Assume that $C$ is the intersection of all of $C_i$'s where $i\in\mathbb{N}$. $C_0=[0,1]$, and $C_i$ is obtained by removing the middle 1/3 of each inverval in $C_{i-1}$.
On an intuitive level I can see how $C=\{\sum_{n=1}^{\infty} \frac{a_n}{3^n}\mid a_n\in\{0,2\}\}$. But how I can prove it?
General idea behind a proof: Consider the ternary (i.e. base $3$) expansion of all the numbers removed in the first iteration. Do you see something they have in common? How about in the second iteration? Keep going. If we remove all such numbers, which are we left with?