While reading Rudin's proof that the Cantor set is Perfect, I realised that it assumes the endpoints of each interval survive the construction. I did not find this to be obvious to me. I tried proving it through induction, but I don't think that's allowed since there are infinite induction steps.
So how to prove that the endpoints survive construction?
The Cantor set $C$ construction being $$C_n = \frac{C_{n-1}}{3} \cup (\frac{2}{3} + \frac{C_{n-1}}{3})\;\;\; \text{ for $n \ge 1$, } \;\;C_0 = [0,1] $$ $$C:= \bigcap_{n = 1}^{\infty} C_n$$
My attempt at a particular case (proving $0 \in C$):
$$\text{a) } 0 \in C_0$$
$$\text{b) For a particular } n \in \Bbb N , \text{ assume } 0 \in C_n$$
$$\implies0 \in C_{n+1} \text{ (as $\frac{0}{3} = 0)$ }$$
$$\therefore 0 \in C_n \;\forall\; n \in \Bbb N$$
But this does not prove that $0 \in C$ as $C$ is the intersection of infinite sets.
Best Answer
To show $0 \in C$ you exactly have to show $$\forall n: 0 \in C_n$$ because that's what the intersection means. This fact can perfectly well be shown by induction on $n$ using the recursive/fractal nature of the $C_n$. So there is no problem at all with infinite intersections.
In fact all other endpoints (they have finite ternary expansion with only zeroes and twos, like $1$, $\frac{1}{3}$, $\frac{2}{3}, \frac{1}{9}, \frac{2}{9} \frac{1}{27}, \ldots$ and countably many more) also all survive, and many others besides. These endpoints even form a countable dense set in $C$.