Why if 1 occurs somewhere in ternary expansion then that number is not belong to Cantor Set?

I had read some articles about Cantor set and ternary expansion.
I come across method of Construction of Cantor set by removing middle third and taking infinite intersection of that set.
and there are at most 2 ternary exapsnsion of any number SO and in case there are 2 exapansion then there is one with no 1's.
Cantor Set as : $$C_0=[0,1]$$
$$C_1=[0,1/3]\cup [2/3,1]$$
$$C_2=[0,1/9]\cup [2/9,1/3]\cup[6/9,7/9]\cup [8/9,1]$$
Cantor Set : $$C=\cap^{\infty}_0 C_n$$

But I had one problem .If some element has only one ternary expansion and contain 1 then why it does not belong to Cantor set?

I am missing some argument . I am very thankful if some one help me to find where I am missing?

• How do you define the Cantor set? – José Carlos Santos Sep 28 '18 at 12:45
• @JoséCarlosSantos Sir, I had written the definition of Cantor Set that I followed – SRJ Sep 28 '18 at 12:54

A number with a $$1$$ in the ternary representation of a number corresponds to a a "middle third" of some interval...
Look at how the ternary expansion works: $$0.1a_1\dots a_n\dots$$ is necessarily in the middle third.
$$0.01a_2\dots$$ in the middle third of the first third.
$$0.21a_2\dots$$ in the middle third of the third third, etc...
This is because, in ternary, $$0.a_1\dots a_n\dots=\sum_{i=1}^\infty \frac{a_i}{3^i}$$, where $$a_i=0,1$$ or $$2\,\forall i$$.