19.6.5 ** Given positive integers $m, a, b, (a, b)=1$. $A$ is a non-empty subset of the set of positive integers, such that for any positive integer $n$, either $a n \in A$ or $b n \in A$. For all sets $A$ with the above property, find the minimum value of $|A \cap \{1, 2, \cdots, m\}|$.


When $a=b=1$, for any positive integer $n, 1 \cdot n=n \in$. Therefore, $A$ is the set of positive integers, $|A \cap\{1,2, \cdots, m\}|=m$.

Assume $a$ and $b$ are not both 1. Without loss of generality, let $a>b$. Consider the set $A_{1}=\left\{n \mid a^{2 i-1} \| n, i=1,2, \cdots\right\}$. For any integer $n$, if $a^{2 i-2} \| n$, then $a^{2 i-1} \| a n$, i.e., $a n \in A_{1}$. If $a^{2 i-1} \| n$, then $a^{2 i-1} \| b n$, i.e., $b n \in A_{1}$. Therefore, $A_{1}$ has the property mentioned in the problem.
By the principle of inclusion-exclusion,
$$
\left|A_{1} \cap\{1,2, \cdots, m\}\right|=\sum_{i=1}^{\infty}(-1)^{i+1}\left[\frac{m}{a^{i}}\right] .
$$

Let $A$ be any set with the mentioned property. If there is $a n \notin A$ while $a \nmid n$, then $b n \in A$. Removing $b n$ from $A$ and adding $a n$ to $A$ results in a new set $A^{\prime}$ with the same number of elements. $A^{\prime}$ still has the property mentioned in the problem: for any positive integer $m$, if $a m \in A$, then $a m \in A^{\prime}$. If $b m \in A$ and $b m \notin A^{\prime}$, then $a m \in A^{\prime}$. Clearly, $b n < a n$. $|A \cap\{1,2, \cdots, m\}| \geqslant\left|A^{\prime} \cap\{1,2, \cdots, m\}\right|$.
Therefore, we can assume that $A$ contains all $n$ such that $a \| n$.
Assume $A$ contains all $n$ such that $a^{2 i-1} \| n, i=1,2, \cdots, k$.
If there is $a^{2 i+1} n \notin A$ while $a \nmid n$, then $b a^{2 i} n \in A$. Removing $b a^{2 i} n$ from $A$ and adding $a^{2 i+1} n$ to $A$ results in a new set $A^{\prime}$ which still has the property mentioned in the problem: for any positive integer $m$, if $a m \in A$ and $b m \notin A$, then $a m \in A^{\prime}$. If $b m \in A$ and $b m \notin A^{\prime}$, then $a m \in A^{\prime}$. And (1) holds.
Therefore, we can always assume $A \supseteq A_{1} .|A \cap\{1,2, \cdots, m\}|$ has the minimum value of $\sum_{i=1}^{\infty}(-1)^{i+1}\left[\frac{m}{a^{i}}\right]$.