On Entropy of Graph Maps That Give Hereditarily Indecomposable Inverse Limits
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Abstract
We prove that if \(f:G\rightarrow G\) is a map on a topological graph G such that the inverse limit \(\varprojlim (G,f)\) is hereditarily indecomposable, and entropy of f is positive, then there exists an entropy set with infinite topological entropy. When G is the circle and the degree of f is positive then the entropy is always infinite and the rotation set of f is nondegenerate. This shows that the AnosovKatok type constructions of the pseudocircle as a minimal set in volumepreserving smooth dynamical systems, or in complex dynamics, obtained previously by Handel, Herman and Chéritat cannot be modeled on inverse limits. This also extends a previous result of Mouron who proved that if \(G=[0,1]\), then \(h(f)\in \{0,\infty \}\), and combined with a result of Ito shows that certain dynamical systems on compact finitedimensional Riemannian manifolds must either have zero entropy on their invariant sets or be nondifferentiable.
Keywords
Topological entropy Attractor Hereditarily indecomposable continuum PseudocircleMathematics Subject Classification
37B40 37B45 37C701 Introduction
2 Preliminaries
2.1 Entropy
Topological entropy is one of the most common measures of complexity of dynamical systems. The reader not familiar with this notion is referred to [1, 47] for more details. Below we recall only the main definitions related to this notion.
Entropy pairs were introduced by Blanchard in [12] and later studied by various authors. The main feature related to entropy pairs is that the dynamical system has positive topological entropy if and only if it has an entropy pair. An important generalization of this concept are entropy sets, introduced in [22]. A set \(K\subset X\) with at least two points is an entropy set if for any finite open cover \(\mathcal {U}\) of X such that K is not contained in the closure of any of its elements we have \(h_{\text {top}}(f, \mathcal {U}) > 0\).
It is much easier to detect if a set is an entropy set, if we use the concept IEtuples (the terminology comes from [35], but the concept appeared earlier in the literature, e.g. [29, Theorem 8.2]). We call a tuple \(x=(x_1, \ldots , x_k)\in X^k\) an IEtuple if for every product neighborhood \(U_1\times \ldots \times U_k\) of x, the tuple \((U_1, \ldots ,U_k)\) has an independence set \(T\subset \mathbb {N}\) of positive upper density, that is for every finite subset \(P \subset T\) and function \(\zeta :P \rightarrow \left\{ 1, 2, \ldots , k\right\} \) we have \(\bigcap _{j\in P} f^{j}(U_{\zeta (j)})\ne \emptyset \). Recall that the upper density of a set \(A\subset \mathbb {N}\) is the number \( \limsup _{n\rightarrow \infty }\frac{1}{n}\# (A\cap [0,n)). \) It is known (see [13, 29, 35]) that a set K, which is not a singleton, is an entropy set iff every finite selection of its points \(x_1,\ldots , x_n\in K\) forms an IEtuple.
Note that the closure of an entropy set is an entropy set, so in this paper we will consider only closed entropy sets. By the same observation (and Zorn lemma), every entropy set is always contained in a maximal entropy set (in the sense of inclusion).
Positive entropy implies existence of entropy sets. It may happen however that there is no entropy set with entropy close to entropy of the map. The following simple example shows that entropy can be infinite, while every entropy set has finite entropy.
Example 2.1
Put \(I_n=[1/2^{n}, 1/2^{n1}]\) for \(n=1,2,\ldots \) and let \(f:[0,1]\rightarrow [0,1]\) be defined by \(f(0)=0\) and \(f_{I_n}\) is \((2n+1)\)fold map (see Fig. 2). Observe that f is continuous and \(h_{\text {top}}(f)=\infty \) since entropy of nfold map is \(\log n\).
But every entropy set of f must be contained completely in one of the intervals \(I_n\) and hence its entropy is finite.
2.2 Graphs and Horseshoes
A (topological) graph is a continuum G which can be represented as a union of finitely many arcs, any two of them having at most one point in common. Each topological graph is homeomorphic to a simplical complex embedded in the Euclidean space \({\mathbb {R}}^3\) (see [40]), hence we may consider each graph G endowed with the taxicab metric, that is, the distance between any two points of G is equal to the length of the shortest arc in G joining these points. A set \(I\subseteq G\) is a closed interval if there is a homeomorphism \(\varphi :[0,1]\rightarrow I\) such that the set \(\varphi ((0,1))\) is open in G. A path in a graph G is any map \(\omega : [0,1]\rightarrow G\). If \(\omega :[0,1]\rightarrow G\) is a homeomorphism on its image and \(q,r\in [0,1]\) then we say that \(\omega ([q,r])\) is an arc in G and we denote it by \(\left\langle a,b\right\rangle \), where \(a=\omega (q)\) and \(b=\omega (r)\). In such a case, without loss of generality, we shall always assume that \(\omega \) is an isometry and we shall write \(t_1<t_2\) (resp. \(t_1\le t_2\)) for \(t_1,t_2\in \omega ([0,1])\) if \(\omega ^{1}(t_1)<\omega ^{1}(t_2)\) (resp. \(\omega ^{1}(t_1)\le \omega ^{1}(t_2)\)). We say that the set \((a,b)=\left\langle a,b\right\rangle \setminus \left\{ a,b\right\} \) is an open arc. Note that sometimes (a, b) is not open in G hence it is not interior of \(\left\langle a,b\right\rangle \) in G (however it is an open set in \(\left\langle a,b\right\rangle \)).
Given a map \(f:G\rightarrow G\) and arcs I, J in G we say that I fcovers J if there exists an arc \(K \subseteq I\) such that \(f (K) = J\). Properties of fcovering relation presented below are adapted from [2, p. 590].
Lemma 2.2
 (1)
If I fcovers I, then there exists \(x\in I\) such that \(f(x)=x\).
 (2)
If \(I\subseteq K\), \(L\subseteq J\) and I fcovers J, then K fcovers L.
 (3)
If I fcovers J and J gcovers K, then I \((g\circ f)\)covers K.
 (4)
If \(J\subseteq f(I)\), and \(K_1,K_2\subseteq J\) are arcs such that \(K_1\cap K_2\) is at most one point, then I fcovers \(K_1\), or I fcovers \(K_2\).
Fix an integer \(s\ge 2\). An shorseshoe for f is an arc \(I\subseteq G\) and subarcs \(J_1,\ldots , J_s\) of I with pairwise disjoint interiors, such that \(f( S_j) = I\) for \(j = 1,\ldots , s\). An shorseshoe is strong if in addition all the arcs \(J_i\) are contained in the interior of I and are pairwise disjoint. It is not hard to show that if f has an shorseshoe for \(s\ge 4\) then it has an \((s2)\)strong horseshoe [36].
Definition 2.3
Let \(f:G\rightarrow G\) be a map of a graph G with metric d and let \(\varepsilon >0\). A path \(\omega :[0,1]\rightarrow G\) is \((f,\varepsilon )\)crooked, if there exist s and t with \(0<s \le t<1\) such that \(d(f \circ \omega (s),f \circ \omega (1) ) \le \varepsilon \) and \(d(f\circ \omega (t),f \circ \omega (0)) \le \varepsilon \) (see Fig. 3).
Remark 2.4
It is an immediate consequence of the definition, that if for some \(\varepsilon >0\) and \(n>0\) the path \(\omega :[0,1]\rightarrow G\) is \((f^n,\varepsilon )\)crooked then it is also \((f^{n+1},\varepsilon )\)crooked.
3 Graphs, Inverse Limits, Crookednes and Entropy Sets
The following result of Brown [18, Lemma 3] will be crucial in the proof of our main results.
Lemma 3.1
Let G be a topological graph. If the inverse limit \(\varprojlim (G,f)\) is hereditarily indecomposable then for every \(\delta >0\) there is \(n>0\) such that each path \(\omega :[0,1]\rightarrow G\) is \((f^n,\delta )\)crooked.
We will also need an important extension of the result of Misiurewicz and Szlenk first proved for interval maps, connecting topological horseshoes with positive topological entropy [36, Theorem B].
Lemma 3.2
If \(\left\langle a',b'\right\rangle \) is an arc, \(a'<a<b<b'\) and \(\varepsilon >0\) is such that \(d(a,a')\ge \varepsilon \) and \(d(b,b')\ge \varepsilon \) then, since distance on G is given by arc length, there are points \(a'\le a_\varepsilon \le a\) and \(b\le b_\varepsilon \le b'\) such that \(d(a,a_\varepsilon )=\varepsilon \) and \(d(b,b_\varepsilon )=\varepsilon \). Therefore we will use the following natural notation \(\left\langle a\varepsilon ,b+\varepsilon \right\rangle =\left\langle a_\varepsilon ,b_\varepsilon \right\rangle \).
Lemma 3.3
 (i)
\(g(l_i)\in \left\langle a\varepsilon ,a\right\rangle \), \(g(p_i)\in \left\langle b,b+\varepsilon \right\rangle \) for all \(1\le i\le k\), or
 (ii)
\(g(p_i)\in \left\langle a\varepsilon ,a\right\rangle \), \(g(l_i)\in \left\langle b,b+\varepsilon \right\rangle \) for all \(1\le i \le k\).
Proof
There are \(c\le l_1<p_k\le d\) such that \(g(l_1)=a\varepsilon /2\), \(g(p_k)=b+\varepsilon /2\) or \(g(p_k)=a\varepsilon /2\), \(g(l_1)=b+\varepsilon /2\).
We have now collected enough tools to prove that the entropy of maps with hereditarily indecomposable inverse limits cannot be bounded, because the size of horseshoes grows very fast when the iteration increases.
Lemma 3.4
Proof
Fix any integer \(\tilde{k}>0\) such that \(\frac{1}{n} \log (\tilde{k})>\alpha \). Denote \(g=f^{n}\) and \(r=\tilde{k}^2\). Without loss of generality we may assume that \(g(J)=L\). Therefore, there are \(\varepsilon >0\) and \(a<b\) such that \(J\subseteq (a,b)\subset (a2\varepsilon ,b+2\varepsilon )\subset L\). Fix any \(\delta <\varepsilon /6r\) and let s be provided by Lemma 3.1, that is each path in G is \((g^s,\delta )\)crooked. Arc J for g can be “iterated” for \(g^s\), that is we may replace J by its subinterval \(J'\) in such a way that \(g^s(J')=L\).
Let us apply Lemma 3.3 to \(\left\langle c',d'\right\rangle =J'\subset J=\left\langle c,d\right\rangle \) obtaining numbers \(c'\le l_1<p_1<\ldots <l_r<p_r\le d'\) and for simplicity of notation suppose that \(g^s(l_j)\in \left\langle a\varepsilon ,a\right\rangle \) and \(g^s(p_j)\in \left\langle b,b+\varepsilon \right\rangle \) for all j (the arguments for the symmetric case \(g^s(p_j)\in \left\langle a\varepsilon ,a\right\rangle \) and \(g^s(l_j)\in \left\langle b,b+\varepsilon \right\rangle \) are the same). Then, since we may view \(g^s :J' \rightarrow L\) as a map between subintervals of the real line, a simple argument shows that each arc \(\langle l_j,p_j\rangle \) \(g^s\)covers J. Hence, there are arcs \(J_{1},\ldots , J_{r}\) such that \(g^s(J_i)=J\) and since \(g(J)=L\) we obtain that each \(J_{i}\) \(g^{s+1}\)covers L. But each path in G is also \((g^{s+1},\delta )\)crooked, so we can continue this procedure recursively. After s steps, we obtain pairwise disjoint arcs \(J_{i_1,i_1,i_2,\ldots , i_s}\subseteq J\), \(i_1,\ldots ,i_s\in \left\{ 1,\ldots , r\right\} \), such that each arc \(J_{i_1,i_2,\ldots , i_s}\) \(g^{2s}\)cover L. Then there are at least \(r^s\) pairwise disjoint intervals which \(g^{2s}\)cover \(\left\langle a\varepsilon ,b+\varepsilon \right\rangle \).
Theorem 3.5
Let G be a topological graph. If the inverse limit \(\varprojlim (G,f)\) is hereditarily indecomposable and \(h_{top}(f)>0\) then there exists an entropy set A such that \(h_{top}(A)=\infty \).
Proof
Take any \(\alpha >0\) such that \(h_{top}(f)>\alpha \). By Lemma 3.2 there are \(s,k>0\) such that \(\frac{1}{s} \log (k)>\alpha \) and a strong \((k+1)\)horseshoe \(J_0,\ldots ,J_k\subset int L\) for \(f^s\).
Now we will present a construction from the proof of Proposition II.15 in [15]. While the construction is standard, we will need to make some essential modifications in two places, so we need to recall this construction briefly. For each \(a\in \left\{ 1,\ldots ,k\right\} \) write \(J(a)=J_a\). Since \(J_1,\ldots , J_k\) is a strong horseshoe for \(f^s\), for every \(a,b\in \left\{ 1,\ldots , k\right\} \) we can find an interval \(J(ab)\subset J(a)\) such that \(f(J(ab))=J(b)\).
By the above construction, for any sequence \(\xi \in \left\{ 1,\ldots , k\right\} ^\mathbb {N}\) the set \(A_\xi =\bigcap _{i=1}^\infty J(\xi _1\xi _2\ldots \xi _i)\) is a nonempty interval, because it is an intersection of a nested sequence of closed intervals. Let \(\Lambda \) be a set that consists of endpoints of all possible sets \(A_\xi \). That way we obtain an invariant set \(\Lambda \subset L\) for \(f^s\) and a continuous onto map \(\pi :\Lambda \rightarrow \Sigma _k\) such that \(\pi \) is onetoone for all but countably many points, \(\sigma :\Sigma _k\rightarrow \Sigma _k\) is the full shift on k symbols and \(\pi \circ f^s=\sigma \circ \pi \). Namely, \(\pi \) sends endpoints of \(A_\xi \) to \(\xi \) but it may happen that some (but at most countably many) intervals \(A_\xi \) are nondegenerate.
We have to make one important remark at this point. In the case that \(A_\xi \) is an interval, it could happen that one of its endpoints is isolated in \(\Lambda \). To remedy this situation we can take \(x\in \Lambda \) such that \(\pi (x)\) has dense orbit in \(\Sigma _k\) and \(\pi ^{1}\pi (x)\) is a singleton. Then x is recurrent, \(\hat{\Lambda }=\overline{\left\{ f^{sn}(x):n\ge 0\right\} }\) is a closed \(f^s\)invariant set without isolated points, and \(\pi (\hat{\Lambda })=\Sigma _k\). Therefore in what follows, we assume that sets \(\Lambda \) constructed by the use of horseshoe are without isolated point.
As a consequence of Lemma 3.4 we easily obtain the following result. It is a direct consequence of Theorem 3.5, but proving it this way would be overcomplicated.
Corollary 3.6
Let G be a topological graph. If the inverse limit \(\varprojlim (G,f)\) is hereditarily indecomposable and \(h_{\text {top}}(f)>0\) then \(h_{\text {top}}(f)=\infty \).
Proof
Corollary 3.6 has an important consequence for smooth dynamical systems on compact finitedimensional Riemannian manifolds, when combined with the following result of Ito.
Theorem 3.7
(Ito [31]) Let (M, g) be a compact ndimensional Riemannian manifold and \(F:M\rightarrow M\) a \(C^1\)diffeomorphism. Then \(h_{\text {top}}(F)\), the topological entropy of F, is finite.
Corollary 3.8
Let (M, g) be a compact ndimensional Riemannian manifold and \(F:M\rightarrow M\) be a homeomorphism with an invariant hereditarily indecomposable continuum X; i.e. \(F(X)=X\). If FX is conjugate to a shift homeomorphism on a graph inverse limit \(\varprojlim (G,f)\) then either \(h_{\text {top}}(F)=0\) or F is nondifferentiable.
4 CircleLike Hereditarily Indecomposable Continua and Entropy
There is a well known example of Henderson [27], who proved that shift homeomorphism on the pseudoarc can have zero topological entropy. For the pseudocircle, there is also an important example of Handel [26] who constructed a homeomorphism on the pseudocircle (extendable to the whole plane) with zero topological entropy. Related results in complex dynamics were obtained by Herman [28] and Chéritat [21]. The natural question is whether the example of [27] can be somehow generalized to obtain the pseudocircle as an inverse limit with one bonding map and with zero entropy. So far such an example was not constructed. We will show, that the reason for lack of such an example is very natural. Simply, such an example does not exist. This situation is in some sense similar to the case of Hénontype attractors. First, Williams showed that every hyperbolic, onedimensional, expanding attractor for a discrete dynamical system is topologically conjugate to the induced map on an inverse limit space based on a branched onemanifold [48]. Later Barge showed, however, that certain dynamical systems with Hénontype attractors cannot me modeled on inverse limits [5].

(\(\mathrm{deg}(f)=0\)) Knaster’s pseudoarc: the unique arclike hereditarily indecomposable continuum.

(\(\mathrm{deg}(f)=1\)) R.H. Bing’s pseudocircle: the unique plane separating hereditarily indecomposable circlelike continuum.

(\(\mathrm{deg}(f)>1\)) Nonplanar pseudosolenoids: an uncountable family of continua.
In what follows we will use some standard fact on rotation numbers and sets for maps on the circle. The reader not familiar with this topic is refereed to Chapter 3 in [1].
We will need the following result of Auslander and Katznelson [3], which extends the classification of periodicpointfree circle homeomorphisms, and implies that every periodicpointfree circle map is semiconjugate to an irrational rotation.
Theorem 4.1
 (1)
for every \(x\in {\mathbb {S}}^1\) there is an interval \(J_x=[a_x,b_x]\) such that \(x\in J_x\),
 (2)
\(f(J_x)\) is an interval with endpoints \(f(a_x)\) and \(f(b_x)\),
 (3)
\(f^m(x)\notin f(J_x)\) for \(m>1\),
 (4)
\(f^m(J_x)\cap J_{f^m(x)}\) for \(m=1,2,3,\ldots \),
 (5)
intervals \(J_{f^m(x)}\) for \(m=0,1,2,\ldots \) are pairwise disjoint,
 (6)
if \(f(x)=f(y)\) then \(J_x=J_{y}\),
 (7)
the sets \(\mathcal {J}=\{J_x:x\in {\mathbb {S}}^1\}\) form a decomposition of \({\mathbb {S}}^1\); i.e. \(J_x=J_y\) or \(J_x\cap J_y=\emptyset \) for all \(x,y\in {\mathbb {S}}^1\),
 (8)
for at most countably many x we have \(J_x\ne \{x\}\),
 (9)
\(J_x=J_y\) if and only if \(J_{f^n(x)}=J_{f^n(y)}\) for some (every) \(m=0,1,2,\ldots \).
Theorem 4.2
Suppose \(f:{\mathbb {S}}^1\rightarrow {\mathbb {S}}^1\) and \(\mathrm{Per}(f)=\emptyset \). Then \(\Lambda _f=\varprojlim ({\mathbb {S}}^1, f)\) is decomposable.
Proof
Let f be as above and let \(\mathcal {J}\) be the collection of intervals guaranteed by (7) in Theorem 4.1. Without loss of generality we may assume that there is an \(x\in {\mathbb {S}}^1\) such that \(J_x\ne \{x\}\), since otherwise by (6) in Theorem 4.1 f would be a homeomorphism and \(\Lambda _f={\mathbb {S}}^1\). Additionally observe that \(f^{n}(z)\ne \emptyset \) for every \(z\in {\mathbb {S}}^1\) as otherwise it is not hard to see that f has a fixed point.
First we claim that \(f^{n}(J_x)\) is an interval in \(\mathcal {J}\) for every n. To see this, fix any \(n>0\) and any \(y,z\in f^{n}(J_x)\). By Theorem 4.1 (4) we have \(f^n(J_y)\cap J_x\ne \emptyset \) and \(f^n(J_z)\cap J_x\ne \emptyset \) which by Theorem 4.1(7) gives \(f^n(J_y)=f^n(J_z)=J_x\). The proof of the claim is completed by Theorem 4.1 (9).
Although this is not essential for the reasoning in the present paper, note that periodicpointfree circle maps are of degree 1 (see [23]).
Theorem 4.3
If \(\Lambda _f=\varprojlim ({\mathbb {S}}^1, f)\) is hereditarily indecomposable and \(\Lambda _f\) is not the pseudoarc (i.e. \(\mathrm{deg}(f)\ne 0\)) then \(h_{\text {top}}(f)=\infty \).
Proof
Suppose \(\Lambda _f=\varprojlim ({\mathbb {S}}^1, f)\) is hereditarily indecomposable and \(\mathrm{deg}(f)>0\). Then \(\Lambda _f\) is either the pseudocircle or it is nonplanar. If \(\Lambda _f\) is nonplanar then \(\mathrm{deg}(f)=n\), for some \(n>1\) (see [45, Theorem 9]). Consequently \(h_{\text {top}}(f)\ge \log n\) by [1] and further, by Theorem 3.6, we obtain that \(h_{\text {top}}(f)=\infty \).
Let \({\hat{f}}:{\mathbb {R}}\rightarrow {\mathbb {R}}\) be a lift of f to the universal covering \(({\mathbb {R}},\tau )\), where \(\tau (t)=(\cos (t),\sin (t))\) and \({\mathbb {S}}^1\) is the unit circle in the plane, centered at the origin. It is known that \({\hat{f}}\) can be extended to \(\hat{\mathbb {R}}={\mathbb {R}}\cup \{+\infty ,\infty \}\), with \({\hat{f}}(\pm \infty )=\pm \infty \). Because \({\hat{{\mathbb {R}}}}\) is homeomorphic to [0, 1] and \(\varprojlim ({\mathbb {R}}, {\hat{f}})\) is a universal covering of \(\Lambda _f\) with the covering map given by \({\hat{\tau }}(x_1,x_2,x_3,\ldots )=(\tau (x_1),\tau (x_2),\tau (x_3),\ldots )\) (see [17, Lemma 3.8]), we get that \(\varprojlim ({\hat{{\mathbb {R}}}}, {\hat{f}})\) is the pseudoarc [10]. Since \(\mathrm{Fix}(f)\ne \emptyset \), \({\hat{f}}\) can be chosen in such a way so that \(\mathrm{Fix}({\hat{f}})\ne \emptyset \). Indeed, any lift of f is uniquely determined by specifying where it sends a single point, and if c is a fixed point of f then there is \({{\hat{c}}}\in [0,2\pi ]\) such that \(\tau ({{\hat{c}}})=c\), hence it is enough to set \({{\hat{c}}}={\hat{f}}({\hat{c}})\).
Because \({\hat{f}}\) has a fixed point in \([0,2\pi ]\) we get by (2) that the rotation set \(\rho ({\hat{f}})\) is nondegenerate. This implies in turn that \(h_{\text {top}}(f)>0\) (see [39]) and so, again by Theorem 3.6, we obtain that \(h_{\text {top}}(f)=\infty \) which ends the proof.\(\square \)
The following theorem follows from the above proof.
Theorem 4.4
Suppose \(f:{\mathbb {S}}^1\rightarrow {\mathbb {S}}^1\) is a map with \(\mathrm{deg}(f)=1\). If \(\Lambda _f=\varprojlim ({\mathbb {S}}^1, f)\) is hereditarily indecomposable then the rotation set \(\rho (f)\) is nondegenerate.
In [17] the authors showed that there exists a 2torus homeomorphism h homotopic to the identity with an attracting R.H. Bing’s pseudocircle C such that the rotation set of hC is not a unique vector. It follows from Theorem 4.3 that the topological entropy of this homeomorphism is infinite.
In [46] Rogers considered a special class of circlelike continua which he called selfentwined. Among the many results, he showed that the pseudocircle is selfentwined, any continuum in this class is indecomposable and no selfentwined continuum is a continuous image of the pseudoarc. Adapting the proof of Theorem 4.3 and properties of selfentwined continua sketched above, we get the following.
Theorem 4.5
Suppose \(\Lambda _f=\varprojlim ({\mathbb {S}}^1, f)\) is selfentwined. Then \(h(f)>0\).
Proof
Since \(\Lambda _f\) is indecomposable, by Theorem 4.2, f must have a periodic point. Again, without loss of generality assuming that f has a fixed point we can consider \({\hat{f}}:{\mathbb {R}}\rightarrow {\mathbb {R}}\), a lift of f to the universal covering \(({\mathbb {R}},\tau )\). Then \({\hat{f}}\) can be extended to \(\hat{\mathbb {R}}={\mathbb {R}}\cup \{+\infty ,\infty \}\), with \({\hat{f}}(\pm \infty )=\pm \infty \). Because \({\hat{{\mathbb {R}}}}\) is homeomorphic to [0, 1] and \(\varprojlim ({\mathbb {R}}, {\hat{f}})\) is a universal covering of \(\Lambda _f\) with the covering map given by \(\hat{\tau }(x_1,x_2,x_3,\ldots )=(\tau (x_1),\tau (x_2),\tau (x_3),\ldots )\) (see [17, Lemma 3.8]), we get that \(A=\varprojlim (\hat{\mathbb {R}}, {\hat{f}})\) is an arclike continuum. Now if we assume by contradiction that \(h(f)=0\) then f has a onepoint rotation set by [39]. Therefore, repeating arguments from the proof of Theorem 4.3, there is a closed \(\hat{f}\)invariant arc \(S\subset {\mathbb {R}}\) such that arclike continuum \(\Lambda _S=\varprojlim (S, {\hat{f}}_S)\subset A\) satisfies \(\hat{\tau }(\Lambda _S)=\Lambda _f\). In this case \(\Lambda _f\) would be a continuous image of an arclike continuum. But since every arclike continuum is a continuous image of the pseudoarc [24] this leads to a contradiction with Roger’s result that no selfentwined continuum is such an image and the proof is complete.\(\square \)
Notes
Acknowledgments
The authors are grateful to the anonymous referee for his careful reading of this paper and many valuable suggestions. The authors express many thanks to Henk Bruin for fruitful discussions during Workshops on Complexity and Dimension theory of Skew Products Systems held at the Erwin Schrödinger International Institute for Mathematical Physics (ESI) in Vienna, in September of 2013, as well as Henk Bruin’s visit at the AGH University of Science and Technology in Kraków, in November of 2013. The authors are grateful for the kind hospitality of the institute during their stay in Vienna. The authors acknowledge financial support of the ESI, the DFG network Grant Oe 538/31, OeAD (PL 02/2013) and MNiSW (AT 2/201315).
Boroński’s work was supported by the European Regional Development Fund in the IT4Innovations Center of Excellence Project (CZ.1.05/1.1.00/02.0070). Boroński also gratefully acknowledges the partial support from the MSK DT1 Support of Science and Research in the MoravianSilesian Region 2013 (RRC/05/2013). The research of Oprocha was supported by the Polish Ministry of Science and Higher Education from sources for science in the years 20132014, Grant no. IP2012 0 04272.
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