Transitive sensitive subsystems for interval maps

Transcription

1 STUDIA MATHEMATICA 69 () (2005) Transitive sensitive subsystems for interval maps by Sylvie Ruette (Paris) Abstract. We prove that for continuous interval maps the existence of a non-empty closed invariant subset which is transitive and sensitive to initial conditions is implied by positive topological entropy and implies chaos in the sense of Li Yorke, and we exhibit examples showing that these three notions are distinct.. Introduction. In this paper an interval map is a topological dynamical system given by a continuous map f : I I where I is a compact interval. In the literature much has been said about chaos for interval maps. The point is that the relations between various properties related to chaos are much more numerous for these systems than for general dynamical systems. As a consequence there is a rather ordered scale of chaos on the interval. For example, for interval maps topological weak mixing and topological strong mixing are equivalent [3], and transitivity implies sensitivity to initial conditions [2], which in turn implies positive topological entropy [8]. For more details on this topic see e.g. [6], [5, 6 9] and [2]. Among the different definitions of chaos, a well known one is chaos in the sense of Li Yorke. The definition follows the ideas of [7] but was formalized later. Definition.. Let T : X X be a continuous map on the metric space X with distance d. Then T is called chaotic in the sense of Li Yorke if there exists an uncountable set S X such that, for all distinct x, y S, lim sup d(t n (x), T n (y)) > 0 and lim inf d(t n (x), T n (y)) = 0. n + n + Note that in the definition of chaos in the sense of Li Yorke some people make the extra assumption that for all x S and all periodic points z X one has lim sup n + d(t n (x), T n (z)) > 0. This gives an equivalent definition since this property is satisfied by all points of the set S except at most one [6, p. 44] Mathematics Subject Classification: Primary 37E05. The author has been partly supported by a Marie Curie Fellowship of the European Community programme Human Potential under contract n HPMF-CT [8]

2 82 S. Ruette Li and Yorke showed that an interval map with a periodic point of period 3 is chaotic in the sense of Li Yorke [7]. In [4] Janková and Smítal generalized this result as follows: Theorem.2 (Janková Smítal). If f : I I is an interval map of positive entropy, then it is chaotic in the sense of Li Yorke. Recently, Blanchard, Glasner, Kolyada and Maass [4] proved that, if T : X X is a continuous map on the compact metric space X such that the topological entropy of T is positive, then the system is chaotic in the sense of Li Yorke. The converse of this result is not true, even for interval maps: Smítal [23] and Xiong [25] built interval maps of zero entropy which are chaotic in the sense of Li Yorke. See also [20] (a correction is given in [8]) or [] for examples of C interval maps which are chaotic in the sense of Li Yorke and have a null entropy. Recall that the map T : X X is transitive if for all non-empty open subsets U, V there exists an integer n 0 such that T n (U) V ; if X is compact with no isolated point, then T is transitive if and only if there exists x X such that ω(x, T ) = X (where ω(x, T ) is the set of limit points of {T n (x) n 0}). The map T has sensitive dependence on initial conditions (or simply is sensitive) if there exists δ > 0 such that for all x X and all neighbourhoods U of x there exist y U and n 0 such that d(t n (x), T n (y)) δ. A subset Y X is invariant if T (Y ) Y. The work of Wiggins [24] leads to the following definition (see, e.g., [3]). Definition.3. Let X be a metric space. The continuous map T : X X is said to be chaotic in the sense of Wiggins if there exists a nonempty closed invariant subset Y such that the restriction T Y is transitive and sensitive. The aim of this paper is to locate this notion with respect to the other definitions of chaos. Remark.4. A continuous map T : X X which is transitive and sensitive is sometimes called chaotic in the sense of Auslander Yorke []. If in addition the periodic points are dense, then it is called chaotic in the sense of Devaney [0]. Transitive sensitive subsystems appear naturally when considering a horseshoe, that is, two disjoint closed intervals J, K such that f(j) f(k) J K, because the points whose orbits never escape from J K form a subset on which f acts almost like a 2-shift [5]. For interval maps, positive entropy is equivalent to the existence of a horseshoe for some power of f [9, 7] (see also [6, Chap. VIII]), and one can deduce that a positive entropy interval

3 Transitive interval maps 83 map has a transitive, sensitive subsystem. More precisely, Shihai Li proved the following result [6]. Theorem.5 (Shihai Li). Let f : I I be an interval map. The topological entropy of f is positive if and only if there exists a non-empty closed invariant subset X I such that f X is transitive, sensitive to initial conditions, and the periodic points are dense in X (in other words, f X is Devaney chaotic). In the if part of this theorem one cannot suppress the assumption on the periodic points. In Section 3 we build a counter-example which leads to the following theorem. Theorem.6. There exists a continuous map f : [0, ] [0, ] of zero topological entropy which is chaotic in the sense of Wiggins. In [23] Smítal built a zero entropy map f which is chaotic in the sense of Li Yorke. If one looks at the construction of f, it is not hard to prove that f ω(0,f) is transitive and sensitive to initial conditions. We show the following theorem in Section 2. Theorem.7. Let f : I I be an interval map. If f is Wiggins chaotic then it is Li Yorke chaotic. The converse of this theorem is not true, contrary to what one may expect by considering Smítal s example. The last and longest section is devoted to the construction of a counter-example that proves the following result. Theorem.8. There exists a continuous interval map g : I I which is chaotic in the sense of Li Yorke but not in the sense of Wiggins. From Theorems.5 to.8 it follows that, for interval maps, chaos in the sense of Wiggins is a strictly intermediate notion between positive entropy and chaos in the sense of Li Yorke. Furthermore the examples of Sections 3 and 4 show that the behaviour of zero entropy interval maps is more varied that one might expect. Let us describe the different kinds of dynamics exhibited by these maps. The next result is well known (see, e.g., [6, p. 28]). Theorem.9. Let f : I I be an interval map. The following properties are equivalent: the topological entropy of f is zero, every periodic point has period 2 n for some integer n 0. According to Sharkovskiĭ s Theorem [22] the set of periods of periodic points of a zero entropy interval map is either {2 k ; 0 k n} for some

4 84 S. Ruette integer n, and then f is said to be of type 2 n, or {2 k ; k 0}, and then f is of type 2. There is little to say about the dynamics of type 2 n, and some interval maps of type 2 share almost the same dynamics [9]: every orbit converges to some periodic orbit of period 2 k ; these maps are never Li Yorke chaotic. The interval maps of type 2 that admit an infinite ω-limit set may be Li Yorke chaotic or not, as shown by Smítal [23]. A map f that is not Li Yorke chaotic is called uniformly non-chaotic in [6] and it has the following property: every point x is approximately periodic, that is, for every ε > 0 there exists a periodic point y and an integer N such that f n (x) f n (y)[< ε for all n N. The maps built in Sections 3 and 4 are both zero entropy and Li Yorke chaotic. In the first example there is a transitive sensitive subsystem which is the core of the dynamics; in particular Li Yorke chaos can be read on this subsystem. In the second example this situation does not occur since there is no transitive sensitive subsystem. 2. Wiggins chaos implies Li Yorke chaos. The following notion of f-non-separable points was introduced by Smítal to give an equivalent condition for chaos in the sense of Li Yorke [23]. Note that Theorem 2.2 was proven to remain valid for all interval maps by Janková and Smítal [4]. Definition 2.. Let f : I I be an interval map and a 0, a two distinct points in I. The points a 0, a are called f-separable if there exist two disjoint subintervals J 0, J and two integers n 0, n such that for i = 0,, a i J i, f n i (J i ) = J i and (f k (J i )) 0 k<ni are disjoint. Otherwise they are f-nonseparable. Theorem 2.2 (Smítal). Let f : I I be an interval map of zero entropy. The following properties are equivalent: f is chaotic in the sense of Li Yorke, there exists x 0 I such that the set ω(x 0, f) is infinite and contains two f-non-separable points. In the proof of this theorem, Smítal showed the following intermediate result which describes the structure of an infinite ω-limit set of a zero entropy map. Lemma 2.3. Let f : I I be an interval map of zero entropy and x 0 in I such that ω(x 0, f) is infinite. For all n 0 and 0 i < 2 n, define I i n = [min ω(f i (x 0 ), f 2n ), max ω(f i (x 0 ), f 2n )], L i n = k 0 f k2n (I i n).

5 Transitive interval maps 85 Then f(l i mod 2k k ) = Li+ k for all 0 i < 2 k, and the intervals (L i k ) 0 i<2 k are pairwise disjoint. Lemma 2.4. Let f : [a, b] R be a continuous map. If f([a, b]) [a, b], then f has a fixed point. Proof. There exist x, y [a, b] such that f(x) a and f(y) b. One then has f(x) x a x 0 and f(y) y b y 0, so there is a point c [x, y] such that f(c) c = 0. Lemma 2.5. Let f : I I be an interval map of zero entropy. If J I is a (not necessarily closed) subinterval such that f p (J) = J and (f i (J)) 0 i<p are pairwise disjoint then p is a power of 2. Proof. If J is reduced to one point then it is a periodic orbit and by Theorem.9, p is a power of 2. We assume that J is non-degenerate, which implies that f n (J) is a non-degenerate interval for all n 0. Since f p (J) = J, by Lemma 2.4 there exists x J such that f p (x) = x. According to Theorem.9 the period of x is equal to 2 k for some k; write p = m2 k. If x J then (f i (x)) 0 i<p are distinct and p = 2 k. Suppose that m 3. Then x J; we assume that x = sup J, the case of x = inf J being symmetric. Since x = f 2k (x) f 2k (J) and f 2k (J) J =, one has x = inf f 2k (J). But also x f 2k+ (J), which contradicts the fact that J, f 2k (J), f 2k+ (J) are pairwise disjoint non-degenerate intervals. Therefore m = or 2 and p is a power of 2. The following result is the key tool in the proof of Theorem.7. A rather similar result can be found in a paper of Fedorenko, Sharkovskiĭ and Smítal [2]. Lemma 2.6. Let f : I I be an interval map of zero entropy and x 0 in I such that ω(x 0, f) is infinite and does not contain two f-non-separable points. Then for all ε > 0 there exists δ > 0 such that if x, y ω(x 0, f) and x y < δ, then f n (x) f n (y) < ε for all n 0. Proof. Let X = ω(x 0, f). For all integers n 0 and 0 i < 2 n define a i n = min ω(f i (x 0 ), f 2n ) and b i n = max ω(f i (x 0 ), f 2n ). Define I i n and L i n as in Lemma 2.3; then I i n = [a i n, b i n]. The points a i n, b i n belong to X and () I i n+ I i+2n n+ Ii n for all 0 i < 2 n. Suppose that there exists ε > 0 such that (2) for all n 0 there is 0 i < 2 n with I i n ε. Using () we can build a sequence (i n ) n 0 such that I i n+ n+ Ii n n and I i n n ε for all n 0. Define I = n Ii n n. It is a decreasing intersection of compact intervals, hence a closed interval, and I ε. Write I = [a, b]; then

6 86 S. Ruette a = lim n + ai n n, b = lim n + bi n n, and thus a, b X. One has a, b L i n n for all n 0. By Lemma 2.3 the intervals L i n+2 n+2, f 2n (L i n+2 n+2 ), f 2n+ (L i n+2 n+2 ) are pairwise disjoint, so {a, f 2n (a), f 2n+ (a)} are distinct. Since a b and by assumption a, b are f-separable, there exist an interval J and an integer p such that a J, b J, f p (J) = J and (f i (J)) 0 i<p are pairwise disjoint. By Lemma 2.5, p is a power of 2; write p = 2 k. Consider the interval K = L i k k J; it contains a, and f 2k (K) K because f 2k (L i k k ) = L i k k by Lemma 2.3. Thus K contains {a, f 2k (a), f 2k+ (a)}. These three points belong to ω(x 0, f) and are distinct, so one of them belongs to Int(K) and there exists an integer n such that f n (x 0 ) K. We thus have X = ω(x 0, f) 2 k j=0 f j (K). Let b X be such that f 2k+ (b ) = b and let 0 j < 2 k be such that b f j (K). The points b, f 2k (b ) and f 2k+ (b ) belong to f j (K) and they are distinct (same proof as for a), hence one of them belongs to f j (K), which implies that b f j (K). One has j 0 because b J and K J. But on the other hand b f j (L i k k ) L i k k, which is empty by Lemma 2.3, and we get a contradiction. Therefore, (2) is false. Let ε > 0; the negation of (2) implies that there exists n 0 such that In i < ε for all 0 i < 2 n. Let δ > 0 be the minimal distance between two distinct intervals among (In) i 0 i<2 n. If x, y X with x y < δ then there exists 0 i < 2 n such that x, y In i ω(x 0, f) = ω(f i (x 0 ), f 2n ). Thus for all k 0 one has f k (x), f k (y) ω(f i+k (x 0 ), f 2n ) Ii+k mod 2n n, which implies that f k (x) f k (y) < ε. Now we are ready to prove Theorem.7. Let f : I I be an interval map. If f is Wiggins chaotic then it is Li Yorke chaotic. Proof. Suppose that f is not chaotic in the sense of Li Yorke. By Theorem.2 one has h top (f) = 0. Consider a closed invariant subset Y I such that f Y is transitive. If Y is finite or has an isolated point, then f Y is not sensitive. If Y is infinite with no isolated point, then there exists x 0 Y such that ω(x 0, f) = Y. By Theorem 2.2, Y does not contain two f-non-separable points, and by Lemma 2.6, f Y is not sensitive. 3. Wiggins chaos does not imply positive entropy. We are going to build an interval map of zero entropy which is chaotic in the sense of

7 Transitive interval maps 87 Wiggins. It resembles the maps built by Smítal (map f in [23]) and Delahaye (map g in [9]), but we give full details because this construction will be used as a basis for the next example. Notation. If I is an interval, let mid(i) denote the middle of I. If f is a linear map, let slope(f) denote its constant slope. We write (resp. ) for increasing (resp. decreasing ). Let (a n ) n 0 be an increasing sequence of numbers less than such that a 0 = 0. Define I 0 = [a 0, ] and, for all n, I 0 n = [a 2n 2, a 2n ], L n = [a 2n, a 2n ], I n = [a 2n, ]. Then In 0 L n In = In. We fix (a n) n 0 such that the lengths of the intervals In, 0 In satisfy: In 0 = ( 3 n I n, In = 2 ) 3 n In if n is odd, I 0 n = ( 2 3 n ) In, In = 3 n I n if n is even, This implies that L n = (/3 n ) I n for all n. Note that I n 0, that is, lim n + a n = ; hence n (I0 n L n ) = [0, ). For all n, let ϕ n : I 0 n I n be the increasing linear homeomorphism mapping I 0 n onto I n; the slope of ϕ n is slope(ϕ n ) = I n / I 0 n. Define the map f : [0, ] [0, ] such that f is continuous on [0, ) and f(x) = ϕ ϕ 2 ϕ n ϕ n(x) for all x In, 0 n, f Ln is linear for all n, f() = 0. Note that f I 0 n is linear. We will show below that f is continuous at. I I 0 0 I 0 I 2 0 I I 2 ϕ ( I2 ) ϕ ( I2 0 ) 0= a a 2 a 3 a 4... a 0 Fig.. The first steps of the construction of f (left) and the graph of f (right). This map has a zero entropy and the invariant set ω(0, f) is transitive and sensitive.

8 88 S. Ruette Let us explain the underlying construction. At step n = the interval I 0 is sent linearly onto I (hence f I 0 = ϕ ) and we require that f(i ) I0 (grey area in Figure ). Then we do the same kind of construction in the grey area with respect to I2 0, I 2 I : we rescale I0 2, I 2 as ϕ (I0 2 ), ϕ (I 2 ) I0 (on the vertical axis) and we send linearly I2 0 onto ϕ (I 2 ); in this way f I 0 2 = ϕ ϕ 2. We repeat this construction on I2 (black area), and so on. Finally, we fill the gaps in a linear way to get the whole map, which is pictured on the right of Figure. Let J0 0 = [0, ] and for all n define the subintervals J n, 0 Jn Jn 0 by min Jn 0 = 0, max Jn = max Jn 0 and J n / J i n 0 = Ii n / In for i = 0,. To show that f is continuous at, it is enough to prove that max(f I n ) tends to 0 as n goes to infinity. For all n one has ϕ n (max I 0 n) = max I n = = min I n + I n, ϕ n ϕ n(max I 0 n) = min I 0 n + I n slope(ϕ n ) = min I n 2 + I n slope(ϕ n ), ϕ n 2 ϕ n ϕ n(max I 0 n) = min I 0 n 2 + I n slope(ϕ n 2 ) slope(ϕ n ),. ϕ ϕ 2 ϕ n ϕ n(max I 0 n) = min I 0 + I n Consequently, = n n I 0 i I i = J 0 n. (3) f(max I 0 n) = J 0 n = max J 0 n. slope(ϕ i ) According to the definition of f, one has max(f I n ) = f(max I 0 n), so that max(f I n ) = Jn 0. By definition, J n 0 /3n 2, which tends to 0, and therefore f is continuous at. The next lemma describes the action of f on the intervals (Jn) i and (In) i and collects the properties that we will use later. The interval In is periodic of period 2 n and the map f 2n swaps In 0 and In. However, we prefer to deal with Jn 0 = f(in); this will simplify the proofs because f I n is not monotone whereas f i J 0 n is linear for all i 2 n. Lemma 3.. Let f be the map defined above. Then for all n : (i) f(i n) = J 0 n, (ii) f(i 0 n) = J n,

9 (iii) f i J 0 n is linear for all i 2 n, Transitive interval maps 89 (iv) f 2n (Jn) 0 = In 0 and f 2n (Jn) 0 = In, (v) f i (Jn) 0 k n I0 k for all 0 i 2n 2, (vi) (f i (Jn)) 0 0 i<2 n are pairwise disjoint; and the previous clauses also imply (vii) f 2n (J 0 n) = J n, (viii) f 2n (J 0 n) = J 0 n, (ix) f 2n I 0 n is linear and f 2n (I 0 n) = I n, (x) f 2n (I n) = I 0 n, (xi) (f i (I 0 n)) 0 i<2 n are pairwise disjoint and f 2n (I n) = I n. Proof. According to (3), max(f I n ) = f(max I 0 n+ ) = max J 0 n; moreover f() = 0 = min J 0 n. Thus f(i n) = J 0 n by continuity; this is (i). According to the definition of f, n f(in) 0 = In 0 slope(ϕ n ) slope(ϕ i ) n = In Ii 0 Ii = I n n In I 0 i I i = J n. Moreover f(max I 0 n) = max J 0 n = max J n by (3), and thus f(i 0 n) = J n. This gives (ii). We show by induction on n that (iii) and (iv) are satisfied. This is true for n = because J 0 = I0, J = I, f I 0 = ϕ is linear and f(i 0) = I. Suppose that (iii) and (iv) are true for n. Since Jn+ 0 J n, 0 the map f i J 0 n+ is linear for all i 2 n and f 2n (Jn+ 0 ) I n; moreover the linearity implies that and min f 2n (J 0 n+) = min f 2n (J 0 n) = min I n = min I 0 n+ f 2n (J 0 n+ ) I n = J 0 n+ J 0 n = I0 n+ In. Therefore f 2n (Jn+ 0 ) = I0 n+. Then f 2n (Jn+ 0 ) = J n+ by (ii). Since Jn+ J n, 0 the induction hypothesis applies: f i J n+ is linear for all i 2 n, f 2n (Jn+ ) I n, and by linearity max f 2n (J n+) = max f 2n (J 0 n) = = max I n+ and f 2n (Jn+ ) = I n+, hence f 2n+ (Jn+ 0 ) = f 2n (Jn+ ) = In+. This gives (iii) and (iv) for n +.

10 90 S. Ruette Now we prove (v) by induction on n: This is true for n = because J 0 = I0. Suppose that (v) is true for n. Since J 0 n+ J 0 n it follows that f i (J 0 n+ ) k n In k for all 0 i < 2n. Moreover f 2n (J 0 n+ ) = I0 n+ by (iv), and f 2n (J 0 n+ ) = f(i0 n+ ) = J n+ by (ii). Since J n+ J 0 n we can use the induction hypothesis again to get f 2n +i (J 0 n+ ) k n In k for all 0 i < 2 n. This gives (v) for n +. Next we prove (vi). Suppose that f i (J 0 n) f j (J 0 n) for some 0 i < j < 2 n. Then f 2n j (f i (J 0 n)) f 2n j (f j (J 0 n)). But f 2n (J 0 n) = I n by (iv) and f 2n (j i) (J 0 n) [0, max I 0 n] by (v), so these two sets are disjoint, which is a contradiction. Finally, we indicate how to obtain the other assertions. Assertions (vii) and (viii) follow respectively from (iv)+(ii) and (iv)+(i). Assertion (ix) follows from (iii)+(iv). Assertion (x) follows from (i)+(iv). Assertion (xi) follows from the combination of (i), (iv) and (vi). Define K n = i 0 f i (I n) for all n 0 and K = n 0 K n. According to Lemma 3., K n is the disjoint union of the intervals (f i (J 0 n)) 0 i 2 n. The set K has a Cantor-like construction: at each step a middle part of every connected component of K n is removed to get K n+. However, K is not a Cantor set because its interior is not empty (see Proposition 3.3). Proposition 3.2 states that the entropy of f is null. Next we show in Proposition 3.3 that the set ω(0, f) contains K. Then we prove that ω(0, f) is transitive and sensitive to initial conditions. Proposition 3.2. Let f be the map defined above. Then h top (f) = 0. Proof. By definition the restriction f Ln is linear decreasing and thus f(l n ) [0, f(max In)]. 0 Moreover f(max In) 0 = max Jn 0 by (3), so that f(l n ) Jn 0. Then Lemma 3.(iii) implies that f 2n Ln is linear decreasing. Since f 2n I 0 n is linear increasing and f 2n (In) 0 = In by Lemma 3.(ix), it follows that f 2n (min L n ) = max In = ; moreover f 2n (In) = In 0 by Lemma 3.(x), and thus f 2n (max L n ) In. 0 We deduce that f 2n (L n ) L n In, and by Lemma 2.4 there exists z n L n such that f 2n (z n ) = z n. The period of z n is exactly 2 n because L n In and the intervals (f i (In )) 0 i<2n are pairwise disjoint by Lemma 3.(xi). By definition L n In, hence slope(f 2n Ln ) 2. If the points x, f 2n (x),..., f k2n (x) belong to L n then f (k+)2n (x) z n 2 k x z n ; thus, for all x L n, x z n, there exists k such that f k2n (x) L n. Since In = I0 n L n In and f 2n (In ) = I n by Lemma 3.(xi), this implies that f k2n (x) In 0 In. In addition f 2n (In) 0 = In by Lemma 3.(ix),

11 and therefore Transitive interval maps 9 x I n, x z n, k 0, f k (x) I n. Starting with I0 = [0, ], a straightforward induction shows that, for all x [0, ], if the orbit of x does not meet {z n n } then for all integers n there exists k 0 such that f k (x) In; in particular ω(x, f) K. According to Lemma 3.(xi) the set K contains no periodic point because K i 0 f i (In) for all n ; thus every periodic point is in the orbit of some z n, and therefore its period is a power of 2. Finally, h top (f) = 0 by Theorem.9. The orbit of 0 obviously enters f i (J 0 n) for all n 0 and 0 i < 2 n, and so ω(0, f) meets all connected components of K. We show in the next lemma that ω(0, f) contains K; the proof relies on the idea that the smaller of the intervals J 0 n+ and J n+ contains alternately either min J 0 n or max J 0 n when n varies, so that both endpoints of a connected component of K can be approximated by small intervals of the form f i (J 0 n). Proposition 3.3. Let f and K be as defined above. Then K ω(0, f). In particular ω(0, f) is infinite and contains 0, and f ω(0,f) is transitive. Proof. According to the definition of K, the connected components of K are exactly the non-empty sets of the form n 0 f j n (Jn) 0 with 0 j n < 2 n. Let y be a point in K. For all n 0 there exists 0 j n < 2 n such that y f j n (Jn), 0 and there exists a sequence (y n ) n 0 such that y = lim n + y n and y n f j n (Jn) 0 = {min f j n (Jn), 0 max f j n (Jn)}. 0 Let ε > 0 and N 0. Let n be an even integer such that /3 n+ < ε and y n y < ε, and let k 0 be such that k2 n+ N. First, suppose that y n = min f j n (Jn). 0 Since 0 Jn+ 0 and f 2n+ (Jn+ 0 ) = Jn+ 0 by Lemma 3.(viii), it follows that f k2n+ +j n (0) f j n (Jn+ 0 ). By Lemma 3.(iii) one has min f j n (Jn+ 0 ) = min f j n (Jn) 0 = y n and f j n (J 0 n+ ) f j n (J 0 n ) = J 0 n+ J 0 n = < ε, 3n+ and therefore f k2n+ +j n (0) y n < ε f j n (Jn) 0 ε. Secondly, suppose that y n = max f j n (Jn). 0 The point f k2n+2 (0) belongs to Jn+2 0 and f 2n+ (Jn+2 0 ) = J n+2 by Lemma 3.(vii), so f k2n+2 +2 n+ +2 n +j n (0) f 2n +j n (J n+2). According to Lemma 3.(iii) (vii) one has max f 2n +j n (J n+2)=max f 2n +j n (J 0 n+)=max f j n (J n+)=max f j n (J 0 n) = y n. Moreover f 2n (J n+2) f 2n (J 0 n+) = J n+ J 0 n.

12 92 S. Ruette Thus f j n+2 n (J n+2 ) f j n (J 0 n ) = f 2n (J n+2 ) J 0 n = f 2n (Jn+2 ) f 2n (Jn+ 0 ) J n+ Jn 0 = ( 3 n+2 2 ) 3 n+. Consequently, f k2n+2 +2 n+ +2 n +j n (0) y n f j n+2 n (Jn+2 ) < ε. In both cases there exists p N such that f p (0) y n < ε, hence f p (0) y < 2ε. This means that y ω(0, f), that is, K ω(0, f). The point {0} = n 0 J n 0 belongs to K, so 0 ω(0, f) and f ω(0,f) is transitive. Finally, K n has 2 n connected components, each of which contains two connected components of K n+ ; thus K has an infinite number of connected components, which implies that K is infinite. In the proof of the next proposition, we first show that K contains a non-degenerate connected component B. Proposition 3.4. Let f be the map defined above. Then f ω(0,f) is sensitive to initial conditions. Proof. First we define by induction a sequence of intervals B n = f i n (J 0 n) for some 0 i n < 2 n such that B n B n and B n = ( 2/3 n ) B n for all n. Take B 0 = J 0 = [0, ]. Suppose that B n = f i n (Jn 0 ) is already built. If n is even, take i n = i n and B n = f i n (Jn). 0 The map f i n J 0 n is linear by Lemma 3.(iii) and Jn 0 Jn 0, so that B n B n = J 0 n J 0 n = 2 3 n. If n is odd, take i n = i n + 2 n and B n = f i n (Jn). 0 According to Lemma 3.(vii) (iii) one has B n = f i n (Jn) and f i n J 0 n is linear, so B n B n = J n Jn 0 = 2 3 n. Let B = n 0 B n. This is a compact interval and it is non-degenerate because log B = log B 0 + ( log 2 ) 3 n >. n Moreover B is a connected component of K, so B K. Let b 0 = min B and b = max B; one has b 0, b ω(0, f) by Proposition 3.3. The set ω(0, f) is included in the periodic orbit of J 0 n, and consequently f k (J 0 n ω(0, f)) = f k (J 0 n) ω(0, f) for all k 0. Let ε > 0 and k. There exists n 0 such that J 0 n < ε and there exist x 0, x J 0 n ω(0, f) such that

13 Transitive interval maps 93 f i n+k2 n (x 0 ) = b 0 and f i n+k2 n (x ) = b. Let δ = b b 0 /4. The triangular inequality implies that either f i n+k2 n (0) b 0 ) 2δ or f i n+k2 n (0) b 2δ. In other words, for all ε > 0 and k there exist x [0, ε] ω(0, f) and i k such that f i (0) f i (x) 2δ. Let y ω(0, f) and ε > 0; there exists k 0 such that f k (0) y < ε/2. By continuity of f k there is η > 0 such that f k ([0, η]) [y ε, y + ε]. What precedes shows that there exist x [0, η] ω(0, f) and i > k such that f i (0) f i (x) 2δ, and if z = f k (x), z = f k (0) and j = i n we find that z, z [y ε, y + ε] and f j (z) f j (z ) 2δ. Then the triangular inequality implies that either f j (y) f j (z) δ or f j (y) f j ( z) δ. We conclude that f ω(0,f) is sensitive to initial conditions. Finally, Propositions 3.2, 3.3 and 3.4 give Theorem.6. Remark 3.5. According to Theorem.7 and Proposition 3.4, the map f is chaotic in the sense of Li Yorke. It can be proven directly that b 0, b are f-non-separable, so Theorem 2.2 applies. 4. Li Yorke chaos does not imply Wiggins chaos. The aim of this section is to exhibit an interval map which is chaotic in the sense of Li Yorke but has no transitive sensitive subsystem. This map resembles the one of Section 3: the construction on the set I 0 n is the same except that the lengths of the intervals differ; the dynamics on L n is different. 4.. Definition of the map g. We are going to build a continuous map g : [0, 3/2] [0, 3/2]. Let (a n ) n 0 be an increasing sequence of numbers less than such that a 0 = 0. Define I 0 = [a 0, ] and for all n, I 0 n = [a 2n 2, a 2n ], L n = [a 2n, a 2n ], I n = [a 2n, ]. Then In 0 L n In = In. Fix (a n ) n 0 such that the lengths of the intervals satisfy n, In 0 = L n = ( 3 n I n and In = 2 ) 3 n In. Let a = lim n + a n. Then n (I0 n L n ) = [0, a) and a < because + ( log( a) = log 2 ) 3 n >. n= For all n, let ϕ n : I 0 n I n be the increasing linear homeomorphism mapping I 0 n onto I n. Define the map g : [0, 3/2] [0, 3/2] such that g is continuous on [0, 3/2] \ {a} and g(x) = ϕ ϕ 2 2 ϕ n ϕ n(x) for all x In, 0 n, g is linear of slope λ n on [min L n, mid(l n )] for all n,

14 94 S. Ruette g is linear on [mid(l n ), max L n ] for all n, g(x) = 0 for all x [a, ], g(x) = x for all x [, 3/2], where the slopes (λ n ) will be defined below. We will also show that g is continuous at a. The map g is pictured in Figure 2. 3/2 0 a I 0 L I 0 0 J 0 2 J 2 I 2 I 2 L 2 +x 3/2=+x 0 Fig. 2. The graph of g; this map is Li Yorke chaotic but not Wiggins chaotic. Let J 0 0 = [0, ] and for all n define the subintervals J 0 n, J n J 0 n such that min J 0 n = 0, max J n = max J 0 n and J i n / J 0 n = Ii n / I n for i = 0, ; let M n = [max J 0 n, min J n]. Note that on n I0 n the map g is defined similarly to the map f of Section 3, and therefore the assertions of Lemma 3. remain valid for g, except (i) and its derived results (viii), (x), (xi). Lemma 4.. Let g be the map defined above. Then for all n : (i) g(i 0 n) = J n, (ii) g i J 0 n is linear for all 0 i 2 n, (iii) g 2n (Jn) 0 = In 0 and g 2n (Jn) 0 = In, (iv) g i (Jn) 0 k n I0 k for all 0 i 2n 2, (v) (g i (Jn)) 0 0 i<2 n are pairwise disjoint. (vi) g 2n I 0 n is linear and g 2n (In) 0 = In, (vii) g 2n Mn is linear and g 2n (M n ) = L n, (viii) g(min L n ) = min M n, (ix) g 2n 2 (min L n ) = min L n. Proof. For assertions (i) to (vi) see the proof of Lemma 3..

15 Transitive interval maps 95 According to (ii), the map g 2n Mn is linear because M n is included in Jn 0. Since M n = [max Jn, 0 min Jn] and L n = [max In, 0 min In], assertions (i), (ii) and (iii) imply that g 2n (M n ) = L n, which is (vii). The map g I 0 n is increasing and min L n = max In, 0 hence, according to (i), one has g(min L n ) = max Jn = max Jn 0 = min M n ; this is (viii). Finally, (vii) and (viii) imply (ix). n+ /3i. It is a For all n 0, define x n = mid(m n+ ), that is, x n = 3 2 decreasing sequence and x 0 = /2; therefore, for all n 0, g( + x n ) is well defined and equal to x n. For all n 0 let t n = slope(g 2n J 0 n ); by convention g 0 is the identity map so t 0 =. Fix λ = 2x / L and for all n 2 define inductively λ n such that (4) L n 2 n n 2 λ i i=0 t i = x n. By convention an empty product is equal to, so (4) is satisfied for n =. The slopes (λ n ) n satisfy g 2n ([min L n, mid(l n )]) = [, + x n ], as proven in the next lemma. This means that under the action of g 2n the image of L n falls outside of [0, ] but remains close to. We also list properties of g on the intervals L n, I n and [, + x n ]. Lemma 4.2. Let g be the map defined above. Then: (i) g 2n [,+xn ] is linear and g 2n ([, + x n ]) = [min I 0 n+, mid(l n+)] for all n 0, (ii) g 2n [min Ln,mid(L n )] is linear and g 2n ([min L n, mid(l n )]) = [, + x n ] for all n, (iii) g 2n+ [,+xn ] is and g 2n+ ([, + x n ]) = I n+ [, + x n+] for all n, (iv) g(i n) [0, mid(m n )] for all n, (v) g 2n ([min I n, +x n ]) [min I n, +x n ] and g i ([min I n, +x n ]) [0, ] for all i 2 n, n. Proof. The map g [,+xn ] is linear and g([, + x n ]) = [0, mid(m n+ )] Jn, 0 so g 2n [,+xn ] is linear by Lemma 4.(ii). Moreover g 2n (0) = min In+ 0 and g2n (mid(m n+ ))=mid(l n+ ) by Lemma 4.(iii)+(iv), which gives (i). Before proving (ii) we show some intermediate results. Let n 2 and 2 k n. Then n λ n 2 i λ n λ k t n 2 t k 2 = k λ i i=0 t i k 3 i=0 t i = x n/ L n x k / L k by (4)

16 96 S. Ruette and so = = n+ i=k+ 3 i 3 n k+ n i=k n i=k 2/3 i 3 n 3 k 3 i 2 (5) λ n λ k t n 2 t k 2 <. By definition, g(mid(l n )) = g(min L n ) + λ n L n /2, and by (4), λ n L n 2 = x n t n n 2 λ n 3 i i=0 t i = M n 3 n+ 2 < M n. 2 = x n L n 2x n t n 2 because t n 2 = L n M n Moreover g(min L n ) = min M n by Lemma 4.(viii), hence by Lemma 4.(vii) (6) g([min L n, mid(l n )]) [min M n, mid(m n )] for all n 2. We show by induction on k = n,..., 2 that the map g 2n 2 +2 n k 2 is linear of slope λ n λ k t n 2 t k 2 on [min L n, mid(l n )] and maps min L n to min L k, g i ([min L n, mid(l n )]) [0, ] for all 0 i 2 n n k 2. By (6) one has g([min L n, mid(l n )]) M n J 0 n 2, so g 2n 2 [min Ln,mid(L n )] is linear of slope λ n t n 2. According to Lemma 4.(ix) one has g 2n 2 (min L n ) = min L n. Now (6) and Lemma 4.(iii)+(iv) imply that g i ([min L n, mid(l n )]) [0, ] for all i 2 n 2. This is our statement for k = n. Suppose that the statement is true for k with 3 k n. By (5) one has λ n λ k t n 2 t k 2 L n /2 L k /2 so that g 2n 2 +2 n k 2 ([min L n, mid(l n )]) [min L k, mid(l k )]. The map g is of slope λ k on this interval, g(min L k ) = min M k 2 by Lemma 4.(viii), and g([min L k, mid(l k )]) M k 2 by (6). Since M k 2 Jn 0, the map g2n 2 +2 n k 2 +2 k 3 is linear of slope λ n... λ k t n 2... t k 3 on [min L n, mid(l n )], and it maps min L n to min L k 2 by Lemma 4.(ix). Moreover g i ([min L n, mid(l n )]) [0, ] for all 0 i 2 n 2 +2 n k 2 +2 k 3 by Lemma 4.(iv) and the induction hypothesis. This is the statement for k.

17 Transitive interval maps 97 For k = 2 we conclude that g 2n = g 2n is linear of slope n i=2 λ n 2 i i=0 t i on [min L n, mid(l n )], with g 2n (min L n ) = min L and g 2n ([min L n, mid(l n )]) [min L, mid(l )]. The map g is of slope λ on this interval, hence, according to the definition of λ n, assertion (ii) holds for all n 2; it also trivially holds for n =. The inductive statement for k = 2 also shows that (7) g i ([min L n, mid(l n )]) [0, ] for all 0 i 2 n, n. Assertions (i) and (ii) and Lemma 4.(vi) imply (iii). One has I n = k n+ (I0 k L k) [a, ]. From the definition of g one can see that max{g(x) x I 0 k L k} = g(mid(l k )), and therefore g(i 0 k L k) [0, mid(m k )] by (6). Hence g(i n) [0, mid(m n )] = J 0 n [min M n, mid(m n )], which is (iv). By Lemma 4.(iii)+(vii), g 2n (J 0 n)=i n and g 2n ([min M n, mid(m n )]) = [min L n, mid(l n )], and by (ii), g 2n ([min L n, mid(l n )]) = [, + x n ]. Combining with (iv) we get (8) g 2n (I n) I n [, + x n ]. Moreover g i (J 0 n) [0, ] for all 0 i 2 n 2 and g i ([min M n, mid(m n )]) [0, ] for all 0 i 2 n 2 according to Lemma 4.(iv). In addition, g 2n +i ([min M n, mid(m n )]) = g i ([min L n, mid(l n )]) [0, ] for all 0 i 2 n by (7). Therefore (9) g i (I n) [0, ] for all 0 i < 2 n. Finally, g([, + x n ]) = [0, mid(m n+ )] J 0 n, and (i) implies that g 2n ([, + x n ]) I n. Combined with (9), (8) and Lemma 4.(iv), this gives (v). Now we show that g is continuous at the point a, as claimed at the beginning of the section. Lemma 4.3. The map g defined above is continuous. Proof. We just have to show the continuity at a. It is clear from the definition that g is continuous at a +. According to Lemma 4.2(iv) one has g(i n) J 0 n. This implies that g is continuous at a since lim n + max J 0 n = 0.

18 98 S. Ruette To end this subsection, let us explain the main underlying ideas of the construction of g by comparing it with the map f built in Section 3. The map g and f are similar on the set n I0 n (which is the core of the dynamics of f); the only difference is the length of the intervals. For f we showed that 2 n i=0 f i (J 0 n) has a non-degenerate connected component B and K = n 0 it can be proven that the endpoints of B are f-non-separable. The same remains true for g with B = [a, ] = n 0 I n (the fact that a, are g-nonseparable will be proven in Proposition 4.4). For f we proved that K ω(0, f), hence B ω(0, f); for g it is not true that {a, } ω(0, g) because the orbit of 0 stays in [0, a]. The construction of g on the intervals L n allows one to approach from the outside of [0, ]: we will see in Proposition 4.4 that ω( + x 0, g) contains both a and, which implies chaos in the sense of Li Yorke. On the other hand, the proof showing that f ω(0,f) is transitive and sensitive fails for g because ω(0, g) does not contain {a, } and ω( + x 0, g) is not transitive. We will see in Proposition 4.7 that g has no transitive sensitive subsystem at all g is chaotic in the sense of Li Yorke Proposition 4.4. Let g be the map defined in Section 4.. Then the set ω(+x 0, g) is infinite and contains the points a,, which are g-non-separable. Consequently, the map g is chaotic in the sense of Li Yorke. Proof. Lemma 4.2(iii) implies that g 2n+ ( + x n ) = + x n+ for all n 0. Since x n 0 as n goes to infinity, this implies that ω( + x 0, g). Moreover Lemma 4.2(i) implies that g 2n () = min I 0 n+ = a 2n for all n, hence a ω(, g) ω( + x 0, g). Suppose A, A 2 are two periodic intervals such that a A and A 2, and let p be a common multiple of their periods. Since g(a) = g() = 0, it follows that g p (a) = g p () A A 2, and A, A 2 are not disjoint. This means that a, are g-non-separable. It is well known that a finite ω-limit set is cyclic. Therefore, if y 0, y are two distinct points in a finite ω-set, the degenerate intervals {y 0 }, {y } are periodic and y 0, y are g-separable. This implies that ω( + x 0, g) is infinite. We deduce that the map g is chaotic in the sense of Li Yorke by Theorem g is not chaotic in the sense of Wiggins. The main result of this subsection is Proposition 4.7 stating that g has no transitive sensitive subsystem. The next lemma is about the location of transitive subsystems. Lemma 4.5. Let g be the map defined in Section 4. and Y [0, 3/2] a closed invariant subset with no isolated point such that g Y is transitive. Then

19 Transitive interval maps 99 (i) Y [0, a], (ii) Y 2 n i=0 g i (Jn) 0 for all n, (iii) g i (Jn 0 Y ) = g i (Jn) 0 Y = g i mod 2n (Jn) 0 Y for all i 0, n 0. Proof. By transitivity there exists y 0 Y such that ω(y 0, g) = Y ; in particular the set Y = {g k (y 0 ) k 0} is dense in Y and y ω(y, g) for all y Y. Let n 0. By Lemma 4.2(iii), g 2n+ ([, +x n ]) = I n+ [, +x n+] and hence Lemma 4.2(v) shows that for all integers k, g k2n+ ([, + x n ]) I n+ [, + x n+] and g i ([, + x n ]) [0, ] for all i > 2 n+, i 2 n+ N. This implies that g i (( + x n+, + x n ]) [0, + x n+ ] for all i 2 n+. Consequently, there is no y (, 3/2] = n 0 ( + x n+, + x n ] such that y ω(y, g), hence Y (, 3/2] = and by density Y (, 3/2] =. Since g 2n (0) = a 2n by Lemma 4.(ii)+(iii), the point 0 is not periodic, hence g k (0) [a, ] for all k. If y (a, ) then g(y) = 0 and g k (y) [a, ] for all k, which implies that y ω(y, g). Consequently, Y (a, ) =. We find that Y [0, a] {}, and Y because Y has no isolated point; this gives (i). Let n. Since min L n = max In 0 and max L n = min In+ 0, it follows that g(min L n ) = max Jn and g(max L n ) = min Jn+ by Lemma 4.(i). Moreover g [min Ln,mid(L n )] is and g [mid(ln ),max L n ] is linear, so there exists c n in [mid(l n ), max L n ] such that g(c n ) = g(min L n ). Since g([c n, max L n ]) = [min Jn+, max J n] is included in Jn 0, the map g 2n [cn,max L n ] is linear by Lemma 4.(ii). Moreover M n g([c n, max L n ]), hence g 2n ([c n, max L n ]) contains L n by Lemma 4.(vii). Consequently, there exists z n [c n, max L n ] such that g 2n (z n ) = z n (Lemma 2.4) and slope(g 2n [cn,max L n ]) 2. Then for every x [c n, max L n ], x z n, there exists k such that g k2n (x) [c n, max L n ]. By Lemma 4.2(v) one has g 2n (I n [, + x n ]) I n [, + x n ], which implies that (0) x [c n, max L n ], x z n, k, We show by induction on n that g k2n (x) I 0 n [min L n, c n ] I n [, + x n ]. () n 0, Y I n. This is true for n = 0 because Y [0, ] = I0 by (i). Suppose that there exists y Y In. Write I n = In = I0 n L n In; to prove that Y In we split into four cases.

20 00 S. Ruette If y I n there is nothing to do. If y I 0 n then g 2n (y) I n by Lemma 4.(vi) and g 2n (y) Y. If y [min L n, c n ] then g(y) g([min L n, mid(l n )] and g 2n (y) [, + x n ] by Lemma 4.2(ii), which is impossible because Y [0, a] by (i). If y [c n, max L n ] then y z n because Y is infinite. In addition g j (y) [0, ] for all j 0 according to (i). Then (0) says that there exists j such that g j (y ) I 0 n [min L n, c n ] I n and one of the first three cases applies. One has g(i n) J 0 n [min M n, mid(m n )] by Lemma 4.2(iv) and also g 2n ([min M n, mid(m n )]) = g 2n ([min L n, mid(l n )]) = [, + x n ] by Lemmas 4.(vii) and 4.2(ii). Together with (i) this implies that (2) g(y I n) J 0 n. Now () and (2) combined with Lemma 4.(i)+(iii) imply that Y 2 n i=0 g i (J 0 n) for all n, which is (ii); furthermore Y g i (J 0 n) = Y g i mod 2n (J 0 n) for all i 0. Since g(y ) = Y it is clear that g i (J 0 n Y ) g i (J 0 n) Y and that g 2n (g i (J 0 n) Y ) g 2n +i (J 0 n) Y ; thus g i (J 0 n Y ) = g i (J 0 n) Y = g i mod 2n (J 0 n) Y for all i 0. The next lemma is the key tool in the proof of Proposition 4.7. It relies on the knowledge of the precise location of g i (J 0 n) in k n I0 n. Lemma 4.6. Let g be the map defined in Section 4.. For all n and all 0 k 2 n one has slope(g 2n k g k (J 0 n)). Proof. A (finite) word B is an element of N n for some n N. If B, B are two words, BB denotes their concatenation and B = n is the length of B. We define inductively a sequence (B n ) n of words by: B =, B n = nb B 2... B n, and we define the infinite word ω = (ω(i)) i by concatenating the B n s: ω = B B 2 B 3... B n.... A straightforward induction shows that B n = 2 n ; thus B + B B k = 2 k and the word B k+ begins at ω(2 k ), which gives (3) (4) ω(2 k ) = k +, ω(2 k + )... ω(2 k+ ) = B... B k = ω()... ω(2 k ).

21 We prove by induction on k that Transitive interval maps 0 (5) g i (J 0 n) I 0 ω(i) for all n k, i 2 k. Case k = : Jn 0 I 0 = I0 ω() for all n. Suppose that (5) holds for k and let n k +. Since Jn 0 Jk+ 0, Lemma 4.(iii) implies that g 2k (Jn) 0 Ik+ 0, and thus g2k (Jn) 0 Jk 0 by Lemma 4.(i). By induction one has g i (Jk 0) I0 ω(i) for all i k, and (4) yields ω(i) = ω(2 k + i) for all i 2 k. Consequently, g 2k +i (Jn) 0 I 0 ω(2 k +i) for all i 2k. Together with the induction hypothesis this gives (5) for k +. Let µ n = slope(g I 0 n ). By definition of g one has slope(ϕ n ) µ n = n slope(ϕ i). It is straightforward from (5) that for all 2 k 2 n, (6) slope(g k J 0 n ) = k µ ω(i). By Lemma 4.(ii)+(iii) the map g 2n J 0 n is linear and g 2n (Jn) 0 = In, thus slope(g 2n J 0 n ) = I n n Jn 0 = 2/3 i /3 i. Since slope(ϕ i ) = I i I 0 i = 2/3i /3 i, we get (7) slope(g 2n J 0 n ) = 2 n µ ω(i) = n slope(ϕ i ). We show by induction on n that for all k 2 n, k n (8) µ ω(i) = slope(ϕ i ) ε i for some ε i = ε(i, k, n) {0, }. µ ω() = µ = slope(ϕ ); this gives the case n =. Suppose that the statement is true for n. Then 2 n 2 n µ ω(i) = µ ω(i) µ n+ by (3) = n slope(ϕ i ) slope(ϕ n+) n slope(ϕ i) = slope(ϕ n+ ). by (7) This is (8) for n + and k = 2 n with ε(i, k, n) = 0 for i n and ε(n +, 2 n, n + ) =.

22 02 S. Ruette Next, ω(2 n + )... ω(2 n+ ) = ω()... ω(2 n ) by (4), so if 2 n + k 2 n+ then k 2 n k k 2 n µ ω(i) = µ ω(i) µ ω(i) = slope(ϕ n+ ) i=2 n + = slope(ϕ n+ ) n slope(ϕ i ) ε(i,kn 2,n) µ ω(i) That is, (8) holds with ε(i, k, n + ) = ε(i, k 2 n, n) for all i n and ε(n +, k, n + ) =. This concludes the induction. Equations (6) and (8) show that for all k 2 n, k+ n (9) slope(g k Jn ) = µ ω(i) = slope(ϕ i ) ε i for some ε i {0, }. Since slope(g 2n k g k (Jn)) 0 = slope(g2n J 0 n ) slope(g k, J 0 n ) (7) and (9) imply that slope(g 2n k g k (J n )) is a product of at most n terms of the form slope(ϕ i ). This concludes the proof of the lemma because slope(ϕ i ) for all i. Proposition 4.7. The map g defined in Section 4. is not Wiggins chaotic. Proof. Let Y [0, 3/2] be a closed invariant subset such that g Y is transitive. We assume that Y has no isolated point, otherwise g Y is not sensitive. The sets (g i (Jn 0 Y )) 0 i 2 n are closed and by Lemma 4.(v) they are pairwise disjoint; let δ n > 0 be the minimal distance between two of them. If x, x Y and x x < δ n then there is 0 i 2 n such that x, x g i (Jn) 0 and g k (x), g k (x ) g i+k mod 2n (Jn) 0 for all k 0 by Lemma 4.5(ii)+(iii). Let ε n = max{diam(g i (J 0 n) Y ) 0 i < 2 n }. By Lemma 4.6, diam(g k (J 0 n) Y ) diam(g 2n (J 0 n) Y ) for all 0 k 2 n. By Lemma 4.(iii) one has g 2n (J 0 n) = I n and by Lemma 4.5(i) one has I n Y [a 2n, a]; thus ε n diam(i n Y ) a a 2n, which implies that lim n + ε n = 0. This shows that g Y is not sensitive. Theorem.8 now follows from Propositions 4.4 and 4.7.

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24 04 S. Ruette [24] S. Wiggins, Introduction to Applied Nonlinear Dynamical Systems and Chaos, Texts in Appl. Math. 2, Springer, 990. [25] J. C. Xiong, A chaotic map with topological entropy [zero], Acta Math. Sci. (English Ed.) 6 (986), Laboratoire de Mathématiques Topologie et Dynamique, Bâtiment 425 Université Paris-Sud F-9405 Orsay Cedex, France Sylvie.Ruette@math.u-psud.fr Received September 7, 2004 Revised version January 2, 2005 (5493)