# For cantor set, any element of $[0,1]$ can be written as the difference from cantor set $C$.

A cantor set is given by \begin{align} &C_0 = [0,1] , \quad C_n = \frac{C_{n-1}}{3} \cup \left( \frac{2}{3} + \frac{C_{n-1}}{3} \right), \quad \textrm{for } n \geq 1. \end{align} Explicitly, \begin{align} &C_1 = \left[0, \frac{1}{3} \right] \cup \left[\frac{2}{3} , 1 \right], \quad C_2 = \left[ 0, \frac{1}{9} \right] \cup \left[ \frac{2}{9}, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, \frac{7}{9} \right] \cup \left[ \frac{8}{9}, 1\right]. \end{align} Define $$C = \cap_{n=1}^{\infty} C_n$$. I want to show that any $$x \in[0,1]$$ can be written as the difference of two elements from $$C$$.

I know $$C$$ is closed [arbitrary intersection of closed set is again closed] and $$C$$ has no interior points and no isolated point.

• You want the description of elements of Cantor's set as the real numbers that are sum of a subsequence of the powers of $1/3$, plus the concept of balanced ternary expansion of real numbers.
– user562983
Commented Sep 28, 2021 at 7:32
• If you use base 3, the Cantor set corresponds to those numbers such that 1 does not appear after the point. E.g. 0.220200020 is in the Cantor set, but 0.1 is not. Maybe this can lead to some elementary arithmetic proof Commented Sep 28, 2021 at 7:49

Recall the construction of $$C$$ as the unique fixed point of the map $$T:A\mapsto \frac13 \big( A+\{0,2\}\big)$$, which is a contraction of constant $$1/3$$ w.r.to the Hausdorff distance on the complete metric space $$M$$ of all nonempty closed subset of $$[0,1]$$.
In general, the standard method to show that an object $$c$$, found as a fixed point of some contraction $$T$$ on some complete metric space $$M$$, has a certain property, is to show that the subset of $$M$$ of all points with that property is a non-empty closed $$T$$-invariant set (then, of course, $$T$$ must have a fixed point in it, which is $$c$$ by uniqueness).
Here we have $$T(A)-T(A)=\frac13\Big(A-A+\big\{0,- 2, 2\big\}\Big)$$, whence it follows that the (non-empty, closed) subset of all $$A\in M$$ such that $$A-A=[-1,1]$$ is a $$T$$-invariant set.