$7 \cdot 58$ For the set $S=\left\{\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right) \mid a_{i}=0\right.$ or $1, i=1,2,3$, $4,5\}$, for any two elements $\left(\bar{a}_{1}, \bar{a}_{2}, \bar{a}_{3}, \bar{a}_{4}, \bar{a}_{5}\right)$ and $\left(\bar{b}_{1}, \bar{b}_{2}, \bar{b}_{3}, \bar{b}_{4}, \bar{b}_{5}\right)$, define the distance between them as $\left|\bar{a}_{1}-\bar{b}_{1}\right|+\left|\bar{a}_{2}-\bar{b}_{2}\right|+\left|\bar{a}_{3}-\bar{b}_{3}\right|+\left|\bar{a}_{4}-\bar{b}_{4}\right|+$ $\left|\bar{a}_{5}-\bar{b}_{5}\right|$. Take a subset of $S$ such that the distance between any two elements in this subset is greater than 2. What is the maximum number of elements this subset can contain? Prove your conclusion.
［Solution］This subset can contain at most 4 elements.
For convenience, in the set $S$, we call $a_{i}(i=1,2,3,4,5)$ the $i$-th component of the element $A\left(a_{1}\right.$, $\left.a_{2}, a_{3}, a_{4}, a_{5}\right)$. Clearly, any two elements in the set $S$ with a distance greater than 2 can have at most 2 components with the same index.

First, we can construct a 4-element subset $S^{\prime}=\left\{A_{1}, A_{2}, A_{3}\right.$, $\left.A_{4}\right\}$ of $S$. Where $A_{1}=(1,1,0,0,0), A_{2}=(0,0,0,1,1), A_{3}=(1,0,1,0,1)$, $A_{4}=(0,1,1,1,0)$.
Clearly, the distance between any two elements in $S^{\prime}$ is greater than 2.
Next, we prove: In the subset of $S$, to ensure that the distance between any two elements is greater than 2, the number of elements cannot exceed 4.

Assume there is a 5-element subset $\left\{B_{1}, B_{2}, B_{3}, B_{4}, B_{5}\right\}$ of $S$, such that the distance between any two elements is greater than 2.

Since there are 5 elements, and each component can only be 0 or 1, by the pigeonhole principle, at least three elements have the same first component. Without loss of generality, assume the first components of $B_{1}, B_{2}, B_{3}$ are all 1.

Since $B_{1}, B_{2}, B_{3}$ have at least two elements with the same second component, without loss of generality, assume the second components of $B_{1}$ and $B_{2}$ are both 1.

At this point, if $B_{1}$ and $B_{2}$ have another component the same, it would contradict the conclusion that at most 2 components with the same index are the same. Therefore, it must be that
\begin{tabular}{crrrrl} 
& $a_{1}$ & $a_{2}$ & $a_{3}$ & $a_{4}$ & $a_{5}$ \\
$B_{1}$ & 1 & 1 & 1 & 0 & 1 \\
$B_{2}$ & 1 & 1 & 0 & 1 & 0 \\
$B_{3}$ & 1 & & & &
\end{tabular}

At this point, consider the 3rd, 4th, and 5th components of $B_{3}$. They must have two components with the same index as $B_{1}$ or $B_{2}$, which leads to at least three components with the same index, thus leading to a contradiction.

Therefore, in the subset of $S$, to ensure that the distance between any two elements is greater than 2, the number of elements cannot exceed 4.