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I am reading this paper and have a question related to the following line:

Quote: Let $D'$ be the collection of all closed discs in the $xy$ plane whose diameters are the intervals in the $x$-axis $$\left[\frac{n+1/3}{3^m},\frac{n+4/3}{3^m}\right],$$ where $0\leq n\leq 3^m$ is an integer which admits a triadic expansion free from $1$'s;...

Now, to construct the Cantor set, we first divide $[0,1]$ into three closed intervals of equal length and discard the interior of the middle one, so we have two closed intervals $C_0, C_2$. Then in $C_0, C_2$ we do the same procedure: divide each of them into three closed intervals of equal length, and discard the interior of the middle one; so we have four closed intervals $C_{0,0}, C_{0,2}, C_{2,0}, C_{2,2}$. And so on.

Question: In the quoted line, are the diameters nothing but all closed intervals that appeared in the $m$-th stage of construction of the Cantor set?

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No, since the interval $[1, 3^{-m}]$ appears in the $m-$th stage of the construction of the Cantor set, however, $\frac{n + 1/3}{3^m} \ne 1$ for any $n, m$ since otherwise we have $$ 3n + 1 = 3^{m+1} $$ which is impossible since $3$ does not divide $1$.

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  • $\begingroup$ Can you explain the part *integer which admits a triadic expansion free from 1's, I mean what are first few diameters then? $\endgroup$
    – Random
    Commented Jan 9, 2022 at 18:19
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    $\begingroup$ Start with $m=1$, write each non-negative integer $n$ not exceeding $3^m$ in base 3, and if this representation contains no 1's, you can compute the endpoints of the interval as in your post. Set $m=2$, repeat and so on. $\endgroup$
    – fwd
    Commented Jan 9, 2022 at 18:23

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