8- 23 Given real numbers $\alpha, \beta$, such that $\alpha \beta>0$. Find the set of real numbers $r$ with the following property: there does not exist an infinite sequence $\left\{x_{n}\right\}$ satisfying
$$x_{0}=r, x_{n+1}=\frac{x_{n}+\alpha}{\beta x_{n}+1}, n=0,1,2, \cdots$$
Let the required set be $M$, and the function
$$f(x)=\frac{x+\alpha}{\beta x+1} \quad\left(x \neq-\frac{1}{\beta}\right)$$

If $\alpha \beta=1$, then for any $x \neq-\frac{1}{\beta}$, we have
$$f(x)=\frac{x+\alpha}{\beta x+1}=\alpha=\frac{1}{\beta} \neq-\frac{1}{\beta},$$

Thus, for any $r \not \neq-\frac{1}{\beta}$, there exists a sequence $\left\{x_{n}\right\}$ that satisfies the given conditions. Therefore, $M=\left\{-\frac{1}{\beta}\right\}$ is a singleton set.

Assume $\alpha \beta \neq 1$. First, consider the case where $\alpha>0, \beta>0$. Let
$$f^{[0]}(x)=x, f^{[1]}(x)=f(x), f^{[n+1]}(x)=f\left(f^{[n]}(x)\right), n=1,2,$$
...

Clearly, $M=\left\{r \in R \mid\right.$ there exists a non-negative integer $k$ such that $\left.f^{[k]}(r)=-\frac{1}{\beta}\right\}$. The inverse function of $f(x)$ is
$$g(x)=\frac{x-\alpha}{1-x \beta} \quad\left(x \neq \frac{1}{\beta}\right) .$$

Since $\alpha \beta \neq 1$, it is easy to see that $\frac{1}{\beta}$ does not belong to the range of the function $f(x)$, so
$$f^{[k]}(r)=-\frac{1}{\beta} \Leftrightarrow r=g^{[k]}=\left(-\frac{1}{\beta}\right) .$$

Thus, $M=\left\{r_{0}, r_{1}, r_{2}, \cdots\right\}$,
where $\quad r_{0}=g^{[0)}\left(-\frac{1}{\beta}\right)=-\frac{1}{\beta}, r_{n}=g^{[n]}\left(-\frac{1}{\beta}\right), n=1,2,3, \cdots$ Since $\alpha, \beta>0$, for $x \leqslant 0$
we have $g(x)=\frac{x-\alpha}{1-x \beta}0$, then
$$\begin{aligned}
\frac{r_{n+1}+b}{r_{n+1}-b} & =\frac{r_{n}-\alpha+b-b r_{n} \beta}{r_{n}-\alpha-b+b r_{n} \beta}=\frac{r_{n}(1-\sqrt{\alpha \beta})+b(1-\sqrt{\alpha \beta})}{r_{n}(1+\sqrt{\alpha \beta})-b(1+\sqrt{\alpha \beta})} \\
& =\frac{r_{n}+b}{r_{n}-b} \cdot \frac{1-\sqrt{\alpha \beta}}{1+\sqrt{\alpha \beta}}=\cdots=\left(\frac{1-\sqrt{\alpha \beta}}{1+\sqrt{\alpha \beta}}\right)^{n+1} \frac{r_{0}+b}{r_{0}-b} \\
& =\left(\frac{1-\sqrt{\alpha \beta}}{1+\sqrt{\alpha \beta}}\right)^{n+1} \cdot \frac{-\frac{1}{\beta}+\sqrt{\frac{\alpha}{\beta}}}{-\frac{1}{\beta}-\sqrt{\frac{\alpha}{\beta}}}=\left(\frac{1-\sqrt{\alpha \beta}}{1+\sqrt{\alpha \beta}}\right)^{n+2} .
\end{aligned}$$

Let $\lambda=\frac{1-\sqrt{\alpha \beta}}{1+\sqrt{\alpha \beta}}$, clearly $|\lambda|<1$, thus
$$r_{n}=-\sqrt{\frac{\alpha}{\beta}} \frac{1+\lambda^{n+1}}{1-\lambda^{n+1}}, n=0,1,2,3, \cdots$$

If $\alpha \beta \neq 1$ and $\alpha<0, \beta<0$, similarly, we can prove
$$M=\left\{r_{0}, r_{1}, r_{2}, \cdots\right\}$$

where $\quad r_{n}=\sqrt{\frac{\alpha}{\beta}} \frac{1+\lambda^{n+1}}{1-\lambda^{n+1}}, n=0,1,2,3, \cdots$
$$\lambda=\frac{1-\sqrt{\alpha \beta}}{1+\sqrt{\alpha \beta}} \text {. }$$