How do I prove the middle-$\alpha$ cantor set is perfect? Let $\alpha\in (0,1)$ and $\beta=\frac{1-\alpha}{2}$.
Define $T_0(x) = \beta x$ and $T_1(x) = (1-\beta) + \beta x$ , $\forall x\in [0,1]$.
Recursively define $I_0 =[0,1]$ and $I_{n+1}= T_0(I_n) \cup T_1(I_n)$.
The Middle-$\alpha$ Cantor Set is defined as $\bigcap_{n\in \omega} I_n$.
I have proved that $I_n$ is a disjoint union of $2^n$ intervals, each of length $\beta^n$. That is, $I_n=\bigcup_{i=1}^{2^n} [a_i,b_i]$
My question is that how do i prove that every endpoint $a_i,b_i$ in $I_n$ is in $\bigcap_{n\in\omega} I_n$?
It seems trivial, but i don't know how to prove this..

Define $I_n^* = I_0 - I_n, \forall n\in \omega$.
Note that (i);
$I_{n+1}^*\\=I_0 \setminus I_{n+1} \\=I_0 \setminus (T_0(I_n)\cup T_1(I_n)) \\=(I_0\setminus (T_0(I_0)\setminus T_0(I_n^*)))\cap (I_0\setminus (T_1(I_0)\setminus T_1(I_n^*)) \\=T_0(I_n^*)\cup I_1^* \cup T_1(I_n^*)$.
Also(ii), it can be found that, $\forall x\in I_n, \beta x\in I_n$ and $(1-\beta)+\beta x \in I_n$.
Let $E_n$ be a set of endpoints of $I_n$.
Let $G=\{n\in \omega | n<m \Rightarrow E_n\subset E_m\}$
Then, it can be shown that $n<m\Rightarrow E_n \subset E_m$.
