The Cantor set $\mathcal{C}$ is defined as follows: $$\mathcal{C}:=\bigcap_{n=0}^{\infty}C_n$$ where $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.
From Wikiwand's page, The explicit formulas of Cantor sets are
$$\mathcal{C} = \bigcap_{n=0}^{\infty} \bigcup_{k=0}^{3^n-1} \left(\left[\frac{3k+0}{3^{n+1}}\,,\, \frac{3k+1}{3^{n+1}}\right] \cup \left[\frac{3k+2}{3^{n+1}}\,,\,\frac{3k+3}{3^{n+1}}\right] \right) \space (1)$$ and $$\mathcal{C} = [0,1] \setminus \bigcup\left\{\left(\frac {3k+1}{3^n}, \frac{3k+2}{3^n}\right) \,\middle\vert\, k,n\in \mathbb Z^+\right\} \space (2)$$
I have tried for several days to get $(1)$ and $(2)$ from the definition of Cantor set, but to no avail.
Could you please help me derive formulas $(1)$ and $(2)$? Thank you so much!
Best Answer
The finite union in (1) is just the expression for $C_n$; you can prove its truth by induction. And $C = \bigcap_n C_n$ by definition.
Then (2) follows by de Morgan: the complement of $C_n$ in $[0,1]$ is just a finite union of open intervals and the complement of $C$ is just the union of the complements of the $C_n$. You then take the complement of that to get $C$ back.