Hi everyone! I have asked a question about a closed formula of Cantor set here. It seems to me that nobody is interested in this question. Could someone help me resolve the question? Thank you so much!
@LeAnhDung I feel like i've done this problem a while ago. Usually when showing two sets are equal I will also some time resort to the following tactic: Let $A_n = \bigcap_{m=0}^{n}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]$... Suppose $x \in A_n$? Is $x \in C_n$? Then consider the other way round.
In case somebody is interested, the question is: [How to prove the formula $C_n=\bigcap_{m=0}^{n}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]$ for Cantor set?](math.stackexchange.com/q/3081687)
Let $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.
Theorem: $$C_n=\bigcap_{m=0}^{n}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]$$
I have tried to prove this assertion by induction on $n$, but to no avail. I...
Let $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.
Theorem: $$C_n=\bigcap_{m=0}^{n}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]$$
I have tried to prove this assertion by induction on $n$, but to no avail. I...
