5. $f$ is a one-to-one mapping on $\mathbf{R} \rightarrow \mathbf{R}$, $y=f(x)$ is strictly increasing, the solution set of the equation $x=f(x)$ is $P$, and the solution set of the equation $x=$ $f(f(x))$ is $Q$, then
A. $P \subset Q$
B. $P=Q$
C. $Q \ni P$
D. $P \nsubseteq Q$, and $Q \nsubseteq P$
5. B Let $x$ satisfy: $x=f(x)$. Since $f(f(x))=f(x)=x$, we have $x \in Q, P \subset Q$; if $Q \nsubseteq P$, then there exists $x_{0} \in Q, x_{0} \notin P, f(f(x_{0}))=x_{0}, f(x_{0}) \neq x_{0}$. If $f(x_{0})x_{0}$ also leads to a contradiction!