6. Let $A$ be a set of numbers, satisfying that if $a \in A$, then $\frac{1}{1-a} \in A$, and $1 \notin A$.
(1) If $2 \in A$, then how many more elements are there in $A$ at least? Find these elements.
(2) Can $A$ be a singleton set? Discuss separately in the set of real numbers and the set of complex numbers.
(3) If $a \in A$, prove that $1-\frac{1}{a} \in A$.
6. Solution (1) $2 \in A \Rightarrow -1 \in A \Rightarrow \frac{1}{2} \in A \Rightarrow 2 \in A$,

Therefore, $A$ contains at least two more elements: -1 and $\frac{1}{2}$.
(2) If $A$ is a singleton set, then $a=\frac{1}{1-a}$,
which means $a^{2}-a+1=0$,
this equation has no real solutions, hence in the real number range, $A$ cannot be a singleton set.
However, the equation has two complex solutions: $a=\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$,
thus in the complex number range, $A$ can be a singleton set, $A=\left\{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right\}$ or $A=\left\{\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right\}$.
(3) $a \in A \Rightarrow \frac{1}{1-a} \in A \Rightarrow \frac{1}{1-\frac{1}{1-a}} \in A$, which means $1-\frac{1}{a} \in A$,