# Subsums of Null Sequences: A Survey and Open Questions

Justin Jacob

Abstract

We give a survey on the different results involving the topological structure
of subsums of null sequences.

##  1 Achievement sets and Null Sequences:

Consider a convergent sequence $x(n)=[x_{0},x_{1},{\ldots},x_{n},{\ldots}],$
such that $\displaystyle\lim_{n\to\infty}{x_{n}}=0$. Denote the set of subsums
of this convergent sequence as:

$E(x)=\sum_{i=1}^{\infty}\epsilon_{i}x_{i},\epsilon_{i}\in\left\\{0,1\right\\},\forall
i\in\mathbb{N}$

where $E(x)$ called the achievement set of $x(n)$. Define the $n$-th tail as:

$R_{n}=\sum_{i=n+1}^{\infty}x_{i}$

For example, we can see that the sequence
$x(n)=[\frac{1}{2},\frac{1}{4},{\ldots},\frac{1}{2^{n}},{\ldots}]$ is a
convergent sequence. Notice that $\sum_{n=1}^{\infty}x(n)=1$. Thus we know the
set $E(x)\subseteq[0,1]$. To show $E(x)$ is equal to $[0,1]$ we use the
following proof from Nitecki [25]. Observe that to express a digit in
base-$2$, or binary, we can use the following notation. Consider
$\epsilon_{i}\in\\{0,1\\},\forall i\in\mathbb{N}$. Then for any $x\in E(x)$ we
have that $x=\sum_{i=1}^{\infty}\frac{\epsilon_{i}}{2^{i}}$ for some sequence
$[\epsilon_{0},\epsilon_{1},\epsilon_{2},{\ldots}]$ Since every number between
0 and 1 has a binary expansion we can see that $E(x)=[0,1]$.
For another example, consider the sequence
$x(n)=[\frac{1}{3},\frac{1}{9},{\ldots},\frac{1}{3^{n}},{\ldots}].$ Again take
$\epsilon_{i}\in\\{0,1\\},\forall i\in\mathbb{N}$. We first give the
definition of the standard Cantor set.

######  Definition 1 (Standard Cantor Set).

The standard Cantor set $C$ is defined as removing the middle third from the
interval $[0,1]$, removing the middle third from each of the remaining
connected components, and so on, iteratively continuing this process. Thus:

$C=[0,1]\setminus\bigcup\limits^{\infty}_{k=1}{C_{k}},\text{where}\;C_{k}=\frac{C_{k-1}}{3}\cup\left(\frac{2}{3}+\frac{C_{k-1}}{3}\right),n\geq
2\;\text{and}\;C_{1}=\left[\frac{1}{3},\frac{2}{3}\right].$

######  Theorem 1 (Nitecki [25]).

The set $E(x)$ for the sequence $x(n)=\frac{1}{3^{n}},n\in\mathbb{N}$ is
homeomorphic to the standard Cantor set.

###### Proof.

First notice that any subsum must belong to the interval $[0,1/2]$. However,
unlike the previous example, every sum excluding the first term will belong in
the interval $J_{0}=[0,\frac{1}{6}]$, while $J_{1}=[\frac{1}{3},\frac{1}{2}].$
So thus subsums are contained in $C_{1}=J_{0}\cup J_{1}$. Now we consider the
exclusion of the second term, $\frac{1}{9}$, we find that the subsum is
contained in four subintervals, two subintervals of $J_{0}$,$J_{00}$ and
$J_{01}$ and two subintervals of $J_{1}$, $J_{10}$ and $J_{11}$. Notice these
correspond to all possible sequences of $\epsilon_{i}$ of length 2. Then
continuing this process for all possible sequences of $\epsilon_{i}$ for
length $n$, or where $1\leq i\leq n$, we have that:

$E(x)\subset
C_{n}={\bigcup_{{\epsilon_{1},\epsilon_{2},{\ldots},\epsilon_{n}\in\\{0,1\\}^{n}}}}J_{\epsilon_{1},{\ldots},\epsilon_{n}}$

With $J_{0^{n}}=[0,R_{n}]$. Notice that at the next step, we have two
possibilities for the next subintervals:

${J_{{\epsilon_{1},{\ldots},\epsilon_{n},0}}}\,\text{and}\,{{J_{\epsilon_{1},{\ldots},\epsilon_{n},0}}}+\textstyle\frac{1}{3^{n+1}}={{J_{\epsilon_{1},{\ldots},\epsilon_{n},1}}}$

So at every step, our intervals are disjoint. We can notice that passing from
the union $c_{n}$ to $C_{n+1}$, a gap is created. Furthermore notice that
$R_{n+1}=\frac{1}{3}R_{n}$, so at each step the size of this gap is the middle
third of each component. Thus we have $C_{\infty}=\bigcap_{n=1}C_{n}$, which
is a version of the standard Cantor set, just scaled down by a factor of
$\frac{1}{2}$. In our construction, our Cantor set contains every number whose
ternary expansion contains digits of only zeroes and ones rather than the
zeroes and twos in the standard Cantor set. ∎

Figure 1: The construction of subsum sets for
$x(n)=\frac{1}{3^{n}},n\in\mathbb{N}$ [1]

So far for the current examples considered, one may notice that in our
investigation of the achievement sets for the geometric series
$x(n)=\frac{1}{3^{n}}$ and $x(n)=\frac{1}{2^{n}},n\in\mathbb{N}$, either the
intervals at each step will continue to overlap or be completely disjoint.

We can now generalize to obtain the following results about subsums of
geometric series. The proof of the theorem found in Aguero’s paper [1] is very
similar to Nitceki’s proof. The key insight is that if $r\geq\frac{1}{2}$, we
will have that the intervals constructed at each step will never be completely
disjoint and thus when we take the intersection we will have that
$C_{\infty}=\bigcap_{n=1}C_{n}$ must be an interval. However if
$r<\frac{1}{2}$ then each interval will indeed be disjoint and thus
$C_{\infty}=\bigcap_{n=1}C_{n}$ will be homeomorphic to a Cantor set. Of
course for a more rigorous notion one can consult [1].

######  Theorem 2 ([1]).

For $0\leq r<\frac{1}{2}$ we have $E(x)$ for $x_{k}=r^{k}$ is a Cantor set and
$\frac{1}{2}\leq r<1$ $E(x)$ is an interval.

Figure 2: Subsum sets for $\frac{1}{2}\leq r<1$ [1]

##  2 General Results on $E(x)$ and Cantorvals

We now shift our attention towards more general results regarding $E(x)$ for
arbitrary null sequences, not just geometric ones. The first prominent result
was done in 1914 by Kakeya [19] showed that $E(x)$ is:

  1. 1.

A perfect set,

  2. 2.

The union of finite intervals if $x_{n}\leq R_{n}$ for $n$ sufficiently large.

  3. 3.

Homeomorphic to a Cantor set if $x_{n}>R_{n}$ for $n$ sufficiently large.

Kakeya also conjectured that $E(x)$ is nowhere dense and homeomorphic to the
Cantor set if $x_{n}>R_{n}$ for infinitely many $n$. However this conjecture
has been shown to be false with many such counterexamples. Ferens [14] was the
first to give the following example with proof:

$\sum_{i=1}^{\infty}\left[\epsilon_{5i-4}\left(\frac{7}{27^{i}}\right)+\epsilon_{5i-3}\left(\frac{6}{27^{i}}\right)+\epsilon_{5i-2}\left(\frac{5}{27^{i}}\right)+\epsilon_{5i-1}\left(\frac{4}{27^{i}}\right)+\epsilon_{5i}\left(\frac{3}{27^{i}}\right)\right]$

with $\epsilon_{i}\in\left\\{0,1\right\\},\forall i\in\mathbb{N}$ as usual.
Gutherie and Nymann [28] gave a much simpler example in 1988:

$T=\left\\{\sum_{i=1}^{\infty}\left[\epsilon_{2i-1}\left(\frac{3}{4^{i}}\right)+\epsilon_{2i}\left(\frac{2}{4^{i}}\right)\right]\right\\}$
(1)

Figure 3: The set of subsums of
$T=\left\\{\sum_{i=1}^{\infty}\left[\epsilon_{2i-1}\left(\frac{3}{4^{i}}\right)+\epsilon_{2i}\left(\frac{2}{4^{i}}\right)\right]\right\\}$

Gutherie and Nymann were also able to come up with a very important theorem
regarding the classification of sets of $E(x)$ for arbitary null sequences.
Before revealing their result, we define a Cantorval, dubbed by Mendes and
Oliviera in the following paper [23]. However, we give more comprehensive
definition regarding them than the one in their paper:

######  Definition 2 (Cantorval [6]).

Let $K$ be a compact subset of $\mathbb{R}$. A gap of $K$ is a bounded
connected component of $\mathbb{R}\setminus K$; an interval of $K$ is a non-
trivial connected component of $K$. A perfect subset of $\mathbb{R}$, such
that any gap is accumulated on each side by infinitely many intervals and gaps
is called an M-Cantorval. A perfect subset of $\mathbb{R}$, such that any gap
has an interval adjacent on its right and is accumulated on the left by
infinitely many intervals and gaps is called an L-Cantorval. The definition of
an R-Cantorval is analogous.

An analgous definition of an M-Cantorval is also any subset of $\mathbb{R}$
that is homeomorphic to $C\cup\left(\bigcup_{n=1}^{\infty}C_{2n-1}\right)$.
Observe that when dealing with Cantorvals we have the following set equality:

$C\cup\left(\bigcup_{n-1}^{\infty}C_{2n-1}\right)=C_{2n}\setminus\left(\bigcup_{n-1}^{\infty}C_{2n}\right)$

Many of the topological properties of Cantorvals have been considered in the
literature, see [8] for some of these results. For special types of
M-Cantorvals found in [9],their Lebesgue measures have been shown to be the
sum of the measures of their interval components. However, Cantorvals do not
only occur in the topic of achievement sets. Bartoszewics et. al. [10] showed
that Cantorvals can occur as attractors of iterated function systems
consisting of affine functions.
We state a lemma attributed to Nymann and Saenz [28]:

######  Lemma 1.

If (a,b) is a gap of $E(x)$, then for some $\epsilon>0$ and
$\epsilon^{\prime}>0$,

$b+([0,\epsilon]\cap E)=[e,b+\epsilon]\cap
E\;\text{and}\;[s-\epsilon^{\prime},s]\cap
E=(s-a)+([a-\epsilon^{\prime},a]\cap E)$

Now we are ready to state the Gutherie-Nymann-Saenz Classification Theorem:

######  Theorem 3 (Gutherie-Nymann Saenz [28]).

The achievement set $E(x)$ for a general null sequence
$x(n)=[x_{0},x_{1},{\ldots},x_{n},{\ldots}]$ is homeomorphic to one of the
following:

  1. 1.

A Cantor set,

  2. 2.

A union of finite intervals,

  3. 3.

Homeomorphic to the set $T$ defined in equation 1 (which is also an
M-Cantorval).

###### Proof.

We have seen examples of null sequences whose achievement sets $E(x)$ are
either a union of finite intervals or a Cantor set. Now suppose that E is
neither a finite union of intervals nor homeomorphic to the Cantor set. Then
it is clear the complement of E must contain infinitely many intervals. E must
contain infinitely many intervals as well, for there were only finitely many
then either $E\cap\left[0,\epsilon\right)$ is an interval or
$E\cap\left[0,\epsilon\right)$ contains no interval for some $\epsilon>0$. If
the former is true, then $\exists n$ such that $R_{n}$ is an interval and thus
by Kakeya’s result $E$ is a finite union of intervals. In the latter case,
$\exists n$ such that $R_{n}$ is a Cantor set, so $E$ would be a Cantor set.
In either case we arrive at a contradiction, so $E$ must have infinitely many
intervals. In fact $E\cap[a,b]$ cannot be homeomorphic to the Cantor set
$\forall a,b\in E$ since every tail $R_{n}$ has intervals in its set of
subsums. Then for $x\in E$ we have $E\cap(x,x+\epsilon)\neq\emptyset$, and
since $E$ is perfect we have $E\cap(x-\epsilon,x)\neq\emptyset$ for some
$\epsilon>0$. Thus we can see that there are intervals dense in $E$.

Now we define a strictly increasing map $f$ from the union of all intervals in
$T$ to the union of all intervals in $E$. We do this inductively: first map
the longest interval in $T$ in a strictly increasing way to the longest
intervals in $E$. There can be at most finitely many intervals of the same
length in either set, so if no one interval is the longest we proceed by
choosing the leftmost one.

After the $n$-th step, $\frac{4^{n}-1}{3}$ intervals of $T$ will have been
identified, in a strictly increasing way, with $\frac{4^{n}-1}{3}$ intervals
of $E$. In addition, $\frac{2(4^{n}-1)}{3}$ gaps of $T$ will be identified in
a strictly increasing way onto $\frac{2(4^{n}-1)}{3}$ gaps of $E$. We repeat
the inductive process above to each of the spaces between any two adjacent
intervals and gaps (of which there are $4^{n}-1$) and to the space between 0
and the leftmost of the gaps and to the space between the rightmost of the
gaps and $\frac{5}{3}$. Lemma 1 guarantees this is possible, since there will
be infinitely many intervals and gaps in $T$ and $E$ in $[0,\epsilon]$ and
$[\frac{5}{3}-\epsilon,\frac{5}{3}]$, respectively.

When $f$ is defined this way, it is a strictly increasing mapping from the
union of all intervals in $T$ to the union of all intervals in $E$. As $T$ and
$E$ are both perfect it is possible to extend $f$ continuously to all of $T$
and then onto all of $E$. Since the extension is strictly increasing, and $T$
is compact, $f$ is the desired homeomorphism. ∎

######  Theorem 4 ([26]).

Any two M-Cantorvals (symmetric Cantorvals) are homeomorphic.

We introduce the following notation introduced by Bartoszewicz in [11]:

######  Definition 3.

For any $0\leq q<\frac{1}{2}$ we use $(k_{1},k_{2},{\ldots},k_{m};q)$ to
denote the sequence:

$\left(k_{1},k_{2},{\ldots},k_{m},k_{1}q,k_{2}q,{\ldots},k_{m}q,k_{1}{q^{2}},k_{2}{q^{2}},{\ldots},k_{m}{q^{2}},{\ldots}\right)$

In 2011, Jones [18] gave a result that yielded a continuum of Cantorvals:

######  Theorem 5.

The set of subsums of
$\left(\frac{3}{5},\frac{2}{5},\frac{2}{5},\frac{2}{5};q\right)$ is a
Cantorval for all $q$ such that

$\frac{1}{5}\leq\sum_{i=1}^{\infty}q^{i}<\frac{2}{9}$

Kenyon and Nitecki [26] also paved the way for another continuum of Cantorvals
with the following result in 2015:

######  Theorem 6.

Suppose we are given $n\in\mathbb{N}$ and $n$ integers
$d_{0},d_{1},{\ldots},d_{n-1}$ such that $d_{i}\equiv i\pmod{n}$. Then the set
of generalized base $n$ expansions using these “digits”:

$S=\left\\{\sum_{i=1}^{\infty}\frac{a_{i}}{n^{i}}\,|\,{a_{i}}\in\left\\{d_{0},d_{1},{\ldots},d_{n-1}\right\\}\right\\}$

has nonempty interior.

Using the notation described in Definition 3, Bartoszewics [11] established
the following result:

######  Theorem 7.

Let $k_{1}\leq k_{2}\leq\cdots\leq k_{m}$ be positive integers and
$K=\sum_{i=1}^{m}{k_{i}}$. Assume that there exist positive integers $n_{0}$
and $n$ such that each of the numbers $n_{0},n_{0}+1,{\ldots},n_{0}+n$ can be
obtained by summing the numbers $k_{1},k_{2},{\ldots},k_{m}$ in some
combination. Then we have:

  1. 1.

If $q\geq\frac{1}{n+1}$ then $E(k_{1},k_{2},{\ldots},k_{m};q)$ has nonempty
interior.

  2. 2.

If $q<\frac{k_{m}}{K+{k_{m}}}$ then $E(k_{1},k_{2},{\ldots},k_{m};q)$ is not a
finite union of intervals.

  3. 3.

If $\frac{1}{n+1}\leq q<\frac{k_{m}}{K+{k_{m}}}$ then
$E(k_{1},k_{2},{\ldots},k_{m};q)$ is a Cantorval.

Ferdinands et. al. [13] have generalized this result to the following:

######  Theorem 8.

Let $k_{1}\leq k_{2}\leq\cdots\leq k_{m}$ be positive integers and
$K=\sum_{i=1}^{\infty}{k_{i}}$. Assume that there exist positive integers
$a$,$d$ and $n$ such that the set:

$\sum_{i=1}^{m}\epsilon_{i}k_{i},\epsilon_{i}\in\\{0,1\\}$

contains each of the numbers $a,a+d,a+2d,{\ldots},a+nd$. Then we have:

  1. 1.

If $q\geq\frac{1}{n+1}$ then $E(k_{1},k_{2},{\ldots},k_{m};q)$ has nonempty
interior.

  2. 2.

If $q<\frac{k_{m}}{K+{k_{m}}}$ then $E(k_{1},k_{2},{\ldots},k_{m};q)$ is not a
finite union of intervals.

  3. 3.

If $\frac{1}{n+1}\leq q<\frac{k_{m}}{K+{k_{m}}}$ then
$E(k_{1},k_{2},{\ldots},k_{m};q)$ is a Cantorval.

Notice, using Definition 3, we have the Gutherie-Nymann example
$T=(3,2,\frac{1}{4})$. Agüero [1] also generalized the Gutherie-Nymann example
as following:

######  Theorem 9.

Using the notation in Definition 3,
$(\underbrace{{3,3,{\ldots}}}_{k\;3^{\prime}s},\underbrace{{2,2,{\ldots}}}_{k\;2^{\prime}s},\frac{1}{3k+2k^{\prime}+1})$
is a Cantorval for $k,k^{\prime}\in\mathbb{N}$.

##  3 Sums of Achievement Sets:

We now turn our attention towards sums of achievement sets, ie:

$P=E_{m}(x)=\sum_{i=1}^{\infty}\epsilon_{i}x_{i},\epsilon_{i}\in\\{0,1,2,{\ldots},m\\},\forall
i\in\mathbb{N}$

with tail:

$R_{z}=\sum_{i=z+1}^{\infty}x_{i}$

In 1995, Nymann [27] gave a result regarding these sums:

######  Theorem 10.

There is a positive integer $m$ for which $P$ is a finite union of intervals
if and only if $\limsup{\frac{x_{n}}{R_{n}}}<\infty$.
Moreover, the smallest positive integer for which $P$ is a finite union of
intervals is the smallest integer $m$ such that $\frac{x_{n}}{R_{n}}\leq m$
for all but a finite number of integers $n$.

We now give a result that uses the proof Gutherie and Nymann used in their
classification theorem:

######  Theorem 11 (Generalized Guthrie-Nymann).

For all $m\in\mathbb{N}$, $E_{m}(x)$ is homeomorphic to one of the following:

  1. 1.

A Cantor set,

  2. 2.

A union of finite intervals,

  3. 3.

A Cantorval.

###### Proof.

First notice that $E_{m}(x)$ is a perfect compact set since it is a sum of
perfect compact sets. Then if the interior of ${E_{m}(x)}=\emptyset$,
$E_{m}(x)$ must be homeomorphic to the Cantor set.
Now suppose the interior of $E_{m}(x)$ is nonempty. Then all but finitely many
tails $R_{n}$ must contain an interval as its set of subsums. Furthermore, we
also have that $E(x)\subseteq E_{m}(x)$, so in this case we know at the very
least $E_{m}(x)$ will contain a finite union of intervals. Now, we can use the
same proof given in [28] to conclude the result. ∎

##  4 Conjectures and Open Questions

###  4.1 Achievement Sets of Compact Sets with Decreasing Diameter:

Now we consider a much more general problem: Define:

${diam(A)}=\displaystyle\sup_{x,y\in A}\\{d(x,y)\\}$

where $d(x,y)$ denotes the usual metric on $\mathbb{R}$. Suppose we are given
a sequence of compact sets in $\mathbb{R}$,
$K_{1},K_{2},{\ldots},K_{i},{\ldots}$ where $0\in K_{i}\,\forall i$, $\forall
x\in K_{i}\,x\geq 0$ and ${diam(K_{i})}>0\;\forall i\in\mathbb{N}$.
Furthermore let ${diam(K_{i})}\to 0$ as $i\to\infty$. Now pick $x_{1}\in
K_{1},x_{2}\in K_{2},{\ldots},x_{n}\in K_{n},{\ldots}$ and consider:

$S=\\{\sum_{j=1}^{\infty}x_{j}|\,x_{j}\in K_{j}\\}\,\,\text{with
tail}\,\,R_{z}=\\{\sum_{i=z+1}^{\infty}x_{i}|\,x_{i}\in K_{i}\\}$

######  Lemma 1.

S is a perfect set.

###### Proof.

Clearly by Tychnoff’s Theorem $S$ is compact. Pick $x\in S$. Then we have
$x=x_{1}+x_{2}+{\ldots}+x_{n}+{\ldots}$. Notice that since $diam(K_{i})\to 0$
as $i\to\infty$, $\forall\epsilon>0,\,\exists K_{y}$ with
$diam(K_{y})<\epsilon$. Then $z_{y}\in K_{y}$ has the property that
$\left|z_{y}-x_{y}\right|<\epsilon$, so the sum
$x_{1}+x_{2}+{\ldots}+x_{n}+{\ldots}+z_{y}+{\ldots}\in(x,x+\epsilon)$. Then
pick $z_{z}\in K_{z}$ where $diam(K_{z})<\epsilon_{1}<\epsilon$, so
$\left|z_{z}-x_{z}\right|<\epsilon_{1}$. So then
$x_{1}+x_{2}+{\ldots}+x_{n}+{\ldots}+z_{y}+z_{z}+{\ldots}\in(x,x+\epsilon_{1}).$
Inductively doing this approach, we get a sequence in $S$ with limit point
$x$. Thus $S$ is perfect. ∎

However, it is unknown what the structure of $S$ is in general. It is trivial
to show that if the interior of $S$ is empty, then $S$ must be a Cantor set,
however we believe that the classifying these sets completely in terms of the
tail $R_{z}$ is far too general of a problem. Instead, we list the following
open questions:

  1. 1.

What are the possible sets $S$ can be homeomorphic to?

  2. 2.

Under what conditions on the tail $R_{z}$ will this set be an interval or a
finite union of intervals?

  3. 3.

Under what conditions on the tail $R_{z}$ will this set be an M-Cantorval? An
L or R Cantorval?

###  4.2 Achievement Sets of Cantor Sets:

Sums of finite Cantor sets appear frequently in the literature and show up in
spectral theory and homoclinic bifuractions. In addition, many of their
topological properties have been considered for sums of more general types of
Cantor sets. Thanks to the Newhouse Gap Lemma [24], we also know if
$\tau(C_{\alpha})\tau(C_{\beta})>1$, then the sum of the two Cantor sets
contain an interval. Here $\tau(C)$ denotes the thickness of the set, see [24]
for a more precise notion. It is well known (see chapter 4 in [21]) that if:

$\frac{\log{2}}{\log{(\frac{1}{\alpha})}}+\frac{\log{2}}{\log{(\frac{1}{\beta})}}<1\Rightarrow\dim_{H}(C_{\alpha})+\dim_{H}(C_{\beta})<1$
(2)

then the Lebesgue measure of the set $C_{\alpha}+C_{\beta}$ is zero and the
sum be a Cantor set. Here $\dim_{H}$ denotes the Hausdorff dimension of a set.
By result 2 and the Newhouse Gap Lemma [24], we have knowledge of many of the
sums of Cantor sets depending on the parameters of $\alpha,\beta.$ However,
many open questions still remain. These results reveal a “mysterious region”
in the space of parameters $\alpha,\beta.$ We denote the area where the sum is
an interval as “large”, the area where the sum has Lebesgue measure zero as
“small”, and the “mysterious region” as R in the figure below. Solomyak [32]
found that if:

$\dim_{H}(C_{\alpha})+\dim_{H}(C_{\beta})>1,$

the set $C_{\alpha}+C_{\beta}$ has positive Lebesgue measure for almost
everywhere $\alpha,\beta\in(0,\frac{1}{2}).$ Furthermore, he and Peres also
found this is true for any compact set $K$ and any middle $\alpha$ Cantor set
$C_{\alpha}$ [29]. This implies that for almost every parameter in the
“mysterious region,” the sum of the corresponding homogeneous Cantor sets has
positive Lebesgue measure.

Figure 4: A graph of the space of parameters and the ”Mysterious Region” R.
[32]

All of these results,however, concern finite sums of Cantor sets. In addition,
many of the results are quite specific and cannot be verified in full
generality, i.e. for every possible class of Cantor sets. However we believe
many of the results will still hold even in our generalization of the problem:
Suppose we have a null sequence
$\alpha_{1},\alpha_{2},{\ldots},\alpha_{n},{\ldots}$ such that $\alpha_{i}\to
0$ as $i\to\infty$. Furthermore suppose we have a sequence
$\lambda_{1},\lambda_{2},{\ldots},\lambda_{n},{\ldots}$ with $\lambda_{i}\to
0$ as $i\to\infty$. For all $i$ let $K_{i}$ be a middle-$\alpha_{i}$ Cantor
set with first step $[0,\lambda_{i}]$. Now consider $S$ for these sets. We
pose the following open questions when considering $S$:

  1. 1.

Under what conditions of $R_{z}$ and $\alpha_{i}$ for $i\in\mathbb{N}$ will
this set be a Cantor set? What type of Cantor set will it be?

  2. 2.

Under what conditions of $R_{z}$ and $\alpha_{i}$ for $i\in\mathbb{N}$ will
this set be a finite intervals or a union of finite intervals?

  3. 3.

Under what conditions of $R_{z}$ and $\alpha_{i}$ for $i\in\mathbb{N}$ will
this set be an M-Cantorval? An L or R Cantorval?

  4. 4.

Denote the Haussdorf dimension of $S$ as $\dim_{H}(S)$. Is it true that
$\dim_{H}(S)=\sum_{i=1}^{\infty}\dim_{H}(K_{i})$ generically? If not, in what
specific cases will this be true?

###  4.3 Achievement Sets in the Complex Plane:

Our last generalization of the problem involves making each set $K_{i}$ a
subset of the complex plane and then investigating it. To do this we give a
brief overview of previous results relating to the problem at hand.

In 1957, Rényi [30] was the first to consider representations of numbers with
radix in $\mathbb{R}$, i.e. the set of points that can be represented as

$x=\sum_{i=1}^{\infty}\frac{\epsilon_{i}}{q^{i}}$

with $q\in\mathbb{R}$ and $\epsilon_{k}\in A$, where A is some finite set of
positive real values called the alphabet. Call this set $H$. This led to many
more developments, including the study of canonical number systems and radix
representations in algebraic number fields. See [2],[4] and [3] for a more
comprehensive summary. The first people to study complex radixes were Daróczy
and Kátai [12], who showed for which $q^{i}=\theta^{i}$ for some
$\theta\in\mathbb{C}$ with $\left|\theta\right|<1$ does $H=\mathbb{C}$. These
conditions were expanded upon by Komornik and Loreti in [20]. The study of
these complex bases has found various applications in computer science and
cryptography, see [17] and [31]. Gilbert [16] was the first to consider the
fascinating geometry of these expansions in 1981. One such example of these
fractals is the Lévy dragon. See [7] for a very thorough analysis on this
figure.

Figure 5: The Lévy Dragon [7]

Lai [22] studied the geometry of these sets with base $q=pe^{\frac{2\pi
i}{n}}$ and characterized their convex hull. Furthermore, the topological
properties of these structures, called self-affine tiles, were done by Akiyama
and Thuswaldner [5]. There are many open questions regarding this subject
already posed in the literature, mostly regarding the representability of
numbers in this radix system as well as the properties of the sets they form.

Finally we give the generalization of our problem: Again, define:

${diam(A)}=\displaystyle\sup_{x,y\in A}\\{d(x,y)\\}$

where $d(x,y)$ denotes the usual metric on $\mathbb{C}$. Suppose we are given
a sequence of compact sets in $\mathbb{C}$,
$K_{1},K_{2},{\ldots},K_{i},{\ldots}$ where $0\in K_{i}\,\forall i$, $\forall
x\in K_{i}\,x\geq 0$, ${diam(K_{i})}>0\;\forall i\in\mathbb{N}$ and
${diam(K_{i})}\to 0$ as $i\to\infty$. Now pick $x_{1}\in K_{1},x_{2}\in
K_{2},{\ldots},x_{n}\in K_{n},{\ldots}$ and consider:

$S=\left\\{\sum_{j=1}^{\infty}x_{j}|\,x_{j}\in K_{j}\right\\}\,\,\text{with
tail}\,\,R_{z}=\left\\{\sum_{i=z+1}^{\infty}x_{i}|\,x_{i}\in K_{i}\right\\}$

Now we pose the following questions, inspired from the results regarding
complex radix expansions and the sets they form:

  1. 1.

Under what conditions on $K_{i}$ and $R_{z}$ will $S=\mathbb{C}$?

  2. 2.

Several authors have investigated the fractal structures of radix
representations using a complex base [15]. Under what conditions on $K_{i}$
and $R_{z}$ will $S$ have fractal structure? What will the Haussdorf dimension
of $S$ be?

  3. 3.

Motivated by results in algebra concerning representations in algebraic number
fields [2], besides the trivial case, what specific conditions on $K_{i}$ and
$R_{z}$ will we able to have a radix representation of all algebraic numbers?

  4. 4.

In the case of real numbers, we can see Cantor sets and Cantorvals arise as
achievements sets of null sequences. Besides the trivial case of real numbers
already examined, under what conditions of $R_{z}$ and $\alpha_{i}$ for
$i\in\mathbb{N}$ will $S$ be a Cantor subset of $\mathbb{C}$? A 2-dimensional
Cantor set (Cantor Dust)?

  5. 5.

Motivated by the results in [5], under what conditions of $R_{z}$ and
$\alpha_{i}$ for $i\in\mathbb{N}$ will this set be a disk or a union of finite
disks?

  6. 6.

Denote the Haussdorf dimension of $S$ as $\dim_{H}(S)$. Is it true that
$\dim_{H}(S)=\sum_{i=1}^{\infty}\dim_{H}(K_{i})$ generically? If not, in what
specific cases will this be true?

##  5 Acknowledgements

This project was done for the Summer Undergraduate Research Experience (SURP)
Program at UC Irvine. I am grateful to the Undergraduate Research
Opportunities Program (UROP) at UCI for providing a stipend to support the
project. In addition, I would like to thank my advisor, Professor Anton
Gorodetski, for guiding me throughout the entire process and supporting me. I
would also like to thank Lucas Booher for his contributions towards the
project as well.

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