Example 3 (China Girls Mathematical Olympiad) Find all positive real numbers $a$ such that there exists a positive integer $n$ and $n$ pairwise disjoint infinite sets $A_{1}, A_{2}, \cdots, A_{n}$ satisfying $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=\mathbf{Z}$, and for any two numbers $b>c$ in each $A_{i}$, we have $b-c \geqslant a^{i}$.
Analysis We look for the most familiar model, and find that classifying integers into odd and even, and then further classifying the even set based on the odd or even nature after dividing by 2, can be extended similarly. Although it requires infinite classification to meet the condition, it provides useful information: Is 2 the boundary? Considering special cases and conjectures are important and commonly used methods in math competitions.
Solution If $0a^{n}$, let
$A_{i}=\left\{2^{i-1} \mathrm{~m} \mid \mathrm{m}\right.$ is odd $\}, i=1,2, \cdots, n-1$,
$A_{n}=\left\{2^{n-1}\right.$ multiples $\}$, then this partition satisfies the requirement.
If $a \geqslant 2$, suppose $A_{1}, A_{2}, \cdots, A_{n}$ satisfy the requirement, let $M=\left\{1,2, \cdots, 2^{n}\right\}$, we will prove $\left|A_{i} \cap M\right| \leqslant 2^{n-i}$.
Assume $A_{i} \cap M=\left\{x_{1}, x_{2}, \cdots, x_{m}\right\}, x_{1}x_{m}-x_{1}=\left(x_{m}-x_{m-1}\right)+\left(x_{m-1}-x_{m-2}\right)+\cdots+\left(x_{2}-x_{1}\right) \geqslant(m-1) 2^{i}$.
Thus, $m-1<2^{n-i}$, i.e., $m<2^{n-i}+1$, hence $m \leqslant 2^{n-i}$.
$A_{i} \cap M, i=1,2, \cdots, n$ is a partition of $M$, so
$2^{n}=|M|=\sum_{i=1}^{n}\left|A_{i} \cap M\right| \leqslant \sum_{i=1}^{n} 2^{n-i}=2^{n}-1$, a contradiction.
Therefore, the required $a$ is all positive real numbers less than 2.