Connected neighborhoods in Cartesian products of solenoids

TL;DR

This paper proves that certain subcontinua in products of co-prime m-adic solenoids are ample, meaning they can be contained in open connected neighborhoods, contrasting with properties of Cartesian squares of fixed solenoids.

Contribution

It establishes that subcontinua with onto projections in products of co-prime solenoids are ample, answering a question and contrasting with known properties of fixed solenoid squares.

Findings

Subcontinua with onto projections are ample in product solenoids.

Such subcontinua can be contained in open connected neighborhoods.

Contrasts with non-ampleness of diagonals in Cartesian squares of fixed solenoids.

Abstract

Given a collection of pairwise co-prime integers $% m_{1},\ldots ,m_{r}$, greater than $1$, we consider the product $\Sigma =\Sigma _{m_{1}}\times \cdots \times \Sigma _{m_{r}}$, where each $\Sigma _{m_{i}}$ is the $m_{i}$-adic solenoid. Answering a question of D. P. Bellamy and J. M. \L ysko, in this paper we prove that if $M$ is a subcontinuum of $\Sigma $ such that the projections of $M$ on each $\Sigma _{m_{i}}$ are onto, then for each open subset $U$ in $\Sigma $ with $M\subset U$, there exists an open connected subset $V$ of $\Sigma $ such that $M\subset V\subset U$; i.e. any such $M$ is ample in the sense of Prajs and Whittington [10]. This contrasts with the property of Cartesian squares of fixed solenoids $\Sigma _{m_{i}}\times \Sigma _{m_{i}}$, whose diagonals are never ample [1].

Equations

$$f(\mathrm{Im}({\gamma }^{(n+1)}))\subset \mathrm{Im}({\gamma }^{(n)})$$

$$f(f^{-1}(\mathrm{Im}({\gamma }^{(n+1)})))\subset \mathrm{Im}({\gamma }^{(n+1)})\subset f^{-1}(\mathrm{Im}({\gamma }^{(n)})).$$

$$v_{N_0+N_1-1}\in L_{N_0+N_1-1},$$

$$v_{N_0+N_1-2}\in L_{N_0+N_1-2}$$

Full text

Key words and phrases: solenoids, connected neighborhoods, continuum, fupcon property, product.

Mathematics Subjects Classification: primary 54F15, secondary 54F50.

Introduction

In the present paper we study Cartesian products of Vietors solenoids. Recall that a (Vietoris) $n$-adic solenoid is a compact and connected space (i.e. continuum) given as the inverse limit of circles, with $n$-fold covering maps as bonding maps ($n>1$). Solenoids are indecomposable; i.e. they cannot be given as the union of two proper subcontinua. Since solenoids are topological groups they are homogeneous. Solenoids arise in the theory of dynamical systems as suspensions of odometers, and therefore minimal sets of suspension flows. Solenoids were also used by Smale to provide one of the first examples of hyperbolic attractors [11]. Williams [12] generlized these examples to higher dimensional analogues of solenoids, first constructed by McCord [8]. Here we study a condition under which subcontinua of products of solenoids have arbitrarily small connected open neighborhoods, called full projection implies arbitrary small connected open neighborhoods, or fupcon for short. Recall that, given a family of metric continua $\{X_{\alpha}:\alpha\in J\}$, the product $X=\mathop{\textstyle\prod}\nolimits_{\alpha\in J}X_{\alpha}$ has fupcon property, provided that for every subcontinuum $M$ and open subset $U$ of $X$ such that $M\subset U$ and $\pi_{\alpha}(M)=X_{\alpha}$ for each $\alpha\in J$ ($\pi_{\alpha}$ is the $\alpha^{\text{th}}$-projection), there exists an open connected subset $V$ of $X$ such that $M\subset V\subset U$. Clearly, each product of locally connected continua has fupcon property. A subcontinuum $M$ of a continuum $X$ is *ample *provided that for each open subset $U$ of $X$ with $M\subset U$, there exists a subcontinuum $L$ of $X$ such that $M\subset$ int${}_{X}(L)\subset L\subset U$. So $X$ is connected im kleinen at a point $p\in X$ provided that $\{p\}$ is ample. By [3, Lemma 1], the product $X=\mathop{\textstyle\prod}\nolimits_{\alpha\in J}X_{\alpha}$ has fupcon property provided that each subcontinuum $M$ of $X$ projecting onto each $X_{\alpha}$ is an ample subset of $X$. Ample subcontinua where introduced in [10] and they have been useful to improve the understanding of homogeneous continua. It is easy to show that if $M$ is an ample subcontinuum of a continuum $X$, then the hyperspace $C(X)$ of subcontinua of $X$ (with the Hausdorff metric) is connected im kleinen at $M$. Thus, if $X$ is a product with fupcon property, then it is possible to find subcontinua $M$ of $X$ at which $C(X)$ is connected im kleinen. This is something remarkable, since in some of the examples considered in this area (products of pseudo-arcs, solenoids and Knaster continua), of products with fupcon property, the properties of local connectedness are very rare. Below we list some of the known results related to fupcon property.

any product of Knaster continua has fupcon property [1, Theorem 4.1 and Observation in p. 230], 2. 2.

any product of pseudo-arcs has fupcon property [1, Theorem 4.4 and Observation in p. 230], 3. 3.

a product of a solenoid with itself does not have fupcon property [1, Corollary 4.3], 4. 4.

a product of the pseudo-arc with any product of Knaster continua has fupcon property [3, Corollary 9], 5. 5.

there exist indecomposable chainable continua whose product does not have fupcon property [3, Example 11], 6. 6.

if a product of continua has fupcon property, then each factor is a Kelley continuum [3, Theorem 10], 7. 7.

there is a complete characterization of chainable continua $X$ for which the diagonal in $X\times X$ has arbitrary small connected neighborhoods [4, Corollary 3.2] 8. 8.

every product of homogeneous continua having the fixed point property has fupcon property [5, Theorem 2.1], 9. 9.

any product of a solenoid and any Knaster continuum has fupcon property [2, Theorem 1.2] and [5, Theorem 3.1], 10. 10.

there exists a Kelley continuum $X$ such that $X\times[0,1]$ does not have fupcon property [5, Example 4.1], and 11. 11.

the product of a chainable Kelley continuum and $[0,1]$ has fupcon property [5, Theorem 5.4], 12. 12.

the product of two Smith’s nonmetric pseudo-arcs has the fupcon property [2, Theorem 3.1].

Answering a question of D. P. Bellamy and J. M. Łysko [1, Question (4)], in this paper we prove the following theorem.

Theorem. Let $m_{1},\ldots,m_{r}$ be a finite sequence of pairwise co-prime integers greater than $1$. For each $i\in\{1,\ldots,r\}$, let $\Sigma_{m_{i}}$ be the $m_{i}$-adic solenoid. Then $\Sigma=\Sigma_{m_{1}}\times\cdots\times\Sigma_{m_{r}}$ has the fupcon property.

Covering spaces

A mapping is a continuous function. A covering space of a space $Y$ is a pair $(X,\varphi)$, where $\varphi:X\rightarrow Y$ is a mapping and for each $y\in Y$, there exists an open neighborhood $U$ of $y$ in $Y$ such that $\varphi^{-1}(U)$ is a disjoint union of open sets of $X$ each of which is mapped homeomorphically onto $U$ by $\varphi$. Given a covering space $(X,\varphi)$ of the space $Y$ and a mapping $\psi:Z\rightarrow Y$, a *lifting *of $\psi$ to $X$ is a mapping $h:Z\rightarrow X$ such that $\psi=\varphi\circ h$. In the case that there exists such an $h$, we say that $\psi$ can be lifted. By $S^{1}$ we denote the unit circle (centered at the origin) in the Euclidean plane. By $\mathbb{R}$ we denote the real line. We consider the covering space $(\mathbb{R},e)$ of $S^{1}$, where $e$ is the exponential mapping $e:\mathbb{R}\rightarrow S^{1}$ defined by $e(t)=(\cos(2\pi t),\sin(2\pi t))$. We consider also covering spaces of the form $((S^{1})^{r},g)$ and $((S^{1})^{r},g^{n})$, where $(S^{1})^{r}$ is the Cartesian product $S^{1}\times\cdots\times S^{1}$ (with $r$ factors), $g:(S^{1})^{r}\rightarrow(S^{1})^{r}$ is a mapping of the form $g(z_{1},\ldots,z_{r})=(z_{1}^{k_{1}},\ldots,z_{r}^{k_{r}})$, where $k_{1},\ldots,k_{r}\in\mathbb{N}$ and $g^{n}$ is the composition $g\circ\cdots\circ g$ ($n$ times). The following lemma is an immediate consequence of Theorem 5.1 of [9].

**Lemma 1. **Let $U$ be a connected locally arcwise connected space, $u_{0}\in U$ and $g:U\rightarrow S^{1}$ a mapping that cannot be lifted to a mapping $\tilde{g}:U\to\mathbb{R}$, with $g=e\circ\tilde{g}$. Then there exists a loop $\gamma:[0,1]\rightarrow(U,u_{0})$ such that $g\circ\gamma:[0,1]\rightarrow(S^{1},g(u_{0}))$ is a non-trivial loop.

Loops in $(S^{1})^{r}$

From now on we fix a finite sequence of integers $m_{1},\ldots,m_{r}$ such that if $i\neq j$, then $m_{i}$ and $m_{j}$ are relatively prime, and $m_{i}\geq 2$ for each $i$. We also fix the mapping $f:(S^{1})^{r}\rightarrow(S^{1})^{r}$ given by

$f(z_{1},\ldots,z_{r})=(z_{1}^{m_{1}},\ldots,z_{r}^{m_{r}})$.

We denote by $1^{-}$ the point in $(S^{1})^{r}$ given by $1^{-}=((1,0),\ldots,(1,0))$. For each $i\in\{1,\ldots,r\}$, we denote by $\rho_{i}:(S^{1})^{r}\rightarrow S^{1}$ the $i^{\text{th}}$-projection.

A general construction

Given $k_{1},\ldots,k_{r}\in\mathbb{N}$, consider the mapping $g:(S^{1})^{r}\rightarrow(S^{1})^{r}$ given by $g(z_{1},\ldots,z_{r})=(z_{1}^{k_{1}},\ldots,z_{r}^{k_{r}})$. Then $((S^{1})^{r},g)$ is a covering space of the space $(S^{1})^{r}$. Given a loop $\gamma:[0,1]\rightarrow((S^{1})^{r},1^{-})$, define $\gamma^{\ast}:[0,\infty)\rightarrow(S^{1})^{r}$ by

$\gamma^{\ast}(t)=\gamma(t-i+1)$, if $t\in[i-1,i]$.

Since the fundamental group of $[0,\infty)$ is trivial, by Theorem 5.1 of [9], there exists a unique lifting $\gamma(g):[0,\infty)\rightarrow(S^{1})^{r}$ of the mapping $\gamma^{\ast}$ (in the covering space $((S^{1})^{r},g)$) such that $\gamma(g)(0)=1^{-}$. Then $\gamma(g)$ satisfies the following properties.

(a) For each $t\in[0,\infty)$, $g(\gamma(g)(t))=\gamma(t-k+1)$, if $t\in[k-1,k]$,

(b) $g(\mathop{\mathrm{I}m}(\gamma(g)))\subset\mathop{\mathrm{I}m}(\gamma)$.

For each $k\in\mathbb{N}$, $g(\gamma(g)(k))=\gamma(1)=1^{-}$. Then $\gamma(g)(\mathbb{N})\subset g^{-1}(1^{-})$. Thus, $\gamma(g)(\mathbb{N\cup}\{0\})\,$ is finite. Moreover, $\gamma(g)([k-1,k])=$ the image of the lifting of $\gamma$ that starts at the point $\gamma(g)(k-1)$. Thus:

(c) $\mathop{\mathrm{I}m}\gamma(g)$ is a subcontinuum of $(S^{1})^{r}$.

Notice that for each $n\in\mathbb{N}$, we can apply the above construction to the mapping $g=f^{n}$. Then for each loop $\gamma:[0,1]\rightarrow((S^{1})^{r},1^{-})$ and for each $n\in\mathbb{N}$, we can define the mapping $\gamma^{(n)}:[0,\infty)\rightarrow(S^{1})^{r}$ by

$\gamma^{(n)}=\gamma(f^{n})$.

Given $s_{0}=(s_{1},\ldots,s_{r})\in(\mathbb{Z}\smallsetminus\{0\})^{r}$ and $n\in\mathbb{N}\cup\{0\}$, let $\sigma(s_{0},n):[0,\infty)\rightarrow(S^{1})^{r}$ be given by

$\sigma(s_{0},n)(t)=(e(\frac{s_{1}t}{m_{1}^{n}}),\ldots,e(\frac{s_{r}t}{m_{r}^{n}}))$.

For each $s\in\mathbb{Z}\smallsetminus\{0\}$, let $\lambda_{s}:[0,1]\rightarrow S^{1}$ be given by

$\lambda_{s}(t)=e(st)$.

**Lemma 2. **Let $n,k\in\mathbb{N}$, $s_{0}=(s_{1},\ldots,s_{r})\in(\mathbb{Z}\smallsetminus\{0\})^{r}$ and $\gamma,\zeta:[0,1]\rightarrow((S^{1})^{r},1^{-})$ loops. Then

(d) $f\circ\gamma^{(n+1)}=\gamma^{(n)}$,

(e) $(\sigma(s_{0},0)|_{[0,1]})^{(n)}=(\sigma(s_{0},0)|_{[0,1]})(f^{n})=\sigma(s_{0},n)$,

(f) If $\gamma$ and $\zeta$ are homotopic, then $\gamma^{(n)}|_{[0,k]}$ and $\zeta^{(n)}|_{[0,k]}$ are homotopic, and $\gamma^{(n)}(k)=\zeta^{(n)}(k)$,

(g) If $\gamma:[0,1]\rightarrow((S^{1})^{r},1^{-})$ is a loop such that for each $i\in\{1,\ldots,r\}$, $\rho_{i}\circ\gamma$ is homotopic to $\lambda_{s_{i}}$, then $\gamma$ is homotopic to $\sigma(s_{0},0)|_{[0,1]}$.

**Proof. **(d) Given $t\in[0,\infty)$, suppose that $t\in[i-1,i]$ with $i\in\mathbb{N}$, then $f^{n}(f\circ\gamma^{(n+1)})(t)=f^{n+1}(\gamma^{(n+1)}(t))=f^{n+1}(\gamma(f^{n+1})(t))=\gamma(t-i+1)=\gamma^{\ast}(t)$ and $(f\circ\gamma^{(n+1)})(0)=f(\gamma(f^{n+1})(0))=f(1^{-})=1^{-}$. Thus, $f\circ\gamma^{(n+1)}$ is the unique lifting of $\gamma^{\ast}$ (in the covering space $((S^{1})^{r},f^{n})$) such that $(f\circ\gamma^{(n+1)})(0)=1^{-}$. Hence, $f\circ\gamma^{(n+1)}=\gamma^{(n)}$.

(e) Given $t\in[0,\infty)$, suppose that $t\in[i-1,i]$ with $i\in\mathbb{N}$. By the properties of the exponential mapping, we have that$\ f^{n}(\sigma(s_{0},n)(t))=f^{n}(e(\frac{s_{1}t}{m_{1}^{n}}),\ldots,e(\frac{s_{r}t}{m_{r}^{n}}))=(e(s_{1}t),\ldots,e(s_{r}t))=(e(s_{1}(t-(i-1))),\ldots,e(s_{r}(t-(i-1)))=((\sigma(s_{0},0)_{[0,1]})^{\ast})(t)$.

Then the mapping $\sigma(s_{0},n)$ is a lifting of the mapping $(\sigma(s_{0},0)|_{[0,1]})^{\ast}$ (in the covering space $((S^{1})^{r},f^{n})$) such that $\sigma(s_{0},n)(0)=1^{-}$. So $(\sigma(s_{0},0)|_{[0,1]})(f^{n})=\sigma(s_{0},n)$.

(f) Notice that $\gamma^{\ast}|_{[0,k]}$ and $\zeta^{\ast}|_{[0,k]}$ are homotopic loops in $((S^{1})^{r},1^{-})$. Notice also that $\gamma^{(n)}|_{[0,k]}$ and $\zeta^{(n)}|_{[0,k]}$ are liftings of $\gamma^{\ast}|_{[0,k]}$ and $\zeta^{\ast}|_{[0,k]}$ (in the covering space $((S^{1})^{r},f^{n})$), respectively such that $\gamma^{(n)}(0)=1^{-}=\zeta^{(n)}(0)$. By Lemma 3.3 in [9], $\gamma^{(n)}|_{[0,k]}$ and $\zeta^{(n)}|_{[0,k]}$ are homotopic and $\gamma^{(n)}(k)=\zeta^{(n)}(k)$.

(g) Is immediate.

**Theorem 3. **Let $s_{0}=(s_{1},\ldots,s_{r})\in(\mathbb{Z}\smallsetminus\{0\})^{r}$. Then there exists $N\in\mathbb{N}$ such that for each $n\geq N$ and every loop $\gamma:[0,1]\rightarrow((S^{1})^{r},1^{-})$ satisfying that for each $i\in\{1,\ldots,r\}$, $\rho_{i}\circ\gamma$ is homotopic to $\lambda_{s_{i}}$, we have that $f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(n)}))=\mathop{\mathrm{I}m}(\gamma^{(n+1)})$ and $f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(n)}))$ is a subcontinuum of $(S^{1})^{r}$.

**Proof. **Let $\alpha_{1},\ldots,\alpha_{r}\in\mathbb{N}\cup\{0\}$ be such that for each $i\in\{1,\ldots,r\}$, $s_{i}=m_{i}^{\alpha_{i}}q_{i}$, where $q_{i}$ and $m_{i}$ are relative prime. Let $\alpha=\max\{\alpha_{1},\ldots,\alpha_{r}\}$, $N=m_{1}^{\alpha}\cdots m_{r}^{\alpha}$ and $n\geq N$. Let $\gamma:[0,1]\rightarrow(S^{1})^{r}$ be a loop such that for each $i\in\{1,\ldots,r\}$, $\rho_{i}\circ\gamma$ is homotopic to $\lambda_{s_{i}}$.

By Lemma 2 (d), $f\circ\gamma^{(n+1)}=\gamma^{(n)}$. Thus, $\mathop{\mathrm{I}m}\gamma^{(n+1)}\subset f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(n)}))$.

By Lemma 2 (g), $\gamma$ is homotopic to $\sigma(s_{0},0)|_{[0,1]}$. By Lemma 2, (f) and (f), for each $k\in\mathbb{N}$, $\gamma^{(n+1)}(k)=(\sigma(s_{0},0)|_{[0,1]})^{(n+1)}(k)=\sigma(s_{0},n+1)(k)$. Since $\gamma^{(n+1)}(0)=1^{-}=\sigma(s_{0},n+1)(0)$, we conclude that $\gamma^{(n+1)}(k)=\sigma(s_{0},n+1)(k)$ for each $k\in\mathbb{N}\cup\{0\}$.

Notice that $f^{-1}(1^{-})=\{(e(\frac{j_{1}}{m_{1}}),\ldots,e(\frac{j_{r}}{m_{r}})):$ for each $i\in\{1,\ldots,r\}$, $0\leq j_{i}<m_{i}\}$. We claim that $f^{-1}(1^{-})\subset\gamma^{(n+1)}(\mathbb{N})$. For proving this, take $i\in\{1,\ldots,r\}$ and $j_{i}\in\{0,\ldots,m_{i}-1\}$. Set $j^{-}=(j_{1},\ldots,j_{r})$.

For each $i\in\{1,\ldots,r\}$, let $\beta_{i}=n-\alpha_{i}$. Set $u=m_{1}^{\beta_{1}}\cdots m_{r}^{\beta_{r}}$ and $u_{i}=\frac{u}{m_{i}^{\beta_{i}}}$. Then $\frac{us_{i}}{m_{i}^{n+1}}=\frac{um_{i}^{\alpha_{i}}q_{i}}{m_{i}^{n+1}}=\frac{u_{i}q_{i}}{m_{i}}$. Notice that $u_{i}q_{i}$ and $m_{i}$ are relative prime. By the Chinese Remainder Theorem, there exists $x\in\mathbb{N}$ such that for each $i\in\{1,\ldots,r\}$,

$u_{i}q_{i}x\equiv j_{i}(\mathop{\mathrm{m}od}m_{i})$.

Then $e(\frac{us_{i}x}{m_{i}^{n+1}})=e(\frac{u_{i}q_{i}x}{m_{i}})=e(\frac{j_{i}}{m_{i}})$ for each $i\in\{1,\ldots,r\}$. Set $k=ux$. Then $k\in\mathbb{N}$ and

$\gamma^{(n+1)}(k)=\sigma(s_{0},n+1)(k)=\sigma(s_{0},n+1)(ux)=(e(\frac{s_{1}ux}{m_{1}^{n+1}}),\ldots,e(\frac{s_{r}ux}{m_{r}^{n+1}}))=(e(\frac{j_{1}}{m_{1}}),\ldots,e(\frac{j_{r}}{m_{r}}))$.

We have shown that $f^{-1}(1^{-})\subset\gamma^{(n+1)}(\mathbb{N})$.

Take a point $v\in f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(n)}))$. Then $f(v)=\gamma^{(n)}(t_{0})$ for some $t_{0}\in[0,\infty)$. Consider the path $\gamma^{(n)}|_{[0,t_{0}]}$ in $(S^{1})^{r}$. Since $((S^{1})^{r},f)$ is a covering space of $(S^{1})^{r}$, there exists a lifting $\zeta:[0,t_{0}]\rightarrow(S^{1})^{r}$ of $\gamma^{(n)}|_{[0,t_{0}]}$ such that $\zeta(t_{0})=v$. Since $f\circ\zeta=\gamma^{(n)}|_{[0,t_{0}]}$, we have $f(\zeta(0))=\gamma^{(n)}(0)=1^{-}$, so $\zeta(0)\in f^{-1}(1^{-})\subset\gamma^{(n+1)}(\mathbb{N})$. Then there exists $k\in\mathbb{N}$ such that $\zeta(0)=\gamma^{(n+1)}(k)$.

Consider the mapping $\omega:[0,k+t_{0}]\rightarrow(S^{1})^{r}$ given by:

\omega(t)=\left\{\begin{array}[]{cc}\gamma^{(n+1)}(t)\text{,}&\text{if }t\in[0,k]\text{,}\\ \zeta(t-k)\text{,}&\text{if }t\in[k,k+t_{0}]\text{.}\end{array}\right.

Given $t\in[0,k+t_{0}]$, if $t\leq k$, by the definition of $\gamma^{(n+1)}$, $f^{n+1}(\omega(t))=f^{n+1}(\gamma^{(n+1)}(t))=f^{n+1}((\gamma(f^{n+1}))(t))=\gamma(t-j+1)$, if $t\in[j-1,j]$ ($j\in\mathbb{N}$). If $k\leq t$, by the definition of $\gamma^{(n)}$, $f^{n+1}(\omega(t))=f^{n+1}(\zeta(t-k))=f^{n}(\gamma^{(n)}(t-k))=\gamma(t-k-j+1)$, if $t-k\in[j-1,j]$. That is, $f^{n+1}(\omega(t))=\gamma(t-k-j+1)$, if $t\in[k+j-1,k+j]$.

We have shown that for each $t\in[0,k+t_{0}]$, $f^{n+1}(\omega(t))=\gamma(t-i+1)$, if $t\in[i-1,i]$ ($i\in\mathbb{N}$). Since $\gamma^{(n+1)}$ also satisfies that for each $t\in[0,k+t_{0}]$, $f^{n+1}(\gamma^{(n+1)}(t))=\gamma(t-i+1)$, if $t\in[i-1,i]$ and $\omega(0)=\gamma^{(n+1)}(0)$, we conclude that $\omega=\gamma^{(n+1)}|_{[0,k+t_{0}]}$.

In particular, $\gamma^{(n+1)}(k+t_{0})=\omega(k+t_{0})=\zeta(t_{0})=v$. Therefore, $v\in\mathop{\mathrm{I}m}(\gamma^{(n+1)})$. This completes the proof that $\mathop{\mathrm{I}m}(\gamma^{(n+1)})=f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(n)}))$.

Finally, by (c), $\mathop{\mathrm{I}m}(\gamma^{(n+1)})$ is a subcontinuum of $(S^{1})^{r}$. $\blacksquare$

Fupcon property

For each $i\in\{1,\ldots,r\}$, we consider the solenoid $\Sigma_{m_{i}}$ which is the inverse limit of the inverse sequence $\{S^{1},f_{i}\}$, where $f_{i}:S^{1}\rightarrow S^{1}$ is defined as before. That is $f_{i}(z)=z^{m_{i}}$. We consider the Cartesian product

$\Sigma=\Sigma_{m_{1}}\times\cdots\times\Sigma_{m_{r}}$

For each $i\in\{1,\ldots,r\}$, let $\eta_{i}:\Sigma\rightarrow\Sigma_{m_{i}}$ be the $i^{\text{th}}$-projection. We identify $\Sigma$ with $\lim_{\leftarrow}((S^{1})^{r},f)$, where $f$ is defined as before. That is

$f(z)=(f_{1}(z),\ldots,f_{r}(z))$.

For each $n\in\mathbb{N}$, let $\pi_{n}:\Sigma\rightarrow(S^{1})^{r}$ be the projection given by:

$\pi_{n}(z_{1},z_{2},\ldots)=z_{n}$, where each $z_{i}\in(S^{1})^{r}$.

**Lemma 4. **Let $M$ be a subcontinuum of $\Sigma$ such that the point $p_{0}=(1^{-},1^{-},\ldots)$ belongs to $M$ and $\eta_{i}(M)=\Sigma_{m_{i}}$ for each $i\in\{1,\ldots,r\}$. Then for each $n\in\mathbb{N}$ and each $i\in\{1,\ldots,r\}$, $(\rho_{i}\circ\pi_{n})|_{M}:M\rightarrow S^{1}$ cannot be lifted (in the covering space $(\mathbb{R},e)$).

**Proof. **Suppose to the contrary that $(\rho_{i}\circ\pi_{n})|_{M}$ can be lifted for some $n\in\mathbb{N}$ and $i\in\{1,\ldots,r\}$. Then there exists a mapping $h:M\rightarrow\mathbb{R}$ such that $e\circ h=(\rho_{i}\circ\pi_{n})|_{M}$. Since $(\rho_{i}\circ\pi_{n})(p_{0})=(1,0)$, we assume that $h(p_{0})=0$.

Suppose that $i=1$, the other cases are similar. For each $j\in\{1,\ldots,m_{1}\}$, define $g_{j}:M\rightarrow\mathbb{R}$ by $g_{j}(p)=\frac{h(p)+j-1}{m_{1}}$. Let $K_{j}=\{p\in M:\rho_{1}(\pi_{n+1}(p))=e(g_{j}(p))\}$. Then $K_{j}$ is closed in $M$.

Given $p=(p_{1},p_{2},\ldots)\in M$, where for each $k\in\mathbb{N}$, $p_{k}=(p_{1}^{(k)},\ldots,p_{r}^{(k)})$, we have $f(p_{n+1})=p_{n}$, so $(\rho_{1}(\pi_{n+1}(p)))^{m_{1}}=(p_{1}^{(n+1)})^{m_{1}}=p_{1}^{(n)}=(\rho_{1}\circ\pi_{n})(p)=e(h(p))=(e(\frac{h(p)}{m_{1}}))^{m_{1}}$. This implies that $\frac{\rho_{1}(\pi_{n+1}(p))}{e(\frac{h(p)}{m_{1}})}$ is an $m_{1}$-root of the unity in the complex plane $\mathbb{C}$. Thus, there exists $j\in\{1,\ldots,m_{1}\}$ such that $\rho_{1}(\pi_{n+1}(p))=e(\frac{h(p)}{m_{1}})e(\frac{j-1}{m_{1}})=e(\frac{h(p)+j-1}{m_{1}})$. Hence, $p\in K_{j}$. We have shown that $M=K_{1}\cup\cdots\cup K_{m_{1}}$.

Take $j,l\in\{1,\ldots,m_{1}\}$ such that $j\neq l$. If there exists $p\in K_{j}\cap K_{l}$, then $e(g_{j}(p))=\rho_{1}(\pi_{n+1}(p))=e(g_{l}(p))$. This implies that $\frac{h(p)+j-1}{m_{1}}=\frac{h(p)+l-1}{m_{1}}+k$ for some $k\in\mathbb{Z}$. Then $h(p)+j-1=h(p)+l-1+km_{1}$, and $j-l=km_{1}$. This contradicts the fact that $j\neq l$ and $j,l\in\{1,\ldots,m_{1}\}$. We have shown that $K_{1},\ldots,K_{m_{1}}$ are pairwise disjoint.

Since $\rho_{1}(\pi_{n+1}(p_{0}))=(1,0)=e(\frac{h(p_{0})}{m_{1}})$, we obtain that $K_{1}\neq\emptyset$. The connectedness of $M$ implies that $K_{1}=M$. Thus, for each $p\in M\,$, $\rho_{1}(\pi_{n+1}(p))=e(\frac{h(p)}{m_{1}})$. Let $h_{1}=\frac{1}{m_{1}}h$. Then $h_{1}$ is a lifting of $\rho_{1}\circ\pi_{n+1}$ such that $h_{1}(p_{0})=\frac{h(p_{0})}{m_{1}}=0$. If we repeat the argument with $h_{1}$ instead of $h$, we can obtain that for each $p\in M$, $\rho_{1}(\pi_{n+2}(p))=e(\frac{h(p)}{m_{1}^{2}})$.

Proceeding this way we can obtain that for each $a\in\mathbb{N}$, $\rho_{1}(\pi_{n+a}(p))=e(\frac{h(p)}{m_{1}^{a}})$.

Since $M$ is compact, there exists $a_{0}\in\mathbb{N}$ such that diameter$(\{\frac{h(p)}{m_{1}^{a_{0}}}:p\in M\})<1$. Thus, $\rho_{1}(\pi_{n+a_{0}}(M))=\{e(\frac{h(p)}{m_{1}^{a_{0}}}):p\in M\}\neq S^{1}$. Therefore, there exists a point $w_{n+a_{0}}\in S^{1}\smallsetminus\rho_{1}(\pi_{n+a_{0}}(M))$. Take $w_{0}\in\Sigma_{m_{1}}$ be any point of the form

$w_{0}=(w_{1},\ldots,w_{n+a-1},w_{n+a},w_{n+a+1},\ldots)$.

Then no point $p$ in $M$ satisfies $\eta_{1}(p)=w_{0}$. This contradiction finishes the proof of the lemma. $\blacksquare$

**Lemma 5. **Let $M$ be a subcontinuum of $\Sigma$ such that the point $p_{0}=(1^{-},1^{-},\ldots)$ belongs to $M$ and $\eta_{i}(M)=\Sigma_{m_{i}}$ for each $i\in\{1,\ldots,r\}$. Then for each $n\in\mathbb{N}$, and each open connected subset $U$ of $(S^{1})^{r}$ with $\pi_{n}(M)\subset U$, there exists a loop $\gamma$ in $(U,1^{-})$ such that for each $i\in\{1,\ldots,r\}$, the loop $\rho_{i}\circ\gamma:[0,1]\rightarrow S^{1}$ cannot be lifted in the covering space $(\mathbb{R},e)$.

**Proof. **By Lemma 4, $(\rho_{1}\circ\pi_{n})|_{M}:M\rightarrow S^{1}$ cannot be lifted. This implies that $\rho_{1}|_{\pi_{n}(M)}:\pi_{n}(M)\rightarrow S^{1}$ cannot be lifted, and then $\rho_{1}|_{U}:U\rightarrow S^{1}$ cannot be lifted.

By Lemma 1, there exists a loop $\gamma_{1}:[0,1]\rightarrow(U,1^{-})$ such that $\rho_{1}\circ\gamma_{1}:[0,1]\rightarrow(S^{1},(1,0))$ is a non-trivial loop. Thus, there exists $s\in\mathbb{Z}\smallsetminus\{0\}$ such that $\rho_{1}\circ\gamma_{1}$ is homotopic to $\lambda_{s}$.

Suppose, inductively, that $1\leq k<r$ and we have constructed a loop $\gamma_{k}:[0,1]\rightarrow(U,1^{-})$ such that for each $i\in\{1,\ldots,k\}$, the mapping $\rho_{i}\circ\gamma_{k}:[0,1]\rightarrow(S^{1},(1,0))$ is homotopic to $\lambda_{s_{i}}$ for some $s_{i}\in\mathbb{Z}\smallsetminus\{0\}$. Suppose that $\rho_{k+1}\circ\gamma_{k}$ (which is a loop in $(S^{1},(1,0))$) is homotopic to $\lambda_{s_{k+1}}$ for some $s_{k+1}\in\mathbb{Z}$.

Using Lemma 1 again, we have that there exists a loop $\zeta:[0,1]\rightarrow(U,1^{-})$ such that $\rho_{k+1}\circ\zeta:[0,1]\rightarrow(S^{1},(1,0))$ is homotopic to $\lambda_{s_{0}}$ for some $s_{0}\in\mathbb{Z}\smallsetminus\{0\}$.

For each $i\in\{1,\ldots,k\}$, let $t_{i}\in\mathbb{Z}$ be such that $\rho_{i}\circ\zeta$ is homotopic to $\lambda_{t_{i}}$. Let $l\in\mathbb{N}$ be such that $\max\{\left|s_{1}\right|,\ldots,\left|s_{k}\right|,\left|s_{k+1}\right|\}<l\leq l\left|s_{0}\right|$.

Let $\gamma_{k+1}=\zeta\ast\cdots\ast\zeta\ast\gamma_{k}:[0,1]\rightarrow U$, where $\zeta$ appears $l$ times in $\zeta\ast\cdots\ast\zeta$. Then $\gamma_{k+1}$ is a loop in $U$.

Given $i\in\{1,\ldots,k+1\}$, $\rho_{i}\circ\gamma_{k+1}=(\rho_{i}\circ\zeta)\ast\cdots\ast(\rho_{i}\circ\zeta)\ast(\rho_{i}\circ\gamma_{k})$ is homotopic to $\lambda_{u_{i}}$ for some $u_{i}\in\mathbb{Z}$.

If $i\leq k$, then $u_{i}=lt_{i}+s_{i}$. In the case that $t_{i}=0$, then $u_{i}=s_{i}\neq 0$, and in the case that $t_{i}\neq 0$, then $l\left|t_{i}\right|\geq l>\left|s_{i}\right|$, so $u_{i}\neq 0$.

If $i=k+1$, then $u_{i}=ls_{0}+s_{k+1}\neq 0$.We have shown that $u_{i}\neq 0$ for every $i\in\{1,\ldots,k+1\}$. This finishes the induction and the proof of the lemma. $\blacksquare$

Theorem 6. $\Sigma=\Sigma_{m_{1}}\times\cdots\times\Sigma_{m_{r}}$ has the fupcon property.

**Proof. **Let $M$ be a subcontinuum of $\Sigma$ such that for each $i\in\{1,\ldots,r\}$, $\eta_{i}(M)=\Sigma_{m_{i}}$. Let $\varepsilon>0$. We are going to show that there exists a connected open subset $V$ of $\Sigma$ such that $M\subset V$ and $V$ is contained in the $\varepsilon$-neighborhood of $M$ in the space $\Sigma$.

Since $\Sigma$ is homogeneous, we may assume that $p_{0}=(1^{-},1^{-},\ldots)\in M$. Let $N_{0}\in\mathbb{N}$ be such that $\frac{1}{2^{N_{0}}}<\frac{\varepsilon}{2}$.

By the uniform continuity of $f$, there exists $\delta>0$ such that $\delta<\varepsilon$ and the following implication holds: if $i\in\{1,\ldots,N_{0}\}$ and $z,w\in(S^{1})^{r}$ are such that $\left|z-w\right|<\delta$, then $\left|f^{i}(z)-f^{i}(w)\right|<\frac{\varepsilon}{2}$.

Let $W$ be the $\delta$-neighborhood of $\pi_{N_{0}}(M)$ in the space $(S^{1})^{r}$.

By Lemma 5, there exists a loop $\gamma_{0}$ in $(W,1^{-})$ such that for each $i\in\{1,\ldots,r\}$, the loop $\rho_{i}\circ\gamma_{0}:[0,1]\rightarrow(S^{1},(1,0))$ cannot be lifted in $(\mathbb{R},e)$. Therefore, there exists $s_{i}\in\mathbb{Z}\smallsetminus\{0\}$ such that $\rho_{i}\circ\gamma_{0}$ is homotopic to $\lambda_{s_{i}}$.

Let $N_{1}\in\mathbb{N}$ be as in Theorem 3 such that $N_{1}>N_{0}$. Let $Y$ be the component of $f^{-N_{1}}(W)$ that contains $\pi_{N_{0}+N_{1}}(M)$. Then $Y$ is a connected open subset of $(S^{1})^{r}$ and $1^{-}\in Y$.

Let $V=\pi_{N_{0}+N_{1}}^{-1}(Y)$. Then $V$ is open in $\Sigma$ and $M\subset V$. We are going to show that $V$ is connected.

Take $v=(v_{1},v_{2},\ldots)\in V$. Then $v_{N_{0}+N_{1}}\in Y$. Let $\eta:[0,1]\rightarrow(S^{1})^{r}$ be a lifting (in the covering space $((S^{1})^{r},f^{N_{1}})$) of the loop $\gamma_{0}$ such that $\eta(0)=1^{-}$. Then $f^{N_{1}}(\eta(1))=\gamma_{0}(1)=1^{-}$. Since $f^{N_{1}}(\mathop{\mathrm{I}m}(\eta))\subset\mathop{\mathrm{I}m}(\gamma_{0})\subset W$, we have $\mathop{\mathrm{I}m}(\eta)\subset f^{-N_{1}}(W)$, and since $1^{-}\in\mathop{\mathrm{I}m}(\eta)\cap\pi_{N_{0}+N_{1}}(M)$, we conclude that $\mathop{\mathrm{I}m}(\eta)\subset Y$.

Since $Y$ is arcwise connected, there exists a path $\mu:[0,1]\rightarrow Y$ such that $\mu(0)=\eta(1)$ and $\mu(1)=v_{N_{0}+N_{1}}$.

Let $\omega$ be the mapping $\eta\ast\mu\ast\mu^{-1}:[0,1]\rightarrow(S^{1})^{r}$ and let $\gamma:[0,1]\rightarrow(S^{1})^{r}$ be the loop $f^{N_{1}}\circ\omega:[0,1]\rightarrow((S^{1})^{r},1^{-})$. Notice that $\mathop{\mathrm{I}m}(\gamma)\subset W$, $\gamma$ is homotopic to $f^{N_{1}}\circ\eta=\gamma_{0}$. By the choice of $N_{1}$, for each $n\geq N_{1}$, $f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(n)}))$ is a subcontinuum of $(S^{1})^{r}$.

Notice that $\omega$ is a lifting of the loop $\gamma$ (in the covering space $((S^{1})^{r},f^{N_{1}})$) satisfying $\omega(0)=\eta(0)=1^{-}$. On the other hand, $\gamma^{(N_{1})}|_{[0,1]}$ is a lifting (in $((S^{1})^{r},f^{N_{1}})$) of $\gamma$ with $\gamma(f^{N_{1}})(0)=1^{-}$. Thus, $\omega=\gamma^{(N_{1})}|_{[0,1]}$. Therefore, $\mathop{\mathrm{I}m}(\omega)\subset\mathop{\mathrm{I}m}(\gamma^{(N_{1})})$. In particular, $v_{N_{0}+N_{1}}\in\mathop{\mathrm{I}m}(\gamma^{(N_{1})})$.

By (b) $f^{N_{1}}(\mathop{\mathrm{I}m}(\gamma(f^{N_{1}})))\subset\mathop{\mathrm{I}m}(\gamma)$, so $\mathop{\mathrm{I}m}(\gamma^{(N_{1})})=\mathop{\mathrm{I}m}(\gamma(f^{N_{1}}))\subset f^{-N_{1}}(W)$. Moreover, since $(\gamma^{(N_{1})})(0)=1^{-}$, we have that $\mathop{\mathrm{I}m}(\gamma^{(N_{1})})\subset Y$.

Consider the sequence $\{L_{n}\}_{n=1}^{\infty}$ of subcontinua of $(S^{1})^{r}$ defined in the following way.

$L_{1}=f^{N_{0}-1}(\mathop{\mathrm{I}m}(\gamma))$, $L_{2}=f^{N_{0}-2}(\mathop{\mathrm{I}m}(\gamma))$, … , $L_{N_{0}-1}=f(\mathop{\mathrm{I}m}(\gamma))$, $L_{N_{0}}=\mathop{\mathrm{I}m}(\gamma)$,

$L_{N_{0}+1}=\mathop{\mathrm{I}m}(\gamma^{(1)})$, $L_{N_{0}+2}=\mathop{\mathrm{I}m}(\gamma^{(2)})$, $L_{N_{0}+3}=\mathop{\mathrm{I}m}(\gamma^{(3)})$, … , $L_{N_{0}+N_{1}}=\mathop{\mathrm{I}m}(\gamma^{(N_{1})})$,

$L_{N_{0}+N_{1}+1}=f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(N_{1})}))$, $L_{N_{0}+N_{1}+2}=f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(N_{1}+1)}))$, $L_{N_{0}+N_{1}+3}=f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(N_{1}+2)}))$, …

Claim.

(1) $\{(L_{n},f)\}_{n=1}^{\infty}$ is an inverse sequence of continua,

(2) the inverse limit $L_{0}$ of the sequence $\{(L_{n},f)\}_{n=1}^{\infty}$ is a subcontinuum of $\Sigma$ containing $v$, contained in $V$ and intersecting $M$.

(3) $V$ is contained in the $\varepsilon$-neighborhood of $M$ in the space $\Sigma$.

Proof of (1). By (c), $L_{1},\ldots,L_{N_{0}+N_{1}}$ are subcontinua of $(S^{1})^{r}$. The choice of $N_{1}$ implies that $L_{N_{0}+N_{1}+1}$, $L_{N_{0}+N_{1}+2}$, … are also continua.

Clearly, for each $1<n\leq N_{0}$, $f(L_{n})=L_{n-1}$. By (b), $f(L_{N_{0}+1})\subset L_{N_{0}}$.

By Lemma 2 (e), for each $n\in\mathbb{N}$,

[TABLE]

and

[TABLE]

So, for each $n\in\{N_{0}+1,\ldots,N_{0}+N_{1}-1\}\cup\{N_{0}+N_{1}+1,N_{0}+N_{1}+2,\ldots\}$, $f(L_{n+1})\subset L_{n}$. Clearly, $f(L_{N_{0}+N_{1}+1})\subset L_{N_{0}+N_{1}}$. We have shown that for each $n\in\mathbb{N}$, $f(L_{n+1})\subset L_{n}$. This completes the proof of (1).

**(2). **By the choice of $N_{1}$, $L_{N_{0}+N_{1}+2}=f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(N_{1}+1)}))=f^{-2}(\mathop{\mathrm{I}m}(\gamma^{(N_{1})}))$, $L_{N_{0}+N_{1}+3}=f^{-3}(\mathop{\mathrm{I}m}(\gamma^{(N_{1})}))$, $L_{N_{0}+N_{1}+4}=f^{-4}(\mathop{\mathrm{I}m}(\gamma^{(N_{1})}))$,…

Since $v_{N_{0}+N_{1}}\in\mathop{\mathrm{I}m}(\gamma^{(N_{1})})$ and $v\in\Sigma$, $v_{N_{0}+N_{1}+1}\in f^{-1}(\mathop{\mathrm{I}m}(\gamma^{(N_{1})}))=L_{N_{0}+N_{1}+1}$, $v_{N_{0}+N_{1}+2}\in f^{-2}(\mathop{\mathrm{I}m}(\gamma^{(N_{1})}))=L_{N_{0}+N_{1}+2}$ and so on. Moreover, by (1),

[TABLE]

[TABLE]

and so on. Therefore, $v\in L_{0}$.

By (b), $f^{N_{1}}(\mathop{\mathrm{I}m}(\gamma^{(N_{1})}))=f^{N_{1}}(\mathop{\mathrm{I}m}(\gamma(f^{N_{1}})))\subset\mathop{\mathrm{I}m}(\gamma)\subset W$. So, $\mathop{\mathrm{I}m}(\gamma^{(N_{1})})\subset f^{-N_{1}}(W)$. Recall that $1^{-}\in\mathop{\mathrm{I}m}(\gamma^{(N_{1})})$. This implies that $L_{N_{0}+N_{1}}=\mathop{\mathrm{I}m}(\gamma^{(N_{1})})\subset Y$. Thus, $L_{0}\subset\pi_{N_{0}+N_{1}}^{-1}(Y)=V$.

Since $f(1^{-})=1^{-}$ and $\gamma(0)=f^{N_{1}}(\omega(0))=1^{-}$, it follows that $1^{-}\in L_{n}$ for all $n\in\mathbb{N}$. Therefore, $(1^{-},1^{-},\ldots)\in L_{0}\cap M$.

**(3). **Take an element $w=(w_{1},w_{2},\ldots)\in V$. Then $w_{N_{0}+N_{1}}\in Y$ and $w_{N_{0}}=f^{N_{1}}(w_{N_{0}+N_{1}})\in W$.

Since $W$ is the $\delta$-neighborhood of $\pi_{N_{0}}(M)$ in $(S^{1})^{r}$, there is $z=(z_{1},z_{2},\ldots)\in M$ such that $\left|w_{N_{0}}-z_{N_{0}}\right|<\delta<\varepsilon$. This implies that for each $i\in\{1,\ldots,N_{0}-1\}$, $\left|f^{i}(w_{N_{0}})-f^{i}(z_{N_{0}})\right|<\frac{\varepsilon}{2}$. Thus, for each $i\in\{1,\ldots,N_{0}\}$, $\left|w_{i}-z_{i}\right|<\frac{\varepsilon}{2}$. Since $\frac{1}{2^{N_{0}}}<\frac{\varepsilon}{2}$, we conclude that the distance in $\Sigma$ from $w$ to $z$ is less than $\varepsilon$.

Property (2) implies that $V$ is connected. By property (3) we conclude that $\Sigma$ has the fupcon property. $\blacksquare$

Open problems

Below we list some related open problems.

• (Q1)

Does the product of two chainable Kelley continua have the fupcon property [3, Question 12]?

• (Q2)

Suppose $X$ and $Y$ are $1$-dimensional continua such that $X\times Y$ has the fupcon property, does the product $X_{P}\times Y_{P}$ have the fupcon property [2, Question 4]?, $X_{p}$ is a $1$-dimensional continuum that admits a continuous decomposition into pseudo-arcs, and whose decomposition space is homeomorphic to $X$ [7]?

• (Q3)

Does the product of $[0,1]$ and the pseudo-circle have the fupcon property [2, Question 5]?

• (Q4)

Does the product of a pseudo-arc and pseudo-circle have the fupcon property [2, Question 6]?

• (Q5)

Does the product of two pseudo-circles have the fupcon property [2, Question 7]?

• (Q6)

Cartesian products of which matchbox manifolds have the fupcon property?

Acknowledgments

The research in this paper was carried out during the 12th Research Workshop in Hyperspaces and Continuum Theory held in the city of Querétaro, México, during June, 2018. The authors would like to thank Jorge M. Martínez-Montejano who joined the discussion on the topic of this paper.

This paper was partially supported by the project ”Teoría de Continuos, Hiperespacios y Sistemas Dinámicos III” (IN106319) of PAPIIT, DGAPA, UNAM. The first author was also supported by University of Ostrava grant lRP201824 ”Complex topological structures” and the NPU II project LQ1602 IT4Innovations excellence in science.

References

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[2] J. P. Boroński, D. R. Prier, M. Smith, Ample continua in Cartesian products of continua, Topology Appl. 238 (2018), 54-58.

[3] A. Illanes, Connected open neighborhoods in products, Acta Math. Hungar. 148 (1) (2016), 73-82.

[4] A. Illanes, Small connected neighborhoods containing the diagonal of a product, Topology Appl. 230 (2017), 506-516.

[5] A. Illanes, J. M. Martínez-Montejano and K. Villarreal, Connected neighborhoods in products, Topology Appl. 241 (2018), 172-184.

[6] A. Illanes and S. B. Nadler, Jr., Hyperspaces, Fundamentals and Recent Advances, Monographs and Textbooks in Pure and Applied Math., Vol. 216, Marcel Dekker, Inc., New York, Basel, 1999.

[7] W. Lewis, Continuous curves of pseudo-arcs, Houston J. Math. 11 (1985), 91-99.

[8] C. McCord, Inverse limit sequences with covering maps, Trans. Amer. Math. Soc. 114 (1965),197–209

[9] W. S. Massey, Algebraic Topology, An Introduction, Graduate Texts in Mathematics, v. 56, Springer-Verlag, New York, Heidelberg, Berlin, 1967.

[10] J.R. Prajs and K. Whittington, Filament sets, aposyndesis, and the decomposition theorem of Jones, Trans. Amer. Math. Soc. 359 (2007), 5991-6000.

[11] S. Smale, Differentiable dynamical systems, Bull. Amer. Math. Soc., 73 (1967), 747–817.

[12] R.F. Williams, Expanding attractors, Publ. Math. I.H.E.S. 43 (1974), 169–203

Addresses

Boroński: National Supercomputing Centre IT4Innovations, Division of the University of Ostrava, Instituto for Research and Applications of Fussy Modeling, 30. Dubna 22, 701 03 Ostrava. Czech Republic, and Faculty of Applied Mathematics, AGH University of Science and Technology, al. Mickiewicza 30, 30-059 Kraków, Poland.

Illanes: Instituto de Matemáticas, Universidad Nacional Autónoma de México, Circuito Exterior, Cd. Universitaria, México, D.F., 04510, México.

Márquez: Departamento de Matemáticas, Facultad de Ciencias, Universidad Nacional Autónoma de México, Circuito Exterior, Cd. Universitaria, México, D.F., 04510, México

Boroński: jan.boronski@osu.cz

Illanes: illanes@matem.unam.mx

Márquez: emanuelrmarquez@outlook.com