On backward attractors of interval maps

Special $\alpha$-limit sets ($s\alpha$-limit sets) combine together all accumulation points of all backward orbit branches of a point $x$ under a noninvertible map. The most important question about them is whether or not they are closed. We challenge the notion of $s\alpha$-limit sets as backward attractors for interval maps by showing that they need not be closed. This disproves a conjecture by Kolyada, Misiurewicz, and Snoha. We give a criterion in terms of Xiong's attracting center that completely characterizes which interval maps have all $s\alpha$-limit sets closed, and we show that our criterion is satisfied in the piecewise monotone case. We apply Blokh's models of solenoidal and basic $\omega$-limit sets to solve four additional conjectures by Kolyada, Misiurewicz, and Snoha relating topological properties of $s\alpha$-limit sets to the dynamics within them. For example, we show that the isolated points in a $s\alpha$-limit set of an interval map are always periodic, the non-degenerate components are the union of one or two transitive cycles of intervals, and the rest of the $s\alpha$-limit set is nowhere dense. Moreover, we show that $s\alpha$-limit sets in the interval are always both $F_\sigma$ and $G_\delta$. Finally, since $s\alpha$-limit sets need not be closed, we propose a new notion of $\beta$-limit sets to serve as backward attractors. The $\beta$-limit set of $x$ is the smallest closed set to which all backward orbit branches of $x$ converge, and it coincides with the closure of the $s\alpha$-limit set. At the end of the paper we suggest several new problems about backward attractors.


Contents
1. Introduction 2 2. Terminology 3 3. Maximal ω-limit sets and their relation to special α-limit sets 4 3.1. Solenoidal sets 6 3.2. Basic sets 8 4. General properties of special α-limit sets for interval maps 10 4.1. Isolated points are periodic 10 4.2. The interior and the nowhere dense part of a special α-limit set 11 4.3. Transitivity and points with sα(x) = [0, 1] 12 4.4. Special α-limit sets containing a common open set 13 5. On special α-limit sets which are not closed 17 5.1. Points in the closure of a special α-limit set 17 5.2. Properties of a non-closed special α-limit set 21 5.3. Maps which have all special α-limit sets closed 22 5.4. Example of a non-closed special α-limit set 23 6. Open Problems 24

Introduction
Let a discrete dynamical system be defined as an ordered pair (X, f ) where X is a compact metric space and f is a continuous map acting on X. To understand the dynamical properties of such a system it is necessary to analyze the behavior of the trajectories of any point x ∈ X under the iteration of f . Limit sets of trajectories are a helpful tool for this purpose since they can be used to understand the long term behavior of the dynamical system.
The ω-limit sets (ω(x) for short), i.e. the sets of limit points of forward trajectories, were deeply studied by many authors. For instance, one can ask for a criterion which determines whether a given closed invariant subset of X is an ω-limit set of some point x ∈ X. The question is very hard in general, however the answer for ω-limit sets of a continuous map acting on the compact interval was provided by Blokh et al. in [4]. A closely related question is that of characterizing all those dynamical systems which may occur as restrictions of some system to one of its ω-limit sets. These abstract ω-limit sets were studied by Bowen [5] and Dowker and Frielander [10]. It was also proved that each ω-limit set of a continuous map of the interval is contained in a maximal one by Sharkovsky [17].
α-limit sets (α(x) for short) were introduced as a dual concept to ω-limit sets and they should be regarded as the source of the trajectory of a point. While for invertible maps α-limit sets are well defined, for noninvertible maps there are many possibilities how to construct the limit of the backward trajectory. One possibility is to take as an α-limit set the set of all accumulation points in X of the preimage sets f −n ({x}). This approach was used by Coven and Nitecki [7], who showed that for an interval map, a point x is nonwandering if and only if x ∈ α(x). A recent result about α-limit sets of unimodal interval maps is due to Cui and Ding [9]. Another approach used by Balibrea et al. [1] proposes instead of looking at all possible preimages to pick one backward branch and check accumulation points of this sequence. The union of the sets of accumulation points over all backward branches of the map was called a special α-limit set (sα(x) for short) by Hero [11].
While α-limit sets and special α-limit sets seem to be similar to ω-limit sets, they were not much explored so far. The reason for this is that they may have very rich structure, and also it is very hard to control the dynamics backward. For instance, it is clear that α-limit sets or ω-limit sets are always closed, but the situation of special α-limit sets is unclear. By definition [11], those sets are in general uncountable unions of closed sets, so a priori their structure may be very complicated. A recent study by Kolyada et al. [12] provided some answers. In particular, they showed that a special α-limit set need not be closed in the general setting. They investigated special α-limit sets of interval maps and proved that for interval maps with a closed set of periodic points, every special α-limit set has to be closed. This result led to the following conjecture: Conjecture 1. [12] For all continuous maps of the unit interval all special α-limit sets are closed.
We disprove the conjecture by showing a counterexample of an interval map with a special α-limit set which is not closed and give the properties of interval maps that determine if all special α-limit sets are closed in Sections 5. On the other hand, we show that for all continuous maps of the unit interval all special α-limit sets are both F σ and G δ . We give further topological properties of special α-limit sets of interval maps. If sα(x) is not closed, then it is uncountable and nowhere dense. If sα(x) is closed, then it is the union of a nowhere dense set and finitely many (perhaps zero) closed intervals, and in Section 4.2 we prove some amount of transitivity of f on those intervals. Since sα-limit sets need not be closed, we propose a new notion of β-limit sets to serve as backward attractors in Section 6. The β-limit set of x is the smallest closed set to which all backward orbit branches of x converge, and it coincides with the closure of the sα-limit set.
Kolyada et al. also made the following conjecture.
Conjecture 2. [12] The isolated points in a special α-limit set for an interval map are always periodic.
We verify this conjecture in Section 4.1. We also show that a countable special α-limit set for an interval map is a union of periodic orbits. These results are opposite to the case of ω-limit sets. The ω-limit sets of a general dynamical system do not posses any periodic isolated points unless ω(x) is a single periodic orbit [16].
The authors of [12] also investigated the properties of special α-limit sets of transitive interval maps and stated the following conjecture: We slightly correct these conjectures in Section 4. 3. We show that f is transitive if there are three distinct points x, y, z ∈ [0, 1] with sα(x) = sα(y) = sα(z) = [0, 1] . If f has one or two points with special α-limit sets equal to [0, 1], but not more, then [0, 1] is the union of two transitive cycles of intervals.
It is known that if two ω-limit sets of an interval map contain a common open set, then they are equal. The last conjecture in [12] suggested that a similar property holds for special α-limit sets: We correct this conjecture by showing that at most three distinct special α-limit sets of f can contain a given nonempty open set in Section 4.4.
The paper is organised as follows. Sections 1 and 2 are introductory. Section 3 investigates the relation of maximal ω-limit sets to special α-limit sets and provides tools necessary for proving the main results. It also contains a simple example showing that, unlike ω-limit sets, the special α-limit sets of an interval map need not be contained in maximal ones. Section 4 is devoted to the above mentioned results on various properties of special α-limit sets of interval maps. Section 5 studies properties of special α-limit sets which are not closed. The paper closes with open problems and related questions in Section 6.

Terminology
Let X be a compact metric space and f : X → X a continuous map.
• a backward orbit branch of a point x if x 0 = x and f (x n+1 ) = x n for all n ≥ 0. A point y belongs to the ω-limit set of a point x, denoted by ω(x), if and only if the forward orbit of x has a subsequence {x n i } ∞ i=0 such that x n i → y. A point y belongs to the α-limit set of a point x, denoted by α(x), if and only if some preimage sequence of x has a subsequence {x n i } ∞ i=0 such that x n i → y. And a point y belongs to the special α-limit set of a point x, also written as the sα-limit set and denoted sα(x), if and only if some backward orbit branch of x has a subsequence {x n i } ∞ i=0 such that x n i → y. If we wish to emphasize the map, we will write ω(x, f ), α(x, f ) and sα(x, f ).
To summarize, the ω, α, and sα-limit sets of a point x are defined as follows. The set ω(x) is the set of all accumulation points of its forward orbit and α(x) (resp. sα(x)) is the set of all accumulation points of all its preimage sequences (resp. of all its backward orbit branches).
Let T : X → X and F : Y → Y be continuous maps of compact metric spaces. If there is a surjective map φ : The forward orbit of a point, regarded as a subset of X rather than a sequence, will be denoted by Orb(x) = {f n (x) : n ≥ 0}. The forward orbit of a set is Orb(A) = {f n (A) : n ≥ 0}. We call f transitive if for any two nonempty open subsets U, V ⊂ X there is n ≥ 0 such that f n (U )∩V = ∅. We call f topologically mixing if for any two nonempty open subsets U, V ⊂ X there is an integer N ≥ 0 such that f n (U ) ∩ V = ∅ for all n ≥ N . Now let f : [0, 1] → [0, 1] be an interval map. We write Per(f ), Rec(f ), and Ω(f ) for the sets of periodic points, recurrent points, and non-wandering points of f , respectively. We say that x is preperiodic if x / ∈ Per(f ) but f n (x) ∈ Per(f ) for some n ≥ 1. We write Λ 1 (f ) = x∈[0,1] ω(x) for the union of all ω-limit sets of f and SA(f ) = x∈[0,1] sα(x) for the union of all sα-limit sets. Following [21] we define the attracting center of f as Birkhoff center of f is the closure of the set of recurrent points Rec(f ) and coincides with Per(f ) [6]. If the map f is clear from the context, we may drop it from the notation. The relation of these sets is given by the following summary theorem from the works of Hero and Xiong [11,21].
If K ⊂ [0, 1] is a non-degenerate closed interval such that the sets K, f (K), . . . , f k−1 (K) are pairwise disjoint and f k (K) = K, then we call the set M = Orb(K) a cycle of intervals and the period of this cycle is k. We may also call K an n-periodic interval. If f | M is transitive then we call M a transitive cycle for f .

3.
Maximal ω-limit sets and their relation to special α-limit sets An important property of the ω-limit sets of an interval map f is that each ω-limit set is contained in a maximal one. These maximal ω-limit sets come in three types: periodic orbits, basic sets, and solenoidal ω-limit sets.
A solenoidal ω-limit set is a maximal ω-limit set which contains no periodic points. Any solenoidal ω-limit set is uncountable and is contained in a nested sequence Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · of cycles of intervals with periods tending to infinity, also known as a generating sequence [3,Assertion 4.2]. Here is the theorem relating sα-limit sets to solenoidal ω-limit sets; the proof is given in Section 3.1.
Theorem 6 (Solenoidal Sets). Let Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · be a nested sequence of cycles of intervals for the interval map f with periods tending to infinity. Let Q = Orb(I n ) and S = Q ∩ Rec(f ).
A basic set is an ω-limit set which is infinite, maximal among ω-limit sets, and contains some periodic point. If B is a basic set then with respect to inclusion there is a minimal cycle of intervals M which contains it, and B may be characterized as the set of those points x ∈ M such that Orb(U ) = M for every relative neighborhood U of x in M , see [3]. Conversely, if M is a cycle of intervals for f , then we will write and if this set is infinite, then it is a basic set [3]. Here is the theorem relating sα-limit sets to basic sets; the proof is given in Section 3.2.
(1) If α(y) contains an infinite subset of a basic set B = B(M, f ), then y ∈ M and sα(y) ⊃ B.
The sharpness of the second claim of Theorem 7 is illustrated in Figure 1. The first map has two basic sets B([0, 1], f ) and B(M, f ), where M is the invariant middle interval. The sα-limit set of 1 does not contain the basic set B(M, f ) although it includes the left endpoint of M , which is preperiodic. The second map shows that we cannot weaken the assumption to α(y). The α-limit set of 1 includes the preperiodic endpoint of M but sα(1) does not contain any basic set. Periodic orbits may also be related to sα-limit sets. The following result is one of the main theorems in [12]. Moreover, it holds for all periodic orbits of an interval map, even those which are not maximal ω-limit sets. One additional observation is appropriate in this section. Unlike ω-limit sets, the sα-limit sets of an interval map need not be contained in maximal ones.
Example 9. Fix two sequences of real numbers 1 = a 1 > b 1 > a 2 > b 2 > · · · both decreasing to 0 and consider the "connect-the-dots" map f : and f is linear on all the intervals [a i+1 , b i ], [b i , a i ]. The graph of such a function f is shown in Figure 2. The sα-limit sets of this map are sα(x) = {a 1 , . . . , a n } for x ∈ (a n+1 , a n ] and sα(0) = {0}.
In particular, we get a strictly increasing sequence of sα-limit sets and no sα-limit set containing them all.

Figure 2.
A map with an increasing nested sequence of sα-limit sets not contained in any maximal one.

Solenoidal sets.
This section is devoted to the proof of Theorem 6.
We start with a broader definition of solenoidal sets, taken from [3]. A generating sequence is any nested sequence of cycles of intervals Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · for f with periods tending to infinity. The intersection Q = n Orb(I n ) is automatically closed and strongly invariant, i.e. f (Q) = Q, and any closed and strongly invariant subset S of Q (including Q itself) will be called a solenoidal set. Two examples described in [3] which we will need later are (1) the set of all ω-limit points in Q, denoted S ω = S ω (Q) = Q ∩ Λ 1 (f ), and (2) the set of all recurrent points in Q, denoted S Rec = S Rec (Q) = Q ∩ Rec(f ). Blokh showed that Q contains a perfect set S such that S = ω(x) for all x ∈ Q [3, Theorem 3.1]. Clearly S = S Rec . We refer to S Rec as a minimal solenoidal set both because it is the smallest solenoidal set in Q with respect to inclusion, and because the mapping f | S Rec is minimal, i.e. all forward orbits are dense.
If ω(x) is a maximal ω-limit set for f and contains no periodic points (what Sharkovskii calls a maximal ω-limit set of genus 1), then it is in fact a solenoidal set [3,Assertion 4.2]. Thus it has a generating sequence Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · of cycles of intervals and belongs to their intersection Q. If Q = n Orb(I n ) is formed from another generating sequence for f , then it is well known (and an easy exercise) that Q and Q are either identical or disjoint. This means that given any solenoidal set S, there is a unique maximal solenoidal set Q which contains it (so Q is uniquely determined, even if the generating sequence is not).
One can use a translation in a zero-dimensional infinite group as a model for the map f acting on a solenoidal set Q = Orb(I j ). Let D = {m j } ∞ j=0 where m j is the period of Orb(I j ) and let H(D) = {(r 0 , r 1 , . . .) : r j+1 = r j (mod m j ), for all j ≥ 0} where r j is an element of the group of residues modulo m j , for every j. One lemma which we will need several times throughout the paper is the following Lemma 11. If A is invariant for f and α(x) ∩ Int(A) = ∅, then x ∈ A. In particular, if sα(x) ∩ Int(A) = ∅, then x ∈ A.
Proof. Choose a ∈ α(x) ∩ Int(A) and choose a neighborhood U of a contained in A. There is n ∈ N and a point x −n ∈ U such that f n (x −n ) = x. Since U ⊂ A and A is invariant, x must belong to A. We get the same conclusion when sα(x) ∩ Int(A) = ∅, because sα(x) ⊂ α(x). Now we are ready to give the proof of Theorem 6.
Proof of Theorem 6. (1): Fix z ∈ Q ∩ α(y) and let S = S Rec = Q ∩ Rec(f ) be the minimal solenoidal set in Q. Then by [3, Theorem 3.1] S = ω(z) and since α(y) is a closed invariant set it must contain S. In particular, α(y) contains infinitely many points from each cycle of intervals Orb(I n ), and so by Lemma 11 y ∈ Orb(I n ), for all n. Therefore y ∈ Q.
(2): Fix y ∈ Q. Since f (Q) = Q we can choose a backward orbit branch for y which never leaves Q. Therefore it has an accumulation point in Q, and so sα(y) ∩ Q = ∅. Let w ∈ sα(y) ∩ Q. By [3, Theorem 3.1], ω(w) is the minimal solenoidal set S. According to [11,Lemma 1], if sα(y) contains a point, then it contains the whole ω-limit set of that point as well. Therefore sα(y) ⊃ S and sα(y) ∩ Q ⊃ S. To finish the proof it is enough to show the opposite inclusion sα(y) ∩ Q ⊂ S.
We will assume otherwise. Suppose there is a point z ∈ sα(y) By Theorem 10, φ −1 (r) has to be an interval and z has to be one of its endpoints, say, the right endpoint, and φ −1 (r) = [x, z], x ∈ S. Since φ is a semiconjugacy we have f i ([x, z]) ⊂ φ −1 (τ i (r)) for all i ≥ 0. But the intervals φ −1 (τ i (r)) are pairwise disjoint. This shows that [x, z] is a wandering interval.
Claim z ∈ Int(Orb(I j )), for every j ≥ 0. We will assume otherwise. Let K be the connected component of Orb(I N ), for some N ≥ 0, where z is an endpoint of K. Let v be a point such that z ∈ ω(v). By [3, Theorem 3.1], S = ω(z) ⊂ ω(v), we have ω(v) ∩ Orb(I N ) infinite and necessarily Orb(v) ∩ Int(Orb(I N )) = ∅. This implies f k (v) ∈ Orb(I N ) for all sufficiently large k. It follows that Orb(v) accumulates on z from the interior of K and we can find k > 0 such that f k (v) ∈ (x, z). But [x, z] is a wandering interval, so Orb(v) cannot accumulate on z which contradicts z ∈ ω(v).
Let {y n } ∞ n=0 be a backward orbit branch of y with a subsequence {y n i } ∞ i=0 such that lim i→∞ y n i = z. Since z ∈ Int(Orb I j ) it follows from Lemma 11 that y n ∈ Orb(I j ) for all j, n ≥ 0. Therefore {y n } ∞ n=0 ⊂ Q. For every n ≥ 1, denote φ(y n ) = r n . Then by Theorem 10, φ −1 (r n ) are connected, pairwise disjoint sets, each containing an element s n ∈ S. Since y n ∈ φ −1 (r n ), we have lim i→∞ s n i = z. But S is a closed set and z / ∈ S, which is impossible. Therefore sα(y) ∩ (Q \ S) = ∅ and sα(y) ∩ Q ⊂ S. Corollary 12. A sα-limit set contains at most one solenoidal set.
Proof. Let Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · and Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · be nested sequences of cycles of intervals generating two solenoidal sets Q = Orb(I n ) and Q = n Orb(I n ). If sα(y) ∩ Q = ∅ and sα(y) ∩ Q = ∅ then, by Theorem 6, y ∈ Q ∩ Q . Since two solenoidal sets Q and Q are either identical or disjoint we have Q = Q . Then the only solenoidal set contained in sα(y) is S = Q ∩ Rec(f ).

Basic sets.
This section is devoted to the proof of Theorem 7. Let f be a continuous map acting on an interval I. We say that an endpoint y of I is accessible if there is x ∈ Int(I) and n ∈ N such that f n (x) = y. If y is not accessible, then it is called non-accessible. The following Proposition is derived from [14, Proposition 2.8].
Proposition 13. [14] Let f be a topologically mixing map acting on an interval I. Than for every > 0 and every x ∈ I such that the interval [x − , x + ] ⊂ I does not contain a non-accessible endpoint and for every non degenerate interval U ⊂ I, there exists an integer where U ranges over all relatively open nonempty subsets of M . It is known that E is finite; it can contain some endpoints and at most one non-endpoint from each component of M . If g is topologically mixing, then By [14,Lemma 2.32], every point in E is periodic and therefore g(E) = E. By the definition of non-accessible points, We use m-periodic transitive maps of cycles of intervals as models for maps acting on basic sets. We recall again Blokh's definition. If M is a cycle of intervals for the map  Proof. Let x ∈ B. There is > 0 such that φ| (x,x+ ) is not constant and φ((x, x + )) ∩ End(M ) = ∅ or φ| (x− ,x) is not constant and φ((x − , x)) ∩ End(M ) = ∅. Otherwise x has a neighborhood N such that φ| N is constant which is in a contradiction with x ∈ B by Theorem 14. We can assume φ| (x,x+ ) is not constant and φ((x, x + )) ∩ End(M ) = ∅, and denote The set E was defined as a union of non-accessible endpoints of a topologically mixing map g m (resp. g 2m ), therefore we can use Proposition 13 for the map g m (resp. g 2m ) acting on M . There is an ). But we have seen that it is not an endpoint. Therefore y ∈ f N (V ) and we can find By the same procedure, we can find y 2 ∈ (x, x + /2) ∩ M and N 2 ∈ N such that f N 2 (y 2 ) = y 1 . By repeating this process we construct a sequence {y n } ∞ n=1 converging to x which is a subsequence of a backward orbit branch of y. Since x ∈ B was arbitrary, this shows B ⊂ sα(y).
Proof. Let M, φ, g, M , E be as in the previous proof. Since the map φ| B is at most 2-to-1 and E is a finite set, there are uncountably many points y ∈ B such that φ(y) / ∈ E ∪ End(M ). The result follows by Lemma 15.
Before we proceed to the proof of Theorem 7 we need to recall the definition of a prolongation set and its relation to basic sets. Let M be a cycle of intervals. Let the side T be either the left side T = L or the right side T = R of a point x ∈ M and W T (x) be a one-sided neighborhood of x from the T -hand side, i.e. W T (x) contains for some > 0 the interval (x, x + ) (resp. (x − , x)) when T = R (resp. T = L). We do not consider the side T = R (resp. T = L) when x is a right endpoint (resp. left endpoint) of a component of M . Now let where the intersection is taken over the family of all one-sided neighborhoods W T (x) of x. We will write P T (x) instead of P T [0,1] (x). The following Lemma 17 and Lemma 18 about properties of prolongation sets are taken from [3].
is a closed invariant set and only one of the following possibilities holds: • There exists a wandering interval W T (x) with pairwise disjoint forward images and There is a close relation between prolongation sets and basic sets. If M is a cycle of intervals for f then we define Proof of Theorem 7. (1) Let φ and g be the maps given in Theorem 14 and E be the set of exceptional points of the map g acting on M . Since φ| B is an at most 2-to-1 map, φ(α(y)∩B) is an infinite subset of M . But E ∪ End(M ) is a finite set, so we can find a point z ∈ α(y) ∩ B such that φ(z) / ∈ E ∪ End(M ) and therefore z / ) is a union of finitely many, possibly degenerate, closed intervals in M .
. Then y ∈ M by the invariance of M . By Lemma 15 applied to y , sα(y ) ⊃ B. But the containment sα(y) ⊃ sα(y ) is clear from the definition of sα-limit sets, and so sα(y) ⊃ B.
(2) Let {y i } ∞ i=0 be the backward orbit branch of y accumulating on x. Since x is not a periodic point, it is not contained in {y i } ∞ i=0 more then one time and we can assume that accumulates on x from one side T . Consider the prolongation set P T (x). Clearly is closed and invariant, x and Orb(x) belong to P T (x), we see that P T (x) contains both periodic and non-periodic points. By Lemma 17, there is only one possibility P T (x) = M , where M is a cycle of intervals. The other possibilities are ruled out -Orb(p), where p is a periodic point and ω(x), where x is a preperiodic point, can not contain a non-periodic point; and a solenoidal set Q can not contain a periodic point. Since Let φ and g be the maps given in Theorem 14 and E be the set of exceptional points of the map g acting on M . By Lemma 18, φ is not constant on any T -sided neighborhood of x and since {y i } ∞ i=0 accumulates on x from the source side T , we can find a pre-image y ∈ M of y such that φ(y ) / ∈ E ∪ End(M ). Then sα(y ) ⊃ B by Lemma 15, and sα(y) ⊃ sα(y ) since y is a preimage of y. We conclude that sα(y) ⊃ B.
We record here one corollary which we will need several times in the rest of the paper. Proof. In this case M is itself a basic set, so we may apply Theorem 7.
4. General properties of special α-limit sets for interval maps 4.1. Isolated points are periodic.
Unless an ω-limit set is a single periodic orbit, its isolated points are never periodic [16]. The opposite phenomenon holds for the sα-limit sets of an interval map.
Theorem 20. Isolated points in a sα-limit set for an interval map are periodic.
Proof. Let z ∈ sα(y) such that z is neither periodic nor preperiodic. Then z is a point of an infinite maximal ω-limit set, i.e. a basic set or a solenoidal set. This follows from Blokh's Decomposition Theorem, that Λ 1 (f ) is the union of periodic orbits, solenoidal sets and basic sets, and from Theorem 5, z ∈ SA(f ) ⊂ Λ 1 (f ). According to [11,Lemma 1], when sα(y) contains a point z, it contains its orbit Orb(z) as well. If z is in a basic set B then, Orb(z) ⊂ B ∩ sα(y) is infinite and by Theorem 7 (1), B ⊂ sα(y). Then the point z is not isolated in sα(y) since B is a perfect set. If z is in a solenoidal set Q = n Orb(I n ) then, by Theorem 6, sα(y) ∩ Q = S. Again, the point z is not isolated in sα(y) since S is a perfect set.
Let z ∈ sα(y) such that z is a preperiodic point. By Theorem 7 (2), there is a basic set B such that z ∈ B ⊂ sα(y). Then the point z is not isolated in sα(y) since B is a perfect set.
In the previous proof, we have shown that if a point z ∈ sα(y) is not periodic then sα(y) contains either a minimal solenoidal set S or a basic set B. In both cases, sα(y) has to be uncountable. Therefore we have the following corollary.
Corollary 21. A countable sα-limit set for an interval map is a union of periodic orbits.

4.2.
The interior and the nowhere dense part of a special α-limit set.
A well-known result by Sharkovsky says that each ω-limit set of an interval map is either a transitive cycle of intervals or a closed nowhere dense set [17]. What can we say in this regard for sα-limit sets of interval maps? When Int(sα(x)) is nonempty, Kolyada, Misiurewicz, and Snoha showed that M = Int sα(x) is a cycle of intervals containing x, see [12,Proposition 3.6]. We strengthen this result by showing that the non-degenerate components of sα(x) are in fact closed, and the rest of sα(x) is nowhere dense. We also get some amount of transitivity.
The following lemma is simple and we leave the proof to the reader.
Lemma 22. Let M be a cycle of intervals for f of period k and let K be any of its components. Then Theorem 23. A sα-limit set for an interval map f is either nowhere dense, or it is the union of a cycle of intervals M for f and a nowhere dense set. Moreover, M is either a transitive cycle, or it is the union of two transitive cycles.
Proof. Consider a limit set sα(x) for an interval map f : [0, 1] → [0, 1]. Let M be the union of the non-degenerate components of sα(x). If M = ∅ then sα(x) is nowhere dense. Otherwise M must be a finite or countable union of closed intervals, and since M contains the interior of the closure of the sα-limit set we know that sα(x) \ M is nowhere dense. Let K be any component of M . By Theorem 5 we have K ⊂ Per(f ), and therefore periodic points are dense in K. Let n ≥ 1 be minimal such that f n (K) ∩ K = ∅. Since f n (K) is connected and K is a component of the invariant set sα(x), we know that f n (K) ⊆ K. Since periodic points are dense in K we must have f n (K) = K. Therefore Orb(K) is a cycle of intervals, and by Lemma 11 we get x ∈ Orb(K). Since this holds for every component K of M , we must have Orb(K) = M , i.e. M is a cycle of intervals and x ∈ M .
From now on we take K to be the component of M containing x. Put g = f n | K . Then g : K → K is an interval map with a dense set of periodic points. There is a structure theorem for interval maps with a dense set of periodic points [14, Theorem 3.9] 1 which tells us that K is a union of the transitive cycles and periodic orbits of g, K = L : L is a transitive cycle for g ∪ Per(g) By Lemma 22, sα(x, g) contains a dense subset of K. By Corollary 19 each transitive cycle L ⊆ K for g must contain x. Since transitive cycles have pairwise disjoint interiors, g has at 1 In fact, [14,Theorem 3.9] tells us that all the transitive cycles for g have period at most 2, and the periodic orbits not contained in transitive cycles also have period at most 2. Some of this extra information can be shown quite easily; it comes up again in our proof of Theorem 25. most two transitive cycles. If their union is not K, then K must contain a non-degenerate interval of periodic points of g. But by an easy application of Lemma 11, no sα-limit set can contain a dense subset of an interval of periodic points. Therefore K is the union of one or two transitive cycles for g. By Lemma 22, M is the union of one or two transitive cycles for f .
Finally, if L is one of the (at most two) transitive cycles for f that compose M , then by Corollary 19 we have sα(x) ⊇ L. Therefore M ⊆ sα(x). Proof. Suppose that f is not transitive. We will prove that at most two points have a full sα-limit set. Suppose there is at least one point x with sα(x) = [0, 1]. By Theorem 23 there are two transitive cycles L, L for f such that [0, 1] = L ∪ L , and by Corollary 19 every point with a full sα-limit set belongs to L ∩ L . We will show that the cardinality of L ∩ L is at most two.
Let A 1 , A 2 , . . . , A n be the components of L, numbered from left to right in [0, 1]. Let σ : {1, . . . , n} → {1, . . . , n} be the cyclic permutation defined by f (A i ) = A σ(i) . If n ≥ 3 then there must exist i such that |σ(i) − σ(i + 1)| ≥ 2, so there is some A j strictly between A σ(i) , A σ(i+1) . Let B be the component of L between A i and A i+1 , see Figure 3. By the intermediate value theorem, f (B) ⊃ A j . This contradicts the invariance of L . Therefore L has at most 2 components. For the same reason L has at most two components. Moreover, L, L cannot both have 2 components; otherwise the middle two of those four components have a point in common, but their images do not, again see Figure 3.
There are two cases remaining. L, L can have one component each, and then L ∩ L has cardinality one. Otherwise, one of the cycles, say L, has two components, while L has only one, and then L ∩ L has cardinality two. In the course of the proof we have also shown the following result: Corollary 26. If f : [0, 1] → [0, 1] has one or two points with full sα-limit sets, but not more, then [0, 1] is the union of two transitive cycles of intervals.
Corollary 26 corrects the second part of Conjecture 3 (originally [12, Conjecture 4.14]) to allow for two points with a full sα-limit set 2 . Both possibilities from the corollary are shown in Figure 4. One of the interval maps shown has exactly one point with a full sα-limit set, and the other has exactly two. (1) Each point x ∈ M \ (E ∪ End(M )) has the same sα-limit set.
(2) At most three distinct sα-limit sets of f contain M .
We will see in the course of the proof that if sα(x ) ⊇ M is distinct from the sα-limit set described in part (1), then x belongs to a periodic orbit contained in End(M ). Since there are at most two such periodic orbits in End(M ) we get part (2).
Before giving the proof we discuss some consequences of this theorem.
Corollary 28. At most three distinct special alpha-limit sets of f can contain a given nonempty open set.
Proof. Let U be a nonempty open set in [0, 1]. If a sα-limit set of f contains U , then by Theorem 23, U contains a whole subinterval from some transitive cycle M . Applying Corollary 19, we see that any sα-limit set which contains U also contains M . By Theorem 27 there are at most three such sα-limit sets.
This corollary corrects Conjecture 4 (originally [12, Conjecture 4.10]), in which it was conjectured that two sα-limit sets which contain a common open set must be equal. For comparison, note that if two ω-limit sets of an interval map contain a common open set, then they are in fact equal, since an ω-limit set with nonempty interior is itself a transitive cycle [17]. We also remark that the number three in Corollary 28 cannot be improved, as is shown by the following example.  In what follows it is necessary to allow for a weaker notion of a cycle of intervals for f . An interval is called non-degenerate if it contains more than one point. If U is a non-degenerate interval (not necessarily closed) such that U, f (U ), . . . , f n−1 (U ) are pairwise disjoint nondegenerate intervals and f n (U ) ⊆ U (not necessarily equal), then we will call Orb(U ) a weak cycle of intervals of period n. The next lemma records one of the standard ways to produce a weak cycle of intervals. Similar lemmas appear in [3] and several other papers, but since we were unable to find the exact statement we needed, we chose to give our own formulation here.
Lemma 30. If a subinterval U contains three distinct points from some orbit of f , then is a weak cycle of intervals for f . Proof. Let x, f n (x), f m (x) be three distinct points in U , 0 < n < m. Clearly Orb(U ) is invariant. Since the intervals U, f n (U ) both contain f n (x) we see that U ∪f n (U ) is connected, i.e. it is an interval. Then also f n (U ) ∪ f 2n (U ) is connected, and so on inductively. Therefore the set A = And since x, f n (x), f m (x) are distinct we get n = j 1 k, m = j 2 k with 0 < j 1 < j 2 . If any f i (B) is a singleton, i ≤ k, then so also is f k (B). Then using B ⊇ f k (B) ⊇ f 2k (B) ⊇ · · · , we see that all the sets f jk (B), j ≥ 1, are the same singleton. But in that case f j 1 k (B) = {f n (x)}, f j 2 k (B) = {f m (x)} are not the same singleton, which is a contradiction. For the rest of the proof we suppose that N ∩ M is finite. We no longer need transitivity and the sets N , M will play symmetric roles. Clearly each nonempty intersection N i ∩ M j is at common endpoints. This shows that N ∩M ⊆ End(M ). It also shows that each component

Proof of Theorem 27.
We have already seen that the set of exceptional points in M may be characterized as Combined with Lemma 11 this shows that Our task is to compare the sα-limit sets of the various points x ∈ M \ E. Since they all contain M it is enough to check whether or not they coincide outside of M . For any point y ∈ [0, 1] let us write sαBasin(y) = {z : y ∈ sα(z)}.
Step 1: If x ∈ M \ E and y ∈ sα(x) \ M , then there is a weak cycle of intervals N for f such that x ∈ N ⊆ sαBasin(y). To prove this claim, let (x i ) be a backward orbit branch of x accumulating on y. The sequence (x i ) cannot contain y twice, for otherwise y would be a periodic point containing x in its orbit, contradicting the invariance of M . Therefore there is a subsequence (x i j ) which converges to y monotonically from one side. We will suppose x i j y from the right. The proof when x i j y from the left is similar. For k = 1, 2, . . . let U k = (y, y + 1 k ). Then U k contains the points x i j for large enough j. By Lemma 30, the set V k = Orb(U k ) is a weak cycle of intervals. Since V k is invariant and contains points x i for arbitrarily large natural numbers i, we must have the whole backward orbit branch (x i ) contained in each V k . Moreover, we have nesting V 1 ⊇ V 2 ⊇ · · · . Letting v k denote the period (i.e. the number of connected components) of V k this implies v 1 ≤ v 2 ≤ · · · . But each V k contains the point x ∈ M , so by Lemma 31 (2) each v k ≤ 2m, where m is the period of M . A bounded increasing sequence of natural numbers is eventually constant. So fix k such that v k = v k +1 = · · · .
For each k let V 0 k be the connected component of V k which contains U k . Choose some i such that x i ∈ V 0 k , x i > y. For k ≥ k , V k is a weak cycle of intervals contained in V k and with the same number of components. It follows that V 0 k is the only component of V k which meets V 0 k . Therefore V 0 k must contain x i as well. In particular, setting δ = x i − y we find that ∀ >0 , Orb((y, y + )) ⊇ (y, y + δ).
Since (y, y + δ) contains x i j for sufficiently large j, Lemma 30 also implies that N is a weak cycle of intervals, and since it is forward invariant we have x ∈ N . This concludes Step 1.
Step 2: If x, x ∈ M \ E and y ∈ sα(x) \ M , then sα(x) ⊆ sα(x ). To prove this claim, fix an arbitrary point y ∈ sα(x). We need to show that y ∈ sα(x ) as well. If y ∈ M then there is nothing to prove, since M ⊂ sα(x ). So suppose y / ∈ M . By Step 1 there is a weak cycle of intervals N such that x ∈ N ⊆ sαBasin(y). Now we apply Lemma 31 (1), noting that M ∩ N contains the point x ∈ End(M ). Therefore M \ E ⊆ N . We now have x ∈ M \ E ⊆ N ⊆ sαBasin(y), from which it follows that y ∈ sα(x ). This concludes Step 2.
Step 3: Each point in M \ (E ∪ End(M )) has the same sα-limit set. Suppose x, x ∈ M \ (E ∪ End(M )). Then we may apply Step 2 to get both containments sα(x) ⊆ sα(x ) and sα(x ) ⊆ sα(x). This concludes Step 3 and the proof of Theorem 27 (1). From now on we will refer to this common sα-limit set as S.
Step 4: If the sα-limit set of x contains M and is distinct from S, then x belongs to a periodic orbit contained in End(M ). The hypothesis M ⊆ sα(x ) implies that x ∈ M \ E. Fix x ∈ M \(E ∪End(M )) and apply Step 2 to conclude that S = sα(x) ⊆ sα(x ). Our hypothesis is that this containment is strict. So choose y ∈ sα(x ) \ sα(x). Clearly y / ∈ M . Applying Step 1 to x we get a weak cycle of intervals N such that x ∈ N ⊆ sαBasin(y). Moreover x ∈ N since y ∈ sα(x). Now M ∩ N contains x but not x, so we may apply Lemma 31 (1) and conclude that M ∩ N is a periodic orbit contained in End(M ). This concludes Step 4.
Step 5: At most three distinct sα-limit sets of f contain M . One of these sets is S from Step 3. By step 4, the only other sets to consider are the sα-limit sets of those periodic points whose whole orbits are contained in End(M ). Since M is a cycle of intervals it is clear that End(M ) contains at most two periodic orbits. And from the definitions it is clear that all points in a periodic orbit have the same sα-limit set. This concludes Step 5 and the proof of Theorem 27 (2). 5. On special α-limit sets which are not closed 5.1. Points in the closure of a special α-limit set. We start Section 5 with two theorems which are important for understanding any non-closed sα-limit sets of an interval map. Theorem 32 relates the closure of a sα-limit set to the three kinds of maximal ω-limit sets. Theorem 33 determines precisely which points do or do not have a closed sα-limit set, and which points from the closure are not present in the limit set. In Section 5.2 we apply these results to establish some topological properties of non-closed sα-limit sets for interval maps, showing that they are always uncountable, nowhere dense, and of type F σ and G δ . In Section 5.3 we address the question of which interval maps have all of their sα-limit sets closed. Most importantly, this holds in the piecewise monotone case. In Section 5.4 we give a concrete example of an interval map with a non-closed sα-limit set.
Recall that a generating sequence is any nested sequence of cycles of intervals Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · for an interval map f with periods tending to infinity. In light of the discussion at the beginning of Section 3.1 we may define a maximal solenoidal set as the intersection Q = Orb(I n ) of a generating sequence. Recall that a solenoidal ω-limit set is an infinite ω-limit set containing no periodic points, and a solenoidal ω-limit set is always contained in a maximal solenoidal set. Recall also that the Birkhoff center of f is the closure of the set of recurrent points.
Theorem 32. Let f be an interval map, let x ∈ sα(y), and suppose that any of the following conditions holds: (1) x is periodic, (2) x belongs to a basic set B, or (3) x is a recurrent point in a solenoidal ω-limit set.
Then x ∈ sα(y). The rest of Section 5.1 is devoted to the proofs of these two theorems. For Theorem 32 the main idea is that sα(y) ⊂ α(y), so we can apply Theorems 6, 7, and 8. However, some extra care is needed for preperiodic points. Recall that x is preperiodic for f if x ∈ Per(f ) but there exists n ≥ 1 such that f n (x) ∈ Per(f ).
Proof. We may suppose without loss of generality that x < f (x). A set U is called a righthand neighborhood (resp. left-hand neighborhood ) of x if it contains an interval (x, x+ ) (resp. (x − , x)) for some > 0. By hypothesis either x ∈ sα(y), or every right-hand neighborhood of x contains points from sα(y), or every left-hand neighborhood of x contains points from sα(y). In the first case there is nothing to prove. The remaining two cases will be considered separately.
Suppose that every right-hand neighborhood of x contains points from sα(y). We will construct inductively a sequence of points y n → x and times k n > 0 such that f k 0 (y 0 ) = y and f kn (y n ) = y n−1 for n ≥ 1. We also construct points a n ∈ sα(y) and open intervals U n a n that are compactly contained (i.e. their closure is contained) in (x, f (x)). The points a n will decrease monotonically to x, the intervals U n will be pairwise disjoint, and the inequalities sup U n < y n−1 < sup U n−1 will hold for all n ≥ 1.
For the base case, choose any point a 0 ∈ sα(y) with x < a 0 < f (x). Choose a small open interval U 0 a 0 which is compactly contained in (x, f (x)). There exists y 0 ∈ U 0 and k 0 > 0 such that f k 0 (y 0 ) = y. Now we make the induction step. Suppose that x < y n−1 < sup U n−1 and choose a n ∈ sα(y) with x < a n < min{y n−1 , inf U n−1 , x + 1 n }. Choose a small open interval U n a n compactly contained in (x, f (x)) with sup U n < min{y n−1 , inf U n−1 }. By Theorem 5 we know that a n is a non-wandering point, so there exist b, c ∈ U n and k n > 0 such that f kn (b) = c. Thus f kn ([x, b]) contains both f (x), c, but c < y n−1 < f (x), so by the intermediate value theorem there is y n ∈ (x, b) with f kn (y n ) = y n−1 . Clearly y n < sup U n , so we are ready to repeat the induction step. This completes the proof in the case when every right-hand neighborhood of x contains points from sα(y). Now suppose that every left-hand neighborhood of x contains points from sα(y). Consider the set For > 0 the set W is invariant for f . It is connected because the intervals f t ((x − , x]) all contain the common point f (x). Now choose a point a ∈ sα(y) ∩ (x − , x). Since a is non-wandering there are points b, c ∈ (x − , x) such that f t (b) = c for some t > 0. Thus c ∈ W , so by connectedness we get [x, f (x)] ∈ W . Since these properties hold for arbitrary , we see that W is invariant, connected, and contains [x, f (x)]. We claim that W contains a left-hand neighborhood of x. Assume to the contrary that Therefore f t (b) > z, so it follows that z ∈ W . Since > 0 was arbitrary, this shows that z ∈ W . Again fix > 0 arbitrary and let U, a, b, c, s be as they were before. We have f (b) ∈ (x, z) ⊂ W and W is invariant, so f s (b) = c ∈ W , contradicting the assumption that x = min W .
Finally, we construct inductively a monotone increasing sequence of points y n → x and a sequence of times k n > 0 such that f k 0 (y 0 ) = y and f kn (y n ) = y n−1 for n ≥ 1. Let δ > 0 be such that W contains the left-hand neighborhood (x − δ, x). For the base case we find a ∈ sα(y) ∩ (x − δ, x). Then there is y 0 ∈ (x − δ, x) and k 0 > 0 such that f k 0 (y 0 ) = y. For the induction step, suppose we are given y n−1 ∈ (x − δ, x). Choose a positive number < min{|x − y n |, 1 n }. Since (0, δ) ⊂ W ⊂ W , we see from the definition of W that there exist y n ∈ (x − , x] and k n > 0 such that f kn (y n ) = y n−1 . Clearly y n = x, since y n−1 = f (x). Therefore y n ∈ (x − δ, x) and the induction can continue.
Proposition 35. Let f be an interval map. If x is preperiodic and x ∈ sα(y), then x ∈ sα(y).
Proof. Find n such that x = f n (x) = f 2n (x) and apply [12, Proposition 2.9], which says sa(y, f ) = n−1 j=0 sα(f j (y), f n ). Since the closure of a finite union is the union of the closures, we find j such that x ∈ sα(f j (y), f n ). By Lemma 1, x ∈ sα(f j (y), f n ). Again applying [12, Proposition 2.9] we conclude x ∈ sα(y, f ).
Proof of Theorem 32. Let x ∈ sα(y). Since α(y) is a closed set containing sα(y), we have x ∈ α(y) as well. If x is periodic, then by Theorem 8, x ∈ sα(y). If x is a recurrent point in a solenoidal ω-limit set, then by Theorem 6 we again have x ∈ sα(y). If x belongs to a basic set B, then it must be periodic, preperiodic, or have Orb(x) infinite. In the periodic case Theorem 8 applies. In the preperiodic case Proposition 35 shows that x ∈ sα(y). And when Orb(x) is infinite, then it must be contained in both B and α(y) since those sets are both invariant. So by Theorem 7 we again have x ∈ sα(y).
Proof of Theorem 33. Suppose first that sα(y) is not closed. Pick any point x ∈ sα(y)\sα(y). By Theorem 5 we have x ∈ Rec(f ) and in particular x ∈ Λ 1 (f ). Let ω(z) be a maximal ωlimit set of f containing x. By Theorem 32, x is not recurrent and ω(z) is a solenoidal ω-limit set. Then ω(z) is contained in some maximal solenoidal set Q. By Theorem 6 we know that y ∈ Q. But Q also contains x, so we have shown that y belongs to a maximal solenoidal set which contains a nonrecurrent point from the Birkhoff center. Now if x is any other point from sα(y) \ sα(y), then by the same argument, x is a nonrecurrent point in the Birkhoff center and belongs to a maximal solenoidal set Q which also contains y. Since two maximal solenoidal sets are either disjoint or equal, we get Q = Q . This shows that To prove the converse part of the theorem, suppose that Q is any maximal solenoidal set for f which contains a nonrecurrent point from the Birkhoff center, and fix y ∈ Q. We will show that sα(y) is not closed, and that To that end, let x be any point from Q ∩ (Rec(f ) \ Rec(f )). Choose a generating sequence Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · with Q = Orb(I n ). Without loss of generality we may assume that x ∈ I n for each n. Since x is not recurrent we have x ∈ sα(y) by Theorem 6. To finish the proof it suffices to show that x ∈ sα(y). We do so by finding periodic points arbitrarily close to x which are in the α-limit set of y, and then applying Theorem 8.
Let K be the connected component of Q containing x. By Theorem 10 we know that the singleton components of Q are recurrent points, and in the non-degenerate components of Q, only the endpoints can appear in ω-limit sets. Since x is in an ω-limit set but is not recurrent, we know that K is a non-degenerate interval with x as one of its endpoints. Without loss of generality we may assume that x is the left endpoint of K and write K = [x, b].
Again using Theorem 10 (and the fact that the set S Rec referred to there is perfect) we know that each component of Q has at least one recurrent endpoint, and the endpoint of a component is recurrent if and only if it is the limit of points of other components of Q. Since x is not recurrent, we know that b is. We conclude that Q does not accumulate on x from the left and it does accumulate on b from the right.
Since x is in the Birkhoff center x ∈ Rec(f ) = Per(f ) and K contains no periodic points, we know that the periodic points of f accumulate on x from the left. In particular, x is not the left endpoint of [0, 1]. That means we can find non-empty left-hand neighborhoods of x in [0, 1], i.e. open intervals with right endpoint x.
Since Q does not accumulate on x from the left, we can find a left-hand neighborhood of x which contains no points from Q, and then there must be some n such that I n does not contain that left-hand neighborhood. It follows that x = min(Q ∩ I n ). At this moment we do not know if x is the left endpoint of I n or not, only that it is the left-most point of Q in I n .
Let m be the period of the cycle of intervals Orb(I n ) and let g = f m . Then I n is an invariant interval for g. Also f and g have the same periodic points (but not necessarily with the same periods). Since g i (x) ∈ I n for all i and x ∈ Q is neither periodic nor preperiodic, there must be i ≤ 2 such that g i (x) is not an endpoint of I n . By continuity there is δ > 0 such that g i ((x − δ, x]) ⊂ I n . Then by invariance we get Choose an arbitrary positive real number < δ. Choose a periodic point p in (x − , x). Let k be the period of p under the map g. Then g ik (p) = p and by (*) g ik (p) ∈ I n . This shows that p ∈ I n . (Incidentally, we see that x is not the left endpoint of I n .) The set of fixed points of g k is closed, so let p be the largest fixed point of g k in the interval [p, x]. Let h be the map h = g k = f mk . We know that h(x) is an element of Q in I n , and therefore it lies to the right of x. We also know h(x) / ∈ [x, b] because no component of Q returns to itself. So the order of our points is p = h(p ) < x < b < h(x), and the graph of h| (p ,x] lies above the diagonal, since there are no other fixed points there, see Figure 6.  Figure 6. The graph of h| In . Since Orb(L, f ) ⊃ Q and each right-hand neighborhood of p eventually covers L, we see that p ∈ α(y). Note: We do not claim that h is monotone on (p , x], only that the graph stays above the diagonal. Since Q accumulates on b from the right we can find n > n such that one of the components L of Orb(I n , f ) lies between b and h(x). By the intermediate value theorem, h([p , x]) contains L.
Let U be any right-hand neighborhood of p , that is U = (p , z) with z−p > 0 as small as we like. We consider the orbit of U . Since the graph of h lies above the diagonal on (p , x] we get a monotone increasing sequence (h j (z)) l j=0 with l ≥ 1 minimal such that h l (z) ≥ x (if there is no such l, then h j (z) converges to a fixed point of h in the interval [z, x], which contradicts the choice of p ). Then h l+1 (U ) ⊃ L. In particular, Orb(U, f ) ⊃ Orb(L, f ) ⊃ Orb(I n , f ) ⊃ Q.
We see that for every right-hand neighborhood U of p there is a point y ∈ U and a time t such that f t (y ) = y. As the neighborhood U shrinks, the value of t must grow without bound, because p is periodic. This shows that p ∈ α(y) = α(y, f ). By Theorem 8 it follows that p ∈ sα(y).
Since p was chosen from the -neighborhood (x − , x) and > 0 can be arbitrarily small, we conclude that x ∈ sα(y). This completes the proof.

5.2.
Properties of a non-closed special α-limit set.
As an application of Theorems 33 and 32, we get the following results.
Theorem 36. If sα(y) is not closed, then it is uncountable and nowhere dense.
Proof. Suppose sα(y) is not closed. By Theorem 33 we have y ∈ Q for some solenoidal set Q = Orb(I n ). By Theorem 6 we know that sα(y) contains Q∩Rec(f ). This set is perfect [3, Theorem 3.1], and therefore uncountable. Let M be a transitive cycle of intervals for f . If Q ∩ M = ∅, then by Lemma 31, each cycle of intervals Orb(I n ) has period at most twice the period of M . This contradicts the fact that the periods of the cycles of intervals tend to infinity. Therefore Q does not intersect any transitive cycle of intervals M for f . In particular, y does not belong to any transitive cycle of intervals M . By Corollary 19 we conclude that sα(y) does not contain any transitive cycle. By Theorem 23 it follows that sα(y) is nowhere dense.
A subset of a compact metric space X is called F σ if it is a countable union of closed sets, and G δ if it is a countable intersection of open sets. These classes of sets make up the second level of the Borel hierarchy. Closed sets and open sets make up the first level of the Borel hierarchy and they are always both F σ and G δ . The next result shows that the sα-limit sets of an interval map can never go past the second level of the Borel hierarchy in complexity.
Theorem 37. Each sα-limit set for an interval map f is both F σ and G δ .
Proof. We write Bas(f ) for the union of all basic ω-limit sets of f and Sol(f ) for the union of all solenoidal ω-limit sets of f . We continue to write Per(f ) for the union of all periodic orbits of f .
To prove that sα(y) is of type F σ we express it as the following union and show that each of the three sets in the union is of type F σ .
The set Per(f ) = n {x : f n (x) = x} is clearly of type F σ . By Theorem 32, sα(y) ∩ Per(f ) is a relatively closed subset of Per(f ), and is therefore of type F σ .
Since an interval map has at most countably many basic sets [3, Lemma 5.2], their union Bas(f ) is of type F σ . By Theorem 32, we know that sα(y) ∩ Bas(f ) is a relatively closed subset of Bas(f ), and is therefore of type F σ .
By Theorem 6 and Corollary 12, we know that sα(y) ∩ Sol(f ) is either the empty set, or a single minimal solenoidal set S, and minimal solenoidal sets are closed. Closed sets are trivially of type F σ .
To prove that sα(y) is of type G δ it is enough to show that sα(y) \ sα(y) is at most countable. By Theorem 33 we know that this set is of the form sα(y) \ sα(y) = Q ∩ (Rec(f ) \ Rec(f )) for some maximal solenoidal set Q. But by Theorem 10 the only points in Q which can be in the Birkhoff center but not recurrent are endpoints of non-degenerate components of Q. Since Q ⊂ [0, 1] has at most countably many non-degenerate components, this completes the proof.

5.3.
Maps which have all special α-limit sets closed.
In [12] it was proved that all sα-limit sets of f are closed if Per(f ) is closed. This is a very strong condition; it implies in particular that f has zero topological entropy. In this section we give necessary and sufficient criteria to decide if all sα-limit sets of f are closed. We show in particular that this is the case for piecewise monotone maps. Note that an interval map f : [0, 1] → [0, 1] is called piecewise monotone if there are finitely many points 0 = c 0 < c 1 < · · · < c n = 1 such that for each i < n, the restriction f | [c i ,c i+1 ] is monotone, i.e. non-increasing or non-decreasing. (3) =⇒ (4): Suppose SA(f ) = Λ 2 (f ) = Rec(f ). Then we can find x ∈ Rec(f ) \ SA(f ). By Theorem 5 we know that x ∈ Λ 1 (f ). Because x is in an ω-limit set, it must belong to a periodic orbit, a basic set, or a solenoidal ω-limit set.
Each periodic orbit is contained in the sα-limit set of any one of its points. Each basic set is also contained in a sα-limit set by Corollary 16. Since we supposed that x is not in any sα-limit set, we must conclude that x is not in a periodic orbit or a basic set. Now we know that x belongs to a solenoidal ω-limit set. Since x is not in any sα-limit set we may use Theorem 5 to conclude that x is not recurrent.
(4) =⇒ (1): Suppose a solenoidal ω-limit set ω(z) contains a non-recurrent point x in the Birkhoff center x ∈ Rec(f ). Since ω(z) is solenoidal it has a generating sequence, i.e. a nested sequence of cycles of intervals Orb(I 0 ) ⊃ Orb(I 1 ) ⊃ · · · with ω(z) ⊆ Q = n Orb(I n ). By Theorem 33 for any y ∈ Q the set sα(y) is not closed.
Corollary 39. If f is a piecewise monotone interval map, then all sα-limit sets of f are closed.
Proof. By [3, Lemma PM2] each point in a solenoidal ω-limit set for a piecewise monotone map f is recurrent, so condition (4) of Theorem 38 can never be satisfied.
We remark that in general the conditions of Theorem 38 may be difficult to verify. Even condition (4) is difficult, since the non-recurrent points in a solenoidal ω-limit set need not belong to the Birkhoff center. For an example, see [2]. But for maps with zero topological entropy, the whole picture simplifies considerably. For the definitions of topological entropy and Li-Yorke chaos we refer the reader to any of the standard texts in topological dynamics, eg. [14]. Proof. For an interval map f with zero topological entropy, the set of recurrent points is equal to Λ 2 (f ) by [21]. Now apply Theorem 38.
Proof. When f is not Li-Yorke chaotic, T. H. Steele showed that Rec(f ) is closed [18,Corollary 3.4]. Moreover, it is well known that such a map has zero topological entropy, see eg. [14]. Now apply Corollary 40.

5.4.
Example of a non-closed special α-limit set.
In 1986 Chu and Xiong constructed a map f : [0, 1] → [0, 1] with zero topological entropy such that Rec(f ) is not closed [8]. This example appeared 6 years before the definition of sαlimit sets [11], but by Corollary 40 it provides an example of an interval map whose sα-limit sets are not all closed.
In this section we give a short direct proof that one of the sα-limit sets of Chu and Xiong's map f is not closed. Here are the key properties of the map f from [8].
(1) There is a nested sequence of cycles of intervals [0, 1] = M 0 ⊃ M 1 ⊃ M 2 ⊃ · · · for f , where M n = Orb(J n ) has period 2 n . (2) For each n ∈ N the interval J n is the connected component of M n which appears farthest to the left in [0, 1]. (3) For each n ∈ N we can express J n = A n ∪ J n+1 ∪ B n ∪ K n+1 ∪ C n as a union of five closed non-degenerate intervals with disjoint interiors appearing from left to right in the order A n < J n+1 < B n < K n+1 < C n . (4) For each n ∈ N the map f 2 n : J n → J n has the following properties: • f 2 n | An : A n → A n ∪ J n+1 ∪ B n is an increasing linear bijection, • f 2 n | J n+1 : J n+1 → K n+1 is surjective, • f 2 n | Bn : B n → K n+1 ∪ B n ∪ J n+1 is a decreasing linear bijection, • f 2 n | K n+1 : K n+1 → J n+1 is an increasing linear bijection, and • f 2 n | Cn : C n → B n ∪ K n+1 ∪ C n is an increasing linear bijection.  Figure 7. Chu and Xiong showed that the left endpoint x of J ∞ is not recurrent, but it is a limit of recurrent points. We give a short direct proof that sα(x) is not closed.
Theorem 42. Let f be the interval map defined in [8] and let x be the left endpoint of J ∞ as defined above. Then x ∈ sα(x) \ sα(x), and therefore sα(x) is not closed.
Proof. Fix n ∈ N and let a n be the left endpoint of the interval A n . Property (4) tells us that f 2 n : A n → A n ∪ J n+1 ∪ B n is linear, say, with slope λ n . By property (5) we have x ∈ J n+1 . Therefore there is a backward orbit branch {x i } ∞ i=0 of x such that x k·2 n = a n + x−an λ k n for all k ∈ N. This shows that a n ∈ sα(x). But n ∈ N was arbitrary. If we let n → ∞, then a n → x. This shows that x ∈ sα(x). Now we will show that x ∈ sα(x). Let {x i } ∞ i=0 be any backward orbit branch of x. Let Q = ∞ n=0 M n . We distinguish two cases. First suppose that x i ∈ Q for all i. For any given i ≥ 1 we can choose n with 2 n > i. Since x i ∈ M n , f i (x i ) ∈ J n , and M n is a cycle of intervals of period 2 n , we know that x i ∈ J n . Since J n is the left-most component of M n in [0, 1], it The graph of f 2 n | Jn . Figure 7. A map with a sα-limit set which is not closed.
follows that x i > y > x (recall that J ∞ = J n = [x, y] is non-degenerate). Since this holds for all i ≥ 1 we see that the backward orbit branch does not accumulate on x. Now suppose there is i 0 with x i 0 ∈ Q. For each i ≥ i 0 there is n(i) ∈ N such that x i ∈ M n(i) \ M n(i)+1 . Since f (x i+1 ) = x i and each M n is invariant, we get n(i + 1) ≤ n(i). A non-increasing sequence of natural numbers must eventually reach a minimum, say, n(i 1 ) = n(i 1 + 1) = · · · = n. For i ≥ i 1 , x i ∈ M n+1 , so in particular x i ∈ J n+1 . But by properties (3) and (5) we know that J n+1 is a neighborhood of x. This shows that this backward orbit branch does not accumulate on x either.
We have seen that even for interval maps, sα-limit sets need not be closed. If we want to work with closed limit sets, then there are several possible solutions. The first one, suggested to us by L'. Snoha, is to answer the following question: What are some sufficient conditions on a topological dynamical system (X, f ) so that all of its sα-limit sets are closed? In this regard we state one conjecture which we were not able to resolve.
Another possibility is to ask whether the "typical" interval map has all sα-limit sets closed. Let In support of Conjecture 51, we show that the conclusion holds for n = 1.
Proof. Let sα(x) contain the periodic orbit {a, b, c} for the interval map f with a < b < c. We may assume without loss of generality that f (a) = b, f (b) = c, and f (c) = a. By continuity we may choose a closed interval U = [b− , b+ ] with > 0 small enough that f 2 (U ) < U < f (U ), that is to say, max f 2 (U ) < min U < max U < min f (U ). Now find x 1 ∈ U and n 1 ≥ 1 such that f n 1 (x 1 ) = x. By the intermediate value theorem we can find x 2 ∈ (max U, min f (U )) such that f (x 2 ) = x 1 . Again, by the intermediate value theorem we can find x 3 ∈ (x 1 , x 2 ) such that f (x 3 ) = x 2 . In the next step we find x 4 ∈ (x 3 , x 2 ) such that f (x 4 ) = x 3 . Continuing inductively we find a whole sequence (x i ) such that f (x i+1 ) = x i , i ≥ 1, arranged in the following order, x 1 < x 3 < x 5 < · · · < · · · < x 6 < x 4 < x 2 . Since a bounded monotone sequence of real numbers has a limit, we may put x − ∞ = lim i→∞ x 2i+1 and x + ∞ = lim i→∞ x 2i+2 , and we have x − ∞ ≤ x + ∞ . Then f (x − ∞ ) = lim f (x 2i+1 ) = lim x 2i = x + ∞ and similarly f (x + ∞ ) = x − ∞ . This shows that {x + ∞ , x − ∞ } is a periodic orbit contained in sα(x). The period is 2 if these points are distinct and 1 if they coincide.