Arithmetic sums of cantor sets

[Proof of the Lemma] Let $w=u+v\in W$ be given, and assume without loss of generality that we know the $j^{th}$ letter $u_j$ of $u$ . We partition $w$ into $p$ $q$ -groups (that is, $p$ groups of $q$ letters) and into $q$ $p$ -groups. Because $p$ and $q$ are relatively prime, the initial letter of a $q$ -group will only coincide with the initial letter of a $p$ -group at the first letter of $w$ . Similarly, the final letter of a $q$ -group will only coincide with the final letter of a $p$ -group at the last letter of $w$ .

Since we know $u_j$ , it follows immediately that $v_j=w_j-u_j$ , and in general, if we know the $k^{th}$ letter of $u$ , we also know the $k^{th}$ letter of $v$ , and vice versa.

Take $n$ such that $nq\le j<\left(n,+,1\right)q$ . Because $u$ is of the form of $p$ groups of $q$ repeated letters, and because we know $u_j$ , we also know the letters $u_{nq}$ through $u_{\left(n,+,1\right)q-1}$ , allowing us to fill in $v_{nq}$ through $v_{\left(n,+,1\right)q-1}$ . We repeat this process now on $v$ ; take $m_0$ such that $m_{0}p\le nq<\left(m_0,+,1\right)p$ and $m_1$ such that $\left(m_1,-,1\right)p\le \left(n,+,1\right)q-1<m_{1}p$ . This gives us $v_{m_{0}p}$ through $v_{m_{1}p-1}$ , which in turn gives us the corresponding letters for $u$ . Because of the condition described above about the initial and final letters of $p$ - and $q$ -groups, we will always be able to extend the range of uniquely determined letters at each step. We repeat this process until all of $u$ and $v$ are determined. Because $u$ and $v$ are finite, this process will terminate at a finite step.

Now, we can apply this Lemma to the claim above. In the case where $w\ne 111\cdots 1$ , there is some $j$ for which $w_j=0$ or $w_j=2$ . In either case, this uniquely determines the corresponding $u_j$ and $v_j$ : They must be both 0 in the former case and both 1 in the latter case. Applying Lemma 4.2, we get a unique pair $\left(u,,,v\right)$ with $u+v=w$ .

In the case where $w=111\cdots 1$ , there are no 0s or 2s, so we must make a choice. We can choose either $u_0=1$ and $v_0=0$ , or $u_0=0$ and $v_0=1$ , but nothing else. With either choice, by Lemma 4.2, the rest of $u$ and $v$ are uniquely determined. These two choices correspond to the two pairs given above.

Thus, for all but one $w\in W$ , there is a unique pair $\left(u,,,v\right)$ for which $u+v=w$ . Since there are $2^{p+q}$ unique pairs of $u$ and $v$ , there must be exactly $2^{p+q}-1$ elements $w\in W$ , as desired.

This Proposition leads us to several results regarding the set $S\left(\frac{p}{q},,,\lambda \right)$ , presented here as Corollaries.

Corollary 4.3 The set of offsets $A+B$ of $S\left(\frac{p}{q},,,\lambda \right)$ contains exactly $2^{p+q}-1$ elements for all but finitely many $\lambda$ .

This is an easy consequence of the Proposition and the construction of $S\left(\frac{p}{q},,,\lambda \right)$ above.

Corollary 4.4 The set $S\left(\frac{p}{q},,,\lambda \right)$ has zero Lebesgue measure for all $\lambda <{\left(2^{p+q},-,1\right)}^{-1/p}$ .

This is an improvement of the result of Lemma 3.2. It is clear that a set $K$ in this form (defined by a set of $N$ offsets and a scaling factor $\sigma$ ) has length ${\mu }_j\left(K\right)$ at the $j^{th}$ stage bounded by

so that its Lebesgue measure $\mu \left(K\right)$ must be bounded by

implying that $\mu \left(K\right)=0$ when $N\sigma <1$ . In this case, we have $N=2^{p+q}-1$ and $\sigma ={\lambda }^p$ , so that $\mu \left(S,\left(\frac{p}{q},,,\lambda \right)\right)=0$ when $\lambda <{\left(2^{p+q},-,1\right)}^{-1/p}$ , as desired. To see this result graphically, see [link] .

The next result from this study involves the results from Theorems 3.4 and 3.5. Using the notation developed above, we can compute algorithmically the value $\lambda \left(\theta \right)$ for any rational $\theta$ . Recall that this value is already known for irrational $\theta$ . To do this, we need the following proposition.

Proposition 4.5 The set $S\left(\frac{p}{q},,,\lambda \right)=\left[0,,,1\right]$ if and only if there are no gaps in the first stage $S_1$ of its construction.

[Proof of the Proposition] $\underline{\Rightarrow }$ : Remember that $S_1$ is the union of intervals of length ${\lambda }^p$ with left endpoints at the offsets in the set $A+B$ . We have

so that

In particular,

as desired.

$\underline{\Leftarrow }$ : Assume that $S\left(\frac{p}{q},,,\lambda \right)\ne \left[0,,,1\right]$ . Thus, $S\left(\frac{p}{q},,,\lambda \right)$ must contain a gap of positive length, and this gap must show up in a finite stage, so it must be a scaled-down copy (or the overlap of multiple scaled-down copies) of a gap/gaps in the first stage of the construction $S_1$ . Thus, $S_1\ne \left[0,,,1\right]$ , as desired.

This Proposition tells us that $\lambda \left(\theta \right)$ is the unique point for which

This point can be computed by finding all the offsets of $S\left(\frac{p}{q},,,\lambda \right)$ , ordering them from least to greatest, and checking the differences of adjacent offsets against ${\lambda }^p$ . This has been done for several values of $\theta =\frac{p}{q}$ in [link] .

The question of what happens below these points and above the points given by Corollary 4.4 is still open. We conjecture that strictly between these points, $S\left(\theta ,,,\lambda \right)$ is an M -Cantorval, and that at the points given by Corollary 4.4, $S\left(\theta ,,,\lambda \right)$ where $\theta >1$ is a Cantor set with Hausdorff dimension 1 and zero Lebesgue measure.

Acknowledgments

This Connexions module describes work conducted as part of Rice University's VIGRE program, supported by National Science Foundation grant DMS–0739420.

We would also like to thank Dr. Danijela Damjanović and Dr. David Damanik for leading this study.