A NEW TYPE OF NON-LANDING EXPONENTIAL RAYS

. In this paper, we will construct a new type of non-landing exponential rays, each of whose accumulation sets is bounded, disjoint from the ray and homeomorphic to the closed topologist’s sine curve.


1.
Introduction. Let E λ (z) = λ · e z (λ = 0). An escaping point is by definition a point z such that the iterates E n λ (z) → ∞ as n → ∞. It is known that all the escaping points of E λ are organized in rays, one end of which are toward infinity [16]. This result has been extended to transcendental entire functions in the Eremenko-Lyubich class with finite order [15]. For these rays, an interesting topic is the landing problem. On one hand, as an analogue of Douady-Hubbard's theorem on external rays in polynomial dynamics [7,Theorem 18.10], all periodic rays land for exponential maps with non-escaping singular value [11,17]. On the other hand, people are also interested in the non-landing rays of exponential maps. For example, the authors in [3,4,5,14] constructed some non-landing rays, each of whose accumulation sets is unbounded and an indecomposable continuum containing some ray in the extended complex plane. Recently, the first author and G. Zhang constructed some non-landing rays, each of whose accumulation sets is bounded and can be an indecomposable continuum containing part of the ray, an indecomposable continuum disjoint from the ray or a Jordan arc [6].
One common feature of the non-landing exponential rays constructed above is that each of the corresponding accumulation sets is an indecomposable continuum or a Jordan arc. Hence, it is natural to ask whether there are some non-landing exponential rays with accumulation sets whose complexities of topology are "between the two extremal cases", such as the closed topologist's sine curve {(x, y)|0 < x ≤ 2/π, y = sin(1/x)} ∪ {(0, y)| − 1 ≤ y ≤ 1}, which is decomposable but not locally connected [8,9].
In this paper, we will give a positive answer to this question by considering the case of post-singularly finite exponential maps. For simplicity, rather than deal with the general case, we shall deal only with the case where λ = 2πi. Theorem 1.1. For post-singularly finite exponential map 2πi · e z , there exist nonlanding rays each of whose accumulation set is bounded, disjoint from the ray and homeomorphic to the closed topologist's sine curve.
We would like to mention that for transcendental entire functions of disjoint type, Rempe in [12] proved that for each function with bounded slope, the Julia continuum is arc-like. Conversely, they also proved that each arc-like continuum with at least one terminal point can be realised as a Julia continuum. In particular, the closed topologist's sine curve can occur as Julia continuum of a disjoint type entire function. While we consider the case of post-singularly finite exponential maps, which is not of disjoint type. In fact, in this case, the Julia set is the whole complex plane, which can be regarded as the closure of all the escaping rays. Theorem 1.1 says that the closed topologist's sine curve can occur as the accumulation set of one of the escaping rays.
The organization of the paper is as follows. In §2, we recall some basic knowledge on the escaping rays of exponential maps, hyperbolic expansion lemma and Bounded-wiggling lemma appeared in [6]. In §3, we show how to determine the itineraries of the non-landing rays whose accumulation sets will be homeomorphic to the closed topologist's sine curve and define the folding points of the rays. In §4, we study several properties of the non-landing rays determined in §3 and prove Theorem 1.1.

2.
Preliminaries. In this section, we first briefly sketch some basic knowledge on the escaping rays of exponential maps (see [2,13] for a thorough background in this aspect). Then, we recall the hyperbolic expansion lemma and the Bounded-wiggling lemma.
2.1. The escaping rays of exponential maps. Here, the construction of the non-landing rays described in Theorem 1.1 rely on the parameter λ = 2πi. In fact, the construction can be adapted to all post-singularly finite exponential maps. Let E(z) = 2πi · e z . Following [1] or [5], there is a unique ray γ 2πi of E landing at 2πi with E(γ 2πi ) = γ 2πi . Let η 2πi be the pre-image of γ 2πi which lands at the origin. Then E −1 (η 2πi ) consists of infinitely many disjoint curves t j , j ∈ Z, with t j+1 = t j + 2πi (we assume that 0 lies in the strip bounded by t 1 and t 0 ). Note that every t j extends from the left to right across the entire plane. Let T j , j ∈ Z, be the union of t j and the open strip bounded above by t j+1 and below by t j . Then the itinerary of z ∈ C is defined to be the sequence of integers s(z) = (s 0 s 1 s 2 · · · ), where E j (z) ∈ T sj for all j ≥ 0.
In our case, the itineraries we involved always satisfy that |s j | is uniformly bounded. For ζ > 0 large enough and M > 0, set Denote by ω s (ζ) the set of escaping points z ∈ C with orbits in H ζ satisfying s(z) = s and Re(E n+1 (z)) > Re(E n (z)) for all n ≥ 0. We call it a tail associated to itinerary s. In fact, the following lemma shows that ω s (ζ) is nonempty and a continuous curve toward infinity.
are the branches of the inverse of E. Then for given s = (s 0 s 1 s 2 · · · ), the set is called the escaping ray with itinerary s, where σ(s) = (s 1 s 2 · · · ). Note that the ray γ s can be parameterized as a continuous curve γ is called the accumulation set of γ s . If the set A s consists of a single point, then we say γ s is landing; otherwise, γ s is non-landing.
As the starting point of our proof, we now recall the "folding phenomena" which was observed by authors in [3,4,5,14]. That is, if a ray is close enough to the origin, then the preimages of the ray extend far away to the negative infinity, and the further preimages will fold somewhere. In our case, suppose ω is a tail in the strip T 1 . We pull back ω by L k 1 with k large enough, so that the endpoint of L k 1 (ω) (⊂ T 1 ) is close to 2πi. Then pulling back L k 1 (ω) by L 0 , we get that the endpoint of L 0 • L k 1 (ω) (⊂ T 0 ) is close to the origin. Next pulling back L 0 • L k 1 (ω) by L ±1 , we get the endpoint of L ±1 • L 0 • L k 1 (ω) (⊂ T ±1 ) extends to the far left. At last, we pull back L ±1 • L 0 • L k 1 (ω) by L 1 and set As a result, the curve G ± k (ω)(⊂ T 1 ) will fold at some point contained in a bounded subset of the plane independent of k. Moreover, if k is large, the endpoint can reach to the right half-plane as far as we wanted. It is worth pointing out that for the folding direction, G + k folds ω up and G − k folds ω down. For later use, we sometimes denote G ± k by G k for simplicity. Denote by I ± k the corresponding itinerary block (1, ±1, 0, 1 k ) of G ± k , where 1 k is the block of 1 of length k. For our purpose, we will be only concerned with the itineraries of the following form (I − k1 ; I + k2 ; I − k3 ; I + k4 ; · · · ; I − k2n−1 ; I + k2n ; · · · ). 2.2. Hyperbolic expansion and bounded-wiggling Lemma. Let us recall the following hyperbolic expansion lemma. Denote by X the hyperbolic Riemann surface C \ {0, 2πi} and ρ X the corresponding hyperbolic density function on X.
Lemma 2.2 (Cf. [6]). There is a δ 0 ∈ (0, 1) such that for all k ≥ 1 and z ∈ X, we have Notations. In the remainder of the paper, we fix some notations for simplicity. First, we always use hyperbolic metric in X = C \ {0, 2πi} rather than Euclidean metric in C unless otherwise stated. In particular, we use dist(·, ·) and diam(·) to denote respectively the hyperbolic distance and the hyperbolic diameter in X. For any R > 0 and a set A, we denote by U R (A) the R-neighborhood of A with respect to hyperbolic metric in X. In addition, the symbol δ 0 always represent the number which is defined in Lemma 2.2.
In the following, we give the concept of folding point and recall the Boundedwiggling Lemma.
For r 0 > 0, let Let Γ a,b ⊂ T 1 ∩ Ω be an injective curve with two endpoints a and b. We say that the map G ± k folds Γ a,b into two pieces with respect to a positive number ζ if min{Re(G ± k (a)), Re(G ± k (b))} > ζ and there is a point c ∈ G ± k (Γ a,b ) such that Re(c) ≤ ζ. Note that if one endpoint of Γ a,b is at ∞, then we put G ± k (∞) = ∞. The point c, together with the endpoints G ± k (a) and G ± k (b), is called the folding points of G ± k (Γ a,b ). Now we claim that for is also bounded away from 2πi by some constant which is independent of k and r 0 . On the other hand, if r 0 is small enough, it is clear that for any z ∈ T 1 , Re(G ± k (z)) > −1/r 0 . This completes the claim. For integers k 1 , · · · , k n (n ≥ 2), let In the following, we will denote g ± i by g i for simplicity. For an injective curve Γ ⊂ T 1 ∩ Ω (finite or infinite), let us define the folding points of g 1 • · · · • g n (Γ) by induction. First, the folding points of g n (Γ) are defined to be the union of the point c (if it exists) and the endpoints of g n (Γ) as above. Suppose the folding points of g i+1 • · · · • g n (Γ) have been defined. We say that z is a folding point of g i • · · · • g n (Γ) if either of the following two conditions holds: 1. There is a folding point w of g i+1 • · · · • g n (Γ) such that z = g i (w).
2. There is a subarc Λ of g i+1 • · · · • g n (Γ) with the endpoints being two adjacent folding points of g i+1 • · · · • g n (Γ) such that z is the folding point of g i (Λ). For any two points x, y ∈ Γ, let Γ x,y ⊂ Γ be the subarc within x and y. The wiggling of Γ is defined by where the sup is taken over all Γ x,y with x, y ∈ Γ and Re(x) = Re(y).
Then there exists a K depending only on K 0 such that for any arc Λ between two adjacent folding points of g i • · · · • g n (Γ), we have where i, n are integers with 0 < i < n < ∞.
3. Realization of the prescribed ray. The goal of this section is to pick out the itineraries elaborately from the following form s * = (I − k1 ; I + k2 ; I − k3 ; I + k4 ; · · · ; I − k2n−1 ; I + k2n ; · · · ), so that the corresponding accumulation sets of non-landing rays are homeomorphic to the closed topologist's sine curve.
3.1. The choice of k n . For any itinerary s and integer n > 0, let γ 0 n (s) be the tail ω s and z 0 n (s) be its endpoint (note that Re(z 0 n (s)) = ζ). For 0 ≤ i ≤ n − 1, denote the tail of g i+1 • · · · • g n (ω s ) by γ 0 i (s) and the endpoint of Let γ i (s) be the escaping ray containing γ 0 i (s) and [z j1 Figure 1 for a sketch of the above definitions. Now let us discuss how to choose the sequence {k n }. We first take k 1 ≥ 1 large enough. For an integer n > 0, suppose k i (1 ≤ i ≤ n − 1) have been defined. Let us define k n . Note that γ −1 n−1 (= g n (γ 0 n )) extends γ 0 n−1 and folds at some point with bounded hyperbolic distance from z 0 n−1 . Moreover, the endpoint z −1 n−1 can reach to the right half-plane as far as wanted provided that k n is large enough. On the other hand, we claim that [z 0 n−1 , z n−1 n−1 ] is independent of k n . In fact, [z 0 n−1 , z n−1 n−1 ] belongs to the tail associated to the itinerary (1, −1, 0, 1, 1, · · · ). This tail is unique since γ 2πi is unique as we mentioned above. Then we show that the point z n−1 , this finishes the claim. Therefore, if we take ξ n−1 ∈ [z 0 n−1 , z n−1 n−1 ], then k n can be defined as follows. ( * ) Let k n be the least integer such that Re(z −1 n−1 ) ≥ Re(ξ n−1 ).

3.2.
The choice of ξ n . In this subsection, we will choose the sequence {ξ n } to realize the non-landing rays whose accumulation sets are homeomorphic to the closed topologist's sine curve.
Lemma 3.1. Let n ≥ 1. Then for any itinerary s with γ 0 n (s) ⊂ T 1 , there is a homeomorphism where D 1 is a universal constant.
Proof. By Lemma 3.1 and a similar argument as in the proof of Lemma 5.3 in [6], we get that there is a universal constant D 4 such that dist(z −1 n−1 (s), ξ(s)) < D 4 . By Lemma 2.3, it follows that the wiggling of both [z 0 n−1 (s), ξ(s)] and [z 0 n−1 (s), z −1 n−1 (s)] is bounded by some universal constant. This implies that the existence ofφ n−1 so that (1) holds for some universal constant D 3 . The inequality (2) then follows from (1) and Lemma 2.2.
3.3. The folding points of γ s * . Recall the definition of the folding point c appeared in § 2.2, one can also take any point c 0 to be the folding point provided that the diameter of the subarc between c and c 0 is bounded by some universal constant. From this, we can also define the set of folding points of by the following inductive way, and denote it by Z i . Let n ≥ 1 be fixed and Z n = {z 0 n }. For any 0 ≤ i ≤ n − 1, suppose Z i+1 has been defined. Let us define Z i as follows. Let W 0 = {z 0 i }. Suppose W j has been defined for j ≥ 0. Let W j+1 be the set consisting of the well defined points of where z ∈ W j andφ l (i + 1 ≤ l ≤ n − 1) are the maps defined in Lemma 3.2. Define In fact, the set of folding points is defined in accordance with geometric visualization, and hence is readily comprehensible.
4. The accumulation set of the prescribed ray. Let γ 0 = γ s * be the escaping ray with itinerary s * obtained in § 3. In this section, we will prove that the accumulation set of γ 0 is bounded, disjoint from γ 0 and homeomorphic to the closed topologist's sine curve. Let i ≥ 0 and j ≤ i. From now on, we redefine the notations γ i , γ j i and z j i as follows. Let γ i = f i • · · · • f 1 (γ 0 ). We denote the tail of γ i by γ 0 i and the endpoint of γ 0 i by z 0 i . Then we define γ j i and z j i by Proof. For k ≥ 1, let It is sufficient to prove that γ 0 \ γ 0 0 is contained in D 5 -neighborhood U D5 (I 1 ) of I 1 with respect to hyperbolic metric in X for some constant D 5 > 0 defined later. For any compact subsets A, B ⊂ X, recall that the semi-Hausdorff distance is defined bỹ Since ξ n−1 ∈ [z 0 n−1 , z n−1 n−1 ], Lemma 2.2 together with Lemma 3.2 implies that d H (I n , I 1 ∪ · · · ∪ I n−1 ) Then, we havẽ d H (I n , I 1 ) ≤d H (I n , I 1 ∪ · · · ∪ I n−1 ) +d H (I n−1 , I 1 ∪ · · · ∪ I n−2 ) + · · · +d H (I 2 , I 1 ) holds for all n ≥ 1. Hence, the proof of the proposition can be finished by taking provided that the first inequality of (5) holds. Now we prove the first inequality of (5) by induction. We first prove the case n = 3, that isd H (I 3 , I 1 ) ≤d H (I 3 , I 1 ∪ I 2 ) +d H (I 2 , I 1 ).
Combining with the two inequalities above, we complete the proof of the first inequality of (5) for n = k + 1.

4.2.
The topology of accumulation set of γ 0 . For any k ≥ 1, set whereφ k is the map defined by Lemma 3.2.
To get the accumulation set of γ 0 , we first give a parameterization of I k = [z −k+1 0 , z −k 0 ] for k ≥ 1. In fact, according to the choice of ξ k−1 , we can parameterize I k inductively as follows (see Figure 2). Let ω 1 (t) (0 ≤ t ≤ 1/2) be any parameterization of I 1 with ω 1 (0) = z 0 0 and ω 1 (1/2) = z −1 0 . We parameterize Then we can define Next, for s ≥ 1, we can parameterize For t ∈ [0, 1/2], the parameterization ω 4s+1 and ω 4s+2 are defined by and Then we define This completes the parameterization of I k for all k ≥ 1.  Proof. Note that for any t ∈ [0, s 0 ], we have and ω 4s+3 (t) If we set by Lemma 3.2. This proves the lemma.
See Figure 2 for an illustration of Lemma 4.2.
On the other hand, by Lemma 2.2, we have and η(t 1 ) = η(t 2 ). Then for any > 0, by taking s large enough, we can get that contains no folding points of γ 4r−1 in its interior, this contradicts with Lemma 2.3. Therefore, η is one-to-one. Proof. We first prove that {η(s − t)} s≥1 is uniformly convergent to an arc α(t) on t ∈ [0, 1/2]. Let , and Note thatg Then by Lemma 3.2, we get that Therefore, we have Here and subsequently, the symbol "⇒" means uniform convergence. In addition, Lemma 4.2 implies that for any > 0 and s 0 ≥ 1, there exists an s * depending only on such that for all s ≥ max{s * , s 0 }, we have Then for the above > 0, we obtain that for all s ≥ s * , Hence, by (8) and (9), dist(η(s + 1 − t), η(s − t)) ⇒ 0 as s → ∞.

JIANXUN FU AND SONG ZHANG
Finally, we prove ∩ n>0 η((n, ∞)) = α. Let {t n } be a sequence with t n → ∞ as n → ∞. Then for any subsequence {t n k } of {t n } such that {η(t n k )} converges to a point z, it suffices to prove that z ∈ α. Suppose z / ∈ α. Since α is compact, we get that dist(z, α) > 0. It is worth pointing out that for any t n k , there exist an x k ∈ [0, 1/2] and a positive integer N k such that t n k = N k ± x k . Then by the conclusion proved above, we get dist(η(t n k ), α(x k )) → 0 as k → ∞. This implies that dist(z, α) → 0, a contradiction.
The proof of the lemma is finished.
Proposition 4.6. The accumulation set A s * of γ 0 is exactly the setη.