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European Journal of Mathematics

, Volume 4, Issue 2, pp 687–698 | Cite as

On the center of distances

  • Wojciech BielasEmail author
  • Szymon Plewik
  • Marta Walczyńska
Open Access
Research Article

Abstract

We introduce the notion of a center of distances of a metric space and use it in a generalization of the theorem by John von Neumann on permutations of two sequences with the same set of cluster points in a compact metric space. This notion is also used to study sets of subsums of some sequences of positive reals, as well for some impossibility proofs. We compute the center of distances of the Cantorval, which is the set of subsums of the sequence \(\frac{3}{4}, \frac{1}{2}, \frac{3}{16}, \frac{1}{8}, \ldots , \frac{3}{4^n}, \frac{2}{4^n}, \ldots \), and for other related subsets of the reals.

Keywords

Cantorval Center of distances von Neumann’s theorem Set of subsums Digital representation 

Mathematics Subject Classification

40A05 11B05 28A75 

1 Introduction

The center of distances seems to be an elementary and natural notion which, as far as we know, has not been studied in the literature. It is an intuitive and natural concept which allows us to prove a generalization of von Neumann’s theorem on permutations of two sequences with the same set of cluster points in a compact metric space, see Theorem 2.1. We have realized that the computation of centers of distances—even for well-known metric spaces—is not an easy task because it requires skillful use of fractions. We have only found a few algorithms which enable us to compute centers of distances, see Proposition 3.2 and Lemma 5.1.

We present the use of this notion for impossibility proofs, i.e., to show that a given set cannot be the set of subsums, for example see Corollary 5.5. We refer the readers to the paper [14] by Nitecki, as it provides a good introduction to facts about the set of subsums of a given sequence. It is also worth to look into the papers [1, 2, 3, 8] as well as others cited therein.

In several papers, the set of all subsums of the sequence \(\frac{3}{4}, \frac{1}{2}, \frac{3}{16}, \frac{1}{8}, \ldots , \frac{3}{4^n}\), \(\frac{2}{4^n}, \ldots \), i.e., the set \(\mathbb {X}\) consisting of all sumswhere A and B are arbitrary subsets of positive natural numbers, is considered. Guthrie and Nymann, see [5] and cf. [15] and [14, p. 865], have shown that Open image in new window But, as it can be seen in Corollary 4.2, we get that For these reasons, we have an impression that the arithmetical properties of \( \mathbb {X}\) are not known well and described in the literature. Results concerning some properties of \(\mathbb {X}\) are discussed in Propositions 4.14.3 and 4.4; Corollary 4.5; Theorems 5.25.35.46.1 and 6.2; and they are also presented in Figs. 12 and 3.

2 A generalization of von Neumann’s theorem

Given a metric space X with the distance d. Suppose that sequences \(( x_n)_{n\in \omega } \) and \((y_n)_{ n\in \omega } \) in X have the same set of cluster points C. For them, von Neumann [13] proved that there exists a permutation \(\pi :\omega \rightarrow \omega \) such that Open image in new window Proofs of the above statement can be found in [6, 18]. However, we would like to present a slight generalization of this result. To prove it we use the so-called “back-and-forth” method, which was developed in [7, pp. 35–36] and is still used successfully by many mathematicians, for example cf. [4, 16] or [17], etc. It is also worth mentioning the modern development of classical works of Fraïssé by Kubiś [11].

Consider the setwhich will be called the center of distances of X.

Theorem 2.1

Suppose that sequences Open image in new window and Open image in new window in X have the same set of cluster points \( C \subseteq X\), where (Xd) is a compact metric space. If \(\alpha \in S(C)\), then there exists a permutation \(\pi :\omega \rightarrow \omega \) such that Open image in new window

Proof

Given \(\alpha \in S(C)\), we shall renumber Open image in new window by establishing a permutation \(\pi :\omega \rightarrow \omega \) such thatPut \(\pi (0)= 0\) and assume that values Open image in new window and inverse values Open image in new window are already defined. We proceed step by step as follows.

If Open image in new window is not defined, then take points \( x_m, y_m \in C\) such that Open image in new window and \(d(x_m, y_m)=\alpha \). Choose Open image in new window to be the first element of Open image in new window not already used such that Open image in new window

If Open image in new window is not defined, then take points \( p_m, q_m \in C\) such that Open image in new window and \(d(p_m, q_m)=\alpha \). Choose Open image in new window to be the first element of Open image in new window not already used such that Open image in new window

The set \(C\subseteq X\), as a closed subset of a compact metric space, is compact. Hence the required points \(x_m,y_m,p_m\) and \(q_m\) always exist and alsoIt follows that Open image in new window \(\square \)

Let us note that von Neumann’s theorem mentioned above is applicable for some other problems, for example cf. [9] or [10], etc. As we have seen, the notion of a center of distances appears in a natural way in the context of metric spaces. Though the computation of centers of distances is not an easy task, it can be done for important examples giving further information about these objects.

3 On the center of distances and the set of subsums

Given a metric space X, observe that \(0\in S(X)\) and also, if Open image in new window and \(0\in X\), then \(S(X) \subseteq X\).

If Open image in new window is a sequence of reals, then the setis called the set of subsums of Open image in new window In this case, we have \(d(x,y)=|x-y|\). If X is a subset of the reals, then any maximal interval Open image in new window disjoint from X is called an X-gap. Additionally, when X is a closed set, then any maximal interval Open image in new window included in X is called an X-interval.

Proposition 3.1

If X is the set of subsums of a sequence Open image in new window , then \(a_n\in S(X)\), for all \(n\in \omega \).

Proof

Suppose Open image in new window If \(n\in A\), then \(x-a_n\in X\) and \(d(x, x-a_n)= a_n\). When \(n\notin A\), then \(x+a_n\in X\) and \(d(x, x+a_n)= a_n\). \(\square \)

In some cases, the center of distances of the set of subsums of a given sequence can be determined. For example, the unit interval is the set of subsums of the sequence Open image in new window So, the center of distances of the subsums of Open image in new window is equal to Open image in new window

Proposition 3.2

Assume that Open image in new window , for a number \(\lambda >0\) and a set Open image in new window If Open image in new window and \(n \in \omega \), then \(\lambda ^n x\notin X\).

Proof

Without loss of generality, assume that \(X\cap (0,b)\not =\varnothing \). Thus \(\lambda \leqslant 1\), since otherwise we would get \( b < \lambda ^m t \in X\), for some \(t\in X\) and \(m\in \omega \). Obviously Open image in new window Assume that \(\lambda ^nx\notin X\), so we getTherefore \(\lambda ^{n+1}x\notin X\), which completes the induction step. \(\square \)

Using Proposition 3.2 with \(\lambda = \frac{1}{q^n} \) and \(b=1\), one can prove the next theorem. In fact, this proposition explains the hidden argument in the next proof.

Theorem 3.3

If \(q>2\) and \(a> 0\), then the center of distances of the set of subsums of a geometric sequence Open image in new window consists of exactly zero and the terms of this sequence.

Proof

By Proposition 3.1, we get \(\frac{a}{q^n} \in S(X)\) for \(n>0\). Without loss of generality we can assume \(a=1\). The diameter of the set X of subsums of the sequence Open image in new window equals Open image in new window Putting Open image in new window , since \(\frac{1}{q-1} < 1\), we getSo, \( \frac{1}{q} \in X \) witnesses that no \(t > \frac{1}{q} \) belongs to S(X). Indeed,
$$\begin{aligned} \frac{1}{q} -t <0 \quad \text{ and }\quad \frac{1}{q} +t> \frac{2}{q} > \frac{1}{q-1}=\mathscr {X}(0). \end{aligned}$$
If \(t\in I\), where I is an X-gap, then \(t\notin S(X)\). Indeed, then \(t\notin X\) and Open image in new window , i.e., \(0\in X\) witnesses that \(t\notin S(X)\). But the intervals Open image in new window are X-gaps, hence Open image in new window is disjoint from S(X).

Now, assume that \(n>0\) is fixed. Suppose that Open image in new window and Open image in new window Thus Open image in new window witnesses that \(t \notin S(X)\). Indeed, Open image in new window implies Open image in new window , and Open image in new window implies that the X-gap Open image in new window has to contain Open image in new window

If Open image in new window and Open image in new window , then the interval Open image in new window is included in the interval Open image in new window No X-gap of the lengthis contained in the interval Open image in new window Therefore, one can findwhich witnesses that \(t \notin S(X)\). Indeed, we get Open image in new window hence \(x-t<0\); and we have that \(x+t \) belongs to the X-gap Open image in new window

By Proposition 3.1, we get \(\frac{1}{q^n} \in S(X)\) for \(n>0\). \(\square \)

Note that, when we put \(a = 2\) and \(q = 3\), Theorem 3.3 applies to the Cantor ternary set. For Open image in new window and \(q=4\) this theorem applies to sets \(\mathscr {C}_1\) and \(\mathscr {C}_2\) which will be defined in Sect. 4.

4 An example of a Cantorval

Following [5, p. 324], consider the set of subsumsThus, Open image in new window , where Open image in new window and Open image in new window Following [12, p. 330], because of its topological structure, one can call this set a Cantorval (or an \(\mathscr {M}\)-Cantorval).
Before discussing the affine properties of the Cantorval \(\mathbb {X}\) we shall introduce the following useful notions. Every point x in \(\mathbb {X}\) is determined by a sequence \((x_n)_{n>0} \), where Open image in new window The value \(x_n\) is called the n-th digit of x and the sequence \( (x_n)_{n>0}\) is called a digital representation of the point \(x\in \mathbb {X}\). Keeping in mind the formula for the sum of an infinite geometric series, we denote the tails of series as follows:Since \(\mathscr {X}(0) = \frac{5}{3}\) we get Open image in new window The involution \(h:\mathbb {X}\rightarrow \mathbb {X}\) defined by the formula
$$\begin{aligned} x\mapsto h(x)=\frac{5}{3} - x \end{aligned}$$
is the symmetry of \(\mathbb {X}\) with respect to the point \(\frac{5}{6}\). In order to check this, it suffices to note thatand also thatSo, we get \(\mathbb {X}=\frac{5}{3} -\mathbb {X}\) and \(\mathbb {X}=h[\mathbb {X}]\).
Fig. 1

An approximation of the Cantorval Open image in new window

In Fig. 1, there are marked gaps Open image in new window and Open image in new window , both of the length \(\frac{1}{12}\). Six gaps Open image in new window and Open image in new window have the length \(\frac{1}{48}\). The rest of gaps are shorter and have lengths not greater than \(\frac{1}{192}\). To describe intervals which lie in \(\mathbb {X}\), we need the following. LetWe get Open image in new window and Open image in new window Keeping in mind \(\mathscr {C}_1(0)=\frac{2}{3}\) and Open image in new window , we check that Open image in new window is the smallest real number in \(K_n\). Similarly, using Open image in new window and \(\mathscr {C}_2(0)=1\), we check that Open image in new window is the greatest real number in \(K_n\). In fact, we have the following.

Proposition 4.1

Reals from \(K_n\) are distributed consecutively at the distance \(\frac{1}{4^n}\), from Open image in new window up to Open image in new window , in the Open image in new window Therefore Open image in new window and Open image in new window

Proof

Since \(|K_1|=1\) and \(|K_2|=5\), the assertions are correct in these cases. Suppose that \(K_{n-1}= \{ f^{n-1}_1\!, f^{n-1}_2\!, \ldots , f^{n-1}_{|K_{n-1}|}\},\) wherefor \(0<j < |K_{n-1}|-1\); in consequence Open image in new window Consider the sum
$$\begin{aligned} K_{n-1} \cup \biggl (\frac{2}{4^n} + K_{n-1}\biggr ) \cup \biggl (\frac{3}{4^n} + K_{n-1}\biggr ) \cup \biggl (\frac{5}{4^n} + K_{n-1}\biggr ), \end{aligned}$$
next remove the point \(\frac{5}{4^n} + \), and then add points Open image in new window and Open image in new window We obtain the set
$$\begin{aligned} K_n=\Bigl \{ f^{n}_1, f^{n}_2, \ldots , f^{n}_{\frac{1}{3}(4^n-1)}\Bigr \} , \end{aligned}$$
which is what we need. \(\square \)

Corollary 4.2

The interval Open image in new window is included in the Cantorval \( \mathbb {X}\).

Proof

The union Open image in new window is dense in the interval Open image in new window \(\square \)

Note that it has been observed that Open image in new window , see [5] or cf. [14]. Since \(\mathbb {X}\) is centrally symmetric with \(\frac{5}{6}\) as a point of inversion, this yields another proof of the above corollary. However, our proof seems to be new and it is different from the one included in [5].

Put Open image in new window , for \(n\in \omega \). So, each \(C_n\) is an affine copy of \(\mathbb {X}\).

Proposition 4.3

The subset Open image in new window is the union of pairwise disjoint affine copies of \(\mathbb {X}\). In particular, this union includes two isometric copies of Open image in new window , for every \(n>0\).

Proof

The desired affine copies of \(\mathbb {X}\) are \(C_1\) and Open image in new window , \(\frac{1}{2} + C_2\) and Open image in new window , and so on, i.e., Open image in new window and Open image in new window \(\square \)

Proposition 4.4

The subset Open image in new window is the union of six pairwise disjoint affine copies of Open image in new window

Proof

The desired affine copies of Open image in new window lie as shown in Fig. 2. \(\square \)

Fig. 2

The arrangement of affine copies of D

Corollary 4.5

The Cantorval Open image in new window has Lebesgue measure 1.

Proof

There exists a one-to-one correspondence between \(\mathbb {X}\)-gaps and \(\mathbb {X}\)-intervals as it is shown in Fig. 3.

In view of Propositions 4.3 and 4.4, we calculate the sum of lengths of all gaps which lie in Open image in new window as follows:Since \(\frac{5}{3} - \frac{2}{3} =1\) we are done. \(\square \)
Fig. 3

The correspondence between \(\mathbb {X}\)-gaps and \(\mathbb {X}\)-intervals

If we remove the longest interval from Open image in new window , then we get the union of three copies of D, each congruent to Open image in new window . This observation—we used it above by default—is sufficient to calculate the sum of lengths of all \(\mathbb {X}\)-intervals as follows:Therefore the boundary Open image in new window is a null set.

5 Computing centers of distances

In case of subsets of the real line we formulate the following lemma.

Lemma 5.1

Given a set Open image in new window disjoint from an interval Open image in new window , assume that Open image in new window Then the center of distances S(C) is disjoint from the interval Open image in new window , i.e., Open image in new window

Proof

Given Open image in new window , consider Open image in new window We get
$$\begin{aligned} x \leqslant \frac{\alpha }{2} \leqslant \alpha - x< t < \beta -x. \end{aligned}$$
Since \(\alpha< x+t < \beta \), we get \(x+t \notin C\), also \(x<t\) implies \(x-t \notin C\). Therefore \(x\in C\) witnesses \(t\notin S(C)\). \(\square \)

We will apply the above lemma by putting suitable C-gaps in the place of the interval Open image in new window In order to obtain \(t\notin S(C)\), we must find \(x <t\) such that Open image in new window and \(x\in C\). For example, this is possible when \(\frac{\alpha }{2}< t < \alpha \) and the interval Open image in new window includes no C-gap of the length greater than or equal to \(\beta -\alpha \). But if such a gap exists, then we choose the required x more carefully.

Theorem 5.2

The center of distances of the Cantorval \(\mathbb {X}\) is equal to Open image in new window

Proof

The diameter of \(\mathbb {X}\) is \(\frac{5}{3}\) and \(\frac{5}{6} \in \mathbb {X}\), hence no \(t>\frac{5}{6}\) belongs to \(S(\mathbb {X})\). We use Lemma 5.1 with respect to the gap Open image in new window Keeping in mind the affine description of \(\mathbb {X}\), we see that the set Open image in new window has a gap Open image in new window of the length \(\frac{1}{12}\). For Open image in new window , we choose x in \(\mathbb {X}\) such that Open image in new window So, if Open image in new window , then \(t\notin S(\mathbb {X})\). Similarly using Lemma 5.1 with the gap Open image in new window , we check that for Open image in new window there exists x in \(\mathbb {X}\) such that Open image in new window Hence, if Open image in new window , then \(t\notin S(\mathbb {X})\). Analogously, using Lemma 5.1 with the gap Open image in new window , we check that if \(\frac{5}{24}< t \leqslant \frac{29}{96} < \frac{5}{12}\), then \(t\notin S(\mathbb {X})\).

For the remaining part of the interval \([0,+\infty )\) the proof uses the similarity of \(\mathbb {X}\) with \(\frac{1}{4^n}{\cdot } \mathbb {X}\) for \(n>0\). Indeed, we have shown that the \(\mathbb {X}\)-gaps Open image in new window and Open image in new window witness that Open image in new window For \(n>0\), by the similarity, the \(\mathbb {X}\)-gaps Open image in new window , Open image in new window and Open image in new window witness that Open image in new window

We have by Proposition 3.1. \(\square \)

Denote Open image in new window Thus the closure of an \(\mathbb {X}\)-gap is a \(\mathbb {Z}\)-interval and the interior of an \(\mathbb {X}\)-interval is a \(\mathbb {Z}\)-gap.

Theorem 5.3

The center of distances of the set \(\mathbb {Z}\) is trivial, i.e., \(S(\mathbb {Z})= \{0\}\).

Proof

If \(\alpha >1\), then \( \cap \mathbb {Z}= \varnothing \), hence \(1\in \mathbb {Z}\) implies \(\alpha \notin S(\mathbb {Z})\). If Open image in new window , then Open image in new window implies \(\alpha \notin S(\mathbb {Z})\). Indeed, the number \(1+\frac{11}{24}\) belongs to the \(\mathbb {Z}\)-gap Open image in new window and the number \(\frac{1}{4}+\frac{11}{24}\) belongs to the \(\mathbb {Z}\)-gap Open image in new window and the number \(\frac{11}{24} - \frac{1}{4}\) belongs to the \(\mathbb {Z}\)-gap Open image in new window Also \(0\in \mathbb {Z}\) implies Open image in new window since Open image in new window is a \(\mathbb {Z}\)-gap. For the same reason \(\frac{1}{4} \in \mathbb {Z}\) implies \(\frac{2}{3} \notin S(\mathbb {Z})\) and \(\frac{1}{3} \in \mathbb {Z}\) implies that no Open image in new window belongs to \(S(\mathbb {Z})\). Since \(\frac{1}{6}<\frac{17}{32} -\frac{1}{3} <\frac{1}{4}\) and \(\frac{2}{3}<\frac{17}{32} +\frac{1}{3} <1\), then \(\frac{17}{32} \in \mathbb {Z}\) implies \(\frac{1}{3} \notin S(\mathbb {Z})\). But, if Open image in new window , then \(\frac{1}{2} \in \mathbb {Z}\) implies \(\alpha \notin S(\mathbb {Z})\). Indeed, \(\frac{1}{6}<\frac{1}{2} - \alpha <\frac{1}{4} \) and \(\frac{3}{4}<\frac{1}{2} + \alpha <\frac{5}{6}\). So far, we have shown Open image in new window In fact, sets Open image in new window and \(S(\mathbb {Z})\) are always disjoint, sinceTherefore Open image in new window implies Open image in new window , a contradiction. Finally, we get \(S(\mathbb {Z}) = \{0\}\). \(\square \)

Now, denote Open image in new window Thus, each \(\mathbb {X}\)-gap is also a \(\mathbb {Y}\)-gap, and the interior of an \(\mathbb {X}\)-interval is a \(\mathbb {Y}\)-gap.

Proof

Since the numbers \(0, \frac{1}{4}, \frac{1}{2}, \frac{17}{32}, 1\) and Open image in new window are in \(\mathbb {Y}\), we getas in the proof of Theorem 5.3. We see that \(1\in S(\mathbb {Y})\), because
$$\begin{aligned} \mathbb {Y}\cap \biggl [0,\frac{2}{3}\biggr ] + 1 = \mathbb {Y}\cap \biggl [1, \frac{5}{3}\biggr ]. \end{aligned}$$
Moreover
$$\begin{aligned} \biggl (\mathbb {Y}\cap \biggl [0,\frac{1}{6}\biggr ] +\frac{1}{4}\biggr ) \cup \biggl ( \mathbb {Y}\cap \biggl [0, \frac{1}{6}\biggr ] + \frac{1}{2}\biggr ) \subset \mathbb {Y}, \end{aligned}$$
so \(\frac{1}{4} \in S(\mathbb {Y})\). Similarly, we check that \(\frac{1}{4^n} \in S(\mathbb {Y})\). \(\square \)

Corollary 5.5

Neither \(\mathbb {Z}\) nor \(\mathbb {Y}\) is the set of subsums of a sequence.

Proof

Since \(S(\mathbb {Z})=\{0\}\), Proposition 3.1 decides the case of \(\mathbb {Z}\). Also, this proposition decides the cases of \(\mathbb {Y}\), since Open image in new window \(\square \)

Let us add that the set of subsums of the sequence Open image in new window is included in \(\mathbb {Y}.\) One can check this, observing that each number Open image in new window , where the nonempty set \(A\subset \omega \) is finite, is the right end of an \(\mathbb {X}\)-interval.

6 Digital representation of points in the Cantorval \(\mathbb {X}\)

Assume that Open image in new window and Open image in new window are digital representations of a point \(x\in \mathbb {X}\), i.e.,where Open image in new window We are going to describe dependencies between \( a_n\) and \( b_n \). Suppose \(n_0\) is the least index such that Open image in new window Without loss of generality, we can assume that \(a_{n_0}\!=2 <b_{n_0}\!=3\), bearing in mind that Open image in new window And then we say that A is chasing B (or B is being caught by A) in the \(n_0\)-step: in other words, Open image in new window and \(a_k=b_k\) for \(k<n_0\). If it is never the case that \(a_k=5\) and \(b_k=0\), then it has to be \(b_k+3=a_k\) for all \(k>n_0\). In such a case, we obtain Open image in new window since Open image in new window

Suppose \(n_1\) is the least index such that \(a_{n_1}\!=5\) and \(b_{n_1}\!=0\), thus B is chasing A in the \(n_1\)-step. Proceeding this way, we obtain an increasing (finite or infinite) sequence \(n_0< n_1 < \cdots \) such that Open image in new window and Open image in new window for Open image in new window Moreover, A starts chasing B in the \(n_k\)-step for even k’s and B starts chasing A in the \(n_k\)-step for odd k’s, for the rest of steps changes of chasing do not occur.

Theorem 6.1

Assume that \(x\in \mathbb {X}\) has more than one digital representation. There exist a finite or infinite sequence of positive natural numbers \(n_0< n_1 < \cdots \) and exactly two digital representations Open image in new window and Open image in new window of x such that:

Proof

According to the chasing algorithm described above in the step \(n_k \) the roles of chasing are reversed. But, if the chasing algorithm does not start, then the considered point has a unique digital representation. \(\square \)

The above theorem makes it easy to check the uniqueness of digital representation. For example, if \(x \in \mathbb {X}\) has a digital representation \((x_n)_{n>0}\) such that \(x_n=2\) and \(x_{n+1}=3\) for infinitely many n, then this representation is unique. Indeed, suppose Open image in new window and Open image in new window are two different digital representations of a point \(x\in \mathbb {X}\) such that \(a_k =b_k\), whenever \(0<k<n_0\) and \(a_{n_0}\!= 2 < b_{n_0}\!=3\). By Theorem 6.1, the digit 3 never occurs immediately after the digit 2 in digital representations of x for the digits greater than \(n_0\), since it has to be Open image in new window for \(k>n_0\).

The map Open image in new window is a continuous function from the Cantor set (a homeomorphic copy of the Cantor ternary set) onto the Cantorval such that the preimage of a point has at most two points. In fact, the collection of points with two-point preimages and its complement are both of the cardinality continuum. By the algorithm described above, each sequence \(n_0< n_1 < \cdots \) of positive natural numbers determines exactly two sequences Open image in new window and Open image in new window such thatand vice versa. In the following theorem, we will use the abbreviation Open image in new window

Theorem 6.2

Let \(A\subset B\) be such that Open image in new window and A are infinite. Then the set of subsums of a sequence consisting of different elements of A is homeomorphic to the Cantor set.

Proof

Fix a nonempty open interval \(\mathbb {I}\). Assume that Open image in new window is the digital representation of a point Open image in new window Choose natural numbers \(m>k\) such that numbers Open image in new window and Open image in new window belong to \(\mathbb {I}\). Then choose \(j>m\) such that Open image in new window , where \(a =2\) or \(a=3\). Finally put \(b_{n} = a_{n}\), for \( 0<n \leqslant k\); Open image in new window and \(b_{i} = 0\) for other cases. Since \(b_m=0\), we get Open image in new window Theorem 6.1 together with conditions Open image in new window and Open image in new window imply that the point Open image in new window is not in the set of subsums of A. Thus, this set being dense in itself and closed is homeomorphic to the Cantor set. \(\square \)

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© The Author(s) 2017

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  • Wojciech Bielas
    • 1
    • 2
    Email author
  • Szymon Plewik
    • 1
  • Marta Walczyńska
    • 1
    • 2
  1. 1.Institute of MathematicsUniversity of SilesiaKatowicePoland
  2. 2.Institute of Mathematics of the Czech Academy of SciencesPraha 1Czech Republic

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