1. The function $y=f(x)$, with domain and range both $R$, is an increasing function. Let the solution set of the equation $f(x)=x$ be $P$, and the solution set of the equation $f[f(x)]=x$ be $Q$. Then, we have () .
A. $P=Q$
B. $P \subset Q$
C. $Q \subset P$
1). $P \not \subset Q, Q \not \subset P$
1. A

If $t \in P$, then $f(t)=t$, so $f[f(t)]=f[t]=t$, hence $t \in \mathbf{Q}$. Therefore, $P \subseteq \mathbf{Q}$. If $t \in \mathbf{Q}$, let $y=f(t)$. Then $f(y)=t$. Suppose $y>t$, then $f(y)>f(t)$, i.e., $t>y$, a contradiction. Similarly, $y<t$ also leads to a contradiction, so $y=t$, i.e., $t \in P$. In summary, $P=\mathbf{Q}$.