Example 1 (Question 3 of the 2nd Test for Grade 11 in the 1st "Hope Forest" Competition) $f$ is a one-to-one mapping from $\mathbf{R} \rightarrow \mathbf{R}$, and the function $y=f(x)$ is strictly increasing. The solution set of the inequality $x>f(x)$ is $P$, and the solution set of the inequality $x>f[f(x)]$ is $Q$. Then
A. $P \varsubsetneqq Q$
B. $P=Q$
C. $Q \varsubsetneqq P$
D. $P \varsubsetneqq Q$ and $Q \varsubsetneqq P$
Solution: For any $x_{n} \in P, x_{0}>f\left(x_{n}\right)$, we have $x_{n}>f\left[f\left(x_{0}\right)\right]$, so $x_{n} \in \mathbf{Q}$. This shows that $P \subseteq \mathbf{Q}$. On the other hand, for any $y_{0} \in \mathbf{Q}$, then $y_{0}>f\left[f\left(y_{0}\right)\right]$. We will prove $y_{0}>f\left(y_{0}\right)$ by contradiction.

Assume $y_{0} \leqslant f\left(y_{0}\right)$, since the function $y=f(x)$ is strictly increasing, then $f\left(y_{0}\right) \leqslant f\left[f\left(y_{0}\right)\right]f\left(y_{0}\right)$, which implies $y_{0} \in P$, i.e., $Q \subseteq P$.
By $P \subseteq Q$, and $Q \subseteq P$, we know $P=Q$. Choose (B).
Analysis: To determine the relationship between set $P$ and set $Q$, we must prove it using the definition of sets. The key to solving this problem is to guess from $y_{0}>f\left[f\left(y_{0}\right)\right]$ that $y_{0}>f\left(y_{0}\right)$, and then use mathematical ideas and methods such as proof by contradiction and case analysis to solve it.
Note that this is a multiple-choice question, so we can also use special techniques for solving multiple-choice questions to eliminate incorrect options.
Solution: Take a special function, let $f(x)=2 x$, then $f[f(x)]=4 x$, the solution set $P$ of the inequality $x>f(x)$ and the solution set $Q$ of the inequality $x>$ $f[f(x)]$ are equal, so $P=Q$, thus eliminating A, C, D. Choose B.
The result of this problem is too amazing! It naturally inspires us to think about the following issues:
(1) Can the condition “$f$ is a one-to-one mapping from $\mathbf{R} \rightarrow \mathbf{R}$, and the function $y=f(x)$ is strictly increasing” be weakened? Does the conclusion still hold when the function $y=f(x)$ is not strictly increasing?
The answer is no. Consider the following counterexample:
Counterexample: When $f(x)=x^{2}-2$, the solution set of $f(x)f[f(x)]$ is $Q$, which is different from $P$.
(2) Extension: If we change the condition “the solution set of the inequality $x>f(x)$ is $P$, and the solution set of the inequality $x>f[f(x)]$ is $Q$” to the general form “the solution set of the inequality $x>$ $f\{f \cdots[f(x)]\}(n$ times $f$) is $Q$”, the conclusion remains unchanged.
(3) Extension: If we change the condition to “the solution set of the inequality $x \geqslant f(x)$ is $P$, and the solution set of the inequality $x \geqslant f[f(x)]$ is $Q$”, the conclusion remains unchanged, i.e., $P=Q$. The proof process can be modified accordingly.

Proof: For any $x_{0} \in P, x_{11} \geqslant f\left(x_{0}\right)$, we have $x_{0} \geqslant f\left(x_{0}\right) \geqslant f\left[f\left(x_{0}\right)\right]$, so $x_{0} \in \mathbf{Q}$. This shows that $P \subseteq Q$. On the other hand, for any $y_{0} \in \mathbf{Q}$, then $y_{0} \geqslant f\left[f\left(y_{n}\right)\right]$. We will prove $y_{0} \geqslant f\left(y_{n}\right)$ by contradiction.

Assume $y_{0}x$;
(3) When $f(x)$ is an increasing function, the system $\left\{\begin{array}{l}f(x)>g(x), \\ f^{-1}(x)=g(x)\end{array}\right.$ is equivalent to $g(x)>x$.
The result of this problem is really amazing!
(5) Let $f(x)=\left(1+\frac{x}{2}\right)^{2}-2, x \in[-2,+\infty)$, then the solution set of the inequality $f(x) \geqslant f^{-1}(x)$ is
Similarly, we can obtain: the solution set of the inequality $f(x) \geqslant f^{-1}(x)$ is the same as the solution set of $f(x) \geqslant x$, which is $[-2,2]$.