According to Wikipedia, each step in the construction of the Cantor Set can be expressed recursively.
Let $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.
Can someone explain how this even works?
To get $C_1$ form $C_0=[0,1]$ what does it even mean to write:
$C_{1} = \dfrac{[0,1]}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{[0,1]}{3}\right)$?
Once I understand how this recursive formula even works, I would appreciate being shown how to prove that that this accurately describes the $n$th iteration of the construction of the Cantor set.
Notation $$ \frac{[0,1]}{3} = \left\{\frac{t}{3}\;:\;t \in [0,1]\right\} , \\ \frac{2}{3} + \frac{[0,1]}{3} = \left\{\frac{2}{3}+\frac{t}{3}\;:\;t \in [0,1]\right\} . $$ So $$ \frac{[0,1]}{3} = \left[0,\frac{1}{3}\right] , \\ \frac{2}{3} + \frac{[0,1]}{3} = \left[\frac{2}{3},1\right] , \\ C_1 = \frac{[0,1]}{3} \cup \left(\frac{2}{3} + \frac{[0,1]}{3}\right)= \left[0,\frac{1}{3}\right]\cup \left[\frac{2}{3},1\right] . $$