10. (1st "Hope Cup" Senior High School Grade 2nd Test, Question 3) $f$ 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. $P \ni Q$
B. $P=Q$
C. $Q \varsubsetneqq P$
D. $P \varsubsetneqq Q$ and $Q \varsubsetneqq P$
10. For any $x_{0} \in P, x_{0}>f\left(x_{0}\right)$, we have $x_{0}>f\left[f\left(x_{0}\right)\right]$. Therefore, $x_{0} \in Q$. This shows that $P \subseteq Q$; on the other hand, for any $y_{0} \in Q$, we have $y_{0}>f\left[f\left(y_{0}\right)\right]$. We will use proof by contradiction to show that $y_{0}>f\left(y_{0}\right)$.

Assume $y_{0} \leqslant f\left(y_{0}\right)$, since the function $y=f(x)$ is strictly increasing, then we have $f\left(y_{0}\right) \leqslant f\left[f\left(y_{0}\right)\right]$. This implies $y_{0} \in P$. Thus, $Q \subseteq P$.
Given $P \subseteq Q$ and $Q \subseteq P$, we conclude that $P=Q$. Choose B.
Explanation: To determine the relationship between set $P$ and set $Q$, we must use the definitions of the sets. The key to solving this problem is to infer from $y_{0}>f\left[f\left(y_{0}\right)\right]$ that $y_{0}>f\left(y_{0}\right)$, and then use proof by contradiction, case analysis, and other mathematical ideas to solve it.

Solution: Take a special function, let $f(x)=2x$, then $f[f(x)]=4x$. 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 excluding A, C, D, choose B.