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 $0_3.1t_2t_3t_4..$?
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.