Arithmetic sums of cantor sets  (Page 5/6)

In general, we will write $S\left(\frac{p}{q},,,\lambda \right)=\frac{1}{2}\left(C,\left(\lambda \right),+,C,\left({\lambda }^{p/q}\right)\right)$ , where $p$ and $q$ are relatively prime.

Now we will present the construction of $S\left(\frac{p}{q},,,\lambda \right)$ for general $\frac{p}{q}$ . Again, the idea will be to write both $C\left(\lambda \right)$ and $C\left({\lambda }^{p/q}\right)$ as homogeneous Cantor sets with the same scaling factor, in this case, ${\lambda }^p$ . For convenience, we will define $\ell ={\lambda }^{1/q}$ .

We start with the set $C\left(\lambda \right)$ . Remember that we may write this set in the form

so that an offset $a$ at the $p^{th}$ stage of the construction may be represented as

Then, we factor out a $\left(1,-,\lambda \right)$ and substitute $\lambda$ with ${\ell }^q$ to get

Rewriting $\left(1,-,{\ell }^q\right)$ ,

Then we combine the sums to get

where $\lfloor n/q\rfloor$ is the standard floor function. Note that $u=u_0{u}_1\cdots u_{pq-1}$ is a binary word of length $pq$ composed of $p$ groups of $q$ repeated letters. There are $2^p$ such words corresponding to the $2^p$ offsets of $C\left(\lambda \right)$ at this stage, so that all $2^p$ possibilities for $u$ are obtained. Similarly, we may write a $q^{th}$ -stage offset $b$ of $C\left({\lambda }^{p/q}\right)$ in the form

so that $v=v_0{v}_1\cdots v_{pq-1}$ is also a binary word of length $pq$ composed of $q$ groups of $p$ repeated letters. Then, finally, we let $A$ be the set of $p^{th}$ -stage offsets of $C\left(\lambda \right)$ , $B$ be the set of $q^{th}$ -stage offsets of $C\left({\lambda }^{p/q}\right)$ and write

where we look at $A+B$ in the following way: given $a\in A$ and $b\in B$ , we have

and we can write a ternary word $w=w_0{w}_1\cdots w_{pq-1}$ to describe this offset $a+b$ . With a slight abuse of notation, we write $w=u+v$ and call $W\left(\frac{p}{q}\right)$ the set of all such $w$ .

It is clear that this sum $S\left(\frac{p}{q},,,\lambda \right)$ is self-similar in the sense discussed previously with homogeneous Cantor sets. This, along with the fact that $0,1\in S\left(\frac{p}{q},,,\lambda \right)\subseteq \left[0,,,1\right]$ , allows us to consider this set as the infinite intersection

Here, $S_j$ is the $j^{th}$ stage of the construction of $S\left(\frac{p}{q},,,\lambda \right)$ , in the sense that it is the union of $N^j$ intervals of length ${\lambda }^{pj}$ where $N$ is the number of offsets in $A+B$ . Note that some of these intervals might overlap. The next proposition is concerned with the computation of the value of $N$ .

Proposition 4.1 For every $\frac{p}{q}\ge 1$ where $gcd\left(p,,,q\right)=1$ , the set $W\left(\frac{p}{q}\right)$ has exactly $2^{p+q}-1$ elements.

For ease of reading, we will write $W$ in place of $W\left(\frac{p}{q}\right)$ . [Proof of the Proposition]Let such a fraction $\frac{p}{q}$ be given. Clearly, we have that $\left|W\right|\le 2^{p+q}$ since each element $w\in W$ can be expressed as the "sum" of two binary words $u$ and $v$ . Since there are $2^p$ such $u$ and $2^q$ such $v$ , there are at most $2^{p+q}$ possibilities for $w=u+v$ . Additionally, since $p$ and $q$ are relatively prime, it is easy to see that $u=v$ if and only if $u=v=000\cdots 0$ or $u=v=111\cdots 1$ . What is left to show is that there are exactly two pairs $\left(u,,,v\right)$ and $\left(u^{\text{'}},,,v^{\text{'}}\right)$ , with $u\ne u^{\text{'}}$ or $v\ne v^{\text{'}}$ , such that $u+v=u^{\text{'}}+v^{\text{'}}$ . In fact, these pairs are $\left(111,\cdots ,1,,,000,\cdots ,0\right)$ and $\left(000,\cdots ,0,,,111,\cdots ,1\right)$ ; in both cases, we have

First, we will show that for all $w\in W$ with $w\ne 111\cdots 1$ , there is a unique pair $\left(u,,,v\right)$ such that $u+v=w$ , and then we will show that when $w=111\cdots 1$ , the only pairs $\left(u,,,v\right)$ for which $u+v=w$ are the two given above. Both of these statements rest on the following lemma.

Lemma 4.2 Given $w=u+v$ , where $u$ and $v$ are initially unknown, and one letter of $u$ or $v$ , the remaining letters of $u$ and $v$ are uniquely determined.