Example 5 Let $S_{n}$ denote some subsets of the set of positive integers $\{1,2,3, \cdots, 100\}$ that satisfy the condition: no number is twice another. What is the maximum number of elements such a subset can contain?
$$
\text { Solution: Let } \begin{aligned}
A_{1} & =\{51,52,53, \cdots, 100\}, \\
A_{2} & =\{26,27, \cdots, 50\}, \\
A_{3} & =\{13,14, \cdots, 25\}, \\
A_{4} & =\{7,8, \cdots, 12\}, \\
A_{5} & =\{4,5,6\}, \\
A_{6} & =\{2,3\}, \\
A_{7} & =\{1\},
\end{aligned}
$$

Then $A=A_{1} \cup A_{3} \cup A_{5} \cup A_{7}$ satisfies the condition. This set has $50+13+3+1=67$ (elements). Next, we prove that $A$ is the set with the maximum number of elements that satisfies the condition.
If $B \subseteq\{1,2,3, \cdots, 100\}$, and no element in $B$ is twice another, then when $a \in B$, $2a \notin B \cap A_{2}$.
Therefore, $\operatorname{Card}\left(B \cap A_{2}\right)+\operatorname{Card}\left(B \cap A_{1}\right) \leqslant 50$.
Similarly, $\operatorname{Card}\left(B \cap A_{4}\right)+\operatorname{Card}\left(B \cap A_{3}\right) \leqslant 13$.
$$
\operatorname{Card}\left(B \cap A_{6}\right)+\operatorname{Card}\left(B \cap A_{5}\right) \leqslant 3 .
$$

Thus, Card $B \leqslant 50+13+3+1=67$.
Comment: Proof and construction are two methods to solve this type of problem.
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