Let $J_{0,1}:=[0,1]$.
Step 1. We remove the central open interval $I_{0,1}=\big(\frac{1}{3},\frac{2}{3}\big)$. We denote with $J_{1,1}:=\big[0,\frac{1}{3}\big]$ and with $J_{1,2}:=\big[\frac{2}{3},1\big]$.
We define $K_1:=J_{1,1}\cup J_{1,2}$ then $\complement{K_1}=I_{0,1}.$
Step 2. We remove from $J_{1,1}$ and $J_{1,2}$ the central open interval of length $\frac{1}{9}$. Then we define what remains with $K_2:=J_{2,1}\cup J_{2,2}\cup J_{2,3}\cup J_{2,4}$, where $J_{2,1}=\big[0,\frac{1}{9}\big]$, $J_{2,2}=\big[\frac{2}{9},\frac{1}{3}\big]$, $J_{2,3}=\big[\frac{2}{3},\frac{7}{9}\big]$, $J_{2,4}=\big[\frac{8}{9}, 1\big]$. At this point we define with $I_{1,1}=\big(\frac{1}{9}, \frac{2}{9}\big)$, $I_{1,2}=\big(\frac{7}{9},\frac{8}{9}\big)$ the two open interval just removed. Therefore, $\complement{K_2}=\complement{K_1}\cup \big(I_{1,1}\cup I_{1,2}\big)$
At the end of step $n$ we will have $2^n$ close interval $J_{n,k}$ for $k=1,\cdots, 2^n$ of length $1/3^n.$
For all $n\in\mathbb{N}$ we define $$K_{n}:=\bigcup_{k=1}^{2^n} J_{n,k}$$ We define the Cantor set $K$ in the follow way $$K:=\bigcap_{n=1}^{+\infty} K_n.$$ In general, as discussed above, $$\complement{K_n}= \complement{K_{n-1}}\cup\bigcup_{k=1}^{2^n/2} I_{n-1,k}\tag1$$
Question 1. It's correct (1)?
We determine, $$[0,1]\setminus K=[0,1]\cap \bigg[\bigcup_{n=1}^{+\infty}\complement{K_{n-1}}\cup \bigcup_{n=1}^{+\infty}\bigcup_{k=1}^{2^n/2} I_{n-1,k}\bigg]\tag2$$
Question 2. How can I proceed in (2)? I would like to prove that $$[0,1]\setminus K =\bigcup_{n=0}^{+\infty}\bigcup_{k=1}^{2^n}I_{n,k}.$$
My answer at question 2.
\begin{equation} \begin{split} [0,1]\setminus K &= [0,1]\cap\bigg[\bigcup_{n=1}^{+\infty}\complement{K_n}\bigg]\\ =&\bigcup_{n=1}^{+\infty} \complement{K_n}\\ =&\bigcup_{n=1}^{+\infty}\complement{K_{n-1}}\cup\bigcup_{n=1}^{+\infty}\bigcup_{k=1}^{2^n/2}I_{n-1,k}\\ \color{BLUE}{=}& \bigcup_{n=1}^{+\infty}\bigcup_{k=1}^{2^n/2} I_{n-1,k}\\ \color{RED}{=}&\bigcup_{n=0}^{+\infty}\bigcup_{k=1}^{2^n} I_{n,k}. \end{split} \end{equation} For the blue equality I used the fact that $$\bigcup_{n=1}^{+\infty}\complement{K_{n-1}}\subseteq\bigcup_{n=1}^{+\infty}\bigcup_{k=1}^{2^n/2}I_{n-1,k} $$ and the red equality it is a simple rewriting of the indexes.
Question 3. It's correct my answer?
Thanks!