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?
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 $2\,\forall i$.