Describe the interior of Cantor set

Describe the interior of Cantor set

I think the interior is empty because the cantor set of nowhere dense, but as I write it correctly?

Solution 1:

The Cantor set is the intersection of the sets $F_i$, where

$F_1=[0,1/3] \cup [2/3,1]$,

$F_2=[0,1/9]\cup [2/9,1/3]\cup [2/3, 7/9]\cup [8/9,1]$,

and, in general: if $F_{n}$ is defined and consists of the union of $2^n$ disjoint intervals of the form $[k/3^n, (k+1)/3^n]$, then $F_{n+1}$ is obtained by removing the "open middle third" of each of these intervals.

Note the total length of $F_n$ is $(2/3)^n$.

Suppose the non-empty open interval $I=(a,b)$ is contained in the Cantor set. Then for each $n$, $I$ is contained in $F_n$, and thus in one of the intervals defining $F_n$. But then $0&lt b-a\le (2/3)^n$ for all $n$. This leads to a contradiction, as $b-a\gt0$ and $\lim\limits_{n\rightarrow\infty} (2/3)^n=0$.

So the Cantor set contains no non-empty open interval; thus the interior of the Cantor set is empty.