Center of Distances and Central Cantor Sets

Abstract

We study a recently discovered metric invariant - the center of distances. The center of distances of a nonempty subset A of a metric space $(X,\,d)$ is defined by $S(A) :=\{ \alpha \in [0,\,+\infty ):\ \forall \ x\in A\ \ \exists \ y\in A d(x,\,y)=\alpha \}$. Given a nonincreasing sequence $(a_{n})$ of positive numbers converging to 0, the set $E(a_{n})\ :=\ \left\{ x\in {\mathbb {R}}:\ \exists A\subset {\mathbb {N}} \ \ x=\,\sum _{n\in A}a_{n}\right\}$ is called the achievement set of the sequence $(a_{n})$. This new invariant is particularly useful in investigating achievability of sets on the real line. We concentrate on computing the centers of distances of central Cantor sets. Any central Cantor set C is an achievement set of exactly one fast convergent series $\sum a_{n}$, and consequently $S(C)\supset \left\{ 0\right\} \cup \left\{ a_{n}:n\in {\mathbb {N}}\right\}$. We try to check which central Cantor sets have the minimal possible center of distances and which have not.

1 Introduction

Let A be a nonempty subset of a metric space $(X,\,d)$. Bielas, Plewik and Walczyńska in [8, before Thm 2.1] have introduced a new interesting invariant of subsets of X. The center of distances S(A) of the set A is defined by

$$\begin{aligned} S(A)\ :=\ \left\{ \alpha \in [0,\,+\infty ):\ \forall \ x\in A\ \ \exists \ y\in A\ \ d(x,\,y)=\alpha \ \right\} . \end{aligned}$$

In the case where $X={\mathbb {R}}$ and $A\subset {\mathbb {R}}$, $\alpha \in S(A)$ if and only if $x+\alpha \in A$ or $x-\alpha \in A$ for any $x\in A$. This notion has interesting properties and applications. First observe that the operator S is not monotonic. For example, $S\left( \left\{ 0,1\right\} \right) =\left\{ 0,1\right\}$ while $S\left( \left\{ 0,1,3\right\} \right) =\left\{ 0\right\}$. Evidently, $0\in S(A)$ always. A nice application of the center of distances is a generalization of von Neumann theorem on permutations of sequences.

This well-known theorem says that two sequences of elements of X, $(a_{n})$ and $(b_{n})$ have the same set of cluster points if and only if there is a permutation $\pi :\,{\mathbb {N}}\rightarrow {\mathbb {N}}$ such that $d(a_{n},\,b_{\pi (n)})\rightarrow 0$ ([12, 20]). It is known that if A is a closed subset of a compact metric space $(X,\,d)$ and if $(a_{n})$, $(b_{n})$ are two sequences with the set of cluster points equal to A, then

$$\begin{aligned} S(A)\ =\ \left\{ \alpha \in [0,\,+\infty ):\ \exists \,\text { a permutation }\,\pi :\,{\mathbb {N}}\rightarrow {\mathbb {N}}\ \ d(a_{n},\,b_{\pi (n)})\rightarrow \alpha \right\} . \end{aligned}$$

More precisely, the nontrivial inclusion $\subseteq$ has been proved in [8, Thm 2.1] using the ”back and forth” method. The reversed inclusion $\supseteq$ is easy. Indeed, take any $x\in A$ and let $d(a_{n},\,b_{\pi (n)})\rightarrow \alpha$. The set $L\left( a_{n}\right)$ of limits points of the sequence $\left( a_{n}\right)$ is equal to A and so there exists a subsequence $a_{n_{m}}\rightarrow x$. Since X is compact, the sequence $(b_{\pi (n_{m})})_{m\in {\mathbb {N}}}$ has a subsequence $\bigl (b_{\pi (n_{m_{k}})}\bigr )_{k\in {\mathbb {N}}}$ convergent to some $y\in A$. Then $d(x,\,y)=\lim _{k\rightarrow \infty }d(a_{n_{m_{k}}},\,b_{\pi (n_{m_{k}})})=\alpha$.

In particular if A is a closed subset of a compact metric space $(X,\,d)$ and if $(a_{n})$ is any sequence such that $L\left( a_{n}\right) =A$, then

$$\begin{aligned} S(A)\ =\ \left\{ \alpha \in [0,\,+\infty ):\ \exists \text { a permutation }\,\pi :\,{\mathbb {N}}\rightarrow {\mathbb {N}}\ \ d(a_{n},\,a_{\pi (n)})\rightarrow \alpha \right\} . \end{aligned}$$

Further, given any $x\in X$, the function $f_{x}:\,A\rightarrow [0,\,+\infty ):\,y\mapsto d(x,\,y)$ is continuous and hence $f_{x}(A)$ is compact. Hence the equality $S(A)=\bigcap _{x\in A}f_{x}(A)$ justifies the implication: if A is nonempty and compact, then S(A) is compact. We will use also the easy observation that if $A\subset [0,\,+\infty )$ and $0\in A$, then $S(A)\subset A$.

The concept of center of distances was introduced as a very useful tool for proving that certain subsets of reals are not sets of subsums of any series [8, Cor. 5.5], [4, Prop. 3.4 and Thm 6.1] (see also the final paragraph of introduction of the latter paper). Of course, it amounts to showing that certain sets are not ranges of any purely atomic finite measures. Many authors prefer to emphasize this connection (see, for example, [4, 9] and [10]).

Given a nonincreasing sequence $(a_{n})$ of positive numbers converging to 0, the set

$$\begin{aligned} E\ =\ E(a_{n})\ :=\ \left\{ x\in {\mathbb {R}}:\ \exists A\subset {\mathbb {N}}\ \ x=\,\sum _{n\in A}a_{n}\right\} \end{aligned}$$

is called (see [13]) the achievement set of the sequence $(a_{n})$ or the set of subsums of the series $\sum a_{n}$. Throughout this note, by a series $\sum a_{n}$ we will always understand a convergent series $\sum a_{n}$ with positive and nonincreasing terms, although, in general, the convergence of the series or even the convergence of $(a_{n})$ to 0 are not necessary to define and investigate the set $E(a_{n})$ ([18, 19, 22, 5, Thm. 21.3]). The k-th remainder of the series will be denoted by $r_{k}:=\sum _{n>k}a_{n}$. In particular, $r_{0}=\sigma$, where $\sigma$ denotes the sum of the series. We will use the symbol $\sigma$ for denoting the sum of the series and for no other purpose. The set of subsums of the k-th remainder series $\sum _{n=k+1}^{\infty }a_{n}$ will be denoted by

$$\begin{aligned} E_{k}\ =\ E_{k}(a_{n})\ :=\ \left\{ x\in {\mathbb {R}}:\ \exists \ A\subset \{k+1,\,k+2,\,\ldots \,\}\ \ x\,=\,\sum _{n\in A}a_{n}\;\right\} . \end{aligned}$$

The set of k-initial subsums of the series will be denoted by

$$\begin{aligned} F_{k}\ =\ F_{k}(a_{n})\ :=\ \left\{ x\in {\mathbb {R}}:\ \exists \ A\subset \{1,\,\ldots \,\,k\,\}\ \ x\,=\,\sum _{n\in A}a_{n}\;\right\} . \end{aligned}$$

The set $F=F(a_{n}):=\bigcup _{k\in {\mathbb {N}}}F_{k}$ is then the set of all finite subsums or, more precisely, of all sums of finite subseries. We agree to write $F_{0}:=\{0\}$. It is obvious that

$$\begin{aligned} E\ =\ F_{k}\;+\;E_{k}\ =\ \bigcup _{f\in F_{k}}(f+E_{k}) \end{aligned}$$

and

$$\begin{aligned} S\left( F_{k}\right) \subset S\left( E\right) \end{aligned}$$

for any $k\in {\mathbb {N}}$. Clearly, $E_{k+1}\subset E_{k}$ and $F_{k}\subset F_{k+1}$ for all k. Every set of subsums $E=E(a_{n})$ has a natural representation $\,E\,=\,\bigcap _{k}I_{k}$, where $I_{k}:=\bigcup _{f\in F_{k}}[f,\,f+r_{k}]$. [5]. The connectivity components of $I_{n-1}{\setminus } I_{n}$ are called E-gaps of order n.

In general, given a nonempty bounded and perfect set $P\subset {\mathbb {R}}$, the bounded components of its complement are called gaps or P -gaps. Since the sets of subsums are bounded and perfect ([14, 15, 22]), understanding their gaps amounts to understanding their geometric properties. There are three very useful facts about the gaps and we will formulate them below for reader’s convenience.

TheFirstGapLemma

([5, Fact 21.9]) If $a_k>r_k$ for some index k, then the open interval $(r_k,\,a_k)$ is a gap of E.

Sometimes, like in the following fact, it is convenient to arrange all elements of $F_k$ into an increasing finite sequence $(f_j^{(k)})_{j=0}^{m(k)}$.

TheSecondGapLemma

([21, Lemma 2], [5, Fact 21.11]). Let $(a,\,b)$ be an E-gap and let $k=\max \{n:\ a_n\ge b-a\,\}$. Then $b\in F_k$. Moreover, if $b=f_j^{(k)}$, then $a=f_{j-1}^{(k)}+r_k$.

TheThirdGapLemma

([4, Lemma 2.4], [3, Lemma 4]). Suppose that $(a,\,b)$ is an E-gap such that all E-gaps lying to the left of it are shorter than $b-a$ . Then $b=a_n$ and $a=r_n$ for some $n\in {\mathbb {N}}$.

The complete characterization of all possible topological types of sets of subsums is known. A set $P\subset {\mathbb {R}}$ is said to be a multi-interval set if it is the union of a finite family of closed and bounded intervals. A set $P\subset {\mathbb {R}}$ is said to be a Cantor set if it is nonempty, bounded, perfect and nowhere dense, that is, if it is homeomorphic to the classic Cantor ternary set. A set $P\subset {\mathbb {R}}$ is said to be a Cantorval if it is homeomorphic to the set

$$\begin{aligned} GN\ :=\ {\mathbf {C}}\ \cup \ \bigcup _{n}G_{2n-1} \end{aligned}$$

where ${\mathbf {C}}$ denotes the classic Cantor ternary set ${\mathbf {C}}=E( \frac{2}{3^{n}})$ and $G_{2n-1}$ is the union of all $4^{n-1}$ ${\mathbf {C}}$ -gaps of order $2n-1$. It is known that a Cantorval is exactly a nonempty compact set in ${\mathbb {R}}$ such that it is the closure of its interior and both endpoints of every nontrivial component are accumulation points of its trivial components. Other topological characterizations of Cantorvals can be found in [17] and [5]. The well-known Guthrie-Nymann classification theorem ([11, Thm. 1], [22]) asserts that the set E of subsums of a series $\sum a_{n}$ is either a multi-interval set or a Cantor set, or a Cantorval. An analytic characterization of the above cases is known only in the case of multi-interval sets. Namely, the set of subsums $E(a_{n})$ is a multi-interval set if and only if $a_{n}\le r_{n}$ for all sufficiently large n ([14, 15, 22]).

A series $\sum a_{n}$ (or a sequence $\left( a_{n}\right)$) is said to be fast convergent if $a_{n}>r_{n}$ for all n. Fast convergence of $\sum a_{n}$ is a sufficient condition for E to be a Cantor set ([14, 15, 22]). For any series $\sum a_{n}$ and any $n\in {\mathbb {N}}$

$$\begin{aligned} \frac{a_{n}+r_{n}}{2}=\frac{r_{n-1}}{2}\text {.} \end{aligned}$$

If $\sum a_{n}$ is fast convergent it means that a middle point of the gap $\left( r_{n},a_{n}\right)$ is a middle point of the interval $\left( 0,r_{n-1}\right)$. Thus, fast convergent series are closely related to central Cantor sets.

A central Cantor set is constructed in the following way. At the initial stage we remove the middle open interval $\left( y_{1},x_{1}\right)$ from a closed interval $K_{0}=\left[ 0,r_{0}\right]$ (a first order gap). The remaining set, which is the union of two closed intervals with the same length, we denote by $K_{1}$. From these two intervals we delete the middle open intervals with the same lengths (second order gaps), obtaining the set $K_{2}$. Denote the left of removed interval by $\left( y_{2},x_{2}\right)$. Proceeding inductively, we remove from each of $2^{n}$ closed components of the set $K_{n}$ the n-order gaps of the length $x_{n}-y_{n}$. The central Cantor set C is given as the intersection of $K_{n}$.

If $\sum a_{n}$ is fast convergent then its achievement set is a central Cantor set. On the other hand, each central Cantor set is an achievement set of exactly one fast convergent series (see f.e. [5, 10] or [22]). However, as we observe later, it is possible to obtain some central Cantor sets as an achievement sets of another (of course not fast convergent) sequences.

Some other sufficient conditions for E to be a Cantor set can be found in [1, 2] and [7]. Only a few sufficient conditions for E to be a Cantorval are known ([1, 7] and [18]) and they all are formulated for multigeometric series. A series $\sum a_{n}$ will be called multigeometric if there are positive numbers $l_{1}\ge l_{2}\ge \cdots \ge l_{m}$, and a number $q\in (0,\,1)$ such that

$$\begin{aligned} a_{km+i}\ =\ l_{i}q^{k} \quad \text {for all }\, k\in {\mathbb {N}}_{0} \,\text { and all }\,i\in \{1,\ldots ,m\,\}. \end{aligned}$$

It will be denoted by $\sum (l_{1},\,\ldots ,\,l_{m};q)$. The multigeometric series play a great role in the rapidly developing study of algebraic and topological properties of achievement sets because their sets of subsums are, in most cases, relatively easy to build and analyze from the computational point of view ([1, 7, 19]). For example, to find out if the series $\sum (l_{1},\,\ldots ,\,l_{m};q)$ is fast convergent it is enough to check if $l_{i}>r_{i}$ for $i=1,\ldots ,m$.

2 Central Cantor Sets with Minimal Centers of Distances

For the reader’s convenience we recall a basically known general fact. The symbol $d(x,\,A)$ denotes the distance of the point x from the set A, that is, the number $\inf _{y\in A}d(x,\,y)$.

Lemma 2.1

Given a convergent series $\sum a_{n}$ of positive terms and of the sum $\sigma$, the following inclusions are true

$$\begin{aligned} \{0\}\cup \{a_{n}:\ n\in {\mathbb {N}}\,\}\ \subset \ S(E)\ \subset \ [0,\, \tfrac{\sigma }{2}+d(\tfrac{\sigma }{2},\,E)]\cap E. \end{aligned}$$

Proof

The left inclusion is well known [8, Prop. 3.1] and so it remain to prove the right one. Setting $\alpha :=\max \{x\in E:\ x\le \frac{\sigma }{2 }\,\}$ and $\beta :=\min \{x\in E:\ x\ge \frac{\sigma }{2}\,\}$, we get $\max _{x\in E}|\alpha -x|=\sigma -\alpha$ and $\max _{x\in E}|\beta -x|=\beta$. Thus

$$\begin{aligned} S(E) \subset [0,\,\sigma -\alpha ]\cap [0,\,\beta ] = \bigl [0,\, \tfrac{\sigma }{2}+\min \{\tfrac{\sigma }{2}-\alpha ,\,\beta -\tfrac{\sigma }{2}\}\bigr ] = [0,\,\tfrac{\sigma }{2}+d(\tfrac{\sigma }{2},\,E)]. \end{aligned}$$

$\square$

In the case of a fast convergent series we have $\frac{\sigma }{2}+d(\tfrac{\sigma }{2},\,E)=a_1$.

It is well known that for a given fast convergent series $\sum a_{n}$ and an $x\in E$, there is a unique set of indices $I_{x}\subset {\mathbb {N}}$ such that $x=\sum _{n\in I_{x}}a_{n}$ [22, Fact 3.3], [10, Prop. 3.1]. In particular,

$$\begin{aligned} x\in E{\setminus } F \quad \text {if and only if} \ I_{x}\text { is infinite.} \end{aligned}$$

(1)

Lemma 2.2

If a series $\sum a_{n}$ is fast convergent, then $S(E)\subset F\cap [0,\,a_{1}]$.

Proof

By the Lemma 2.1 and (1), it suffices to show that $(E{\setminus } F)\cap [0,\,a_{1}]\cap S(E)=\emptyset$. Take any $\alpha \in (E{\setminus } F)\cap [0,\,a_{1}]$. We search for an $x\in E$ such that $x+\alpha \not \in E$ and $x-\alpha \not \in E$. Let $I_{\alpha }$ be the unique set of indices such that $\alpha =\sum _{n\in I_{\alpha }}a_{n}$. Define $m:=\min I_{\alpha }$. Since $\alpha <a_{1}$, we get $m\ge 2$. Further, define $k:=\min \{n\in I_{\alpha }:\ n>m\ \text {and}\ a_{n}<a_{m-1}-r_{m-1}\,\}$, $I:=\{k\}\cup {\mathbb {N}}{\setminus } \bigl ( \{1,\,2,\,\ldots ,\,m-1\}\cup I_{\alpha }\bigr )$ and $x:=\sum _{n\in I}a_{n}$ . Clearly, $x-\alpha<r_{m}-a_{m}<0$ and hence $x-\alpha \not \in E$. Finally, we get $r_{m-1}<r_{m-1}+a_{k}=x+\alpha <a_{m-1}$ where the last inequality is a consequence of our choice of k. Thus, $x+\alpha \not \in E$.

Bielas, Plewik and Walczyńska have proved in [8, Thm. 3.3] that for any $q<\frac{1}{2}$ $S(E(aq^{n}))=\left\{ aq^{n}:n\in {\mathbb {N}}\right\} \cup \left\{ 0\right\}$.

We adapt the nice geometric method of the proof of the above result to prove a generalization of it:

Theorem 2.3

If E is a central Cantor set generated by a fast convergent sequence $\left( a_{n}\right)$ such that for any $n\in {\mathbb {N}}$ all gaps of E, starting from $n+2$ order, are shorter then $a_{n}-r_{n}$ (i.e. shorter then gaps of n-order), then

$$\begin{aligned} S\left( E\right) =\left\{ a_{n}:n\in {\mathbb {N}}\right\} \cup \left\{ \emptyset \right\} . \end{aligned}$$

(2)

Proof

Suppose that a positive number $t\!\in \! S\left( E\right)$ does not belong to $\left\{ a_{n}:n\!\in \! {\mathbb {N}}\right\}$. We can assume that $t\le r_{1}$ so there is $n\ge 2$ such that $a_{n}<t\le r_{n-1}$. Hence, $r_{n}+t>r_{n}+a_{n}=r_{n-1}$.

If $r_{n}+t<a_{n-1}$ then $r_{n}+t\notin E$ (because $\left( r_{n-1},a_{n-1}\right)$ is a gap) and $r_{n}-t\notin E$ (because $r_{n}-t<0$ ). It means that $t\notin S(E)$.

If $r_{n}+t\ge a_{n-1}$ we consider a translation

$$\begin{aligned} I:=\left( r_{n-1},a_{n-1}\right) -t=\left( r_{n-1}-t,a_{n-1}-t\right) \subset \left[ 0,r_{n}\right] . \end{aligned}$$

The interval I is longer than any gap in $\left[ 0,r_{n}\right]$, so there is a point $x\in I\cap E$. Then, $x+t$ belongs to the gap $\left( r_{n-1},a_{n-1}\right)$ and $x-t<0$, so $t\notin S\left( E\right)$. $\square$

To formulate the assumptions of the Theorem 2.3 in the language of inequalities between the terms of the series we need the following observation.

For the lenghts $L_{n}$, $L_{n+k}$ of the gaps of the order n and $n+k$, respectively, $L_{n}>L_{n+k}$ is equivalent to

$$\begin{aligned} a_{n}>\left( \sum _{i=n+1}^{n+k}a_{i}+a_{n+k}\right) \end{aligned}$$

Indeed, $L_{n}>L_{n+k}\Leftrightarrow \left( a_{n}-r_{n}\right)>\left( a_{n+k}-r_{n+k}\right) \Leftrightarrow a_{n}>( a_{n+k}+r_{n}-r_{n+k}) \Leftrightarrow a_{n}>\left( \sum _{i=n+1}^{n+k}a_{i}+a_{n+k}\right)$. The equivalences remain true if we change the symbol > to <, $\ge$, $\le$ or $=$.

Hence, the conjuction of two conditions

$$\begin{aligned} \forall _{n}\;a_{n}>a_{n+1}+2a_{n+2} \end{aligned}$$

(3)

and

$$\begin{aligned} \forall _{n}\;a_{n}>a_{n+1}+a_{n+2}+2a_{n+3} \end{aligned}$$

(4)

mean that the gaps of orders $\left( n+2\right)$ and $\left( n+3\right)$ are shorter the ones of order n. By our observation and an obvious recursion we obtain that assumptions of Theorem 2.3 are equivalent to the conjuction of (3) and (4).

Note that none of these conditions implies the other but both of them follow from

The last condition says that gaps of subsequent order are getting shorter. It is more convenient to check than conditions (3) and (4) but there are interesting sequences which do not satify ($\varDelta 2$) but satisfy (3) and (4) - see the Example 2.8 below. We will also show that there are fast convergent series $\left( a_{n}\right)$ such that $S\left( E\left( a_{n}\right) \right)$ are bigger than $\left\{ a_{n}:n\in {\mathbb {N}}\right\} \cup \left\{ \emptyset \right\}$ (Example 2.7), and that the Theorem 2.3 cannot be reversed (Example 2.7).

Of course, the earlier recalled Theorem of Bielas, Plewik and Walczyńska is a straightforward corollary from the Theorem 2.3. In fact, for the geometric series $\sum a_{n}=\sum aq^{n}$ we have

$$\begin{aligned} a_{n}-r_{n}=aq^{n}\frac{1-2q}{1-q} \end{aligned}$$

and the function $f:{\mathbb {N}}\rightarrow {\mathbb {R}}$ given by $f(n)=aq^{n} \frac{1-2q}{1-q}$ is decreasing.

The inclusion

$$\begin{aligned} \left\{ a_{n}\right\} \subset S\left( E\left( a_{n}\right) \right) \text {.} \end{aligned}$$

(5)

is true for any sequence $\left( a_{n}\right)$. It was generalized in [4] in the following way

Lemma 2.4

[4, Lemma 2.5]. If

$$\begin{aligned} a_{n}=a_{n+1}=\cdots =a_{n+2j-2} \end{aligned}$$

then $ja_{n}$ belongs to $S\left( E\left( a_{n}\right) \right)$.

For the reader’s convenience we present the brief proof of it. If we replace the terms $a_{n},a_{n+1},\ldots ,a_{n+j-1}$ in the sequence $\left( a_{n}\right)$ by one term $ja_{n}$, then in the modified sequence we can obtain any number $ma_{k}$ where $m=1,2,\ldots ,n+2j-2$. Consequently, $E\left( a_{n}\right)$ is equal to the achievement set of the modified sequence. Therefore, by (5), we obtain that $ja_{n}\in S\left( E\left( a_{n}\right) \right)$.

The following corollary is a direct consequence of the Lemma 2.4.

Corollary 2.5

If $\left( a_{n},a_{n+1},\ldots ,a_{n+k}\right)$ forms an N -segment with respect to $a_{n+k}$ that is if

$$\begin{aligned} \left\{ \varepsilon _{n}a_{n}+\ldots +\varepsilon _{n+k}a_{n+k}:\varepsilon _{j}\in \left\{ 0,1\right\} \right\} =\left\{ 0,a_{n+k},2a_{n+k},\ldots ,Na_{n+k}\right\} \text {,} \end{aligned}$$

then

$$\begin{aligned} \left\{ ja_{n+k}:j=1,2,\ldots ,\left\lceil \tfrac{N}{2}\right\rceil \right\} \subset S\left( E\left( a_{n}\right) \right) \text {.} \end{aligned}$$

In particular, if

$$\begin{aligned} a_{n}=a_{n+1}=\ldots =a_{n+k}=a \end{aligned}$$

then $\left\{ a,2a,\ldots ,\left\lceil \frac{k}{2}\right\rceil \cdot a\right\} \subset S\left( E\left( a_{n}\right) \right)$.

Example 2.6

The Theorem 2.3 resulted from our failed attempt to prove that the equality (2) holds for every fast convergent series. We know now that the hypothesis was false. Indeed, consider the multigeometric series $\sum (4,\,2,\,1;q)$. This series is fast convergent if $4>3+\frac{7q}{1-q}\$, $2>1+\frac{7q}{1-q}\$and $1>\frac{7q}{1-q}$ i.e. for $q<\frac{1}{8}$. Note that $\left( 4,2,1\right)$ is a 7-segment with respect to 1. Hence, by the Corollary 2.5 we have $3\in S\left( E(4,\,2,\,1;q)\right)$ .

Note, in a similar manner, that every number of the form $3q^{k}$, $k\in {\mathbb {N}}$, belongs to S(E), not being a term of the series. A question raises if $S(E)=\{0\}\cup \{a_{n}:n\in {\mathbb {N}}\}\cup \{3q^{k}:\ k\in {\mathbb {N}}\,\}$ for any $q<\frac{1}{8}$. We will be able to answer it positively later in the paper.

The series from the previous example does not fulfill the principal assumption of the Theorem 2.3, that is,

$$\begin{aligned} \max _{k\ge n+2}(a_{k}-r_{k}) <\ a_{n}-r_{n} \quad \text {for all}\, n . \end{aligned}$$

(6)

More precisely, this inequality fails exactly for any n divisible by 3. In fact, the condition (6) is not necessary for the equality (2).

Example 2.7

Let us consider the fast convergent multigeometric series $\sum (11,\,6,\,3;\,q)$ with q such that $q<\frac{1}{21}$, or equivalently $\sum _{n=1}^{\infty }q^{n}<\frac{1}{20}$. For every $k\in {\mathbb {N}}$, we get

$$\begin{aligned} r_{3k-2}=9q^{k-1}+20\sum _{i=k}^{\infty }q^{i},\quad r_{3k-1}=3q^{k-1}+20\sum _{i=k}^{\infty }q^{i},\quad r_{3k}=20\sum _{i=k}^{\infty }q^{i}. \end{aligned}$$

Further,

$$\begin{aligned} a_{3k-2}-r_{3k-2}= & {} q^{k-1}\left( 2-20\sum _{i=1}^{\infty }q^{i}\right) , \\ a_{3k}-r_{3k}= & {} a_{3k-1}-r_{3k-1}\ =\ q^{k-1}\left( 3-20\sum _{i=1}^{\infty }q^{i}\right) , \end{aligned}$$

for any $k\in {\mathbb {N}}$, and hence $a_{n}-r_{n}<\max _{k\ge n+2}(a_{k}-r_{k})$ for any $n\in 3{\mathbb {N}}-2$ which means that (6) does not hold for the series.

It remains to show that the center of distances of the series does not contain anything but 0 and all terms of the series. Take any $\alpha \in \bigl (F\cap [0,\,11]{\setminus } \bigl (\{0\}\cup \{a_{n}:n\in {\mathbb {N}} \,\}\bigr )$. There exists a finite set $I_{\alpha }\subset {\mathbb {N}} {\setminus } \{1\}$ with $\alpha =\sum _{n\in I_{\alpha }}a_{n}$. Clearly $I_{\alpha }$ has at least two elements. Define $m:=\min I_{\alpha }\ge 2$, $M:=\max I_{\alpha }$ and $J:=\{1,\,2,\,\ldots ,\,m-1\,\}\cup I_{\alpha }$. Then define $x:=\sum _{n\in ({\mathbb {N}}{\setminus } J)\cup \{M\}}a_{n}$. The inequalites $x-\alpha<r_{m}-a_{m}<0$ yield $x-\alpha \not \in E$ easily.

In order to show that $x+\alpha \not \in E$, observe first that

$$\begin{aligned} x+\alpha \ =\ r_{m-1}+a_{M}. \end{aligned}$$

(7)

We have to analyse three cases depending on the form of the remainder $r_{m-1}$:

• (a) $\displaystyle r_{m-1}\ =\ 9q^{k-1}+20\sum _{n=k}^{\infty }q^{n}\ \text {for some}\, k\in {\mathbb {N}}\,(\text {when}\, m=3k-1),$

• (b) $\displaystyle r_{m-1}\ =\ 3q^{k-1}+20\sum _{n=k}^{\infty }q^{n}\ \text {for some}\, k\in {\mathbb {N}}\,(\text {when}\, m=3k),$

• (c) $\displaystyle r_{m-1}\ =\ 20\sum _{n=k}^{\infty }q^{n}\ \text {for some}\, k\in {\mathbb {N}}\,(\text {when}\, m=3k-2).$

In the case (a) $a_{M}=3q^{k-1}$ or $a_{M}\in \bigl \{11q^{k-1+i}, \,6q^{k-1+i},\,3q^{k-1+i}:\ i\in {\mathbb {N}}\,\bigr \}$. If $a_{M}=3q^{k-1}$, then the inequalities

$$\begin{aligned} a_{3k-2}+r_{3k}&=\ 11q^{k-1}+20\sum _{n=k}^{\infty }q^{n}<\ x+\alpha \ \overset{(7)}{=}12q^{k-1}+20\sum _{n=k}^{\infty }q^{n} \\&<\ 11q^{k-1}+3q^{k-1}\ =\ a_{3k-2}+a_{3k} \end{aligned}$$

imply that $x+\alpha \in (a_{3k-2}+r_{3k},\,a_{3k-2}+a_{3k})$, that is, $x+\alpha \not \in E$. If $a_{M}$ is of the form $11q^{k-1+i}$ or $6q^{k-1+i}$ , or $3q^{k-1+i}$ for some $i\in {\mathbb {N}}$, then

$$\begin{aligned} r_{3k-2}=9q^{k-1}+20\sum _{n=k}^{\infty }q^{n}\ \overset{(7)}{<}\ x+\alpha \ \overset{(7)}{<}\ r_{m-1}+q^{k-1}<11q^{k-1}=a_{3k-2} \end{aligned}$$

and hence $x+\alpha \in (r_{3k-2},\,a_{3k-2})$ which implies that $x+\alpha \not \in E$.

In the case (b) $a_{M}=11q^{k-1+i}$ or $6q^{k-1+i}$, or $3q^{k-1+i}$ for some $i\in {\mathbb {N}}$. Thus

$$\begin{aligned} r_{3k-1}< & {} 3q^{k-1}+20\sum _{n=k}^{\infty }q^{n}+a_{M} = x+\alpha< 3q^{k-1}+q^{k-1}+q^{k-1}\\ {}< & {} 6q^{k-1}=a_{3k-1} \end{aligned}$$

and hence $x+\alpha \not \in E$.

In the case (c) we get $a_{M}=6q^{k-1+i}$ or $3q^{k-1+i}$ for some $i\in {\mathbb {N}}$ or $a_{M}=11q^{k-1+i}$ for some $i\ge 2$. Thus

$$\begin{aligned} r_{3k}<\ 20\sum _{n=k}^{\infty }q^{n}+a_{M}<\ q^{k-1}+q^{k-1} <\ 3q^{k-1}\ =\ a_{3k} \end{aligned}$$

and hence $x+\alpha \not \in E$.

We have proved that $\alpha \not \in S(E)$ and it must be $S(E)=\{0\}\cup \{a_{n}:\ n\in {\mathbb {N}}\,\}$ by the Lemma 2.1.

Example 2.8

The multigeometric series $\sum (2,\,1;q)$ is fast convergent for $q<\frac{1}{4}$. It is easily seen that it does not fulfill the condition ($\varDelta 2$) . However, it satisfies conditions (3) and (4) and hence the center of distances is minimal, i.e., contains only zero and the terms of series.

The problem of characterizing those fast convergent series for which (2) holds remains open. A complete characterization of (2) will be even more difficult, because there are series that satisfy (2) and are not fast convergent ([8, Thm. 5.2]).

3 Central Cantor Sets with Large Centers of Distances

Recall that a set E is a central Cantor set if and only if it is an achievement set of a fast convergent sequence $\left( a_{n}\right)$. Moreover, a correspondence between fast convergent sequences and central Cantor sets is one to one. However, there are sequences which are not fast convergent but their achievement sets are Cantor sets (i.e. homeomorphic to the classic ternary Cantor set). Some of them even generate central Cantor sets ([6]) .

A monotonic sequence $\left( a_{n}\right)$ with positive terms converging to 0 (or a series $\sum a_{n}$) is called semi-fast convergent if for all $n\in {\mathbb {N}}$ we have

$$\begin{aligned} a_{n}>\sum _{a_{k}<a_{n}}a_{k}\text {.} \end{aligned}$$

A strictly decreasing sequence $\left( a_{n}\right)$ is semi-fast convergent if and only if it is fast convergent.

Let $\sum a_{n}$ be a semi-fast convergent series. Then there exist two uniquely determined sequences, $\left( \alpha _{k}\right)$ of positive numbers decreasing to zero and $\left( N_{k}\right)$ of positive integers such that

$$\begin{aligned} a_{i}=\alpha _{k}\ \text { for }\sum _{j=0}^{k-1}N_{j}<i\le \sum _{j=0}^{k}N_{j} \end{aligned}$$

(where $N_{0}=0$). The numbers $\alpha _{k}$ are the values of the terms of the series $\sum a_{n}$ and $N_{k}$ is the multiplicity of the value $\alpha _{k}$ in the series. Thus, we can identify

$$\begin{aligned} \sum a_{n}=\sum \left( \alpha _{k},N_{k}\right) \qquad \text {or}\qquad (a_{n})=(\alpha _{k},\,N_{k}) \end{aligned}$$

and the sum of the series is $\sum a_{n}=\sum \left( \alpha _{k},N_{k}\right)$=$\sum _{k=1}^{\infty }\alpha _{k}N_{k}$.

It was proved in [6] that for any semi-fast convergent series $\left( a_{n}\right)$ its achievement set $E\left( a_{n}\right)$ is a Cantor set ([6, Thm.16]). Moreover, $E\left( \alpha _{k},N_{k}\right)$ is a central Cator set if and only if for any k the number $N_{k}$ is equal to $2^{m_{k}}-1$ for some $m_{k}\in {\mathbb {N}}$ ([6, Thm.19]). If this condition holds, then the fast convergent sequence

$$\begin{aligned} \left( 2^{m_{1}-1}\alpha _{1},2^{m_{1}-2}\alpha _{1},\ldots ,\alpha _{1},2^{m_{2}-1}\alpha _{2},\ldots ,\alpha _{2},\ldots \right) \end{aligned}$$

and the semi-fast convergent sequence $\left( \alpha _{k},N_{k}\right)$ have the same achievement sets.

It follows from Corollary 2.5 that the center of distances of a semi-fast convergent sequence $\left( \alpha _{k},N_{k}\right)$ contains the values $\alpha _{k},2\alpha _{k},$...,$\left\lceil \frac{N_{k}}{2} \right\rceil \alpha _{k}$ for all $k\in {\mathbb {N}}$. So, if we want to exhibit a family of achievement sets being Cantor sets but possesing centra of distances larger than the set of values of the terms of the generating sequence with zero attached, it is reasonable to consider semi-fast convergent sequences and fast convergent sequences with the same achievement sets. From the previous considerations and Lemma 2.2 it follows that if $E=E(\alpha _{i},\,N_{i})$ is a central Cantor set, then $S(E)\subset F(\alpha _{i},\,N_{i})$. Indeed, by Lemma 2.2$S(E)\subset F\left( a_{n}\right)$ where $\left( a_{n}\right)$ is a fast convergent series for which $E=E\left( a_{n}\right)$, and each $a_{n}$ is a finite sum of $\alpha _{i}$. In the next Lemma we show a bit more. We hope that it could be a step in the study on centers of distances for achievable Cantor sets, not necessarily central.

Lemma 3.1

If $\sum (\alpha _{i},\,N_{i})$ is a semi-fast convergent series, then $S(E)\subset F$.

Proof

Observe that $S(E)\subset [0,\,{\hat{\sigma }}]$ where ${\hat{\sigma }} \!:=\!\inf \left( E\cap (\frac{\sigma }{2},\,\sigma ]\right) <\sigma \!:=\!\sum (\alpha _{i},\,N_{i})$ by the Lemma 2.1. Moreover,

$$\begin{aligned} {\hat{\sigma }}\ \le \ \left\lceil \frac{N_{1}+1}{2}\right\rceil \alpha _{1}. \end{aligned}$$

(8)

Take any $y\in (E{\setminus } F)\cap [0,\,{\hat{\sigma }}]$, We are going to show that $y\not \in S(E)$. There is a unique sequence $(m_{i}^{y})_{i\in {\mathbb {N}}}$ of nonnegative integers such that $0\le m_{i}^{y}\le N_{i}$ for all i and $y=\sum _{i}\alpha _{i}m_{i}^{y}$ (see the first part of the proof of Lemma 14 from [6]). By our assumption on y, $m_{i}^{y}>0$ for infinitely many i’s. Denote $k:=\min \{i:\ m_{i}^{y}>0\,\}$. In particular,

$$\begin{aligned} y>\alpha _{k}. \end{aligned}$$

(9)

The case $m_{k}^{y}=N_{k}$. Then $k>1$, because $m_{k}^{y}\alpha _{k}<y\le {\hat{\sigma }}$. Indeed, in the case $m_{1}^{y}=N_{1}$, these inequalities imply $N_{1}<\left\lceil \frac{N_{1}+1}{2}\right\rceil$ which is false for all positive integers $N_{1}$. Define

$$\begin{aligned} p\ :=\ \min \left\{ i>k:\ m_{i}^{y}>0\ \text {and}\ \alpha _{i} <\ \alpha _{k-1}-\sum _{j=k}^{\infty }\alpha _{j}N_{j}\,\right\} \end{aligned}$$

(10)

and

$$\begin{aligned} \mu _{i}\ :=\ {\left\{ \begin{array}{ll} 0\qquad \qquad &{} \text {if}\, i<k; \\ N_{p}-m_{p}^{y}+1 &{} \text {if}\, i=p; \\ N_{i}-m_{i}^{y} &{} \text {if}\, i\ge k, i\ne p. \end{array}\right. } \end{aligned}$$

Clearly, $\mu _{k}=0$ and $\mu _{p}\le N_{p}$, because $m_{p}^{y}\ge 1$. Thus, $x:=\sum _{i}\alpha _{i}\mu _{i}$ is an element of E. Further,

$$\begin{aligned} x-y\ \overset{(9)}{<}\ \sum _{j=k+1}^{\infty }\alpha _{j}N_{j}\,-\,\alpha _{k} <\ 0 \end{aligned}$$

Hence,

$$\begin{aligned} x-y\ \not \in \ E. \end{aligned}$$

(11)

Moreover,

$$\begin{aligned} \sum _{j=k}^{\infty }\alpha _{j}N_{j}<\ x+y\ =\ \sum _{j=k}^{\infty }\alpha _{j}N_{j}\ +\ \alpha _{p}\ \overset{(10)}{<}\ \alpha _{k-1}. \end{aligned}$$

Hence $x+y\not \in E$ which together with (11) implies $y\not \in S(E)$.

The case $m_{k}^{y}<N_{k}$. Define

$$\begin{aligned} p\ :=\ \min \left\{ i>k:\ m_{i}^{y}>0\ \text {and}\ \alpha _{i} <\ \alpha _{k}-\sum _{j=k+1}^{\infty }\alpha _{j}N_{j}\,\right\} \end{aligned}$$

(12)

and

$$\begin{aligned} \mu _{i}\ :=\ {\left\{ \begin{array}{ll} 0\qquad \qquad &{} \text {if}\, i\le k; \\ N_{p}-m_{p}^{y}+1 &{} \text {if}\, i=p; \\ N_{i}-m_{i}^{y} &{} \text {if}\, i>k, i\ne p. \end{array}\right. } \end{aligned}$$

We put $x:=\sum _{i}\alpha _{i}\mu _{i}$. Using the same argument as before one can show that $x-y\ \not \in \ E$. The fact that $x+y\not \in E$ follows from

$$\begin{aligned} m_{k}^{y}\alpha _{k} + \sum _{j=k+1}^{\infty }\alpha _{j}N_{j}< x+y = m_{k}^{y}\alpha _{k} + \sum _{j=k+1}^{\infty }\alpha _{j}N_{j} + \alpha _{p} \overset{(12)}{<} (m_{k}^{y}+1)\alpha _{k}. \end{aligned}$$

And again $y\not \in S(E)$. $\square$

Lemma 3.2

Suppose that $\left( a_{n}\right)$ is a fast convergent sequence, $E=E\left( a_{n}\right)$, $k,m\in {\mathbb {N}},$ and $k>m$. If

$$\begin{aligned} \left( a_{k-\left( m-1\right) },\ldots ,a_{k-1},a_{k}\right) =\left( 2^{m-1}a_{k},\ldots ,2a_{k},a_{k}\right) \end{aligned}$$

then $ja_{k}\in S\left( E\right)$ for any $j=1,\ldots ,2^{m-1}$. Equivalently

$$\begin{aligned} \left\{ \varepsilon _{k}a_{k}+\ldots +\varepsilon _{k-\left( m-1\right) }a_{k-\left( m-1\right) }:\varepsilon _{i}\in \left\{ 0,1\right\} \right\} \subset S\left( E\right) \text {.} \end{aligned}$$

Proof

Note that $\left( 2^{m-1},\ldots ,2^{0}\right)$ forms $\left( 2^{m}-1\right)$-segment. Hence, by Lemma 2.4 and Corollary 2.5 we obtain the conclusion. $\square$

The next theorem describes the typical situation when the sum of more than one terms of a fast convergent series $\sum a_{n}$ belongs to the center of distances of the central Cantor set E which is an achievement set generated by this sequence. We suppose that the set $S\left( F_{n_{2}}\right)$ in (i) and (ii) can be replaced equivalently by S(E) but we believe that our, a bit weaker theorem is also interesting.

Theorem 3.3

Let $\sum a_n$ be a fast convergent series and let $n_1,\,n_2$ be positive integers sastisfying the condition $2\le n_1<n_2$. Then the following conditions are equivalent:

• (i) $\displaystyle \left\{ \,\sum _{i=n_{1}}^{n_{2}}\varepsilon _{i}a_{i}:\ \varepsilon _{i}\in \{0,\,1\}\,\right\} \ \subset \ S(F_{n_{2}})$ ;

• (ii) $\displaystyle \sum _{i=n_{1}}^{n_{2}}a_{i}\in S(F_{n_{2}})$;

• (iii) $\displaystyle a_{n_1-1}\,=\,\sum _{i=n_1}^la_i\,+\,a_l$ for $n_1\le l\le n_2$;

• (iii’) the gaps of orders $n_{1}-1,n_{1},\ldots ,n_{2}-1$ are of the same length;

• (iv) $\displaystyle a_i=2^{n_2-i}a_{n_2}$ for $n_1-1\le i\le n_2$.

Proof

(i) $\Rightarrow$ (ii). It is obvious.

(ii) $\Rightarrow$ (iii). If we assume that $\sum _{i=n_{1}}^{n_{2}}a_{i}\in S(F_{n_{2}})$, then

$$\begin{aligned} \sum _{i=n_{1}}^{n_{2}}a_{i}+a_{n_{2}}\in F_{n_{2}}. \end{aligned}$$

It is easy to see that the following inequalities hold

$$\begin{aligned} \sum _{i=n_{1}}^{n_{2}}a_{i}<r_{n_{1}-1}+a_{n_{2}}<a_{n_{1}-1}+a_{n_{2}}. \end{aligned}$$

Observing that $F_{n_{2}}\cap \left( \sum _{i=n_{1}}^{n_{2}}a_{i},a_{n_{1}-1}+a_{n_{2}}\right) =\{a_{n_{1}-1}\}$, we can conclude

$$\begin{aligned} \sum _{i=n_{1}}^{n_{2}}a_{i}+a_{n_{2}}=a_{n_{1}-1}. \end{aligned}$$

(13)

Further, since $\sum _{i=n_{1}}^{n_{2}}a_{i}\in S(F_{n_{2}})$ we have

$$\begin{aligned} \sum _{i=n_{1}}^{n_{2}}a_{i}+a_{l}\in F_{n_{2}}\text { for }n_{1}\le l\le n_{2}-1. \end{aligned}$$

Moreover for $l\in \{n_{1},\ldots ,n_{2}-1\}$

$$\begin{aligned} a_{n_{1}-1}+\sum _{i=l+1}^{n_{2}-1}a_{i}\overset{(13)}{=} \sum _{i=n_{1}}^{n_{2}}a_{i}+ \sum _{i=l+1}^{n_{2}}a_{i}<r_{n_{1}-1}+r_{l}<a_{n_{1}-1}+a_{l} \end{aligned}$$

with the convention $\sum _{i=l+1}^{l}a_{i}=0$.

Then since

$$\begin{aligned} F_{n_{2}}\cap \left( a_{n_{1}-1}+\sum _{i=l+1}^{n_{2}-1}a_{i},a_{n_{1}-1}+a_{l}\right) =\left\{ a_{n_{1}-1}+\sum _{i=l+1}^{n_{2}}a_{i}\right\} \end{aligned}$$

we have

$$\begin{aligned} \sum _{i=n_{1}}^{n_{2}}a_{i}+a_{l}=a_{n_{1}-1}+\sum _{i=l+1}^{n_{2}}a_{i} \end{aligned}$$

and therefore

$$\begin{aligned} \sum _{i=n_{1}}^{l}a_{i}+a_{l}=a_{n_{1}-1}. \end{aligned}$$

(iii)$\Longleftrightarrow$(iii’) This equivalence follows directly from the observation of the lengths of the gaps after Theorem 2.3 (using the symbol $=$ in place of >).

(iii) $\Rightarrow$ (iv). It follows immediately from that $a_{i}=2^{n_{2}-i}a_{n_{2}}$, $n_{1}\le i\le n_{2}$, is the solution of the system of equations

$$\begin{aligned} \left\{ a_{n_{1}-1}=\sum _{i=n_{1}}^{l}a_{i}+a_{l},\text { where }n_{1}\le l\le n_{2}\right. . \end{aligned}$$

(iv) $\Rightarrow$ (i). Let $t:=\sum _{i=n_{1}}^{n_{2}}\varepsilon _{i}a_{i}$ for some$\ \varepsilon _{i}\in \{0,\,1\}$ and $x\in F_{n_{2}}$. Then

$$\begin{aligned} x=g+\varepsilon a_{n_{1}-1}+f \end{aligned}$$

where $g\in F_{n_{1}-2}$, $\varepsilon =0$ or $\varepsilon =1$ and $f\in E\left( a_{n_{1}},\ldots ,a_{n_{2}}\right)$. If $\varepsilon =0$ then $x+t\in F_{n_{2}}$ and if $\varepsilon =1$ then $x-t\in F_{n_{2}}$. $\square$

Theorem 3.4

Let $\sum a_{n}$ be a fast convergent series and let k and $n_{1}<n_{2}<\cdots <n_{k}$ be positive integers greater or equal to 2. If $\sum _{i=1}^{k}a_{n_{i}}\,\in \,S(E)$, then $\sum _{i=n_{1}}^{n_{k}}a_{i}\,+ \,a_{n_{k}}\,=\,a_{n_{1}-1}$.

Proof

If we assume that $\sum _{i=1}^{k}a_{n_{i}}\in S(E)$, then

$$\begin{aligned}&\sum _{i=1}^{k}a_{n_{i}}+\sum _{i\in \{n_{1}+1,\ldots ,n_{k}\}\backslash \{n_{2},n_{3},\ldots ,n_{k-1}\}}a_{i}=\sum _{i=n_{1}}^{n_{k}}a_{i}+a_{n_{k}}\in E \end{aligned}$$

(14)

$$\begin{aligned}&\text { and } \nonumber \\&\sum _{i=1}^{k}a_{n_{i}}+\sum _{i\in \{n_{1}+1,\ldots ,n_{k}\}\backslash \{n_{2},n_{3},\ldots ,n_{k-1}\}}a_{i}+r_{n_{k}}=r_{n_{1}-1}+a_{n_{k}}\in E. \end{aligned}$$

(15)

Suppose $\sum _{i=n_{1}}^{n_{k}}a_{i}+a_{n_{k}}<a_{n_{1}-1}$. Then since $r_{n_{k}}<a_{n_{k}}$, the element $\sum _{i=n_{1}}^{n_{k}}a_{i}+a_{n_{k}}$ belongs to a gap $\left( r_{n_{1}-1},a_{n_{1}-1}\right)$ and therefore it does not belong to E, contrary to (14).

Now suppose that $\sum _{i=n_{1}}^{n_{k}}a_{i}+a_{n_{k}}>a_{n_{1}-1}$. Then

$$\begin{aligned} a_{n_{1}-1}+r_{n_{k}}<r_{n_{1}-1}+a_{n_{k}}<a_{n_{1}-1}+a_{n_{k}}. \end{aligned}$$

Hence $r_{n_{1}-1}+a_{n_{k}}\not \in E$. This leads to a contradiction with (15).

Therefore

$$\begin{aligned} \sum _{i=n_{1}}^{n_{k}}a_{i}+a_{n_{k}}=a_{n_{1}-1}. \end{aligned}$$

$\square$

Since $a_{n_{1}-1}-r_{n_{1}-1}=a_{n_{k}}-r_{n_{k}}$ is equivalent to $a_{n_{1}-1}=\sum _{i=n_{1}}^{n_{k}}a_{i}+a_{n_{k}}$, the above theorem can be equivalently formulated in the following way: If $a_{n_{1}}+\ldots +a_{n_{k}}\in S\left( E\right)$ then the $\left( n_{1}-1\right)$-gap of E has the same length as the $\left( n_{k}\right)$-gap. As a conclusion we get a useful property, that extends our Theorem 2.3.

Corollary 3.5

Let $\sum a_{n}$ be a fast convergent series such that the gaps of not neighboring orders of its sets of subsums $E=E(a_{n})$ are of different length. Then

$$\begin{aligned} S\left( E\right) =\left\{ a_{n}:n\in {\mathbb {N}}\right\} \cup \left\{ \emptyset \right\} . \end{aligned}$$

(16)

Now let us return to Example 2.7 and multigeometric sequences of the form $\left( 11,6,3;q\right)$. It is not difficult to check that a sequence $\left( 11,6,3;q\right)$ is fast convergent for $q<\frac{1}{11}$, so $\sum q^{n}<\frac{1}{10}$. On the other hand, such sequence fullfiles the assumption on the above corollary. That is why the tedious computations in the Example 2.7 were successful. Moreover, comparing the conditions for q we can see that the result obtained from Theorem 3.4 is stronger then the one obtained in Example 2.7.

The next theorem describes the centers of distances of the central Cantor sets which can be obtained as achievement set of semi-fast convergent series.

Theorem 3.6

Let $E=E(\alpha _{i},\,N_{i})$ be a central Cantor set given by a semi-fast convergent series $\sum (\alpha _{i},\,N_{i})$ satisfying $\alpha _{i}>(N_{i+1}+1)\alpha _{i+1}$ for all $i\in {\mathbb {N}}$. Then

$$\begin{aligned} S(E)\ =\ \left\{ 0\right\} \cup \,\left\{ k\alpha _{i}:\ i\in {\mathbb {N}} ,\,k\in \{1,\,2,\,\ldots ,\,\tfrac{N_{i}+1}{2}\,\}\,\right\} . \end{aligned}$$

Proof

Since $E(\alpha _{i},\,N_{i})$ is a central Cantor set, each $N_{i}$ is of the form $N_{i}=2^{m_{i}+1}-1$ for some $m_{i}\in {\mathbb {N}}_{0}$ by the Thm. 19 of [6] . Define $M_{i}\,:=\,\sum _{j=1}^{i}(m_{j}+1)$, $M_{0}:=0$ and

$$\begin{aligned} a_{n}\ =\ 2^{M_{i}-n}\alpha _{i}\qquad \text {for}\, i\in {\mathbb {N}}, M_{i-1}<n\le M_{i}. \end{aligned}$$

We obtain a fast convergent series $\sum a_{n}$ such that $E(a_{n})=E(\alpha _{i},\,N_{i})$. Now since the equalities (iii) of Theorem 3.3 hold for the terms $2^{m_{i}-1}\alpha _{i},\,2^{m_{i}-2}\alpha _{i},\ldots ,\,2^{1}\alpha _{i},\,2^{0}\alpha _{i}$, we can use the statement (i) of the theorem and conclude that $k\alpha _{i}\in S(E)$ for all $i\in {\mathbb {N}}$ and $k=0,\,1,\,\ldots ,\,2^{m_{i}}$. Moreover all terms $a_{n}$ belong to S(E) and thus $2^{m_{i}}\alpha _{i}$ are in S(E) for all i.

We are now going to exclude from S(E) all other elements of F which will conclude the proof by the Lemma 3.1. Let $\sum _{i=1}^{m}a_{n_{i}}$, $m\ge 2$, $n_{1}<\ldots <n_{m}$, be any element of S(E) that is not of the form $k\alpha _{i}$ for some $i\in {\mathbb {N}}$ and $k\in \{0,\ldots ,2^{m_{i}}\,\}$. Clearly, $n_{1}>1$, because $S(E)\subset [0,\,a_{1}]$ and $m\ge 2$. Therefore, there exists $i\in {\mathbb {N}}$ such that $n_{1}-1\le M_{i}<n_{m}$. Let $r<p$ be positive integers such that $n_{1}-1\in [M_{r-1}+1,\,M_{r}]$ and $n_{m}\in [M_{p-1}+1,M_{p}]$. Then

$$\begin{aligned} a_{n_{m}}\ -\ \sum _{j=n_{m}+1}^{\infty }a_{j}\ =\ \alpha _{p}\ -\ \sum _{j=p+1}^{\infty }\alpha _{j}N_{j} \end{aligned}$$

(17)

and

$$\begin{aligned} a_{n_{1}-1}\ -\ \sum _{j=n_{1}}^{\infty }a_{j}\ =\ \alpha _{r}\ -\ \sum _{j=r+1}^{\infty }\alpha _{j}N_{j}. \end{aligned}$$

(18)

Now, the assumption $\alpha _{i}>(N_{i+1}+1)\alpha _{i+1}$ implies that

$$\begin{aligned} \alpha _{i}\ -\ \sum _{j=i+1}^{\infty }\alpha _{j}N_{j} >\ \alpha _{i+1}\ -\ \sum _{j=i+2}^{\infty }\alpha _{j}N_{j}\qquad \text {for all}\, i \end{aligned}$$

and hence by induction

$$\begin{aligned} \alpha _{r}\ -\ \sum _{j=r+1}^{\infty }\alpha _{j}N_{j} >\ \alpha _{p}\ -\ \sum _{j=p+1}^{\infty }\alpha _{j}N_{j} \end{aligned}$$

which by (17) and (18) means that

$$\begin{aligned} a_{n_{1}-1} >\ \sum _{i=n_{1}}^{n_{m}}a_{i}\ +\ a_{n_{m}}. \end{aligned}$$

Thus, by the Theorem 3.4, $\sum _{i=1}^{m}a_{n_{i}}\not \in S(E)$. $\square$

Let us observe that if $\sum \left( \alpha _{k},N_{k}\right)$ is semi-fast convergent, and for some i we have

$$\begin{aligned} \alpha _{i}=\left( N_{i+1}+1\right) \alpha _{i+1}\text {,} \end{aligned}$$

then we can replace two segments $\left( \alpha _{i},N_{i}\right)$ and $\left( \alpha _{i+1},N_{i+1}\right)$ by the one segment $\left( \alpha _{i+1},N_{i}\left( N_{i+1}+1\right) +N_{i+1}\right)$.

Example 3.7

Let $\alpha _{i}=2\alpha _{i+1}$, $N_{i}=3$ and $N_{i+1}=1$. Then, instead of two segments

$$\begin{aligned} \left( \alpha _{i},\alpha _{i},\alpha _{i}\right) \left( \alpha _{i+1}\right) \end{aligned}$$

we can equivalently use one segment

$$\begin{aligned} \left( \underset{\text {6 times}}{\underbrace{\alpha _{i+1},\ldots ,\alpha _{i+1}}},\alpha _{i+1}\right) \text {.} \end{aligned}$$

Hence $3\alpha _{i+1}\in S(E)$. This shows that we cannot take the weak inequality in the assumption of Theorem 3.6 instead of the strong one.

If for all i we have $\alpha _{i}\ne \left( N_{i+1}+1\right) \alpha _{i+1}$ then we call the semi-fast convergent series $\sum \left( \alpha _{i},N_{i}\right)$ irreducible.

Example 3.8

Let $a_{1}=9$, $a_{2}=5,a_{3}=2$, $a_{4}=1$ and $a_{n}=\frac{1}{3^{n}}$ for $n>4$. Then $a_{1}=9<10=2a_{2}$ and it is easy to check that $6\in S\left( F_{4}\right) \subset S\left( E\right)$. Hence the assumption $\alpha _{i}>(N_{i+1}+1)\alpha _{i+1}$ of Theorem 3.6 cannot be replaced by $\alpha _{i}\ne (N_{i+1}+1)\alpha _{i+1}$.

Example 3.9

Let $k\in {\mathbb {N}}$ and a series $\sum a_{n}$ satisfies $\frac{ a_{n+1}}{a_{n}}<\frac{1}{2^{k+1}}$ for all n. Denote $E:=E(a_{n})$. Then $S_{2^{k+1}-1}(E)$ (that is, the algebraic sum of $2^{k+1}-1$ copies of E) is a central Cantor set with the center of distances

$$\begin{aligned} S\bigl (S_{2^{k+1}-1}(E)\bigr )\ =\ \{0\}\,\cup \,\bigl \{ma_{n}:\ n\in {\mathbb {N}},\,m\in \{1,\,2,\,\ldots ,\,2^{k}\,\}\,\bigr \}. \end{aligned}$$

Indeed, the assumption $\frac{a_{n+1}}{a_{n}}<\frac{1}{2^{k+1}}$ for all n implies that $a_{n+i}<\frac{a_{n}}{2^{(k+1)i}}$ for all $n\,i\in {\mathbb {N}}$ . Hence $r_{n}<\frac{a_{n}}{2^{k+1}-1}$ and thus

$$\begin{aligned} \frac{a_{n}}{r_{n}}\ +\ 1 >\ \ 2^{k+1}\qquad \text {for all}\, n\in {\mathbb {N}}. \end{aligned}$$

(19)

Denoting $\lambda _{n}:=\frac{r_{n}}{r_{n-1}}$, we see that

$$\begin{aligned} \lambda _{n}\ =\ \frac{1}{1+\frac{a_{n}}{r_{n}}}\ \overset{(19)}{<}\ \frac{1}{2^{k+1}}. \end{aligned}$$

Since $E=C_{(\lambda _{n})}$ [6, p. 1525], the Corollary 20 of [6] implies that $S_{2^{k+1}-1}(E)$ is a central Cantor set generated by the semi-fast convergent series $\sum (a_{n},\,2^{k+1}-1)$ with the property $a_{n}>2^{k+1}a_{n+1}$ for all n. Thus, all assumptions of the Theorem 3.6 are fulfiled by the set $S_{2^{k+1}-1}(E)$ and a direct application of the theorem yields the desired conclusion.

Example 3.10

Let k be a positive integer and let $q\in (0,\,1)$. Define

$$\begin{aligned} x_{i}\ =\ 2^{j}q^{n} \end{aligned}$$

for $i=(k+1)n-j$ where $(n,\,j)\in {\mathbb {N}}\times \{0,\,1,\,\ldots ,\,k\,\}$. If $q<\frac{1}{2^{k+1}}$ then the series $\sum x_{i}$ is fast convergent. Observe that $E(x_{n})=S_{2^{k+1}-1}(C_{q})$ where $C_{q}=E(q^{n})$. A direct application of the Example 3.9 to the series $\sum a_{n}$ with $a_{n}=q^{n}$ yields

$$\begin{aligned} S\bigl (E(x_{i})\bigr )\ =\ \{0\}\,\cup \,\bigl \{mq^{n}:\ n\in {\mathbb {N}} ,\,m\in \{1,\,2,\,\ldots ,\,2^{k}\,\}\,\bigr \}. \end{aligned}$$

Example 3.11

Let $\sum x_{i}$ be the multigeometric series $\sum (4,\,2,\,1;\,q)$ with $q< \frac{1}{8}$ from our Example 2.6. Then $\sum x_{i}$ is fast convergent and - by the previous example - the center of distances of its set of subsums is

$$\begin{aligned} \{0\}\ \cup \ \bigcup _{n\in {\mathbb {N}}}\{q^{n},\,2q^{n},\,3q^{n},\,4q^{n}\}. \end{aligned}$$

It would be interesting to find a characterization of all central Cantors with minimal centers of distances. Other possible direction of research could be to investigate the relationshops between centers of distances and the uniqueness of purely atomic probability measures with prescribed range. Powerful results, recently published by Marchwicki and Miska in [16] demonstrate that it is impossible to precisely describe the topological type of achievement sets using only the Kakeya conditions: $a_{n}>r_{n}$ and $a_{n}\le r_{n}$.