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?

  • 1
    $\begingroup$ How do you define the Cantor set? $\endgroup$ – José Carlos Santos Sep 28 '18 at 12:45
  • $\begingroup$ @JoséCarlosSantos Sir, I had written the definition of Cantor Set that I followed $\endgroup$ – 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 $2\,\forall i$.

  • $\begingroup$ Yes. Inspect the ternary expansion. $\endgroup$ – Chris Custer Sep 28 '18 at 13:32

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