Example 7 On a plane, there exist $n$ points, no three of which are collinear, and when these $n$ points are arbitrarily labeled as $A_{1}, A_{2}, \cdots, A_{n}$, the broken line $A_{1} A_{2} \cdots A_{n}$ does not intersect itself. Find the maximum value of $n$.


【Analysis】When $n=2,3$, it is obviously true.
When $n=4$, if the convex hull of the four points is a triangle, then it satisfies the condition.
Next, we show that when $n \geqslant 5$, it is impossible to satisfy the condition.
Consider only five points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ among these $n$ points.

If the convex hull of these five points is a pentagon $A B C D E$ or a quadrilateral $A B C D$, then the broken line $A C B D$ does not satisfy the condition.

Assume the convex hull of points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ is $\triangle P_{1} P_{2} P_{3}$, i.e., points $P_{4}, P_{5}$ are inside $\triangle P_{1} P_{2} P_{3}$. The line connecting $P_{4}, P_{5}$ does not pass through $P_{1}, P_{2}, P_{3}$, so it must intersect two sides of $\triangle P_{1} P_{2} P_{3}$. Assume without loss of generality that $P_{4} P_{5}$ intersects $P_{1} P_{3}$ and $P_{2} P_{3}$. Then, quadrilateral $P_{1} P_{2} P_{4} P_{5}$ is a convex quadrilateral, which does not satisfy the condition.

Therefore, $n \geqslant 5$ does not satisfy the condition, i.e., the maximum value of $n$ is 4.

Using the convex hull to analyze problems involving finite point sets is a common method in combinatorial geometry.