$2 \cdot 56$ Let $p \in N, n=2^{p}$. Consider the subset $A$ of $E=\{1,2, \cdots, n\}$ with the following property: if $x \in A$, then $2 x \in A$. Find the maximum number of elements in such a subset $A$.
[Solution] For any $e \in E$, we write $e=2^{s} t$, where $s$ is a non-negative integer, and $t$ is an odd number. According to the given condition, if $2^{s} t \in A$, then $2^{s-1} t \notin A, 2^{s+1} t \notin A$. Therefore, for each $t<n$, the number of elements in the intersection of set $A$ and the set $\left\{t, 2 t, 2^{2} t, \cdots\right\}$ is no more than the number of elements in the set $\left\{t, 2^{2} t, \cdots, 2^{2 m} t\right\}$, where $m$ is the largest non-negative integer such that $2^{2 m} t \leqslant n$. Thus,

summing the right-hand side of the above expression, we get
$$
|A| \leqslant\left\{\begin{array}{l}
\frac{1}{3}\left(2^{p+1}-1\right) \\
\frac{1}{3}\left(2^{p+1}+1\right)
\end{array}\right\}=\left[\frac{2}{3} n\right]+\varepsilon_{p},
$$

where
$$
\varepsilon_{p}=\left\{\begin{array}{ll}
0, & \text { when } p \text { is odd, } \\
1, & \text { when } p \text { is even. }
\end{array}\right.
$$

It is easy to see that the equality in inequality (1) can be achieved, hence the maximum value of $|A|$ is $\left[\frac{2}{3} n\right]+\varepsilon_{p}$, where $\varepsilon_{p}$ is given by (2).