Show that end points of the Cantor Set $C$ other than $0$ and $1$ can be written as $0.a_{1}a_{2}a_{3}...a_{n+1} (base 3)$ where each $a_{i}$ is $0$ or $2$ except $a_{n+1}$ which is either $1$ or $2$.
I know that the end points of the set are the ends of the interval of the form $((2k-1)/3^n, 2k/3^n)$ which are removed from the set $[0,1]$. So, I only have to show that numbers of the form $(2k-1)/3^n$ and $2k/3^n$ have form as mentioned above. I am not able to proceed from here.
This follows from the construction of the Cantor set. Namely, at each step $n\ge 0$ we remove $2^n$ open intervals with endpoints $\left(\tfrac{3k+1}{3^{n+1}},\tfrac{3k+2}{3^{n+1}} \right)$ from the closed intervals $\left[\tfrac{k}{3^{n}},\tfrac{k+1}{3^{n}} \right]$ surrounding them.