{i}\right| \leqslant C_{k-1}^{r-1}$.
On the other hand, the intersection of $A_{i}$ with any $r-1$ other sets chosen from $S^{\prime}$ contains at least one element, so $\left|A_{i}\right| \geqslant C_{k-1}^{r-1}$. This shows that when $|A|$ takes its minimum value $C_{k}^{r}$, each set $A_{i}$ has size $C_{k-1}^{r-1}$.