Please see my Edited version at the end of the post.
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http://en.m.wikipedia.org/wiki/Cantor_set
My definition of Cantor Set is just like that of wikipedia.
That is, $C=[0,1]\setminus \bigcup_{i=1}^{\infty} \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i}, \frac{3k+2}{3^i})$.
With this definition, i have shown that $C$ is compact, perfect, equipotent with $2^{\aleph_0}$ and contains no openset. (i.e. Basic properties of Cantor set)
I preferred this definition to another since this definition is simple and strictly written in first-order logic.
Let $C_n = [0,1]\setminus \bigcup_{i=1}^n \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i},\frac{3k+2}{3^i})$.
Then $\bigcap_{n\in \mathbb{N}} C_n = C$.
Here, how do i prove that $C_n$ is a disjoint union of $2^n$ intervals, each of length $3^{-n}$?
(To make it clear, intervals here refer to closed connected sets)
=========================== EDIT:
This is not actually i meant, but this is exactly the same as what i wanted to prove anyway..
Let $A_0=B_0=[0,1]$. Define $\{A_n\}$ recursively such as; $A_{n+1}=\frac{A_n}{3} \cup (\frac{2}{3} + \frac{A_n}{3})$.
Now, define $B_n=[0,1]\setminus \bigcup_{i=1}^n \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i},\frac{3k+2}{3^i})$, $\forall n\in \mathbb{Z}^+$.
How do i prove $A_n=B_n$?