Limit points and interior points of the cantor set Please help to prove that every point of the cantor set is a limit point and no point is an interior point ( i.e. it has empty interior ) . 
 A: Hint 1: If there is an interior point, then the Cantor set contains an interval. What is the measure of the Cantor set?
Hint 2: Points in the Cantor set are either end points of the excluded intervals or limit points of endpoints. The Cantor set is closed.
A: The cantor set is all real numbers between $0$ and $1$ with no $1$s in the ternary representation, i.e.
$$x=\sum_{n=1}^\infty {d_n\over 3^n},\quad d_n\in\{0,2\}$$
that's exactly what it means to take out the "middle third" at each step.
If we had an interval, say $(x_0-\epsilon, x_0+\epsilon)$ for some $x_0$ in the Cantor set and $0<\epsilon<1$, then we note that this includes triadic rationals with a digit of $1$. In fact if $3^n\le \epsilon^{-1}<3^{n+1}$, and the first $2$ in the expansion of $x$ is at the $N^{th}$ digit of the ternary expansion then clearly $n>N$ and the number
$$x_1=x_0+3^{-n-1}\in (x_0-\epsilon, x_0+\epsilon)$$
hence this interval cannot be contained in the Cantor set since $x_1$ has a $1$ in it's triadic expansion.
Closure is simple, it is the intersection of the closed sets which compose the steps of the process of removing the middle thirds, step $n$ is just the closed set
$$C_{n}=\{x\in [0,1]: \text{there are no $1$s in the ternary expansion until at least digit }n\}$$
Any intersection of closed sets is closed, so we have the desired result.
