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It is known, that the Cantor set is the subset of points of the segment $[0,1]$ allowing ternary expansion $0.a_1a_2\ldots$, where $a_i=0$ or $2,\; i\in {\mathbb N}$, neither of those values ($0$ or $2$) in period.

In English edition of V.I. Bogachev's fundamental work Measure Theory vol 1, we read.

1.12.60$^◦$. Let $C$ be the Cantor set in $[0, 1]$. Show that $C + C := \{c_1 + c_2 : c_1, c_2 ∈ C\}$ $ = [0, 2]$, $C − C := \{c_1 − c_2 : c_1, c_2 ∈ C\} = [−1, 1]$.

Hint: the sets $C + C$ and $C − C$ are compact, hence it suffices to verify that they contain certain everywhere dense subsets in the indicated intervals, which can be done by using the description of C in terms of the ternary expansion.

In ternary expansion $1/2=0.(1)$ (1 in period). The question is: How to show that there exist Cantor set poins $c_1=0.a_1a_2\ldots$, and $c_2=0.b_1b_2\ldots$ (where both $a_i,b_i$ are $0$ or $2$, for all $i\in {\mathbb N}$, none of them in period, such that $c_1-c_2=0.(1)$?

Jose Avilez
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  • It is enough to show that $C+C=[0,2]$ for $t\in[-1,1]$ iff $1+\lambda\in[0,2]$. Is if $C+C=[0,2]$, then $1+\lambda=c_1+c_2$ where $c_j\in C$ and so $\lambda= c_1-(1-c_2)$ which is in $C-C$ as $C$ is symmetric around $1/2$. Not that the steps can be reversed – Mittens Jun 09 '21 at 16:55
  • Notice that any point $x\in [0,1]$ is the midpoint between two cantor sets, that is $x=\frac{1}{2}(c_1+c_2)$ form some $c_i\in C$ – Mittens Jun 09 '21 at 17:01

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