Example 5 (Canada Olympiad) A set of real numbers is called "simple" if it contains an element $z$ such that $x+y=z$. Given the set $\{1,2, \cdots, 2 n+1\}$, find the maximum number of elements a simple subset can contain.
Consider the two subsets of $\{1,2, \cdots, 2 n+1\}$: $A=\{1,3,5, \cdots 2 n+1\}, B=\{2,4,6, \cdots, 2 n\}$. Since the sum of any two numbers in $A$ is even and thus not in $A$, $A$ is a simple subset, and the number of elements in $A$ is $n+1$. We will now prove that $n+1$ is the maximum number of elements in any simple subset of $\{1, 2, \cdots, 2 n+1\}$. Suppose $C$ is a subset of the original set with the maximum number of elements being $n+2$, and let $C$ contain $p$ numbers from $A$ and $n+2-p$ numbers from $B$. Clearly, $p \geqslant 2$. Let the numbers from $A$ in $C$ be $a_{1} < a_{2} < \cdots < a_{p}$, then $0 < a_{2} - a_{1} < a_{3} - a_{1} < \cdots < a_{p} - a_{1}$ are $p-1$ distinct even numbers, which should be part of subset $B$. Let the $n+2-p$ numbers from $B$ in $C$ be $b_{1} < b_{2} < \cdots < b_{n+2-p}$. Since $(p-1) + (n+2-p) = n+1$, but $B$ contains only $n$ even numbers, there must be indices $i$ and $j$ such that $a_{i} - a_{1} = b_{j}$, where $2 \leqslant i \leqslant p$ and $1 \leqslant j \leqslant n+2-p$. Thus, $a_{i} = a_{1} + b_{j}$, indicating that $C$ is not a simple subset. This shows that the maximum number of elements in a simple subset cannot exceed $n+1$. However, we already know that the simple subset $A$ has $n+1$ elements, which completes the proof.