# Arithmetic sums of cantor sets

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For each $\theta \ge 1$ there is a unique $\lambda \left(\theta \right)\in \left[{\lambda }_{1},\left(\theta \right),,,{\lambda }_{2},\left(\theta \right)\right]$ such that $C\left(\lambda \right)+C\left({\lambda }^{\theta }\right)=\left[0,,,2\right]$ if and only if $\lambda \ge {\lambda }^{\theta }$ . Moreover,

• if $\theta$ is irrational, then $\lambda \left(\theta \right)={\lambda }_{2}\left(\theta \right)$ ;
• if $\theta =n$ or $\theta =1+\frac{1}{n}$ where $n$ is a positive integer, then $\lambda \left(\theta \right)={\lambda }_{1}\left(\theta \right)$ .

The lines ${\lambda }_{1}\left(\theta \right)$ and ${\lambda }_{2}\left(\theta \right)$ can be seen in [link] as the blue and red lines, respectively.

A partial to this result was given by Cabrelli, Hare, and Molter in 2002, regarding the possibilities that $\lambda \left(\theta \right)={\lambda }_{1}\left(\theta \right)$ or $\lambda \left(\theta \right)={\lambda }_{2}\left(\theta \right)$ [link] .

Theorem 3.5 (Cabrelli, Hare, and Molter 2002) Using the notation from above, we have the following results:

• $\lambda \left(\theta \right)={\lambda }_{1}\left(\theta \right)$ if and only if $\theta =n$ or $\theta =1+\frac{1}{n}$ for some positive integer $n$ ;
• Let $\theta =1+\frac{p}{q}$ where $gcd\left(p,,,q\right)=1$ and $p=1,2,\cdots ,8$ or let $\theta =n+\frac{1}{2}$ with $n\in \mathbb{N}$ . Then, $\lambda \left(\theta \right)<{\lambda }_{2}\left(\theta \right)$ .

Theorem 3.4 proved that (i) is sufficient, and this theorem proves that it is also necessary. As for (ii) , we now know that there are some rational $\theta$ , but not necessarily all, for which $\lambda \left(\theta \right)$ must lie strictly below ${\lambda }_{2}\left(\theta \right)$ . It is conjectured, but not known, that this is true for all rational $\theta$ .

A summary of the known results presented in this section is given below in [link] .

As mentioned in "Sums of Mid-α Cantor Sets" , we now know that $C\left(\frac{1}{3}\right)+C\left(\frac{1}{3}\right)=\left[0,,,2\right]$ because $\frac{1}{3}\ge \lambda \left(1\right)=\frac{1}{3}$ and that $C\left(\frac{1}{5}\right)+C\left(\frac{1}{5}\right)$ is a Cantor set because the point $\left(1,,,\frac{1}{5}\right)$ in the $\left(\theta ,,,\lambda \right)$ -plane lies in Region I.

## Results from the study

In our study, we decided to focus on rational $\theta$ . We found a nice way to write the sum set $C\left(\lambda \right)+C\left({\lambda }^{\theta }\right)$ when $\theta =\frac{p}{q}$ , where $gcd\left(p,,,q\right)=1$ , that is similar to the notation used for homogeneous Cantor sets. Sets of this form may be homogeneous Cantor sets, but in many cases they are not. We will present this process first for $\theta =2$ and then for general rational $\theta$ .

The idea behind this process is to write the two mid- $\alpha$ Cantor sets as homogeneous Cantor sets with the same scaling factor. For $\theta =2$ , remember that we may write $C\left(\lambda \right)$ as a homogeneous Cantor set with scaling factor $\lambda$ , as in

$C\left(\lambda \right)=\sum _{n=0}^{\infty }{\alpha }_{n}{\lambda }^{n},\phantom{\rule{1.em}{0ex}}{\alpha }_{n}\in A\left(\lambda \right)=\left\{0,,,1,-,\lambda \right\}.$

The second stage of the construction of $C\left(\lambda \right)$ , as described in "Mid-α Cantor Sets" , is the union of 4 intervals, each of length ${\lambda }^{2}$ , starting at the points $0,\lambda \left(1,-,\lambda \right),1-\lambda ,\phantom{\rule{0.166667em}{0ex}}\text{and}\phantom{\rule{0.166667em}{0ex}}\left(1,+,\lambda \right)\left(1,-,\lambda \right)$ . With this in mind, we can rewrite $C\left(\lambda \right)$ as a homogeneous Cantor set with scaling factor ${\lambda }^{2}$ and 4 offsets. That is,

$C\left(\lambda \right)=\sum _{n=0}^{\infty }{\alpha }_{n}{\lambda }^{2n},\phantom{\rule{1.em}{0ex}}{\alpha }_{n}\in A=\left\{0,,,\lambda ,\left(1,-,\lambda \right),,,1,-,\lambda ,,,1,-,{\lambda }^{2}\right\}.$

One might notice here that $A=A\left(\lambda \right)+\lambda ·A\left(\lambda \right)$ . And since $\theta =2$ , we have also that

$C\left({\lambda }^{2}\right)=\sum _{n=0}^{\infty }{\beta }_{n}{\lambda }^{2n},\phantom{\rule{1.em}{0ex}}{\beta }_{n}\in B=\left\{0,,,1,-,{\lambda }^{2}\right\}.$

Now, we can take the sum

$\begin{array}{cc}\hfill C\left(\lambda \right)+C\left({\lambda }^{2}\right)=\sum _{n=0}^{\infty }\left({\alpha }_{n},+,{\beta }_{n}\right){\lambda }^{2n},\phantom{\rule{1.em}{0ex}}& {\alpha }_{n}\in A=\left\{0,,,\lambda ,\left(1,-,\lambda \right),,,1,-,\lambda ,,,1,-,{\lambda }^{2}\right\},\hfill \\ & {\beta }_{n}\in B=\left\{0,,,1,-,{\lambda }^{2}\right\}.\hfill \end{array}$

Or equivalently,

$C\left(\lambda \right)+C\left({\lambda }^{2}\right)=\sum _{n=0}^{\infty }{\gamma }_{n}{\lambda }^{2n},\phantom{\rule{1.em}{0ex}}{\gamma }_{n}\in A+B$

where $A+B$ is considered as the normal set-wise sum. In this case, we have

$A+B=\left\{0,,,\lambda ,\left(1,-,\lambda \right),,,1,-,\lambda ,,,\left(1,+,\lambda \right),\left(1,-,\lambda \right),,,\left(1,+,2,\lambda \right),\left(1,-,\lambda \right),,,\left(2,+,\lambda \right),\left(1,-,\lambda \right),,,\left(2,+,2,\lambda \right),\left(1,-,\lambda \right)\right\}.$

This sum $C\left(\lambda \right)+C\left({\lambda }^{2}\right)$ is contained in the interval $\left[0,,,2\right]$ . However, due to our notation, it is much simpler and more convenient to consider a scaled-down copy of this set so that it is contained in the interval $\left[0,,,1\right]$ . We do this by halving the offsets. This does not change any properties of the set. Also, each offset contains a factor of $\left(1-\lambda \right)$ . This is not an accident, and we will use it here to condense the notation, writing

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