$4 \cdot 25$ Introduce a relation “ $\rightarrow$ ” among the elements of set $S$, such that
(1) For any $a, b \in S$, either $a \rightarrow b$ or $b \rightarrow a$, but not both;
(2) For any $a, b, c \in S$, if $a \rightarrow b, b \rightarrow c$, then $c \rightarrow a$.

How many elements can set $S$ contain at most?
[Solution] Clearly, when $S=\{a, b, c\}$ and the relationships among them are $a \rightarrow b, b \rightarrow c$, $c \rightarrow a$, the conditions of the problem are satisfied. Therefore, the maximum number of elements sought is no less than 3.

If $S$ contains 4 elements $a, b, c, d$, then by the pigeonhole principle, the arrow relationships between $a$ and $b, c, d$ must include at least two that are the same, i.e., either two arrows point to $a$, or two arrows leave $a$. Suppose $b \rightarrow a, c \rightarrow a$. By symmetry, we can assume $b \rightarrow c$. Thus, we have $b \rightarrow c, c \rightarrow a$. According to the given (2), there should be $a \rightarrow b$, which is a contradiction.
In summary, $S$ can contain at most 3 elements.