# $y-x=a$ in the Cantor set

Given any $a\in [0,1]$, I would like to prove that there are $x,y$ in the Cantor set such that $y-x=a$. I need a hint or something, this question seems a little bit unintuitive for me.

Thanks a lot.

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## 3 Answers

Hint: what is the base 3 representation of points in the cantor set?

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these numbers have a ternary expansion with only 0,s and 2's, as such 0,2202202 – user42912 Feb 23 '13 at 23:00

Have you seen the ternary expansion formulation of the Cantor set? The Cantor set is precisely those real numbers that have a ternary expansion consisting entirely of $0$s and/or $2$s. Your task, then, is equivalent to determine how to add such a number ($x$) to some non-Cantor element ($a$) of $[0,1]$ such that the resulting sum ($y$) is a Cantor element.

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how can I do this? for example if I have 0,7977, how can I add a cantor number to have a cantor number in this case? it seems impossible, thank you for your answer. – user42912 Feb 23 '13 at 23:03
Don't forget to put it into ternary form, not decimal. – Cameron Buie Feb 23 '13 at 23:41
ha yes of course, yes, you're right – user42912 Feb 24 '13 at 9:13

First remember that every number $\in [0,1]$ can be expressed in base 3 by $\sum_{k = 1}^{\infty}{\frac{\beta_k}{3^k}};\; \beta_k \in \{0,1,2\}$.

Notations:

$A + B = \{a + b;\; a \in A,\; b\in B\}$

$A - B = \{a - b;\; a \in A,\; b\in B\}$

$|A - B| = \{|a - b|;\; a \in A,\; b\in B\}$

The question asked is equivalent to prove that $[0,1] = \{|x - y|; x, y \in K = Cantor\;set\}$.

We know that the elements of the Cantor set are the numbers $\in [0,1]$ that only have 0 or 2 in their base 3 representation.

The trick is to add 1 to x - y, and remember that because of the Cauchy series, $1 = \sum_{k=1}^{\infty}{\frac{2}{3^k}}$ in base 3:

$x - y + 1 = \sum_{k = 1}^{\infty}{\frac{\sigma_{xk}}{3^k}} - \sum_{k = 1}^{\infty}{\frac{\sigma_{yk}}{3^k}}+\sum_{k = 1}^{\infty}{\frac{2}{3^k}}; \sigma_{xk}, \sigma_{yk} \in \{0, 2\} \text{ Therefore:} \sum_{k = 1}^{\infty}{\frac{\sigma_{xk} - \sigma_{yk}+2}{3^k}}; \sigma_{xk} -\sigma_{yk} + 2 = \alpha_k \in \{0, 2, 4\}$

putting $t_k = \frac{\alpha_k}{2}: \sum_{k = 1}^{\infty}{\frac{\alpha_k}{3^k}} = 2\sum_{k = 1}^{\infty}{\frac{t_k}{3^k}}; \;t_k \in \{0, 1, 2\}$. In other words, every number $\in [0,1] \text{ is expressed by }\sum_{k = 1}^{\infty}{\frac{t_k}{3^k}}$. However, since we're multipling it by 2, every number $\in [0,2] \text{is expressed by} \sum_{k = 1}^{\infty}{\frac{\sigma_k}{3^k}}$ In conclusion:

$K - K + 1 = [0,2] \rightarrow K- K = [-1, 1] \rightarrow |K-K| = [0, 1]$

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