Let $C_0$ be the segment $[0,1]$ $C_2$ will be $[0,1]$, with the middle third, an open set removed, so $[0,1/3]\cup[2/3,1]$
First, if we removed closed sets would the cantor set, the limit of what remains from this process, be the same?
I was able to show that the limit of the sequence of partial sums of the pieces converges to $1$. And this happens if the removed bits are open or closed, it doesn't matter. Don't know if that helps or not.
Second, I've been trying for two hours and I still can't find a sensible way to write $C_n$, in some compact notation.