Example 7 In a simple graph with 8 vertices, what is the maximum number of edges in a graph that does not contain a quadrilateral? (A simple graph is one where no vertex is connected to itself by an edge, and there is at most 1 edge between any two vertices)

As shown in Figure 7-10, the graph has 8 vertices and 11 edges, and there is no quadrilateral, so the maximum number of edges sought is no less than 11.
Next, we prove that if the graph has 12 edges, then it must contain a quadrilateral.
First, we point out two obvious facts:
(i) If $A \neq B$ are two vertices, and if point $A$ is connected to points $C_{1}, C_{2}, \cdots, C_{k}$, and $B$ is connected to at least two points in $\left\{C_{1}, C_{2}, \cdots, C_{k}\right\}$, then the graph must contain a quadrilateral.
(ii) If 4 points are connected by 5 edges, then the graph must contain a quadrilateral.
Assume that in a graph with 8 vertices and 12 edges, there is no quadrilateral, and let point $A$ be the vertex with the most edges.
(1) Suppose $A$ has $m \geqslant 5$ edges, and the set of vertices connected to $A$ is denoted as $M$, and the set of all other vertices except $A$ and $M$ is denoted as $T$. By (i) and (ii), the number of edges between points in $M$ is at most $\left[\frac{m}{2}\right]$, the number of edges between points in $T$ is at most $C_{|T|}^{2}$, and the number of edges between $M$ and $T$ is at most $|T|=7-m$. Therefore, the total number of edges in the graph is at most
$$
m+\left[\frac{m}{2}\right]+|T|+C_{|T|}^{2}=7+\left[\frac{m}{2}\right]+C_{|T|}^{2}.
$$

When $m=5,6,7$, $T=2,1,0$, the value of the above expression is 10 in all cases, meaning the graph has only 10 edges, which is a contradiction.
(2) Point $A$ has exactly 4 edges: $A A_{i}(i=1,2,3,4)$. Let the other 3 points be $B_{1}, B_{2}, B_{3}$. Then, the set $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ has at most 2 edges, the set $\left\{B_{1}, B_{2}, B_{3}\right\}$ has at most 3 edges, and the set $\left\{B_{1}, B_{2}, B_{3}\right\}$ and $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ have at most 3 edges between them. Since the graph has 12 edges, the number of edges in these 3 groups is exactly 2, 3, and 3. Assume the 3 edges in the third group are $A_{j} B_{j}(j=1,2,3)$. Since the first group has 2 edges with no common endpoints, there is an edge among $\left\{A_{1}, A_{2}, A_{3}\right\}$, say $A_{1} A_{2}$. Thus, $A_{1} A_{2} B_{2} B_{1}$ forms a quadrilateral, which is a contradiction.
(3) Point $A$ has exactly 3 edges. Since the graph has 12 edges, each point must have 3 edges. Suppose there is no edge between points $A$ and $B$, and the 3 edges from each are $A A_{j}, B B_{j}(j=1,2,3)$. Then, by (i), the sets $\left\{A_{1}, A_{2}, A_{3}\right\}$ and $\left\{B_{1}, B_{2}, B_{3}\right\}$ have at most one common point.

If they have no common point, the 3 points in each set can have at most one edge between them, and the two 3-point sets can have at most 3 edges between them, so the total number of edges in the graph is at most $3+3+3+1+1=11$, which is a contradiction. If the two 3-point sets have exactly one common point, then consider the 8th point $C$. By the pigeonhole principle, at least two of the 3 edges from $C$ must go to the same 3-point set, which leads to the formation of a quadrilateral, which is a contradiction.

In summary, we have proven that a graph with 8 vertices and 12 edges must contain a quadrilateral, so the maximum number of edges sought is 11.