Cantor Set ternary representation Let $I=[0,1]$ be an interval in $\mathbb{R}$
We represent all numbers in $I$ in the ternary form as $t=0_3.t_1t_2t_3...$ where $t_j=0,1$ or $2$
As per the usual construction of the Cantor set, we remove the middle third which in our case would be $(\frac{1}{3},\frac{2}{3})$
When we remove the middle third $(\frac{1}{3},\frac{2}{3})$, why does this means that we remove all numbers of the form $0_3.1t_2t_3t_4..$?
What is $0_3.1t_2t_3t_4..$?
 A: Because $0_3.1t_2t_3t_4...$ is the general (ternary digital)
form of a number $\ge \frac{1}{3}$ and $\le \frac{2}{3}$   
Why? Because 
$0_3.1t_2t_3t_4... \ge 0.1 = \frac{1}{3}$  
and in the same way   
$0_3.1t_2t_3t_4... \le 0.2 = \frac{2}{3}$ 
In fact, to be exactly correct it should say that you do not remove
$\frac{1}{3} = 0_3.1000000...$ and $\frac{2}{3} = 0_3.1222222...$ which are the ends of this interval.
But indeed you remove all the others.     
A: Imagine a decimal version of the Cantor set, where instead of removing the middle third, we remove a tenth of the numbers. Let's say we take out the third tenth (so that two tenths remain on the left, and seven on the right).
In other words, we're removing all numbers in the interval $(\frac2{10},\frac3{10})$, i.e., those whose decimal representation is $0.2t_1t_2t_3\ldots$ (except for the endpoints, $0.2\bar0$ and $0.2\bar9$).
In the case of the actual Cantor set, we use ternary because that base corresponds to the fraction of the numbers being removed.
