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

Question

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 $C$.

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

Answer 1

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.