TREE-LIKE CONTINUA THAT ADMIT POSITIVE ENTROPY HOMEOMORPHISMS ARE NON-SUSLINEAN

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1 TREE-LIKE CONTINUA THAT ADMIT POSITIVE ENTROPY HOMEOMORPHISMS ARE NON-SUSLINEAN CHRISTOPHER MOURON Abstract. t is shown that if X is a tree-like continuum and h : X X is a homeomorphism such that the topological entropy of h is greater than 0, then X must be non-suslinean. 1. Introduction In dynamics, topological entropy is a number in [0, ] that gives a measure of the chaotic behavior of a function on a space. The connection between entropy and the dynamics of a continuous function is well documented. Many simple functions such as the tent map have positive entropy. However, for a homeomorphism to have positive entropy, it appears that the local structure must be complex. For example, in [5] it is shown that no homeomorphism of a regular continuum can have positive entropy. A continuum X is a compact connected metric space. A continuum is regular if it has a neighbor base of open sets with finite boundary. In [*] it is shown that every chainable continuum that admits a positive entropy homeomorphism must contain a nondegenerate indecomposable subcontinuum. A continuum X is decomposable provided there exist proper subcontinua H and K such that X = H K. A continuum is indecomposable if it is not decomposable. Indecomposable continua are known to have a very complicated local structure. However, the Cantor fan is a continuum that admits a positive entropy homeomorphisms but does contain an indecomposable subcontinuum. The Cantor fan is the Cartesian product of the Cantor set with an arc with the arcs identified at one endpoint. However, the Cantor fan is non-suslinean and tree-like. A continuum is non-suslinean if it contain an uncountable number of disjoint nondegenerate subcontinua. It will be shown that every tree-like continuum that admits a positive entropy homeomorphism is non- Suslinean. If U is an open cover, the mesh of U is defined as mesh(u) = sup{diam(u) : U U}. The nerve of U, denoted N U is a geometric simplex where each element U i U is represented by a vertex v i N U and there exists a arc from v i to v j if and only if U i U j. A cover T is a tree cover if is nerve is a tree. A continuum is tree-like if for every ɛ > 0 there exists an open tree cover such that mesh(t ) < ɛ. A tree cover T is taut provided that for every A, B T if A B =, then A B =. Finally, a chain C is an indexed collection of open sets {C 1, C 2,..., C n } such that C i C j if and only if i j 1. Here, C 1 and C n are the endlinks of the chain. Let X be a compact metric space, f : X X be a map, and U be a finite open cover of X. Define N(U) be the number of sets in a finite subcover of U with smallest 2000 Mathematics Subject Classification. Primary: 54H20, 54F50, Secondary: 54E40. Key words and phrases. entropy, tree-like continuum, non-suslinean. 1

2 2 C. MOURON cardinality. If U and V are two open covers of X, let U V = {U V U U, V V} and f 1 (U) = {f 1 (U) U U}. Also, define n 1 i=0 f i (U) = U f 1 (U)... f n+1 (U), where f 0 = id and Ent(f, U, X) = lim n (1/n) log N( n 1 i=0 f i (U)). Then the topological entropy of f is defined as Ent(f) = sup{ent(f, U, X) : U is an open cover of X}. If U and V are finite open covers of X such that for every V V there exists a U U such that V U, then V refines U. If for every V V there exists a U U such that V U, then V closure refines U. The following propositions are well known and can be found in several texts such as [3] and [6]. Proposition 1. If V, U are open covers of X such that V refines U, then Ent(f, V) Ent(f, U). Proposition 2. For each positive integer k, Ent(f k ) = kent(f). 2. The Entropy of Words and Collections of Open Sets For a more complete treatment on the entropy of words see [4]. An alphabet A is a finite set of symbols. A word is a sequence composed from the elements of A. Words can be finite, W n = A i n i=1, or infinite, W = A i i=1. The length of a word is the number of elements in the sequence. Define π k ( A i i=1 ) = A k and Π k ( A i i=1 ) = A i k i=1. Likewise, define π k( A i n i=1 ) = A k and Π k ( A i n i=1 ) = A i k i=1 if the word is finite and k n. If W is a word, then Π k(w ) is called the prefix of length k of W. If W n is a word of length n then (W n ), A is a new word of length n + 1 formed by adding the symbol A to the end of W n. W(A) (W when there is no confusion) will represent some collection of infinite words on A. If B A, then W(B) = { B i i=1 B i B}. Likewise, W n will represent some collection of words of length n. Next, define Π k (W) = {Π k (W ) W W}. Π k (W n ) can be defined in a similar manner provided n k. Let B A. A collection of words W is complete on subalphabet B provided that every possible word composed from B is in W. That is, if B i B, then B i i=1 W. Likewise, W k is complete on B provided if B i B, then B i k i=1 W k. Clearly, if W is complete on B, then W(B) is complete on B. Also notice that if W(B) is complete on B, then W(B) takes on the structure of a B ary tree. In this section, we will use the structure of words to aid in finding results on the entropy of self maps. Let f : X X be an onto continuous function and let A be a finite open cover of X. Cover A can be thought of as an alphabet where the elements of A are the symbols. Inverse images of f will then be used to create the words. Let f : X X and A be an open cover on X. A finite sequence A 1, A 2,..., A i of elements A k A is a (f, A)-word of length i provided A 1 f 1 (A 2 )... f i+1 (A i )

3 TREE-LIKE 3 Let W i (f, A) be the set of all (f, A)-words of length i and W(f, A) = {W Π i (W ) W i (f, A)}. Next, suppose that B is a finite collection of open sets that do not necessarily cover X. Let G = {g(i)} i=1 be an increasing sequence of non-negative integers. Then define B i k i=1 to be a (f, B, G)-word of length k provided B i B and f g(1) (B 1 ) f g(2) (B 2 )... f g(k) (B k ). Let W k (f, B, G) be a collection of all (f, B, G)-words of length k. Then define W W(f, B, G) = (W(f, B)) G provided Π k (W ) W k (f, B, G) for all k. Let ENT(f, B, G) = lim sup i and define the entropy of (f, B) by log Π i (W(f, B, G)), g(i) ENT(f, B) = sup{ent(f, B, G)}. G Also, (W(f, B)) G and W(f, B, G) are themselves collections of words and ENT(W(f, B, G)) ENT(W(f, B), G) = ENT(f, B, G). Proposition 3. If ENT(f, {A, B}) > 0 and B C, then ENT(f, {A, C}) > 0. Proof. Let G be an increasing sequence of non-negative integers. Define r(a) = A and r(b) = C. Then given any A 1,..., A k W(f, {A, B}, G) it follows that r(a 1 ),..., r(a k ) W(f, {A, C}, G). Hence Thus, W(f, {A, C}, G) W(f, {A, B}, G). 0 < ENT(f, {A, B}) ENT(f, {A, C}). Let K n = (k 1,..., k n ) be a collection of positive integers (not necessarily distinct). Define R Kn (B) recursively in the following way: Let D 0 = B and given D j 1 define D j = D j 1 f kj (D j 1 ). Let R Kn (B) = D n. Notice that if B is disjoint, then R Kn (B) must be disjoint. Also R Kn (B) = B 2n. The next result is the Theorem 19 of [*]. Theorem 4. Suppose that A, B are open sets such that ENT(f, {A, B}) > 0. Then for each positive integer n, there exists a finite set of positive integers K n = {k i } n i=1, r 1, and G = {g(i)} i=1 such that g(i) ri and W(f, R K n {A, B}, G) is complete on R Kn ({A, B}). Furthermore, for every distinct A, B R Kn ({A, B}), ENT(f, {A, B }) > 0. The following theorems are Corollaries 24 and 25 respectively from [*]. Theorem 5. Suppose ENT(f, {A, B}) > 0, B is a collection of open sets such that B B. Then there exists B B such that ENT(f, {A, B }) > 0. Theorem 6. Suppose ENT(f, {A, B}) > 0, A, B are collections of open sets such that A A and B B. Then there exists A A and B B such that ENT(f, {A, B }) > 0.

4 4 C. MOURON The next theorem is a generalization of Theorem 27 in [*]. To prove the generalization just replace the phrase chain cover with the phrase tree cover in the proof of Theorem 27. Theorem 7. If T is a tree cover such that ENT(f, T ) > log(2), then there exist disjoint open sets A, B of C such that ENT(f, {A, B}) > 0. Let U be collection of open sets. C U is a cover component of U if for every A, B C there exists a chain from A to B completely contained in C. Let C(U) be the collection of cover components of U. Corollary 8. Let A, B T such that Ent(h, {A, B}) > 0 and let T be a refinement of T such that T = T. Then there exists B T such that B B and Ent(h, {A, B}) > 0. Proof. Let B = {B T B B}. Then B = B so the corollary follows from Theorem 5. Theorem 9. Suppose that W is an collection of infinite words of the form α i i=1, where α i = {1, 2}. If for every W k Π k (W) there exists a n > k and W 1 n, W 2 n Π n (W) such that Π k (W 1 n) = Π k (W 2 n) = W k, π n (W 1 n) = 1 and π n (W 2 n) = 2, then W is uncountable. Proof. It can easily be shown that there exists a subset W of W that can be mapped in a 1-1 way with the Cantor set i=1 {1, 2}. Lemma 10. Suppose that U is a finite open cover of X, A, B, C and D are disjoint elements of U and there exist subchains {C 1, C 2, C 3, C 4 } of U such that 1) C 1 has endlinks A and B, 2) C 2 has endlinks C and D, 3) C 3 has endlinks A and C, 4) C 4 has endlinks B and D. Furthermore, suppose that C 1 C 2 = and that C 3 C 4 =. Then U cannot be a tree cover. Proof. Let 1) C 1 = [C1, 1 C2, 1..., Cn 1 1 ] where A = C1 1 and B = Cn 1 1, 2) C 2 = [C1, 2 C2, 2..., Cn 2 2 ] where C = C1 2 and D = Cn 2 2, 3) C 3 = [C1, 3 C2, 3..., Cn 3 3 ] where A = C1 3 and C = Cn 3 1, 4) C 4 = [C1, 4 C2, 4..., Cn 4 4 ] where B = C1 4 and D = Cn 4 4. Notice that if there exist indices i < j < k such that C p i, Cp k C q and C p j C q for p q, then U must contain a circle-chain. So suppose that this cannot happen. Let p 1,3, p 2,3, p 1,4 and p 2,4 be indices such that (1) Cp 1 1,3 C 3, Cp 2 2,3 C 3, Cp 1 1,4 C 4 and Cp 2 2,4 C 4 (2) p 1,3 p 1,4 and p 2,3 p 2,4 are minimized. It follows that p 1,3 p 1,4 and that p 2,3 p 2,4. Define C 3 to be subchain of C 3 from Cp 1 1,3 to Cp 2 2,3 and C 4 to be subchain of C 4 from Cp 1 1,4 to Cp 2 2,3. Then [Cp 1 1,3,..., Cp 1 1,4 ] C 4 [Cp 1 2,3,..., Cp 1 2,4 ] C 4 is a circle-chain. Hence U cannot be tree-like. Let T be a tree-like continuum and T be a taut tree-cover of T such that there exists A, B T such that

5 TREE-LIKE 5 (1) ENT(h, {A, B}) > 0 (2) The unique chain from A to B has at least 5 links. Since T is taut, it follows that A B. Let U be an element in the chain from A to B such that U (A B) =. Suppose that T 1, T 2 are collections of open sets that both refine T and whose nerve contains no simple closed curve. Then we say that T 1 T2 if (1) T 1 T 2 = (2) There exists A T 1, B T 2 such that A B A B and Ent(h, {A, B }) > 0 (3) If T C(T 1 ) C(T 2 ) then T U. We say that T 2 splits T 1 into T 1,1 and T 1,2 if there exists refinements T 1,1, T 1,2 of T 1 such that (1) T 1,1 T 1,2 = (2) T 1,1 T 2 (3) T 1,2 T 2 (4) mesh(t 1,1 T 1,1 ) < (1/2)mesh(T 1 ) Let T (T i ) = {A T there exists B T i such that A B}. Proposition 11. Suppose that X is a tree-like continuum, T is a taut tree cover of X and U T. Let T be any taut tree-like refinement of T. If T 1 is a component of T (T {U}), then T1 U. Proof. Let T 1 be a component of T (T {U}). Let A T 1 and B T such that B U. Since T is a connected tree cover, there exists a unique chain [C 1, C 2,..., C n ] from A to B in T where C 1 = A and C n = B. Let k be the maximal index such that C k T. It follows that C k+1 U. Otherwise C k+1 T (T {U}) and since C k C k+1 it would follow that C k+1 T 1 which contradicts the maximality of k. Thus, C k U. So T1 U. Lemma 12. If T 1 T 2, then one of the following must be true: 1) T 2 splits T 1 into T 1,1, T 1,1. 2) T 1 splits T 2 into T 2,1, T 2,1. Proof. Suppose T is a tree-cover of X, h : X X is a homeomorphism and T 1, T 2 T such that T 1 T 2. Then there exists A T 1 and B T 2 such that ENT(h, {A, B}) > 0. Let k be such that D 1 = A h k (A), D 2 = A h k (B), D 3 = B h k (A), and D 4 = B h k (B) are all nonempty and ENT(h, {D i, D j }) > 0 for all i j as prescribed by Theorem 4. Let λ be the Lesbegue number for T. Since h is uniformly continuous, there exists a γ > 0 such that if d(x, y) < γ then d(h i (x), h i (y)) < λ for all 2k i 2k. Let T be a taut tree-cover that refines T and mesh(t ) < min{γ, (1/2)mesh(T ). Then by Theorem 6, for each i j there exist disjoint D 1,4, D 2,3, D 3,2, D 4,1 T such that D i,j D i and ENT(h, {D 1,4, D 4,1 }) > 0 and ENT(h, {D 2,3, D 3,2 }) > 0. Furthermore it follows from Proposition 3 that ENT(h, {D 1,4, B}) > 0, ENT(h, {D 2,3, B}) > 0, ENT(h, {D 4,1, A}) > 0, and ENT(h, {D 3,2, A}) > 0. Also, ENT(h, {h k (D 1,4 ), B}) >

6 6 C. MOURON 0, ENT(h, {h k (D 2,3 ), A}) > 0, ENT(h, {h k (D 4,1 ), B}) > 0, and ENT(h, {h k (D 3,2 ), A}) > 0. Case 1) Suppose D 1,4 and D 2,3 are in different cover components of T (T 1 ) say T 1,1 and T 1,2. Then T 1,1 T 2 and T 1,2 T 2 since B T 2. Case 2) Suppose D 4,1 and D 3,2 are in different cover components of T (T 2 ) say T 2,1 and T 2,2. Then T 2,1 T 1 and T 2,2 T 1 since A T 1. Case 3) Suppose h k (D 1,4 ) and h k (D 3,2 ) are in different cover components of h k (T )(T 1 ) say T 1,1 and T 1,2. Then T 1,1 T 2 and T 1,2 T 2 since B T 2. Case 4) Suppose h k (D 4,1 ) and h k (D 2,3 ) are in different cover components of h k (T )(T 2 ) say T 2,1 and T 2,2. Then T 2,1 T 1 and T 2,2 T 1 since A T 1. Case 5) Suppose that D 1,4 and D 2,3 are in same cover component of T (T 1 ), D 4,1 and D 3,2 are in the same cover component of T (T 2 ), h k (D 1,4 ) and h k (D 3,2 ) are in the same cover component of h k (T )(T 1 ) and h k (D 4,1 ) and h k (D 2,3 ) are in the same cover components of h k (T )(T 2 ). Then there exist unique disjoint subchains C 1, C 2 of T from D 1,4 to D 2,3 and from D 4,1 to D 3,2 and unique disjoint subchains C 3, C 4 of h k (T ) from h k (D 1,4 ) to h k (D 3,2 ) and from h k (D 4,1 ) to h k (D 2,3 ). Hence, there exists a unique disjoint subchains h k (C 3 ), h k (C 4 ) of T from D 1,4 to D 3,2 and from D 4,1 to D 2,3. It follows from Lemma 10 that T is not tree-like. Hence, Case 5 cannot happen so at least one of the Cases 1-4 must hold. {Ti α}n i=1 splits T β into {T β j }m j=1 provided (1) 2 m (2) Each T β j refines T β (3) For each i {1,..., n} there exists j {1,,.m} such that T α i (4) For each j {1,..., m} there exists i {1,..., n} such that Ti α (5) mesh( m i=1 T β j ) < (1/2)mesh(T β ). T β j T β j. Lemma 13. Suppose that T1 α splits T β into T β 1 and T β 2 and Ti α T β for i {1,.., n}. Then {Ti α}n i=1 splits T β into {T β j }m j=1 for some m 2 Proof. For each i > 1 there exists A i Ti α and B i T β such that Ent(h, {A i, B i }) > 0. Let T be a refinement of T β such that T = T β and T β 1, T β 2 are components of T and mesh(t ) < mesh(t β ). By Corollary 8, there exists B i T such that Ent(h, {A i, B i }) > 0. Let T β i+1 be the component of T that contains B i. Then T β i+1 T β i+1. Remove all repeated elements and the lemma is shown. Proposition 14. Suppose that P = {T i } n i=1 is a collection such that for every i {1,..., n} there exists j {1,..., n} such that T i Tj. Let T β P and Q = {T α T α T β }. If Q splits T β into {T β j }m j=1, then P = (P {T β ) {T β j }m j=1 has the property that for every T i P there T j P such that T i T j.

7 TREE-LIKE 7 Proof. Pick T P there are 3 cases to consider: Case 1 T Q. Then by definition, there exists j {1,..., m} such that T T β j. Case 2 T {T β j }m j=1. Then by definition, there exists T Q Q such that T T Q. Case 3 T P (Q {T β j }m j=1 ). Then T P Q thus there exists T γ P such that T T γ. Since T Q, T γ T β so T γ P Let P be some collection of elements of the form T α1,...,α n where α i {1, 2}. Let s n (α 1,..., α n ) = n max{k Π k ( α 1,..., α n ), 1 Π k+1 (P n ) and Π k (α 1,..., α n ), 2 Π k+1 (P n )} If no such k exists, then take s n (α 1,..., α n ) = n. Let P n be and ordering on P n defined by T α1,...,α n < T β1,...,β n if and only if (s n (α 1,..., α n ), α 1,..., α n ) > L (s n (β 1,..., β n ), β 1,..., β n ) where > L is the lexicographical ordering. Notice that the inequality is reversed. Lemma 15. Suppose that T Suslinean. T but T does not split T. Then X is non- Proof. Then by Lemma 12, we may assume that T splits T into T 1 and T 2. Continuing inductively, there suppose that {T α1,...,α n } α1,...,α n n i=1 {1,2} has been found where each T α1,...,α n T but no T α1,...,α n splits T. Then T splits each T α1,...,α n into T α1,...,α n,1 and T α1,...,α n,2. Then T α1,...,α n+1 T but no T α1,...,α n+1 splits T for all α 1,..., α n+1 n+1 i=1 {1, 2}. Notice that in the construction, Tα 1,...,α m 1 Tα 1,...,α m 1,α m. So let T αm m=0 = T α 1,...,α m and let W be the collection of all α m that can be formed in this way. It follows that if α m γ m, then T αm T γ m =. Also, in each T α1,...,α m there exists a chain from A B to U. Hence, each T αm contains a continuum that intersects both A B and U and therefore is nondegenerate. Furthermore, by Theorem 9, the collection {T αm } α m W is uncountable. Thus X is non-suslinean. Theorem 16. If X is a tree-like continuum and h : X X is a positive entropy homeomorphism, then X is non-suslinean. Proof. If Ent(h) > 0, then there exists an integer k such that kent(h) > log(2). Hence by Proposition 2, Ent(h k ) > log(2). So without loss of generality, we may assume that Ent(h) > log(2). Hence by Theorem 7, we may assume that there exists a tree-cover T such that 1) there exist A, B T 2) A B =, 3) ENT(h, {A, B}) > 0.

8 8 C. MOURON 4) the chain from A to B has at least 5 links 5) there exists a link U in the chain from A to B such that U (A B) =. Let T 1 be the cover component of T {U} that contains A and T 2 be the cover component of T {U} that contains B. Then T 1 T 2. By Lemma 12, one of the following must be true: (1) T 2 splits T 1 into T 1,1, T 1,1. 2) T 1 splits T 2 into T 2,1, T 2,1. Suppose 1) is true. (Proof is similar if 2) is true.) Let T 2,2 = T 2. Then T 1,1 T 2,2 and T 1,2 T 2,2. Let P 1 = {T 1,1, T 1,2, T 2,2 }. Notice that if T α is any element of P 1 then there exists T β P 1 such that T α T β. Furthermore, T β must split T α otherwise by Lemma 5, X is non-suslinean and we are done. Suppose that P m has been found and P m has the property that for every T α P m there exists a T β P m such that T α T β. Let P m be an ordering of P m as defined earlier. Let T α1,...,α m be the first element of P m. Then there exists a T β1,...,β m P m that splits T α1,...,α m into T α1,...,α m,1 and T α1,...,α m,2. (Otherwise, by Lemma 5, X is non-suslinean and again we are done.) Let Q = {T β T β T α1,...,α m }. Then by Lemma 13, Q splits T α1,...,α m } into {T α1,...,α m,j } m j=1. Then define P m+1 = {T γ1,...,γ m,γ m = T γ1,...,γ m if γ 1,..., γ m α 1,..., α m } {T α1,...,α m,j } m j=1. Then by Lemma ***, P m+1 has the property that for every T α P m+1 there exists a T β P m+1 such that T α T β. Notice that if T γ1,...,γ m is the first element of P m, then T γ1,...,γ m,1 and T γ1,...,γ m,2 are the last and next to last elements of P m+1. Also if T γ1,...,γ m, is the nth (n > 1) element of P m, then T γ1,...,γ m,γ m is the (n 1)th element of P m+1. Hence, for every T γ1,...,γ m P m there exists a k > m such that T γ1,...,γ m,...1, T γ1,...,γ m,...2 P k. Furthermore, from in the construction, T α T 1,...,α m α. So let T 1,...,α m 1 α m = m=0 m=0 T α 1,...,α m and let W be the collection of all α m that can be formed in this way. It follows that if α m γ m, then T αm T γ m =. Also, in each T α1,...,α m there exists a chain from A B to U. Hence, each T αm is a continuum that intersects both A B and U and therefore is nondegenerate. It follows from Theorem 9 that W is uncountable. Therefore, X is not Suslinean. Department of Mathematics and Computer Science, Rhodes College, Memphis, TN address: mouronc@rhodes.edu