1 Introduction

In [2], the authors examine a family of Mahavier products \(M_{r,\rho }\) of closed relations \(L_{r,\rho }\) on [0, 1]. These relations \(L_{r,\rho }\) are the unions of two line segments joint in (0, 0) that have positive slopes r and \(\rho \) and extend to the opposite boundary of \([0,1]\times [0,1]\). More precisely, let r\(\rho >0\). It is shown in [2] that

  1. 1.

    if \(r=\rho \) and \(r\le 1\), then \(M_{r,\rho }\) is an arc from \((0,0,0,\ldots )\) to \((1,r,r^2,r^3,\ldots )\).

  2. 2.

    if \(r=\rho \) and \(r>1\), then \(M_{r,\rho }=\{(0,0,0,\ldots )\}\).

  3. 3.

    if \(r\ne \rho \), the following possibilities are considered.

    1. 1.

      If \(r>1\) and \(\rho >1\), then \(M_{r,\rho }=\{(0,0,0,\ldots )\}\).

    2. 2.

      If \(r\le 1\) and \(\rho \le 1\), then \(M_{r,\rho }\) is a Cantor fan.

    3. 3.

      If \(r=1\) and \(\rho > 1\), then \(M_{r,\rho }\) is as pictured in Fig. 1. See [2] for a detailed description of the fan.

    4. 4.

      If \(r<1\), \(\rho >1\) and r and \(\rho \) never connect (see Definition 11), \(M_{r,\rho }\) is homeomorphic to the Lelek fan, i.e., the fan with the dense set of end-points, see Fig. 2 and [7] for more information about the Lelek fan.

Fig. 1
figure 1

The fan \(M_{r,\rho }\) for \(r=1\) and \(\rho > 1\)

Fig. 2
figure 2

The Lelek fan

In this paper, we study the final case where \(r<1\), \(\rho >1\) and there are integers k and \(\ell \) such that

  1. 1.

    \(k\ne 0\) or \(\ell \ne 0\), and

  2. 2.

    \(r^k = \rho ^{\ell }\).

Explicitly, we prove that the Mahavier product \(M_{r,\rho }\) of \(L_{r,\rho }\)’s in this case is a fan with certain additional properties. More precisely, let \(E(M_{r,\rho })\) be the set of end-points of \(M_{r,\rho }\) and for each \(e\in E(M_{r,\rho })\), let \(A_e\) be the straight line segment in \(M_{r,\rho }\) from its top v to e. Then \(M_{r,\rho }\) is the union of a countable family \(\mathcal {C}\) of Cantor fans such that

  1. 1.

    for each \(C_1,C_2\in \mathcal {C}\), if \(C_1\ne C_2\), then \(C_1\cap C_2 = \{v\}\) and

  2. 2.

    for each \(e\in E(M_{r,\rho })\), the set of limit points of the set \(E(M_{r,\rho })\) in \(M_{r,\rho }\) forms a harmonic sequence in \(A_e\).

This gives a solution to the open problem from [2].

The paper is divided into the following parts: In Section 2, we give definitions and notation that we use later in the paper and in Section 3, among other things, our main result is presented and proved.

2 Definitions and Notation

In this section, we introduce definitions and notation that is used later in the paper. These follow [8, 9].

Definition 1

For each positive integer n, let \(X_n\) be a compact metric space and let \(f_n:X_{n+1}\rightarrow X_n\) be a continuous function. We then define the inverse limit as the subspace of \(\prod _{n=1}^{\infty }X_n\):

$$\underleftarrow{\text {lim}}\{X_n,f_n\}_{n=1}^\infty = \{(x_1,x_2,x_3,\ldots )\ \mid \ \forall n\in \mathbb {N}: x_n = f_n(x_{n+1})\}.$$

We will shorten the notation to \(X_\infty \) to denote the inverse limit \(\underleftarrow{\text {lim}}\{X_n,f_n\}_{n=1}^\infty \).

Definition 2

Let X be a compact metric space. We use \(2^X\) to denote the set of all non-empty closed subsets of X. Now, let \(G \in 2^{X\times Y}\) and \(x\in X\). Then we define

$$G(x) = \{y\in Y\ \mid \ (x,y)\in G\}.$$

Definition 3

Let XY be compact metric spaces. A function \(F:X\rightarrow 2^Y\) is called a set-valued function from X to Y. We will use \(F:X\multimap Y\) to denote F as a set-valued function from X to Y. Now, let \(F:X\multimap Y\) be a set-valued function. We say that F is upper semicontinuous at a point \(x_0\in X\) if for each open set \(U\subseteq Y\) such that \(F(x_0)\subseteq U\), there is an open set V in X such that

  1. 1.

    \(x_0\in V\) and

  2. 2.

    for each \(x\in V, F(x)\subseteq U\).

A set-valued function is upper semicontinuous, abbreviated to USC, if it is upper semicontinuous at every point in X.

We give a characterization of USC set-valued functions, which makes detecting them much more intuitive. This characterization can be found in [1, Prop. 11, page 128] as well as [5, Theorem 1.2, page 3].

Theorem 1

Let \(F:X\multimap Y\) be a set-valued function. Then F is USC if and only if the graph of F, defined as

$$\Gamma _F = \{(x,y)\in X\times Y\ \mid \ y\in F(x)\},$$

is closed in \(X\times Y\).

Definition 4

For each positive integer n, let \(X_n\) be a compact metric space and let \(F_n:X_{n+1}\multimap X_n\) be a USC set-valued function. We define the generalized inverse limit of the double sequence \(\{X_n,F_n\}_{n=1}^\infty \) as the subspace of the product space \(\prod _{n=1}^{\infty }X_n\):

$$\mathop {\lim }\limits _ {\circ \hspace{-1.8pt}-\!\!-\!\!-}\{X_n,F_n\}_{n=1}^\infty = \{(x_1,x_2,x_3,\ldots )\ \mid \ \forall n\in \mathbb {N}: x_n \in F(x_{n+1})\}.$$

We use \(X_\infty \) to denote the generalized inverse limit \(\mathop {\lim }\limits _ {\circ \hspace{-1.8pt}-\!\!-\!\!-}\{X_n,F_n\}_{n=1}^\infty \). Notice that this is a generalization of the inverse limit as defined before in Definition 1.

The following is a well-known result for generalized inverse limits. Thus, the proof is omitted. Further information can be found in [6, Theorems 5.1 and 5.2, page 125].

Theorem 2

For each positive integer n (Fig. 3), let:

  1. 1.

    \(X_n\) and \(Y_n\) be compact metric spaces,

  2. 2.

    \(F_n:X_{n+1}\multimap X_n\), \(G_n:Y_{n+1}\multimap Y_n\) be USC set-valued functions and

  3. 3.

    \(\phi _n:X_n\rightarrow Y_n\) be a continuous function such that for each \(x\in X_{n+1}\):

    $$\phi _n(F_n(x)) \subseteq G_n(\phi _{n+1}(x)).$$
Fig. 3
figure 3

A diagram from Theorem 2

Then the function \(\phi : X_\infty \rightarrow Y_\infty \) where for each \((x_1,x_2,x_3,\ldots ) \in X_\infty :\)

$$\phi ((x_1,x_2,x_3,\ldots )) = (\phi _1(x_1),\phi _2(x_2),\phi _3(x_3),\ldots ),$$

is continuous. Furthermore, if the function \(\phi _n\) is a homeomorphism for each positive integer n and it also holds that for each \(x\in X_{n+1}:\)

$$\phi _n(F_n(x)) = G_n(\phi _{n+1}(x)),$$

then the function \(\phi \) is also a homeomorphism.

Next, we define Mahavier products. First, we introduce the following notation.

Definition 5

Let XY be compact metric spaces and \(G \in 2^{X\times Y}\). We define the inverse of G, denoted as \(G^{-1}\), as

$$G^{-1} = \{(y,x) \in Y\times X\ \mid \ (x,y) \in G\}.$$

Definition 6

Let X be a compact metric space, \(G \in 2^{X\times X}\). We use

  1. 1.

    \(\star _{i=1}^{\infty }G = \{(x_1,x_2,\ldots )\in \prod _{i=1}^{\infty }X\ \mid \ \forall i\in \{1,2,\ldots \}: (x_{i},x_{i+1})\in G\}\) to denote the Mahavier product of G.

  2. 2.

    \(\star _{i=1}^\infty G^{-1} = \{(x_1,x_2,\ldots )\in \prod _{i=1}^\infty X\ \mid \ \forall i\in \mathbb {N}: (x_{i+1},x_i)\in G\}\) to denote the Mahavier product of \(G^{-1}\).

Observation 3

Let X be a compact metric space, let \(F: X \multimap X\) be a USC set-valued function. Then

$$ \mathop {\lim }\limits _ {\circ \hspace{-1.8pt}-\!\!-\!\!-}\{X,F\}_{n=1}^\infty = \star _{n=1}^\infty \Gamma _F^{-1}. $$

Definition 7

For each positive integer i, let \(X_i\) be a compact metric space. We use

  1. 1.

    \(p_n\) to denote the nth standard projection, \(p_n:\prod _{i=1}^\infty X_i \rightarrow X_n\),

    $$p_n(x_1,x_2,\ldots ) = x_n.$$
  2. 2.

    \(p_{1,n}(\underline{x})\) to denote the projection \(p_{1,n}:\prod _{i=1}^\infty X_i \rightarrow \prod _{i=1}^nX_i\)

    $$p_{1,n}(x_1,x_2,\ldots ) = (x_1,x_2,\ldots ,x_n).$$
  3. 3.

    \(p_{n,\infty }\) to denote the projection \(p_{n,\infty }:\prod _{i=1}^\infty X_i \rightarrow \prod _{i=n}^\infty X_i\),

    $$p_{n,\infty }(x_1,x_2,\ldots ) = (x_n,x_{n+1},\ldots ).$$

The following definition is needed in Theorem 14.

Definition 8

Let \((x_1,x_2,\ldots ,x_{n-1},x_n)\) be a finite sequence and \((x_n,x_{n+1},x_{n+2},\ldots )\) be a sequence. We use \((x_1,x_2,\ldots ,x_{n-1},x_n)\star (x_n,x_{n+1},x_{n+2},\ldots )\) to denote

$$(x_1,x_2,\ldots ,x_{n-1},x_n)\star (x_n,x_{n+1},x_{n+2},\ldots ) = (x_1,x_2,\ldots ,x_{n-1},x_n,x_{n+1},\ldots ).$$

Definition 9

A continuum is a non-empty compact connected metric space. A subcontinuum is a subspace of a continuum, which is itself a continuum.

Definition 10

Let X be a continuum.

  1. 1.

    The continuum X is unicoherent, if for any subcontinua A and B of X such that \(X = A\cup B\), the compactum \(A\cap B\) is connected.

  2. 2.

    The continuum X is hereditarily unicoherent provided that each of its subcontinua is unicoherent.

  3. 3.

    The continuum X is a dendroid, if it is an arcwise connected, hereditarily unicoherent continuum.

  4. 4.

    Let X be a continuum. If X is homeomorphic to [0,1], then X is an arc.

  5. 5.

    A point x in an arc X is called an end-point of the arc X, if there is a homeomorphism \(\phi :[0,1]\rightarrow X\) such that \(\phi (0) = x\). The set of all end-points of the arc X is denoted by E(X).

  6. 6.

    Let X be a dendroid. A point \(x\in X\) is called an end-point of the dendroid X, if for every arc A in X that contains x, x is an end-point of A. The set of all end-points of X will be denoted by E(X).

  7. 7.

    A continuum X is a simple triod, if it is homeomorphic to \(([-1,1]\times \{0\})\cup (\{0\}\times [0,1])\).

  8. 8.

    A point x in a simple triod X is called the top-point or, briefly, the top of the simple triod X, if there is a homeomorphism \(\phi : ([-1,1]\times \{0\})\cup (\{0\}\times [0,1])\rightarrow X\) such that \(\phi ((0,0)) = x\).

  9. 9.

    Let X be a dendroid. A point \(x\in X\) is called a ramification point of the dendroid X, if there is a simple triod T in X with top x. The set of all ramification points of X will be denoted by R(X).

  10. 10.

    The continuum X is a fan, if it is a dendroid with at most one ramification point v, which is called the top of the fan X (if it exists).

  11. 11.

    Let X be a fan. We say that X is a Cantor fan, if X is homeomorphic to the continuum

    $$\bigcup _{c\in C}A_c,$$

    where \(C\subseteq [0,1]\) is the standard Cantor set and for each \(c\in C\), \(A_c\) is the straight line segment in the plane from (0, 0) to (c, 1).

Observation 4

Note that a space is a Cantor set if and only if it is a non-empty, compact, completely disconnected space with no isolated points. See [9, Theorem 7.14, page 109] for details and more information.

3 Connected Numbers and a Family of Fans

In this section, we answer the open problem from [2]. First, we introduce the notation that is needed later.

Definition 11

Let \(r\in (0,1)\) and \(\rho \in (1,\infty )\). We say that r and \(\rho \) never connect if for all integers k and l,

$$r^k = \rho ^l \iff k=l=0.$$

We use \((r,\rho )\in \mathcal{N}\mathcal{C}\) to denote that r and \(\rho \) never connect.

Definition 12

Let \(r\in (0,1)\) and \(\rho \in (1,\infty )\). We say that r and \(\rho \) connect if there exist positive integers MN such that

$$r^M\rho ^N = 1.$$

We use \((r,\rho )\in \mathcal {C}\) to denote that r and \(\rho \) connect.

Observation 5

Let \(r\in (0,1)\) and \(\rho \in (1,\infty )\). Then r and \(\rho \) connect if and only if there exist positive integers MN such that

$$r=\rho ^{-\frac{N}{M}}.$$

Theorem 6

Let \(r\in (0,1)\) and \(\rho \in (1,\infty )\). Then \((r,\rho )\in \mathcal C\) if and only if they do not never connect.

Proof

First, let \((r,\rho )\in \mathcal C\). By Definition 12, there exist positive integers MN such that

$$r^M\rho ^N = 1.$$

Let \(k = M\) and \(l = -N\) be integers, then

$$r^k = \rho ^l.$$

Since k and l are non-zero, it follows that r and \(\rho \) do not never connect.

Now, let r and \(\rho \) be such that they do not never connect. There must then exist integers kl such that

$$r^k = \rho ^l$$

and at least one of them is non-zero. It is clear that neither can be zero, otherwise the other would also have to be zero or either r or \(\rho \) would have to equal 1, all of which are contradictions. Since r is less than 1 and \(\rho \) is greater than 1, it is also clear that one of the integers kl is negative and the other is positive. Without loss of generality let k be negative. It follows that for positive integers \(M = -k\) and \(N = l\),

$$r^M\rho ^N = 1.$$

This means that r and \(\rho \) connect. \(\square \)

Definition 13

Let r, \(\rho >0\). We define the following sets:

  1. 1.

    \(L_r = \{(x,y)\in [0,1]^2\ \mid \ y = r\cdot x\}\),

  2. 2.

    \(L_\rho = \{(x,y)\in [0,1]^2\ \mid \ y = \rho \cdot x\}\),

  3. 3.

    \(L_{r,\rho } = L_r \cup L_\rho \),

  4. 4.

    \(M_{r,\rho } = \star _{i=1}^\infty L_{r,\rho }\).

Example 1

In Fig. 4, an example of such an \(L_{r,\rho }\) for \(r = \frac{1}{2}\) and \(\rho = 2\) is given.

Fig. 4
figure 4

\(L_{\frac{1}{2},2}\)

Definition 14

Let A be a non-empty set. We use \(A^\mathbb {N}\) to denote the set

$$\{{\textbf {a}}\ \mid \ {\textbf {a}}:\mathbb {N}\rightarrow A\}$$

of all sequences in A.

Definition 15

Let \((r,\rho )\in \mathcal {C}\) and let \({\textbf {a}} = (a_i)_{i=1}^\infty \in \{r,\rho \}^\mathbb {N}\).

  1. 1.

    We define the set \(L_{\textbf {a}}\) as follows:

    $$L_{\textbf {a}} = \star _{i=1}^\infty L_{a_i} = \{(x,a_1x,a_2a_1x,\ldots )\ \mid \ x\in [0,1]\}\cap M_{r,\rho }.$$
  2. 2.

    We say that \({\textbf {a}}\) is a useful sequence if \(L_{{\textbf {a}}}\) is an arc in \(M_{r,\rho }\).

  3. 3.

    We define \(\mathcal {L}_{r,\rho } = \{L_{\textbf {a}}\ \mid \ {\textbf {a}}\in \{r,\rho \}^\mathbb {N}\}.\)

First, we prove the following theorem.

Theorem 7

Let \((r_1,\rho _1), (r_2,\rho _2)\in \mathcal {C}\) such that \(r_{1}^{M}\rho _{1}^{N} = r_{2}^{M}\rho _{2}^{N} = 1\) for some positive integers MN. Then \(M_{r_1,\rho _1}\) and \(M_{r_2,\rho _2}\) are homeomorphic (Fig. 5).

Proof

We prove this using Theorem 2. To do this, we define the following:

For each positive integer n, let

$$f_n:[r_1^{\frac{n}{N}},r_1^{\frac{n-1}{N}}]\rightarrow [r_2^{\frac{n}{N}},r_2^{\frac{n-1}{N}}]$$

be such that it’s graph is the line segment from \((r_1^{\frac{n}{N}},r_2^{\frac{n}{N}})\) to \((r_1^{\frac{n-1}{N}},r_2^{\frac{n-1}{N}})\). Explicitly, for each \(x\in [r_1^{\frac{n}{N}},r_1^{\frac{n-1}{N}}]\),

$$f_n(x) = \left( \frac{r_2}{r_1}\right) ^{\frac{n-1}{N}}\cdot \frac{1-r_2^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}\cdot x + r_2^{\frac{n-1}{N}}\cdot \frac{r_2^{\frac{1}{N}}-r_1^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}.$$

Next, we define \(f:[0,1]\rightarrow [0,1]\) as follows:

$$\forall x\in [0,1]: f(x) = {\left\{ \begin{array}{ll} 0; &{} x = 0\\ f_n(x); &{} x\in [r_1^{\frac{n}{N}},r_1^{\frac{n-1}{N}}]; \end{array}\right. }$$

see Fig. 4.

Fig. 5
figure 5

The graph of the function f

It is clear that such f is a homeomorphism. Now, for each positive integer n, let

  1. 1.

    \(X_n = Y_n = [0,1]\),

  2. 2.

    \(F_n:X_{n+1}\multimap X_n, G_n:Y_{n+1}\multimap Y_n\) be set-valued functions such that their graphs are \(L_{r_1,\rho _1}^{-1}\) and \(L_{r_2,\rho _2}^{-1}\), respectively, and

  3. 3.

    \(\phi _n = f\).

To use Theorem 2, we show that for each \(x\in [0,1],\)

$$\phi _n(F_n(x)) = G_n(\phi _{n+1}(x)).$$

To prove this, we need to prove that

$$ \{f(r_1\cdot x), f(\rho _1\cdot x)\} = \{r_2\cdot f(x), \rho _2\cdot f(x)\}, $$

since \(\Phi _n(F_n(x)) = \{f(r_1\cdot x), f(\rho _1\cdot x)\}\) and \(G_n(\Phi _{n+1}(x)) = \{r_2\cdot f(x), \rho _2\cdot f(x)\}\). It is enough to prove the two following claims:

  • Claim 1. For each \(x\in [0,\rho _1^{-1}] = [0,r_1^{\frac{M}{N}}]: f(\rho _1\cdot x) = \rho _2\cdot f(x)\).

  • Proof: We treat the following possible subcases.

    • (i) \(x = 0:\) This is clearly true.

    • (ii) \(x \ne 0:\) Then there exists some positive integer n such that

      $$x\in [r_1^{\frac{n}{N}},r_1^{\frac{n-1}{N}}]$$

      and because of that (noting that \(n>M\)),

      $$\rho _1x\in [r_1^{\frac{n-M}{N}},r_1^{\frac{n-M-1}{N}}].$$

      It follows then that

      $$f(\rho _1x) = f_{n-M}(\rho _1x) =$$
      $$= \left( \frac{r_2}{r_1}\right) ^{\frac{n-M-1}{N}}\cdot \frac{1-r_2^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}\cdot r_1^{-\frac{M}{N}}x + r_2^{\frac{n-M-1}{N}}\cdot \frac{r_2^{\frac{1}{N}}-r_1^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}.$$

      On the other hand,

      $$\rho _2f(x) = \rho _2f_n(x) =$$
      $$= \rho _2\cdot \left[ \left( \frac{r_2}{r_1}\right) ^{\frac{n-1}{N}}\cdot \frac{1-r_2^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}\cdot x + r_2^{\frac{n-1}{N}}\cdot \frac{r_2^{\frac{1}{N}}-r_1^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}\right] $$

      and we can quickly see that these two values are the same.

  • Claim 2. For each \( x\in [0,1]: f(r_1\cdot x) = r_2\cdot f(x)\).

  • Proof: We treat the following possible subcases.

    • (i) \(x = 0:\) This is clearly true.

    • (ii) \(x \ne 0:\) Then there exists some positive integer n such that

      $$x\in [r_1^{\frac{n}{N}},r_1^{\frac{n-1}{N}}]$$

      and because of that,

      $$r_1x\in [r_1^{\frac{n+N}{N}},r_1^{\frac{n+N-1}{N}}].$$

      It follows then that

      $$f(r_1x) = f_{n+N}(r_1x) =$$
      $$= \left( \frac{r_2}{r_1}\right) ^{\frac{n+N-1}{N}}\cdot \frac{1-r_2^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}\cdot r_1x + r_2^{\frac{n+N-1}{N}}\cdot \frac{r_2^{\frac{1}{N}}-r_1^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}.$$

      On the other hand,

      $$r_2f(x) = r_2f_n(x) =$$
      $$= r_2\cdot \left[ \left( \frac{r_2}{r_1}\right) ^{\frac{n-1}{N}}\cdot \frac{1-r_2^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}\cdot x + r_2^{\frac{n-1}{N}}\cdot \frac{r_2^{\frac{1}{N}}-r_1^{\frac{1}{N}}}{1-r_1^{\frac{1}{N}}}\right] $$

      and we can quickly see that these two values are the same.

Having proven this, it follows that

  1. 1.

    if \(x\in [0,\rho _1^{-1}]\), then for each positive integer n,

    $$\phi _n(F_n(x)) = f(L_{r_1,\rho _1}^{-1}(x)) = f(\{r_1\cdot x, \rho _1\cdot x\}) = \{r_2\cdot f(x), \rho _2\cdot f(x)\} = G_n(\phi _{n+1}(x))$$

    and

  2. 2.

    if \(x\in (\rho _1^{-1},1]\), then for each positive integer n,

    $$\phi _n(F_n(x)) = f(L_{r_1,\rho _1}^{-1}(x)) = f(r_1\cdot x) = r_2\cdot f(x) = G_n(\phi _{n+1}(x))$$

and, therefore, by Theorem 2 and Observation 3, \(M_{r_1,\rho _1}\) and \(M_{r_2,\rho _2}\) are homeomorphic. \(\square \)

Observation 8

Let \((r,\rho )\in \mathcal {C}\). For each point \({\textbf {x}}\in M_{r,\rho }\) there are a point \(x\in [0,1]\) and a sequence \((a_1,a_2,a_3,\ldots )\in \{r,\rho \}^\mathbb {N}\) such that

$${\textbf {x}} = (x, a_1x, a_2a_1x,\ldots ).$$

If \(x\ne 0\), then there is only one such sequence \((a_1,a_2,a_3,\ldots )\).

The following lemmas are needed in the proof of our main result.

Lemma 9

Let \((r,\rho )\in \mathcal {C}\) and let \({\textbf {a}} = (a_1,a_2,\ldots )\in \{r,\rho \}^\mathbb {N}\) be such that the set

$$A = \left\{ \prod _{i=1}^{n}a_i\ \mid \ n\in \mathbb {N}\right\} $$

has no maximum in \(\mathbb R\). Then the set A has no upper bound in \(\mathbb R\).

Proof

Suppose the set A has no maximum, but does have a supremum, let

$$M = \sup \left\{ \prod _{i=1}^{n}a_i\ \mid \ n\in \mathbb {N}\right\} .$$

Then, for each \(\varepsilon >0\) there exists a positive integer n such that

$$M-\varepsilon< \prod _{i=1}^{n}a_i < M.$$

It follows that there is a strictly increasing sequence \((m_n)_{n=1}^\infty \) of positive integers such that

  1. 1.

    for each positive integer n: \(a_1a_2\cdot \ldots \cdot a_{m_n} < a_1a_2\cdot \ldots \cdot a_{m_n+1}\) and

  2. 2.

    \(\displaystyle \lim _{n\rightarrow \infty }a_1a_2\cdot \ldots \cdot a_{m_n} = M\).

Note that for each positive integer n, from

$$\prod _{i=1}^{m_{n+1}}a_i = \left( \prod _{i=1}^{m_n}a_i\right) \cdot a_{m_n+1}\cdot \ldots \cdot a_{m_{n+1}} > \prod _{i=1}^{m_n}a_i,$$

it follows that

$$a_{m_n+1}\cdot \ldots \cdot a_{m_{n+1}} > 1.$$

We see also that

$$\frac{\prod _{i=1}^{m_{n+1}}a_i}{\prod _{i=1}^{m_n}a_i} = a_{m_n+1}\cdot \ldots \cdot a_{m_{n+1}}$$

and applying the limit as n approaches infinity, we get

$$\lim _{n\rightarrow \infty }(a_{m_n+1}\cdot \ldots \cdot a_{m_{n+1}}) = \lim _{n\rightarrow \infty }\frac{\prod _{i=1}^{m_{n+1}}a_i}{\prod _{i=1}^{m_n}a_i} = \frac{M}{M} = 1.$$

It follows that for each \(\varepsilon > 0\) there exists a positive integer n such that

$$1< a_{m_n+1}\cdot a_{m_n+2}\cdot \ldots \cdot a_{m_{n+1}} < 1+\varepsilon .$$

Since \((r,\rho )\in \mathcal {C},\) there exist positive integers MN such that \(r^M\rho ^N = 1\) or, equivalently, \(r = \rho ^{-\frac{N}{M}}\), so every such product can be written as

$$a_{m_n+1}\cdot a_{m_n+2}\cdot \ldots \cdot a_{m_{n+1}} = \rho ^{A_n-B_n\cdot \frac{N}{M}},$$

where \(A_n\), \(B_n\) are non-negative integers. For each positive integer n, let \(A_n,B_n\) be such integers.

It follows then that

$$\inf \left\{ A_n-B_n\cdot \frac{N}{M}\ \mid \ n\in \mathbb {N}\right\} = 0$$

and 0 is not the minimum of this set. Note that for each positive integer n,

$$A_n-B_n\cdot \frac{N}{M} = \frac{A_nM-B_nN}{M},$$

but since the integer M is fixed and \(A_nM-B_nN\) is an integer, for each positive integer n, these numbers cannot get arbitrarily close to 0, which is a contradiction. \(\square \)

Lemma 10

Let \((r,\rho )\in \mathcal {C}\). Then \({\textbf {a}} = (a_1,a_2,\ldots )\in \{r,\rho \}^\mathbb {N}\) is a useful sequence if and only if there exists

$$\max \left\{ \prod _{i=1}^{n}a_i\ \mid \ n\in \mathbb {N}\right\} .$$

Proof

Let \({\textbf {a}}\) be a useful sequence. By Definition 15, \(L_{{\textbf {a}}}\) is an arc. Since

$$L_{{\textbf {a}}} = \{(x,a_1x,a_2a_1x,\ldots )\ \mid \ x\in [0,1]\}\cap M_{r,\rho },$$

it follows that there is \(m\in (0,1]\) such that

$$L_{\textbf {a}} = \{(x,a_1x,a_2a_1x,\ldots )\ \mid \ x\in [0,m]\}.$$

Suppose the set \(\{\prod _{i=1}^{n}a_i\ \mid \ n\in \mathbb {N}\}\) has no maximum. By Lemma 9, \(\{\prod _{i=1}^{n}a_i\ \mid \ n\in \mathbb {N}\}\) then has no upper bound, meaning that for any \(x\in (0,m]\) there exists a positive integer n such that,

$$a_{n-1}\cdot \ldots \cdot a_2a_1 > \frac{1}{x}$$

or, equivalently, that

$$a_{n-1}\cdot \ldots \cdot a_2a_1x > 1.$$

The only value of x where such \((x,a_1x,a_2a_1x,\ldots )\) could be in \(M_{r,\rho }\) is therefore \(x=0\), but then \(L_{\textbf {a}} = \{(0,0,0,\ldots )\}\) is not an arc \(M_{r,\rho }\). Then \({\textbf {a}}\) is not a useful sequence, which is a contradiction. This proves the first implication.

To prove the second implication, let

$$m = \max \left\{ \prod _{i=1}^{n}a_i\ \mid \ n\in \mathbb {N}\right\} .$$

It is clear then that \(L_{{\textbf {a}}} = \{(x,a_1x,a_2a_1x,\ldots )\ \mid \ x\in [0,\min \{1,\frac{1}{m}\}]\},\) showing that it is an arc. Therefore, \({\textbf {a}}\) is a useful sequence. \(\square \)

Remark 1

Note that for any useful sequence \({\textbf {a}}\),

$$(0,0,0,\ldots ) \in E(L_{{\textbf {a}}}).$$

In the following two theorems, the properties of arcs \(L_{{\textbf {a}}}\) are studied.

Theorem 11

Let \((r,\rho )\in \mathcal {C}\), let \({\textbf {a}}\) be a useful sequence and let \(x\in (0,1]\). Then

$$ (x,a_1x,a_2a_1x,\ldots ) $$

is an end-point of \(L_{{\textbf {a}}}\) if and only if either \(x=1\) or there exists a positive integer n such that \(a_{n}\cdot \ldots \cdot a_2a_1x = 1\).

Proof

First, suppose that \(x=1\) or that there exists a positive integer n such that \(a_na_{n-1}\cdot \ldots \cdot a_1x = 1\). We prove that \((x,a_1x,a_2a_1x,\ldots )\) is an end-point of \(L_{\textbf {a}}\). If \(x=1\), then obviously, \((x,a_1x,a_2a_1x,\ldots )\) is an end-point of \({\textbf {a}}\). Now let \(x<1\). Let \((x,a_1x,a_2a_1x,\ldots )\), where \(x\ne 0\), be such that there exists a positive integer n so that \(a_{n}\cdot \ldots \cdot a_2a_1x = 1\). Since

$$ L_{\textbf {a}} = \left\{ (t,a_1t,a_2a_1t,\ldots )\ \mid \ t\in \left[ 0,\frac{1}{a_1a_2\cdot \ldots \cdot a_n}\right] \right\} $$

is an arc from \((0,0,0,\ldots )\) to \((x,a_1x,a_2a_1x,\ldots )\) where \(x=\frac{1}{a_1a_2\cdot \ldots \cdot a_n}\), it is clear then that \((x,a_1x,a_2a_1x,\ldots )\) is the end-point of \(L_{{\textbf {a}}}\) which is different from \((0,0,0,\ldots )\).

Next, let \((x,a_1x,a_2a_1x,\ldots )\), where \(x \ne 0\), be an end-point of \(L_{{\textbf {a}}}\). Since \({\textbf {a}}\) is a useful sequence, and \(x>0\), it follows that

$$ m = \max \{a_n\cdot \ldots \cdot a_2a_1\ \mid \ n\in \mathbb {N}\} $$

exists and it follows that

$$ \sup \{a_{n}\cdot \ldots \cdot a_2a_1x\ \mid \ n\in \mathbb {N}\} = x \cdot \sup \{a_n\cdot \ldots \cdot a_2a_1\ \mid \ n\in \mathbb {N}\} = $$
$$ = x \cdot \max \{a_n\cdot \ldots \cdot a_2a_1\ \mid \ n\in \mathbb {N}\} = x\cdot m. $$

It is clear then that

$$x\cdot m = \max \{a_{n}\cdot \ldots \cdot a_2a_1x\ \mid \ n\in \mathbb {N}\}.$$

We consider the following possible cases.

  1. 1.

    If \(m\le 1\), then

    $$L_{{\textbf {a}}} = \{(t,a_1t,a_2a_1t,\ldots )\ \mid \ t\in [0,1]\}$$

    and since \((x,a_1x,a_2a_1x,\ldots )\) is an end-point of \(L_{{\textbf {a}}}\), it follows that \(x=1\).

  2. 2.

    Now let \(m>1\). Then

    $$L_{{\textbf {a}}} = \left\{ (t,a_1t,a_2a_1t,\ldots )\ \mid \ t\in \left[ 0,\frac{1}{m}\right] \right\} $$

    and since \((x,a_1x,a_2a_1x,\ldots )\) is an end-point of \(L_{{\textbf {a}}}\), it follows for some n, \(xa_1a_2\cdot \ldots \cdot a_n = 1\).

\(\square \)

Theorem 12

Let \((r,\rho )\in \mathcal {C}\) and \({\textbf {a}}, {\textbf {b}}\) be two different useful sequences, then

$$L_{{\textbf {a}}} \cap L_{{\textbf {b}}} = \{(0,0,0,\ldots )\}.$$

Proof

Suppose that this is not true and there is a non-zero point \((x,a_1x,a_2a_1x,\ldots ) = (x,b_1x,b_2b_1x,\ldots ) \in L_{{\textbf {a}}} \cap L_{{\textbf {b}}}\). It immediately follows that for each positive integer n: \(a_n = b_n\) and therefore \({\textbf {a}} = {\textbf {b}}\), a contradiction. \(\square \)

Remark 2

Let \((r,\rho ) \in \mathcal {C}\). From previous theorems we see that \(M_{r,\rho }\) is the union of a collection of arcs, joint only in \((0,0,0,\ldots )\). Also, it is easy to check that the dimension of \(M_{r,\rho }\) is 1. To see that, one can use the main result from [3].

Our next goal is to prove that \(M_{r,\rho }\) is a fan. To see this, we introduce the following notation about end-points.

Definition 16

Let \((r,\rho )\in \mathcal {C}\). For every positive integer n, we use \(E_n(r,\rho )\) to denote the set

$$E_n(r,\rho ) = \{(x_1,x_2,x_3,\ldots )\in M_{r,\rho }\ \mid \ x_n = 1\ and\ \forall i\in \{1,2,\ldots ,n-1\}: x_i \ne 1\}.$$

Definition 17

Let \((r,\rho )\in \mathcal {C}\). For each useful sequence \({\textbf {a}}\), let \(e_{{\textbf {a}}}\) be the non-zero end-point of \(L_{\textbf {a}}\) and let

$$E(r,\rho ) = \{e_{\textbf {a}}\ \mid \ {\textbf {a}}\ is\ a\ useful\ sequence\}.$$

Observation 13

$$E(r,\rho ) = \bigcup _{n\in \mathbb {N}}E_n(r,\rho )$$

Next, we prove some properties of the above defined sets.

Theorem 14

Let \((r,\rho )\in \mathcal {C}\). For each positive integer n, \(E_n(r,\rho )\) is an uncountable set.

Proof

Let \((r,\rho )\in \mathcal {C}\) and let MN be positive integers such that

$$r^M\rho ^N = 1.$$

The proof is by induction on n.

  1. 1.

    \(n = 1\). First, we define two finite sequences:

    $${\textbf {x}} = (1,r,r^2,\ldots ,r^M,r^M\rho ,\ldots ,r^M\rho ^{N-1},1)$$
    $${\textbf {y}} = (1,r,r^2,\ldots ,r^{2M},r^{2M}\rho ,\ldots ,r^{2M}\rho ^{N-1},r^M,r^M\rho ,\ldots ,r^M\rho ^{N-1},1)$$

    It follows then that

    $$\{\star _{i=1}^\infty {\textbf {z}}_i\ \mid \ \forall i\in \mathbb {N}: {\textbf {z}}_i\in \{{\textbf {x}},{\textbf {y}}\}\}\subseteq E_1(r,\rho )$$

    and since this subset is uncountable, so is \(E_1(r,\rho )\).

  2. 2.

    Let \(n\in \mathbb {N}\setminus \{1\}\) be some positive integer and \(E_{n-1}(r,\rho )\) be uncountable. We prove that \(E_n(r,\rho )\) is uncountable as well. Let \((x_1,x_2,\ldots ,x_{n-2},1,x_{n},\ldots )\in E_{n-1}(r,\rho )\). We see then that

    $$(\rho ^{-1}x_1,x_1,x_2,\ldots ,x_{n-2},1,x_{n},\ldots )\in E_n(r,\rho )$$

    because it’s nth coordinate is 1 and \(\rho ^{-1}x<1\) since \(\rho >1\) and \(x<1\). Thus there exists an injection from \(E_{n-1}(r,\rho )\) to \(E_n(r,\rho )\). Since \(E_{n-1}(r,\rho )\) is uncountable, it follows then that \(E_n(r,\rho )\) is also uncountable.

\(\square \)

The following lemma is needed in Theorem 16.

Lemma 15

Let \((r,\rho )\in \mathcal {C}\) and let \({\textbf {x}}=(x_1,x_2,x_3,\ldots )\in M_{r,\rho }\) be any point. Then for each positive integer k, there is a point \({\textbf {y}}_k\in M_{r,\rho }\) such that

  1. 1.

    \(p_1({\textbf {y}}_k) = x_k\) and

  2. 2.

    \(p_{k,\infty }({\textbf {x}}) \ne {\textbf {y}}_k\).

Proof

Let k be a positive integer. We treat the following possible cases.

  1. 1.

    There exists a positive integer l such that \(x_{k+l} = \rho \cdot x_{k+l-1}\). Then let

    $$ {\textbf {y}}_k = (x_k,x_{k+1},\ldots ,x_{k+l-1},r\cdot x_{k+l-1},r^2\cdot x_{k+l-1},\ldots ). $$

    Then \(p_1({\textbf {y}}_k) = x_k\) and \(p_{k,\infty }({\textbf {x}})\ne {\textbf {y}}_k\).

  2. 2.

    For each positive integer l, \(x_{k+l} = r\cdot x_{k+l-1}\). Then there is a positive integer \(m>k\) such that \(\rho \cdot x_m < 1\). Then let

    $$ {\textbf {y}}_k = (x_k,x_{k+1},\ldots ,x_m,\rho \cdot x_m,r\cdot \rho \cdot x_m,r^2\cdot \rho \cdot x_m,\ldots ). $$

    Then \(p_1({\textbf {y}}_k) = x_k\) and \(p_{k,\infty }({\textbf {x}})\ne {\textbf {y}}_k\).

\(\square \)

Theorem 16

Let \((r,\rho ) \in \mathcal {C}.\) For each positive integer n, \(E_n(r,\rho )\) has no isolated points.

Proof

Let \({\textbf {x}} = (x_1,x_2,\ldots ,x_{n-1},1,x_{n+1},\ldots ) \in E_n(r,\rho )\). Since \((r,\rho )\in \mathcal {C},\) let MN be positive integers such that

$$r^M\rho ^N = 1.$$

We construct a sequence \(({\textbf {x}}^k)_{k=1}^\infty \) in \(E_n(r,\rho )\) such that it converges to \({\textbf {x}}\) and for each positive integer k, \({\textbf {x}}^k \ne {\textbf {x}}\).

For each positive integer k, \(k>n\), let \({\textbf {y}}_k\in M_{r,\rho }\) be a point such that

  1. 1.

    \(p_1({\textbf {y}}_k) = x_k\) and

  2. 2.

    \(p_{k,\infty }({\textbf {x}})\ne {\textbf {y}}_k\).

Such \({\textbf {y}}_k\) does exist by Lemma 15. For each positive integer \(k>n\), let

$$ {\textbf {x}}_k = p_{1,k}({\textbf {x}})\star {\textbf {y}}_k. $$

Then \(\displaystyle \lim _{k\rightarrow \infty }{} {\textbf {x}}_k = {\textbf {x}}\) and for each such k, \({\textbf {x}}_k\ne {\textbf {x}}\). \(\square \)

Theorem 17

Let \((r,\rho )\in \mathcal {C}\). For each positive integer n, \(E_n(r,\rho )\) is compact.

Proof

Let \((r,\rho )\in \mathcal {C}\) and let MN be positive integers such that

$$r^M\rho ^N = 1.$$
  1. 1.

    \(n = 1:\) Let \(({\textbf {x}}^k)_{k=1}^\infty \) be a sequence in \(E_1(r,\rho )\):

    $${\textbf {x}}^1 = (1,x_2^1,x_3^1,\ldots )$$
    $${\textbf {x}}^2 = (1,x_2^2,x_3^2,\ldots )$$
    $${\textbf {x}}^3 = (1,x_2^3,x_3^3,\ldots )$$
    $$\vdots $$

    and let \({\textbf {x}} = (x_1,x_2,\ldots ) = \displaystyle \lim _{k\rightarrow \infty }{} {\textbf {x}}^k\). Then, obviously, \(x_1 = 1\) and \({\textbf {x}}\in E_1(r,\rho )\) follows.

  2. 2.

    Let \(n\in \mathbb {N}\setminus \{1\}\) be some positive integer and suppose that \(E_{n-1}(r,\rho )\) is compact. We prove that \(E_n(r,\rho )\) is compact. Let \(({\textbf {x}}^k)_{k=1}^\infty \) be a sequence in \(E_n(r,\rho )\) and let \({\textbf {x}} = (x_1,x_2,\ldots ) = \displaystyle \lim _{k\rightarrow \infty }{} {\textbf {x}}^k\). Then

    $$(p_{2,\infty }({\textbf {x}}^k))_{k=1}^\infty $$

    is a sequence in \(E_{n-1}(r,\rho )\) such that

    $$\lim _{k\rightarrow \infty }p_{2,\infty }({\textbf {x}}^k) = p_{2,\infty }({\textbf {x}}).$$

    It follows that \(p_{2,\infty }({\textbf {x}})\in E_{n-1}(r,\rho )\). All that is left to see is that \(x_1\ne 1\). For each positive integer k, we denote the elements of \({\textbf {x}}^k\) by

    $${\textbf {x}}^k = (x_1^k,x_2^k,x_3^k,\ldots ).$$

    Note that since all \({\textbf {x}}^k \in E_n(r,\rho )\), it follows that for each positive integer k, \(x_1^k < 1\) and \(x_n^k = 1\). Suppose now that \(x_1 =\displaystyle \lim _{k\rightarrow \infty }x_1^k = 1\). For each positive integer k, there exist \(a_{n-1}^k,\ldots ,a_2^k,a_1^k\in \{r,\rho \}\) and positive integers \(A_k,B_k\) such that

    $$1 = x_n^k = a_{n-1}^k\cdot \ldots \cdot a_2^ka_1^kx_1^k = x_1^k\cdot r^{A_k - B_k\cdot \frac{N}{M}}.$$

    It follows that for each positive integer k,

    $$x_1^k = \left( \frac{1}{r}\right) ^{A_k - B_k\cdot \frac{N}{M}}.$$

    Because of this,

    $$1 = \lim _{k\rightarrow \infty }x_1^k = \lim _{k\rightarrow \infty }\left( \frac{1}{r}\right) ^{A_k - B_k\cdot \frac{N}{M}} = \left( \frac{1}{r}\right) ^{\displaystyle \ \lim _{k\rightarrow \infty }\left( A_k - B_k\cdot \frac{N}{M}\right) }.$$

    This would imply that there exists a double sequence of positive integers \(\{A_k,B_k\}_{k=1}^\infty \) such that

    $$\lim _{k\rightarrow \infty }\left( A_k - B_k\cdot \frac{N}{M}\right) = \lim _{k\rightarrow \infty }\left( \frac{A_kM-B_kN}{M}\right) = 0$$

    and for each positive integer k,

    $$\frac{A_kM-B_kN}{M}\ne 0.$$

    Since M is fixed and \(A_kM-B_kN\) are all integers, this is not possible. This leads to a contradiction.

\(\square \)

Theorem 18

Let \((r,\rho )\in \mathcal {C}\). For each positive integer n, \(E_n(r,\rho )\) is a Cantor set.

Proof

Let n be any positive integer. Every element of \(E_n(r,\rho )\) can be uniquely represented by a useful sequence

$${\textbf {a}} = (a_1,a_2,a_3,\ldots ).$$

This representation is unique because different elements of \(E_n(r,\rho )\) are elements of different arcs in \(M_{r,\rho }\). With this in mind, let \(f:E_n(r,\rho )\rightarrow \{r,\rho \}^\mathbb {N}\) be a function such that for each \((x,a_1x,a_2a_1x,\ldots )\in E_n(r,\rho )\),

$$ f(x,a_1x,a_2a_1x,\ldots ) = (a_1,a_2,a_3,\ldots ). $$

This function is injective. To prove that it is continuous, let

$$ {\textbf {x}}^1 = (x_1^1,x_2^1,x_3^1,\ldots )\in E_n(r,\rho ) $$
$$ {\textbf {x}}^2 = (x_1^2,x_2^2,x_3^2,\ldots )\in E_n(r,\rho ) $$
$$ {\textbf {x}}^3 = (x_1^3,x_2^3,x_3^3,\ldots )\in E_n(r,\rho ) $$
$$ \vdots $$
$$ {\textbf {x}} = (x_1,x_2,x_3,\ldots )\in E_n(r,\rho ) $$

such that \(\displaystyle \lim _{k\rightarrow \infty }{} {\textbf {x}}^k = {\textbf {x}}\). We prove that \(\displaystyle \lim _{k\rightarrow \infty }f({\textbf {x}}^k) = f({\textbf {x}})\). For each positive integer k, there is a useful sequence \({\textbf {a}}^k = (a_1^k,a_2^k,a_3^k,\ldots )\) of r’s and \(\rho \)’s such that for each positive integer k,

$$ {\textbf {x}}^k = (x_1^k, a_1^kx_1^k, a_2^ka_1^kx_1^k,\ldots ) $$

and there is a useful sequence \({\textbf {a}} = (a_1,a_2,\ldots )\) of r’s and \(\rho \)’s such that

$$ {\textbf {x}} = (x_1,a_1x_1,_2a_1x_1,\ldots ). $$

It follows that \(\displaystyle \lim _{k\rightarrow \infty }x_1^k = x_1\). It follows from

$$ \lim _{k\rightarrow \infty }x_2^k = \lim _{k\rightarrow \infty }(a_1^kx_1^k) = a_1x_1 $$

that \(\displaystyle \lim _{k\rightarrow \infty }a_1^k = a_1\). Now, let \(n>1\) be a positive integer and for each positive integer \(l<n\), let

$$ \lim _{k\rightarrow \infty }a_l^k = a_l. $$

It follows from

$$ \lim _{k\rightarrow \infty }x_{n+1}^k = \lim _{k\rightarrow \infty }(a_n^ka_{n-1}^k\cdot \ldots \cdot a_2^ka_1^kx_1^k) = a_na_{n-1}\cdot \ldots \cdot a_2a_1x_1 $$

that \(\displaystyle \lim _{k\rightarrow \infty }a_n^k = a_n\). It follows that

$$ \lim _{k\rightarrow \infty }(a_1^k,a_2^k,a_3^k,\ldots ) = (a_1,a_2,a_3,\ldots ). $$

Therefore, \(\displaystyle \lim _{k\rightarrow \infty }f({\textbf {x}}^k) = f({\textbf {x}})\). So, f is continuous.

Since \(E_n(r,\rho )\) is homeomorphic to a subspace of \(\{r,\rho \}^\mathbb {N}\), it is also totally disconnected. Therefore, \(E_n(r,\rho )\) is a Cantor set. \(\square \)

Theorem 19

Let \((r,\rho )\in \mathcal {C},\) then \((0,0,0,\ldots ) \in Cl (E(r,\rho )).\)

Proof

For each positive integer n, let

$${\textbf {x}}^n = \left( \left( \frac{1}{\rho }\right) ^n,\left( \frac{1}{\rho }\right) ^{n-1},\ldots ,\left( \frac{1}{\rho }\right) ^2,\frac{1}{\rho },1,\ldots \right) \in E_{n+1}(r,\rho ).$$

It follows that \(\displaystyle \lim _{n\rightarrow \infty }{} {\textbf {x}}^n = (0,0,0,\ldots )\). \(\square \)

Observation 20

Let \((r,\rho )\in \mathcal {C}.\) The set \(E(r,\rho ) = \cup _{n\in \mathbb {N}}E_n(r,\rho )\) is not compact. This follows from Theorem 19. To show this using a different approach (proving also that \(E(r,\rho )\cup \{(0,0,0,\ldots )\}\) is not compact), let \((r,\rho )\in \mathcal {C}\) and let MN be positive integers such that

$$r^M\rho ^N = 1.$$

First, we define the following:

$${\textbf {x}} = (r^M,r^{M+1},\ldots ,r^{2M},r^{2M}\rho ,\ldots ,r^{2M}\rho ^{N-1},r^M) $$
$${\textbf {y}} = (r^M,r^M\rho ,\ldots ,r^M\rho ^{N-1},1,r,r^2,r^3,\ldots )$$

It is clear that each coordinate of \({\textbf {x}}\) is less than 1 since \(r<1\), \(\rho >1\), M is a positive integer and for each positive integer \(k<N\),

$$ r^{2M}\cdot \rho ^k< r^{2M}\cdot \rho ^N = r^M < 1. $$

For each positive integer k, let

$${\textbf {x}}^k = (\star _{i=1}^k{\textbf {x}})\star {\textbf {y}} = {\textbf {x}}\star {\textbf {x}}\star \ldots \star {\textbf {x}}\star {\textbf {y}} \in E(r,\rho ),$$

where \({\textbf {x}}\) is repeated k-times. It is clear then that

$$\lim _{k\rightarrow \infty }{} {\textbf {x}}^k = \star _{i=1}^\infty {\textbf {x}} = {\textbf {x}}\star {\textbf {x}}\star {\textbf {x}}\star \ldots \notin E(r,\rho ).$$

In the following definition, special sets \(F(M_{r,\rho })\) are defined. They play an important role in the proof of Theorem 21.

Definition 18

Let \((r,\rho )\in \mathcal {C}\) and let M,N be such positive integers that \(r^M\rho ^N = 1\). Let also \(\mathbb {Z}_0^-\) denote the set of all non-positive integers. We use \(F(M_{r,\rho })\) to denote the set

$$F(M_{r,\rho }) = \{(x_1,x_2,\ldots )\in M_{r,\rho }\ \mid \ x_1\ne 0,\ \forall z\in \mathbb {Z}_0^-:\ x_1 \ne \rho ^{\frac{z}{M}}\}.$$

Theorem 21

Let \((r,\rho )\in \mathcal {C}\) and let \({\textbf {x}}\in M_{r,\rho }\) be such a point that

$${\textbf {x}}\ne (0,0,0,\ldots )\ and \ {\textbf {x}}\notin E(r,\rho ).$$

Then \({\textbf {x}}\in M_{r,\rho }\setminus F(M_{r,\rho })\) if and only if there exists a sequence \(({\textbf {x}}^k)\) in \(E(r,\rho )\) such that

$${\textbf {x}} = \lim _{k\rightarrow \infty }{} {\textbf {x}}^k.$$

Proof

Let \((r,\rho )\in \mathcal {C}\) and let M, N be positive integers such that \(r^M\rho ^N = 1\), or, equivalently, that \(r = \rho ^{-\frac{N}{M}}\). We may assume that the greatest common divisor of M and N, denoted as \(GCD (M,N)\), equals 1.

First, let \({\textbf {x}} = (x_1,x_2,\ldots )\in M_{r,\rho }\setminus F(M_{r,\rho })\). By Definition 18 and since \({\textbf {x}}\ne (0,0,0,\ldots )\) there exists a non-positive integer \(z_1\) such that

$$ x_1 = \rho ^{\frac{z_1}{M}}. $$

Since for each positive integer k, \(x_{k+1}\in \{r\cdot x_k,\rho \cdot x_k\}\), it follows that \({\textbf {x}}\) is of the form:

$$ {\textbf {x}} = \left( \rho ^{\frac{z_1}{M}},\rho ^{\frac{z_2}{M}},\ldots \right) $$

where for each positive integer k, \(z_{k+1} \in \{z_k-N,z_k+M\}\) is a negative integer. Since \(GCD (M,N) = 1\), it follows from Euler’s theorem that there exist integers x,y such that

$$Mx+Ny = 1.$$

Since both M and N are positive integers, it is clear that either x or y is positive while the other is negative. Without loss of generality, let \(x>0\) and \(y<0\). Then for \(u = x\) and \(v = -y\), it follows that

$$Mu-Nv = 1.$$

Multiplying by \(-\frac{z_k}{M}\), we see that

$$-uz_k - vz_k\cdot \left( -\frac{N}{M}\right) = -\frac{z_k}{M}.$$

Note that \(-uz_k\) and \(-vz_k\) are both positive integers. With this in mind, for each positive integer k, let

$$ {\textbf {y}}^k = \left( \rho ^\frac{z_k}{M}, r\cdot \rho ^\frac{z_k}{M}, \ldots , r^{-vz_k}\cdot \rho ^\frac{z_k}{M},\rho \cdot r^{-vz_k}\cdot \rho ^\frac{z_k}{M},\ldots ,\rho ^{-uz_k-1}\cdot r^{-vz_k}\cdot \rho ^\frac{z_k}{M}, 1\right) $$

and

$$ {\textbf {x}}^k = p_{1,k}({\textbf {x}}) \star {\textbf {y}}^k \star (1,r,r^2,\ldots )\in E(r,\rho ). $$

It is then clear that

$$\lim _{k\rightarrow \infty }{} {\textbf {x}}^k = {\textbf {x}}.$$

This proves the first implication.

To prove the other implication, let \({\textbf {x}} = (x_1,x_2,\ldots )\) be the limit of a sequence \(({\textbf {x}}^k)\) in E and for each positive integer k, let \({\textbf {x}}^k = (x_1^k, x_2^k, x_3^k,\ldots )\). For each positive integer k, \({\textbf {x}}^k\in E\), therefore for each positive integer k there exists a useful sequence

$$ {\textbf {a}}^k = (a_1^k,a_2^k,\ldots ) $$

such that

$$ {\textbf {x}}^k = (x_1^k, a_1^kx_1^k, a_2^ka_1^kx_1^k,\ldots ) $$

and either \(x_1^k = 1\) or \(a_n^k\cdot \ldots \cdot a_1^kx_1^k = 1\) for some positive integer n. Also, for each positive integer k, let

$$ m_k = \max \left\{ \prod _{i=1}^na_i^k\ \mid \ n\in \mathbb {N}\right\} . $$

Since for each positive integer k, \({\textbf {a}}^k \in \{r,\rho \}^\mathbb {N}\) and \(r = \rho ^{-\frac{N}{M}}\), there exist positive integers \(A_k\),\(B_k\) such that

$$ m_k = \rho ^\frac{A_kM-B_kN}{M}. $$

Since \({\textbf {x}}^k\), for each positive integer k, is the end-point of \(L_{{\textbf {a}}^k}\), different from \((0,0,0,\ldots )\) it follows that

$$ x_1^k = \frac{1}{m_k} = \rho ^{\frac{B_kN-A_kM}{M}}. $$

It then also follows that

$$ x_1 = \lim _{k\rightarrow \infty }x_1^k = \lim _{k\rightarrow \infty }\rho ^{\frac{B_kN-A_kM}{M}} = \rho ^{\displaystyle \ \lim _{k\rightarrow \infty }\left( \frac{B_kN-A_kM}{M}\right) }. $$

Since \({\textbf {x}}\notin E(r,\rho )\), \(x_1\) must be less than 1 and so the limit of integers \(B_kN-A_kM\) must be some negative integer. By Definition 18, \({\textbf {x}}\notin F(M_{r,\rho })\). \(\square \)

Observation 22

Let n be a positive integer. For each \({\textbf {e}}\in E_n(r,\rho )\) let \(K_{\textbf {e}}\) be the straight line segment in the Hilbert cube from \((0,0,0,\ldots )\) to \({\textbf {e}}\). Note that

  1. 1.

    \( \displaystyle \bigcup _{{\textbf {e}}\in E_n(r,\rho )}K_{\textbf {e}} \subseteq M_{r,\rho } \) and \(\bigcup _{{\textbf {e}}\in E_n(r,\rho )}K_{\textbf {e}}\) is a Cantor fan. We denote \(F_n = \displaystyle \bigcup _{{\textbf {e}}\in E_n(r,\rho )}K_{\textbf {e}}\).

  2. 2.

    \( M_{r,\rho } = \displaystyle \bigcup _{n\in \mathbb {N}}F_n. \)

  3. 3.

    \(M_{r,\rho }\) is a fan which is a union of a countable family \(\mathcal {C}\) of Cantor fans such that

    1. 1.

      for each \(C_1,C_2\in \mathcal {C}\), if \(C_1\ne C_2\), then \(C_1\cap C_2 = \{(0,0,0,\ldots )\}\) and

    2. 2.

      the set of limit points of the set of end-points of F, forms in each arc from \((0,0,0,\ldots )\) to an end-point, a harmonic sequence.

    and \( E(M_{r,\rho }) = E(r,\rho ). \)

Finally, we are ready to state and prove Theorem 24, which is the main result of the paper. First, we introduce the following definitions and present some examples.

Definition 19

Let F be a fan with top v and let

$$ \mathcal {A}(F) = \{L\subseteq F\ \mid \ L\ is\ an\ arc\ in\ F\ from\ v\ to\ some\ point\ {\textbf {e}}\in E(F)\}. $$

For each \(L\in \mathcal {A}(F)\), we use \(S\!_{\!F}(L)\) to denote the set

$$ S\!_{\!F}(L) = \{x\in L\ \mid \ \exists (e_n)\ in\ E(F)\ such\ that\ \lim _{n\rightarrow \infty }e_n = x\}. $$

For each \(L\in \mathcal {A}(F)\), we call the elements of \(S\!_{\!F}(L)\) the end-point-significant points in L.

Example 2

Let F be the Lelek fan, then for each \(L\in \mathcal {A}(F),\)

$$S\!_{\!F}(L) = L.$$

Note that F is a Lelek fan if and only if for each \(L\in \mathcal {A}(F)\), \(S\!_{\!F}(L) = L.\)

Example 3

Let F be the Cantor fan with top v, then for each \(L\in \mathcal {A}(F),\)

$$S\!_{\!F}(L) = E(L)\setminus \{v\}.$$

Definition 20

Let \(X\subseteq [0,1]\) be closed and non-empty and let F be a fan with top v. We say that F is X-end-point-generated if for each \(L\in \mathcal {A}(F)\) there exists a homeomorphism \(\phi _L:[0,1]\rightarrow L\) such that

  1. 1.

    \(\phi _L(0) = v\), \(\phi _L(1) \in E(F)\) and

  2. 2.

    \(\phi _L(X) = S\!_{\!F}(L)\).

Observation 23

Let F be a fan. Then F is a Lelek fan if and only if it is [0, 1]-end-point-generated.

Finally, we prove our main result.

Theorem 24

Let \((r,\rho )\in \mathcal {C}\). Then \(M_{r,\rho }\) is a fan having the following two properties.

  1. 1.

    \(M_{r,\rho }\) is the union of a countable family \(\mathcal {F}\) of Cantor fans such that for all \(F_1\) and \(F_2\) from \(\mathcal {F}\),

    $$ F_1\ne F_2 \implies F_1\cap F_2 = \{(0,0,0,\ldots )\}, $$

    and

  2. 2.

    \(M_{r,\rho }\) is \(\left\{ \frac{1}{n}\ \mid \ n\in \mathbb {N}\right\} \cup \left\{ 0\right\} \)-end-point-generated.

Proof

Let \((r,\rho )\in \mathcal {C}\) and let MN be positive integers such that

$$r^M\rho ^N = 1.$$

We have seen from Observation 22 that \(M_{r,\rho }\) is a union of countably many Cantor fans satisfying 1. It follows (also using Theorem 21), that for each \(L\in \mathcal {A}(M_{r,\rho })\),

$$ S\!_{\!F}(L)\!=\! L\setminus F(M_{r,\rho })\!=\!\left\{ {\textbf {x}}\in L\ \mid \ there\ is\ z\in \mathbb {Z}_0^-\ such\ that\ p_1({\textbf {x}})\! =\! \rho ^{\frac{z}{M}}\right\} \cup \{(0,0,0,\ldots )\}. $$

Note that for each \(L\in \mathcal {A}(F)\), \(S\!_{\!F}(L)\) is homeomorphic to \(\left\{ \frac{1}{n}\ \mid \ n\in \mathbb {N}\right\} \cup \left\{ 0\right\} \). Therefore, \(M_{r,\rho }\) is \(\left\{ \frac{1}{n}\ \mid \ n\in \mathbb {N}\right\} \cup \left\{ 0\right\} \)-end-point-generated. \(\square \)

Observation 25

Let \((r,\rho )\in \mathcal {C}\) and let \(F:[0,1]\multimap [0,1]\) be the set-valued function such that its graph equals to \(\Gamma _F = L_{r,\rho }^{-1}\). F is then a USC set-valued function and the generalized inverse limit \(\mathop {\lim }\limits _ {\circ \hspace{-1.5pt}-\!\!-\!\!-}\{[0,1],F\}_{n=1}^\infty \) is homeomorphic to \(M_{r,\rho }\).

We conclude this paper by stating the following open problems.

Problem 26

Let F and G be fans, which are unions of countable families \(\mathcal {F}\) and \(\mathcal {G}\), respectively, of Cantor fans such that for all \(F_1, F_2\in \mathcal {F}\) and \(G_1,G_2\in \mathcal {G}\),

$$ F_1\ne F_2 \implies F_1\cap F_2 = \{(0,0,0,\ldots )\}\ and\ G_1\ne G_2 \implies G_1\cap G_2 = \{(0,0,0,\ldots )\}. $$

If F and G are both \(\left\{ \frac{1}{n}\ \mid \ n\in \mathbb {N}\right\} \cup \left\{ 0\right\} \)-end-point-generated, are then F and G homeomorphic?

Problem 27

Let X be a non-empty closed subset of [0, 1] and let F and G be fans, which are unions of countable families \(\mathcal {F}\) and \(\mathcal {G}\), respectively, of Cantor fans such that for all \(F_1, F_2\in \mathcal {F}\) and \(G_1,G_2\in \mathcal {G}\),

$$ F_1\ne F_2 \implies F_1\cap F_2 = \{(0,0,0,\ldots )\}\ and\ G_1\ne G_2 \implies G_1\cap G_2 = \{(0,0,0,\ldots )\}. $$

If F and G are both X-end-point-generated, are then F and G homeomorphic? If not, find all such X for which they are homeomorphic or find a non-trivial subset of such subsets \(X\subseteq [0,1]\).

Problem 28

Let X be a non-empty closed subset of [0, 1] and let F, G be fans such that they are X-end-point-generated. Are F and G homeomorphic? If not, find all such X for which they are homeomorphic or find a non-trivial family of such X.

Problem 29

Characterize all non-empty closed subsets X of [0, 1] for which there exists a fan F such that F is X-end-point-generated or find a non-trivial family of such sets X.

Problem 30

Does there exist a fan which is \(\left\{ \frac{1}{2}\right\} \)-end-point-generated?