How come Cantor set is 'the' Cantor Set? 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)
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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 $A_n=B_n$?
 A: I don't have enough reputation to post a comment, so i'll write it here.
There's a little typo in the definition of the Cantor-set (first formula), the interval should be $\left(\frac{3k+1}{3^i},\frac{3k+2}{3^i}\right)$
I think you can prove it by induction using the recursive formula of the wikipedia article :
$C_n=\frac{C_{n-1}}{3}\cup\left(\frac{2}{3}+\frac{C_{n-1}}{3}\right)$
I  hope this is helpful.
EDIT : I was typing when other comments came up. So here are some details you need.
By induction, assume that $C_{n-1}$ is the disjoint union of $2^{n-1}$ intervals in $[0,1]$ each of them with lenght $\frac{1}{3^{n-1}}$. 
On one hand, $A_n:=\frac{C_{n-1}}{3}$ is a subset of $[0,\frac{1}{3}]$ and is again a disjoint union of $2^{n-1}$ intervals but of lenght $\frac{1}{3}\cdot\frac{1}{3^{n-1}}=\frac{1}{3^{n}}$
On the other hand, $B_n:=\frac{2}{3}+\frac{C_{n-1}}{3}$ is a subset of $[\frac{2}{3},1]$ and is as well a disjoint union of $2^{n-1}$ intervals of lenght $\frac{1}{3^{n}}$. 
Conclusion : $C_n$ is the union of $A_n$ and $B_n$ which is disjoint and hence composed of $2\cdot2^{n-1}=2^n$ intervals of the desired lenght.
A: In $\mathbb{R}^2$, for any two Cantor sets $A$ and $B$ we can find a homeomorphism $h \colon \mathbb{R}^2 \to \mathbb{R}^2$ which carries $h(A)$ onto $B$. This essentially says they're equivalently embedded. I'm not sure this directly answers your question but in $\mathbb{R}^2$, talk of `the' Cantor set isn't necessarily correct, however upto homeomorphism of the ambient space we do only have one equivalence class of Cantor set.
