the elements of Cantor's discontinuum Let $(A_n)_{n \in \mathbb{N}}$ the sequence of subsets of $\mathbb{R}$, given by $A_0 := \bigcup_{k \in \mathbb{Z}}[2k, 2k + 1]$ und $A_n := \frac{1}{3}A_{n-1}$ for $n ≥ 1$.
Also, we define
$$ A := \bigcap_{n=0}^\infty A_n, \,C:= A \cap [0, 1]$$
We call $C$ Cantor's discontinuum.
Given this definition, I now want to prove that $C$ consists of all the real numbers of the form $x = \sum_{k=1}^\infty \frac{a_k}{3^k}$, where $a_k = 0$ or $a_k = 2$.
Thanks in advance! I'm beginning to understand what $C$ actually looks like, but I don't really know how to show this. My idea was to maybe use the three-adic representation of numbers?
 A: Here is an image of the Cantor set:

Image File:Cantor5.svg by User:sarang
Imagine, you want to describe a number $x$ of the cantor set uniquely. Here you have to specify at each iteration, whether you have to go to the left or
the right subinterval in order to finally reach $x$:

Because each number of the cantor set is uniquely defined by a path of the form
$$( \text{left, right, left, left,}\ldots)$$
The numbers of the cantor sets are bijective to $\{\text{left}, \text{right}\}^\mathbb N$ or $\{0,1\}^\mathbb N$ (0 may mean "go to the left subinterval" and 1 may mean "go to the right interval"). So we see, that the Cantor set has uncountable many objects.
Now take the starting point $s=0$. To reach a final number $x$ of the Cantor set you add $\frac{2}{3^k}$ to go the right subinterval in the $k$-th iteration and you add $0$ to go to the left subinterval:
 
This shows that the Cantor set consists of all real numbers with a 3-adic representation only containing 0 and 2.
Okay, this is no rigorous proof, but I think you get the basic idea... ;-) 
